Week 3 Chapters 6 - موقع المهندس حمادة ... · Chapters 6 Numerical Methods ... 6.3...

June 2016

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يتوقعون الحصول عليه.

Open Methods

Chapters 6

Numerical Methods

النوميركالنوت

. أجزاءعشرة تتكون النوت من

على كل أسبوع نوتيحتوي

من ألمثلة وتمارينشرح وحلول

سابقة.هوموركات وامتحانات

Week 3

شرح النوت فيديو متوفر على قناتكم

HS Engineers

تابعونا ليصل لكم كل جديد

حقوق النوت

حقوق النسخ أتيحلتعميم الفائدة فإني

لكل وشروحات الفيديو وإعادة اإلصدار

لبعض األجزاء أو الكل. سواء البشر

(00965-26094444واتساب )

يتم تنقحة النوت وتحديثها

وإضافة الجديد إليها بشكل دوري.

.نتشوق النتقادك الهادف

June 2016



إذا قدمت معروفا ألحد، ال تنتظر إيصاالا

بعلم الوصول أو خطاب شكر.

For bracketing methods, the root is located within an interval

prescribed by lower and an upper limits. Repeated application of those

methods always results in closer estimates of the true value of the root.

Such methods are said to be always convergent because they move

closer to the truth as the computation progresses as seen in figure(a).

In contrast, the open methods are based on formulas that require only

a single starting value of or two starting values that do not necessarily

bracket the root.

They sometimes diverge or move away from the true root as the

computation progresses (fig(b)). When the open methods converge

(fig(c)), they usually do so much more quickly than the bracketing

methods.

Chapters 6

Open methods

تعتمدددل نددد ددد

واحدددلر بة مددد ددد

الجذر.

بددل تمتعددل دد ال دد

وال تصدددددددد ل دددددددد

Divergent.

ذا ابتةبددددددد ددددددد

ال دد تنددسر ددة

Bracketing.

Bracketing

Methods

تمدددل التنمددد لنجدددذر

بدددددل ن ددددد تددددد

.ب نهما الجذر

ت تةب ال كثدة

كنمدا اات الن ددسا

Convergent.

ت تدددددةب ددددد ال ددددد

بمعددددددل ب دددددد دددددد

Open methods.

June 2016



في النهر ال تصنعن لنفسك معبراا

ثم تجاهد بعد ذلك لتجمع أجره.

6.1 Simple Fixed-Point Iteration

Rearranging the function or simply adding to both sides of original

equation so that is on the left hand of the equation:

(6.1)

For example,

–

can be simply rearranged:

Whereas sin could be:

The utility of Eq. (6.1) is that it provides a formula to predict a new

value of as a function of an old value of :

(6.2)

Examples of simple fixed-point can be found on pages (14-20).

Where one formula of ( ) can be dive gent, another formula

can be convergent.

Using calculator enables you to fill the table of results quickly (learn it

well)

طريقة الحل:

( من أحد حدود المعادلة. صفرية ثم نفصل ) نجعل معادلة

( من الحد الذي يحتوي على أكبر يفضل فصل ) معاملل، وكلذلك الحلد اللذي يعطلي أكبلر

ع الحل.ي( وذلك لتسر قيمة عند التعويض بقيمة )

.تعلم برمجة اآللة البيضاء إلنجاز الحسابات بأقل من نصف الوقت

June 2016



األمر ببسلاطة: إن أهلم سلبع لعلد

.تحقيق األهداف هي عد وجودها

6.2 The Newton-Raphson Method

The most used formula for finding roots is the Newton-Raphson.

If the initial guess at the root is , a tangent can be extended from

the point . The point where this tangent crosses the axis

usually represents an improved estimate of the root.

The Newton-Raphson method can be derived as follows:

which can be rearranged to yield:

Which is called the Newton-Raphson formula.

Raphson-Newton

هذه الطريقة هي أسرع الطرق في

كان الحلالوصول إلى الجذور في حال

Convergent.

June 2016



من اإلنسانية أال يتم التضحية

بإنسان في سبيل غاية.

EXAMPLE 6.3 Newton-Raphson Method

Problem Statement:

Use the Newton-Raphson Method to estimate the root of – ,

employing an initial guess of .

Solution:

The first derivative:

–

which can be substituted:

Starting with an initial guess of , this iterative equation can be

applied to compute new estimated values

0 0 100 1 0.500000000 11.8 2 0.566311003 0.147 3 0.567143165 0.0000220 4 0.567143290

The true percent relative error at each iteration decreases much faster

than it does in other method.

اسلللللتادا اآلللللللة لبرمجلللللة

المعادلة ومأل الجدول بلدون

عمل أية حسابات.

June 2016



مللن علللو أخأقللك أن تمللنح معارضللك

فرصة جيدة لأنسحاب دون إحراجه.

6.3 The Secant Method

A potential problem in implementing the Newton-Rapshon method is

the evaluation of the derivative because there are certain functions

whose derivatives are extremely difficult. For these functions, the

derivative can be approximated by a backward finite divided

difference, as in the following way:

–

This approximation can be substituted into Eq. (6.6) to yield:

Notice that Secant method requires two initial values of . However,

because is not required to change signs between the two initial

values, it is not classified as a bracketing method.

تكرر باالمتحاني سؤال

طرق ضمن( Secant Methodلماذا تم تصنيف )

( مع أنها تتطلع نقطتين مثل Open methodsالـ )

(Bracketingطرق الـ )

والجواب

June 2016



إذا لم تاطط ألهدافك، فليس من

حقك أن تند على عد تحقيقها.

EXAMPLE 6.6 Secant Method

Problem Statement:

Use the secant method to estimate the root of – .

Start with initial estimates of and .

Solution:

Recall that the true root is 0.56714329

First iteration:

Second iteration:

(Note that both estimates are now on the same side of the root.)

Third iteration:

–

ال يتم تصنيفها ضمن لذلك

(.Bracketingطرق الـ )

June 2016



في حال عد وجود أهداف يكون لدينا والء

.ألمور تافهة لكن بشكل منظمغريع

Problem 6.2

Determine the highest real root of

(a) Graphically.

(b) Fixed-Point iteration method (three iterations, ). Note: Make

certain that you develop a solution that converges on the root.

(c) Newton-Raphson method (three iterations, ).

(d) Secant method (three iterations, , )

(e) Modified secant method (three iterations, , ).

Compute the approximate percent relative errors for your solutions.

Solution

a) Graphical Method:

Root 3.58

b) Fixed Point Method: The equation can be solved in numerous ways. A simple way that

converges is to solve for the that is not raised to a power to yield:

June 2016



قد يكون مهما أين أنت اآلن،

لكن األهم إلى أين تتجه.

Continue

The resulting iterations are

| | 0 3 1 3.180791 5.68% 2 3.333959 4.59% 3 3.442543 3.15%

(c) Newton-Raphson:

| | 0 3 -3.2 1.5 1 5.133333 48.09007 55.68667 41.56% 2 4.26975 12.95624 27.17244 20.23% 3 3.792934 12.57%

(d) Secant Method:

| | 0 3 -3.2 4 6.6 1 4 6.6 3.326531 -1.9688531 20.25% 2 3.326531 -1.96885 3.481273 -0.7959153 4.44% 3 3.481273 -0.79592 3.586275 0.2478695 2.93%

(e) Modified Secant Method ( = 0.01):

| | 0 3 -3.2 0.03 3.03 -3.14928 1.6908 1 4.892595 35.7632 0.048926 4.9415212 38.09731 47.7068 38.68% 2 4.142949 9.73047 0.041429 4.1843789 10.7367 24.28771 18.09% 3 3.742316 2.203063 0.037423 3.7797391 2.748117 14.56462 10.71%

June 2016



من علت همته،

ه. طال َهمُّ

Problem 6.3

Use:

(a) Fixed-Point iterations.

(b) The Newton-Raphson method to determine a root of

using .

Perform the computation until is less than . Also perform

an error check of your final answer.

Solution

(a) Fixed Point Method:

The equation can be solved in two ways.

The way that converges is

√

The iterations are shown in table.

(b) Newton-Raphson Method:

| |

0 5 -13.5 -8.2 1 3.353659 -2.71044 -4.90732 49.09% 2 2.801332 -0.30506 -3.80266 19.72% 3 2.721108 -0.00644 -3.64222 2.95% 4 2.719341 -3.1E-06 -3.63868 0.06% 5 2.719341 -7.4E-13 -3.63868 0.00%

| | 0 5 1 3.391165 47.44% 2 2.933274 15.61% 3 2.789246 5.16% 4 2.742379 1.71% 5 2.726955 0.57% 6 2.721859 0.19% 7 2.720174 0.06% 8 2.719616 0.02%

June 2016



ما يمكن تايله يمكن تحقيقه، وما يمكن

ا للوصول إليه. تحقيقه لن نعد طريقا

Problem 6.9

Determine the highest real root of

(a) Graphically.

(b) Using the Newton-Raphson method (three iterations, )

(c) Using the secant method (three iterations, , and )

(d) Using Modified Secant method (three iterations, , ).

Solution

(a) Graphical Method:

Highest real root 3.3

(b) Newton-Raphson Method:

| | 0 3.5 0.60625 4.5125 1 3.365651 0.071249 3.468997 3.992% 2 3.345112 0.001549 3.318537 0.614% 3 3.344645 7.92E-07 3.315145 0.014%

June 2016



وسيلة النقل والارائط غير هامة

إذا لم تكن تعرف وجهتك.

Continue

(c) Secant Method:

| | 0 2.5 -0.78125 3.5 0.60625 1 3.5 0.60625 3.063063 -0.6667 14.265% 2 3.063063 -0.6667 3.291906 -0.16487 6.952% 3 3.291906 -0.16487 3.367092 0.076256 2.233%

(d) Modified Secant Method ( = 0.01):

| | 0 3.5 3.535 0.60625 0.76922 4.6563 1 3.3698 3.403498 0.085704 0.207879 3.6256 3.864% 2 3.346161 3.379623 0.005033 0.120439 3.4489 0.706% 3 3.344702 3.378149 0.000187 0.115181 3.4381 0.044%

ا ـــــــه

الشللللكور الرئيسللللية مللللن امتحانللللات

النيوميركال هي ضيق وقت االمتحان،

اآلللة البيضلاء لتلوفير لذا تعلم برمجلة

أكثر من نصف وقت الحسابات.

June 2016



المنافسة الواقعية تكون بين ما تقو بعمله

قادر على عمله، قارن نفسك مع وما أنت

نفسك وليس مع أي شاص آخر.

Problem 6.03

You are designing a spherical tank (like the following fig), to hold water

for a small village in a developing country. The volume of liquid it can

hold can be computed as

Where = volume [ ], = depth of water in tank [ ], and = the

tank radius [ ].

If , what depth must the tank be filled to so that it holds 30 ?

Use three iterations of the Newton-Raphson method to determine your

answer. Determine the approximate relative error after each iterations.

Note that an initial guess of will always converge.

June 2016



كون البعض ناجحون يثبت أن اآلخرين

ا. يمكنهم أن يكونوا ناجحين أيضا

Solution

The equation to be solved is:

(

)

Because this equation is easy to differentiate, the Newton-Raphson is

the best choice to achieve results efficiently. It can be formulated as:

( )

substituting the parameter values:

The iterations can be summarized as

iteration | |

0 3 26.54867 28.27433

1 2.061033 0.866921 25.50452 45.558%

2 2.027042 0.003449 25.30035 1.677%

3 2.026906 5.68E-08 25.29952 0.007%

Thus, after only three iterations, the root is determined to be 2.026906

with an approximate relative error of 0.007%.

( )

في حال عد ذكرها يتم

افتراض قيمة منطقية.

June 2016



6.5 Multiple Roots

إلى نفسك.ال تنسع أفكار اآلخرين

A multiple root corresponds to a point where a function is tangent to

the x axis. For example, a double root results from

or, multiplying terms, .

The equation has a double root because one value of

makes two terms in equation (6.11) equal to zero.

Graphically, this corresponds to the curve touching the axis

tangentially at the double root.

A triple root corresponds to the case where one value

makes three terms in an equation equal to zero, as in:

Or, multiplying terms,

Figure (b) indicates that the function is tangent to the axis at the root,

but that for this case the axis is crossed.

In general, odd multiple roots cross the axis, whereas

even ones do not. For example, the quadruple root

Fig(c) does not cross the axis.

Multiple roots have two main difficulties:

1. The fact that the function does not change sign at even multiple

roots prevents the use of the bracketing methods.

2. The fact that not only but also goes to zero at the root.

This causes problems for both the Newton-Raphson and secant

methods.

June 2016



Modified Newton-Raphson Method for Multiple Roots

البعض يسلافر لليس ليصلل بلل لمجلرد

السفر، أغلبنا يعاني من أجل ال شيء.

A slight change in the Newton-Raphson formula can be used for solving

multiple root:

Where m is the multiplicity of the root (that is, for a double root,

for a triple root, etc.). Of course, this may be an unsatisfactory

because it depends on foreknowledge of the multiplicity of the root.

Another alternative formula:

EXAMPLE 6.10 Modified Newton-Raphson Method for Multiple Roots

Use both the standard and modified Newton-Raphson methods to

evaluate the multiple root of Eq. (6.11), with an initial guess of .

Solution:

The first derivative of Eq. (6.11) is , and

therefore, the standard Newton-Raphson method for this problem is

Which can be solved iteratively for:

0 0 100 1 0.4285714 57 2 0.6857143 31 3 0.8328654 17 4 0.9133290 8.7 5 0.9557833 4.4 6 0.9776551 2.2

June 2016



شيئا بصدق، تآمر الكون إذا أردت

كله لمساعدتك على تحقيقه.

As expected, the method is linearly convergent to the true value of 1.0.

For the modified method, the second derivative is , and:

(

)(

)

(

)

Which can be solved for:

0 0 100 1 1.105263 11 2 1.003082 0.31 3 1.000002 0.00024

Thus, the modified formula is quadratic ally convergent.

We can also use both methods for the single root at . Using an

initial guess of gives the following results:

standard Modified 0 4 33 4 33 1 3.4 13 2.636364 12 2 3.1 3.3 2.820225 6.0 3 3.008696 0.29 2.961728 1.3 4 3.000075 0.0025 2.998479 0.051 5 3.000000 2 × 10-7 2.999998 7.7 × 10-5

Thus, both methods converge quickly, with the standard method being

somewhat more efficient.

June 2016



إذا لم تعلم أين تذهع، فأي

طريق يفي بالغرض.

Non – Linear Systems of Equations

, and

are two simultaneous nonlinear equations with two unknowns, and .

They can be expressed in the form of Eq.(6.14) as

(6.19a)

(6.19b)

Fixed-Point Iteration

The fixed-point-iteration can be modified to solve two simultaneous,

nonlinear equations as illustrated in the following example.

EXAMPLE 6.11 Fixed-Point Iteration for a Nonlinear System

Use fixed-point iteration to determine the roots of Eq. (6.19). Note that

a correct pair of roots is and . Initiate the computation with

guesses of and .

Solution:

Equation (6.19a) can be solved for

and Eq.(6.19b) can be solved for

On the basis of the initial guesses, Eq. (E6.11.1) can be used to

determine a new value of :

This result and the initial value of can be substituted into Eq.

(E6.11.2) to determine a new value of :

June 2016



ركز على أهدافك، كثير من الناس

حاربوا وماتوا لغير قضية.

Thus, the approach seems to be diverging. This behavior is even more

obvious on the second iteration:

Obviously, the approach is becoming worse.

Now we will repeat the computation but with the original equations set

up in different format. An alternative formulation of Eq. (6.19a) is:

√

and of Eq. (6.19b) is

√

Now the results are more satisfactory:

√

√

√

√

Thus, the approach is converging on the true values of and .

June 2016



EXAMPLE 6.12 Newton-Raphson for a Nonlinear System Use the multiple-equations Newton-Raphson method to determine roots of Eq. (6.19). Note that a correct pair of roots is and . Initiate the computation with guesses of and .

Solution: First, compute the partial derivatives and evaluate them at the initial guesses of and :

Thus, the determination of the Jacobian for the first iteration is

The values of the functions can be evaluated at the initial guesses as

These values can be substituted into Eq.(6.24) to give

Thus, the results are converging to the true values of and . The computation can be repeated until an acceptable accuracy.

Just as with fixed-point iteration, the Newton-Raphson approach will often diverge if the initial guesses are not sufficiently close to the true roots.

Week 3 Chapters 6 - موقع المهندس حمادة ... · Chapters 6 Numerical Methods ... 6.3...

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