Chapter 5 Bracketing Methods - موقع المهندس حماده شعبان (b) shows the case where a single root is bracketed by negative and positive values of f(x). However, fig(d)

May 2014

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Chapter 5

Bracketing Methods

May 2014




Numerical Methods

Roots of Equations

Chapter 5 Chapter 6

Bracketing Methods Open Methods

ا يصرر اامهندسونررالاامهورر عثالا مرر ع ا رر ا ررا اكثيرر ما ررpolynomialsيصررمحاعادرر اا

ا:ب هط قامهنمت ةا ثل

امهـ(Numerical Methodsا)تمنرلا،اأيامهصموةامهنم ع تس عوافياإيج اجذارا ثلاهذها

.مهتياتجملاقينةامهنم هةا س ايةاهاصف اxقيماقينةاأااعاىاإيج ا

تسقسماط قامهحلاإهىاقسني ارئيسيي اعسحامهشكلاأعاله،اكلاقسرمايحتراياعارىا رالقاطر قا

.مهنم ع إليج اجذارا

ا

كررلا جناعررةاألامهنجناعررةامباهررىاتورروأامهحسرر ا رر اقينترري ااطرر قمهفرر قامه ئيسرريابرري ا

ةامهث يرةافتوروأابقينرةاامعروةاق يورةا ر امهجرذراتق يويتي ابيسدن امهجرذرامهنطارا ،اأ ر امهنجناعر

ا.مهنطاا

Graphical Method

Bisection Method

False Position Method

Simple Fixed Point Method

Newton Raphson Method

Secant Method

May 2014




This chapter on roots of equations deals with methods that use the

fact that a function changes sign around its root.

These techniques are called Bracketing Methods because two guesses for

the root are required, these guesses must “bracket” the root. These

methods will reduce the width of the bracket.

5.1 Graphical Methods

A simple method for obtaining an estimate of the root of the

equation f(x) = 0 is to make a plot of the function and observe where it

crosses the x axis. This point, which represents the x value for which

f(x) = 0, provides a rough approximation of the root.

x x root root

ا.اقيمامهومهةاقولاابموامهجذراهد اإش رم ا ختافةا-

ينك امهحصالا اهذهامهط يقةاعاىاقيماتق يويرةاا-

ا.هاجذاراتستخوماكسقطام طالقاهو قيامهط ق

. القاط قاإليج اجذارامهنم ع (ا5) س قشافياش بت ا -

.تق يويتي اهاجذراع اينيسهاايس رهتووأاكلاط يقةا سد ابقينتي ا -

كررلاوطرراةا رر اوطررام اهررذهامهطرر قامهررثالقاتمنررلاعاررىاتقاررييامهنسرر فةابرري امهقينترري امهاترري ا -

ا.تحص ملامهجذرابيسدن

الذين يسعون دائما نحو التميز واألفضل .يصبحون تلقائيا مثال لآلخرين

May 2014




EXAMPLE 5.1 The Graphical Approach

Problem Statement:

Use the graphical approach to determine the drag coefficient c needed for

a parachutist of mass m = 68.1 kg to have a velocity of 40 m/s after

free-falling for time t = 10 s. Note: The acceleration due to gravity is

9.8 m/s2.

Solution:

using the parameters t = 10, g = 9.8, v = 40, and m = 68.1:

or

Various values of c can be substituted into the

right-hand side of this equation to compute the

values of f(c).

These points can be plotted as shown in the

figure.

f(c) = g m

c (1 – e –(c/m)t) - v

f(c) = 9.8 (86.1)

c (1 – e –(c/68.1)10) - 40

f(c) = 667.38

c (1 – e -0.146843c) - 40

دت أن تختلف مع إنسان، رإذا أ

.قل له ما يجب عمله

ا.ا.ا.اا21،اا21 امه نماينكر امنرتست وااجرا اجرذرابري ا

ا. اهذهامهط يقةامهنطاا اهذماهاا

May 2014




Special Cases

Graphical techniques are of limited practical value because they are

not precise. However, they can be used to obtain rough estimates of roots.

These estimates can be employed as starting guesses for the other

numerical methods discussed in this and the next chapter.

They are also important tools for understanding the properties of the

functions and guessing the pitfalls of the other numerical methods.

Figure (b) shows the case where a single root is bracketed by negative and

positive values of f(x).

However, fig(d) where f(xl) and f(xu) are also on opposite sides of the x

axis, shows three roots occurring within the interval.

In general, if f(xl) and f(xu) have opposite signs, there are an odd number of

roots in the interval.

As indicated by fig (a) and (c), if f(xi) and f(xu) have the same sign, there are

either no roots or even number of roots between the values.

غالبا ما تكون النهاية

.حسب الغاية

عاألامهست ئجامهتق يويةامهتيا حصلاعايد ا اهذهامهط يقةاغي ا قيقةاإعاأ د اتنثلاقيمام طالقاهور قيا

ا.مهط ق،ااأ ام ا دنةاهفدماطويمةاهذهامهوامل

May 2014




Although previous generalizations are usually true,

there are some cases where they do not hold.

For example, functions that are tangential to the x

axis as in figure(a) and discontinuous functions as in

fig(b) can violate these principles.

An example of a function that is tangential to the

axis is the cubic equation f(x) = (x – 2)(x – 2)(x – 4).

Notice that x = 2 makes two terms in this polynomial

equal to zero. Mathematically, x = 2 is called a

multiple root.

The existence of such cases makes it difficult to develop general

computer algorithms guaranteed to locate all the roots in an interval.

عسوام قط امهومهةااعومامنتن مريةامهنسحسىامهخ صابد ،ايحتنلااجا اجرذري ابري ا

ا.قينتي اتختافاقينةاإش رةامهومهةاهدن ،اأااعومااجا اأيةاجذاراعاىامإلطالق

المنافسة الحقيقيةة دائمةا مةا تكةون ةين

.وما أنت قادر على فعلهما قمت فعله

عسو اتكالاقيمامهومهةابرسف ا

مإلشررررررر رةاعسررررررروا قطتررررررري ا

يحتنررررلاعرررروماا تجرررر ارتي

اجا اجذاراأاااجرا اعرو ا

ا.جذارازاجي

عسرواتيير اإشر رةاقينررةا

مهومهررررررةاقوررررررلاابمرررررروا

ا قطتررررري ا تجررررر ارتي

يحتنررررلااجررررا اجررررذرا

امعرررواأااعرررو اجرررذارا

ا.ف ي

May 2014




EXAMPLE 5.2 Use of Computer Graphics to Locate Roots

Problem Statement:

Computer graphics can fasten and improve your efforts to locate roots of

equations. The function f(x) = sin 10x + cos 3x

has several roots over the range x = 0 to x = 5.

Use computer graphics to know about the behavior of this function.

Solution:

Packages such as Excel and MATLAB software can be used to generate

plots. The following figure(a) is a plot of f(x) for x = 0 to x = 5.

This plot suggests the presence of several roots, including a possible

double roots at x = 4.2 where f(x) appears to be tangent to the x-axis.

In fig(c), the vertical scale is narrowed further to f(x) = -0.15 to

f(x) = 0.15 and the horizontal scale is narrowed to x = 4.2 to x = 4.3.

This plot shows clearly that a double root does not exist in this region, but

there are two distinct roots at about x = 4.23 and x = 4.26.

لمسة األخيرة على عملك، وضع ال

. مثا ة التاج الذي يكلل نجاحك

May 2014




-8

-4

0

4

8

-5 0 5 10

Problem 5.1

Determine the real roots of f(x) = -0.5x2 + 2.5x + 4.5:

(a) Graphically.

(b) Using the quadratic formula.

Solution

(a) A plot indicates that roots occur at about x = –1.4 and 6.4.

(b)

40512.6

40512.1

)5.0(2

)5.4)(5.0(4)5.2(5.22

x

الكثير من العمل الجيد يضيع سبب

.العجز عن ذل ما هو أكثر قليال

May 2014




5.2 The Bisection Method

When applying the graphical technique in example 5.1, you have

observed that f(x) changed sign on opposite sides of the root.

In general, if f(x) is real and continuous in the interval from xl to xu and f(xl)

and f(xu) have opposite signs, that is,

f(xl) f(xu) < 0

then there is at least one real root between xl and xu.

The bisection method, which is alternatively called binary chopping, is

one type of incremental search method in which the interval is always

divided in half. If a function changes sign over an interval, the function value

at the midpoint is evaluated. The location of the root is then suggested to lie

at the midpoint of that new subinterval. The process is repeated to obtain

refined estimates. The figure shows the steps for the bisection method.

Step 1: Choose lower xl and upper xu guesses for the root such that the function changes sign over the interval. (ensure that f(xl) f(xu) < 0)

Step2: An estimate of the root xr is determined by

Step3: Make the following evaluations to determine in which subinterval

the root lies: (a) If f(xl) f(xr) < 0, the root lies in the lower subinterval.

set xu = xr and return to step 2.

(b) If f(xl) f(xr) > 0, set xl = xr and return to step 2.

(c) If f(xl) f(xr) = 0, the root equals xr, terminate.

xr = xl + xu

2

ا.إهىامهسصفااxuاااxl قامابتضييقامهف قابي اااiterationأن ساهذهامهط يقةاأ هافياكلا

ا ذل كل ما تستطيع، لكل

.شيء أنت مسئول عنه

May 2014




The following figure shows a graphical description of the bisection

method. This plot conforms to the first three iterations from Example 5.3.

Example 5.3: Bisection Method

Use bisection method to solve the same problem approached graphically

in example 5.1.

Solution:

The first step in bisection is to guess two values of the unknown (xr) that

give values for f(xr) with different signs. We can see that the function

changes sign between values of 12 and 16. Therefore, the initial estimate

of the root xr lies at the midpoint of the interval

This estimate represents a true percent relative error of εt = 5.3%

(note that the true value of the root is (14.7802).

Next we compute the product of the function value at the lower bound

and at the midpoint:

f(12) f(14) = 6.067 (1.569) = 9.517

= 14 xr = 12 + 16

2

نقطةةة اطنطةةالل لكةةل اتنجةةا ات هةةي

.و استمرارالرغبة، تذكر هذا دائما

May 2014




Continue

which is greater than zero:

Since no sign change occurs between the lower bound and the midpoint,

the root must be located between 14 and 16.

Therefore, we create a new interval by redefining the lower bound to 14

and determining a revised root estimate

which represents a true percent error of εt = 1.5%.

f(14) f(15) = 1.569 (-0.425) = -0.666

Therefore, the root is between 14 and 15. The upper bound is redefined to

15, and the root estimate for the third iteration is calculated as:

which represents a percent relative error of εt = 1.9%. The method can be

repeated until the result is accurate enough to satisfy some standard error

(εs).

= 15 xr = 14 + 16

2

= 14.5 xr = 14 + 15

2

ا رع ذرة الرغبة في عقلك، وسوف

تشكل نواة ذات قوة تجذب إليه كل

.شيء تحتاجه لتحقيق نجاحه

May 2014




In the previous example the true error does not decrease with each

iteration. However, the interval is halved with each step in the process.

The interval width provides an exact estimate of the upper bound of the

error for the bisection method.

5.2.1 Termination Criteria and Error Estimates

We must now develop an objective criterion for deciding when to

terminate the method. An initial suggestion might be to end the

calculation when the true error falls below some pre-specified level.

For instance, in Example 5.3, the true percent relative error dropped from

5.3% to 1.9% during the computation. We might decide that we should

terminate when the percent error drops below, say, 0.1% percent.

This strategy is not useful because the error estimates in the example

were based on knowledge of the true root of the function. This would not

be the case in an actual situation. Therefore, we require an error estimate

that does not depend on the knowledge of the root. An approximate

percent relative error εa can be calculated as

where xrnew is the root for the present iteration and xr

old is the root from

the previous iteration. The absolute value is used because we are usually

concerned with the magnitude of εa rather than its sign. When εa becomes

less than a pre-specified stopping criterion εs, the computation terminates.

األمر ط ينتهي أ دا، فهناك دائما

.الغاية التالية، والهدف التالي

100% εa = xrnew – xr

old

xrnew

(5.2)

May 2014




EXAMPLE 5.4 Error Estimates for Bisection

Continue Example 5.3 until the approximate error falls below a stopping

criterion of εs = 0.5%. Use the error approximate equation.

Solution:

The result of the first two iterations for example

5.3 were 14 and 15. Substituting these values

into Eq. (5.2) yields

Recall that the true percent relative true error for the root estimate

of 25 was 2.5%. Therefore, εa is greater than εt. This behavior is clear for

the other iterations:

Thus, after six iterations εa = 0.442% and the computation can be terminated.

في أي مكان ترى فيه عمال ناجحا، ستجد

.شخصا اتخذ ذات مرة قرارا شجاعا

100% εa = 15 - 14

15 = 6.667%

May 2014




The plot shows that εa εt thus, when εa falls below εs. the

computation could be terminated with confidence that the root is known

to be at least as accurate as the pre-specified acceptable level.

Although it is always dangerous to draw general conclusions from a

single example, it can be demonstrated that εa will always be greater than

εt for the bisection method. This is because each time an approximation

root is located using bisection as xr = (xl + xu)/2, we know that the true root

lies somewhere within an interval of (xu – xl)/1 = Δx/1.

Therefore, the root must lie within ± Δx/1 of our estimate.

xr = 14.5 ± 0.5

Because Δx/1 = xrnew – xr

odd, Eq. (5.2) provides an exact upper bound

on the true error. For this bound to be exceeded, the true root would have

to fall outside the bracketing interval, which, by definition, could never

occur for the bisection method.

Although bisection is generally slower than other methods,

the neatness of its error analysis is certainly a positive aspect that could

make it attractive for certain engineering applications.

وصةةةل إليهةةةا ت التميةةةز هةةةو النتيجةةةة التةةةي ي

.من أجل التحسن متدريجيا عد الكفاح الدائ

May 2014




The following figure shows three ways in which the interval may

bracket the root. In (a) the true value lies at the center of the interval,

whereas in (b) and (c) the true value lies near the extreme.

Notice that the difference between the true value and the midpoint

of the interval never exceeds half the interval length, or Δx/1.

افعل الشيء الصحيح، في الوقت

.الصحيح، الطريقة الصحيحة

اΔxأكث ا ا صفااrtع ااxrعسوا ستصفامهنس فة،اب هت هياه ايوموااxrب س افياكلا ةا خت را

May 2014




The following figure shows a graphical description of why the error

estimate for bisection (Δx/1) is equivalent to the root estimate for the

present iteration (xrnew) minus the root estimate for the previous iteration

(xrodd).

We should note that the relationships

can be substituted into Eq.(5.2)ا to develop an alternative formulation for

the approximate percent relative error

xu - xl xrnew – xr

old = ا2

xu + xl xr

new = ا2

نجز عض األعمال شكل أفضل أ

.مما تم إنجا ها ه من قبل

May 2014




This equation yields identical results to Eq. (5.2) for bisection, in

addition, it allows us to calculate an error estimate on the basis of of our

initial guesses, that is, on our first iteration. For instance, on the first

iteration of Example 5.2, an approximate error can be computed as

Another benefit of the bisection method is that the number of

iterations required to reach an absolute error can be computed

before starting the iterations. This can be seen by recognizing that before

starting the technique, the absolute error is

Ea0 = xu

0 – xl0 = Δx0

where the superscript designates the iteration. Hence, before starting the

method, we are at the “zero iteration.” After the first iteration, the error

becomes

Because each succeeding iteration halves the error, a general formula

relating the error and the number of iterations, n, is

100% = 14.29% εa = 16 - 12

16 + 12

Ea1 = Δx0

2

نحةةةن مجةةةانين إذا لةةةم نسةةةتطع أن

نفكةةةر، متعصةةةبون إذا لةةةم نةةةرد أن

.نفكرنفكر، عبيد إذا لم نجرؤ أن

Ean = Δx0

2n

(5.4)

May 2014




If Ea,d is the desired error, this equation can be solved for

Let us test the formula. For Example 5.4, the initial interval was

Δx0 = 16 – 12 = 4. After six iterations, the absolute error was

We can substitute these values into Eq. (5.5) to give

Thus, if we know in advance that an error of less than 0.0625 is

acceptable, the formula tells us that six iterations will yield the desired

result.

إنك ط تقوى على أسر فكرتي،

ألنهةةةةةةا حةةةةةةرة كالنسةةةةةةيم فةةةةةةي

.الفضاء، ط حد له وط مدى

= 6 n = log (4 / 0.0625)

log 2

= 0.0625 Ea = |14.875 – 14.75|

ا2

= log2 n = log (Δx0 / Ea,d)

log 2 (5.5)

Δx0

Ea,d

May 2014




-8

-4

0

4

8

-1 0 1

EXAMPLE 5.2

Determine the real root of f(x) = 5x3 – 5x2 + 6x - 2

(a) Graphically.

(b) Using bisection to locate the root, Employ initial guesses of xl = 0 and

xu = 2 and iterate until the estimated error εa ≤ εs = 10%

Solution

(a) A plot indicates that a single real root occurs at about x = 0.42.

(b) First iteration:

5.02

10

rx

%100%10001

01

a

75.0)375.0(2)5.0()0( ff

Therefore, the new bracket is xl = 0 and xu = 0.5

The process can be repeated until εa falls below 10%. As summarized

below, this occurs after 5 iterations yielding a root estimate of 0.40625.

iteration xl xu xr f(xl) f(xr) f(xl)f(xr) a

1 0 1 0.5 -2 0.375 -0.75 2 0 0.5 0.25 -2 -0.73438 1.46875 100.00% 3 0.25 0.5 0.375 -0.73438 -0.18945 0.13913 33.33% 4 0.375 0.5 0.4375 -0.18945 0.08667 -0.01642 14.29% 5 0.375 0.4375 0.40625 -0.18945 -0.05246 0.009939 7.69%

عجيةةةةب أن نمجةةةةل مةةةةا وجةةةةب

.عمله اآلن ثم نصر على الندم

May 2014




-1.5

-1

-0.5

0

0.5

0 0.5 1 1.5

Problem 5.5

Locate the first nontrivial root of sin x = x3, where x is in radians.

Use a graphical technique and bisection with the initial intervals from 0.5

to 2. Perform the computation until εa is less than of εs = 2%. Also, perform

an error check by substituting your final answer into the original equation.

Solution A graph indicates that a nontrivial root (nonzero) is located at about 0.93. Using bisection,

75.02

15.0

rx

092067.0)2597638.0(354426.0)75.0()5.0( ff

Therefore, the root is in the second interval and the lower guess is

redefined as xl = 0.75.

All the iterations are displayed in the following table:

i xl f(xl) xu f(xu) xr f(xr) a

1 0.5 0.354426 1 -0.158529 0.75 0.2597638 2 0.75 0.259764 1 -0.158529 0.875 0.0976216 14.29% 3 0.875 0.097622 1 -0.158529 0.9375 -0.0178935 6.67% 4 0.875 0.097622 0.9375 -0.0178935 0.90625 0.0429034 3.45% 5 0.90625 0.042903 0.9375 -0.0178935 0.921875 0.0132774 1.69%

Consequently, after five iterations we obtain a root estimate of 0.921875

with an approximate error of 1.69%, which is below the stopping criterion

of 2%. The result can be checked

0132774.0921875.0)921875.0sin()sin()( 33 xxxf يخدع المظهر

ا.األول الكثيرينا

May 2014




5.3 The False-Position Method

Although bisection is a perfectly valid technique for

determining roots, it's "brute-force" approach is relatively inefficient.

False position is an alternative based on graphical knowledge.

A shortcoming of the bisection method is that, in dividing the

interval from xl to xu into equal halves, no account is taken of the

magnitudes of f(xl) and f(xu). For example, if f(xl) is much closer to zero

than f(xu), it is likely that the root is closer to xl than to xu.

An alternative method that uses the graphical knowledge is to join

f(xl) and f(xu) by a straight line. The intersection of this line with the

x-axis represents an improved estimate of the root. The fact that the

replacement of the curve by a straight line gives a "false position" of the

root is the origin of the name, method of false position.

من األفضل أن تكون مكروها

لشيء فعلته، من أن تكون

.محبو ا لشيء فعله غيرك

May 2014




Using similar triangles,

Which can be solved for

This is the false position formula. The value of xr computed with

the previous equation then replaces whichever of the two initial guesses xl

or xu, yields a function value with the same sign as f(xr). In this way, the

values of xl and xu always bracket the true root.

The process is repeated until the root is estimated adequately. The

algorithm is identical to the one for bisection with the exception that the

previous equation is used for step 2. In addition, the same stopping

criterion is used to terminate the computation.

= f(xl)

xr – xl

f(xu)

xr – xu (5.6)

أسةةاا النجةةاح فةةي اتدارة أن تسةةتفيد

أقصةةى مةةا يمكةةن مةةن قةةوة اآلخةةرين،

.وتتضرر أقل ما يمكن من ضعفهم

xr = xu - f(xu)(xl – xu)

f(xl) – f(xu) (5.7)

May 2014




EXAMPLE 5.5 False Position

Use the false position method to determine the root of the same equation

in Example 5.1.

Solution:

Initiate the computation with guesses of Xl = 12 and xu = 16

First iteration:

xl = 12 f(xl) = 6.0699

xu = 16 f(xu) = - 2.2688

which has a true relative error of 0.89 percent.

Second iteration:

f(xl) f(xu) = - 1.5426

Therefore, the root lies in the first subinterval, and xr becomes the upper

limit for the next iteration, xu = 14.9113:

xl = 12 ا f(xl) = 6.0699

xu = 14.9113 f(xu) = - 0.2543

xr = 14.9113 -

which has a true and approximate relatives errors of 0.09 and

0.79 percent. Additional iterations can be performed to refine the

estimate of roots.

xr = 16 - -2.2688 (12-16)

6.0669 – (-2.2688) = 14.9113

-0.2543 (12 – 14.9113)

6.0669 – (-0.2543) = 14.7942

له، فال تفع إذا لم يكن ما ستفعله صحيحا

.فال تقله صادقاوإذا لم يكن ما ستقوله

May 2014




A feeling for the relative efficiency of the bisection and false-position

method can be appreciated by plotting the true percent relative errors for

Examples 5.4 and 5.5.

Recall that in the bisection method- the interval between xl and xu

grew smaller during the course of a computation. The interval, as defined

by ∆x /2 = │xu – xl│/2 for the first iteration, therefore provides a

measure of the error for this approach. This is not the case for the method

of false position because one of the initial guesses may stay fixed

throughout the computation as the other guess converges on the root.

For instance, in Example 5.6 the lower guess xl remained at 12 while xu

converged on the root. For such cases, the interval does not shrink but rather

approaches a constant value.

الخيال هو داية ات داع، إنك تتخيل

يما تتخيله، ما ترغب فيه، وترغب ف

.فيه وأخيرا تصنع ما ترغب

May 2014




5.3.1 Pitfalls of the False-Position Method

Although the false-position method would seem to always be the

bracketing method of preference, there are cases where it performs poorly. In

fact, as in the following example, there are certain cases where bisection yields

superior results.

EXAMPLE 5.6 A Case Where Bisection Is Preferable to False Position

Use bisection and false position to locate the root of

f(x) = x10 - 1

between x = 0 and 1.3.

Solution:

Using bisection, the results can be summarized as

Thus, after five iterations, the true error is reduced to less than 2 percent.

For false position, a very different outcome is obtained:

.لئيمأذل الناا معتذر إلى ا

May 2014




Continue

After five iterations the true error has only been reduced to about

59 percent. In addition, note that εa ‹ εt. Thus, the approximate error is

misleading.

As in the figure, the curve violates the idea which false position was

based on, that is, if f(xl) is much closer to zero than f(xu), then the root is

closer to xl than to xu. Because of the shape of the present function, the

opposite is true.

القدر من الخيال يمكن أن يحقق نفس

.النجاح أو الفشل وفقا لتوجيهك له

May 2014




-8

-4

0

4

8

0 0.5 1 1.5

EXAMPLE 5.3

Determine the real root of f(x) = -25 + 82x – 90x2 + 44x3 – 8x4 + 0.7x5

(a) Graphically.

(b) Using bisection to determine the root to εs = 10%.

Employ initial guesses of xl = 0.5 and xu = 1.0.

(c) Perform the same computation as in (b) but use the false-position

method and εs = 0.2%.

Solution

(a) A plot indicates that a single real root occurs at about x = 0.58. (b) Bisection:

First iteration:

75.02

15.0

rx

%33.33%1005.01

5.01

a

063213)072362(478131)750()50( ....f.f

Therefore, the new bracket is xl = 0.5 and xu = 0.75. The process can be

repeated until the approximate error falls below 10%. As summarized

below, this occurs after 4 iterations yielding a root estimate of 0.59375.

ط يمكنك أن تصنع ما هو أكثر حماقة من أن تجلس على

.جانب الطريق حتى يأتيك أحدهم ويحاول مساعدتك

May 2014




Continue

iteration xl xu xr f(xl) f(xr) f(xl)f(xr) a

1 0.50000 1.00000 0.75000 -1.47813 2.07236 -3.06321 2 0.50000 0.75000 0.62500 -1.47813 0.68199 -1.00807 20.00% 3 0.50000 0.62500 0.56250 -1.47813 -0.28199 0.41682 11.11% 4 0.56250 0.62500 0.59375 -0.28199 0.22645 -0.06386 5.26%

(c) False position:

First iteration:

xl = 0.5 اf(xl) = –1.47813

xu = 1 f(xu) = 3.7

64273.07.347813.1

)15.0(7.31

rx

35808.1)91879.0(47813.1)64273.0()5.0( ff

Therefore, the bracket is xl = 0.5 and xu = 0.64273.

Second iteration:

xl = 0.5, f(xl) = –1.47813

xu = 0.64273, f(xu) = 0.91879

58802.091879.047813.1

)64273.05.0(91879.064273.0

rx

%304.9%10058802.0

64273.058802.0

a

The process can be repeated until the approximate error falls below

0.2%. As summarized below, this occurs after 4 iterations yielding a root

estimate of 0.57956.

iteration xl xu f(xl) f(xu) xr f(xr) f(xl)f(xr) a

1 0.5 1.00000 -1.47813 3.70000 0.64273 0.91879 -1.35808 2 0.5 0.64273 -1.47813 0.91879 0.58802 0.13729 -0.20293 9.304% 3 0.5 0.58802 -1.47813 0.13729 0.58054 0.01822 -0.02693 1.289% 4 0.5 0.58054 -1.47813 0.01822 0.57956 0.00238 -0.00351 0.169%

إن المعةةةةةةةةةاذير

.ي ش و ها الكذبا

May 2014




-2

0

2

4

6

0 1 2 3 4 5

Problem 5.7

Determine the real root of f(x) = (0.8 – 0.3x) / x

(b) Graphically.

(c) Using three iterations of the false-position method and initial

guesses of 2 and 3. Compute the approximate error εa and the true

error εt after each iteration. Is there a problem with the result?

Solution

(b) The graph of the function indicates a root between x = 2 and 3.

Note that the shape of the curve suggests that it may be ill-suited for

solution with the false-position method

(c) Using false position, the first iteration is

875.2)03333.0(5.0

)31(03333.03

rx

%81.7%10066667.2

875.266667.2

t

01087.0)02174.0(5.0)875.2()1( ff

إن كل فكرة من أفكارك هي لبنة إما تضيفها

.إلى قصر أحالمك أو إلى قبر مشاريعك

May 2014




Continue

Therefore, the root is in the first interval and the upper guess is

redefined as xu = 2.875. The second iteration is

79688.2)03333.0(5.0

)875.21(03333.0875.2

rx

%793.2%10079688.2

875.279688.2

a

%88.4%10066667.2

79688.266667.2

t

00698.0)01397.0(5.0)79688.2()1( ff

Consequently, the root is in the first interval and the upper guess is

redefined as xu = 2.79688. All the iterations are displayed in the

following table:

i xl xu f(xl) f(xu) xr f(xr) f(xl)f(xr) a t

1 1 3.00000 0.5 -0.03333 2.87500 -0.02174 -0.01087 7.81% 2 1 2.87500 0.5 -0.02174 2.79688 -0.01397 -0.00698 2.793% 4.88% 3 1 2.79688 0.5 -0.01397 2.74805 -0.00888 -0.00444 1.777% 3.05%

Therefore, after three iterations we obtain a root estimate of 2.74805

with an approximate error of 1.777%. Note that the true error is greater

than the approximate error. This is not good because it means that we

could stop the computation based on the erroneous assumption that

the true error is at least as good as the approximate error.

This is due to the slow convergence that results from the function’s

shape.

دينيةة، : لتكن أهدافك منوعةة

أسةةةةةةةةةةةةةرية، اجتماعيةةةةةةةةةةةةةة،

.اقتصادية، مهنية، رياضية

May 2014




Problem 5.8

Find the positive square root of 18 using the false-position method within

εs = 0.5%. Employ initial guesses of xl = 4 and xu = 5.

Solution

The square root of 18 can be set up as a roots problem by determining the

positive root of the function

018)( 2 xxf

Using false position, the first iteration is

22222.472

)54(75

rx

34568.0)17284.0(2)22222.4()4( ff

Therefore, the root is in the second interval and the lower guess is

redefined as xl = 4.22222. The second iteration is

24096.4717284.0

)522222.4(722222.4

rx

%442.0%10024096.4

22222.424096.4

a

Thus, the computation can be stopped after just two iterations because

0.442% < 0.5%.

إننةةي ط أستسةةلم أ ةةدا، حتةةى عنةةدما

.يقول لي الناا أنني لن أنجح أ دا

Documents

Chapter 5 Bracketing Methods - موقع المهندس حماده شعبان (b) shows the case where a single root is bracketed by negative and positive values of f(x). However, fig(d)