Numerical Analysis (MATH 292)

Lecture 5- Numerical Integration and Differentiation

1- Numerical Integration

Newton - Cotes formula

b b

na a

I f x dx f x dx Where fn(x) is a polynomial of the form

1 20 1 2 ...n

n nf x a a x a x a x Where n is order of the polynomial

The Trapezoidal rule:

.2

f b f aI b a

Error in Trapezoidal rule

True error: Et = Itrue - Iapprox

True percent relative error: 100true approx

t

true

I I

I

Simple application of the trapezoidal rule

2 3 4 50.2 25 200 675 900 400f x x x x x x From a = 0 to b = 0.8

The exact value of the integral is 1.640533

Solution:

0 0.2

0.8 0.2323

f

f

0.17282

f b f aI b a

1.640553 0.1728 1.46775tE

1.46775

100 89.5%1.640533

t

In actual situations, we would have no knowledge of the true value. Therefore, an

approximate error estimate is required

3

2

1''

12aE f b a

n

where

''''

b

af x dx

fb a

In the example above:

2 3'' 400 4050 108000 8000f x x x x

0.82 3

0'' 400 4050 108000 8000

''0.8 0

60

b

af x dx x x x dx

fb a

3 3

2

1 1'' 60 0.8 2.56

12 12aE f b a

n

Multiple application of the trapezoidal rule

1

0

1

22

n

i n

i

b aI f x f x f x

n

Example:

Use the trapezoidal rule to obtain an approximate value of the integral

2

315

1

d xwhere n

x

Solution:

x 1 1.2 1.4 1.6 1.8 2

3

1

1x 0.707107 0.605449 0.516811 0.442981 0.382583 0.33333

2 1

[0.707107 2(0.605449 0.516811 0.44298110

0.382583) 0.33333] 0.493609

b

af x dx

Simpsons 1/3 Rule

1 2

0

1,3,5 2,4,6

4 23

n n

i i n

i i

b aI f x f x f x f x

n

Example:

Use the Simpsons 1/3 rule to obtain an approximate value for the integral from 0

to 0.8 (n=2). 2 3 4 50.2 25 200 675 900 400f x x x x x x

0

1

2

0.2

0.4 2.456

0.8 0.232

f x f a

f x f

f x f

0.8

0.2 4 2.456 0.232 1.3674672 3

I

Recall that 1.640533trueI

1.640533 1.367467 0.2730667tE

16.6%t Error in Simpsons rule:

5

4

4180a

b aE f

n

where 4f is the average fourth derivative

0.8

2

4 024 900 120 400

24000.8 0

x x dxf

5

0.82400 0.273

2880a

Example:

Use Simpsons 1/3 rule with n=4 to estimate the integral

2 3 4 50.2 25 200 675 900 400f x x x x x x

Solution:

0 4 23

n

b aI f x odd even f x

n

x 0 0.2 0.4 0.6 0.8

y = f(x) 0.2 1.288 2.456 3.464 0.232

n = 4 0.8

0.24

h

0.2

0.2 4 1.288 3.464 2 2.456 0.232 1.6234673

I

5

4

0.82400 0.017067

180 4aE

1.04%t

Simpsons 3/8 Rule: In the similar manner to the derivation of the trapezoid and Simpsons 1/3 rule, a third order Lagrange polynomial can be fit to four points and integrate

3b b

a aI f x dx f x dx

To yield

0 1 2 33

3 38

I h f x f x f x f x

Where b a

hn

Use Simpsons 3/8 to integrate the same function for n = 3

2 3 4 50.2 25 200 675 900 400f x x x x x x

0.8 0

0.26673

b ah

n

0 0.2 0.5333 3.487177

0.2667 1.432724 0.8 0.232

f f

f f

0.2 3 1.432724 3.487177 0.2320.8 1.519170

8I

5

2400 1.640533 1.51917 0.1213636480

a

b aE

7.4%t

For n = 5, use the same function to integrate the first two segments using Simpsons 1/3 and the last 3 segments using Simpsons 3/8

0.80.16

5h

0 0.2f 0.16 1.296919f 0.32 1.743393f 0.48 3.186015f 0.64 3.181929f 0.8 0.323f

0.16 0.32 0.48 0.64 0.80

Simpsons 1/3 rule

Simpsons 3/8 rule

The integration for the first two-segments is obtained using Simpsons 1/3 rule.

0.16

0.2 4 1.296919 1.743393 0.38032373

I

For the last 3 segments, the Simpsons 3/8 rule can be used to obtain

3

0.16 1.743393 3 3.186015 3.181929 0.323 1.2647548

I

The total integral

0.3803237 1.264754 1.645077I

0.00454383tE 0.28 %t

Errors in numerical integration:

t t appE I I

100true app

t

true

I I

I

Approximate Error Estimate:

1) Trapezoidal rule

3

2''

12a

b aE f

n

2) Simpsons 1/3 rule

5

4

4180a

b aE f

n

3) Simpsons 3/8 rule

5

4

480a

b aE f

n

Numerical Differentiation

Mathematically, the derivative represents the rate of change of a dependent variable with

respect to an independent variable.

The mathematical definition of the derivative begins with a difference approximation

( ) ( )i if x x f xy

x x

If x is allowed to approach zero, the difference becomes a derivative

0

( ) ( )lim i ix

f x x f xdy

dx x

The derivative dy

dx is the slope of the tangent to the curve ( )y f x at

ix .

Taylor Series:

Taylor series is of great value in the study of numerical methods.

By using the Taylor method we can predict a function value at one point in terms of the

function value and its derivatives at another point.

2

1 1 1

( )

1

''' ...

2!

!

i

i i i i i i i

nni

i i n

f xf x f x f x x x x x

f xx x R

n

where nR is the reminder term.

It is usually convenient to simplify the Taylor series by defining a step size 1i ih x x

Let 1i ix x h , then

( )2

1

'''

2! !

n

ni i

i i i n

f x f xf x f x f x h h h R

n

Finite divided difference approximation of derivative

First derivative

Taylor series can be solved for

1

1

( )

2

1

' ( ) (1)

where ( )

''and

2! !

i i

i

n

ni i

n

f x f xf x O h

h

RO h

h

f x f xR h h R

n

1( ) ( )i if x f x is the first forward difference and h is called the step size ( length of

interval)

There are three ways to approximate the first derivative form Taylor series

1- Forward difference: 1' i ii

f x f xf x

h

2- Backward difference: 1' i ii

f x f xf x

h

3- Centered difference: 1 1'

2

i ii

f x f xf x

h

Example:

Use forward, backward and centered difference approximation to estimate the first

derivative of

4 3 20.1 0.15 0.5 0.25 1.2f x x x x x

at 0.5x using a step size 0.5h . Repeat the computation using 0.25h , compute the

true percent relative error

Solution:

Exact first derivative 3 2' 0.4 0.45 0.25f x x x x ' 0.5 0.9125f

For h = 0.5

1 1

1 1

0 1.2

0.5 0.925

1.0 0.2

i i

i i

i i

x f x

x f x

x f x

Forward divided difference:

0.2 0.925

' 1.45 58.9%0.5

i tf x

Backward divided difference:

0.925 1.2

' 0.55 39.7%0.5

i tf x

Centered divided difference:

0.2 1.2

' 1.0 9.6%1.0

i tf x

For h = 0.25

1 1

1 1

0.25 1.10351

0.5 0.925

0.75 0.6363281

i i

i i

i i

x f x

x f x

x f x

Forward divided difference:

' 0.25 1.155 26.5%tf

Backward divided difference:

' 0.5 0.714 21.7%tf

Centered divided difference:

' 0.75 0.934 2.4%tf

For both step sizes, the centered difference approximation is more accurate than the other.

Finite difference of higher derivative:

Taylor Series expansion can be used to derive numerical estimates of higher derivatives.

To do this, we write a forward Taylor expansion of 2( )if x in terms of ( )if x

2

2

''' 2 2 ...

2!

i

i i i

f xf x f x f x h h (2)

The Taylor expansion of 1( )if x is

2

1

''' ...

2!

i

i i i

f xf x f x f x h h (3)

Multiply equation (3) by 2 and subtract from equation (2)

22 12 '' ...i i i if x f x f x f x h

2 122

'' i i iif x f x f x

f x O hh

(4)

Equation (4) is called the second forward finite divided difference.

High-accuracy differentiation Formulas

High-accuracy forward divided-difference

We have

1 '''

2

i i i

i

f x f x f xf x h

h

(5)

Substitution of equation (4) into equation (5), gives

1 2 1

2

2 1

2' .

2

4 3

2

i i i i i

i

i i i

f x f x f x f x f x hf x

h h

f x f x f x

h

High-accuracy backward divided-difference

1 23 4( )

2

i i i

i

f x f x f xf x

h

High-accuracy centered divided-difference

2 1 1 28 8( )

12

i i i i

i

f x f x f x f xf x

h

Example:

Find the first derivative of 4 3 20.1 0.15 0.5 0.25 1.2f x x x x x at 0.5x , use high accuracy differentiation formulas with 0.25h

Solution:

2

1

1

2

0 0 1.2

0.25 0.25 1.103515

0.5 0.5 0.925

0.75 0.75 0.636328

1.00 1.0 0.2

i

i

i

i

i

x f

x f

x f

x f

x f

Forward difference:

0.2 4 0.636328 3 0.925' 0.5 0.859375

2 0.25f

5.82 %t

Backward difference:

3 0.925 4 1.103515 1.2' 0.5 0.878125

2 0.25f

3.77 %t

Centered difference:

0.2 8 0.636328 8 1.103515 1.2

' 0.5 0.912512

fh

0 %t