t 5 open methods

8/10/2019 t 5 Open Methods

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Motilal Nehru National Institute of Technology

Civil Engineering Department

Computer Based Numerical Techniques

CE-401

Roots of the equations: Open Methods


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Open Methods

Bracketing methodsare based on assuming aninterval of the function which brackets the root.

The bracketing methodsalways convergeto theroot.

Open methodsare based on formulas thatrequire only a single starting value of x or twostarting values that do not necessarily bracket

the root. These method sometimes divergefrom the true

root.


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1. Simple Fixed-Point Iteration

Rearrange the function so that x is onthe left side of the equation:

)(

)(0)(

1 ii xgx

xxgxf

Bracketing methods are convergent.

Fixed-point methods may sometime diverge,depending on the stating point (initial guess) andhow the function behaves.


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Simple Fixed-Point Iteration

Examples:1.

2. f(x) = x 2-2x+3 x = g(x)=(x2+3)/2

3. f(x) = sin x x = g(x)= sin x + x

3. f(x) = e-x- x x = g(x)= e-x

x

xg

or

xxg

or

xxg

xxxxf

21)(

2)(

2)(

02)(

2

2


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Simple Fixed-Point Iteration Convergence

x = g(x) can be

expressed as a pair of

equations:

y1= x

y2= g(x). (component

equations)

Plot them separately.


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Simple Fixed-Point Iteration Convergence

Fixed-point iteration converges if :

( ) 1 (slope of the line ( ) )g x f x x

When the method converges, the erroris roughly proportional to or less than the

error of the previous step, therefore it is

called linearly convergent.


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Simple Fixed-Point Iteration-Convergence


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Steps of Simple Fixed Pint Iteration

1. Rearrange the equation f(x) = 0 so that x ison the left hand side and g(x) is on the righthand side.

e.g f(x) = x2-2x-1 = 0x= (x2-1)/2

g(x) = (x2-1)/2 2. Set xiat an initial guess xo.

3. Evaluate g(xi)

4. Let xi+1= g(xi) 5. Find a=(Xi+1xi)/Xi+1, and set xi at xi+1 6. Repeat steps 3 through 5 until |a|


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Example: Simple Fixed-Point Iteration

1.f(x)is manipulated so that we get

x=g(x)g(x) = e-x

2.Thus, the formula predicting the

new value of x is: xi+1= e-xi

3.Guess xo = 0

4.The iterations continues till theapprox. error reaches a certain

limiting value

f(x)

Root x

f(x)

x

f(x)=e-x- x

g(x) = e-x

f1(x) = x

f(x) = e-x

- x


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i xi g(xi) % t%

0 0 1.01 1.0 0.367879 100 76.3

2 0.367879 0.692201 171.8 35.13 0.692201 0.500473 46.9 22.14 0.500473 0.606244 38.3 11.85 0.606244 0.545396 17.4 6.896 0.545396 0.579612 11.2 3.83

7 0.579612 0.560115 5.90 2.28 0.560115 0.571143 3.48 1.249 0.571143 0.564879 1.93 0.70510 0.564879 1.11 0.399


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i xi g(xi) % t%

0 0 1.01 1.0 0.367879 100 76.3

2 0.367879 0.692201 171.8 35.13 0.692201 0.500473 46.9 22.14 0.500473 0.606244 38.3 11.85 0.606244 0.545396 17.4 6.896 0.545396 0.579612 11.2 3.83

7 0.579612 0.560115 5.90 2.28 0.560115 0.571143 3.48 1.249 0.571143 0.564879 1.93 0.70510 0.564879 1.11 0.399


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Example


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Flow ChartFixed Point

Start

Input: xo, s, maxi

i=0

a=1.1s

1


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Stop

1

while

amaxi

xn=0

x0=xn

100%n oan

x xx

Print:xo, f(xo) ,a, i 01

nx g x

i i

False

True


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2. The Newton-Raphson Method

Most widely used method.

Based on Taylor series expansion:

)(

)(

)(0

g,Rearrangin

0)f(xwhenxofvaluetheisrootThe

...!2)()()()(

1

1

1i1i

2

1

i

iii

iiii

iiii

xf

xfxx

xx)(xf)f (x

x

xfxxfxfxf

Solve for

Newton-Raphson formula


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The Newton-Raphson Method

A tangent tof(x)at theinitial pointxiis extended

till it meets the x-axis at

the improved estimate of

the rootxi+1.

The iterations continues

till the approx. error

reaches a certain limiting

value.

f(x)

Root x

xixi+1

f(x) Slope f/(xi)

f(xi)

)(

)(

)()(

/

/

i

ii1i

1ii

ii

xf

xfxx

xx

0xfxf


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Example: The Newton Raphson Method

11)(

)(/1

x

x

ix

x

i

i

iii

e

xex

e

xex

xf

xfxx

Use the Newton-Raphson method to find theroot of e-x-x= 0f(x) = e-x-xand f`(x)= -e-x-1;thus

Iter. xi

t%

0 0 100

1 0.5 11.8

2 0.566311003 0.147

3 0.567143165 0.00002

4 0.567143290


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Flow ChartNewton Raphson

Start

Input: xo, s, maxi

i=0

a=1.1s

1


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Stop

1

while

a>s &

i


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M-file for Newton Raphsons Method

function [root,ea,iter]=newtraph(func, dfunc, xr, es, maxit, varargin)

% newtraph: Newton-Raphson root location zeroes

% [root,ea,iter]=newtraph(func,dfunc,xr,es,maxit,p1,p2,...):% uses Newton-Raphson method to find the root of func

% input:

% func = name of function

% dfunc = name of derivative of function

% xr = initial guess% es = desired relative error (default = 0.0001%)

% maxit = maximum allowable iterations (default = 50)

% p1,p2,... = additional parameters used by function

% output:

% root = real root% ea = approximate relative error (%)% iter = number of iterations


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M-file for Newton Raphsons Method

if nargin


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Pitfalls of The Newton Raphson Method

Cases where Newton Raphson method diverges or exhibit poor

convergence.a) Reflection point b) oscillating around a local optimum

c) Near zero slope , and d) zero slope


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Example Newton Raphsons Method

Evaluate 29 to five decimal places by Newton-Raphsons iterative method


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Solution

Evaluate 29 to five decimal places by Newton-Raphsons iterative method


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Solution


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3. The Secant Method

The derivative is

sometimes difficult to evaluateby the computer program. It

may be replaced by a backward

finite divided difference

)()(

))((

i1i

i1iii1i

xfxf

xxxfxx

Thus, the formula

predicting thexi+1is:

/ 1

1

( ) ( )( ) i ii

i i

f x f xx

x x

/ ( )i

x


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The Secant Method

Requires two initial estimates of x , e.g, xo, x1.However, because f(x) is not required to change

signs between estimates, it is not classified as a

bracketing method.

The scant method has the same properties asNewtons method. Convergence is not

guaranteed for all xo, x1,f(x).


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Secant Method:Example

Determine a root of the equation sinx + 3 cos x

2 = 0using the secant method. The initial approximations

x0 and x1are 0 and 1.5.


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Comparison of convergence of False Position and

Secant Methods

False Position Secant Method

Use two estimatexland xu Use two estimatexiand xi-1

f(x) must changes signs betweenxl

and xu

f(x) is not required to change signs

betweenxiand xi-1

Xrreplaces whichever of the original

values yielded a function value with

the same sign as f(xr)

Xi+1replace xiXi replace xi-1

Always converge May be diverge

Slower convergence than Secant in

case the secant converges.

If converges, It does faster then False

Position

1

1

1

( )( )

( ) ( )

i i i

i i

i i

f x x xx x

f x f x

( )( )

( ) ( )

u l u

r u

l u

f x x xx x

f x f x


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Comparison of convergence of False Position and

Secant Methods

Use the false-position and secant method to find the root off(x)=logex. Start computation with xl=xi-1=0.5, xu=xi= 5.

1. False position method

2. Secant methodIter xi-1 xi xi+1

1 0.5 5.0 1.8546

2 5 1.8546 -0.10438

Iter xl xu xr

1 0.5 5.0 1.8546

2 0.5 1.8546 1.2163

3 0.5 1.2163 1.0585


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Comparison of

the true percentrelative Errors Et

for the methods

to the determine

the root off(x)=e-x-x


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Flow ChartSecant Method

Start

Input: x-1, x0,s, maxi

i=0

a=1.1s

1


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Stop

1

while

a>s &

i < maxi

Xi+1=0

Xi-1=xiXi=xi+1

1

1100%

i ia

i

x x

x

Print:xi, f(xi) ,a, i1

1

1

( )( )

( ) ( )

1

i i ii i

i i

f x x xx x

f x f x

i i

False

True


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Modified Secant Method

Rather than using two initial values, an alternative

approach is using a fractional perturbation of the

independent variable to estimate

1

( )

( ) ( )

i i

i i

i i i

x f xx x

f x x f x

is a small perturbation fraction

/ ( ) ( )( ) i i iii

f x x f xxx

/ ( )i

f x


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Modified Secant Method:Example

Use the modified secant method to find the root of

f(x) = e-x-x and, x0=1 and =0.01

0 0

0 0 0 0

1 1

1 1

1 1 1 1

2 1

1 0.63212

1.01 0.64578

( )0.537263

First Iteration

Second Iteration

5.3%( ) ( )

0.537263 0.047083

0.542635 0.038579

( )

(

i ii i t

i i i

i ii i

x f x

x x f x x

x f xx x x

f x x f x

x f x

x x f x x

x f xx x x

f x

0.56701 0.0236%

) ( ) t

i i ix f x


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Multiple Roots

x

f(x)= (x-3)(x-1)(x-1)= x3- 5x2+7x -3

f(x)

1x

3

Doubleroots

f(x)= (x-3)(x-1)(x-1)(x-1)

= x4- 6x3+ 125 x2- 10x+3

f(x)

1 3

tripleroots


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Multiple Roots

Multiple root corresponds to a pointwhere a function is tangent to the x axis.

Difficulties- Function does not change sign with double

(or even number of multiple root), therefore,

cannot use bracketing methods.

- Both f(x) and f(x)=0, division by zero withNewtons and Secant methods which maydiverge around this root.


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4. The Modified Newton Raphson Method by

Ralston and Rabinowitz

Another u(x)is introduced such that u(x)=f(x)/f/(x);

Getting the roots of u(x) using Newton Raphsons

technique:

)()()(

)()(

)]([

)()()()(

)(

)(

)(

//2/

/

1

2/

/////

/1

iii

iiii

i

iiiii

i

iii

xfxfxf

xfxfxx

xf

xfxfxfxf

xu

xu

xuxx

This function has rootsat all the same locations

as the original function


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Modified Newton Raphson Method: Example

Using the Newton Raphson and Modified Newton

Raphson evaluate the multiple roots off(x)= x3-5x2+7x-3with an initial guess ofx0=0

)106)(375()7103(

)7103)(375(

)()()()()(

2322

223

//2/

/

1

iiiiii

iiiiii

iii

iiii

xxxxxx

xxxxxx

xfxfxfxfxfxx

7x10x33x7x5xx

xfxfxx 2

i

i

2

i

3

ii

i

ii1i

)()(/

Newton Raphson formula:

Modified Newton Raphson formula:

M difi d N t R h M th d E l


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Newton Raphson Modified Newton-Raphson

Iter xi t% iter xi t%

0 0 100 0 0 100

1 0.4286 57 1 1.10526 11

2 0.6857 31 2 1.00308 0.31

3 0.83286 17 3 1.000002 00024

4 0.91332 8.7

5 0.95578 4.46 0.97766 2.2

Newton Raphson technique is linearly converging towards thetrue value of 1.0 while the Modified Newton Raphson is

quadratically converging.For simple roots, modified Newton Raphson is less efficientand requires more computational effort than the standard

Newton Raphson method

Modified Newton Raphson Method: Example


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Systems of Nonlinear Equations

Roots of a set of simultaneous equations:

f1(x1,x2,.,xn)=0

f2 (x1,x2,.,x

n)=0

fn (x1,x2,.,xn)=0

The solution is a set of x values that

simultaneously get the equations to zero.


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Systems of Nonlinear Equations

Example:x2

+ xy = 10& y + 3xy2

= 57u(x,y)= x2+ xy -10 = 0

v(x,y)= y+ 3xy2 -57 = 0

The solution will be the value of x and y which makesu(x,y)=0and v(x,y)=0

These arex=2and y=3

Numerical methods used are extension of the openmethods for solving single equation; Fixed point

iteration and Newton-Raphson. (we will only discuss

the Newton Raphson)


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Systems of Nonlinear Equations:

2.Newton Raphson Method


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x 2+ xy =10 and y + 3xy 2 = 57are two nonlinear simultaneous equations with two unknown x and y they canbe expressed in the form: use the point (1.5,3.5) as initial guess.

2

2 ,

3 , 1 6

u u

x y xx y

v vy xy

x y

i xi yi Ui Vi ui,x ui,y vi,x vi,ya,x

a,y

0 1.5 3.5 -2.5 1.625 6.5 1.5 36.75 32.5

1 2.03603 2.84388 -.06435 -4.7560 6.91594 2.03603 24.26296 35.74135 26.3 23.1

2 1.9987 3.00229 1.87 5.27




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i xi yi Ui Vi ui,x ui,y vi,x vi,y a,x a,y

0 1.5 3.5 -2.5 1.625 6.5 1.5 36.75 32.5

1 2.03603 2.84388 -.06435 -4.7560 6.91594 2.03603 24.26296 35.74135 26.3 23.1

2 1.9987 3.00229 1.87 5.27




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Problem

The polynomial f (x) = x3 -6x2 +11x- 6.1 has a real root

between 3 and 5. Apply the Newton-Raphsonsmethod to

this function using an initial guess ofx0= 3.5. Explain your

results

Problem


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Problem



this function using an initial guess of x0= 3.5. Explain your

results

Problem


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Problem




results

Problem


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Problem




results

Problem


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Problem

Find the root of polynomial f (x) = x3 -6x2+11x- 6.1. Apply

the Secantsmethod to this function using an initial guess

of x-1= 2.5 and x0= 3.5.

Problem


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Problem



of x-1= 2.5 and x0= 3.5.

Problem


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Problem



of x-1= 2.5 and x0= 3.5.

t 5 open methods

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