numerical analysis 2210

8/9/2019 Numerical Analysis 2210

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Numerical Analysis

Determination of roots of polynomials and transcendental equations byNewton Raphson, Secant and Bairstow's method.

Motivation:

Let us look at a set of problems of scientific and engineering interest toget a feel of what is root finding and why to find roots. Later we learnhow to find them.

Problem 1:Suppose we are asked to cut a rectangular sheet with one of its sides

1.25mts longer than the other and the area being 0.875 mts from a

thin iron sheet of 5 mts area.What will be length of the 'smallest side'?

Say, length of the smallest side =

Length of the other side

Area of rectangle

mt

i.e.

i.e. say

(1)

So you need to solve a quadratic equation to find the required quantity.i.e. we have to find the roots of a quadratic equation.

We know that the roots of a quadratic equation

are given by

(3)


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Problem 2:Concepts of thermodynamics are used extensively in their works by sayaerospace, mechanical and chemical engineers. Here, the zero-

pressure specific heat of dry air say KJ/(KgK) is related totemperature 'T' by

Now, determine the temperature that corresponds to a specific heat of1.2 KJ/(KgK). So, here we have to solve or find the roots of

=1.2

i.e. find roots of

(

Problem 3:The concentration of pollutant bacteria 'C' in a lake decreases as per the model:

where 't' is the time variable. Determine the time required for the bacteria

concentration to be reduced to 9.

Here, we have to find the roots of

(5)

Problem 4:

The volume of liquid in a hollow horizontal cylinder of radius r and


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length L is related to the depth of the liquid h by

Determine h given thatHere we have to find the roots of

(6)

So we have seen that finding roots of ..................(7)

is very important in finding solution to several scientific and engineeringproblems.

The equation may be a polynomial equation or atranscendental equation.

Polynomial Equations:Polynomial equations in one independent variable 'x' are a simple classof algebraic equations that are represented as follows:

The degree polynomial has roots. These roots may be real orcomplex.Examples:

Transcendental Equation:The equations include trigonometric or exponential or logarithmicfunctions.Examples:


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We may note that the examples are nonlinear functions.

Method of solution:

Some of the ways of finding the roots or solution of are:

Direct analytical methods Graphical approach Iterative methods etc.

Direct analytical methods: One may be able to find a mathematical

expression for the solution (root) of

.For example, for quadratic equation (2), we have solutions given by (3).However a large number of equations cannot be solved by directanalytical methods.

Graphical Method: This approach involves plotting the given function anddetermining the points where it crosses the x-axis. These points,

extracted approximately from the plot, represent approximate values ofthe roots of the function.Example:

Find the positive roots of

Rewrite as

Now consider and and plot them .


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The x-co-ordinate of the point of intersection of , and

gives the required positive root of the given function. Clearlythis approach is cumbersome and time consuming.

Iterative Methods:Starting with an initial guess solution these methodsgenerate a sequence of estimates to the solution which is expected toconverge to the true solution. They are grouped into two categories :

(a) Bracketing methods (b) Open methods

(a)Bracketing Methods:These methods exploit the fact that a functiontypically changes sign in the vicinity of a root. They start with two initialguesses that bracket the root and then systematically reduce the widthof the bracket until the solution to a desired accuracy is reached. Thepopular bracketing methods are: (a) Bisection Method, (b) FalsePosition (or) Regula Falsi method, (c) Improved or modified Regula

Falsi Method.(b)Open methods: These methods are based on formulas that require onlya single starting (or guess) value of solution or two starting values thatdo not necessarily bracket the root. They may sometimes diverge ormove away from true root as the computation progresses. Howeverwhen the open methods converge , they do so much more quickly thanthe Bracketing methods. Some of the popular open methods are: (a)Secant method, (b) Newton-Raphson method, (c) Bairstow's method (d)Muller's method etc.

Bracketing Methods:


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(a) Bisection Method:This is one of the simplest and reliable iterative methods for the solutionof nonlinear equation. This method is also known as binary chopping or

half-interval method. Given a function which is real and continuous

in an interval and and are of opposite sign i.e.

, then there is at least one real root of

.

Algorithm:

Given a function continuous on a interval satisfying theBisection method starting criteria, carry out the following steps to find a

root of

(1) Set(2) For n=1,2,...until satisfied do

(a) If

(b)If

otherwise

Note:

1) The subscripts in etc denote the


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iteration number. is the interval for the zeroth or starting iteration.

is the interval for the n iteration.

(2) An iterative process must be terminated at some stage. 'Untilsatisfied' refers to the solution convergence criteria used for stoppingthe execution process. We must have an objective criteria for decidingwhen to stop the process.We may use one of the following criteriadepending on the behavior of the function (monotonous/steepvariation/increasing /decreasing)

(i) (Tolerable absolute error in

(ii) (Tolerable relative error in

(iii) (Value of function as )

(iv) (difference in two consecutive iterationfunction values)

Usually are referred to as tolerance values and it is fixed byus depending on the level of accuracy we desire to have on the solution.

For example etc.

Example:

Solve for the root in the interval [1,2] by Bisectionmethod.

Solution: Given on

There is a root for the given function in [1,2].

Set


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Set

and

Set

Details of the remaining steps are provided in the table below:

Bisection


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Method

Iterationno.

0 1.0000000000 2.0000000000 1.5000000000 -2.0000000000

1 1.5000000000 2.0000000000 1.7500000000 1.3437500000

2 1.5000000000 1.7500000000 1.6250000000 -0.4804687500

3 1.6250000000 1.7500000000 1.6875000000 0.3920898438

4 1.6250000000 1.6875000000 1.6562500000 -0.0538940430

5 1.6562500000 1.6875000000 1.6718750000 0.1666488647

6 1.6562500000 1.6718750000 1.6640625000 0.0557680130

7 1.6562500000 1.6640625000 1.6601562500 0.0007849932

8 1.6562500000 1.6601562500 1.6582031250 -0.0265924782

9 1.6582031250 1.6601562500 1.6591796875 -0.012913236410 1.6591796875 1.6601562500 1.6596679688 -0.0060664956

11 1.6596679688 1.6601562500 1.6599121094 -0.0026413449

12 1.6599121094 1.6601562500 1.6600341797 -0.0009283243

13 1.6600341797 1.6601562500 1.6600952148 -0.0000717027

14 1.6600952148 1.6601562500 1.6601257324 0.0003566360

15 1.6600952148 1.6601257324 1.6601104736 0.0001424643

16 1.6600952148 1.6601104736 1.6601028442 0.0000353802

17 1.6600952148 1.6601028442 1.6600990295 -0.0000181614

18 1.6600990295 1.6601028442 1.6601009369 0.0000086094

Example:

Solve for the root in the interval byBisection method.

Bisection Method

Iterationno.

0 0.5000000000 1.5000000000 1.0000000000 3.1720056534

1 0.5000000000 1.0000000000 0.7500000000 0.6454265714

2 0.5000000000 0.7500000000 0.6250000000 -1.0943561792

3 0.6250000000 0.7500000000 0.6875000000 -0.1919542551

4 0.6875000000 0.7500000000 0.7187500000 0.2357951254


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5 0.6875000000 0.7187500000 0.7031250000 0.0240836944

6 0.6875000000 0.7031250000 0.6953125000 -0.0834089667

7 0.6953125000 0.7031250000 0.6992187500 -0.0295295101

8 0.6992187500 0.7031250000 0.7011718750 -0.0026894973

9 0.7011718750 0.7031250000 0.7021484375 0.0107056862

10 0.7011718750 0.7021484375 0.7016601562 0.0040097744

11 0.7011718750 0.7016601562 0.7014160156 0.0006612621

12 0.7011718750 0.7014160156 0.7012939453 -0.0010144216

13 0.7012939453 0.7014160156 0.7013549805 -0.0001766436

14 0.7013549805 0.7014160156 0.7013854980 0.0002420362

15 0.7013549805 0.7013854980 0.7013702393 0.0000326998

16 0.7013549805 0.7013702393 0.7013626099 -0.0000715650

17 0.7013626099 0.7013702393 0.7013664246 -0.0000194324

18 0.7013664246 0.7013702393 0. 7013683 319 0.0000069206

Exercise: - Find the solutions of the following problems accurate

to within using Bisection Method.

(1) for

(2) for

False Position or Regula Falsi method:

Bisection method converges slowly. Here while defining the new interval

the only utilization of the function is in checking whether

but not in actually calculating the end point of the

interval. False Position or Regular Falsi method uses not only in

deciding the new interval as in bisection method but also incalculating one of the end points of the new interval. Here one of end

points of say is calculated as a weighted average defined on

previous interval as


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( have opposite signs).

The algorithm for computing the root of function by this method isgiven below.

Algorithm:

Given a function continuous on an interval satisfying the

criteria , carry out the following steps to find the root of

in :

(1) Set(2) For n = 0,1,2.... until convergence criteria is satisfied , do:

(a) Compute

(b) If , then set

otherwise setNote:Use any one of the convergence criteria discussed earlier underbisection method. For the sake of carrying out a comparative study wewill stick both to the same convergence criteria as before i.e.

(say) and to the example problems.Example:

Solve for the root in the interval [1,2] by Regula-Falsi method:

Solution: Since , we go ahead in finding the root ofgiven function f(x) in [1,2].

Set .


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set

, proceed with iteration.Iteration details are provide below in a tabular form:

Regula Falsi

Method

Iterationno.

0 1.0000000000 2.0000000000 1.4782608747 -2.2348976135

1 1.4782608747 2.0000000000 1.6198574305 -0.5488323569

2 1.6198574305 2.0000000000 1.6517157555 -0.1169833690

3 1.6517157555 2.0000000000 1.6583764553 -0.0241659321

4 1.6583764553 2.0000000000 1.6597468853 -0.0049594725

5 1.6597468853 2.0000000000 1.6600278616 -0.0010169938

6 1.6600278616 2.0000000000 1.6600854397 -0.0002089010

7 1.6600854397 2.0000000000 1.6600972414 -0.0000432589

8 1.6600972414 2.0000000000 1.6600997448 -0.0000081223

Note : One may note that Regula Falsi method has converged fasterthan the Bisection method.

Geometric Interpretation of Regula Falsi Method:

Let us plot the polynomial considered in the above example and trace

, its movement and new intervals with iteration. From thefigure(), one can verify that the weighted average


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is the point of intersection of the secant to , passing through points

and with the x-axis. Since here is concaveupward and increasing the secant is always above . Hence,

always lies to the left of the zero. If were to be concave downwardand increasing, would always lie to the right of the zero.

Example:

Solve for the root in the interval [0.5,1.5] byRegula Falsi method.

Regula FalsiMethod

Iterationno.

0 0.5000000000 1.5000000000 0.8773435354 2.1035263538

1 0.5000000000 0.8773435354 0.7222673893 0.2828366458

2 0.5000000000 0.7222673893 0.7032044530 0.0251714624

3 0.5000000000 0.7032044530 0.7015219927 0.0021148270


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4 0.5000000000 0.7015219927 0.7013807297 0.0001767781

5 0.5000000000 0.7013807297 0.7013689280 0.0000148928

6 0.5000000000 0.7013689280 0.7013679147 0.0000009526

Exercise:1)Solve for the root in the interval [2,3]

by Regula-Falsi Method.

2)Find the solution to , in the interval [1,2] accurate

to within using Regula-Falsi Method.

Modified Regula Falsi method:

In this method an improvement over Regula Falsi method is obtained byreplacing the secant by straight lines of even-smaller slope until falls

to the otherside of the zero of . The various steps in the method aregiven in the algorithm below:Algorithm:

Given a function continuous on an interval satisfying the

criteria , carry out the following steps to find the root of of

in :

(1)Set(2) For n=0,1,2...., until convergence criteria is satisfied, do:

(a) compute

(b) If then


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Set

Also if SetOtherwise

Set

Also if SetExample:

Solve for the root in the interval [1,2] by ModifiedRegula Falsi method.

Solution: Since we go ahead with finding the

root of given function f(x) in [1,2]. Setting and followingthe above algorithm. Results are provided in the table below:

Modified Regula Falsi Method

Iterationno.

0 1.0000000000 2.0000000000 1.4782608747 -2.2348976135

1 1.4782608747 2.0000000000 1.7010031939 0.5908976793

2 1.4782608747 1.7010031939 1.6544258595 -0.0793241411

3 1.6544258595 1.7010031939 1.6599385738 -0.0022699926

4 1.6599385738 1.7010031939 1.6602516174 0.0021237291

5 1.6599385738 1.6602516174 1.6601003408 0.0000002435

The geometric view of the example is provided in the figure below:


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Example:Solve for the root in the interval[0.5,1.5] by Modified Regula Falsi Method.

Modified Regula Falsi Method

Iterationno.

0 0.5000000000 1.5000000000 0.8773435354 2.1035263538

1 0.5000000000 0.8773435354 0.7222673893 0.2828366458

2 0.5000000000 0.7222673893 0.6871531010 -0.1967970580

3 0.6871531010 0.7222673893 0.7015607357 0.0026464546

4 0.6871531010 0.7015607357 0.7013695836 0.0000239155

5 0.6871531010 0.7013695836 0.7013661265 -0.0000235377

6 0.7013661265 0.7013695836 0.7013678551 -0.0000003363


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Secant Method

Like the Regula Falsi method and the Bisection method this method

also requires two initial estimates of the root of f(x)=0 butunlike those earlier methods it gives up the demand of bracketing theroot. Like in the Regula Falsi method, this method too retains the use of

secants throughout while tracking the root of f(x)=0. The secant joining

the points is given by

Say it intersects with x-axis at , then

If (say) then replace with

and repeat the process to get and so on . Themethod is algorithmically described below:Algorithm:

Given a , two initial points a, b and the required level of accuracy

carry out the following steps to find the root of f(x)=0.

(1) Set(2) For n=0,1,2... until convergence criteria is satisfied, do:Compute

Example:

Solve for the root with by secant

method to an accuracy of .

Solution:

Set


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Repeat the process with and so on till you

get a s.t. These results are tabulated below:

Secant Method

Iterationno.

0 1.0000000000 2.0000000000 1.4782608747 -2.2348976135

1 2.0000000000 1.4782608747 1.6198574305 -0.5488323569

2 1.4782608747 1.6198574305 1.6659486294 0.0824255496

3 1.6198574305 1.6659486294 1.6599303484 -0.0023854144

4 1.6659486294 1.6599303484 1.6600996256 -0.0000097955

Geometrical visualization of the root tracking procedure by Secantmethod for the above example.


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Exercise: Find the solutions accurate to within for thefollowing problems using Secant's Method.

(1)

(2)

Convergence of secant method:

Definition: Say, where is the root of

, , are the errors at n and (n+1) iterations and

are the approximations of at , (n+1) iterations. If

where is a constant, then the rate of convergence of

the method by which is generated is p.Claim: Secant method has super linear convergence.Proof: The iteration scheme for the secant method is given by

Say and i.e the error in the n iteration in

estimating .

Using (iii) in (ii) we get

By Mean value Theorem, in the interval and s.t.


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We get

i.e.Using (iii)above, we get

using (v), (vi), in (iv) we get

By def of rate of convergence , the method is of order p if

>From (vii) and (viii) we get

i.e

i.e

From (viii), (ix) we get

i.e.


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Hence the convergence is superlinear.

Example:

Solve for the root in the interval [0.5,1.5] bysecant method.

Secant Method

0 0.5000000000 1.5000000000 0.8773435354 2.1035263538

1 1.5000000000 0.8773435354 0.4212051630 -4.2280626297

2 0.8773435354 0.4212051630 0.7258019447 0.32987323403 0.4212051630 0.7258019447 0.7037572265 0.0327354670

4 0.7258019447 0.7037572265 0.7013285756 -0.0005388701

5 0.7037572265 0.7013285756 0.7013679147 0.0000009526

Newton-Raphson Method:

Unlike the earlier methods, this method requires only one appropriate

starting point as an initial assumption of the root of the function

. At a tangent to is drawn. Equation of thistangent is given by

The point of intersection, say , of this tangent with x-asis (y = 0) istaken to be the next approximation to the root of f(x) = 0. So onsubstituting y = 0 in the tangent equation we get

If (say) we have got an acceptable approximate root

of , otherwise we replace by , and draw a tangent to

at and consider its intersection, say , with x-axis

as an improved approximation to the root of f(x)=0. If , we

iterate the above process till the convergence criteria is satisfied. This


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geometrical description of the method may be clearly visualized in thefigure below:

The various steps involved in calculating the root of by NewtonRaphson Method are described compactly in the algorithm below.

Algorithm:

Given a continuously differentiable function and an initial

approximation to the root of , the steps involved in

calculating an approximation to the root of s.t. are:

(1) Calculate and set(2) For n = 0,1,2... until convergence criteria is satisfied ,do:Calculate


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Remark (1): This method converges faster than the earlier methods.In fact the method converges at a quadratic rate. We will prove thislater.

Remark (2): This method can be derived directly by the Taylor

expansion f(x) in the neighbourhood of the root of . The

starting approximation to is to be properly chosen so that the first

order Taylor series approximation of in the neighbourhood of

leads to , an improved approximation to . i.e

, neglecting and its higher powers, we get

i.e.

Now the successive approximations etc may becalculated by the iterative formula:

Remark(3) : One may also derive the above iteration formulationstarting with the iteration formula for the secant method. In a way this

may help one to visualize Newton-Raphson method as an improvementover the secant method. So, let us consider the iteration formula for thesecant method i.e.

Add and subtract to the numerator on the R.H.S. to get


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or ,

Clearly, is the slope of the secant to the curve

through the points , . This alsorepresents slope of the tangent to f(x)=0 parallel to the secant

intersecting x-axis between and . If is differentiable one

may as well approximate this slope by and thus arrive at theiteration formula.

Example:

Solve for the root in [1,2] by Newton Raphsonmethod.

Solution:Given

Take and


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Since,

Therefore repeat the process.

Results are tabulated below:

Newton Rahpson Method

Iteration no.

0 2.0000000000 1.7209302187 0.8910911679

1 1.7209302187 1.6625729799 0.0347661413

2 1.6625729799 1.6601046324 0.0000604780

3 1.6601046324 1.6601003408 0.0000002435

Example:

Solve in [0.5,1.5] for the root by Newton-Raphsonmethod.

Solution: Given

Say,

The results are tabulated below:


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Newton Raphson Method

Iteration no.

0 0.5000000000 0.6934901476 0.1086351126

1 0.6934901476 0.7013291121 0.0005313741

2 0.7013291121 0.7013678551 0.0000003363

Exercise: Find the solutions accurate to within for thefollowing problems using Newton-Raphson Method.

(1) for and

(2) for and

Convergence of Newton-Raphson method:

Suppose is a root of and is an estimate of s.t.

. Then by Taylor series expansion we have,

for some between and .By Newton-Raphson method, we know that

i.e.

Using(2*) in (1*) we get


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Say

where denote the error in the solution at n and (n+1)iterations.

Newton Raphson Method is said to have quadratic convergence.

Note:Alternatively, one can also prove the quadratic convergence of Newton-Raphson method based on the fixed - point theory. It is worth statingfew comments on this approach as it is a more general approachcovering most of the iteration schemes discussed earlier.

A Brief discussion on Fixed Point Iteration:

Suppose that we are given a function

on an interval for which we need to find a root. Derive , from it, anequation of the form:

Any solution to (ii) is called a fixed point and it is a solution of (i). Thefunction g(x) is called as "Iteration function".

Example:

Given , one may re-write it as:

or ,

or ,

where g(x) denotes possible choice iteration function.


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Fixed point Iteration:

Let be a root of and be an associated iteration function.

Say, is the given starting point. Then one can generate a sequence

of successive approximations of as:

...............

...............

.................


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This sequence is said to converge to iff as.

Now the natural question that would arise is what are the conditions on

s.t. the sequence asHere, we state few important comments on such a convergence:

(i)Suppose on an interval is defined and .i.e. g(x) maps I into itself.

(ii) The iteration function is continuous on I=[a,b].

(iii)The iteration function g(x) is differentiable on and

s.t.

Theorem :Let g(x) be an iteration function satisfying (i), (ii) and (iii) then g(x) has

exactly one fixed point in I and starting with any , the sequence

generated by fixed point iteration function converges to .

(iv) If then . For rapid

convergence it is desirable that . Under this condition for the

Newton Raphson method one can show that (i.e.quadratic convergence).

Remark 1: One can generalize all the iterative methods for a system

of nonlinear equations. For instance, if we have two non-linear

equations then given a suitable starting

point , the Newton-Raphson algorithm may be written as follows:

For i=1,2... until satisfied , do


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Exercises: Solve the following systems of equations by NewtonRaphson Method.

(1)

Use the initial approximation

(2)

Use the initial approximation


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Bairstow Method

Bairstow Method is an iterative method used to find both the real andcomplex roots of a polynomial. It is based on the idea of syntheticdivision of the given polynomial by a quadratic function and can be usedto find all the roots of a polynomial. Given a polynomial say,

(B.1)

Bairstow's method divides the polynomial by a quadratic function.

(B.2)

Now the quotient will be a polynomial

(B.3)

and the remainder is a linear function , i.e.

(B.4)

Since the quotient and the remainder are obtained by

standard synthetic division the co-efficients can beobtained by the following recurrence relation.

(B.5a)

(B.5b)

for(B.5c)


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If is an exact factor of then the remainder is zero

and the real/complex roots of are the roots of . It may

be noted that is considered based on some guess values

for . So Bairstow's method reduces to determining the values of r

and s such that is zero. For finding such values Bairstow's methoduses a strategy similar to Newton Raphson's method.

Since both and are functions of r and s we can have Taylor series

expansion of , as:

(B.6a)

(B.6b)

For , terms i.e. second and higher order

terms may be neglected, so that the improvement over guess

value may be obtained by equating (B.6a),(B.6b) to zero i.e.

(B.7a)

(B.7b)

To solve the system of equations , we need the partial

derivatives of w.r.t. r and s. Bairstow has shown that these partial

derivatives can be obtained by synthetic division of , which


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amounts to using the recurrence relation replacing

with and with i.e.

(B.8a)

(B.8b)

(B.8c)

for

where

(B.9

The system of equations (B.7a)-(B.7b) may be written as.

(B.10a)

(B.10b)

These equations can be solved for and turn be used to

improve guess value to .

Now we can calculate the percentage of approximate errors in (r,s) by

(B.11)


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If or , where is the iteration stopping error, then

we repeat the process with the new guess i.e. .

Otherwise the roots of can be determined by

(B.12)

If we want to find all the roots of then at this point we have thefollowing three possibilities:

1. If the quotient polynomial is a third (or higher) orderpolynomial then we can again apply the Bairstow's method to the

quotient polynomial. The previous values of can serve as thestarting guesses for this application.

2. If the quotient polynomial is a quadratic function then use

(B.12) to obtain the remaining two roots of .

3. If the quotient polynomial is a linear function say

then the remaining single root is given by

Example:Find all the roots of the polynomial

by Bairstow method . With the initial values

Solution:Set iteration=1

Using the recurrence relations (B.5a)-(B.5c) and (B.8a)-(B.8c) we get


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the simultaneous equations for and are:

on solving we get

and

Set iteration=2

now we have to solve

On solving we get


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Now proceeding in the above manner in about ten iteration we get

with

Now on using we get

So at this point Quotient is a quadratic equation

Roots of are:

Roots are

i.e

Exercises:

(1) Use initial approximation to find a

quadratic factor of the form of the polynomial equation

using Bairstow method and hence find all its roots.

(2) Use initial approximaton to find a quadratic

factor of the form of the polynomial equation

using Bairstow method and hence find all the roots.


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numerical analysis 2210

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