DK Jain M3

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S h r e e R a m F r a k a s h a n

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NGINEERING MATHEMATICS-III!. t f ' r i ; I 1 n . t h i h u s i h ' i ' i i n U i l ! I M; . , ; ‘

SHREE RAM PRAKASHAN

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Published by

>i i -j ; (•. : I'R a Patankar Bazar. Gwalior (M.P.)

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r I Al l r i g h t s r e s e r v e d . N o p a r i (it t hi > b o o k m a \ b e r e p r o d u c e d o r t r a n s m i t t e d i n a n \ f o r m or by j

m e a n s , e l e c t r o n i c o r m e c h a n i c a l , i n c l u d i n g p h o t o c o p y i n g , r e c o r d i n g , o r an y i n f o r m a t i o n storage;

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mailto:[email protected]

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REFACE OF THE FlRST EDITION

We are very happy to present to our readers, students, and teachers alike, this book on EngineeringMathematics-Ill. The present book is to meet the requirements of the students of B.E. Second Year, the

eed of which was being felt very anxiously. In the treatment of subject matter, we have tried toMaintain die same style as used in the other books of Engineering Mathematics (namely I, II andItecrete Structure).

All the topics have been covered comprehensibly not only with clarity but in a lucid and easy wayd be grasped. There are a good number of fully solved examples with exercises to be worked out at theHid of each unit. B esides, there are fu lly solved questio n papers o f previous year RGPV xamination a t th e end o f th e bo ok. We are sure that this book will receive the same warm

welcome by its users, as in the case of our previous volumes.

We shall feel highly obliged for any suggestions and constructive criticism for the improvement of

he book.The book is fully on accordance to the syllabi provided for the Bachelor of Engineerging Fourthemester students of Rajiv Gandhi Prodyogiki Vishwavidyalaya, Bhopal.

Lastly, we thank our Publisher, M/s Shree Ram Prakashan, Gwalior for publishing this valuablewok.

AUTHOR For suggestions, write toDr. D .K. Jain,B-3 Dwarika Puri, Behind Prem Nagar,

Gwalior (M.P.) 474002Phone:9826086788Email: [email protected]

........................................ .......... —— ............ 11 ...................

P r e f a c e o f t h e F o u r t h E d i t io n

We are pleased to bring out the fourth edition of this book. This new edition is throughly revisednd updated in accordance with the demands of the modified syllabi and that of the technical studentsnd the interdisciplinary applications that they need. The new revised edition includes modified units

ealing with statistics adding up topics namely, correlation and regression, theoretical distribution andsting of hypotehsis, with particular references to mathematical statistics.

Suggestion for the further improvement and upgrading of the book are heartily welcome and theditors and the authors will be thankful to the readers for responding with the same.

AUTHOR

Dr. D. K. JainGwaliorJanuary 2012



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IV

C o n t e n t s

UNIT-1

FUNCTIONS OF COMPLEX VARIABLES1.1 Introduction l

*2 Complex Variables 1

1.3 Polar Forms ofAComplex Number ~ I

0 4 Functions o f A Complex Variable 2

1.5 Limit ofA Function 2

1.6 Continuity of /( z ) 2

1.7 Differentiability of f(z) 2

J.8 Analytic Function 3

•+.9 Câuchy-Riemann Equations 4

1.10 Cauchy-Riemann Equations In Polar Coordinates 6- t. l l Harmonic Function 7

1.12 Orthogonal System 7

1.13 Methods for Constructing an Analytic Function 15

1.14 Bilinear Transformation 32

1.15 Conformal Transformation 40

1.16 Some Standard Transformations 40

1.17 Some Definitions 48

1.18 Complete Integration 52

1.19 Some Definitions 53

-120 Complex Line Integral 5442 1 Cauchy's Integral Theorem or Cauchy's Theorem 58

H-22 Cauchy's Integral Formula 59

123 Residues o f/( z ) at Pole (A

J2 4 Methods of F inding Out Residues of / (z) at A Pole (A

125 Cauchy's Residue Theorem 65

>+26 Application of Residues To Evaluate Real Integrals 79

127 Taylor's Series 99

128 Laurent's Series 99

129 Circle and Radius o f converge nce 105

Practice Questions 106

UNIT-2 NUMERICAL ANALYSIS-I

2.1 Introduction 115

22 Approximation And Errors 115

23 Some Important Rules For Absolute and Relative Errors 118

2.4 Algebraic and Transcendental Equations 119

2.5 Bisection or Bo lzano Method 119

2.6 Iteration M ethod or Successive Approxim ation Method 123

2.7 Method of False Position or Regula Falsi Method 128

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Derivatives ofy =f(x) Based on Newton’s Forward Interpolation Formula 257Derivatives of y = /( x) Based on Newton’s Backward Interpolation Formula 258Derivatives ofy =f(x) Based on Stirlings or Central Difference Formula “ 258 Numerical Integration 269

Newton-Cote’s Quadrature Formula 269Trapezoidal Rule 270Simpson’s One-third Rule (or 1/3) Rule 270

H Simpson’s Three-Eighths (3/8) Rule 27132 Weddle’s Rule 272

Practice Questions 290UNIT-4

NUMERICAL ANALYSIS-III, CORRELATION AND REGRESSION

Numerical Solution of Ordinary Differential Equations 299Picard’s Method 299Taylor Series Method 307Euler’s Method 311

Improved Euler’s Method 314Modified Euler’s Method 315Runge-Kutta Method 318Milne’s Predictor-correct or Formulae 326Working Rule for Milne’s Predictor Corrector Method 327

4# Concept of Correlation 333ill Coefficient of Correlation 333.12 Types of Correlation 333S3 Properties of Coefficient of Correlation 335

Covariance 336Methods to Study Correlation 336Scatter or Dot Diagram Method ' 336Karl Pearson's Coefficient of Correlation or Covariance Method 337Rank Correlation Method 344Lines of Regressions 351

Regression Coefficients 351Properties of Regression Coefficients 352Properties of Lines of Regression 353Relation Between Regression Analysis and Correlation Analysis 354Curve Fitting 366Principle of Least Squares 366Fitting ofA Straight Line 368Fitting ofA Parabola (or Second Degree Curve) 369Fitting of A Curve of The Type^ = ab* 370

Fitting of The Exponential Curve y = a(*>x 370b

Fitting ofA Curve of The Type y = ax2+ ~ 370

Practice Questions 380

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' F u n q t j o n s o f C o m b w x V a r d w i . e s

— ..■■■"■■" — ■......................■■■■.................................. w fo M -wfw i m iilit . >■——

i n t r o d u c t i o n ^ ! I 'w - - .- - ;* * * * ■ -

We havife 'fafafl that there exist one-one correspondences between the set of real

numbers and me points on the real line. The concept of ^ (- l j is meaningless, since there is no

real number whose square is a negative real ^70t*l783) was the first mathemati

cian who iiôellticed the symbol i with the property that ? = - l l i was Gauss (1777-1855) who

first studied thatan algebraic equation with re&f <io#1$iehfs HafecbmpleX roots of the forma+ ib, where a and b are real numbers. 3 '/Si IMWO > A Kf ?M\i : *

ftp ii:.; i"

&]^npM v iab le . Ateb i a r i t f y ^ respectively called real and

imaginary patfsrafa SonKtimes we express sr as z ~ x+ iy * (*,31).

° ^ ' a l » M - v ! r t ; ' ' :"r ’

•' Thecomplex conjugate, ar briefly conjugate o f z=x+iy is I =x-iy.

For exgmp&4ao)iigitic of tx m - 3 - 5 i is z * r 3f t # •i i ‘

- .f.*i

It is easy to verify that: R(z)=x =Z- ? * , and l ( z )=y=^~ . ! 'v ’ '; i •- i , ■ . 2,1

POLARFORMS OF A COMPLEX NUMBER ' . . ' i t - -

Consider a point /* in the complex pl{(he corresponding to a complexnumber z=x+iy. From the

adjoining figure,

putting x =rcos#, andy=fgin^

Then r = ^ (x s + >*) , j

' w ' ■ k If follows that r = x + iy K 1

? iA(^W +il8mff)=rei& ...(1)

It is called/?o/<^ybrw dfthe coirt^T®i<i»«mber*;rand $ arecafted polar coordinates o f z. ^

Hsaaited modutuserabsGkitevaiueofz.Theaagge & r ' \ which line OP makes wi&Die positive jc-îs, i eaiied argument or amplitude of z. It is also

written as <9= am p(i) or 0=arg(4:). t --' ra

y k ; \ j O ' i'i i ■f { t .••< ' f- ■

; ; - XP(x,y)

r /

•' ' '■ j r ■ y V * * s ■j '*

y S \ D

A L-: x M X

(Argand diagram)

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represents

2 | Eng ineh u ng MAmEMAncs il

Remarks :

1. |*,*2H*,| | *2| * A

'*>2£*. , ' | 4 ^ %MI * i m * $ & C - --D ,

■ 3r - " - ' ■ •■■ - 7-4. Equation of circle | r - a j= r where a is center and r is a radius,

5. Re (z)=x and Im (z)=y

6. . Jzj2=zz

-ij.- <•.7. The equation ‘T ~ ~

* “ * 2 ^

(i) ac to je if A 0 and (ii)» straight line If A»i..

1.4 FUNCTIONS OF ACOMPLEX VARIABLETo each value of complex variable z, if there correspond*4tatfMr more than one value of acoppkat vari*We H^then we say that w is a function sof a ccjpjplex variable z. Also we write

w - f ( z ) = u + i v , ..... .

w is said to be single valued or many valued functions of z according as for a given value of z,there corresponds one or more than one valye o f w, and u and v are ftmctiom of x aftd>For examples:

1. f( z)= z2 2. f(z)»&D.z 3. /(^ « lo g * 4. f ( z ) * z 2

1.5 LIMIT OF AFUNCTIONA number / is die limit of a function f(z) as z tends to a and write

lim f{z ) - l.

1.6 CONTINUITY O F/ft)

A function f{z) is called continuous at z = a if 1

l im/(a)=/(o).

A function f(z) is said to be continuous in a domainD if it is continuous at every point of D.1.7 m f f e r e n h a b i l i t y o f / ^

A function fin) in domain D, is called differentiable at a poiptz =f a, if

tim— <L—— — is exist and unique limit

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1 4 : I f f( z)= z * 0

"4,

Prove that ->0 as g -> o ekmtgltirf m lto 4ettM kdnat 4» jt ■-»%xj»**y X w

manner. >tf

i M Suppose *>-* 0, along the radios vectef y » « * , then we have

x2y(y -ix ) '

m M z M . i J i i - u !2*-to z - 0 *-*o (x+*y)-0

[Since y ~ m x then « -► 0=>x+iy -* 0 =».*+*rox -► 0=> x -+ 0]

. I j J - = lim[ , ----- 1

= lim - g f c & : * 0. bl!6 n {.< f.-v. **C(x7+M2)$+im) ■ ■ -!•

• i ' i. ' * I . . . • .Further, if g -* 0 along the eurve y = m x2 (ie., in any manner) then we have

x 2y(y~ix)

<£±£1*-*« * - 0

Ans.

(*+yr ) -0

( !

[Since y = mx2, then z ->,0 => x -* 0]

_ Hni x2mx2(mxl -ixy fj. ljm mxi(mx-i)

(x4 + m2x*)(x+imx2) z~*4>x5(1 +nt2Xl+iinx)

- m i= :----- 2 ~ does not exist, [because limit depends on m i.e. not unique].

1+ rn •• t ': u : < \ r ' Proved.

J ANALYTIC FUNCTION. _A single valued function ffz) in a domain P is said to be analytic at.a point z = a if there exists a

neighbourhood \ z - a \ < 6 at all points of which /"(zjeadsts.

If f ( x ) exiits at every point of a d&main D exceptar a flnite^tanber of exceptional points then f(z) is said to be analytic in Z > .T h e « exceptional points are calledsingula- points or singularities of the function jjjft).If f { z ) exists at every point of D, then we say that f( t) is regular in D.

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The terms regular and holomorphic are also sometimes used as synonyms fo r analytic.

For examples : 1. f ( z )= z7 2. f ( z ) -e* t*

R w ir l : '■ ■ H V' : £!-> :

Singular point: A point z =zn is stid to be a singutotgoint o f a fraction f(z) if f ( z Q) does flat

exist. •;

A necessary condition Ifcat w = f(z)=u(x,y)+i v(x,y ) y^- - vv

be analytic in a donwin'OiMljat options, ^ ««*.-.?<;■

du dv ,d u dv•*—- **-—an d ——=— (■<■ - jv j 1dx dy dy dx

or ux =vy and uy ~-vx . , iiiui - - • • -u., at- - i ti

Theorem [1.1]; Necessary condition fo r f(z) to be analytic:t f fi(z) = u+iv is analytic in adomain D, then u, v satisfy the C-R equations

• , -I . -J

dx 8$ „«***; _ r’s->-

/ W / J = «■+ iv JfMQanalytic functionIn S f f i g i g ? d f e n S _ mvr..f* where « and v are the function of * and y. Let, S u oftL the H)c*enients of wand v

or

f(z + Se) - (u + Su)+i(v + Su) f(z + s j ) - f ( z ) (u +Sy)+*iv~±$$^(u + iv)

We have JZ s * * ''> .. .. 1 1, ........... "'r^~ ' *'“■■*“ . ti. i - • '

_ Su + iSv _ 8 ’u ' .Sy

Sz 5z Sz

^ ^ f j «

Sz-*0 j., ^ 2* , . S'9.p ».

i: ‘ •? fito * >«,.N v.i’.- tV V -.i’

.fes/o'

or . f ( z ) - S z ) *I'jJ; : no «SR3Q9b”mn u>. p?/•? • . ' •

Sino*, & M »ajjftrwtch zrw&aiongany path.

w Along real axis : On real axis y=0 so that z =x, an^ d'y=^, ( t r s f

“ -& U * !te tR ,-dri: •,.V

1*0 'M- -f-o i>norr

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FUN C | f

=(foln

-3 * .• vi

n:L& i Su

J 1 dy dy

, k f(z) is dilYertmiaWeêi* two values o f f! (i) must be the same i.e., • 'tf , r - \ r

du . dv _ .d u d v . ^ -'-equating (2) and (3) we get, + l dx~~* dy*dy ,■

■■ ; ;S ■v: * - ; . • — - i- du dv j do \ -du

Equating real and imaginary pats : ^ - ^ a n d — = - — ■

J-Jence the necessary condition fbrj^T’t^'be analytic is thatthe C-R equations must besatisfied. v Proved.

.t The function w = f (z)=u+ iv is analytic in a domain D, if

|f> U, v are dtfferentiaple in D and =t>,, ur =-,%*iti .fi ’ .. V\ -.I!, ■ i y,U- v , . ;

(it) The partial derivatives ux, vx, uy, vy all are continuous in D.

du du dv dv Let,f(z) x=u + iv be a single-valued function possnG&iqgpaftial derivatives Qy

at each point ofa region Z) and satisfyirtgOR equations.

du dv , du dv ,

’ dx dy dy dx

We shall prove that f(z) is analytic, i.e., f\z) exists at every (idiot of the

region/). :By Taylor's thtorem 4or functions of two variables,; we have

f(z + Sz) = u(x + Sx, y + S y ) f i v(x + Sx, y + Sy)

[on omitting seoon^ and higher degree terms of Sc and Sy]

' [V/< « )« »+ »wj

=' ( s + i s ) & + ( " i +‘i ) ' Sj' [Using C-R equations]

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6 | Enginebr mo MftnENAncs-Ili

dv

dx dx

f(z + S z )- f (z ) du | . dv

Ss dx dx<

du

dx

du

(du . 5uV-i — <5cc+ — +i— \i$y d x) t a* d x)

7 w &-*> & d x d x '

du dvThus f'(z) exists, because — >— exist.

Hence f(z) is analytic.

1.10 CAUCHY-RIEMANN EQUATIONS IN POLAR COORDINATES

Proved,)

Let w = f( z) => u + iv=f(x + iy)Put x=rcos0, y =rsin0,'then

u+iv=f(rcosd+ir&wff)=f(rel9)

u+iv=f(re10)-

Differentiating (1) partially w.r.t r, we have

+ *— = / '( r e " ) . * "dr dr

Differentiating (1) partially w.r.t 6, we havedu . dv

+ i — = f '(r ew).ire* = irld Q d d 1

dv . du = - r — + i r — .

dr dr

. ( du t .dv'

V5r dr,

[RGPV June 2609]\

[ve* = cos#+isin0)

•••(I)

...(2)

[Using (2)]

Equating real and imaginary parts, we get

du dudv , dv = - r — and — = r

dQ dr d6 dr du _ 1 dv , du _ 1 du

dr ~ rdO ° dr ~ r d9

Which is the polar form o f C-R equations. Example 1.1 ; Test the analytic behaviour o f log z.

Solution. Let /(z) = legz => u+iv = \og(x+iy)

zz> u +iv = log(rcos.f? + irsinff)

=> u+iv = log(re*)

=> v + iv = lo%r + id

[•.•elB ~ co8#+isin#]

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u ~ logr and v ~ 0

Differentiating partially w.r.t. r and 6.

du 1 , dv ^ — = — and — = 0,

or r or

— = 0 and — =1.dO 39

du Id v . dv 1du It is clear that - = and - =

=> The function log z is analytic except at origin. Ans.

Ill HARMONIC FUNCTION

d2u d2uA real valued function u = u(x, y) is called harmonicJunction, if 7 T +7 T

» ' ox oy

2 d2u d2u i.e., u satisfy the Laplace’s equation : V u = = 0*

qgftmark : Harmonic Conjugate Function : I f the function f(z) - u(x, y) + iv (x>y), is analytic in a domain

u" D, then the Junctions u and v are said to conjugate Junctions.J& Alpit 1.2 : Prove that, u + tvis analytic function in domain D, then « and v are harmonic.

M ention : Since f(z) = u + iv is analytic => Cauchy-Reimann equations are satisfied

du dv d u __ dv

L e - ~ t e ~ d y m d - ( 1 )

d2u d2v d2u d2vDifferentiate: 7 7 - . ~ and a

dx dxdy dy~ dydx

d2u d2u _ /Adding: ~ ^ u *s harmonic.

Again from (1), differentiate w.r.t. y and x respectively, we get

d2u d2v d2u d?v

Fu n c t io n s o f Co m pl e x Va r ia b l e s | 7

and dydx dy dxdy , <dx

d2v d2vOn adding: T T +T T => *>is harmonic.

ox oy

i.e., this shows that u and v satisfy the Laplace's equation. Proved.

ORTHOGONAL SYSTEM

Two families o f curves u(x ,y)=c]and v(x,y)=c2are said to form an orthogonal system i f they intersect at right angles at each o f their points o f intersection.

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u = logr and v = 0

Differentiating partially w.r.t. r and Q.

du 1 , 5d _

TT = " «nd T - = 0,or r or

du dv , — =0 and — =1.80 d9

du Idv , dv 1du It is clear that - = — and - = - -

=> The function log z is analytic except at origin. Ans.

HARMONIC FUNCTION

d2u d2uA real valued function u = u(x,y) is called harmonic function, if T T +T Tox oy

d*u <?ui.e., u satisfy the Laplace's equation : V u =

rk : Harmonic Conjugate Function : I f the function f(z) - u(x, y) + iv (x, y), is analytic in a domain D, then the functions u and v are said to conjugate functions.

1.2: Prove that, ifflz )* u + iv is analytic function in domain D, then u and v are harmonic, n : Since f(z) = u + iv is analytic => Cauchy-Reimann equations are satisfied

du _ du du dv

i e " ~dx ~~dy ~dy~~~dx'

d2u d2v d2u d2vDifferentiate : T T ’ T r and r r =" -“T"

dx2 dxdy dy~ dydx

d2u d*u , n Adding: = ^ u is harmonic.

Again from (1), differentiate w.r.t. y and x respectively, we get

d2u d2v c?u d?v

Fu n c t io n s o f Co m pl e x Va r i a b l e s | 7

and dydx dy2 ™ dxdyi dx*

d2v d2vOn adding: T T +T T =* v *s harmonic.

ox oy

i.e., this shows that u and v satisfy the Laplacefs equation. Proved.

ORTHOGONAL SYSTEM

Two families o f curves u (x ,y )-c i and v(x,y)=c2are said to form an orthogonal system if they

intersect at right angles at each o f their points o f intersection.

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81 ENOiNaaaNG M n m u n c s4 li

Theorem f l. 3 |; If f(z) = u + ivisan analytic function in a domain D, prove that the curves u = const and v = const form two orthogonal families.

Proof Let f(z) = u + iv be an analytic functions of z, so that Cauchy Riemannequations

ur ~vx uy = -vx are satisfied.To prove that the curves u(x,y) =c,, and L’(.T,.y) - c2 are orthogonal.

Let, m, = slop of the tangent to the curve m= c , and

m 2 = slop of the tangent to the curve v =c2.

Taking differential of and v(xty)-c2

or

or

du . du , _ dv j dv .and T x d * ^ d y - «

dy ,dx u.

m1(say) and dy _ -vx _

dx"h (say)

rmtm2—

u. V '~ v x

V J

- Ux Vx _ u x v x _ |

u =c, and v =c2 form an orthogonal system.

[v by C - R equations]

Proved.

Example \J ; I f u(x, y) and vfx, y) ure harmonic functions in a region R, prove that the function

Solution.

r du 51;") .v rk» * + ldy dx)

f du ( dvy {dx dy. is an analytic function o f z= x + iy.

fRGPV DEC 2004J

du dv . ^ du dvLet $ = ----------and t =-— + — .

dy dx dx dy

Then we shall to prove that s + it is analytic, i.e., s and t satisfy the C-Requations.

ds _ dt ds _ dt

The C-R equations are : ^ ^ anti ^ ■

ds dt We have, = XT

d_

dx

f du dvs d fdu dv>

Kdy dx; dy Kdx dy j

d2u d2v ^ d2u d2v

dxdy dx2 dydx dy2

d2v d2v _ nr T +7 J “ 0 => v is harmonic.oy ox

Sinced u

dxdy dydx

...d)

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Function* of Co m pl e x Va r i a b l e s | 9

da dt d (d u ____ d (du dvSim ilarly,^ ^ ==> fa ) cbe^d* dy

d2u d2 v _ d2u d2v

^ dy2 dydx dx2 dxdy

d2u d2u n~TT + ~rT = 0 => k is harmonicoy ox

- . (2 )

It is clear that Cauchy-Riemann equations arc satisfied, because from equation (1) andv (2), u and v satisfy Laplace's equation.

i Hence s + it is analytic. Proved.

Example f .4 ; Show that the function f(z ) ~ >/| xy | is not analytic at the origin, although the

Cauchy-Riemann equations are satisfied at that point Solution. Given, f (z) = - 1 xy |=> u + iv =» yj\ xy\ + i.O

=> u(x >y) - a/1 *y t and = 0At the origin:

xÔ x

Similarly,

and

du- lim

dx (0.0) x- °

du

= lim[dy (0,0) '-*0

i = limdx (0,0) *-*

'd v

_dy\ [0.0) ^

y-+o y

x -* 0 x

Hence C-R equations are satisfied at the origin.Further,

A O ) .]im /fr)- f i 9 . 1 i m ^ T~ ° - Um*-»o z *-+Q z *+*y-*o (x + iy)

Along y=mx, then we have

Vlmx21 J\m\ , f (0)= lun ----- ;— - =-—:— = depends on m he., not unique.

*->0 (X + fWMC) 1+tflt

=> AO) does not exist.Hence function is not analytic at origin. Proved.

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T

Fu n c t io n s o f C o m pl e x Va r ia b l e s I 11

Cauchy-Riemann equations arc:

du dv , du

dx dy*n $y dx

=> 2x =0 => x » 0 and 2y = 0 => y=0.Hence Cauchy-Riemann conditions are satisfied only at origin.Thus the given function is now here differentiable except at the origin. Proved.

1.7 : Show that the function f( z) * e '* * , andffO) - 0 is not analytic at z « 0,

although Cauchy-Riemann equations are satisfied at this point [RGPV 2001]

Here, f(z) = e ’ 4=«-<**>'4 ...(1)

— U J z M " 7 j h ^ ixUyt~ 6xV) ~a- e (* ♦ <* - «y)‘ e l(** ♦/>* J _ e <**+/>*

6xV]

* I JU) = P L ' -*

Equating real and imaginary part, we get

«4+y*- (kV1

( ^ + y s>4

and

At* = 0,

u - e

v= e

00B

‘byOic2-y 2) . . 4ay(* * -/ )006— -;----- r-r ---- K S i n

t f + f ) *

‘fap|(* 2-y 2)( x 8 + / ) 4

z w d ]8ini ^ V )

f o g | im «*(*»<>)-« (0 ,0 )

dx *-*<> x

r e-*~4- 0 v 1= lim ----- — = hm — “ — 3 ux4

[Using (1)]

lim-*-*o

= l im-*->o i 1 11H-----r + ----=- +

x 4 2 jc

= lim*->o

1

1 1X + — + ---- - +

JT

= 0

and du

- limdy y-*0

Now,dv

=limdx x-+0

and dv

=lim4y y-fO

~- j ~4= lim I f -----

*-*° y

lim—= 0x->0*

lim—= 0. y-O y

[Using (l)]

8/11/2019 DK Jain M3


12 I ENGINEERINGMMHEMAnoMtf

Hence Cauchy-Riemann equations are satisfied at 2 = 0.

Further , / ' (0 ) = l im ~ lim ----

z

e -- ; if z -> 0 along z ~ rei7t,A

e~r \ e= lim--- r*77- = «>

r_*®re^

.*. f \ z ) does not exist at z = 0 and hence f(z) is not analytic at z

a* a* . a* Example 1.8: Provelkat: + -* }£ £

Solution. Since z - x + i y and z = x - i y , then

x = ~ (z + z) and y = ~ ( z ~ z )2 2i

dx _ 1 dx _ 1 dy __ 1 ffy _ J_

dz ~ 2 ' dz ~ 2' dz~ 2i and d z ~ ~ 2 i '

Since, f = f (x ,y ) so that / = f(z, z ) .

ax Sz $y’dz 2 dx 2* 2^5x

Sf df dx | df dy _ i <y 1 a / iY a/ ( .a/~j

5? 3* 65 dy Sz 2 dx 2idy 2â * 3yJ

dzd? SsvcSjJ 4 \d * dy.

and

or

dzdz

d*f I

dzdz 4

Y s .eO — + 1—{dx dyt

2‘ d ^

{dx2*dy2 f-

d2 t d2 d2

dx2 dy2 dzdz

Example 1.9 ; Ifffz) is a regular function o f z, prove that

<£_ a

la * 2 + d y ' )

i/r*>i*=4i f w r .

- 0. Proved.

. . . ( 2 )

[by (1) and (2)]

Proved.

[RGPV June 2004J

8/11/2019 DK Jain M3


F u n c t i o n s o f C o m pl e x Va r i a b l e s ) 13

Solution. Recall that :

L.H.S. =

•2 + - -2 ~*Szdz' dx dyJ

' a2 a2

cbc2 dy*I/(*>!*

= 4-î-1 /(*) |J= 4 -£_(/(*)/<?))dzdS

= 4—dz &

m m

dzdz

oz

= 4f' (z) f'(z) = 4\ fX z) \2

= R.H.S.Example 1.10 : if a regular function o f z» prove that

I L J L Y 3* , + V ;

[v| 2p=2z]

[As differentiate w.r.t. £ ]

[v \ z \2= zz]

Proved.

Solution. Recall that:

L.H.S.

= 4-dx2 dy2 dzdz

d2i,dx dy J

I —i

logl/'(*)!

= 4 - ^ - -Mog|/'<*)dzdz [2

[logtfWX*)}]

- 2-^~=:[log/'(2) + log /'(*)] = 2 AA * )

= 0.

[ v | z | 2= :r* ]

[Since f \ z ) and / #(z) treated as constants w.r.t. to z] ~ R.H.S. Proved.

Example 1.11 : Show that a harmonic fun ction sa tisfies the formaI differential

d2uequation =u. ‘

ozoz

Solution. Recall th at:a 2

dx2 dy2 dzdz. . . (I)

.(>*<

d2u d2u _Since u is harmonic function => T T + T T “

• ox oy

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14 | En g s*«?w g H*t hc>mjbc*4II

( d2 s2 A

dx2+ dy2u = 0 => 4

<?u

dzSs= 0

&u

dzdz = 0.

[Using (1)]

Proved. Example 1.12 .* Ifffz) is an analytic function ofz, prove that

| i H / w i * - 2 i /r * ; i *.

Solution, Let /(z) = u+iv. Then R f{ z )~ u => i/=s

du2 _ 5u Now, differentiate, we get -r— = 2u—

6 cte dbc

a vAgain differentiating, we get — j- = 2

dxr

( d u ) 3 0*u

[ a j +lV

2..S

Similarly,3ya

= 2( d u *

<sy.+ u-

a2u

3y2...(2)

Adding (1) and (2), we get

*r2 By2 [ W

f 32u 52u

or a V a2u3

i ? - + ^ =2

Since /(z ) is analytic => u is harmonic =>

o x ) \ o y )

d2u ^d2u Q

a *2 + dy2 ”

...(3)

v 0u; aw an . dt> ' x ,

Also, Since / ( * ) * “aT * TC * aT+l<32? Kt GX OX

dll t

dx dx

or i rw f -(£ ) ’ * 0Hence, from (3) and (4), we get

aV aVdx2 + '*~>

a2or

s2 "N

{dx2* f t 2)

= 2| /'(z) I

! U/Tz) |2= 2 1/'(*) I2 Proved.

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< a

t»13: Prove that the function ffz)= xy+ iy is everywhere continuous but not analytic.

Given: f (z) = xy+iy => u + iv = xy+iy Hence u = xy and v = y.

du dv . du dv .Differentiate: - y - ^ - 1 . - 0.

du dv , du dv cleafly- * _ a n d - * - -

=> Cauchy-Riemann equations are not satisfied.=> ffz) is not analytic.Evidently, u and v both are polynomials in x and >>and so they are continuous everywhere.

Aas.

1.14; Determine whether ~ is analytic or not? [RGPV June 2003] X

Let f (z) = u + iv =>u + iu = — x

U+W~lT ily = x 2+y2 [Rationalize]


x - yU = — ------f> U = 2 2 x + y* x l + y

du (x2 + y2) . l - jc .2x y 2 - x 2 du _ - 2 xy

dx (x 2+ y2)2 (x2+ y2)2 ^ dy ( x 2 +y2)2

dv _ 2xy dv y 2- x 2

dx (x2+y2)2 8,1(1 dy {x2+y2)2

du dv du _ dv

Clearly' a T *

Fu n c t io n s o f C a a nA x Va r ia b l e s | 15

Thus C-R equations are satisfied. Also partial derivatives are continuousexcept at (0,0). Therefore, 1/z is analytic everywhere except at z = 0. Proved.

M3 METHODS FOR CONSTRUCTING Al* ANALYTIC FUNCTION

M ethod!; If u is given function then to find :

, dv j d v * , ' ' dv » — d x + — dy

dx dy

du .. du ,mH ? ~di y [By C-R equations]

= M d x + N dy ...(I)

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16 I EhGMBBUNG JtaMBAIttKS-111

M _ du duwhere ™ and ^ ~ ~ r ~

oy dx

dM d2u dN d2u Differentiate : - r “ = ” T T and — = zTT oy dy* dx dx2

U U, U U ASince u is harmonic r > r y + y 58 0

a7u dht

dx2 + dy‘

dM _ dN => dy ~ dx [By C-R equalions]

Hence equation (1) is exact differential equation.\ dv can be integrated to get v.

Method t i : MILNE'S THOMSON METHOD:Type (i) : To construct analytic function J{z) in terms of z, when real part is given by

the following formula:i

/ (* )» | i M z ’ 0)- i^(2 , 0 ))d z+C ...(1)

. , . du i / x du where $ (*»?) - — and ri (*»y) - ~r~

dx oy

Type (ii) : To construct analytic function f(z) in terms of z, when imaginary part v is given by the following formula,

where $ (*»y) “ and (*>y) ~ ~ -oy dx

Type (iii) : To construct analytic function f(z) in terms of z, where u - v is given :Let U ~ u - v then

(1+ i) f( z)= J [&(z, 0)]dz+Ct . ...(3)

. , , dXJ v dU where and f i (*»?)55

Type (iv) : To construct analytic function^ in terms <kz, when u + v is given :Let V = u + v, then

(l + i ) / (z )= J [^(z f0) + i 2(z,0)]dz+C ...(4)

j r . dV . , . dV where fS (x,y) - — and (*.30 = — .

oy ox

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t

1.15 .*Use Cauchy-Riemann equation to fin d v, where u = flo?y-y?. [RGPV Dec. 2001]

u = 3x*y - y*

Fu n c t io n s b n n o n n c * s ¥ a — im g s | 1 7

Here

Differentiating (1) partially w.r.t x and respectively.

We have — = = 4 (x>y)>sayvx

^ = l x 2 -3y 2=t2(x,y), say

Cfu &U a / n \Also ==*>y+v-oy; = u=> u is a harmonic function.

Now putting x » x, y * jCUft (2) and (3), we get

(z , 0) = 0, and# 1(ar, 0) = 3**i

Hence by Milne-Thomson method, we have

/ ( * ) =J [^(z,0)-i^2(zt0)]dz+c

...(2)

...(3)

[.*. Real part u is given]

or

or

f(z)= J -i3 z2dz+c= -12!*+c

u+iv = -4(x+iy)3+ci- ~*(x8 +3ixuy - 3x0? - 1i f ) +ci [v c=ic]

or u+iv = (3 x* y- f ) + #3 x^ -#* +c)

Equating real and imaginary parts, we have

u=3x2y - y 3, u=3j^2- x 3+ c. Ans.

Example 1.16 .*Show that u = ~ log(x* +y*) is harmonic and Jtod Us harmonic conjugate.

Soiuthn. Here

du , du and —

and

Clearly

dx x 2 + y2 ~ dy x2 + y 2

d*u y* - x 1 } d%u _ x* -y *

d x2 (x2+y2)2 an dy2 (r* + ya)*

S?u n= u ^ « is harmonic.

Let the hgqnonic conjugate of u be v. Then

, dv , dv , du j du dv = — dx +— ay = -dx dy dy

[By C - R equations]

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1 8 1 En q b« b M M 4 M !maiio »4II

or

Where

dv = —

Af = — * * + /

a w / - * *

x*+y*

and dN y*-x*

dx (x* + y2)*dy (x* + / ) *

dM _ diV _ Clearly ~ljy "Idx^* ®Quat*on (1) is an exact

Its solution is

- 1 ? + ? jao*at <3X+ J + C

{krwmof N uhkhan fm fivmx)

= -y ~ t a n ' ' ( ^ ) + 0 +

or i; = -ta n~ ,(^)+ c. Ans.

Example \M : Show that u - 2 x - x l +3 xy* fr harmonic and determine its conjugate.- [XGTV June 2002J

u-2x~x*Solution. Given

| ^ = 2 -3 * * +3 / anddr dx

- 6x

Then

~ = 6^y and ^ ~ = 6x

d*u d*ud ? + dy*~ ^ t t l s ™ur,nonic*

Let the harmonic conjugate of wbe v, then we know that

, dv * du ,dv = - - d x + —-dy

dx dydu . du , $

=~— ax+—-a y dy dx

dv~-6xy cb+(2-3x*+3;y2)dSy

du = Mdx+ AWy,(say)

or

[v B y C - R equation]

...d)

or

dM dN

8/11/2019 DK Jain M3


R w c m n tg t C o w w y m m » 119

Clearly * — => equation (1) is an exact.dy dx

/. Its solution v - | Mdx+ J (terms of N which are free from x)dy + c* * •-< jreouM

or

u = -6 y j xdx+J (2+3,y:i)cj(y + c

_ x* 0 3y3v=s"6^,Y +2;y+ 3 +c

or i/ = -3 b c * ,y + ^ y + d Ans.

f 1.11 x) ; Prove that u • x2 - y 2 - 2ap - 2x + 3y is harmonic. Find a function v such that

ffz) = u + tv is analytic. Also express f(z) in terms o f & fRGPV June 2005J » Given : u =x2- y 2-2ay-2x+3y

^DUaferetrtiate: ^ = 2^ _2x+3 - d )

i ? = 2’ ^ - ' 2

52u d2u

G N*V-

Adding, —r- = 0 => wis harmonic.dx2 dy

Further, let v be harmonic conjugate of u, then we know that

, dv , dv , d v ~ — dx + — ay

dx dy

du , du ,

= -{ - 2y-2x+3)dx+(2x-2y-2)dy

- M dx+ Ndy, (sty)Here, Af = 2 y + 2 x -3 and N - 2 x - 2 y - 2

[v By C-R equations]

[Using (1)]

...(2)

dM

dyand

So clearly, = dy dx

dN .

I T 2

equation (2) is an exact

Its solution,

- J (2y+2x-3)dx + J ( - 2y-2)d[y+c

= 2 xy + x? -3x-*at* - 2y+c • Ans.

8/11/2019 DK Jain M3


Finally, by Milne-Thomson method, analytic function

/(*) “ J 0) - i^(z , 0)]dz + c ...(3)

A

where, M x>?) = -r- = 2 x - 2 y - 2 => (z , 0) = 2z ~ 2 - "ox

and M x> = 2y-2x+S=> ^(z, 0) = - 2z + 3.

Hence (3) becomes:

f(z) ~ J [2z - 2 - i ( - 2z + 3)]<iz+c

= z 2 - 2z - i ( - z 2 + 3z) + c

= z 2 - 2 z + iz2 ~3iz + c= (l + i)*2 -2 z -3 iz + c. Ans.

Example 1.18 (b) : Show that the function u mx? - Sxy3 is harmonic and fin d the corresponding analytic function o f this as the real part [RGPV, Dec. 2011 J

Solution. Given u - x3-3xy*

d2u- 3x2- 3y* and —y ~ 6x,

dx

&U

= - 6xy and ~ - 6*.

2 0 1

Again, we /rove /o f in d f (z) = u + *v.By Milne Theorem, we have

/O ? )* J[î(^0) + /<J>2[2,0]]flfe+iC ... ( 1)

du dvHere <J>t (x, y) = ^ and ^ (*, y) *

We have

♦i (*..>') = | j “ = 3X2 - 3y2

Replacing x -> r and >>-> 0, we get*, (z, 0) = 322.

Again, we have• • #*•' i

dv

du

dx

du

dy

d2u

dy2

8/11/2019 DK Jain M3


t

FuNC|W .^ .q 9 W ^ J f |}S?|W5s, | 21

du du dv= [ : by C-R equation ^ = ~ ^ ]

” - [0 - 6 xy] = 6 xy Replacing x -* z and y 0, we get

♦i <*, 0) « 0.Hence (1) becomes

/(z) * J(3z2 +i.0)dz + iC = z* + #C. Ans.

jip/e 1.19 .* Finrf /A* im aginary part o f the ana lytic fu n ctio n whose real pa rt is

x? - 3 x f +3x* -3 y f . fRGPV Dec. 2005, 2008(N) and June 2011]

tion. Given : u - x * - 3 xy2 + 3x2-3 y 2

=> — = 3x2- 3 y 2 +6x and ~ = ~ 6 xy -S y.dx dy

t

Let v be the imaginary part of the function u.

, dv dv , du , du .'^c 'dy 'dy 'dx [By C-R equations]

= - ( - 6xy - 6 y )d x + (3x* - 3 y* + 6 x)dy

« ( 6 j ^ + 6y)<i3c+(3*2^ 3y2 + 6x)dy = M dx+ Ndy ...(1)

»• dM _d N L is an exact differential equation because ^T

.*. Its'solution :

l>= J (6ry + 6y)<2x + J (terms in N which do not contain x)dy + c y const.

( 2 > X ----+ X

<2 J - 6 y J(x + l)d* + J - 3 y 2dy+c = 6y

3 y3-£ - + c

3

= 3x2y+6jo r~<y3+c. Ans*•J

1.20 .* I f u ~ x ? - y * b o t h u and v sa tisfy Laplace's equation but

lVi u + tv is not an analytic function o f z>Or

y •<•<: Prove that -y * and v = ^ +y t ^ torm onkfunctions ofxan dy but are not

harmonic conjugates.

\

8/11/2019 DK Jain M3


Solution, Let u - x 2 - y 1

22

du _ d2u * du ~ d*u nDifferentiate: ^ - = 2 * ^ — = 2 m i — ‘ - Z y => ^ = - 2

d 2u 3 2w ,,T T + T T = z - z = u => t tjs harmonicSbr 3>r

/.e.„ u(x, y) satisfies Laplace's equation.

Again let, * * x 2 +y*

dv -2 xy d2v 6x *y -2 y3Differentiate: ( ^ 7 7 7 =>

5u x 2 - y 2 d2v - 6 x 2y+2y*

8 y ~ (x 2+y2)* ^ dy2= (*’ V ) 3

j + T 7 = ® =* vis harmonicaJu a2u

i ? +î.e., v satisfies Laplace's equation.

du dv dv du

But* Imd

Therefore « and v are not harmonic conjugates. Henoe ar +iv is not an analytic function. Proved.

Example \2 \ : Show that the function e* (coa y + i t i n y ) is holomorphic (Le. analytic) and fin d

its derivative. fltiGPV, June 2008]

Solution. Let, f(z) = ex co sy+ iex sia y => u + it; * e* co sy + iexs iny

Then, u « ex cosy, v ~ e x sin y

Differentiate:

du dv , . du x . dv x — = excoa.y, —- = e erny and *£r~~e “ “ y* t ~ = e cosydx dx oy oy

du _ d v du _ dv

Clearly, and * = " * •Hence, C -R equations are satisfied. Therefore f(z) is analytic.

Now, we know that, / #<*) * ~ + ^ * « * c o s y + i V 8in ySbe dx

* eJf(cosy + iBiny) * e*.e* = ex+ = e*. Ans.

Example 1.22 .*Sfcw tta* the function f (z )=z\ *\ is not analytic anywhere.

Solution. We have, f(z) * (x +ty) \x + iy \ [v z = x + iy]

8/11/2019 DK Jain M3


f i s t in g real and imaginary parts, then

; ■' u = xyjx2 + y2, and v * y ^ x 2 +y2.

F u n c t i o n s or Co»fe*£X V&s a b l Es | 23

du x 2 r ~ ~ i xy

and

Differentiate: + +Jx‘+y‘’ a n d ^ "

du _ .*y . a ; y2

dx dy J x 2+y yjx +y2=> x - y .

du dvWhich is satisfied only for x * y and ^ * ~" T*

Bk f; Hence C -R equations are not satisfied.

/fa) is not analytic everywhere. Proved.

: Shaw that the Junction v m sin kx cosy is harmonic and fin d its harmonic conjugate.

jfc Gives: v = ainhx.cosy ...(1)

^ dv , tfu . ,Differentiate: — = coehx cosy => ^ - j = sinhx cosy ...(2)

and ~ = s-s in hxsin y => - sinhx cosy ...(3)

d2v &v

dx2 + dy:

. du . du , dv . dv .We know that ®u ~ — a x +— ay - [v B y C -R

or du = (-sinhx sin y)dif -(coshx cosy)dfy = Mdx + Ndy, say ...(4)

Here, M = - (sinhx siny) and N - - (coshx cosy)

dM . . dN

... r—- = -sin h x cosy ^ - sinhx cosy**» cfe

dM _d N So clearly => equation (4) is an exact.

8/11/2019 DK Jain M3


M | C u n m — i - w J U

.*. Its solution:

u = J Mdx + J (terms of 2V which are free from x)dx + c jC&W t.

= ~ s in y j s in h x d r+J Odx+c

= -ainy coshx+c. Ans.

Example 1.24 ; Show that ex( x c o a y -y sin y) is harmonic function. Find the anaiytic'function

fo r which ex(x cosy - y s iny) is imaginary part [RGPV June, 2004]

Or Show* that ike function e* (x cosy —y siny) is harm&nlc and fin d its conjugate.

[RGPV, Feb. 2010!

Solution. Given imaginary part, v ^ e x(x oos y-y s iny) . . ■•( 1)

Differentiate: “ - ax(xcosy - ysin y)+ ex(cosy) ...(2)

i

and — = e*(- xsiny-einy-ycosy) (3>

Again differentiate (2) and (3), we get

^4 r * e*(zcosy - y siny)+ e*cosy+ ex cosydx2

a-ex(-x cosy - cosy - cosy + y siny)

V V UV AT T + T T = u v is harmonic function.

= e*(jccos<y--y8iny+2oosy) ...(4)

8y‘

- ex (-x cosy -2 cosy + y siny) ...(5)

Adding equations (4) and (5), we get

d2v &v

dx2 dy2

Now, MilneThomson. method, If vis given:

A * ) 5* / W ( * ,0 ) + i ^ ( a r ,0 ) } d e + c ...(6 )

where M x »y) “ t - “ ex(*o os y-y siny )+ eJC(cosy)1ox

Then ^(* ,0) = e*(z - 0)+«*(!) * ex(z +1).

dvand M**y) m -rr= e*(-xsiny -siny - ycosy),

dy

8/11/2019 DK Jain M3


!

jfe then 4(* ,0) = e* (0 -0 -0 )* 0 .

Hence (6) becomes: /(*) 35 J *(* + Oe*dz+c

= i£(* + l)e* - J e*darj+c = i[(2 + l)e* - e* ] + c

- i z e 9 +c. Ans.

0 5 : Determine the analytic function f(x) = u + iv, If

v^ log b f +y*)+x-fy .

Given imaginary part, v = log(*2 + jy2) + x -2 y .

dv 1 s .Differentiate: ^ = ”i"” T ^ + 1 --(0

1 /n V „and ~T~ — f7 T '(2 y )-2 , ...(2)

qy r + y

By M:foe Than son method, if v is given:

/ ( * ) = / W(^,0) + i^(z,0)]d2r + c ...(3)

. z v 2 y 0where s ^ ~ ^ ^ (z»0) * -2.

Fw cH ft—r t » I t iw W f iH M ii iM | a t

and

Hence (3) becomes : /(* )* J dz+c

= - 2* + i (2 log 2 + z) + c

=±*21ogs£+(*-2)js+c. Ans.

Example 1.26; T/' u = ----------------------, /bwf the corresponding analytic function.

coah2y+coa2x■* fRGPV Dec, 2003/ Solution. Here, given real part of/fo) is , ,

sin2xu~-

cos/i2y+ c o s2 jc

du _ (coe/i2y 4-co82x)>2co62x - s in 2x (-2 sin 2s)Differentiate. (cosA2^+ooe2x)*

2cos2jcco8/i2y+2(co83 2 jc + sin2 2x)3 — . - - - — - - J - — — ----------------

(cosft2y+ cos2x)

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26 I FrmngTnwini MinM~iiivnrfii IH

2 cos2xcosft2y +2 _ x ( \

(coah2y +cos2x)2 1 X*^ ’ ^

dtt_ (co8ft2y+coa2x)(0) - 8in 2x(2gin/t2y)

31x1 c>y (cos^2y+cos2x)2

28in2xsin/i2y~ 7I7Z 7^r~^~ ^77^2-r2\x>y)> say ...(2)(cos^2y+cos2x)

Now putting x = z, y '» 0 in (!) and (2), we get

2cos2sr +2 2 2 2------------------------- ------ - —=sec ,(l+cos2z) l + cos2if 2cos 2

and fa(z , 0)=0.

Hence by Milne-Thomson method, if u is given :

f( z)= J [<}>j (z,0)-«|>j (z,Q)]dz+ci= J sec2 zdz+ic=t8inz+ic. Ans.

y Example 1.27: In a two-dimensionalfluid flow , the stream function is V = ~^ j ~ y » finding the

velocity potential f .

Solution. Given: y/= — ——- ...(1) x 2 +y2

_ ^ 2rv y 2 - x 2Differentiate: "aT= 7 1 ---- ITT

cb (x + y ) & < * + y )

We know that, cty = — dx+~d[y=— dx - — d|ydr §y dy dx

by C -R equations— and— = - — d* $y dy cbc

Of equation (2) is an exact

O'8-**)Its solution / ^ + j taj2°ê+ | Oc^y+c

8/11/2019 DK Jain M3


,.28 s i f w ~ f + i\k represents the com plex poten tial fo r an electric fie ld and

T * x * fin d the function o f +

Here, imaginaiy part of w, “ >* +^ T ~ T

Differentiate: ^ « - 2 y - ; - f rr *r t <*»*). say§y (x2+y2)2

| 2 7

. . 0 )

,x},k:ag' - (s 2 +y*)fl)-*(2s)

6 bc (x2 + y2)2

Now putting x - z , y = 0 we get A (*»0)«0, an d 4%(z,0)-2z ~ ~ Z

1

Hence by Milne - Thomson metho djf yf is imaginary, then

/(*)= J W<A0)+ i#2 (z,0)ck+c

= J [0+ i(2z — ~ ) ] d z +c=i j 2 z — y j c £ z + c = ^ z 2+ —

„ (x+iy)2+—-^ -|+ c = i x2 - y 2 +2i ^ + - ^ y \+cL x+ iy] I x + y J

vEquating real parts, we get # = - 2*y+>

+ c ‘

+ c

Ans.

^ . I 29 ; Mow that the fun ction u = e~**8in (x* - y * ) is harmonic. Find the

~ conjugate function v and express u + iv as an analytic function o f z,[RGFV Dec. 2006 and June 2007]

to u ,vHere, u = e “2ao's in (x*-y*)

; Differentiate: ~ * - 2 y e '1* ain(x2 - y 2)+2xe ~2* coe(x2 - y2)ox

-~-r = 4yae~** sin(x* - y2) - 4 xye~** cos(x* - y i )+2e~**f cos(x2 - y 2)

oar -4jg'e~2v co8(x*-y* )-4x, e~**ain(x2 -y 2)

= - 2x « '2j5f sin(x* - y2) - 2ye~i*y cos(x* -y 2).3y

...(1)

\

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^ ( * , y > » ^ - » ( ^ + 4 a g i + / ) + ( * - y ) ( 2 * + 4 y ) [by (I)]

=> ^ ( z yO)-z +2z =3z [':x-*zandy-*Q \

dU and *2 (*>?)=— = - ( * 2 + 4ji?y+ jy2)+(x - y)(4x + 2y)

ys => 4 (s ,0 )= -z 2 +4z2 = 3z2. [vx-»zatt<i;y->0]

Hence (2) becomes:

(1+ i)f(z )= j [îZfOy-ifaiz, 0)]dz+c= J (3*2 -<tiz2)dz+c=(l-i)z3+ c

' ■_r' ' / / \ lj**”l 3 C

r t * ) = f - r * + r - r 1+ 1 l+ l

‘nifi ; , c , 1- i .or f(z) = -is r +C,, where c* and ^ - r = -» . Ans.

e ^ - c o s j c + s m xis 1 3 1 : / / W “ « + & an analytic function o f z - x + iy, and **-v - c o a h y - c o a x ' *

| 2 9

flndf(z) subject to the condition # ¥ •

y 7 e^-cosx+sinx

“ * ^ “ ~ U' coshy-coax ’ " ( l)

By Milne’s - Thomson method, if i/ ~ - v is given, then

(l + i)/ (* )= J 0,0)] dz+c ...(2)

, / x SU , / v 517where ^ ^

. , . 3 1 / (sinx+©08x)(oosh)y-co8x)-(ey-cosx+sinxXsinx)

Now, ----------- ------------- (coBhyleos*? ------------------------ '

Put x-frzandyÔ, then

(sinz+cos z)(l-cosz)-(l-cos2r+sinz)8in 2r

(1-COS2T)2

sin z-s inz cosz + cosz-cos2z-s inz+ sinz cosz-s in2z

(1 - cos*)2

coez-(cos22+sin2 z) cosz-l -122 ............... mm f — =?' » — —

(1-COBZ)2 (l-c o sz )2 1-C 0S2

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9 0 1riwn—iMM iUnaiiimrniiil

^ / „ _ \ dU ey(ooBhy-Q98 x)-(ey -cosx+ainaOsinhy.and ti v^y/*”^*”0* 11 ‘/ v \2 • 1 1 - ’

dy (coshy-cosx)

Put *->2randy->0, then

(1-cos*) 1(I-cosa?) l-cos*

Hence (2) becomes:

(1+ i)/w ' / [ o i - ' a ^ j ] ^ ^

=-(1+i)J a ^ )+c""Pr)/co*<c,(f)*+c*(1+i)cot(f)+<:[v (l-c o s0 )= 2 sinJ(0/2)]

=> /(* )= co tf |J+ where ...(3)

But given ^ f y ] =~ 2^ ’ **len 0 ) becomes at *= -j.

3 -i .ff 3- i t 3~i - 1 -i — - e o t ^ c , ^ — - l+ e , = > c ,= — - 1 = — .

Hence (3) becomes: /(* )* cotf—j+ —j - . Ans.

Example 132 : iff(z) mu + tv is an analytic function o f z, findf(z) ( f

u - v - e * (c o s y-s ln y ).

Solution. Let £/=u-i> =ex(cosy~8in y) —(1)

By Milne's - Thomson method, \ f U ~ u - v is given, then

<i+i)/(*)=J U M y i M z M d z + c , ‘ ...(2)

where ****

Now, #i <*,*)»—“««* (coey-s iny).ox

Put * -> z and y-»0, then $(z ,0) *e* (l-0)= e*.

dU and h (x >y) = — * e *(-« n y-c oe y)

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Fu n c t jJMs W C6WH& VWSfitts I 31

Put and ;y-»0, then ^ ( 2,0)= e* (0- 1) = -e*.

Hence (2) becomes : (l+i)f(z)= J [V -i(-e*j^dz+c

k or (1 + i) f{ z) = (l + t)e*+c

f(z) = e* +-^-r - e +cr Ans.- 1 + t

^jfiit 133 .* Ifffz) m u + iv is an analytic function o f z, flnd ffz) i f

2a&n2xu + v =

(e* + e ^ -2cos2x)

T/ 2sin2jif Letimaginaiypart, v - “ + t' - (ej , + e-2, _ 2cog2x)

or V - , . 8in2* ' ...(1)e2' + e"3' cosh2y - c o s 2x

-C082JCV 2

By Milne - Thomson method, \ fV = u + v is given, then

(1 + i)f(z) = J [^(z.O) + ifc(*,0)Jdz +c ...(2)

5 V 5 Vwhere A(x >y) ~ and

, , v dV -28w2xsiiih 2y

Now, M x .y ) - a y - (c08h 2 j,_eos2 x)2 => * (2' 0>= °- [v 8111110 = °1

. . . d V 2co82x(co8h 2y - c o s2x )“ 28in2 2*= — = ----------- -- -------------

(cosh2y-cos 2x)

. , „ 2 cos22(1- cos2^) -2 si»8 22

A ( 2 ' 0 ) = ------------

( I - c m W --------

“

_ 2oo82z - 2în>2g+coe*2s) 1

(1-cos22)* 1

2 (cos22- 1) - 2 - 2

(l- co s2 z )2 (l-*cos2z) 2s in 22 00860 **

Hence (2) becomes r 0 + 0 /(*) = J [0+ i(-cose c2z)Jck + c = i cot z + c

i c tor /(*) = -—rCotz+-— 7*-(l-£)cotz+cl. Ans.

1 + t l + i 2

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m

Example 134 / Let f ( z ) ~ u ( r , 0) + iu (r, $) be am analytic function, whete u ~ -r fism 38. Con

struct the corresponding analytic function ') in terms o f z.

Solution. Here u = - r 8 sin 30 => — = - 3r* sin 30, and ~ = -3 rs cos 3#dr 58

By total differentiation formula:

dv = ~ d r + ~ d e = ( “ + (r f : ) cW [By C-R equations]

or di> = 3r2cos30 d r - 3 r 8 sin 30 dS

or d v - (3r2dr)cos30 + r3{-sin30.(3d0)} = d (r 3 cos 30)

On integrating, v = r 3 cos30+c.

f (z ) = u + it; = - r 3 sin 30 + ir3cos 30+ic

= ir3(cos30+ i sin 3 0)+ ic ir 8 8**+ ic = iz3 + ic. [v z = re**] Ans.

Exam ple 1.35 : i f n is real, M«f u> = r* (c osnB+ ts in nO ) is analytic except

possibly when r —0.

Solution. Given: w = r*[cosn0 + isin n0]« u + iv

=> u(r,^ ) = r* cosnl, and v(r,0) = r* sinn0 .

du «_i ^ dv . j .Differentiate: — = n rR ‘costttf;-—= nr* sinn0,

<t t or

and t t - " n r^ sin n ^ ;-rr * w**cosn^.a# a*

. . du 1 „ _ i a u . 1 » • /i 1 duCtariy ¥ = 7 nr « osn6 = ~ . and F = ; nr « . n f i .— .

Thus Cauchy-Riemann equations (polar form) are satisfied.Hence w = rn (cos 0+ i sin nQ is analytic except r = 0. Proved.

1.14 BILINEAR TRANSFORMATION

02 + 6The transformation : w = f(z) -

------ —, with ad - 6c # 0 ...(1)cz+a

is called bilinear transformation or MobiuS transformation, where a, b, c, d arecomplex constants.

Remarks :

(i) The transformation (1) is said to be normalized if ad -b e = I.(ii) In bilinear transformation (1), the (ad - be) is called determination of the transformation.

aar +6 -dw + b v .(iii) If w = ----- then w(cz+ d) = az+b =0- z ~ -------------- * / say is known as in-

cz + a cw — a

verse transformation.

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u~-r^sh%SSLU

.

Sr* cos 3#

Cor rfiu a d - b e]point : Differentiating (1) with respect to 2, we get “ j“ * ^ + 2 ■

d . dw du) . — >then ~r~ = 00 and if z * <», then t - = 0.

c dz dz

F u n c t i o n s o p Comkex Va r ia b l e s | 33

[By C-R equations]

0830)

* r d 0 '

Proved.

#*™ c, are

**y is known as in-

'j*

1*

?=? — and z~<x> are the critical points.C u

or Fixed points: The invariant points or fixed points o f a transformation (1) potting w - z .

: M>= — ,..(l) is a transformation,

fixed points, put w * z in (1), we get

—^ V a -1 =>z * ±i Fixed points are z « -* and /.£

rotation, magnification and inversion are special type o fbilinear transformations.

-Ratio : If there are four points z ]tz2iz3 and z4 taken in order, then the ratio

Wr J—, ” *4) - II J , -

cfsss-ratios are invariant under a bilinear transformation; i.e.

■ WK'W.j:’ - u ; 4) (g, - * 2)(23-Z4)

(w2 ~w 3)(w4 ^tu, ) (*2 - z 3)(z4 - z x)

We may use (1) to find a particular bilinear transformation which transforms one giventilfcle into another given circle or straight line. Thus to obtain the required transformation,

We fake any three distinct points z i, z2,z3 on the first given circle and any three distinct

inti wuw2,w3 on the second given circle or straight line and substitute them in (1)..*Find the bilinear transformation which transform the pbitHs z m 2,1, 0 into w « /,

A I respectively. [RGPV 2001/Ue< required bilinear transformation be :

(wL~w2)(w3-w 4)h (zl - z 2)(zi - z 4)

^ t o f 2^ w 3)(w4 -m , ) (z2- z 3)(z4 -* , )

*Nrting m/j = 1, tt?2 = 0, w3- i ,w 4 - iv, z, = 2,z2 = 1, z3 - 0, z4 =z, we get

...(1)

( 0 - i ) ( w - \ ) ~ ( ) -< h( z - 2 )'

=> zu>-iz-2w + 2i = -ziw + zi

=> to)z~$w+wz**i*-2i+zi

=> w( iz- 2 + z) = 2iz -2i

2t ( * - l )

w - i

i ( w - 1)

~ z

z - 2

w =( l + 0 * ^ 2

, which is a bilinear transformation. Ans.

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z =-3±^/9^H -3±1

Fu n c t i o n s o p C owi if cxV aB aiaa* jf 55

2i 2it ' *

- 2 - 1

=> z = -T- and 9 ~~J~ i-e >z ~ 2<# and / are fixed points. Ans.139 ; Find the bilinear transformation which maps the points » = <*>, z=i, z - 0 on to

the points wÔ, ip*4 and ic=oa (RGPV June 2004 am i June 2007]

Let tiie required bilinear transformation b e :

( t ^ - w 2)(w3 ~w 4) (zl - z 2)(z3- z A)

(w2- w i)(w4 -it?,) (z2 - * 3) ^ 4 - z {)

Putting, zl =<*>, z t = if zs = 0, z4 = z , and

u\ - 0, k>* « i, » 00, w;4 = u;, we get(0-t)(qo-i6>) (oo-i)(0-;g)

(i-oo)(u>-0) (i-O)^-oo)

ao[v Remove the factors, which containing » i.e., in—form]

00

- (H)- (B)

-£ - z -2=> - i = - z ww i

=> n> = - —.which is the required bilinear transformation. Ans.

1.40; Find the bitintar tnwsformatioft which maps z - 1 , 1,-1 re^tectivefy on to w ~i,0 ,~L[RGPV Dec. 2005, Feb. 2010 and Dec. 2011]

Also fin d : (i) the image o f |z| <1,

(U) the invatfant points o f this transformation. Let the required bilinear transformation be :

(Wj ~W2)(W3 ~W4) (Zy ~ z2)( zy - z A)

(f v2- w 3)(w4 -u>,) ~ (z2 ~ z 3)(z4 - z x)

Putting, z} - 1, z 2 = i, z3 = -1, \z4 - z, and

wx ~i, wt = 0, wn = - it w4 = wt we get

( i - O K - i - w ) _ ( l ~ i )( - l- z )

(0+i)(w-i) ( i+ l ) (z- l )

- (w+i) ^ ( t - 1) (z +1)

^ (w~i) (»+1) ( z - i )

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36 | SMQMKMWQItoB M IIG lJI

~(w + i) i( z +1)

( w - i ) ( z - i )

~(w+i) ( z -1) = (i z+0(w - 0

- (wz + i$ -w - i ) = izw ~ i2z + iw ~ i2

-w z - i z+ w + i = wtv+ 2 + iu; 1 1.

-iu2-i2u>+u;-iu;*=*+l+i3”ri

- w z + w - i z w - iw = z ( 1 + 0+ (l - 0

u ?( -2 + l-« -0 = ^ ( i + l )+ ( l -0 => u>=2 (“i -!)+ (!-» )

; .( t - D _ ( t - i ) > -& .1

i + 1 i* - i2 -2 J

or

8 )lt> = -

- 2 +

Z - l

- z - i

1- i (1 -0 2 ~2i

‘ 1+ i ” 1 - i2 2

or 1 -2

w - - — , which is the required bilinear transformation.i+z

(i) Image q f\z\<\: From (1), solve for z, we get - w z ~ i w = z - i

~«wz - z ^ i w - i => z *•iw - i i - iw

~w -1 MJ+1

Since given image in Z - plane is jzjcl

i(l -w )

...(I)

...(2)

1+ w<1

|i(l-t^)|<|l + t^ => (l-u -w j-cj l+u+ u^

7(1 ~u)2 +u2 < ^(l + ii)2 +u2' *

l + w8 - 2u + v2 < l+ u a +2u+va

0<4u u>0.

[Using (2)]

[•i*MJ=u+iu and |if=l]

jv |x + iy|=</ j x2 + y 2

This shows that the interior of the circle x 2 + y 2 =1 in the 2-plane is mapped onto the

entire half of the w-plane to the right of the imaginary axis. Ans.

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!

Functions g* J 37

(ii) To find the invariant points : Put w=z in (1), we get

z = -—- => 2 = —^ z - i z 2~ l- i z= 0 => iz* + ( i- l)z + l=0i + z 1 - i z

f t

2* 2 t 2 4

These are the required invariant points or fixed points. Ans.

2z+ 31.41: Show that the transformation to - — — maps the circle i t + 3? -*X = 0 onto the

straight Une 4u+ 3= 0, and explain why the curve obtained is not a circle,

(RGPV, June 2002, Dec. 2003, June 2005 and June 2011J

Given ...(1) z - 4

Solve above for z, we get wz-4w^2z+3 z - ...(2)w - 2

Now the given equation of the circle x* +y* - Ax * 0 in 2-plane can be written as

2 2 —2(2 4-2 )=0 ...(3)

jêinces - x + iy ,z= x~ iy , x = —* Z and z z = x 2 + y2j

Substituting for z and z from equation (2) in equation (3), we get

=04u>+3^ ( 4w +3A ~{4w+3 4^+3^

2 ----------+■v + 3 ) ( V . - 2 ) \ w - 2 J 4w - 2 ) \ w - 2 J ^ w - 2 w - 2 j

=> \6ww + I2w+I21v + 9-2(4wuJ + 3w -%w-6 + 4ww+3w-%w-6) =0

=> 22(u> + «;)+33=*0 => 44u + 33 =0 [vw=u + iuand w=u~iv]

=> 4u + 3=0,

which is equation of straight line in w-plane i.e., not a circle. Ans.

1.42 .*Find the bilinear transformation which maps the points z m0, - / , I onto w=i, 4 <xxv Also, fin d the image o f the unit circle fzj * /•olution. Let the required bilinear transtbnp}tion be

(W] - w 2)(w3~ w a) (*1 - ^ 2X^3 ~zi )

r (U>2-W{)(WA- W X) (z2 - 23X*4 -* | )

Putting, 2 j = 0, 22 = -I, 23 = i, z4 as2, and

u>, ss j, =0, u>3 = oo, w4 = u;, we get

(i-0)(oo-u>) (0 + l) (i“*2)(0-oo)(w;-t) (-1 -i )( 2 -0 )

(co-u >) *1s ince ------- => -1(O-oo) J

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7— => ( z i + z ) i = w ( i - z ) - i ( i - z ) ( w - i ) -2 (i + l)

=> - z + zi~wi~wz+l+iz => u;z - u;i = z +1

2 + 1= -------- z - Iwhich is the required bilinear transformation.

ii£/+1Solve above (1) for z, we get z -

Now, since in z-pJane, |z |= l

iw + 1

w - 1...(2)

=1

=>

=>

:=>

=>

[Using (2)J

[yu; =u+ty=>u7= u-ii;]

[vl a + ib \~ â2 +b2]

w - 1

I + iw H w - 1 1

II + i(u + iv) j=| u+ iv - 11

| l - u + i t t | « | u - l + i i ;|

(1-u)2 + u 2= (u -l) 2+u2

1+ u2 - 2 y+w2 =»u2+1-2u+i;2

u - v - 0 or i.e.» straight line in vv-plane.

Hence the image of unit circle |z |= l in z~p!ane is a straight line making an angle */A to u- axis and passing through origin in w-plane. A as.

S - 4 x Example Show that the transformation w -

4 z - 2radius unity In w-plane and find the centre o f the circle.

5 - 4 z

transform the circle \z =1 into a circle of

Solution* Given: w -4 2 -2

=> Solve above for z, then z =2w + 5

4o>+4Since inz-plane, |z |= 1=> zz = 1

f 2w + 5 Y 2ElT+ . ' '

^ 1,410 + 4 y v + 4 J=> 4ww + 25 + \0(w + ttF) = 16(ww +1 + + 57)

=> 4(u2 + u2) + 25 + 20u = \6[u2 + u2 +1 + 2tt]

2 2 3

u +v + « - —= 0 , which is equation of circle in w-plane.

[Using (1)]

[v w = u + iv)

...(2)

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g m *

* Comparing this with equation of circle: u2 +u2 +2 £u'+2 /v+c = 0

=> g = 1/2, / a 0 and c = - 3/4

Centre = ( - £ ,- / ) = (- 1 /2 ,0 ), and Radius = V#2 + f 2 ~ c = ^ “ +0 + ~‘ = 1-

Hence (2) represents circle with its centre at (-1/2,0) and of radius unity in w-plane. Ans.

az+b \ M : Find the condition that the transformation w -------— transforms a straight Une o f

cz+d the z-plane into 0te unit circle o f the w-plane►

Or

Find the condition that the transformation w = make the circle |ic|= 1 cow -

to a straight line in the z-plane,We know that the equation of a straight line in the z-plane is

F u n c t io n s o p C o * p t£ x Va w w u ® | 3 9

z - z 2= 1

The equation of the unit circle in w-plane is | w |= 1

az + b a z + (bJa)>——,---- ----- -The given transformation is w = wcz + d c z+ (d lc )

Now, the transformation (3) transforms the unit circle (2) into the curvea

c

-.(I)

...(2)

...(3)

l- |u>i= z + (bta) —s z+(jbla) c

z + {d1c) z + (dfc) a

1=

[v zx = -bla, 2 j =d/a]

Which is the required condition.Example 1.45 .*Find bilinear transformation whose fixed points are 1 and 2. olution. Let the required bilinear transformation be

az + b 1w -

cz+d

[Using (1)]

Ans.

...(1)

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\

Since, z - 1 and z - 2 are fixed points, then

( z - \ ) ( z - 2 ) = 0 => z 2 - } z +2 = 0

Comparing (2) and (3), we get

d _ d

— b

------ --3 and — - - 2 => b - -2c and d = a -3 cc c

az - 2cHence the transformation (1) becomes : w - - ------- r~-

' cz + a -3 c

In particular, taking a = 1, c - - 1, then bilinear transformation

z + 2

...(3)

w =4 - z

Ans.

1.15 CONFORMAL TRANSFORMATION

Let Ci and Ci be two curves in the z-plane intersect at the point P and the corresponding curves

C, , C2 in the w-plane intersect at point P' under transformation w - f (z ). If the angle of

intersection of the curves at point P is the same as the angle of intersection of the curves at pointP ' , both in magnitude and sense, then the transformation is said to be conformed at point P.

Definition. A transformation, which preserves angles both in magnitude and sense between every pair o f curves through a point, is said to be conformed at the point.

1.16 SOME STANDARD TRANSFORMATIONS

(1) Translation : w ^ z + c , where c is a complex constant, i.e., c =a+ib Let z - x + iy and U) —U + , then w —z+c becomes z

u + iu= (x+ iy) + (a +jb) - ( x + a) + i(y + 6)

=> u —x+ a and v —y + b

This transformation represents a simple translation of the axes. Accordingly, the transformation maps a curve in die z-plane into a curve in the w-plane of the same shape and size.

(2) Rotation and Magnification : w = cz, where c is complex constant.

Let x= rcos 0, y= rsin 0, u = R c qb#, v - R s i n ^

then z —X +iy - r e l0 and w=u + iv

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let, c=peia, then the transformation w ~cz becomes :

R 8 ^ * /7 e w.re ,<? => Re* « pre*0+a)

=> R = p r a n d ^ = ^ + a .i Thus the transformation maps a point, />(r,0) in the z-plane into a point p ' (pr,0+a) in the w-

«, plane. Hence the transformation consists of magnification of the radius vector of P by p~\ c | and

^ + rotation through and angle a - amp (c).

10 ) Inversion : w = — v ' z

Let z =re10 and w =Re‘*

then the transformation w =— becomes : Re* * - e * => R * — ^ - - 9 . z r r

g 1Thus under the transformation w; , a point P(r, 0) in 2-plane is mapped into the point

z

Fu n c t io n s o f C o m k j e x Va r ia b l e s ( 4 1

in w~] plane.

Example 1.46: Under the transformation w=—t find the image o f \z-2i\= 2.

[RGPVDec. 2006]Solution. The given transformation is

1 1 1 u - ivw * - or or x + i y - ----- —= -^— f-

z u; u + i u u +v

[v z - x + iy and w = u +iv JEquating real and imaginary parts, we get

u - v x s s~ — 2 and y = ~~ 2— 2 - 0 )

u +v i r + i r

The given curve is \ z - 2 i \~ 2 or |x + i(y-2)|=*2or x * + 0 ' -2)2 =4 or x 2+ y2- 4 y - 0 ■♦•(2)

ri': 1 which is a circle in the jr-plane with center (0, 2) and radius 2.Substituting the values of x and y from (1) ki equation (2), we get

(u2 +V2)* (u2 + 1 )* u*+v

ua +u* 4t> A 1 4t» A=> 7-3----- — r * 0 = > - y — j-+ —

(i r+ ir ) a +i> u +u u +u

l+ 4t> =0, a straight line which is the required image of the given curvein w-plane. Ans.

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Example 1.47 ; Find the image o f the infinite strip under One transformation e c - ~ .4 X z

Also show the regions graphically. Solution. The given transformation is :

[RGPV Dec. 2002, Dec. 2008(N)J

1 1w or * = —z w

or 1

u + iv u2 +v2[v z - x + iy and w - u +u>]


x =u - v

2 2 y “ 2 2W +ITU +V

1 - v 1

When J '* - , then ---- r = T4 u + ir 4

=> u2 +i>2+4v = 0 => i*2+(t;+2)2 =4,which is the equation of circle in w-plane whose center is (0, - 2) and radius 2.

When1 - v 1

^ then —— 7 = ”2 i r + i r 2

=> u2+i>2+2v = 0 => u2 +(i>+l)2 =1,which is the equation of circle in w-plane whose center is (0, - 1) and radius 1.Graphs:

u

w-plane

Hence the infinite strip i £ y $, i is transformed into the region between the two circles

: u2 + (u + 2)2 = 4 . center( 0, - 2) radius 2

and u2 + (y + 1)2 ■ 1, center (0, -1 ) radius 1. Ans.

x ~ i Example 1.48^ Show that unfter the transformation w - — the real axis in z-plane is mapped

4H*I

into the circle J 1. What portion o f the z~p!ane corresponds to the interior o f the

circle?

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0M. Given: “' = 7 7 7 •••(«)* TlSince, given the real axis in z-plane (Le., y - 0) i.e., z - x transforms into (1),

x - i (x - i )2 x 1 - l - 2ix x2- l . ( - 2x)we get w * -------------------------- =------- 5 —r + l“ 2 —T-x + 1 x2 +1 x +1 x +1 x +1

Fu n c t io n s o p g b # & r V A a ta u s t 4 3

IW 1 => | W |**1.

Hence the real axis in z-plane transforms into the circle Iw |= 1. Proved.

Further, put z * / in (1), we get w * 0. Hence the half z-plane above the real axis corre

sponds to the interior of the circle I u; 1*1. Ans.

1Vj&mmple 1.49; Show that under the transformation » ■ -» circle ^ - 6 x = 0 Is transformed

([ i... into a straight line in w-plane.

Solution. Given: w = — /.e., z = — ...(1)w

Again, given in z-plane: x 2 + y2 - 6x = 0

or g - * f e ± I > , 0 z * = x 2 + y2andx = y

or zz -3 ( z+ z ) ~ 0

1 1 , [ 1 t l _ => — - 3 —+ r r = 0 [from (1)]

W W lw w j

=> 1 - 3(w + i^ ) = 0

=> 1- 3(u+ iu +u - iv) = 0 [v u; = u - iu]

=> 1 - 6 u = 0 .Which is equation of straight line in w-plane. Proved.

Example 1.50; Show that the transformation pj = x* maps the circle \ z - l \ = l into the cardioids

p = 2(1+ cou<p), where pet* in the w-plane.

Solution. Given: w - z 2 ...(1)

put x = rcos0 , y ^ r s in Q in z~ x+ iy= > z~ rew

and u = pcos^ , y » p sin ^ in w ~u + iv = pe* .

Hence (1) becomes : pe* ■ rV *

=> p * r 2 and ^ = •♦•(2)

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Now the given equation of circle in the z-plane is

| z - l | = l =5. I x + O' - 1 1= 1

=> |(x -l )+ iy | = l => (x -l ) 2 + j 2 = l

=> x 2 +y2 - 2x = 0 => r 2- 2rcoe0 *O [v put x = rcosl, y = rsinfl]

r = 2 cos# => r2 —4oos2 0

=> r 2 = 2(1+cos20) => /? = 2(l+coa^). [using (2)]

The circle |x - l |* l in z-plane transforms into the cardioids p - 2(l + cos^) in

w-plane. Proved

i - z

441

Example 1.51 .•Show that w = j —~ maps the real axis o fthe z-plane into the circle | i c |= l and the

ha(f-ptane y > 0 into the interior o f tke unit circle | i c |= l in the w-pfana.

Solution. Given: w = *——. * ...(1)Given: w - l ~ Z i + z

■ .....

Since |u;|ss 1=>i - z

i + z= 1

=> \ i - z \= \ i+ z \

=> | i -x - i ,y i =U+x+’iy|

x 2 +(1 - y ) 2 = x 2 -n(l+y)2

=> \+ y 2 - 2 y = l + y 2 +2y => 4^ = 0 ,y = 0 .

[from (1)]

[v z - x + iy]

This shows that transformation maps the real axis of z-plane into the unit circle |ia>|= 1.

Proved.

Further, since | m-' | < 1=>i - z

< 1 [from (1)]i + z

=> | i -x - i ,y |< l i+x+<y | => x2 + ( l - y )2 < x 2 + ( l + j )2

=> x 2 + l+ y2 - 2 y < x 2 +I+y2 +2y => 4^>0=^y>0.Hence the given transformation maps the halffllanej^ 0 of Die z-plane into Die interior of

the unit circle I w {= 1 in the w-plane. Proved.

Example 1.52 : Discuss the conformal transformation w m fRGPV Dec. 2005]

Solution. Given: w = J z

Squaring both sides, we get w2 ■ z

=> (u+ iv)2 = x+iy [v w = u + iv and z - x + iy]

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FutCTQWSQ^^PIjEX,V4ÂejUEa |* S

#*i!

It).

=> u 2 - v 2 + 2 iu v = x +iy


u2 - v2 = x

and 2 u v - y

Now following two cases arise :Case I. When x = constant say, (a and b).i.e.y p u t x = a a n d x ~ 6 (where a> b > 0 and b < a) in ( 1), we get corresponds to the

hyperbolas u2 - v 2 - a and u2 - v2 * b respectively in the w-plane.

—(2)

w = Jz

w - plane

Case IL When y = constant, say (a and b)' i.e., Put^ = a and>> = b (where a, b > 0, andb > a), (2), we get the area included betweenthe rectangular hyperbolas 2uv = a and 2uv ~ b.

v

w = 4 z

O

Ans.•*> u

w -plane

Sxample\.S$: Under the traatfpm atbn n? = —, show that the image o fthe hyperbola = J'Ex. *

is the lemniscate p* * cos 2 4 •.tn

Solution. Given:1 1

w = — => u + w - ----- — z x + i y

\ p c o s j+ ip s m j1

rcoefl + irs in tf

[v x « rcos l , y « rs in# and u * pcoaf, v ~ p em f ]

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46 ) Ei i w h w w i w I I w m k i h c i M

*>C* !S—Wrelv

P j *

{v e“ =cosr 4-isinjj^

Given hyperbola in2-plane: x 2- y 2 -1

=> (rcoatf)* - ( r o n ^)2 =*1 =3 r a{ooe20 - 8in20] = 1

=> r 2 ooe20 = I => W co«2(-#) «1 [Using (1)]

=> cos2^ m p 2.

Which is a tenmiaeafe in w-plane. Proved

Example IM ; Determine Ufa region in the w-plane In which the rectangle bounded by the lines x™Q ,y m0 , x * 2 , y - 1 Is mapped under the transformation w = >[2ei(K/4) z-

Solution. Given,

Since e** scostf+iain#

u+iv = ^2 coe-j+isin-jj(ac+ iy) = (!+ i ) (x+ gr) * (x - y )+ i(x+y)

=> u = x - y , and v**x+y

Given :(i) Line x ■ 0 is mapped into u = -> \ v = y => u = -v

(ii) Line>>~ 0 is mapped into u~x, v = x => u = v

(ii i)L m ex *2 is mapped into u = 2 - y , v = 2+ y => u + u = 4

(iv)Line,y*l is mapped into u = * -1, v = x +l => v - u = 2

Y

*= 0

u+v=4

v—u

Ans.

w-plane

Example 1.55; Determine the region o fthe w-plane Into which the region and

is moped by the transformation u>~z*.

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Then

and

=> tt + u> = c(co80 + i8in0) + i(co80-icos0 ) fv Euler theorem®c 1

U+i„ = (c +i ) costf+ - i je in0

a = fc + ijco80 —(1)|x ) *

U= ( C~ c ) 8in e "•(2)|

which are the parametric equations of an ellipse.

When c - 1 then (1) and (2) becomes: u = 2cosd and v - 0.

Since |coa#|<c 1, we get ~ 2 ^ u £ 2,and v = 0. fv - l £ c o s 0 £ l]

The transformation gives a segment of the v-axis of length 4. Ans.

1.17 SOME DEFINITIONS

Zero o f an analytic function : A zero o f cat analytic function f(z) is the value of z for which

/ ( * ) = 0 .

/ / \ z + *

For example : / W - z (z _ 2)'

For zero of f(z), we put z +1 *=0 =>z = -1 .

Order o f zero: If, f(z) ~ ( z - a)mf(z),

Such that /(a ) = 0, f*(a) = 0, /' (a ) = 0, ..... / <m_ 1)(a) = 0, f m(a) * 0.

then yf(z) has a zero of order m at z = a.

For example: /(*) - - 72 +1

For zero of f(z), put (z - i)3 = 0 => z - i of order m = 3.Singularity o f an anafytic function :

A singular point of function f(z) is the point at which the function not analytic In etfMr words a point at which function/^ is not defined.

For example : f( z ) 35 has a singularity at z * -yti.

Remark: For finding the Singular point of analytic functions put deaom&itor = 0.

fiinzF .r e t* .p k : «*>= (z_1)(*_ )r/2)

For singular points, we put (z - /) (z - n/2) = 0 => z = /, n il i.e., i and n il are singular points of

. f(z).

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1

0d*1td and Non-isolatd singularity:i * a is said to be an isolated singularity of the funetiofr f(z) rf f (t) is analytic at each

in some neighbourhood of the point a defined by | z - a |< 6 , except at the ptfteit a ftstff,

ise it is called non-isolated singularity. fc if 2 ~ a is a. singularity of the function f(z) such that there exists no othergularity within a small circle of circumference with centre at theôint a, tftfn

* a is said to be an isolated singularity. .example: '

\ _ Z + 1Consider the function / W - z (z + 2) '

is analytic everywhere except at z ~ 0 and z - - 2. Thus z - 0 and z = - 2 are the onlyttngfS&ftfe* of this ftmcfion. There are no other singularities ofyfcj toili#neigftbdurtood of t *=

Z Hence 2 * 0 and 2 ~ 2 are the isolated singularities of this

f(z) = cot (#72) ------ \ —-Let ta n. — I

It is not analytic at the points where tan! — = 0 = tan nit i.e., =? — = fin \ z ) z

=> 2 = —, where (w = 1.2, 3«..)n

Thus, z = 1, ........ are the isolated singutarties of f(z ) excejjf 2 = 0 because in thez. 3

J neighbourhood of z ~ 0, there are infinite number of other singwlartfes.

f 1 , J.e„ z = — if n is very large => 2 = 0 is the non-isolated singularity of f(i).

types o f singularity :

^luppose f(z) is analytic within a region R except at i * it, whtett to in isolated

singularity. Let C be a circle, whose centre is *a’ sueh that 0 <] !< R.

00 eoThen by Laurent's series, f (z ) * V aK (z*a)n + 6*(2 - o ) " \

!!«« »«i

oc; . •The part 2h (z ~ <*) *...(2) is called principal part off(z), at z -a .

n*\

There afe three distinct possibilities:

(i) Removable singularity:All are zero i.e., fto firth in principal part of ffz).

Fu n c t io n s o f C o n R£X VMBttLes [4 9

•40

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BA | CMmMaarida U m uc Iim um JII■OPf j B J R J N W n M IW lV B V 8 J M IIC W * tll

i.e., '« . (* - « ) * . |* - o |< f i jM

Thai z * a is said to be removable singularity o f f(z).

Alternatively, jjj} /(*) exist finitely, then z « a is a removable singularity,

v sinzFor example: f\*> --------has removable singularity at z = 0 since

£

00 .

sinz 1^ z3 z5z —

3! 5! ,

t z 2 z4= 1----- - H---- - — ...

3! 5!

It has no term containing negative powers of z Le.t principal part of ffz) has no term.

However^ the singularity z - 0 can be removed and the function be made analytic by defin lsinz . . .------= 1at z ss 0.

z r

(ii) Isolated essential singularity:

If the principal part of f(z) atz = a contains an infinite number of terms, then z - a is called)isolated essential singularity of f(z).

i.e., if there exists no finite value ofm such that lim (z -a )m/’(z) * (finite non-zerox-* a

then z = a is called an isolated essentiaI singulatity of f(z).

For example: The function eu* has essential singularity atz = 0, since

i/g t 1 1 1elZ = 1+ —+ ---- - + ---- r- + ...

* 2 \ t 3lz

has infinite number of terms in negative powers of z.(HQ Pole: If the principal part of/^z) atz * a consists of a finite number of terms, saym, thenthe singularity at z - a is called a pole o f order m off(z).

A pole of onler 1 is called a simple pble.

Alternatively, If lim /(z) * oo, then z - a is a pole of fiz).I - W

Remark :For finding the pole of f(z), we put denominator tjp zero.

\

(2- l ) 3 (*-3)

of order m = 6.

Rem ark:

If lim /(z) does not exist, then z = a is an essential singularity.

For example: Let f(z) - - — ------- - . Then z = I is pole of order m = 3 and z * 3 is a pole

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L P

FucnOW^CONM£XVA8>ABLE8 |# 1

i l te- cat** v IS 7 : Find Ike kind c f singularity o f the fknctUm -— at z - a and * = » .

[RGPV Dec. 2002]

f t \ c0*** oo s j cG‘vcn: * (* -a )2 rim teO r-a)2

For the pole of f(z), taking denominator to zero.

i.e., sin/nr. ( z - a ) 2 =0

=> ( z - a ) 2 =0 *®d ain*z=0

=> z = a is a pole of order 2. and ain*z=8inrwr

=> z =»**(where » = 0, ± I,±2.............. .)

=> z=0, ±1, ±2, are simple poles.,\ If z *ao is a limit of these poles hen z=*oo is a non-isolated essential singularity.

Ans.138 : Find the kind o f singmlnrity o f the function

f(x ) = ^ -j j [RGPVDec. 20041

For Poles o f f(z ) denominator of f(z) = 0 i.e., ** = 0 i.e., z * 0 is a /><?/* o/order 2.

t*n

For zeros of/fzj, the numerator of/fz) = 0 te., (z - 2) sin |

N*(a

z = 2 and ~^— =nn\ te.y z - 2 ,z= — +1. (wheren»±1, ±2,.... ). z —1 n x

Thus z = 2 is a simple zero. The limit point of the zeroes given by

z =— +1, X» = ± 1, ±2,...) is z • 1. [v lf n is large]„• rur

V0 ‘ Hence z * 1 is an isolated essential singularity. Ans.

1 3 9 : Show that the fitnction e* kns nn M ated essential singntdrity a t *s ®.eh*- [RGPVDec. 2003]

Given f ( z ) - e M . ; . .(1)

>, Putting 4 to ( l) b ec o r ae s :

n.....

8+ ......00

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aZ -J EIWSMBStfNG nXMMVKMH

Which contains infinite number of terms.• * •• •• v ' . . • » . - • • ■

Hence / = 0 is an isolated essential singularity of => z -oo is an isolated essentia

singularity of ez. Ans

Example 1.60 : Find the kind o f the singularity o f f(» ) * s i n z ~ l .

f i >sSolution. Let /(* ) = sin

\ l - z )

For zeros of f(z), put numerator of f(z) - 0

i.e., sin7:— r = 0 or — [ V 8 i n ^ = 0](1- z ) 1 -2 L J

or 2 =1 — !—(where n -± I, ± X .... )♦ Ifw is huge, Clearly 2 - 1 is a limit point of thesen it

zeroes:Hence 2 - 1 is isolated essential* singularity. • Ans.

iJLV Example 1.61 ; Show that e '■**' has no singularities.

V7 0 -Solution. (i) f(z)=e

For zeros e ^ ^ = 0 = e~*y =>—- = » [v Num erator of f(z) =0]

=> z2 =0 => z= 0 is a zero of order 2. [v e^° =0]

Since there is no limit point of zeros, then/ft) has no singularity.

(ii) For poles e~iU*2) *0 [ v Denominator of f(z) = 0]

, Which is impossible for any value of 2, râl and complex. Hence no singularity. Proved.

Iv l8 CO M PLEX INTO GH Am W ? h

Historically, Euler was the first to obtain the value of a definite integral by replacing a real variable by a complex variable, whereas P.S. Laplace (1749 - 1827) in vestig a^ (J J82 - thevalidity of such a process. S.D. Poisson (1781 —1840) is believed to have been the first to use lineintegral in the complex plane. , ...It is interesting to note that the concept of indefinite as the process of inverse differentiation in caseof a function of a real variable is extended to a function of a complex; variable if the complexfunction f(z) is analytic. It means that \ff[z) is an analytic function of a complex variable 2, and if

f f( z)dz=F(, is » >J * / '• V '

then differential ofF(z) is equal to f[z);i.e’.9

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f

\ I I #

rver, the cofK^pt of defmrte integral of a fimcntkm, of a real variaWfdpesnot^xteo^cd out, rightlydomain of complex variables. For example, in the case of real variaMe, the path of integration of

6J f(*)dxa

Iways along the real axis from x = a to x — b. But in the case of a complex function f(z), the

| may be along any curve joining the points z - a and z * b; so that its value depends upon the pathv (curve) of integration. However, we shall see that this variation in the value of definite integral willF'Uisappear in some special circumstances.

For instance, the variation in values can be made to disappear if the different paths (curves) joining z = tftoz = 6are regular path? (corves).

} SOME DEFINITIONSDomain (Region): A set S of points in the Aigand plane is said to be connected set if any two of

' its points can be joined by a continuous curve, till of whose joints belong to S.-*»An open connected set is called an open domain. If the boundary points ofS are also added to an

open domain, then it Ts called cloied domain,

Jordan Arc : Let x(t) and y(t) be continuous function of a real variable / in the interval a s t <,fi.

Then the set of points x in the Argand plane given by die equation

is called a continuous arc if, corresponding to one value of t, there exists more then one value forf z, then z is said to be a multiple point A continuous arc with no multiple point on it is called a

Jordan curve.$ Contour : By contour, we mean a continuous chain of a finite number of regular arcs. If the

contour is closed and does not intersect itself, then it is called closed contour. h Example: Boundaries of triangles and rectangles.* Sfrrtpiy and Multiply Connected Domains :

f/ A domain in which every closed curve can be shrunk to a point without passing out of Die regionis called a simply connected domain. If a domain is not simply connected, then it is called multiplyconnected domain.

path of the definite integral

b

J f(z )d z

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I I | &wg«mi»3 Mxmounc»4H

IM COMPLEX LINE INTEGRAL

Suppose,/?# is continuous at every point of a closed curve C having a finite length, Le.; C is a

rectifiable curve. Divide C into n parts by means of (it + 1) points z9

Let a = zQ,b * zH,

We choose a point £* on each arc joining z*_j to z k.

Form the sum

Sn=t. /(*•)(*,~*r,l).\ rml

Suppose maximum value of (zr - z r_,) 0 as n -* « .

Then the sum S„ tends to a fixed limit which does not depend upon the mode of subdivision and

i denote this limit byhi : .

£ f(z )d z or I f{ z)d z

L

which is called thecomplex line integral or line integral offfz) along C. An evaluation of integral by such method is alsocalled ab-initio method i

r * :

By the symbol f(z )d z we mean the along a boundaryC intfce positive sense.

In case of closed paths, the positive direction in anticlockwise. The integral along C is oftencalled a contour integral.

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F u N r t i O H i w C n w H f l f Vm b w u s | 5 5

1.62 ; Evaluate ](3x* +4xy + 3y* )d x +2 (x* + 3xy + 4y* )d y]

(I) a lo n g y = x* (U) a lo n g y = x .

Does the value o f the Integral depend upon the path ?

(i) Along curve, y = x 2 => d y - 2x d x and limit jc -► 0 to 1.

I * £ [(3x2 + Ax3 + 3 x 4)d x +(:2x2 + 6x3 + 8x4)2 xdx]

= £ (3x2 + 4x3 + 3x4 + 4x3 + 12x4 + 16xs)dx =

(ii) Along curve, y * x x> dy “ dx and limit x -> 0 to 1.

I » £ (3x2 + 4x2 + 3x2)dx + (2x2 + 6x2 + %x2)d x =

Clearly, both values are the same, hepce the value of integrationdoes not depend upon the path o f integration. Am.

Rppfc 1.63 ; Evaluate J [ (z)2dz, along

(I) the real axis from z~ 0 to z~ 2 and then along a line parallel to y-axisfrom z m 2 to

Zm2 + L

(ii) along the line 2 y -x . IRGPV June 2905 and June 2007]

Since (z)2 = (x - iy)2 - (x2 - y 2)~2ixy

(i) Along the path OMP where M is (2,0) and P is (2, 1).

i2+i .

Qc) d t

M o n .

= (x2 - y 2-2 ixy)dz+ (x2- y 2 -2ixy)dz ...(0

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M |

Now, along OM, y**Q =>dz = dx and x varies from 0 to 2.

x

3Jo* ~ y2 ~ 2“ 0f)cfe =

Also* along MP, x~ 2 ,d z~ id y and>- varies from 0 to 1

I mp “ y2 " 2ixy^dz = I (4 “ y2 ~ Aiy ^ y

4 iy~ i^Y + 2y7 ~ 4 i i + 2 = 2 + i i f 3 3

Hence 0 ) becomes,

r2+tr r * 8 _ II . 14 11./ = 1 ( z ) d z = —+ 2 + — 1 = — + —i.

* 3 3 3 3 Example 1.64: Evaluate f * (2x+/;y + t)d z along the two pa ths:

(i) x - t + 1, y - I t2-- I(ii) The straight line joining (1 - i) and (2 + i) ' '

Ans.

Solution* Let £** (2x+iy+ l)dz

fftGPV Dec. 2009 (N)f

.-(1)

(i) Alongpath: x = / + 1and y - 2t2 - 1=> dx ~dt and dy = At dt.Then dz = dx + idy.Also from path x ~ t + 1 :when x = 0, then t ~ 0when x = 2, then t = 1i.e.. limit = / -> 0 to I.Hence (1) becomes :

i / = j[2(t +1) + 1(2*8-1) + l](<fc +i4tdt )

o j

= j(2t + 2 + 2it2- i + l)(\ + 4it)dt o

. 25.= 4 + —-i.

3(ii) Along path is straight line joining (1 - i) and (2 + f)'i.e., point (I, - 1) to (2, 1)

1+1Equation of path; .y + 1 = — ~(r -1 ) => +1 = 2 jc - 2

2^1=> y+ 1 = 2 x - 2=> y = 2x - 3

=> dy ~ 2dx

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Fuwcttowi<ibCm i » ¥ m m u 8 [17

Also dz - dx + idy and limit x -> I to 2.Hence (I) becomes :

/ = J2[2x + i(2x - 3)+1)(dz+ i2dx)

= j^[2x +1 + i(2x - 3](2i + l>ijc» 4 + 8i. Ans.

mxample 1.65 : Evaluate £ (** + 3z+ 2)dzt where C is the arc o f the cycloid x - a (0 +sin 0),f ■ y = a (l-c o s6) between the points (0, 0) to (no, 2a).

'Solution. Let us consider the path of integration along a curve C consisting o f :

I (i) the part ofthe real axis from the point (0,0) to the point (*0,0). On this line: z = x ,d z = dx

and x goes from 0 to na.

(ii) followed by a line parallel to the imaginary axis from the point (m , 0) to the point(na, 2a). On this line:

z ~ m + iy, d z = idy an^y goes from 0 to 2a.

Let, / = £ (z 1+ lz + 2)dz 'a

or

or

or

“ F + 3x+ 2)c£x + £ ° {(>ro + iy)2+ 3 (fro+iy) + 2]idy

I=f"i x*+-x2+2x1 +i|"-(?ra +iy)s + (>ra +iy)2+21ylL4 2 J0 [_3 2 J0

/ = | j ( W + | (^ a ) 2 +2*aJ

+i[i(â+i2o)s +^(>ra+i2a)2+4o-i(xa)8- (/ra)2l|_3 2 3 2 J

1 3 I - 2 ; ra + - ( * a + i2a)4 + — (tra+ i2a) + 4ia . Ans.

3 2

Example 1.66: Integrate z2 along the straight line OA and also along the path OBA consisting o ftwo straight line segments OB and BA where O is the origin, B is the point z = 3 and A the point z * 3 + L fRGPV Dec, 2004]

Solution. Since, z = x+iy => d z - d x + i d y .Along Die curve C, we have *

£ z 2d z = j c (x + iy)2(dx+i dy)= £ (x2 - y 2 +2ixy)(dx+idy) ...(I)

(O'ThOpoint Afe z ** 3 + /, i.e:%A fa (3,1).The equation of the line OA is :

y - 0 = y ^ ( x - 0 ) le.%x = 3y

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5 S | FmrnO Ti^Kdi rattATjca-MI

Hence, x*=3y=>dx « 3dy and .y varies from 0 to 1.

Now from (1), we have

JL *2<3k = £ ( ^ 2 - ^ 2 + 2i3y.y).{3dy+icfr) = £ (&+6i)0+i)y2dy

= 0 8 + 2 6 0 ^ j = i(l8 +26 .-)= 6 + j ^ . ...(2)

00 Again, z 'd z = z 'd z = J ^ ^ + *2<fc

(x 2 - y 2 +2wy)(dx + M f y ) + ( x 2 - y 2 + 2ixy)(dx+ idy) ...(3)

On the line OB, y = 0=>dy = 0 and x varies from 0 to 3.On the line BA, x = 3=>dx = 0 and >>varies from 0 to 1.Hence, (3) becomes:

z2dz * x2dx+ £ ( 9 - y 2 +6iy).idy *

= 9 + i r 9 - I + 3 i j= 6 + i^ .

Thus equations (2) and (4X show that:

L z 2dz = I z2dz. JOBA

x 33

+ i 9^--^-+3i,y23 0 3

...(4)

Am .

U l CAUCHY’S INTEGRAL THEOREM OR CAUCHY’S THEOREMStatement: i

I f f ® & m anotytic function and f( x ) k continuous at each point wtthim an# on a simple closed curve C, then

£ f(z)dz=0. [RGPV Dec. 2002 and Dec. 2011]

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FuNCTfONS^rCo t f f t ex Va r ia b l e s |* W

fc*•6)

afqrr

Let R be the region bounded by the curve C.

Let f(z) - u (x ,y)+ iv(x,y) = u + iv and z - x + iy

=> dz = dx+idy, then

£ f( z)dz = (u + iv)(dx + idy)

= £ (udx-vdy)+i £ (vdx+udy)

du du dv dvSince f ( z ) is continuous, the partial derivatives are also continuous in region R.

So using Green's Theorem Mdx + Ndy » JJ

1) becomes:

I f(z )d z = JJ / / ...(2)

Since f(z) is analytic so using Cauchy-Rientann equations, we get

d u _ d v d u __ du

8 x~ By and Sy~ 3x

Thus, (2) becomes : jj, /(*)<& = 0.

wit •r n *

Extortion o f Cauchy's Theorem: If ffz) is analytic in the region R between

two simple closed curve C\ and C* then /(*)<** = k f (z )dz .

CAUCHY'S INTEGRAL FORMULAStatement: If f ® is anotytic within and an a etaaed curve C and *a * is any point within C, then

^ " 5 5 1: Je -a )dX' °T i o eC [RGPVDec. 2007, June 08/

f(z)roof. Consider the function------ >which is analytic at every point within C except at z - cl

2-0

Draw a circle C\ with a as centre and radius r such that C\ lies entirely inside C. ie., C\ :\ z-a\ = r.- »

f Thus function ——- is analytic in the region between C and Ci. z - a

By using extension of Cauchy's theorem, we have

f f J V L d z mJc ( z - a ) Jc, ( z - a ) *(1)

Since, C, : |* - a |= r or * - a = rew or z = a + new

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=> dz ssi r e^dS and 0 -► o to 2 n .

Hence (1) becomes:

r j w _ <fe= [ J ^ - d z

k (Z '-a) k ( z -a )

= f ^ ’ - )'.irei9d& * i f t a + re ^) d 0 ...(2)

In Die limiting form, as the circle C\ shrinks to the point a, i.e., r -* 0 , then (2) becomes:

Jc <** * * f / W * - * » r 60 a < a >

Hence /( ° ) = < J e ...(3) froved.Deduction :Differentiating both sides of (3) w.r.t. a, we get

/ ’(a) = — — f f .. ftfcK <fe (4|' 1 2 x i d a * ( z - a) 2« (z - a f ~ (4)

Similarly, r (a ) = l J ^ f dz' -(5)

\

In general, /<' >(<,) = t ( I - a p * * 1 -<6)

Results (4) to (6), are Cauchy's integral formula fo r the derivative qff(z). Remarks:

(i) If point a ©/<then fc

(ii) Converse of CMefcy'i Theorem (Morera's Theorem ):Let f(z) be continuous in a simply connected region R and let for every simpleclosed curve C, such that

£ f( z )dz = 0. Theft fl(z) is analytic,

(iii) Cauchy Inequality :

I ff(z) is analytic within a circle C given by |s -a |s = jR an d if l/ (* } |£ M on C, then

\ f f(z)dz

Proof. Since / W = Jc ^ , qy»*i P*y Cauchy integral formula]

6 0 1 Encmhebw h q

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v.

FuNciidJferWGbiAfcEXVariables | (fcl

| /»(o)l» J l L f J W ± ,' ' 2in. Jc (z —nY**2xi Jc (z - a )*+l

[Since, z ~ a = Re* =5 dz - iRa'8dOaxi&\dz\-\ iRewd6\= R do\

n\ e !/(*)! \dz\ [vUI-1]2 j t j c K ^ -o )**1I

n! M r2*

2* f lB+l

fix

I

* * J £ - 2 * r 2n R n+i

Mn\ Proved.

(iv) Liouvllle's Theorem :0 -y If a ftinctionjfc) is analytic for all finite values of z and is bounded, then f(z) is a constant

e**1.67: Using Cauchy's integral formula, evaluate

| * | - 3 .Uiwftm. By derivative of Cauchy's integral formula:

Here a = - I, n = 3, and f( z )~ e 2*

JLr * + i /d z, where C is the circle

[RGPV, June 2011 J

-.(1)

,2 i

= 1

2>riJc (z+1)4

Since at pole z = a = - t , so that | z |=| - 1 1= 1< 3 i . e point z = -1 He in circle C.

/(*) = e2z => A * ) * 2e2x => f \ z ) = 22e2*

=> f m(z) = 23e2* => / '( - I ) = 8«“2.‘ • i ) ‘

_2 3 f e2* ,Hence (2) becomes : 8e ~~r L 7— TT®*

M ** (Z + l)

*♦•(2)

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62 1 M il Mill—IIIiTlB III

Example 1.68 : Using Cauchy's integral formula, evaluate I

d x

*(s+*i)* where C is the circle

Solution.

| *+3I |=1 .

For poles (or singular points),

put z (z+m) = 0 => z = 0, «*-» *At pole z = 0

|*+3i|=|0+3i 1=3*1, te., pole a tz ~ 0 out side of C.

At pole z - -tri :

|z+ 3 i|=|-* * + 3« M (3 -; r) i |= 3 -; r <1, /.e., pole at z = -jd, insideofC.

L 11 -]* * " & z (z + m) m * I z z + n J

= ± f ± _ ± f L - d zm k z m *c (z + » )

[Using partial fractions]

= -~[03-A l2* ./ ( -» )]sa m

m —~.2*X lm

1 Because f ° eC■ Jc { z - a ) \ 0 , a « C

[v / (z ) = 1 / (-m ) = 1)

A bs.

Example 1.69 ; I f /(S)® £ fMiww C Iv the circle x? +y* =4, fin d the values of

Solution.

= -2.

3 ^ +7»+l z - \

f ( 3 ) , f ' ( l - l ) and fX l+ i ) .

The given circle C is x 2 + y2 * 4 or C :|« |= 2.

The point z » 3 lies outside the circle \ z \ - 2 while z= 1 - / and z - 1 + / lie inside the

circle. [Because!*M31*34:2, \ z |* | l - i | - V 2 < 2 , \ z \ - \ \ + i \ - j 2 <2]

(0 / (3) = X

3z2 + 72 +1 j j3 * 2+72 + 1 . j ^dz a n 4 ---- - —-— in analytic within and on C.

*~3By Cauchy's integral theorem,2 - 3

ic3z2 +72 + 1

dz = 0,2 - 3

(ii) By Cauchy's Integral formula:

f(z) I - ^ r d z k (z - g)

~ \ C = | z | = C OI I

+ • MI I N

\ 0 z=i-l)

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Ftacfflotf &CowtoVAMAMJir I 69

Hence

(iii)Also,

£ ^LIJJLLL c I z * 2xi[3z* +7z +1]^

/ (# ) - fcri(3£*+7#+l)

f \ 0 - 2**(6£+7)

/ ,a - 0 = 2jri{6a-0 + 7>

* 2*1 (13-60 = 2* (6 + 13i).

A » - 2« (6) « 12*i

r a + o * i 2 « .

Ans.

[From (2)]

Ans.

mfii 1.70:1/F(t) - £ z -% &***’ where* is the ellipse:

Fi*rf the values o f ft) F(3.5), M F®, F f - /) ami F 'f - i). [RGPV June 2009]

ution : The given ellipse c 1

Here the point z - 3.5 lines out side of the ellipse c, while z * j, - 1 and - i lie inside the

ellipse c as shown in figure.t (4z* + z + 6) j

0) ^3 .5) = 1 (z _ 3 5) d* •••(»

since z = 3.5 e c ie.t outside, by using Cauchy integral formula

i(4z2 + z + 5)

z - 3.5dz * 0

Hence (1) becomesF{3.5) = 0.

(ii) By Cauchy integral formula:

f(z)

I

“ Ini/lt), if / € c 2

(4z* + z + 5) z —t

*>

and Putting * = /, in (2), we get

•dz as2«/[4z2 + z + 5Jx- ,

FX/) *2w i(4^ + / + 5)F'(f) ~ 2fti (8/ + 1)

F" (0 -2**(8 )

F( 0 * 2*/ (- 4 + i + 5) = 2*/ (1 + 0 * 2*(/ - 1).

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Putting / - - 1 in (3), we getF \ - 1) « 2*/ ( - 8 + 1) * - 14*/.

Putting t = - / in (4), we getF"(-0=2itf(8) -167«. Ans.1

1J3 RESIDUES OF/jy AT POLESince analytic function fiz) can be expanded in a Laurent's series

/ (z) = y a „ (* -a )»+ T 6 . (»-o )-n-0 n-1

_ _ I ,

Then the coefficient of (z _ ay i.e., bx is called the residues off(z) at pole z = a. It is denoted by

[Res f ( z ) \ ^ or Res. f(a).

fa68 f (z )\-a = fr| - ~ r f( z)dz , where C is the closed curve around the point 2 - a.

fcj |

For example:

2m

x / v sin* 1

Z* *

*3 *5 z 72——H---- ------+ .

3! 5! 7S

1 1 1 1 1 3= —j———+ —2 “ —* + .... ZS »3 3! ' 5! 7!

... [ R e s^ z )^ 4-0 = Coefficient of = ~ =

1.24 METHODS OF FINDING OUT RESIDUES OF/fcJ AT A POLE(i) Residue off(z) .at simple pole z - a (i.e., pole o f order I) :

[Res /(z)L« = lim (z- a)/ (z) .*-KJ

(ii) Residue off(z) has pole o f order m at z = a :

(iii) Residue o f f(z) at infinity :

Le., Residue oifiz) at z = a> = lim{-z./(z)}

A«s.

= - [Coefficient of — in the expansion of fi z) .for values of z in the neighbourhood z j

by z = oo ]

- 3 < l t

where C is the closed contour enclosing alt the singularities of fiz) except at infinity.

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..... -Him ^ i r u r i j a w r " 16I

rCAUCHY'S RESIDUE THEOREMStatement :

r If f(z) is analytic function, except at a finite number of poles a,, a2, a3,....an within a

> dosed contour C and continuous on the boundary C, then

J f ( f z ) d z =2/ri.ERe s/faJ nt the pole a, at , .... ,o„ in C

£= 2xi (sum of re sidue s nt the poles with in C).

>41

Let c,, c2,.... ,cH be the circles widi centres at 04, 02*03.......o»*respectively and radii so

small that they lie entirely within the closed curve C and do not overlap.Then f(z) is âlytic within the multiply connected region enclosed by

Q O'

the curve C and circles Cu C2,.... C ,. So using Cauchy's theorem i Q Q ,

- for multi connected regions, we have /

£ f ( z)dz =* £ f i z )dz+ £ f { z )dz+....+ £ f{z)tdz

A = 2m[Re8f(z)]gma) + 2m[R£*f(z)}^2+.... + 2»IRes f ( z ) \ ^ [By definition of residues]

= 2j»fsum of residues 0# /(z ) a t poles alta2, .......a* i*kC]. Proved.

• V i - 2 z1.71: Find the outer o f each pole and restore at it i f z ( z - l ) ( z - 2 ) ' IR**PV Dec. 2001]

f / _ \ _ 1- 2zLet A*/* ", rr*

z ( z - l ) ( z - 2 )

For the poles of/Ts$> : put *(z ~ l)(a? - 2) « 0

=> 1,2 aH are simple poles.

We know that [R e s. /f a )^ « lim (» -o )/(z )

(i) - *£<*-«»«*> = lim = i Ans.

(ii) Ans-

(Bi) [lte .rt« > |0 -lm < * -2 )r t2) = lim = Abs.

and the residue at each

w

umple 1.72 .* Determine-the pots o f the function /fe>= f y - x f f c + 2 )

pole. [RGPV June 2002]

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Solution. Let f ( z )~

661

*2

(z - 1)2(z + 2)

For the poles of f(z) : put (z -1 )2(a + 2)-0

=> 2 = -2 is simple pole and * = 1is pole of order 2,

We know that [Res. f { z ) \ a = lim (z - o) f(z) M-H I

(i) + A b s .

We know that [R m / W L . . = ~ o)'" / (z )! '

(«) [R«s ;r« ] , . ,m. ! = ^ ^ 7 lim £ « z - l ) 7 M ) .

;lim i L * L U «-■ d*[* + 2j “

lim(z + 2)(2z) -z *(l) | (z2+ 4z) 5

W “ «• Ans.(z + 2) j *-*» (z + 2) 9

** Example 1.73 ; Evalute the residues o f (z _ j)( z _2 )(* -3 ) <**2 = 1* 2, 3 and infinity and show

that their sum is zero. M [RGPV Dec. 2002]

z 2Solution. Let f(z) =( z - l ) ( z - 2 ) ( z - 3 ) '

For the poles of f(z) : put (z 1) (z - 2) (2 - 3)=0

=> z =1,2, and 3 all are simple poles i.e., each pole of order m = 1.

We know that [Ree./Xz)]^ * lim (z-a)/(z)

(i) [Res. A*)],., * Km (z - l)/(z ) = limz8

*-* 1 , «-»i ( z - 2 ) ( z - 3 ) 2

(io [ * « . / ( C - 5 9 c»- » / » = Hg ( 7 : ^ - 3 ) '° a l i l j — 4-

(BO [ R « . / W L = S (*-3)/(*>“ ten (zi 4 - 2 ) = ^ ) 4

(iv) Residue of f(z) at z = 00= lim {-2 f{z)} [by definition]

=liml “MO

- zz2

(z-1) (z-2 ) (z

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flim

Fu n c t i o n s o f C o m p i e x V a m a m j e s | 6 7

- z 8

z \ l -X ) ( i- ^ ) ( l - ^ )

-1

a-o>a-o)d-o)=-i.

j 9 i _g + 9_ 2Thus, sum of residues of /(*)■ —-4 + —-1 ------- -------=0. Proved.

11.74; Find the poles, order o f the poles and residue at It fo r the function

fRGPV June 2003]

Let /(*) = ~T—r zA+1

For the poles off[z) : put z 4 +1 * 0 =» z 4 * -1

i.e., z* = eoe?r + £8in*, = cos(2n + l);r + i8in(2n + l);r

z = [oos<2n + un + i sin(2n+ l)jr]l/#

^ z 4 *[cos(2^ + l ) ^ + i8in(27i + l ) ^ ]

=> z - et<2n+v/ i , where n - 0,1,2,3. [y e* = coed + tsintf]

or z - ei*f4le3i*,4fe5i*,*fen*u are four simple poles.We know that [Ree./X*)]^ =Um (z -a) f (z )

(i) [Res./X*)] ^ = lim ( z - e ix'*).-}— ,w " 4 * +1

[which is — form, so using L*hospital rule]

I 1 -3 /4 1r 3 x . .=5lim —r - —e = — cos----- ism—

Az >A 4[ 4 4 .- I . 1

v r ' - ^ j

i ( i + » )

* ~ 1 T ' Ans.

(ii) [Re8./0z)l w . * lim ( z - e 3m!4).— — , now using V hospital ruleW " 4 * +1

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68 I ENGlJte£Hl MArrtEMATlC&-llI

(iii) [Res./(2:)je_eJ«/4 = lim ( z - e 5jn/4).—p— , using L’hospital rulez 4 + 1

4 4

(iv) [R es ./^ l^T.M = lim ( z - e 7j”/4) . - - — , using L'hospitairuJe Z +1

= lim —- = lim z z-H?*'* 4Z Z e1*'*

1 1 1 1

UJ

Ans.

3 IV z = —

- z

An*.

v s in z . . Example 1.75; Find the sum o f residues o f the function t(z) = —— — in the circle \z = 3.

- X COS Z 1 1

Solution. For the poles of f(z), : put zcosz = 0 => z **0 and cos r = 0** '■ 1

=> r = 0 (simple pole) and z = ±^ /2, ± 3 ^ /2 ,± 5 ^ /2 .... are simple pole.

At the pole, z ~ 0, then | z |=| 0 1- 0 < 2 inside the circle | 2.

Atthepoles , z = ±a72, then |z |= i ±»72)= /r/2 = 1.57 < 2, inside the circle |* |= 2,At the poles z = ±3jt/2, then |z |= | ±3;r/2 |=3*/2 = 4.71 <%

Which is outside the circle |z j= 2

Hence poles z = 0, z = /r /2 and z - - t / 2 lie within the given circle C. 'V\v V

(j) [Res./(z)] - l im (^ -0 )- ^ ^ -= l im 8*n ? ~0.^ ' 1 «-*<> ZCOSZ <X)SZ

(ii) [^®s• f <*)Lrf - Hjg (* ~ M) zc~ ” • [which is “ from, using I *Hospital rule]

= Hm (sin z +(z - ^ycos z) 1 2*-*'A cosz-zsin z (0 -f ) ;r*

(iii) [Res./(«)]*»-*/ = lim (z + ) - -------, so using L ’ Hospital rutezcosz

sinz+(z + )c 08z -1* 2= am -------------------- ——-r ---- r-——.

*—% cosz-zsinz j * ) n

r * j

2 2Sum of all residues of f ( z ) =0 — +— =0. Ans,

7T K

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M f# 1*76: Determine the po/e and residue at each pole. I f f (z) -

m m - . . Let

1-e*2

F u n c tio n s o f Co m d e x V /W a k -e s j 69!i,

V

i -‘ (2 z) (2 z)2 (2 z f I + + ....

I! 2! 3!

4 •> 2 -»(2 + 2z + - z + 3 .... ^

-.(I)

V*

n:1

n*

Hence f(z) has pole at z = 0 of order 3./. Residue of f(z) at z = 0 of order m = 3

i i »•r Ilim

(3-1)! *-x> dz[ c - o f . S z f l

2 - 2 z - ~ z 23

lim2 z-*Q

8 2 *--------.6z~3 3

[Using (1)]

Ans.

1*77 .* Determine the poles and residue o f the function sinz

thm, Let f (z)=— sm z

*v IvihK U'A poles : put sin z=0 =?sin z - sin tin

=>z = nn (where, n »0,±1, ±2 ..... )

Hence z = «/rare simple poles.

[Res./Xz)]^, = lim ( z - a)f (z ) [By definition]

[Res.ftz)]^, = lim { z - n n )

- lim

z-*mr

(2 Z - T17Z)

sinz

nn

[Since — form, using L Hospital rule]

nn

z-*n/r COS Z coann (-1)*= (- l) nnn. Ans.

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70 | E N G i c ^ l i i t ^ t ic s 4 I I

Example 1.78 : Find the residue at z ~ O of the following function :

J + e(i) f(*) - ~

a m z +z c o s z

Solution. [R«* • /(*)],_ - tim (z -a ) /(*)

(i) [Res./Xz)! =lim ■r-^—— —u 0 sinz+zoosz

- limr-»fl

1 + e*

+ C0S2

1 +1

1+11.

x ^ x*(ii) Now, since c o sx * l- — +— ....... . then

/ (z)=2 C oaf i ]=z[ l -—^ -+ T-i-T - ...... \ z ) |_ 2! z* 4! z J 2z 24 z8

This is the Laurent series about z 530.

Residue of / (z )a tz » 0 =coefficient of — ^( z - 0)

[By definition]

sin zv lim ------=

*-*o z

...d)

[By definition]

Ans.

Example 1.79; Find residue o f f(z)= z a in (m z)

Solution. Given /(*)

a/ Me origin.

z 1 z* 1+* + —+—-+.

2! 3!

3! 5!

H+jrjr jmz* 1 -

m’z* m424

3! 5!

[Since (1 - x)~l =1+ x + x 2 + .......]

1

j»2

22 23

1 + * + — r + - T + .

2! 3!1+

3!

[Because for all sufficiently small value of 2]

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// X 1 1 1 1 ;... /(*)=—.-Y+—.-+ .......... ...(1)m z m z

which is Laurent series about z = 0

1

I 7 1

Hence, Residue of /(*) a t z =0 = Coefficient of

Ans.

( z - 0)J_

m

(4-3z)dzr ( * - 4 z )a z1.80; Evaluate Jp z (z - i ) ( z ~2)* w*tere ^ k the circle | 2 1=3/2. [RGPVJune 2008J

Let f(z)= " lZ z ( z - l)(*-2)

[ Using residue theorem:

^ jc f(z)dz = 2ni [S Res.of f(z) at each pole with in C.] ...(1)

For the poles of f(z) : put z (z - 1)(2 - 2)=0

=> 2 =0, 1, 2, are simple poles.

At pole z = 0, then 12r|~| 0 1—0< 3/2, which lie inside the circle C.

r' Atpole 2 = 1, then |ar J=| 11— 1< 3/2, which lie inside the circle C.

At pole z - 2, then |* |=121= 2< 3/2, which is not lie inside the circle C.

We know that : [Res f(z))gma= lim (z -a ) f(z)." Then

(i) [R * f ( z ) U 0) =1 =2 .

[Res/(2)1, i =lim (2 —I ) — —— — =—-—=— 1.1 2(2 —1)(2 - 2) (-1)

Hence (I) becomes : | c z ( J l i ) ( z !_2) ^ = 2*-i (2 + (-1)] = 2ni. Ans.

z*mmple 1.81 .* Determine the pole o f the function f(z)=—~ — rrrr — -7 and the residue at each

r ( z - i y ( z + 2 )

f Z * pole. Hence evaluate I -------- =-------- d z where C :\ z |=3.

* (z - 1) (z + 2)

WnHon. For the poles of f(z) : put (z - 1)2 (2 + 2)=0. f(z) has a pole of order m ~ 2 d X z - 1and a simple pole at z = -2.

d f / __\i d

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72 I

= = U m l i l H i l = £*-*11 (z + 2) J 1(«+2) J 9

*2 4(ii) /( z )U 2 * Um [(2: + 2) /(*)]= lim -— - j = - .z~*“2 *-*-2 (2; —1) 9

Using residue theorem:

Jk f (z)dz~2m [I Res. of f(z) at each pole with in C], ...(1)

At pole «=l,thenlr|=|l|=l<3, inside in C.

At pole 2= -2 ,then |ar|= |-2 != 2< 3, inside in C.

Hence (1) becomes :

f 7----- %-------- d z ±2x i \—+^-\ =2ni Ans(* - l) 2(* + 2) [9 9] iia Ans*

Example 1.82: Evaluate f ----- - f ------ where C Is the circle I z + * l=>/3.Jc (Z - t )* & Z + 3) .

12z-7Solution. /(*)= '

( z - \ ) 2 (2z +3)'


f(z)dz=2m [IRes.of f{z) at each pole with in C]. ...(1)

For poles off(z) : put (z - 1)2(2z + 3) -0

3=>z = 1 (order 2)andz=— (simple pole)

2

At pole z - 1, then |« + i |= |l + i |« -J\2+12 =<J2<J3, inside in C.

At pole 2 - - 3 / 2 , then |« + i| = |-3/2+ 11*^9/4+ 1= 713/4 <^3 out side in C.

Thus, only pole z = 1 lie inside the circle C : ,| z + i j=^3 .

'• We kn°w «“ t S S ^ . T ^ _ a)m /(2 )]

=> [R®«/(a)],.i. « .i =7f l*m — { W 7 . 11! *-»i a z [ ( z —l) (2 z+ 3)J

=Bin ± ( U * z l) = K * r (2 £ ± 3 m z Q 2 jz Z H*-* cfe\ 22+3 J [ (2z+3)2

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6 ° - 10= 50=225 25

Hence (1) becomes:

122-7

(2 - I ) 2 (2* + 3)

z - 3

[ ----- i g - L — cb =2m[2]=4m.k (2 -1 ) (22 + 3)

pxttmpie 1.83 : Evaluate - j — - — —dz, where C Is the circle |2 + l+i|= 2,

Ans.

U t f(z)= z 322 + 22 + 5


£ /( z )d z= 2;n[IRes. of /(z) at each pole with in CJ. ...(1)

-2± >/4 -20 - 2 ± 4iFor the poles of/fe) : put z +2z + 5 = 0 => z -------- ----------— -—

=> z ~1 ±2i => simple poles are z = -1 + 2i and - l - 2 t .

At pole z = - l+ 2 i, then 12+1+ t|= |-- l + 2i + t + £H 3i 1=3*2, outside in C.

At the pole z - - 1- 2i, then 12 +1 + i M -1 - 2i +1 + i H - i 1=1 <2, inside in C.

Thus only pole z =-1 - 2i lie inside in C.

• [Res /(*))*— 1-2. = ^ <* + *+ 2i)f(z)

- te + l + 2Q (g -3 )_ Km (z + 1 + 2i)(z-3 )2Z+ 22 + 5 (z+ l + 2i)(2 + l -2 i)

.. 2 -3 —1 —2£ —3 4 + 2i 1 .= li m ----------------------------------------------- s=----------- ---— i

2 + 1- 2 t - l - 2 i + l - 2 i 4i 2 Hence (1) becomes :

Jc 7 ^ E dz = 2m [ v i h (2 + i )- Ans- Z *

Example 1.84 .* Find the residue of f(z)= -------- ------------------ at its pole and hence( x - l ) ( z - 2) ( z - 3 )

evaluate Jj, f (z)dz, whereC: |* |=6/2.

Solution. For poles of ffz) : put (2 - 1)4 (z- 2)( z-3 )= 0

=> 2 =1 (order 4) and z

=2, 3 (simple poles)Residue of f(z) at z “ 0 “ jj®} ~ [By definition]

8/11/2019 DK Jain M3


(i) [Res f(z)\z- i = lim (z -2 )-— — - —-— -w *-»2 ( z - \ ) a( z -2 ) ( z -3)

7 4 1 ...... T

=lim8

(2 -I)4(2 -3) (-1)

~-8

(ii) [Res/(2)]z_3=lim(2-3). ■= lim-27

*-3 (* - 1)4 (z - 2 ) ( z - 3) «-5(*-1)4 (*-2) 16'

[By definition](Hi) (Res « * ) U . ■=— Um {(z-a.)" f(z)\

... [Res A * ) U i , « - 4 =(4-0! tfe3

(*-D4.(*-1)4(z - 2 ) ( 2 - 3 )

Z*l

15— 6

1

6

J_

6

6

J_

6

J_ 6

' d3I

dz3' (z-2)(z-3)J*-i

— I<fes i2 + 5 +

(19« - 30) z* - 52 + 6 *=»

d3f , 27 8 1 — r <j* + 5+ -------------- ^cfe 1 2 -3 2 -2 J

f 27 8 1«fe2l (z-3)J+(*-2)JJ

_ d [ 54 16 ]]

d z \ ( z - Z f (z-2)3jJi.1

A

[v as dividon algorithm]

[v resolve partial fractions]

Z*1

-162 48•+ -16216J 2 -3 )4 (2 - 2)4

Further, using residue theorem :

f(z)dz-2ni[ZUes.oi f (z) at each pole with in C].

At pole z= 1, then 12 |—11 |=l<5/2, inside in C.

At pole z - 2, then | z |^| 2 1=2<5/2, inside in C.

At pole z - 3, then 12 |-j 3 l=3*5/2, outside in C.

+48U J l = i ° i .J 16 16

~.(!)

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Fu n c t i o n s OFÔNrtJix V a r ia b l e s I 75

Clearly, only the poles z = 1and 2 2 2 lie inside the circle C.Hence (1) becomes :

I 1

Wfamtple 1.85 : Evaluate Jc where C i t the circle |z - l |= 2 .

Natation. f(z) =1

z 2 sin hz z2 (e* - e"*)2

* - e °v sin hO =

*2 *3

1+ * + 2T+ 3!+ '

f, *2 *3

T ‘ *+2T‘ 3f+-

1

- 2zJ 2 Xs

^ i r i T -

, *2 *41 +3T+ I T +- •

= z 3 1+( z 2 z*---+-----+.

,6 120=z -3 , z 2 _ z *

6 120 + ‘ "

-I z _ J ___ 1 ___ *_+

120*.... z 3 6* 120which is the Laurent's series.

By Residue Theorem:£ f(z)dz=2m[2.Res / ( z ) at each pole in C].

From series (1 \ f(z) has pole at z - 0.

At pole z = 0, then |z - l |= ! 0 - l 1=1 <2, lie inside the circle C.

1 1 Resf(z) at z = 0, * coefficient of in (senes) •

Hence (2) becomes :

f ___ l ___ dz = .Jc (z2 s ta te ) I 6) 3

Example I M : Evaluate f -------- dz, where C is the unit circle | *|= L Jc COS 71 z

...(2)

Ans.

Solution. Let /(*)=-COSKZ


£ /(z)dz=2/»[£Re8.of /(z ) at each pole with in C). ..(I)

\

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For the poles offtz) : put cosn t * 0

=> Jt z=± —,± — ,±— ........2 2 2

7 6 |

1 3 5=> z = ± - , ± - ,2 2 2 are all simple pole?.

At poles 2=±“ , then | z|=I ±-~ < 1, inside the circle C.2 Z Z

3 3 3 At poles z~ ±—, then |* H ±—1=—* 1» outside the circle C.

2 6 /

Hence only two poles * = ~ and ~ ~ He inside the circle C.

• [Res/(z)] | =lim ( z -1/2).*

which is — from, using V Hospital rule

=limi Z~*2

Similarly,

-a inxz .n tf.(-sin;r/2) it

[Res f (z)] , = lim (z + 1/2 ). ____ usingV Hospital rule

= lim

C08XZ

erX

-i (-sin nz)n (-;r)si n (-* / 2 ) (->r )(-l ) n

Hence (1) becomes :

r ez dz = 2nicos/rz

e^ e ^K 7t

( e%V

- -A is ia h—.2

mple 1.87 («) ; Evaluate J ^ _ j ^ z _ ^ C is the circle { z \ * ■?.

_2r

fon. Let ■**>“ 771

[v sin ^ /2 = 1]

Ans,

Fef 20101

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We know that 1

| f(z)dz * 2ni * [Z Res. of f(z) at each pole in C] ... (1)c

ft.- For the poles of J(z) : put ( z - 1) (2 -2 ) = 0=> z = l and z = 2 are simple poles.

At pole z * I, then |z | = | 1 | = 1 < 3, inside the circle C. At pole z - 2, then 12 1* | 2 }* 2 < 3, inside the circle C.Hence both simple poles are He inside of C.

2x

FuNcrtbxg Vdw»ai a s | 77

e2x e2

= lim------- = ----- = — e2 z-*i (z-2 ) (-1)

e2*V

J ) - (ii) [Res-/(z)]...2 = Ji» (^ -2 ).( i _ 1)(2_ 2)

[ - j s ^ o - s - *Hence (1) becomes :

• 1 (z -I )( z -2 )A = 2t u [e* - e2]. Ans.

t cos k z * , , ,Mxsmple 1.87 (A) : Evaluate Jc l ) ( z - 2 ) where C is the circle \ z \ - 3.

[RGPV June 2003 and Dec. 2011J

cos x z2n.

M\‘

Let ^ z) = U -l) (z -2 ) '

£ f(z)dz=2m[ZJte8.of f(z) at each pole with in CJ. •■•0)t

For the poles of f(z) : put (z -1 )( z -2 ) = 0

=> z ~ 1and z = 2 are simple poles.

At pole z - 1, then |z |= | 1h i < 3, inside the circle C.

At pole z = 2, then | z |=121=2 < 3, inside the circle C.

Hence both poles are He inside of C.

ro VI v / COSJTZ2 CQ8/TZ2 COB7T -1 ,(0 [R -/W L , =lun <* - Im

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78 | EwnnmwiQ>fcBW|Mnc»m

(ii) .[Res/’(«)]„, *Um (z -2 )(*-1X2-2) (2-1) I U

cos**3 cos4* 1

Hence (1) becomes:

d*« 2*i (1 + 1) *4*i.

Solution. Let

By residue theorem:

Jc /■(*)<&=2*i[E Res. of/(*) at each pole with in C].

For the pole of f(z) : put (* -* /6 ) s * 0

=> z * * /6 is a pole of order m = 3

At the pole *= */6, then |* |s |* /6 | * 0.523 <1, inside the circle C.

We know that:

(m-1)! *-** dz"1'*{(*-o)m/(*)}

^ [Re8 f (*)I,«,/8(M«3' (* -* /6 )a

sin**

lim [30 sin4 * cos* * - 6 sin6 z)2 «-**/•

Hence (1) becomes:

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FuNC^WolaH*VW dAB^s I79

Example 1.89 : Evaluate f t a n z d z , where Cis the circle |* | = 2.

„ sin zn. Let /(z) = tanz = - — .

COS 2

By residue theorem:

f /(z)dz=2*»[ERe8.of /(z) at each pole with in C]. ...(1)I*

l t;y For the poles of/fe): put coaz = 0 => z * ±*72, ± 3*72, ±5*72..........

all are simple poles.

\ At the poles z= ± */2 , then | e |=| ±*721= jt/2 = 1.57 < 2, inside the circle C.

At the poles z=±3*/2, then | z |=l ±3*/21=3*72 = 4.71 < 2 , outside the circle C.Hence only two poles z * *72 and z * -*72 lie within circle C.

[Res/ Xz)],.* ,, = lim (z->r / 2 )^ ^ ,z-wr/l COS 2

0[which is ~ form, using V hospital mle]

- Jim Qinz + (z-nl2)ca9>z _ 1 ^ ^

. -sin 2 -1

^ y(- and /(«)],._,/* “ ,55/, (z + *f 2) using L’hospital mle.

+v' = lim *MZ + (z + ffl2)cQ&z _ -1 _ L-sin 2 1

Hence (1) becomes:

f tanzdz = 2*t[-l + (-1)] = -Am. Ans. j c

APPLICATION OFRESIDUES TO EVALUATE REAL INTEGRALSW n f2*

Type (I) : Integrals of the type I i?(cos0,8in0)<20, where F (cos#, sin0 ) is a&89V

rational function of cos 9 and sin£, such integrals can be reduced to complex line integrals by

the substitution,

dz put z=ei0>so that dz=i eie dO, => “ ^ “ 77

U " Also, o o . , = ^ l = i ( , + i ) and Sin*

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As 0 varies from Oto 2?r, then z moves one round the unit circle in the anti clockwise direction,

i.e., \z \ = \

f F(cos 0, sin B)d8~ = £ « * ) & ,

where C is the unit circle I z 1=1.

Then above integral on the rigit can be evaluated by using the residue theorem.

C f (x) —— tfcc, where f(x) and F(x) are polynomials in x such

xf(x)that >0 as x oo, such that F(x) has no zeros on the real axis.

Consider the integrals / = ~ ^ d z , over the closed contour C consisting of the real axis

from -R to R and the semi circle C, of radius R in the upper half plane. We take R large «*mgh,

so that all the poles of

m

f(z) F(z) in the upper half plane lie with in C. By Residue theorem we have

/(*) .= [ sum of the residue of *n ^ u P P e r pi®0*]

r f f (z )dz cR f ix ) , „ . f (z)

k ' F(z) "+ 1/? 'F(x) m sum of residue of f H ) in

plane] ...(1)

If we put z ~ Re*e in the first integral on the left side then R is constant on C\ and a» z moves

along C| and 8 varies front 0 to n.

• H •» F (R e i e )c F(z) R e ‘’ id 0

For large R,

r

W e '

F (R e l

f f(R) 1 R e ia&\ is of the order of real part of

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. tFu n c t i o n s o t C o m p u e x Va r i a b l e s |0 1

" r F(f leT ^ R e ‘ ) t d 0 - * ° when « - » «

Hence from (I), we have

tv -» f ( z ) E f{X) , ^- ^~ ^dz -2m [ sum of the residue of in the upper half plane]

f2* d8'e 1.90: Using contour integration evaluate the integral

LetH

2 + cose*

[RGPV June 2002, Dec. 2003 and June 2011/

2* dO

2+cos£

Put z ~ e iB => dz= et9 .id $ = > d 0 ^ -^ - =» d 0 ~ êl9.i iz

«** +e~ie 1Since eos£*— — -----=> co&0=^ z + —\, then we have

3 -

/ * f2# r 1 dz f 2z dz

*# 2 + co80 g + 1 ( z + 1 '\ i z $4z+ z^ * l i z

IR )dz

t-

I

_ 2 f azor I - j )c where C is unit circle |z |*1 ...(1)

Let /(*)=■z2+4* + l

- 4 ± V l6 ~ 4 -4 ± 2 > /§For poJes : put z+ 4 2 -f 1 =0 =>z « ------ ------- o z -

2 2

=> z =-2±V?=> z = - ? + V ian d z -- 2 -V 3 are simple poles.

4 / pole z = - 2+ V?, then | ? |=1 -2 + VJ 1-0.267< 1, inside the circle C.

ifc/>e/e z a - 2 - V5, then U N - 2 - V II*2.73 ft 1 i.e., out side of G

Thus, only the pole z =. -2 + V3 fws inside the unit circle |z |= l but z= -2 -V 3 * s

outside the circle C.Hence by Residue theorem :

Jc f(z) dz =2*x [Ree /(-2 + V3)]

=2** f lim [ z - (-2 + V§) /(«)]]

8/11/2019 DK Jain M3


$2 | En6mzkmu M xn n w ics ^ f l

=2xi

=2m r (»+2 -Vi>) 1 r * * «~<-«^>(*+2-,/3)(*+2+v/3)J L ~ +

Hence (1) becomes:

fi* dO 2T f** a v z f 2 m 2x

/ = I 2 ^ 0 ’ 1 1 m d *=m * n2+CO80eU dQ

Example 1.91 •* Evaluate

pin

rSolution. Let

H

5+3sia0

dO

5+3sin0

Put z =ei6 => dz=i ei0d&=> ~ = d 0vz

e i0 -e~ i9 i f OSince, s in 0 = — —:— => sin 0=—• z —2i 2i\, z j

- (2* r Jo 5 + 3sin0 i :

-lie

- H

dz

5+3

dz

[ S R I "

5 + 3. J 72s

<z} - l '

* / j

dz

lOtar+BQg* -1)2iz

2ick

a0w+32*-3)

2dz

or

(lbw + Ssr* -3 )

/ = 2 ^ /(z)dz

(z + 2-V 3)(z+ 2 + /3)]

Am,

4 fRGPV Dec. 20011

8/11/2019 DK Jain M3


Fu n c t i o n s o p C o s W l ex V a r ia b l e s

where 'C is the unit circle, Le., C : |zj = 1 and — t : — r 3z2 + 1012-3

For poles: put 3z2+ 10 iz-3=0-10*±>M 00+36 -10f±/8

6

-6i±4i - -K+4i . 0 — 4i *=— 3 — • ^ en a " — 3 — ond — 3 — (say)

ar=-~ and p *-3 i are simple pole3

At pole z= a= -i/3 ,th en 1z |=| —i / 31=1/3 < 1, inside the circle C.

At pole z *£ - -3 i ,then |zH -3£ | = 3 < 1 outside the circle C.

'• f Z)=3 ? + lOi* -3 ~ 3(* - a)(* - £) " (2)

We know that:

1

3{z -P)[Ree/(^)=La = Km (z - ar)/(*)-limJ-*«f !«*«

i i _2 .

By using residue theorem:

Jc f(z)dz * 2xi (IR es /(z)at pole z =« in C).

0 . 1 2*=2bnx—*— =— .

8t 8 4

Hence (1) becomes : /■ 2 Jc f(z)dz=2. *r/4=*72 Ans.

Example 1.92.*Evaluate £* (g + « e y »,a> 6 > 9 ' [RGPVJune2004/

olution. Let 7= I** ------ — ---- r ■*> (O + &CO80)2

Put zê* => dz = e* idO = ~~iz

e +e z+ l /z 2*+l.. CO80 -------------- ------------- ----------- .

2 2 2z

8/11/2019 DK Jain M3


8 4 -I

and limit 0-+O to 2/r reduces to C:\z \ - \

j - f 2* ^ f 1(a + fccos)2 *c

a + b

(z2+1)

2 Z

IZ

\

■ i t z d z

(62; +2az+b)z *6* Jc ( , 2a V

I2+T 2+,J

£ f(z)dz, where C:|«|=l

dz

11 Here f(z)~ 2 2o

2 + — *+1

...d)

For the poles o f/f e j: put *2 + ~ * + I= 0

or I

Z ~ 2 ~P 4

- a ± J ( a 2 - b2)

6

—a+-v/(o* - 62) —a —J(a2 -b*)Let <*=---- — --------- and /?=------ ------------ are two poles of order m = 2.

0 o

Since, a >6>0,l/? |< l and that |a/?|= l, it follows |a |< l . Hence 2 = a is the only pole

of order two with in circle C. Hence pole 2 = $ outside o f C.

f(z) can be written as : ^ *

Now, Residue ^t a pole 2 = 0 of order m ;

d"

(m-l)t dz*- [ (2 -a r/(2 )}

.<*)•(*- f f \

=lim ± \ J L -,} -Km*-** d z \ ( z - f l ) J *-** ( z - P )

=1-m z - P - 2 z (a + P)*-* (2-^)3 “ (a- /?)3

8/11/2019 DK Jain M3


I

F u n c tio n s o p C o m p le x V a r ia b le * | 85*

-2 a

*(-i)x-ab2

By residue theorem:

** Jc (Sum of the residues of the poles which lit: with in C)

r ab2 1~2 j A

\ 4 (a2 - b 2)3/2 J

Hence (I) becomes : f (z )dz

-..(2)

[Using (2)]

=—~x2mib

ab2 1 2jta

• f —*• 1,0 (a +

4 (a2 -& 2)3/2

d0 2 xa

(a%-b2),3/2

(a + fc coe0)2 (a* - 62)3'* •

f*« cos 26xample 1.93 .* Evaluate J, 5 +4 00*0 Avcontour Integration.

olution. Put z=e‘s => dz= ie'e => ~ = d S • iz

„ e^ + e '" i f O l | V + l \ 1Since oorf ------- — ‘ 2{2 + z h { — ;

1 ' 2 1 'j I= +? j =I

/= f2' 00826 i c f 2 v *2 J <fe_ Jo 5 + 4cos0 Jc (*2 + 1 iz * where C is the circle \z I~1.

5 + 2 l — J

- h i e S & l L i * mh l c *«>*

Ans.

[RGPV June 200SJ

and cos20=V * J

...(0

8/11/2019 DK Jain M3


8 6 1 E m w m pm M

zA+ 1where /(*)=- 2/_ 2 - —„v

z (2z +5z +2)

For the poles of f(z) : put z2(2z2 +5s+2)=0

-5±V25-16i.e., 2 = 0, pole of order 2, and z - ---------------

4

-5+3 -5-3 Le.t z=-

4

z ~~ \ anc* -2, ^ simple poles.

At pole2 =»0, then | z |- | 0 |=0<1, inside the given circle C.

At pole 2=-1/2, then |z |= |-1/21=1/2 <1, inside the given circle C.

At pole2 = 2, then I z 1=1 -21=2 <1, outside the given circle C.

... [Re»rt2)ko,m_2, =— Um — {(*-0)2./(*)}

(*4 + 1)- i nn— ■i 1■ 1■

zôdz[ z2(2z +5ar+2)

[B*>/(*)Jz-x = J j k + Y d ' f W

= lim (z+1/2)-^---- & t.fj------ L --------i--------------------*--------- 1* .2(*+l/2 )(z +2) L **(2**+fo + 2) 2«*(*+l/2)(« + 2)J

- M m . J Z . ‘ .2z*(s + 2) 12

By using residue theorem;

Jc f(z)dz=2xi* (Sum of residue at each pole in C)

I 4 12 J 12 3

Hence (1) becomes:

r2* cob 2 # j q 1 m x

Jo 5 + 4cos0 _ 2 j X3 ~ 6 Am*

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FuNcnbNroFCo mmJex Vahabl e s | 87t

r(l + 2co«0) ,(S+4©m8) [RGPV Dec, 2004/

T ‘[(L + 2CQ80) 1 l + 2coe0 r * * * ” ' j (5 + 4co8g) 2^° [v f l» r- « - /< « ]

Sincc = co 80+ iein0

/ = Real part (R.P) of i f

Put z - p*6 => ~r- = d$t and cos 0 = — * ~ e t2 2z

7 = R.P of i f 1+2* 4* where, C is the unit circle I z |= 1.

= R . p . o f i j r - ^ p -2i Jc ^2Z+

(1 + 2z)dz

52 + 2

f = R.P.of — f <i+3 g > ± -2i Jc (2 + 2)(l+ 22)

- R ^ o f ^ r f r - i r r c fe2i Jc (2 + 2)

7 = RP*of^ Jc f(*)dz where, f(z) = ^ - ...(1)

For the pole o f : put 2 + 2 = 0 =>2 = -2.

z = -2, is a pole but it does not lie within C le. 12 1= 1,

because (2 1=4 —2 1= 2*1.Hence by Cauchy's residue theorem

£ f{z)dz = 2m(0) * 0.

Hence (1) becomes: J = R.P. of ~ x 0 = 02i

rl + 2CO80 A

(5 + 4cos0) A"5-

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Example f .95 .* Evaluate by contour integration, £ TTJdd

(5-3«in0)2 *

dzSolution, Put Z = d z - i e ^ d O => -r- = d0

iz

Since sin 9 -e - e-is z - z -I

2i

- r

2i 2i

1

(5-3s in^) i

dz

5 -3

where C is the unit circle | z\= 1.

1 dz

- I 5 -

3<*2 -1)

az r

iz Jc

4iz"T"

2iz J

- L

iz JC (10w -3z2 + 3)2

4 izdz

iz

4 izdz

(3 « + l)J(w+3)*

where f(z) = -

(3z- i )2(z-3i )

4iz

(3z- i)2(z-3i)2

For the poles off(z) : put (3z- i )2(z -3 i? = 0

=> z - if 3 and z - 3 i are poles, each of order 2.

At pole z - i/3, then | z | ={i / 3 1= 113 < 1, inside the circle C.

At pole z - 3/, then |z|=J 3i |*=3 < 1, outside the circle C.

Thus, pole z ® i/3 of order m = 2, inside the circle C.

[R ao /M U s..., <*&

= lim .£ { . (^ - ‘ /3)24« 1

4i d — lim ------- ~ l im

9

(z-3i)2( \) -2z(z-3i)

(z - 3£)4

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Functions or ft*

By residue theorem: i.

f(z)dz = 2» (sum of residue at each pole in C.)

* . f SH 5 r = 2 « x ^ - — J = — .

r f2» dO _ 5 jt Hence (1) becomes . 28 J, (5 -3sin0 )2 * 32* An8‘

Example 1.96; Apply calculus o f residue to prove that

r co*2Q 2u a 1

( l -2 ac o a 0 + a* ) ( 1 -a 2)* <1 *

[RGPV June 2003, June 2009 and Feb. 2010]

Solution. Put, z = et& => dz = et0.idO => dO - —

- e*+ e* 1 ( 1)Since, «»* = — — - j r + -J

v i l I - S ? J dz 1 f ( * 4 + l) < fe

, " TT T 2i tT T ” —l+O

2» 22(z -o )( l~ aar) 2i

*4+l

z ( z - a z - a + a z)

...(1)

where C is the unit circle, C:| z i= 1 and f(z) - — w, . •2 ( z —a)(L-az)

For the poles offfz) : put z 2(* -a )( l-a z ) = 0

=£>z = 0 is a sole of order 2 and z = a, z = — are two simple poles.

At pole z - 0, then | 2 |H 0 1= 0 < 1, inside the circle C.

At pole z = a, then | z |=| a 1=a < 1(v a 2 < 1), inside the circle C.

At pole z - \fa, then | z |-1 1/a |< 1(v a 2 < I), outside the circle C.

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90 1 I

Thus only pole z = 0 (order 2) and z = a (simple pole) are inside the circle C.

[Ree* f ( z )\z-a =lim[(z-o)]/(z)

= lim*-KI

and

'(Z-„) (£ l± lL _]=liin (£l±iL„ toliu2: (2 - a ) (1- cup) J *-** z*(l-az) a2( l - o s ) ‘

1! dz [ Z ‘J i z - o H l - a z ) ) ^ { d z [ z - a f -a + a 4z JJ^

3) - ( z 4 + l ) ( l - 2az + a2)] (1 + a2)

- o + o V “ o 2

f( * -o z 2- a + a 2;g)(4z3)-»

| ( z -az2

By Residue theorem: /(z)dz = 2*x[ERes/(z)at each pole in C.]

= 2mo 4+ l l + o 2

o 2( l - o 2) o2

Hence (1) becomes:

r*» cos 2#

4wro32 \ *

1 4iô* 2na%/ « f * ____ —"■■■ d0 = — x(l-aocosd +o* ) 2i ( l- o 2) l-o *

f * (id 4k

(2 + cos 8)2 ~ 3^3 *

Solution. Put z = e*, then cosO = — z + i j = j

• J _ f2* rifl f 1 dz■*° (2 + cos0)2 «

Proved.

2+i £ ± l2 s

7- , where, C is circle |z | = 1.

...(i)

where, /(*) “(z2 + 4z + l)2 *

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For the poles of f(z) :

/ . . „ i , , „ -4 ± x /l6 -4 -4±2>/3 put, + 4 z+ l) = 0 =>2r + 4ar+l=0=>z = ---------------* ----- ------

i.e. z - -2±V 3 are poles of order two.Here, pole * = -2 + ^ 3 Hes inside the ciicle |*|= 1,

because [2 1=|-2 + >/3|=-2 + 3 <1.

But pole * = -2 - V3 outside of C.

Residue at (2 * -2+ i/3 ) of order m = 2 is

Fu n c t io n s o f C o m m a Va r i a b l e s | 91

(2 + 2->/§)*.« j

i#»-s+wS

k f * ,-11<& (2 + 2-Va^(2+2+V3)*J dz 1(2+2+$)* J]

(z + 2+V 3)2.l -g .2 (g + 2 + V3)

(z+ 2+ ^ 3)4 -2+V3

2 + V5 + 2->/3 = 4 1 1

~ (-2 + + 2!+«$)* * 8.3%/3 “ 2.3^3 “ i j s

By Residue theorem:

1 ffi£ f (z)dz » 2 » [£ R et/(z ) at pole in C ] s ^ x *

r r** 4 *i 4*Hence (1) becomes: ^ = J, (2Ve0itf)r = T Xi ^ = i ^ ' A“ -

#st

MVlr 1.M r AmfeMir j, ^ resUuethMfem'

Mutton. Put z = e*8 =» coa6a^ (* + ;z) anc*S’n ^ = 2i (Z - zJ

/= (*' --------- *£------------f «» (3 -2 co s0 + sin 0) Jc

d$ t I dz

2co80+ein0) ” *(*2+l) (*2- l ) 'w

3” 2 2z 2*2

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921En g iM ^ |I # I 9 I M |qbA

- i c *1(l-20+6* e-( l + 20

2 r 1

dz

L( l -2 i ) *(1-2© (1 - 2f) j

3 6iz (1 + 2*)For pole of/fe>: put * + j ^ — f T ^

=» «2(l -2 i)+6i* -(l + 2i) = 0

£ b = ( r H ) J c ...d)

- 6 i ± J - 36+4(5) - 6i ± 4t -2 i , -10* z ~ ------- — ----- -------- --------------------------a n d --------

2(1-2 i) 2(1-2*) 20 -2* ) 2(1 -20

“ j~ 2 ^,ap l , # i e *z ~ a y arc S*®P ®P°l®s*

—IAt pole z = a = — , then |* |-

which is inside the circle C.

- i

1-2*

-5 iAt pole z - p = ——— outside the circle C.

( l - 2 t )

[R esf(z)]Zma = lim (* - a).f(z) = lim (z - a ). 1*-K* 9-UX

v/(*) =6w (1+20 ( z - a ) ( z - 0 )

1 F ^ a T T ^ T

(1-20( -4 , 5i } 4iU - 2 i 1 - 2 * J

By Residue theorem:

f ( z)dz ~ 2m (£R es/(z)at each pole in C]

- . (1-2 0 = 2 j b x X— i = - ( l - 2 0 .

4t 2Hence (1) becomes:

Ans.

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C<&lf*fcVARIABLE8 | 93)

sin*6 JO %n. p iT T E *mmpie 1.99: Prove that J8 (a + b coa 9) = jJT»a ~ v 'a )> wAcrr a > b > 0.

[RGPV June 2008/

m d n u t / . p - J S S * * , <!». p . g r P g g L ^“www- Jo o + 6coe0 2(a+6cos0)

f*' 1-V* ^= Real part of J0 2a + 2*ooe*

Putting z = e10 => and cos# =* -2 we get

/= Real part of f - — rr»T“ >where C is the unit circle \z \= 1Jc 2o+b(z+afl) i z 1 1

-Realpartof j £ <** = Real part of £ /(*)<&,

*t \ _ (I"**)where f M m w ? 7 £ i 7 i )

For the poles of/fc): put i(bz2 +2az+b) = 0 => bz1 +2az +b = 0

ten-

-2a±nJ{4a2 -4 6 2} - o ± V(o2 -&2)Z = . . . . . - . - . - - - . s . . --------------

26 6

I) Let or ------- Jti--------- and P * ------ — - ------ are simple poles.o o

u Since a > 6 > 0, so that I p |> 1, hence pole z * p is outside the circle C.

S' Also I ay? I = 1, so we have I or |< 1. Hence z *=a is the only pole which lies within

C: |* | - l .

[Res f(z ))tma = Urn (z~a) f (z ) [By definition]

(I-**)

r f t \ 1” ^ I - 2 */(*)= T

i(bz* + 2oz + b) ib(z - or)(z - P)

.H m = , g , - . l (g I - . . l £ . [ v ^ = l]*-•« i6(z->£) ib (a -p) , ib (a- f i ) i b ( a- p) ib

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94 | En g b o &un g M/mtEMAncs-IIl

Hence by Cauchy's residue theorem (1) becomes :

/ = Real part of £ f(z)dz - 2m [sum of residues at poles within C]

Example 1.100: Using the calculus o f residue, prove that

r x ai n( ax )J. 2

Proved.

Solution. Let /(*) =

2.***

where C is the contour consisting of

(i) Real axis from -R to R

(ii) Semi-circle C r of radius A:! z \= R

p R

y

ydz, y S

I 1

Ssv\ ^ 11

R ' 0 +R

Hence £ f (z)dz = J ^ f(x)dx + Jc#/(*)<&

To prove that, as R -+<*>, f(z)dz -► 0.

On the real axis z = x and on the semi-circle z ^ R . e * and limit 0 varies from 0 to n.

Now consider £ f{R»J0) .B . e P i d 0 * £ f( z )dz

R eie eia(Rco*e+iR'toe) Reie M

...d)

—(2)

-r-r

v « * ) - ze

z*+k2

R .e ^ .e * M B

R 2e2Uf +k2

[Now Divide N r and Dr by R2e**]

~ r

Sincel« . e lit I

11+ lr ./2 (cos20 - i sin 20) |

g-sfirin0.1

I J l+ k 2j r 2(coe2?-imn20? I

[.• e“2W= cos20-isin20]

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' fak ing real pamoniy]

Fu n c t io n s o f C o m pI e x Va r ia b l e s | 9 §

-afteinff

e-aRnnB( l- fc’ i r 1)

Therefore from (2) and (3) we have

...(3)

e * . i d 6

( J l V W )

...(4)

[ v io r0 £ 0 iir l2 ,^ ~ r -'z.‘l l x e

r Jo ^ ^ • d ea -f c ’/ r V 0

2 r*/2 -alt— £ -------- T----5-1 e * d6

a - f c ,R s) »

[-

...(5)

By Jordan inequality — £ ^ 1or sin# £ ~ It & ft J

(1- k l R * )

T#/* ,-alHUVir

~aR.~ ft J

l-Jfe2i r J [ 'a R

-] *72,-aR(2*)/jr

- * ___ (e*^ -1)= * 0 “ «““*)...(6)of l( l- fe2i r 2) af l ( l - * 2j r 2)

/.A s ft -> oo, we get Jc</(2)<fe -+ 0

Therefore (1) gives

Jc f(z)dz = j* f(x)dx + 0

f(x)dx = f (z )dz - 2m (Sum of Residues) [vBy residue theorem]

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The pole off(z) are »**±ik of which z = +ik ties m upper tetfptane 1*1=1.

£ f (x)dx - £

' 2 4:#4 . j^ j * ^ \ 2ia j

2m-z~e * » *2

»«**

Henee- £ x ’ h k r d x =

Equaling (he imaginary parts

£xsinax ,

or

or r*2 +**

x sin ax , jr _<*

7 T F T ‘ '

f* cosmx . Example / ,101: iy contour integration |

Solution. Let J * |* BE?*1*,0 X* + 1

dx.

Taking* /(*)==.mix

(* +0

in: .

/= f /(«)<&- f *4— ^* * (*2 +i) - * o

Where C is the con**sr consisting 6f the ».(i) Real axis from -R to R

(ii) Semi-circle CR of radius R in the upper half plane.

Then £ /(*)<£* - £* / ( x ) d x + / ( * ) d *

To prove that as £ -► *>, : f f ( z )dz -> 0" r- •/.'• 5

...(7)

fRGPV Dec. 2*06]

m

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N

F u n c t io n s o f QomnJExVABUWGS | 9 7

z |2 -I |

[.-. U 2+ l | -U 2-( - t ) | a | z |2 - l]

On semi-circle z * R ^ 6 and Q varies from 0 to it. ... ..T

\t\=R=>\dz\-\R.t?°.ide\=RdO

and I M j_j g-foJfsInPgfatRcosff |_ e~mRnn6

to •

2fl r /J

- s € -mR20lx40

j By Jordon’s inequality — or — ^sintfÔ

2R

(«*-1) m itt

ir .

Ktl

n 0

-2 R x x r , ,

( t f - D M l

Jt

xR

Hence

Hetce

m(fi2-I)f l

/(z)<2z|-»0 ae fl -> °o

;i) becomes:

£ f (z)dz - £ f (x)dx => £ f (x)dx « 2m [$um ofresidue]

Nowj fe) has simple pole at z * ±* of which only z = +i lies in the upper half plane;

1*1-1.

...(2)

imt .-it(z-Qe

(z- i ) ( t+iy 2i

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9 # 1BkmmmHwoMxnmucnc*ms

Hence (2) becomes:

£ —r — dx =■2m x — —- m x 2+l 2i

Equating Real parts on both side, we get

Ccosmx , r° cosmx , k

Example 1.102 : Evaluate by contour integration £ y ^ » i /o > 0 .

Solution. Let /(«) = — — 4 and L /(*)<** = f ; - 4 dz, z*+a* Jc z +<y

where C is contour consisting of (i) Real axis from -R to R and

(ii) Semicircle C* of radius R.Therefore

£ f (z)dz = /(*)d*+ f(z)dz

For fee poles f[z) put z 4 + a 4 * 0 => z 4 * - a 4 => 2 = o (-I)

■mAns.

...(1)

(- i) !/4 = (eos(2n + l)/r + i s in ( 2 * + i» l/4]

/.e. z4 « o 4. e ^ l)' => 2 = a w h e r e w = 0, 1, 2, 3

le . The poles are at z = oe*>/4,aei3' /4, ae*5*M, aeilx!4 of which first two lies in the

upper half plane 1*1=1.

Let a denote any one of these poles then a * = [v z4+a* - o]

Now [Res. f(z)]gma =h m ( z - a ) f ( z ) >l-W

= lim (z-or)—r——r ar F— z4+a4 LO

V t *= lun—-7- = — -*-** 4z 4a'

(i) W**- /<*>]«**"

Of

4 a 4

i*/4

formusingL’ hospital rule J

a »

4o4

ae,»3jr/4

4a

...(2)

[By (2)1

[By (2)3

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F u n c t io n s o p C o m p l e x Va r ia b l e s ( J l i

- 1 r . iwU . IXmU j __ I |c**^4 t C*3* ^ ].•. Sum of the Residues = r t 0 ®*** +ae

- U 2 i \ e i" * - t i "*' \ i i J 2

j - i r

Now, |jc. A * H “ | k 7 ^ ^ .

...(3)

B;

r* i?d<9 D= —3--------r -»0 a s I f -t o o

Jo J f - a 4

Hence (1) becomes:

£ f ( z ) d z = f t f ( x ) d x + 0

=» f° —T— -T- f /(*)<far»2*ix(suiii of residues at each poles in C)*-■ r 1 j./** JC* + a Jc

=:2»4a

dx

x

l 2 o > '

7 r a - f Jo x 4 + a4 * a3

r dx x

jc4 + a 4 i j l a 1 4a 3Ans.

4? TAYLOR'S SERIESIf a function^ is analytic at all the points with in a circle C with its centre at the points 'd

* ’ and radius r, then

/( * ) = £ a* (*-«)* ; where£ * , n !

/ W - £ with 12 - 0 1<#*-0 n '

or f(z)=f (a)+^— ^(z - o ) + ^ p (2 - d f + ...... + (z - a)* +1! Z! A!

LAURENTS SERIES

If/(^ is analytic in a region A, bounded by two concentric circles C\ and C2 of radius R\ and

Ri (jR, >R2) with center V then for all z in &

/(z ) = 2 a,(*-a)"+ 2 ] 7 -^ x r, with |z-o|<JZ *-0 *«i \Z ~ a)

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8/11/2019 DK Jain M3


1.105 ; Expand in the series the/unction f (z )

FbNcnQNS o p C o m pl e x Va r i a b l e s | 101

1in the regions

V - 3 * + 2)

(i) 0 < \ z \ < l (U) 1 < \ z \ < 2 (W ) \ z \ > 2 (tv) 0 < \ z - l \ < l[RGPV Dec. 2005]

or +

Expand A* ) * ( *- 2) ** *** reS^>n \z\> 2.

Resolving into partial fractions, we get

1 I 1 1

[RGPV June, 2009]

/(*)=*2 -3 z + 2 ( z - 2 ) ( z - \ ) z - 2 z - 1

(i) When 0<UI<1: we have \z \<\ or

F ro m (l):/W = — - 2^ - -

•< -3

— - i h itv a - * r , = £

= 1 x 4 i f lt-0 ta*0

’" S j 1 2**1}**• if 0<|*|<1.

(ii) When 1<|*|<2: i.e, \z\>\ and |z |< 2 => | j | <1 and |j < 1

Ans.

Using binomial expansion, we get

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(Hi) WhenUf>2: Since \ z |>2:=>

1 I

1 1<x=>

1<1 and

2

2 2 z z< h

From (1): /(*) =

2r —2

- K ' - r - K ' - i r -Using binomial expansion, we get

m = - t ( i f - t ( 2 " - l) - i r . if |z |> 2. z t 0 \ Z ) *£* W f t 3? 1 1

(iv) When 0 < |* -1 |< I:1 1

From (1): /(*) *ar-2 ar-1 z - 1-1 z -1 ,

1 t

- f l - ( * - l ) ] 2 ~ l

= - Z (2 -1 )* -<*-! )-*

-i

Ans.

J*»#

[‘.•(I- * ) " 1 =S**J. Ans.

(a?-2) ( z + 8) Example 1.106; Obtain the expression fo r ( , + ! ) (*+ 4) which me voUd when

a) u i < / m i < \ z \ < 4 m u i > *

z 2- 4 , 5*+8 , 1 4Solution. Let, / ( * ) - — ----------= > - ------ —— s - - l — = 1

z 2 +5*+4 (*+4)(*+l) 1+* 4+2

, {v Reaolving partial fractions}

(i) For |* | <1: we have

8/11/2019 DK Jain M3


Funct ions cwCompiex Variables j 103

= l - l - £ (-!)»*• - 1 - j ; (-1)” ^ " .H»1

= - l + £ (~1 )*+1[l + 4~*]z*; when |ar|<l.M l

1 \z I(ii) For 1<| z |< 4: we have 1 —: < 1 and — < 1 i.e.,

\z | 4

, - > - K 4 r - H F - 4 [ , - w - -

1<1 and

z

z 4<1.

z f z

‘ - 4 +U ' -

= S ( - o ’Ml

which is valid in 1<\z |<4.

4 1 1(iii) For I z |> 4: we have 7—7 < 1 and 7—7 < —< 1

I z I \z \ 4

, I ) "

•4 [4

= i + i ^ a + 4 * * 1)•-0 z

= 1 + £ (-1)*(1 + wk®n \ z\>4.

„ * + i

A+l

A bs .

z + 3mmpie 1.107; Expand f(z ) « — -—— in powers o f s.

(Z —Z —Z /Z

(!) within the unit circle about the origin,

(ii) within the onnutus region between the concentric circles about the origin having radii 1 and 2 respectively.(Ill) the exterior to the circle with centra as origin and radius 2 Le. fo r |z|>2.

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z + 3

1 0 4 | E N 6 ffe £ R r aM M H m n c » in

(* + 3 )t / \ _ 46 *' _ > * j )Solution. Let « « * t f _ z _ 2) z ° *(* + l)(z-2 )‘

Hence resolving into partial fractions.

„ . 3 2 5

-i

we have

2 z 3(2+1) 6(*-2)

(i) For 0 <1 z |< 1, we have

- t t [§<-»■-$)■]*•-I

(ii) For 1 < |2 1<2 => and M < 2 = > |||< 1 .

, 3 2 ( , 1Y‘ 5 (. zY ‘

/ w " - * +5 l 1+; J - l a ^ - a j

= _ i . + 2 f , . i 4 4 J _ ± f i + £ + J + J2z &z I z s? £ J 12|_ 2 2 2

\ **■

~ i — 2-f (-i)*-L—l y r*r ** 3 * a ** “ S ' W

i i 121 ,(iii) For \ z |>2, => j T p J . |z | we have

3 2 r. 1 X 1 I 5 r, 2 l 2 23 1= +-^jK+-*+“5-+“5'+...

2z 3z L * *2 z3 J &*[_ * * z3 J

[v |z |<1]

Ans.

Ans.

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Fu n c t i o n s o f C o m p i£ x Va r i a b l e s 1 1 0 5

>1.108; Obtain the Taylor's or Laurent's series which represents the function.

f<z>= lf /+ V > 7*+ i> '

(l) | r | < / m / < | z | < ZResolving/fo) into partial fractions, we get

5|_* + 2 z2 + l j

(i) For | z | < 1, we have

10

/w 4 i [ !+f ) ' 4 <2- 2)0” ! r '

-[ (1 -z2 + z4 - z 6 + ... + (-I)**2" +...]2"

10 Ans.

This series being in the positive powers of z represents ’Itoylor’s expansion for ffz).(ii) For 1 <) z |<2, we have

- - i t “ r & t CIRCLE AND RADIUS OF CONVERGENCELet power series about z = a.

fr m '

, / (* > = £ ...(1)m*

The interior circle |z - a\ < R which include all the values of z fotf (z), is called the circle of convergence o f series (1). The R is called radius of convergence of series (1).

ar*:

Let z = a be centre of circle and f{z) be analytic function in the circle j s - o je i? , except the poles

a,,da,a, soon.

Then, R = mm. distance of poles a ,, o2>. _ from the centre z = a.

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1

106 | En g in e e r in g Ma t h e m a t ic s -H I

Example 1.109 .* f(z) =

Solution. Let f{z) =

( z - l ) ( z 3 )

indicate the circle o f convergence.

1 1

. Expand the function in Taylor's series about z **2 ani

( * - 1 ) 0 - 3 ) z 2 - 4a:+ 3

t

( z - 2)z~l= -11 - ( z - 2)2]'1= - [1 + (2 - 2)s + (z - 2)4 +...]

or / ( z ) - (~ 2) , whjch is Taylor’s series about z = 2.

Circle o f convergence:

Here centre is z = 2 i.e., C ente r* (2,0)

The poles f(z ) are z = 1 and 3 i.e., z = (1, 0) and (3, 0).distance of pole z = (1,0) from the centre z = (2, 0)

d, =V( 2 - l ) 2 +0J =1Also distance of pole z = (3, 0) from the centre z - (2, 0)

di = yj(3 -2 )z+02 =1

Radius of convergence; /? = min {d,, } = I.

Thus circle of conveiênce = | z - a | < z - 2 J<1. Ans.

1.

2.

3.

Exercise-l(A)

Prove that the function | z \2 is continuous every where but no where differentiable except at theorigin.

Determine analytic function f(z) = u + iv, if u = log(x2+ y 2) + x ~ 2y.

If iv = A log

' z - o '

, where a is real, show the lines u ~ constant, v - constant, form two \ z + a )

orthogonal families of coaxial circles. Prove that this function may be used to represent a sourceand sink combination.

Show that f (z) = z + 2z is not analytic anywhere in the complex plane.Show that the function f(z) defined by

x + y

= 0 z = 0.

is not analytic at the origin even though it satisfies C-R equations at the origin.

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F u n c tio n s o f C o w i£ X V a r ia b le s | 107

L If f(z) = u + iv is a regular function of z in any domain, prove that

(0

(ii)

! t + t t ] i /<*> !p= p 2-1a* ) r 2 -im i2 •dy )

i L + i L j i M r= p (p _ o.i u r M f \ z) i2 .

V.5x dy

I Determine a, b, c ,d so that the function

' /(z) = (*2 + <My + 6y2) + »(c*2 +dxy+y2) is analytic,

i Detennine k such that the function

f(z) = |lo g (* J + y J) + it a n 'l^ y j be an analytic.

|i Detennine the analytic function f(z) in terms of z whose real part is

(0 cosxcoshy (ii) e_Jt(x 8iny-y co8y) (iii) e* cosy

sin 2*(iv) e~x {xcosy + y&\ny)\f{0) = 1 (v) ^ j ^ y + cos2*‘

10. Find the analytic function f(z) in terms of z whose imaginary part isi

x - yit- (0 x 2+y2 0*) sin hx eos^

(fii) &ry-5x + 3 (iv) ^ p - + coshlcos *

Prove that u = x 2- y 2 - 2 x y - 2 x + 3 y is harmonic. Find a function v such that f(z) = u + iv

^ . is analytic. Also find f(z) in tenns of z.

An electrostatic field in the jy-plane is given by the potential function f - x 1~ y 7, find the

stream function.

cos* + simc — L' If f ( z ) ~ u + iv is an analytic function of z and u - v - ------------------ — .prove that

2coax - 2co8hy

f* f(z ) = - when = 0-

(4. Uffz) ~ v + iv is an analytic function find ffz) if u + v - —•, when /(I ) = 1.x +12

[5. Find the analytic function f (z) = u + iv, given

t? (i) u = a( l+cos0) (ii) r* 0 .

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16. Prove that y = lo g[(*-I)2 + (y -2 )2] is harmonic in every region which does not

include the point (1, 2). Find a function ^ such that # + it// is an analytic function of the

complex variable z - x + iy. Also find f(z) in terms of z.-

Answers-1(A)

2. f(z) = -2z + 2ilog(iz) + c. 7. a = 2, 6 = -1, c = -I, d = 2. 8. k * - l

(ii) iz e' z +c (iii) e* +c (iv) cot z + c

1 0 8 | En g in e e r in g M a t h e m a t ic s -H I

9. (i) co sz + c

1+ i10. ( i ) ------+ c

z(ii) is inhz + c (iii) 3z2 - 5iz + c (iv) —+ icoehz + c.

11. v - x 2- y 2 + 2 x y - 2 y ~ l x and f(z) = (]+i)z2 ~(2 + 3i)z. 12. t// = 2xy + c.

14. iz i +c.

15. (i) a(1+co80 + isin01og r) (ii) ( 0r + — ( 0cos <9+ r — I- rj I r)

sin0 + c.

16. ^ - - 2 tan and /(z) = 2 ilo g (z -l-2 i) + c.

»•«* «* K * '* (• C;

1. Prove that bilinear

2. Prove that every bi

Exercise-1 (B)

transformation maps circles or straight lines into circles or straight lines.Itnikar tronrfnrmotirtn 11/itk hi/A(inita nnmte r t /? /vanKa mrt In fir

' v ) - a ' z - a '|

\ w ~ P ) U - p )- P J

bilinear transformation which has only one fixed point a can be put in the formProve that every1 1 5------- = ---- --■f A.

w - z z - a

Find the fixed points of the following transformations :

z 3 z -4 3i’z + l(’) W ~ T ~ 7 (") w = T T W W = -7 7 T 2 - Z Z — \ Z +l

(2 + i ) z - 2(iv) w = -------- :— .v 7 z + i

iz+2Show that the transformation w = ~^Z+J maps the real axis in the z-plane onto a circle in the »-

plane. Find the centre and the radius of the circle and the point in the z-plane which is mapped onthe centre of the circle.

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F u n c tio n s o f C o m p le x V a r ia b le s | 109

p ds6. Show that J ~ in an inverse with respect to the transfonnation

az + bw -------- j . ad -6c = 1, where ds = J d x 2+ d y .cz + a

5 - 4 z7. Show that the relation : w = ——- transforms the circle 1 z 1= 1 into a circle of radius unity in

the w-plane and find the centre of this circle.8. Find the Mobius transformation which makes the set of points in the z-plane

(i) a, b, c (ii) 2, l+2i, 0 correspond to the points 0, 1, oo of the w-plane.

1+ iz9. Show that w = maps the part of the real axis between z - 1 and z = - I on a semi-circle in

the w-plane.10. Find the Mobius transformation which maps.

(i) 1, —f, 2 onto 0, 2, respectively.(ii) I, - I, oo onto 1+ i, 1- i, 1 respectively. •

1+ z11. Show that the bilinear transformation w = ----- maps the region I z |S 1onto the half R(w) > 0.

I ' z

z - i12. Find what regions of the w-plane correspond by the transformation w - to

(i) the interior of a circle of centre z = -j.(ii) the region y > 0, * > 0, |z + i|< 2.

3. Show that inverse of point a with respect to a unit circle is i/o .

i f z + 2^Show that the relation u>= - . - - - j transforms the real axis in z-plane to a circle in w-plane.

V Find the centre and radius of circle and the point in z-plane which is mapped onto the centre of the‘ circle.

i$. For the transformation w = z 2, show that the circles I z - a 1= c (where a, c being real) in the z-

plane correspond to the limacons in the w-plane.

\6. Show that the mapping z = 4w transforms the family of circles | tt>-l |= X into the family of

leminscates.

1z - 1 1| z +11= A, with focal points at z = ±1.

7. Show that the transformations w.(z + i)2 =1 maps the interior of the circle I z 1= 1 in the z-plane

*' 1on the exterior of the parabola — = 2(1 - cos^) where w - pe'*, in the w-plane.

P

1$. Show that the transformation w = - ^ z + — 'j maps the upper half on the circle I z |< 1 on the

upper half of the w-plane.

8/11/2019 DK Jain M3


(

j 4- , ft yr \ ft ----- I represents I ar |< 1 on the strip ~ y ( u ; > < -

of the w-plane.

20. If w = 2z + z 2, prove that the circle | z |= 1 corresponds to the cardioid in die w-plane.

1 1 0 | En g b e e r m o M athem atics -! !!

4. (i) z = 0 and 1

Answers-l(B)

(ii)z = 2 (iii)z = i (iv) z = 1± t.

7 1 ( 7 95. In vv-plane u1 +v7+—v — = 0, centre 10* ~ —I i.e. w = -~ -i and radius ~-

4 2 V *>/ 8 ®

In z - plane, z ~ —.4

7. Centre is • 8. (i) ut - b ~ c ( b - a { z - c j

(ii) w - i. ( 2 - z z j

10. (i) w =2 ( z - »

(1 + 0 (2 -2 )

.... z + 1(it) w -------

z

m m

' W

'Jr M W K iT

Exercise-1(C)

3.

4.

1+2 z 1 1 , 2 JProve that ---- T ” - T + — t+ z - z +z .... where 0< |z |< 1.

z ‘ +z z z

Find Laurent’s series expansions in powers of z of the function^,) given by

Z(\ + Zl )

Represent a function f(z) = -<*-!)<*-3)

which converges to f(z) when 0 <1 z - 1 1< 2.

(z-2)(z + 2)

by series of positive and negative powers of (z -1)

Obtain the expansions for

(i) UI<1

(z +1)( z + 4)

(ii) !< ]z|< 4 (iii) |z |> l.

which are valid. When

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F u n c tio n s o f C o m p le x V a r ia b l e s | 111

2* 25Prove that: ta n -1 z = z - — + when |z |< 1.

(z — (z —|)Show that: log z - ( z - l ) ---- —— + — ------ when I z - 1 1< 1.

Expand f{z) ~( * - 1)2

about 2= 1 , as a Laurent’s series.

K I

Find the expansion of ^ 2 + 5)^2 + 2) 'n P°wers of 2, when

(i) !*!<• (ii) l< |z |<>/2 (iii) )z |> 42.

9. Find expansion of z 2 + 3

valid for:

(i) |ar-l |< 2 (i i)j* + l |<2

0. By considering the Laurent expansions of —^ -r about the points z - ±1, show that the function1- z

; /(2) = 74

, 0 <11 -JT t< 2 .

l.

Q*'

3.

4.

5.

6.

7.

Discuss the nature of singularities of the following functions and also calculate the residues at thesingularities.

1 sing

® z ( i - z 2) ^ ( z - x )Find the zeros and discuss the nature of the singularities of

Discuss singularity of

(z - l )

1

Also calculate the residue at the poles.

71 — at z ~ —.

(sinz-cosz) 4

Discuss is the nature of the singularity at z = 00 of the function f(z) = cosz - sinz.

Discuss the singularity of the function f(z) - —

“ *1*

at z = 0.

nat z - 0.Discuss singularity of the function f(z) = tan —

\ zj Find the nature of singularities of the following functions

(i) -—— at z - 00 (ii) coaec] —) at z = 0.l + e* \ z j

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112 I E ng ineer ing M athem at ics-!! !

Answers-! (Q

2. f{z) = — z + z —z +..., when 0 <| z |< 1 z

t( \ ~ * 3 ^ f z - l V3‘ /(Z)~ 2 ( z - l ) I 2 .

e2 2e2 2e2 4e2 2e2 .------- r-+------- T + ----- + -----+ -----(z —1) +..( z - l) 2 (z -1 )2 z —I 3 3

n-0

,2n «

»2»+2 j»»0 pt-1

M ) n( l- 2 " )

_ 2»+2

n-0

I 111. (i) The points 0, 1,-1 are three simple poles with residues 1. - — — respectively.

(ii) ;ris a simple pole with residue -1.

12. 2,1 + — are zeros; 0 is a pole of order 2 and 1is an essential singularity. The residue at z = 0 iinn |

2 cos 1 - sin 1.13. Simple pole. 14. Isolated essential.15. Non-isola!ed essential. 16. Non-isolated essential.

17. Non-isolated essential, Non-isolated essential.

»

pExercise-! (D)

Use Cauchy integral formula to evaluate :

L

sin t o 2 +cos® 2

( z - l ) ( z - 2 )

Evaluate the following integral

d z , where C is the circle | z | = 3.

ic “ Cos z dz t where C is the ellipse 9x2 + 4y2 = 1.

Use Cauchy integral formula to evaluate :

£3z + z

(z2-1)dz , where C is the circle (z - 1| = 1.

4. Evaluate by using Cauchy's integral formula

z - l

L <z + l)2(z -2 )d*, where C is |z - tl = 2 .

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f .

: Evaluate f — dz, where C is I z + i 1=!.R k z 2 + 1 1 1

r r sin3 z Evaluate by using Cauchy integral formula : I --------dz, where C:| z |= 5.

Z + —

r 2Evaluate by using Cauchy integral formula

f * +Z+ dz, where C is the ellipse 4x2+ 9 y2= 1.* z 2 -3 z + 2

, Evaluate by using Cauchy integral formula

j* COS7TZ { — ----- dz , around a rectangle with vertices 2 ±i and - 2 ±i.)c z2- \

, Evaluate by using Cauchy integral formula M QQg jgp

} ------ dz, where C is the circle [ z - l [= 3 . Z~~7T *

6. Evaluate f \ -dz, where C is the circle | z \ ~ 4.K sm hz

I. Evaluate f --------- dz, where C is the triangle with vertices (0, 1) (2, -2) (7, 1). fc z" sins:

Using residue theorem to solve the following integrals :

2. where C : |2 - l |* l .

- £ w f * ? 'where C : |z “ i,=2

F u n c t io n s o f C o m p i e x V a r ia b le s | 113

f 3 z 2+2

' Jc (z~\}(z2+9)

Zz + l

dz, where C : \ z - 2 \ ~ 2 .

( 22 + 1 ,5- Jc (2z-])2 C C :|2 ,= L

* I ( ? T “ h ere C: U, = I-5 '

i. I <fe, where C : Iz 1= 1.* (2-2X 4 + 22)

*. £ ~ T ^ Z ‘

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19. £ z 2eUxdz;C:\z \=\.

20. Prove that

114 | E ng ineer ing M athem at ics - II I

sin20 ,a I n . / I TT.I r ---- ~ d d ~ —y { a - s a - b ), where 0 22‘ " 5 + 4cos0de

r _ d t > _ [“ z s - Jo 17 - 8cos# 1 ■ Jo S-4cos0

dx25. Prove that £ + “ 4^ ' ° > ° '

26. Prove that f” —^ ^ — 5 — -r-d* = — (a > 0, 6 > 0)J-« (* + a )(x +6 ) a + 6

27. C ctr r *sinax , ., 3 ■ 28. L , —, » <**. (o>0).

» (x +1) x*+fc*

Answer*-1(D)

1.4ai 2. 2«i 3. 4 jb2m

4 ' T5. m

7. 0 8.0 9. -2 m 10. -2*»2i{ - \y

11. 2n «•

12. - 2«i 13. «716 14. m 15. » 16. 0

17.0 18. - 2 m8m

19 3 ?

7t

2L i r ^ f

VI+ 0

22.0

23. /r/15 24. * /4 27. 3*/8 28. | e - “ .

6. 2»

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N u m e r ic a l A n a l y s i s - I

LI INTRODUCTION

Numerical Analysis play an important role in solving many real life mathematical, physical andengineering problems. In the eighteen and nineteenth centuries, great mathematicians like Gauss

Newton, Lagrange, and many others developed the numerical techniques which are still widelyused. The digital computer, however, has enhanced the speed and accuracy of the numericalcomputations. Numerical computations is an approach for solving complex mathematical problems

using only simple arithmetic operations. The approach involves formulation of mathematicalmodels of physical situations that can be solved with arithmetic operations.

One of the most useful tool ofNumerical Analysis is interpolation. According to their ''Interpolation is the art of reading between the lines o f the table Interpolation is the technique to obtain thevalue of a function for any given functional information for it*

Let y - f (x ) be a functional relation between the independent variable x (i.e., argument) and the y be dependent variable (i.e., entry). Until otherwise stated we shall consider / (or)to be a polynomialof all discussions.

L2 APPROXIMATION AND ERRORS [RGPV Dec. 2003]

In any numerical computations :

Error = (True value) - (Approximate value)M** Approximate Numbers :

9The numbers such as 6, 9, —> 15, 36 are exact numbers. But these are numbers such as

4

i i r — ,n/2, it, e which can not be expressed by a finite numbers of digits. We may approximate

them by numbers — **3.6667, >/2»1.4142, n * 3.1416 and e * 2.7183 respectively. Such3

numbers are called approximate numbers.

Significant Figures :

The digits that are used to express a numbers are called significant digits or figures. The numbers

89653,2.8659,0.33857,119.36 each contains five significant figures. But the numbers 0.00869,0.00000108, 0.098500 have only, three significant digits (869, 108, 985), since the zeros serve**»’ only to fix the position of the decimal point.

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1 1 6 | Engineering Mathematics-111

Rounding O ff :

20We know that — = 2.8571428 .....with is never ending. In practice, we have to cut it down#

7 ; ’

a manageable form such as 2.86,2.857,2.85714 and so on. This process ol nttiing off superfluousdigits and retaining as many as desired is called rounding off.

Truncation Err o r;

They are caused, when an infinite process is replaced by a finite one.

jg3 ^5 jp7For example: a im = x~ — + ..too° = T (say).

If it is replaced by * - — + — = Tx (say).

Then, the truncation errors- is | r~ 7|(.

Absolute, Relative and Percentage Errors :

Ldt *i be the approximate value of the exact number x, then :

The absolute error Ea is defined as E„ = \x - x \ j

The relative error is define as: Er ~ X-Xi

- — » and

EaThe percentage error is define as : Ep - —- x 100.

X

Example 2.01 : lf 0.333 is the approximate value o f 1/3, find the absolute, rehtivc and percem# errors.

Solution.HtK x = 1/3 and xa = 0.333.

(i) Absolute error : The absolute error Ea is given by

i R d P V J m e m

ga _ | x xa I -

1000-999

- - 0 . 3 3 3

3

1 333

3000 3000

3 1000

= 0.00033.

(ii) Relative error ; The relative error En is given by

£, = — = ■ = 0.00099.* 1/3

(iii) Percentage error : The percentage error Ep is given by

Ep = Er x 100 = 0.00099 * 100 = 0.099.

At

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i* •. » > Numerical Anaiysm-I | 117

fRGPV June 2006/ &&82 :Find the relative error o f 2/3 is approximate to 0.667.

Here x -- 2/3 and xa = 0.667

Relative error E, x x 2/3

2 667

= 3 1000 12000-20011

2/3 2/3x3000

Ans.

2,9# :tVhat are different types o ferrors and safeguards against them T fRGPV Julie 2003J

of erro rs:

The procedures are adopted to eliminate all mistakes whether these are human Le„ due tocomputer of technical. To the calculating device, when all such mistakes have been eliminatedfrom a computation, the solution is not generally exact oh account of inherent errors ofvarious types. They are classified as follows :

(t) ftttflcation errors :

They are caused, when an infinite process is replaced by a finite one.

For example,

If it is replaced by x - — + — = T, (say).o! 5!

Then, the truncation errors is |T - Ti|.

(J) ft mindin'; off error The errors are unavoidable in most of die calculations will be non-terminating decimals andfor practical reason only certain numbers of figures will be carried in a calculation. If anumber is rounded of with decimal place a change in the number of 7 * 10~(" + mayincrease <>i decrease.

sin j c = Xs x& x 1

X ~ 3! + 5! ~ 7l

x3 *5

EaPerceirt;»>;0 error E P = — *100.

xA n s .

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1 1 8 f E w b— ew w b Ma t h e m a t ic s -111

2.3 SOME IMPORTANT RULES FOR ABSOLUTE AND RELATIVE ERROR!

Then, we have the following rules regarding the absolute and relative errors :Rule I : Absolute error because of truncation. Ifxa is the approximate value of x after truncato m digits, then

Rule II : Relative error because of truncation. If xa is the approximate value of jc afltruncation to m digits, then

Rule III : Absolute error because of rounding-off. If xa the approximate value of x aftrounding-off to m digits, then

[r-xa | <0.5 x io»~*

The above type of analysis is useful when we want to keep track of the errors that occur if tiinumbers in a calculation are truncated or rounded-off to a certain number of decimal digits.

Example 2.04 :Round~off the number 537.261 to four significant digits and then calculate absohn error, relative error and percentage error.

Solution. Given number is 537.261 (= x say).

After rounded-off to four significant figures, the given number would be 537.3 (= X\ say)

.\ (a) Absolute error, Ea - i i -X i| - 1537.261 - 537.300j = 0.039. Am

Example 2.05: Evaluate the sum S = -Js + 5 + J l to fo ur significant digits andfind its absoiuu

and relative errors.

Solution. J s - 1.732, Vs = 2.236, 77 = 2.646 (rounded-off four significant figures)

Hence sum, S = 6.614 and Ea = 0.0005 + 0.0005 + 0.0005 = 0.0015.The total absolute error shows that the sum is correct to three significant figures only.We take, 5 = 6.61

If x be any number expressed as

jc = O.î di di ... x 10", (where d\, di, dj , ... are decimal digits).

jir-*d < i 0 " -m

Rule IV : Relative error because of rounding-off. If x„ is the approximate value of x aftrounding-off to m digits, then

<0.5 x 10-"+I.

x - x x .039 _ oc _ (b) Relative error, Er = — ----- 5 3 7 2 6 I _ X

x - x xAm,

(c) Percentage error, Ep = Er * 100 = 7.25 x 10“i. Ans.

then Ans.

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N u m e r ic a l An a l y s is -! 1 1 1 9

Overflow and Underflow:An overflow is said to have occurred when the sum of two n digits occupies n + 1 digits. Thenumber is larger than the largest number that can be stored in word. This condition is called anoverflow. On the other hand the result is smaller than the smallest number which couid be stored

in a word. This condition is called underflow.For example : 0.9998E1 + 0.1000E - 99 = 9.99980E100

~0.9998E101This is known as overflow.0.9998E (-5) + 0.1000E98 = 0.9998E (- 104)This is known as underflow.

ALGEBRAIC AND TRANSCENDENTAL EQUATIONSAn equation of the form/fr) = 0 is called algebraic and transcendental equations depending uponwhether flx ) is purely a polynomial of jc or contains exponential, trigonometric or logarithmic

functions. For example; 2Jc3 +3jc2+ 17 = 0 is an algebraic equation white Sx3 + 4 log x + 2sin x =0 is transcendental equation. The process of finding die roots of an equation (or zeros of anequation) is called the solution o f j(x) = 0. We shall discuss some numerical methods for thesolutions of algebraic and transcendental equations,rla :0) lf/jc) is divisible by jr - a, then jc = a is a exact root of the equation/*) = 0.00 An equation of the n* degree has onlyn roots.(iii) Every equation of odd degree has atieast one real root

BISECTION ORBOLZANO METHOD:Bisection method based on die repeated application of the intermediate value theorem. This method

one of die simplest iterative methods. Let/jc) = 0 algebraic or transcendental equation and let/a)and /i) are opposite sign i.e.,fid^b) < 0 then root of equation/jc) = 0 lie between a and b. Then

a + bthe first approximation to the root is x\ = ^ •

Now ifX ri)= 0. Then x\ is a root offlx) = 0. otherwise the root lie between a and xi or x\ and b according asflx) is positive or negative. Then we bisect the interval as before i.e., second

o + jc, x, +b .

approximation to the not root x2

- —^— °r *2 = —g— and continue the process until the root isfound to desired accuracy.

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a + bFrom the figure root lie betwen a and b i.e., first approximation to the root *1 = —-—.

a + XiThen the second approximation to the root lie between a and X[ i.e., *2 = — „— •

I S O | E n g o s e r in g M / u ic o y m c s -I lI

Similarly, the third approximation to the root He between X| and i.e., *3 - ~ and so

Exmmpie 2.06: Find a root o f the equation xi ~ x - 4 =0 between I and 2, to four places q fdecim bisection method.

Solution. Let f{x) = jc3 - x - 4 = 0

Initialisation: M x - Ij{1) = (l)3- I - 4 = -4 and atx=2 ,/2)= (2)3-2-4=2.C!early1j

= - 8 < 0. Therefore, the root o f/x ) = 0 lies between I and 2.

a + b 1 + 2First iteration : xj = —- — = —-— = 1.5,

Since XO = - 4,^2) = 2, fa ) = (1.5)3 - 1.5 - 4 = -2.125.

Since/1.5) is negative andX2) is positive ».<?., XI-5) X2) < 0The root lies in interval (1.5,2).

1.5 + 2Second iteration : X2 = — ~— = 1.75,

Now, X 1.75) = (1.75)3 _ 1.75 _ 4 = _ 0.39062.

Clearly/1.75) is negative and/2) is positive./. The root lies in interval (1.75, 2).

1.75 + 2Third iteration : x$ = ---- ----- = 1.875,

Now, / I -875) = (1.875)3 - 1.875 - 4 = 0.71679.

Since/1.75) is negative and/1.875) is positive.

/. The root lies in interval (1.75, 1.875).

1.75 +1.875Fourth iteration : x\ = ---------------= 1.8125,

Now, /1.8125) = (1.8125)3- 1.8125-4 = 0.14184.

S ince/1.75) is negative a n d /1.8125) is positive.The root lies in interval (1.75, 1.8125).

1.75 + 1.8125Fifth iteration : x5 = ----------------- = 1.78125.

Now, / 1 . 78125) = (I.78I25)3 - 1.78125 - 4 = - 0.12960.

SinceXL78125) is negative andXl-8125) is positive./. The root lies in interval (1.78125 1.8125).

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N l m b u c a l An a l y § » 4 | J 2 1

1.7126 + 1.8125Sixth iteration : -------------------- - = 1.79687,

Now, / l .79687) - (1.79687)* - 1.79687 - 4 = 0.00477.

S ince/1.78125) is negative an d /l.79687) is positive..*. The root lies in interval (1.78125, 1.79687).

1.78125 + 1.79687 ,Seventh iteration : xy - ----------- ------------ = 1.78906,

Now, X I.789.6) =(1.78906)3- 1.78906-4 =-0.0672.

S ince/1.78906) is negative and/1.79687) is positive.The root lies in interval (1.78906, 1.79687).

1.78906 + 1.79687

Eighth iteration : xg = ----------------------- = 1.79296,

Now, / 1 . 79296) = (1.79296)3 - 1.79296 - 4 = - 0.02913.

Since/1.79296) is negative an d /l.79(687) is positive.

The root lies in interval (1.79296,1.79687).

1.79296 +1.79687 Nineth iteration : xg ~ ----------- ------------ = 1.79491,

/ l . 79491) =(1.9491)3- 1.79491 -4 =-0.02513.

Since/l.79491) is negative and/1.79687) is positive./. The root lies in interval (1.79491, 1.79687).

1.79491 +1.79687Tenth iteration : xio ------------- £----------- = 1.79589.

Now, / 1 . 79589) = (1.79687)3 - 1.79589 - 4 = -0.00375,

S ince/1.79589) is negative and/1.79687) is positive.

1.79589 + 1.79687 Eleventh iteration : xn = ----------- ------------ --- 1.79638

Now, / 1 .79638) =0.0005Since/1.79589) is negative a n d /1.79638) is positive.

The root lies in interval (1.79589, 1.79638).Here, we see that the digits in the first three places of decimal are the same in the interval(1.79589, 1.79638). Therefore, the value of the root to three places of decimal is t.796.

Ans.

Example 2.07 : Find the real root o f the equation x logto x “ 1.2 by Bisection method correct to three decimal places.

Solution : Let / x ) = x logjo x - 1.2 = 0 ...(1)

Here / l ) =-1 .2X2) = - 0.59 an d/3) = 0.23, clear ly/2)/3 ) < 0.

Hence a root lies between 2 and 3.

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1X2 | E n o m e e h n g M a t h e m a t i c s -H I

First approximation to the root is,

x, = 1 ( 2 + 3) =2.5

Now, fo x ) =/2 ,5) = - 0.205 (-ve)

Hence root lies between 2.5 and 3.Second aproximation to the root is,

*2 = ±(2.5 + 3) =2.75

Now, ftx2) =/2.75) = 0.008(+ve)Hence root lies between 2.5 and 2.75

Third approximation to the root is,

*3 = ^ (2.6 + 2.75) = 2.625

Now, f a ) =/2.625) = - 0.099 (-ve)Hence root lies between 2.625 and 2.75.

Fourth approximation to the root is,

* = | (2.625 + 2.75) = 2.6875

Now, ftxA) =/2.6875) = - 0.046 (-ve)Hence root lies between 2.6875 and 2.75.

Fifth approximation to the root is,

X5 = | (2.6875 + 2.75) = 2.71875

Now, ftxs) 2.71875) = - 0JI9(-ve)Hence root lies between 2.71875 and 2.75.

Sixth approximation to the root is,

| (2.71875 + 2.75) = 2.734375

Now, ftx6) =.*2.734375 = - 0.00546 (-ve)Hence root lies between 2.734375 and 2.75.

Seventh approximation to the root is,

x, = | (2.734375 + 2.75) = 2.7421875

Now, fix-,) =/2.7421875) = 0.0013 (+ve)Hence root lies between 2.734375 and 2.7421875.

Eighth approximation to the froot is

= |(2.734375 + 2.7421875) « 2.73828125

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Now, j{xg) =/2.73828125) = - 0.002 (-ve)

Hence root lies between 2.73828125 and 2.7421875.

.\ Nineth approximation to the root is

** = i (2.73828125 + 2.7421876) = 2.740234

Now, / x 9) = /2 .740234) = - 0.00035 (-ve)

Hence root lies between 2.7402 and 2.7421.

/. Tenth approximation to the root is,

x,0 « \ (2.7402 + 2.7421) = 2.74115mt

Now, j ix io) =/2.74115) = 0.00043 (+ve)

Henê root lies between 2.7402 and 2.74115

.’. Eleventh approximation to the root is,

x„ - = 2.740675

Now, / * „ ) - /2.740675) = 0.00025 (+ve)

Hence root lies between 2.7402 and 2.740675

.\ Twelth approximation to the root is

x X2 = - (2.7402 + 2.740675) = 2.74042

Now, Xx,2) =/2.7404) = - 0.00021 (-ve)Hence root lies between 2.7402 and 2.740675

.’. Thirteenth approximation to the root is

xo = \ (2.7404 + 2.740675) =2.7405

Now, fi? n ) =X2.7405) - - 0.00012 (-ve)Hence the root lies between 2.7405 and 2.740675.

Since X12 = xI3 = 2.741 upto three decimal places, hence the root of given equation is 2.741.Ans.

ITERATION METHOD OR SUCCESSIVE APPROXIMATION METHODTo find the real root of at equation/x) = 0...(1)

Which can be expressed in the form x - j ( x ) ...(2)

Firet we find an initial approximate value xo for (1). A better approximation xi for the root is obtained by replacing x by xq in R.H.S. of (2) I.e.. xj = (*o)

A still better approximation x2 for the root is obtained by putting x = xj in the R.H.S. of (2). Thus x2

= <*(*,)■

NUMERICAL ANALYSIS'! | 123

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This procedure is continued and we get

x3 ~ 4 (*2>

M = 4 fo )

12 4 j En GINEQONG MAnftMATTCS-UI

XnIf the sequence xq , x \, x ^ ,......... x„ ..........of approximate roots converges to a limit a, thefi a istaken as the root of the equation/*) = 0.

Remarks:The sufficient condition for convergence of iterations :(i) If I is the interval in which the root a of the equation x = f (x) lies, then < t

for all x in the interval L

(ii) The initial approximation xq for die root lies in I.

Example 2.08 ; Find a real root o f the equation x* + x* -100 * 0, by iteration method,

Solution.Gwm fix) = r 3 +X2 - 100 = 0 , . .yC1) ~ —98,/2 ) = - 88,/3 ) = - 64/ 4 ) =64 + 16-.100 = -2 0 (-ve). ... .«/ 5 ) * 125 + 25 - 100 = 50 (+ve).

Hence /4 ) and /5 ) are opposite sign.So that a root of equation lie between 4 and 5.The given equation can be written as,

10 xH? +1) =100=>x = => x =${x)

10where j ( x ) = J x + [ '

<f>'(x) = 10 x Y * (* + I)3' 2 = (x + lp/z

5=> |^'(*)| ** j(x + 1)3/* j < 1 in flw interval (4, 5).

So that the iteration method can be applied. Since the root is nearer to 4. 4-v

Taking xq = 4.2. Then approximations are :• «*1*f '

10 .. - . , - -t (*o) = J 4 2 + i = 43852

10 x 2 = *(*i) = 3852 + i = 4 3092

10

*3 ~ # fo ) * ^4,3092 + 1 " 4 33990

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N u m e r ic a l An a l y s is -1 1 12 5

« - * < * > - A 3 3 9 9 0 + 1 ’ 4'32750

10

*s * V4.32760 + 1 " 4 3325610

JC6 = S(x5) = 33266 + 1 = 4-33050

10*7 *= = 3 3 0 6 o + T ~ 4 33 10 6

10

xg (x7) - ^ “ 4:33,06

Since xy-x% = 4.3311 correct up to four decimal places./. root of given eqution is x = 4.3311. Ans.

frawyiV 3 J& : Find the positive root o f the equation x 3 - Sx +8 - 0, other than 2, by the method o f simple iteration, correct to 3 places o f decimals.

Let f{x) - x3 - 8x + 8. Then/0) - 8 ;/ l) = l ;/ 2 ) = 0i.e., J(0) > 0 ;/ l) > 0;/2) = 0; (i.e., exact root at x = 2),so that / l . 5) =-0 .625 < 0.

A positive root other than 2, lies between 1 and 1.5.

We rewnte the equation asx = 1 + I — I , ...(I)

10

1 Vs3i.e., in the form x = $ (x) where (x) = 1 + (fl

3x2=* 4>'(x) * “i " =>1*>'(*)1=

3{Because j*'(125)| = - (1-25)2 = 0.5859 < 1.]

O

So that iteration method can be applied.

Let us take xo = 1 {which is near to root)

3 2 — x 2 < 1, V * e (1, 1.5)

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1 2 6 | En g in e e r in g M a t h e m a t ic s - II I

= 1.2043

= 1 +

I 2 )

- 1.2183

i f 1-*5 = 1 + { —

0%= l + ( -

1.2261

*7

= 1 + ^1.230 4?

(Vj

2183 f =

2 J "

2261 f

2 ~ -J = 12304

j = 1.2328

2355

= 1.2357

, f 1-2342 ' f X9 = 1 + [ — — J - 1.2350

. ( 1.2350 )*x . 0 - 1 ^ — J - 1 .

, ( 1.23651

+ J. f 1.2357 f

x'2 = 1 + { — — j = 12359

. ( 1.2359 f

*l3 = { 2 ~ I = 1,2360

. ( 1.2860?X l 4 = 2 ~ J = 1 2 3 6 0

Since *13 = X]4 = 1.2360 correct up to four decimal places,

root of given equation is x - 1.2360 Ans.

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Example 2.10 .* By iteration method find the value of (48)I/3, correct to three decimal places.

IRGPV Dec. 2003/

Nu m er ic a l An a l y s»4 | 117

••■(I)

Solution. Let, X = (48)l/3=>x3 = 48Suppose A x) = x3 - 48 = 0Put x = l,th e n /l) = -4 7 = -ve

Put x = 2, then/2) = - 40 = -vePut x = 3, then/3) = - 21 = -vePut x = 4, then/4) = 64 - 47 = + 16 = +ve

Rewrite the equation x3 = 48

48 48 jc = - 5- => x = ^(x), where ^(x) - - j

t \ x )

!#'(*)!

x 1

-96

x3

-96

x3= < l,Vx € (3,4)

XsBecause

Hence adding 50 x both sides of equation (I), we get5Qx = - x3 + 48 + 5Qx

=> x ~ TR (48 + 50x - x3)00

=> x = ^ (x), where $ (x) = — (48 + 50x - x9)

Clearly |^'(x)f < 1 for all x e (3, 4).Hence iteration method can be applied.

Taking initial rootx© = 3.6, then iterations are:

, _ 48 + 50xo - x* 48 + 50(3.6) - (3.6)3* ! - # « > - -------- go------^ = ------------50---------------1627

...(3)

48 + 50x! - x? 48 + 50(3.627) - (3.627 y>X2 - 4 M ---------- jg --------= ---------------- 50------------------3'633

48 + 50x2 - x | 48 + 50(3.633) - (3.633)8 „ _ *3 = *(*2) = -------- ------------------------------------------------ =3-634

48 + 50x 3 - xf 48 + 50(3.634) - (3.634)8** = #(x 3) = -------- g0---------- ---------------- 50--------------- = 3*634

Since x3 = X4 - 3.634 correct up to three decimal places.Hence root of given equation is 3.634. Ant.

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2.7 METHOD OFFALSE POSITION ORREGULAFALSI METHOD

To find the root Ax) = 0 ...[

SinceAxo)Ax\) < 0, then root of equation (1) lie between xq and x>. Let xi = —- .

In this method x-±will be taken as the point where a straight line or chord AB through the point fixo) a nd /x i, /x i) ) intersect the x-axis i.e.. intersect at point fo, 0).

.\ Equation of straight line AB is

/ ( * i ) - / ( * o )y~Ax o) =

1 2 8 | E n g in e e r in g M xruSNA ncs-iH

*i

Since straight line intersects at fo, 0) so that

/(* i) - (*o)

( j c — JCo)

0-Axo)

=

*2

(* 2 - X o )

*, - x 0

(* 1 - X o )/ (X q )

/ ( * i) - A*o)

or may be used the following formula :

(*i -X o )

*>X

B(xt MX

X 2 = X j - f ( x , ) - / ( x 0 )

* f ( x , )

Which is an approximation to the root of equation (I). If/xo) and/x 2) are opposite signs then the

root lie between x q and x2. So that replacing X| by x i in (2), we obtain next approximation x }.

However, ifAx\)Ax 2) < 0 i.e., root lie between X| and X2 then we obtain next approximate* x$ accordingly. We continue the iterations till the root is found to the desired accuracy.

Remark:

This method is also known as method o f linear interpolation.

Example 2.11 : Find a real o f the equation x? - 9x + 1 * 0 by the method offalse position.

fRGPV June 20O7j

Solution. Let, Xx) = x3 - 9x + 1here A 2) - - 9 ,/3 ) - 1 = > / 2 )/3 ) < 0root lies between 2 and 3.

Taking xo = 2, x\ - 3 so that/xo) = - 9, and/xj) = 1.By method of false position, First approximation :

(*I - *o) x

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N u m e r ic a l An a l y s is -! | 129

Now, fo r ) = /2.9) = (2.9>» - 9(2.9) + 1 = - 0.711

Clearly/xi).X*2) < 0, therefore, the root lies between 2.9 and 3.

Takingx0 = 2.9, xt = 3,/xo) = - 0.71 l.j fo ) = 1 in (1), then second approximation:

root xi - 2.9 - ~ —- —— (-0.711) = 2.94161 + 0.711

Now, / x 3) = /2 .9 4 16)

= (2.9416)3- 9(2.9416) + 1= - 0.0207

Clearly/xO/xj) < 0, so the root lies between 2.9416 and 3.

Taking x0 = 2.9416, xj = 3 ,/x 0) « - 0.0207,/x O = 1 in (1), we get third approximation

root x4 = 2.9416 - 2 ^ 1 (-0.0207) = 2.9428

Now, y(x4) - - 0.0003so that root lies between 2.9428 and 3

Taking xo = 2.9428, xi = 3, we get

x} - 2.9428 - (-0.0003) = 2.9428,

Clearly x4 = x5 = 2.9428.Hence the root is 2.942 correct to three decimal places. Ans.

2.12 : Find a real root o f the equation xi - 2 x - 5 ’*0by Regula-Falsi method correct to

three decimal places. fRGPV, Dec. 2011f Let /( x ) =x3-2x — 5 = 0 ... (1)Here / (0) = - 5,

/ ( I ) = - 6,/ ( 2 ) = - l ,/ (3 ) = 16,

Clearly root lies between 2 and 3.First approximation :

Taking x0 = 2, Xj = 3,/ (x0) = - l, /( x i) = 16.

By Regula-Flasi method an approximate root:(X) Z lo }

f ( x \ ) - f ( x 0) x = X*~~77Z ~ 777Tx f ( xo) .. .(2)

- 2 —— x ( - 1 ) = 2 + — =2.0 5916+1 17

Now, from (1), we have/ (2.059) - - 0.389 and / (3 ) - + 16.

=> next approxim^root lies between (2.059, 3).

Second approximation :Taking x„ = 2.059, xj = 3 ,/( x 0) = - 0.389,/( x ,) = 16.

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1 3 0 1 En g in e e r in g M a t h e m a t ic s -111

( 3 - 2 f l 5 9 ) x H ^

16 + 0389 1 y ' j|

i.e., x = 2.081From (1), / (2.081) = - 0.150 an d/(3 ) = +16.=> next approx ima te root lies between (2.081,3)

,\ Third approximation:

Taking x0 = 2.081, xt = 3,/ (x 0) = - 0.150,/xt) = 16

, , 2 .q 8 i - <3 - 2 0 8 I> - ^ ! 5,)> [b s f f l16 + 0.150 1 y ( n

= 2.090

Now, from ( I ), we have

/ ( 2 . 0 9 0 ) = - 0 . 0 5 ! a n d / ( 3 ) = + 16

=> next approximate root lies between (2.090, 3).

Fourth approximation:Taking x0 - 2 .090, x , = 3 , / ( x 0) = - 0 .51 ,/ ( x , ) = 16.

o w _ f l z 2 j 9 0 ) x ( - a 0 5 0 :

16 + 0.051 i y w i

= 2.093

Henc e root o f the given equation is 2.093. Ans.

Example 2.13 : Find the real root o f the equation x log/g x - 1.2 = 0 by the method o f false position coned to three decimal places. {RGPV Dec. 2002, Dec. 2005 & Dec. 2001)

Solution. Let f ix ) = x logio x - 1.2

Here f i 2) - 2 log,0 2 - 1.2 = - 0.59794

f i 3) = 3 log ,0 3 - 1.2 = 0.23136

Clearly root lies between 2 and 3.

For the first approximation,

Taking xq = 2 , jq = 3 , /x o )= - 0 .59794 a n d /x j ) = 0.23136 .

By the m ethod o f Regula Falsi or false position method

0 (3 - 2) x (-0.59794)(.23136 + .59794) ” 2-72102

N ow, f ix 2) =y(2.72102) = - 0 .01709 and / 3 ) = +ve

i.e., The root lies between 2.72102 and 3.

Taking xo = 2.72102, X| = 3,y(xo) = 0.01709 a n d /x j) = 0.23136 in equation (1) we get.

Second approximation,

x = 2.72102 + -------------------------- x 0.01709*3 (0.23136 + 0.01709)

= 2.74021

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N u m e r ic a l A n a l v s is -I ( 131

clearly A *t) ./ ta ) < 0root lies between 2.74021 and 3.

Taking xo = 2.74021, X| = 3,/*o) = - 0.00038, fix \) = 0.23136, we get

0 25979.■. Third approximation x^ = 2.74021 - ■ (-0.00038) = 2.74064

Now, Xx4) =-0.00001

Then fourth approximation is

, 5 . 2.74064 - (-0.0001) - 2.74065

Since X4 = xs = 2.7406 correct up to four decimal places./. Hence root of given equation is 2.7406. Ans.

XmpU 2.14 : Find the root o f equation xe* * cos x, using Reguia Falsi method, correct to four decimal places. fRGPV June 2009J

Given cosx-xe*=0

Let /x ) = cos X- xe* ...(1)/(O) = cos 0 - 0 = 1 [taking mode of calculator is Radian]XI) =-2.1779

HenceX0) a n d /1) are opposite sign.Root of equation (I) lie between (0,1).

Taking xo = 0, x\

= l,X*fc) = UX*i) = - 2.1779By Reguia falsi method:First approxination:

(Xj - x„) v

' - ' - u M - f a r ™ ' (2)

= ° “ (- 2.1779-1) Xl =0 -3147

Now, from (1),

X0:3147) = 0.5198ClearlyX0.3147) andXl) are opposite sign, so that next approximate root lie between (0.3147,

I)-Taking xo = 0.3147, x, = l,X*>) - 0.5198,/x i) - - 2.1779Second approximation : from (2), we get

(1 - 0.3147) x .5198

(-2.1779- .5198)

From (1) XO-4467) = 0.2036

Clearly next approximate root lie between (0.4467,1).Taking xq = 0.4467, x, = 1,/xo) = 02036,/ x ) = - 2.1779

Now, / x 3) = X2.74021) = - 0.00038

x « 0.3147 - - ---- J - - = 0.4467

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132 |£«eB«i*saNaM*rHEMAiTcs-UI

Third approxunate : from (2), we get

(1 - 0 .4467 ) x 0 .2036r = U . 4 4 b 7 ------------------------------------- — a 4Q40

x ( - 2 . 1 7 7 9 - 0 . 2 0 3 6 ) U4y4U

From (1), we have/0 .4 9 4 0 ) = 0 .0709

Clearly root lie between (0.494, I)

Taking x0 = 0 .4940 , x , = 1 , /xo ) - 0 .0709, f a , ) = - 2.1779Fourth approximation • Prom (2) becomes

a AriAn (1 - 0 -4940) x 0 .070 9= i).4y4U------------------------------ = n cfiQi( - 2 .1 7 7 9 - 0 . 0 7 0 9 ) UM,yi

Now from (I) , we have

/0 .5 0 9 I ) = 0.0261

Hence nex t approxim ate root lie between (0.509 1, 1)

Taking = 0.5091, x, - l.yfo) = 0.0261 and/xt) = - 2.1779

Fifth approximation : From (2), becomes

_ 0.509i _ £ 1 ^ 0 0 ^ * 0 0 2 6 1

( - 2 . 177 9 - 0 . 0261) u 'M 4y

From (I), we have

■ .. /0 .5 1 4 9 ) =0 .0087

Clearly root lie be tye en (0.5149, J).

Sixth approximation : From (2), becomes

(1 - 0 .5149) x (0 .0087 )- u .5 i4y -------------------------------- = a si a s

A ( - 2 . 1 7 7 9 - 0 . 0 0 8 7 )

From (1), we haveX0.5168) =0 .0029

Clearly root lie between (0.5168, 1).

Seventh approximation : From (2) becomes

( 1 - 0 . 5 1 6 8 ) x 0 .0029r = U.OlbO -------------------------- :-------- = n 517 A. x (-2.1779 - 0.0029) 101/4

We have / 0 . 5 174) = 0.0010

Hence root of given equationx = 0.5174. Am

Example 2.15 ; Find the root o f the equation cos x ~ 3 x - l f using the Regula falsi method correct i three decimal places* .

Solution. le t , f[x) = c o s x - 3x + 1Hence J(0) - 2 [taking mode o f calculato r in radian

7(1) = - 1 .45969. H en ce /0) a n d / l ) are opposite signs./.The root lies between 0 and 1.

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Nu m er ic a l A n a l y s i s -! | 133

/. Taking x0 = 0, x, = l , /x 0) = 2-X*i) = - I-45969. we getFirst approximation:

or x 2 = 0 “ f r.M fln --- n7 * 2 = 0.5781(1 - 0)

[-1.46969 - 2]

Now, J(x2) =/0 .578I) = 0 .1032098 = +ve

root lies between 0.5781 and I.

Taking xo = 0.5781 and X| = 1 ,/x o ) = . 1032098,/x i ) = - 1.45969, then

Second approximation:

Xx = 0.5781 + ------- ( - ~ - 5 — ------ - x 0.1032098[1.458969 + 0.1032098]

= 0.606 • (Using (1)]

Now, / x j ) = / .6 0 6 ) = 0.003 95 5 = +ve

root lies between 0.606 and !.

Taking xo = 0.606, X| = 1 ,/xq ) = 0.003955,/x j ) = - 1.45969, then

Third approximation:

= 0 .606 + -------- (1 ~ ° - 0 6 )------- - x 0 .003955 x* {1.45960 + 0.003955]

= 0.6071

Now, / x 4) = /0 .6 0 7 1 ) = 0.000 005 9 = +ve

/. root lies between X| = 0.6071 and I .

Takingx0 = 0.6071, x, = l , / x 0) = 0.000059, fix) = - 1.45969, then

Fourth approximation:

x = 0 .6071 + ------- -1- — -------- x 0 .0 0 0 0 0 5 9 x [1.45969 + 0.000059]

= 0.607

Since x4 = X5 = 0.607 correct up to 3 decimal places.

Hence, the root o f given equation is 0.607. An s.

Example 2 . 1 6 : Find a root o f the equation 3x + sin x - e * * 0 by Reguia fa lsi method correct to four decimal places.

Solution. Let j(x) = 3x + sin x - e* ...(1)

Then / (0 ) = - 1, / 0 . 1) = - 0 .705 . /0 .2 ) 0 .423

[taking mo de o f calculator in Radian]/0 .3) = - 0 .1 54 . /0 .4 ) = 0.0975

Clearly/0.3) and/0.4) are opposite s igns.

The root o f / x ) - 0 lies between 0.3 and 0.4. . 1 ■

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1 3 4 | En g in e e r in g M a t h e m a t ic s -111

Taking xo = 0.3, x, = 0.4, / ( x q ) = - 0.154,/x[) = 0.0975By Regula falsi method :

(*i ~ *o)*2 = X0 - f(*o)

First approximation:

[ f i x , ) - f ( x o)]

(0.4) - (0.3)

(0.0975) - (-0.154)

0.1 x 0.154

(-0.154)

0.3612

Now, f lx2) = X0.3612) = 0.0019 = (+ve)The root lies between 0.3 and 0.3612

Taking x q « 0.3, jc, = 0.3612,/ x q ) = - 0.154,/x,) = 0.0019, thenSecond approximation:

‘ (0-3 , + O < 0 i5 4 )" o 36<m

Now, fix )) =/0.3604) = - 0.00005 = (-ve)

root lies between 0.3604 and 0.3612.Taking x0 = 0.3604, x { = 0.3612, /x o) = - 0.00005, /x j) = 0.0019, then


= 0.3604 +

(0.3612 - 0.3604)

(0.0019)-(-0.00005).(-0.00005)

[Using (2)]

0.0008

0.00195(0.00005) = 0.36042

Ans.Since X3 = X4 = 0.3604 correct to four decimal places,

root of given equation is 0.3604.

2,8 SECANT METHOD OR CHORD METHOD

The secant method is an improvement over the method of false position, such that condition/jc0)

J(x 1) < 0 are not required.The general formula for successive approximations is given by

(*n ~ %a-1 )/(*«)

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N u m e r ic a l A n a l y s is -1 I 1 3 5

r - (• *! x 0 ) f / Y \ i.e. for n - 1 « - *>

(x2 - x,) n

for n = 2, x3 = x 2 ~ , , , _ 77" A ** J and so on.[ / ( * 2) - / (* k ) l

2.17 : Find the root o f cosx-xe* —0 by using Secant or Chord method.

Let, A x) ~ cosx-xe* ...(1)/(0) - cos 0 - 0 = !(+ ve)/ I ) = -2.177979 (-ve)

root of equation (I), lies between 0 and 1.Hence/0) a n d /l) are opposite sign.Using Secant method :

r ~ *i»-l) f (Y \ i / d . ) - / ( * „ ) ] f i , ) • "-(2)


(■ i ~ *o) \

f0Tnml ' x^ ^ X l " 7 ( x ly - - f ( x 7 ) fiXl )-

Taking xq = 0, xx= 1, jfa>) ~ l,X*i) = - 2.177979, we get

, (1 - 0)(-2 .177979)

X2 = [2.177979 + 1] = 03147

Now, f ix2) = /0 .3 ! 47) = 0.51797Second approximation : for n - 2, in (2), we get

■ ** ' ( / ( S - / ( i , ) i /(X2> * 04467

.•. Now, / f o ) = /0 .4 4 6 7 ) = 0.2036

Third approximation : for n - 3, we get

(x3 - x2) ,/ ,

f e W f e » / ( l a ) ' 0'53' 8 Now , / [xA) = /0 .53 )8 ) = - 0.0432.

Fourth approximation : for n = 4, we get

(x4 - x3) .. .

" °'5173'

Now , f [xs) = /0 .5173) = 0 .0014.

Fifth approximation : for n = 5, we get

/. Now, f[xb) = /0 .5 l77 ) = 0 .0001 .

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Sixth approximation : for n = 6, we get

v - (-r 6 _ * 5 ) f t \

*?= 6 _ [/(*6)- /( *5 )] 6 a5177‘

Since xg = *7 = 0.5177 correct to four decimal places.Hence root of given equation is 0.5177. Example 2.18 : Find a real root o f x* - 5x + 1 = 0, by using Secant Method. S o l u t i o n Let / * ) = x 1 - Sx + 1

/ (0) = + I/ ( ! ) - 1 - 5 + 1 = - 3

Clearly/( 0 ) and/(I) are opposite sign.Hence root of equation (1) lie between (0,1).Using Secnat Method :

136 | EwoiwttHHto Ma t h e m a t ic s -!!!

for « = 1, 2, 3.....

.. F/ntf iteration : Put n - ! in (2), we get

T <*»-■ *<>> ■; f ( r >

*2 ' ' 1 /(* > )-/(* 0)

Here x„ = 0, jc, = 1,/x 0) * /(0 ) * ! ,/( x t) = /( !) = - 3

* ■ ' ' - { ^ x<- 3) ■ ' “ -0.25.

Thus x2 = 0.25 and / (jr2) = /(0.25) = - 0.2344.Second iteration : Put n = 2 in (2), we get

x f ( x )

(0.15 - I) x (-0 23 44 )= 0 25 - ----------£— ----------- -

" (-0 2 34 4 + 3)

= 0.1864.Hence x, = 0.1864 a n d /(r3) = / (0.1864) = 0.0743.

Third iteration : Put« = 3 in equation (2), we get

(Xj - x 2)

( / (*a ) - / (*2) )

= 01S61 (Q-1864- 0-25) * ^ 0743)(0.0743 + 0.2344)

= 0.2017.Hence x4 = 0.2017 and /(.r4) = /(0 .20I7) = - 0.0005

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Fourth iteration : Put n = 4 in (2), we get

' < / ( * . ) - / < * » / M

. 0 2 0 1 7 - (^ 0 1 7 - ° l8 6 4 ) .( -0 .0 00 5 )(-0.0005 - 0.0743)

= 0.2016.Hence = 0.2016 and/(x5) =/(0.2016) = 0.0002.Fifth iteration : Put n = 5 in (2), we get

(x5-*4)

_ 0 2 0 1 6 _ (0 2 0 2 6 Z 0 20 17)

0.0002 + 0.0005

= 0.2016Since xj = X 6 = 0.2016Hence root of given equation is 0.2016. Ans.

NEWTON-RAPHSON METHODLet xo be an approximation value of a root of the equation/x) = 0.Let X| be the exact root closer to xo, so that x\ = xo + h, where h is small.Since xi is the exact root of/x ) = 0, we have/xj) = 0, i.e., J[x0 + h) = 0

h h2i.e., /(x0) + — f'(xo) + — f ' (xo) + ... = 0 [By Taylor’s theorem]

Since h is small, h 2 and higher powers ofh may be omitted.

/(*o)Hence / x 0) + hf\xQ) = 0 =>h = _ ^ j ...(I)

/(x0)x i= x 0 + /r=*o ^ j [Using (1)]

Taking xt as an approximation value of the root, a still better approximation x 2 can be obtained by

f(* i)

Nu m e r ic a l A n a m s »-I (>137

... * *i -continued un

The iterative formula is

*2 ‘ A x ,)The iterative process is continued until we get the required accuracy.

/(*„) , x„+1 = x* ~ (for n = 0, 1 , 2 , .....)

is called the Newton-Raphson formula or Newton’s iteration formula.

Remarks ;

(i) When the derivative o f /x ) can be easily found and is a simple expression, then the real root

o f the eq u a tio n /x ) = 0 can b e computed rapidly by the Newton-Raphson method.

(ii) New ton’s formu la conve rges provided the initial approx imation xo is chosen sufficiently to

close to the root.

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138) Engineer Mathematics-HI

(iii) Condition fo r Convergence o f Newton-Raphson M ethod:

fix*)The Newton-Raphson formula is

The iteration formula is

From (1) and (2), we get

f '(xn)

*»+i =<f>ixn)= x -

f ix )

f i x )

...(2)

[Using (2)]

...(1)

1 , f ix ) . f ‘(x)r 

f'Hx)< l

We know that the condition for convergence of the iteration method is |<#'(x)! < 1.The corresponding condition for the Newton-Raphson method is

, f / ,2(*) - /(*)• / ' (*) 1

I f ' 2(x)

=>\ f(x ).f\ x)\ < {fix)}2, for all x in the interval in which the root lies.2.10 GEOMETRICAL INTERPRETATION OF NEWTON-RAPHSON METHOD

Let x0 be a point near the root a of equation fix) = 0, then tangent at /*0 {(x0, _/fa0)} is

f (x0) y - fi xo) = /{x0X* - Xo). It cuts x-axis at (xu 0) then x i ~ *o “ Y’(x^) *

Which is first approximation to root a. IfP\ corresponds to X| on the curve, then tangent at P\ will

cut x-axis at X2, nearer to a and is therefore second approximation to root a.

Repeating this process, we approach the root a quite rapidly. Hence the method consists in replacingthe part of the curve between PQ and x-axis by the means of the tangent of the curve at P q.

Example 2.19: Solve by Newtons-Raphson method :x* -3 x+ / * ft /RGPV Dec. 200/ & Dec. 2007j

Solution.Let, fix ) - x3 - 3* + 1 ...(I) fiO) - + 1= + ve/ ! ) = I - 3 + 1 = 2 - 3 = - 1 = - ve.

Clearly/0) and f i \) are opposite signs.

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root of equation (1) lies between 0 and 1. Since |/(0)| =

N u m e r ic a l Ah a l y s is -I | 139

Taking initial approximation xq = 0.5

Using NewtorvRaphson method.

0 + 1 n*i.e.,*© = —-— = 0.5

/(*»)

x"* ' ~ *" " T o o - (2)

Since fix ) = x3 - 3x + 1=> /t x ) =3*2-3Hence (!) becomes:

- 3x, +1) 3xf - 3xn - xg + 3x, - 1

11 3 x 2 - 3 ~ 3 ( 4 - D

2xg - 1

x" +1 ~ 3(xf -1 ) " <3)

Putting n = 0, in (3), then first approximation :

2xp3 - 1 _ 2(.5)3 - 1 _

JC| “ 3(x$ - 1) ~ 3(,52 - X) _ 0 3333

Putting n - 1 in (3), then second approximation:

2V - 1 X2 ~ 3(x,2 -1) = 0,3472

Putting n - 2 in (3), then third approximation:

Hence x2 = X3 = 0.347 correct to 3 decimal places.Thus root of given equation is 0.347. Ans.

xample 2.20 : Solve by Newton-Raphson method : x* - 3 x - 4 - 0. fRGPV June 2002/olution. Let, / x ) = x 3- 3 x - 4 ...(1)

Then fiO) = - 4 = - ve;/ l ) = l - 3 - 4 = - 6 = -ve f i l ) = 8 —6 - 4 = —2= —ve, fi} ) = 27 - 9 - 4 = 14 = + ve

.'. fil) and/3) are opposite signs.Hence root lies between 2 and 3.Also since /2)( < jfl3)| i.e., root near to x = 2, so taking xo = 2.By Newton-Raphson formula:

* A x .) ~(2)

Since fix ) = x3 - 3x - 4

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=> /'( * ) = 3x2-3

(xf - 3x„ - 4) 3x% - 3xn - x* + 3x„ + 4Hence (2) becomes : x„ + , = - ... i --------------= ------- 3 ( x f - l )


or x„+ , 3(X2 _ w ...(312 x 2 +4-»+1 ~

Putting u = 0 in (3), we get

2xg + 4 2(2)3 + 4 20First approximation: x t = j ) “ ^(22 - 1) ~ IT = ^-222

Putting n= I, in (3), we get

Second approximation: x2 = ^ _ J} - ^ . 2 2 2 ) * ' - 1] ' T l s T T b = 2,197

(2x f + 4) 2(2 .222 )3 + 4 25 .9472*2 = =

Putting n = 2, in (3), we get

2xf + 4 25 .209Thin!approximation: x3 = 3^ 2 ^ ) “ 1 1 4 8 0 4 = 2,196

Putting >? = 3, in (3), w e ge t

2 x f + 4 2 5 . 1 8 0Fourth approximation: X4 = ^ ~ 21467 ~ 2.196

Hence x3 = x 4 = 2 .196 co rrect to 3 decimal places.

.'. root o f given equation is 2.196. Ans.

Example 2.21 : Find the real root o f x 4 - x - 10 ** 0 by Newton-Raphson method.[RGPV Dec. 2003 & Dec. 20///

Solution. Let f(x) = x4 - x - 1 0 = 0 ...(I)T hen 7 (0 } = - 1 0 = - v e

I ) = I - 1 - 10 = - 10 = ( - ve)7 (2 ) = 1 6 - 2 - l0 = 4 = (+ve) .

Clearly fi 11 and 7(2) are opposite signs*

root ot equation ( 1 ) lies between 1 and 2 .

Taking initial approxim ationx o = 1.5

By Newton-Rephson m eth od :

. /(*«>•*»+1 = *» f ' (xn) ’ " = 0‘ ’’ 2.......

Since f i x ) - x * - x - 10

=> f '(x) = 4x3 - 1.

Hence (2) beco m es:

(x* —xn — 10) 4x* - x n - x< + xn + 10

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Putting, n = 0 in (3), we get

3 * * + 1 0 3 ( 1 . 5 )4 + 10 25 .1875

First approximation: x, = 4 x ^ - 1 4 (1 .5 )3 - 1 1 2.5 _ 2 0 1 5


3*j4 + 10 3(2.015)4 + 10 Second approximation: x2 - 4 i3 “ 4(2.015)3 - 1 = 1,874


N u m e r ic a l An a l y s is -! 1 1 4 1

3x4 + 10 3(1.874)* + 10x3 = “

Putting, n = 3, in (3), we get

Third approximation: xj = 4jcg _ j " 4(1.874)3 - I = 1 8 56

3x$ + 10 3(1.856)4 + 10Fourth approximation : = 4^3 _ j ~ 4^ g gg )3 - 1 = 1

Since, X3 = X4 = 1.856 correct to 3 decimal places.root of given equation is 1.856. Ans.

xsmple 2.22 : Solve the equation 3x = cosx+ 1,by using Newton-Raphson method. fRGPV Dec. 2002, June 2004, Dec. 2006 and June 2011 f

olution. Let X*) = 3x - cos x - I ...(HX0) = 0 - cos 0 - I = - 2 = (- ve)/ I ) = 3 -co s (I) - I = 1.4596 = (+ ve)

ClearlyX0) andXO axe opposite signsroot of equation ( 1) lies between 0 and 1.

Taking initial approximation xo=0.5By Newton-Raphson formula:

) x „ ^ ^ xn ~ J ~ , (« = 0, 1,2, 3......) ...(2)

SinceXx) = 3x - cos x - 1 =>/'(x ) - 3 + sin x

[3x„ - cos(.tn) - 1]Hence (2) becomes: x„+1= g + ^ ...(3)


(3x0 - cos Xq - 1)

First approximation : x, = *0 " 3 + 8m(Xo)

[3(0.5) - cos(.5) - 1}= 0 .5 -

3 + sin(.5)

Putting n ~ 1, in (3), we get.'. Second approximation:

n _ (-0.37758)

' a 6 - T J 7 9 5 T ' 0'608

[3^,

Xl 1 3 + sin x,

(0.0032079)

3.57123

= 0.607

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Putting n = 2, in (3), we getThird approximation:

_ [3*2 - COS *2 -1] _ 0 g07 _ (- 003629) Xi 2 3 + sin *2 ' (3.57041)

= 0.607

Since x 2 = = 0.607 correct to 3 decimal places,root of given equation is 0.607. Ans.

Example 2.23 ; Find by Newton’s method correct to five places o f decimals the root o f the equation x loge x = 4.772393. [RGPV Dec. 20041

Solution. Let J{x) - x log x - 4.772393 ...(I)A O = - 4.772393 =(-ve) f{2) = - 3.386099 = (- ve)/ ( 3 ) = - 1.476556 = (-ve ) f{ 4) = 0.772784 = (+ ve).

Clearly/3) and/4) are opposite signs.

.'. root of equation (1) lies between 3 and 4.Taking initial approximation x$ = 3.5By Newton-Raphson method:

f ( x n)

*« + 1 = x» “ "f'(xn) ’ (" = °» *» 2......... ) - ( 2)

Since f[x) - x log x - 4.772393=> /'( x ) = togx +1

.. _ I*„ log(xn) - 4.772393]Hence (2) becomes : x„+ j = * „ ---------- (i T iog a ")--------

Putting n = 0, in (3), we get. . . . .. „ [x o lo g x o -4.772393]

.. First approximation: x\ = x0 --------- —— ------- --------(1 + log x0)

0 _ [-0.3877226]

- 3 8 - 1 2.282763) - 3 6721098'Putting tt = 1, in (3), we get


[x\ log X\ -4.772393]

X2 X* (1 + log*!)3 67-1098 ( ° 04163956) , r71n,nr

- 3.6721098 ^300766374) = 3'6739196

Putting n - 2, in (3), we get

, [X 2 log X 2 - 4.772393]Third approximation: xx = x 2 -------------- — --------- ----------

(1 + log x j

(.008328323)■ 3.6739196 - i - ■ 3.6703006

1 4 2 | E n g j n e £RING M a t h e m a t ic s -III

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Putting n = 3, in (3), we getFourth approximation:

Putting n = 4, in (3) we get

N u m e r ic a l An Al y s i s -I I 143

xa = 3.6703006 - - ° Q- 0{ 4— = 3.67029979(2.300273566)

[*< log x4 - 4.772393]Fifth approximation : x$ - x* (1 + log *4) = 3.67029

Hencex* = x$ = 3.67029 correct to 5 decimal places,root of given equation is 3.67029. Ans.

2.24 : Find the real root x4 - x - 9 ~ 0 by Newton-Raphson method correct to 3 decimal places. IRGPV June 2005 & June 2008/Let, /(x )=x4 - x - 9 ...(1)

7(0) = - 9 = (- ve)7{l) = ] - l - 9 = -9 = -ve7(2)= l 6 - 2 - 9 = 5 = + ve.

Clearly/1) and/2) are opposite signi.root of equation (1), lies between 1 and 2.

Taking initial approximation *o= 1.5By Newton-Raphson formula:

/(*.) f ’( x j

( * 2

*« + 1= x* ~ * (» = o, 1, 2 , .......)

(4*3 -1 )

I v fix) = X* - X - 9 then /(x) = 4x3 -

4*« - ** - *2 + ** + 9

4 x 2 - 1

3** +9or x„+ j — 3 _ j ‘ •••(2)


3x^ + 9 3<1.5)4 + 9 24.1875 First approximation : *> = 4xg - 1 ~ 4(1.5)3 - 1 = 12.5 = 1,935

Putting n ~ 1, in (2), we get

~ 4*? - 1 " 4(1.935)3 - 1 27.9803

3*? + 9 3(1.935)4 + 9 51.0577Second approximation: x 2 = ir3 _ i" “ 4/1 _ 1 ~ 97 QftDS ” 18248

Putting n = 2, in (2 \ we get

3sj+9 42.2646Third approximation: Xj - _ j - 23 3056 = 1,8135

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Putting n - 3, in (2), we get

3x^ + 9 41.4483Fourth approximation : X4 = 4 j,| _ j = 228568 = I-8*34

Putting n = 4, in (2) we get

3x< +9Fifth approximation : xs = ^ - 1.8134.

Hence x4 = X5 = 1.8134 correct to 4 decimal places.root of given equation is 1.8134. Ans.

Example 2.25 : Find the root o f 4 x - e * ’m0 that lies between 2 and 3, by Newton-Raphson method. Solution. Let, / { x )= 4 x - e x ...(I)

Then / ( l ) - 4 - e = + ve/(2) * 8 - e 2 = 0.6109 = + ve

J{3) = 12 - e3 = - 8.0855 = - ve.Clearly root lies between 2 and 3 and root closer to 2 because ]/[2)\ < J/(3)|, Taking initialapproximation x q - 2 .By Newton-Raphson formula:

fix*)1 11 f*(X ) * (" _ .......)

(4x„ - ex») r [ v / ' W = 4 - e» ]

4x„ - x .e1" - 4x. +

144 |i£Nq*ifflppa#l«HEMAncs-IH

or x" +l = ( 4 ^ T j ‘ -(2»

4 —e*"('i _

_ e*n~lt f I

Putting n = 0 in (2), we get

e**{\-x0) «2(1 - 2)First approximation : Xj = (4 __gxo) ~ (4 _ e2) ~ 2.18026963


e1’ (1 - Xj) 10.443822Second approximation: x2 = ( 1 _ ^ ) ~ 4.87019074 = 2 ' 14444213


e ^ d - X j ) 9.77041972Third approximation : x3 = (4 _ g n ) - 4 53727722 = 2.15336627

Putting n = 3 in (2) we get

e*3(l - xs) 9.93487335Fourth approximation : x* = ^ _ gX3^ ~ 4 61280604 = 2-15329237.

Hence x3 = X4 = 2.1533 correct to 4 decimal places.root of given equation is 2.1533. Ans.

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N u m e r ic a l Attu,V8(&4 | 145

I. '

2.26 : Find the smaller root to the equation e~x - sin x = 0 by Newton-Raphson method.

Lei; f(x) =e~*~ sinx ...(1)/(O) = I — 0 = 1 — ve)/(I) = e~1- sin (I) = -0.4736 = (- ve).

/. Clearly/O) a n d /1) are opposite sign.Hence root lies between 0 and 1.Taking initial approximation xq - 0.5.By Newton-Raphson formula:

/(*«) xn + 1 ~ « f ' M

(e~Xn - sin x„)~ xn —

( -e~xn - cos xn)

(e-*n _ sinx*)*„+i = *» + {er** + cos xn )


.\ First approximation: x\ = xo +(e -*# - sinx0)

(e~*o + cosx0)

..-(2)

(e~5 + cos(0.5))

ne 0.12711= 0.5 + _ = 0.58565.

Putting n = I in (2), we get



/. Third approximation:

x2 = +

1.48411

(e xi - sin Xj)

(e x>+ c o b xx)

= 0.58853.

= 0.58565 +

058853 +

(0.00400)

a.39010)

(0.0000)

1.38690 Xi ‘ (c‘x* + cos**)

= 0.58853.

Since x%= X3 = 0.5885 correct to four decimal places./. root of given equation is 0.5885. Ans.

227 ; Find the real root o f the equation x sin x + cos x * 0 near to x - it correct to three decimal places by Newton-Raphson method.Let ^ x ) = xs in x + cosx ...(1)Given that initial root xo= *=3.14159By Newton-Raphson formula:

/ (* , )

f X x J

[xn sin x„ 4 c o a x j

[ *•’ f ( x ) = x cos x + sin x - sin x]

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146 | E n g in e e r s ^ M a t h e m a t ic s -! 11

or x„ + i = x *


/. First approximation:


Second approximation: xi = x\ ~

(x„ sinx* + cosxj

.. (*osinx0 + cos*o) ____ __ lx, = *0 ---------------------------= 3.14159 -

..Oil

COSXq

= 2.82327

(jtj sin xr + coajq)

AT] COS


Third approximation: x 3 - x 2 ~

. 2 . 8 2 M 7 - (- 08? 15.49>. . 2 79859(-2.681433) 279859

(x% sin x 2 + cosx2)

xz cos x 2

= 2.79859 — (^000537)_ (-2.635569)

Hence x 2 ~ x3 = 2.798 correct to three decimal places.root of given equation is 2.798. Atis,

Example 2.28 •• Using Newton's iteration method, fin d the real root o f x logio x = 1.2 correct to fix decimal places. [RGPV, Feb. 20/0/

Solution. Let, fi x) = x logio x - 1.2 ...(I) J { \ ) = - 1.2 = (-ve)

J{2) = 2 logio 2 - 1.2 = - 0.59794 = (- ve)A3) = 3 logio 3 - 1.2 = 0.23136 = (+ ve).

Clearly/2) and/3 ) are opposite signs.root of equation (1) lies between 2 and 3.

Since f[x) ~x logio x - 1.2

v logio x = log* x log* 10log* 10

= x logio e- x - 1.2 = .43429 x loge x - 1.2

[v logô e = -43429]

f ( x ) = 0.43429 (1 + log, x) = 0.43429 + log10xBy Newton-Raphson formula:

Xn+ 1 = x» f j x j (s , logl0 xn - 1.2)

f \ x n) n (.43429 + logio *n)

or ■43429xw+ 1.2

Xn+ 1 " .43429 + logio xn...(2

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Taking initial approximation : xq = 2.5 Putting n = 0 in (2), we get

.43429x0 + 1.2 2.28575

.43429 + logl0 x0 .83223

.43429xL+1.2 2.39278

.43429 + log10 x, .87307

.43429x2 + 1.2 2.39026


First approximation: x\ = AOAOn , i _ ~ “ QOooo = 2.74651

Putting n - 1 in (2), we get

.'. Second approximation: xi = AOAn _ ~ = ononn ~ 2.7407


Mird approximation: x , - .43429 + ',ogl0 ^ - 2.7406

Since x 2 = *3 = 2.741 correct to 3 decimal places..*. root of given equation is 2.741. Ans.

2.29 ; Using Newton-Raphson method, evaluate to four decimal places.

Let, x = 728 =>x2 = 28=*Jta - 2 8 = 0

and let J(x) = x2 - 28 ...(1) AO) = - 28,/H) = - 27,/2 ) = - 24,/3 ) = - 19, M ) = -1 2 , /5 ) = -3 , / 6 ) = 8.

Clearly/5) an d /6) are opposite signs.

Hence root lies between 5 and 6.Taking initial root: xq = 5.5By Newton-Raphson formula:

fi*n) Xn ~(xj ~ 28)

2x„

or + i “


x? +28

2x.» (« = 0, 1, 2 , ....)

iv m = 2 x]

...(2)

Firj/ approximation:


.*. Second approximation:

Putting h = 2 in (2), we get

xg + 28 (5.5)2 + 28 58.25

Xl " 2xo 2(5.5) 11 ~ 5-29545

x? + 28 _ (5.29S45)2 + 28 x2 =


Clearly jc2 = x 3 = 5.2915root of given equation is 5.2915.

2Xj 2(5.29545)

xf + 28 _ (5.2915)2 + 28

2x2 2(5.2915)

Clearly JC2 =X3 = 5.2915 correct to four decimal places.

= 5.2915.

= 5.2915

Ans.

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.148 j E n g i n e e r i n g M a t h e m a t i c s -111

Example 2.30 : Find the cube root o f 2 approximately by Newton-Raphson method correct to ft decimal places.

Solution. Let x - (2),/3 => x3 = 2 => x3 - 2 = 0.

Let f{x) = x3 - 2, then /’tr) = 3X2.

Now/1) = - 1and/2 ) = 6 => root of equation (1) lies between 1and 2.By Newton-Raphson method:

■*n+ I f '(xn)

( 4 - 2 )

3 4

2 * 3 + 2 2 ( 1 1

3 4 = 3 P + 4 >

Taking initial approximation :jco= 1.5







Putting n ~ 3 in (2), we get

Fourth approximation:

2x2 = 3

2xj = 3

1.5 +(1.5)2 J

1.29630 +

= 1.29630

1

1.26093 +

1.25992 +

(1.29630)2

1

(1.26093)2

1

(1.25992)2

= 1.26093.

= 1.25992.

= 1.25992.

Since jc3= x4 = 1.2599 correct to four decimal places.

root of given equation is 1.2599. A

Example 2.31 .*Find the negative root o f the equation x 3 - 21 x + 3500 * 0 correct to three decimal places using Newton-Raphson method.

Solution. Let, j(x) - x3 - 2 Ix + 3500 ...(1)

/ - 5) - + 3485 = (+ ve) A - 10) = + 2710 = (+ ve)A - 15) = + 440 = (+ ve)

A - 16) = - 260 = (- ve).Clearly/- 15) a n d /- 16) are opposite signs.

root lies between (-15) and (-16).

Taking initial root: xq = - 15.5

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?fcBy Newton-Raphson formula:

*« +t = x * ~

fix*) (* | - 21x„ + 3500)

N u m e r ic a l A n a l y s i s -1 | 149

(3x2 _ 21)

2x3 _ 35QQ(3x2 _ 21) - ( 2)


.'. First approximation:





xi =

x2 =

2xg - 3500 _ -10947.75

(3x§ - 21) 699.75

(2x? - 3500) _ -11158.729

= - 15.645.

3xf -21 713.298

= - 15.644

(2*2-3500) -11157.260xj = — 7 T* = = - 15.6443xf - 21 713.204

Since x2 = x3 *=- 15.644 correct to three decimal places.

.‘. Negative root of given equation is - 15.644. Ans.2J2 ; Find the Newton-Raphson iterative formula to fin d the pth root o f a positive number

/Vand hence fin d the cube-root o f 17.

. Let N up = x. Then xP ^ N or fix) = xP - N = 0. Newton-Raphson iterative formula:

xn+i = * » A x J

(*£ - N )

p x F

(p - l)x£ + N or

p x V .••(I)

The cube root of 17 lies between 2 and 3.

Taking W= 17, xo = 2 and p = 3 in (1), we get

2 x 23 + 17

3 x 22

2 x (2.75)3 + 17

= 2.75,

= 2.58264. X2~ 3 x (2.75)2

Similarity x3 = 2.57133; x4 = 2.57128; xs = 2.57128.

Since x* = X5 = 2.5713.

/. The cube root of 17, correct to 4 places of decimals = 2.5713. Ans.

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1 5 0 j E n q n s o b n g M a t h e m a t ic s -111

Example 2-33 .*Find the root o f the equation x* **100, coned to 4 places o f decimals, using Nem Raphson method.

Solution. Let, f [x) ~ x* - 100 = 0 =>J\x) ~ x* (1 + log x)Since / 3 ) = - 73 and/4) = 156.

The root of the equation lies between 3 and 4. f (xH)

Newton-Raphson iteration formula : x„ +\ = “ ~fXx~)

Let us take xq ~ 3,/xq) = - 73; /'(*&) = 56.66253

*' = 3 + 8 6 ^ 5 3 * l 2 m i * <3' 4>

Since the value of the root has to lie with in (3,4), our choice of x$ is wrong. Now taking)

= 4»/*o) = 156 and f ' f o ) = 610.89136, we get

156/. First approximation : x\ = 4 - ^ 39136 = 3.74464

Now, / * , ) = 40.3506 l ; / ’(x1) = 325.65909

Second approximation: jc2 “ 3.62074.

Now, f t x 2) = 5.50! 80; f '( x 2) - 241.24870

.*. Third approximation : xj = 3.59793.

Now, / x 3) = 0.14718;/y(jcJ) = 228.37149

Fourth approximation : X 4 = 3.59728.

N ow, / x 4) = - 0.00115;/X*4) = 228.05119,\ Fifth approximation : x$ = 3.59729.

Since *4 = jc5 = 3.597 correct to 3 decimal places.Required root of the equation is 3.597. Aw

2.11 ORDER OF CONVERGENCE OF AN ITERATIVE METHOD

Let,/*) = 0 be given equation, and let xq, *i, x i ,...... x„ be the successive approximations of tht

root a of/x) = 0. Let e„ be the error in the root n = 0, 1,2 ,........Since a is the exact root

fix) = 0, then e„ =x„ - a and en+ j = *„ + 1 - a.

For any iterative method p is said to be order o fconvergence, if \e„ +1! S k\etf, where * is a positive constant.If p - 1 then the convergence is linear and \ fp = 2, then it is quadratic.

2.12 CONVERGENCE OF BISECTION METHOD

According to bisection method, we have seen that, the original interval is divided into half intervalin each iteration and bracketing the root in new interval. If we take the mid points of the successiveintervals to be the approximations of the root, then the one half of the current intervals is theupper bound to the error.Thus, ife„ is the error in nAiteration, then e„ +1 is the error in (n + 1)* iteration, which is given by

en +1 ~ 0.5 en

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N u m e r ic a l An a l y s is -! J 151

'Now comparing (1) with order o f convergence p :

M £ t y X Hence p = 1 and k = 0.5The shows that, the bisection method is of 1“ order convergent or linear order convergent.

CONVERGENCE OF SECANT METHOD

® X|, X2.......are successive approximation obtained by SECANT method :

(*n ~ *»-t)f(xn)

/(*») ~ /(*„_,)tod a be the exact root of equation/(.t) = 0, then

Let x„ = a + e wandx„+i = a + e„+\

where en,e„+] being errors in n* and (n + 0 th approximation respectively.Hence (1) becomes:

(e„ —

...d)

a + e „ + i = a + en

or

f (a + e„) - f (a + en_r)■/(« + O

en_i f (a + en) - ew/(g + e».i)

e" + 1 _ / ( a + e j - / ( a +

f(a) + e j ' ( a ) +£ f ' ( a ) +.... ~«n /(«)+«,-! A«)+-|f-/•(«)+....

/(a)+€kA«)+^/'(«)+.... -

<e„.i - e„ )/ (a ) + ^ < e „ - * -!>/*(«) + ....

(e„ - <Vi)/ '(«) + / '( a ) +

/ '(a ) + / '(a) +... . tv /«) =0]

-.(2)or eB+i 2! /' ( a )

[Neglecting high powers of em e„ _ i due to very small]

Since by order of convergence p is en+j = k e% where k is constant

=* en = k e£_, [v « - > « - ! ] -..(3)

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=> e)lp ~ kvf e„_ i => e„. t = k~l'P e^/p

1 52 | E n g o ^ A n g M a t h e m a t ic s -II)

e,Vrk-Vp p(a)(2) becomes : e„ +\ = ----- 5----- ei

* / ’(«)

'n

f'(a)

ff-np -+i f ( a )or ke£ = ~ z - < ~ t [Using (3)1

Comparing both sides, we get

1 , , . . k -"P f " (a )

^ = p 2 f \ a ) = constant

=> p2 -p -1=0

1 ± ^ T 4 1 + y/E /n=> p = -------------= — -— (Remove -ve sign because p > 0).

/. Order p - 1.618.Thus order of convergence p = 1.618.

2.14. CONVERGENCE OF REGULA FALSI METHOD

Let f (x) - 0 be given equation. If the ftuiction/(x) in the given equation/(r) = 0 is convex in theinterval (*o> *1) that contains the root, then one of the points x0 or *i is always fixed and other

points varies with n. If the point Xo is fixed, then the function / (jc) is approximated by the straightline passing through the points (x0, / (x0)) and (xm f (x„)), n - 1, 2, 3,....

Then the error equation (2) see in above Secant method, becomes :- 1 /"(<*> - „

2 / ' ( a ) 0’ "

Here e0 = x0 + a is independent of n.6„ + j — k e„ ... (A)

1 / " ( a )

where k = ^ f ' ( a ) e° iS tl,e asymptotic error constant.

Now by the definition of order of convergence p is

e „ +1 <; k\e„\p ...(B)Comparing (A) and (B), we get p - 1=> Regula - Falsi method has linear rate of convergence.

2.15 ORDER OF CONVERGENCE OF NEWTON-RAPHSON METHOD

Let a be the exact root of the equation fix) = 0. Let x„ and x„+ 1 be the approximations for the root

after and (n + I)**1iterations respectively. Let e„ and e„+ 1 be the corresponding errors.

Then e„ = x „ - a and e„+ ( = x„ +1- a /.e., x„ = + a andx„+ 1~e„+J + a.

f(*n)

By Newton-Raphson formula, x„ +1 = xn f ’M

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N tM BubM 'iAm nrsfr'I | l5 S

>•

9

....

<* + e» + I = a + c«/(g + <*n)

f \ a + en)

_ f(g + e j

e"+ >“ * A * + 0[Since expanding X<* + e*) and/ '( « + e„) by Taylor’s series, we have

f ia + «.) - /(«) + / ' (« ) + = e*A«) + f | / '<«) + ....

and f \ a + e„) -/Xflf) + e/"(a) + ....Putting these values in (I), we get

-•(I)

(since/or) = 0]

e*+i - e» -

or en+ i

% A «) + g /■<«> + . . .

A«) + «,/ * («) + -

/ ' ( a )

2[ A « ) + e . m i

e* T(«)

or

2 A « )

e l f \ a )

1 - e .

2f'(a) [

- ] f \ a )

f\cc)

f \ a ) v e. 1---- -^ A«)._

[v (1 + x) '1 = 1 - X + X2 - ....]

...(2)e„+1 = *^*y7^y [Neglecting small quantities]

Comparing by definition of order of convergence e„+\ < k \e„\P.

h _ 1 /'(«)Hence p = 2 and “ *j f '(a) *

This show that order of convergence of Newton-Raphson method is 2, i.e., Newton-Raphsonmethod is quadratic convergent

Example 234 : What do you mean by convergence o f the method and its important in numerical analysis. fRGPV June 2003]

Solution. A few method which are applicable to both algebraic and transcendental equation/*) = 0to

finding the real root of an equation, we first find an approximate value of the root of the gfvenequation and then successively improve it to some desired accuracy. The general technique to(. finding the root of equationX*) - 0 is that we start with an initial approximate value, say xq

then find the better approximations successively xj, x2, ...... x„.By repeating the same method the root more and more closely, i. e., is called method converges.LetXx) = 0 be given equation, and let x& xj, x -l ......x„ be the successive approximations tothe root aofXx) **0. Let e„ be the error in tire root xn ; n ~ 0 ,1,2, ......Since a is the exact

root of/x) = 0, then e„=x„- a and e„+ i =x„ + 1- aFor any iterative method p is said to be order o fconvergence, if

|e„+l| £ k [ejf , where k is a positive constantIfp = 1 then the convergence is linear and i f p ~ 2 then it is quadratic.

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154 | En g i n e e j^ n s Ma t h e m a t i c s -III

2.16 COMPARISON STUDY OF ITERATIVE METHOD

Nameof Method

Iterative formula Order ofConvergence

P

ConvergenceCondition

Reliability ofConvergence

1. Bisection *" +! 2

where x„ and x„. tare enclose roots

Gain of one bit

per iteration

/(d) and/(a)

must be oppositesign i.e.,

Ad)Ab)< o.

Guaranteed

convergence

2 Regula Falsi xn*i xn _ j

Or

(x„ X f(x ) X"+,~ " f(x„) — f ( xn~\ ) / U )

Xa and i are enclosed roots

m m <o Guraranteed

convergence

2. Secant(xn ~ *«-l) u f(x )

*"+1 " / ( * > - / < * -■ > n n)

x„ and x„_ , are enclosed roots

i+Vs p - ----- —or y 2

p~ 1.618

f(Xn)*f(x„- |) No guarantee

of convergenceif not near the

root.

4. Newton-

Raphson

f(x„)

***' *" f'(x„) P= 2 f(x„)<f(xn) Convergent

fast, if startingvalue near die

root.

2.17 GRAEFFE’SRGOI’SQUARING METHOD

In all the previous methods we require prior information about die root, but in this method we donot required any previous information and it is capable of giving alt the roots of algebraic equation

fix) ~ 0 at a time.

Consider the polynomial:

fix) =xfl + + aix” ~ 2 + .... + a„-\x + a„ = 0 ...(1)

Now, separate even and odd degree powers of x and then square of it, we get

( xn + 02*" “ 2 + —¥ = (<*!■*" - 1+ “ 3 + —-J2-

Now, substitute x2 =y and simplifying, we get

y + b ^ y - 1+.... + bn — o —(2)

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Nu n e r jc a IÂn a l y s i s-1 1155

where, b\ - -of + 2o2

bz - of - 20]0a + 2a,

A* - (-D»a».If x,, X2, ....x* be the roots of equation ( I X then the roots of (2) are xf , x f ....... jc*. Now, after

squaring m times, let the new transformed equation be«" + ifcja*- 1+ kit /1" 2 + .... + k itH2 + iii-1« + ^ = 0 ...(3)

whose roots are c$|, <5j,....4 a« such that

4 = *?".(» = 1,2,...si.)

Now, from (3), we get1 4 = - * ,

£<Si cJj * kj.

=(-1)"^ Jand from these equations we can obtain the result for <5as

«?i « * ~T-» ............... .«! «2

f k Y/aMand since Sl = xfm *i = (^ )1/2j" - ~ T ~ *»-i /

Hence, we can determine the actual

roots xi, X2, ....x* of the equationX*) * 0.xample 2 3 5 : Apply Greffe's root squaring method to solve the equation

x i - t x 2+ 17x-10 - 0 fRGPV Dee. 2003/

olution. Given x3-8x2+ !7 x -10 =0 ...(1)Rearranged given equation by separating even and odd degree powers of x, we get

x* + 17x = 8x? + 10

Squaring both sides, we get (x3 + Mx')1 * (8X2 + 10)2=> x* + 34** + 289x2 = 64x4 + 160c2 +100.

Now, let x 2 - y, we get/ + 3 4 / + 289y * 6 4 ^ + 160y+ 100.

=> / + 129v = 30/+ 100.[As rearranging by separating even and odd degree terms]

Again squaring both sides, we get( / + 129y)2 - (30V2 + 100)2

=> / + 25 8 / + 16641 / = 9 0 0 / + 600 0 / + 10000

=> /+ 1 0 6 4 1 /= 6 4 2 /+ 1 0 0 0 0

Now, let / = z, then we get

r3 + 1064 Iz =642^ + 10000

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Again, squaring both sides, we get(z* + 10641 z f = (642Z2 + 10000)2

=> z6 + 21282z4 + I13230881Z2 = 41 21 64z 4 + 128 400 00z 2 + 100000000

Put z2 = h, then we get

- 3 90 88 2m2 + 100390881 « - 10s = 0 . ..(2)

Comparing (2) with m3 + k\u2 + k2u +k% = 0 and hence

*, = - 390882, k2 - 100390881, *3 = - 108.

Let the roots of equation (2) are <%, then we get

*i = = ( - ki)m (3908 82)1'8 = 5.004 « 5

156 | Enginhwno M a t h e m a t ic s -IH

*2

* 3 ( ^ 1 0 0 3 9 0 8 8 1 J

Thus, roots of given equation are 5 , 2 and 1. Ans. Example 2J6 : Solve the equation x 3 - 6 x* + l l x - 6 “ 0 by Graeffe’s root squaring method.

[RGPV June 2006J

Solution. Given x3 - fix2 + 1lx - 6 = 0 ...(1)

Rearranged given equation by separating even and odd degree powers of x, we get x3 + I I t

= 6x2 + 6

=> x(x2 + U ) = 6(x2 + i )Squaring on both sides, and putting x2 = y,we get

y(y+ 1 1 ) 2 = 3 6 ^ + 1 ) 2

=> Xy2 + 22 y + 121) = 360^ + 2 y+ 1)

=> y3 + 22y2 + 121y — 3 6 )^ + 72y + 36

=> y* + 49y - My2 + 36

=> yiy 2 + 49) —(My2 + 36)

Again squaring and putting y2 = z, we get

z(z + 49)2 —(14 ^ + 36)2=> r3 + 98z2 + 2 4 01 z = 1 9 6 z2 + 1 0 0 fe + 1 29 6

=> z3 + I393z = 98Z2 + 1296

=> z(z2 + 1393) - (98Z2 + 1296)

Further, squaring and putting z 2 - u, we get

=> «(« + 1393)2 = (98 « + 1296)2

=> m3 + 2786m2 + 1940449m = 9604m2 + 2 54 01 6a + 1679616

=> «3 -6 8 1 8 m2 + 1686433m - 1679616 - 0 . . .. (2 )

Comparing (2) with m3 + k\u* + k$u + *3 = 0 , so thati ] ——6818, k 2 = 1686433, k3 * - 1 67 96 16 .

- f - S s s r )

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N u m e r ic a l A n a l y w *! 1157

Let the roots of equation (2) are S\, <%, then we get

x, = ($)>*==(-*i),/8 = (6818)l/8 = 3.01444 * 3

/ rvw« f kz)Ua (1686433Y/8- ( - j g i g - J - 1 . 9 9 1 4 . 2

/ J t u / s - f feaV/8 f 1679616f ' 8,• ^ x* ^ U 686433J 0.9999 « 1.

Thus, roots of given equation are 3,2 and 1. Ans.

^KACRSTOW’S METH OD O R METHOD FO R CO MPLEX R OO T

i 3Rbs method is useful to obtain complex roots of an algebraic equation/*) = 0. The complex roots

-Of such an equation occur in pairs a ± ib. Each such pair corresponds to a quadratic factor, which j® as follows:

‘ ’ t { x - ( a + ib)} { x - ( a ~ ib)} =x2 ~2ax + a2 + b2 - x 2 + px + q,

where p and q are real.

’Let f i x ) - x” + a\x"~ 1 + . .. . + On - \x + a„ be given polynomial . . .(I)

If we divide/*) by ( jc2 + px + q), we obtain a quotient

Qn-2 = X n ~ 2 + b ]Xn- i + „.. + b„-2

and a remainder R„ = Rx + S

Hence (1) becomes:

f ix ) = (x2+px + q) (x"-2 + - 3 + .... + b„-j) + Rx +S •—(2)

IfX2+px +q divides/x) completely, the remainder Rx + 5 = 0 i.e., R - 0, S’= 0. Obviously R and S both depend upon p and q.

So our problem is to find p and q such that

Rip. q) = 0, S{p, q) = 0. ...(3)

Let p + Ap, q + Aq be the actual values of p and q which satisfy (3) then,

Rip + Ap, q + Aq) = 0; Sip + bp, q + Aq) = 0

To find the correction Ap, Aq, we get the following equations :

Cn- 2*P + Cn-S&I * K -l \

(Cn~l - V l W + * , -2 ^ = K J " W

After finding the values ofbp and c,s by synthetic division scheme, we obtain approximate values of

Ap and Aq say, Apo and Aqo

Ifpo, q0 be the initial approximations then their improved values are

P\ ~P 0 + Apo, q\ - qo + Aqo. Now taking p\ and q\ as the initial values and repeating the process, we can get better values of p

and q.

This method works well only if the starting values of p and q are close to the correct value.

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158 | EN O osaaNoM A nouncs-n i

Remark : The synthetic division procedure are as follows:

a* : a0(=l) <»1

- p b o

a2

- P b i

-flbo

- P ^ 2 .... - f * n-3

-<£1 .... -<jbn-i

On-1

-P*>n-2

-P<V3

~pb„. i

- p b n_2

- p

~Qho(=D

- p c 0

b 2

“ PC t

-<JCo

-PC2.... - P C ^

-QCj.... -<JC^4

*PCn-2

-<*n-3- P~Q

c0(—1) Cl c2 C*-l

Example 2.37 : Find a quadratic factor o f the polynomial X 4 + ix* + 3x? - 5* - 9 ® ft Starting with

pg~3 ,qo~~ 5 by using Bairstow s method. fRGPV June 2003]

Solution. Synthetic division scheme is as follows :

c*_ j — b„ _ j = —36 + 1 ——35.

The corrections Apo and Aqo are given by

Cn- 2&P0 + C»-3^0 = V !=> 10 Apo - = ~ 1 —(1)

and (c*.| - b „ - ]yipo + c„ 2 ^!o = l>n=> -3 5 Apo + 10 = 4 ...(2)

On solving (1) and (2), we get, Apo = - 0.09, A?o = 0.08

Thus pi, q\ the fjgp approximations ofp and q are given by

pi =Po + Apo = 2.91, = qo + ~ ~ 4-92 .

Repeating same process Le., dividing/x) by x2 + 2.91x - 4.92, we get

1 5-2 .91

3-6 .084.92

- 5-5 .3510.28

- 90.209.05

-2.91

4.92

1 2.09 1.84 -0 .07 0.25 -2 .91-2 .91 2.37 -26.57 4.92

4.92 -4 .03

1 -0 .82 9.13 -30 .67

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At this step, the corrections Ap, and Aî are given by,9.13 Ap) - 0.82 Aqi =-0.07

- 3 0 . 6 0 ^ + 9.13 Aqt =0.25=> Ap, = - 0.00745,

Aq\ - 0.00241Hence second approximation of p and q are given by,

Pi + Api - 2.91 - 0.00745 = 2.90255

q 2 = q\ + A?, = - 4.92 + 0.00241 = - 4.91759.Thus a quadratic factor is, x 2 + 2.90255 x - 4.91759.Dividing the given equation by this factor, we can obtain the other quadratic factor. Ans.

2.19 SOL UTION OFSIMULTANEOUS LINEARALGEBRAIC EQUATIONSWe come across very often, simultaneous linear algebraic equations in various fields of science andengineering. We have solved such equations by mathematical methods such as Cramer’s rule

(determinant method) and matrix method. These methods involve a great amount of labour, whenthe number of equations exceeds four. On die other hand, the numerical methods, which involveless amount of labour and are also best suited for computer operations.These numerical method are of two types namely ; (i) Direct method (ii) Iterative method.In the direct method the given system of equations will be transformed into an equivalent uppertriangular system.In the iterative method, the solution of given system of equations is obtained by successiveapproximations.

2.20 GAUSS-ELIMINATION METHOD (DIRECT METHOD)

Consider the n-linear equations in n unknowns,

a u X i + O j 2* 2 + + ° \ n x n ~ ^1

021*1 + <*22*2 + •—+ °2nxn = 2

+ °»2*2 + ■■■*+ a«rt*n = K

where dy and b, are known constants and x,’s are unknowns.The system (1) is reduce to matrix form :

AX = B

.-(I)

...(2)

On <*12 •• °1n *1 Vwhere A = 021 °22 °2n . x= *2 and 5 = h

_anl ®rt2 ••ann . .Xn. A. Now our ob

operations.

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160 | Fit i th a m m lU n g m n c s - m

On Now take the pivot an, so that multiply the first row of (3) by “ ~ and add to the ith row of

11

[A : B], where i = 2 , 3 , n. So that all elements in the first column of [A : J?] except a\ yare made

to zero, then (3) reduces to

[A : B] ~

°ii °iz *•°21 a22 ••

a»l ar >2 "îw

..(4)

Now taking 622 as the pivot, using elementary operations, we make all elements below bn as zero,then (4) reduces to

...(5)

Again taking C33 as the pivot, using elementary operations, we make all elements below C33 as zeros.Continuing the process, all element below the leading diagonal elements of A are made to zero.Hence

Oil °12 a 13 *• ° l n

0 ^2 ^!3 C2[A : B] ~ 0 0 C33 •> <h

0 0 C*3 •• Cnn

Oil °12 °13 „ a ln fcl‘

0 ^22 kt 3 .. ^ 2 n C2

0 0 C33 c34... ^3n <h

0 0 0 0 a « .

[A :B]~

From, (6), the given system of linear equations is equivalent to

a„*! + 0x2*2 + 0(3*3 + ™+ “i^%2*2 + Sl3*3 + ••• + = C2

C33*3 + —CstXn = dj

-(6)

<*nnXH =

By using back substitutions, we get values ofxm xn. 1, x„_ 2, .... *2, *1. Example 2.38 : Solve the following system o f equations by Gauss-eUmlnatktn method :

2 x -y + 3z » 9,

x + y + z * 6 , x - y + z = 2 fRGPV Dec 2001}

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elution.

N u m e r ic a l An a l y s is *! j 161

The given system in matrix form : AX = B

where, A =

2 - 1 3

1 1 11 -1 1

Augmented matrix: [A : 5] =

2 - 1 31 1 11 -1 1

..(I)

[Taking «n - 2 as pivot]

Operating j and - 3 **3 “ ( 2 ) we get

[A : B]

'2 -1 3 9

0 3 /2 - 1 / 2 3 /20 - 1 /2 - 1 /2 - S /2

(1 Ôperating: ^3 + I & 2 , we get

[Now taking 622 = 3/2 as the pivot]

' 2 - 1 3 90 3 /2 -1 / 2 3 /20 0

- 2 /3 -2 J

[A:B\

Which is in upper triangular form.From (2), we get

2jc - y + Sz = ®

( 3 / 2 ) y - ( l / 8 k = 3 / 2

( - 2 / 3 ) 2 = - 2

Using back substitution, we get z = 3, y= 2, x = I.

,\ The requked solution is x = 1, >>* 2,2 = 3. Example 2*39: Solve by Gaim-elimiHOtkm method:

lO x + y + 2 z - 13

3 x + lO y + z = 14

2 x + 3 y + lOz = IS

Solution. Given equations write in matrix form :AX~ B.

.(2)

Ans.

(R G lty June 2002]

...d)

where, A =

1 e * 0 r H' X 13*

3 10 1 , X= y » 14

2 3 10 z _ 15

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1 6 2 | En g in e e r in g M a t h e m a t ic s -!!!

Augmented matrix : [A : B\ =

Operating: R2 -* Rz

‘ 10 1 2 13

3 10 1 14

2 3 10 15

— | and

[A:B]

^ - j “ j ^ i . we get

" 10 1 2 13

0 9 7 / 1 0 2 / 5 1 0 1 / 1 0

0 1 4 / 5 4 8 / 5 6 2 / 5

Operating : _ ^ - ^ 2* we get

[A:S\

10 1 2 13

0 9 7 / 1 0 2 / 5 1 0 1 / 1 0 '

0 0 460 460

Which is in triangular form.

From (2), we get

IOjc + y + 2z = 13 (97 / 10)y + (2 / 5)z = 101/10

4602 = 460

Using back substitution, we get

2 = l , y = l ,x » 1/. The required solution is x=l,>’= l >z = l .

Example 2.40 : Solve by Gauss-eliminatioH method :3x + 4y + 5i - 18 2 x -y +8 z = 13

Sx-2y + 7z =20.

Solution. Aug mented matrix [A : B] =

3 4 5 18"

2 - 1 8 13

5 - 2 7 20

fl,1 ■ we get

3 4 5 18 '

[A : B) ~ 0 - 1 1 / 3 1 4 / 3 1

0 - 2 6 / 3 - 4 / 3 - 1 0

Ans.

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164 | En g in e e r in g M a t h e m a t ic s -III

f 6 . 7 3 3 5 )Operating : -“ 3 n 3 ~ I 6 4453 I™2’ we get

[A:B]~

From (1). we get

3.15* - 1.96y + 3.85z = 12.956.4453y - 5.4933* = -17.3666

0.6534* = 0.6853

Using back substitution, we get

0.6853

3.15 -1.96 3.85 12.95

0 6.4453 -5.4933 -17.3666 ...(10 0 0.6534 0.6853

0.6534 ‘ , 048S-

5.4933 x 1.0488 - 17.3666 y =

x =

6.4453 l'80°5,

1.96 x (-1.8005) - 3.85(1.0488) + 12.95

3.15= 1.7089

/. The required solution is x = 1.7089, y - - 1.8005, z = 1.0488. Example 2.42 : Solve by Gauss-elimination method:

lOx- 7y + 3z + 5u = 6 , - 6 x + 8y~z~4u = J,3x +y + 4z + l lu = 2, S x - 9 y - 2 z + 4u = 7.

Ans.

Solution. [A: B] =

‘ 1 0.7 0.3 0.5 0 .6'

-6 8 -1 -4 53 1 4 11 2

5 -9 -2 4 7

[Since dividing first equation by 10

Operating : R 2 + 6 R\> R 3 Rj - 3/?i, R4 R* - 5R\, we get

[A :B)~

Operating: Ri

[A : B] ~

"l 0.7 0.3 0.5 0.6

0 3.8 0.8 -1 8.6

0 3.1 3.1 9.5 0.2

0 -5.5 -3.5 1.5 4

.l)/?2 and R4 -> (3.8)&i + (5.5

1 0.7 0.3 0.5 0.6 ‘

0 3.8 0.8 -1 8.6

0 0 9.3 39.2 -25.9

0 0 -8.9 0.2 62.5

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N u m e r ic a l A n a l y s is -! | 165

Operating : / ? 4 —► (9.3) R4 + (8.9) # 3, we get

1 0.7 0.3 0.5 0.6

0 3.8 0.8 -1 8.6

0 0 9.3 39.2 -25.9

0 0 0 350.74 350.74

[A:B]

From (1), we get

x + 0.1 y + 0.3z + 0.5 u = 0.63.8,y + 0.8z - u ~ 8.6

9.3 z + 39.2u = -25.9350.74u = 350.74

Using back substitution, we getm= 1, z = - 7,>> = 4, x = 5

/. The Solution of given equations is x = 5, y - 4, z = - 7, u - 1.xampie 2.43 ; Solve the following system o f equations by Gauss-ellmination method:

Sxj - x 2 + x 3 ~ 10

2xt + 4 x 2 = 12 x , + x 2 + Sx 3 - -1

’elution. The given system in matrix form is AX - BAugmented matrix:

'5 -1 1 10'

[A : B) =2 4 0 12

1 1 5 -1

Operating : R\ R}, we get

1 1 5 - l ‘

[A :B]~ 2 4 0 125 -1 1 10

Operating : R 2 -* Rz~ 2R\, R-} -> R 3 - 5R,, we get

Operating: ^2

1 1 5 -1

[A: B] ~ 0 20 -6

-10-24

1415

f1 ]UJ we get

1 1 5 - l '

[A: B ]~ 0 10 -6

7

3 1 1

7

15

.-(I)

Ans.

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Operating : Ri -» Ri - R\, Ry -» Ry~ Ru R a R a ~ 5/?i, we get

[A : B)

"l 1 1 4 -6 '

0 6 0 -3 18

0 0 5 -3 10 -4 -4 -19 34

Operating: ^2( V

6 jwe get

1 1 1 4 -6

0 1 0 -0.5 3

0 0 5 -3 1

0 -4 -4 -19 34

1 0 1 4.5 -9

0 1 0 -0.5 3

0 0 5 -3 1

0 0 -4 -21 46

[A : 5] -

Opera ting : / ? [ -» - R 2 , then Ra -* Ra + 4/?2, we get

[,4:2?]

Operating : Ra j - ^ 3 >we get

[A:B]

Operating : R\ -> R\ - Ri, Ra R\ + 4R}, we get

1 0 1 4.5 -90 1 0 -0.5 3

0 0 1 -0.6 0.2

0 0 -4 -21 46

[A:B] ~

1 0 0 5.1 -9.2’

0 1 0 -0.5 30 0 1 -0.6 0.2

0 0 0 -23.4 46.8

Operating: - ^ 4

' 1 N

{ 23.4 R* >we get

[A : B]

“l 1 1 5.1 -9.2 ‘

0 1 0 -0.5 3

0 0 1 -0.6 0.20 0 0 -1 2

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168 | E n g in e e r in g M a t h e m a t ic s -H I

Operating: —► JR3 - (0.6) -* R? - ( 0 . 5 ) R i , R l -* R\ + (5.1)/L», we get

1 0 0 0 1

0 1 0 0 20 0 1 0 -10 0 0 -1 2

{A :B]~

Using back substitution, we getxi = l , x 2 - 2, X3 = - 1, jt4 = - 2.

Example 2.45 ; Apply Gauss-Jordan method to find the solution o f the following system: I 8 x +y + 2 ™ 12 ;

2x+ I0y + z a 13; x +y + Sza 7.

Solution. Interchange the first and the last equation, we get

Am

[A : B) =

Operating : R 2 -» R 2 - 2R]> R)-+ R 3 - 10/?i, we get

" 1 1 5 7 '

2 10 1 13

10 1 1 12

1 1 5 7 ‘

[A : B] ~ 0 8 -9 -1

0 -9 -49 -5 8

Operating : Ri [ g w e 8et

1 1 5 7

[A : B) ~ 0 1 - 9 / 8 - 1 / 80 -9 -49 -58

Operating : R -»• R$ + 9R2, we getr l 1 5 7

[A : B) ~ 0 1 - 9 / 8 - 1 / 8

0 0 -4 73 /8 - 47 3/ 8

Operating: R9 1 ^ ’ weget

‘l 1 5 7

[A : B] - 0 1 - 9 / 8 - 1 / 8

0 0 1 1

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N u m e r ic a l A n a l y s is -1 1 16 9

Operating R} - R?. we >yi

I u 4 9 / 8 5 7 / 8 '

[A : B] - o 1 - 9 / 8 - 1 / 8

0 0 1 1

Operating: ^2 •fta + [ g ] - 3*- 1 ^ I we get8

' l 0 0 I

[A : B) ~ 0 1 0 1

0 0 1 1

Using back substitution, we get x = l ,y = I, z = 1. Ans.

CROUT’S TRIANGULARIZATION METHOD OR CHOLESKEY’S METHODOR LU FACTORIZATION METHOD (DIRECTM E T H O D )

This method is based upon the fact that any square matrix can be expressed as the product of alower and a upper triangular matrices (All the principle minors of the matrix must be non-singular).i.e., if A ~ [o,y], Then \A\ * 0.Let us consider the equations :

0)1*1 + a J2*2 + °13*3 ~ &!

<*21*1 + 022*2 + °23*3 = &2

° 3 l x l + ° 32 x 2 + °33*3 = ^3

Which can be written in matrix form : A X - B

...d)

•••(2)

«1! °12 °13 X\ Vwhere, A « 021 °22 023 > x= H and B = b 2

.°3I °32 O33 _ _*s. bs

Since

where

A = LU

L =

' 1 0 O' Uj2 “131 0 and t / = 0 °22 23

hi 1 0 0 33 _

...(3)

are lower and upper triangular

matrices./. System (1) becomes :

We assume UX = V, where V -

Then system (4) becomes :

LUX - B l()

-.(5)»>2y3

LV = B -(6)

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170 | En g in e e r in g Ma t h e m a t ic s -!!!

1 0 0' V!1 i 1 l>2 = *>24i 2 1 3 .

k

h \V\ + V2 = &2

hivi + httVi + 3 =

Solve these for Vi.e., vj, v2, v3 and putting in (5), we get

U;i.T| + U,2*2 + «J^X3 = V,

U 22 X 2 + 11*3X3+ =“ 33X 3 = V3 j

Using back substitutions, we have obtain x\, xi, xj.

Example 2.46 : Apply Crout's method to solve the equations :

3x + 2y+7z = 4;

2x + 3y + z ~ 5;

3x + 4y + z —7.

> ’tution. Given system can be written in matrix form : AX = B

' fRGPV Dec. 2004 & Dec. 2007}

-(I)

' 1 0 o '1 “ ii Ui2 «%

Sincc A ~ LU where L = hi i 0 and ( / = 0 !i22 °23

M h i 1 0 0 “ 33 .

' 1 0 O' ui\ Ui2 Wia 3 2 7'

h i 1 0 0 “ 22 «23 = 2 3 1

hi h i 1 0 0 “3 3 . 3 4 1[v LU = A]

T>

Hence

un^!ultk l uU

U 124 lu12 + **22

^ lu12 + ^ “22

w13

41 13 + °2S

3l“l3 + S2°23 +

3 2 72 3 13 4 1

«n - 3,

21 11 = ^21 —2 / 3>

/3]Uu = 34n =

u, 2 = 2

21ui2 + u22 = ^=> U-22 = 5 / 3,

^!U32 + 32°22 = 4=> 42 = 6 /5 ,

u!3 - 7

21u13 + 23 = 1^ lio3 ~ ” 11 /3

h l Ul2 + ^32^23 + °33 ~ 1=> U-33= -8 / 5

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N u m e r ic a l A n a l y s is -1 | 17 1

Thus,

Suppose,

where

1 0 0 "

2 / 3 1 0and {/ -

1 6 / 5 1

UX = V

V =

V 3

3 2 7 ]

0 5 / 3 - 1 1 / 3 |

0 0 - 8 / 5 j

then A X - B => LUX = B=>LV= B

1 0 0 " ii 4

2 / 3 1 0 u2 =5 5

1 6 / 5 1 _U3_ 7

2 7 6=> V! = 4; - i'i + i '2 = 5 => v2 = - ; u, + - v2 + vs = 7

o .i 5

Hence (2) i.e.%UX = V becomes :

L*3 —— .3 5

'3 2 7 4

0 5 / 3 - 1 1 / 3 > = 7 / 3

0 0 - 8 / 5 2 1 / 5

3% + 2y + I z ~ 4^ ( 5 / 3 ) y - ( l l / 3 ) z = 7 / 3

- (8 / 5 )z = 1 / 5

Using back substitution, we get z - - 1/8, y = 9/8, x - 7/8.

Example 2.47 : S'o/ve Ay Crout’s m eth od :

10x + y + z = 12;

2x + lOy + z ~ 13;

2x + 2y + lOz = 14.

olution. Given system in matrix form : AX - B

' 1 0 0 ‘ u n u t2 UI3

Let, A = LU, where L = h\ 1 0 and t / = 0 «22 ^23

3! 32 1 0 0 ^32 _

Since,

«jt = 10,

hiuu ~ 2> 21 = 1/5,

*3l“ ll - 2> *31 = 1 / 5 ,

LU = A, we get

u12= 1

*2IU12 + 22 =:> U q 2 = 49 / 5,

41 12 + 32 22 = 2^ /g, = 9 / 49,

••(2)

Ans.

(I)

u13 = 1+ u23 = 1

^ u 23 = 4 / 5

^31*% + ^ 2 U23 + u 33 ” 10

u 33 = 473 / 49

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1 7 2 | E n g in e e r in g M a t h e m a t ic s -III

Suppose,

where V =f-’i

3

UX = V

. then AX = B => LUX = 5 => LK = B

- (2)

1 0 0 ‘ 12

1 /5 1 0 U2 _ 13

1 /5 9 /4 9 1 -y3_ 14

=> V[ = 12; /2 [V| + vj = 13

=> v2 = 53/5

/31 vj + /32V2 + V3 = 14

=> v3 = 473/49.

Hence (2) i.e., UX = V becomes :‘10 1 1 x 12

0 4 9 /5 4 /5 y 53/50 0 473/49 z 473/49

lOx + y + z = 12(49/5);y + (4/5)2 = 53/5

(473/49)2 = 473/49

By back substitution, we getz = l,.y = l ,x = 1. Ans.

2.23 ITERATIVE METHOD

The iterative method is not always successful to all systems of equations. In order for the iteration procedure to give the solution of a system of equations, each equation of the system must containone coefficient much larger in magnitude than the others in that equation and the large coefficientmust be that of a different unknown in each equation. In other words, after rearranging the equationsif necessary, the large coefficients must be along the leading diagonal of the coefficient matrix.i.e., the system of equations :

will be solvable, if

I Oj 1 I> I I + I <*13 I

I O02 l>l @21 I + I ®23 I ’I O33 I> I £*31 I + I O32 I

2.24 GAUSS-JACOBI METHOD

Consider the system of equations of three variables in three unknowns :

Oj* + l\y + cxz = dt

02* + f^y + CjZ = dz

a^x + bsy + c3z = tf3if , .(1)

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N u m e r ic a l A n a l y s is -! | 173

11*! I > I bi 1+ I c, |I fife !> | Oj I + I c2 I

| Cs |> | <*3 | + | 63 I

then, iterative m ethod can be used for the system (I )Hence we write x, y, z as follows :

x =~(dt - b y - c i z ) >,<*1

y = (dj - oax - c^z)

z = — (da - agx - b^y).

If x<0>, > 0), z<°> are th e initial values o f x, y, z respectively, then

First iteration :

-.(2)

^ l ) = — (d[ - 6|y° > - c,z<°>) ^«1

yl) = “ (A - O2X{0) - C Z™)0*

2<1) = — (dg - <%x(m - b ^ y ^ ) ) c3

Again using these values j in (2), we get

Second iteration:

y( 2) = — (d, -<*1

1y2) = — (<4 - - C2Z<‘>)

■(3 J

..(4)

2(2) = — (di - 0 3X<1> - y « ) J

Continue the proced ure till the con vergence is assured co rrect to required decimal places.

emark : In the absence o f initial values o f x , y , r w e take, usually x = y - z ~ 0.

xample 2.48 ; Solve the following system by Gauss-Jacobi method :

lOx - 5y-2z = 3;

4x - lOy + 3z = - 3; x + 6y + lOz - - 3 . fRGPV Dec. 2008(N)j

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Solution. Since |tO| > |- 5| + |~ 2|, |- I0| > |4| + |3| and |10| > |1| + \ 6 \.Then Gauss-Jacobi method can be used for the given system of equations.

Given system can be written as :

x — [3 + 5^ + 2 z]

1 7 4 | E n g in e e r in g M a t h e m a t ic s -!II

v = — [3 + 4x + 3 z] y 10

Let the initial values be x = y = z = 0, we getFirst iteration :

* 1, = — [3 + 5(0) + 2(0)] = 0.3

yi) = - 1 [3+ 4(0)+ 3(0)] - 0.3

^ [-3 - (0) - 6(0)] = - 0.3

Second iteration : Putting the values o f x ^ . y 1>, z*1*in (I), we get

*2) = i - [3 + 5(0.3) + 2(-0.3)] = 0.39

y 2) = -L [3 + 4(0.3) + 3(—0.3)] = 0.33

= ^ [- 3 - (0-3 ) - 6(0-3)] ——0.51

Third iteration : Putting the values of xP \ y 2\ z<2) in ( I), we get

& = ^ [ 3 + 5(0.33) + 2(—0.51)] = 0.363

^ = ^ t 3 + 4<03 9>+ 3(~°-51)} = 0.303

aP> = ^ [ ~ 3 - (0.39) - 6(0.33)] = - 0.537

Fourth iteration : Putting the values of / 3), in (1), we get

= j^ [3 + 5(0.303) + 2(-0.537)] = 0.3441

= ^ [ 3 + 4(0.363) + 3(—0.537)) = 0.2841

^4) = j L[-3 -(0.363) -6(0.303)] =-0.5181

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Fifth iteration :

Sixth iteration :

y5)

x<(>)

y(6)

t<6)

Seventh iteration :

Eighth iteration :

Nineth iteration :

X<7)

y7)

*8)

^8)

2<*>

x<*>

= TX

10

J_ 10

JL_

10

J_

10

J_ 10

J_

10

J_

10

J_

10

J_

10

J_

10

J_

10

J _

10

JL_

10

J_

10

110

3 + 5(0.284) + 2(-0.5181)] =0.33843

3 + 4(0.3441) + 3(-0.5181)] = 0.2822

-3 - (0.3441) - 6(0.2841)] « _ 0.50487

3 + 5(0.2822) + 2(-0.50487)] = 0.340126

3 + 4(0.33843) + 3(-0.50487)] = 0.282911

-3 - (0.33843) - 6(0.2822)] = _ 0.503163

3 + 5(0.283911) + 2(-0.503163)| .. 0.3413229

3 + 4(0.340126) + 3(-0.503163)] = 0.2851015

-3 - (0.340126) - 6(0.283911)] = _ 0.5043592

3 + 5(0.2851015) + 2(-0.5043592)J - 0.34167891

3 + 4(0.3413229) + 3(-0.5043592>] = 0.2852214

-3 - (0.3413229) - 6(0.2851015)] = - 0 . 5 0 5 1 9 3 1 9

3 + 5(0.2852214) + 2(-0.50519319)] = 0.34157

3 +4(0.34167891)+ 3(-0.50519319)] =0.28511

-3 - (0.3416789) - 6(0.2852214)] = - 0.5053

Nu m e r ic a l An a i,y s is -I | 175

Since eight and nineth iteration are same correct to 3 decimal places.

Hence, x = 0.342, y » 0.285, z - - 0.505. Ans.

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Example 2.49 ; Solve the following system o f equations by using Gauss-Jacobi method correct to} decimal places:

8 x - 3y + 2z * 20;

4x + H y - z -33;

6 x + 3y

+12z ~ 35.

Solution. Since |8| > {- 3| + |2|, (11| > |4| + j- !j and |i2| > |6| + |3|.Then Gauss-Jacobi method can be used for the given system of equations.

Hence, we w rite jc, y, z as follows

x = ±[20 + 3y - 2 z] 'o

J j [ 3 3 - 4 1 + *J

i [ 3 5 - 6 x - 3 > U

Let the initial values be x = 0, y - 0, z = 0, we get

First iteration :

*0) = I [20 + 3(0) -2(0)] =2.5O

y n = _L [33 - 4(0) + 0] - 3.0

2(1) = A-[35 - 6(0) - 3(0)] = 2.916666

XzSecond iteration : Putting the values o f 1 \ >, r *1 >in (1), w e get

x<2) = I [20 + 3(3.0) - 2(2.916666)] = 2.895833O

yt2j = l - [3 3 - 4(2.5) + (2.916666)] = 2.356060

176 [ ILn g in l e k in g M a t h e m a t ic s -111

Third iteration:

p . ) = ± - [ 3 5 - 6 ( 2 . 5 ) - 3 ( 3 .0 ) ) = 0 .9 1 6 66 6X

* 0) = i [ 2 0 + 3 ( 2 .3 5 6 0 6 0 ) - 2 ( 0 . 9 1 6 6 6 6 ) ] = 3 .1 54 35 6O

y 3 ) = - ± - [3 3 - 4 ( 2 . 8 9 5 8 3 3 ) + (0 . 9 1 6 6 6 6 ) ] = 2 .0 3 03 0 3

2(3) = — [ 3 5 - 6 ( 2 . 8 9 5 8 3 3 ) - 3 ( 2 . 3 5 6 0 6 0 ) ] = 0 .8 7 97 3 512

"Fourth iteration :

x<4) = ± [ 2 0 + 3 ( 2 . 0 3 0 3 0 3 ) - 2 ( 0 . 8 7 9 7 3 5 ) ] = 3 . 0 4 )4 3 08

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Fifth iteration :

Sixth iteration .

Seventh iteration

Eighth iteration :

Nineth iteration:

y 4) = jy [3 3 - 4 (3 .1 5 4 3 5 6 ) + (0 .8 7 9 7 3 5 ) ] = 1 .93293

2(4) = — [35 - 6(3 .154 35 6) - 3(2 .0303 03)J = 0 .831913

12

x<5) = I [20 + 3(1.932 937 ) - 2(0 .831 91 3)] = 3.016873

y j) = JL [3 3 _ 4 (3 .04 143 0) + (0 .831913)] = 1 .96965^

i s ) = — [ 35 - 6 ( 3 .0 4 1 4 3 0 ) - 3 ( 1.9 3 2 9 3 7) ] = 0 .9 1 2 71 712

= - [ 2 0 + 3 (1 .9 6 9 6 5 4 ) - 2 ( 0 . 9 12 7 1 7 )] = 3 .0 1 0 4 48

y«> = -L [3 3 - 4(3 .016 87 3) + (0 .91 27 17 )] = 1 .985930

* 6) = — [35 - 6 (3 .0168 73) - 3 (1 .969654)] = 0 .91581712

jfrt = - [ 2 0 + 3 (1 .9 8 5 9 3 0 ) - 2 (0 .9 1 5 8 1 7 ) ] = 3 .0 15 77 08

yP) = j j [ 3 3 - 4 (3 .0 1 0 4 4 1 ) + (0 9 1 5 8 1 7 ) ] = 1.988550

2(7) = J - [35 - 6 (3 .01 044 1) - 3 (1 .985930)] = 0.91496412

X») = - [ 2 0 + 3 (1 .988550) - 2 (0 .9149 64)] = 3 .0169468

yP) = JL [3 3 - 4 (3 .015 77 0) + (0 .91496 4)] = 1.986535

2(8) = — [ 35 - 6 ( 3 . 0 1 5 7 7 0 ) - 3 ( 1 .9 8 8 5 50 ) ] = 0 .9 1 164 I-12

*9) = —[20 + 3(1 .98653 5) - 2(0 .9 116 44 )] = 3 .0170398

^ 9) = I T 133 ” 4 <3(>1 694 6) + <° 9 11 64 4 )I = 1-985805

N u m e r ic a l A n a i y s i% | | 1 7 7

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1 78 | B M » S B K > i U n e u n c s* IIl

Tenth iteration

4* ) = i - { 3 5 - 6(3.016946) - 3(1.986535)] = 0.911560

jfm = i ( 2 0 + 3(1.985805) - 2(0.911560)] = 3.016786

8

yiO) = _L[33 _ 4(3.017039) + (0.911560)] = 1.985764

2(10) = J - J 3 5 - 6(3.017039) - 3(1.985805)] = 0.91169612

Since nineth and tenth iterations are same correct to 3 decimal places.Hence, x = 3.017,^= 1.986, z = 0.912. Ass,

2.25 GAUSS-SEIDEL METHOD

This is the modification of the Gauss-Jacobi method.

Let us consider a system :

ai* + 6|.y + cxz = d, 1

OiX + + C 2 Z - dz OgX + l^y + CgZ = dz

Transform the above system in the following form :

* = — [dy - 6)y-CizI )<h

y = T-{<h ~ ChX-^z]

Z = - O3 X JC3

We start with the initial values y = yi°> and z = z<0\ we get x*1*from the first equation, i.e.,

*(1) = — (dy - - c ^ 05)°i

Now using the second equation, we use z = 2*°> and x = x<l), we get

,0) = ^~(<k~ OjxW - Ca ^ )02Finally, having known jK1) a n d ^ 'l use x = x<1) and y = y ■>in the third equation, we get

C3

In finding the values of the unknowns, we use the latest available values on the right hand side. This process of iteration is continued until the convergence is assured. As the current values ofthe unknowns at each stage of iteration are used in getting the values of unknowns.

Remark :The convergence in Gauss-Seidel method is very fast when compared to Gauss-Jacobi method.

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Example 230 .*Solve the following system by Gouss-Seidel method:27x + 6 y - z = 85

6 x + 15y + 2z - 72 x +y + S4i * 110

[RGPV Dec. 2002, Dec. 2006, June 2008, June 2011 and Dec. 2011] Solution. Since in each equation one of the coefficient is larger than the other, satisfying the condition

for Gauss-Siedal method, we write the given equations in the following form :

x - ~ j ( M - 6 y + z)

y = — (72 - 6x - 2 z) y 15

2 = i - d l O - x - ^ )54

we start with y = 0 and 2 = 0 and using the most recent values of x. y, z, thenFirst iteration :

XH> = ~ (85 - 0 - 0 ) =3.15

yi) = — (72 - 6 x 3.15 - 2 x 0 ) = 3.5415

z(i) = — (110 - 3.15 - 3.54) = 1.9154

Now using the most recent values of x, y. z, thenSecond iteration:

*<2) = — (85 - 6 x 3.54 + 1.91) = 2.4327

y2> = ± ( 7 2 - 6 x 2.43 - 2 x 1.91) = 3.57

e ± ( n o - 2.43 - 3.57) = 1.9254

Third iteration:

xf.1) = — (85 - 6 X 3.57 + 1.92) = 2.42627

yf}) = — (72 - 6 X 2.426 - 2 x 1.92) = 3.57215

20) = — (110 - 2.426 - 3.572) = 1.925.54

Since second and third iterations are same correct to 3 decimal places.

The required solution is : x = 2.426, y = 3.572, z = 1.925. Ans,

Nu m b o c a l An a l y s » I | 1 79

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Example 2.51 : Solve the equations I Ox + 2y + z~ 9,

~ 2x + 3y + lOz = 22; x + lOy - z = - 22

by Gauss-Seldel method. {RGPV June, 2003]

Solution. S ince in each equation one of the coefficient is larger than the other, satisfying the conditionfor Gauss-Siedel method, we write the given equations in the following form :

1

1 8 0 | En g in e e r in g M a t h f m a t ic S'111

y = — (-22 - x + z)

z -

10

J_

10(22 + 2 x - 3 y) J

•••(I)

We start with y ~ z - 0 and using the most recent values of x, y, z, we get

First iteration:

y n . 1 . 0.9

y > = ( - 2 2 - x <" ) = - 2.29

^ ^ (2 2 + 2.t “ > - 3 y f)) = 3 .0 67

Second iteration :

Third iteration

* 2) = T o * 9 ' 2yW ~ zW) = 1 0 5 , 3I

- Yo (" 22 _ JC‘2’ + z<1>) = “ ' " 84

* 2> = ^ (2 2 + 2x<» - 3y(2)) = 3

jc<3>= — (9 - 2yW - z<2>) = 0.99968

y 3> = ( - 2 2 - * (3) + z(2)) = - 1.9999

z<3) = ^ ( 2 2 + 2 x< 3> - 3 / 3 )) = 2.9 9 9

Fourth iteration :

*<4>- ^ (9 - 2y 3>- 2(3)) = i

y 4) = (-2 2 - *<« + z<3>) = - 2

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2<4 ' = ^ ( 2 2 + 2 *< 4 > - 3 vv<4 ) ) = 3

a<5> = ^ [ 9 - 2 < - 2 ) - 3 ] = 1

^ - ^ [ - 2 2 - 1 + 3 ] = - 2

~ ( 2 2 + 2 (1 ) - 3 ( - 2 ) ] - 3 .

The required solution is • x = l ,y = - 2, z - 3. Ans.

Example 2.52 .*Solve the following equations by Gauss-Seidel method:

83x + Hv -4 z =957x + 52y + /i z - 1043x + 8 y + 29z = 7/ Ztec. 200i a«rf 2006}

Solution. Since in each equation one of the coefficient is larger than the other, satisfying the conditionfor Gauss-Siedel method, we write the given equations in the following form :

x ^ J - ( 9 5 - U y + 4 * )Oo

y - — ( 1 0 4 - IX - 1 3 2 )52

r = (71 - 3* ~ 8^We start with y = z - 0 and using the most recent values of x, y, z, we getFirst iteration :

^ 1) = - L ( 9 5 - l ly « » + 42<°>)8 3

= i < 95 - 0 + °> = 1145O O

y j ) = J - ( 1 0 4 - 7x"> - 1 3 2 '° ’)52

= — ( 1 0 4 - 7 x 1 . 1 5 - 1 3 x 0 ) = 1 . 8 4 652

Ah = — ( 7 1 - 3 * t f > - 8 y > )2 9

N u m e r ic a l A n a l y s is -! | >181

Fifth iteration :

Second iteration :

= — (71 - 3 x1 .14 5 - 8 x 1.846) = j.821^9

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Fifth iteration

V* = ^ [9 - 2(-2) - 3] =1

>*5) ’ {"22 - 1 + 3 ] ~ ~ 2

& = ^ [22 + 2(1) - 3(—2)] - 3.

The required solution is x = 1, v = •• 2, 2 = 3. Ans.

Example 2.52 : Solve the following equations by Gauss-Seidel method:

83x + llv - 4 z = 9 S 7x + 52y + 13z - 1043x + 8 y + 29z - 71 fRGPV Dec. 2003 and June 2006}

Solution. Since in each equation one of the coefficient is larger than the other, satisfying the conditionfor Gauss-Siedel method, we write the1given equations in the following form :

x - -rr<95 - + 4z)O O

y , ± ( 1 0 4 - 7 x - 13z)

Z= 2 ^ (71 ~ 3 x ~ 8 y )We start with >>= 2 = 0 and using the most recent values of x, y. t. we getFirst iteration :

*1) = ± ( 9 5 - liy°> + 4z<0’)83

= ± ( 9 5 - 0 + 0) = 1.145oa

y(i) = ± (1 0 4 - 7x(n - 13z‘0’)52

= ± (1 0 4 -7 x 1 .1 5 -1 3 x 0 ) = 1.84652

4i) » ± (71 - 3xlI) - 8 / » )29

= ± (71 - 3 x 1.145 - 8 x 1.846) = 1.82129

N u m e r ic a l An a l y s w -I | 181

z<4>= ^ (22 + 2x (4) - 3 y 4)) = 3

Second iteration :

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1 82 |f e i6 w a w > M « iB M iic s-lll

Third iteration:

jp> = ^ - ( 1 0 4 - 7 * ^ ~13*W> = 1.412

2® = (71 - 3x«> - s y 2>) = 1.956

*<3) = - ^ ( 9 5 - l i y 2> + 4z«>) = 1.051O O

y3) = i - ( 1 0 4 - 7x<3> - 13*«>) = 1.34452

Fourth iteration:

* 3) = ^ ( 7 1 - 3 x < 3 > - 8 ^ 3 ) ) = 1.969.

* 4) = ^ ( 9 5 ~ H y 3> + 4*<3)) = 1.051o o

A d s .

y4) = — (1 0 4 - 7x<«> - 1 3 *« > ) = 1.3452

~ 3l<4) " 8^ 4>> = 1-9694&V The required solution is :

x = 1.051, y = 1 .34, z = 1 .969.

Example 2.53 ; Solve the following equations by Gauss-Seidel method :1 0 X { - 2 X 2 ~ X 3 ~ X 4 * 3

- 2 x] + 10 x 2 - x j - x j = 15 - x , - X 2 + IOX 3 - 2 X 4 = 27 —Xj —Xj — 2 x 3 + IOX 4 ——9 fRGPV June 2004, Dec. 2006 A June 2009}

Solution. Since in each equation one of the coefficient is larger than the other, i.e., satisfying thecondition for Gauss-Siedel method, we write the given equations in the following form :

1=

*2

10

JL10

(3 + 2x2 + xs + x4)

(15 + 2xj + X3 + x4)

*3 = 10 *27 + x* + x* + 2X4

1X4 = — ( - 9 + Xi + Xj + 2x3)

we start with xi = x3 = x4 = 0 and using the most recent values of xj, X2, x3, x4, we getFirst iteration :

x,(l) = X o (3 + 2X2(01 + x3f0' + **(0)) = 0 3

*2’ 10 (15 + 2x,;i> + x3<°> + x ^ ) = 1.56

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Second iteration

Third itera tion:

Fourth iteration

Fifth iteration :

x,a> = 10

- 10

X4<!> = 10

,(5) = —1 10

■*2,(5> = —10

27 -r *,<»' + x j" ^ 2 x 4<°>) = 2.886

-9 + *i<‘>+ x2‘l>+ 2r3‘l>) = -0.1368.

3 + 2x,<‘>+ *3<»> + x4<” ) = 0.887

Nu m e h c a i. An a l y s i s*! | 183

X2<2>=1

1015 + 2x,(2> + X3W + x4(1)) = 1.952

I I % 1

1027 + x,‘2) + x*<2>+ 2x/») = 2.957

X& =1

10-9 + Xi<2>+ x2<« + 2x3(2)) = - 0.025

X,,3>=1

103 + 2x2<2>+ x3«> + x 4‘2>) =0.984

x,(3>=1

1015 + 2x,<3) + x 3(2) + x 4(2)) = 1.99

I I 1

10 27 + x,<3) + x 2<3>+ 2x4<2) ) = 2.99

Jt4<3) =1

10-9 + x,'3>+ X2<3>+ 2x3‘3') = -0.0046

X,W =1

103 + 2x2(3) + X3<3) + x4(3)) ==0.997

I I S '

1

1015 + 2x,(4> + x3(3) + x 4<3>) = 1.998

H U J i u 1 1 110

27 + x,<4>+ x2<4>+ 2x 4<3>) = 2.999

x*<4>=1

10-9 + x,<4>+ x2<« + 2x 3(4)) = - 0.0007

3 + 2x.<4>+ x3<4>+ x4<4’) = 0.999

15 + 2x,«» + x3«> + x4«>) = 1.999

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1 he required solution is :

X] * 1, xi » 2, X3 « 3,

x4 * 0. An Example 2.54 : Solve by Gauss-Seidel method, given the following system :

28x + 4 y - z »32 x +•Sy + lOz = 24 2x + 17y + 4z * 35

Solution. Since in each equation one of the coefficient is larger than the other, i.e., satisfying t!condition for Gauss-Siedel method, we write the given equations in the following form

x = ^ (3 2 - 4 y + z)

1

y jy (35 - 2x - 4z)

z = ~ (24 - x - 3 y)

Start with y = 0, z = 0 and using the most recent values of x, y, z, we get/ irst iteration :

= ~ (32 - 4(0) + 0] = 1.1429

1y*> = j f f35 - 2(1.1429) - 4(0)] = 1.9244

^ (24 - (1.1429) - 3(1.9244)] = 1.8084

t'econd itera tion :

*<2) = ~ (32 - 4(1.9244) + 1.8084] = 0.9325

1^ = y j (35 - 2(0.9325) - 4(1.8084)] = 1.5236

1^ * j0 124 " <0-9325^ ' 3(1.5236)] = 1.8497Third iteration :

J<3) = ~ (32 - 4(1.5236) + 1.8497] = 0.9913

y 3t = (35 - 2(0.9913) - 4(1.8497)] = 1.5070

z<3) - ^ (24 - 0.9913 - 3(1.5070)] = 1.8488

I ourth iteration :

1& “ ^ 3 (32 - 4(1.6070) + 1.8488] = 0.9936

1 6 4 | &NGlNE£JON6 MXTHfcMATICS-III

,r4^ - ~ (- 9 + x ,(8> + * 2<5' + 2 x 3<5)) = - 0.00044

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N u m e r ic a l A n a l y s e ! 1 1 8 5

y 4) = — [35 - 2(0.9936) - 4(1.8488)] = 1.5069

z<4>= [24 - 0.9936 - 3(1.5069)] = 1.8486

Fifth iteration

** = 28 132 ” 4<L5069>+ (1-8486)] = 0.9936

y s> = 7 7 [35 - 2 (0 .9936 ) - 4 (1 .8486)] = 1.5069

* 5) = “ [24 - 0 . 993 6 - 3 (1 .5069 )] = 1-8486.

Since 4th and 5 ,h iterations are sam e, we getx - 0 .9936, y = 1.5069, z = 1.8486. Ans.

Example 2,55 : Solve by Gauss - Seldal method 5x + 2 y + 3 -1 2 x + 4y + 2z ” 15 x + 2y + Sz =*20 [RGPV, Feb. 2010}

ibttioit. Since in each equation one of the coefficient is iaiger then the other, i.e., satisfying the

cond ition fo r Gauss-S eidat m ethod, we write the given equations in the following form :

x = ±[12-2 y -z ]

y = ±[15- j r -2^}4

2 = ±[20-x~2 y]

Start with y = 0. z = 0 and using the most recent values o f x, y , z we get

Fist iteratio n:

jd » = “-[1 2 - 0 - 0] = 2.4

= ~[15- (2 4)-0] = 3.15

Second iteration

2<» a y [2 0 -2 ,4 -2 x 3.15] - 2.26

x<« =• - [ 1 2 - 2 x 3 . 1 5 - 2 . 2 6 ] = 0.688

y(2> = - [ 15 - 0.688 - 2 x 226] = 2.4484

*2) = j[ 2 0 -0 .6 8 8 -2 x 2.448] = 2.8832

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186 j En g m s h n g Ma i h e k a t j c s-HI

Third iteration .

v,.<i = -(1 5 - 0.8442 - 2 * 2.8832] = 2 .0 9 '4

4

= j (20 - 03442 - 2 * 2 .0974] =2 .9922

Fourth iteration:

= i[I2 - 2 *2.09^4-2.9922] =0.9626

= -^[15-09626-2x29922] =20133

r J- = 4(20-0.9626-2 2.0133] = 3.0022

Fifth itera tion :

*.<< = -{12-2x2.0333-3.0022] = 0.99425

vt5< = -(15- 0.9942 - 2 x 3.0022] =2.00034

HM= -[20 - 0.9942 - 2 x 2 0003] =3 00105

Since 4* an d 5 * iterations are pa rtially same , hence solution o f given sy stem is

x * 1.000, y a 2.000. z ~ 3 .000 Ans.

2.26 RELAXATION M ETH OD :

Consider a system of equations :a\X * b\y - C;z - i/j

OiX + £> 2 }' + c 2- “ 2

ay.x t bxv cp = d\.

Let, Rx, Ry, R. be ‘residual’ defined by

Rj = d\ ~ - b-j- --\-and Ry = d2 - o^x - - c;z ...(1)

R. = d} - GjX - b£ ' - cxz

Now we start initially x = y = r = 0 and obtain the initial residuals, (i.e. the initial residuals are Rx

= d i, R> = </2. = d}). Then the residuals are reduced step by step by giving increments to thev ariables. For example, if the variable x is increased by 5t = 1f keeping y and z constants) then thecorresponding change in residuals Rx. /?v. R. are SRx = - o;, 6 Ry - - a>. SRz = - ay i.e.. Rx, Rt,

R. are reduced by aj. a2, ay respectively. Similarly, we can find the effects on the residuals whenv and z are given increasements dy = 1 (keeping x and z constants) and S: = 1 (keeping x and y constantns). The operation table is :

x*'' = ^ [1 2 -2 * 2 448-2.8832] =0.8442

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N u c r ic a l A n a l y s s 4 | 1 8 7

Incrment in Variable

Change in Residuals

6R Z SRf 8R t

Sx = 1 —®i ~ a 2

*y = 1 - * > 3

S z - 1 - c . - c 2 ~ C 3

At each step then we reduce to almost zero the numerically largest residual. In order to reduce a particular residual, then the value of the corresponding variable is changed, for example if residual

Rx is to be reduced by a quantity a, then

a

J LSimilarly if residual Ry is to be reduced by a quantity # then Sy = and so on. Continued such

processes till al! the residuals become zero or negligible, Hence the solution is obtained by addingthe increments Sx, Sy, 8 z in x ,y , z separately.

(Nm i* ;

The convergence conditions that were given for iterative methods must be satisfied for convergence, of relaxation method also.

hnfrff 156 : Solve the following equations by relaxation method :

1Qx-2 y-3z = 2Q5;

~2x + lOy ~2z~ 154;

- 2 x - y + lOz = 120.

p tetafc Since, |10{ > h 21 + r 31; |10| > h 2| + f- 2| and |10) > h 2| + | - 1|.

So we can use relaxation method.

The residuals of given equations are :

Rt = 205 - IQr + 2y + 3r. R, = 154+ 2 r- I0 y + 2r,

R: = l20 + 2* + y-t0 z.

The operations table is :

Increment in varia ble j 6 R t 6R , SR,

o1

«-i 1 1 X2 2

S y = 1 | 2 - 1 0 1

S z = 1 j 3 2 - 1 0

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18* | f tatN anteM m siM ncs-ni

Relaxation table is :

R , K. x = y = z = 0 205 154 120

S x = 20.5 0 195 161

195 1Q c' y = ¥ a l 9 S

39 0 180.5

_ 180.5 10 ,<>* = — — = 18 (approx.) 93 36 0.5

5 x = 9 3 54 18.5

6 y - 5.5 14 -1 24

S t = 2 20 3 4

<5x = 2 0 7 8

J* = l 3 9 -2

<5y = 1 6 -1 -1

<?x = 0.5 0 0 0

Nowx = H 6 x 3 32, y = Z 6 y = 26, z = E£z = 21

The required solution is : x - 32, ’ y = 26, 2 = 21. Ans.

Example 2.57 ; So/we tike following equations by relaxation method:

10 x - 2 y - 2 z - 6 ,

—x + lOy —2z ** 7,

- x - y + l O z - 8.

Solution. Since, |I0| > (- 2| + h 2|; |I0| > |- lf+(~ 2| and |10| > (- i| + h 1|!

So we can use relaxation method

The residuals of given equations are :

Rx = 6 - IGr + 2y + 2z/^= 7 + JC -l0y + 2z

=8 +x + y - IQz.

The operations table is :

Increment In

variable5*, 5Rt

6x = 1 -10 1 1

¥ I I 2 -10 1

6z = l 2 2 -10

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Nu m e r k n >An a l y s e 1189

tlaxation table is ;

H.

x = y - z - 0

0.8« , - J L10

. 8.6 .6 y =----= 0.86

10

&c = £ 2 1 ^ 0 .9 3 210

& = 0.1792

8y = 0.129

8r = 0.062

& = 0.019

Sy = 0.0102

&r = 0.005

52 = 0.0016

By = 0.0008

&x = 0.0009

6.0

7.6

9.32

0.0

0.3584

0.6164

-0.0036

0.0344

0.0548

0.0048

0.0080

0.0096

0.0006

7.0

8.6

0.0

0.932

1.2904

0.0004

0.0624

0.1020

0

0.0050

0.0082

0.0002

0.0011

8.0

0.0

0.86

1.792

0.0

0.129

0.1910

0.0010

0.0112

0.0162

0.0002

0.0010

0.0019

Hence, x= I.S y = 0.9999 « 1,5 y = 'LSy m 1,

x = Z 8 z = 0.9998 * 1The required solution is : x - y ~ z = 1.

ample 2.58 : Solve by relaxation method :3x + 9 y - 2 z - l l ;

4x + 2y+ I3z ° 24;4x -4 y + 3i 'm- 8

ption. Since in each equation one of the coefficient is not larger than the

the condition for relaxation (iteration) method.

Ans.

fRGPV Dec. 2005] other, i.e., not satisfying

Ans.

Exercise-2(A)

1. An approximate value of misgiven by 3.1428571 and its true value is 3.1415926. Find absoluteand relative errors.

4 jc2v32. If u = — — and errors in x, y, z be 0.001, compute the relative max. error in u when x = y =

z - l .

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3. Let x = 0.5998 * 10 “ 2. Find the relative error, if x is truncation to three decimal digits.

4. Let x = 0.458529 * 10-2. Find the absolute error if x is rounded - off to three decimal digits.5. Explain underflow condition and overflow condition of error in floating points addition and

subtraction.

1 9 0 1f t g i e B n w r .lt o M B M iiC 5 -ll l

1. Ea - 0.0012645,2. 0.009

Answer-2(A)

Er = 0.000402,3. 0.133 x 10-2, 4. 0.471 * I0-3

Exercisc-2(B)

Use bisection method to find the roots o f the following equations as indicated, correct to three

places o f decimals :

1. Find the smallest positive root of the equation x3 + x2 - 1 = 0.

2. Find the root of the equation x3 - 4x - 9 = 0, that lies between 2 and 3.

3. Find the root of the equation x3 + Zr2 + 2.2x + 0.4 = 0 that lies between 0 and - 1.

4. x4 - x - 1 0 = 0. 5. x - cos x = 0. 6. 3 x -e I = 0.

7, 9X2- 1—sinx = 0. 8. x6 -x 4- x 5- 1 =0. 9. tanx +x = 0.

10. xe* = 1. t l. Compute cUbic root of 36.

Use simple iteration method to fin d the roots o f the following equations, correct to four places of decimal:

12. Find the root of the equation x3+ x 2 - 100 - 0, that lies between 4 and 5.

13. Find the root of the equation x? + x - 1 = 0, that lies between 0 and 1.

14. Find the root of the equation 3x - logic x - 6 = 0, that lies between 2 and 3.

15. Find the root of the equation 3x + sin x - e* = 0, that lies between 0 and I.

116. x = 17. x = (5 -x )w 18. sin x - 10(x- 1)

21. x1- 1^sin2*

(X + 1)2

19. x sin x = 1 20. e* cotx22. Compute square root of 30.

Use Regula Falsi method to find the w ots o f the following equations, correct to four places of decimals:

23. x2 + x2 + x - 100 = 0 (Root lying between 0 and I).

24. x3 - 8x +■40 = 0 (Root lying between - 5 and - 4).25. x tan x + 1= 0 (Root lying between 2.5 and 3).

26. x6 - x 4 - x 3 - 1 =0 27. xeJt= 2.Use secant method to find the roots of the following equations, correct to three places ofdecimals:

28. x3 + x2 + x + 7 = 0 29. x - e r x ~0.

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NumacAL Ax m.yws-1 | 191

Use Newton-Raphson method to fin d the roots o f (he follow ing equations, correct to fo i plac.

o f decimals : !

X4- x ~ 10 = 0 (Root lying between 1and 2).

I!. xJ - x 2 + x + 100 = 0 (Root King between - 5 and - 4V

x - e~x = 0 (Smallest positive root).

xe* - cos x = 0 (Smallest positive root)

[M. 2 x - logio x = 7. 35. tog x = cos x 36. x4 + 4 sin x = 0.

xs n x = — 38. tan*=x 39. Mx - sin x) = 1.

e* = x3 + cos (25)x which is near x = 4.5.

41. 10* + x - 4 = 0. 42. Compute ^ 9 .

•43. Using N-R method, obtain formula for and find ^20 correct to 2 decimal places.

44. Obtain the cube root of 120 using Newton-Raphson method starting with xo - 4.5,

45. Apply Newton-Raphson method to find an ap proximal ton solution of the equation

e* = ¥ correct to three significant figures assuming x = 0.4 as an approximate root of theequation.

46. Find the root of the equation f i x ) = sin x - ^ = 0 near x = - 0.4..

Compute cubic root of 42, by Newtoo-Raphson Method Develop an algorithm using Newton-Raphson method to find the fourth root of a positive mcnber

N and hence find ^ 32.

f, 49. Find cube root of 3 correct to three decimal places by Newton's iterative method.

/1_

56. Prove the recurrence formula x* + j - 3.V

2xj. + —

\

for finding the cube root of N. Hence find cube root of 63.

Solve the following equations by Graffe's root squaring method :

51. x4- lflfcr3 + 35x*- 5Qx + 24 = 0.

52. xJ -4x3 -3 x + 18 = 0.

Extract the quadratic factor o f the form x 2 + px + q from each o f the following polynomials, using Bairstow i method and assuming indicated initial values o f p and q.

53. X4~x3 + 6x* + 5x + 10;p= 1.14; = 1.42.

54. x* + xi + 2x 2 + x+ \',p = q = 0.5.

55. x4 - 3x3- 4 X 2 - 2x + %\p = 0.95; q = 1.0554. x4 - 2 x 3 + x 2 - 4 x + 4;/j = 0.5; q = 1.5.

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I92r\ F fry cnwtirHf hihbw ics-H I

Answers-2(B)

1. 0.755. 2. 2.707. 3. 0.222. 4. 1.813.S. 0.739. 6. 0.619. 7. 0.392. 8. 1.404.9. 2

.020

.10.

0.567.11.

3.301912.

4.3311.13. 0.6823. 14. 2.1080. 15. 0.3604. 16. 0.4655.17. 1.516. 18. 1.088. 19. 1.068. 20. 0.5314.21. 1.404. 22. 5.4772. 23. 0.4264. 24. -4.1891.25. 2.7980. 26. 1.4036. 27. 0.853. 28. -2.0625.29. 0 567. 30. 1.8558. 31. -4.2644. 32. 0.5672.33. 0.5178. 34. 3.7892. 35. 1.303. 36. -1.934.37. 1,896. 38. 4.4934. 39. 1.171. 40. 4.545.41. 4.4346. 42. 5.3852. 43. 4.47. 44. 4.9324.45. 0.619. 46. - 0.42036. 47. 3.4760

48.1

x"+1 43x„ +^-j, 2.3784. 49. 1.442. 50. 3.979.

51. 4,3.21. 52. 3, 3, - 2. 53. jc2 +1.1446*+1.4219.54. x2+x +•1 55. x2 +2x +2. 56. x3+x +2.

Exercise-2(C)

Solve the follow ing systems by using (I) Gauss-elimination (ii) Gauss-Jordan

(iii) Crout’s methods :

1. t + 3y + 6 z = 2, jc - 4y + 2z - 7, 3x- y + 4z = 92. 2x + 2y + z = 12, 3x + 2y + 2z = 8, 5x + IOy - Sz - 103. 2.1 Ox - 4.50y - 2.00 z = 19.07. 3.0Qr + 2.50y + 4.30z - 3.21, - 6.00x + 3.50y + 2.50z = - 18.25

4. x - v + 2 = 1,- 3x + 2 y - 3z - -6 , 2x - 5y + 4z = 55. x + 3y + lOz = 24, 2x + 17 y + 4z = 35, 28x + 4 y - z - 326. x - 3y~z = -3 0 , 2x - y - 3z = 5, 5 x - y - 2 z ~ 1427. 5x -9 y~ 2z + 4w - 7, 3x +y + 4z + 1lw = 2,

IQr - ly + 3z + 5w = 6, - 6 x + %y - z - 4w = 5.8. 10 x + y + z = 12,x + 10>' + z = 12, x + y+ 10z = 12.9. lOx + y + z - 18.141, x + 10 y + z = 28.140, x + y+ lOz = 38.139.

10. 3x + y - z = 3,2* - 8y + z = -• 5, x - 2y + 9z - 8.11. 3x-.y +2z = 12,x + 2y + 3z = II , 2x -2 ;y -z = 2.

12. 2x - 3y + z = - 1, x + 4y + 5z = 25, 3x - 4y + z = 2.13. x + 2y + 3z = 6, 2x + 4y + z - 7,3x + 2y + 9z = 14.

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N u m e ric a l A naly sis* ] | 193

tH. 2v 'S v - t v

|5 Iy 1 ■ '

LJA x \ p7. 2> •" V18. x - :>•

' »• 9. vt r y - 4" t = 3,

- _ hw — 2 y — r A - —2.U 5 ■ t-. >7 H. ?x + 3y +r - 7.

4. ' V - v 3.t - - 4. 2v - 3y - 5z = - 5.■=- 12. 5v y r = M, Ax- y + 3z = 10.

vf -- - 2, Z.v +3y — z + 2w - 7,

til 6. < f 1 j + w - 2.19. 4 . i 2 v y.his) 2 O lr *• 4 . 9 3 , 1 ,88x - 4.62y + 5.50z = 3 . 1 1

1 . 1 0 v - 0 . 9 6 v J - 2 . 7 2 r =- 4 . 0 2

20. 6 y - v +z = 13. x + y + z - 9, lOx + y - z = 19

21. x + 2 y - i 2- -f- 8m’ - 27. 5x + 4 y + Iz - 2w - 4,6 r - 12-v - 8; *• 3w = 49 , 3x - l y - 9z - 5w = - 11

,22. x + 0.5 v- + 0.33z = 1, 0.33x + 0.25y + 0.2z = 0, 0.5x + 0.33y + 0.25z = 0

23. 2v + 4y + r “ 3, 3.t ^ 2y - 2z = - 2, x -y ° + r = 6

24. x + y +: - h>= 2, 7.x ■+• y + 3z + 2w = 12,8x —y + r - 3m-= 5, lOx + 5y + 3z + 2w = 20.

25. 2 x -r 4y - 8r - 4 1 . 4.v + 6y + lOz = 56, 6x + 8y + lOz = 6426. 2v -r 2y - z + «• = 4, 4x +■ 3y - z + 2w = 6,

S.v * 5;.' - or +- 4iv = 12, 3x + 3y - 2s + 2w = 6 .

Ansners-2(C)

1. 2. - 12.75, 14.375, 8.75 3. 1.34,- 4.76, 2.58.

4. - 2, 3, 6 5. 0.99, 1.50, 1.84 6. 39.2, 16.7, 19.

> 7. I. 4, 5. - 7. 8. 1, 1, 1 . 9. 1.234, 2.348, 3.455.

10. 1 , 1. i. 11. 1. 2, 3. 12. 8.7, 5.7, - 1.3.

13. 1 , 1. 1 . 14. 1. 2 , 2, 2. 15. 1, 1.2.J6. 1,- 1,2. 17. 1.64, - 2.49, 0.32. 18. 1 ,0,- 1,2.19. 4.2075. 1.3327, 0.2468 20. 2, 3, 4. 21. 3, - 2, 1, 5.

22. 55.56, - 277.78,. 255.56. 23. 2 . - 1,3 24. 1, 1, 1, 1.

25. 1.5, 2.5, 3.5. 26. 1. 1,~ 1,- 1.

Exercise-2(D)

,SV;/\v’ e/,‘c following system o f equations by (i) Gauss-Jacobi method 00 Gauss-Seidel method ( t i l ) and relaxation method.

1 . 5.v -■ 2y + r ~ - 4 . x + 6y - 2z - - I , and 3x + y + Sz = 13

2. 8.<r <y +- = 8, 2 r +-4y + z = 4, x + 3y + 3z = 5.

3. 8.c ■6y + r - 13.67, 3.v + y - 2: = 17.59, 2x - 6y + 9z - 29.29.4. 30.v - 2v ' 3r - 75, 2 r * 2v + 18z = 30, x + 1 7 y - 2z = 48.

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5. y ~ x + lOz = 35.61,* + z + 10y = 20.08, y - z + lOx = 11.19.6. 3.I22* + 0.5756y-0.1565;-0.0067/ = 1.571.

0.5756* * 2.938v + 0.1103z - 0.0015/ - - 0.9275.- 0.1565x + 0.1l()3y + 4,127- + 0.2051/ = - 0.0652.

- 0.0067*-0.0015y + 0.205 lr + 4.133/= -0.0178.7. 10* - 2y + z = 12, * + 9v - z - 10, 2.r Hr = 20.8. 10*-2 y - z ~ / = 3, - 2* + \ 0 y - z - / = 15

- * - y + 10; - 2/ = 27, - * - y - 2z + 10/ = - 9.9. S x - y + z = 18, 2x +5y-2z = 3 .x + y -3 z - 16.

10. 2x + y + z = 4, .r + 2y + z = 4. * + y + 2z ~ 4.11. 4* + 2y + ; = 8, * + 5y - z = 10, * +y + 8s = 20.12. 8* + y + z = 8, 2* + 4y + z = 4, * + 3y + Sz - 5.13. 14* - 5y= 5.5, 2r + 7v = 19.3.

14. * - 2y + JOr « 30.6, Ix + 5y - z = 10.5, 3* + y + z = 9.3.15. 8* - 6 y + z = 13.67. 3* + 11 v - 2z = 17.59, 2x - 6y + 9z = 29.2916. 7.6x - 2.4y + 1.3z = 20.396, 3.7* + 7.9y - 2.5z - 35.866,

1.9* - 4.3y + 8.2z = 32.514.17. 10* - 2 y + z = 12, * + 9 y - z = 10, 2x ~y + 1Iz = 2018. 5*- y - z = 3, - * + 10y- 2 z ~ l , - x - y + lOz = 8.19. 2 r - y +z - 3, 2z + y - z - 1, ,r +y + z = 0.20. 50* + 2y - 3z = 196, 3* + 65 y + 2z = 8 1 ,- * + y + 33z = 6321. 8 * + y -z = 8,*-7>> + 2z = - 4 , 2x+ y + 9z=12.

22. 27* + 6 y - r = 85, 6* + 15y + 2z = 72, * + y + 54z =110.23. 9* - 2y + z = 50, * + 5y - 3z = 18, - 2* + 2y + Iz = 19.24. 2 * - 3 y + I Or - 3, - * + 4y + 2z = 20, 5* + 2y + z = - 12.

25. 6*( + *2 - *3 = 14, *i + 5*2 - *3 = - 18, 2xj + *2 + 9*3 = 68 .

26. 3* + y - z - w - 0. * + 3y - z + 2w + 3 = 0,- 2* + 2y + 3z - 2w - 4 = 0, * + 2y + z - 5w + 1=0.

1. - 1.001,0.999, 3.04. 2.5796, 2.7976, 1.0693

Answers-2(D)

2. 0.83,0.32, 1.075. 1.321, 1.522, 3.541.

3. 2.45, 1.62, 3.79

6. 0.5835, - 0.4307, 0.0181, - 0.0044.7, 1.262, 1.159, 1.694.8. 0, 1,2,3

11. 2, 2,2

9. 2,0.999,2.999

12. 0.876, 0.419, 0.574.10. 1, 1. 1.13. 1.25, 2.40.15. 2.45, 1.62, 3.79.14. 1.232, 2.262, 3.382, Actual 1.2, 2.3, 3.4.

16. 3.23, 4.85, 5.76.19. I,-1.0.22. 2.473, 3.256, 1.931.

25. 4, -3, 7.

17. 1.3, 1.2, 1.7.20. 4, 1,2.23. 6.15,4.31,3.24.

26. 2,-1,4, 1.

18. 1, 1, 1.21. 1, I, 1.24. -4, 3, 2.

a

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N u me r ic a l A n a l y s i s - I I

,-^jt

CALCULUS OF FINITE DIFFERENCES^*The calculus of finite differences, in contras^fo infinitesimal calculus, deals with the changes ofthe functional value, the dependent variitte'due to finite change in the independent variable(argument). The finite variation arguments may either be equal or unequal. Consequently, thefinite variation (the differencing interval) is discrete in nature. On the other hand, in infinitesimal

calculus, we study the variations which occur when the aigument changes continuously with a passage to the limit.Suppose that y - f i x ) has set of (n + 1) tabulated values with equally spaced

x : xo, = *0 + ft, xi = .to + 2h, X}— xq + 3h, ,...x„=xo + nh andCorresponding value of y are

y y(b yuy 2i ...-yn To determine the values of f ix) , f (x) or/ ”(*) for some intermediate value ofx,then the various types of different opereators are found useful.

DIFFERENT OPERATORSWe will study the following operators :1. The Shifting Operator (E ):

E f ix )= f ix + h) i.e.,

=fix + 2h) i.e.,

EPfix) = f ix + nh) i.e.,

Here, n takes up integral or fractional, positive or negative values.

For examples : E~ lfix) =f i x - h) i.e.,

E * f ( x ) = f ( x + \ h ) i.e..

Properties o f Operator E :(i) Operator E is distributive.(ii) Operator E is commutative with respect to constant.(iii) Operator E obeys laws of indices.2. Forward Difference Operator (A):

If x0, x \ , x i , x „ are equally spaced with interval of differencing h and if y =fix), then A/fr) =fix

+ h) ~A*) or Ay0 =yt ~yo

i.e., Ay, =y,+ i -y , for i ~ 0, 1,2, 3.....n - I,

The symbol A is called forw ard difference operator and Ay, is called first forward differences.

Similarly, the second forward differences are A2yt - Ay, +1- Ay,For example : A2y 0 = Ayi - Aya ~ (y j ~ y i ) - (yi -yo ) = yi - 2y\ +y0.

Clearly any higher order difference can easily be expressed in terms of ordinates.

Eyo = y i

Ezyo = y 2

£nyo =y n

E~[yi =y\

E\ayx

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Forward difference table :

1 9 6 | En g in e e r in g M athem atics-1 11

Arugments x

Entryy~ f ( x )

Firstdifference

Ay

Second difference

A*y

Third difference

A3y

Fourthdifference

Fifthdifference'

A5,y

* 0 yo

* 1 y\

x2 y2

Xi ys

*« y*

*5 y5

Ay0Ay,

A y 2

Ay3

A y4

A2y0

a V,A2ya

AV3

AVo

a V i

a V 2

a V o 5

A4y, A *

Here the first entry yo is called leading term and Ayo, A2y& .... are called leading differences.

Properties o f A :

1.A[/to±#x )] = Afa)±A #*)2. A[c/x)] ~ c A /(x); c is constant.3. AmA"fix) - Am +"fix), where m, n being positive integers.4. Ac = 0

5. If given «-observations (x„ y,), then A" /= 0.

Example 3.01 :Evaluate when A = /.

Solution. A 2 x3= A[(x + 1}3 - x3] = A[x3 + 3x2 + 3x + 1- x 3]

“ Atfx2 + 3x + 1) = 3AX2 + 3Ax + A1

- [3(* + l)2~3x2] + [3{x + l)-3 x ] +0 [ v A1 =

= 3*2 + 6* + 3 - 3x2 + 3* + 3 - 3x =6 x + 6 - 6(x + 1). Ans.

Example 3.02 : Prove that

n-\

= Ay* - Aye .

*-oSolution, We have

ji-l n-l

+1 - y*>. now open the summation

= A(yi ~yo) + A(yz-yi) + Afo -yz) +... . + A(y„ _ i - y*_ 2) + A (y„ ~y„~i)

- Ay» - Ayo + Ays - Ayt + Ay3 - Ayz +.. .. + Ay«_ i - Ay„_2 + Ay„- Ay„_ t

= Ay* - Ay0. Proved. Example 3.03 : Prove that

A logf{x) « /o^ | J + } .

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N u m e r i c a l Analysis-11 | 197

Nation. We haver

. f f ( x + h) log^x + h) - log/w = log |- y (x)-

,„„j ' f ( x + h ) - f ( x ) + f ( x ) \ - log j [ ( v As add and subtract]

Proved. f ix ) J { fix)

3. Backward Difference Operator (V) :

!f*o, X |.x2, .... x„ are equally spaced with interval o f differencing h and y =fix) then Vfix) =fix)

- f i x - h )Vvi ~y i~y i~ i> f o r »'= J, 2 , 3 ...... , n

The symbol V is called the backward difference operator and Vy, is called fi rst backward

differences.

Also, the second backward differences a re V2y/ = Vy, - Ay,_ 1? i = 2, 3, 4 ,..., n

For example : V2}^ = V(V>2) = - ^ i ) = Vy2 - Vy,

= iyi ~y\) ~ Cvi - o)= >'2-2>’ j +/0

Similarly, the higher order backward differences are defined as follows :

Vky, 1-

Backward difference ta ble :

Argument

XEntry

* = /(*)

Firs tdifference

Vy

Second difference

V*y

Th ird j Fo urthdifference i difference

V3y i V4y

Fifthdifference

Vsy

*0

*1

x2*3

*4

*5

Vo

Vi

l>2V3

V4

Vs

Vyj =y, -y 0

Vy2

Vy3

Vy4

Vys

^ y ^ V y g - V y !

v 2y3v zy4

v 2ys

v 3y3

V3y4

v 3y5

— ———„i

..............

V4y4

V4ysv 5y5

Example S. 04 :Find the first backward difference o f x? + 2x,

Solution.v Vfix) -fix ) -fi x - /i)

=> V(x2 + 2x) = [*2 + 2x] - [ ( jc -A)2 + 2( jc - / i)]= (x2 + 2x\ - [x2 + - 2xh + 2* - 2h]

- 2/tx + 2h - fi1. Ans,

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froof: (i) By definition o f A :

Hence

(ii) By definition o f V :

Hence

(iii) By definition of S :

Hence

(iv) By definition p :

or

Afix) ^ fix + h) ~A x)- E fix) - f ix )

A = E - 1.

Vfix) - f i x ) - f (x - ft)V/.r) -fix ) -E ~ ]fix)

Vfix) = ( l -£ -» )Xx)V = 1

= f ^ / x ) - E- m fix )

» ( £ '« - £ - 1/2)At)

= ^(S»^+£?-«'2)/(x)

[ v Efix) - fix + /;)]

[v £ - • / * ) = / * - * )]

N u m e r i c a l Analysis-11 | 199

Hence

(v) By definition of E :Using Taylor's series, we get

W = /(* )+ hf '( x ) + — f \ x ) + .......

= f ( x ) + kD f ( x ) + l- ~ D 2f (x )- ^. .

I , urt h2D' = 11 + hD + —^ - + . /(*)

= e*Pfix)

Hence, E = e*0.

or eM) = I + A

(vi) Taking iog both side o f (I ), we get

log ( I + A) = h D

Also, V = I - => V = i - <r«>

=> - hD = log (1 - V)= > hD - - log ( I - V)

D - i

v ex = 1 + x + ■+ .2!

... (1) Proved.

Proved.

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Remarks:

1. A = £ - 1=> A = ehD- 1=> A + 1 = ehD.

2. V = 1 - 1 V = 1 -e ~hD.

3. Since V = I -e~hD =>e~hD~ 1- V=> - hD = log (I - V)=> hD = - log (1 - V)

2 0 0 | E n g in e e r in g M a t h e m a t ic s -IH

|R G P V Dec. 20051

1

D = h

y2 V3 y4V + — + — + — + .

2 3 4

4. Since A + 1= ehD

hD = log (1 + A).

1 D = h

. A2 A3 A4A----- + ---------- + .2 3 4

|R G P V Dec. 2005|

-(2)

<3()* 0.

(where h = interval)

N ote : Results (1) and (2) will be used in numerical differentiation.

3.4 DIFFERENCE OF A POLYNOMIAL

Ifflx) is a polynomial o f degree n in jc,

i.e., fi x ) = aoxn + a\x? ~ 1 + azx"~2 +.. . . . + a„.

Then &nfix) = constant = ao n\hn

and A" + 1 fix ) = 0.

Theorem [3.1 J : Show that the n'h difference o fa polynomial o fdegree n will be constant and all (n - 11"1and higher order difference are low.

fR GPV June 20091

OR

Iff (x) be a polynomial o f degree n in x, then the nth difference off(x ) is constant and the (n + l)th and higher differences are zero.

Solution : Let fix ) = a0 + a t. t + . . . . + a„x". ...(!)

be a poly nomia l o f degre e n, where a„ * 0 .

By definition o f forward difference op era tor:

4 / 1*) =f ix ^ h ) - f i x )- [do + <3 | (x + h) + ...... + a„ (x + /j)n] - [oq + «|T + ........+ 0,,v"]

[using (I))

= a\[(x + h ) - x \ + an [(* + h f - ~ xz ] + ...... + a„ [(.v + h f - x"]

= a\h + aJpCixh + h2] + .......... + a„[nC\ .v" _ 1/; + nCi .x"~ 2 h2

+ ...... + "Cn h"\

- [a\h + a^h2 + ........ + a j f 1] + j rpcj crh + h'l a^h2 ~ .....

+ a j i " - 1] ^ ...... + nc\ anhxn~i

= b\ + b x + ------ + b„_ i jc"-2 + najixn~1 ...(2)where, b\, b i ........... . 6„_ | are con stant coefficients. This show s that, first difference of fix)

i.e., Afix) is a polynomial o f degree n - 1 .

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N u m e ric a l An a l y s is -11 | 2 0 1

Again, we have

A~/(x) = A [A /x)] = A[/(x + h)-f ix)]

= Afix + h) ~ Afix)

w’ 1 = 0 i +b 2 (x + h) + ..... + b„_ | (x + f t y - 2 + na„ h (x + h y -* ]- [61 + bjx + ........+ 6„_ 1 xn~2 + naj?x n~l]= hi [ (x + h ) - x ] + by [ (x + A ) 2 - x 2] + ........+ 1 \{x + h y * ' 2 - jc " - 2 ]

+ na j t [(x + /»)*~ ’]

= b2 + bypCi xh + h2} + ......+ 6„_ 1[n~ 2C\ x * - 1 h + .... + h "~ ']

= d2 + dyx + .....+ d„_ 1 x"-* + n (n - 1)h2a„x^~2 . . .(3)

where, d2, d^.........d„- \ are constant coefficients.Th i s sh o w t h a t , s e c o n d d i f f e r e n c e o f f{x) i.e. , A2 f {x ) i s a po lynomia l o f

degree n - 2 .

There fo re , con t inu ing above b rocess w- t imes , so we ge t a po lynomia l o fdegree n ~ n = 0 i.e, constant polynomial is

A"f{x) ~ n ( n - 1) ( w - 2 ) ....... 2.1 h"a„

= n \ h n cin ...(4)

= constantAlso A" ■*"1 f[x ) = 0, An + 2 f{x) = 0 , ..... so on.

/.e A" + mJ(x) = 0, when m > I .

tie 3.05 :Evaluate A2 (cos 2x)

>n. We have

A2 (cos 2y) = ( £ - 1)2 (cos 2x)

= (A*2 - 2 E + 1 ) cos 2x

= E1 cos 2x - 2E cos 2x + cos 2x

= cos 2( r + 2hi) - 2 cos 2 (x + h) + cos 2x

= cos (2x + 4h) - 2 cos (2x + 2ft) + c os 2x.

1 pie 3.06 :Evaluate A (e^.log bx)

blution. We have

A(c?°-'.log bx) = e°lx + U og b(x + h )~ e^ . log bx= e*x +h) |0g fjfa + ga(x +h) ]0g i x + gafx + h) |0g bx-e™ log bx

’ [v add and subtract]

x + h

Proved.

[ v A = E - \ ]

f v ET'ftx) = / x + nh)} Ans.

[RGPVDec. 2001}

e<i(x+A) Jog . ( ax+ah _ ea*)l0g6x

_ eax.ah log^1 + — j+ €*“ (6°^ -I) l0g6x. Ans.

3.07 :Show th at : hD - sin h~1 (fiS).

blution. We have

[.. £ = eAD]

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2 0 2 I En g in e e r in g M athem atics-111

Hence,

Exampie 3.08 :Evaluate A

Solution. We have

pS = sinh(hD)

hD = sinfr 1

v s in h 6 -e° *

2

Proved.

cos 2 x

i — 1I, cos2x J

(x +h )2

cos 2 (x + h) cos 2 x

(~ + h)- cos 2.r - x2 cos 2 (x + h)

cos 2(x + h). cos 2 x

[(x -f /t)2 - x2] ct>8 2 x +x2[cos 2 x - cos 2{x + /t)]

cos2<x + /i)-cos2x

( 2hx + h2)cos2x + 2x- sin(/i) sin(2x + h)

cos 2(x + h). cos 2xAns.

a9 [ 5x + 12 ] Example 3.09 :Evaluate A x"s + 5 * + 6 ’

= A2 |

fRGPV June 2006]

Solution. We have

5x + 12

x2 + 5x + 6

5x + 12

= A2

[ (x + 2)(x + 3)

2 3

(x + 2) (x + 3) [ V Resolving into partial fractions]

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N u m eric a l A n a ly s is -H | 2 0 3

6

(x + 2)(jc + 3)(x + 4) (x + 3)(* + 4)(x + 5)

2(5x + 16)

(x + 2)(x + 3)(x + 4)(x + 5) Example 3.10 : Show that:

8* / 8 * 1

A = T + S r + t (i)

Solution. Since S = £ l/2 -E ~ m

(0 ,H.S.

+ (E 1/2 - E~v 2)^1 + < ^ L J( £ 1 / 2 _ £ - 1 /2 )2 / p i / 2 r l/2 x jt ( £ 1 / 2 _ £ - 1 / 2 )2

(g + g " 1 2) + /2 _ jp_U 2 J T + E + E 1 2

(E + - 2)

2+ (E 1'2 - E - y 2) J - ( E ' /2 + E ~1'2)2

Ans.

= jE + y -----? ) + £l/i _ £ - i / 2 ) i ( £ l / 2 + £ -1 /2 )

(E + JS-1 - 2) (E - S - J)

2 + .2

£ + £ -> - 2 + £ - 2E - 2 _ , A= .......... ■■ ■ = --------- - iS —1 = A.

2 2= R.H.S. Proved.

(ii) We have

L.H.S. = - ~r(E 1'2 + j^-v 2)(E i'2 - E 'y~) {*,• By definition of p and d'jm

- - ( £ - £~l) = i l l + A - (1 - V)J [v A - E - I and V = 1- £ 1j

= I [A + V] = R.H.S. Proved.

Example X I I : Evaluate (V+ Ap (x 2 + x), taking h “ I. Solution. We have

(A + V)2 (x2 + ac)= [£ -1 + 1- £ - I]2(x2+x)

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= ( £ - £ - 1)2(x2 +x)

~ (E? ~2 + E~2) (x 2 + x) [ v £ £ - * “ lj= (£2-2 + E - 2)x* + (E2- 2 + F 2) x

= (x 2) - 2X 2 + E~ V ) + £*(x) - 2x + 2(x)

= (r + 2A)2 - 2X2 + (x - 2A)2 + (x + 2h) - 2 x + ( x - 2/») [ v E"fix)^fix + nh)}

= ( x + 2 ) 3 - 2 x 2 + (x - 2 ) 2 + (x + 2 ) - 2 x + (x - 2 ) [ v A * 1]

- x2 + 4x + 4 - 2x2 + x2-4 x + 4+ x + 2 - 2x + x - 2= 8. Ans.

Example 3.12: Evaluate (I + 4) (I - V).Solution. We have

(1 + A ) ( l-V ) = ( ! + £ - l ) [ l - ( |- £ - i ) ]

= £(1 - 1+ E~)] = EE~i = 1. Ans. Example 3.13: Evaluate

Al0 f ( l - a x ) ( l - b x * )( l - cx3)(1 - dx<)]

Solution. We know that, if fix ) = a^x” + a\x”~1+.... + a„.

Then Anfix )=aonlhn. ...(1)

A10[(l - axX 1- ~ c*3Xl - <&*)] = A10[abcdx'0 + ......]= abed A10(x10)= abed 10!. [v h= 1] Ans.

204 | E n g in e e r in g Mathematics-111

f /iM Ee*

{ E Y '^ e * Example3.14: Evaluate | - j fJ« • {RGPVJune 2002, Dec. 2004; Dec. 2006, Dec. 2008(H)]

Or

(A3) , Ee*

k e y ' a V

?*Provt that e* * ~p \e *7O ’ {RGPV, Feb. 2010 and June 2011}

Solution. We have

i?e*a M £«x —rl e 1. — — =Aa.£-£ J A V le*

A2ex

[Since *_/(x) =^x - A) and £^x) - fix + h)]

ox+h px+A= A V ' * . - ------ = e~hA2ez.A V A2€x

= e~h.ex+h = e*. Ans. Example 3.15: Find the second difference o f x? -5 x +6 the Interval o f difference being /.Solution. Let fix) - x 2 - 5x + 6

Afix) = A(x2 - 5x + 6)

= Ax2 - 5 Ax + A6

= [(x+ l)2-x 2]- 5 [( x + l) -x ] [v AC = 0]= 2x - 4. Ans.

Now, A? fix) ~ A(2x - 4) = 2Ax - A4= 2(x + 1 - x) - 0 = 2. Ads.

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(i )EV= VE = A (II) F= S E ~1/2

(Hi) 6=4(1 + A>-,/2= V(1 - V rm (iv) VA = A - V= 62

x m A v

Solution, (i) We have, £V = £(l - E~ ') - E - I = A

and S/E = ( l - £ - > ) £ = £ ~ 1 = AThus EV = A E ~ A.(ii) R.H.S. =<SE-i/2

= (£ l/2 - £ - I/2) £~1/2

= 1- £ - ' = V = L.H.S.

(iit) A(1 + A)~!/2 = ( £ - 1)(1 + £ - l )~ ,/2 = ( £ - l ) £ " l/2

= £1 21 £ - i/2 = sAlso V(1 - V)-I/2 = (1 - £ _l)[l -(1 - £ - i ) J - 1/2

= (1 ~E~l)E l/2 = E 'K -E - ]/2 = 5.

(iv) VA =(1 - £ - |X £ - l ) = £ + £ - , -2

= (£l/2- £ - ,/2)2= £2.Also A -V = (£ - 1)-(1 -£ -!) = £ + £ -* -2

= (£ '/2- £ ~ 1/2)2 = £2

- N u m e r i c a l A n a l y s is -11 | 2 0 5

(v) R.H.S.

* + /* r

Example 3.17: Prove that fi = ~ \JJ +

A V A2 - V 2 ( E - l ) 2 - ( l - f i 1)2

_ V ~ A ~ VA “ (E - D(1 - E - 1)

(E 2 - 2 E + 1) - (1 + E~2 - 2 E ~ l )

(e - d (i - e - ’ )

E 2 - 2 E - E ~2 + 2E~*

£ - 1 - E E - 1 + E _1

_ ( E 2 - E ~2) - 2 ( E - E~»)

E + E - 1 - 2

_ ( E - E -* ) ( E + E - 1 ) - 2 (E - E - 1)

( E + E -* - 2 )

_ ( E - E - 1 ) (E + E~* - 2 )

( E + E - ‘ - 2 )

= =(} + A) - (1 - V)= A + V = L.H.S. Proved.

_ £ + f*

2 j l + A

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2 + A

206 | En g in e e r in g M a t h e m a t ic s -111

2 + E - lSolution. We have

2 ^ + A

1 E

J E + y[E

l + E

2 yfE

.-(I)

Also f + ~ + ±( E * * - E - " * y

= ~yj4 + (E + E - 1 - 2 ) = - J E + E - 1 + 22 2

= I ^( E l '2 + £ - 1 /2 ) 2 . A ( £ 1 /2 + £ - 1 / 2 ) = ^2 2

...(2)

Proved From (1) and (2), L.H.S. « R.H.S. Example 3. IS : Prove that

(E 1(* + E - " ' ) ( l + A)” » =2 +A

Solution. L.H.S. » (£ ,/2 + E ~l/2) (1 + E - 1)m

= (£I/2+ £ -• * ) E lf2 = E + I= £ - l + 2 = A + 2 = R,H.S.

Example 3.19 :Given uj » 13, u j - 28, u4 - 49, us “ 76fin d A? u j and A? «>

Solution, (i) A2h2 = ( £ - 1)2^

= (£? —2£ + l)u2 = £2u2 —2£«2 + «2= a4 - 2«j + »2 = 49 - 2(28) + 1 3 = 6 .

(ii) A3k 2 = (£ - I)3 «2

= ( f 3 - 3 ^ + 3 £ - I )«2 = £ 3«2 - 3 £ 2»2 + 3£ «2 - «2

= «5 - 3«4 + 3«3 - «2

= 76 - 3(49) + 3(28) - 13 - 0 . Ans.

Example 3.20 .'Given y^ * 580, yi » 556, y 2 ~ 520 andy4 - 385. flndy>Solution. Given four observations, thus we get fourth difference will be zero, i.e., A y = 0

fRGPV Dec. 2005/

[v A = £ - 1J

Proved.

[v A - E - 1]

Ans.

=> AVo ~ 0 [As puty = yo]

=> oI I £ 1 * * »

[v A = E - 1]=> (1 -£)*yo = 0=> (E* - 4£3 + 6 E 2 - 4£ + 1)yo =0 [Using binomial expansion]=> y* ~ 4K3 + _ 4Vi +>5o =0 [Enyx ~ yx+n\

385 -4 y ,+ 6(520) - 4(556) + 580 = 0

yj =465. Ans.

Example 3.21: Prove that y 4 = y s + A y , + A*y, + Asy,.

Solution. R.H.S. = y 3 + &y 3 + (A2 + A3)^

= E 2yi + A Eyx + Az(l + A)^ ['•’ E nyx = jy +n]

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= 2?2y, + AEyi +A2Ey j [v A = E - 1]

= (E2 + A E +A2E)yj = [ E 2 + A # (l + A)]y,

= [ E 2 + A S2]*, = £ 2(1 + A)yt= - y4 = L.H.S. Proved.


Jo + yi + yt + ....+ yn - "*JCiyf +

N u m e r ic a l A n a l y s is - ! ! | 2 0 7

Solution. L.H.S. = y0 + + ^ 2 y0 + .... + E"y0

= ( 1 +E +E'1 +......+ E ") v0

( 1 ~ En+l >

[ 1 - E J* > - ( _A .

V 1 + r + r ' +

(v E*vr = y ^ J

1 . ,.».i 1

J

Provwl.

= — [1 + n+1 C j A + rt+1 CoA2 t .... +rt+1 C„+iA '1*5 - lbnA

= BtIC,y0 + "+,C2Ay0 + .... + *+,C^,A"y0

Example 3.23: Prove that

y tx + y t x * + y 3 x* + .......... - + (J - x)* A y* + < T ^ * F * ' y ' + .....

Solution. L.H.S. = y,x + y2x2 + y3x3 + .....

= yj.x + EyiX 2 + E 2 yxx 3 + .......= x[l + Ex + E~x- 4 ........ ]yj

[Herey is not function of x but it is a fun ction o f some other variable]

= x(l - Ex)~ly 1 [ v As binnnval exp an sion ]

= x[l - (1 + A)x]-Jyi

- x[l - x - Ax]-1 y, (*.* A = E - 1]i

1 - x

l - x

1 -

yi +

Ax

l - x yi = l - x

Ax A2x21 + -------- + --------- - +

(1 -JC) ( l - x )2 yl

Ayi + A2yt + .... = R.H.S. Proved.


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2 0 8 | E n g in e e r in g M*nsMAncs>IU

Solution. We have

= A"'1

(

_1 ___ 1

x + h x

1)A*-2

ln-1x(x + ft)

A l l - «x x + ft

[l X + ft X J V* + 2ft * + A JJ

* (-l)A "-2

= (-DA*-2

* (-1)2A"-2

= (-1)3A"'3

1

x + h x jc + 2ft

?/ 2 x(x + hVr + 2ft)

____ 2 !A*__

x(x + h)(x + 2ft)

________ M W ________

x(x + A)(x + 2ft)(x + 3ft)

In general,

A» f —1 = (-1 )" -------------------------- .V x J x(x + ft).......(x + nft)

[As similar processes]

Proved.

3.5 FACTORIAL POLYNOMIAL

The factorial polynomial of degree n denoted by the symbol or aW. Thus

xM = x(x - h)(x - 2ft)......... (x - n ^ l ft).

Here ft ?c difference and n is positive integer.In particular ft = 1, we have

jrf‘1 = x J&l = x ( x - l )

xPJ =x(x- IX*- 2 ) and so on.

If a is negative integer, then reciprocal factorial function of order n and denoted b y^-") or.tl-,,i.Thus.

In particular ft = 1 we have

1

(x + ft)(x + 2ft)(x + 3ft).........(x + nh)

1

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WwEWCtu^tew rw s-II j 209

________ 1 ________

**~3} ~ (x +XKx+2Xx + 3) 30(150 on- Remarks:

1. AxW = (x + /r)W -*MI

- [(x + AX*X*- h).......(* - n - 2 ft)l- [ x(x - h)....... (x - n -1 h)]

~ x (x - h).......(x - n - 2 A)I(x + h ) - ( x - n - 1 A)]

= nh x(x - A)........(x - n - 2 A)

“ nh x**"lJ.

For example : Axî = ^ = 3x1*1.

Similarly A2xM ”=n(n - I) A2 xi" ~ 21A«jrl»l = «!/i»

2. A*xW = 0 for k > « and Ax1*1= — x1*1= ux^*1).dx

X f 3. — x**l =* ----- t + C, where C is constant

A J n. + 1 Example 3.25 (a) : Represent the function :

f(x) «=x* - 12x?+24x*-36x + 9mud Its successive differences in factorial notation. fRGPV June 2004}

Solution. Given/x) = x* - !2xJ + 24c2 - 30* + 9.

Letjfc) * + lftcPJ +Oc£l + Hafii + E in factorial notation

=>x*~ 12x3 + 24x2-3Qx + 9= Ax(x- IX *-2 Xx-3) + BxOc- IX*- 2 ) +C x(x -l) + Dx + E ...(1)

Now, putting x = 0 on both sides of (1), we get E = 9

Again putting x - 1, on both sides of (1), we get1 -12 + 24 -3 0+ 9 =D + £=>/> + £ = -* = > £ - - 17

Further, puttingx = 2, on both sides of (1), we get16 - 96 + 96 - 60 + 9 = 25 + 3C + Z)

2C + 2D + E = 35 =>C = - 5Similarly, putting x = 3, we get

6fl + 6C + 3i>+£ « -I0 8 = » 3 = - 6Finally, equating the coefficient of x4 on bolh sides of (IX we get A = 1.Thus, the required polynomial in factorial notation is

f e ) - xW - 6 - SxPl - 17xt‘l + 9. Ass.

I

** 2J = (x + IX * + 2)

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'210 | Engbveebng H m e u n c sJ II

Hence, 4/1*) = 4xiJl - 6 x 3xl2>- 5 x 2xW - 17 x I + 0

= - 18brt2l - 10xW- 17,

Also ^ A x) ~ 12xt2l —36xt11—10,A3/x ) = 24*m - 36,

A4/x ) = 24.AsJ{x) = 0. A b s .

Example 3.25 (b) : Express y *=2xJ - 3x 3 + 3x - 10 in factorial notation form. / RGPV, Dec. 20J1] Solution. Given f (x ) = 2x3- 3x 2 + 3x - 10

Let f(x) - A xW + fltf2! + Cx 1*1+ D ... (1)in factorial notation.=> 2x} - 3xJ + 3x - 10 -~Ax (x - I) (x - 2) + Bx (x - 1) + Cx + D ... (2)

Now, putting x = 0 on both sides of (2), we getZ) — 10

Again, putting x = I in (2), we get

C + D = - 8 =*C = 2Finally, putting x = 2 in (2), we get

B = ~(-2C - D ) ^ s - 3.a*

Also, equating the coefficient of x3on both sides, we get A =2

Thus, polynomial in factorial notation (2) becomes ; f ix ) = 2rP) + 3xt2J + 2xf J - 10. Ans.

Example 3.26: Obtain the function whose first difference is Xs + 4x? + 9x + 12.

Solution. Let fix) be the required function, we haveA/tx) - x3 + 4x* + 9x + 12

or 4/1*) -AxW + + CxI'l +- D in factorial notation ...(1)

=> x3 + 4x2 + 9x + 12 =Ax(x - lX x-2) + Bx (x- I) + Cx + D ...(2) Now, putting, x = 0,1, 2 on both sides of (2), we get

D * 12C + D = I + 4 + 9+ 12 = 26 o C - 14

2fl + 2C + £> = 8 + 1 6+ 18 + 12= 102 D ^ 1

and equating the coefficients ofx3, we get A - I.

Hence (1) becomes: Aflx) = xt3l + 7x 21+ 14xt’l + 12, on integrate, we get

2fHl xW mi fix) = —— + 7. —- +14. —— + 12x'1’ + k, [k being constant]

4 o a

= *<* - j ) + + 7*< » - 1) + i 2 r + *

= —(x4 - 6 l 3 -7 x 2 ~Gx) + -~(x 3 - ix ‘ *2x ) + 1(x' 2 - x ) + 12x + k 4 3

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Nu m e r a l An a l y s is -!! | 211

*4 ( 6 7 1 3 ( 7 _ , f 6 14 _ . A . — + — ■+— x3 + ------ 7 + 7 x2 + — + ------7 + 12 x + fc

4 I 4 3J i 4 J I 4 3 J

49Ans.

Ans.

/RGPVDec. 2004/

-(I)[ V * l ! J = X, x*2l *=x ( x - 1)]

= — x* + —x ?— r *2 + — x + ft.

4 6 4 6 Example 3.27 : Obtain the function whose first difference is 2x* + 3x* - Sx + 4 fRGPV,.Feb, 20l0 f Solution. Proceed as above example, we get the function

y - ( = i x ‘4' + 3x,J1 + 4x111+ k.

Example 3.28 : Find the function whose fir st difference is Px2 + / /x + 5.

Solution. Lety(x) be the required function, we haveA/{x) = 9x2 + llx + 5

or A/(x) = A x^

+ fltM + C in factorial notation=> 9x2 + 11x + 5 = Ax(x-‘1) + B x + c

Puttingx = 0 we get C = 5.Putting x - 1, we get 9+11 + 5 = 27+ C = ?iJ ;=20.Equating die coefficient of xJ, we get A - 9Hence (1) becomes:

A/W = 9x ^ + 20xni + 5

v(3] y\X\ => Ax) - 9 ~ + 2 0 ~ + 5x W + fe

=> /x ) =3 x(x-lX x-2)+ 10 x(x -l ) + 5x + /k Ax) =3x3 + x2 + x + *, where k is constant. Ans. Example 3.29 : A second degree polynomial passes through (0,1) (I, 3), ( 2 , 7) and (3,13). Find the

polynomial.

Solutbn. Let the polynomial beX*) = + + c. ...{1)The difference table for the given values of x and/x ) is as follows :

0

1

2

3

f(x)

1

3

7

13

Af(x)

2

46

As f(x)

2

2

A*f(x)

From (1) Ax) - + + c&Ax) = 2ax + b =s>A/(0) = b=> b ~ 2

and A2/(x) =2a=> A2jt0) = 2a=s>2a = 2=>a = 1Again putx = 0,^ 0) = c ^ c = 0Hence (1) becomes:

Ax) =x<2>+ 2x<»+ 1 = x(x- l) + 2x+ I = x2 + x + 1.

[v Mx) = 2]

[v A2/0 ) = 2][v /0 ) = 0]

Ans.

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F xample 3.30: Find A* fix), i f

f(x) * (2x + 1) (2x +3) (2x + 5) _ (2x + 19). blution. Given J[x) = (2x + 1X2* + 3X2x + 5)....(2x + 19)

2 1 2 { En g w e j m w g MAWtomcs-lH

[ v There are 10 factors]

- - R ) K ) K ) ..

(-4) Ax) - 210 I

, r 19 f o1+ » [whereh «=1and by definition o f factorial polynomial.]

( 19 V95Hence &ftx) = 2tox lO lx + — I

[81

\f«1

tffix) =2^x90^ + !

A*) =2*°x72o(x + y j

=210x6040 +y j

, 2>«*504o(*+ ^ } ( * + ^ ] ( * + H } ........

=80640(2* +9) (2x + ll)(2x +13) ....(2x+ 19).Frampfe 3,31 .'Prove that

* y x « j w - HCiyx+n-t + *c,yx.n t — + ( - i y i y x .

Solution. R.H.S. = ys„ - nClE ^ y x^ + "C ,£-2y«»-.. +(-1)*C»-E-*yx+»

= {1- "C,£-J+ KC2E~i - ....+(-l)»* (1 - £?_l )*>*♦„ [As binomial expansion]

= { (I-# * 1)* .£*}>* * ((X-E~l)Eyyx

= [ E - \ Y y x =&*yx -L.H.S.

L <ample 3.32: Given that y s m4 ,y tm3ty r m4 ,y tm 10 andy9 - 24, find the value o f4 y f :

(i)By using the difference tabie and (U) Without using the difference table. Solution, (i) By using the difference table :

Ans,

Proved.

X y A y * y Sy *’f (x)

5 4 16 3

—I 1 2 Q

7 4I614

5Oo 0

. . 8 . 1 0 8o

9 24

From the difference table AVs = 0. Ans.

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Nu m b k c a l An a l y s i s -11 | 213

A*ys = ( £ - l)*ys

= (£< -*C ,£3 + 4C j£2_4C8£ + 4C 4)yb

= > 9-4 ^8 + 6y7-4 y 8 +y8

= 24 - 4* 10 + 6 x 4 -4 * 3 + 4-24 - 40 + 24-12 + 4 *0

Example 3.33 : Given x : 1 2 3 y : 2 5 10

Find the value ofV*y$

Solution, (i) By using the difference table to find V2y5 :

(ii) We have

From the difference table, we get V2 = 2.

(ii)Without using the difference table to find V2 :We have V2 = (1 - E~ 'J2 ys

= (l + E - i - 2 E~')ys

=ys +y3~2y4

= 26 + 4 0 -2 x 17 = 3 6 -3 4 -2 . Example 3.34 : Estimate the missing terms in the following table,

x : 0 1 2

y = f ( x ) : I 3 9 Explain why vmtue differs from 3* or 27.

SolutioH.Wen number of known data = 4 values A*y = 0

=> ( £ - 1)4* = 0

=> (£* - 4C |£ 3 + aC2E2- *CjE + *C4)y = 0

=> (£* -4£ 3 + 6£2- 4 £ + l) v = 0

Putting .y -yo> ( E 4 - 4 Ei + 6£? - 4£ + 1 ^ = 0

=> r4-4>»3 + 6>»2-4yi +yo =0

[Since y0 = t.yi =3,>^ = 9,>^>=?,y4 = 8t]

4 17

3?

[v A = £ - I)

Iv E*yx -

Ans.

526

[RGPV June 20(hf

1 X1 > Vy

i 3 k m >

V3y

I u

i i y» * 5 3

2 2

3 J ? I i 1 & 5

20

4 yt = & 7

20

5 >1 = 17 9

y, = 26

Ans.

[v V = 1-&'■ ')

[v E~*yx = yx.„ j

Anv

4 81

(RGPVDec. 2002, Dec. 2008(N)J

[By definition]

{•• E nyx = y x+«1

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=> 81 - 4 * >>3 + 6 * 9 - 4 > 3 + 1 0

=> 4 y} = 124

=> >*3=31=> fl3) = 31. Ans,Explanation : This differs from 33 i.e., 27 because differencing values arise due to theassumption that/x) = 3 is a polynomials of degree 3 (Since four values are givefi) so thatfourth and higher order differences vanish. Ans.

Example 3.35 :Flnd the missing values in the following data. x : 2.0 2.1 2.2 2,3 2.4 2.S 2.6

y=f(x ): 0.135 - 0.111 0.100 - 0.082 0.074[RGPV June. 2004}

Solution. Let yo,y] ,y 2 ,yhy*,y$< V6 be given values, in which missing values are y\ and y4

Here, number of known data = 5 values Asy = 0 => (E ~ I )5y = 0

=> ( f 5 - 5C,F» + 5C2& - SCyE 2 + 5C 4 E - 5Cs)y - 0=> (E 5 - 5E* + 10£3 - 10/s2 + 5£ - l)y = 0 ...(1)

(i) Putting .y = .yo in ( I), we get

(£s -5 £ 4+lO£3-lO£2 + 5£ -l)yo =0

=> ys-$y4+ 'Q w - iQyz + Sy, -yo =0 [v E ny0 =y0+J

=> 0.082 -5>-4 + 10*0.100-10*0.111 + 5yi-0 .l3 5 =0

=> 5^|-5>»4 =0.165

=> y \ - y* - 0.0326 ...(2)

(ii) Again putting y - y\ in (1), we get

(£ 8 - 5 E* + 10F8 -1 0 E* + 5 E - I)*, = 0

=> - y6 - 5y6 + lOy* - 10y3 + 5y2 - >i = 0

=> 0.0 74 -5x 0.0 82 + 10 ;^ -1 0 x (0.100) + 5(0. I ll ) -j y , =0

=> , 10^4 -^ i =0.781 ...(3)

Solving equations (2) and (3), we get9y4 = 0.8163 =>>4 = 0.0904 i.e.,/2 .4) = 0.0904. Ans.

and y\ =0.123 i.e.,fil.\)~ 0.123. Ans.

Example 3.36 :Find the first term o f the series whose second and subsequent terms are 8 , J, 0, -1,0.[RGPV June2903, Dec. 2007}

Solution, Letyo = ? and y\ = 8,>»2 = 3,y j = 0,^4 = - l.ys = 0.

Here number of known data = 5 values .'. &y = 0

=* ( £ - l)5,y = 0

=> (Es ~ 5C ,^+ 5C 2£;s -6C 3£ ? + 5 C 4 « -6C5)y = 0

=> (£ 6- 5 £ 4 +10 F>-10 £2 +5 £-l) ;y “ 0 ...(1)

(i) Putting y = >5q in ( I). we get f

{E* - 5£’4 + 10 £3 - 10 £2 + - I)y0 =0

2 1 4 j En g w e e b w g M m m x n c s -U I

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(i) Puttingy ■ yo in (IX we get

(E*-4E*+GE 3 ~4 E + l)y0 =0

=> > < +>0 - 0 [v£»y0 = y ^ J

=> >4 -4(17) + 6(y3) -4(10)+6 -0

=> 6 yz+yi = l02 ...(2)(ii) Puttingy « y j in (1), we get

(E* — 4E^ + 6JE® — 4E + l)>i = 0

=> -4 y 4 + 6 -3 -4>2+>i =0

=s> 8 1 -4 y 4 +6(17)-4ya +10 = 0

=» 4> + 4^4 = 143 ...(3)Solving equations (2) and (3), we get

y 2 * 13.25 and y4 - 22.5.i.e., 'ftlQ)= 13.25 9ni 22.5. > Ans.

Example 3.39 {interpolatef(2) from Ike following table: x : 1 2 3 4 5 f(x) 7 - 13 21 37 and explain why the value obtained is different from that obtained by putting x = 2 in the expression 2 *+$.

Solution. Letya,y\,y 2 ,y},y4 be given data in which y( is missing.

Number of known data = 4 values, so that A*y = 0

=> ( £ - l ) V = 0=> (£*-4£3 + 6£2- 4 £ + l) y “ 0 ...(1)Putting y = yo in (1), we get

=> y 4 -4 y 8 +6y3 -4 y ,+ y 0 =0 (v £*yo=ycn-„]

=> 37-4 (21)+ 6(13)- 4 y i+ 7 =0

=> 4yi *38

=> y\ “ 9.5 orX2) = 9.5. A b s .

Also since j{x) = 2* + 5.*. Exact value at x m 2 Le.,J(2) m 9.But interpolate value ofJ(x) is 9.5 i e , 9.5 differs from the actual value 9 obtained by theexpression 2* + 5.The reason for this is : we have assumed/x) to be a polynomial of degree 3. Ans.

Example 3.40 .Find logie 7 and logic U from the following table:

x : 6 7 9 9 10 11 12 y - h g n x : 0.77815 - 0.90309 0.95424 1.000 - 1.07918

Solution. Letyo, yj.y2.y3.y 4.y 5.y6 be given data in which yi and >>4are missing.Here number of known data = 5 values, so that A5y = 0=> (£- l)5y=0

=* (E5^6C t£4+6C2£ s -6 Q S 2 +6C4£ -e C 8)y =0

2 1 6 |

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Nn—atKWM w w u fl j 217

=> ( ^ - S ^ + l O E ’ - lO ^ + S g - D y =0 ...(1)

(i) Puttingy * yo in (IX we get

(£ 8 - 5 E* +10 E * -10 E* + 5 E - l)y0 = 0

=> y&-b y* + 10>3 - 10>J + 5yj - y0 = 0 fv & y a = y ^ J

=> y 5 - 5(1.000) + 10(0.95424) - 10(0.90309) + 5>, - 0.77815 - 0

5y,+y5e 5.26615 ...(2)

(ii) Puttmgy = yi in (IX we get

y« - 5 % + i o ^ - i 0 y 8 + 5 y2 - y i *o

1.07918 - 5ys +10(1) -10(0.95424) + 5(0.90309) - yx =0

=> >i + 5ys • 6.05223 ...(3)Solving equations (2) and (3), we get

y, = 0.84494 Le., tog,0 7 = 0.84494

and y$ = 1.04146 i.e., ^ 1 0 11 = 1.04146. Ans.

3.6 I N T E R P O L A T I O N W I T H E Q U A L I N T E R V A L S

interpolation is the process of finding the intermediate value of a functiony-fix) from a set of itsvalues at specific points given in a tabulated form i.e.,Suppose we are given the following values of y ~j(x) for a set of values of x :

x : xo x\ X 2 — xH

y : yo y\ n ... y»Thus the process of fmding the value of y corresponding to any value of x = x, between xo and x„ is called interpolation.Hence interpolation is the technique of estimating the value of a function for any intermediatevalue of the independent variable while the process of computing the value of the fistction outsidethe given range is called extrapolation.

3.7 G R E G O R Y - N E W T O N ’S F O R W A R D I N T E R P O L A T I O N F O R M U L A F O R

E Q U A L I N T E R V A L S

Statement:Let yo, y \, y-i. ..... y„ are the value of y - J(x) corresponding to equidistant values of

x ~ jcq, a'2, .... xm such that x~x$ + uh^>u~ — j— [A = length of interval]

■ me n, , -M ) » ....

• ........(I)n!

Proof:

Let y = J[x) =/x o + uh) [v x = xo+uh]

= E?j(xo) [Using Shifting operator]

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= (1 + A)“>>o [vo =X*o)]

= U j f o . . . .

^ ( y - l ) :- ( u - ^ i )n!

[Since A" + % , A"+2>b. etc are zero]. Proved. Remarks:

(i) Since the formula derived involves the forward differences of y at >>0, it is called

Newton's forward interpolation formula.

(ii) If three value of y , namely, yo, y i and y% corresponding to x - xo. *1 and xi are

given, then Newton's forward interpolation formula is called parabolic interpolation

formula.3.8 GREGORY-NEWTON'S BACKWARD INTERPOLATION FORMULA FOR

EQUAL INTERVALS (RGPV Dec. 20021

Statement:

y\> y i t ...... yn are the value of y = ftx ) corresponding to equidistant values of

x =*X&, x \, X 2......... xm such that x, - x, _ j = h for i = 1, 2, 3 , ...... n, and x - x„ = uh

x - x n =s>u = —^— . then

u „ , u(u-»-l> u(u * l )....(u h 1) y=Ax)= .......+ — ^ ...(])

Proof: Let y ^J{x)

=> y ~A*n +uh) Iv X - Xn = ii/l]= HyjiXn) [Using shifting operator]

= (£ ‘ l )-" /(x?) = (1 - V Y u y n [ v 7 = l -E ->]

[l t “ V >U(U+1) V* i u& + 1'>(u+ ?) V3 ,I 1! 2! 3! .......

218 | Cn q b« ^ b *o MxmEMAncs-IlI

n !

[Since V” + lym V"+ 2 y„ etc. are zero]

t u _ u(u + l ) u(u + l )....(u + n -l )y =f(x) » yn + - + - 1 - - V2 yn +.... + ----- !— i----------i V » y „ .

1! 2! n !

Un

proved.

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Numer ica l Ana l ys is-# j 219

Remark :Though it is usually suggested that the forward formula should be used for interpolating orextrapolating y at points near xq and the backward formula should be used for interpolating or

extrapolating near xm it is not necessary. Either of the formula can be used for interpolating orextrapolating.? at my point In feet that the interpolating polynomials of fix) that occur in theR-H.S.'s of both the forward and backward formulas are identical.

3.9 ERROR IN NEWTON’S INTERPOLATION FORMULAE

1. Error in Newton's forward interpolation formula:

™ - l) ( u - 2)....■■(»-n )^ > = (n •+-1)! ( y>™ '

X—Xq where xq <c < xBand u - — t — •

2. Error in Newton's backward interpolation formula:

r , , «(« + 1)(« + + nl m»+i V)(n +1)! * 9}~ *

where x q < c < x „ and«:h

Example 3.41: Estimate the sale fo r 1966 using Newton forward interpolation fo rm ula :Year: 1931 1941 1951 1961 1971 1981

Sale iu thousand: 12 15 20 27 39 52Solution. Here given interval is equal and h ~ 10.

The forward difference table is :Year

<x)Sa le (y) Ay A*y A*y A*y A&y

1931 123

1941 155

20

1951 207

23

3-1 0

1961 2712

5- 4

- 7

1971 39 13 11981 52

We know that Newton forward interpolation is

u . u (u - l) . u(u - l) (u~2 )

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220 I ep * NW ^J4ro «M t f iC84 i I

1966-1931A t,* - 1966, taking *o® 1931 and A - 10, so that u * ------^ ---------3J.

Hence (1) becomes:

- * * « > -

. (3.5)(2.5K1.5X.5) . „ . (3.S)(2.5X1.5K.5)(-.6).. --------- -- --------- H ( '

= 12 + 10.5 + 8.75 + 0.8203 + 0.2734 - 32.34Thus, the sale for the year 1966 * 3234 thousands. Ans.

Example 3.42 .’Estimate the sale fo r the year 1966 using Newton backward interpolation formula :

Year : 1931 1941 1951 1961 1971 1981

Sale in thousand: 12 15 20 27 39 52

Solution. Here given interval is equal Le., h ■=*10.The backward difference table is :

Year (x) S a le(y) Vy V*y Vsy V5y

1931 12 o1941 15

U2 *

5 01951 20 2 3

7 3 -1 0

1961 275 - 7

12 - 41971 39 1

131981 52

We know that Newton backward interpolation is

....... - 0 )

1966-1981Here, « = — -

------ ^ ----- - 1 . 5/

Hence (1) becomes:

>1966 =/(1966)= 62 + (-1.5) *13 + (~L^ ~ 5)- x 1 + ( 15)L x (-4)

( (-1.5)(-.5)(.5)q.5) : ( 7)[ (-1.6)(-.5K.5Xl-5)(2.5)„( ^

4! 6!

= 52 - 19.5 + 0.375 - 0.25 - 0.1641 - 0. 11 72 = 32.34Thus, the sale for the year 1966 - 32.34 thousands. Ans.

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NtowErttUtAMmlvs»41 t 221Example 3.43 :The table gives the distances in nautical miles o f the visible horizon fo r the jfcven

heights in fe e t above the earth’s surface :x - height t 100 ISO 200 2S0 300 350 400

y -d is ta n ce: 10.63 13.03 15.04 16.81 18.42 19.90 21.27 Find the values ofy , when x ~ 218 f t and 410f t

Solution. Here h - 50 [equal interval]The difference table is :

X y Ay A2y A*y A4y

100160200260300

350400

10.6313.0316.0416.8118.42

19.9021.27

2.402 . 0 1

1.771.61

1.481.37

-0.39-0.24-0.16-0.13

-0.11

0.160.080.030.02

-0.07-0.06- 0 . 0 1

(0 At x** 218: Which is near to x&= 200, so using Newton forward interpolation formula:

u . u(w~l) . . « (« -!) (« -2 )yz =AX) =yo + Avo + '2\ y°+ — fi----- y° +..............(I)

x-xo 218-200Here u = —j-— = -----— ---- =0.36 and

yo = 15.04, Ayo * 1.77, &?yo * - 0.16, A3>»o = 0.03 etc.Hence (1) becomes:

K ,, -^ 2 1 8 )- 1 5 .04 +0 .36 0 .77 )+2 iS± «£ (-0.16)

+ 0.36(- 0.&4X-1.64) (003)+

= 15.04 + 0.637 + 0.018 + 0.001 +.... = 15.696 i.e. 15.7 nautical miles. Ans.QS)Atx~ 410: Which is near the end of the table, so using Newton backward interpolationformula:

» - /X ) . * V i v * + v= ,„+...................(2,

x -x „ 410 - 400 _ "here. -------gj---- 02Here,x“ 410, x„ -400 ,yn - 21.27, Vy„ - 1.37, V2 ,, = -0.11, V3# , = 0.02 etc.

Hence (t ) becomes:

y* 10 =*410) = 2 1 . 2 7 + 0 . 2 ( 1 . 3 7 ) + ^ ^ ( - 0 . 1 1 ) + ....

21-53 nautical miles. Ans.

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Example 3.44 :The following table gives the population o f a town during ike last six censuses. Estimate using Newton’s interpolation formula the increase in (he population during 1946 to 1948. '

1911 1921 1931 1941 1951 1961

12 15 28 27 39 52

{RGPV June, 2003]

2 2 2 | E m m tam a M ashesm xics-II I

fear;

Pop. in thousand:

Solution. Here h ~ 10. (equal intervals)The difference table is :

Year xPop. in

thousand

y

A > A*y a V A4 f(x) A 5y

1911 123

1921 15 2 A1931

19412027

07

1213

25

II3

3- 7

- 10

1951 39 1 — •*1961 52

1-------------------.

x - x Q 1946-1911 A tx~1946i v «= - y 1 -------- ^ ------=3.5.

Using Newton’s forward interpolation formula :

. u ( u - 1) fi x) = +«&*> + ■ 2 j - A2y0 + .......

=^1946)» + +

-(1)

2! 4!

+ (3-5)(2.5)(1.5)(0.5)(-0.5) (_1q)

5!

or fi\946) = 12 + 10.5 + 8.75 + 0.8203 + 0.2734 = 32.34.

Again,

*-*o 1948-1911 At x ~ 1948: v «= —r 2- “ ------“ ------=3.7.

10Hence (1) becomes

10 (3.7)(2.7) 0 . (3.7)(2.7)(1.7)(0.7) .>1948 - / 1 9 4 8 ) = 12+{3.7)(3) + - — ~ — 'x 2 + 0 + ~ — —— —— - — - x 3

2 \ 4!

| (3.7)(2 .7)q.7)(0 ,7)(~0 ,3)5!

Hence, poiulation increase during 1946 and 1948 19 4 8 )1 9 4 6 )

= 34.87 - 32.34 = 2.53 thousands = 2530.

x (-10 ) = 34.87

Ans.

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N u m e r ic a l An a l y s i s -11 | 223

Example 3.45 :Find the cubic polynomial which takes the following values : x : 0 1 2 3

f(x) 1 2 1 10 Hence or otherwise evaluate ff4). fRGPV Dec. 2001 and Feb. 2010}

Solution. T he difference table is :

X y = f(x) Ay A2y A3y

0 1 t1 2

1-19

-212

2 1 103 10

x - x o x - 0 Here, h = 1and u = —r— “ —;— - x

n 1

/. By Newton’s forward difference formula,

u . u (u -l ) .» Ax) = yo + J7 Ay0 + —gj— A o+...

x ( x - l ) x ( x - l K x - 2 ) Ax) - yo + *Ay0 + - A2y„ + ------- J j------- A3y0 [v u = x]

, ! + , ( ! ) . £<* ^> (-2) + ^ X- S ^ Z ?1 (12)2 6

[v y0 = 1, Ay0 = 1, A2y0 = -2, A8y0 = 12]

= 1 + x -x ( x - 1) + 2*(x-1X*~2)= 1 + x - x2+x +2 x3 -6 x2 + 4x

= 2x3- 7** + dx + 1

.Ax) =2x3~7x2 + 6 x + 1. .—(I) Ans.Also, put x = 4 in (1), we get

A*) = 2* 43- 7* 42+ 6 * 4 + 1= 128-112 + 24 + 1=41. Ans.

Example 3.46 :Given sin 45* * 0.7071, sin 50* m 0.7660, sin 55* m 0.8192, sin 60* m 0.8660, find sin S2* using Newton*s forward Interpolation formula.

Solution. Lei Ax) = sin x°.The difference table is :

X y A y &*y A‘y

46505560

0.70710.76600.81920.8660

0.05890.05320.0468

- 0.0057-0.0064

- 0.0007

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X —Xt) 5 2 - 4 5 A tx “ 52 : Here h **5, *8 - 45, then a * —^— * — g— = I 4

Using Newton forward interpolation formula:

Ax) - yo.+«Ay0 + A2y0 +.....

or sin52° - 0.7071 + (1.4)(0.0589) + (-0.0057)

(1.4)(0.4)(-0.6)(-0.0007)+ * 6

= 0.7071 + 0.08246 - 0.001596 + 0.0000392 - 0.788. Ans. Example 3.47 -.Form the table, estimate tke number of students who obtained marks between 40 and

45.

Marks 30-40 40-50 50-40 60-70 70-80So. o f students: 31 42 51 35 31

[RGPV 20011Solution. To find/4 0 andX45).

The difference table is :

M a rks le ss th en

(x) y= f(x ) A y A ' y Aay A<y

40 8142

5135

31

50 73 9

- 2 61260 124 - 1 6 3770 15^ - 480 190

From the table: A tx * 40=>fl40) = 31. ...(1)

X —Xa 4 5 - 4 0 Now, Atx * 45, then u = —^ ~ — = 0.5

/. Using Newton’s forward interpolation formula:

M ......

=> A 4S) . 3 U (0.5)» 42 , , 9 » - ’-a? , (-25)2 6

t (.5K-.5)(-l.S)(-2.5)j

24

= 47.87 * 48. ...(2)Hence the number of students getting marks between 40 and 45

=X45)-A*0) ~ 48 - 31 = 17. Ans,

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' ’ NowHBCAt *0*L Y sis-fl1 2 2 5

Example 3.48 :ln an examination the number of candidates who secured marks between certain limits were as foUows :

Mdrks : $-19 20-39 40-59 60-79 No. of Candidates: 41 62 65 50

Estimate the number o f candidates getting marks less than 70.The difference table is :

80-9917

Solution.

M arks lean then (x)

1939597999

Num ber o f candidates}? f (x )

41103168218235

Ay

62655017

A 'y

S

- 15 —33

A 'y

-1 8-1 8

A4y

We have, « 19, h * 20, x * 70 f To fmdX70)]

*~«o 70-19u=> h ^ ° 2.55.

By Newton’s forward difference interpolation formula :

u (u -1 ) Ax) - +“4y0 +

/7 0) * 41 + 2.55x62 +

21 * * +

2.55x1.55

*<3) +

2.55x1.55x0.55 x (-18) + 02x1 ^ 3 x2 x1

= 41 +■158.1 + 5.92 -6 .5 2«205*2-6 .5 3 - 19*49 * 198. Aas.Example 3.49 :The following are the marks obtained by492 candidate* in a certain examination :

Marks: 0-40 40-45 45-89 $0-55 55-60 60-65 No. of Candidates: 219 43 54 74 32 79Find out the number of candidates:(I) Who secured more than 48 but not more than 50 marks.(Ii) Who secured less than 48 Nit not less thaw 45 mark.

Solution. The difference table is :

M arks less then (x)

No. o f candidates

(y)

A y &*y * y * 4y A 'y

40 21043

45 253 1154 9

50 307 20 -7 122274 —62

55 381 -4 289

15132

60 413 47

7965 492

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226 | E n g w e e k h K j M a t h e m a t ic s >111

From the table: Here/40) = 210 and/50 ) = 307.y y AO J A

To fbid f(48) i Here xo = 40, * - 48, h •= 5, .*. u = — r— = — r**- = 1.6.n o

By Newton’s forward interpolation formula,

Ax) » + “ A y0 + U(| ~ 1- A2y0 + .....

=> /48) *210+ 1.6x434 y >((ll)^ 6M ^ x(9)6 *

(1.6)(.6)(-.4)(-1.4) q.6)(.6)(-.4X-1.4X-2.4)

24 120*210 + 68.8 + 5.28 - 0.576- 1.5964 - 2.3869 * 280.

(i) Number of candidates who secured marks more than 48 but not more than50 marics = /50) ->(48) =,307 - 280 * 27. Ans.

(ii) Number of candidates who secured marks less than 48 but not less than45 marks = /4 8) -X45) - 280 - 253 = 27. Ans.

Example 3.50 .’The following table gives the marks obtained by 100 students in Mathematics in a certain examination:

Maris : 30-40 40-50 5040 60-70 70-80 No. o f Candidates: 23 35 22 11 7 How many students got more than 55 marks T


M arks less than (x)

No. o f candidates

(y>

A» A*y A*y A4y

40 25352211

50 60 -1 860 82 -11

A

n 570 93 -4

7

• 80 >007

To And f(55) : Here x - 55, xo “ 40, h ~ 10

x-xp 55-40

U ~ h 10/. Using Newton forward difference formula :

Ax) = yo+jfoo* - ^y1- A2y0+.....

=> X55) - 25 + x 3 5 + ^ 5^x(-13) + )(- )f c f )-x 21 2 6 ' ‘

(1.5)(.6)(-.5)(-1.5)jt5

24

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I

Numer ica l An a l y s k -11 I 227 »

- 25 + 52.5 - 4.875 - 0.125 + 0.117 = 72.6 » 73'students.Hence number of students getting marks less titan 55 - 73Therefore, number of students getting marks more than 55

- 100 - 73 - 27. Ans.

Example 3.51: From the following data, estimate the number o f persons having incomes between<1)2000-2500, (U)1000-1700, (U) 3500-400*

Income: Below 500 SOO-IOOA 1000-2000 2000-3000 3000-4000 No. o f Persons: 6000 4250 3600 1500 650

Solution. First we prepare the cumulative frequency table make equal interval (A - 1000), as follows: Income less than (x): 1000 2000 3000 4000 No. o f persons (y): 10250 13850 15350 16000

Now the difference table is :

X y A y A A 'y

10002000

30004000

1025013850

1535016000

36001500

650

- 2100- 8 5 0

1250

(i) From the table :/2000) « 13850 ...(1)To find f(2500) : Here x - 2500, xq - 1000, A- 1000.

_ X - X o 2500-1000

^ en U h 1000 “ K5'

.". Using Newton’s forward interpolation formula :

Ax) = y o + ~ Ay 0 7 1} A>*>+ ........

=> /25O0) - 10250 -t- (1.5)x 3600 + x (-2100) + ^ x (1250)2 6

- 10250 + 5400 787.5 - 78.125 - 14784 students ...(2)Hence the number of persons having income between 2000 and 2500

-X2500)-X2000)-1 4 7 8 4 - 13850

= 934. Ans.(ii) From the ta ble /1 000) = 10250 and ...(3)

/4000) = 16000 ...(4) Now to fin d/ft ) at x - I7 O 0 Le.f(l700) :

Here x = 1700, xo - 1000, h - 1000

x-xo 1700-1000

“ = h 1000 ~°*7Using Newton's forward interpolation formula,

Ax) ~ yo +«Avo+ A2y0 + -g> t f y<>+ .........

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228 | ENQiNEcnNo MiWTamncs-IH i

*1700) = 10250+(0.7)(3600) +(0.7)(-0.3)

( -1 2 0 0 ) +(0.7)(-0.3X-1.3)

2 6

= 10250 + 2520 + 220.5 + 56.875= 13047.375 * 13047.

Hence the number of persons having income m between 1000 - 1700=/1700) -^1000) - 13047 - 10250 = 2797.

(tii) Again to flBdJfr) a tx - 3500Le.,ff3500) :Here x = 3500 which is near 4000, so that taking x„ - 4000, h = 1000

x - x n 3500 - 4000

...(5)

r "•Ans.

(1250)

u = = -0.5h 1000

Using Newton's backward interpolation formula:

„ tt(u + l ) _ , u(u + 1Xk + 2 )_ . Ax) = yn + wVvB+ ------- ***• + •

2!

/3500) = 16000 + (-0.5)(650) +

3!

(-0.5)(0.5)( -850) (-0.5)(0.5)(1.5)(1250)

...(6)

Ans.

2 6

- 16000 - 325 4 106.25 - 78.125 * 15703.125 « 15703.Hence the number of persons whose income in between 3500 -4000 :

=/4000) -X3500) - 16000 - 15703 = 297.h xomple 3.52 :To obtain the interpolating polynomialf[x) satisfying the following data:

x : I 2 3 4

y ~ f ( x ) : 26 18 4 1 I f another point x ~ 5, f ( x ) m 26 is added to the above data, will interpolation polynomial be the same as before or different T Explain why ?


X y - f ( x ) * ’y A*y

1 26' - 8

- 1 4O

2

3

184

- 6

1117

4 1—

To find f(x) : Here jco = I, h ~ 1

. t - * 0 i - lI hen it - —r— => » = x - I

h 1

Using Newton’s forward interpolation formula .

Ax> - ‘v* f * .......

= J to . 2 6 - ( * ~ i x -8) + ( - 6 a7)

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N um eric al An a l y s is -11 | 2 2 9

= 2 6 -8 (x - 1) - 3(x* - 3x + 2) + — (x3 -6 x 2 + l lx -6 )6

*7 , on 2 193 , , Ax) - — x3 - 2 0* 2+ - ^ - x + l l . Ads

Further, if another point x - 5, Ax) “ 26, is added to the data, then ihe difference table a given a s :

12

345

y = / «

261841

26

Ay

- 8

-1 4- 3

- 2 5

A*y

- 6

11

28

A*y

1717

A*y

Since, the fourth differences is zero i e.t AV “ 0. hence die interpolating polynomial will i ethe same as before. Ans.

Example 3.53 : A third degree polynomial passes through the points (0,-1), (1,1), (2, I) and (3, -21, find the polynomial.

Solution. The forward difference table is :

X y « / W Ay A*y A5y

0 - l 91 l

it

n - 2

2 l uQ - 3 —13 - 2

u

X - Xn X ~ 0 Here xo = 0, h ~ 1 u~ —r— = —— - x

« 1By Newton’s difference formula for interpolation,

. x (x -l ) x (x -l ) (x -2) A*) - >o ■*-*Ay0 — — A2y0 +

2! 3*A3y0 ( v u = ? j

2 6

- -1 + 2x - i 2 + x — x3 - 3 x 2 +2x

6- l + | l x 2 + f 2 +1 - | ) x - l = - - x 9- -

l 6 ; I, 6 j 6 2x 2 + - x - l . A ns.

Example 3.54 :Flnd the cubic polynomial y(x) which takes the following values: y(0) - / , y(D - ft y(2) - /, y(3) - 1$, Also fin d y(4).

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230 t ENGM3&ING MATHENAT)CS~III


X y=/W Ay A*y

0 l-1

1 0 21 62 l 8

93 10

x-x^ x - 0 u - h ~ i " *

By Newton’s difference formula for interpolation, we have

. * . *(* -1 ) *9 x (x - l Xx-2 ) . a y(x) - y0 + *Ay0 + ot Aay0 +— — ~ 7---- -A3y0

2! 3! [■:«=*]

1+ X<-1)+j f c V (2) + (6)2 ' ' 6

= 1 -x + x2-x + x3“ 3x* + 2x “ xj - 2x2 + I .

Also,y(4) = 43 - 2 * 4* + I = 6 4 -3 2 + 1 “ 33.

Example 3.55 : Given 229249, H x)* 175212 an4

A b s .

A bs ,

M

j f(10) -40365, fin d f( l) .

Re-write the above data:

4 J

x ; 1- 10 4-10 7-10

l f ( x ) : 500426 329240 175212

I i i

(1) (2) (3)

By subtracting 0M2)» (2)—(3> and (3 )^10), we get

x : 1-3 4-6 7-9

y = f(x) : 171186 154028 134847

Now prepare cumulative frequency table:

x : 3 6 9

y = f(x): 171186

The difference table is :

325214 460061

X y Ay A?y3

6

9

171184

325214

460061

154028

134847-19181

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x - x o 1 -3To Rod/ If I) : Here x - 1, xo - 3, h = 3 u = —^— =^~s~ ~ ~

Now using Newton i forward interpolation formula :

Ax) = y0 +«Ayo+—^ y 1}A2yo+ ........

= 171186+ (-.667X154028)+ (~’66T}, (-19181)

= 171186 - 102736.7- 10663.5 ■=57785.7. Ans. Example 3.56 : The following data :

x : 0.51 0.52 0.53 0.54 0.55 y - f ( x ) : 0.5292 0.5379 0.5465 0.5549 0.5633

gpves the values of the probability Integral f ( x ) = £ e~** d x fo r certain equidistant

1 ' {*.333 .values o fx. Find the value o f ^ e ~*dx , using Newton’sforward interpolation

formula.Solution. The difference table is :

' Nw u » c a i. A »* y s »-II j 231

(x) y ~f(x) A y &'y &*y A4y

0.510.52

0.530.540.55

0.52920.5379

0.54650.55490.5633

0.0087

0.00860.00840.0084

i

i

p

P

° 8

§

I

s -0.00010.0002 0.0003

\

To f ind y at jc - 0.533 Le.,f(0.533) ;Here x = 0.533 , xo = 0.51, h = 0.01

x - V 0.533-0.51 _ “ h 0.0J. 3

Using Newton’s forward interpolation formula :

u a u ( u - l ) . ,

Ax) = y0 + Y7Ay° — 2! ° ......

X0.S33) - 0.5292 ^ ^ > , g ^ x ( - 0 0 01)1 *4

! = 0.5292 + 0.02001 -0.0001495 - 0.00001495 - 0.000007848= 0.549037702 * 0.5490.

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r

NtMHOCAlA*ALYSW-II | 2 8 3

To find/fc> * tx - 42 :Here x* 42 which Is near to 45, so taking %

x -x , 42 -45 -3x„ -45 , / .i f - * g “ 5

a Using Newton’s backward interpolation formula:

- 0.6

f/O ) - 204+(-.6)(-27)+(~ 6*-4 )><(2)+

31

(-.6X.4)q.4)

6x(0)

( (-.6X.4XI.4)(2.4) :(S) t (-.6X.4X1.4X2.4X3.4),.(ie)

24 120219. Ans.

Example 3.59 :From the following data, find f(7 .5) x : I 2 3 4 5 6 y : 1 9 27 64 125 216

Solution. The difference table Is :

7343

8512.

X y A y A*y * 'y A4y

1234

567

x» —8

l82764

125216343512

71937

6191127169

121824

303642

66

666

00

00

To find/ft) at jc • 7.5U.,f(7.5) :

Here x * 7.5, which is near to x „ -8, *■x -x . 7 .5-8

= -0.5.

.-. Using Newton’s backward interpolation formula:

fix) « ym+ uVym+ v2>» + •21

, ______ _ (-0.6X0.6) , 0 (-0.6X0.6X1.5) _ , nX7-5) * 612+(-0.5)(169)+-— — £x42+------ a , —1x6 + 0

3!= 512 - 84.5 - 5.25 - 0.375 - 421.875. Ann.

3.10 GAUSS’S CENTRAL DIFFERENCE FORMULA FOR EQUAL INTERVALSWe shall develop central difference formulae which are best suitable for interpolation near themiddle value of the tabulated set (table).

x : X -2 x -i xi x2y -f(* ) y~2 y -i y« yi yi

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234 f En g in eer in g Ma t ^ ca tk -'s -III

Then difference table is :0

X l -d i f ference ll-Dif f . I I I D i f f . rV-Diff.

x0 ~2h = x_2

x0 - h = x .l*0

x0 + h = x 1

x0 +2h = x 2

y-2

y-iyoV\

\>2

Ay_z(~ - 5 )

Ay0(=<^)

AVi(=^)

Azy_ 2(=<?2y-i)aV i<= <?2y0)

A2y0(=«f 2y1)

A3y .2{= * 3y.^)

A3y_i(= <?3y^)

A4y _2

(=<?4y0)

3.11 GAUSS’S FORWARD INTERPOLATION FORMULA FOR EQUALINTERVALS

U u (u - l) , 2 (u + l)(u)(u - 1) 3 Ax) = *> + Ay0 + ~ 2 ! A + ------- 31-------- A y ~l

. (u + l)u(u - l)(u -2 ) a4„ . , x-Xo-----------4!----------- y~2 + ........where, u= —^—•

Remark:This formula is applicable when u lies between 0 and 1 i.e., (0 < « < 1).

Example 3.60 :Using Gauss’s forward interpolation formula to evaluate yj$ given thatyj j «*18,4708,

y2s - 17.8144, y n ~ 17.1070, y}3 - 16.3432 andy 37 - 15.5154.

Solution. The difference table is ? x A 'y A4y

21=x _2

25 = x _ i

29 = x0

33 =x,

37= x2

18.4708

17.814417.1070

16.3432

15.5154

- 0.6564

-0.7074

r 0.7688- 0.8278

-0.0510

- 0.0564

-0.064

-0.0074

- 0.0076- 0.0022

To find y - f f x ) at x - 30, Le.,f(30) :

x-xn 30-29Taking xo = 29, h = 4, x = 30, then u = —^— = — - — = 0.25.

Using Gauss’s forward difference formula.

u 4. u (u - l) . . (u+l )u( t t - l )

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X30) *>>30= 17.1070 + (0.25X-0.7638) +

NuW-ioCAL Al«LYS*s4l | 235

f f c S f c a & H w n e o

a j g w b z a (-o.i6 24

= 17.1070 - 0.19095 + 0.00529 + 0.0003 - 0.00004or .fto =■ 16.92

3.12 GAUSS’S BACKWARD INTERPOLATION FORMULA

V ' u . (u+l )u *o (u+l)u<u-l) .«* ~ J (x )« % + J j A y - i + i - ^ A » y . 1+ i ------ ^ --------A*y _2

(m + 2)(u + 1)u (u -1)

Ans.

4!A4y-2 +.

where u -X - X q

Remark:This formula is applicable when u lies between - I and 0.

Example 3.61 :Find by Gatos's backward interpolation formula the sales o f a concern for the year 1936, given that

Year: 1961. 1911 1921

Sahfln thousands rupees): 12 IS 20 Solution. Taking the origin at 1941 and h * 10, also x = 1936

193127

1941 1951

39 52

x-*o 1936-1941

------- h ~ ----- 10---------- 05 [v -1 < u < 0]The i ? 'tral difference table Is :

X y A y A * y A * y A * y A ‘ y

1901=x_412

1911=x_3 153

2

1921=x_2 205

20

3

1931=x_! 27719 5

3 —i - 7

- 1 0

1 9 4 1 = x 0 89

A2

13 1

■“ *k

1951=Xj 52

Using Gauss backward formula:

Ax) = y o + j y Ay-i + ^ ~ A > y - i A»* »+■(o + l)u (u + l)u(u —1)

2! 31

A , m - 39 , ( - . S)« (I2) , M x a ) * « > f c & a . M ) A D

= 39 - 6 - 0.125 - 0.25 * 32.625 thousands. Ans.

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3.13 STIRLING’S FORMULA FOR EQUAL INTERVAL '

2 3 6 |

Let*-*o

U = —

Then Gauss’s forward interpolation formula is

u“ a u( u- l) (u + l)u(u -l) A*) - y0 + j j Av0 + 2~ & 2 y~i + ------- ~ ------- A*y-1

3!

(u + l)u(u - l)(u - 2) .+ ------ — , — A<y_, + .... ...(i)

and Gauss backward interpolation formula is

u- * (u+l)» ,, (u + l)u (u -l) Ax) = yo + Jj Ay-! + " i ^ i A3y_2

3!

. (w + 2)(« + l)u(u -1) lA ......................+ -------------------------A y.2 + ..... ...(2)

4!

Taking the mean of (1) and (2), we get

. , U [Ay 0 + Ay-i 1 u2 u (u -l) (u + l) \ A* y.t +A3y - 2 '

Ax) - yo+IT[— g— J "iT y“l+ ----- 3 !----- — ■*------- .

u2(ua -1 )

which is called Stiriling formula.

4!■A*,y_2 +. -(3)

It is used when « lies between - -7 and ~ .4 4

Example 3.62 :Employ Stir lin g fo rm ula to compute y i 2.2 fr om the follow ing table

(yx ~ I + log M.stox)

x ° : 10 II 12

10syx : 23967 28060 31788

13 14

35209 38368

{RGPV Dec. 2005/ Solution. Rewriting:

jP: 10 11

y,: 0.23967 0.28060

Taking the origin at xo “ 12°, A= I, x “ 12.2,

x - X q 12.2-12

h 1

12 13 14

0.31788 0.35209 0.38368

then u - 0.2,

which is lies between - •7 and —4 4

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The central difference table is

it NtMEBtCAl.AMALYSIS-II | 237

X y ~ f(x) Ay A 'y A3y A4y

X_2 =10

X_j =11

x0=12 %

Xj =13

x2 =14

0.23967

0.28060

0.31788

0.35209

0.38368

0.040930.03728

0.03421

0.03159

- .00365

-.00307

- .00262

.00058

-.00045- .00013

v Stirling formula is

yx=fix)= yo+l iAyo+Ay-i

2u2

+ i T 1

u(u2 - 1)

3!

A3y_j + A3y_2

2

U2(uz - 1) 4

Hereto = 0.31788, Ayo =0.03421, A>_ i = 0.03728

A V t = - .00307, A3y - ) = - .00045, A3>_2 = .00058, A4>_2 = - .00013.

.03421+ .03728' y,2 z - 0.31788 + — ^

(■2)[(.2): - lj -.00045+ .00058 (,2)2[(.2)2 Y\ .00013)24

= 0.31788 + 0.00715 - 0.00006 - 0.000002 + .0000002

= 0.32497. Ans.

Example 3.63: Given

0: CP 5° 10° IS0 2tf> 25° 3<P

tan 9 : 0 0.0875 0.1763 0.2679 0.3640 0.4663 0.5774

Find the value o f tan 16°, using Stirling formula.

Solution. Take origin at 15°

i.e., xo = 15°, and h - 5 and here/fr) = tan x

x - x 0 16-15u = 0.2.

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238 I ENGMEEHNe Mathemat ics-III

The central difference table is :

X y A y A*y A 'y A4y A 'y A‘ y

X_3=0

x_2 =5X_! =10

x0 =15

xj =20

x2 =25

x3 =30

00.08750.17660.26790.36400.46630.6774

0.08750.08880.09160.09610.10230.1111

0.00130.00280.00450.00620.0088

0.00150.00170.00170.0026

0.000200.0009

- . 0 0 0 2

0.0009 0 .0 0 11

Using Stinting formula,

« TAyo+Ay-il a 2

+ ~ ^ T ~ ^ [ A~ ' ' 2~ '3 j,~ ] + — ^ A4y_2 + ••••

, JÎS) —tan 15° —M C T + f f l g g j J ™ ] +< f-( .0 0 46 )

( (1.2)(.2)(-.8) ^.0017 + .0017~j | (.04)(-.96) (Q)

= 0.2867 (Approx.) A3.14 BESSEL’S FORMULA FOR EQUAL INTERVAL

The Gauss’s forward formula is

m - ? o + Z * y o + ^ A ’>., , ^ 1w » - d A V |

, (u +l)u(u-l)(u- 2)+ -----------J j ----------- A^.2+. .„

and Gauss’s backward formula is

M - * + + & |2 t 4 V - ,

, (u +2)(u+l)u(u-l)4I y~2 + ........- ( 2)

In equation (2) shift the origin to 1 by replacing a by (« - l) and adding 1 to each ajgument 0,•i- * i ......men wc get

Ax) =yt+(«-l)u&y0+ ^ +u(u - l)(u - 2)(u - 3) 2 ! 3 1

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240 | Ewoii a iBgjilijiip iin ir f i#

Putting all values from the table, we get

^25) = 3162 + (.25) x (382) + [ 74 + 66+ b 2 5 ) (.2 5 K -7 5 )x (_8)

2 L'

or >>25 = 3162 + 95.5 - 6.5625 - 0.0625 = 3250.875. Ans.

Example 3.65 : Apply Bessel’s formula to find the value off(27.4) from the table : x : 25 26 27 28 29 30

f f x ) : 4.000 3.846 3.704 3.571 3.448 3J33 Solution. To find y =fix)vtx = 21A : Taking origin at x = 27, h - I

x - x * 27.4-27U = h ~ = 1 ~ 0-4

1 , 3which is lies between — ~r •

4 4


X y * ‘y A*y A*y A4y 4 *y

x_2 = 25

x_j =26

*0 =27

-2 8

x2 =29

x3 =30

4.000

3.846

3.704

3.571

3.448

3.333

—.154

- .142

- .133

- .123

- .115

.012.009

.010

.008

- .003

.001-0 .002

.004

- .003- .007

:. The Bessel’s formula is

A x) = y 0 +uAyQ+u(u~ 1) I A ^ + A ^ q ]

2!

u(u

3! [A3Jl J +

(u + l)(u)(u - l)(u * 2)

/ 2 7.4) = 3.704 + M x ( - .l 3 3 ) + <-4)<°-6) x

4!

.009

r A<y.2 + A4y.i

L 2+ .01]

2 J

-.003)(.4X-.6X-.1) r i (1.4)(.4)(-.6)(-l .6) r .004+ (-.

6 24 I 2

(-.l)(1.4)(.4)(-.6)(-1.6) r + 1 — ----- x[- .0 07j = 3.6497. Ans.

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Numerical An a l y s i s-11 j 241

3.15 EVERETT’S FORMULA FOR EQUAL INTERVALS

x - x QLet u = — 7 — and u + v = 1 i.e., v = 1- u

h

viv2 - ! 2) ^ uiv2 - l 2) ^ - 2 2) (Then, fix ) ~ Wo + ---- — ---- A2 + ---------- — --------- A y .2

u(u 2 -1 2) a9 u(u2-12)(u 2~22) 4 + m +- - 3t a V o + — -------^ -------- - A4 -i + - •

This is known as Everett’s formula.

1 a 3It is applicable when u lies between — an(1 •j •

Example 3.66 :Apply Everett's formula to find the value o f log 337.5, given th at:

x : 310 320 330 340 350 360log,ox: 2.49136 2.50515 2.51851 2.53148 2.54407 2.55630

Solution. Let y - fix) = logj0 x

Tofindffx) a t x = 337.5 : Taking the origin atJCQ= 330 and h - 10,

337.5-330then ------------ j j ---------0.75

1 J 3Which is lies between — -7 •4 4


X y = f ( x ) A y A*y A4y A*y

x_2 =310

x_j = 320

x0 = 330

x l =340

x 2 =350

x3 -360

2.491362.505152.518512.531482.544072.55630

.01379

.01336

.01297

.01259

.01223

-.00043-.00039-.00038-.00036

.00004

.00001

.00002

-.00003.00001

.00004

Using Everett’s formula is

vW - 12) A, v(v* - l 2)(u2 - 22) . 4fix) = Wo + 2, ^ ^ --------- tfy-z

u(u2 - I2) 2 u(us - l 2)(u2 - 2*) A.+ m & y *+---------------------- A y- i +— -C)

where v = 1- u = 1 - .75 = .25

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Putting the values from the table in (1), we get

y(337.5) = .25 x (2.51851) + k 2j M ? g5 Z 1). x (-.00039)

+ C2aC^625_-l)(.062_5-4)120

+ (,75)x(2.53148) + — x(-.000386

t (.75)(:5 6 2 5- l ) ( .56 2 5 - .4 )x(00001

6= . 6 2 9 6 3 + . 0 0 0 0 2 - .0 0 0 0 0 0 2 + 1 .8 9 8 6 1 + . 0 0 0 0 2 + .0 0 0 0 0 0 1

/ ( 3 3 7 .5 ) = 2 . 5 2 8 2 8 .

Thus fi x) ~ log x^ lo g io ( 3 3 7 . 5 ) = 2 . 5 2 8 2 8 . Ans.

3.16 INTERPOLATION WITH UNEQUAL INTERVALS

If the values of x are given at unequal intervals, then the definition of differences for equal intervalsis not applicable.In such situations, we make use of a new kind of differences, called divided differences.

3.17 DIVIDED DIFFERENCESLet y = fix) be a function. Suppose/xn), fix\), fix -i),..... fix,,) are the (« + I) values affix)

corresponding to (n + I) values xo, xi, x2........ x„ of the argument x, where X| -x<),x2- x ] , ...... x»-x„_ | are not necessarily equal, in other words there is a case of unequal intervals.

The first order divided ’-fi'-v te o f fix) for the arguments xq, x\ is denoted by y[x0, X|] or by

f ( x i ) ~ f ( x 0)A f ( x 0) and is defined by “ 3 •xj Xj - x 0

Tl>us,*k X) - x0

Similarly , / x t,x 2) = J

The second order divided difference ofy(x) is

A 2 t t ~ \ f ( Xl > ^ 2) — ^ l )/ x o , x , , x 2) = ^ / < * 6 > = ----------- — -----------

_. .. - . A Z — \ /(^2j X3 ) —/(Xj, Xj)Similarly, f i x \ ,x 2, x3) = xf xs / W = x3 - x , etc'

Hence nth divided difference is

A n t t ~ \ f ( . x l * x 2 .......... X „ ) — / (X g , X j .......... X n _ i ).... *> = Xii Xn n * 0) = ------------------— -----------------

242 | En g in e e r in g Ma t h e m a t i c s-HI

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P roof. By definition :

/ ( * ) - / ( * o ) A*, xo) = ^ 7 ^

A x) =Ax o) + (X - * o ) /x , Xo) •••( I

/ ( X o > x , ) - / ( x , X o )Also Ax, xo, xt) = ^

f(x,x0)-f(x0,xj)or Xx , xo, xi ) =

=> Ax, Xo) = jfa o , X,) + (x - X| )y(x, Xo, X,)

Using this value o f^ x , xo) in (1), we get

Ax) =Axo) + (x-xo)Xxo,xl) + (x~xo)(x-x,)Ax,xo,x i) ...(2)Proceeding in this manner, we can get

/ X ) = f (X0) + ( X - X 0)Af(x0) + ( X - X 0) ( X - X l )A2f ( x 0) +.............+ ( x - x 0)(x - * i ) ......( x - x „ _ 1 )A“/(jc0)+ 0 ...(3)

Since (n + 1 ) values o f f ix) are given, => An*lf{x) = 0. Proved.

3 20 LAGRANGE’S INTERPOLATION FORMULA FOR UNEQUAL INTERVALS

Statements : Let A xq ), A x 0 , ..... A x*) be the values o f the function fix ) corresponding to the {n + 1)

aiguments xo, x j, X2, ..... . x„ not necessarily equally spaced.

( x ~ x i ) ( x - x z ) ..... ( x - x „ )Then y - fix) = 7} c }-----------r x

' yw ( x o - x j i x o - x ? ) .....( X o - X * )

, ( x -X q K x - X z ) ..... ( x - x „ ) K

(Xi - X0)(xl - X 2)......( X i~X„) 1

+ , ( x - X o ) ( X - X i ) ..... ( x - x n_x)

(*»-XoXx„-xt)......( x „ - X n. l )

Proof. Since (n + 1) values of y =AX) are given, so that we can expressed Ax) by polyno mial o f the nth

degree, which is as follows :

y =Ax) - - X]XX - x2) .. .. (x - x„) + A j(x - xoXx - x2) .... (x - x„)

+ ..... + A ix - x0Xx - X |) . .. . (x - x , _ (Xx -x ,+ j ) ... (x -x„ )

+ .....+ / f ^ x - x 0X x - x , ) .......(x -x ^ i ) . ...(1)Equa tion (1) is satisfied by the given pairs o f values (xo, yo). (x j.y i) , {x„,y„).

i.e., w hen x = x0, y -yo- Using this in (1), we get

yo = M x o - x iXx q - x 2) ...... (xo - x„)

214 | En g in e e r in g Ma t h e m a t i c s-1 JI

_____________ 1 ____________

A° ~ (Xo-X1)(xo-X2)..... (Xo-X,)^0

Similarly putting x = X| andy =y\ in (1), we get

___________ 1 ___________

A ' = (Xj-XoXXi-Xz)......(X l~Xn) yi 311(150 on-

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N u m e r ic a l A n a l y s is - I I | 245

Finally An -1

y *-(*» - X*n - *1)......(*n ~ **-» )

Using these values of Ao,A}........ A„ in (I), we get Lagrenge's interpolation formula:

_ ( * - * l X * ~ * a ) - ( * ~ * n )

y ~ W “ (Xo - X , X * 0 -*2)....(Xo - Xn)xô

) (x -jq l))(x -3 fe ). . . . (x -x ,l) ;c

(jfl -*o)(^ l -* * )—(*j - x n )

t (x-acb)(x-*i)....(x-x,,.1) }

(x, -Xo)(*n - x , )....(*,, -x„_,)

Example 3.67 :Given the values

x : 5 7 11 13 17

f ( x ) : 150 392 1452 2366 5202

Evaluate f(9), using (I) Lagrange’s formula, (ii) Newton's divided difference formula.[RGPV, June 201!]

Solution, (i) Lagrange's Formula :

= (« -X i) (x -x2) (x-x 8) (x -x4) 3 .

(Xo - xx)(x0 - x3 )(x0 - x3 )(x0 - x4) 0

+... , (* -* oX *- *iX *- *2X *-*3) :(xi ‘-x0)(xi ~x i)(x4 - x 2)(x4 - x 3) 4

Here xq = 5,xi = 7,X2 = 11, xj * 13, X4 * 17

and yo = 150, y\ =»392, y%- 1452, y3 = 2366, y4 = S202.Putting x = 9 and substituting the above values is (1), we get

*01 = ( 9 -7 X 9 - H X 9 - 1 3 X 9 - 1 7 ) 15Q ( 9 - 5 ) ( 9 - ll )( 9 - 1 3 ) ( 9 - 1 7 ) ^ ,

A } (5 - 7)(5 -11X 5 - 13X5 -17 ) (7 - 5)(7 - 1 1)(7 - 13)(7 -17 )

-xl452(9-5)(9-7)(9-13X9-17)(11-5X11-7X11-13X11-17)

(9-5)(9- 7 ) 0 - 1 1)(9-17)

(13 - 5)(13 - 7)(13 - 11)(13 -17)x2366

(9-5 )(9 -7 )(9 -l 1)(9-13) ,

(17 - 5X17 - 7X17 -1 1)(17 -13)

50 3136 3872 2366 578 aiA — + ------+--------------- +----- = 810.

3 15 3 3 5

An*.

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246 | Eng i neer ing Ma t h e m a t i c s-1 II

(ii) The divided difference table is :

X V = f(x) Ay * 'y A4y

x0 = 5150

xj =7 392121

24

x 2 = l l 1462 265457 32 11 0

X3 —13 2366709

42

x4 =17 5202

By Newton’s divided interpolation formula :

Ax) - f(x0) + ( x - x 0)Af(x0) + ( x - x 0) ( x - x :)A2f(x 0) +...... ...(2)Here, x - 9 and putting the values from the table in (2), we get

A9) = 150 + (9 - 5X121) + (9 - 5X9 - 7X24) + (9 ~ 5X9 - 7X9 - 11X1) + 0= 810. ' Ans.

Example 3.68 :Using Newton’s divided difference formula, flnd/(8 ), ff9) andf(15) i f x : 4 5 7 10 11 13

y = f ( x ) : 48 100 294 900 1210 2028

Solution. The divided difference table is : fRGPV Dec. 20061

x y = f(*) Af(x) A‘f(x) A'fix) A4f(x)

X0 =4

Xj =5

x2 - 7

x3 =10

x4 - 11

xs =13

Vo =48

yj =100

y2 = 294

ya =900

y4 = 1210

y5 - 2028

100 - 48 = 52

5 - 4

294-100 97 - 5

900 - 294

10-7

1210 - 900 = 31011-10 *

2028-1210---------------= 40913-11

202-97 01

1 0 -5310-202

11-7409 - 310

13-10

21-1510-427-2111-5

33-27

13-7"

l ~ l =011-4

^ =01 3 -5

v Newton’s divided difference formula : Ax) = f (xQ) + ( x - x 0)Af(,xQ) + ( x - x 0)( x - x 1)A2f(x0) +....... ...(1)

(i) At x = 8, then (1) becomes : A*) = 48 + (8 - 4X52) + (8 - 4X8 - 5X15) + (8 - 4X8 - 5X8 - 7X1) + 0

= 48 + 208+180+12 = 448. Ans.(ii) Atjc = 9 ,then(1 )becomes:

/ 9 ) = 48 + (9 - 4X52) + (9 - 4X9 - 5X15) + (9 - 4X9 - 5X9 -7 X 0 + 0= 48 + 260 + 300 + 40 * 648. Ans.

(iii) At x - 15, then (1) becomes^1 5 )= 48 +(15-4X52)+ (15-4X15-5X1 5)+ (15 -4X 15 -5X 15 -7X 0 + 0

= 48 + 572+ 1650 + 880 = 3150. Ans.

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v Using Newton’s divided difference formula :

Ax) = f(x0)+(x-x0W(x<))+(x- x0)(x - x, )A2 /(*o)

+(x~ x0)(x - xt )(x -xz )A3/(x0) + ........ (1)

Putting these values from the table in (1), we get J(x) = -11 + (x - 0)(3) + x{x - 1){-1) 4-xix- l)(x - 2) ( - i j

+x(x - l)(x - 2)(x -4K1 /4) +x(x - l)(x -2)(x -4)(x - 5) f- — \ 24;

= 11 + 3 x - x 2 + x - —(x2 -3 x + 2) + ~ ( x - l) ( x 2 -6 x + 8)4 4

- — x(x - l)(x - 2)(x2 - 9x + 20)24

.. . , x8 3x2 x x* 3X3 2 x3= l l + 4 x - x 2 ----- + ---------- + ------------+ 2x2 ------

4 4 2 4 24 4

3x2 0 x6 x4 49 , 39 , 5x+ -------2 x ------ + ---------- x3 + — xz -----

2 24 2 24 12 3

1 3 . 97 ,, 13 , 1 „= - —-x * + —x4 ----- x3 + — x2 — x +11

24 4 24 2 6

or Ax) = “ ^ (*6 - 18*4 + 97*3 ~ + 4* - 264>- Ant Example 3.71: F in d f(10) from the following data:

248 | Eng ineer ing Mathemat ics -!!!

x : 3 5 11 27 34

f ( x ) : -13 23 899 17315 35606

Solution. Divided difference table is :

X y=f(x) A/(x) A3f(x)

A4/M

xo =3

xj =5

x2 =1 1

x3 = 27

x4 =34

-18

23

899

17315

36606

23 + 13 tJ85 - 3 =18

899-23 = 14611-5

17315-899

27-11

35606 - 173 15 =26133 4 - 2 7

146-18 , ,

11-3 =161026-146

2 7 - 52613-1026 ,Q

34-11 =69

40-16 ^

27-369-40

34-11

0

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By Newton’s divided difference formula.

Ax)= /(jc0) + (*-JC0)A/(x0) + (x-jc0)(x-x,)A2/(^)+ (x - Xo )(* - xt )(x - x2 ) A*/(xo J+ ........... ( 1)

Putting above values from the table in (1). we get Ax) = - 13 + (j r- 3)18 + (x -3 X x -5)16 + (x-3 X x-5 X x-11)1

= x3-3 x 2-7 x + 8Differentiate, we get

f i x ) = 3x2- 6 x - 7 ...(2)

Put jc = 10, in (2) we get/XlO) = 3(10)2 -6(10) - 7 = 233. Ans.Example 3.72 :Findf(x) a polynomial in the power o f (x -5 ) from the following data:

N u m e ric a l A n a ly s js -II | 2 4 9

x : 0 2 3 4 7 9

f i x ) : 4 26 58 112 466 922Solution. The divided difference table is :

X y*/M A/W A*/W A‘/W

Xq =0

xi = 2

Xj = 3

x3 = 4

x4 = 7

*s -9

4

26

58

112

466

922

26-4-11

2-0

s * " 26- *3 - 2

112-584 - 3466-112 „ 0

7-4 =118

922 “ 466 =2289 -7

32-11 _ ----------=73-0

54 - 3 2 . , ,

4 -2118-54 ,,7 - 3 ' 16

228-118 ^9 -4

n " 7=l4-0

16-11

7 -222-16 j9 -3

0

0

Using Newton’s divided formula,

Ax) = f(x0) + (x - x0)hf(xQ) + (x - Xo )(x - x, )A2/(x0)

+(x - Xo)(x - Xj )(x - Xj )A3/(Xo) + ........(1)

Putting the values from the table in (I), we get Ax) =4 + (x-0 X H ) + (x - OXx- 2X7) + (x - OXx - 2Xx - 3X0 + 0

=> Ax) - x* + 2X2 + 3x + 4. -.(2)We know that Taylor’s series in the power of (x - a)

Ax) - ««>+^ « < ■ )+ 6 ^ n o )+ ^ 2 . A o)+.... ...(3)

[Here a = 5].

From (2), Ax) =x3 + 2** + 3x + 4 => /5 ) - 194then f i x ) = 3X2 + 4x + 3. => f ( 5 ) = 98.

8/11/2019 DK Jain M3


f i x ) = 6x + 4 => /"(5) = 34. f i x ) =6 => f" ($ ) = 6./*(*) =0

Hence (3) becomes :J(x) = 194 + 98(x - 5) + 17(x - S)2 + (x - S'p. Example 3.73 : Find the missing term by using Newton’s divided difference form ula:

250 | E n g in e e r i n g M a th e m a tic s !II

x ; 0 1 2 3 4

y : 1 3 9 — 81

Solution. The divided difference table is : (As given known data)

Ans.

X y=/W A*/W *S/W

o * I I O l2

Xj =1 3 1 -0

9 ' 3 =6

6 - 2 =22 -0

-10- 2-=2x2 =2

X3 =4

9

81

2 - 1 ’

T l - 364 - 2

^ - ^ io4 -1

4 - 0

Using Newton’s divided difference formula,

Ax) = ) +(x - x0 W( xq) +(x~xc)(x-xl )A2f(x0)+(x - X0 )(X - Xj )(X - Xo )A3f(x0) + ...

=> Ax) - 1+(X - 0X2) +(x - OX*- IX2) +(x - ox* - 1x* - 2X2)or Ax) = l+ 2x + 2x (x-l) + 2x(x-lX *-2)

= 2x3 - Ax2 + Ax + 1Put* - 3, we get/3) = 2(3)3 - 4(3)2 + 4(3) +1=31.Hence missing term is 31. Ans.

1 Example 3.74 :Show that f a ( £ ) - “ o6c<T

Solution. The Newton divided difference table is :

X i i 5 A*/W As/W

a

b

c

d

i

a1

b

1

c1

d

1 1

b a _ 1 b - a ba

1 1

c b _ 1

c - b be 1 . 1

d c _ * d-c dc

( -1)2 - J -abc

abed

From the table, we get g d ( £ ) - - • Ans.

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Example 3.75: Use Newton's divided difference formula to find the form off(x), given :

N u m e ric a l An a l y s i s -11 J 251

x : 0 2 8 6

f ( x ) : 648 704 729 792

[RGPV June 2006, Dec. 20M(N)jSolution. The Newton’s divided difference table is :

X y = /W 4/W **JW

k * 0 = °648

704-6482 5 - 2 8 _ 704 2 - 0

729-704 o- 3 - 0 0

x2 = 3

x3 =6

729

792

3 - 2 '792 -729 01

6 - 3

21-25 -

6 - 2

Using Newton’s divided difference formula,

X*) = /(*o) + (* - *o W O o ) + (x -x 0)<•* ~ x\ W f i xo)+•

= 648 + (jc-0X28) + (*-0X*-2X- I)= 648 + 28x - x 2 + 2 x

= - x 2 + 3Ctr + 648. Example 3.76 :Flnd the polynomial o fdegree three which takes the values given below :

Solution.

X 0 1 2 4 y 1 1 2 5

Using Lagrange’s formula,

Ans.

(x - l)(x - 2)(x - 4)

^ = (0 - l ) (0 -2 ) (0 -4 )

x l + £ (£Z 2K x-4 )x l + x - l ) ^ x2

l ( l - 2 ) a - 4 ) 2(2 - 1)(2 - 4)

x (x - l ) (x -2 )

4(4- l ) (4-2)x5

= — (-x3 + 9x2 - 8 x +12).12 Ans. Example 3.77 '.The following table Is given below. Find the Lagrange’s Interpolating polynomial flx) :

x : 0 1 2 5

f ix ) : 2 3 12 147

Solution. Here,

xo = 0 ,xt = l,x 2 = 2,x3 = 5 a n d ^ = 2,^ i=3,> ^= 12 ,^3 = 147.By Lagrange’s interpolation formula,

(x -x 1)(x -x 2)(x -x3) t t (x -x 0)(x -x i)(x -x 2)

■ x) “ (x0- x 1)(x0-x 2)(x0- x 3) 0 ...... (*s ~Xo)(*3 -Xj)(x3 -x 2) 3

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2 5 2 | E n g i n e e r in g M a th e m a tic s -! 1!

_ ( * - l ) (* -2 ) (* -5 ) (x -0 )(x -2 ) (* -5 )

(0 - l ) (0 -2 ) (0 -5 ) (1 -0X1-2 ) ( l -5 )

( x - 0 ) ( * - l ) (* - 5 ) „ ( * - 0 ) ( * - l ) ( * - 2 ) :

(2 -0 ) (2 - l ) (2 -5 ) (5 -0X5-1X5-2)= (x- lX *-2 X *-S ) 3x(x -2) (x-5) 2x(x _l)(x_ 5)

-5 4

. 49 (x)( x-l)( x-2 )

20

= x3 + x2~x + 2. Ans. Example 3.78 :The following values o f the function ffx) fo r values o fx are given :

f f l ) - 4,f(2) - S,f(7) * 5,f(8) - 4.Find the value of f( 6) and also the value o f x fo r which/fx) Is maximum or minimum.

Solution. Here, we haveXo = I j Xj = 2, X2 —7, X3 = 8 andyo “ 4,_yi ” 5 ,^ = 5 ,^ = 4,

Using Lagrange's interpolation formula, we get

( * - 2 X * - 7 ) ( x - 8 ) ( x - l ) ( x - 7 ) ( x - 8 ) : : 6

" x ) ~ (1- 2)(1- 7 ) ( l - 8 ) ( 2 - l ) ( 2 - 7 ) ( 2 - 8 )

(x - l )(x - 2 )(x~8) . - (« - l ) (» -2 ) (x -7 )

(7 - l ) (7 -2X7-8 ) (8 - l ) (8 -2 ) (8 -7 )

On simplification, we get

-x 2 3 8

* > - - r +r " s •(1)

6s 3 „ 8 34Put,x = 6, the n /6)= — +—x® + - = ~ =5.66. Ans.

Again, for maximum or minimum of /(*). - - 0 , gives

dy -2x 3 _

dx 6 2 ^

( d2f( x) )=> x =4,5 and . 2 = - 2 < 0

v /*-4.B

Hence, fix) is maximum at the point x = 4.5. Abs. Example 3.79: Given loga 654 « 2,8156, log ft 658 - 2.8182, log 19 659 “ 2.8189 and logio 661 m

2.8202. Find logit 656.

Solution.Given, xo * 654, Xi = 658, X2 = 659, X3 “ 661 and yo = 2.8156, y\ = 2.8182,

y2 —2.8189,y j = 2.8202.

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N u m er ic a l An a l y s h J I I 253

By Lagrange’s interpolation formula, we have

(656 - 658)(656 - 659)(656 - 661) /0 Qt cg,X656) -log ,0 (656) - (664 _ 65s)(654 - 659)(654 - 661)

t (656-654X 656-6S9X656 - 661)

(658 - 654)(658 - 659)(Go8-661) '

t (656 - 654)(656 - 658)(656 - 661) g

+ (659 - 654)(659 - 668)(659 - 661) '

(656 - 654)(656 - 658)(656 - 659) 82Q

(661-654X661-658)(661-659) V'

(-2X -3X -5) ^ . s i s e ) ^ 2^ ) (2.8182) + (2-8189)(-4)(-5)(-7)

log io (656) = 2.8169. Example 3.80 .-Express the function

4(-l)(-3) 5(l)(-2)

2(-2)(-3)

7(3)(2)(2.8202)

Ans.

m3x* + x + J

Solution. Given

( x - l ) ( x - 2 ) ( x - 3 )

as a sum o f partial fractions by using Lagrange’S interpolation formula.

3x2 + x +1

= (x - l ) (x -2 ) (x -3 )

Consider (x) = 3X2 + x + 1and tabulated its values for* = 1,2, 3, we get

x : 1 2 3

3x2 + x +1: 5 15 31

Using Lagrange’s interpolation formula, we get

# w . M ( 5) + f c M ( 1 5 ) + i ^ M <31)( l -2 ) ( l -3 ) (2 -1) (2 -3) (3—1) (3 —2)

t ( jc) = | (x - 2)(x - 3) -1 5 (x - lXx - 3) + ~ (x - l)(x - 2)it

m

2

15 31

&x) ~ (x - l ) (x -2) (x-3 ) *" 2 (x - l ) x -2 2 (x-3)

Example 3.81: By means o f Lagrange’s interpolation formula , show that :

Ans.

0) y<>~ ~{yi - * / ) - ( > - / -y-s)]

(U) y i ~y» - 0.3 (ys -y . j) + 0.2 (y.j -y ^ ) .

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Solution. Here yz - f ( x ) i.e.Argument * : - 3, - I, 1,3.By Lagrange’s interpolation formula:

( j c + IX* - IX* ~ 3) ( * + 3X* - IX* - 3) y = f t x \= ---------- —— —£ -------- — x v , + -----------— ------ — ---------- x y ,

y (-3+1X -3-1X -3-3) 3 (-1 + 3X -1-1X-1-3)(x + 3X * + 1 X * ~ 3 ) + ( * 1 3 X x + 1 X £ - 1 )

+ (I + 3X1 + W - 3 ) ” (3+ 3X3+1X3-1)

(x + lX * -lX * -3 j (* + 3X* -1 Xx -3 )

y (-4S) ,_3 (16) '

(x + 3X* + 1X *~3). . | (* + 3X* + lX * - l) .+ (-16) 1 (48) ^

Put x ~ 0, we get

1 9 9 1 y o - i6 -f ~3+ i 6 >'- ' + 16>'> l 6 y *

or yo = + y_i) —^[O 'j - y i) - O '- ! -y- i)] Proved.

(ii) The argument arex : - 5 , - 3 , 3 , 5 .

Using Lagrange’s furmula:

(jc-t-3X*~3X*~S) v | (* + 5X* ~ 3X* ~ 5) ,

* ( -5 + 3X-5-3X-5-5) 5 (-3 + 5X-3-3X-3-5) 3

(x + 5X* + 3X*-5) (x + 5X* + 3X *-3)+ ------ — ----- — ----- - * y-i + ------ — --------------x

+ (3 + 5X3+ 3X3-5) (5 + 5X5 + 3X5-3)Putx= I, we get

yi = - 0.2 y-s + 0.5 y - i+ y i - 0.3y} or yi - y 3 - 0.3 (y5-> ,3) + 0.2 (yj - jlj)- Proved.

3.21 INVERSE INTERPO LA TION JRGPV June 2003}

The method of determining the value of the argument xt corresponding to the given value of the

function when the function lies between two tabulated values, is called inverse interpolation. Thus My =J{x) be the given function then to find the value of x corresponding to the value of y, when y lies between two tabulated values ofy, is called inverse interpolation.The method o fdetermining the inverse interpolation is as follows :

3.22 LAGRANGE’S INV ERSE INTERPOLATION FORM ULA [RGPV June 20031

The Lagrange’s inverse interpolation formula is given by

( y - y i X y - y j )- ( y - y . ) _ . ( y - y o X y - y i M y - y » ) „ ,


8/11/2019 DK Jain M3


N u m e r ic a l An a l y s is -11 | 255

Example 3.82 :Apply Lagrange’s inverse interpolation to find to one decimal place, the value o f x

when y = 13.6, given the following. [RGPV Dec. 2002]

X 30 35 40 45 50

y = f ( x ) •• 1S.9 14.9 14.1 13.3 12.5

Solution. The Lagrange’s inverse interpolation formula is as given below :

(j'~yi)(y->2>... ( y - y n) „ . (y-jbXy-jfc).... i y - y*)_ - ■■i■■ II —■ ■■-Ja -f- ■■ .... mm I ■ . ^ 1

(yo-y\)(yo - t t ) .... (yo -y «) U -yo)(y1 - vt) .....Cxi - y . )

j ( y - y 0)Qy-y i) .... x

(;y»- 0 )(y, - yi)... (y* - yn-i)We have, xq = 30, x\ = 35, x 2 = 40, xj = 45, X 4 = 50 and

yo = 15.9, yj = 14.9, w = 14.1, j-j = 13.3, yA= 12.5. Also y= 13.6Putting the above values in equation (1) we get

(13.6 - 14.9)(13.6 -1 4 .1)(13.6 - 13.3)(13.6 -12 .5 ), 3Q

* " (15.9-14.9)(15.9-14.1)(15.9-13.3X15.9-12.5)X

: (13.6 - 15.9X13.6 - 14.1)(13.6 - 13.3)(13.6 -12.5) „ g5

+ (14.9 - 15.9)(14.9 -1 4 .1)(14.9 - 13.3)(14.9 -12.5 ) X

t (13.6 - 15.9)(13.6 - 14.9X13.6 - 13.3)(13.6 -12.5) ^

+ (14.1-15.9)(14.1-14.9)(14.1-13.3)(14.1-12.5)X

(13.6-15.9)(13.6-14.9)(13.6-14.1X13.6-12.5) ;15

+ (13.3 - 15.9X13.3 - 14.9>(13.3 - 14.1)(13.3 -12.5) X(13.6 - 15.9XI3.G - 14.9)(13.6 - 14.1)(13.6 -13.3)

-.(I)

x50(12.5 - 1 5.9)(12.5 - 14.9)(12.5 -1 4 .1)(12.5 -13.3)

= 0.45833 - 4.32373 + 21.41276 + 27.7954 - 2.14700=> x = 43.195. Am.

Example 3.83 :Applv Lagrange’s formula inversely to find a root o f the equation f(x) “ 0, when f (30) = - 30, f(34)= - I3,f(38) - 3 andf(42) «=18, Or given table:

X 30 34 38 42

f ( x ) : -30 -13 3 IS

Solution. We havexoÔ,*] = 34, x2 = 38, xj =42,yo = -3 0, y, = - 13, >^ = 3, 3 = 18.We are to find-the value of x when y =fix) - 0. Using Lagrange’s inverse interpolationformula

(0 + 13)(0 - 3)(0 -18) OA , (0 + 30)(0 - 3)(0 -18)x = (-30 + 13)(-30 - 3)(-30 -18)

■x30 +(-13 + 30)(-13-3)(-13-18)

x34

(0 + 30)(0 + 13)(0-18) £ (0 + 30X0 + 13X0-3) . g

(3 + 30)(3 +13)(3 -18) (18 + 30)(18 + 13)(18 - 3)

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2 5 $ ^ En g in e e r in g M a ih e m a t ic s -I II

13x3x18x30 30x3x18x34 30x13x18x38

17x33 x 48 + 17x16 x 31 + 33x16x15x = - 0.7821 + 6.5323 + 33.6818-2,2016 - 37.2304.

Hence the required root of the equation/*) = 0 is 37.2304.

Example 3.84 :Fbut the value o f x when y - 5, given that:

30x13x3x42

48x31x15

A n s.

X 1.8 2.9 2.2 2.4 2.6

y = f ( x ) : 2.9 3.6 4.4 S.5 6.7

Solution. We have, xo = 1.8, Xi = 2.0, x2 = 2.2, xj = 2.4, x* = 2.6, y o = 2.9, y \ = 3.6, >5 = 4.4, y^ = 5.5, y 4 = 6.7 and given y = 5.By Lagrange’s inverse interpolation formula

( y - y i X y ~ y 2) ( y -y a ) ( y -y <) .. ^ (y - y0)(y- y2)(y- ysXy- y*)

x ~ (yo-yi)(yo -y*)(yo-ysXyo- y*)

(y-yo )(y-yi)(y-y3Xy - y4), ~ ~ ..................................... , ( y - J b X y - r iX y - ^ X y - t t)(yi - jbX** - yiXy2- - y*) 2 Cya- y<>)(ys ~ y \)( s - yz)(j3 - y*)

x3

1 (y -y o X y -flX y - jfrX y - jfr)

(y4 - yo)(y* -yi)(y* -y2)(y* -y*) 4“ 1}Putting above values in (1), we get

(5-3 .6)(5 - 4.4X5 -5.5)(5 - 6.7)

* (2.9 - 3.6)(2.9 - 4.4)(2.9 - 5.5)(2.9 - 6.7)

(5-2 .9)(5 -4.4X 5 - 5.5)(5 - 6.7)

(1.8)

(3.6 - 2.9)(3.6 - 4.4X3.6 - 5.5)(3.6 - 6.7)

(5 - 2.9)(5 - 3.6)(5 - 5.5)(5 - 6.7)

(2.0)

(4.4 - 2.9)(4.4 - 3.6)(4.4 - 5.5)(4.4 - 6.7)

(5 - 2.9)(5 - 3.6)(5 - 4.4)(5 - 6.7)

(2.2)

(2.4)

(2-6)

(5.5 - 2.9)(5.5 - 3.6X5.5 - 4.4)(5.5 - 6.7)

(5 - 2.9)(5 - 3.6)(5 - 4.4)(5 - 5.5)

+ (6.7 - 2.9)(6.7 - 3.6X6.7 - 4.4)(6.7 - 5.5)

(1.4)(0.6)(-0.5)(-1.7)(1.8) (2. l)(0.6)(-0.5)(-1.7)(2.0)(1.5)(0.8)(-l.l)(-2.3) + (0.7)(-0.8)(-1.9)(-3.1)

t (1.4)(0.6)(-0.5)(-1.7)(1.8) ] (2.1)(0.6)(-0.5X~ 1.7)(2.0)

(1.5)(0.8)(-L l)(-2.3)

x = 1.285

3.036 3.29842.617. (Approx).

2.142 5.498 7.1971. + —-— + - 2.293

3.036 6.5208 32.5128

(0.7X-0.8)(-1.9)(-3.1)

(2.1)(1.4X0.6)(-0.5)(2.6)

(3.8)(3.1)(2.3)(1.2)

Ans.

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Example 3.85:Apply Lagrange's method to fin d the value o fx when f(x) “ IS from the given data: fRGPV June 2005 and Dec. 2011 J

N u o m c a i. Am a l t h s -H { 2 5 7

* 5 6 9 11

y = f ( x ) : 12 13 14 16 Solution. We have, jro = 5, X| - 6, v> = 9, xj * II,

Vo = 12,yi = 13, 2 = 14,>>3 = 16 and given y - 15. Tofindx:By using Lagrange’s inverse interpolation formula

( y -y i ) {y -y ' i ) { y -yz ) x , (y-y0Xy-yaXy-ya)

x ~ (yo-yiXyo-yzKyo-ys) 0 (* -yo)(yi -ysXyi -ya) 1

, (y-ypX y-yiX y-ya) , (y-y o)(y -yi) (y-y z)

(y2 -yo)(y2-yi)(y2-y3) (ys-yo X ys-yiX ys-ya) 3

Putting given values from the table in (I), we get(15-13)(15-14)(15-16) t (15-12)(15-14)(15-16)

•c “ (12-13)(12-14)(12-16) (13-12X13-14X13-16)

( (15-12)(15-13)(15-16) (15-12)(15-13)(15-14) y U

(14 - 12)(14 - 13)(14 -16) (16 - 12X16 - 13X16-14)

16 27 11=> x = 9 + — + — =9.125. Ans .o * 4

3.23 NUMERICAL DIFFERENTIATION

Numerical differentiation is a process where we differentiate a suitable interpolation polynomialin place of actual function when we are given a set of values of that function and find the approximate numerical value of the derivative (or derivatives) of a function.Working Rule : (i) Fit up an interpolation polynomial to the given set of values of die functionusing an appropriate interpolation formula.(ii) Then substitute the values of x to find the derivative (or derivatives) at that point, if desired.

dy Remarks: (i) If the values of x having an equal interval and ^ is required near the beginning of the

data, we use Newton’s forward formula, and if it is near the end of the data, we use Newton’s

dy backward formula, and for the values near the middle of the table, ^ is calculated by means of

Stirling’s or Bessel’s formula.(ii) If values of x have not an equal interval, then we use Newton’s divided formula to representthe function.

324 DERIVATIVES OF>>*= fix) BASED ON NEWTON’S FORWARD INTERPOLATION

FORMULA

(01

" hAy0 - ~ A2.y0 + ~ A Vo - \ A4y 0 + £ A V o - ~ A6;y0 + .

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(ii)

258 | E n g i n e e r in g M a th e m a tic s - II I

{*2L\ 1[dx2jr=xo =h2

(d*y) 1

" * 3 .

A2y0 - “ A4y0 - | A5y0 + A6y0 - .

AVo A4;y0 + ^-A3;y0 ~.

325 DERIVATIVES OF y - J{x) BASED ON NEWTON’S BACKWARDINTERPOLATION FOR MU LA:

(i)1

[ d x , ~ h

d2y ) 1dx2 j *=*n = h?

v 3 + | v ^ n + l v ^ „1(iii) {dx>l__Xn -

326 DERIVATIVES OE.y /.x) BASED ON STIRLINGS OR CENTRAL DIFFERENCEFORMULA:

' & ] = I

dy

(i)

(ii)

(iii)

( Ay0 +Ay-1 1 ( A3y_t +A3y_2 I l f A*y_2 + A6 _3 "

I 2 j “ 6 2 J 301 2 J

dx L

dx , ft3

f A3y .t + A3y_2

Idy d*y

Example 3.86 : Find and a t x - 4 , using

x : 1 2 4 8 10

y : 0 1 S 21 27

Solution. Divided difference table is :(RGPV Dec. 2901 A June 2006J

X y Ay A 'y A*y A4y

*0 =1Xj =2

x2 =4

x3 =8

x4 =10

015

2127

1243

1/31/3

- 1/6

0- 1/16

-1/144

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By using forward divided difference formula :

Ax) = y0 + (* - i)Ayo +(* - 1)(* - 2W y 0 + ( x - l) (x - 2)(x - 4)A3>0

+(x - 1)( jc - 2){x - 4)(x - 8)A4y0 + ....

=* Ax) = 0 + < * - l ) .l+ < x - l ) < x - 2 ) |+ 0 + ( x - l ) ( x - 2 ) (x - 4 ) ( x - 8 ) ( ^ }

= ( x - l ) + i ( x 2 -3 x + 2) + - ^ ^ ( x 2 + 1 2 x + 32)3 144

= x —1 + —— x + 4 + x4 -15X3 +70x + 120x + 643

211or y x4 -15 x3 + - x 2+120x + 67 ...(1)

V

Differentiate : — = 4x3 - 45X3 - 45x2 + —- x +120 ...(2)ax <3

Nu m e r ic a l An a l y s i s -II | 259

= 252.8. Ans.Ar-4

d?v „ 422Again, differentiate (2) : - 12x2 - 90x +

^ = 12(4)2 - 90(4) + “ = - 27.33. Ans.

Example 3.87: Values o fy as plotted against x are :

x : 0.5 0.75 LOO 1.25 1.50

y : 0.3521 0.3011 0.2420 0.1827 0.1295

Find ~ at x = 0.5 dx

olution.

[RGPV June 2002}

To find f'(x) ai x - xo = 0.5: Using derivative of forward difference formula.

f& \ , d x )

X~XQ

Ay0 - - A2 jy0 + “ A3>0 - -7 A4y 0 + .

The difference table is : Here h ~ 0.25

••(I)

X y A 'y A’ y A4y

0.50.751.00

1.251.50

0.35210.30110.2420

0.18270.1295

- 0.051-0.0591

-0.0593- 0.0532

- 0.0081-0.0002

0.0061

0.00790.0063

- 0.0016

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Hence {I) becomes :

260 | E n g i n e e r in g M a th e m a tics-I II

f dy] 1

«-o.» ' 0.25

0.0081 0.0079 0.0016- 0.051 + --------- + ---------- + ----------

Hence

0.25

= -0.17567

(- 0.051 + 0.00405 + 0.002633 + 0.0004)

= -0.17567

0.25

- 0.0439167

Ans.

Example 3.88: Find — a tx = 1.5 from the following table :

dx

x : 1.5 2.0 2.5 3.0 3.5 4.0

y • 3 .375 7.0 13.625 24.0 38.875 59.0 IRGPV, June 2008 & June 2011 j

Solution. To find f \ x ) at x = xo - 1.5 :Using derivative of forward difference formula.

The difference table is : Here h - 0.5

X ... y Ay A2y AJj A f y15 3375

3.62S2.0 7.0 3.0

6.625 0.7525 13.625 3.75 0

10375 0.753.0 24.0 45 0

14-875 0.7535 38.875 2.25

4.020.125

59.0

From the table,A>>o= 3.625, Atyo = 3.0, A3y0 = 0.75, A^0 - 0Hence (I) becomes:

-ii o-sL3.625 - - x 3.0 + - x 0.75 - - x 0

2 3 4

_ 237505

= 4.75 Ad s .

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Numerical Anal ys is-! I | 2$1

Example 3.89 :Findf%x) at x - 0.1 from the following table :

x : 0.1 0.2 0.3 0.4

y = f ( x ) : 0.9975 0.9900 0.9976 0.9604

Solution. Tofindf' fx) at x **0.1 : So taking xo = 0.1

The forward difference table is :

X y ~ f(x) Ay A*y A*y

x 0 =0.1

0.20.30.4

.9975

.9900

.9976

.9604

- .0075-.0124- .0372

- .0049- .0048

0.0001

Using derivative of forward difference formula

( * L ........ ]

Here h = .1, Ay0 = - .0075, A2y0 = - 0049, A*yo = 0.0001, and x0 = 0.1

/'(0.1) = 7TT - 0075- i (-.0049) + 1 (.0001)O.l A o

Example 3.90 (a) : find f%x) and f% x) a t x m6, given that

= -0.0502. Ans.

x .* 4.5 5.0 5.5 6.0 6.5 7.0 7.5

y ~ f ( x ) : 9.69 12.90 16.71 21.18 26.37 32.34 39.15

olution. To fin d f%x) andf 'f x) at x * 6: So taking xq = 6

X y

4.5 9.695.0 12.95.5 16.71

x„ = 6.0 21.18

6.5 26.37

7.0 32.34

7.5 39.15

Ay

3.213.814.475.195.976.81

A*y A*y

0.60 0.060.66 0.06 00.72 0.06 00.78 0.06 00.84 0.06

A4y

Using derivative of forward difference formula :

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2 6 2 | EngemesmNg M athem at ics - !II

or

and

or

5.19 - i (0.78) + (0.06)12 3 J™ = a e

= 2(5.19 - 0.39 + 0.02) = 9.64. Ans.

d 2ydx2 h2

A2y0 - A3y0 + A*yo +••••

/"(«) = T T ^ t0-78 - 0-06] =4(0.72) = 2.88. \j.Ab A d s .

To fi n d f ’ (x) a n d f" (x) a t x - 1.6: So taking x„ = 1.6 and using derivative of backwardformula.

Here Vy„ =0.281, ^ y „ = -0 .018, VV„ = 0.005, V*yn = 0.002, V% = 0.003,

V*y„ = 0.002Then, we have

( £ L . - i K + i v ' * 4 v V * + + i v V ' +- ]

S L . ' w [ “ 8 , + r ^ 0 , 8 ) t r (0005)]

+ - x (0.002) + - x (0.003)+- X (0.002)4 5 6

= 2.7476.

Also, = - y f v2>«+ v V (I+ — V4y » + -V 5>'B+— VV),^ A-*. A L 12 6 ” 180 J

(0.1)2 L-0.018 + 0.005 + — X(0.002) + ~ x (0.003) + — x (0.002)

12 6 180

= -0.638.

Hence / " (1.6) = -0.638.

Example 3.90 (b): Given that:

Ads .

x : 1.0 1.1 1.2 1.3 1.4 1.5 1.6

y ♦* 7.989 8.403 8.781 9.129 9.4S1 9.750 10.031

dv d y Find -j- and — r at x - 1.1.

dx dx2[RGPV, Dec. 2011 J

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Solution. The difference table isgiven below :

N u m eb jc a l A naiy& is-U | 2 6S

X y A y A*y A ^ _ A4 y A5 j > A*y1.0 7.989

0.414

1.1 8.403

0378

-0.036

0.006

12 8.781

0348

-0.030

0.004

-0.002

0.001

13 9.129

0322

-0.026

0.003

-0.001

0.003

0.002

1.4 9.451

0299

-0.023

0.005

0.002

15 9.750

0.281

-0.018

1.6 10.031

To f in d f ’(x) and f" (x) atx = 1.1 : So that taking xo - 1.1Thenyo = 8.403, AVo= 0.378, 5A*y« = - 0.03. A^0= 0.004, A4>’o = - 0.001. and A5y0= 0.003Then, we have

(& !.„ ■ 4 ^ + j aV«■ ? + ■ i aV,+"".($L\ = 0378 + - x 0.03+ - x 0.004+ - x 0.001+-x0.003U J „ n 0.1 L 2 3 4 5 J

= 3.9518. Ans.

and 1 r*2 *3 11 a4 5 s 1 3 7 . 1

- r I . ¥ + - 4 » - ? a * — 4' *«■*<) W

= _ L . r - 0 . 0 3 - 0.004 + — x (-0 .0 01 ) - 1 x (0.003)]

* L u 12 6 J

1-------T . . . 0.011 0.015= -----1—0.03 —0.004---- — ------- - —

0.01 L 12 6 ,

= - 3.742 Ans.

dy d*y

x : 1.0 1.2 1.4 1.6 1.8 2.0 2.2

y : 2.7183 3.3201 4.05S2 4.9530 6.0496 7.3891 9.0250

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2 6 4 | E n g m e e r »*g M a t h e m a t ic s -111

Solution. To findf% x) andf"{x) at x « 2.2 : So taking x„ - 2.2 and using backward difference formula:

X y Ay A*y 4*3- A4y A*y 4 'y

1.01.2

1.41.61.82.0

Xn = 2.2

2.71833.3201

4.05524.95306.04967.38919.0250

0.6018

0.73510.89781.09661.33951.6359

0.1333

0.16270.19880.24290.2964

0.02940.03610.04410.0535

0.00670.00800.0094

0 00130.0014

0.0001

Derivative of backward difference formula :

(* L]

2 ‘"* 3 4 *"* 5 6

[Here h = 0.2, x„ = 2.2, Vy„ = 1.6359, V*yn = .2964. V*y„ = .0535, V*y„ - .0094, V ^„ =.0014 and V€y„ = .0001]

■■ A 2-2)= £ 2 1.6359 + - (0.2964) + - (0.0535) + - (0.0094)2 3 ‘1

= 9.0228. Example 3.92 : Find the value offfx ) at x

-(0.00 14) + -(.0001)5 6

A n s.

.04 from the following table

x : 0.01 0.02 0.03 0.04 O.OS 0.06 f ( x ) : 0.1023 0.1047 0.1071 0.1096 0.1122 0.1148

{RGPV Dec. 2005]

Solution. To fin d f'(x ) at x —.0 4 : So taking x„ = 0.04 and using backward difference formula

x y= f(x ) Ay A* y A 'y A4y

0.010.020.03

=0.040.050.06

0.10230.10470.1071

0.10960.11220.1148

0.00240.0024

0.00250.00260.0026

00.0001

0.00010

0.0001

0-0.0001

-0.0001

-0.0001

Using derivative of backward difference formula

( S L - i [V y „ + ^ V 2 y „ + - V ^ „ + .

2 3

[Here h - .01, Vy„= .0025, V*y„ = .0001, V - 0.0001, and = .04]

f ( 0 M ) = -^ .0 0 2 5 + | ( .00 0 1 ) + ~(0.0001) = 0.255. Ans.

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NdMMCKk'MM^MB-n | 265

Example 3.93 :Flnd the first and second derivatives o fthe function given below at the point x - 1.2 :

Solution.

x : i 2 3 4 5

y «* 0 1 5 6 8

[RGPV June 2003]

To find f%x) and f fx) a t x m 1.2 : Which is near to xq - 1. So using Newton’s forwarddifference formula:

u+ •r -.(I)

where X - X q

u = —j-— => x * xo + uh. (i.e., xq + uh = 1.2)

X ~~

:

-• yi

$

I I Ay A 'y A* y A*y

1 0 12 1

1 A 3

3 541 -3

—o A 10

4 612

15 8

From (1), rewriting

A*o +uh) = f (xo)+ j Aft*o) + A* rt*o) 3 ,u . , , x u ( u - l) . u ( u -l) (u - 2 )

A3/(x0)

^ u (u - lX « - 2 )( u r 3) (2)

Differentiate (2), w.r.t. u, we get4!

(3u2 -6 u + 2)

6A3/(*o)

(4u3 - 12u2 + 22u - 6) . . . , ,+ i ------------- — -------------- A4/( x 0) ...(3)

24Again differentiate w.r.t u, we get

h>r(*o + uh) = A2/( x0) + (u - l)A3/’(x0) + (e“2- A V(*o)12

••(4)

x - x 0 1.2-1Putting, A = l ,x 0 = l ,u = —^ j

and (4), we get

(0.4 - l )x 3 [3(0.2)2 - 6 x (0.2) + 2]

= 0.2; and x = xq + uh = 1.2, in equations (3)

f( 1.2) = 1+ -6

x ( - 6 )

(4 x (0.2)3 -1 2 x (0.2)2 + 22 x (0.2) - 6} x 10

24= 1- (0.9) - (0.92) - (0.853) = 1 - 2.673 = - 1.673. Ans.

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And /" (1 .2) = 3 + ( 0 .2 - l) (- 6 ) + -— (— ^ 2—(— ^ xlOXa

= 3 + 4.8 + 0.33 = 8.13. Ans. Example 3.94 :FUut the 1st and Ilnd derivative of the function tabulated below at point x = 1.1:

2 6 6 | E n q m e e b jn g M /u h e n a tk s - I II

* 1.0 1.2 1.4 1.6 1.8 2.0 f i x ) 0 0.128 0.544 1.296 2.432 4.000

[RGPV June 2009 and Feb. 2010] Solution. Since at point j c = 1.1 lies near the 1.0, so we shall use Newton’s forward formula.

j w - m * ^ * ......2! 3!

•-(I)

x - awhere u - — r — =>uh= x - a=>x = a + uh = 1.1

hThe Difference Table is :

X I I I Af(x) A’ fix ) 1 A3 f(x) A4fix)

1.0 00.128

1.2 0.1280.416

0.2880.048

1.4 2.5440.752

0.3360.048

0

1.6 1.2960.136

0.3840.048

0

1.8 2.4321.568

0.432

2.0 4.000From (I)

f a + uh) = f(o) + uAf(a) + — A2/(o) + — ---- --- - 2 — A3f(a) Z o

Differentiating twice w.r.L V we get

h f\ a + uh) = A / (a) + A2f(a) + A*f(o) Z D

h?f ' (a + uh) - A2/ ( a ) + (u - l) A 3/ ( a )

Here, h = 0.2, a = 1.0, so that u = 0.5 and a + uh = 1.1Hence (2) and (3) become :

'••(2)

A M ) - a *

l

and /'(1 .1) =

0.2

1

o.i28+ x o.288 + (g ^ ^ a s ^ ) k ao48

[0.128-0.002] =0.63.

(0.2)![0.288 +<0 .5-1)(0.048)] = 6.60. Ans.

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Nomemcal An a l y s is-!! (267

Example 3.9S :A slider in a machine moves along a fixed straight rod. Its distance x cm. along the rod is given below fo r various values fo r the time t seconds. Find the velocity and acceleration o f the slider, when t» 0.3 second:

t: 0 0.1 0.2 0.3 0.4 0.S 0.6

x : 3013 31.62 32.87 33.64 33.95 3X81 33.24

[RGPV Dec. 2004J

Solution. Here the velocity (dx / dt) and acceleration ((fix i dt *) are to be determined at the point t =0.3 sec. which is near the middle of the table. Therefore, centra! difference formula will beused.The central difference table is :

X y Ay 4*y A*y A4y A 'y A(y

t 3 =0

t .2 =0.1 L i =0.2

to =0.3

tx = 0.4

t2 =0.5

t3 =0.6

30.1331.6232.8733.6433.9533.8133.24

1.491.250.770.31

-0.14-0.57

o o o o o

! 1 1 1 t

-0.240.020.010.02

0.26-0.010.01

-0.270.02

0.29

0

We know that central difference formula for derivatives ;

d y ) 1 r (Ayp+Ay-i) l (a 3 + A8y_2) [ l (A5y_2 + A5y_3)

2 ” 6 2 30 2

Here h = . 1, tQ= 0.3, A y<>= .31, Ay_ j = .77, 2 “ *02,

A^l j = .01, AV-2 = .02, A5>_3 = - .27.

1 f~(,31 + .77) 1 <.02+ .01) | 1 (.02 +(-.27))]^ U o t..s "C1)L 2 6 2 + 30 2 J

Velocity - / '( .3 )= + jjjj (”*125) j = 5.34 cm/sec. Ans.

Also- ( ^ L , -

Putting h = .1, A2y~i -- .4 6 , A*y~2 ~ - .01, A^y _3 = .29.

- / ”< • » - c F [ - 46- S (- 01)+^ <i!9)]

= - 45.6cm/sec2. Ans.

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268 I En g in e e r in g MATHKM*ncs-III

Example 3.96 :Find x for which y is maximum and fin d this value o f y

x : 1.2 1.3 1.4 1.5

y •' 0.9320 0.9686 0.9855 0.9975 0.9996

Solution. The difference table is as follows :

X y Ay A*y A*y 44y X q =1.2

1.31.41.51.6

0.93200.96360.98550.99750.9996

0.03160.02190.01200.0021

-0.0097-0.0099-0.0099

- 0.00020

0.0002

Let, y0 = 0.9320 and x0 = t .2, Ayo = .0316, A2yo = - .0097By Newton’s forward difference formula :

u(u -1) . „ y =f{x) = y0 + «Ayo t — y° + •*“

y =0.9320+ (0.0316m)+ ^ ( - 0.0097)2

Differentiate, ^ =0.0316 + —-— j ( - 0.0097)

[Neglecting higher differences]

dyFor a maximum put — - 0

0.0316 =j u-|j(0.0097) =>,, = 3.76

X — Xn x-1 .2u = — 7 ~ ^ 3.76 = — - — => j c = 1.576. Ans.

n .1

(which is near to x„ - 1.6, so using backward difference formula)

To fin d y m we use backward difference formula, x - x K 1.576-1.6

u = —r— -------- :------=> u = - 0.24n .1

JM * ♦ “V ,. + v y . * 2 < ! £ ± » * *

X I -5 7 6 ) - 0 .9 9 9 6 - ( 0 . 2 4 k 0 . 0 0 2 1 ) t ~ ° 2 4 ) ( - 0 .0 0 9 9 )2

= 0.99% - 0.000504 + .0009028

= 0.9996 + 0.0003988 = .9999 approx.Vmax- =0.9999. Ans.

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N u m e r i c a l A n a l y s is - ! ! | 269

3.27 NUM ERICAL INTEGRATION

fb f(x )dx can be evaluated exactly, if/(x) is explicitly defined as a mathematical function and if

Joit is integrable. But sometimes, the function y~f{x) is defined by means of (w + I) paired values

(xh y/); i = 0, 1,2, 3,...., n. In such situations, we resort to numerical integration, in which we

f *replace fix) by any convenient interpolating polynomial P{x) and calculate J P{x)dxt that is

f approximately equal to I f(x)dx. The process of numerical integration of a function j{x) is

Jacalled quadrature.

3.28 NEW TON -COTE’S QUADRATURE FORMULA

Let y<h yu yi>..... . yn be the values of y - J{x) corresponding to x = x q , X | , x 2.......... . x„ which areequally spaced with step-size h. Then, by Newton’s forward interpolation formula.

u„ u (u - l ) _ u (u - l )(u - 2) , y = yo+Y jAyo+-" 2! -A 2y0 +------- r ;-------A3y0 +.

where,X-X0

u - — r— =>x = xf, + uh =^dx = hdu n

fXQ+nh

3!

f*n , r*0+nft f» u u (u - l) Now ydx = y dx = j o y0 +^ j4yo — A

u(u - l)(u - 2)

-(I)

A3y0 +-

= h

3!

(u2 - u) .2 (u3 -3 u 2 +2u) , y0 + uAyQ+ - A2y0 +------------ --------------A3y0 +

hdu [By (l)j

du

*" “ + T + I ( t ' “3 + 1 )■&3yo

fn 6 3n4 lire3

24- 3 a 2 A*y0

1 f n e

'120 [ 6

0 . 35n* 50n3 22n9 + ---------------- + 12n2 A5y0 + ....0 ...(2)

Result (2) is called Newton-Cote i quadrature formula. We can obtain a variety of special quadratureformulas by putting n = 1,2, j ........ as discussed below :

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270 | E m m m vw M a j h o u t i c s -IH

3.29 TRAPEZOIDAL RULE

By putting n = ! in Newton’s-Cote’s quadrature fonnula and omitting terms with A yo and higherorder differences, we get

Similarly, f * y dx = £ [y, + y21......... so o n . ...(2)J*i it

Finally, P ‘‘ y d x = £[y„_ i + y„ ] ...(3) Jxn a

f^ ydbc = P y d x + P ydx+ ......+ (*“ y d x Jx0 *cq Jxi

= ^ 0 'o + y i ) + | -( y i + y a ) + - + - |( y ). - i+ y J .

[Using (1), (2) and (3)]

= j [(y0 + y„>* 2<>i +V i+ ....+ y*-i >] ,..(4)

Which is known as general trapezoidal rule. Remarks:

(i) It is to be noted that Trapezoidal rule is applicable for any number of intervals or any numberof ordinates.(ii) The total truncation error is Trapezoidal formula is

E = - ^ ( x n - x 0 ) W ( i \ )

where h - ** and/" (? ) = 1^*^» where x o < t j< x „.

i.e., The errors in the trapezoidal rule is the order o f A2.

3.30 SIMPSON’S ONE-THIRD RULE (OR US) RULE:

By putting n = 2 in Newton-Cote’s quadrature fonnula and omitting terms with AJyo and higherorder difference.

Assuming that y - fix ) is a polynomial of degree two, we get

J % d b c 2y0 +2Ay0 + ^ | - 2 A2y0 j .

= M2y0 + 2(y, -y 0) + ^ 0*2 -2 yt +y0)

[Since A*yo = (E - l)2^ -2y, +y0]

=> f x y d x = ^ (y o+ ^y i + y*). ...(l)

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Similarly, J J y dx = ^ [ y 2 + 4y3 + y4 ] ........so on

Finally,

Numer ica l Ana l Vs is -II | 271

3 - - ------------------- "#>

J ! = -r[5 '»-2 + 4yfl.1+yR] ...(3)**n-t a

f^yctc = f* y d x + P ydx + ....+ j*" ydx . Jxg «2 «»-*

h h h = J (>0 + 4yi + y2) + - (y2 + 4y3 +y4) + .... + - (yju-2 + 4y2j,_1+ y2J

p y d x = ^ [(y 0 +y2*) + 4(yl +y3 +... + y2(l_I)

+ 2(y2 +y4 + .....+y2n-2)] -(4)Which is known is general Simpson's 1/3 rule.

Remarks:

<i) To apply Simpson’s rule, the interval of integration must be divided into an even number o f sub-intervals each of width h i.e., number of known ordinates is odd.

(ii) The total truncation error is Simpson’s 1/3 or Simpson’s formula is

E - ~TZf; (x 2n -*b)fc4/ (K,,(il)> Since ~ r *° = h, whereloU

/">(?) -

3 31 SIMPSON’S THR EE -E IGH TH S (3/8) RULE

By putting n = 3 in Newton’s - Cote’s quadrature formula and omitting terms with andhigher order differences, we get

= h 9 9 33y0 + - (yi - y0) + ^ (y2 - 2yi + y0) + - (y3 - 3y2 + 3yj - y0)

[Since A2y0 = (E - 1 yo and A3yo = (£ - I )3 yoJ

. p y d * = -^(y(,+3 y, +3y2+y3) ...(1)J*o o

3*Similarly f y dx = — (y3 + 3y4 + 3y5 + y6), and so on

o

Finally f*“ y d x = ^ (y „ _ 3 + 3y„_2 + 3y„_, + y jJ%-s 8

-.(2)

••■(3)

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y d x = [** ydx+ f*8 y d x +..... + f *" y d x JXQ J xq Jj* J**.3

= — [(y0 + 3y, + S>2 + y3) + 0-8 + 3y4 + 3>5 + y6) + .....

+ CK3«—3 + 3>^_2 + tyin- ) +>’3«)] [Using(I), (2) and (3)]

j ^ y d x = ^ K y 0 +J'3.) + 3(yi+ y4 + y7 +--) + 3(y2+>'5+>'8 + ....) Jxo o

+ 2(^3+ y e+y9+ ....+ yjn)] —(4)which is known as general Simpson i 3/8 rule.

Remark :

To apply this rule, the interval of integration must be divided into a 3-multiple number of sub

intervals each of width h.33 2 WEDDLE’S RULE

By putting n = 6 in Newton’s-Cote’s quadrature formula and omitting terms with d?yo and higherorder differences, we get

P y d x = I 6y0 + 184y0 + ^ ( 7 2 - 18)A2y0 +~ (324 - 216 + 36)A3y0L 2 6

27 2 | rN W frM M M*W«tlitrK*-Hl

+ J- ( H 2 1 _1944+ 792 - 1082 4 { 5 A4y0+ -

6y0 + 18Ay0 + 27A2y0 + 24A3y0 + A*y0 + A8y0 + A6y010 10 140

41 , 3 -If we replace 10 °’ ^ lus ^ em>r comra‘tte^ will be negligible and using

Akya = (E~ l)*)^ we get

I T y(i r = B ’ ,0+5;>,1+^ +6y3+y4+5;y6 + y6) - d )

Which is called weddie i rule for n * 6.

Similarly y d x . | ~ ( y 6 + 5 ^ + yt + 6y9 + y,0 + 5yu +y2) and so on ...(2)

Finally f* =* ~ (y»-« + By,.* + yM + 6 y ^ + + 5y„_, + yB) ...(3)' 1»-» 1U

f ** y d x = ydx+ [*** y d x +...... + (*" y d xJjfc Jxo J*n-«

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- jq [^ 0 + y») + 5^ l + ^7 +>ia + —) + 02 + /8 + > ')4 + ••••)

+ 6(^3 +.V9 +y\2 + ....) +0>4 + >’10 + y\6 + •••■)

+ 5(vs +yu +>>17--) + 20'6 +y n +yw + ••••)] •■•(*)which is known as Weddle's rule.

emark:To apply this rule, the interval of integration must be divided into a 6-multipie number {i.e., 6, 12.18, etc.) of sub-intervals each of width h.

Example 3.97 :Fhtd £ j + x 7 *** Av Trapezoidal rule, Simpson’s 1/3, Simpson’s 3/8 and Weddle

rule, where the interval is divided into 6 equal parts. [RGPV Dec. 2001/

1 -0 1

olution. Here, we have n = 6 and h = —g—= —•Since x„ = x0 + nh for » = I, 2, 3 , .....„6,

N u m e ric a l An a l y s i s -1! | 273

3h

X * 0 1 1 0 * 1 1

1 O ' ! | » — • 2

Xz=6

3X3=6

4X« = 6

5

* 5 =6x6 =1

y=/W 1

l + x *

Vo =1 y, =.97297y2 = 9 y3 = .8

yx =0.6923 Vb =.59016 1 1

Using Trapezoidal ru le :

= ^ [( y o +>6) + 2(y ,+ ^ + .... + >5)]J o l + x2 2

= — [(1 + .5) + 2(.97297 + .9 + .8 + .69230 + .59016)]12

9.41086= — —— - .78424. Ans.

Simpson’s 1/3 rule for n ~ 6 :

f1— 1— dx = - [(;yo +y6) +4(yi +y3+y5) +2(3'2+y4)lJ o l + x2 3

= — [(1 + .5) + 4(.97297 + .8 + .59016) + 2(.9 + .69230)]6x3

.I

14.13712--------- — ----- - .78539. Ans.

Simpson’s 3/8 rule for n = 6 :

• i l , 3hf ----- j d x = — [(>'o+ ^ ) + 3 0 ,1+>'2+3'x+3'5) + 2(y3)]Jo I + jc2 8

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274 | En g in e e r in g Ma t h e m a t i c s-!!!

= f i [(1 + .5) + 3(.97297 + .9 + .69230 + .59016) + 2(.8)}8 6

12.56575

16= .7853593. At

Weddle role for n = 6 :

1 , 3hJo 1+** *** = l o ^ 0 +5:Vl +yi + 6y* +y* +5y* + * ]

= + 5(.97297)+ .9 + 6(.8) + .69230 + 5(.59016) + .5]

= 0.7854

x r 1 d x

AO!

x r • ax Example 3.98: Compute the value o f xfrom the formula ~ = j + using trapezoidal rule wit,

JO sub-intervals.Solution. When the interval (0, 1) is divided into 10 sub-intervals, n = 10 am

, 1 -0h = = 0.1, since x„- x0 + nh for n = 1,2........ . 10.

X oI I x, =0.1 x2 =0.2 x3 =0.3 x4 =0.4

1

1 + x

y0 =1.0000 yt =0.9901 y2 =0.9615 y3 =0.9174 y4 =0.8621

x ; x5 =0.5 x6 =0.6 ! • ?

1 i i

. * n o o o x9 =0.9 x» =i-o

y ■ ys =0.8000 y6 =0.7353 yT =0.6711 y8 = 0.6098 y» =0.5525 yi0 = 0.5000

By Trapezoidal rule fo r n** 10: j ^

I J'd x= -[ (y 0 +>-10)+2(y l +y2+......+y,)]« *

i.e., [(1-0000 + 0.5000) + 2(0.9901 + 0.9615 + 0.9174 + 0.8621

- 0.78498.

r* dx

+ 0.8000 + 0.7353 + 0.6711 + 0.6098 + 0.5525)]...(I)

Now, exact value j o j — £ = (tour1x)£ - j

Hence

n * 3.13992.Since die actual value of it is 3.14159

The error =3.13992-3.14159 = -0.00167.

[Using (I)]

Ans.

Ans.

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N u m e r ic a l A n a l y s is -11 I 2 7 5

f t jExample3.99 '.Evaluate J p by using (i) Trapezoidal rule (ii) Simpson’s 1/3 rule (iii)

Simpson *s 3/8 rule (iv) Weddle’s rule and compare the result with its actual value (Taking h “ I). [RGPV Dec. 2002]

Solution. Here,6-0

h — 1 i.e., h -------- =>n —6

Since x„ = xq + nh for n = 1,2 ,3...... . 6.

x : Xq = 0 x ,= l *2 = 2 x 3 = 3 x« = 4 xs =5 x6 =6

1

,= J + x*Vo=l Vi ~ 0.5 VZ = 0.2 V3= 01 y, = .0588 y5 = .0385 V6 = 027

(i) By Trapezoidal rule, for n =*6 :

r6 dx h= ~ K o + >e) + 2(yt + y2 + 3 + + ya)]

Jo 1+x2 2

= - [(1 + 0.027) + 2(0.5 + 0.2 + 0.1 + 0.0588 + 0.0385)]2

= 1.4108.(ii) By Simpson’s 1/3 rule, for n = 6 :

C 7— 7 = ~ f(y0 + Ve) + 4(y, + y3 + y6) + 2(y2 + y4)]Jo l + *z 3

- -1(1 + 0.027) + 4(0.5 + 0.1 + 0.0385) + 2(0.2 + 0.0588)]3

= 1.3662.(iii) By Simpson’s 3/8 rule, for n=> 6 :

f 6 = — [Oo + y8) + 3(yi +y2 + y i + y6) + 2y3]Jo l + x2 8

= - ((1 + 0.027) + 3(0.5 + 0.2 + 0.0588) + 0.0385 + 2(0.1)]8

= 1.3571.(iv) By Weddle’s rule, for n = 6 :

f 6-r ——r = |^Lyo + 5yi+ y2+6y3+y4+5y5 + y6]Jo l + x2 10

= 0.3(1 + 5(0.5) + 0.2 + 6(0.1) + 0.0588 + 5(0.0385) + 0.027]= 1.3735.

fG dxAlso Jo f +~g ~ [tan-1 x]3 = tan- 16 = 1.4056 which is exact value.

This shows that the value o f the integral found by Weddle's. rule is the nearest to the actual value. Ans.

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Example 3.100: Evaluate x l ' *e~*dx, approximately by using a suitable formula.

{RGPV June 2003 and Dec. 2006,

Solution. Suppose, we divide the interval (0.5, 0.7) into four equal parts by taking h = :

40.05. Now the va lues o f given function y - x ia e~x.

276 | En g i n e e r i n g Ma t h e m a t i c s-111

X y ~ x l1e~*

x0 = .50

x, = x 0 + h - .55

x 2 = x 0 + 2h = .60

x 3 = x 0 + 3h = .65

x4 = x 0 + 4h = .70

y0 - 0.4288819

Vl = 0.4278774

y2 = 0.4251076

y3 = 0.4208867

y4 * 0.4154730

By Simpson's ‘1/3’ rule, forn -

4 :f 07 x 1'2e~xdx = ~r[(y0 + y4) + 4(yj + y3) + 2 y2 ]J 0.5 d

= [(0.4288819 + .415473) + 4(.4278774 + .4208867)

+ 2(.425!076)|

or f V v ^ d x = ^ [ 5 . 0 8 9 6 2 6 5 ] =0,084271. AmJ0.5 3

Example 3.101: Find an approximate value o flog 5 by calculating to 4 decimal places, by Simpson's i/1

e5 dxrule, J o - , dividing the range into 10 equal parts. / RGPV June 2004, Dec. 2008(N)

5 -0Solution. Here range of integration (0, 5) is divided into 10 equal parts, i.e.. h = ------= 0.5. Giver

10

function y = 77 — - , then(4* + 5)

X v - 14 x + 5

0 1 1

* y0 = 0.20000

x, = x0 +fc = 0.5 y x = 0 . 1 4 2 8 6x 2 = x 0 + 2h = 1 y 2 = 0 . 1 1 1 1 1

X3 - Xq + 3/l = 1.5 > 3 = 0 . 0 9 0 9 1

x 4 = x0 + 4h = 2 y4 = 0 .07692

X5 = Xq +5 ft = 2.5 y B - 0 06667

Xg = Xq + 6h = 3 y 6 ~ 0 .05882

:= Xq + 7ft = 3.5 y 7 = 0.05263

x8 = x0 + 8/1 = 4 y 8 = 0.0476 2

x9 =Xq + 9ft = 4.5 y 9 = 0 .0 4 3 4 8

X10 = X0 + 10ft = 5 y 10 = 0 . 0 4 0 0 0

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From Weddle jr rule for n - 6 :

2 7 8 | En g in e e r in g Ma t h e m a t ic s -111

■ \yd x = — [>0 + 5 }', +y2 + 6 yy +y4 + 5 ys + y6]

= A x JLrj + 5 x 0.98281 + 0.9306 + 6 x 0.84089 + 0.727110 12

= —-x 15.14079 = 1.189240

>5.2

+5 x0.50874 +!

Ai

Example 3.103 :Evaluate loge x d x by (l)Simpson’s 1/3 rule, (ii Simpson’s 3/S rule. Af

finding the true value o f the integral, compare the errors in the both cases.. 5 .2-4 . .

Solution. Suppose, we divide the interval (4, 5.2) into six equal parts by taking * = — “— = U.2.

Now the values of given function y = log, x are as follows :

X y - log. x o * I I 4 5 k ©

111 Ly0 =1.3862944

Xj =x0 +h =4.2 yj =1.4350845

x2 =x0 +2h =4.4 y2 =1.4816045x3 =x0 +3h =4.6 y3 =1.5260563

x4 = x0 + 4h = 4.8 y4 =1.5686159

x5 = x0 + 5h = 5.0 y5 =1.6094379

Xe =Xq +6h — 5.2 y6 =1.6486586

(i) By Simpson's 1/3 rule, for it a 6 :

L yd x = -j[(yo+ ^6) + 4(>, +y3 +ys) + 2(y2+y4)]

J * 2 log x dx = ^127.4 17706] = 1.827847. Ans.

(ii) Simpson’s 3/8 rule, for » = 6 :

f ** y d x = ^ • [ (y 0 + 6) + 3(y1+y2 +>'4+ 3’5)-,-2(y3)] jjoq a

f 5 log x dx = [24.371294] = 1.827847. Ans. J 4 o 44

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Now exact value:

f 6.2 <<1.2log x dx = I log x. Id x = [x(log x - 1)152

= [5.2 (log, 5.2 - l ) “ 4(loge 4 -1 )] = 1.827847.Hence the errors are :(i) Due to Simpson’s 1/3 rule = 0.000000.(ii) Due to Simpson’s 3/8 rule = 0.000000. Ans.

r * dxExample 3.104 :Find the value o f J f — by Simpson fcrule. Hence obtain approximate value ofloge 2.

(RGPV June 2006)

Solution. Suppose, we divide the interval (1,2) into 6 equal parts.

l 2 - 1 1

Taking, A = —— = — • Since x„ = x0 + nh, for n = 1,2, 3 , 6 .

N u m e r ic a l A n a l y s is -11 | 2 7 9

X i i ~ H x

Xq = 1 Vo = 1

Xy = 7 /6 Vi = 6 /7

x2 = 8/6 j?2 = 6 /8

x3 = 9/6 Vz = 6 /9

x4 = 10/6 y4 =6/10

x5 = 11 /6 y s = 6 / 1 1

x 6 = 2 * 6 = 1 / 2

By Simpson’s 1/3 rule for « = 6 : We have

P y d * = -7[Cyo+^6)+4(3'i+>'3+y8)+2(>2+y4)]

or

Now,

r^ = ± r r 1 + n + 4 f 6 + 6 + M + 2 r 6 + o i

Ji x 18 \ \ 2 ) U 9 l l j U 10J

1 |~3 5736 | 216~

_ 18L2+ 693 + 80 _ _L18

3 (1434) (108)

2 693 80

= — (12.477056) =0.693169.lo

Ans.

r2dx

x

log, 2 = ^593169. [by using (I)]

r2dxJj — = [log, x]? * log, 2 - log, 1= log, 2 [v log, 1=0]

Ans.

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Example 3.105 :A solid o f revolution is formed by rotating about the X-axis, the area between the X- axis, the line x - 0 and x 1 and a curve through the points with the following coordinates : \

280 | E n g i n e e r i n g M a i t c m a t i c S'I I I

X 0,00 0.2S 0.50 0.75 1.00 y 1.00 0.9896 0.9589 0.9089 0.8415

Find the volume o f the solid formed, using Simpson *s rule.

Solution. Here, h = 0.25 and n = 4 (i.e., 5 ordinates are given)

Required volume V of the solid generated is given by

hV = f 7i y2dx = Jt — [ ( ^ + y |) + 4(> f+>!) + 2>’|]

■0 i)

[ v By simpson’s rule for n = 4]

0.25x3.141[{1 + (0.8415)*} + 4{(0.9896)2 +(0.9039)2} +2(0.9589)2J

= 0 . 2 6 I 8 [ 1 .7 0 8 1 + 7 . 2 2 1 6 + 1 .8 3 9 0 ] = 0 . 2 6 1 8 * 1 0 .7 6 8 7 = 2 . 8 1 9 2 . Ans.

Example 3.106 :The speed, v meters per second o f a car, t seconds after it starts, is shown in the following table :

t : 0 12 24 36 48 60 72 84 96 108 120

v : 0 8.60 10.08 18.90 21.60 18.54 10.26 5.40 4.50 5.40 9.00

Using Simpson's rule, find the distance travelled by the car in 2 minutes. Solution. If s meters is the distance covered in t seconds, then

dsas cimSince, - v => [sjjfg = vdt [y / = 2 min. = 120 sec.]

Here 11ordinates are given i.e., n = 10. Using Simpson’s 1/3 rule : (Here h = 12)

i.e., s = I(t»o + t»j0) + 4(11! + v3 + v6 + v7 + u9) + 2(v2 + v4 + l>6 + u8)]

12= — {(0 + 9.00) +4(3.60+ 18.90+ 18.54+ 5.40+ 5.40) ,3

+ 2(10.08 + 21.60 + 10.26 + 4.50)]

= 1236.96 meters. Ans.

Example 3.107: The following data give the velocity v o f a particle at time t :

t(sec): 0 2 4 6 8 10 12

v( m / sec): 4 6 16 34 60 94 136

Find the distance moved by the particle in 12 seconds and also the accerelation at t =2 seconds. fRGPV, Feb. 2016J

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Solution, (i) Let S’= distance

N u m e r ic a l A n a l y s is - ! ! | 281

12dS

dt " v

12

s= j v d t

i.e. distance S = jv d t ... (1)o

Here 7 ordinates ae given i.e., number of intervals are n = 6.

. b - a 1 2 - 0 „Using Simpson’s 1/3 rule : I Here h = ——- = —- — = 2.

From the table, we have:v0 = 4, Vj = 6, v2 = 16, v3 = 34, v4 = 60, vs = 94, v* = 136.

f hS = J vdt = j [ ( v 0 + v6) + 4(vj + v3 + v5) + 2(v2 + v4)]

o

= y [(4 +136) + 4(6 + 34 + 94) + 2(16 + 60)]

1656 _ ____ _ meters. Ans.

We know that

Velocity vAccerelatin a = — -------= — ... (2)

time t K

Here at time t = 2 sec. die velocity is 6 m/sec. from the table, hence (1) becomes :

6Accerelation = — = 3 m/sec2. Ans.

Example 3.108 :Estimate the length o f the arc o f the curve 2y = x 2 from (0, 0) to (1,1/3), using Simpson *s 1/3 rule taking 8 sub intervals.

Solution. The length of arc is given by curve 2y~x?-=$y = x2/2.

dyDifferentiate, — = *

dx

We know that length of die given are is

S - f ^jl + (dy/dx)2dx = f ^ l + x 2dx [v x = 0 / o x = 1] Jxâ J 0

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282 | En g in ee r in g Ma th em a t i c s-!II

The values of fix) = J \ + x2 since (0, 1) divided into 8 sub-intervals :

x : y = V l + x*

oi

i X

y0 =iX j = .125 y> =1.008

x2 = .25 y2 = 1.031

x3 = .375 y3= 1068x4 = .5 y4 =1.118

x5 = .625 y5 = 1.179* * ? i

i y6= 1-25 Xy =.875 y7 = 1.329

x8 =1 y8 = 1.414

Here, h » - £ - = > h =.125 and« = 8.

By using Simpson's rule : we get,

[ 1-Ji+x2dx = ^-[(>0 +>8> + 4Cyi + Js +^5 +yy) + 2(y3 +y4 +y6)] Jo 3

= — ((1 + 1.414) + 4(1.008 + 1.068 + 1.179 + 1.329)

+ 2(1.03 J + 1.118+1.25)]= 1.14783. Ans.

Example 3.109 :A reserviour discharging water through sluices at a depth h below the water surface has a surface area A fo r various values o fh as given below :

h( f t ) : 10 11 12 13 14

A ( ttq.fi) :950 1070 1200 1350 1530

l f t denotes time In minutes, the rate o f fa ll o f the surface is given bydh -48jh

dt A

Estimate the time taken fo r the water level to fa ll from 14 to 10 ft. above the sluices.

dh -4 RJh A

dt Solution. Given that -rr *

—48 -Jh ** j !

— r - ^ d , ' ^ h dh

C - C - A dh

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N u m er ic a l A n a ly sk -M j 2 83

i * ■ £ h * w *

= / “ /( A ) * w here/*)- ...(I)

h : 10 11 12 13 14 A : 960 1070 1200 1350 1530

yo= -6.259 yt= -6 .7 2 y1 = -7 .217 C O t > 1 I I > > y i- -8 .52

Here, n =4 i.e., 5 ordinates are given,

Using Simpson’s 1/3 rule, fo r « - 4 :

f U f(h )d h = ~ [(>0 + J j + 4(ji + Js) + 23-2]J10 3

or (<)}$ = |[ ( -6 i2 6 9 + (-8.52)) + 4 (-6 .7 2 -7.8) + 2(-7.217)]

= -[-87.293]3

= - 29.09 min. [- ve sign indicate the fall of water level]/. The time of water level to fall from 14 to 10 f t = 29.00 min. (approx.). Ans.

Example 3.110 :A river is 80 meters wide. The depth d (In meters) o fthe river at a distance xfrom the bank is given by the following table:

x : 0 10 20 30 40 50 60 70 80

d: 0 4 7 9 12 15 14 8 3

Find approximately the area o f cross-section o f the river.\

, 80-0

Solution. By Simpson’s ‘1/3’ rule, here 9 ordinates are given i.e., n = 8 and » = —g— = 10.

/. Area = P f(x)dx Jio

»80 h= I ydx = - [(y0 + ys) + 4(>1 + y3 + y8 + » +2(y^+y4 + y6)]

Jo 3

= ~ [ ( 0 + 3) + 4(4 + 9 + 15 + 8) + 2(7 + 12 + 14)] [Herej = d, h = 10]u

= — [213] = 710 square meter,t)

Ans.

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Example 3.111 '.Calculate by Simpson’s rule an approximate value o f x 4dx by taking seven

equidistant ordinates. Compare it with the exact value and the value obtained by using

the Trapezoidal rule. , 3 - ( - 3 ) .

Solution. Here the given interval (-3, 3) is subdivided into six equal parts, taking «•------~— - L

2 8 4 { En g in e e r in g Ma t h e m a t i c s -111

x :* X1 ! a .

x0 = -3

x1 =x0 +h=-2

Xj = x0 + 2h = -1

x3 = x0 +3h=0x4 =x0 +4h =1

x5 =x0 +5h=2

Xg =x0 + 6h=3 1

Vo = 81

Vi =16

Vz =1

y3 =0J/4 =1

Vs =16V6 -81

(i) By Simpson's 1/3 rule, fo r n - 6 :

P y d x * ■^[(3'o+y6) + 4(yi+>3+>’5) + 2(>2+>4)I•>*0 o

• f ^ d x - i [ 2 9 4 ] .-3 3

(ii) The exact value :

98.

(iii) By Trapezoidal rule fo r n - 6 :

i * y d x = f[^> ,o+y6)+2Cy1+>-2 + y3+y4+y5)]

fx*dx - -i[230] = U5.- 3 ^

...(1) Ans.

...(2) Ans.

...(3) Ans.

From (I), (2) and (3) this shows thatSimpson’s rule gives the value which is much nearer to the exact value. Ans.

f x / i Example 3.112 .’Calculate | e* *dx correct to four decimal places, by taking four ordinates.

• 0

Solution. Divide the range (0, n t 2) into three equal parts because given 4 ordinates.

n / 2 - 0 i t v Taking h --------------* - -

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v By Simpson *s 3/8 fo r n = 3.

r*'2 • , 3h ,

N u m e ric a l A n a l y s is - 0 | 2 8 5

f* gsinxfa = — [(yQ+y3) + 3(yx + y2)]Jo 8

= - , - [(1 + 2.71828) + 3(1.64872 + 2.3774))

8 6

= [3.71828+ 12.07836]

= ^[15.79664]

= 3.1029. Ans.

Example 3.113 :Find J e ^ d x , by taking seven ordinates using simpson’s 1/3 rule.

[RGPV June 2909]

Solution. Given seven ordinates i.e., number of sub-intervals n = 6. x6 - *o 0 . 6 - 0 „

Then h = — g - = — t — = 0.1 b , b

X y - f(x) = e *2

1 ! O % i i f t ) 1 o k I I > —

* 1

Xj = X q + h - 0.1 y, » e~i = e-° 01 = 0.9900

x2 = x0 + 2h - 0.2 y2 = e-OM = 0.9608

Xg = 0.3 y3 = «-«•» = 0.9139

x4 = 0.4 y4 = e-°16 = 0.8521

x5 « 0.5 y5 = €-°25 = 0.7788

x6 = 0.6 y6 = <r« 36 = 0 6977

By Simpson *s1/3 rule for n = 6 :

10° V -^d x = ^ [(y 0 + ye) + + M) + 2( 2 + ^ ) ]

= — [(1 + 0.6977) + 4(0.9900 + 0.9139 + 0.7788) + 2(0.9608 + 0.8521)]3

= ~ [1.6977 + 10.7308 + 3.6258]

= 0.5351. Ans.

2 1f „-x / 2

Example 3.114 : Evalute: J e dx, by Simpson’s 1/3 rule, taking four intervals.i

Solution. Here n = 4, x0 = 1,*4 = 2.

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286 | En g in e e r in g Ma t h e ma t ic s -IU

h =

xt =x0 + h= 1.25,x2=jto + 2h= 1.5,

x3=x0 + 3h = 1.75

X 4 Xn 2 “ I I- ^ = — - ^ = 0 . 2 5

X y - f ( x ) - e - * i

X q ~ I y0 = 0.60653Xj = 1.25 y, = 0.53526 x2= 1.5 y2 = 0.47237

xj - 1,75 yj = 0.41686x4 = 2.0 yA= 0.36788

By Simpson’s 1/3 rule for n = 4, we have

7 AJydx = -[ 0 'o+>'4) + 4(>'1+>'3) + 2>2]*0

f e-*/2dx = — [(0.60653 + 036788) + 4(053526 + 0.41686) + 2(0.47237)]. 3

= 0.4773. Ans.

r* s inx , Example 3.115:Using Simpsoni 1/3 rule, fin d J -------by dividing (0, x) in 6 equal parts.

JCSolution. Here, Sub-intervals is n = 6, and

h =

x :ainx

v - —X

x0 =0 Vo =1Xj =ft 16 j>! =3 In

x2 =2*76 y2 =2.59009 in

x3 =3*76 y3 = 2/*x4 =4*/6 y4 =1.29904/*

x5 =5*/6 ys =0.6/*.<Jc

x6 =*• Ve =0

Using Simpson’s 1/3 rule, fo r n *=6:

C - ^ d x = ^ [(y0 + y6) + 4(yj + y3 + ) + 2(y2 + )] X o

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N u m e r ic a l A n a ly s is - !! J 2 8 7

7t f , 22.4 7.7426'I

= — 1 + ------+ ---------18 Jl It ;

= 1.852. Ans.

fK H Example 3.116:Calculate the approximate value o f s in x d x

(i) by Trapezpidal rule

(ii) by Simpson*5 rufa using 11 ordinates.Solution. Here given 11 ordinates i.e. number of sub-intervals is n = 10

, w'2-0.*. Taking h = ——— = n! 20.

X y <=atnx oI f X

Vo =0

Xj = 7t /20 Vi =0.1564

xz =*710 y2 =0.3090

x3 =3?r/20 y3 =0.4540

x4 =*75 y4 =0.5878

x5 = n ! 4 y5 =0.7071x6 =3*710 y6 =0.8090

x7 =7*720 y7 =0.8910

x8 =2*75 yt =0.9511

x9 =9*720 y9 =0.9877

x10 =*72 =1.0000

(i) Using the Trapezoidal rule, fo r n *» 10:

«**/2 hJo s i n x d x =-[(yo+yio) +2(yt+y2+ .....+*»)]

= ~ [ ( 0 + l) + 2(0.1664 + 0.3090 + 0.4640 + 0.6878 + 0.7071

+ 0.8090 + 0.8910 + 0.9511 + 0.9877)]= 0.9981. , Ans.

(ii) Using the Simpson’s rule, fo r n m 10:

*>2 h

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n

2 8 8 | E n g in e e r in g M a t h e m a t ic s -III

[(0 + 1) + 4(0.1564 + 0.4540 + 0.7071 + 0.8910 + 0.9877)60

+ 2(0.3090 + 0.5878 + 0.8090 + 0.95 II)}1.0006. Abs.

f l . 4 Example 3.I I 7 .-Evaluate (s inx~ lo g t x ^ e x)dx approximately using Weddle’s rule correct >1.4

to 4 decimals. (Taking h * 0.1). Solution. Let J[x) = sin x - log,.* + e*.

1.4-.2Since, h =

.1 =

n

1.4-.2

n

* = 12X y = sin x - l o g ' * + e x

X q =0.2 y0 =3.0295

Xi =0.3 Vi =2.8494

x2 =0.4 y2 =2.7975

x3 =0.5 y3 =2.8213

x4 =0.6 y4 =2.8976

x5 =0.7 y5 =3.0147

x6 =0.8 y6 =3.1661

x7 =0.9 y7 =3.3483

x8 =1.0 ys =3.5598

x9 =1.1 y9 =3.8001

x10 = 1-2 y10 =4.0698

X„ =1.3 yn =4.3705

xI2 =1.4 yl2 =4.7042

Using by Weddle’s fifle, for n =*12 :

f ' f(v)dx = ~ b 0 +5>i +yt +6ys +y4 +5y8 +2>6 + 5y7 + y8 +6y9 J0 .2 10

+îo + 5^n + yu]

= ^ (0.1)[3.0295 +14.2470 + 2.79J6 + 16.927#;*B.8976 +15.0735

+ 6.3322 + 16.7415 + £.55$ + 22.8006 + 4.069*4; ?1.8525 + 4.7042]- (0.03) [ 135.0335] = 4.051, ’ \ " Am.

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f u i t - ------------------Example 3. 118 :Using Simpson’s rule, evaluate I s i + 3 tin * 9 d9 , taking n = 6.

* ff

N u m e r i c a l A n a l y s is -1 I J 2 8 9

7 1 / 2 -0 71 ,--------------------------

Solution. We have, h = — - — = — and y= fl,0 ) = >/l + 3sin2 0

Let x ~ 19 y = J l + 3sin *e o

i i oX y0=1

X! = x / 1 2 y, =1.095

x 2 =2^/12 y2 =1.323

x3 =3/77 12 y3 =1.581

x4 =4>r / 12 y4 =1.803

x5 =5^/ 12 >5= 1-949

x 6 - n i l ye =2

Using Simpson’s 1/3 rule, for « " 6 :

f* 2>/l + 3sin2 0 dQ = ~ [(y0 + _)'6) + 4(.y1+ y3 +yi) + 2(y2 +>4)]Jo o

n

= — [(1 + 2) + 4(1.095+1.581 +1.949) + 2(1.323 +1.803)]36

= ^ ( 3 + 18.5 + 6.252) =2.422.36

Example 3.119 :The following table gives the values o f a function at equal intervals :

Ad s .

x : 0.0 0.5 1.0 1.5 2.0

f i x ) : 0.3989 0.3521 0.2426 0.129S 0.0540

Evaluate (i)f(1.8) (U) f f l .S ) (IB) J * f (x )d x , stating me formula used.

fRGPV June 2005}

on. The difference table is :

X y = f(x) Ay A’y A*y A*y

0.00.51.01.52.0

0.39890.35210.24200.12950.0540

-0.0468-0.1101- 0.1125- 0.0755

-0.0633-.00240.087

0.06090.0394

- 0:0215

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290 | E n g in e e r in g M a t h e m a t ic s -U I

(i) To fi nd ff l. 8) : At x = 1.8, so using Newton’s backward interpolation formula

_ u(u +1) u(u + l)(u + 2) „ Ax) = yn + uVy« + - gT 7 + — - ------ v + •

here

2!

* -* n 1 .8-2 .0u = — ;— = — ~ — - - 0.4

3!

0.5

A 1.8) = 0.0540 + (- .4 )(- 0.0755) + (0.037)A

, (.0394) + M X q-g g-6 )gL 6) 0 m6 24 ^

or / 1 . 8) = 0.078133. Am

(ii) To findf' (1.5 ): Taking** = 1.5 and using derivative of Newton’s backward interpolateformula:

l

0.5- 0.1125+- ( - 0.0024)+- (0.06091)

2 3

Am= -0.18 54

(iii) Given : y0 = 0.3989, y x= 0.3521,^ * 0.2420, y3 = 0.1295, .>>4= 0.540

Using Simpson 3r 1/3 rule for n = 4.

C = +y*^ + 4(>l + + 20^2)]

f 2 f{x) dx - — [(0.3989 + 0.0540) + 4(0.3521 + 0.1295) + 2(0.242)1Jo 3

0.5(2.8633) =0.4772. Ass

Exercise-3(A)

1. Given «o + u&= 1.9243, u\ + «7= 1.959, U 2 + «6 = 1.9823,1/3 + u$ = 1.9956, find 1/4.

2. if «o = 3, u\ = 12, U 2 = 81, V} = 2000, U 4 - 100, calculate A4ho.

3. Form a table of differences the function/*) = x3+ 5x - 7 for x = -1 ,0 ,1 ,2 ,3 ,4 , 5 continue dutable to obtain/6).

4. Find the missing values in the following table :x : 45 50 55 60 65

y : 3.0 — 2.0 — -2.4

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5. Obtain the missing terms in the following table :

x : 2.0 2.1 2.2 2.3 2.4 2.5 2.6

Ax): 0.135 — 0.111 0.1 — 0.08 0.07

6. Express 3x* - 4jc3 + 6x2 + 2x + 1as a factorial polynomial and find differences of all orders.7. Obtain the function whose first differences is 2x3 + 3X2 - 5x + 4.

8. Locate and correct by means of differences the error in the following table of values.

x : 1 2 3 4 5 6 7 8

y : 3010 3424 3802 4105 4472 4771 5051 5315

9. Find the function whose first difference is e*.

10. Given log 100 = 2, log 101 = 2.0043, log 103 - 2.0128. log 104 = 2.0170, find log 102.

11. What is the lowest degree possible for a polynomial ux which takes the following values :

u\ = 0, u2 = 3, «3 = 8, u4 = 15, u$ ~ 24, u6 - 35, also find the value of ux.

12. The following table is given. Find ux

x: 0 I 2 3 4

y : 3 6 11 18 27

If for another point x = 5, ux - 38 added to above data, will the function be a same as before ordifferent.

13. Prove th at:

(0 /4 ) = /3 ) + A/^2) + A ftl) + Aftl)

(ii)X4) =A®) + 4A/(0) + 6A2A~ 1) + 10A3 - I) as for as third differences.

14. Evaluate (£2 - 3 E + 2) (2 x/h + x).

15. Estimate the production of cotton in the year 1935 from the given data :

Year*: 1931 1932 1933 1934 1935 1936 1937

Production f(x) 17.1 13 14 9.6 12.4 18.2

(in millions)

16. Given u ^ = 0.2884, = 0.5856, U49 = 0.6313,1/50 = 0.7620, find the value of U47.17. Prove that:

(i) E '" = M + l s (ii) fxS =~(A + V) (Hi) l + y = Vl + V 2

(iv) tf(«* + £ -! « )= ft2£)2 _ hzDa + _L h*D* - ......

18. Prove that: W4 = u0 + 4A«o + 6 A2u. 1+ I0A3u_ |.

19. Evaluate :(i) (E ~1A)r3; Takingh = \

N u m e r ic a l A n a l y s is -11 j 2 9 1

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A2*2

2 9 2 | E n g i n e e r i n g M a t h e m a t ic s -111

(ii) E(x + log x) ’

(iii)e*

e* + e~

(iv) A cos (a*);

Taking h - I

Taking h = 1

Taking h - 1

Answers-3(A)

. u4 = 0.9999557 2. - 7459

3. 239 4. ;-5o = 2.925, yM = 0.225 5. J[2.1) = 0.123,X2.4) = 0.09

16. y = 3(.v ]4 + 14 [x ] 3 + 15M2 + l[x) + I; k *y - 72 7. - *[4) + 3*t31 + + c

8.4151 9. f(x) = - r —.-e* -1

12. No change and equal to 2

16.0.4147 19. (i)3x2- 3 x + 1,

(iii)

eh _ g-A

10. 2.0086

14. - h

(ii) *!

(iv) 2 sin

II. x2- I

15. 6.6

' a x(l + a /,) > 'crr ( l - a h)'|

t o I 2 J

5 t c 's * i m n f T T F S

Exercise-3(B)

I. The area A of a circle of diameter d is given for the following values

d: 80 85 90 95 100 A : 5026 5674 6362 7088 7854

Calculate the area of a circle of diameter 105.

2. Given:

x : 1 2 3 4 5 6 7 8

f ix ): 1 8 27 64 125 216 343 512

Find/7.5) using Newton’s Backward difference formula.

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N u m e ric a l A n a ly s is -II | 2 9 3

3. From ihe following table of half yearly premium for policies maturing at different ages, estimatethe premium for police maturing at age 46 and 63.

Age: 45 50 55 60 65

Premium: 114.04 96.16 83.32 74.48 68.484. From the following table of values of x an A fix), determine

(i)X0.23)(ii)A0.29)

x : 0.20 0.22 0.24 j 0.26 0.28 0.30

f ix ): 1.6596 1.6698 1.6804 1.6912 1.7024 1.7139

5. The probability integral /(■*) = e ' z ^d x has following values ;

x: 1.00 1.05 1.10 1.15 1.20 1.25 I

f(x): 0.682689 0.7062S2 Q.728668 0.749856 0.769861 0.78S700 |

Calculate XI.235).6. Find the number of men getting wages between Rs. 10 and Rs. 15 from following table :

Wages (in Rs.): 0 -1 0 10-20 20-30 30-40

Frequency: 9 30 35 42

7. In an examination, the number of students who obtained marks between certain limits are asfollows :

M arks: 0 -1 9 2 0-39 4 0 -5 9 6 0 -7 9 8* -9 9 |

No. of s tudents: 41 62 • ' 65 50 17

Estimate the number of candidates who obtained fewer than 70 marks.8. Using Newton backward differences formula to the given data, find a polynomial J(x);

x : 1 2 3 4 5 £ t i

1 -1 1 -1 "l

9. From the following table, find y, when x = 1.84 and 2.4 by Newton’s interpolation formula :

* : 1.7 1.8 1.9 2.0 2.1 2.2 2.3

I I U 5.474 6.050 6.686 7.389 8.166 9.025 9.974

10. Find the polynomialX*)» if

x: 3 5 7 9 11

y = f i x ) : 6 24 58 108 174

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11. The following table gives the values of e* for certain equidistant values of x. Find the value of <?when x - 0.644 using(i) Stirling’s formula, (ii) Bessel’s formula, (iii) Everett’s formula

2 9 4 | E n g i n e e r in g M a t h em a ti c s- ] II

x : 0.61 0.62 0.63 0.64 0.65 0.66 0.67 y - e x : 1.840431 1.858928 1.877610 1.896481 1.915541 1.934792 1.954237

12. From the following table, find y when x = 1.45

x: 1.0 1.2 1.4 1.6 1.8 2.0

y- 0.0 - 0.112 - 0.016 0.336 0.992 2.0

13. Find ><0.543) from the following values of x and y

x : 0.1 0.2 0.3 0.4 0.5 0.6 0.7 y ix ) : 2.631 3.328 4.097 4.944 5.875 6.896 8.013

14. Evaluate sin (0.197) from the data given below :

x : 0.15 0.17 0.19 0.21 0.23

sin ,v: 0.14944 0.16918 0.18886 0.20846 0.22798

15. Interpolate by means of Gauss’s backward formula, the polulation of a town for the year 1974,given that:

Year: 1939 1949 1959 1969 1979 1989

Population

(in thousands)12 15 20 27 39 52

16. Using Gauss’s backward formula estimate the number of persons earning wages between Rs. 60and Rs. 70 from the following data :

Wages (Rs.) : Below 40 4 0 -6 0 6 0 -8 0 80-100 100-120

No. o f persona: (in thousands)

250 120 100 70 50

Answers-3(B)

1. 8666 2. 421.875 3. (i) 110.52567(ii) 1.7081 5. 0.783172 6. 15

8* f ix ) = \ x i - 8x3+ - ^ x 2 - 56x + 31.

1 0 . / x) = 2t2- 7 x + 9. 12. 0.046 13. 6.303

II. (i) 1.90408214.0.19573

(ii) 70.5851527. 194

9. 6.36, and 11.02

(ii) 1.90408215.32.345 (thousands)

4. (i) 1.6751

(iii) 1.904082.16.54000.

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N u m e r ic a l A n aly s is -U | 2 9 5

M M U r m f r I* 5* * r mi M j r n x i *

Exercise-3(C)

1. Using the following table, find^x) as a polynomial in x :

x: - I 0 3 6 7 Ax): 3 - 6 39 822 1611

2. If I) = —3,>{3) = 9,j{4) = 30, and><6) = 132, find the Lagrange’s interpolation polynomialthat take the same values as the function^ at the given points.

3. Given the table of values* : 150 152 154 156

12.247 12.329 12.410 12.490

Evaluate Vl55 using Lagrange’s interpolation fonnula.

4. Applying Lagrange’s formula, find a cubic polynomial which approximates the following data :x : - 2 - 1 2 3

.yCO: - 12 - 8 3 55. Find/(5X by Lagrange’s fonnula for the following data :

x : 1 3 4 6 10 y(x): 0 18 48 480 900

6. The following table gives the normal weights of babies during the first eight months of life :

4ge in months 0 2 5 8Weight in lbs: 6 10 12 16

Estimate the weight of the baby at the age of seven months :7. Find by Lagrange's formula the value of tan 33°, if tan 30“ = 0.5774. tan 32° = 0.6249, tan 35°

= 0.7002 and tan 38° = 0.7813..8. Determine by using Lagrange’s formula the percentage number of criminals under 35 years :

Age (Year): Under 25 Under 30 Under 40 Under 50% number o f criminals: 52.0 67.3 84.1 94.4

x 2 + 6 x - l

9. Express the function (x2 - i)(x - 4)(x - 6) ** a sum fract>ons-

1#. The following table gives the viscosity of an oil as a function temperature. Use Lagrange’s

formula to find viscosity of oil at a temperature of 140°.

Temp0 : 110 130 160 190

Viscosity: 10.8 8.1 5.5 4.8

11. Certain corresponding values of x and logio x are given below :x : 300 304 305 307

l o g i o x : 2.4771 2.4829 2.4843 2.4871

Find logio 310 by (i) Newton’s divided difference formula (ii) Lagrange’s formula.Given XO) = - 18,^1) = 0 ,# ' ^ 0,^5) = - 248,/6 ) = 0,^9) = 13104, find/x).

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13. The following table gives the values of x and y : jc : 1.2 2.1 2.8 4.1 4.9 6.2 y : A2 6.8 9.8 13.4 15.5 19.6

Find the value ofx corresponding to 12 using Lagrange’s technique of inverse interpolate14. Obtain the value of t when A = 85 from the following table using Lagrange’s method

t: 2 5 8 A: 94.8 87.9 81.3

15. Find the value of x when y = fix ) = 19, given thatx : 0 1 2

y=Ax)-- 0 1 2016. Find the value of x when y - 18600, given that

x : 52 53 54 y : 19231 18868 18519

17. Find the value of x when y -fix) - 6, given that

x : 168 120 72 y -A x ) : 3 7 9

1468.7

55IS 182

6310

561-855

l .y(x)= x4 -3 x 3 + 5x2 - 6

1 , 3 , 241 _ 4 . ----- x3 ------ x 2 +------ x - 3.9

15 20 60

Answers-3(C)

2. x3- 3.*2 + 5x - 6

8. 77.4

1

5. 160

13

6. 14 lbs

3. 12.45

7. 0.6494

71

5 (x - l) 35(x +1) 10(x- 4) 70(.r -6 )(ii) 2.4786 12. *s - 9x* + I&c3 - x 2 + 9r - I'i 13. 3.5516. 53.7646 17. 147.

10.7.03 11. (i) 2.478614. 6.5928 15. 2.9

1.

2.

Exercise-3(D)>

Givenx : 3.0 3.2 3.4 3.6

y ~AX) : -14.000 -10.032 - 5.296 0.256Find first and second derivative of/x) = at x - 3.0,Given

x : 7.47 7.48 7.49 7.50.195 .198 .201

3.86.672

7.51.203

4.014.000

7.52.206 y ~AX) '• .193

find dyldx atx = 7.50.3. Using Newton’s divided difference formula, find/XlO) from the following data :

x : 3 5 * H 27 34 A x ) : -1 3 23 899 17315 35606

7.53.208

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Numerical A nal ysis-!! { 297

4. From the table below, for 'vhat value of x, y is mini muni ? Af.;o find ilv.y ■fc 'j •x : 3 4 5 6 *7 s

y : 0.205 0.240 0.259 0.262 '•.250 \J . _ 5. Calculate y'(0) and v'’(0) from the following table I

x : 0 " I 2 J 4

y : 4 8 15 7 66. Find >'(0-5) from the following table :

x: 0.35 0.40 0.45 0.50 0.55 0.60 y r 1.521 1.506 1 . 4 8 8 i .4o~ : 11t ! .4 IL

0 .65

7. A rod is rotating in a plane. The following table give s the 0; rad ians ) through which the rcxi

turned for various values of the time i second t : 0 0.2 0.4 0.6 O.S i.O i.2

6 . U. 0.12 0.49 M 2 2 .0 2 3 .2 0 4 .6 7

Calculate the angular velocity and angular acceleration i\- the :^d w h en / = 0 . 6 seco n d

8. Find the values of cos 1.74 using values given in the ihie below :x : 1.70 1.74 1.78 1 S2 i.So

sin x • 0.9916 0.9857 0.9781 U.v69[Hint Find first derivative at x = 1.74]

9. The following data gives corresponding values of pressure and specific vo:ume of a super heiueosteam.

v : 2 4 6 . 8 1 0

p : 105.0 42.7 25.316 1 3 . 0

Find the rae of change of (i) Pressure with respect to volume when v = 2.

(ii) Volume with respect to pressure when p=

105.10. Find /'(4) from the following data using Lagrange’s formula.

Ax)00

2

8

5125

r i.

Answers-3(D)

t. 3, 18 2. 0.22666 3. 232.87 4. 5.6875,0.2628.

6.0.44 7. 4,08 radian/sec; 7 radian/sec2 8.-0.19833(ii)-0.0191. 10.48

5. - 27.1>. 17.-,.

9. (i) - 52.4

S3SPT3G7 JT~..

Extrcise-3(E)

log(l T x ~)1. Use Simpson’s rule with ten equal to prove that J0 ” (i'+x2}

2. Evaluate e ' ^dx by Simpson’s ‘1/3’ rule (taking h = 0.1).

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ft)3. Use an approximate integration formula to find the vaJue of Jo yx dx , , using the following

2 9 8 | E n g i n e e r i n g Ma t h e m a t ic s -111

6

0

values o fyx : x: 0 1 2 3 4 5 6

yx : 0.146 0.161 0.176 0.190 0.204 0.217 0.230o.s

of

4. Use Simpson’s *3/8* rule to obtain an approximate value of Jq (1 -8 x 2)l/2c£r.*

5. Evaluate J(ein x + cos x)dx. Correct to two decimal places using seven ordinates.

6. Calculate yj(l+x~+ x2)dx. Correct to five places of decimals using eleven ordinates.

7. Find the value of J 3 fe + x 2) *** ^ dividing thp range into eight equal parts.

fS 8. Use Simpson’s ‘ 1/3’ rule to find value of J f(x) dx given

x : I 2 3 4 5 fix): 10 50 70 80 100

9. Dividing the interval in 8 equal use Simpson’s 1/3 rule to evaluate <Jsin x dx .

10. Find the value of x2 log xdx by taking 4 strips.

11. The velocity v of a bike which starts from rest, is fixed intervals o f time /, as follows :/min 2 4 6 8 10 12 14 16 18 20

vkm/min 10 18 25 29 32 20 11 5 2 0Estimate approximately the distance covered in 20 minutes.

12. A curve is given by the table:x : 0 1 2 3 4 5 6

y : 0.0 2.0 2.5 2.3 2.0 1.7 1.5

The x co-ordinate of center of gravity of the area bounded by the curve, the end ordinates and thef 6

-Y-axis is given by A x ~ xy dx, where, A is the area. Find x by using Simpson’s rule.

Answers-3(E)

2. 0.746832 3. 1.13625 4. 0.29159 5. 1.13935 6. 2.970497. 0.4478 8. 256.667 9. 1.61 10. 177.483 11. 309.33

12. 3.032

□

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N u me r ic a l A n a l y s is - I I I ,

Co r r e l a t io n a n d R e g r e s s io n

4.1 NUMERICAL SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS

Many problems of science and engineering can be formulated in terms of ordinary differentialequations with appropriate initial and boundary conditions. When the analytical methods are notof any help to solve differential equations, we shall go in for numerical methods to solve them.It is step-by-step method. The given interval [a, A] over which the independent variable x, is

divided into sub-intervals i.e.,a = x0 <JC1<JC2 < ......< Jt„ = b

The length (h) of each sub-interval is called a step, such thath - x i +] Xf, (/ = 0, 1, 2 , ....h - I)

The numerical solution of initial value problem :

dy — ~ fix , _y) with yixo) = yo are obtained one of the following two forms :

(i) An approximate series for y is ascending powers of x or (x - X q), from which thevalues of y corresponding to specified values of x can be obtained by direct

substitution. This category is known as single step method.For example : Picard’s and Taylor series methods.(ii) Approximate value of y. corresponding to only specified value of j c ,

This category is known as multi-step method.For example : Euler and Runge-Kutta methods.

4.2 PICARD’S METHOD

Consider first-order differential equation :

dyt o =fix>y) - 0 )

with the initial condition, jô ) “yo *•«•>y = yo at x - jtoIntegrating (1), we get

= £ / (* , y)dx

=> y = + ^ f ( x , y ) d x ...(2)

Now, we solve (2) by the method of successive approximations, we assure yo as an initialapproximation of y and obtain a better approximation _vi by replacing/ in the integrand byyo-

Then first approximation _vi is : y] - yo + f f (x ,y0)dx.

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3 0 0 | E n g in e e r in g M a t h e m a t ic s -!H

Similarly, The second approximation y? is given b> :

y 2 >» S'o + £ f ( x , y i ) dx.

By repeated this process, the nth approximation y„ is given by :

y„- yo+£ f(x,yn.i)dx.TJie equation (3) is called Picards iteration formula.

Example 4.0} : I \ in g Picard method to obtain y fo r xÔ .l , i f

dy— = / + X)', with y (0) = I.

Solution. Here. Ax. y) = 1+ xy, xq = 0 and yo = 1.First approximation :

V; -Vo + f f ( x , y 0)dx = 1 + f (1 + x. 1) dx J*o Jo

approximation :

v>- y0 +f f(x.y})dx J.tu

r JO

1 + x I + x + * L '

2 ;

dx

=1 +I'ji+X +X +yJ<£X

_ X2 X3 X4=I +X+--+-- +--.2 3 8

Third approximation : .

y 3 = yo + f X f ( x , y 2) d x JXQ

..(I)

v yx- 1 + x + c-1

...(2)

rJo

I t X2 X 3 X41 + X 1 + X + — + — + —

2 3 8 Jdx

. x- X3 X4 X5 X6= 1 + X + --- + ---- + --- + --- + ---

2 3 8 15 48

[using (2)]

When, x = 0.1, then

t , , , ( .l)2 (.l)3 (.l )4 (.l )5 (.l)6J<0!) - 1 + <■•) * 2 * ~ * g + I T * I T

><0.1) - 1.105.

= 1.105.

Ans.

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N u m e r ic a l A n a l y s i s -111, Correlation& Rckession j 3 0 1

Example 4.02 .*Use Picard’s method to find approximate value o fy when x = 0.1 given that y = 1

when x —Oand *®3x + y 2. {RGPV June 2003j

r

\ (T.ySolution. Here,/*, y) = 3x + y1, xo = 0, yo = 1 ! v ^ 7 = f ( x*-v)

The first approximation :

vi = yo + f f(x,yn)dx -- ,vn + f (3.v + y-W \ Jro *v«»

= 1 -V f X (a r + D d r ^ \ * x f — .Jo 2 ...(0

Second approximation

y-> = >’o + f•>xo

J Ir , . . . 3.v- v ! . r . 3x2

3x + I 1 + x + —— ] idx i i - 1 + x *•

= 1 + j " f f 3x' ~ 4.v2 - 5.v * i ; •

o ,, 4 3 "*=1 + X -»•— X- +—.V-r —.V: • —.. ..12 '2 3 4 2-.

When,x = 0.1, then (2) becomc:.:

n = ><0.1) = 1 * (-1) * | u ) ; - ~ l-ly‘ * ~ 1.1264

Hence, M0.1) = 1.1264. Ans.Example 4.03 : Use Picard’s method to approximate the value o f y when x = 0.2 given thaty ~ / when

dy x —0 and = x + y , (three iterations). Also find error.

i .. dy _ // \ !Solution.Here, fix.y) = x+y, jc0 = O.yo = 1 j - ~

The fir st approximation ;

y\ = > ' « + [ f ( x ,y0) d x = 1 + f (x + 1)dx

= 1 + .*+*^- . ...(I)

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302 | E n g i n e e r i n g M a t h e m a t ic s - ! 11

Second approximation :

y i ~ 1 + J0 f (x ,y \ )d x x + ^1 + x + ■—j

= 1 + X + X 2 + — Xs .6

Third approximation :

y } = 1 + \ j ( x , y 2)dx = 1 + J* * + j l + x + x2 + ^ x 3 j dx

[Using (2)]

= 1 + x + x2 + - x3 + — x4. (3)

3 i 24 JWhen, x ~ 0.2, then (3) becomes :

yi = y (0.2) = 1 + (-2) + (.2)2 + + — - - 1.2427o 24

i.e., y{ 0.2) = 1.2427 ...(4)

dy Exact solution : Given — ----y = x

dxwhich is linear differential equation in y.

I.F. = e 'l ldx = e~J.

Its solution: y{I.F.)~ Jx (I.F.) dx + C

=> ye~x = jxe-* dx + C =~xe-x-e ~ x + C

=> y = - x - \ + C e xUsing initial condition y = 1 whert x = 0, then C = 2

Exact solution, y ~ - x - 1+ 2e*

when, x = 0.2 then ><0.2) = - 2 - 1+ 2e° 2 = 1.2428 ...(5) Error = 1.2428 - 1.2427= 0.0001. Ans.

Example 4.04 :Obtain Picard’s second approximate solution o f

dy x*dx = y T l »M y(0) = 0.

x2Solution. Here. fix , y) = ' ! " £ - x0 « 0, y0 = 0

First approximation :

y i = J o + f * f(x,y t)dx

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Cx X 2 , e x *3=*>+ LTr~TT4* = I *2dx =—. ...(1)J o ( ^ + l ) Jo 3 " y >

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yi = yo + f' f (x ,yi) dx

= 0+ I* -? -— ** - W * 6 + iJ o j f + i T + iJ,

*

I 9

dx[Using (1)]

Putting t =—=>dt =x2dx and t -> 0 to3 J

* = Jo**/a? T I A = [ tan ' l * I ,/ * = tan_1(*3/3)-

0* 9 8Since tan-1 0 =0 - -—+—-3 5

Xs l ( x 3 yi ■ 3 31 3 Ans.

dy Example 4.05: Using Picard's method of successive approximations to ^ = y* ~ **» x**0,y~ I

obtainy in the interval 0 £ x £0.5.(taking h <*=0./).

Solution : Here, y) - y 2- * 2* xo~0,yo~ I.First approximation :

>>i = >o + J* / (x ,l) dx

I** X®

= 1 + f (l-x2)dx « l +x-=-. Jo 3

Second approximation

dx

. Q A M C * 7= 1 + x + x — —— ——x5 + ——x .

6 15 63We obtain value of><x) using (2) at x = 0.1, 0.2,0.3,0.4, 0.5 are :

x : 0 0.1 0.2 0.3 0.4 0.5 y : 1.00 1.11 1.24 1.39 1.56 1.74

tv)’o=U

...d)

[Using (1))

...(2)

Ans.

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Example 4.06 :Use Picard’s method to find approximate y when x * 0.1 given that

dy y - x y = /,* * e n x = 0 a n d - £ = — -

304 J En g i n l e k in g Ma t h e m a t i c s -Ui

Solution. Here,

•: - - £ A*, jbxfc - 1 * £ I n r I * - 1)

r [ 2 - a + x)i= 1+ (X-±~dx-(Xl d x

( l + x ) J o U j c Jo

- t - (2 !og(l + x) - a*)0‘ - I - x + 2 log (I + x) ...(J)

SvconJ <jc; :o;:

r v „ , , , f * (I - 2x + 2 log(l + x)J j■■ •" - £ - 1 * Jo 1 + 2 log(l + jc ) *

- 1 - r : 1-

1 + x - 2 f* —Jo 1 +

_____ 2x _____

1 + 2Iog(l + x)

x

dx

dx2 log(l + x)

whici-. is very difficult to integrate. Hence, we have considered only first approximation.Vvnen * ~ >!. then (1) becomes:

Vi - y (0.1) = 1-(0.1) + 2!og*(l +0.1) = 0.983. Ans.rU.,rf!'s method, find.approximate values o fy and z corresponding to x a 6.1,

■ ■ . , iff: - 2. z (0) - } and

d v d zr - a* + s, ~r~ * x - j r .

< ! X dx

......*’ ft. ji) 2, zq —*I,

A *■— - fix.y, z) = x + z, and

-7~. - 4>U,y,z) = x - ) t-( f-\

tn : U>iiig Picard’s method

n - .To *- f f(*,yo>zo)<&'' [Since-jfo>,> o) = + 2q = x + 1)**0

- 2 •*- f (x + l)dx = 2 + x + ~ x2.•»o 2

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and

N u m e r ic a l A n a ly s is -IH , C o r r e l a t i o n f t R e g re s s io n | 305

z\ = zo + f 4(x,y0,z0)dx [Since #(x,yo, zo)=x+yo2 = x - 4 )

=1+jj(x - 4)dx =1- 4x +|x 2.Second approximation :

y2 = yo+ P f ( x ,y l , z l )dx

c(*= 2 + | U + 1-4 X + -X 2 !dx [v■ /•(Xly „* 1) = x + 2,]

•d)

z2 = 26 + J* t ( x , y l , 2 i)dx [v ^ (x,yi,z) = * - y,2]"v

=1 + f x- | 2 +x+ — x2h { 2 J

=1- 4x +—x2 -x3 ---- - —.2 4 20When, x = 0.1, then (1) and (2) becomes :

n = 2.0852 i.e., ><0.1) = 2.0852

and Z2 = 0.5840 /.e., 2(0.1) = 0.5840,

...(2)

Ans.Example 4.08 :V$lng Picard's method, find approximate values o fy and z corresponding t o x - 0.1,

given thaty(O) - 1 and z(0) - 1/2 and ^ = z > - x 3( y + z )

Solution. Here, *o =0,>o= l»zo = 1/2

dyLet, =/(x ,y,z) = z

dz

dx

First approximation :

fa = t ( x ,y t z) = x*(y + z).

> 1 = ^ 0 + f ( x , y 0 ,Z i})dx = 1 + = 1 +Jo =1+f •

2«= 20 + /* ** ■ *■ *> * - \ + / '* * (* + j ) * = »3x4

+ -----2 8

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306 ) En g in e e r in g M a t h e m a t ic s -] 11


12 + 8

dx

1 x 3x6

= 1 + —+ -----2 40 - .(I)

and z2 = *o + r »*!><** = 77 + + Zi)dx Jxo 2

, i + f V2 Jo

- 2. ^x * x8 3*B _ 2 + ~ 8 ~ + 1 0 + ~64~

When,x = 0.1, then (I) and (2) becomes :n = 1.0500008 Le., >(0.1) * 1.05and z2 - 0.5000385 /.e„ z(0.1) * 0.5. Ans.

Example 4.09 :Uslng Picard 's method, fin d approxim ate y when x - 0.1 given that

“ 7 + f i ~ + y = (»4«f> = W, = «.l whenx~0.

Solution. Let,d2y dz

dx zz> dx 2 dx

dzThus the given equation reduces to — + 2xz + y = 0; y(0) = 0.5, z(0) = 0.1

dy dzLet, = J(x, y, z) = z\ and ^ = $ (x, z) = - (2xz + j>) with the conditions

yo = 0.5, zo = 0.1 atxo = 0.First approximation:

y\ = yo + f* f (x ,y0,Zo)dx JXQ

y x = 0.5 + J o Zq dx = 0.5 + J (0.1)dx = 0.5 + (0.1)x. ...(1)

zi * *o + f* ^ J o - ^ b )■>*0

= 0 . 1 - J o ( 2 x z q + y 0 ) dx [ v ^ x , y 0 , Z 0> = - 2 ( x z q + y 0 ) ]

= 0.1 - J J (0.2x + 0.5) ctx » 0.1 - 0.5x - (0.1) x2. ...(2)

or

and

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I1

yj = J o + f * f (x ,y i ,z , ) dx

or y 2 = 0.5 + zx dx [v f ( x ,y u z{) = zv\

= 0.5 + f * (0.1 - 0.5* - 0. lx 2]d* [Using (2)]Jo

= 0.5 + (0.1)* - x 2 - t ^ - x * . ...(3)2 3

and z2 = *o + J* = 0.1 - £ (2*Zj + y,)dx

= 0.1 - J*[2*(0.1 - 0.5* - 0.1*2) + (0.5 + 0.1*)}dr

[by (1) and (2)]

= 0.1 - (0.5)* - ^ *2 + ^ *4. ...(4)Z O 4

When, X= 0.1, then (3) becomes:^ = 0.50746667 ».e.,j<0.1)* 0.5075. Ans.

4J TAYLOR SERIES METHOD

Taylor series method is the simplest method for obtaining the solution of an initial value problem

provided all the conditions prescribed are valid at the initial point x - x q .

Consider a first order differential equation :

dy fa =fa , >0 with y(xo) =y0. -•■(')

Let the y(x) is the exact solution of (1) with ><xo) * 0. then y(x) can be expanded into a Taylorseries about the point jc = jc0 is given by

. . . . . . . . ^ . ( * " * o > 2 . . . . (* " *o)3 . y(x) = yo +{x - x0)y0 + ---- — ---- y0 + ----- — yo + — ...(2)

which is convergent series in* within the interval [*&, *„] the values of y£„ yj, y j , ... are known.

From (1), we can write / =Ax,y) from which/o can be found i.e., / 0 -flxo, yd)- Now, using the fonnula for total derivatives, we can write

/ ' =f(x>y) =fx +yfy =fx +ffy IV f= A ^y))when the suffixes denote partial derivatives with respect to the variable concerned.Similarly,

/ " =» f ' (x ,y ) = f„ + f ^ f + f[ fys + fyyf] + fy[fx + fyf]

~ fxx + If fyy + P f y y + fx fy + f fy an d SO On.

Now, by substituting* = x q andy =yo i n / , / ' , / " , .... we can obtain the constants y , y'0t y%,...

Substituting these values in (2) we can obtain the desired solution of (1).

N u m e r i c a l A n a l y s is -111, C o r r e l a t io n & R e g r e s s io n | 3 0 7

Second approximation :

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3 0 6 | En g in e e r in g Ma t h e j k a t k s -III

dy , „ Example 4.10: Solve - 1 - 2 xy, given that y(0) ^ 0 by any method. /RGPV Dec. 2

Solution : Given,

dy - 1- 2xy =>y' - \ -2 xy and xq ~ 0,yo = 0.

By Taylor series : ><x) = yo + (* ~ ■*o)>o + -—— Jo + —

Since / ( x ) - 1- 2 x y ^ / o ~ 1-Ix^yo = i [vyo=}<

Differentiate : / '( * ) ~ - 2 y ~ 2 x /, then yo" - - 2y0 - 2xayo = 0.Differentiate : /" (* ) = - 2 / - 2xy" - 2? = - 4 / - 2xy”,

then yo"’ ~ - 4yd - 2 xq / ' q - - 4.

Differentiate : /% x) « - 4y>’ - 2jy”' - 2/ ' , then y0tv = - 6y0" - 2xQy0"' = 0.Differentiate : yvW, = - 6 /” - 2xy* - 2 /", then

yov = - Syo'" - 2xdyolv = 32 .... and so on.Hence (1) becomes:

Mx) = 0 + x.(l) + 0. ~ < - 4 ) + 0 + ^ - (32) + ...

2x? 4 .

><jc) = x - - r + i 6 ”d y

Example 4.11 :Vsing Taylor’s series, fin d y(2.1) correct to 5 decimal places. Given * = *

withy (2) - 2.

x — y ySolution. Here xo - 2,yo = 2, / --------- = 1 ------

X X

ySince / = 1 - — J then yo' =0,

Differentiate: y" = — + —£ I then yd' = ^

y " 2 y ' 2 yDifferentiate : / " ---------- + — ---------3-; then yd " = “

X X X

y" 3y* 6y' 6y 3Differentiate : y* = - — + — - ; then y ^ = •

Taylor series expansion about xo = 2 is

< n

|

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N u m e ric a l A n a ly s k - I I I , C o r r e l a t i o n f t R e g re s s io n | 3 0 9

When, x = 2.1, then (1) becomes :

*2 .1) = 2 + (2.1 - 2)(0) + - •I ~ 2)2 ( l /2 ) + ( — ~ 2)3 ( -3 /4 ) + ^ 1 ( 3 / 2 ) .

><2.1) =2.00238. Ans.■4.12: U sing Taylor’s series method to obtain approximate value o f y at

x - 0.2fo r the differential equation - 2 y + 3e* ,y (0 ) = 0. Compare the numerical

solution with the exact solution. [RGPV June 2004jGiven,

dy f a = 2y + 3ex; xo = 0,yo = 0

Since / = 2y + 3e* => yo ~ 2yo + 3ex = 3.

Differentiate : y" = 2 / + 3ex => y b" = 2yo + 3e° = 2(3) + 3 = 9.Differentiate : / " = 2y" + 3e* => y0'" —2yo" + 3e° = 2(9) + 3 = 21.

Differentiate: / ’'= 2 / ' ' + 3e* => yiv = 2yo‘" + 3e° = 2(2!) + 3 = 45and so on.Using Taylor’s series :

x , ) - * + < » - xM ♦ * -

=> a ») - o + *<3)+ f r <9>+ | r <21) + i t <45>+•••9x2 7ac3 15x4

=> ><*) * 3x + ~2~ + ~2~ + ~ 8 ~ +

When, x = 0.2, then (1) becomes :

9(0.2)2 7(0.2)3 15(0.2)4>(0.2) = 3(0.2) + + ~ 1 —

or ><0.2) =0.6 + 0.18 + 0.028 + 0.003 = 0.8110. ...(2)To find exact value : Since,

which is linear differential equation in7 ,

I .F _ = e - JM * = ^ 2 *

Its solution : X®-21) = J3e*.(e_2r)dx

=> y / r2* = 3je~xdx =-3e~x + c

y =~3<*+ce2x

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310 | Eng ineer ing Mm w n a h c s -111

Give initial conditions : y - 0, when x - 0, then0 = - 3 +c=>c~ 3

Hence y = 3c2* - 3e*

v Exact value at x - 0.2 is>(0.2) = 3e%02) - 3e° 2 = 0.8112 ...(2)J£ m r = 0.8112 - 0.8110 = 0.0002. Aa

dy Example 4.13 :Solve by numerical method ^ = * + y* , y (0 ) - J. Find y a tx =»0.1.

[RGP V Ju ne 2002ft

Solution. Here, xo =0,>^= 1a n d / = x + .y2.Sin ce y +x => >>o' = 1.

Differentiate : / ' = 2>/ + 1 => yo" - 2 + 1=3Differentiate : / " = 2(y')2 + 2>y' => yo’" = 2 + 2 * 3 = 8.Differentiate : y» = 4 /y ' + 2 / / ’ + 2>y" = 6yy ' + 2>y”

=> >0^ = 6 * 3 + 2 * 8 = 34.Using Taylor’s series :

(X - X n ) , (X - X n)2 , (x - Xo)3 „* x ) = J o + ■ •-y !" - -J o + v- - 2-^ -yg + -v- - y p— J o + -

><x) = J o + ~ JS + § y J o + j j - J g ' + - [ v xo = 0]

= 1 + x + -—(3) + ~-(8 ) + -—(34) + ...2 6 24

3 , 4 , 17 ,« 1 + X + — X2 + — Xs + --- X4 + ...2 3 12

When x = 0.1, then (1) becomes

X0.1) =f 1 + 0 .1 + | (0.1 )* + i (o . i)8 + i | ( 0 .1 ) 4 = 1 .1165 . Ans.

dy d z Example 4.14:Solve by Taylor’s series method - x + z , — = x - y* with y(0) * 2, z(0) * /.

Flndy (0.1) and z(0.1).

Solution. Here,xo = 0 , = 2, zo ~ 1an dy = x+>, t - x - y 1. Theny =x + z=i*>^'=xo + 2o= 1, and / = x - y 2=>zo' = -Jco->'o2 = - 4

y ' = I + z7=> yo" = 1+ z&' = 1- 4 = -3, and

z" « I-2>y =*Zo" « 1-lyay* = “ 3y " = z" => >o'" = zo" - - 3 and so on. and z"' = - 2(>y' + y 2)

=> zo'" » - 2j>oVo" +>'o'23 = 10 and so on.’)T-J»

Using Taylor i ieriei y am/ z :

(X - Xn) , (X - Xn)2 . (X - Xq )3 » ><x) - Jo + Jo + 2 ? " J o + ------3 j Jo + - ( I )

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(X - X n ) , (X - X * )2 , ( x - Xn )3 ,

and zfr) = *o + - ! ! 2 ! 15 3 ! + "( }

When x = 0.1, then (1) and (2) becomes : ,

*0.1) = 2 + (0.1X1) + (-3) + - ( - 3 ) + .... [v xo = 0]

= 2 + 0.1 - 0.015 - 0.0005 * 2.0845. Ans.

and 2(0.1) = 1 + (0.1X-4) + - ( - 3 ) + ^ - ( 1 0 ) + ....

= 1- 0.4 - 0.015 + 0.001667 = 0.5867. Ans.

d*y dyxample 4.15 :Solve - y + * — > given y(0) - l t yV) = 0 and calculate y (0.1).

olution. Here, xq = 0,>t> = l,yo' = 0 and / ' = y + x y x> yQ” = >>o+ xq>>o' - 1 .

Differentiate : / " = / + / + xy" = 2 / + x y ' => yo" = 2ytf + xqvo" = 0Differentiate : /* - 2 / ' + y" + xy"' => yd* = 3yo" + xqVo"' = 3.Differentiate : y v = 4 /" + xy** =>yov = 0.

Differentiate : y* = Sy*+ xyv => yo*1= 15, and so on.Using Taylor's series :

(X - Xn ) , ( X - X n )2 ,

yQ c)-* o+ - { i Ly° + 21 + **'

When x = 0.1 and putting these values, we get

K0 , , . 1 + 0 + C ^ a ) + 0 * ® j f ( 3 ) t 0 + i ( I 5 )

= 1.005013. Ans.

14 EULER’S METHOD

Euler's method is one of the oldest and simplest method.Let the first order differential equation be

dy

~AX»y) with yixo) - yo- ...(1)

Suppose we are to find the value / at n-points besides xo, which are equally spaced with a step-length h.

Then, Xj =xo + h, X 2 = xo + 2h,..... . x„ - xo +nh.First approximation:

p d y = P f(x ,y)dx [From (1)] J jo *>*0

Replaclng/x, y) by the approximation/xo, yo), we get

yi -yo = *** f(x*>yo)dx = f(Xo,yQ) p dx

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=> y\ =>^ + (jfi -xo)A*<hyo)

y \ =yo + ¥(xo>yo) [vA=*i-*o] ...(2Similarly second approximation:

yz +/»/(X|,>'i)

«* approximation:y„ = y„-

j+hj(x„- ,,y„ .

,)■ - (3 Remark: To get more accurate result, h should be very small. Example 4.16:Using Euler's method, fin d an approximate value ofy corresponding to j c * 0.1, given

d y y - x that - yJr X withy(O) * /.

y - xSolution : Here,./fo y) = . x0 = 0, y0 = 1.

We divide the interval (xo, x) =(0,0.1) into n - 5 steps i.e.,

X —Xq 0.1 ~ 0h -

------ — =>A = — -— =0.02.n 6

Then, xi = xo + h - 0.02, x2 - xo + 2h = 0.04,X3 = 0.06, X4 = 0.08 and X5 = 0.1.

By Euler's method :

yn ~yn-1 + A/(x„-1, yn- 1), where n » 1,2,3,4,5.

y \ =>0+Wxo,yo) * 1+(0,02 (lTo, = I 02‘ ' = jj "+'*£

^ =>1 + M î.Ji) « 1-02 + (0.02) / 0.02, 1.02). ,n nn J 1.02 - 0 .0 2 )

. 1.02 + (0.02)^ ^ — j , 1.0392

« - » * 1 03 92 + (0 0 2)( i S ^ h ) * u m

* - « - « f e n > - U W T ♦ - 1.0756.

» + « « . * > - » 0756 ♦ W ( ^ S S ) - 1.0928.

Thus, ys ~y(x =0.1)= 1.0928. Am Examplf .ft.V7 : Using Euler’s method solve the differential equation In six steps :

* ! ~dx ^ x + y* y(0) ** 0, choosing h m 0.2. [RGPV Dec. 2003}

Solution. Here,/x,y) = x + y, xo ~ 0,yo - 0, h - 0.2 and n - 6 .By Euler’s method:

yn=yn- 1+ hAxn- i,y n- 1). where, n = 1,2,3 ,4, 5 ,6.Then, xj = xq + h~ 0.2,X2 “ 0.4, xj = 0.6,X4 = 0.8, X5 = 1.0,x6 = 1.2.

312 | E n g b s e k n g Ma t h e m a t i c s -III

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N u m e r ic a l A n a l y s is -111, C o r r e l a t io n & R e g r e s s io n | 3 1 3

y\ =yo + a/xo,/o) = o + (0.2X0 + 0) = o. y i= y i+ hfixuyO - 0 + (0J2X0.2 + 0) =■0.04.

y3=yi + h fe 2,yi ) = 0.04 + (0.2X0.4 + 0.04) = 0.128.>4 =>3 + hftx3, / j ) = 0.128 + (0JX0.6 + 0.128) = 0.2736.

/5 =/4 + hfix4,y*) - 0.2736 + (0.2X0.8 + 0.2736) = 0.48832.= /5 + AXx5, / 5) - 0.48832 + (0.2X1.0 + 0.48832) = 0.785984.

Hence, y$ =><1.2) = 0.785984. Ans. Example 4.18: Findy (2,2) using Euler methodfrom the equation

yi =yo + h.f(xo, yo) = 1+ (0.04) [- x0y<?) = 1+ 0.04 (-2 * P)

= 0.92 y i = yx + h.f(xu * ) = 0.92 + (0.04) [- (2.04) (0.92)*]= 0.8509

yy =y2 + fif(x2,y2)= 0.8509 + (0.04) [- (2.08) (0.8509?]= 0.7907

y* =/3+ h.f(xi,yi)= 0.7907 + (0.04) [- (2.12) (0.7907)2]- 0,7377

>5 =/4-»-A./(jf4,>4)= 0.7377 + (0.04) [- (2.16) (0.7377)2]= 0.6907.

Solution. Here, xo = 0, = 0,Xx, y) “ I - 2xy, h “ 0.2 and x - 0.6.

— — xy2 wttky(2)~ 1. [RGPV, June 2011[

Solution. Here f (x ,y ) = -x y t

Xo - 2, /o “ 1,

7b f indy(x) a t x = 2.2 i.e., y (2.2) : •

x - x0 2.2 -2Taking rc = 5, h - ------- -- — -— = 0.04

f t J

x0 = 2, Xj = Xo + h = 2.04, x2= xo + 2A = 2.08xj = xb + 3h - 2.12, X4 =xo + 4h - 2.16,xj = xo + 5h = 2.2.

By Euler's Method: y„ =y»-1+ hf(x„-uy0. i), Where n = 1,2,3,4, 5.

Then

2xy, y(0) =0, toke h- 0.2.

[RGPV June 2006}

Ans.

X - X q h --------- - => 0.2 =

0 . 6 - 0

n n=>n- 3 (steps)

Then, xj = xo + h = 0.2, x2 = 0.4, and xj = 0.6.

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314 | E n g i n e e r i n g Ma ih e m a t j c s -II!

By Euler’s method:

yn = /« - j + hj{x„_ u y » - l), where, « = 1,2, 3.

y\ =>o + M*o,yo) = yo + AO - 2xq)>d) = 0 + (0.2X0 = 0.2.Yi =y\ +AX*J,yi)=>i + W -2 ri /i)

= 02 + (0.2X1 ~ 2 * 0.2 x 0.2) = 0.384.>3 + ft fix2t + - 2X2 2)

= 0.384 + (0.2X1 - 2 x 0.4 x 0.384) - 0.52256.Hence, -yixy) =><0.6) = 0.52256. Ans.

Example 4.20 : Solve the equation $x + y* - 2 ~ 0 ,y (4 ) = 1 fo r y(4.1), taking h “ 0.1 and

using Euler's method.

dy 2 - y3 2 - y 2Solution. Given : — = — ----- , i.e.,Ax, y) = — ----- . xo = 4, /o = 1. A = 0.1, then

ax dx ox x j = xo + A —4 + 0.1 = 4 .1 .

By Euler’s or Simple Euler’s method :

y\ =yo + hfiM,ya)

2 -

5x0

Hence, y\ =XJci ) =y(4-0=: 1.0050. Ans.

4.5 IMPROVED EULER’S METHOD

The Improved Euler’s fonnula is :

y„<'> = y«-i + ~ [ f i x K_l ,y n. x) + f(xK, y J \ , wherey„=yn_ i + hf(xn~ h y„ . i)

or yn<» = yK-1 + [/(*»-!,y ,-i) + f{x«,y*-\ + /i/(xB_„y,_i)}] for n = 1,2 ,3,...

dy . Example 4.21 : I f = x + y* andy ® 1 a t x mO. Find an approximate value o fy a tx « 0.2 by

improved Euler’s method (taking h m 0,1).Solution. Here, /x ,y ) = x + y 2,xo = 0,yo= 1, A= 0.1.

Then, X] =xo + h - 0.1, andx2 = xt>+ 2h = 0.2.By Improved Euler’s fonnula :

/ ,«) = y0 + [/(xo.yo) + /f a .f t) ] ...(1)

where y x =yo + = I + (0.1)(0 + 12)= 1;1From(l), we get

/(<'> = yo + ^ ^ [/(* o » > o ) + /(x1(l . l) ]

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N u m e r i c a l A n a l y s is -III , C o r r e l a t i o n & R e g r e s s io n | 3 1 5

where

Now,

= 1 + ~ [ / ( ° . 1) + /(0.1, 1.1)]U

= 1 + 0.05[(0 + 12>+ (0.1 +(1.1)2)]= 1.1155.

n (» - y,(,) +~[f(xl,y ^ ) + f(xi ,y^] , ...(2)

><2 = y,(,>+ M * ,IJ l(,)) != 1.1155 + (0.1)x0.1. 1.1155)= 1.1155 +(0.1) [0.1 +(!. 1155)2]= J.2499.

From (2), we get

^(D = 1.1155 + 1 / (0 .1 , 1.1155) + /(0.2, 1.2499)]2

= 1.1155 + (0.05)[{0.1 +(1.1155)2} + (0.2 + (1.2499)2}]

= 1.2708.Hence, >2(I) = X*2) = X0.2) = 1.2708.

dy

Ans.

Example 4.22 ••Solve the equation $ x - 2 - y*; y(4) - I, fo r y (4.1), taking h - 0.1 and using

improved Euler’s formula.

2 -y *Solution. Here, J[x,y) =

5x■. xo = 4,>o= 1, and A= 0.1.

Then, x\ =xo + A= 4 + 0.1 =4.1.

By Improved Euler’s formula :

where

>i(l) “ y<>+ + / ( * l J i )]

y\ =yo +hfixo,>0) = l + (0.1 M4, i)

...(l)

1.005

From (1), we get

3l(i) = 1 + M [ / ( 4 , i ) + /(4.1, 1.005)]

,(i) = 1 + 0.05(2- l2) ] [2 - (1.005)2]5x4 5x4.1 = 1.0049.

Hence, yjW =y(;ri) =>(4.1) = 1.0049.

4.6 MODIFIED EULER’S METHOD

Consider the differential equation:

dy

4 i

Ans.

...(1)

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316 | En g in e e r in g M a t h e m a t ic s -IU

The approximation are:

yn

i.e., y j

+ , = j* +~ » j , + -/<* *, j , ) j ; f0rn = o, 1, 2,

= Jo + * / {* « + ~ » Jo + | / ( ^ J o ) J

yz = Ji + h / jx , + y, + | / ( x , ,y ,) j and,1so on.I z z j

Example 4.23: So/ve = J “ “ » >(ft>“ 1 f»ry(0.1), taking h * 0.1 and using modified Euler’s

method.

2x

Solution. Wen, fix>y) = J “ »*o = 0> Jo = I, A = 0.1, then xj = xo + /r = 0.1.

By Modified Euler’s formula :

= Jo + j*o + “ »Jo + Jo )|

(0.1)/ jo.05, 1+ f / (0, i)j ^ s i n c e /(0,1) = 1- — - = 1

(0.1) f{0 .05, 1.05} = 1 + (0.1) |l .0 5 -

y\

= 1 +

= 1 +

= 1+(0.1X0.95476)= 1.09548.

Hence, y\ =><xj) =><0.l) = 1.09548. Ans.

A'V Example 4.24 : Solve the equation 5x — + y* = 2; » 7 fory(4.1), taking h “ f t / ««</ using

modified Euler’s method.

2 - y 2Solution. Here,y(x, y) = — ----- , xa~ 4,><o = 1, /i = 0.1, then Xj = xq + A= 4.1.

OX

By Modified Euler’s formula :

y, = Jo + h f

= 1+

h x o + - , y o + | /(aco,Jo)j

(0.1)/|4 + H , i + M /(4 , i)J

= l + ( 0.1 )/ |4 .0 5 , l + 0 . 0 5 ^ ^ ^ - J j

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N u m e r ic a l An a l y s is -111, C o r r e l a t i o n & R e g r es sio n | 317

= 1+(0.1)/{4.05, 1.025}

2 - (1.025)2= 1 + (0.1)

5 x 4.051.0047.

Hence, y \ =><xt)“ >(4-1)= 1.0047. Ans. Example 4.25 : Use m odify E u ler’s m ethod to compute y fo r x — 0.05. Given that

dy * , — = * + y , x0 = O, y0 = J, result correct upto three decimal places.

[RGPV Dec. 2002]

Solution. Here,./fa y) = x + y, with xq = 0, = I

Since compute ><0.05), so we take h - 0.05, then xi=xo + h ~ 0.05.By Modified Euler’s formula :

y\ =y0+ A/ | *o + yo + f(*o>yo)J

(0.05) / |o + 1+ /(0, 1)J [Since X0, 1)= 0 + I = 1]= 1 +

- 1+ (0.05)f[0.025, 1.025} = 1 + (0.05X0.025 + 1.025)« 1.0525.

Hence, y\ =><*)) = .K0.05) = 1.0525. Ans.

Example 4 . 2 6 : Solve by Euler’s modify method ^ = iog,(x + y ), y ( i ) =2 at x ^ 1.2 andatx= 1.4

with h “ 0.2. [RGPV Dec. 2004 A June 2009]

Solution. Here,X*, y) = log* (x + y)f x0 = l,yo = 2 and h = 0.2,then *i = xq + h= 1+0.2 = 1.2 and X 2 = xo + 2h - 1.4.By Modified Euler’s formula:

+A/ ' j*b + ~ , yo + ^ / ( ^ . > o ) J

= 2 + (0.2 )/j l + 2 + /(l, 2)|

[Sin ce/l , 2) = log, (1 + 2) = 1.0986]

= 2 + (0.2)/{1.1,2.10986} =2 + (0.2) log,(1.1 + 2.10986)= 2.2332.

Hence, yi =y(xi) =y( 1.2) = 2.2332.

Again, yi = * + | / ( * j . y , ) |

= 2.2332 + (0.2) f { 1.2 + , 2.2332 + /(1 .2, 2.2332)J

[Since/1.2, 2.2332) = 1.2335]

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= 2.2332 + (0.2)/{1.3,2.35655} = 2.2332 + (0.2) log, (1.3 + 2.35655)= 2.4925

Hence, yi = yix2) =>(1.4) = 2.4925. Ans.

4.7 RUNGE-KUTTA METHOD

Runge-Kutta method is one of the earliest and the most widely used in numerical methods ofsolving ordinary differential equations of the first order.The fourth order Runge-Kutta method is most commonly used and is reffered as 'Runge-Kutta method Consider first order differential equation :

dy fa ~A*) with ><xo) = yo.

To find >i atxi =X q + h \y iH x i~ ao + 2h and so on.Then we compute, = h Axo, Jo);

*4 = HA*o+h, y0+h).

k - + 2*2 + 2*3 + k4) wherek is weighted mean of *i, k2, k$ and fcj.6

Hence y\ = yo + ki.e .,y(x\ )=yo + k. Now to compute>2 i.e., >2 ~y\ + £: we start from (jci,>i) and replacing the above process, we get

(*2» yi) and so on- Remarks:

(i) MThe Runge-Kutta method of first order is known as Euler ’s method (ii) The Runge-Kutta method of second order is :

k\=hfa>,yQ)

*2 - v ( * o + f ’» + y )

Then,>i =>0 + where k - k 2. Similarly we can compute yi, >3 ....and so on.“The Runge-Kutta method of second order is known as modified Euler’s method ’.

(iii) The Runge-Kutta method of third order:*1 =hAxo,yo)

* * = * / ( * > + £ . * > + 7 )

*3 = I + *. >D + 2 * 2 -* t)

* = “ (&l +4*2 +*})• b

Then, y\ - yo + k, where k = k2. Similarly we can compute yi, >3, .... and so on.

“The Runge-Kutta method of third order is known as Range’s method'.

318 j En g in e e r in g Ma t h e m a t i c s -III

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Example 4.27 .* Apply Runge-Kutta method o f fourth order to solve :

dy10 d x +y2>'y(°> " 1 f° r x * 0 L fRGPV Dec. 2002}

*

x2 + y2Solution. Here, fix,y) = — rr — > XQ-OtyQ- 1, and x = 0.1,

N u m e r ic a l A n a l y s is -111, C o r r e l a t io n & Regression | 319

Then,

We have

Hence,

=> y(0A) **1 +0.0101345 - 1.0101345. Ans.Example 4.28 .*Using Runge-Kutta method of fourth order to solve:

y '~ x y fo r x - 1.2. Initially x - l , y - 2 ( t a k e h - 0.1) fRGPV June 2003f

Solution. Here, fix,y) ~xy,xo = !,>>o = 2,A = 0.1.

Then = xq + h= M ,andx2 = *o + 2A = 1.2.Hence we compute^) and /2 (i.e., two steps)

Step 1: Starting (x6,>»o):We have ky = hfixo, y0) = (0.1 Xô/c) = (0.1X1 *2) = 0.2

* » + ^ -j - (0.1)^1.05,2.1)

= (0.1X1.05^2.1) = 0.2205.

xu

X - X q 0 . 1 - 0h = — — = — ----- =0.1

n 1 x\ =x q + h = 0 . 1 .

*i = M ^ > o ) = ( o . i) m i ) = 0*1

[As taking n = 1 i.e., one step]

( 0 + l2 ''I

n rJ"0-0k2 = h f \ x o + - , y0 + —J =(0.1)/0.05, 1.005)

= 0.1(0.05)2 + (1.005)21

1 0 J- 0.01012525

*3 = h f [ y Co+ | , Jo + y j =(0.1)/0.05, 1.0050626)

(0.05)2 + (1.0050626)* 1

= 0.1 10 0.010126508

*4 = hfixo + h, y0 + *3) = (0.1 )X0.10, 1.010126508)= 0.010303555

k = ^(A, + 2*2 + 2*3 + A*)D

= 1(0.060807071) =0.0101345 b

y\ ~yo +ki .e . ,y(x\)=yQ + k.

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3 20 | E n g in e e r in g M athem atics-III

= h f ^ X o + ~ , J o + ~ j = ( 0 . 1 ) X 1 . 0 5 , 2 .1103) = 0.2215

and *4 = hftx}i-h,yo + * 3) = (0 .1 )/l .l , 2.2215) = 0.2443

* = ^(*i + 2*2 + 2*3 + *4) • 0.2583.

Hence, y t -yo + k i.e.,y\ =X*i) =X l- 0 ~ 2.2583.Step 2 : Starting (X), >j) :We have

*i = M x ,,y ,) = (0.1)X U,2.2583)=(0.!X U * 2.2583) = 0.2484

h = h f{ *\ + f >J» + ^ j = (0.1)/(1.15,2.3825) = 0.27399

*3 - * /( * » + f ’ Ji + r ) -(0.1V(1.I5,2.395295) = 0.27546and *4 = A/x, + h,y\ + * 3 ) * (0.1)ffl.2, 2.53376) = 0.3041.

k = !<*i + 2*2 + 2*3 + *4) » A. (1.6514) = 0.27523.

Hence, =>1 + * *.e..><1.2) = 2.2583 + 0.27523 = 2.53353. Ans.

dy y* - x* Example 4.29; Using Runge-Kutta method offourth order, solve “j~ - with y(0) • l a

x - 0.2 and 0.4. (RGPV June 2004 A June 200*1

y2 - x2Solution. Here, / x , y) = y + JC2 **0 = 0, yo = I, taking h = 0.2.

.-. We compute y\ and y2. Then xj = 0.2 and x2 = 0.4.Step 1 : Starting from (x&, >0) •

We have *, = hfix<>,yo) = (0.2)/0 , 1) = 0.2 v /(0. 1) = « 1

k2 = h f ^ + | , Jo + y j =(0 2^ 0.1, 1.1)**0.19672

*3 = h f [* 0 + ~ , Jo + y j = ( 0 ^ 0 .1 , 1.09836) = 0.1967

aad k4 • A/x0 + *, + * 3) = (0.2 )/0 .2 ,1.1967) = 0.1891

* = |( * i + 2*2 + 2** + *<)

- i [ 0 . 2 + 2 x ( 0 . 1 9 6 7 2 ) . + 2 x ( 0 . 1 9 6 7 ) + 0 . 1 8 9 1 ] = 0 . 1 9 5 9 9 6

Hence, y\ =yo +k ^ y(0.2) ~ 1+0.19599= 1.19599 * 1.196.

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Step 2 : Starting (xt, / | ) :

We have *, = h f a uy i) = (0.2)tf0.2, 1.196) = Oil891.

*2 = A /(* 1 + ! , y i+ ! } = (0.2)/0.3, 1.2906) = 0.1795

k} - h f { x i + £ . yi + ~ j = (0 .2 )M l 1.2858) = 0.1793

and *4 = hj{x] +h,y\ + kj) = (0.2)/0.4, 1.3753) = 0.1688.

**=!■ (*i + 2/% + 2*3 + *4) = 0.1792.D

Hence, >-2 = y t + k ^ v(0.4) = 1.196 + 0.1792 = 1.3752. Ans.

Example 4 J O : Apply Runge-Kutta fourth order method to fm d an approximate value o fy when x =dy •

0.2 in steps o f 0.1 i f ~ = x + y 2 gi\>en thaty = / where x**0.

fRGPV Dec. 2004, Feb. 2010 and June 2011} Solution. Here, fix , y) - x + y2, x0 = 0,>-0 = 1, taking h = 0.1.

Then xi = xo + * “ 0.1 and x i^xo + 2h = 0.2.Hence we have to compute y\ and y%in two steps :

Step 1: Starting from (xo, >o) :We have *, = h f o o, v0) = (0.1 )tf0, 1) = 0.I [v /0 , 1) = 0 + 1= = 1]

( h fc ^k2 = h f j^xo + - , y0 + — J =(0.1)/0.05, 1.05) = 0.11525

u A h Mk} = h f xo + - , y0 + — I = (0.1)/0.05, 1.057625) = 0.11686

aad *4 = hflx0 + h,y() + kJ) = (0.1 )/0 .1 , 1.11686) = 0.1347

k = i ( * i + 2*2 + 2*3 + *4) =0.1165.o

Hence, y\ =yo + *=>_y(0.1) = 1.1165.Step 2 : Starting(xi,/i), hereX| =0.1,>1 = 1.1165 and * = 0.1.

We have *1 = M * ,,/ i) = (0.1)X0.1,1-1165) = 0.13466

k2= +|, +yj =(0.1)/0.15, 1.18383) =0.15514

u A h Mki = h f l*i + yt + y l =(0.1)/0.15, 1.19407) = 0.15758.

and *4 = hAx \ + h,y \ + *3) = (0.1)/0.2, ! .27408) = 0.18233.

k = |-(*i + 2*2 + 2*3 + *4) = 0.1571.o

Hence, y^ = i| + k i.e.,y(0.2) = 1.1165 + 0.1571 = 1.2736. Ans.

N u m e r ic a l A n a l y s is -111, C o r r e l a t i o n & R e g r e s s io n | 321

L

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Example 4.31; Use Runge-Kutta method to obtain y when x = /./ , given thaty = 1.2 when x = / oji y satisfies the equation :

322 | E n g in e e r in g M a t h e m a t i c s -111

Solution. Here,

so that

Then

We have

IRGPV June 200$

[As n = l]

dy

Ac ~ 3x+S - Ax,y) = 3x + y2, xo = l,yo= 12,x= 1.1,

X - x 0 1 . 1 - 1h --------- - =>* = — -— = 0.1.

n 1

X) = x0 + A= 1 +0.1 = 1.1.

i, = hAxo, y0) = (0.1) (3x0 + >o2) = (0.1 ){3 + (1,2)2] = 0.444

k2 = h f ^ x o + | , Jo =(0.1)^1.05, 1.422)

= (0.1 )[3 x 1.05+ (1.422)23 = 0.517.

*3 = * / ( x 0 + | , y 0 +~*- = (0.1 )/(1 .05 ,1.459)

and

Hence,

= (0 .1)[3 x 1.05 + (1,459)2] = 0.528.

*4 = ^ o + A ,Jo + ^ ) = ( 0 . 1 ) X l .l , 1 -728)

= (0.1)[3 x l.i +(1.728)2] = 0.629.

k = i ( * , + 2*2 + 2*3 + *4) = i (3.163) =0.527. b b

yi =>o +ki.e., >(1 .1)- 1.2 + 0.527= 1.727. Ans.

Example 4.32; Use Runge-Kutta method to approximate the value o f y when xÔ .l given thaty(O)

[RGPV Dec. 2008 (N)J

Solution. Proceed as above example.

Taking x0 = 0, yo = 1, h **0.1 . Then we get

>l =>o + k = 1 + 0.12725= 1.12725. Aw.

dy Example 4.33 ; Solve fa

x + y fo r x = 0.5 by using Runge-Kutta Method with x0 - 0,

..>0 = / (take ft *=0.5).

Solution. Here, xo = 0, >o = 1, h = 0.5 and/x, >) =

Then xj = xo + * = C.5.

(* + >)

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(0.5^0.25, 1.25)

‘ (0 5>( o.25 ’ <0 '6) i 1 * I ‘ 0J3333

*3 = h f \ x o + | , y0 + y ) = (0-5)/^0.25, = 0.35294

*4 = >o + *J) = 0.5/0.5, 1.35294)

(0 .5+”T35294) = °-26984.

/t = •“ (*i + 2*2 + 2*j + *4) = —(2.1423) = 0.35705.6 ” 6

Hence, yi = yo + k i.e.,y(Q.S) = 1.35705. Ans.

dyExample 4.34 : The unique solution o f the problem ^ - ~ xy ,y (0 ) = 1 is y = e~x i a .

Find approximate value ofy(0.2) using Runge-Kutta Method.

Solution. Here, xo = 0, yo ~ 1 ,fix , y) = -xy, taking h = 0.2,

Then X] = xo + h = 0 + 0.2 = 0.2

We have *1 = hfipco, yo)= (0 .2)/0, 1) = 0

Nu m er ic a l An a l y s is -11!, C o r r e l a t io n & R e g r e s s io n | 323

= (0.2X- 0.1 x I) = -{0.2X0.1) = - 0.02

= k f \ x o + | . y0 + -~ j = (0 .2 )/

- (0.2)/0.1, 1-0.01) = (0.2 )/0.1 ,0.99)

= 0.2(- 0.1 x 0.99) = 0.2(- 0.999) = 0.198.*4 - hfix0 + h,y$ + *3) = 0.2 /0 .2 ,1 - 0.0198) = 0.2A0.2,0.9802)

= 0.2(- 0.2 x 0.9802) = 0.2(- 0.19604) = ^ 0.039208

* = - ( * , + 2*0 + 2A3 + *4) = -(-0.118808) = -.019801.6 6

Hence/1 -yo * ki.e.yy(0.2)= 1-0.019801 = 0.9802. Ans.

Since y = er*m solution of given problem,

Then atx = 0.2, is exact value = e~<'2',2/1 = 0.98012.

Hence, error = 0.9802 - 0.98012 = 0.00008. Ans.

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dy - Example 4.35 : Apply Runge-Kutta method to — = xya, >(1)= 1 to obtain y(1.1).

dx

Solution. Here, - 1, yo ~ 1and A= 0.1 then X| = x<>+ A = 1+ 0.1 = 1.1.

We have kx = Afco, y0) = 0.1(1X1),/J = 0-1l

3 2 4 | E n g in e e r in g M a t h e m a t ic s -111

h = A / jx 0 + | , Jo + j ] = 0 . l( l + ~ ) ( l + j 3 =0.10672

i

1.A h h ) n A* 0.1V, 0.10672 >3ky = A/1 x0 + —,y0 + — J * 0.1 1 + — | 1 + --- ---- j -0.10684

*4 =W.x0 + h,y0 +ki) = 0.1(1 +0.1X1 + 0.t0684),/3 = 0.11378.

k = i( 0 .1 + 2x 0.1 06 72 + 2 x0 .10 68 4+ 0.11378) = 0.106826 '

Hence y } = yo +k i .e.,y(lA) = I +0.10682 = 1.10682. Ans.

dy Example 4-36 ; Solve the initial value problem ~ ^ = -2xy* withy(O) * / and A = 0.2 on the intern!

[0, IJ. Use Runge-Kutta fourth order method.

Solution. Step 1 : Here XQ= 0,yo = 1, A = 0.2 andjfc, y ) ~ - 2xy2

*i = hfao, y o ) - - 2(0.2X0X 1y = 0 [v y0) = - 2x<>v02

k2 = hf^Xo + | , y0 + = - 2 ( 0 . 2 ) (l)2 = _o.04

*3 = A/J Xo + | , >-o + y j = -2(0.2) ^pj(0.98)2 = _0.38416

*4 = Aftr0 + A, y0 + h ) = - 2(0.2X0.2X0.961584)2 = - 0.0739715

* = - (*i + 2*2 + 2As + *4) = - 0.0384672. b

Hence y } =y(0.2)=y0 + * = 1-0.0384672 = 0.9615328.

Step 2 : We have x\ = 0.2, yi = 0.9615328, A = 0.2, we have*1 = A/xj.y ,) - - 2(0.2X0.2X0.9615328)2 = - 0.0739636

u A h kAk2 = A/^Xj + - . J 1 +-J-J =-2(0.2X0.3X0.924551)2 = -0.1025754

{ h> ho ^*3 = V ^ X , + - . J 1 + y j = -2 (0 .2X 0.3 X0 .9 10 24 51)2 = -0.0994255

*4 = A /x i + A, y, + * 3) = -2(0.2X0:4X0 .8621073)2 = -0 .1189 166 .

* = ~ (Aj + 2*2 + 2*3 + *<) = _ 0.0994803. b

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Hence y2 = MO-4) = y, + * = 0.9615328 - 0.0994803 - 0.8620525.

Step 3 : y3 = y(0.6) - 0.7352784

Step 4 : y4 = y(0.8) = 0.6097519

Step 5 : y5 = y( 1.0) = 0.5000073. Ans.Example 4.37 ; Using Runge-Kutta method, solve y"** xy'2- y 2 for x = 0.2 corrected to four decimal

places. Initial conditions are x =■0, y = /, y '= ft

N u m e r ic a l A n a l y s is -111, C o r r e l a t i o n & R e g r e s s io n j 325

Solution. Letdy

y = =z=Ax ,y , z ) . say.

dzSo that y" = = xz2 - y 2 = <f>(x, y, z), say.

Here x0 = 0, yo ~ 1. *o ~ 0, h = 0.2Using Runge-Kutta formulae : we have

*l = tyfro, yo, Zo) = 0.2(0) = 0

[vy '=2]

[v y -Z =>=o = yo' = 0]

/i = h# (xo, yo, 20) = 0.2(- 1) = -0 .2

i J h h U ' h *, nk2 = y o + f>Zo + J h = **(*> + 2 ,y° + 2 ’Z° + 2,

= 0.2( - 0 . 1) = - 0.02

*r ( * *2 ^*3 = h f ^ x 0 + - , y o + ^ , z 0 + ^ )

= 0 . 2 ( - 0 .0 99 9) = - 0 . 0 2

*4 = hj[xo + h, y0 + *3, z0 + /3)

= 0 .2 ( -0 .1958 ) = -0 .039 2

k - ^ (* t + 2*2 + 2*3 + fe4) b

= 0.2 ^(0.1, 1, -0.1) = 0.2

[(0.1K-O.1)2 - 1] = - 0.1998

h - h * { *

= 0.2 (-0.9791) = -0.1958

l4 = h<f>(xo + A, yo + *3,20 + 6 )= 0.2(^0.9527) = -0.1 90 5

/ = —(/j + 2fj + 2/3 + /4) b

= -0 . 1 9 7 0= - 0 .0 1 9 9

Hence, for x = 0.2, we have

y =>’0 + *= 1 ~ 0.0199 = 0.9801 /.e.,y(0.2) = 0.9801.and y' = z = z0 + / = 0 - 0.1970 = - 0.1970. i.e., y'(0.2) = - 0.1970. Ans.

Example 4 3 8 ; .S’o/ve ~ = y z + x; — = xx + y , g/ven /Aa/ y(0) ■* /; z(0) = - / fo r y(0.l),

z(0.l).

Solution. Here ^ = y z + x = / i(x,y, z), s a y and ^ = x z + y = f2(x , y , z ) , say

Given : xq = 0, yo = I, Z q = - 1, and h = 0.1, we have

k\ = hMx0,y 0, zq) = (0.1)[(1) (- 1) + 0] = - 0.1,

l\ = hf2(xo, y0, 20) - (0.1 )[(0X- 1) + 1)] = 0.1.

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uf f h A ) ,.2 = hfi i + - , Jo + ~ . *o + ~ . J = h/](0.05,0.95, - 0.95)

=(0.1)[(0.95X- 0.95) +0.05} =- 0.08525,

( H h h \ /2 « hf 21x0 + -- ,y 0 + -± , zo + * , J * h/2(0.05, 0.95, - 0.95),

= (0.1)[(0.05)(- 0.95) + 0.95] = - 0.09025.

, , ( h *2 Iq )kj = */, I x0 + - , y0 + y >*o + J -I - A/i(0.05, 0.957375, - 0.954875)

= (0. !)[(0.957375)(- 0.954875) + 0.05] = - 0.0864173,

/3 = */2 I x0 + - , y 0 + y > z o + 2“! =*6(0.05,0.957375,-0.954875)

- (0.1 )[(0.05X-0.954875) + 0.957375] * 0.0909631.h = */i(xo + h,y0 + kh 20 + 13) = */i(0.1,0.9135827, - 0.9090369)

=(0.1 )((0.9 T35827X- 0.90903&) +0.1] =- 0.073048U = V i ix o + h , y o + k h z 0 + / j ) = */2(0.1.0.9135827,- 0.9090369)

= (0.1)[(0.1X- 0.9090369) + 0.9135827] = 0.0822679.

k = ~(k , + 2*2 + 2*g + *4) b

= A [-0.1 + 2(-0.08525) + 2(-0.0864173) - 0.073048] =-0.08606376

and I = —(.h + + 2lg + l4) b

= A [0.1 + 2(0.09025) + 2(0.0909631) + 0.0822679] = 0.0907823.6

Hence, y , = y(0 .1) = y0 + k = 1- 0.0860637= 0.9139363and z, =z(0.1) =z0 + / =- 1+0*0907823 =-0.9092176. Ans.

4.8 MILNE’S PREDICTOR-CORRECTOR FORMULAE

It is a simple and accurate method for solving the differential equation. If we want to find the

value of>> using this method, we require four values of x and y namely, X0,xi,X2,xj and yo,y\,yi,

y$ such that h = x<-x<_ j.Consider the differential equation,

dy

y' = dx

or Euler method we calculate y\ - Xxi'), y j - y(x2), y i - .yfo), by using Picard’s method orTaylor’s series method or Runge Kutta method.

326 | E ng inee r ing M athem at ic s - IH

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Next we calculate,

/ o = X * o > M V\ =A* uy\ ), y'2=A* 2,y2\ / } =Ax* yi)- Then to find >>4= y(x4) by :

4hPredictor formula : >>4 = >o + ~ [2>! - >2 + 2^3] ...(1)o

Equation (I) is known as Milne ’ 5 Predictor formula.

Now calculate / 4 =Xjc4, y4) and apply :

hCorrector formula : y+ = y-i + —[y'2 + 4y'a + yi] ...(2)

Equation (2) is known as Milne’s Corrector formula.Again apply corrector formula till we get the difference between any two corrected value lessthan the desired accuracy. Continue this process till we get the required result.

emark:To apply Milne method, we require four starting values of y which are calculated by means ofPicard’s method or Taylor’s series method or Eultfr’s method or Runge-Kutta method.

9 WORKING RULE FOR MILNE’S PREDICTOR CORRECTOR METHOD

Step-1: First we have to calculated : yo =f(xo,yo) ,yi ' = f ( x h y l) , y 2l = /(x2, y 2)

y i =f(xi,yi)Step-2: To find :

y\ =/(*<)•

Sten-i: Apply Predictor formula :

y* =yo+ — [2y ' \ - y \ +2y'i \

Then obtain the value of v4' = / (x4, yi).'Step-4 : Again apply Corrector formula :

yi =j’2+

which is the corrected value o f y* - y (X4).

Step-5 : If required, again apply corrected fonnula till to get the difference between any two

corrected values less than the desired accuracy.&SP-6: Continue this process til! to get the required result,

i.e., to find y s =f(x,)For apply Predictor fonnula :

ys =yi+ y [ 2 / 2- / j + 2 / 4]

Then obtain the value of y i - f ( x j, y 5)

Again apply Corrector fonnula :

ys = yj + | [ y ,3+4y,4+y'j]

which is the corrected value z . y5=y (x5) and so on.

N u m e r i c a l A n a l y s i s -III, C o r r e l a t i o n & R e g r e s s io n | 327

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3 2 8 | E n g in e e r in g M athem atics-111

dy Example A M : The differential equation ^r ; = 1 + y* satisfies thefollowing sets i f values o fx and y

0.6

0.6841

Solution.

x : 0 0.2 0.4

y : 0 0.2027 0.4228 Compute y (0.8), using Milne’s method.

Here, x0 “ 0, xt = 0.2, x2 = 0.4, x3 = 0.6,

yo = 0, y }= 0.2027, f t = 0.4228, yy = 0.6841,and h = 0.2, then *4 = x$ + h = 0.8.

Since,/0 =A*o,yo) = I + j*>2 = I y ’l = I +yi2 = 1.0411

y’z = 1+y22 = 1.1787 y ' i = I + y s 2 = 1.4681.

Now using Milne predictor formula :

= 0 +

4h

3

4x0 .2

y A = y0 +Y 2y i ~ y ' 2+ 2y^

[2 x 1.0411 - 1.1787 + 2 x 1.4681] = 1.0239

Then, >>4' =J{x4,y4) = I +y A2 = I + (1.0239)2 * 2.0480.Using Milne corrector formula :

yA = y-i+ |Lyo + 4 9 + y \ ]

= 0.4228 + — [1.1787 + 4 x 1.4681 + 2.0480] = 1.0294.3

Hence, y4 = y(x4) = X0.8) = 1.0294.

dy

Am.

Example 4.40 : Apply Milne's method to find a solution of the differential equation ^ = x ~ T in

the range 0 £ x £ I, fo r the boundary condition y - 0 a tx = 0. [RGPV Dec. Solution. First we compute y lf y j, yy Using Picard’s method :

Here, / =/{x,y) - x - y 1, x0 = 0,y0 = 0, taking h = 0.2.

So that x 1 = x0 1- h = 0.2, x2 = 0.4, x j = 0.6, X4= 0,8, X5 = 1.0..•. Using Picard’s method :

= y0 + £ f (x ,y 0)dx = 0 + £ x d x = y .

v(2) = y0 + £ f { x ,y ^ )d x = 0 + £ * “ ( ^ r ) dx20

x2e.

y = T ~ 20 for computing >>,{x) :

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N u m e r ic a l Analvsis-111, Corrioation & Regression |3 2 9

At, x, = 0.2, then y x =><0.2) = - 0 .02 .

At, x2 = 0.4, then y i =>’(0.4) = — r ----------- — — - <J-0j9;>

2 20

(0.4)* i0 .4 )r'2 20

(0.6)* ,i0 .6)5

04# 20

, 2 - ( 0 . 0 2 r - 0.19

At, jf3 = 0.6, th en y j =><0.5) - — ----------- = 0.170.

Hence,

y i ~Axi- yi) ~X 2 ~ yz 2 ~ 0-4 - (0.0795y = 0.393'

y i ~ A x h yi) = *3 - j ’32 = 0*6 - (0 .1 ?6 )2 = 0.5690


Ah

3 y 4 = yo + — [2^1 - y i + 2^ 3 ]

= 0 + 4 * ° ‘2 [2 X 0 .199 6 - 0 .393 7 + 2 x 0 .5690 ] = 0 .3049.3

Hence, / A = / x 4) y A) = 0.8 - (0.3049)2 = 0.7070.

Using MiJne corrector formula :

h, y 4 = y-> + - b i + 4 ,V3 + y U

= 0 .079 5 + — [0 .3937 + 4 x 0 .5690 + 0 .7070 ] = 0 .30463

Hence. y Aat x = 0.8, is y(0 .8) = 0.3046.

Then, y 4’ = / * 4, y A) = 0.8 - (0.3046)2 = 0.7072.

Again, using M ilne predictor formula, w e get

y s - y , + Y l 2 A ~ y> + 2y' t]

= 0 .02 + 4 X 0-2 [2 X 0 . 3 9 3 7 - 0 . 5 6 9 0 + 2 x 0.7072] = 0.45543

y i = / x 5, ys) = I - (0,4554)2 « 0.7926.

Using M ilne corrector formula :

ys = J s + j b s + 4 y ; + J s 1

= 0 .1 7 60 + — [0 .5 69 0 + 4 x 0.70 72 + 0 .79 26 ] = 0 .4554.3

Hence, ><*5) =><•)“ 0,4554. Ans.

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dyExample 4.41: Solve numerically the differential equation — ■* x + y with y (0) = /,

h] Wine’s predlctor-corrector method from x m 0 to x m 0.4.

Solution. G >'pn that y1=x+y andx<> = 0 , yo = I.Since / =1 +y => yo1 = *o+yo= l»yit = j ^ ,0it = 1 + yfli = 2,

> = y => yo"1 = yo" = 2,

> v = => y0IV = 2 and so on.Taking/» “ 0.1, so thatX) =xo + A= 0.1, x 2 = 0.2, X3-0.3, x4 = 0.4.First we have to find y i,y 2 , ya by Taylor’s series :


(x -x 0)2 it (x -x 0)3 lU y

(*)=yo + (x-xo)yol

+ — — >o + — ^ — >o +•■

/ x _ . / \ i . (* | -- * 0 ) .. H . (.X, Xq ) III y(x ,) = y o + (x < -* o )y o ' + — ~ t — y 0 + — j ,— x >o + ...... . - 0 )

where /= 1,2,3.(i) For/=1:

Put Xj = 0.1, and Xo, y0 yo1. V , yom and so on, we get

y(0.1) = 1+ (0.1) * 1 + ~ “ *2 + ^ - x 2 = 1.1103

Hence y, = 1.1103(ii) For i - 2, Put x* = 0.2 in (1), we get

(02)2 „ (02)3 „ y (0.2) « 1 +(0.2) * I + ^ - x 2 + — - x 2 = 1.2427

« J*

i.e., ^ « 1.2427(iii) For i - 3, Put x3 - 0.3, in (1), we get

COJ'i2 COJ'I1y ( 0 . 3 ) = 1 + ( 0 . 3 ) x 1 + L J T x 2 + ^ f ' x 2 “ 1 3 9 9 0i.e., y} = 1.3990Since y1 - f (x,y) = x + y.Then, we have

y,» = x ,+yi = 0.l + 1.1103 = 1.2103 y ix =*2 +yi = 0.2 + 1.2427 = 1.4427yj' =x3 +>>3 = 0.3 + 1.3990 = 1.6990


Ah

>4 =yo +-r[2>'i1 ~y\ +2yj]3

= 1+ f ( | l ) [2 x 1.2103 -1.4427 + 2 x 1.6990]

= 1.5835

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N u m e r ic a l A n a l y s is -111, C o r r e l a t i o n & R e g r e s s io n | 3 3 1

Hence y4' = x4 + y 4 - 0.4 + 1.5835 = J.9835

Using Milne corrector form ula

hy 4

=y? +'

(0.1)

values and =y2-xK

= 1.2427 + 1.4427 + 4 x 1.6990 + 1.9835J = 1.5834

Hence y (0.4) - y ( x 4) = 1.5834 Ans .

Example 4.42: Using M ilne predictor-corrector method, obtain y (0.4) from the given set o f tabulated

f&Ldx

x ; 0 0.1 0.2 0.3 y : 1 l . U 1.25 1.42

y ' : 1 1.22 1.52 1.92Solution. Given that x0 = 0, x t = 0.1, x2 - 0 .2, xj = 0.3 and h ~ 0 . l

Th en Xj = xq + 4h. Then we have to fin d y4 - y (0.4).

Again given that

yo = U = 1-11, y2 = 1-25, y j = 1.42, y0' = h y , ' = 1.22, ^ * = 1.52, *> = 1.92.

and )S =f(x ,y)=y2-x> . . . ( 1 )

Using Milne predictor form ula :

4h\ y* =jo + *[2 y\ - y \ +2y]]

= I+ ^ M [ 2 x 1 2 2 - U 2 + 2 x l .9 2 ] = j .635

* ~ u? —r? =(1.635)2 - (0 .4 y = 2 .513

Since y = / ( x , y) = y2- x 1

=> y*1=y4-xiAgain using M ilne corrector formula

h

3

= 1 .2 5+ — [132 + 4 x1 .9 2 + 2 5 13 ]= 1 .64

Hence y (0 .4 ) = y 4 ~ 1.64 Ans.

Example 4.43: Using Milne’s predictor-corrector method, obtain y (4.4), fro m the given set o f tabulated values and Sxy’ + y * - 2 “ 0.

x : 4.0 4.1 4.2 4.3 y : 1.0 1.0049 1.0097 1.0143 y ' : 0.05 0.0483 0.0467 0.0452

Solution. Given

x0 = 4.0,

yo = i.o, yo 1= 0.05,

y . 2 ^ y 5xx , =4 .1 ,

yx = 1.0049,

V = 0.0 48 3,

0 )*2

yi y l

4.2 , • x3 = 4.3 the n x4 = 4.4 an d h - 0 . I

1.0097, y i —1.0143 th en find y4 = y (0.4)

= 0.04 67, y 3' = 0.0452.

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3 3 2 | E n g i n e e r i n g M a t h e m a t i c s -111

Using the Milne p redictor formula, we have

y4 =>'o+ ~r[2y* ~y\

= 1.0 + x A.0483 - 0.0167 + 2 x 0.0452]

*= 1.0187

2 - r 2 - .v ; 2 -(1 .0 l8 7 )2Since v' — r '— => v* - --- ------ - -----r ------------

5.t 3Jf4 5 x 4.4

i.e.. v / - 0.0437

New . vising M ilne-c orrecto r fo rm uia , we have

hi_j3

I j - V> I- + 4 .1*1 + v ’ j

-- . -i ® [0.0467 + - !0 .0 45 2 -r 0.0437] j

- 1.0187

H cn ce >- (0 .4 ) ~ - 1.01 87. Ans.

Example 4 . 44 : Using Milne Vmethod to fin d y (0.3) from ~” - x * + y 2 .* y ( 0 ) - J. Find the initial

values \m(—0. l) ,y( 0. l) an dy(0.2) from the Taylor’s series method. fRGPV June 2006 and June 2007 /

Solution. First we co m pute y \ , y i , >'3.

Here. >*' - f ix , y ) = x2 + y2, x q = 0 , y 0 = 1 . taking /; = 0 .1.

Since / = x2 + v2 => yo’ = !.

Differentiate : v" = 2x + 2yy‘ =>>0” = 2 x 0 + 2 * l x l = 2 ;

y ‘" = 2 + 2(yy" + / 2) = *^ ’"(O) - 2 + 2 {1 * 2 + I2) = 8 .

.*. By Tay lo r's serie s :

(x - x0) , ( * - *o)2 « (x - x 0)3 *) ix) - y0 + yo + 2 1 ^ + 3 ! -v° + -

, 4 3or it v ) -- 1 + X + X- + - X 3 •+■....

O4

Put x = - 0.1, we get>< - 0.1) = 1 - 0 .1 + (0.1)2 + - ( - 0 .1 ) 3 = 0 9087O

4

Put X = 0.1, we getjv(0.1) = 1 + 0.1 + (0.1)- + - ( 0 . 1 )3 = 1.1113.O

Put x = 0.2, we get;< 0.2) = 1 + 0.2 + (0 .2 )2 + i (0 .2 )3 = 1 .2506.

Hence points a r e :

x _i = - 0 . 1 , x 0 = 0 ,x i = 0 . 1 ,X 2 = 0-2 ,X 3 = 0.3, a n d /j = 0.1

y- 1 = 0 . 9 0 8 7 , > 0 = l ,yi = 1.1113, >5 = 1.2506.

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y d =X*o, yo) = *1 + y i - o 2 + i 2 = i .

y i - / x t . t t ) - * ? + y f =(0 .1 )= + (1.1113)2 =1 .24 49 .

y2 =AX2*>'2) = X- + y f « (0.2)2 + (1.2506)2 = 1.6040.Using Milne predictor formula :

y i = J-i + y l-J'o ~ Jt ■*"“ '2J

= 0 .90 87 + - X —-1.[2 x 1 - 1 .2449 + 2 x 1.6040] = 1.4371.

y ’i =f(xy v3)= xt * v- (0.3£ • (1 .4 '7 1): = 2.1552.

Using Milne corrector formula :

Vj = y\ + ^f .v i + 4v j + v , ]

= 1 .1113 + y - { l 244 9 + 4 > 1 .6040 ^ 2 .155 2] = 1 .4385.

Hence. yj - ) i x j ) =y(0.3)z 1.4385. Ans.Remarks:

(i) The truncation error in Euler’s method ib of O (ir)

(ii) The truncation error in modified Euler’s method is o!

(iii) The truncation e rror in Ru nge-Kutta method o f fourth order is of Q(h~)

(iv) The truncation error in M ilne's method is of 0( /;s)

4.10 CONCEPT OF CORRELATIONThe correlation is a statistical tool which studies the relationship between variables, say x and v,

and correlation analysis involves vario us mctht«dv u n i tech niqu es used for studying and m easuring

the extent o f the relationship between two variables.

In other words, correlation analysis is a stc . '•cu. pv c. -i:.. b. • v- .* can determ ine the

degree of relationship between two or more \ar:?.r*!e--.

4.11 COEFFICIENT OF CORRELATION

Th e deg ree to w hich th e tw o variab les are inter-re hi led i* m easured by a coe fficient which is calledthe coefficient o f correlation.

The coefficient o f correlation between the two variables x s ; i d i s generally denoted by r or rn. or

p or p (x, y).

4.12 TYPES OF CORRELATION1. Positive or Direct Correlation : If the increase (or de crease) in one variable, say x results in a

corresp ond ing increa se (or dec rease ) in the oilier variable say then the correlation is said to be

positive or direct.

For example, the height and weight of a growing child.

2. N egative o r In ve rse co rre latio n : If the increase (or decrease) in one variable, sav x results ina correspon ding decrea se fo r increase) ir, the other variable say v. then th e correlation is said to be

N u m e r ic a l A n a l y s i s -III, C o r r e l a t i o n & R e g r e s s io n j 3 3 3

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3 3 4 | E n g in e e r in g M athematics-111

For example, the consumption of electricity and its bill.

3. Linear correlation : If the plotted points (x,,y,) are approximately on or near about a straighline, then the correlation between x and y is said to be linear and see the following figure 4.2 (a)

yA

o ----------------- **

Fig. 4.2 (a) : Linear and Perfect Positve Correlation

For example, the correlation between the saving and wealth of a man.4. Perfectly Linear Correlation : If all the plotted point ( x,, yi) lie exactly on a straight line, then thecorrelation is said to be prefectly linear and see the following figure 4.2 (b):

Fig. 4.2 (b) : Perfect Linear Correlation

5. Perfect Correlation : If the correlation is one variable x is followed by a corresponding and proportional deviation in the other variable y, then the correlation is said to be perfect correlation

6. Perfect Positive Correlation : If the equal proportional changes in the two variates x and arein same direction, then it is a perfect positive correlation and see figure 4.2 (a). In this case r = 1.

7. Perfect Negative Correlation : If the equal proportional changes in the two variates x and y arein opposite direction, then it is a perfect negative correlation and see figure 4.2 (b). In this case

r = - I.8. High Degree Positive Correlation : If the plotted points (x,, y,) fall in a narrow band and the

points are rising from lower left hand comer to the upper right hand comer, then there will be ahigh degree postive correlation between x andy, and see the following figure 4.2 (c).

y

Fig. 4.2 (c): High Degree Positive Correlation

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N u m e r i c a l A n a l y s is *!!!, C o r r e l a t i o n & R e g r e s s io n j 335

9. High Degree Negative Correlation : If the plotted points (x,, yt) fall in a narrow band fromupper left band comer to the lawer right comer, then there will be ahigh degree negative correlation

between x and y and see the following figure 4,2 (d).

v*

Fig. 4.2 (d) : High Degree Negative Correlation

10. Low Degree Positve and Negative Correlation : If the plotled points (x#, y,) are not clustereda round a straight line but are widely scattered over die diagram, then there is a very low degree of

positive and negative correlation between the variables x and y.

11. No Correlation : If the plotted point (x,, yi) lie on a straight line parallel to thex-axis or in a haphazard manner, then the variables are not correlated and see the following figure4.2 (e). In this case r - 0.

I

Fig. 4.2 (e) : No Correlation

4.13 PROPERTIES OF COEFFICIENT OF CORRELATION1. It is measure of the closeness of* fit in a relative sense.

2. Correlation coefficient lies between -1 and +1, i.e., -1 <, r <, I.3. The correlation is perfect and positive, ifr = 1and it is perfect and negative if r = -1.4. Ifr = 0, then there is no correlation between two vaiables and thus the variables are said to beindependent.

5. The correlation coefficient is a pure number and is not affected by a change o forigin and scale.

6. It is a relative measure of association between two or more vaiables.7. Following are the diagrams for various value of r.

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336 | E n g in e e r in g M a t h e m a t ic s -! 11

4.14 COVARIANCELet (x , , y, ). i = 1,2 . 3.......n be a hivariate distribution, when X|, X 2 .........x„ are the values of variable

x and y i , y i ..... ,y„ those of y. Then Covarience between x and y is denoted by Cov. (x, y) and isdefined a s :

Cov. (x. y) = i 2 ( x , - x ) ( y , - y ) ...(1)n ivsj

Svwhere x - and y - are means of x and y. respe ctively

Sipcc the above f<'r;r.i.tia lo r the calculation s o f covarianc e is com plicated so we use altenativeformula as below;

4.15 METHODS TO S Tl DY CORRELATIONThe various methncK to determ ine whether two variables are correlation or not are as follows :

(i) Sca tter ;>r Do t Diaeram method ,^ 1(ii) Karl Peam-.i;’? in ef fic ie n t o f Corre lation or Cov ariance method,

(iii) Spearm an's Rank Method.

We shall discuss each o f the methods in tetaii in the following ahed sections.

4.16 SCATTER OR DOT DIAGRAM METHODThis is the sim p ler method to ascertain whether two variables are correlated or not and in casethey are, what is the nature of correlation? In this method we use the rectangular coordinate axesto mark a dot corresponding to each pair of x and y values and thus obtain as many points as thenumber of ordered pairs in the given bivariate distribution. This diagram of dots is called thescatter diagram. B v looking to the trend of the plotted points in the scatter diagram we caR ftmn

an idea as to u he: her the variables are correlated or not.

The scattcr diagram may indicate both degree and the type of correlation. From scatter diagram wecan form a fairly good, though rough, idea about the relationship between the two variables. The

different types o f correlation are depicted by means of scattered diagrams as shown by figures 4.2(a), (h). (c), (d). and (e).

The advantages and disadvantages of scatter diagram method are as follows.Advantages :

(i) It is readily comprehensible and enables us to form a rough idea of the nature ofrelationship between the two variables x and y .

(iii It is not affected by extreme observations.

(iii) ,‘t is not influenced by the size o f exterme items.

Disadvantages :

(i) It i> run a suitable method if the number* of observations is very large.1ii' Ii enables us to oh: tin an approximate estimating line or line of best fit1by this method.

: 1nuijh incûr- or corre!atioto where the exact magnitude cannot be know.

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4,17 KARL PEARSON’S COEFFICIENT OF CORRELATION OR COVARIANCEMETHODThe correlation coefficient r between two variables x and y is given byFirst Form :

r (x, y) or r ~

r {x, y) or r -

Cov (*,;>>)

JV3r(x ) x J V ar(y)

Cov (x, y)

or

Whereand

Also

Second Form ,

ax x c y •d)

Var(x) =o,2,Var (y) = c r /

Var (*) =

r or r(x, y) -

n

'L{.xi~x)Z(yi-y)

Where x and y are means of x and y respectively.Third Form:

r or r(x,y )nZxiy i -( Z x i)(I .y i )

.(2)

(3)^{n£*i2 - (SXi )2 } x {nSy - ( l y i ) 2 J

This formula (third form) saves a lot of computational labour and it reduces the error due to

computation and rouding off. Again if the values of x i 's oryi’s are large or involve fractions, thennone of (1), (2) and (3) is convenient to use. In such a case use another formula : i.e., short cut method which is given by the following form ula:

n'Zuivi -(£ u i)('Zvi)_______

r (x, y) or r =^{»£«<2 -(E u;)2} * ^ ^ 2 ~(£v,)2}

{S«)(Zv)luv--

or r = •(4)

where u, = xf - a and v , = y , - b and o and by being assumed means of x & y series respectively.

Note : It is advised to calculate the correlation coefficient by using short-cut method* fonnulaonly for easier calculations.Remarks: \ . r ( x , y ) = r(u, v) = r 2. r2is called cofficient ofdetermination.

3, S.D. awhere u = x -A , A being assumed mean.

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Example 4.45 : Find the Kart Pearson's coefficient o f correlation between x and y fo r the following

data:

x : 6 2 4 9 I 3 5 8

y : 13 8 12 15 9 10 11 16.

Solution : Let u, = x, - a = x<-5 and v, =yt - b =yt- 12, where 5 and 12 are assumed means o f x and y respectively.


Xi y> 5 vj —12 Vi2

6 13 1 1 l 1 1

2 8 -3 -4 12 9 16

4 12 - 1 0 0 1 0

9 15 4 3 12 16 9

1 9 -4 -3 • 12 16 9

3 10 -2 -2 4 4 4

5 11 0 -1 0 0 t

8 16 3 4 12 9 16

Ik, =-2 1 ? i i * L > Iw,v, =53 la,2= 56 Ev,2=56

Here n ~ 8

We have

nSujUj - (£u,-) (Su*)

yjnZuf -(E U j)2 -(Ei>;)2

8(53 )-(-2) (-2)

^8 (56) - (-2)2x ^8 (56) - (-2)5

420V444 V444

= 420~ 444

= 0.946

Example 4.46: Calculate the coefficient o f correlation between x and y :

x :

y •

2 3

18

2 7

20

28

20

28

27

2 9

21

30

29

31

27

Ans.

33

29.

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N u m e r ic a l A nalysis-1 11, C o r r e l a t i o n & R e g re s s io n j 3 3 9

Nation: Letw, = x, -2 8 and v ^ y , -2 7

Xi y> u i = Xi - 28 V i - y i - l l UiVt u? V/2

23 18 -5 -9 45 25 83

27 20 -1 -7 7 1 4928 20 0 -7 0 0 49

28 27 0 0 0 0 0

29 21 1 - 6 - 6 1 36

30 29 2 2 4 4 4

31 27 3 0 0 9 0

33 29 5 2 10 25 4

£«, = 5 Ivy = -25 Iu,v, = 60 Sw/2= 65 Ev,2 = 223

We have n = 8,

r = r (x, y)nZtfjVj -(2ui)(Lvi)

V n l u ' f - a u , ) 2 jn -L vf - (ZVi)2

8 ( -6 0 ) - ( 5 ) ( - 2 5 )

Vs (65) - (5)2 ^8 (223) - ( -25 )2

= = 0.80 Ans.V495 V1159

ample 4.47: Calculate the Kart Pearson’s correlation coeftclent between x andyfor the following data :

x : ISO 153 154 155 157 160 163 164

y : 65 66 67 70 68 53 70 63

tation : Let ut = x, - 155 and v,=y,~ 68.

Xi y> Ui = xt - 155 v ^ y i - 6 8 u(v, ui1 Vi2

150 65 -5 -3 15 2 5 9

153 66 - 2 - 2 4 4 4

154 67 -1 -1 1 1 1

155 70 0 2 0 0 4

157 68 2 0 0 4 0

160 53 5 -15 -75 25 225

163 70 8 2 16 64 4 i164 63 . 9 - 5 -45 81 25

Eh, = 16 Ev, = -2 2 Eu,v, = -84 «2 = 204 = 272

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Here n = 8,

nZuivi -<£ui)(l.vl)We have r ~r(x, y) = ---------------------------------

,J n l u f -( Z u £)2 x -y j nZv f - (Zu*)2

= 8 W M * 6 ) (- 2 2 )^8 (204) - (16)2 xyjs (272) - (-22)2

-6 7 2 + 352

Vl376 Vl692

- 3 2 0

= Vl376 V1692

320

1525.84= -0.20 97 . An

Example 4.48: From the following data obtain the value o f the correlation coefficient: n = 10,2>c-140, Ey - 750, £(x - 70>> -- 7*0, S(y - 75^ - 2 / Jand £ (x - 10)( y-15)~ 60.

Solution : Let w= x - 10 and v = y - 15. ThenZw= Z (a - - 10) = Z x -n (10) = 140- 10 (10) = 40. [ v Z10=I0Zv = Z (y - 15) = Z y-« (15)= 15 0- 10(15) = 0. [ v £15 = 15It is given thatZ (jr- 10)2= 180,Z (y - 15)2 = 21 andZ (x- 10)(y- 15) = 60-

i.e., Zu1 = 180, Zv2= 21 and Zwv = 60., ritu.v, — (Zu;) (Zu,)

We know that r(x ,y ) = r= ■— L- *— p - — ------ ■—■■■ yjnZuf -( Z u , )2 x ,JriZvf -(Z u ,)2

10 (6 0) -(4 0) (0)

/ i 0 (108) - (40)2 x yjlO (215) - (0)2

600

V200 >/2150600 6

V430000 V43= 0.9151. An

Example 4.49 : From Me following data obtain the value o f the correlation coefficient: n = 10 ,£x = S S ,fy = 40, 2k1 = 385, = 792 anrf +y)2 » 947.

Solution : We have Z (x + y¥ = 947=> Z (x2 +2 xy +v2)= 947=> Zx2 + 2Zxy + Zy2 = 947

=> 385 + 2Zxy= 192 = 947=> Zxy = 185

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We know that r = (by third form)

1 0 x 1 8 5 - 5 5 x 4 0

^ 1 0 ( 3 8 5) - ( 55 )2 x ^1 0 (1 9 2 ) - ( 4 0 ) 2

1 8 5 0 - 2 2 0 0

V 3 85 0 - 3 0 25 x ^ 1 9 2 0 -1 6 0 0

- 3 50 - 3 5 0

V 8 2 5 > / 3 2 0 V 8 2 5 x 3 2 0

Ans.

ample 4.50 : From the following data compute the coefficient o f correlation between x and j>.(i) Arithmetic Mean o fx series is 25 ?nd that o fy is 18.(ii) Sum o f the products o f deviations o f x and y series fro m their

respective means = 122.(iii) Sum o f the squeares o f deviation fr om their respective means are

136,138 respectively fo r x series and y series.(iv) Number o f pain o f values ■= IS.

station : We are given the following information

Sample 4.51 : From the following data, find the number o f items n. r = 0.5, 2xy * 120, o> “ He2 * 90,

where x andy are deviations from arithmetic mean.

ifatkra: Let arithmetic mean o fX and Y be X and Y respectively. Then x and y are deviation from

arithmetic mean, i.e. x - X - X , y ~ Y - Y

«= 15, x =2 5, y = 18.

and 2 (x - x ? = 136, E ( y - £ ) 2 = 13 8 ,2 ( x - x ) ( y - y ) = 122.

~ 136.996 ~ 0 89-

Hence the coefficent of correlation is 0.89. Ans.

Now

Also

an d

2 (Y - Y Y = 64 n.

U X - X y = Er* = 90,

Z(X- X ) ( Y - Y) = Zxy= 120.

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342 | En g in e e r in g Ma t h e ma t ic s -1 II

We know that correlation coeficient (Second form) :

t ( X - X ) ( Y - Y )r =

=> 0.5 =

Squaring both side, we have

0.25 =

V i (x - x ) 2 x V z (y - y )2

120

VsiOx V64s

120x12090 x 64n

120x120^ ” 90 x 64 x 0.25 IU-Hence number of items are n = 10. Ans.

Example 4.52 : A computer operator while calculating the coefficient between two variates x and

for 25 pairs o f observations obtained the following constants, n - 25, Ex ~ 125, £x* * 650, - 100, 2 p - 460, £xy - 508. It was, however later discovered at the time o f checking that he had copied down f< pairs as (6, 14) and (8, 6). while the correct pairs were (8, 12) and (6, 8). Obtain <1correct value o f the correlation coefficient.

Solution : Given:

Incorrect data Correct data

Xt 6 8

Xi 8 6

y\ 14 12 y2 6 8

Incorrect date are: Ex = 125. Ey = 100, Ex2 = 650, Ey2 = 460, E xy =508.ThenCorrected Ex = Incorrect Ex - (6 + 8) + (8 + 6) - 125 - 14 +14 = 125Corrected Ey = Incorrect Ey - (14 + 6) + (12 + 8) = 100 - 20 + 20 = 100Corrected Ex2 - Incorrect Ex2- (62+ 82) + (82 + 6 ) = 650 - 100 + 100 = 650Corrected Ey2 = Incorrect Ey2- (142+ 62) + (122+ 82) = 460 - 232 + 208 = 436Corrected Ery = Incorrect E xy - (6 x 14 + 8 x 6) + (8 x 12 + 6x8)

= 508 - (84 + 48) + (96 + 48) = 520.We know that, correlation coefficient

•>' ‘ • _ ________ flEyy-(Ejf)(Ejr)

- (Ex)2} x -(E .p)z]

= 25x520-125x100 .

^{25 x 650 -(125 )2} x 25 x 436 - (100)2}

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Example 4.53 : A computer white calculating the correlation coefficient between the variables x and y obtained the following constants, n -30, 23c « 120,22c2 = 600, fy = 90,ty2'=2S0, Hey = 356.

It was, however, later discovered at the time o f checking that it had copied down two pairs o f observations as

X z

X y6 id while the corrected values were 8 12

12 7 10 8

Obtain the correct value o f correlation coefficient between x andy. Solution : Given : Incorrect data

Zx = 120, Ey = 90, Zx* = 600, Zy2= 250, Ixy = 356Also Incorrect values of x and y Correct val ues of x and y

X y X y& i i 8 \2

12 7 10 8

Then :Corrected Zx = Incorrect Zx - (8 + 12) + (8 + 10) = 120 —20 +1 8 —118Corrected Zy = Incorrect Zy - (10 +7) + (12 + 8) = ° ’ j - 17 + 20= 93Corrected Zx2 = Incorrect X*2- ( 82 + 122) + (82 + 10') = 600-208 + 164 = 556Corrected Zy2 = IncorrectZ^-( 102 + 72) + (12?4 82) = 250- 149 + 208 = 309Corrected Zxy = Incorrect Zxy-(8 x 10 + 12 * 7) + (8 * 12 + 10 * 8)

= 356 - 164 + 176 - 368.

We know that, correlation coefficient

r =nZxy - (Zx) (Zy)

^{nZx2 -( Z * )2} x J{nI .y2 - ( X y f }

30 x 368 -118 x 93

^ 3 0 x 556 - (118)2} x {30 x 368 - (93)2}

66 66= 0.0257 Ans.a/2756 x >/2391 2567.02

Sxample 4.54 : Prove that the Karl Pearson's coefficient o f correlation independent o f change o f origin and scale.

•ohition : We know that coefficient of correlation :

S ( * - * ) ( y - y )

Vz(x-x)2x^Z (y-y)2r = r(x,y ) = .. ■iVT.

Let u = x - a

and y - b

k

- ( I )

(h, Jfc> 0)

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and a, b being assumed means of x and y series, respectively.

=*• x - a + uh and y = b + vk

n> x = a + uh and y ~ b + vk

Putting these values in equation (1), we get

£{(a + u h - a - uh) (b +v k - b - vk)}

3 4 4 | En g in e e r in g Ma t h e m a t t c s - I II

>-= r(x,y) y/lCa + u h - a - u h ) 2 x i]z(b +v k ~ b - v k ) 2

_ h k Z { ( u - u ) ( v- v )}

iJh2Z(u - u )2 x yjk2Z ( v ~ V ^

J z ( u - u ) 2 x j z ( v - v ) 2

- r ( u , v) = r This shows that r (x, y) is independent of the change of origin and scale. Proved

Example 4.55: Prove that the Karl Pearson's coefficient of correlation r lies between - 1 and 1. Le; - 1 & r £ 1. or \ r | £ L

Solution : We know that coefficient of correlation

, - r t e , ) - - — 2 2 ^ 2 < 2 L a L = _ f o r i- 1,2.

V£(*«— *)2 x y j U y i - y ) 2

Let x, ~x - a, and>-; - y - b,: i = 1,2 ,...... n,

Then r = Zoibi

■Jza2 x J z b 2

r, _ ( s° A )2

(Za,!>ab,2)

By Cauchy- Schwarz's inequalitv, we have(IaA)2 *(Za?) (Ebfl

(ZQi b j ) 2 £ l

( 16 , )

=> r2 <: 1(Using I)=» -1 £ r :£ 1. Prove

4.18 RANK CORRELATION METHOD

Rank correlation is the coefficient of correlation between different ranks attained by a groupindividuals in two characteristic or attributes.

The coefficient o f rank correlations is given by Spearman’s by the following formula.

, , _ 1 ,r{x,y) = r =1n (n2 ~ 1)

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where n is the total number of pairs of observations, and d} is the square of the difference ofcorresponding ranks and d, = (Rank x, - Ranky) or d = Rx - R y TYPE I : Wfaea the R u k are Given :Working Rale:Step 1. Calculate the differences of the ranks, i.e.,

d —Rx —Ry

Step 2. Square the difference d and write it under column headed by d1.Step 3. Apply the formula, coefficient of rank correlation.

Example 4.56: Two gents X and Yare asked to rank 7 different types o f shirts. The ranks assigned by them are given below :Shirts A B C D E F G

Rank given by X 2 I 4 3 5 7 6

Rank given by V I 3 2 4 5 6 7

Calculate Spearman’s rank correlation coefficient

Solution : We know that rank correlation coefficient:

Where d - Rank X - Rank Y

Here n = 7.

Rank o fXLe; Rx Rank o f Y Le; R, d — Rx — Ry d 2

2 I I 1

1 3 -2 4

4 2 2 4

3 4 -1 1

5 5 0 0

7 6 1 I

6 7 -1 1

lcP= 12

Hence (1) becomes

6x1 2 = , 3 = 11

7(4 9-1 ) 4 14

i.e.; r = 0.786 Ans.

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Example 4.57 : Ten competitors In a musical test were ranked by the three judges X, Y and Z in the following order :

3 4 6 j E n g in e e r in g M a t h e m a t i c s *III

Ranks by X : 1 6 5 10 3 2 4 9 7 8

Ranks by Y: 3 5 8 4 7 10 2 1 6 9

Ranks by Z : 6 4 9 8 1 2 3 10 5 7

Using rank correlation method, discuss which pair o f judges has the nearest approach to common likings in music.

Solution:

Ranks

byX

(Rx)

l l

^

! Ranks

byZ

(RJ

d i m

Rjf-Rf

d i m

Ry~Rt

d s ~

Rx~Ri

d ,> d f d ?

1 3 6 -2 -3 -5 4 9 25

6 5 4 1 1 2 1 1 4

5 8 9 -3 -1 ^4 9 1 16

10 4 8 6 -4 . 2 36 16 4

3 7 1 -4 6 2 16 36 4

2 10 2 -8 8 0 64 64 0

4 2 3 2 -I I 4 1 1

9 1 10 8 -9 -I64 81 1

7 6 5 1 1 2 1 1 4

8 9 7 -1 2 1 1 4 1

Total Id ,2 =200 I # = 214 S< # = 60

Here n = 10, = 200 , td 22= 2 ! 4 and W 32 = 60

(Y — _ . 6 x 200 40 _^£7 = p* r (X Y) ~ 1 " n ( n 2 - l ) ~ 1" 10(I6b-T) "" 33 33 0212

r(Y’ Z ) ~ l ~ n ( n 2 ~l ) 165 0297

Since r ( £ 2 ) is maximum, therefore the pair of judges X and Z has the nearest approach

to common likings in music. Ans.TYPE I I : When the ranks are not given :

In this case we assign the ranks to both the series x and y by giving the rank 1to highest values in boththe series (or to the lowest values in both the series) and so on.

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'Working Rule:

Step 1 : Assign ranks to each data of the both series.

Step 2 : Calculate d = Rx - R y

Step 3 : Square the difference d and write it under the column headed by d1.

Step 4 : Apply the formula :

, = 1-

N u m e r ic a l Analysis-111, C o r r e l a t i o n & R e g r e s s io n | 3 4 7

n (n2 -1 )

Example 4.58 : The marks obtained by 9 students in chemistry and Mathematics are given below:

Marks in Chemistry : 35 23 47 17 10 43 9 6 28

Marks in Mathematics : 30 33 45 23 8 49 12 4 31

Compute their ranks in the above two subjects and the coefficient o fcorrelation o franks.

Solution : We first assign ranks to X and Y series giving rank 1 to the highest value in both the series.

Marks in

Chemistry (X)

Marks In

Maths (Y)

Ranks in

X (RJ

Ranks in

Y (Ry)

d = R ^R , <p

35 30 3 5 -2 4

23 33 5 3 2 447 45 1 2 -1 1

17 23 6 6 0 0

10 8 7 8 1 1

43 49 2 1 1 1

9 12 8 7 1 I

6 4 9 9 0 0

28 31 4 4 0 0

. .lcP= 12

Here n = 9, = 12

61d 2 , 6 x 12 _.w„r - I --------5----- = 1------ -5 — - = u.9 Ans.

n (n -1 ) 9 (9 - 1)

Example 4.59 :10 students got the following percentage o f marks in Mathematics and Physics.

Mathematics (X) 8 36 98 25 75 82 ' 92 62 1 65 35

Physics (Y) 84 51 60 68 62 86 58 35 49

Find the coefficient o f rank correlation.

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3 4 8 | En g in e e r in g Mathematics-111

Solution : We first assign ranks to X and Y scries giving rank 1to the highest value in both the series.

Marks in

Maths (X)

Marks in

Physics (Y)

(RJ Rank

in X

(Rj) Rank

in Y

d**Rx- R , <P

8 84 10 3 7 49

36 51 7 8 -1 198 91 1 I 0 025 60 9 6 3 975 68 4 4 0 082 62 3 5 -2 492 86 2 2 0 062 58 6 7 -1 165 35 5 10 -5 2535 49 8 9 -1 1

Total U 2- 90

1.1

Here, n = 10, and 90

Rank of correlation : r = >1----- --------- = 1- - \n(n - 1) 1 0(100- 1)

= 1 - ^ = 1- 0.545 - 0.455. Ans.

TYPE I I I : When Equal Ranks are given Le.; Tie Case :

If two or more items are placed together in any classification with respect to an attribute, i.e; there

are more than one item with the same value in either or both series. Then such type problem issolved by assigning average rank to each of these individuals who are put in tie.

For example : Suppose an item is repeated at ranks 7, (i.e; the 7* and 8* items are having same

7 + 8values), then the common rank assigned to avarage of 7th and 8* is g— = 7.5, the ranks which

these items would have been assigned if they were different, and the next rank assigned will be 9.

Again, if an item is repeated thrice at rank 3, then the common rank assigned to each value will be

= 4.5 i.e; mean of 3, 4 and 5, and the"next rank assigned will be 6.

In this way, to find the rank correlation coefficient of repeated rank by the following formula :

6 (Id2+correction factor)r'*- n(n2-l)

Where, correlation factor = m (*2- 1)—

Here m is the number of times an item is repeated.

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i.e.; The modified formula for Tie rank correlation coefficient is given by

6 [Z<i2 + 0 mi + ( " I - D + ..... ]

r ’ l ~ • n (»2 - l )Where, mlt m 2 , ...... are the number of times an item is repeated.

Example 4.60: Front the following table find the rank correlation coefficient X : 48 33 40 9 16 16 65 24 16 57 Y : 13 13 24 6 15 4 20 9 6 19

Solution : Here in the series X, the value of 16 is repeated thrice say, mi = 3.In series Y the value of 6 and 13 are repeated twice say, m2 = 2 and mi = 2.

Tie rank correlation coefficient is given by

n (n2 -1 )

Here n = 10, m\ = 3, m2- 2, m3= 2To prepare the table :

...(2)

...(1)

X Y Rank in X

Rx

Rank in Y

Ry

d mRx —Ry <P

48 13 8 5 ± 6 = 5 52 3 2.5 6.25

33 13 6 5.5 0.5 0.2540 24 7 10 -3.0 9.0

9 6 1(2+3)

2 -1.5 2.25

16 15(2 + 3 + 4)

3 i7 -^.0 16.0

16 4 3 1 2.0 4.065 20 10 9 1.0 1.024 9 5 4 1.0 1.016 6 3 2.5 0.5 0.2557 19 9 8 1.0 1.0

Total •V 1 ^ = 41

Hence (1) becomes : j-

6[41 + i 3 ( 9 - l ) + 2 ( 4 - l ) + i 2 ( 4 - l ) ]

r = 1 10(100-1)

6(41 + 2 + 0.5 + 0.51 _ 264990 990

i.e; r - 0.733 Ans.

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3 5 0 | E n g in e e r in g M a t h e m a t i c s -!!!

Example 4.6! : Find the rank correlation coefficient fo r the following data ;

X : 68 64 75 50 64 80 75 40 55 64

¥ : 62 58 68 45 81 60 68 48 50 70

Solution ; Here in the series X, the value of 75 is repeated twice and vaiue of 64 is repeated thrice, (saym i = 2, m2= 3).

In the series Y, the value of68 is repeated twice, (say m3 = 2).Tie rank correlation coefficient is given by

r = 1-6 [ ^ 2 + U my - 1 ) + ]]2m 2 (m 2 - l ) + j 7 > m 3 ( " t f - 1 ) ]

n ( n 2 - 1)•••(I)

Here, n - 10, m\ =2, = 3, my = 2

Table for Calculation Rank Correlation :

X Y Rank in X Rank In Y ( V

f t . I

> 1

tp

68 62 7 6 1 i

64 58(4 + 5 +6) ,

3 54 1 i

75 68 (8 ; 9 ) = 8.5 (7 + 8) _ 7 ^2 1 i

50 45 2 1 1 i

64 81 5 10 -5 2580 60 10 5 5 25

75 68 8.5 7.5 1 1

40 48 I 2 -1 1

55 50 3 3 0 0

64 70 5 9 -A 16

©i i a 2 ^ = 7 2

Hence (1) becomes:

6^72 + i 2 ( 4 - l ) + i 3 ( 9 - l ) + j | . 2 ( 4 - l ) j

10 (100- 1)r = J -

. , [72 + 0.5+2 + 0.5] _ 1 ~ 6 990

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4.19 LINES OF REGRESSIONS

A line of regression is the line which gives the best estimate of one variable 'x' for any given valueof the other variable y.(i) Line of Regression of x

on y :

It is the equation of line which gives the best estimate for the values of x for a specified value of y. It is given by

x - x = r — ( y - y ) ...(I)

wherexandyare means ofx and y series, respectively; axanday are standard deviation ofx andy series,respectively and r is the correlation coefficient Which are calculated as follows.

mean o f x : * = ~ o r n = ~ - where u= x -A (A being assumed mean)

- Ey Elf y = — or y = ~ »where v =y - 3 (B being assumed mean)

\ 2

N u m er ic a l An a l y s i s -111, C o r r e l a t io n & R e g r e s s io n | 351

S.D. of x is a , = -J —— | — J (where u = x - A )

(Eu) (Iv ) E uv-

and coeff. of correlation isn

r — ----------------------------------

The above regression line x on>> i.e; equation (1), can be written as :x = ay + b,

where a is the slope and b is intercept of the line.(ii) Line of Regression of y and x :It is the equation of line which gives the best estimate for the values ofy for any specified valuesofx.Regression line ofy on x is given by

y ~ y = r°x

The above regression line (2), can be written as : y - a x + b

where a is the slope and b is intercept of the line.

4.20 REGRESSION COEFFICIENTSCoeffidept of Regression ofy on x :

The regression coefficient of y and x is denoted by and is given by

a v ...(1)

where o„ crv are S.D. of x and y series respectively, and r is the correlation coefficient.

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Coefficient of Regression of x on y :

The regression coefficient of x on y is denoted by and is given by

. _ b*y = r ctv

3 5 2 | E n g i n e e r i n g Ma t h e m a t i c s -]II

4.21 PROPERTIES OF REGRESSION COEFFICIENTS(1) The coefficient of correlation is the geometric mean of the coefficients of regression Le.;

r = ± X b y''yx

Proof: We know that

and K ~ r n> <*X

-d )

...(2)

Hence

, , SU. °y b*>x b» = r c r * r T x =r

r —± ijb^y x byx Proved.

(2) If one of the regresion coefficients is greater than unity, then the other is less than unity Le.;

If

Proof: Suppose that

We know that

Since

by* > 1=>bv < 1or

if bxy > 1 => byx< ! .

byx > 1 .

— yjbxy x byx

r2 = bxy* b^

- 1 S r S 1r1 < 1

bxy * b^ <1 (by 2)

bxy S p < luyx

..•(1)

..-(2)

Hence, if byx <1 bxy<\.

[becauseb^ >

1=> r^ - < 1]°yx

Proved.

(3) Both regressioH'coefneients and correlation coefficient are of the same sign Le.;

r > 0, ifbxy and > 0,

r < 0. rfbxy and b^ < 0.

For example : If b„ ~ - 0.4 and bxy-- 0.8, find r.

Solution : v r = ± Jbyx *bxy = ± V0.4x 078

[because by, and b , < 0] Ans.= ±0 32r =-0.32

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N u m e r i c a l Analysis-111, C o r r e l a t i o n & R e g r e s s io n | 3 5 3

(4) Regression coeficrents are independent of change of origin but Rot of scale.

(5) Arithemetic mean of the regression coefficient is greater than the correlation coefficient.Remark: To find b ^ and b^ by the following formulae :

(i) b „ - ------or 4 , - -----------------------------a ----

( * * - ¥ )

a » - g a g a ________ n ___ l - ___ _ ____ n __

n n

22 PROPERTIES OF LINES OF REGRESSION

1. Two regression lines x o n y andy on x always intersect at the means ( x , y ) .

For example : Find mean values of x and y, of two regression lines are :4 x - 5 y + 33 =0

and 20* - 9y - 107 =0.Solution: Given 4 x - 5 y = - 33 ...(1)

20x- 9y = 107 ...(2)Eliminate x, from (1) and (2), operate: 5 * (1) - (2), we get20*-25 y =-16520*- 9y = 107- + -

- 16y *-2 7 2 y = 17

Putting the value of y in (1). we get

* = 13

Hence mean of x andy are 13 and 17 respectively.2. If r = 0. then the regression coefficients are zero, i.e; regression lines are perpendicular to eachother.3. If r = ± I , then the regression lines become identical, i.e; both pass through the Common point

( * , y ) i . e ; jc = * andy = y .

4. The angle between two regression lines is given by

\

tan 0a x. Oy

_ 2 , _ 2a x + ° y )

If r = 0 => tan 0 = «>=>0= t i/2 => lines are perpendicular

If r = ± I = ^ t a n 0 = O >0 = O=5 lines are identical.

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4.23 RELATION BETWEEN REGRESSION ANALYSIS AND CORRELATIONANALYSIS

3 5 4 | En g i n e e r in g M a t h e m a t ic s -I 11

Correlation Analysis Regression Analysis

1. It is relationship between two or 1. Regression means stepping back or returningmore variables. to average value.

2. Correlation coefficient r between x and 2. bxy and are mathematical measures y is a measure of direction and degree of expressing the average relationship betweenlinear relationship between x and y. the two variables.

3. It is symmetric in x and y i.e; 3. The regression coefficient and by* are notr (x. y) = r(y, x) symmetric in x and y i e.,biy* b>x

4. The correlation coefficient does not 4. Regression coefficients reflect on the nature of reflect upon the nature of variable variable i.e., which is dependent variable and (independent or dependent variable).

• which is independent variable.5. It does not imply cause and effect 5. It indicates the cause and effect relationshiprelationship between the variables between the variables. The variabJeunder study. corresponding to cause is taken as independent

variable. Whereas corresponding to effect istaken as dependent variable.

6. It is a relative measure and is 6. Regression coefficients are an absoluteindependent of the units of measure of finding out the relatinshipmeasurement. between two or more variables.

7. It indicates the degree of 7. It is used to forecast the nature of dependent

association. variable when the value of independent variableis known.8. It has limited application as it is 8. It has wider applications as it also studies

confined to the study of linear non-linear relationship between therelationship between two variables. variables

Example 4,62 : Find the regression equation o f y and x and the coefficient o f correlation from the following data : Zx~60, Zy-40 , X xy -1 ISO, I k 2 - 4160, 2 y ~ 1 7 2 0 ,n - 10.

Solution: We have. _ Zx _ 60 _ ,

x " n "TO " 6: _ Iy 40 .v ' - ' i o ' 4'

Zxly n _

n

bxy =1160- 60x40

10

ZytJME

1450-240

1720-160

1720- (40)10

910

1560 = 0.583

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1150-gP*40

N u m e r i c a l A n a ly sis-1 11 , C o r r e l a t i o n & R e g r e s s io n | 3 5 5

and b„ = -------------- — -------------------— £jc2 _(E x)2 4 1 6 0 - ^ ^

n lu910

3800= 0.239

Since r = + x [v b^Sc. by, > 0, then r >0]

r * + -v/0.583 x 0.239= 0.373 Ans.

Again, the regression equation ofy on x :

J ' - j ' -&>*(*- *)

=> y - 4 = 0.239 (x -6 )=> _y = 0.239.x + 2.566. Ans.xample 4.63 ; / f regression equation o f x on y : 5x - y - 22 and regression equation o f y on

x : 64x - 4Sy —24 are two lines o f regression. Find (i) the mean values o f x andy,(ii) the regression coefficients,(iii) the coefficient o f correlation between x andy,(iv) the standard deviation ofy, i f the variance o f x is 25.

olution : (i) The mean values ofx andy , i.e; (jc, y ) lie on the both lines.

S x - y = 2 2 ...(1)

64 x - 45 y =24 ...(2)Multiplying the equation (I) by 45 and equation (2) by 1, we get

2253c -4 5 y =990

64 if -4 5 y =24

on subtract, we get 161 x =966

=> jc = 6.Putting the value of jc in (1), we get

3 0 - J? =22=> y =8,

Hence means: jc - 6, y = 8. Ans.(ii) The regression equation o fy on x is

64x - 45 y ~ 24=> 45 y = 64x - 24

_ 64 24^ y 45* 45

64*>i “ 45 [*•’ R.L. y on x is y = by,, x + c]

Again R.L. ofx on y i s 5x-y=22=> 5x = y + 22

1 • 22

x “ 5 T

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=> b„ = ~ [ v R.L. x in y is x = bzy y + c] Ans

(iii) We have, coefficient of correlation

r = ± ,Jbyx x bxy Since and b„ both are positive, then r will be postive, so that

3 5 6 | E n g in e e r in g M a t h e m a t i c s -III

- +V45 5

= jjr = 0.533. Ans.

(iv) Since variance of x = 25o*2 = 25 o a, - 5.

We know that by, = r — x .

6 4 8 <*y —S ----- — -*— X —~

45 15 5

40 .1 «<Sy ~ "jj* = 13.33

=> S.D. o f y = 13.33 Ans. Example 4.64 : I f 4x - 5y + 33 ~ 0 and 20x -9 y —107 are two lines o f regression. Find

(i) the mean values o f x andy,(ii) the regression coefficients,(iii) the correlation coefficient,(iv) the standard deviation o f x i f the variance o f x is 9,(v) the value o f y for x 15J.

Solution : (i) The mean values of jc and y, i.e; ( j , y ) lie on both lines.

4* - S y =-33 ...(I)

20* - 9 y = 107 ...(2)Multiplying (1) by 5 and subtracting from (2), we get

- 16 y = -2 72 => y = 17

Putting the value of y in (1) we get x =13.

Hence the mean values of x and y are 13 and 17.(ii) Let us assume that R. L. o f y on x be 4 x - Sy + 33 = 0=> 5y = 4 x + 33

4 33^ y 5 X 5 *•* . . I/. Regression coefficient yon.c :

h - ib>r~ 5

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2 0 c -9y = 10720* =9 'y + 107

9 107* 20y 20

h - - 1Oxy 20

4 9Clearly by, x bv = g x — < l

Hence our assumption is correct.

4 9by, — g and = 2q * Ans.

(iii) Since r = ± J b ^x ~ b ^

N u m e r ic a l A n a l y s is -111, C o r r e l a t i o n & R e g r e s s io n | 3 5 7

Also the R.L. of jc on y be

(iv) Since variance of x = 9=>We know that

[ v bn & are positive]

3= + g = + 0.6. Ans.

0*2 = 9 => ox ~ 3

± - *

5 5 TCTj. = 4

=> S.D. ofy = 4. Ans.(v) Since the R.L. y on x gives the best estimated value of y for given x,i.e; Put* = 3 in 4 * - 5y + 33 = 0, we get

I2-5y+ 13 =0=> y - 9 . Ans.

Example 4.65 : l/ 2 x + 3 y a:7 and 5x + 4ym9 are two lines o f regression. Find

(i) the mean values o f x andy,

(ii) the regression coefficients,(iii) the correlation coefficient between x andy.

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Solution : (i) the mean values of x and y, i.e; (it, y ) lie on both lines.

2x +3y =7

5 x + 4 y =9

Solving equations (1) and (2), we get

- I? ^ - 5x = -y and y - y

17 5Hence mean of x is -y and mean of y is y .

(ii) Let us assume that R.L. of y on x be2x + 3y = 7

=> 3y ~ - 2 x + 1

- 2 x 7

_ 2V 3

Also, the R.L. of x on y be5x + 4y = 9

=> 5x = - 4y + 9

4 9x = - g y + 5

h = ib*y ~ 5

3 5 8 | E n g i n e e rin g M a th e m a tic s -H I

< 1

Hence our assumption is correct.Hence of regression coefficients are

I. - 2 A k _ — O yx g 8 n t * w jy g

(iii) v r - ±

=> r - - ^6,* x [ v & by, both are negative]

— f i l l A .f 15

= -0.73.

Hence correlation coefficient between x and y is - 0.73.

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Now ,

\

or

Hence (1) becomes

o , 2.5

= 0-8 x a s

y - 6 5 = y (x -6 7 )

y = 0 . 5 7 J ( x - 6 7 ) + 65

y = 0.57Lt + 26.71

Pu t.r = 70 in (1), we get estimated height of sister

y =0.57! « 70 + 26.71

= 66.68 cm. Aw

Example 4.68 : Calculate the coefficients o f regression lines and find the two lines o f regression from the following data: x : 78 89 , 97 69 59 79 68 61 y : 125 137 156 112 107 136 123 108.

Solution : Let u = x - 78 and v = y - J 25, where 78 and 125 are assumed means o f x and y respectively

To obtain the regression lines, prepare the following table :

x y u - x -7 8 v “ y -1 25 u v w1 V2

78 125 0 0 0 0 0

89 137 11 12 132 12 1 144

97 156 19 31 589 361 961

69 112 - 9 -1 3 117 81 16959 107 -1 9 -1 8 342 361 324

79 136 1 11 11 1 12 1

68 123 - 1 0 -2 20 100 4

61 108 -1 7 -1 7 289 28 9 289

Zx = 600 Z y= 1004 Zu = -24 Zv = 4 Zwv = 1500 Z«J= 1314 Zv2 = 20 J 2

We have n = 8 ,

Zx 600= 75 Mean o fx : x - - a

n o

.. , - zy 1004 _ Mean o fy : y = — * — 5 — - 125.5

Coefficent o f regression y on x :

n

by, f n 8

o1 3 H -

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N u m eric a l An a l y s i s -111, C o r r e la t io n & R e g re s sio n | 361

1500 + 12 _ 281314-72 23 l-2,7>and

Coefficient o fregression x on y :

Zut>- Z uZu1500-

(-24)(4)8

Z.2 - < ^n

2012- &L.8

1512 756= 0.752.

2010 1005

The line o f regression y o n x :

y - y = by, (x - x )

=> y - 125.5 = 1.217 ( j c - 75)'

=> y = 1.217* + 34.225.

The line o fregression x a n y :

x - x = bv ( y - y )

=> x - 75 = 0.752 iy - 125.5)=> x =0.752>'- 19.376. Ans.

Example 4.69 : Find the two lines o f regressions from the following data :x : 158 160 163 165 167 170 172 175 177 181

y : 163 158 167 170 160 180 170 175 172 175.

Estimate y, when x ~ 164.

Solution : Let A = 170, B= 175 be the assumed forx and y series, so that u =x - 170, and v = y - 175

X u =x-170 i»* y v=y-175 V-2 U V

158 -12 144 163 -12 144 144

160 -10 100 158 -17 289 170163 -7 49 167 -8 64 56

165 -5 25 170 -5 25 25

167 -3 9 160 -15 225 45

170 0 0 180 5 25 0

172 2 4 170 -5 • 25 -10

175 5 25 175 0 0 0

177 7 49 • 172 -3 9 -21

181 11 121 ; 175 0 0 0

Zx = 1688 Zw = -12 Iw2 = 526 I y= 1690 Zv = -60 8 I I W

Zkv= 409

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Here n - 10.

We have,

- Z* 1688 , , 0<) Mean o fx : * ~ ~ ~ ~J q ~ = 168.8

. . , ~ ly 1690 Mean ofy : y = „ = " jq " = 169-

Regression coefficient y on x :

E u a - S a J i 1

* - - - ------------ ---------------■----- »fi-

362 | En g i n ee r in g M a t h e m a t i c s - I I I

[a.*-fi£] [526- t f i ]

409 - 72 _ 337 =M J9_ 526-14.4 51L6

Regression coefficient x on y :

4og_ (-12) (-60)

- f § ' 0 - 7 5 6 . Regression line ofy on x :

y-y =M*~* )=> >>—169 = 0.659 (x - 168.8)

=> y = 0.659*+ 57.761. ...(1)

Regression line ofx on y :

x = ^ 0 - y )

=> * - 168.8 =0.756 ( y - 169)

=> x =0.756 >>+ 40.236. ...(2)

Estimated value o fy when x —164 :

From equation (1), we have

y =0.659 x 164 + 57.761

i.e. _y(164) = 165.837 Ans.

Example 4.70 : From the following data, obtain the two regression equations :

Sales : 91 97 108 121 67 124 51 73 111 57

Purchases : 71 75 69 97 70 91 39 61 80 47.

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N u m e r ic a l A n a l y s is -111, C o r r e l a t io n & R e g r e s s io n j 3 6 3

Solution : Let the variables x and y, respectively denote the sales and purchases. We prepare the

following table : Let u = x ~ 90 and v = y - 70

X u x-90 v -y -7 0 u1 V? uv

91 71 I 1 1 1 1

97 75 7 5 49 25 35

108 69 18 - 1 324 1 -1 8

12 1 97 31 27 961 729 837

67 70 -2 3 0 529 0 0

124 91 34 21 1156 441 714

51 39 -3 9 -31 1521 961 1209

73 61 -1 7 - 9 289 81 153

1 1 1 80 21 10 441 100 2 10

57 47 -3 3 -2 3 1089 529 759

Zx = 900 Zy= 700 Z« = 0 Zv = 0 Zu2= 6360 Zv2= 2868 Zuv = 3900

Here, n = 10.

We have,

Mean o f x : x = ^ , 900 = 90 x _ lnn 1 0

, , - - 700 - Mean ofy : y - — ------- Jq* -


Z uLv

70,

n

n

3900-0[6360-0]

390636


= 0.613.

ZuZuZu v -

Oxv---------------------3900 -0[2868 - 0] = 1.36.

Regression line o f y o n x :

y - y = M * ~ x )

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3 6 4 | E n g in e e r in g M a t h e m a t ic s - ! II

Regression line o f x on y :

x - x ^ b r y iy -y )

=> x - 90 = 1.36 (y - 70)=? x = 1.36y - 5.2 Ans.

Example 4.71: Find the two lines o f regression from the following data : Age o f husband : 25 22 28 26 35 20 22 40 20 18 Age o f wife : 18 15 20 17 22 14 16 21 15 14.

Hence estimate : (i) the age o f husband, when the age o f wife is 19 and (ii) the age of wife, when the age o f husband is 36. (iii) the correlation coefficient between them.

Solution: Let x = Age o f husband

y = Age o f wife

Let u - x - 26 and v = y - 17, where 26 and 17 are assumed means of x and y series

respectively. __________________________________ ___

X u ~ x - 2 6 i i* y v my - 1 7 uv

25 -1 1 18 1 i -1

22 ~A 16 15 - 2 4 8

28 2 4 20 3 9 6

26 0 0 17 0 0 0

35 9 81 22 5 25 45

20 -6 36 14 . - 3 9 18

22 - 4 16 16 -1 1 4

40 14 196 21 4 16 56

20 -6 36 15 - 2 4 12

18 -8 64 14 -3 9 24

Ex = 25 6 E« = -4 £«* - 450r -I I S k

W < NI I £ Ev2 =78 Euv= 172

Here n = 10.

We have

Mean o f x : _ _ Ex _ 2561 * T - To"’ 256

Mean o f y :


.a »vv

» • v •

y = S = 1 Z2 =172 y n 10 1 • •

Euv- EuEt> 172-n

(-4)(2).10

n4 5 0 - < -4 )2

10

T O U 8 _ 172.8450 -L 6 448.4 = 0 .385.

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N u m e ric a l A n a ly s is -III , C o r r e l a t io n & R e g r e ss io n | 3 65


Zuu-brv

LuZv172-

7 8 -

(-4)(2)10

(2)1 '10

172.877.6

= 2.23.

Regression line o f y on x :

y - y =b}, ( x - x )

=> y - 17.2 =0.385 (x - 25.6)

=> y = 0.385* + 7.34

Regression line ofx on y:

x - x = bv ( y ~ y )

=> x - 25.6 =2.23 0 - 17.2)

=> x = 2.23y- 12.76

The age o f husband (x) when the age o f wife (y) is 19:

i.e. ; puty = 19 in equation (2), we get

x = 2.23 x 19- 12.76 = 29.6 « 30 years.Hence age of husband = 30 years.

The age o f wife (y) when the age o f husband (x) is 30 :

i.e; Put x ~ 30, in equation (1), we get

y = 0.385 x 30 + 7.34 = 18.89 #» 19 years.

Hence age of wife = 19 years.

Correlation coefficient:

- ( I )

- ( 2)

-±$. yX X

= + V0.385x 2.23 by, > 0]

= 0.927 Ans.

Example 4.72 : The following data represents rainfall (x) and yield o f paddy per hectare (y) in a

particular area. Find the linear regression o f x on y.

x : . 113 102 95 120 140 130 125

y : 1.8 1.5 1.3 1.9 1.1 2.0 1.7.

Solution : Let 120 be the assumed mean for x series;?i)d 1.8 be the assumed mean fory series, i.e.. letu = x - 120 and v - y - 1.8.

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366 | E n g o c e r m g M a th em a tic s-1 1 1

We prepare the following table :

X u “ *-120 U1 y v- j>-1 .8 V2 u v

113 -7 49 1.8 0 0 0102 -18 324 1.5 -0.3 0.09 5.4

95 -25 625 1.3 -0.5 0.25 12.5

120 0 0 1.9 0.1 0.01 0

140 20 400 1.1 -0.7 0.49 -1.4

130 10 100 2.0 0.2 0.04 2.0

125 5 25 1.7 -0.1 0.01 -0.5

Ix =825 M « I f £k2 =1523 iy= 11.3 Ev = -0.13 Lv2 = 0.89 £i/v = 18

Here n ~ l . We have

Mean o f x :

Mean o f y ;

—_ _ 825 . . . x — = - s t 117.86.

n r

- Xy 1L3


n

Z .uv -

1.614.

Zulu 1 8 -(-15) (-0.13)

n0 .8 9 - ( 3)- j

18-0.279 17.721= 19.965.

Ans.

0.89 - 0.00241 0.8876 Regression coefficient x on y :

x - x = b iy (y - y )

=> x - 117.86 * I9.965(y- 1-614)=> x = 19 .965^+ 85.637.

4.24 CURVE FITTINGCurve fitting means an expression of the relationship between two variables by algebraic equations.

Suppose that we are given n-values*i, x2, ....... x„ of an idependent variable* and the correspondingvalues yu y i....... . y» of a dependent variable^ depending on x. We try to find a curve that servesas best approximation to the curves ~ f (x). Such a curve is called as the curve of best fit and the

process of determining a curve of best fit is called curve-fitting, i.e., The general problem offinding equations of approximating curves which fit given sets of data is called curve-fitting.

4.25 PRINCIPLE OF'LEAST SQUARES

Principle of least squares provides a unique.set of values to the constants and suggests a curve of best fit to the given data.

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4.26 FITTING OF A STRAIGHT LINELet (*,, y,), i = l , 2....... . n be V sets of observations of related data and

y - a + bx ... (1)

be the straight line to be fitted The residual at x = x, is

e, = y, ~f(x,) = y, - (a + bx,), (= 1 ,2 .......... n.

n n

S - ~ a ~ bx')2‘ -(2)1=1 /=i

By the principle of least squares, S is minimum

55n a 8S n

=0 and — = 0. -da db

Differentiating (2) partially w.r.t. a and b respectively, we get,

— = - 2 Y iy , - a -b x ,) i.e., T ( y , - a - b x , ) =o ...(3)da t t

30 n " and — = - 2 £ x,(y i-a - b x ,) i.e., £ 0 ^ ~ ax>~ bx^ = 0 ■«<♦)

8b ,=1 /=i

From (3) and (4), it follows that

n n n

Y.y> = £ * * * & i.e.,Iy = m> + £>Lx ...(5);=1 (-1

n ft tl

and S r ' v' = aZ x' + x' 2 i-e-, Zxy = olx + * Ir2 ... (6)/=! >=l /=!

These equation given by (5) and'(6) are called Normal&quatiom, can be solved for determining‘a’ and lb' X] and yt are known.

cS dS Differentiating — w.r.t. 'a ' partially and — w.r.t tb' partially, we get

ca oo

368 | E n g i n e e r in g M a th e m a tic s -HI

= 2; and —y = 2 Tx,2 = +ve.Ti

S2S „

3 a 2 * db2

which are posit iv es Kies. Hence, ‘S' bSMwimum.

Now putting the values of ‘a ’1and'b7

get equation of <hostraight line o f best fit theciven dat.i points.

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4.27 FITTING OF A PARABOLA *OR SECOND DEGREE CURVE)L<*» •• =( ' f’r •-rx2 •••(!)

e a , i/u'w. ( -hi . : rVe) to be fitted for the V data ( x,, yt), i - 1 ,2,..... . n.

Rie r -sidual at jc = x, is

e< y, -/(*»)-y , - (<a + bx, + cx,2)

n n

s = z y = Z (^ ~a~bx>- ^ z 2) ... (2)r=l z«l

By the principle of least squares, S should be minimum for the best vaJues of o, b and c.

dS dS j dS = 0, — = 0 and — - 0. ... (3)

da db dc

Using (2), these three conditions given in (3), we get

Zy = na +bZx + cZx2

Lxy = a lx +bLx2 + d r 3 ... (4)

Zx2y = aZx2 +bZxi +cZx*

N u m e ric a l An a l y s is -111, C o r r e l a tio n & R e g re ss io n | 369

here Zx = and so on./=i

These three equations given by (4) are called the normal equations, can be solved for determining

a, b and c.Putting the values a, b, c in (1), we get the equation of the parabola of best fit for the given data.

Remarks:

(i) Normal equations to the straight line y = a + bx are :

Zy ~ ma + bZx, where m is number of data points,and Zxy —aZx + bZx2.

(ii) Normal equations to the parabola y - a + bx + cx2 are :

Zy - ma + b lx + cZx2

£x>> - aZx +bZx2+ clxi

Zx2y = aZx2 + bZxy + cZx*

(iii) If the values in the given data are large, we changing the origin and scale by using substitutions.

If m is odd i.e., number of data is odd, then we assume.

x-(middle term)u ---------------------- > (h = length of interval)

If m is even i.e., number of data is an even, then we assume

_ x - (mean of two middle terms)

M= (A/2) '

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4.28 FITTING OF A CURVE OF THE TYPE y?=ab*

Let the curve to be fitted be y = ab1. ... (1Taking logarithm on both sides, we get

logy = log a + x.\ogj>

or Y=A+Bx. ...{2)where Y = logy, A -\o g a and B = log b. Normal equations of (2) are :

I.Y = nA + BTx (where n is number of data) ...(3) ZxY =AZx + Bl jc2 ...(4)

where n is number of given points.Solving (3) and (4), we obtain A = log a and B - log b i..e, a and b are found by taking antilog.

Note-: Similarly we can fit the curve y = ax*.

4.29 FITTING OF THE EXPONENTIAL CURVE y = ae**

Let the curve to be fitted be y - ae** , ... (1)

Taking logarithms of both sides of (1) to the base 10, we get

togio y = logio a + bx log,0 e ... (2)

This is of the form Y = A + Bx ... (3)

wher Y= logio y, A - log,0 a, B = b log)0 e.

:. Normal equations of (3) are :

Ly = nA +B lx ... (4)

XxY = ALx + B lx2 ...(5) 1

Solving (4) and (5), we get A

and B

from whicha

and ft can be determine. !

430 FITTING OF A CURVE OF THE TYPE y - a*2 + ~

For obtaining normal equations :

J 2 »VThe squared sum of errors S - ~~x) ** niinimum.

dS

= 0 given lx , y? ~ alx,* + iEx,i.e., Ixy2 = aZxt+blx ...(1)

dS v y( 1 — = Ogives = aZx,+ b l 2

* Ou

370 | E n g i n e e r in g M a th e m a ttc s -H I

From the normal equations (1) and (2), a and b are obtained, and as such the desired curve is

fitted. v*

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Example 4.73 ; The weights o f a calf taken at weekly Intervals are given below. Fit a straight line using the method o f least squares, and calculate the average rate o fgrowth per week.

Age: 1 2 3 4 3 6 7 8 9 10

Weight: 52.5 58.7 6.5 70.2, 7S.4 81.1 87.2 95.5 102.2 108.4

Solution. Here number of values is 10, i.e., even, so we changing the origin and scale by put

N u m e ric a l A n a l y s is -111, C o r r e l a t i o n & R e g r e s s io n (, 3 7 1

x - (mean of two middle terms)u -----------

- r a

(h i 2) 1/2

x - 5 5i.e., u = — j - and v = y.

Let the straight line to be fitted bev =a + bu —(I)

Its normal equations are ;Iv = ma + bZu ... (2)I w = d a + bZu2 ... (3)

* v m y*-5 .5

M~ 0.5uv

1 52.5 - 9 - 472.5 81

2 58.7 - 7 -410.9 49

3 65.0 - 5 - 325.0 25

4 70.2 - 3 -210.6 95 75.4 - 1 -1 5 A 1

6 81.1 1 81.1 1

7 87.2 3 261.6 9

8 95.5 5 477.5 25

9 102.2 7 715.4 49

10 108.4 9 975.6 81

Sv = 796.2 Z««0 Ii»v= 1016.8 In 1 = 330

Putting these values in (2) and (3), we get796.2 - 10 a and 1016.8 = b (330)

=> a -79.62 and * = 3.081.The required equation of curve of fit is

v =(79.62) + (3.081) u [From(l)]

or

y .= 6.162*+4£729. 'if!

Thus, the average rate of growth per week is 6.162 units. Ans.

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Example 4.74 : Fit a straight line to the following data, taking y as the dependent variable

x : I 2 3 4 5 6 . 7 8 9

y : 9 8 10 12 11 13 14 16 15

Solution. Since me an of x-s erie s is 5 and me an o f> se rie s is 12. [i.e., use changing the origin]

Let * = * - 5 and Y = y - U Let the straight line to be fitted be

Y = a + b X ...(I)

Then its normal equations are :

ZY = ma + bZX ... (2)

ZXY =aZX+bZX> ...(3)


where m is numbe r o f data i.e., m - 9.

X y X = x - S Y - y - 12 XY X 2

\ 9 - 4 - 3 12 16

2 8 - 3 ' - 4 12 9

3 10 - 2 - 2 4 4

4 12 - 1 0 0 1

5 n 0 - 1 0 0

6 13 1 1 1 1

7 14 2 2 4 4

8 16 3 4 12 9

9 15 4 3 12 16

ZX = 0< 5 & t l w

ZXY =5 7 ZX2 = 60

Putting these values in normal equations (2) and (3), we get

0 = 9a + 0 => a = 0

and 57 = 0 + 60 b = > 6 = 0.95

The required equ ation o f curve o f fit is

Y = 0 + 0 .9 5 *

or y - 12 =0 .95 ( j f -5 ) [ \ ‘ X =x

=> y = 7.25 + 0.95x Example 4.75 : The results o f measurement o f electric resistance R o f a copper

temperature fC are listed below

t: 19 25 30 36 40 45 R : 76 77 79 80 82 83

Find a relation R = a + bt, when a and b are constants to be determine. Solution. Here num ber o f data m = 7.

Let the given relation to be fitted to the data be :

R = a + bt

-5 , Y=y~ 12]

Ans.bar at various

50

85

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its normal equations are :

Z* = ma + bZt ... (2)

ZtR = aZt + bZt2. ... (3)

N u m e r ic a l A n a l y s is -!!!, C o r r e l a t i o n & R e g r e s s io n | 3 7 3

t R tR fi-19 76 1444 36125 77 1925 62530 79 2370 90036 80 2880 129640 82 3280 160045 83 3735 202550 85 4250 2500

14 ~ 245 Z R ^ 562 Irtf “ 19884 Zt1 = 9307

Putting all values from the table in normal equations, we get562 = la + 2456

19884 = 245a + 93076On solving equations (3) and (4), we get

a = 70.06, b = 0.292Hence the requried relation (1) becomes :

R = 70.06 + 0.292fxample 4.76 : Fit a second degree parabola to the following data

. . . (3)

... (4)

Ans.

olution.

x : I y : 1090

Here m - 5 (number of data).Changing of Origin

21220

u = x - 3 and v =

31390

y - 1220

41625

51915

Let the parabola of the curve bev = a + bu + cu2

Its normal equations areZv = ma + bZu + cZu2

Zav = aZu + bZu2+ cZ«3.

Z«2v = aZu21 6Za3+ cZu*.

( 1)

X y u = x - 3(y~1220)

v= 5 u2 u3 u4 uv u2v

1 1090 - 2 -26 4 - 8 16 52 -1042 1220 - 1 0 1 - 1 1 0 03 1390 0 34 0 0 0 0 04 1625 1 81 1 1 1 81 815 1915 2 139 .. 4 8 16 278 556

I I < s >

Zv - 228 Zu3^ 10 Zu3 •=0 Zu* = 34 Zuv = 411 I« 2v ■=533

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Putting all values from the table in the normal equations we get,

5a+ A(0) + c(10) = 228

a(0 ) + /?(lO) + c(0) =411=> A= 41.1c(10) + 6(0) +c(34)=533.Sovling (2) and (4), we get

a - 34.6 and c = 5.5

Heace(l) becomes:

v = 34.6+ 41.14« +5.5a2

3 7 4 { En g in e e r in g M a t h e m a t k s -JII

...(2

...(J

-{4!

or O'-1220)

= 34.6 + 41.1 (x - 3) + (5.5) (x - 3 f

=> y = 1220 + 173 + 205.5 (x - 3) + 27.5 ( x - 3J2

y » 1024 + 40.5x + S7.5*2. Ant

Example 4.77: Fit a parabolic curve o f regression o fy on x to the 7 pairs o f values (orfit a secom degree parabola to the following).

Solution.

x : 1.0 1.5 2.0

y : 1.1 1.3 1.6

Here m = l and changing the origin

x -2 5

2.5

2.0

3.0

2.7

3.5

3.4

4.0

4.1

Suppose u -0.5

= 2x - 5 and v =y.

Let the parabola or second degree parabola of the curve be

v = a + bu + ew2

Its normal equations are

Ev = ma + bZu + cZu2

Zuv = aZu + bZu2 + cZu3

Zu2v = aZu2 + bZu3 + cZu4.

•d)

X y u - 2 x - 5 ym y uv u2v u4

1.0 1.1 - 3 1.1 9 -3.3 9.9 -2 7 81

1.5 1.3 - 2 1.3 4 - 2.6 5.2 - 8 162.0 1.6 -1 1.6 1 - 1.6 1.6 : -1 1

2.5. .2.0 =“ . o ;>o ' 0 0 0 0 0

3.0 2.7 ■ — 1 ---- -2.7 1 2.7 2.7 i 1

3.5 3.4 2 '3.4 4 6.8 13.6 * 8 16

4.0 4.! 3 4.1 9 12.3 36.9 27 81

Z u -0 Iv - 16.2 Zu*-2 8 Zuv - 14.3 Zu2v~69.9 Lu< = }96

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Putting all values from the table in die normal equations, we get

la + 28c = 16.2 ,..(2)

286= 14.3 =>i = G.5i ...(3)

28a + 196c =69.9 ...(4)

Solving equations (2) and (4) we get,a = 2.07 and c = 0.061

Hence (1) becomes:

v =2.07 + 0.5U« + 0.061u*

or y =2.07 + 0.511 (2 x -5) + 0.061 (2x -S? [v « = 2x- 5 and v = y]

or y = 2.07 + 1.022x - 2.555 + 0.061 (4jc2 - 20x + 25)

or y =1.04 - 0.193* + 0.243xJ. Ans.

Example 4.78: The profit o f a certain company in the xf* year o f Os Ufe are give by -

x : 1 2 3 4 5 y : 1250 1400 1650 1950 2300

Hiking u - x - 3 and 50v - y - 1650, show that the parabola o fsecond degree o f y o n x is y - 1140.05 + 7Zlx + 32.1 Sx*.

( y - 1650)

50 ’

Let the parabola of the second degree of the curve be

v - a + bu + cv*. ... (1)

Its normal equation is

' ’ Iv = ma + bZu + cZu2

Zi/v = a lu + bZu2 + cZu3

ZuJv = aZu2 + bZu3 + cZu4.

N u m e r ic a l A n a l y s is -U I , C o r r e l a t io n & R e g r e s s io n | 3 7 5

Solution. Here m = 5, and given that k = x - 3 and v1

X y u m x - 3 y -1650

uv u2 vu2 i t * u*50

1 1250 - 2 - 8 16 4 -3 2 - 8 16

2 MOO - 1 - 5 5 1 - 5 - 1 1

3 1650 0 0 0 0 0 0 0

4 1950 1 6 6 1 6 1 I 1

5 2300 2 13 26 4 52 8 16

Z u -0 Zv-tf Zuv■ 53 Zu3- 10 Zvu2 —21 Z*f . 0 Zu4 ” 34

Putting all values from the table in normal equations, we get,

5a + 6 (0) + c (10) = 6

a (0> + i (10) + c (0) - 53 => b - 5.3

a(10) + 6(0) + c(34' 21 sT

»■• (2)

v-(3)...(4)

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On solving equations (2) and (4), we get

a = -0.086, c — 0.643Hence (I) becomes:v = - 0.086 + 5.3a + 0.643a2

y -1650or — —— = - 0.086 + 5.3 (x - 3) + 0.643 (x - ty

50

y -1650

50= - 0.086 + 5.3 {x - 3) + 0.643 <x2 - 6x + 9)

v =( y - 1650)

50

>- = 32.l5x2+ 72.1x+ 1140.05. Example 4.79: Fit a second degree parabola to the following data

x : 0 1 2 y : 1 1.8 1.3

Solution. Here

m = 5,Changing the origin a = x ~ 2 and v *= y.

Let the second degree parabola of the curve bev = a + 6a + ca2.

.\ Its normal equation isZv = ma + 6Za + cZa2

lav = aZa + AZa2+ cZa*Iw2v = aZa2+ 6Za3+ cZa4.

32.5

and v = x -3

Ans.

46.3

- ( I )

X y u = jc - 21 1 uv u2 U2V u} u<

0 i - 2 1 - 2 4 4 - 8 161 1.8 - 1 1.8 -1.8 1 1.8 - 1 1

2 1.3 0 1.3 0 0 0 0 0

3 2.5 1 2.5 2.5 1 2.5 1 1

4 6.3 2 6.3 12.6 4 25.2 8 16

Z« = 0 Zv = 12.9 Zuv “ 11.3< S b * * * i

i Zm*v = J3.5 Zu* = 0 Zu* = 34

Putting all values from the table in the normal equations, we get.

5a + 10c = 12.9 ...(2)106 =11.3 =>6=1.13 ...(3)

10a + 34c = 33.5. ...(4)

On solving equations (2) and (4), we get

a = 1.48 and c = 0.55

Hence (1) becomes:

v = 1.48+ I.l3u + 0.55w2

or y = 1.48+ 1.13 (x-2> + 0.55 (x -2 )2 [v a = x -2 an d v=y]

=> y = 1.42 - 1.07x + 0.55x2. Ans.

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Example 4.80 : Fit a curvey * ah* to the following data :

x 2 5 4 5 6

y : 144 172.8 207.4 248.8 298.5

Solution. Given y = ab* ... ( t)Taking logm of both sides, we get

logio y = logio a + jc logio b

:. Its normal equations are :

2 logio > = w logio a + [og,o b Zx [here ,7t - 5]

Ix logio y - log,0 a Zx + log]0 b Zx2

N u m e r ic a l A n a l y s is -111, C o r r e l a t i o n f t R e g r e s s io n | 3 7 7

X y X 1 togi t y r ogi,y

2 144 4 2.1584 +.3168

3 172.8 9 2.2 >>5 6.7125

4 207.4 16 ?<168 9.2672

5 248.8 25 2.3959 11.9795

6 298.5 36 2.4749 14.8494

Ex = 20 Zx3 <=90 2logi»y “ 1 1 .5 8 35 Zxlog,9y “ 47 .1254

Putting all values from the table in the normal equations, we get

5 iogio a + 20 logio 6 = n .5835 ... (2)

20 logio a +90log]0 b = 47.1254. ... (3)On solving equations (2) and (3), we get

logio a ~2 and logio b = 0.079

=> a - 100 and b = 1.2 [as taking antilog]

Hence (I) becomes :y= 100(1.2)*. Ans.

Example 4.81: Fit a curve y = ax* to the following data and estimate y a tx m 12:

x : 20 16 10 11 14

y : 22 41 120 89 56

Solution. Given y = ax6 ... (1)

Taking logt of both sides, we get

Ioge y - log. a + b log, x

or Y = A + bX ' ...(2)

where Y = \ogey,A -\o g caandX= \o^cx

Then normal equation of (2) are :

ZY = mA + bZX ... (3)

ZXY~AZX+bZX> ...(4)

m = 5 (No. of data)

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3 7 8 | En g in e e r in g M a t h e m a t k s -III

* y X - log, x Y m iog<y X ? XY

20 22 2.996 3.091 8.976 9.2606

16 41 2.770 3.7135 7.6729 10.2864

10 120 2.30 4.7875 5.29 11.01111 89 2.398 4.4886 5.7504 10.763

14 56 2.64. 4.02535 6.9696 10.627

X X -13.107 Z Y - 20.106 ZX*-34.66 L X Y -51.95

(4)

Putting all values from the table in the normal equations, we getSA + 13.1076 = 20.106

and 13.107/«f+ 34.666 =51,95Solving above, we get

A = 10.254 and 6--2 .37948 « - 2.38=> log* a — 10.254 a ~ 28491.416 « 28491

Hence (1) becomes y =28491 xx-23*

Further, estimates at * = 12 : y =28491x(i2)-2« - 76.956 « 77 A bs.

Example 4.82: By the method o f least squares, find the curve y m ax+ bx? that best fits the followingdata:

x : I 2 3 4 5 y : 1.8 5.1 8.9 14.1 19.8

Solution. Given y = ax + bx2

Let (*1» yi) (*2, k ) (*3, yi) ■••• (** y») be the points.Then error or estimate for f* the points (**, y,) is Ei=(yi~ a x ,- bx?).By principle of least squares the values ofa and 6 are such that

S = ZE? = Ky, -rax, - bx?)1 is minimum..*. Normal equations are given by

£da

Txty im aXx? +

blxj3and

Zx?y, =

dZx? *

bZx,4 or Zxy * aZx2 +6£x3 (2)

1x2^ = o Ix 3 + 6 I^ ...(3)

(1)

dS O . a n d ^ - 0

X y xy X? x* ’ X* x3y

1 1.8 1.8 1 1 1 1.82 5.1 10.2 4 8 16 20.4

3 8.9 26.7 9 27 81 80.14 14.1 56.4 16 64 256 •• 225.6

5 19.8 99 25 125 625 495

I x y - 194.1 I x* -55 Tx> « 255 Ex* m 979 Ix*y * 822.9

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N u m e ric a l A n a ly s re -III, C o r r e l a t io n & R e g r e s s io n | 379

Putting all values from the table in the noimal equations, we get

55a+ 225* - 194.1225o + 979* =822.9

On solving equations (4) and (5), we geta - 1.52 and * = 0.49.

Hence required curve (1) becomes: y - 1.52x + 0.49x2. Example, 4.83 :fita second degree parabola to the following data

x : 1929 1930 1931 1932 1933 1934 1935 y : 352 356 357 358 360 361 361

Solution. Here m = 9 (which is odd number)Changing the origin u ~ x - 1933 and v=y-357.Let the parabola of the curve be

v = a + bu + cu1.

Then its normal equations are :Ev = ma + bZu + cZu2

Zvv - aZu + bZu1 + cZu3

Zu2v - aZu2 + bZu* + cZu*.

X u -x -1 9 3 3 y v “y - 357 uv u* u2v u* u*

1929 - 4 352 -5 20 16 -8 0 -6 4 256

1930 - 3 356 -1 3 9 - 9 -2 7 81

1931 - 2 357 0 0 4 0 -8 16

1932 -1 358 I -1 1 1 -1 1

1933 0 360 3 0 0 0 0 0

1934 1 361 4 4 1 4 1 1

1935 2 361 4 8 4 16 8 16

1936 3 360 3 9 9 27 27 81

1937 4 359 2 8 16 32 64 256

Z u~ 0 £ v « / / Z uv-51 1 m 2-6 0 Zh2v - - 9 Z u*-0 Zu*“ 70

Putting all values from the table in the normal equations, we get

9a + 0 + 60c —11. ...(2)

0 + 60* + 0 + 0 = 51 => * = 0.85. ' ...(3)

60 a+ 0 + 708c =-9 . ^

On solving equations (2) and (4), we get ■ ,.. ; - f

a = 3 and c = - 0.27. o y

Hence fitted parabola (1) becomes: • ;<:• t « ■

v = 3 + 0.85a-0 .27 •••

=> y - 357 = 3 + 0.8j$r - 1933) -0 .27 (x - $33)*=> y =-0 .27x^+ J044.67 x - 1010135.08. Am.

... (4)

...(5)

An.

1936 1937 360 359

...(1)

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3 8 0 | En g in e e r in g Ma t h e m a t ic s -III

Ezercise-4(A)

hod, obtain the <

XO. 1) and y(0.2).

dyI. Using Picard’s method, obtain the solution of — = *(1 + x 3y), y(0) = 3. Compute the value

dy2. Use Picard’s method to approximate the value ofy when * = 0.1, given that — = 3* + y2 with

A O )-I.

dy n y3. Find the third approximation of the solution of the equation ~ x by method

when >>= 2 forx= 1.dy

4. Find the fourth approximation of the solution of the equation -fa = 2 x - y with y( 1) = 3 using

Picard's method. '

dy5. Given — = 1 + y2 with><0) = 0, obtain^ as a series in powers of* in the range 0 <x£ 1, using

Picard’s method.6. Solve the following initial value problem by Picard’s method

^ = xey, withXO) = 0. Compute ><0.1), ><0.2) and XI)-

7. Solve the following simultaneous differential equations, using Picard’s method for

* = 0.1 and 0.2; t r “ ** + = °» and 2(0) = 1.dx dx

8. Using Picard’s method, obtain the second approximation to the solution of

d2y „ dy , 1

= H + y withX0) = ] ,A 0) = 2 ‘

dy — = x2 + y2, y(0) = 0; determine the first three non-zero terms in Taylor’s series expansion

ax

9. Given the differential equation

= xz + y2, y(Q) ~ 0; det<

dxfor X*) and hence obtain the value ofXO-

dy10. Apply Taylor’s series expansion to the differential equation = 2x + 6y with

X0) = 1, obtain X0-1)>XO-2), X0-3) and X°-4) in steps of A= 0.1.

dy11. Solve ^ f=y sin * + cos x, subject to initial condition X0) ~ 0, using Taylor’s series expansion.

12- Given / ~ + ^ and X0) = I, find the Taylor series X°-1) X 0-2)-^ •S'?

13. Using Taylor series method solve y = xy +y*, y(0) =1 at * = 0.1, 0.2,0.3.

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14. Solve by Taylor series method of third order the problem/ = ( x3 + xy2) e~x,y(0) = 1to fmdy forx = 0.1,0.2,0.3. ‘

dy dz15. Solve - z - x , " - y + x withy(0) = 1, z(0) = 1 to get><0.1) and z(0.1), using Taylor’s

method.

Answers-4(A)

1. y(0.1) = 3.005, y(0.2) = 3.020, 2. ><0-1) = 1.12721.

3. y s = 2x - 21og x - ^ (log x)3.

1 5 5 , 7 , 47 , 73 7374. v = ----- x + — x ----- x + — x 2 ------ x + ----- .

' 60 24 6 12 12 120

1 , 2 . 1 ,5 y - X + — x 3 + — Xs + — x 7.

y 3 15 636. y(0.1) = 0.0050125. ><0.2) = 0.0202013, y(l) = 0.6487213.7. ><0.1) = 0.1050, y(0.2) = 0.2200,

z(0.1) = 0.9997, z(0.2) = 0.9971.

1 3 v3 v7

8- ^ 1 + 2 X + X5-

N u m e ric a l A n a ly s is -H I, C o r r e la t io n & R e g re s s io n | 381

9. j r t x ) = y + — + ...;><l) = 0.3502.

10. y(0.1) = 1.355; y(0.2) = 1.8555; y(0.3) = 2.5516; y(0.4) = 3.5109.

11. y = x + — + —— +* 6 120 12. y(0.l) = 1.0665;y(0.2) = 1.1672.

13. X0-1)= 1.1167; y(0.2)= 1.2767; y(0.3)= 1.5023.14. y(0 1)= 1.0047;>(0.2) = 1.01812;><0.3) = 1.03995.15. y(0.l)= 1.003; z(0.1) = 1.1102.

2. Solve — = x2 + y2 subject tox = l,y = 0 atx = 1,3

Exercisc-4(B)Using Taylor series method do the following problem (1 -7 )

1. So lve/ =y sinx + cos x subject tox = 0,y = 0.

dx

dx

3. Find the values of y(l ) and y ( l. l) for the equation f{x, y) I 2 y - 3e“ where *o = 0,yo - 0. Check the values.

4. Find the values ofy(l. l) and y(1.5) for the equation/ = xym withy(l) = 1 correct to threedecimal places.

5. Find the value ofyupto five decimal places atx= 1.02 for the equation — = (xy - 1) with>(I) = 2.

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dx dfi y dy6. Solve 3 7 = x + y + t and - j- r - x - 1 with y - I .3 —- -1 .* = 0 at / = 0.

dt dt 8 dx

dx dy 7. Solve — = xy + 2t and — = 2ty + x with the initial conditions x = l ,y = 1, t = 0. Using

3 8 2 | En g in e e r in g Ma t t c m a t t c s -III

Picard’s method do the following problem (8 - 12).

dx

dy8. Find third approximation for the differential equation ~r~ = 2 y - 2x2 - 3 with

m = 2 .

dy „ y9. Find a solution upto third approximation of the differential equation = 2 - —, x i ) = 2.

dy10. Find upto third approximation to solve ^ = x + y where y = 0 when x = 0.

dy - _ 11. Find the solution of the diflferential equation = 1 - 2 xy correct to four decimal places for

x = 0.2 given that><0) = 0.

dy dz _ 12. Find third approximation of the equation — = 2x + z and ^ = 3 xy + x z with >(0)= 2

and z(0) *»0.

dy y — x13. Use Euler’s method to approximate >>when a: = 0.1 given that ^ = y Z x ’ = by taking

A = 0.1.14. Solve/ = 3X2 +y in 0 <,x <. 1 by Euler’s method h = 0.1 given that><0) = 4.15. Solve/ = x + y, XO) = 0 choosing the step length 0.2 for y (l2 ) by Euler’s method.16. Solve / = —j/; ><0) = 1by Euler’s method for><0.04).17. Solve/ = x+ y + xy, X0) = 1 forXO.l) by taking h = 0.025, using Euler’s method.

dy y - x18. Use improve Euler’s method to find X01X given ^ - y + x *X0) ~ 1-

dy19. We Improved Euler’s method to find X0-5X taking h =0.1 and ^ = y + 8in x >XO)= 2.

20. Use Improved Euler’s method to fineXI-6) if ^ = y2 ~ ^-.>(1) = 1-

21. Using Euler’s Improved method findX0.2),X0.4) given ^ - x+ I \fy I > = 1

22. Find X0.1) given y’ = x2+y, X0) = 1 using Improved Euler method.23. Use modified Euler method to obtain X0-2), X0.4) and XO-6) correct to three decimal places

given th a t/ =y - x 2,Xp) * 1.24. Use Euler’s modified method to get X0.25) given that / = 2 xy, XO) = 1-25. Given / = x2 + y2-, X0) = 1determine X0.0 and j£0.2) by modified Euler’s method.26. Solve / = y + x \ y i0) = 1forX0.02), X0.04) ajid'xO-06) using Euler’s modified method.

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27. Using second order Runge-Kutta method find > at x = 0.1, 0.2 and 0.3 given2 / = (I+jc>2;>(0)=1.

dy 2xy + e*28. If — = x 2 + ygx ' yW = ® solve for> at x.= 1.2, 1.4 using Runge-Kutta method of fourth

order.29. Using Runge-Kutta method of fourth order find y at jc = 1.1, 1.2 given that

2y = 2x3+r>J<l) = 2.

dy30. Use Runge-Kutta method evaluate y at x = 0.2,0.4 given that ^ - xy - 1; y(0) = 2.

31. Solve the following for y(0.1), y (0.2) using Runge-Kutta method of 4th order

~ + 2y = x; y(0) = I.

dy dz32. Solve — = 1 + = ~*y for jc = 0.3 given that > = 0, z = 1 at x = 0 by Runge-Kutta

method.

d*y a dy _ 33. Solve ~ x — ~ 2a?y = 0 given that >(0) = 1, /(0 ) * 0 for ><0.1) using Runge-Kutta

method.34. Use Runge-Kutta method to solve / ' - xy + 4y = 0; ><0) = 3; /(0 ) - 0 at x - 0.1.

cPy35. Apply Runge-Kutta method to find y at x = 0.1 given -j^ r = y3; y(0) = 10; y'(0) - 5.

dy36. Solve by Milne’s predictor-corrector method, the differential equation — = y - x 2 with fee

following starting values : y(0) = 1, >(0.2) - 1.12186, ><0.4) - i .4682, ><6)= 1.7379 and find thevalue of> when jc = 0.8.

dy 137. Given ^ = - (! + x 2)y* and><0) = 1,><0.1) - 106, ><0.2) = 1.12. ><0.3) - 1.21, evaluate

><0.4) by Milne’s predictor-corrector method.38. Solve the differential equation / = jc? + y1 - 2 using Milne’s predictor-corrector method for jc -

0.3, given the inital values x - 0, > = 1. The values of > for jc = - 0.1, 0.1 and 0.2 should becomputed by a Taylor series expansion.

N u m e r ic a l A n a l y s is -111, C o r r e l a t io n & R e g r e s s io n | 3 8 3

Answers~4(B)

L;F=X + T + i20 + ...... 2. 0.4158 3.X i)=13 .91 ,>( i;i)=i7 .87

4. >(1.1) = 1.228, >(1.5) = 1.686 5. 2.02061

6. x(t) = t + — + .... y(t ) = 1 - t + z(t) = -1 + ~ where * =2 4! 3! d t ^

212 ^ ' 47. x(t) =1- t + — +...;>(*) * -1 +t - ~-t2+- 13-...o A o

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384 | E n g in e e r in g M a th e m a tic s -! !!

8. y = 2 + x + x2 - — - — x6. 9. y = 2x - 2 log x - —(log x)3O

X2 X6 X6 X 11

10. y = — + — + 160 + 4400 11. 0.1946

3+ — x

408l i v — 2 + x^ + x® + — x® + — ■ z —3x2 + —x4 + —x^ + — Xs1 2 . y * + x + x ' ' + 2 0 x’' + i 0 > z « + 4 x + 5 x - (- 2 8 x

13. 1.092814. 4.4, 4.843.5.339, 5.9002, 6. 53825, 7.267078, 8.101786, 9.0589647, 9.10396, 10.2573615. 1.1831818. 1.093221. 1.2309, 1.515324. 1.062527. 1.0552, 1.123C, 1.207330. 2.2430, 2.5890

33. 1.00533436. 2.011

16. 0.960319. 3.439422. 1.105525. 1.1105, 1.2502628. 0.1402,0.270531. 0.82342,0.6879.

34. 2.939937. 1.2797

17. 1.144820. 1.176623. 1.218, 1.467, 1.73726. 1.0202, 1.0408, 1.061929. 2.2213, 2.491432. 0.3448, 0.99

35. 17.4238. 0.6148

r i •- *m * itftiilitt ^ ^ pi *•* > > - « * V «

1

4.

5.

6.

Exercise-4 (C)Calculate Karl Pearson coefficient of correlation from the following data, using 20 as workingmean for price and 70 as working mean for demand.Price : 14 16 17 18 19 20 21 22 23Demand : 84 78 70 75 66 67 62 58 60From the following information relating to the Stock Exchange Quotations for two shares A andB, ascertain by using Pearson's coefficient of correlation how shares A and B are correlated, intheir prices ?Price Share (A) iis. : 160 164 172 182 166 170 178Price Share (B Rs.: 292 280 260 234 266 254 230Find the correlior the resrltMonthIncomeExpenditure

coefficient between the income and expenditure of a wage earner and comment

Jan. Feb. March April May June July46 54 56 56 58 60 6236 40 44 54 42 58 54

Calculate correlation coefficient from the following data : N= 10, Ex = 140, Sy f 150,1 (x -1 0)2 = 180, K y - I5)2 - 215, E(x - 10) (y - 15) = 60Calculate Ae correlatin coefficient between the following pairs of values :x : 100 ' 110 115 116 120 120 125 130 135y : 18 18 17 16 16 16 15 13 10From thc&Howing d$t$» examine whether input of oil and output of electricity can be said to becorrelat^f':Input of pit : 6.9 8.2 7.8 4.8 9.6 8.0 7.7Output of Electricity : 1.9 3.5 6.5 1.3 5.5 3.5 2.2

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N u m e r ic a l A n a l y s is -III , C o r r e l a t i o n & R e g r e s s io n | 3 8 5

7. Given the following data :

Sum of deviation from assumed mean: -1 4 18Swn ef sqMm of deviation from assumed mean value : 4308 6308

Sum .of predwet ef deviations from their respective assumed mean: 1SO Nuiaier fff pate of observations : 12CalcVtate the lorrtifaflee coefficient.

8. The coefficient of correlation between two variables x and y is 0.3 Their Covariance is 9. Thevariance of x is 16. Find the standard deviation of y series.

9. Find the coefficient of correlation between the ranks obtained by 10 students in Mathematics andPhysics in a class test are given below :Ranks in Mathematics: 1 2 3 4 5 6 7 8 9 10Kfeks in PHJrsics: 3 10 5 1 2 9 4 8 7 6

10. 'Em Competitors in a beauty contest are ranked by three judges in the following order:

Jwtge x : 1 6 5 10 3 2 4 9 7 8Judgey : 3 5 S 4 7 1* 2 1 6 9Judge z : < 4 ■- 9 • 1 Z 3 10 5 7

11. Calculate fee w flH li Rt ©f ctiffWMion rfktot (fee foBbwim dtta by fee method of rank correlation: x i 7 | |S 95 7* A M SI SO y : 134 45# 115 l i t 4 148 142 100

12. Compute Rank cfWMwen f»m fee fellofcinj HWe #f HMk Wiiwber ®fSupply (x) and price (y)

* : US 134 120 130 124 12ty : 130 „ 132 128 130 127 12$Ten competitors in a Nnwty contest are ranked by thwe judges in fee following order:First Judge : 1 5 4 8 9 6 10 7 3 2Second Judge : 4 8 7 6 5 9 10 3 2 IThird Judge ; 6 7 8 I 5 10 9 2 3 4Use the rank correlation to gaagc which pair of judges have the nearest approach to common taste

14. Following are the scores of ten1sft&nts in a class and their I.Q.Student: 1 2 3 4 5 6 7 8 9 10Secore: 35 40 25 55 85 92 65 55 45 50

I.Q.: 100 100 110 140 150 130 100 120 140 110.Calculate the rank correlation coefficient between the score and I.Q.15. Find the mean values of the variables * and y and correlation coefficient between them from the

following regression lines :2y - jc - 50 = 0, 3 y - 2 x - 10 = 0.

16. Regression equations of two variables x and y are as follows :3ac + 2y - 26 - 0; 6x + y - 3 1 = 0 .Find (i) the means ; (ii) the regression coefficients; (iii) the coefficient of correlation between x andy.

17. The equations of two regression lines obtained in a correlation analysis are ;

3a: + 12y = 9 and 3y + 9* = 46. Find (i) Mean val ues of x and y ,

(ii) the correlation coefficient between x and y.

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3 8 6 ( En g in e e r in g M a t h e m a t ic s -111

18. The lines of regression of y on x and * on y are given to be

y = 5+ 0.2x and x = 3+.y.Find the means for the two variable x and y.

19. In a regression analysis problem, the following data are given. The regression lines are x + 2y - S

= 0 ,2x + 3y = 8 and <sx2 = 12.Find x , y , r and ay\

20. Find the lines of regression x o n y and y on x for the following data : jc : 3 5 6 6 9 y : 2 3 4 6 5

21. From the following data find the regression equations and estimate the likely value ofy when * =100.x : 72 90 76 81 56 76 92 88 49

y : 124 131 117 132 96 120 136 97 85

22. Find out the likely production corresponding to a rainfall of 40 cms. from the following data.Rainfall (in cms) Output (in quintals)Average 30 50S.D. 5 10,r = 0.8. ,

23. Given die following data for the variables x and y.

x = 36, y = 85, a , = 11, a* = 3 and r = 0.66.

(i) Find the two regression lines (ii) Estimate the value of x when y - 75,24. The following data give the correlation coefficient, means and standard deviation of rainfall and

yeild of paddy in a certain tract:

Yield per hectare in Kgs. Annual rainfall in cm.Mean 973.5 18.3S .D. 38.4 2.0Coefficient of correlation = 0.58. -Estmate the most likely yield of paddy when the annual rainfall is 22 cm, other factors beingassumed to remain the same.

25. Given is the following information :x y

Arithmetic mean 6 8Standard deviation 5 40/3.Coefficient o f correlation between x and y = 8/15.Find (i) the regression coefficient of y on x(ii) the regressionequation o f x on y,(iii) the most likely value of y , where x = 100.

r-

*?<>: Answers—4(C)

1. r = - 0.954 2. r = -0 .96

4. r = 0.915 5. r = -0.915

7. r = 0.2949 8. ct, = 7.5 *

10. xandz. 11. r = 0.9

3. r - 0.769

6. r = 0.696

9. r = 0.224

12. r = 0.33

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N u m e r ic a l A n a l y s is -III, C o r r e l a t io n & R e g r e s s io n j 387

13. Second and third judges are in nearest approach to the sense of beauty.

14. r - 0.47.

15. x = 130, > = 9 0 , r = 0.866.

16. x =4, byx = - | , r = -0 .5 .

17.i n1 1 1 * y = 0.333; r = -0.2887.

18. X - 10, y =7.

19. x - \> y = 2 , ay2= 4, r = -0.866

20. y = 0.6x + 0.4; x —1.2y+ 1.2.21. x = 0.64y + 2.63;; y = 0.83x + 51.88.22. 66 quitals.

23. y = 0.9075x + 41.1375, y - 0.48x + 67.75; and x = 26.93.24. 1014.7.25. byx = 1.42; y - 1.42x - 0.532; and y = 141.668

»

*iMMMH ' f Ip* n' r >•**. kto A Vi w j* '# + * yn miWiW

Exercise-4(D)

1. Fit a straight line to the following data :x : I 3 5 7 9

y : 8 15 24 32 402. Taking 1990 as origin for x-series, fit a straight line to the following data showing the production

of a commodity in different years in Gwalior.Yearx: 1988 1989 1990 1991 1992Productiony: 10 12 8 10 14

3. Fit a second degree parabola to the following data :x ; - 3 - 2 - 1 0 1 2 3

y : 4.63 2.11 , 0.67 0.09 0.63 2.15 4.584. Fit a parabola to the following data :

x ; 2 4 6 8 10 y : 3.07 12.85 31.47 57.48 91.295. Fit a parabola of second degree to the following data :

x ; 0 200 300 400 y : 1544 1898 2133 2327

6. Fit a second degree of parabola to the following data :x : - 4 - 3 - 2 -1 0 1 2 3 4

y : 101 104 107 107 108 110 109 110 108

Fit a curve y = ax + — to the following data :

x ;

y-

1

5.43

2

6.28

3

8.23

4

10.32

5

12.63

6

14.84

7

17.27

8

19.51

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3 8 8 | En g in e e r in g M a t h e m a t t c s -III

b8. Fit a curve y = ax1+ —to the following data :

x

x : 1 2 3 4

y : -1 .51 0.99 3.88 7.669. Fit a relation of the form R = a + bV + cV2 to the following data, where, V is the velocity in

km/hours and R is the train resistance in kg/quintal:

V: 20 40 60 80 100 120

R : 5.5 9.1 14.9 22.8 33.3 46.0

Estimate R when V= 90.

10. Fit a curve of the form y~ abx to the following data :

*. 1 2 3 4 5

y : 87 97 113 129 20211. Fit a curve of the form y = ax* to the following data :

x : 2 4 7 10 20

y : 43 25 18 13 ' 8

12. Fit a curve of the form y = ae* to the following data :

x ; 1 2 3 4

v. 1.65 2.70 4.50 7.35.

Answers-4(D)

1. 7 = 3.55 + 4.05x 2. ,y = I0.8 + 0.6x. 3. y =1.243 - 0.004x +0.22 j:2

4. y - 0.34 - 0.78x + 0.99x2. 5. y = 1547.9 + 378.4x- 40x2. -

6. j = 109 + 0.883*-0.2457xJ.. . 3.03

7. y = 2.4x + ----- . 8. y = X

0.509X2- — .X

9. R = -1.5726 + 0.329F + 0.00008K1 andrta tF = 90 is 28.69.

10. jf = (73.97)(1.68)'. 11. = (78)x‘fl*. 12. y-= (l)e0<w*.

a

6 7

195 193

40 60 80

5 3 2

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S t a t i s t i c s

5.1 CONCEPT OF PROBABILITY

When an experiment is performed it may result in one of possibly several outcomes. Because of'chance' factors, we may not be able to predict the outcome precisely. However, we may not beable to offer our opinion about die possible outcomes, by giving a quantitative measure of the’chance' that a particular outcome would occur. Thus, with every event A consisting of one ormore outcomes of an experiment, we may associate a numerical quantity, called the probability of

A, denoted by P(A), which will measure the 'change* that the event A will occur. We take 0 < P{A)

5 1, xi'PiA) - O' implies tint there is no ’change’ that A will occur and if 'P{A) - 1' implies 100%chance that A will certainly occur. IfP(A) ~ H, then it will mean that there is 50% ’chance1that A

would occur.The advantage of defining probabilities of events in quantitative terms is that it now becomesamenable to further mathematical analysis. We can talk of the possible gain to a gambler orecpected profit from a risky venture. We can calculate the insurance premium to be charges tocover a certain risk.Let A be an event, then probability of an event deponds only on the number of sample pointscontained in A. We can easily verify that this probability function defined on subsets of samplespace S, possesses the following properties:

(i) P{A + B) « P(A) + PiB) (ii) if A ^B = > P(A) <; P(B)(iii) P(S) = 1 (iv) 0 5 P(A) <, 1.Let A be the class of events associated with the outcomes of a certain experiment i.e; Ah A2,........eA, and S bethe sure events. Then the probability function P on A is also called probailitydistribution on S. It tells us how the unit probability mass is distributed among different subjectsA of S.For example : A die is thrown, then the probability of getting a prime number can be find asfollows:

Here sample space S = (1, 2, 3,4, 5, 6}n (5) = 6

Let E be the event if selecting a prime number i.e; E = {2, 3, 5}

=? n(E) = 3

n(E) 3 iProbability: P(E) = 7i(S) ” § = 2 '

5.2 RANDOM VARIABLE

Let S be the sample space associted with a given random experiment. Then a real valued function X which assigns to each outcome x e S to a unique real number X{x) is called a random variable.

In other words a random variable is a real valued function having domain as the sample spaceassociated with a given random experiment.

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The random variable are two types :

(i) Discrete Random Variable: A variable, when real valued function defined on a discrete samplespace is called a discrete random variable.

For example : The marks obtained in a paper or in a test, number of telephone calles per unit time,number of successes in n-triais, and so on.(ii) Continuous Random Variable: A random variable* is said to be continuous random variable,if it can take all possible values betwen certain limits.For example : Age, weight, height and so on.PROBABILITY MASS FUNCTION:Let X be a discrete random variable. A probability mass Junction (p.m./.) is given by

fx (x )= P (X = jc), for all*or, f { x ) = P { x : X fo) = x }It is a function with values between 0 and 1, and whose sum is 1, over all values of x.For example; Toss a balanced coin once. Let X be the number of heads that occurs, then probability

mass function (pmf) can be determined as follows :Random Variable :X - { x = 0, 1 }

Sample space S = { H, T } =>n{S) = 2

3 9 0 | En g in e e r in g M a t h e m a t ic s -III

Number of Elementary Probability Mass Function

Heads *= { 0 , 1} Events : (£) f { x ) = P iX = x )

r = 0 T Vi

JC- I H '/ j

Example 5.01 : Toss a balanced coin twice. Let X denote the number o f heads. Find the probability mass function o f X

Solution : Random Variable: X = {x = 0, 1, 2)

Sample space: S = (HH, IT, HT, TH) => n{S) = 4

Number of Heads Elementary Probability Mass Function

*= { 0 , 1,2} Events : E f i x ) = P iX = x)

jc = 0 TT 1/4

x= 1 TH, HT 2/4 x = 2 HH 1/4

5.3 DISTRIBUTION FUNCTION

Let X be a discrete random variable, then distributive function o f* is given by

F(x) =P ( X £ jc)= 2 Pi Xi*X

CO

such that p, k

0, and 2 p, = I, where p, - p

( jc,)

i=l

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St a t is t ic s | 391

Again, Let X be a continuous random variable, then distributive function of A" is given by

X

F ( x ) = P ( X < x ) ^ I f (x) dx. —OO

Remark: A distributive function F ( jc) or Fx (x) is also called cumulative distribution function( c d f ) .

xample 5.02 zA random variable X has the following probability function : X - x 0 1 2 3

/ i * 2 2 p(x) 0 g 5 5 .

Determine the distributive function o f x.olution ; We know that distributive function

F ( x ) = P ( X Z x ) = £ p t XjfiX

F(Q) = P(X^Q)=p(Q) = 0

F ( \ ) = P ( X * ] ) = p ( Q ) + p (\) = 0 + l = |

F(2)=/'(XS 2) = 0+ | + ! = |

F(3) =/>(*< 3) = 0+ | + | + | = r. Ans.

xample 5.03 xA random variable X has the density function :

f ( x ) = | \ x > , 0 £ x £ 3

1 0 , Otherwise.Find the distributive function.

lution : By definition of distributive function for continuous random variable :

X

F (x ) = P ( X £ x ) = J f ( x ) d x

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Properties of Dbtribative fiidiea : If X is the modom variable, then it holds the following properties:(i) F{x)7> 0,( i i ) 05 f (* )S l ,(iii) F ( - 00) = 0 and F (+*>) - I,(iv) F is an increasing and continuous function from the light at every point

5.4 PROBABILITY DENSITY FUNCTION:Let/(x) be a probability function in die interval [a, b], then the probability that the variate value X to lie in the interval [a, b] is given by

b

P(a <, x <, b) - J f i x ) dx.a

The functin f ix ) is probability density function ( p.df.) if and only if, the total probability is unity,i.e., the probability density function (p.df.) possessess the following two properties:

(i) f i x ) 2 0

00(ii) \ f{x)dx - 1, i.e; total probability is unity.

-CO

For example : whether check the following function p .d f ? f i x ) = fix (I - i ) , 0 S x ^ l

1 1

For p.d.f., we have J / (x>& = j fix (1 - x) dx0 0

3 9 2 | E n g d s e h n g M a t h e m a t t c s -H I

H ence / (x) is the p.d.f.

Remarks : [Formulae]

Let/ (x) be th e p.d.f. o f a random variable X, where X is defined in [a, b]. Then

( 1) Arithmetic mean or Expected valve o f x

b

i.e; x or E(x) - j x. f (x)dxa

( 2) The expected value o f x2 :

b

E (x2) = J xV(x)dxQ

(3) First moment about origin :

b

m* =£(x)= J x .fix )d x.

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S ta t i s t i cs [ 3 9 3

* Tm» *b

Hz' = £ ( x 2) ' I x\f(x)dx,

($) >* moment about origi®:r " r

b

Hr' = £ ( * 0 = I xr. f(x)dx, a

(6) Mean = \i\

(7) Variance = »**'- M <1))^

and standard deviation

(8) t* moment about mean :

6

Hr * I (X - x y f '(x)dx

W f t -

= h j’ - (ftO2,

U3 = H j ' - 3 ji2Vi' + 2 (Hi*)3

M4 = fi4 - 4h3’ ( f t1) + 6 ft' (f t1)2 - 3 (f tT and so on.

(10) Median : If M is the median, then

M 1 \ f(x)dx = g

a ~bor f f(x)dx - ~

M £ (11) Mode : mode is the value of x for which/ ( j c ) is maximum.

For finding the mode : / ' (x) = 0 and f ”( jc ) < 0 at point x = a.

Such point is called mode.

(12) Mean Deviation: Mean deviation about mean * isb

M.D. = J | x - x |./(x ) dx,a

(13) If F (x) is a distributive function, then corresponding p.df. is given by

f i x ) = £ ( F (x)}

le 5.04 : I f f (x) * Kx2,0 < x < 1, has probability density function (p.d.f), determine K and

find P ( | < x < ! ) and fin d V i f P (X > a) - 0.05.

lotion : Since / (x) is a p .d f, so thatbi f(x)dx = i

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j Kx*dx = 1

■ k " K= 3.

^ / (x ) is a />.<£/

ro/?nrf/» ( | < x < | )

we have

Since / (x) = 3x2, 0 < x < 1

1/31/2

= J 3 x*dx [by (1)]i/s

MTI -11/31/3

1_J_ = _19^8 27 216 '

To find a :

Since P{X> a) = 0.05(X>

=> I f (x)dx= 0.05o

1=* J 3x*dx = 0.05

=> [*3]> = 0.05

=> I - o3 = 0.05

=> a3 =0.95

=> a = (0.95)I/J

Example 5.05 : /4 random variable X has the p .d f.:ax, 0 £ x £ l

a, l £ x £ 2

-ax + 3a, 2 £ x £ 3

0, otherwise

Find (i) value o f a (U) P(X S 1.5).

Solution : (i) Since / (x) is pd-f, so that<30

J f(x)dx = 1 —«0

1 2 3

=> I ax dx+ j adx + ! (~ax + 3a)dx = 10 1 2

/ M

(i)

Ans.

D>y(D)

Ans.

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3 9 6 | ENaNrEMNG MATHEXATtCS-III

Thus p.df.

| [ 2 7 - 0 ] - I

- I/(*> =

To compute P (1 £ x £2) :

We have

i o, otherwise

025x53

P ( 15 x 5 2) = J f{x)dx i2

= J J x2«4c1 a

•iW= 2 7 18 n

_ . I27'

7b fin d distributive Function F(x) :

We have, / (x) = /> (*5x)X

= / /( X > &

[ t y (O I

[ by definition]

Hence D.F.

F{x) 0 5 x ^ 32_ 271, x £ 3.

Example 5.07: The distributive function of* random variable X is given by0, x£0

= J 4 x2dx o *

- b K 0,

= x 27

x < 0

F(x)

Ans.

/* x > 3t then (I) find corresponding density function of random variable X, (ii) compute P(2 < x *3).

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S t a t is t ic s | 3 9 7

Solution : (i) We know that, p.d.f. :

/ (*>= £ (*-(*)}

i.e; p .d f.

(ii) We have

/(*) =

ft x i 1

A . J dx |

\ x - l ) 4]

, 16 jI I < x i 3

$ < > > • x > 3

0 , x 5 1

- i y 1 <X £ 3

0 , x > 3

P(2<x<;3) = ! f(x)dx2

3 -

= i T ( x - \ydx

_ i F( x - i) 4 '" < 1 4 _

= _Lnfi n = l§16 * ~ ' 16

- ao<x < ooExample 5.08 : Ifp .d .f f ( x ) m k *----- J ,1+x ,

then fin d k and the distributive function F(x).

<30

Solution : For p .d f., we have J f{x )dx = \

- ( I )

[by 0 )]

Ans.

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3 9 8 I E n g in e e r in g M a ih e m a t ic s -III

p.d.f.

Distribution Function

- I 1* ' ( l + x 2 )

F(x) = P ( *£ x)X

= ! f(x)dx

1- C O

X

= 1 *<l + x2) *

= i [tan-1*]*

- ± O n > x + ± Example 5.09 : A frequency distribution is defined by

x }, 6 <Tx £ 1

3(2-xy, IZXZ2.

Prove that f(x) is a p .d f also fin d mean and the standard deviation.

Solution : For p .d f , We have

(

«. 1 2

J f i x ) d x-

\ x 3 d x + J

3(2- x y d x

0 1 r in2

- M + 3( 2 - x ) 4

=>f (x) is a p.d.f.

To find mean : We have

mean x =^l ' =£(x)= / x/ (x)dx-<o

I 2

=! x. x*dx +j x. 3(2 —x)3dx0 l1 2

= J x*dx + 3 I x.(2-xydxo i

-M + 3

\x. ( 2 - x ) 42 2

l -4

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S t a t is t i c s | 3 9 9

To find S.D. We have

5 4

* H H l M

- z £ 5

1 , 3 65 4‘6

11io-

Hz’ = E (X*) = J x*f(x)dx-CO

1 2= J *2. x3<&+ i xJ. 3(2 - x f d x

o i

+ 31 -4

dx

U 3 + 3 [Jv<2 -= 6 + 4 + 2 I U

- i + - ——1(0= 6 4 10 |1

# i ; - h

- M

dx

I + I - J L L i + I o j - o l6 4 10L 6 V J

= 1 + 1 + J l x16 4 10 6

1 3 7= —+—+ — 6 4 2010 + 45 + 21

60

= ! § - ! §~ 60 ~ 15-

Hence S.D .:

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4 001ENOweaiH atlwiWWlWH

- W- s o n

V15~100- H E

V1500

Ans.uu

Example 5.10 : For the distributive function (dF) :

dF ■> y,e~w dx, -co<x<<*>

Prove thatye * ^ , mean ” 0, S.D. “ 4 t , variance * 2, and mean dtviation about mean

1st.

dF -yêr^xdx

% - » * "

f{x )

Solution : Given

[v /(* ) = {F (*)}]...•(!)To find yo :

For p.d f, J f (x ) dx = I

=> ) > 0 fl&f = 1,-00Since (r| is an even function, so that

00

2 \ y0e^*\dx = 10

00

=> 2yo 1 e~* dx = I 0

[by (l)]

[v |x | = jc, jc > 0}

2y0 -1 = 1

=> - 2yo [0 - 1] = 11

=> yo = 2*

Hence p.d.f. : f (x) = e-W

To find mean: By definition00

mean = j*i‘ = E (x) = J x . f (x)dx

...(2)

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[Because x e-« is an odd function and I odd = 0]-cO

To find S.D. and Variance :

We have \i2' = E (x2)

00= j x2f(x)dx

—g o

00 1= J *2 “

and S.D.

= J x2erxdx 0

= X2 e V°- J 2*

/o oCO

0 *+- 2 J xe-*dx o

/ -* V° 00 --r

x 4 t - J £ tv. A o _1

c

~ 2

- 2

dx

0+ le~*dx o

N . — —2 (0 —1} —2

fi2’ = 2.

Variance = p2*- (n,1)* = 2 - (0)2 » 2,

° ~ -/Variance“ >/2<

To Afecw Deviation about mean :00

M.D. = J Jx- x ]/ (x ) dx -0 0

* 1= J | x -0 | . ~ «~wdx

= | J |»|

S t a t i s t i c s 1401

v xVW is an even fun.]

[v \x\ = x, x 2s 0]

[by def.]

[v mean x = Oj

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402 | EnowBERTng Ma t h e ma t ic s-111

1 °°= £ .2 J |x | e-w abc

* 000

= J xer*dx

- k r - i $ *

M

[ v I*) is an even /?.]

[v 1* | = *, x > 0]

= 0 +

— ~ I (0 —1) ==1

Hence M.D. = I

Example 5.11 ; Determine the value o f k, so that the following function represents the p .d f.0, x £ - !

Proved.

fix ) =

Also fin d median*

Solution : For p.d.f. :

*fx + />, - / < x £ 3

4k, 3 < x £ 4

0, otherwise

I / (x) dx = 1

-J 3 4 co

J 0 <&+ J *(* + 1) a!r + J 4fefc + j 0 = 13 4

(x + I) ?

-1

+ 4k{xh* = 1•1

=>

| [16] + 4A(4 - 3) = 1

8* + 4* = 1

k = — * 12'

7b median : Let M be the median of/(*) M

We have j / (*>a6c = ~ [ by definition}

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r

4. A

M 1 13 2 3

M = |A/= 2.5

Median - 2.5Exampte 5.12 : For the distribution Junction (<Lfi) ;

dF • y ac*/*dx, 0<!x<oo Find mean, S.D., variance and r4* moments about origin.

dF

4kM = g + 4*

Solution : Given:

/. p.d.f. :

To find yo : For p.d.f.

=yoe-x'°, Osx<co.

f ( x ) ^yoe -^

=>

=> yo

! f ( x)dx = 1-<30

00

J yo e~xtt,dx = !o

I --x/a'

-1/cr

a

Hence p .d f.


f (x) - — e-x!°, 0 5 jc < co

Mean = Mi’ = E(x) - J x.f(x)dx

S t a t is t ic s ] 403

A n s .

[ v j t o = £ (F(x)]

•••(I)

1

o- J l

,-xlc

1/oL o -1/cr d x

- J e~xladx = - a[e’x/tJ|0 1 Jo

= - o [0 - 1] = a.

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To find Variance and S.D. : We havec0

Hz' = E (x2) = J x2f{x) dx

4 0 4 | ENGffltemNG M A ne nx nc s- l I 1

X 1Let t - — => dt = ~ dx, then

c a

I x2e~*/adx £’**/(* ) - ~> e~xh

H 2' ~ i (t2**2) ^ dt 0

00

= a2 J er*. t2di0.00

= tr2 J dt

0CO

[Since Gama function J e~‘. t^ 1dt - 0

fi2’ = o 2 ((3 ) = or2. 2 ! = 2 ct 2.

Herce H2' = 2crJWe know that

Variance - ^2' - (Mi1)2 = 2o2 - (a)2 = a2,

and S.D. = ^Variance = V ? = ° r 7b (i/ : We have, 1* moment about origin :

«0n; = J xr. f{x)dx

—00

1 00= — J of. er**dx

0 0

[ Put t = ^ di =

Hr = J (toy. e^dt 0

OD

- or. J e~'. r+l~'dt

= 0'. |(r +1) (v £ = J e->r-ldt)

= r ] <r. Adj,

Example 5.13 : For the frequency o f probability a w e :

y mflx) ■ \ sin x, 0 £ x £x .

Find median and mode.

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S t a t is t i c s j 4 05

For p.df. : 1 /(x) dx = j sin x dxo 2

= - 2 (w**)!

=> / ( x ) is a p .d .f To find median : Let m be the median, for which

«

! f(x ) dx = 4 (by definition)2

T 1 • ^ 1J sin x. dx = g

-[cosx]; =1- [cos m - IJ = 1

- cos m + 1= 1- cos « = 0

cos m = 0

j .

2

Hence median =

To find mode : We know that x *=a is mode, i f / (*).*» 0 and/* (x) < 0 at x = a .

Given / (x) = h sin x

and

For mode : Put

/ (x) - g cos x

/ ' (x) = - ^ sinx

/ ( x ) =0

cos x = 0

cos x = 0

...d)

...(2)

[by (i)]

x ~ 2

Put x = in f (x), we have

f ( f ) — 5 si" ( f )

1

~ 2<0

Hence x * t e/2 is the mode of die distribution.

[by (2)]

Ans.

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5.5 THEORETICAL PROBABILITY DISTRIBUTION

A theoretical distribution is the frequency distribution of certain events in which frequencies areobtained by mathematical computation.There are two types of theoretical distributions :(i) Discrete theoretical distributions.

(ii) Continuous theoretical distributions.some of the discrete and continuous theoretical distribution are : Discrete theoretical distributions:1. Binomial distribution.2. Poisson distribution.3. Power series distribution.4. Multinational distribution.5. Negative Binomial distribution.6. Geometric distribution.7. Hyper geometric distribution.Continuous theoretical distribution :

1. Normal or Gaussian distribution.2. Rectangular or Uniform distribution.3. Exponential distribution.4. Gama distribution.5. Beta distribution.6. Weibul distribution.

5.6 BINOMIAL DISTRIBUTION

Binomial distribution is a discrete probability distribution. Let an experiment consisting of n-trials be performed and let the occurrence of an event in any trial be called a success and its nonoccurrence a failure. Lelp be the probability of success and q be the probability of the failure in

a single trial, when q - \ - p , so that p+ q-\.

Probability ofr successes in n-trials

/ ,( r ) = " C / > ', r = 0,1,2........... M

or P[X=r\~ "Cr q*~rpr,r^Qt \,2,................. j i

where P (X =r)or P{r) is called binomial probability distribution o f a random variable X.

If an experiment consists of n-trials and this experiment is repeated N times, then

Binomial frequency distribution or Expectedfrequency o fr successes is

f(r)=NP(r )

or f(r)=N.*Crq”'p rt r= 0,1,2..........n.

5.7 HYPOTHESIS FOR BINOMIAL DISTRIBUTION

1. The variable should be discrete i.e. r=0,l,2,3,4,........etc. and never 1.3, 2.1,3-41etc.

2. The happening of events must be either a success or failure.

3. The number of trials n should be finite and small.

4. The trials or event must be independent.

5. The trials or events must be repeated under identical conditions.

4 0 6 | E n g in e e r in g Ma t h e m a t ic s -111

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S t a t is t i c s | 4 0 7

5.8 CHARACTERISTIC OF BINOMIAL DISTRIBUTION

1. ft is a discrete which gives the theoretical probabilities.

2. It depends on the parameters p or q and n.

3. The distribution will be symmetric if p - q. ll is skew symmetric or a symmetric if p * q.

4. The statistics of the Binomial distribution are mean = np, variance = npq and standard

deviation = Jnpq.

5. The mode of the Binomial distribution is equal to the value of random variable X which hasthe largest (maximum) frequency.

Example 5.14 : For the Binomial distribution, prove that:

From (1) and (2), we have

P(r +1) "Cq'-'-'p"' n\ q"'-'p'1 r?(«-r)!P(r) ~ "Crq"~r pr “ ( r + I ) ! ( w - r - l ) ! « W

_ r ! (w -r ) (» - r -1)! p _ (w - r ) p

~ (r+ \) r\ (n -r -\)\q ~ (r + l ) 7 ‘

Hence / ,(r + l)=y «-'~ ^>(#') Proved.(r +1) q5.9 MEAN, VARIANCE AND STANDARD DEVIATION O F BINOMIAL

DISTRIBUTION

We know that

mean ...(1)

and Variance -(ji,*)2, —(2)

where ji |’ and |i2' are first and second moment about origin, respectively.

We have first moment about origin,

which is known as recurrence relation fo r the Binomial distribution. Solution. By the Binomial distribution, probability of r-successes

P(r) = "C, q*~rpr\ r =*0,1,2 , ..... ji .

=> P(r +1) = " C „ , p r*K. [ v r - * r +1] ... (2)

•••(I)

H,’ = £ r.P(r)

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where />(r)=:*C,,tf*~ p, is Binomial probability distribution.

=>r»0

V’ **• *-<■ f S I r,~j7— >

" v,t= Y --------------- q - ' p '

408 j EnGJNEHMNG MATHeKArtCS-III

*> H, = «p

Thus, mean = jit ~np, i.e., m ~ np.

New, second moment about origin:

' = £ r ^ ( r ) = £ r ' * C r q ~ p ' r~0 r-0

= £ W r-l)+ r} »C f9- > 'r-0

= J > ( r -1) *C, r■*Cr O '

r!(«-r)!

f Hr “ !)»(» -1)0* -2)! _ ^ ,

= «(n-l)p*Y ------ (w-2)! ,.f-& ( r - 2 ) l ( « - 2 - r - 2 ) !

:■ r p(r) = 0 for r = 0]

(*-»)? __(r-1)!(»-l-r-l)

[v ? + />=!]

...(3)

[v rJ=r(r~i)+r]

[By (3)]

: p'~z + np

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I

• ><^Sfctasncs j IW

r-2

= « ( « -i ) p 2( ? + p r 2+^p

= h2p2 - np2 + np [•; q + p - \ ]

S p ' + n p Q - p )

==> fi2 = n2p 2 + npq ['•'? = !-/?] ...(4)

Since, variance = j^ ' -

- n2/>2 + "?*? ~(«p)2 [By (3) & (4)]

= npq.Hence, Variance (ji2) = nP9 (5)

Furhter, standard deviation (S.D.) a = \/Variance

or tr = V«W. {By (5)]

Example 5.15: In a Binomial distribution the mean and standard deviations are 12 and 2 respectively. Find n and p.

Solution. Given: mean =np = 12, ... (1)

S. D. - *Jnpq = 2 => *P? = 4. —(2)

npo 4We have, ~ = [by (I) and (2)]

I=> ? = j-

, 1 2Since q + p = l /> = 1- => /» = ! - - => P = —• Ana.

2 Now, form (1) np = 12 => n.—~\2 => « = 18. Ans.

Example 5.16: Using the Binomial distribution, find the probability o f rotting at most 2 sides in 5 rolls o f a dice.

Solution. Given: « = 5, P=~ [v Probability of rolling six is^J

5

o

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By Binomial distribution: P(r) =Cr q"~'p' ..•(!)

P (at roost 2 sixes) = P(r £ 2) = P(r = 0) + P(r = 1) + P(r = 2)

(? )+’C’(?) (? )+’C’(f) (?) Pw«**d)]

410 | ENGBeEHNGMAneMxnc8-IlI

54 _ 625

= $r 648 ' * “ ■

Example 5.17 : A coin Is tossed 4 times, what is the probability to getting:

0) Two heads,

(li) at least two heads.

Solution. Given n = 4, let />= prob. of getting a head in a sinf' ; throw of one coin

i.e., p = x A and q=Vi

ByBinomialdistribution: P(ryCrq*~'pr ...(!)

(i) P (two heads) = p (r = 2)= “C, [ ]11

1 3. 6.— =4=0.37516 8

(ii) P (at least two heads) = P(r =2 or3 or 4)

= P(r =2)+P(r ss3)+P(r = 4)

6 4 ! 11= — +— + --------=0.6875 Ans16 16 16 !6 AnS'

Example 5.18 : The Incidence o f occupational diseases in an industry is such that the workers have a 20% chance o fsitfjfaingfrom U. Find the chance that out o f 6 workers 4 or more will catch the dhmse.

Solution: Given;

p =20% =*0.2, q=\- p-l-0.2=0.&and n~6, (total workers)

By Binomial distribution

P(r) = • C . q T p ' - (1)

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P (4 or more catch the disease)

= P{r =4 Of 5 or 6)

= P(r=A)+ P(r =5)+P(r = 6)

= 5C4(0.8)5(0.2)4 + ‘C,(0.8)'(0,2)5+ 6C# (0.8)°(0.2)6= 0.01696 Ans.

Example 5.19: A cubical die is thrown in sets o f & The occurrence o f 5 or 6 is called a success. In what proportion o fthe sets you expect three successes?

2 1Solution, p = prob. of throwing 5 or 6 with one die =—= -

1 ^ 2i.e., p - —and q = -

By Bir ..ma] distribution:

p(r) = mc r< r Pr - 0 )

/. P (getting 3 successes)= P(r =3)

-•(tm= 56x^=0.2731

3*

Hence the required percentage = 27.31 % Ans.

Example 5.20 : Ten percent o fscrews produced in a certain factory turn out to be defective. Find the probability that in a sample o f 10 screw chosen at random, exactly two will be defective.

1 9Solution. Given : P ~ ? = —»n = 10, r = 2

By Binomial distribution: P(r) ="CrqT'p'

P (exactly two defective) = P(r = 2) = 10C3 j j

= 10x9 |r 9_>| _ J — = ] =0.1937.2x1 VIO; 10x10 2 U 0J

j St a t i s t i c s | 411

Ans.

Example 5.21: A bag contains 3 red and 4 blackballs, one ball is drawn and then replaced in the bag and the process in repeated. Getting a red ball in a draw is considered a success. Find the distribution o f X where X denotes the number o fsuccesses in 3 drawn, assuming that in each draw each ball Is exactly likely to be selected.

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3 4Solution. Given: P = ~j> ? = ~ .« = 3, and r = 0,1,2,3.

Using Binomial distribution:

412 | En g w q q *#q M * n « u n c 8 - II I

- o )

Then

P(X = 0) = }C0m n i

[v n = 3]

3 T T 4 f 9 f 4 nJ p ( j r = i ) = Jc ,

J ) Vl 717

Thus, the probability distribution of A" is given by :

X = 0 1 2 3

" * > - $ « ) ' f f i ) ' ( « ' Example 5.22 : Out o f $00 fam ilies with 4 children each, how many families would be expected to

have

(i) 2 boys and 2 glris (ii) at least one boy (iii) no girl (iv) at most two girls ? Assume equal probabilities fo r boys and girls.

Solution. Given that equal probabilities for boys and girls :

1 1P ~~2 9 ~2 " [p = probability of a boy]

Also given n = 4, N - 800, using Binomial distribution

- 0 )

Then expected numbers f(r)~N .P(r) ... (2)

(i) The expected number of families having 2 boys and 2 girls,

/ ( r = 2) = 800x4Cz[ j j = 8 0 0 x 6 x ^ = 300. Abs .

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(ii) The expected number of families having at least one boy,

= f ( \ o r 2 or 3 or 4) = / ( r = l) + / ( r = 2 )+ /(r = 3 )+ /(r = 4)

= 800 Mill}) +‘ c,(i)W +‘ Ca(i)(i)* +‘ c*(iJ

S t a t is t i c s | 4 1 3

= 8 0 0 x — [ 4 + 6 + 4 + 1 ] ~ 7 5 0 .1 6 J(iii) The number of expected families having no gjris (i.e., r = 4 boys)

/ ( r = 4 ) = 8 0 0 x 4 C 4 ^ i j = 8 0 0 x ^ = 5 0 .

(iv) The expected number of families having at most two girls, (i.e., at least 2 boys)

= / ( r = 2 or 3 or 4 ) = / ( / • = 2) + / ( r = 3 ) + / ( r = 4 )

Ans.

Ans.

= 800

- 800 x — = 550.

16 ■

Ans.

Example 5.23: Out o f 800families with 5 children each, how many would you expect to have (i) 3 boys (ii) 5 girls (iii) either 2 or 3 boys ?

Solution. Given th at: N = 800, n = 5, and

1 , . 1 1Probability of having a boy, P=— .then q - \ - p - \ - ~ - ~ .

(i) P{ 3 boys out of 5 children) = P(r = 3)

10

32= 03125.

The number of expected families =800 x 0.3125 = 250.

(ii) /*(5 girls out of 5 children) = P( i£., r = 0 boys)

^( i r ( i j4—/. The number of expected families = 800 x 0.03125 = 25,

Ans.

A n s .

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(iii) / ’(either 2 or 3 boys) = P(r = 2 or 3) = P{r - 2) + P(r = 3)

= 1 0 x i x i + 1 0 x i x i = — = 0.625.8 4 4 8 32

The number of expected families = 800 x 0.625 * 500. Ans.

Example 5.24 : The probability o f entering student in chartered accountant wlil graduate is OS Determine the probability that out o f 10 students (I) none (U) one and (iii) at least one will graduate.

Solution. Given that, the prob. of entering student in chartered accountant will graduate is p = 0.5

/. q = 1- p = 1- 0.5 = 0.5 and n - 10, using B.D. P(r)=*C, <f~'p'.

(i) P (none will graduate) = P(r - 0)

4 1 4 | En q m v b n g M a t h e m a t j c &-III

= ,oC0(0.5)'M (0 J ) ° = ^ r.

(ii) P (exactly one student will graduate) = P(r = 1)

= ,flC1(0.5)^,(0.5), = ^ .

(iii) P (at least one student will graduate) = P(r 2 1)= P(r = 1) + P(r = 2)+ ....+ /*(r = 10)

= 1- P(r ~ 0)

Ans,

Ad i

r«0

= 1- Co(0.5) (0.5) = 1 - - ^ = 0.99. Ans.

Example 5.25 : In a bombing action, there is 50% chance that any bomb will strike the target Two

direct hits are needed to destroy the target completely. How many bombs are required to be dropped to give a 99% chance or better o f completely destroying the target

Solution. Given: P =50 1

100 2

Since the probability must be greater than 0.99, if n bombs are dropped, we have

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• ' S t a ik t ic s | 4 1 5

- I r c 2 +" Cj +" C4 + .......+" C J2 :0.99

2 - n - 1

2" 2:.99 "C0 +*C, +"C2+ ....+'CM =2*]

zo — =£0.01.r

=> 100 + 1 0 0 n -2 " s 0 = > r £100« + 100

By trial, n ~ 11 satisfies the inequality.Hence 11 bombs are required. Am*.

Example 5.26: I f 10% o f the bolts produced by a machine are defective, determine the probability that out o f 10 bolts chosen at random

(I) one (ii) none (lit) at most 2 bolts will be defective.

10 1 . 1olution Given p (defective) = — = i.e., P = —<

, 1 9 9q (non-defective) = 1“ — = — i.e., 9 = — •

Also, n = 10, (n is no. of bolts).

By Binomial distribution:

PW =C r p 'q '-

( \ Yf 9 YM(i) /. P(one defective) = Pir = 1) ='°C, j —J

" of e B ) ’ =(9 ,’ = oj874 '

(ii) /'(none defective) = P(r = 0)

\0 / n \10-0

.. ( I ) A n .

u o j b o ,

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(iii) /*(at most 2 bolts will be defective) = P(r 5 2)

= P(r = 0) + P(r = I) + P(r = 2)

=0.3486 + 0.3874 + 0.1937 = 0.9297. Ans.

Example 5.27: Assuming that 20% o f the population o f a city are literate, so that the chance of an Individual being literate is 1/S and assuming that 100 investigators each take IQ Individuals to see whether they are literate, how many investigators would you expect to report 3 or less were literate ?

1 4Solution. Given: p = — and? = —>* = 10,usingB.D., P(r)=nC,<f~*p',r = 0,1,2.....10.

= 0.34M+0J»74+» C , ^ J ( i J [by (1) and (2)]

5 5P(r = 3 or less) = P( r= 0o r\o r2 or 3) ± P(r = 0 ) + P(r = 1)+P(r = 2) + P(r = 3)

10

' , v ,i- " & li ? ' + “Ci

• e r - T B M s ' - i s j ) '

4 ' t= ( j j K0.8)5+ 2(0.8)* +1.8(0.8) + 0.96]

= 0.2097[0.512+1.28+1.44 + 0.96]

= 0.2097 x 4.192 = 0.8791

Required number of investigators = 0.8791 * 100 = 88 (nearly). Aas.

Example S.2S : Find the binomial distribution whose mean is 4 and variance is 3. Also find its mode.

Solution. Given: mean = np = 4 ... (1)

and variance = npq = 3 ... (2)Dividing (2) by (1), we get

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Thus the required binomial distribution is

Now mode of this binomial distribution

= integral part of (np + p)

- integral part of f 4 + —J , integral part of 4.25 = 4.

S t a ti s ti c s { 4 1 7

Ans.

Ans.

Example 5.29 : The probability that a bomb dropped from a plane will strike the target is 1/S. I f six bombs are dropped, find the probability that

ft) exactly two will strike the target,

(ii) at least two will strike the target.

1Solution. Given that P - ~ and n = 6.

, , 1 4q = \ - p = \ — - —. H y 5 5

Using B.D. P(r) ='C, p ' , r = 0,1,2, .., j i

(i) P (exactly two will strike the target) = P(r ~ 2)

(ii) P (at least two will strike the taqjpt) ;= P(r 1 2)

= P(r = 2 or 3 or 4 or 5 or 6)

= P(2) + P( 3) + />(4) + P( 5) P{6)

= 1- [P(r = 0) + P(r = 1)]

Ans.

v £ P(r) = 1r»0

S=l- = 1-[0.26?1 +0.3932]

= 0.345. Ans.

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418 ( En g m e er m g Ma t hemat ic s-! 11

Example 5.30 : The sum and the product o fthe mean and variance o fa Binomial distribution are I and 128. Find the distribution.

Solution. Given: np + npq = 24 and (np)(npq) = 128 [ v mean = np and variance = np

==» «p( l + q ) = 24 and n1pzg ^ 128

24

and (np)2q = 128

~ ~ 1 9 = 128 [by (11+9/

576=> (FT^F 9 ^<*q = 2 ( U q f

=> 9q = 2(1 +2? +•./)

=> 2q2-Sq + 2=0 => (2q-l )(q-2)=0

=> q = [But here = 2 is not possibi

I 1I hen l> i - <7^ i - —-=>/> = —

From (1), we get

24x4„ = _ _ _ ^ „ - 32.

f i i Y

Hence Binomial distribution is (? + P f - 1^ + J J * Am

Example 5.31: Suppose A and B are two equally strong table tennis players. Which o f thefollowk two events is more probable

(a) A beats B exactly in 3 games out o f4 or

(b) A beats B exactly in 5 games out o f 8.

1Solution. Given that P = <?---

Binomial distribution P(r) = *Crq*~rp f.

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(a) Prob. of A beating B in exactly 3 games out of 4

, \3

—(ij(0

S t a t is t i c s I 4 1 9

[V r = 3. » = 4]

= 4 x i x i = i = 25% . .. (I )2 8 4 w

(b) Prob. of A beating B in exactly 5 games ou; of 8

[ v r = 5, n = 8]

= 5 6 x lx — = — = 21.875% . ...(2)8 32 32 K '

Clearly from (I) and (2), that first event Is more probable. Ans.

ample 5.32: A set o f 8 biased coins was tossed 256 times and the frequencies o fheads obtained are given by the following tab le:

Number o f heads (x) 0 1 2 3 4 5 6 7 8

Frequencies (f) 2 6 24 63 64 50 36 JO J

Fit a binomial distribution to this data,

ution. Here n = 8 and x = r ~ 0, 1,2, ......8.

9 *Since Mean =m = — W

_ 2 xQ+6xl+ 24x2 + 63x3 + 64x4 + 50x5 + 36x6 • -Uv /-t-lx8

2 + 6 + 24 + 63 + 64 + 50 + 36 + 10 +1

0 +6 + 48 + 189 + 256 + 250 + 216 + 70 + 8 _ 1043

256 256'

Also I / = W = 256

1043Since mean of B. D. m = np~ — —

6JO

M - S S t " = * l

p - - 0.5093

20489 = l - p = l - 0 .5 0 9 3 = 0.4907

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420 I ENGINEERINGMaTHEMAT1CS-IH

Binomial distribution is (q + p )' = (0.4907 +0.5093)* - = *Cr(0.4907)*'(0.5093)'.

Now expected frequencies for jr=r=0,l,2, ...............,8 are :

f ( x = r)=N. "C, q-'p'-,

for x - 0. / ( 0)=256x 'C0 (0.4907)* (0.5093)° =0.861.

for j : = 1, /(l)=256x*C, (0.4907)’ (0.5093)' =7.145.

for x = 2, /(2)=256x aC2(0.4907)6 (0.5093)2=25.956.

for x = 3, /(3)=256x % (0.4907) 5(0.5093/ =53.88.

for x = 4, /(4)=256x SC„ (0.4907)4 (0.5093)4 =69.903.

for x = 5, /(5)=i56x*C, (0.4907)1(0.5093)5=58.042.

for x = 6, /(6)=256x ’C4 (0.4907)2(0.5093)6 =30.121.

for x = 7, /(7 )=256x8C7(0.4907)' (0.5093)7=8.932.

for x ~ 8, / ( 8)=256x *C„ (0.4907)° (0.5093)* =1.1589. Ads

Example 5.33 : The following data are the number o f seeds germinating out o f 10 on damp fitter f 80 sets o f seeds. Fit a binomial distribution to these data

X : 0 1 2 3 4 5 6 7 8 9 10 Total

F : 6 20 28 12 8 6 0 0 0 0 0 80.

Solution. Here n =10 N - I f - 80.

Ifx 0x6 + 1x20 + 2x28 + 3x12 + 4x8 + 5x6mean ~ m ~ y ~ g o

_ 174

” 80

_ 87

~~40

Since87

mean of B.D, m = n p ^ > — = n p 40

=3* /> = -* ! = 0.2175.400 [• « = 10]

Then q = 1-/?= 1-0.2175 =0.7825.

Hence the binomial distribution to be fitted to the data is = 80(0.7825 + 0.2175)10

[V f ( r ) = N ”CTqn rp'\

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or /( * =r)= 80x ,oCr(0.7825),<>' ' (0.2175)', r = 0,l ,2 .........,10.

Thus, expected frequencies of x = r = 0,1,2,.............10 are respectively

6.9, 19.1, 24.0, 17.8, 8.6, 2.9, 0.7, 0.1, 0, 0, 0. Ans.

Example 5.34; In litters o f 4 mice the number of litters, which contained 0,1, 2, 3, 4 females were noted. The data are given below

No. o f female mice 0 1 2 3 4 Total

No. o f litters : 8 32 34 24 5 103

I f the chance o f obtaining a fem ale in a single trial Is assumed constant, estimate this constant o f unknown probability. Find also the expected frequencies.

Solution. Here n~4. N - I f = 103.

L/x 8x0 + 32x1 + 34x2 + 24x3 + 5x4

mean ~ m ~ ~ 8+32+34 + 24+5

32 + 68 + 72 + 20 192 ,----------------------- — = I .864.

103 103

n = 4 and np = mean = 1.864. [ v mean of B.D., m = np\

=> /> = 1 = 0 .4 6 6 . [v « = 10]4

Then q = !- /> = 1-0.466 = 0.534.

Hence the expected frequencies are the respective terms of the binomial expansion of

= N.(q + p)n ~N.RCt q*~r p '.

or / ( r ) = 103x-'C,(0.534)*" (0.466)' for r = 0,l, 2,3,4 . Ans.

Example 5.35 : Fit a Binomial distribution thefollowing data and compare the theoretical frequencies with the actual ones.

x : 0 1 2 3 4 5

f : 2 14 20 34 22 8.

Solution. Given : n = 5, N = E/= 100.

Zfx 0x2 + 1x14 + 2x20+3x34 + 4x22 + 5x8mean = ">= — =---------------------- —

.’. Since, mean of B.D, m - n p ^ n p - 2.84

St a t i s t i c s | 421

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p ~ 0.568.

i'hen q ~ \ —p ==> ^ = 1-0,568 => q = 0.432,

! lento theoretical or expected frequencies are

A x = r) = N.(q + p y = N. *C, q*~'p'

= 100 xs Cr(0.432)w (0.568)', for r = 0,l ,2 ,5.

Thus, theoretical frequencies of x - r - 0,1,2,3,4,5 are respectively

1.5, 9.89, 26, 34.2, 22.48, 5.9. A

5.10 POISSON DISTRIBUTIONThe Poisson distribution was discovered by a French Mathematics Simesi Denis Poisson i1837. It is a discrete distribution and is very widely used. Poisson distribution is a limiting foof the Binomial distribution in which n, the number of trials, becomes vsiy large and p, t probability of the success of the event, is very-veiy small such thar me.tn m - n p is a fini

quantity.

.-. Probability of r-successes.

e~mmr P(X = r)=— “ , r .v 3....................

Also theoretical or exported frequencies is

g~m (ri f ( X - r ) = N ..... . where m = mean and N is number of trials.

r!

The following are the statistical measures of the Poisson distribution.

(i) Mean - np to m.

(ii) Variance ~ m or np.

(iii) Standard Deviation 0 ^ .

(iv) Moment measure of skewness (?i)= '^ p

(v)Moment measure of kurtosis (Y j ) = ~ .rrt

Some examples o f Poisson distribution :

1. The number of deaths in a city in one year by a rare disease.

2. The number of printing mistakes in each page of the first proof of a book.3. The number of defective screws per box of 100 screws.

4 2 2 | E n g i n e e r i n g M a th e m a ti cs - 1 II

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S t a t i s t ic s | 423

11 CONDITIONS UNDER WHICH POISSON DISTRIBUTION IS USED

1. The random variable X - r should be discrete i.e., r = 0. 1,2,... n. here n i» i.i-jic.

2. The happening of the events must be of two alternatives such as succeo. *>■-' ; jlure.

3. It is applicable in those cases when the number of trials n is very iarj:c u p .! probability ofsuccess p is very small but the mean m is finite.

4. p should be very small (close to zero). If p -* 0, then the Poisson distribution is J-shaped and imimodal.

xample 5.36 : Prove that Poisson distribution as a limiting case o f Binomial distribution,

when n -* oa

olution. Binomial distribution

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424 j Engineer ing Ma mema t ics-UI

Since n is very large so that

v lim fl-—)"*=<;X J

Then (1) becomes :

. e mmr p (x ~ r )= — ~ , r = 0, 1 ,2 , . . . ,«

which is probability function of Poisson distribution. - Proved.

5.12 MEAN VARIANCE AND STANDARD DE VIATION O F TH E POISSONDISTRIBUTION

For the Poisson distribution

e mrn' P ix= r)= ~ 7 r > r =0,1,2..... «

We know that

mean ... ( 1)

and variance = ^ 2' - ( n , ) 2,

where |ij' and u2’ are first and second moment about origin.

Forr=0,1,2,..,,oo fir st moment about origin.

L i m ..... f l - ^ l j = ( l - 0 ) ( 1 - 0 ) ..... ( 1 - 0 ) « 1 a „ d

since = e'm. and

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From (1) mean = jti' = m.

S cMts nc^ 142S

Now, second mom ent abovr origin,

r 2 P(r)r= 0

^ , , .e 'mrnr V ' e~Mmr

mO r ■ >0 ' *

AC

- ‘ - I r=t

= m2e~m

( r - 2)1

__3 _ 4 j W m

, m m1 + — ,1! 2 !

m ‘ + — + — ■•+.1! 2!

+ «

+ m

[Since r 2= r ( r - l ) + r ]

[w = mean]

[by (3) and (4)3

« « " / + «

jij =m +m

From (2), variance

t t

• (5)

-(h, )j

-m 2 +m -(m)2

= m.

Thus, variance = m

[by (4) & (5)]

(RG PV Feb. 2006 A Jun e 2008 (N)]

Further, standard deviation, <r=■/variance =4m.

Example 5 .3 7 : Show that in a Poisson distribu tion with un it m ean, the mean deviation about the

mean, is 2 /e tim es the standard deviation.

Solution. By Poisson distribution

P (X ~ r ) = ? - ^ f - , >"=0,1,2......oo.

r\

Given that mean m =1.

X f \ x - m \

We know that mean deviation about mean,

I / k - 1 !r *0 ______ __

X f

N

- I f and x=r] ... (1)

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Here

426 | En g d e e h n g MxnEMAnc&-lIi

/ —P‘r')-e~m J r\ /•! r\ [ v « = 1]

and Z / = I r-'”> I

Hence (I ) becomes :

u j x - i* > - l |

[■.* total frequencies - 1]

=e

- e

) 0 - I | ) l l - l [ | | 2 - l | | | 3 - l j t

0! 1! 2! 3!

1 + ( y ) + < y ) + ( i z i ) + .

2! 3! 4!

, 2 1 3 1 4 11 + ----- —■-i----------- ^ ---------- K

2! 2! 3! 3? 4! 4!

, 1 1 1 1 1 11 + ——— + — —— + — -— + .

1! 2! 2! 3! 3! 4!

= 2e"

Since standard deviation o f Poisson deviation S.D . =Jnt=-J \ =1.

Hence (2) becomes: M .£).=—.! =—(S.Z>.).e e

Exam ple 5.38 : For the Poisson distribution, pro ve that

P(r+1)*m

P(r).(r + 1)

W hich is known as recurrence relation fo r Poisson distribution .

Solution. By Poisson distribution

e mP ( r ) = _ _ r=0)1>2t>

replacing r-»r+l.

P{r +1)=e mr *1

(r+1)!

... (2)

I v « = 1 ]

Proved.

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Now,P(r +1) e"m mr*> . r \ r! m

S t a t is t i c s | 4 2 7

P{r) (r + 1)! e - m r (r + l)r! r + 1

Hence* Prvoed.

Example 5.3 9: F ind the probui !!'• y that a t m ust 5 defective fuses w ill befound in a box o f 200fuses,

i f experience sho- >that 2 percent o fsuch fuse s are defective.

Solution. G iven : p -2 Vo

- i t r 02- "■ *”

mean m=np=> m= 200 x.02 => m =4 .

Since n i» lt«ge so that using Poisson distribution

g-m

p (r>----- — • r = 0, l ,2 , ........... 200. ... ( 1 )

/. P (at m ost 5 defective) = P (r £ 5)

- / » ( 0 ) + /*(!)+/*(2 )+ P(5)+P(4)+P(5)

- e 4 42 4J 44 4s1 ^— 4--*'— + — + — II 2! 3! 4! 5!

[ by ( 1 )]

=0.0183[1 + 4 + 8+10.6667 +10.6667 + 8.5333]

=0.0183x42.8667=0.7845. Ans.

Example 5.40: In a certain fa ctory turning razor blades, there Is a sm all chance o f 0.002 fo r any

blade to be defective. The M odes are in pa cke ts o f 10. Use Potsson's distribution to calculate the approxim ate num ber o f packets containing no defective, one defective and

two defective b lades respectively in a consignm ent o f 50,000 packets.

Solution. Given, p =0.002, n=10, N =50,000

m - n p = 10x 0.002=0.02

By Poisson distribution

„-K» _ .re m

P{X=r) = — — , , -=0,1 .3 , .......... 10

(i) />(nod ef ic t ive)=/ ’ ( r= 0) = ® 1 = 0 . 9 8 0 2

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4 2 8 | En g i n h e w n g M a x h o c a t t c s -III

Hence number of packets containing no defective blades

=N.P(r=0)=50,000x0.9802=49010. Ans.

(ii) P (one defective) = P(r=l)=e~001.— ^-=0.019604

Number o f packets containing one defective blade

=M />(r=l)=50,000x0.019604 =980. Ans.

(iii) P (two defective) ~ P(r ~2)- e -0® ——=0.00019604

/. Number of packets containing two defective blades

=Ar./>(r=2)=50,OOOx0.00019604=9.8«10. Ans.

Example 5 .41: I f the proba bility that an Individual su ffers a bad reaction from a certain injection is

0.001 , de term ine the probability that ou t o f 2000 individuals

(i) exactly 3 (li) more than 2 individuals

(iii) none (iv) more tha t 1 individuals

w ill su ffer a bad reaction.

Solution. Given p =0.001 (which is very small)

n =2000 (which is large), then m - n p => m* 2000x0.001=2

Using Poisson distribution:

P(X= r )=e m~ ^ , r = 0 , l , 2.................2000 .

(i) P (exactly 3 suffer a bad reaction) ~ P ( r = 3)

8e_I 4r= 0 . 180.

3! 6 3e2

(ii) P (more than 2 suffer a bad reaction) = />(r>2)= l -[ i>(r =0 )+ i>(r = l) + />(r=2)]

= 1 - e~m+ m em V * ‘

= 1 - r i 2 —-i ~. 2 "1J. .1 ! 2 ! 2

_e e V .

=1-4=0.323. Ans.e

(iii) P (none suffers a bad reaction) =P(r= 0)=e '" =l /e 2 =0.135. Ans.

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St a t i s t i c s | 429

(iv) P (more than 1 suffers a bad reaction)

= />(r>l)=l-[/’(r=:0)+i,(r=l)]

= 1-

_m mle 'm■ 1-

' I 2 14 __ _

1! y

= 1 - 4 = 0 . 5 9 4e

Ans.

Exam ple 5.42: A ca r-hire fir m two cars, which it h ires out day by day. The num ber o f demands fo r

a car on each day is distributed as a Poisson distribution with mean 1.5. Calculate the

proportion o f days on which neither car is used and the proportion o f days on which

some dem and Is refused.

Solution. Given mean m = 1.5,

Poisson distribution ^ (r )—

i _rm

r\ r =0,1,2,3,.

(i) The proportion of days when neither car is used

P(r=0) =m

i.e.,0!

e‘l 5 =0.2231.

(ii) Since total cars = 2

demand is.refused when r £ 3.

Hence, the proportion of days on which some demand is refused

= />(rs 3 )= l-[F (r = 0)+ /,(r=l)+P(r=2)]

Ans.

„0 1 2-m rrt ffi .mr7i

e — +e — +e — 0! 1! 2!

1 + 0 - 5 ) + ( I S ?

Ans.=1-0.8087 =0.1913.

Exam ple 5 .43: A telephone sw itch handles 600 ca lls on the average during a rush hour. The board

can m ake a m axim um 20 connections per m inute. U se Poisson distribu tion to estim ate

the probability tha t the board w ill be o vertaxed during any given m inute.

Solution. Given that mean number of calls per minute

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fit'

P (using 0 to 20 callas per minute) =P(r-Z2Q)

=P(r=0)+P(r=l)+...... + /V= 20)

» „<■ 20 _ r -io V m

430 | En c h n e h o n g M /m o w ic s-IIl

Poisson distribution P(r )= e r = 0 .' 2,............600.

“ 5 * 7 T “ * § 7 T - 0 )** r -0

Thus P (the board will be over-taxed during any given minute)

" P (when the calls are more than 20)

= P ( r>20)= l -P( r520 )

f r -*»0 ~\

Ans.

Exam ple 5.44 : I f 3% o f th e elec tric bulbs m anufactured by a com pany are defective, fin d the

probability th a t in a sam ple o f 100 bulbs exactly fiv e bu lbs are d efective.

3Solution. Given that =0.03, n=100

/. mean m=np=100x0.03

=> m = 3.

irf By Poisson distribution ^ ( r )~e r=0,1,2,......... 100.

/. P (exactly five bulbs are defective) =P(r=5)

= e „ ( 3 ) l = 0.0 49 79 > 2 4 3 = 0 1 0 0 8

5! 120 An8,

Example 5.45 : A m anufacture knows that the condensers h e m akes contain on an average 1% o f

defective. H e pa cks them in boxes o f 100. W hat is the prob ab ility tha t a box picked at

random w ill con tain 4 or m ore fau lty condensers T

Solution. Given th a t/> = ^~= 01, »=100. Mean =/n=«p=100x0.01=I.

By Poisson distrib ution:

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S t a t i s t i c s | 431

P (4 or more faulty condensers)

=/>(r£4)=P(4)+/>(5)+....+ iT.,;0)

=1 -[/>(0)+ P(\) + P(2) f /’(.I)]

= 1-

<f' e l e-' ---- + —— + 4- — 0! 1! 2! 3! - • - j ' ;+H

=1 ——=1 -0.981=0.019.3e

Ans.

Example 5.46: A Skilled typ ist on routine work kept a record o f m istakes m ade per day daring 300

working days

M istake p er day : 0 1 2 3 4 5 6

Num ber o f days : 143 90 42 12 9 3 1 F it a Poisson distribu tion to the above date an d calculate expected (or theo retica l) frequen cies

Solution. Mean = «= -

or

143x0+ 90x1 + 42 x2 + 12x3 + 9 x 4 + 3x5 + 1x6

300

0 + 90 + 84 t- 36J- i 5 +6 267

[V N = 300]

m --300

By Poisson distribution:

e~mmr

H ’ r = 0 ’1-2’ ............6>

The expected (or theoretical) frequency for r success

f(r )= N P (r)^N .^-~ ~ - . r = 0,l,2....... 6

for r =0 /(0)= NP(0)=-^.C—”i2. -=300 x 0.411 =123.3 «123

for r = 1, / (l )=A y( l )=^ - —^ =300x0.365 =109.5 «1

300e“°*9(0.89)2

10

for r = 2, /(2)=A7><2)=

for r = 3, /(3 )= ^ (3 )=

2 !

300e~°*9(0.89)*

3!

-=300x0.163=48.9*49

=300 x 0.048 =14.4 «14

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432 | B iQiBBBjiin Mwwamaics-ni

fo rr = 4, /(4)= jW>(4)=-— - ”p ^ -=300x0.011=3.3 *3

f(5)=NP(5)= 3Q ^ ^ .-- ) -=300 x 0.002 = 0.6 »1forr = 5, . x - 5 ,

for r = 6, / ( 6) =M>(6)=— «300 x 0.0003 « 0.09 * 0

Thus expected (or theoretical) frequencies are :

r: 0 1 2 3 4* 5 6

/ , : 123 110 49 14 3 1 0. Ans.

Example 5.47: The frequency o f accidents per sh ift in a facto ry Is given in the fable below

A ccident p er s h # (x) : 0 1 2 3 4 Tpftl

Frequency ( f ) : 192 100 24 3 1 320.

F ind the corresponding Poisson distribution an d compare w ith actua l observation.

„ , ^ w l£c 0x192+1x100+ 2 x 24 + 3x3 + 4x1

Mean =mf = — 1«+fdo724 ; t ; i—

=— =0.503320

. e '"mr By Poisson distribution, ^ \ x ~ r )----- —* r =0, 1,2,3,4.

The corresponding probabilities for r success

o /x -« « (0.503)'P i r )* * * 3” - ~ r=0,l,2,3,4.

i.e., Probabilities are : 0.605, 0304, 0.076, 0.0128, 0.0016. Am .

Now the corresponding expected frequency for r success

/( r ) = JVJ»(r)=320x«-05® i 2 ^ L , r= 0,l ,2,3,4.r\

i.e., Corresponding frequencies are :

193.6, 97.3, 24.5, 4.1, 0.5. A»s.

Example 5 .48 : F it a P obson’s distribution to the follow ing ca lculate theo retica l frequencies;

r : 0 1 2 3 4

f : 122 60 15 2 1

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, , w Zfx 0x 122 + 1x60 + 2x15 + 3x2 + 4x1olution. :. M ean = « = — ------------------------ — ---------------------- N = Z f = 2 0 0 ]

2*/ 2w

60 + 30 + 6 + 4 100 _1 _Qs

200 2002

~ ‘

The theoretical frequencies of r success f { r )= N P (r )

= N . e - ^ = 200x«f<’»x<°^l; r= 0,l,2,3,4.

For r - 0, / ( 0)=200xe-<,J1 2 ^ ! =i21.3*121

For r * 1, / ( I ) - 2 0 0 x ^ s ^ - 6 0 , 6 5 * 6 1 .

For r = 2, / ( 2 ) - 2 0 0 x « ^ £ ~ 9 l - l S . 2 5 * l S• !

For r - 3, / (3 ) = 2 0 0 x e ^ M l = 2 .5 3 * 3

For r = 4, / (4 ) * 2 0 0 x e ^ & 2 l * 0 3 l 6 « 0 .

Thus, theoretical frequencies are :

r: 0 1 2 3 4

/ , : 121 61 15 3 0. Aos.

xam ple 5.49: The m ortality rate fo r a certain disease is 7 in 1900. W hat is the probability fo r ju st 2 deaths on account o f this disease in a group o f400 ?

7olution. Given: p

------- = 0.007 and n * 400.

F 1000

mean, m =np * 400x0.007 - 2.8.

By Feisson distribution, probability of r-success

^ ( r ) = ~ r = 0 , U ........... 400.r\


P{2deaths) = P(r = 2) =~— = 0.2384 = 23.84% A ns .

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434 | En g r c e h in g Ma themat i cs-III

Exam ple 5.50 : I f x is a Poisson variety suck that

P(x = 2 )- 9 P (x 'm4) + 90P (x'*6), fin d the m ean o f x.

Solution. By Poisson distribution with mean m is

P(x) = x = 0,1,2...... .co.

Since, P(x = 2) = 9 P(x = 4 ) + 9 0 P(x = 6)

e~mm 2 e~"m4 e "m 6

— = 9 x — + 9 0 x —

, 9m 2 90m 4

=> 1 = — + .12 6x5x4x3

, 3 2 1 4=> U - w + 2m = > » 4-3 m l - 4 = 0

=> (m2 + 4)(m2 -1 ) = 0

=> m = ±1 [*.* m1 - -4 is rejected]

=> m - 1. Ans.

Exam ple 5 .51 : Six co ins are tossed 6,400 tim es. Using the Poisson distribution , w hat is approximate

probability o fge tting six heads x tim es.

Solution. Given: n - 6400.

v Probability of getting a head in a throw of a coin =

Then, Probability of getting six heads in a throw of six coins, P = ^ ~ ~ -

mean = m - n p = 6400x— = 100.64

By Poisson distribution, probability of r success,

i>( r ) = l ^ l ( r = 0, 1 ,2 ........r!

e mm‘ e-m (\Q0y P (six heads x tim es) = P{r = o r ) ------------------------------- Ans.

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S t a t is t ic s | 435

5.13 NORMAL DISTRIBUTION

It is a theoretical and continuous probability distribution in which the relative frequencies of acontinuous variable are distributed according to the normal probability law. In other words, it isa symmetrical distribution in which the frequencies are distributed evenly about the mean of the

distribution. Normal distribution is a limiting form of Binomial distribution under the following conditions:(i) h , the number of trials is infinitely large, i.e., «-><».(ii) neither p for q) is very sm all I.e., p and q are fairly near equally.A random variable x is said to have a normal distribution with mean ‘n’ and standard deviation‘or’ if its probability density function is given by

/ ( * ) = — «W2«

• 00<X<<X) .

The probability density function with mean zero i.e., h = 0 and standard deviation <r is given by

f i x ) -----

o V 2 rt-00 < x < 00.

Normal distribution was first discovered by British Mathematician De-Moivre in 1733. Normaldistribution is also known as Gaussi an d istribu tion

The total area under the normal curve is equal to unity and the percentage distribution o f area

under the normal curve is given below as shown in the figure.(i) About 68% o f the area fells between h - cr and (i + a .

(ii) About 95.5% of the area fells between - 2o and (i + 2o.

(iii) About 99.7% of the area falls between p - 3cr and n + 3a.

68.27% •

95.S% •

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4 3 6 | En g j n e e w n g M a t h e m a t ic s -1]I

5.14 STANDARD NORMAL DISTRIBUTION

A random variable z which has a normal distribution with m ean - 0 and a standard deviation

cr = 1 is said to have a standard normal distribution. Its probability density function is given by

/ ( z )= -?L=-e',J/2, —oo <2 < oo,V2 rt

It is denoted by N (0, 1). In short, standard norma] variety is written as S.N. V.

Th e area unde r any no rmal curv e is found from the table o fa standard norm al probability distribution

showing the area between the m ean an d any v alue o f the normally distributed random variable.

For a given value o f and f i and <j specific value, X, o f the random variable, the standardized variety Z is derived from the following formula :

CT

The p urpose o f standardization o f normal distribution is to enable us to make use o f the tables o f t

the area o f the standard curve /( * ) = ^ for various points along the x-axis.

The standard normal distribution is also known as Unit Normal Distribution or Z-Distribution. The standard no rma l curve helps us to find the areas w ithin two assigned limits under the curve.

The areas betwe en the standard normal curve draw n at two assigne d limits a and b will give the

pro port ion o f ca ses fo r w hic h th e valu es o f z lie between a and b. Thus the area between two

assigned limits a and b un der the standard normal curve w ill represe nt the probability that Z will

be betw eena

andb.

It is denoted byP ( a £ Z £ b ) .

z = -3 z = - 2 z - 0 z = + l z = +2 z = +3

t4

0.6827

0.9545

0.9973

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S t a t is t ic s | 43 7

5.15 PROPERTIES OF NORMAL CURVE

The normal probability curve with mean p and standard deviation o has the following properties:

1. The equation of the curve is

and it is bell-shaped. The top of die bell is directly above the mean p.

2. The curve is symmetrical about die line x = p and x ranges from - » to +«.

3. Mean, mode and median coincide at x * p as the distribution is symmetrical.

4. .A'-axis is asymptote to the curve.

5. In Normal distribution: Arithmetic Mean = p and Variance * a 2

6. The points of inflexion of the curve are at x = p + a, x = p - a and the curve changes from

concave to convex at x = p + o to x= p~ cr .

7. The mean deviation from the mean in normal distribution is equal to 4/5 of its standarddeviation,

S. All the odd moments about the mean are zero i.e., =0.

9. The maximum ordinate lies at the mean i.e., at x = p.

10. The curve of normal distribution has a single peak, i.e., it is a unimodal.

11 .The two tails of the curve extend indefinitely and never touch the horizontal line.

5.16 METHOD TO FIND THE PROBA BILITY WHEN TH E VARIATE IS

NORMALLY DISTRIBUTED

Let X be a normal variate with mean p and standard deviation or. Suppose we want to find the

probability that a randomly selected value fo r X will lie between a and b i.e., P(a <X <b ).

Step / . Convert X into a standard normal variate by die formula

1, -OOSXÔO.

. . . { 0O

Step 2. Find the limits of Z corresponding to the limits of X.

ry t f - PWhen X = a then Z -------- tP ut X ^ a in (1)]

~ b - pWhen X - b then 2 = ------ [PutX=bin( l ) }

a

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4 3 8 | En g in e e r in g Ma t h e ma t ic s -III

Step3. Thus P {a < X < b ) = p [ ^ ^ < Z < t t ^

Step 4. From the normal table find the probability that Z is between (a - n )/o and (6 - \ i) /a

Exam ple 5.52 : A certain type o f wooden beam has a m ean breaking strength o f 1500 kg s and a standa rd deviation o f 100 kgs. F ind the relative frequency o f a ll such beams whose

breaking strengths are between 1450 and 1600 kgs.

Solution. Let X be the breaking strength. Then we are to find P(1450< X < 1600)

Let A" be a normal variety with mean p = 1500 and standard deviation a = 100.

Then standard normal variety N (0,1) :

^ _ X - y . _ X 500

cr 100

When X = 1450, then Z = !i f f = -0,5

When X = 1600, thenlUU

Thus P( \450<X <1600) = i ,(-0 .5< Z < l)

= P (-0.S < Z < 0) + i>(0 < Z < 1)

= 0.1915 + 03 4 13 = 0,5328 [From normal table]

=> 53% of the beam has the breaking strength between 1450 kgs and 1600 kgs. Ans.

Example 5.53 :Prove that the points o f inflex ion o f the norm al curve are x = ±<r.

Solution. Let the equation of normal curve be

y= y0e*’/2*’ ...d)

We know that points of inflexion are given by

d 2v d*y

Differentiating (1) w.r.t. V, we get

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S t a t is t i c s j 4 3 9

dx3 a * { a 2)

Now , fo r in flex io n poin ts :

d ' y '

dx2 - 0 =>

x J ^

V v /

Also at x = ±o , d 3y ld x>* 0.

Hence this show that points o f inflexion are at x = ±a . Proved.

Example 5.54 : Prove that in a normal distribution, all the moments o f odd order about the mean

are vanish Le., zero.

Solution. By definition o f mom ents about the mean u o f odd order :

H ^, = ) (x-M)*V<x)<fc '-00

W h e re ,/(x) is p.d.f. o f norma l distribution i.e.,

/ « = — ! r - e W '” ‘ - « >c tv 2 ji

' ' [by ( 1 ) ]

Putting,

I <N<2» )

= — p L = f ( x - M y + ' - e ^ r f ' ^ d x

= - J _ ] ( o O ^ '.e ^ . o d t ,

v m i

= 0; since tu *xe~'>12is an od d function, so tha t J oeW = 0

Thus, A d s .

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4 4 0 1 E n g m b h m o M * n tEM x n cs4 U

Exam ple 5.55 .*Prove tha t in a norm al distribution, a ll the m om ents o f even order about the meat is

= (2n - l)(2n - 3)------- 5.3.1 — a u .

Solution. Let

f ( x ) be p.d.f. of normal distribution i.e.,

/ ( * ) =1

We know that moments o f even order about the mean jt is

= } ( x - n ) u f ( x ) d x

= — J — T ( x - p . f ' . e ^ ' ^ d x .rt J

Putting, 1 -- - —— =>dx = a d t

ii - — 7-— f {at)2”.e~*>n.od t ^ o V ( 2 « ) i

= J t 7 Me -1,2dt

2o>

' ^ 0

(1)

[by (1)]

since t ^ e ^ ' 1 is an even function, j even- 2 J even-e ' d z . f z(*+V '<£rV7t ; v k 5

2*o*"rf J l v J e" 'zN' xdz ~ V(N)

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2- ‘a 1 1

^ i ^ r T 2 J

From (1) by (2), we have

Replacing n by (n - 1), we get

S t a t o u c s { 44 1

- (2)

[v r*=(x—i)r(x-i>]

Now proceeding as above processes, we get

m ,=(2 »-l)(2» -3)....5.3 .1.o2j\ Proved.

Exam ple 5.56 : To drive the norm al distribution as a lim iting case o f Binom ial d istribution, when

p - q .

Solution. We know that in a Binomial distribution frequency of r-success is

m = N.*Crq” p r

1 1Since P =9 => P = ~ and ? = -

Suppose that n is an even integer, i.e., n = 2k, k being an integer.

Hence (1) becomes:

... ( 1 )

[because p + q = \]

m

Replacing r - y r + 1, we get

f ( r + 1)^ (2t)l(r)K 2t-r)l _ 2k - r

f i r ) u Cr (2k - r ~ l)!(r + 1)!(2£)! r + 1 ' ... (2)

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442 | En g in e e r in g M a t h e m a t i c s -!!!

The frequency of r successes will be greater than the frequency of (r + 1) successes if

/ ( r + l ) / / ( r ) < l

2 k - r < r + \ => r > k ~ - 2

- (3)

In a sim ilar way the frequencies o f r successes will be greater than the frequencies o f (r - 1)

successes if

. 1r < k + —.

2... (4)

From (3 ) and (4 ), we observe tha t i f * “ < ' , < * + ^ , t h e fr equency correspond ing to r

successes will be maximum. Clearly r - k is the value o f the success corresponding to

which the frequency is maximum. Suppose it is Then from (1),

— ' ( a ' - ' i s a i rLet yx be the frequency o f ( i + jc ) successes then we have

(2*)!

k l k l

x )K k -x ) \

k ( k - \ ) ( k - 2 ) . . . ( , k - x + \)

y0 ( it + x ) ! (£ -x ) ! (k + x) (k + x - ! ) . .. .( £ + 1)

... (6)

[by (5) & (6)]

o r

Taking log of both sides, we get

logf e ) ' K 1' l ) +log( , ' l ) +-"+log( ' - T i l

- Iog(, + i)+log(,.l)+... +Iog(, +i)

Now writing expansion fo r each term and neglecting hig her powers o f x/k as k is very

large), we get

l o g f ^ U - r t f + 2 + 3 + . . . ( * - ! ) } - - i - 0 + 2 + 3 + . . . ( x - I ) + x }

U J * *

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= - j { l + 2 +3 + . ..+ (x - l )}~~

or I O g | ^ y.

X . X X3

y . - y + ~ * n

Since in Binomial distribution, o = -Jnpq

9 * 1 1

=> t r = npq => a2=2k.—.— 2 2

=> 2^ = * .

Hence (7) becomes:

y t =y*e ■

i.e. f { x ) = y0e~’3'2*3.

Which is normal distribution.

Exam ple 5.57 : For the norm al curve

y - M >—o ^ l 2 x

Find m ean and standard deviation.

Solution. Since / (x) is p .d f , then

oo

mean = J x . f ( x )d x

- I J __

.xdx.

— j i — r f e''1lta^/2 + n )a -j2 dt o>/(2n ) i V ’

1

[Putting / =

2a 1 J e~ *tdt+ \iB 'j2 J e~*dt\ -«© *« J

j s in c e J e^tdt--

[By using A.P.]

... (7)

[v n = 2k]

Ans.

— J* dx = o -\/2 dt. icw2

<8

Oand J e~,1dt = Jn

S t a t is t i c s | 443

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Mean = X0 +

4 4 4 | En g in e e r in g MATHFJumrs-III

1

x n o V 2 x-v/ir = j icV 2n

Thus mean of normal curve is p

Further, by definition o f variance

oo

Variance = J (x - mean)5. f (x )dx

A d s .

= f(x-/<)2- 4 — dx a>/27t [v Mean = jx]

[Putting t = ~r-= > dx = a^ 2d l .] a v 2

Variance = , f {to42 \ a^2dtW ( 2n ) i ' '

= —1

—| e t2dl =—n - \

(te * ltdt

ctV2 tc i i ' '

On integrating by parts, we get

2a 3Variance =

>/jr

2a

2 —x>/n V J e'^dt^yfH

= 0 ‘ .

Hence Variance of normal curve is o2. Ans.

Example 5 .5 8 : The m ean deviation fro m the mean o f the norm al distribution is tim es its standard deviation.

Or F ind the m ean deviation from mean fo r norm al distribution.

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S t a t i s t i c # j 445

Solution: By Normal distribution

m - < »tWZJl

We know that, mean deviation from mean V

M.D. = £ \ x - \ i \ f ( x ) d x

IP“ ' * J

M.D. = £ I ' T a , \ . - ^ - e - f . ^ 2 d ,

= Ule- 'd t

duPut u= t2=> du=2tdt => tdt- —

2a >/2 r® du ct f e - T

M D - ~ ~ 7 T ‘° e 7 = 7 S r h l

- 2 o

yj2%

* (0.8)o approx

4 a „~ — approx Proved.

Exam ple 5.59 : In a norm al distribution 31% o f the item s are under 45 and 8% are over 64. F ind the

mean a nd standa rd deviation o f the distribution.

Solution. 3 \% of items are under 45 => Area to the left is 0.31. But area right from this point is(0.50-0.31) =0.J9. (See figure)

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44 6 | En g w e h b n g Ma t h e m a t c s -III

Let X be the random variety, which is normally distributed with mean ji and standard

deviation o .

Z » -Z , 2T-0 Z>

Then, 2 - X ~ \n is a standard norm al variable (S.N.V.)

The S.N.V. corresponding to X - 45 and X - 64 are as below

W hen X = 4 5 , then Z = i l l E = _ z i. (Say)c

W hen X = 6 4 , then z = — ^ = Z , , (Say)

a

From the figure it is obvious that

P ( 0 < Z < Z i ) = 0,42=>Zj = 1.405

and P( -Z , < Z < 0 ) = 0 .1 9 = > / >( 0 < Z < Z , ) - 0 . 1 9 .

=> Z, = 0.4 96

Substituting the values of Z\ and Z j in ( I ) and (2), we get

4 5 - n■= -0,49 6 => 4 5 - - 0 . 4 9 6 c

6 4 - n= 1.405 6 4 - n = 1 . 4 0 5 a

Solving (3) and (4), w e get

a =10, n =4 9.96 = 50 (approx .) i.e., S.D. = 10 and mean « 50 .

•••(!)

.. .(2)

[From the normal table].

[by symmetry]

[From the normal table],

... (3)

... (4)

Ans.

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Example 5.60 : The mean height o f 500 studen ts is 151 cm , and the standard deviation is 15 cm

Assum ing that the heights are norm ally distribu ted, fin d how m any studen ts have heights

between 120 and 155 cm. ?

Solution. N o. o f students = 500 .'. N = 500

M ean, p = 151 cm , and a = 15

S t a t is t i c s | 4 4 7

By standard normal variable z =

x . - n 120-151 -31Stan dard norm al varia ble = -*------= ------— -----= -77 - - ' •

0 15 15

When x2 = 155 cm,

Standard nonnal variable *

xi - \ i _ 153-151 4 _ Q

15 27.o . 15

^(120 < jc < 155) = / ’(-2.07 < z < 0.27)

= P (-2.07 iz £ 0 ) + /»(05zS 0.27)

= P(0Zz£2.Q7) + P (Q lzZ 0 27 )

= 0.4808 + 0.1085= 0.5892.

The required no. o f students =0.5892x500 = 294.

Exam ple 5 .61: The distribution o f weekly wages o f 500 workers in a factory is approxim ately norm al with the means and standard deviation o fR s. 75 and Rs.15. F ind the num ber o f workers

who receive weekly wages:

(i) M ore than Rs. 90

(ii) Less than Rs. 45.

Given: No. o f workers = 500 i.e., N = 500

Mean, ft = 75 and S.D. a = 15

[by norm al table]

Abs.

Solution:

x - \ iBy standard normal v ariable z= ~a

(i) When, x - 90 : Then standard normal variable

j c - * i 90 - 75 15

. . . ( 1 )

z= -o 15

P (more than Rs. 90) = P(x>90)

= P(z> 1)

= 0 . 5 - / >( 0 < z < l )

= 0 .5-0 .3413

= 0.1587

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4 4 9 1 En g in e e r s * ; M a j h e m a t k s -III

Hence number of workers who receive weekly wages more than Rs. 90

■ 0.1587x500

= 79.35*79 workers

(ii) Whem, x = 45 : Then 2 = £ l i U 4 5 -7 5 = _ 2o 15

P (less than (Rs. 45) - P(x < 45)=P(z < - 2)

= 0.S-P(-2<z<0)

= 0.5 - 0.4772

= 0.0228

Hence No. of workers who receive wages less than Rs. 45

= 0.0228 x 500 = 11,4 '» 11 workers Ans,

Exam ple 5.62 : A sam ple o f 100 dry battery cells tested to fin d the length o f life produced the fo llow ing resu lts :

M ean x -1 2 hours, standard deviation a - 3 hours.

A ssu m ing the data to be normality distributed, what percentage o fbattery cells are expected

to have life

(I) M ore than 15 ho urs

(ii) Less than 6 hours

(iii) Between 10 end 14 hours ?

Solution. Here x denotes the length of life of dry battery cells.

x - j f x - 1 2The S.N.V.; * « -

a 3

(i) When x - 15, then z = 1.

P (more than 15) - P (x > 15) = P{z > 1)

- P { 0 < z < c o ) ~ P ( ( ) < z < 1)

= 0.5 - 0.3413 = 0.1587 - 15.87%

6-12 „(ii) When x = 6, then 2 = —~— = - 2

P (less then 6) = P (x < 6) = P (z < - 2)

Z - - 1 X«0

= P ( z> 2 ) - P ( 0 < z < oo) ~ P (0 < z <2)

= 0.5 - 0.4772 = 0.0228 * 2.28%.

Ans.

Z - 2

10-12(iii) When x = JO, then 2 = — -— = -0.67.

when x = 14 then z ~~

3

14-12=0.67.

A bs .

Z=-.67 Z =0 Z =.67

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P (Betw een 10 & 14) = P (10 < x < 14) = P ( -0 .67 < z < 0 .67)

= IP (0 < i < 0.67) = 2 x 0.2487

= 0.4974 = 49.74% An s.

Exam ple 5.63 iAssum ing th at the diam eters o f1000 brass plugs taken consecutively fro m a machine

fro m a norm al distribu tion with mean 0.7515 a n . and standard deviation 0.0020 cm .,

how m any o f the plugs are like ly to be rejected, i f the approved diameter is.

Solution. Given range of diameter are :

x , - 0 .752 - 0 .004 = 0 .748 cm an d

x2 = 0.752 + 0.004 = 0.756 cm

Mean jx = 0 0.7515 and S.D. «r = 0.0020 cm

St a t is t ic s | 449

The S.N.V. is r = x - n

0.748 - 0.7515 Z — 1.75 Z ~« Z -22S A t Xi *= 0.748, then = -------i t t : ------ = -1.75.

Also a t x2 = 0,756, then z2 = ■

0.002

0 .756-0 .7515= 2.25.

[From nonnal table)]

0.002

P (x 1 < X < X2) = P (Z| < 2 < Z2)

= /> ( - 1.75 < z < 2 .25)

= P (0 < z < ~ 1.75) ■+■.P (0 < z < 2.25 )= 0.4599 + 0.4878

= 0.9477.

.’. Nu mb er of plugs likely to be rejected = 1000 (1 - 0.9477)

= 1000 x 0.0523 [ v Shaded are* = (1-0.9477)]

= 52.3 s> 52. A ns .

Example 5.64: A m anufacturer knows fro m experience that the resistance o f resistors be produces is

norm al with m ean ft = 100 Ohms and standard deviation a m 2 O hms. W hat percentage

o f resistors w ill have resistance between 98 Ohm s and 102 Ohm s ?Solution. Given that mean p = 100 Ohms, standard deviation ct = 2 Ohms

x - pBy standard normal variable is 2 = ~

9 8 - 1 0 0 ,Whem x = 09, then z, = ----- :----- = - l .

W hen x = 102, then

2

102-100e l .

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P (98 < x< 102) = P ( z \ - - \ < z < t 2 ~ \)

= P (~ ) < z < 1)

= / > ( - 1 < z < 0) + />(0 < z < 1 )

- 0.3413 + 0.3413 = 0.6826.Thus, the percen tage o f resistors having resistance between 98 ohms and 102 ohms.

= 0.6826 x 100 = 68.26% Ant

Example 5 .65 : In a sam ple o f1000 cases, the mean o f a certain test is 14 and standard deviation is

2.5. Assum ing the distribution to be norm al fin d

(i) How m any stud ents score between 12 and 15 T

(ii) How m any score above 18 ?

Solution. Give n that 1000, mean \i = 14, o,= 2.5.

The S.N.V. is z = -CT

450 | En g in e e r in g Ma t h e m a h c s-III

(i) When x = 12, then.

When x - 15, then, zi =

1 2 - 1 4

o 2 .5

1 5 - 1 4

= - 0.8

2.5= 0.4

P (12 < x < 15) = P ( -0 .8 < z < 0.4 )

« P ( - 0 . 8 < z < 0) + P (0 < x < 0.4)

= 0.2881 +0 .1554

= 0.4435.

.*. Th e required nu m be r o f stude nts = 1000 * 0.4435

= 443.5 * 444.

I O | A

(ii) When x= 18, then, z — = 1.6

[From normal table]

2.5t *

P (score above 18) = P (x > 18) = P (z > 1.6)

- Left area - Area between 0 and 1.6

F 0.05 - i* (0 < z < 1.6)

*= 0.5 - 0.445 2

= 0.0548.

.v The required num ber o f students = 1000 * 0.0548

“ 54.8 * 55.

Z - 0 Z - 1.6

[From normal table]

Ans.

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Exam ple 5.66 :The lifetim e o f a certain kind o f battery has a mean o f 300 hours and a standard

deviation o f 35 hours. A ssum ing tha t the d istribution o f life tim es, which are m easured

to the nearest hour, is n o rm a lfind the percentage o f batteries, which have lifetime o f

more than 370 hours.

Solution. Let x be a random normal variate measuring the life tim e o f batteriesHere mean j i=300 ,cr=35

x —ii x - 3 0 0Th e S.N.V. is z = -----~ = - ~

o 35

170 — 00W hen x = 3 7 0 , t h e n z = ± ^ ^ = 2 .

P (M ore than 370) = P ( x > 370) = P(z >2) z - o z - 2

= left area - area between z = 0 and z = 2 * - 3 7 0 z = 0 and z = 2

* 0 . 5 - P (0 < z < 2) = 0.5 - 0.4772

= 0.0228.

Hence percentage o f batteries having life time m ore than 370 hours

= 0.0228 x 100 = 2.28%

Exam ple 5.67 :F it a norm al curve to the follow ing data

length o f line (in cm ): 8.60 8.59 8.58 8.57 8.56 8.55 8.54 8.53 8.52

frequency: 2 3 4 9 10 8 4 1 1

Ans.

<*-n)3Solution. Let the nonn al curve be : y -

Here iV= 42.

... ( 0

X f u * * x -A Taking

A = 8.56

u^ /« f »2

8.60 2 0.04 0.0016 0.08 0.0032

8.59 3 0.03 0.0009 0.09 0.0027

8.58 4 0.02 0.0004 0.08 0.0016

8.57 9 0.01 0.0001 0.09 , 0.0009

8.56 10 0 0 0 0

8.55 8 - 0.01 0.0001 -0 .08 0.0008

8.54 4 - 0.02 0.0004 -0 .08 0.0016

8.53 1 ^0.03 0.0009 -0 .03 0.0009

8.52 1 -0 .04 0.0016 -0 .04 0.0016

2 /= 42 0.0060 t fu = Olll Tfit 2 « 0.0133

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452 | E ngineer lng M athem atics- II I

and

, 0. 1 1mean (n) = ^ 4" ^ r "*I,56+' 42 _

=8.56 +0.0026 =8.5626 =8.563(approximate)

S.D . (ct) =2/

J IfuY i 0.0133 r0.11Vl^ J j Vl 42 A42 J

= yjUO• 000317 -0 - 0000069}] =0-0175cm .

Hen ce from (1 ) the equation o f non nal curve to be fitted is

y - / (x) = 9,8 « « « < * * » ».

5.17 RECTANGULAR DISTRIBUTION OR UNIFORM DISTRIBUTION;

/ { * )A random variable x is said to be a continuous rectangular or

uniform distribution ove r an interval

(a, b), where b > a, if its p .d f. is given by :

f i x )a < x < b

b - a '0, otherwise

Distribut ion Function o f Rectangular Distribution :

The p .d f. o f rectangular distribution is given by

1

f i x )a < x < b

6 - a ’

0, otherwise

Then distribution function or {C.D.F.) F (x) is given by

F i x ) = P i X Z x )

X

= j f ( x ) d x-<so

. - J 1b - adx

Thus, d.f.

F(x)

b - a

- x ~ a b - a

0 ,

x - a b - a *

1 ,

ixYa

x< ,a

a < x a < x 2

To find S.D. and Variance : T he second mom ent about origin :at>

( b - a ) (6 + a +ab)

We know that

Variance = jij ' - (u i1)2

•(2 )

Mi’ - I x2.f(x )d x

6 1= J x \ ( & T ^ ) dx [ by ( 1 )]

__ L_ [63- a3](6 - o ) 3

2 , _2

“ ( b - a ) ---------------T

= •! (a 2 + 62 + ab). ...(3)

= | (a2 + 6^ + ab) - [by (2)]

= g (a 2 + 6* + ab) - ^ (a2 + 62 + 2ab)

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454 | E n o n b e h n g Ma t h e m a t i c s -1 11

and S.D.

_ ( b - a ) 2 12

o' = VV a r i a n c e

- a2 + b2 - 2 ab12

[v * > a]

yfl2.( b - a ) .

5.19 MEAN DEVIATION OF A RECTANGULAR DISTRIBUTION:The p.d.f. o f rectangular distribution is given by

f i x )

1

b - a

0 ,

a < x I

x -

( b - a )

(a + b)

dx

dx

P u t / = x - (a + b) => d t -dx

( b - a ) ^ ( b -a )and limits are : t => - to

M.D.

>K>- (b-a)

- 1

( b - a ) j Wdt

(b-a)2

(±±)

2 ?( b - a ) J ^

•d )

[by def.]

(o + 6) ,[ v m ean = — ^ — ]

[ v jf| is an ev en /* .]

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(fr-o)

2 ? .

= (6- 0) I tdt

(6-o)

= _ _ 2 _ 1 r a]—( b - a ) ' 2 I Jo

1

[ v |/| = /, if (b - a) > 0 ]


( b - a ) *( b - a ) • 4

, 0 - ° >4 ’

Exam ple 5.68 : A variate X has the rectangular o r uniform distribu tion with p .d f. L given by :

1 r 0 < x <100 100

0 , otherw ise

Com pute (i) P [X > 60} and (ii) P [20 * X *4 0 ]

Solution : (i) We have

too

m -

&

P [X> 60] = J f ( x ) dx 60

100 ^

= I 1 0 0 *

~ 10 0 W

= 0.4

100

100 l*J60

(ii) We have,

40

P [20 5 x 5 40] = J f (x )dx20

40

= 1 A *20 100

= 1 f ]4010 0 ' XJ20

20100

[v by given/(x)]

= 0.2 Ans.

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5.20 GAMA DISTRIBUTION:

( i ) A con t i r ious r andom var iab le X is said to be a gama distribution with parameter

X > 0, if its p.d.f. is given by :

456 ( En g in e er in g M a t h e m a t i c s -H !

/ ( * > «

e - V - '~ W

0, othewise

0 < jc < oo, X > 0

Where [( I j = J e~zxx~'dx.0

(ii) The p.d.f. a gama distributin with two parameters a > 0 and X > 0, is given by

f i x ) o r f {x , a, \ ) =

° gâyX-i • 0 < x < o o , a > 0 , X > 0

I w

0, ‘ oth erw ise

5.21 DISTRIBUTION FUNCTION OF GAMA DISTRIBUTIONThe p .d .f of the gama distribution is

fix) m

0 ; oth

We have, distributive function or C.D.F. of / (x) is given by

0 < x < c o , A_ > 0

F (x) = P iX < x) = 1 f i x ) dx-00

dx x( e - '.x x-'

F i x ) - J - r ^ r -o 1(A)

1

W ) j e~*xM dx

Hence d .f F{x) =

1 x

f (X) / e~* xK~'dx,o

0 ,

0 < x < oo

otherwise.

5.22 MEAN, S.D. AND VARIANCE OF GAMA DISTRIBU TION:

The p .d.f. o f gama distribution is given by

/ ( * ) = .

e~*.xk‘‘

m

0,

0 < x < o o , X > 0

otherwise

•••(J)

[by definition]

[by (l)]

. . ( l)

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Mean = fit’ = E (x)

<30

= J x f ( x ) d x —00

7 e 'V ' '= \ x dx

0 RT)


I 00-T—» J e *x*'- l dx m o

■ W

= X.

7o fin d S.D. and variance :

The second m oment about origin :<50

j i2’ = J x* .f(x ) dx

7 ,

We know that

- i<U *

= X2 + X.

Variance (n 2) = to' - (Mi')2

- K + X - ( V = X

Variance ( |i 2) = X

[by (l)]

CO

[v « = J e*. J f '1<sfc]o

[ v | ( » + 1) = « [(n)3

[by (1)3

[by def. o f Gam a function]

[v |(n + l) = «((w)]

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and S.D. a — -^Variance

This shows that mean and variance are equal fo r the gama distribution.

5.23 BETA DISTRIBUTION OF FIRST KIND:T h e r a n d o m v a r i a b l e X i s said to be Beta distrib ution o f fi r s t k ind , i f i ts p.d .f . is

given by

_ 1

f ( x ) or f i x , a , P ) =

458 | E n g m e e m n g M a tv e m a tk s - III

-5 ( a ,p ) ■xa~l ^ 0 < x < 1 and a , 0 > 0

0, otherwise.

Where B (a, 0) is a Beta function.

I

R e m a r k : B ( a , P ) = f (1 - xp -' dxo

fwl¥)| (a + P)

5.24 COMMULATIVE DISTRIBUTIVE FUNCTION OF BETA DISTRIBUTION OFFIRST KIND:

The p .d f. o f Beta distribution o f first kind is given by :

1

/ ( * ) =5 < a r p) 0 - ^ l. 0 < x < 1, a, p > 0

0 ,

By definition o f distrubtive function

Fi x) =P{X<. X)

otherwise

Hence d.f. is

F{x) =

“ / f i x ) d x-ao

= I T t k $ xa~ ' v ~ xr ' dx

= 5 ( ^ p ) I r -10 *

X

j /*-1 (1 - f )M dt, 0 £ x < I

[by (i)]

1 ; x £ 1 .

N ote : The above cum ula tive dis tr ibutive fu nction f i x ) is called Incomplete Beta Junction, which

is giving by equ ation (2 ).

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5.25 MEAN S.D. AND VARIANCE OF BETA DISTRIBUTION OF FIRST KIND:

The p.d.f. o f Beta distribution o f first kind is given by


/ < * ) =

13 < a , p ) 0 < x < 1 ; a , P > 0

0, otherw ise ... ( 1 )


Mean - fi\ = E (x)

00

= J x/(x) dx

j x**'-» (1 - JL)M dxi ? ( a , p ) 0

^ p j * B ( a + i , P)

[by definition]

[by ( 1 )]

( v J x ~l (1 - x ^ d x - - B ( l , m)]

VB (/, m) =

( a + p ) . a [ (a )[ (a ) (a + p) | (a + P)

a

a + P

|(/ + m)

[v (n + 1 ) - n («)]

- ( 2 )


The second moment about origin

H2 = j x>f{x)dx

*(a,p) J 0 j x2. jf“-' (1 -x )^ - ‘ dx [by (1)]

= 5 (^~p) * B ( a + 2, P) [by def. o f Beta function]

l(a + P + 2)

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4 6 0 | En g in e e r in g M a t h e m a t ic s -U I

IjEH*---- (g+l)«|(5 CT -X,R o) ( a+ ( S + I X a + P ) ^ a + P ) l( ) ' 1

a ( a + 1 )

We have

(a + p)(a + p + l)

Variance (n 2) = M2 - (Mi')2

a (a + l)

••(3)

a(a + p)(a + p + l) Va+p

<*• a +1 aa + p j .a + p + 1 a + p

« f(a + p) ( a + 1)- a (a + p + l)" a+p [ (a + p) (a + p + l)

[by 2 & 3]

a p

(a+ p) (a + p + l)

and S.D. (<i) - -^Variance

1 a p

(a + p) V(a + p +1)

5.26 BETA DISTRIBUTION OF SECOND KIND:

The random variable X is said to be Beta distribution o f second kind, if its p .d f. is given by

/ ( * ) =

1 .0 - 1

S M ' m S ; 00

0 ; otherw ise ...(I)

Re m ark : If we put y ~ r -— , then distribution (I) becomes, Beta distribution o f fir st kind.

Exampla 5.69 : Prove that total area (total probability) o f Beta distribution o f second k ind is 1.

Solution : The p.d.f. o f Beta distribution o f second kind is given by :

f i x ) =

1

B < a , P ) ' ( i + x )a+P

0,

We have Total area = J f ( x ) d x

= 1 }0 B (a, P) (1 + x)a+p

0<x<oo& a , p>0

otherwise ...(1)

[by definition}

dx [by (I)]

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a-1- * - jB (a ,p ) J (l + x)a+P

dx

1 1 1 - y ~dy[Put ting y = - :— - => 1 + x = ~ =>x = — ~r~ =>dx = 2 and limit y~-> 1 (whenx = 0)

1 + x y y yto y —* 0 (when x = 00)]

Total area - _ _ L _ f f i z z ) 1 I - d y B ( a , p ) 1 I y ) V j Y ^ ' U 2 J

p . : J (1 y 5-1 <v B ( a , P) 0

1

b 7 I 1 (I - > ) a"‘ dy B ( a , p ) 0 ,

1

— p (a , P) [ v B (/. * ) = J x '- ‘ (1 - or > -1 dx] B ( a , p ) 0

Hence total area = 1.

5.27 MENA, S.D. AND VARIANCE OF BETA DISTRIBUTION OF SECOND KIND:The p.d.f. o f Beta distribution o f second kind is given by :

/ ( * ) =

1 .'*-1


S ( a , P ) ' ( W . - c f -3

0 ,

mean = j i | ' = E (A)

00

= i x f (x) dx

0 < x < co and a , p > 0.

otherwise . . .( 1 )

[by definition]

1 * x a _ *

* ( a ,p ) [ * ' ( l + x )a ^ *

1 dx

[Putting y =1 + x

X =1 - y

dx = - —7 dy and limits of v -> 1 to 0]v"

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4 6 2 | En g i n e e r s M a t h e m a t ic s - I 11

l ? (i- y ) a.(y)a+>i a ,.2 &

“ e) f 0 -y Y * x~' dy B ( a , p )

= ^ ' p j x B (P - 1, a + 1) [by def. o f Beta function]

W ) W ) ] ( a + p >

Hence meai


The second m om ent about origin is given by :

l(P — 1) - a ( a )

l ( a X ( P - l ) ( P - D ■

a

P - l 'when P > I

a

>*» “ F T

'GO

Hz' = / x » / ( r ) dx-<C

°° 1 ^ a -1

_ | 5 ( a , p ) x2- ( i + x )a+p *

£ ( a , p ) o ( l + * ) a+p

1 l —V _■»[Putting y = — — =>x= —— => a&c = —A ay and limits o f y 1 to 0]

A * y y 2

[by (1) ]

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We have

’ B l k j )

‘ B ( h >


= B ( a,p) xB0 -2 .a + 2)

j f o + P) .. R F 5 j 1( ^ 2 )

[ (* jR M l (a + P )

( a + l ) a l a |(p - 2 )

“ |( a ) 0 - 1) 0 - 2 ) |0 - 2 )

, _ <*(<* + !) mMz ( p - l ) 0 - 2 ) "(3)

Variance (jij) = n2' - (u i1)2

[by 2 and 3]

-(4)

- + f a" < p -l) (p -2 ) U - l J

g f (g + l) a |" ( p - l ) L ( p - 2 ) ( P t f J

_ g(a + p- l )

~ (p- l )z (p-2)

an d S.D . a — ^ V a r i a n c e

_ 1 /a (a + p - 1)~

" 0 - D V ( P “ 2 )

5.28 EXPONENTIAL DISTRIBUTION:

A random variable X is said to be an exponential distribution w ith parameter X > 0, if its p.d f. is

g iven b y :

/«- f Xr": xS00 ; otherw ise ...( 1 )

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464 | En g i n k f r t n g Mathematks-W

The graph o f expone ntai) distribution is

R a m s r k : The total area o f exponential distribution is unity.

<X>

So lutio n : We hav e, total Area = 1 f { x ) d x

= j Xe- dxo

[by definition]

[by distribution]

.-Xx

~x

/ (* > =

5.29 DISTRIBUTIVE FUNCTION OF EXPONENTIAL DISTRIBUTION:

The p.d.f. of exponential distribution is given by

Xe-**; x2Q0 ; otherw ise ...( 1 )

By definition o f distribution function p r C .D .F.:

X

F ( x ) = P ( X < , x ) = ! f( x )d x

0 *

1 f (x )d x + J f (x )d x —eo 0

= 0 + J Xe^xxxix [by 0)1

[■j i

: J _

1]

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S t a t i s t i c s I 4 65

Graph o f distribution Function is

Hence distributive fu nc tion :

F ( *) = 1 - er*', jc > 0, X > 0

1, JC= ao

0, jc < 0

5.30 MEAN, S.D. AND VARIANCE OF EXPONENTIAL DISTRIBUTION:

The p.d f. o f exponential distribution is given by :

m _ j le-**; . x z o , X > 0 .

i. 0 ; otherw ise ...( 1 )

To find mean :

We have, Mean = ji|* = E ( jc )

= / x f ( x )dx [by defh.]

! x. Xe-^dxo

= XJC.

-X

\00 oo -Xx- S ^ - d x

/o-X

[by ( 1 )]

0 + I e -^dx o

i~Xx

-X

- X ( ° - !>= X

Hence1

jii = mean = ^ ••(2 )

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4 6 6 j E n g in e e r in g M a t h e m a t ic s -III

To find. S.D. and Variance :

The second m oment about origin

- J x \ f ( x ) d x

= J x2.Xe~** dx

- X J x2.e~k*dx

= X2 e

x .

- x L 0 ”

; 0 + 2 J xe~u dx 0

= 2

_ 2 _ ■ 00I

0

I *

„ \ / 0 _

Hence

We have

' X* ( 0 ~ ° ~ X2

2

“ X2

Variance fi2' - ^ 2 - (Mi')2

2

-*-«r

and S.D. : a - 1/ V a r i a n c e ~ T

[by (i)I

...(3)

[by 2 & 3]

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5.31 MEAN DEVIATION ABOUT MEAN OF EXPOBEBTIAL DISTRIBUTION:

The p .d f. o f an exponential distribution is given by

Xe-**; x'i.O and X > 0


/ < * ) =0 ; otherwise

We know that mean deviation about mean is given by

M .D . = J l x - m e a n | /( * ) d r

- . ( I )

[by ( 1 ) and mean ~ jj-]

= J |Xjc —1[ er^dx 0

Puting Xx = t => Xdx - dt a nd limit t -> 0 to «>, we have

go

M.D. = f i \ t - \ \e -<dt

Since I '- H = - . ( 2 )

i

X

f 1 °o

] (1 - i) e~ld t + J (* - 1) e ' d t o l

[by 2)

= X [ - ( 0 - D + ( O o - ( 0 - 0) - ( * r O r ]

X [1 + ( <r , - l ) - ( 0 - <H )J

* r , X r ‘

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468 | E n g in e e r in g Mathematics-411

Example 5.70: A random variable X has an exp onen tial distribution w ith p.<Lf. is given by.

m . I[ 0, i f x SO .

Com pute the probab ility th a tX is no t less than 3. A lso fin d , m ean and standard deviation

and prove tha t coefficien t o f variation is unity.Solution : (i) Compute P (X K 3) :

We have P (X * 3) = P (X > 3)

00

= / f ( x ) d x 3 '

00

= J 2 e-2* dx

3

= - [0 - e~*] =

(ii) To fin d m ean :

We have, mean = fi| ' = E (x)

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(iii) To fin d S.D . :

We have,

_ I2

— 5 ( 0 - 1 )

H:’ = 1-C O

t <x>

= J x2. 2e-2* cfr

= 2 J x 'e r^dx 0

= 23- e

-2x V -2x- ! 2x S ~ d x"2J o o ~2

2

( - 2 )0 - 2 J xe 2xd x

o

CO

= 2 / x<r*<ic

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4 7 0 | En gineer ing M athem at ics - !! !

- j er^dx o

e-2*)*

- 2 Jo

= - 1 (0 - 0

We know that,

S . D . : a - ^jj-2 ' - ( m ') 2

f W

= / H IV2 4

- U H = I~ V 4 ‘ 2

and Variance = ^ Ans.

To find Coefficient o f Variation :

We have that, coefficient of variation =

1/2 „ _ ,= 1 7 2 = 1 . ^ v e d .

Example 5.71 : The Income tax X, o f a man has an expo nential distribu tion w ith p .d f. is given by

f ( x )

0 ; x < 0

I f incom e ta x is levied a t the rate o f S%, what is the probability th a t h is Income exceed

Rs. 10,000 ?

Solution : If income exceed Rs. 1 0 , 0 0 0 , then income tax will be :

Income tax = 1 0 , 0 0 0 * = 5 0 0 .

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Hence tax o f a m an will exceed Rs. 500.

required prob ability:


Example i

Solution :

P (X > 500) = \ f ( x ) d x

500

00

= J 7 e~*IAdx [by given p.d.f]600 4

i [ l -*n y4 1 - 1 / 4

aO

I

500

= _ (0 - e-[”)

= e~]2S Ans.

5.72 : Suppose Ike life o f mobile batteries is exponentially distributed with parameter

X —.001 days. What is the probability that a battery will last more then 1200 days ?

The p.d.f. o f exponentia l d istr ibutio n :

f i x ) - Xe~^ \ x > 0

0 ; otherw ise ...( 1 )

where X, = 0.001 days.

P (battery will last more than 1200 days)

= P (X > 1200)

= I f ( x ) dx 1200

J h r ^ d x

1200

( * rV J\2<

[using ( 1 )]'1200

=_ (0 _ {j-XxIMO)

: -0.001*1200 [X = 0.001]

e~12 « 0.301 A ns.

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5.32 TEST OF SIGNIFICANCE

The statistical constan ts for the given po pulation such as m ean (p ), stan dard deviation (a ) etc. are

called the parameter's. Similarly, the statistical constants for the sample drawn from the given

populat ion such as mean (ic ) , standard dev iation (j) etc. ar e ca lled th e 'statistics'. In general, the

populat ion parameters are no t know and their estim ates given by the corresponding sample statisticare used.

It is necessary to test the statistic to see whether their difference is significant or not. Such tests

are called the 'tests o fsignificance'. It can be used to com pare the characteristics o f two samples

o f the same type. For applying the test o f significance we first setup a hypo thesis.

5.33 NULL AND ALTERNATIVE HYPOTHESISA hypothesis w hich is tested for possible rejection u nder the assum ption that is true is called null

hypothesis, and is denoted by Ho-

In other words, a null hypothesis Ho that there is no variation exist between variables or set of

given observations.For example, The null hypothesis H0 m ight be that there is no relationship betw een two measured

phenom ena.

The hypothesis which is accepted, w hen the null hypothesis has been rejected is called the altenative

hypothesis, and denoted by H) o r HA

For example-, in a clinical trial o f a new drug. Then, w e have

Null hypothes is Ho : there is no differenc e betwe en the two drugs on average i.e; the new drug is

no better on average, than the current drug.

Alternative hypothesis H\ : the two drugs have different effects on average, i.e;

the new drug is better on average, than the current drug.The purp ose o f hypothesis testing is not to qu estion the com puted value o f the sam ple statistic but

to make a judgm ent ab out the difference betw een that sam ple statistic and a hypothesized population

parameter.

534 LEVEL OF SIGNIFICANCEThe probability level below which are reject the hypothesis is known as the level o fsignificance.

The level o f significance usually em ployed in testing o f hypothesis are, 5% an d 1%. W ha t, if we

test a hypothesis at the 5% level of significance? T his mea ns that w e w ill reject the null hypothesis,

if the difference betw een the sam ple statistic and population param eter is so large or a larger

difference would occur, on the average, only fine or fewer times in every 10 0 samples when the popu lation para m eter is co rrec t. Assuming th e hypothes is is co rrect, then the sign ifican ce level

indicates the percentage o f sam ple mean tha t is outside certain limits.

In this section we shall study the fo llowing four tests o f significance only, viz :

(i) x2 - test, (i i) t - test, (iii) F - test and (iv) Z - test.

5.35 CHI-SQUARE TEST OR x2 - STATISTICThe C hi-square test is a very pow erful test for testing the significance o f the discrepan cy between

the actual (or observed ) freq uencies and the theoretical (o r expe cted) frequencies in a sample, ft

was given by prof. Karl Pearson in 1900 and is know n a s "Chi-square test o fgoodness offit". The

Chi-square is a letter o f the Greek alphabet w hich is written as %2.

472 | En g in e e r in g Ma t h e m a t j c s -JH

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Let f 0 and f be a set o f observed an d expected frequencies of a class interval or cells. Then Chi-

square distribution is de fined as


X2

Where 'Lf„-'Lft = N= total frequencies

Another Form o f %z distribution :

i f o - f e ) 2

- ( 1 )

Since X2 = 2k / w . r l

I I [/„*

I '•

- I { £ - % ♦ / . [

f 2/o

4

f 2 *0

fe

- 2 N + N

or <<? -AT

[ v i / 0 = i / , = a o

...(2 )

5.36 PROPERTIES OFx2 DISTRIBUTION :

(i) The values o f %2 are always positive, i.e; x 3-c u rv e is alway s positively skewed.

(ii) The lowest value o f %2 is zero an d the highest va lue is infinity.

(iii) If degree o f freedom v = 1 , then x ^ u rv e will be y = y 0e~* , w hich is the exponentail

distribution as show n the following figure

(iv) If v > 1, then x* -curve is tangential to X -axis a t the origin and is positively skewed.

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474 ( E n g i n e e r i n g M a t h e m a t ic s - !! ]

(v) The probability P that the value o f y} from a random sample will exceed xi >s gives by

P = J ydx.

*0

537 CONDITIONS FOR THE VALIDITY OF x*-TEST:

(i) The mem bers o f sample should be independent.

(ii) Co nstrains on the cell frequen cies should be linear.

(iii) Total freque ncies N shou ld be reason ably large say, greate r than 50.

(iv) As x^-tes t is based wholly on samp le data, no assump tion is ma de about the population

values (or parameters).

(v) The expec ted frequencies should not be less than 5, If it is less than 5, than adjacent it.

(vi) x 2-test wholly depen dent on degress o f freedom (do/.) v.

538 USES OR APPLICATION OF x*-TEST:

(i) Test o f goodness o f f i t : Ch i - square s t a t i s t i c may be used to de te rmine whe the r

two curves are fitted good or not.

(ii) Test o f independent o f attributes in a contingency table test for a specified standard

deviation.

(iii) Test fo r a s pec ified standa rd deviation i .e; i t may be used to tes t o f popu lation

variance.

(iv) Com pare a num ber o f frequency distributions.

Remarks on the x*-Test :

1. Cells : I f given data are arranged in the com partments, which is called cells or class interval

and frequency o f a cell is called cell frequency.

2. Contingency Table : Let a group having N persons is divided into two attributes A and B. A

having m classe s say, A lt A 2........A , .........Amand B having n classes say, Bu B2, Bi t ......... B„ and

/oij are observed frequ encies o f celles belonging to classes A, and Bj.

A \ B B , B2 .... ...... Bj ....... ...... B„ Total i

A| f \ l f \ 2 • — ...... A j ......... / .

a 2 fix h i ...... ...... f u ......... ....... f u f i

A fix f a ...... ...... f j .......... .......U f .

A* fm\ fm2 ...... fm

Total Ft Fi .......... Fj ....... .... Fn N

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3. Calculation of Expected Frequencies :

Consider the 2 * 2 contingency table :


A \B B , b 2 Total ^

A, / . h N,

a 2 h U n 2

Total H i N a iV (Grant Total)

r

The exp ected (or theoretica l) frequen cy o f cell 11 is given by

f t \ \ -

Expected frequency o f cell 12 is :

f t \2

Expected freque ncy o f cell 21 is

Expected frequency of cell 22 is :

fe\ \ - N

-------- J T

n 3 * n 2 M i N

f - ^ 4 * ^ 2U 22 N

4. D egree o f freedo m (dof) : T he number o f independent variates is called the number o f degree

of freedom (dof) and denoted by v.

(i) In Binomial distrib utio n:

dof v = n - 1 ,

(where n is total sam pling, for example x : 0, 1, 2, 3 ,4 then n = 5 - 1 = 4 )

(ii) In poisson distribution :

dof v - n - 2

(iii) In No rmal distribution :

dof v = n - 3 .(iv) For m * n contingency table.

do f v = (m - 1 ) ( n - IX

where m = num ber of rows, and n - num ber o f columns.

W orking R ule of x1- Test :

Step 1. Consider the Null Hyp*, s is H 0 : N o association exists between the attributes.

Alternative Hypothsis H \ : An association exists between the attributes.

Step 2. Calculate the expec ted (or theoretic) frequency / „ corresponding to each cell.

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476 | E ng ineer ing M athem atics- IH

Step 3. Calculate x2-statistic by the formula.

and calculate the degree o f freedom v.

Step 4. See the value o f x 2 from the table, i.e; value o f at u.% level o f significance and for thedof v, as calculated in step 3 .

Note : If no valu e o f a % is mentioned, then take 5% i.e; a ~ .05

Step 5. Decision :

(i) If calculated value of x~ < tabulated value o f x i >•

Then accept the null hypothesis H0.

(ii) If calculated value o f x 2 ^ tabulated value of xi,*

Then rejected null hypothesis l lQi.e; accept the alternative hypothesis H\.

Example 5.73 : From the table given below, whether the colour o fso n ’s eyes is associated with that of fa th er's eyes ? Given that the value o f x 2 fo r / do f at 5% level of significance is 3.841.

Eye Colour is son ’s

Not Light Lig ht

Eye colour Mot Light 230 148

in father Light 151 471

Solution : Step 1 : Null Hyothesis H0 : Th e colou r o f the son ’s eyes is not associate d w ith the colour

of father's eyes.

Step 2 : Calculation o f Expected Frequencies f < :

Given the Observed frequ encies/,// are :

Not Light Light Total

Not Light fo\\ = 230 f , n = 148 378

Light foi\ - 151 i v i ! 622

Total 381 619 N « lOUO

The Expected frequencies will be :

Not Light Light

Not Light

Light

3 7 8 x 3 8 1 . . .

A " 1(X)0

3 8 1 x 6 2 2 _

/<21 ' 10 0 " 237

. 3 7 8 x 6 1 9

1000

6 1 9 x 6 2 2 _

} ' n 1000

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Stat is t i cs | 4 7 7

Step 3 : Calculation ol' x : -stutisiu- •

^ (fo fc?We have

or

x- =

2 = < 230 -144 )* (148 - 234 )2 (151 - 237 )2 (471 -3 8 5 )21 A A O O A O l3 ’7 g g g144 234 237

- 51 .36 + 31.61 + 31.21 + 19.21 = 133 .29.

and degree of freedom

V - ( w I ) I )

- ( 2 1 ) ( 2 I ) [ 2 ^ 2 c o n ti n ge n c y ta ble ]

-- I.

Step 4 : The calculate value of X2 at 5% levels o f significance and for 1 do f is 3.841 i.e..

Xno5.i “ 3 84 I

Ste» S : Decision : Cle arly calculate value o f x ’ - 133.39 ^ tabulated value of Xnovi - 3 84

ro the Null hypothsis is rejected i.e.

there is an asso ciation between thfe colours o f eyes o f son's and c olours o f eyes of

father's. Ans.

xample 5.74 : In a sample survey o f public opinion, uswers to the questions fi) Do you drink? (ii)

Are you in favour o f local option on sale o f liquor ? are tabulated below :

Question

Yes No Total

Yes 56 31 87

No 18 6 24

Total 74 37 U I

Can you infer whether or not the local option on the sale of liquor is dependent on

individual drink ? (Given that the value o f z 2 fo r do f at 5% level o f significance is

3.841.)

o lution : Step I : Null Hypothesis : The option on the sale o f liquor is not dependent {or not

associated) with the individual drinking.

Step 2 : Calculation o f expected or theoretical frequencies f e

The exp ected frequenie.s. co rrespond ing to the observed frequencies are calculate as follows

Expected frequencx o f cell 11 is

8 7 x 7 4

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Expected frequenc y o f cell 12 is

4 7 8 | E n g in e e r in g M a t h e m a t j c s -111

f<" = - t i t 5=29

Expected frequency o f cell 21 is

_ 7 4 x 2 4

~ 1 1 1 ~

Expected frequency o f cel! 22 is

x 24x37U i ~ m - 8.

Step 3 : Calculation o f statistic :

We have x : “ ^ ! ^ ~ - (1>e

fa f .i f o - f e f

U

56 58 4/58

31 29 4/29

18 16 4/6

6 8 4/8

Total z ! . m , o _937

Hence Value o f j } = 0.957

Also degree o f freedom (dof) v - (« - 1) (n - 1)

= (2 - 1 ) ( 2 - 1 )

= 1 .

Step 4 : The tabulated value o f %2 at 5% level o f significace and for 1 d o f is 3.841 is 3.841

X0.05,1 = 3.841,

Step 5 : Decision: Clearly calculated value o f x2 = 0.957 < tabulated value o f x2o.os.i =3.841.

=> die Null hypothesis is accepted.

=> Sale o f liquer is no t dependent o r not associated w ith the individual drinking. Ads.

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Example 5.75: SO students selected at randomfrom 500students enrolled in a computer crash programme

were classified according to age and grade points giving the following data :

S t a t is t ic s 1479

Grade Points Age (in years)

Total20 and under 21-30 above 30

up to 5.0 3 5 2 10

5.1 to 7.5 8 7 5 20

7.6 to 10.0 4 8 8 20

Total 15 20 15* l i *

Test at 5% level o f significance the hypothesis that age an d grade po int are Independent

(Given that Xaos,i = 5.99J).

Solution : Step 1 : Nu ll Hy pothe sis H0 : The ag e and grade po int are independent.

Step 2 : Calculation o f expected frequencies f t :

The expected frequencies are as follows :

G rade P o i n ts 20 an d u n d er 2 1 - 3 0 above 30

upto 5 o

* w 1 1 U J 2 0 x 10 .

5 0 4

1 5 x 1 0 .

50 3

5.1 to 7.51 5 x 2 0 ,

50 “ 6

2 0 x 2 0 '

5 0 8

1 5 x 2 0 ,

5 0 = 6

7.6 to 10.0 1 5 x 2 0 ,50 6

2 0 x 2 050

1 5 x 2 0 .50

From the abov e table, we see that some values o f expected frequencies are less than 5, so

that to amalgam ate these cells to its neighbours (i.e; m erge 1 & U rows).

Then new obseve d and expec ted frequenies table are :

Expected frequencies table (ft) :

G rade P o i n ts 20 a n d u n d e r 2 1 - 3 0 above 30 I b t a l

upto 7.5 9 12 9 30

7.6 to 10. 6 8 6 20

Total 15 20 15 Z I I g

Observed f requencies t ab le {f0} :

G rade P o i n ts 20 & u n d e r 2 1 - 3 0 above 30 Total

upto 7.5 11 12 7 30

7.6 to 10.0 4 S 8 20

Total 15 20 15 N = 50

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We have, x 2 ~

n r v» - i ( 1 2 - 1 2 )2 ( 7 - 9 )2 ( 4 - 6 )2 (8 - 8)2 , (8 - 6)2

X 9 12 9 6 8 8

= 2.222.

Also degree of freedom . (dof)

v = (m - 1 ) (n - !)

= (2 - 1) (3 - I) [ v As new table]

i.e; d o.f. • v = 2

Step 4 : The tabulated value of x2 at 5% level o f significance and for d o f v = 2 is 5.991 i.e..

X&.m 2 = 5.991.

StgP-5 : Dec ision : Clear ly ca lcu la ted va lue of x 2 = 2 .222 < tabulated value of

XSoj.2 = 5.991

=> Null hypothesis ts accepted.

=> the grade points and age are independen t o f each other. Ans.

Exam ple 5.76 : In experiments on pea-breading, mendal obtained the fo llowing frequencies o f seeds:

4 8 0 | E n g i n e erin g Ma t h e m a t ic s -111

§ t s n l : Calculation o f x2-statistic :

Round and

yellow

W rinkled and

Yollow

Round and

Green

W rinkled and

Green

Total

315 101 108 32 556

Theory prtod lcts tha t the freque nc ies shou ld be in propo rtion 9:3:3:1. Exam ine the

correspondance between theory and experiment Given tha t the value o f x 1 fo r 3 d o f at

5% level o f significance is 7.81 S.

So lution : Step t : Null Hypothesis H0 : there is a correspondence between theory and experiment.

: Calculation o f Expected frequencies ( fj :

Given f requencies in p ropor tion 9 : 3 : 3 : 1

Total sum o f proportion = 9 + 3 + 3 + 1 = 16, therefore

9(i) Expected frequency o f Round and Yellow seed is = j g * 556 » 313

g(ii) Expected frequency of Wrinkled and Yellow seed is = j g * 556 * 104

3(iii) Expected frequency o f Round and Green seed is - j g * 556 ~ 104

(iv) Expected frequency o f Wrinkled and Green seed is = * 556 * 35.

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Here f 0 = observed or actual frequencies

f e = expected o r theoretical frequencies

r , _ ( 3 1 5 - 3 1 3 ) 2 ( 1 0 1 - 1 0 4 ) 2 ( 1 0 8 - 1 0 4 ) 2 ( 3 2 - 3 5 ) 2

X 31 3 104 104 35

^ 4 9 , 16 , 9

313 104 104 35 c

A ls o th e de gre e o f freed om v = n - l = 4 ~ 1 = 3 .

Step 4 : The tabulated value of x 2 W 5% level o f significance and for do f v = 3 is 7.8 15

X005.3 ~ 7.815

Step 5 : C alculated value o f %2 = 0.5 < tabulated value o f x 5.os,3 = 7.815=> Null hypothesis is a accepted.

there is very high degree of correspondence between theory and experiment. An s.

Example 5.77 : A servey o f 320fa milie s with S children each revealed the fo llowing distribution :

Stat is t i cs | 481

Step 3 '.Calculation o f %2-statistic :

No. o f boys : 5 4 3 2 1 0

No. o f g ir ls : 0 I 2 3 4 5

No. o f families : 14 56 n o 88 40 12

Is this result consistent with the hypothesis that the male and fem ale births are equally

probable ? (Given zks, *-s * 10.07).Solution : Step 1 : Null Hypothesis Hq : M ale and fem ale births are equ ally propable.

S t e a l : Calculation o f Expected frequency f t :

From the Binomial distribution, the expectedfrequency:

f (r) = N. "C, q^'p ”', r = 0, 1, 2, 3, 4, 5.

Here N = 320, p = q = ^ (equal probability) and n = 5.

For r = 0, then (1) becomes :

/ , ( 0 ) = 3 2 0 *5C 0 ( ! ) 5 ( ! ) ° “ 10

For r = 1, then

/ , ( ! ) = 320 * ^ (I ) ' ( I / = 5 0

For r = 2, then

f t (2) = 320 x JC2 ( i ) ( i ) = 100

f (3 ) - 320 x SC3 ( ! ) = 1 0 0

For r = 3, then

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4 8 2 1ENQMemiG M/u t c m/o t c s-III

For r = 4, then

For r ~ 5, then

/ . (5) = 320 x SC5 ( | ) ° ( | ) 5 = 10.

Hence expected frequencies for r = 0, 1,2, 3,4 , 5 or 5,4 , 3,2 , 1,0 a re :

/ , - 10, 50. 100, 100, 50, 10 :

Step 3 : Calculation o f z 2-statistic :

/ , (4) = 320 x !C4 ( i ) 4 ( i ) ' =50

we have = Z f e

/ . / , < / ; - / . ) * ' <f*- f .W

14 10 1.60

56 50 36 1.72

110 100 100 1.00

88 100 144 1.44

40 50 100 2.00

12 10 4 0.40

Totalf( /o- /«)*!

’ = 7.16%2 = Z fe J

Also degree of freedom for BD. v " 6 - 1= 5.

Seto4 : The tabulated value of %2 at 5% level of significance and for dof v - 5 is 11.07

i.e, $ 05,5 ■ it.07

StenS : D ecision: Clearly calculated value of %2 * 7.16 < tabulated value o f xS.05,5 * 11.07

=> Null hypothesis is accepted

=> the male and female births are equally probable. Ans.

Example 5.78: In 60 throw s o f a dice, fa ce one turn ed up 6 tim es, fa ce two or three, 18 tim es, face

fo u r o r fiv e 24 tim es and fa ce six, 12 tim es, Test a t 10% level o f significance, i f the dice

is honest, it being gives tha t P (z * > 6*25) ~ 0,1 fo r 3 degrees o f freedom .

Solution : Step 1 : N ull Hypothesis H q \ The dice is honest

Step 2 : Calculation o f expectedfrequenies f , :

Given that the observed frequency are as follows :

Face o f dice x or (r) : 1 %or 3 4 or 5 6observed frequency f a : 6 18 24 *12 (N = 60)

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Under the assumption of Ho that the dice is honest, so that expected frequency for eachface is

/ . W “ N x g - 6 0 x i = 10 .

[Because prob. of turning up any one of the numbers J, 2,3,4, 5,6 is g]. i.e.

x or r : 1 2 or 3 4 or 5 6expected frequency f t 10 20 20 10Step 3 : Calculation o f£ sta tistic :

Wehave — s -

Tkble for x2-statistic :

S t a t i s t ic s J 483

, - E

Face of dice x or r / . feu

I 6 * 10 16

2 or 3 18 20 0.24 or 5 24 20 0.8

6 12 12 0.4

Total tf-6 0 JV*60 x2= £ ' \ i fo ~ f e )2 f e r 3-0

Thus, Calculated value of %2 * 3.0.Also degrees of freedom v = 4 - 1=3.

Step 4 : Since given that tabulated value of v3 at 10% level of significance and for 3 dof is6.25 i.e. xS.i,v»3 “ 6.25.

: D ecisio n : Clearly calculated value of x2 “ 3.0 < tabulated value ofXS.i.3 - 6.25.=> Null Hypothesis is accepted => The die is honest Ans.

mple 5.79: Calculate the expected frequencies fo r the follow ing data presum ing the two attribu tes

Condition o f home

Gear D irty

Condition d e a n 70 50

o f child F airly clean 80 20 D irty 35 45

UsecM* quaretestat5%lmldf9ignyteancelosm6 whethcrthetwo attributes areindependent

(table value o f£ at 5% for 2 dcf it 5,991, end for 3 dof is 7,915and fo r 4 d o f Is 9.488).

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484 | ENQBoramo Mm h e m m k s -111

Solution i Step 1 : N ull H ypothesis H q : No association between the attributes.

Step 2 : Calculation o f expected frequency ( fj o f each c e ll :

Clean Dirty Total

Clean

Fairly Clean

Dirty

fo : 70

120x185300 1

/0:8 0

185x100*e 300 1

fo : 35

/ • - 1SIZom = 49*33

f o ' 50

, 115x120 „ f t = ~ m " = 4 6

/ . : 20

115x100300

fo *45

, 115x80 ____

300 = 30-67

120

100

80

Total 185 115 N = 300

Step 3 : Calculation o f x 2 -statistic :

_ (7 0 -7 4 )2 . (50 - 46)2 . (8 0 -6 L 6 7 )2 . (20-3& 33)2X2 “ - 7 4 - ’+ — 4 6 - + ~ L67— + “ W ‘

t (35 -49 33 )2 (20 - 38.67)2

4933 + 30.67

•» 0.2162 + 0.3478 + 5.4482 + 8.7657 + 4.1627 + 6.6954

= 25.636.

Also degree o f freedom (d.o.f.):

v = (m „ I) (n - 1) = (3 - 1) (2 - 1) = 2.

Step 4 : Since the tabulated value of %2 ** 5% level of significance and for dof

v = 2 is 5.991, i.e; xfl.ov * 5.991.

Step 5 : D ec ision : Clearly, calculated value of x 2 * 25.636 tabulated value ofX&.05,2 = 5.991=> the Nufl Hypothesis is rejected.

=> there exists and association between the attributes. Ans.Example 5.80 : The follow ing fig u re s show the distribution o f digits in num ber chosen at random

fro m a telephone d ire ctory :

D ig its: 0 1 2 3 4 5 6 7 8 9

F reu ency: 1026 1107 997 966 1075 933 1107 972 964 853.

Test a t 5% level wh ether the digit* r m y b f Ufken to occur eq ually frequ en tly in the

directory. (The tab le w ine o f fo r 9 degree o f freed om m 16.919)

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S t a t is t ic s I 485

Step 1 : Null Hypothesis H o: The digits occur equally frequently in the directory.

Step 2 : Calculation o f expected frequency ( f j : V

Given that total observed frequencies N * l f Q“ 10 ,000.

Expected frequency for each digits 0, V, 2 , ..... 9;

f . { r ) ~ N * p { r ) I *

10,000 x ~ * 1000;

Step 3 : Calculation o f -statistic :

We have, f2 *- I

p (r ) = t o ]

for r = 0, 1 , 2,..., 10.

x o r r f . / . .( f o - f e ) *

fe

0 1026 1000 0.676

1 1107 1000 11.449

2 997 1000 0.009

3 966 1000 1.156

4 1075 1000 5.625

5 933 1000 4.489

6 1107 10Q0 11.449

7 972 1000 0.7848 964 1000 1.296

9 853 J m . . 21.609

N = 10,000 N = 10,000k / w , ) 2 !

■ jk- — 58.542le

Also degree of free& ^ v ^ 10 - f 1* 9.

Steiy 4 : Tfife tabulated vafue o f x* a?5% level of significance and for dof v = 9 is 16.919,

i-e;$o5,9 16.919:

Step 5 : D ecision : Clearly, calculated value of x2 * 58.542 * tabulated value of xios.9 =

16.919.

=> the Null Hypothesis is rejected.

=> the digits do not occur equally frequently.

Example 5.81: A d ice is tossed 120 time* w ith the follow ing resu lts:

N um ber tu rned u p : 1 2 $ 4 5

F reuency: 30 25 18 10 22

Test the hypothesis tha t the d ice is unbiased ( fi ts , s m 11*07)

Ans.

6

15

Total

120

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Solution : Step I ; Nutt Hypothesis H q : The dice is unbiased oat.

Step 2 : Calculation o f expected frequency : Here N 31120.On the basis of H q , the dice is unbiased* so that the expected frequency

/ , ( r ) - N p ( / 0 - 120 » | - 20; for r - 1,2 ,....6

i.e; expected frequencies for each turned up r * 1,2 ,3 ,4 ,5 ,6 is 20.Step 3 : Calculation o f s ta tis tic :

4 8 6 { E f in n rm a ig M a t h e m a t ic s - i n-V

We have

Table for x2-«tatistic

... (1)

xorr / . f . C f . - f # ( f o - fe )

fe

1 30 20 100 5.002 25 20 25 1.25

3 18 20 . 4 0.20

4 10 20 100 5.00

5 22 20 4 0.20

6 15 20 25 1.25

Total AT-120 tf»120 — X2 - 12.90

Also degree of freedom v = 6 - 1=*5.

Step 4 : The tabulated value ofx2 at 5% level of significance and for dof v=5is 11.07, i.e; Xtas = 11.07.

Step 5 : Decision: Clearly, calculated value of x2* 12.90 K tabulated value of xS.os.j * 11 07. =>the Null Hypothesis is rejected.

=> the dice is a biased one. Ans.

Example 5.82 : Front the adult male population c f seven large cities random sample giving 2 x 7

condnguency table cf married andunmarried mem ra given below were taken. Can it be

said that there is a significant variation among the cities In the tendency o f men to

marry T

City A B C D E F G Total

Married 133 164 155 106 153 123 146 980

Unmarried 36 57 40 37 55 33 36 294

Total 169 221 195 143 208 156 182 1274

(At (2 —1) (7—1) d.f. Hike 12.6)

Solution : Step 1 : Null Hypothesis H0 : There is no significant variation among the cities in the

tendency of men to many.

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Step 2 : Calculation of Expected frequency: On the basis of Null hypothesis the expected frequencies are:

980Expected number of married people in City A * J2 7 4 x *69 = 130

980Expected number of married people in City B * J2 7 4 x 221 = 1 7 0

980 “Expected number of married people in City C * J2 7 4 * = 15®

980Expected number of married people in City D ~ x =

980Expected number of married people in City E ~ J2 7 4 x 208 - 160

980Expected number of married people in City/*= 1 2 7 4 x 156 ~ 120

980Expected number of married people in City G m J2 7 4 x 182 = *40

Similarity the Expected frequency are 39, 51,45,33,48,36 and 42.

T a b k for expected freqieacy f4

Married 130 170 150 110 160

Unmmarried 39 51 45 33 48

Step 3 : Calculation of -statistic :

S t a t is t ic s 1 4 8 7

120

36

We have,

140

42

/ . A V . - f # fe

133 130 9 0.069164 - - -THT 36 0.212

155 150 25 0.167106 110 16 0.145153 160 49 0.306

123 120 9 0.075146 140 36 0.25736 39 9 0.23157 51 36 0.70640 45 25 0.55637 33 16 0.48555 48 49 1.021

33 36 9 0.250

36 42 36 0.857

N= 1274. N** 1274 X2 = 5.337

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4 8 8 | En g m e e m n g M a ih e m * i i c s -H1

m m i • Th* tabulat&d value of x2 a* 5% level of significance and for v = (2 - 1) (7 -1) =6 dof is 12.6 i.e., xfl.os,6 * 12.6 .

Step 5 i Decision : Clearly calculate value of x2 * 5.337 < tabulated value of X2 * 12.6=> the null hypothesis is accepted.=> these is no significant variation among the cities in the tendency of men to marry.

Example 5.83: For 2 * 2 contingency table

(o + b + c + d) (ad - be) 8 prove that, z - (a+ b)fr+ ,rf(b+d) ia+ cy

Solution : Given the observed frequency (fa) table :

Total a b i a + b |

c d c + d

Total *+-c b + a N = a + b + c + d

Table for expected frequency f e :

(o + 6)(a + c) a+ b+ c+ d

(c+<t)(a+c)

Wehave,

q + 6 +e+d

(a+6)(6+d)o+6+c+d

(c+d)(]b+d)o + 6 +c+d

a w . ) 2!

r (o+6)(«+c)i2 E l a+ b+ c+ d J

( q + 6 ) ( q + c )

q + 6 + c + d

- V' [q2 + ofr+qe-fad-q2 - a c - a b - ^ (q + 6) ( q+c) (q + 6 + c+ d)

mV (ad-be)2______( q + 6) (q + c) (in- 6 + c + d)

[As four terms]

[As four terms]

Now open the summation for four terms, we have

r 2 -(qd-fcc)

(q+ 6 +c+L_r____ i d) L(q+W

J* b)(c * c) (q + b)(b+ d) ( c+d)(a + c) (c+d)(b+d)

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S t a t is t i c s 1 4 8 9

{ad-be)2 f f b+ d+ a+c[ l ( a + 6Ko+ c)(b+d)

\ . f b+d + a + c 11

r U c + d X o + cXb+d)!}( a + b + c+ d ) Ll(a+&Xa+c)(t>+d)J ' l(c+dXo+c )(6

- ( a d - b e y [ ( a + b ) ( o + c ) ( b + d ) + f c + d X « + c ) ( 6 + d ) ]

_______ c + d + a + b _______ 1

(a +b)(c+d)(b* d)(a + c) J(a+b+c+d) (ad-bc)2

(a + b)(c+d)(b+d)(a+c) Proved

Example 5*84: Prove that fo r a 2 * n contingency table:

x ^ f i + f t where fi, f 2 are two frequencies in a subgroup and Nb are the marginal sums of two rows.

Solution : Table for observed freqaeacy / , :----- 1-----------

Marginal sum (Total)

f \ »••• N,

h - N*

Marginal sum(Total)

f i + f i N, + N2

The expected frequency corresponding to f , :

( / . + / 2) * i

n , + n 2

( / l + / 2) * 2

n , + n 2 '

We have %2 = £ ^ y - -

Taking value of chi-square for the first column are :

{ f i+ f i )N\ Nt +N2

C/i + fi ) ^ 2

n x+ n 2

[v by (I)]

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4 9 0 1 En g m k r m g M athem atics-!!!

= 1 v (a" 6)2 = <*'

iVi(/i+/2)W + N 2) W2(/, +f2XN1+Nt )

,XM M(A+/*)W+iV2) AW,

n , n 2 \&__-^-T1 2 LM N t l

( f l + f l )

Hence the value of chi-square for 2 * n table is

/ T > Proved./ l + / 2

Example 5.85 : /» jfte accounting department of a bank 100 accounts are selected at random and examined fo r errors. The following results have been obtained.

No. o f Errors 0 1 2 3 4 5 6 No. rtf Accounts 36 40 19 2 0 2 1 DoestUs information verify that the errors are distributedaccerdlng to Poissson probability tow 't (tbs, s m 11.07).

Sobttkm ; Stop 1 : Null Hypothesis Ho : The errors are distributed according to Poisson probability law.Stea l : Calculation of expected frequency (£):We know that the expected frequency of poisson distribution :

e~m mr fAr)=N- — 0, 1, 2,3,4, 5, 6 .

where N = y ~ 100

Zfr m = mean =

- Qx 36-flx40-h2xl9 + 3x2 + 4xQ-hSx2-f6xl

’ m ~ 36 + 40 + 19 + 2 + 0 + 2 + 1

_ 100

100 " *•

, 1° 100x0367

Expected frequency:

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For r = 2: / , (2) = 100 * ^ « 18[2

Forr=3: f. (3) - 100 x e->. S i - « 6

(i)4For r~ 4 : / , (4) » 100 x e~K «2

14

o )5

For r ~ 5 : f t (5) * 100 * e~l. "jj" = 0.306 » 0

0)‘For r - 6 : f, (6) * 100 x * 0.051 « 0

Step 3 ; Calculation of £ statistic :

We have X2 = -< l) /*

Since same expected frequencies axe less than 5, so that the frequncies for r =3,4,5 , 6 have

been pooled together (see in table).


For /• = 1 : ^ (1) = 100 x r i. ffi!. * 37

No. of

errors (r )

Obseved

frequency ( f , ) frequency (/ , ) f '

0 36 37 0.027

1 40 37 0.243

2 19 18 0.055

3 2 1 6 1

4 0 2 1.125

[ 5 1 85 2 0

6 1 J 0 J

Total iV* 1 0 0 //= 100 X2 = 1.450

Step 4 : The tabulated value ofy* at 5%level of significance and fordofv-6-1 =5 is 11.07

i.e., .*$.05, j —11.07.

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4921Enumiuumg MxiwMxncs-fll

S b a i : Decision: Clearly calculated value of x2 = 1-45 < tabulated value of xl.os, 5=11.07.

=> the null hypothesis is accepted.

=> the errors are distributed according to Poisson probability law.

539 /-STATISTIC AND/-DISTRIBUTION:

The statistic t was introduced by W.S. Gosset in 1908.Let x\7x2, jc* be the values of random sarnble of size n, drawn from a nonnal population with mean ji and variance a2 (unknown).Then we define the /-statistic as :

* -

where S* = O T l) (X'~ * y

1 £and Jc = sample mean - — Ex,.n i=lHere S is the standard deviation of sample with degrees of freedom v - n - 1, when

n < 30. If we calculate t for each sample, we obtain the sample distribution for t. This distribution known as student's t-distribvtion is given by

....................... H )Where y0 is a constant, so chosen that the total area under the curve is 1.

Remark: When random sample of size n is large, or [when population variance are known] that /-statistic define as:

where a 2 = ^ 2 (xt - x Y -

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<r

5 .4 0 C O M P U T A T I O N O R W O R K I N G R U L E O F l J H S T R f f i U T l O N :

(a) To test the significance of a mean (snail sample)Step 1 : Consider the Null Hypothesis ffo: There is no Significant difference is between the sample means and population means.Alternative Hypothesis H%: There is significant difference in between the sample means and population means.

S m s n c t | 493

Step 2 : Calcualte /-statistic :

where

When variance is known:

then

[when variance is not known]

- ( ^ i )

where o2 ** “ I (x -x )2. fwStep 3 : Level ofSignificance or Tabulated value of t : See the value of / from the table i.e.» value of t at a% level of significance and for the dof v.St£&4 : Decision:(i) If calculated value of |/j< tabulated value off. Then accept the null hypothesis H0.

(ii) If calculated value of J/|Af tabulated value of t. Then reject the null hypothesis Ho.(b) To test the s&rificaace of dtffcreacc hetweea two sample weans :

Let xu x2, xHi and y u yi* «•« y*2 be two independent random samples with means x and y and standard deviation S\ and $ from a normal population with the same variance. Suppose we have to test the hypothesis that the population means |ij and (i2 are the same, for it we calculate /-statistic as follows :

SSzH p* j 2 2 _1 S i ( n 1+n2) ’ "•'l>

where S2 - ------- ---------- s----- *Hi + n 2 - 2

with degrees of freedom v - n\ + t*i - 2 .When variances cr? and aj are known : Then

2 2 „ »i<*i + *2<*2

»1 + fl2 - 2 *

Remarks : When variance b unknown :1. For dof v and 95% confidence limits for p are given by :

cJ1 = 7 ± to.03 v>* “7*»

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2. For dof v and 99% confidence limits for \i are given by:

gM“ 5 ± /o.ei *»x "7“

Vit <When variance is known :

1. For dof v and 95% confidence limits for p are given by :

|i * jp ± 4mh« x 4n - 1

2. For dof v and 99% confidence limits for ji are given by:

ai i ssx ± y i » x

Example 5.86 : Find the student's t-statistic fo r the following variable values in a sample :- 4 , - 2 , - 2 , 6 , 2 , 2 , 3 , 3 ,

taking the mean of the universe to be zero.Solution : We have /-statistic :

/ - ...d)

Where = Z (* -x )2 ...(2)

4 9 4 1 En o o c b b m q M a t h e m a t ic s -III

Here n = 8, and mean of universe p - 0.

Serial No. Xx - X (*- 3c)1

1 - 4 - 4.25 18.0625

2 - 2 -2.25 5.0625

3 - 2 -2.25 5.0625

4 -0.25 0.0625

5 2 1.75 3.0625

6 2 1.75 3.0625

7 3 2.75 7.5625

8 3 2.75 7.5625

Zx« 2 49.5000

v T o x - Mean - ~ = f =0.25

71 o

and

From (1), we have

S ~ = J i p = 2.6591 » - l i 7

(025-0) V8 . 2.659

[by (2)]

Ans.

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Example 5.87 : A sampoo manufacturing company was distributed a particular brand of sampoc through a large number of retail shops. Before a heavy advertisement Campaign, the mean sales per sampoo was 140 dozens.

After the campaign a sample of 26 sampoo was taken and the mean sales figure was found to be 147 dozens with standard deviation 16. Can you consider the advertisement effective 7 (Given /***2 s m1.708)

Solution : Given: n - 26 (small samples)

mean * y * 147; S.D. Le.» ct= 16 (variance are known), and

degrees of feedom v *= n -1 * 26 - 1* 25.

StgflLl : Null hypothesis H q : There is no difference in between the sample means and population means, i.e., the advertisement is not effective Le., Hq : * 140.

\A : Calculation of/-statistic :I

We have t = —— —- [when S.D. a is known]

_ (147-140)^26----------- 16-----

= 2.19.

Step 3 : The tabulated value of / at 5% level of significance and for v * 25 dof is 1.708.

i.e., fo.os,25 = 1.708.

Step 4 : Decision : Clearly calculate value of |f| - 2.19 K tabulated value of

*0.03,25 = 1*708.=> Null hypothesis H q is rejected.

=> The advertisement is effective. Ans.

Example 5.88 : Ten objects are chosen at random from a population and their heigktsare found to be in inches 63, 63, 64, 65, 66, 69, 69, 70, 70, 71.

Discuss the suggestion that the mean height in the universe is 65 inches, given that for 9 dof the value o ft and 5% level of significance is Z262.

Solution : Given n » 10, (small smaples)

Universe mean : \i = 65 and variance is not known and degrees of freedomv * = n - l ” 9.

Step 1 : Null hypothesis H q : There is no difference between the mean height of sample and universe i.e., p » 65.

Step 2 : Calculate of /-statistic :

We have t - ...(1)

Where ' ...(2)71-1

Svftnsnc* (4 9 9

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4 M | En g b c h m g Ma t h e m a t i c s-IR

and mean x ~ Zx

n

67010

67.

X ( x - x r

63 - 4 16

63 - 4 16

64 - 3 9

65 - 2 4

66 - 1 , 1

69 2 4

70 2 470 3 9

70 3 9

71 4 16

670 Z(x- jc)^ = 88

From die table, then equation (2) becomes

* 9

Hence (1) becomes

= 3.13 inches

t = ( 6 7 - g y i Q =2Q24

S sp j : The tabulated value of t at 5%level of significance and dof v = 9 is 2.262 i.e., 0.05,9 = 2.262

Step 4 : Decision: Clearly calculated value of j/ j312 .0 24 < tabulated value ofto.05,9 - 2 .262.

=> the null hypothesis Ho is accepted.

=> die mean height of the universe is 65 inches. Ans.

Example 5.89: A salesman is expected to effect an average sales of Rs. 3500. A sample test revealed that as particular salesman had made thefollowing sales. Rs. 3700,3400,2500,5200,3000and 2000. Conclude whether his work is below standard or not Using 5% level of significance of

t and 5 dof is 2.015.

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Solution : Given n - 6, (small sample)

3700+ 3400 + 2500 + 5200 + 3000 + 2000 Mean of sample: J - ------------------------g------------------------

or 3c = = 3300 and n - 3500

Here variance is not known and degrees of freedom v = w- 1=5.

Step 1 : Null Hypothesis Ho : the salesman is up to standard.

Step 2 : Calculation of /-statistic :

We have t = - — ...(1)

Where S2 * ...(2 )71—1

~ S t a t is t ic s j 497

Sale in Rs. x x — jg 1 * 1

&

3700 400 160000

3400 100 10000

2500 - 800 640000

5200 1900 3610000

3000 -30090000

2000 - 1300 1690000

Ix = 19800 S ( x - x >2 = 6200000

From the table, we have

S1 - = 1240000 [by (2)]o

=> 5 = 1113.55 Rs.

Hence (1) becomes:

(3300 - 3800) s ’ n a 5 5

or / * - 0.44

| / 1= 0.44

Step 3 : The tabulated value of t at 5% level of significance and dof v = 5 is 2.015 i.e., *0.05.5 = 2.015.Step 4 ; Decision : Clarly calculate value of \ t | = 0.44 < tabulated value of *0.05, 5= 2.015.

=> the hull hypothesis is accepted.

=> the salesman is upto standard. Ans.

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Example 5.90 : A random sample of size 16 has as mean is 53. The sum of the squares of the deviations taken from the mean is 1501Can this sample be regarded as taken from the population having 56 as mean ? obtain 95% and 99% confidence Umits of the mean population. (For v * 15, taoi * 2.95 and tAB5 * 2.131).

Solution: Given that n = 16 (small sample)Sample mean 3c = 53 and n = 56.

Also I (x - x ¥ - 150 and variance is not known.

S2 = = io.n ~ l l b - l

=> S = y/lO

=> 5 = 3.162.The degrees of freedom v = n - J= 16 - 1= 15.

Step 1 : Null Hypothesis H q : Population mean is 56 i.e., H0: \i = 56.


We have, / = ...(1)

(53-56) Vl6 _3.162

=> | / 1- 3.79.

Step 3 : Given that tabulated values for do f v = 15 are :

(i) o.os, is = 2.131

(*0 *o.oi, t5 = 2.95.

Sten 4 : Decision:

(i) Since calculated value of |/j = 3.79 K tabulated value of^ 05,15 = 2.131

=> the null hypothesis is rejected.

=> the population mean is not 56 at 5% level of significance.

(ii) Since, calculated value of |/1 = 3.79 K tasulated value of fooi. 15 = 2.95

=> the null hypothesis is rejected.

=> the population mean is not 56 at 1% level of significance.

Again, we known that

95% confidence limits for ft (dof v = 15) are given by

4 9 8 | En g in e e r in g M a t h e m a t ic s -11 I

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H= 53 ± 1.684

50.316 < < 54.684.

H = 53 ±2.131 x

99% confidence limits for p (dof v = 15) are given by

M“ x ± fo.oi x ~T=V/i

= 53 ± 2.95 x3.162

= 53 ±2.33, 50.67 < n < 55.33. Ans.

Example 5.91; A random sample of size 7from a normal population gave a mean of977.51 and a standard deviation of 4.42. Find 95% confidence interval for the population mean, (given that * - 2.447)

Example 5.92 : The nine items of a sample had the following values: 45,47, 50, 52,48,47, 49, 53, 51.

Does the mean of the nine items differ significantly from the assumed population mean of 47.5 ?

Given that

Solution : Given that n * 7, (small sample)

x = 977.51 and a = 4.42 (known) and dof v = « - 1=6.

We known that 95% confidence limits for population mean ji for 6 dof. are given by

442= 977.51 ± 2.447 x

V6

= 977.51 ±4.416

973.09 <n <981.93.

[** fo.05,6 ~ 2.447]

Ans.

Where Pis the area to the left of the ordinate at t

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Solution : n - 9 (small sample)

H- 47.5 and variance is not known

and degrees of freedom v -n - 1=8.


JC x~ x ( x - x ) 245 -4.11 16.8921

47 -2.11 4.4521

50 0.89 0.7921

52 2.89 8.3521

48 -1.11 1.2321

47 -2.11 4.4521

49 -0.11 0.012153 3.89 15.1321

51 1.89 3.5721

Zx ~ 442 E( jc - x P = 54.8889

2* = 442 n 9 x = = — =49.11

n - 1 oand

By /-statistic

S = V6.8611 = 2.62

= (x~\i)yfnS

_ (49.11-47.5) >/9 2.62

= 1.84.From the given data let us interpolate the value of P for / = 1.84 is as follows :

for/=1.9 P = 0.953 / = 1.84

for / = 1.8 P = 0.945 /= 1.8

difference = 0.1 difference = 0.008 difference * 0.04

(Diff. in P) (Diff. in calculated value of t )

Difference -----------------Diff. of t as given--------------

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Thus required value of P for

P « 0.945 + 0.0032

* 0.9482 = 0.95Thus *0.95

t - Area to the right of / = 1.84

= !-/> ,. i.m= 1-0.95 = 0.05.Hence /’ (I/I * 1.84) - Left + Right

= 0.05 + 0.05= 0.1 which is greater than 0.05

| / 1* 1.84 > 0.05=> Chance of getting value of t is greater than 0.05.

The value of / is not significant

=> Sample may be a random sample which is drawn from the normal population with mean

xample 5.93 : The average number of articles produced by two machines per day are 200 and 250with standard deviations 20 and 25 respectively on the basis of records of 25 days production. Can you regard both the machines equally efficient at 1% level of significance ? (Given that t6.0j,*s * 2.58)

S = V533.85 = 23.1 Degrees of freedom v = n\ + n2 - 2 = 48.

Step I : Null Hypothesis Ho : both the machines are equally efficient i.e.,

Hi = H2-Step 2 : Calculate of /-statistic :

47.5.

olution : Given that fli » 25, xl = 200, S.D. C| = 20

n2 ~ 25, x2 ~ 250, S.D. a2 = 25

We have p = "l°l + »2°rl

_ 25 x (20)2 + 25 x (25)2 25 + 25 -2

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5 0 2 | E ng i nee r ing M athem atic s -! 11

-50x 3,54

23.1

- - 7.65=> | / 1= 7.65Step 3 : The tabulated value of / at 1%level of significance and dof v = 48 is 2.58 i.e.,

*o.oi,4* = 2.58.Step 4 : Decision : Since calculated value of |/| = 7,65 K tabulated value of fo.oi, 48 = 2.58.=> the null hypothesis is rejected.=> the two machines are not equally efficient at!% level of significance. Aas.

Example 5.94: Two horses A and B were tested according to the time (in seconds) to run a particular track with the following results: Horse A 28 30 32 33 33 29 34 Horse B 29 30 30 24 27 29Test whether you can discriminate between two horses♦ (Given that

t&os, n “ 2.20).Solution : Given that «i = 7, n2 - 6

_ 2 * _ 2 1 9x ~ n x ~ 7mean of A : 31.29

mean of B : y = 28.17.169

*2 6

Degress of freedom v = «t + »2 - 2 = 11.Step 1 : Null Hypothesis H q : No discriminate between two horses i.e.,

Ho: Hi = *i2.Step 2 : Calculation of /-statistic:

We have

Table for calculation of S

t = € x - y n|»2

X (x - x ) (x - X?;

y y - y ( y - y )2

28 -3.29 10.8241 29 0.83 0.688930 - 1.29 1.6641 30 1.83 3.3489

32 0.71 0.5041 30 1.83 3.3489

33 1.71 2.9241 24 -4.17 17.3889

33 1.71 2.9241 27 - 1.17 1.3689

29 -2.29 5.2441 29 0.83 0.6889

34 2.71 7.3441

Lx - 219 31.4287 t y = 169 26.8334

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Since„ Z ( x - * ) 2 + Z ( y - y ) 2

n l + n 2 ~ 2

3L4287 + 26.83347 + 6 -2

S = VS!296 = 2.30

Hence (1) becomes

/ =(31.29-28.17)

2.30 .'7x67 + 6

5.608 •2.30

= 2.44.

Step 3 : The calculated value of t at 5% level of significance and for dof v = 11 is 2.20 i.e., *0.05, n = 2.20.

Step 4 : Decision : Since the calculated value of |/| 2.44 < tabulated value of t = 2.20.

=> the null hypothesis is rejected

=> there are dicriminate between two horses at 5% level of significance. Ans.xample 5.95 : Below are given the gain in weights in kgs of cows fed on two diets x andy.

Diet x : 13 14 10 11 12 16 10 8 11 12 9 12

Diety: 7 11 10 8 10 13 9.

Test at 5% level of significance whether the two diets differ as regards their effect on mean increase in weight (given that n «* 2.11)

lution : Given that: n\ ~ 12, n2^ l and degrees of freedom v = nx+ n2 - 7 = 17.

= bT = 12' = 11S

Step 1 : Null Hypothesis H 0: two diets do not differ significantly i.e., Ho : Hi:


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504 | En g in e e r in g M a ih e m a h c s -III

Table for calculation o f S :

X x - x ( x - x ? y y - y (y-y )2

13 1.5 2.25 7 -2.71 7.3441

14 2.5 6.25 11 1.29 1.6641

10 - 1.5 2.25 10 0.29 0.0841

11 -0.5 0.25 8 - 1.71 2.9241

12 0.5 0.25 10 0.29 0.0841

16 4.5 20.25 13 3.29 10.8241

10 -1.5 2.25 9 -0.71 0.5041

8 -3.5 12.25

11 -0.5 0.25t

12 0.5 0.25

9 -2.5 6.25

12 0.5 0.25

138 53.00 Sy = 6 8 23.4256

_ S ( * - i ) 2 + S ( y - J ) 'Smce * ------- n ^ n t - 2

5a00+2a425612+ 7 - 2

« 4.4956

From (1), we have

S = V4.4956

= 2.12

>.71) 112x7V12 + 7212

- 1.77.

Step 3 : The tabulated value of * at 5% level of significance and for dof v = 15 is 2.11i.e.,

*0.05.15 = 2.11

Step 4 : Decision : Since calculated value |*| = 1.77 < tabulated value o f

*0.05, 15= 2.11.

=> null hypothesis is accepted.

=> the two diets do not differ significantly. Ans.

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5.41 FISHER’S F-TEST

The F-test was first originated by a great statistician even agriculturist prof. R.A. fisher. The test is also known as Fisher’s F-test or simply F-test It is based on F-districution. The F-test refers to a test of hypothesis concerning two variances derived from two samples.

The distribution of variance ratio F with vt and v2degres of freedom is given by :

where y0 is so chosen that the total area under the curve is unity.

5.42 F-STATISTIC:

Let xu *2, ....* x„u and y x%y i , .... y*i be the values of two independent random samples drawnfrom the two normal populations with mean iii and n2and variances ai2, a22 respectively, then we define variance ratio F as :

and jc, y are the sample means, with vi = n\ - 1, and v2= - 1are degrees of freedom.

Note : The value of F-statistic is always greater then L

5.43 ASSUMPTIONS IN F-TEST

The F-test is based on the following assumptions :

(i) The values in each group should be normally distributed.

(ii) The error should be independent of each value.

(iii) The variances within each group should be equal for all groups.5.44 COMPUTATION OF F-TEST:

We have the following steps :

Step 1 : Null hypothesis Ho : ctj2 = a22

Alternative Hypothesis H \ : <J\2 * a22.

Step 2 : Calculation of F-statistic :

when

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5 0 6 | E ng ineer ing M athem atics- ! 11

or F = — , when S22£ Si2.Sf

Step 3 : Choose tabulated value of / ’-statistic at a% level of significant with degrees of freedoms vj and V2.

Step 4 : Decision : If calculated value of F < tabulated value of F *\, v2)

Then accept null hypothesis Hq.

If calculated value F \ tabulated value of i(vi, V2>

Then reject null hypothesis H q .

Example 5.96 : Random samples are drawn from two populations and the following results were obtained:

Sample X : 20 16 26 27 23 22 18 24 25 19

Sample Y: 27 33 42 35 32 34 38 28 41 43 39 37 \ Find variance of two populations and test whether the two samples have same variance.

(Given that F&0Sfor 11 and 9 dof is 3.112).

Solution : Given that wj = 10 and n2 - 12 with degrees of freedom v = n\ - 1 = 9 andV = - 1 — 11.

Step 1 : Null Hybothesis H q : Let CTj2 = 022 i.e., the two samples have the same variance.

Step 2 : Calculation of F-statistic :

We have to find Si2 andS22 :

S 2

Sample X Sample Y

X x - x (x-x)2 y y - y ( y - y ) 2

20 - 2 4 27 - 8 64

16 - 6 36 33 - 2 4

26 4 16 42 7 49

27 5 25 35 0 0

23 1 1 32 - 3 922 0 0 34 - 1 1

-----18 -4 16 38 3 9

24 2 4 28 - 7 49

25 3 9 41 6 36

19 -3 9 43 8 64

30 - 5 25

37 2 4

Ex = 220 0 120 V : t l - f * . t o 0 0 314

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I

S t a t is t i c s I507

Here n\ = 10,

- = « 2 2 0 = x nx 22*

Si2 =

10SQc - x)2

fl i~ 1

120= 133

42012

£(y-y)*

* 2 - l

wj - 12

- . a ^ »2

= 35

314

11= 28.5

Hence F=£2 _=gas

Si2 " 1032.14 [ V S22 > S , 2 ]

Hence calculated value of Fis 2.14.Step 3 : The tabulated value of F at 5% level of significance for the dof v = 11 and 9 is 3.112 i.e.t Fo.os - 3.112.Step 4 : Decision : Calculated value of F = 2.14 < tabulated value of F0.05 = 3.112 => the null hypothesis H0 is accepted.

=> the two samples have the same variance. Example 5.97: in a test given two groups of students drawtf from two normal populations, the marks obtained were as follows:Group A : 18 20 36 50 49 36 34 49 41 Group B : 29 28 26 35 30 44 46

Examine at 5% level, whether the two populations have the same variance, (Given that Fe.83(*,6) “ 4,15).

Solution : Given that: n\ = 9, and n2= 7

j jc 333

* “ »i 9

37

and 34.- SL 238

Step 1 : Null Hypothesis H %: ctj2 = a^, Le., two populations have the same variance. Step 2 : Calculation o f F-statistic:

Group A Group B

x x - x (x-x)* y y - y ( y - y ?

18 -19 361 29 - 5 2520 -17 289 28 - 6 3636 -1 1 26 -8 6450 13 169 35 1 149 12 144 30 - 4 1636 - 1 1 44 10 10034 - 3 9 46 12 14449 12 14441 4 16

Ex = 333 1134 = 238 386

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Step 4 : Decision : Since calcualted value of F ~ 1.018 < tabulated value of Fo.os ~ 2 . 9 0 .

the null hypothesis //<>is accepted.

the two populations have the same variance. Ans.

5 .4 5 F I S H E R ’S z - D I S T R I B U T l O N

l^t x \ , x 2, ...... .v„i and y i% y2....... yn2 be the values of two independent random samples'with

estimated variance S{2and S22.

Then Fisher’s /-distribution is the statistical distribution of half of the logarithm of an F-distribution variate:


i .e . , z ^ | l o g « F

= o log.S 1

V I; where S\2> S22.2 ^ S^ ,

With degress of freedom vt ~ n\ - 1and i>2 = n2 - I.

It was first described by Ronald Fisher in 1924.

The probability density function of /-distribution is defined as follows :

u\!2

y - ------------- --------- -- X --------------------------- -.----------- r j~ ; - o o < z < OC

a t e . * )

Remark: With the help of equation (1), all the applications of F-distribtution may be regarded as the application of z-distribution also.

Example 5.99 : Two gauge operators are tested for precision in making measurement One operator completes a set of 26 readings with a standard deviation of 134 and the other does 34 reading with a standard deviation of 0.9%. What is the level ofsignificance of this difference ? Given

that for dof v, * 25 and v2 m33, the value of z*os * 0.306 and z+oi *=0.432.

Solution : Given . = 26 and n2~ 34

S.D. at = 1.34 and o2= 0.98

, Z(x~x)2 , U y ~ y ) 2Since Oj2 - ---------- and o22- ----“-----

rt, n2

=> (1-34)2 - and (0.98F =

=> I ( x - x f - (1.34)2 x 26 and 2. ( y -yY = (0.98)2* 34

=> I. (x - x)2 ~ 46.68 and I iy - y? = 32.6536

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5 1 0 | E n g m e h bn g M a t h e m a t ic s -111

Step I : Null Hypothesis H0 : The difference between variances is not significant i.e.,

O!2 = <J22.

Step 2 : Calculation of z-statistic.

Step 3 : The tabulated value z at 5% and i% level of singificance and for dof 25 and 33 are zo os —0.306 and zo.ot360.432.

Step 4 : Desicion :

(1) Since Calculated value of z = 0.3148 K tabulated value of £o.os= 3.06

=> the null hypothesis Ho is rejected.

=> the difference between variances is significant at 5% level of significance.

(ii) Since calculated value of z = 0.3148 < tabulated value of zo.oi ~ 0.432.

=> the null hypothesis H0 is accepted.

=> the difference between variances is not significant at 1%level of significance. Ans.

I : 17 27 18 25 27 29 27 23 17

n : 16 16 20 16 20 17 15 21

Indicate samples drawn from the same universe. (The value of z at 5% level for 8 and 7 degrees of freedom Is 0.6575J.

Since

and

Clearly Si2 >

we have

= | log, (1.8770)

= 0.3148.

Example 5.100 : Test whether the two sets of observations:

Solution : Given that

with dof

/i, =9

V| = «2 - 1 = 8

J»2 = 8

and v2 = *2 - 1= 7.

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Step 1 : Null hypothesis Ho : Two population variances are same i.e., H0 : ai2 = O22.

Calculation of z-statistic:

S t a t is t i c s | 5 l l

I observation 11 observation

X x - X ( x - x Y y y - y 0>- y Y

17 -6.33 40.0689 16 - 1.625 2.6406

27 3.67 13.4689 16 - 1.625 . 2.6406

18 -5.33 28.4089 20 2.375 5.6406

25 1.67 2.7889 16 - 1.625 2.6406

27 3.67 13.4689 20 2.375 5.6406

29 5.67 32.1489 17 - 0.625 0.3906

27 3.67 13.4689 15 - 2.625 6.890623 -0.33 0.1089 21 3.375 11.3906

17 -6.33 40.0689

210 184.6001 141 37.8748

We have

n2 - 1

Since S\2> S22

z = i lo& l ^

_ £ ( * - 3 i f _ 184.0001 = , ,

S' n i - 1 8 23

w . a t B i . 2 z * 6 a - M107

5 ,2ri2

= 1 10&(sfiW )

= \ log, (4.2508)

= 0.7236.

Step 3 : The tabulated value of z at 5% level of significance with dof 8 and 7 is 0.6575. i.e.,

20.05 = 0.6575.

Step 4 : Decision: Since calculated value of z = 0.7236 K tabulated value of zo os= 0.6575.

=> the null hypothesis is rejected.=> the two variances are not same. Ans.

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5 J & I E n q w e e o n g M a t h e m a t t c s -III

Exercise-5 (A)1. Find whether the following function is a p d.f.

r f x _ J *» 0 < ;x s l J W “ |2x, 1 £ x £ 2 ’

2. A continuous random variable A"has />.<£/

/(x ) = 3x2»0 ^x ^ 1Find the value of ‘ a \ if P(X<>a) = P(X> a).

4.

6.

3. If f ix ) -0, x < 2

• (3 + 2x), 2 < x < 4

8 0, * > 4

Prove that/ (x) is a p .df Also compute P (2 < X < 3). A random variable X has the p.d.f.

f i x )

x, 0 £ x < 1_ •{2 - x, 1 £ j c < 2

0, x 2: 2

Find distributive function or cumulative distributive function (c.d.f.) of X. A random variable X has p.df. :

/far3*' *>o j o , i s o

Find (i) the value of k (ii) compute P (1 < X < 2) and P (X> 3).A random variable A'has a p.d.f.

f ix ) =

x2 ’

0 < x £ 1

4 ( 3 - x ) , 1 <x <;241

4 ’2 < x £ 3

4 (4 -x ), 3 < x ^ 440, otherwise

Compute (i) P(XZ 3), (ii) P (1 < X < 3) (iii) P (| X | > 1.5).

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S t a t is t i c s ) 513

7. Find the mean o f the following p .d .f.

f i x )0, otherwise.

8. For the continuous distribution :

dF = y0 (x - jc2) dx, 0 £ x <, 1.

Find mean, median and mode.

9. If a random variable X has p.df.

f i x ) = ^ - a Z x Z a .

Find first, second, third and fourth moments about origin.

10. A p.df. function defined by

f ix ) =

—(3 + x)2, - 3 < *£--!

X ( 6 - 2 * 2 ) , - 1 < x < 1

~ ( 3 - a:)2, 1 < x £ 3.

Find mean, variance and standard deviation.

111. Vox p.df f {x) -

Find mean, median, and mode.

12. A distributive function F (x) is defined as

0 , x <>2

F (x ) -

1

Find p.df. of x. Also find mean.

x > £

Answers-5 (A)

* 71. No., because J f(x) dx = « * 1.

2.

3.

a * (0.5)l/3.

4/9

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514 \ E n g in ee r ing M athem attc s -II I

fo

4. F(x) =

, x < 0

2* 2 l * x < 21 , xZ 2

5. k - 3 , P ( I < A*< 2) = e*3- e** and /> (A' £ 3) = e~9.

6.I 3 158 ’ 8 ’ 32 ‘

135

8 . _y0 = 6, mean = median = mode = ^

9. m' = o, ^ ,M 3’ = o, h4'=

10. mean = 0, variance = 1, S.D. = 1.

11. mean = mode = median - 0 .

, _ x + 312. a = , mean = 5.375

Exercise-5 (B)

For a p.d.f f(x)

/ ( * ) = s - G<* <6[ 0, otherwise

Find the value of k and show that/ (x) is a rectangular distribution. Also fmd mean.

For the rectangular distribution ;

f i x ) ~ fjj~ , - a < x < a

Find \i2, ^3 and m (i.e., 2nd, 3rd& 4thmoments about mean).

For the rectangular distrubiton :

1 0 - f e < x < 1 0 + f c f(x\ = \ 2k

[ 0, otherwise

Find mean and variance. Also find distribution function F ( x ) .

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S t a t is t i c s I 5 1 5

4. Given the Incomplete Beta function:

x 3 ( I f n ) B, (/, m) * J jc 1(1 - *)"-1 dx and lx (/, m) =

then prove that Ix f/, tti) * 1 - / lHr (m, /)5. The income tax of a man is exponentially distributed with the p.d.f is given by

1 --x/3/<*) = 3*

What is the probability that his income will exceed Rs. 17000 assuning that the income tax is

levied at the rate of 15% on the income whose Rs. 15,000 ?

The life time X in hours of a T.V. tube of a certain type obeys an exponential distribution with

X = 0.001 hours. Find(i) P (X> 1000), (ii) P (700 ZX<> 100)

Find the value of *c\ so that / ( j c ) be a p.df. of/ ( j c ) = ce-2*, x > 0, then prove that

(i) fiQ O M (Jffc), (ii) Cwfficient of variation is 1.

Answers-5 (B)

k - b ~ a, mean = ^ (a + b).

2 4

|«2 "jp ® IT*

mean = 10, variance = and

jc- ( 1 0 - £ )d.F.

-100

F(x) =10-/t <x < 10 -£

xZ 10+£.

6. (i) 0.368, (ii) 0.129.

m m

j* **A f* |» >

Exercise-5 (C)

1. Find the Binomial distribution whose mean is 5 and variance is 10/3.

2. Find the Binomial distribution whose sum of mean and variance is 4.8 for 5 trials.

3. In a Binomial distribution, the sum of the mean and variance for 5 trials is 1.8. Determine the distribution.

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4. If 10% of the rivets produced by a machine are defective, find the probability that out of 5 rivets chosen at random :

(i) None will be defective, (ii) One will be defective, (iii) At least two will be defective.

5. If the probability that a newborn child is a male is 0.6, find the probability that in a family of 5 children there are exactly 3 boys.

6. Three percent of a given lot of manufactured parts are defective. What is the probability that in a sample of four items none will be defective.

516 | Engineer ing Mathemat ics -III

7. The average percentage of failure in a certain examination is 40. What is the probability that out of a group of 6 candidates, at least 4 pass in the examination.

8. Five coins are tossed 3,200 times. Find the frequencies of the distribution of heads and tails and tabulate the result. Calculate the mean number of successes and standard deviation.

9. In a particular market 40% of the consumers prefer readymade clothing. A sample of n = 5 consumers is to be drawn. Give the probabilities of having 0,1,2,3,4 and 5 preferring readymade clothes.

[Hint. Probability of r consumers preferring readymade clothes out of a sample of

5= 5CX (0.4)r(0.6)5_r. Put r =0,1,2,3,4 and 5.]

10. The incidence of occupational disease in an industry is such that the workers have a 20% chance of suffering from it. What is the probability that out of six workers chosen at random four or more will suffer from the disease.

[Hint. Here p=0.2, ^=0.8, n=6.

[Hint. />(0) =4 C0(O.O3)°(O.97)4 = 0.885 ]

[Hint. p = 0 A , <7=0.6

/»(*;> 4)=P(x = 4)+/>( jc=5)+P( jc =6).

=C4 (.6)4 (.4)2 +C5(0.6)5(0.4)=C6 (0.6)6 =0.5443].

[Hint.

p(r ;>4)=C4 (0.2)4(0.8)2 +C5(0.2)5(0.8)+C6 (0.2)6].

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11. A sample of 10 pieces was examined out of a large consignment which has 5% defective pieces. Give the probability of 1defective in the sample of 10.

10[Hint Here 0.05, ?=0.95, n=10 />(r= 1)=C,(0.05)(0.95)9].

12. In 100 sets of ten tosses of an unbiased coin, in how many cases should we expect seven heads and three tails.

[Hint Here Probability of head »$=—,»=10 #=100.

ioo V r n 3'Required number of cases »100xC7l —j I—I

13. Fit a Binomial distribution for the following data and compare the theoretical frequencies with the

actual ones:x : 0 I 2 3 4 5

/ : 2 14 i0 34 22 8

14. In a Binomial distribution, for n = 5, if P(r - 1)= 0.4096 and P(r - 2 ) - 0.2048, then find the value of p.

15. Ten coins are tossed 1024 times and the following frequencies observed. Compare these frequencies with the expected frequencies.

x : 0 I 2 3 4 5 6 7 8 9 10

/ : 2 10 38 106 188 257 226 128 59 7 316. Seven coins are tossed and the number of heads are noted. This experiment is repeated 128 times and the following distribution is obtained:

x : 0 1 2 3 4 5 6 7 Total

/ : 7 6 19 35 30 23 7 1 128

Fit the binomial distribution considering the coin is unbiased.

17. Fit a Binomial distribution for the following data and find the expected frequencies of the chance of machine being defective is 1/2.

x : 0 1 2 3 4 5 6 7

/ : 7 6 19 35 30 23 7 1


Answers-5 (C)

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518 | E ng inee r ing M athem attc s - II I

4. (i) 0.5905, (ii) 0.3281, (iii) 0.0815.

5. 0.3456. 6. 0.885. 7.0.5443. 8. Mean = 2.5, S.D. =1.118.

9- C,(0.6)r(0.4)5*', r = 0,1,2,3,4,5.

1°* Ca { 0.2)4+ C5( 0.2)5 (0.8)+(0.2)6.

10 / ] V°11. ,0C, (0.05)(0.95)9. 12. i°0xC7| j J

13. /(r) = 100x5Cr(0.432)5~f(0.668)r, r - 0, I, 2, 3, 4, 5.

I14. P = ?

15. 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1.

16. 128(0.517+ 0.423)7; 1, 8, 23, 36, 33, 19, 6, I

17. 1,7,21,2,5,21,7,1.

Exercise-5 (D)

1. If a random variable has a Poisson distribution such that P (1) -P (2), find

(i) Mean of the distribution (ii) P{ 4).

22. Suppose that X has a Poisson distribution. If P{X - 2) ~~P{X = I), find,

(i) P(X = 0) (ii) A ^ = 3).

3. A certain screw making machine produces on average 2 defective screws out of 100, and packs them in boxes of 500. Find the probability that a box contains 15 defective screws.

4. The probability that a man aged 35 years will die before reaching the age of 40 years may be taken as 0.018. Out of a group of 400 men, now aged 35 years, what is the probability that 2 men will die within the next 5 years ?

5. Suppose a book of 585 pages contains 43 typographical errors. If these errors are randomly distributed throughout the book, what is the probability that 10 pages, selected at random, will be free from errors ?

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Using Poisson distribution, find the probability that the ace of spades will be drawn from a pack of well-shuffled cards at least once in 104 consecutive trials.

If m and pr denote by the mean and central rth moment of a Poisson distribution, then prove that

am

In a normal summer, a truck driver gets on an average one puncture in 1000 km. Applying Poisson distribution, find the probability that he will havie *.

(i) No puncture (ii) Two punctiires in a journey of 3000 kms. j

Six coins are tossed 6400 times. Using Poisson's distribution find approximate probability of getting 6 reads x times.

A manufacturer knows that the razor blades he makes contain on an average 0.5% of defectives. He packs them in packets of 5. What is the probability that a packet picked at'random will contain3 or more faulty blades ?

A source of water is known to contain bacteria with mean number of bacteria per cc equal to 2. Five 1 cc test tubes were filled with water. Assuming that Poisson distribution is applicable, calculate the probability that exactly 2 test tube£ contain at least* 1 bacterium each.

Ten percent of the tools produced in a certain manufacturing process turn out to be defective. Find the probability that in a sample of 10 tools chosen at random

(i) Exactly two will be defective and

(ii) More than one will be defective, by using Poisson distribution.

Between the hours 2 P.M. and 4 P.M., the average number of phone calls per minute coming into the switchboard of a company is 2.35. Find the probability that during one particular minute there

will be at most 2 phone calls.

n/ , e mm' e~235 x(2.35)r[Hint p( r ) =— T ’= ----------------- »r =°»X' 2»3*

P(X 2) = P(X = 0)+ P(X = 1) + P(X = 2),

v2= <T235

, 2.35 (2.35)' 1+ -------- * +

1! 2!= 0.583].

If 5%of the electric bulbs manufactured by a company are defective, use Poisson distribution to find the probability that in a sample of 100 bulbs, 5 bulbs will be defective.

[Hint

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5 2 0 | E n g in e e r in g M a t h e m a t ic s -!!!

15. The number of accidents in a year attributed to taxi drivers in a city follows Poisson distribution with mean 3. Out of !,000 taxi drivers, find approximately the number of drivers with (i) no accidents in a year, and (ii) more than 3 accidents m a year.

16. The probability that a man aged 50 years will die within a year is 0.01125. What is the probability

that out of 12 such men at least 11 wilt reach their fifty-first birthday.17. A machine produces large quantities of hems and past experiment shows that on an average it

produces 1%defective items. A sample of 20 items in drawn from a large number of Hems. Use Poisson distribution to determine the probability of 2 defectives.

18. Fit a Poisson distribution to the following data and calculate theoretical frequencies :

Deaths: 0 1 2 3 4 5 6 7

Frequencies: 305 365 210 80 28 9 2 1

19. Fit a Poisson's distribution for the following data:

x: 0 1 2 3 4 5 6 7 8 9 10

y: 103 143 98 42 8 4 2 0 0 0 020. From records of Prussian Army corps kept over 20 years, the following data was obtained showing

the number of deaths caused by the kicks of a horse. Calculate the theoretical Poisson frequencies :

Number of deaths 0 1 2 3 4 Tbtal

Frequencies 109 65 22 3 1 200

21. Fit a Poisson's distribution for the following data:

x: 0 1 2 3 4 5 Tbtal

f: 229 211 93 35 7 1 576

Answers-5(D)

1. (i)2

2. (i) e-4

(ii) 2/3<?2

3. 0.035.

(ii) 4e~*.

4. 0.01936. 5. 0.4795. 6. 0.865.

8. (i) (ii) (415)<f3.

11. 0.3459.12. (i) 0.18395 (ii) 0.2642,

15. (i) lOOxe-3 = 49.8,

13. 0.583.

(ii) 353 nearly

14. 0.1823.

16. 0.9916.

17. 0.0164.

18. 301.2, 361.4, 216.8. 86.7. 26.0, 6.2, 1.2, 0.2.

19. 107, 141, 93, 41, 14, 4, 1, 0, 0, 0, 0.

20. n = 109, 66, 20, 4, 1.

21. 227.26, 211.35, 98.28, 30.46, 7.08, 1.31.

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S t a t t s t k b i 521

5.

6.

Ex

Find the probability that the standard nonnal variate lies between 0 to I 5.

Find the area under the nonnal curve for (a) Z ~ 1.64, (b) Z *• 132. (c) Z ~ 0.56 (d) Z - J54.

Find the area to the (a) right of Z * 0.25; (b) left of Z * 1.96; (c) left of Z = - 1.96; (d) between Z = 0.4 and Z* 0.6.

A large number of measurements in normally distributed with mean of 65.5 cm and a standard deviation of 6.2 cot. Find the percentage of measurements that fad between 54.8 cm. and 68.8 cm.

In an intelligence test administered to 1,000 students the average score was 42 and standard

deviation 24. Find <a) the number of students exceeding a score of 50; (b) the number of students trying between 30 and 54, <c) the value of score exceeded by top 100 students.

[Hiat (a) Where X ~ 50, Z = =*0.333.

P(Z > .333) = 0.5-0 1304- 0.3696.

Expected number of students exceeding a score of 50 = .3696 x 1000 = 369.6*370.

[Hint (b) When Xhes between 30 and 54, then Z lies between - 0.5 and 0.5. Probability of having students with score between 30 and 54 = 0.1915 + 0.1915 - 0.3830.

Expected number of students to score between 30 and 54 = 1.000x0.3830 = 383.

100[Hint (c) Probability of getting top 100 students ------- = 0.1. Standard normal variate having

1,000

0.1 area to the right = 1.28.

Also X -42

24>X = 72.72 « 73

In a distribution exactly 7% of the items are 35 and 89% are under 63. What are the mean and standard deviations of the distribution ?

[Hint ! 48^35-n = I.48aa

/ ' j

Z = = j.23 63 - n = -1 .23o a

Solve (1) and (2) we get ji = 50.3, o = 10.33 ].

... (0

.. (2)

t

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7. A sample of 100 diy battery cells tested to find the length of life produced the following results:

|A=12 hours, a = 3 hours.

Assuming the data to be normally distributed, what percentage of battery cells are expected to have life (a) more than 15 hours, (b) less than 6 hours (c) between 10 and 14 hours.

[Hint (a)WhenX= IS, Z = ^ j^ * = l

Area to the right of Z = 1 is 0.5-0.3413 = 0.1587.

Percent of battery cells having life more than 15 hours = 0.1587 x 100 = 15.87%.

[Hint (b) When X = 6, Z = = -2

Area to the left of Z is -2 = 0.5 - 0.4772 = 0.0228.

%of Battery cells having life less than 6 hours = 0.0228 x 100 = 2.28%.

[Hint (c) When X= 10, Z = —~ =-0.67 When JIT=14, Z = - 2 =0.67

Area between X - 10 to X~ 14 is twice the area to the left of Z = .67 = 2x0.2487 = 0.4974.

% of Battery cells having life span between 10 hours and 14 hours = 49.74%].

8. The mean lifetime of 100 Watt light bulbs produced by Larsen and Turbo is 200 hours. It is known that the standard deviation is 20 hours. Assuming that the life time of light bulbs are normally distributed, what are the probabilities that a single 100 Watt light bulb extracted from the production lot will (a) bum out between 180 hours and 210 hours ? (b) bum out for a time greater

than 250 hours.

[Hint (a)WhenX= 100,then Z = - — °^ = -land vhen*= 210 then Z = 2- ° = 0.520 20

Area between Z= - 1and z * 0.5 is 0.3413 + 0.1915 = 0.5328.Required probability = 0.5328.

[Hint (b) When X = 250, Z = — - - = 2.5.

Required Probability=Area to the right of Z equals 2.5

= (0.5 - 0.4938) = .0062]

9. In a sample of 1,000 the mean weight is 45 kgs with standard deviation 15 kgs. Assuming the normality of the distribution, find the number of items weighing between 40 and 60 kgs.

= /X-0.333 <; Z £ 1) = P{r0.333 < Z < 0) + P(0 < Z < 1) = 0.1293 + 0.3413 = 0.4706.

Required number of items = 1,000 x 0.4706 = 470.6 » 471].

5 2 2 | En g in e e r in g M a t h e m /u t c s -III

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10. In a sample of 120 workers in a factory the mean and standard deviation of wages were Rs. 11.35 and Rs. 3.03 respectively. Find the percentage of workers getting wages between Rs. 9 and Rs. 17 in die whole factory assuming that the wages are normally distributed.

11. Fit the equation of the best fitting normal curve to the following distribution:

x : 0 \ 2 3 4 5 / : 13 23 34 15 11 4

12. Fit a nonnal distributions to the following data:

Mid point of interval 100 95 90 85 80 75 70 65 60 55 50 45

Frequency 0 1 3 2 7 12 10 9 5 3 2 0

13. Fit a normal distribution to the following data:

Variable 60-62 63-65 66-68 69-71 72-74

Frequency 5 18 42 27 8

S t a t i s ti c s | 520

Answers-5(E)

1. 0.4332

2. (a) 0.4484; <b) 0.4049 (c) 0.2123; (d) 0.4382.

3. (a) 0.4013; (b) 0.9750 (c) 0.0250; (d) 03811.

4. 0.6601 5. (a) 370; (b) 383; (c) 73.

6. Mean = 50.3 Standard deviation a * 10.33

7. (a) 15.87%; (b) 2.28%; (c) 49.74%.

8. (a) 0.5328; (b) 0.0062 9. 471 10. 75.1

II.„ , ioo -(it)1

y = f (x ) = - r r r e ^ - V3.4*

12. Mean (*i) = 71.2, S.D. (a) = 9.95;

Theoretical frequencies : 0.2, 0.6, 1.8, 4.1, 7.3, 10.1, 10.7, 8.9, 5.7, 2.9, 1.1, 0.3.

13. Mean (p) - 67.5, S.D. (cr) = 2.9;

Expected frequencies : 2.55, 13.43, 13.10, 18.64, 4.81.

VMftx mi f. * *r >- X. ' Ml « h* *, '

Exercise-5 (F)1. A sample of 300 students of Under-Graduate and 300 students of Post-Graduate classes of a

University were asked to give their opinion toward the autonomous colleges. 190 of the Under- Graduate and 210 of the Post-Graduate students favoured die autonomous status.

Present the above data in the form of a frequency table and test at 5% levels, the opinions of Under-Graduate and Post-Graduate students on autonomous status of colleges are independent

(Table value of chi-square at 5% level for 1d.f. is 3.84).

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2. A certain drug is claimed to be effective in curing colds. In an experiment on 164 poeple with cold,

half of them were given the drug and half of them given sugar pills. The patient’s reactions to the treatment are recorded in the following table. Test the hypothesis that the drug is not better than sugar pills for curing colds.

Helped Harmed No effectDrug: 52 10 20

Sugar Pills: 44 12 26

(Table value of chi-square at 5% level for d f 2 is 5.99)

3. In a survey of 200 boys, of which 75 were intelligent, 40 had skilled fathers, while 85 of the unintelligent boys had unskilled fathers. Do these support the hypothesis that skilled fathers have intelligent boys. Use Chi-square test. (Values of x2for 1-degree of freedom at 5% level is 3.84).

4. Four dice were thrown 112 times and the number of times 1, 3 or 5 were as under :

Number of dice showing

1,3, or 5: 0 1 2 3 4Frequency: 10 25 40 30 7

(Table value of chi-square at 5% level for 4 d.f. is 9.488).

5. To test the efficiency of a new drug a controlled experiment was conducted where in 300 patients were administered the new drug and 200 other patients were not given the drug. The patients were monitored and results were obtained as follows :

Cured Condition No Total

worsens effect

Given the drug: 200 40 60 300

Not given the drug: 120 30 50 200Total 320 70 no 500

Use x2test for finding the effect of the drug. (x0 o$,2 = 5.99)6. The following table gives the classification of 100 workers according to sex and the nature of

work. Test whether the nature of work is independent of the sex of the work.

Skilled Unskilled

Male 40 20

Female 10 30

(Use xi.05.4 * 3.84)7. There is a general belief that high income families send their children to public schools and low

income families send their children to Government schools. For this 1000 families were selected in a city and the following results were obtained.

Income Public Schools Govt Schools Total

Low 100 200 300

High 500 200 700

Total 600 400 1000

Use Chi-square test to determine whether income level and the type of schooling were associated.

(Xoos. 1 = 3.84).

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S t a t is t i c s J 5 2 5

8. Calculate chi-square for the follow ing data :

Class : A B C D Total

Jn : 8 29 44 15 4 10.0

f '7

24 38 24 7 1009. Calculate the value of t in the case of two characters A and B whose corresponding values are given below:

A 16 10 8 9 9 8

B 8 4 5 9 12 4

10. The table below are for protein tests of the same variety of wheat grown in two districts. The average in District I is 12.75 and in District II is 13.03.

Calcualte / for testing the significance between the means of the two districts :

Protein results

District 1 12.6 13.4 11.9 12.8 13

District II 13.1 13.4 12.8 13.5 13.3 12.7 12.4

[Given that /oos. to ~ 2.228]

11. Ten cartons are taken at random from an automatic filling machine. The mean net weight of the 10 cartons is 11.8 kg and standard deviation is 0.15 kg. Does the sample mean differ significantly from the intented weight of 12 kg ? (Given that for / 0.05, - 2.26).

12. A fertiliser mixing machine is set to give i 2 kg ofnitratefor every quintal bag of fertiliser. Ten, 100 kg bags are examined. The percentages of nitrate are as follows :

II, 14, 13, 12. 13. 12, 13, 14, II, 12.Is there reason to believe that the machine is defective ? Value of t for 9 degrees of freedom is2.2621.

13. Ten students are selected at random from a college and their heights are found to be 100, 104, 108,110,118,120,122,12-4,126 and 128 cms. In the height of these data, discuss the suggestion that the mean height of the students of the college is 110 cms. (Given / 005.9 = 2.26)

14. Talcum powder is packed into tins by a machine. A random sample of 11 tins is drawn and their contents are found to weight in kgs as follows :

0.44. 0.51, 0.49, 0.52, 0.45, 0.48, 0.46, 0.45, 0.47, 0.45 and 0.47Test if the avarage packing can be taken to be 0.5 kgs. (/0 05. 10 = 2.28)

15. Samples of the types of electric bulbs were tested for the length of and the following data were obtained

Type I Type II

Sample No. 8 7

Sample Mean 1234 hours 1036 hours

Sample S.D. 36 hours 40 hours

In the difference in the means sufficient to warrant an inference that type I is superior to type II regarding the length of life ? (Given /0.<)5,15 = 2.131)

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16. The standard deviations calculated to n two random samples of size 9 and 13 are 2 and 1.9 respectively. May the sample be regarded as drawn from the nonnal populations with the same standard deviation. (Given Aw(s. «> * 3.51)

17. TWo sample are drawn from two normal populations. From die followong data test whether the

two samples have the same variance at 5% level:Sample I 60 «S 71 74 76 82 85 87

Sample II 61 66 67 85 78 63 85 86 88 91

(Given f 9 .9 s ( 9 .7) * 3.68)

18. Following results were obtained from two samples, each drawn from two different populations A and B.

Population A B

Sample I II

Sample Size n\ «*25 rt2 = 17

Sampled s.<L s\ = 3 s j - 2

Test the hypothesis that the variance of brand A is more than that of B.

(Given that Fqxh (24, 16) “ 2.24).

19. Two samples were drawn from two normal populations and their values are :

A 66 67 75 76 82 84 88 90 92

B 64 66 74 78 82 85 87 92 93 95 97.

Test whether the two populations have the same variance at 5% level of significance. (Given that Fo.os (10, 8) ” 3.07).

20. Find the value of z-statistic, if two independent samples are:

Sample I 7 2 4 ' 6 4 6

Sample II______ 8 4 5 7 4 2 5.____________________

Answers-5(F)

1. Opinion on autonomous status and level of graduation are independent

2. Computed %2= 1.622, and Drug is not better than sugar pills3. x2- 0- 8, Skilled fathers have intelligent boys.

4. All the four dice are fair. (%2 * 1.845)

5 . x2~ 2.434, Drug is not effective.

6. x2~ 16.666, The nature of work does not seem to be independent of the sex of the worker.

7. The level and type of schooling are associated.

7. x2= 6.77

9. tA= 1.66;/a = 0.85

1 2 6 | E n @m b e h i n g M a t h e m a t ic s -111

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10. / = 0.85, No difference.

11. t - 4; the sample mean differs significantly.

12. t = 1.464, there is no reason to belive that the machine is defective.

13. t - 1.937; the mean height of the students can be taken as 110 cm.14. / = - 3.43 i.e., | f | = 3.43; No

15. / = 9.35, Type I bulb is superior to type II.

16. F ~ 1.15; the samples can be regarded as drawn from the population with the same standard deviation.

17. F= 1.468, the samples have the same variance.

18. F - 2.205; variance of brand A is not more than the variance of brand B.

19. F = 1.415, the two populations have the same variance.20. z = 0.086.

□


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5 2 6 | E n g in e e r in g M a ih e m a iic s -III

TABLEAREA OF A STAMJAMB NORMAL DISTRIBUTION

As entry in At total* Is Dm pmpmrtkm m&r «bt / Iottav am i wfelck ii brtwn t I ml t piMhc / I

valmoft Aw wfcwrf tiw nMnined / I bytymetry. —^ -----£

z .00 .01

. . . —

.02 .03 .04 .05 .06 .07 .08 .09

0.0 .0000 ,0040 .0000 .0120 ,0160 J0199 .0239 .0279 .0319 .0359

0.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753

02 .0793 .8832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141

03 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517

0.4 .1554 1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879

0.5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .222406 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2517 .2549

0.7 .2580 .2611 .2642 .2673 .2703 .2734 .2764 .2794 .2823 .2852

0.8 .2881 .2910 .2939 .2967 .2995 .3023 .305! .3078 .3106 .3133

0.9 .3159 .3186 .3212 .3238 .3264 .3289 .3315\

.3340 .3365 .3389

1.0 .3413 .3438 .3461 .3485 .3508 .3531 .3554 .3577 .3599 .3621

1.1 .3643 .3665 .3686 .3708 .3729 .3749 .3770 .3790 .3810 .3830

1.2 .3849 .3869 .3888 .3907 .3925 .3944 .3962 .3980 .3997 .40151.3 .4032 .4049 .4066 .4082 .4099 .4115 .4131 .4147 .4162 .4177

1.4 .4192 .4207 .4222 .4236 .4251 .4265 .4279 .4292 .4306 .4319

1.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441

1.6 .4452 .4463 .4474 .4484 .4495 .4505 .4415 .4525 .4535 .4545

1.7 .4554 .4564 .4573 .4582 .459! .4599 .4608 .4616 .4625 .4633

1.8 ..4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706

1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767

2.0 .4772 .4778 .4783 .4788 .4793 .4798 .4803 .4808 .4812 .4817

2.1 482! .4826 .4830 .4834 .4838 .4842 .4846 .4850 .4854 .4857

2.2 .4861 .4864 .4868 .4871 .4875 .4878 .4841 .4884 .4887 .489023 .4893 .4896 .4898 .4901 .4904 .4906 .4909 .4911 .4913 .49162.4 .4918 .4920 .4922 4925 .4927 .4929 .4931 .4932 .4934 .4936

Z5 .4938 .4940 .4941 .4943 .4945 .4946 .4948 .4949 4951 4952

2.6 .4953 .4955 .4956 .4957 .4959 .4960 .4961 .4962 4963 .4964

2.7 .4965 .4966 .4967 .4968 .4969 .4970 .497! 4972 4973 .4974

Z8 .4974 .4975 .4976 .4977 .4977 .4978 .4979 .4979 .4980 .4981

19 .4981 .4982 .4982 .4983 .4984 .4984 .4985 .4985 .4986 498fc

3.0 .4987 4987 .4987 .4988 .4988 .4989 .4989 .4989 .4990 >4990

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a e . (Third Stouter) EnMinttM, Ji m , 2008

eacft UniftfrooMlpidsory. AH questions

E x a m . P a p e r s | i

ifnifrlis aiMmatytic function and find its derivative.

Ans. (Refer Page 22)

f ( ° ) = Aw. (Refer Page 59)C

Or(a) Evaluate the following integral using Cauchy’s integral formula;

Ic 4 -3z

z(z - l)(z - 2)

dz

where C is the circle jzj = 3/2. Ans. (Refer Page 71)

l*2x sio 2 0 2 ll i « . #j(b) Prove that: = pr(a-V j -b ) where 0 < 6 < a. Ans. (Refer 93)

A2 , . A2 sin(x+h)(a) Evaluate: ~£ + )+ Esin(x+h)

Ans. (Refer Page 202for a Similar Pmtetm)(b) Use Bessel’s formula to find the value ofy if x - 3.75, given:

x y25 24.1453.0 22.04333 20.225’4.0 1164445 17.2625.0 * 16J047;

Ans. (Refer Page 239for a Similar Problem)

Or(a) Find the first, second and third derivatives of the function tabulated below, at the pintx= 1.5:

X mis 33752D 7-025 13.6253.0 24.035 38*754.0 59.0

Am. (Refer Page 260)(b) A body is in the form of a solid revolution. The diameter D in cm of the sections at distances x cm from one end are given below. Estimate the volume of the solid:

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%| EN GJ^&O NGM AneM /mcs-Ul

X..-. . Di t *. f i *i V/fi: *»*»*$,' v

~ * H * .v . • .-i< bni s VJ im

s 'l S jO 6.0

’ « .%« V. v • W *“ 8.75*'* 10.0 625

12.5 55

15.0 4.0

<i\ *J*'4*L>

sir-1..- ‘ C

nm> r *w j v t maiiwnw rrwmiy(a) By using Newton-Ramphson method, find the root o f x ^ x - 9 * 0 , which is near tox~ 2

correct to three places of decimal. Aas. (Rrfer P*$t 143)

Aw. (X ^r Atfff 179)

2Jx+ 6 y - z m 85 6x + 15y + 2r - 72

x'+<y + 54x* 110Or*

(a) Pafcrm twoiteatioiBofPk^’snie<hods loManapprDMmalB9ofetk3nofthek^v»k»probfcm:

dy '•£; - * + / ;> < 0 ) = i

AM.(gêr 50/ jbr a Similar Problem)

dy y —X*(b) Using Runge-Kutta method o f fourth order, solve ---- r with v(0 )" 1 atx “ 0.2, 0.4.

dx y*+xAm. (Xtfer Page 329)v .

4. (a) Find the rank o f:

Unit-IV

' 6 1 3 8

4 2 % -1

10 3 9 7

16 4 12 15

(b) Show that the set S ® {(1,2, IX (2,1, OX(1, - 1,2)} forms basis of V JX).Or(a) Show that the mapping/: V3 (K) -> defined as below

/(a , A, c) (c, <i + b) is linear.(b) Show that the equations:

3x + Ay + Sz * a 4x + 5 y + 6z ** b 5x + 6y + Iz = c

: do not have a solution unlessa + c * 2b.

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Ex a m. Pa p e i b | 3

5.

[_• Unit-V(a) Prove that every Hermitian matrix #ean be represented in the form P+IQ, where P and Q are

real symmetric and skew-symmetric metric? and phow that: H * H is real if PQ “ - QP,

t* 1

(b) Using Cayley-Hamilton theorem find the inverse 6f the matrix: A

Or

(a) Find the matrix P which transforms the matrix A

3 -I I

- 1 5 -1

1 -1 3

-1 2 -1

1 -1 2

to diagonal form. Hence

calculate if1.(b) Reduce foeanonical form and find the rank imd the signatwe of the quadratic form:

q ■ 1x2-2> ,2-3*2 - 4 yz+ 6zx

B.E. (Third Semester) EiM riM tjuit D c^ 20N (N)Mathematics-Ill

. Note : The question paper ,is divided into five tmits. Each unit carries an internal choice. Attempt one question fiom each unit All questions carry equal marks.

Uaift'I

(a) Find the imaginary part oftheanalytic function/OO whose real part is x?-Ixy* +3x2-3y2.

Ans. (R tfer Page 21 )*2+/

(b) Evaluate J^(2 jc + + 1)<* along the two paths:

. CO * ■ /+ U y - 2 f i - 1. (ii) The straight line joining ( 1 - 0 and (2 + /). Ans. (R efer Pmge 56 )

Or

I.

2. 1 1

3.

(a) Find the image of the infinite strip ^ < y < j under the transformation to * l/z. Show the

region graphically. Ans. (Rtfer Page 42)(b) Apply the calculus of residues to show th at:

f2*— ^ — * ■r ?mmmta> b> 0 Ans. (Refer Page 81) fo r a similarproblem.

Jo a + b cos0 Va2-&2Uatt-II

(a) Use Newton’s divided difference formula to find a polynomial hi jc, given: jc 0 2 3 6

f ( x ) 648 704 729 792 Am*. (Refer Page 251)

(b) Find the approximate value of log. 5 by caicu&ing to four decimal places using Simpson's

f* dxl/3rd rule for dividing the range into 10 equal parts. Aas. (Refer Page 276)

Or

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JE x a m . P a p e h » I 5

-* Machines

Operators

4

Mi m 2 M i- m 4

I 47 32 46 44M 38 48 40 38h i 32 42 36 30 1IV 46 40 42 46

(b) Find the dua] of the following L.P.P:

Minimize: z * x, + x2 + x3

Subject to; xt - 3x2 + 4x3 = i

Xj - 2x2 £ 3

2x2-x3i4X j.X j Ô

x3 is unrestricted in sign.»»i t -V

9. (a) Obtain the steady stale equations for the queueing model (M/M/1): (oo/FCFS).(b) Find the transition matrix corresponding to the following transition diagram:

Or10. (a) One type of aircraft is fqund to develop engine trouble in 5 flights out of a total of 100 and

another type in 7 flights out of a total 200 flights. Is there a significant difference in the two types of aircrafts so far as engine defects are concerned ?

(b) Write short notes on the following:(i) Sampling (li) Random smapling (iii) Factorial design(iv) Tfegpchi loss function

B.E. (Third Semester) Examination, Jine, 2009 Matkematics-ni

Note: Attempt all questions. All questions carry equal marks. One complete question solve at one place.1. (a) Show that the Polar form of Cauchy-Reimann equations are :

du 1 dv dv 1 du

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61 EwawcMBwa m nKM tm xm

A du « ''

3.

?r' toVtfrr+ ~ -~ +

$0? frl 58$ ‘.'-.-ir- • 1i A‘J

421 +z + 5^----------- - (<v{

where C is the Ellipse: (x/2)2 +(y/3)2 =*I ;

Find the value o f:

co f (35> mOr

1(a) Expand:

Abs. (ReferPage 63)

(b) Show that:

m

f _________________*> l-2acos0 +a2 I

(*- !)(*^2) ^ * e ,^ on W> 2. A it (Refer Page 101)

2* cos26d6 2 jw 2 j-(oJ < 1) Am. (Refer Page 89)

2. (a) Show that the nth difference ofa polynomial of degrees will be constant and all (»+ l)thand higher ofder difference arc low. A*%. (Refer Page 200)

lx= 1.1:

X . f i x )1J5 "' 6 ' 12 0.128

U 0544

1 j6 L296U 2432 2J0 4 jOOO Ans. (Refer Page 266)

(a) Find; rOr

e*dx

Ana. (Refer Page 285)by taking seven ordinates using Simpsoffe 1/3 mle.(b) Define forward, backward, central difference operator and shift average operator also.(a) Solve the following by Euler modified method:

^ = log(x+ y), y(0) * 2

at x » 1.4 with h « 0.2. An. (Refer Page 317)

(b) Find the roots of equation xe* = eosx using Regula-Falsi method, correct to four decimal places. Or Abs. (Refer Page 131)

(a) Solve by Gauss-Seidel method:

lOx, -2x2 -*j -x 4 * 3 -2xj + I0x2 -x 3~x4 «15

-x, -x2+ 10*3 -24x4 » 27 -x, - x 2 -2xj + 10x4 m-9 Ana. (Refer Page 182)

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V - v B c mi. Papgbs { 7

(b) Giventhe values of u (x, y ) on the boundary'of the square in bloow. Evaluate foe function‘‘‘i (x,$ safl&fymg foe Laplace equation ^ *0 aMhe Pivotal poWofthis figure.

1000

aooo

aooo

i© i 1000. 1000

500

behB the lower

combined). Tbe belt A requires* fimcy buckle and only 400 per day are available. There are

700 buddes a day available for belt B. Deferable foe optimal product mix by graphical method,(b) Solve the following transportation problem:

Destination

Source

Demand

SyS2S,

Di O i D* Da Supply

19 30 50 10 770 30 40 60 940 8 70 20 18

5 8 7 14

Or (a) Four jobs are to be done on four different machines The cost (in rupees) of producing fth job

on the Jth machine is given below:

Machine

Mx M2 M3 M4

Job

A

Ji

A J a

15 U 13 15

17 12 12 13

14 15 10 1416 13 II 17

Assign the jobs to different machine so as to minimize the total cost

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(b) Solve by Simplex method: Minimize:

| E n g in e e r in g M m h e m a iic s -ID

*rti

Subject to:

x f -3ucj + 2 x 3 .

3jcj ~x2 + 3x3 5 7-2 xj + 4x2 512

~4X| •+ 3x2 + 510 X|, Xj, x3 £0

(a) In a railway marshalling yard goods train arrive ata rate of 30 tram per day. If the distribution of arrivals is the Poisson and that of service time is exponential with an average 36 minutes,1then find:

(i) Mean queue size.

(ii) The probability the queue size exceeds.

If the input of trains increases to an average 33 per day what will be the change in (i) and (ii) ?(b) The state transition matrix’for retentions gains and -losses of firms A, B and C is given below.

Using this matrix determine the steady state equilibrium condition: .

Form To

A B C

A 0.700 0.100 0.200

B 0.100 0.800 0.100

C 0.200 o.ioo 0.700 !

Or

(a) Write the advantage and disadvantage of Robust design method.

(b) The number of units of an item that are withdrawn from inventory on a day to day basis is a Markov Chain Process in which requirement for tomorrow depend on today’s requirement A one day transition matrix is jpven below.

Number of units withdrawn from inventory :

5

Today 10

12

Tomorrow 5 10 12

0.6 0.4 0.0

03 03 0.4

0.1 03 0.6

(i) Construct a tree diagram showing inventory requirements on two consecutive days.

(ii) Develop a two day transition matrix.

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B.E* (Third Semester) Fiawlasrton, Feb, 2010 Matfceraatks-IU

Note : Attempt five questions in all selecting one question from each unit All questions cany equal marks.

Unft-IJ . (a) Show that the function e* (x cos y - y siny) is harmonic and find its conjugate.

Abs. (Refer Rage 24)

2 : - •

Ex a m. Pa f b k } 9

(b) Evaluate f -----*c ( z - lX*-2)

where C is the circle Iz 1*3.

Or

Abs. (Refkr Page 76)

2. (a) Find the bilinear transformation which maps the joints * = l , i , - 1onto points w * 4 0, - i.

1 Abs. (Refer Page 25)r2/% COS20 dB

(b) Evaluate ! ~—r----- r—Jo l-2tfcos6+<zr* (tf2< t.)

Unfc-II

Abs. Page 89)

3. (a) (i) Prove that:

Abs. (Refer Page 204)

(ii) Obtain the function whose first difference is 2X3 + 3x* - 5x + 4 .Abs. (Refer Page 211)

dy(b) Find ~ at x * 1.1 from the following table:

X y’ TF 0

12 0.128

1.4 05441.6 1296

1* 2.432 '

2.0 4.00 Abs. (Refer Page 266)

Or

4. (a) Find the cubic polynomial which takes the following values

x : 0 1 . . 2 3

f(x) : 1 2 1 10 Hence or otherwise evaluate /(4). Abs. (Refer Page 223)

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1 0 1EmMHOHQKwMAncs-ill

(b) The following tabfe gives the vetority vofapartjcleattinK;;;

! v(m W .i v "-

24

6r

8

10

12

■' . * ' '

*616

34

60 '

94

136

Find the distance moved by the particle in 12 seconds and also the acceleration at/-»2 second.

Ass. (Refitr Page 28$)

Unit-III

5 (a) Find a real root of the equation jc log* x ~ 1.2, correct to five decimal places by Newton- Raphson method. Ans. (Refer Page 146)

(b)Sotvetheequdiom:

Sx+2 y + z * 12 x + 4 y + 2 z * 15 x+2^ +5z * 20

byQmm Saidel method. Ans. (Refer Page 185)Or

6. (a) Solve the equations:

3x + y + z » 4 x + 2 y+ 2z * 3 * •

2 x + y + 3 z * 4 byCroutVmethod

(b) Apply Runge-Kutta method to find an approximate value of>»for x * 0.2 in steps of 0.1, if

*dy— “ X+^.^CO)* 1. Ans.(ReferPage321)

Urit-IV

7. (a) Solve the following L.P.P., using Simplex method:

Maximize z*5xj + 3xj Subject to : x\ +x2 £ 2

5xi + 2x2 5 10

3xi + 8x2 5 12

xi,x2 it 0

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(b) Solve the foi *’*L *' v

O m s u i ^ ^ ..A” ‘ *i

i r Available

I 6 a 4 1411 4 9 8 12

, ni I 2 6 5

Required 6 10 15 31

11

(a) Using duality solve die following problem:

Maximize: •z **0-7 x, + 0.5 x2 .Subject to : x i ^ 4 , x2 ^ 6 .

xj + 2xi * 20

2&1+X2 £ 1 8 *>' > .

X|,X2 * 0

(b) A firm plans to begin production of three new products on its. three ptftst*. H eiptt cost of producing / at plant j is as given below. Find the assignment that Minimises the total unit cost:

Plant *

’ 1 2 31 10 8 12

Product 2 18 6 - 143 6 4 2

U n i t -V

. (a) Obtain the steady state equations for the (M j M11) : (oo /FCFS) qticuing model.

(b) Write sfaoit notes on the following:

(i) Taguchi toss function

(K) Markovian process

Or

0. (a) There is congestion on the platform of railway station. The trains arrives at the rate of 30trains per day. The waiting time for any train to hump is exponentially distributed with an average of 36 minutes. Calculate the mean queue size and the probability that queue size exceeds 9.

(b) Write short notes on the following:

0) Factorial design(ii) Robust design method.

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B.E. (Third Semester) Euniaitkm , J u e , 201!Mathemattes-III

Note: Attempt all questions by selecting any two part*\fi*« e*ch Questions. All questions carry equal marks.

(a) Find the imaginaiy part of the analytic function whoê/eal part is :

12 | Enq b«e8b«3 M athem at ics- !! !

1.

2.

(b) Evaluate:

x 3 - 3 ^ - 3 * a - 3 ^

f _ £2z... dzk < * + l ) 4


where C is the circle 12 j * 3.

(c) Show that:

r*« *0__£o 2 +cos0

2it ^ '

Ans. (X ^r 6/)


2*+3(d) Show that the transformation co “ changes the circle x1 +>*-4z * 0 into straight line

4* + 3 ~ 0,

(a) Prove that:

(i) e*>»l+A

09 e*

Ans. (ffgfer 5??


Ans. (Rtfer Page 204)

(b) Find /(9 ) from the following table:

x

5 7

11 13

17

(c) Find — atx * 1.5 from the following table:

x

1.5

2.02.5

3.0 15

4.0

/<*)150

392

1452

2366

5202

y 3.375

7.0

13.625

24.0 38.875

59.0

Ans. (Rtfer Page >

Abs. (Refer Page 260)

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• A EXAK. PAFtMfi | IS

(d) Evaluate: "V *'* *’*'m / 2 fU -*"viV/

. ... : . . t, i: . i u,

\ . . ,.*■ , • • . .

(i) Rising Simpsons - ~ rule.(ii) Using Weddle’s rule. Aim*.(Refer Page 277)

(a) Using Newton-Raphson method find the real root of the equation 3x * cos x + 1 correct to five decimal places. Abs. (Refer Page 141)

(b) Solve the following system of Aquations using Gauss-Seidel method :27x + 6y-* * 85

6x + \Sy + 2z “ 72 x + y + 54z « 110 Ans.(ReferPage 179)

dy ‘ .

(c) Findy (2.2) using Euler’s method from the equation xy*9y (2) * 1.Ass. (Refer Page 313)

(d) Apply Runge-Kutta method to find an approximate value ofy for x » 0.2 in steps of 0.1 if

dy-r£ pcx+â, given thaty* 1whenx~0. Ans. (Refer Page 321)

(a) Solve the following L.P.P. by Simplex method:Max:

z - 3r|+2r2Subject to:

X| + Xj £ 4

xj-xj £i 2 X|,X2 fc'O

(b) Solve the following L.P.P. by Graphical method:Max.: z * 5xi + 7xjSubject to constraints:

xj + x2 S 4 3xt + 8x2 £ 24

10bC| + 7x2 £ 35

xn x2 * 0(c) Find the initial basic feasible solution of the following transportation problem by Vogel’S approximation method:

Wvehouse

F,Factory F2

F,

W, w, w, w4

19 30 50 1070 30 40 60

*40. 8 70 20

5 8 7 14

7918

34

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minutes. If be mpaks scts in oif4fryta which ti&y come maad if ike arrival of sets Is approximately Potoon within averagfrfjats of 10 per 8 hours per day, what is repairman's expected idfe time each day? How manyjobs are ahead of the average set just brought in j —

(c) Write short note on the foUowbgg:

N toetioiiRo^dwgnoKttxxL ?f,

U t (Third Stme&er) fcuwhurtlee, Dee, 2011

ii*fiom each Unit All question carry equal marks!

(Jlitt-I

(a) Shew that the frootiong^x3- la y 1 is harmonic andlind the corresponding analytic function afthisas the*Wpait ( Ass. (Refer Page 20)

fh) jjl lTed flflrivn airTij theorem. Aaa. (3Iq tr Age 5*>

Or

(a) Find the b l t & i e a ^ j s u q a s U iv*-1to /, 0, - / respectively. >

1 Abs. (Ifgfcr Page 35)* A V

f oosxr «integral formula, evaluate the integral1 L 7' ’n ;r .’n where Cis the circle

. *; • . • • • ' . ; ’ .* .. -iy ;vv’ ' U " ‘V- V

UttU-H

(a) Prove with usual notations that:

AD " log (1 + A) * - tog (1 - V)

(b) Expressy-2X3 - 3X2 + 3Lc- 10 in fectoriri notation form.

Or

(a)G|ventiiat:

Ans. (K gtr/ to? * 77)

Aaa-fref&Fage Iff) Abs. (Refer-Page 210)

/*

x

1.0

1.1

U

1.3

1.4

F.5

1.6

y .7.989

8.403

8.?8l

9.129

9.451

9.750

10.031

Find ffe dx

1 j6» Abs. (R tfer Page 242)

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*?«

( b f m * J » d * > fS ttd 4 # k fa e N lf J r S N ljfc e fh w »“r; .v.'.'f.'.>' J £* o*. >:$<k>v,,.w:. • *>

y. ' ijuT • •'?* *‘'f, ‘ . **,v,~ .

' ** ’ f2 ** •*•*'• V* ,r>' •»*•"A/ -c. HJt'".• « , ' * ' ; i f

' r .1;ruc •; *••: v - ^ V

911

13 ‘

14 16

• vU n l t- m

(b) Performvalue profefem-:

efft* M ia

(a) Ffcid a nal coot of fee equation ** - 2x - £ * 0, by ffegnMyai d+cinidptocet. \ y

ofPkwd’s netted to find aaj^xadflnle

*'. - •' t ; ' V>. A • *|V V: -

i . , : J - * ♦ * > < » - *

Am *Or

(a) Solve thesystem ofequations by Qqm*G*toi ftyeffooafeod: .i . ., 27* + 6 > - z » | 5

<k+15>n-2z - 72.?' > + * + 54* « 110, ,, . ^

(b) By using NeftftteRaphson mated, find* rpatf** of*ie S ^ p t t a i i ^ w f e f e f c li.

n ar to*» 2#Mitect to three deota l places, " IMIttMdWW M V

(a) Find th v M v f ftr&ltowing L.P.P.:

Mf*%jr+z-x -3 y + 4z *>.5 '• ..

x + 2y 5 . 3 2 x - i i 4 x,^,z £ 0

(b) Four jobs are to be demon four (Sfiefeat xaacktnes. The cast 0n rupees) oCfKofecmg Alt job to yth machine is gh*a below:

Vmkim — ►

M i* ! * 2 • if#

i i i 30 2i- ^ ------

u IS 33 * 31 1

in 41 25 24 21

IV 13 30 —------ f 'H f I f . .

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r 'W f T - ' "

Stauncei A *>2 *3 o 4

s t 19 30 50 10 7

S2 70 30 40 60 9

Si 40 8 70 20 18

8 7 14 34

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DK Jain M3