Lecture 9 Root Finding:

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Lecture 9 Root Finding:

Solution of Nonlinear Functions

Prof. M. Moness

Minia University

Department of Computers and Systems Engineering

CSE 121, Prog. Lang.2, Prof. M. Moness,

2020 1

Issues

– Definitions

– Classification of Methods

• Analytical Solutions

• Graphical Methods

• Numerical Methods

– Bracketing Methods

– Open Methods

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§ 9.1 Introduction

• Definition :Root Finding Problems

• Many problems in Science and Engineering are expressed as:

These problems are called root finding problems.

A number r that satisfies an equation is called a root of the equation, e.g.,

This image cannot currently be displayed.

) 1 ( ) 3 )(

2 ( 18 15 7

3 . 1 ,

3 , 3 , 2

18 15

7 3

2 2

3 4

2 3 4



















x x

x x x

and

x x

i.e.,

: roots four has

: equation The

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4

Solution Methods

Several ways to solve nonlinear equations are possible:

– Analytical Solutions

• Possible for special equations only – Graphical Solutions

• Useful for providing initial guesses for other methods

– Numerical Solutions

• Bracketing methods

• Open methods

CSE 121, Prog. Lang.2, Prof. M. Moness, 2020

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5

Analytical Methods

Analytical Solutions are available for special equations only.

if f(x) is a polynomial of degree 5 or higher, there is no such formula.

Also for nonlinear functions:

a ac b

roots b

c x b x a

2 4

0 2

2 



 





 :

of solution Analytical

 0

 _e  x x

: for available

is solution analytical

No

• Another example, consider the following function:

F(x) = 48x(1 + x) ⁶⁰ ‐ (1 + x) ⁶⁰ + 1 = 0

• What are the roots (solutions) of f(t), i.e., values x _i , I = 1,, 2, …. , 60 such that f(x _i )=0.

– How would you solve such an equation?

HIGH‐DEGREE POLYNOMIALS:

• As we mentioned before, for a quadratic equation : ax ² + bx + c = 0, there is a well‐ known formula for the roots.

• For third‐ and fourth‐degree equations, there are also formulas for the roots.

– However, they are extremely complicated.

If f (x) is a polynomial of degree 5 or higher, there is

no such formula.

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7

Graphical Methods

Graphical methods are useful to provide an initial guess to be used by other methods.

 0.6



 

root root The

e x Solve

x

] 1 , 0 [

e ^ x

x

Root

1 2 2

1 • Likewise, there is no formula that will enable us to find the exact roots of a transcendental equation such as :

cos x = x.

• APPROXIMATE SOLUTION

• We can find an approximate solution by plotting the left side of the equation.

• Using a graphing device, and after experimenting with viewing rectangles,

we produce the graph shown.

Another examples are the transcendental equations

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• We see that, in addition to the solution x = 0, which

doesn’t interest us, there is a solution between 0.007 and 0.008

– Zooming in shows that the root is approximately 0.0076

• If we need more accuracy, we could zoom in repeatedly.

• That becomes tiresome, though.

Zooming in

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Numerical Methods

A faster alternative is to use a numerical root finder on a calculator or computer algebra system.

Many methods are available to solve nonlinear equations:

Bisection Method

Newton’s Method

Secant Method

– False position Method – Muller’s Method – Bairstow’s Method – Fixed point iterations – ……….

Numerical methods are classified as either

bracketing method or open methods

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11

Bracketing Methods

• In bracketing methods, the method starts with an interval that contains the root and a procedure is used to obtain a smaller interval containing the root.

• Examples of bracketing methods:

– Bisection method – False position method

Open Methods

• In the open methods, the method starts with one or more initial guess points. In each iteration, a new guess of the root is obtained.

• Open methods are usually more efficient than bracketing methods.

• They may not converge to a root.

In both cases, initial guesses are required

§ 9.2 Bracketing Methods

• Both bisection and false‐position methods require the root to be bracketed by the endpoints.

• How to find the interval (endpoints)?

* plotting the function

* incremental search

* trial and error

• We 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 x

_l

to x

_u

and f(x

_l

) and f(x

_u

) have opposite signs, that is f(x

_l

) f(x

_u

) < 0

Then there is at least one real root between f(x _l ) and f(x _u )

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• Given some function, find location where f(x)=0

• Need:

Starting position x

₀

, hopefully close to a solution

Ideally, points that bracket the root

Well‐behaved function f(x

₊

) > 0

f ( x

_–

) < 0

Incremental Search:

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Intermediate Value Theorem

• Let f(x) be defined on the interval [a,b].

• Intermediate value theorem:

if a function is continuous and f(a) and f(b) have different signs then the function has at least one zero in the interval [a,b].

a

b f(a)

f(b)

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15

Examples

• If f(a) and f(b) have the same sign, the function may have an even number of real zeros or no real zeros in the interval [a, b].

• Bisection method can not be used in these cases.

a b

The function has four real zeros

The function has no real zeros

Design a MATLAB function that implemental search (Steven Chapra, page 160)

function xb = incsearch(func,xmin,xmax,ns)

% incsearch: incremental search root locator

% xb = incsearch(func,xmin,xmax,ns):

% finds brackets of x that contain sign changes

% of a function on an interval

% inputs:

% func = name of function

% xmin, xmax = endpoints of interval

% ns = number of subintervals (default = 50)

% output:

% xb(k,1) is the lower bound of the kth sign change

% xb(k,2) is the upper bound of the kth sign change

% If no brackets found, xb = [].

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% Check for input arguments if nargin < 3

error('at least 3 arguments required') end

if nargin < 4 ns = 50;

end %if ns blank set to 50

% Incremental search

x = linspace(xmin,xmax,ns);

f = func(x);

nb = 0; xb = []; %xb is null unless sign change detected

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for k = 1:length(x)-1

if sign(f(k)) ~= sign(f(k+1)) %check for sign change nb = nb + 1;

xb(nb,1) = x(k);

xb(nb,2) = x(k+1);

end end

if isempty(xb) %display that no brackets were found disp('no brackets found')

disp('check interval or increase ns') else

disp('number of brackets:') %display number of brackets disp(nb)

end

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Example :Use the previous M-file incsearch to identify brackets within the interval [3, 6] for the function:

f(x) = sin(10x) + cos(3x) Solution:

incsearch(@(x) sin(10x)+cos(3x), 3 , 6)

>> incsearch(@(x) sin(10x)+cos(3x), 3 , 6) number of brackets:

5 ans =

3.2449 3.3061 3.3061 3.3673 3.7347 3.7959 4.6531 4.7143 5.6327 5.6939

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3 3.5 4 4.5 5 5.5 6

x

-2

-1.5 -1 -0.5 0 0.5 1 1.5 2

f(x)

f(x)=sin(10x)+cos(3x)

missed missed

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If we increase increase subintervals to 200

xb=incsearch(inline('sin(10x)+cos(3x)'),3,6,200) number of brackets:

9 xb =

3.2563 3.2714 3.3618 3.3769 3.7387 3.7538 4.2211 4.2362 4.2513 4.2663 4.7035 4.7186 5.1558 5.1709 5.1859 5.2010 5.6683 5.6834

Find all 9 roots!

3 3.5 4 4.5 5 5.5 6

x -2

-1.5 -1 -0.5 0 0.5 1 1.5 2

f(x)

f(x)=sin(10x)+cos(3x)

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§ 9.1.1The Bisection Method

• It is the simplest root‐finding algorithm.

• requires previous knowledge of two initial guesses, a and b, so that f(a) and f(b) have opposite signs.

• Two estimates are chosen so that the solution is bracketed.

i.e., f(a) and f(b) have opposite signs.

• In the diagram this f(a) < 0 and f(b) > 0.

• The root d lies between a and b!

initial pt ‘a’

initial pt ‘b’

a

root ‘d’

d c

b

• The root d must always be bracketed (be

between a and b)!

• Iterative method.

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 MORE:

• The Bisection method is one of the simplest methods to find a zero of a nonlinear function.

• It is also called interval halving method.

• To use the Bisection method, one needs an initial

interval that is known to contain a zero of the function.

• The method systematically reduces the interval. It does this by dividing the interval into two equal parts,

performs a simple test and based on the result of the test, half of the interval is thrown away.

• The procedure is repeated until the desired interval size is obtained.

The Bisection Method (Cont.)

• The interval between a and b is halved by calculating the average of a and b.

• The new point c = (a+b)/2.

• This produces are two

possible intervals: a < x < c and

c < x < b .

initial pt ‘a’

initial pt ‘b’

a

root ‘d’

d c

b

• If f(c ) > 0, let a _new = a and b _new = c and repeat process.

• If f(c ) < 0, let a _new = c and b _new = b and repeat process.

• This reassignment ensures the root is always bracketed!!

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Bisection Method : Algorithm

Step 1 : Choose and such that and bracket the root, i.e. f( **)*f( ) < 0.**

Step 2 : Estimate the root (bisection),

**= 0.5*(** + )

Step 3: Determine the new bracket, If f( **)*f( ) < 0**

= and still the same else

= and still the same end

Step 4: Examine if is the root, i.e., ) =  , where  is the error (required accuracy), if not continue.

Step 5 : Repeat steps 2 and 3 until convergence

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Flow Chart of Bisection Method

Start: Given a,b and ε u = f(a) ; v = f(b) c = (a+b) /2 ; w = f(c)

is u w <0

a=c; u= w b=c; v= w

is (b-a) /2<ε yes

yes

no Stop

no

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Error

• Convergence rate:

– Error bounded by size of [x _a … x _b ] interval – Interval shrinks in half at each iteration – Therefore, error cut in half at each iteration:

|  _n+1 _| _{= ½ |}  _n _| – This is called “linear convergence”

– One extra bit of accuracy in x at each iteration

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Error Estimates for Bisection

At the iter1:

∗

length of the interval

12 16

‐34.8 17.6

16

‐12.6 17.6

12

14 15

‐12.6 1.5

‐34.8

Change of sign

17.6 Change

of sign

Iter1

Iter2

14

16

∗

True root live inside this interval

∗

∆ ∆

At the iter2:

∗

length of the interval

∗

∆ ∆

At the nth iteration:

∗

length of the interval

∗

∆ ∆

∗

the absolute error in the n-th iteration

Error Estimates for Bisection

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Suppose is a continuous function defined on the interval , , with

and of opposite sign. The Intermediate Value Theorem implies that a number exists in , with

.

This technique is based on the Intermediate Value Theorem

Show that

has a root in [12, 16]

Example:

12 16

. .

Solution:

Since there is a change in sign, then f(x) has a root in the rang [12,16]

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Use Bisection method to find the root of the function

in [12, 16]

Example:

16

‐12.6 17.6

12

14 15

‐12.6 1.5

14.5 151.5

1 14.0000000000 2 15.0000000000 3 14.5000000000 4 14.7500000000 5 14.8750000000 6 14.9375000000 7 14.9062500000 8 14.8906250000 9 14.8984375000 10 14.9023437500 11 14.9003906250 12 14.8994140625 13 14.8999023438 14 14.9001464844 15 14.9000244141

‐34.8

17.6

Change of sign

Iter1

Iter2

Iter3

14

True root: ^∗ 14.9

16

14

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12 16

‐34.8 17.6

16

‐12.6 17.6

12

14 15

‐12.6 1.5

14.5 151.5

‐34.8

17.6

Change of sign

Change of sign

Iter1

Iter2

Iter3

14

16

14

∆

At the n

^th

iteration:

endpoints of the inrteval

∆ ,

Length of the interval

∆

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If

^,

is the desired error, this equation can be solved for

Error Estimates for Bisection

. n must be an integer !

1 14.0000000000 ‐9.0000000000e‐01 2 15.0000000000 1.0000000000e‐01 3 14.5000000000 ‐4.0000000000e‐01 4 14.7500000000 ‐1.5000000000e‐01 5 14.8750000000 ‐2.5000000000e‐02 6 14.9375000000 3.7500000000e‐02 7 14.9062500000 6.2500000000e‐03 8 14.8906250000 ‐9.3750000000e‐03 9 14.8984375000 ‐1.5625000000e‐03 10 14.9023437500 2.3437500000e‐03 11 14.9003906250 3.9062500000e‐04 12 14.8994140625 ‐5.8593750000e‐04 13 14.8999023438 ‐9.7656250000e‐05 14 14.9001464844 1.4648437500e‐04 15 14.9000244141 2.4414062500e‐05 16 14.8999633789 ‐3.6621093750e‐05

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Bisection Method : function file

• What should do?

You pass the function required (func) along with lower (xl) and upper (xu) guesses.

an optional stopping criterion (es) and maximum iterations (maxit) can be entered.

The built‐in function first checks whether there are sufficient arguments and if the initial guesses bracket a sign change.

If not, an error message is displayed and the function is terminated.

 It also assigns default values if maxit and es are not supplied.

Then a while...break loop is employed to implement the bisection algorithm until the approximate error falls below es or the iterations exceed maxit.

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Bisection Method : function file (Chapra, page 151)

function [root,fx,ea,iter]=bisect(func,xl,xu,es,maxit,varargin)

% bisect: root location zeroes

% [root,fx,ea,iter]=bisect(func,xl,xu,es,maxit,p1,p2,...):

% uses bisection method to find the root of func

% input:

% func = name of function

% xl, xu = lower and upper guesses

% es = desired relative error (default = 0.0001%)

% maxit = maximum allowable iterations (default = 50)

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

% output:

% root = real root

% fx = function value at root

% ea = approximate relative error (%)

% iter = number of iterations

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if nargin<3

error('at least 3 input arguments required') end

test = func(xl,varargin{:})*func(xu,varargin{:});

if test>0

error('no sign change') end

if nargin<4||isempty(es) es=0.0001;

end

if nargin<5||isempty(maxit) maxit=50;

End

iter = 0; xr = xl; ea = 100;

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while (1)

xrold = xr; xr = (xl + xu)/2; iter = iter + 1;

if xr ~= 0

ea = abs((xr ‐ xrold)/xr) * 100;

end

test = func(xl,varargin{:})*func(xr,varargin{:});

if test < 0 xu = xr;

elseif test > 0 xl = xr;

else ea = 0;

end

if ea <= es || iter >= maxit break

end end

root = xr; fx = func(xr, varargin{:})

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The Bisection Method Stopping Criterion : version 1

function [xr,err,yc,iter,x]=bisect_ver1(f,a,b,es)

%Input: f is the function, a, b are endpts

% es is the tolerance, imax is max iter

%Output: c is the zero, yc= f(c) ya=f(a); yb=f(b); iter =0;

if **ya*yb > 0,return,end for k=1:1000**

iter = iter +1;

xr=(a+b)/2; yc=f(xr); x(k)=xr;

if yc==0

a=xr; b=xr;

elseif **yb*yc>0 b=xr; yb=yc;**

else

a=xr; ya=yc;

end

if b‐a < es, break,end end

xr=(a+b)/2; err=abs(b‐a); yc=f(xr);

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Example: Use Bisection method to find the root of the

function in the interval [1,16]

Using bisect_ver1 function :

**>> f ‐@(x)(10x^6‐149x^5+10x‐149)/((10(x^4+1)))**

>> xl=12;xu=16;es=0.001;

>> [xr,err,yc,iter,x]=bisect_ver1(f,xl,xu,es) xr =

14.8999 err =

9.7656e‐04 yc =

‐0.0015 iter =

12 x =

Columns 1 through 11

14.0000 15.0000 14.5000 14.7500 14.8750 14.9375 14.9063 14.8906 14.8984 14.9023 14.9004

Column 12

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The Bisection Method Stopping Criterion : version 2 function

[xr,err,yc,iter,x]=bisect_ver2(f,a,b,es)

%Input: f is the function, a, b are endpts

% es is the tolerance, imax is max iter

%Output: c is the zero, yc= f(c) ya=f(a); yb=f(b); iter =0;

if **ya*yb > 0,return,end**

max1=1+round((log(b‐a)‐log(es))/log(2));

for k=1:max1 iter = iter +1;

xr=(a+b)/2; yc=f(xr); x(k)=xr;

if yc==0

a=xr; b=xr;

elseif **yb*yc>0 b=xr; yb=yc;**

else

a=xr; ya=yc;

end

% if b‐a < es, iter=k; break,end end

xr=(a+b)/2; err=abs(b‐a); yc=f(xr);

Stopping Criteria

>>

[xr,err,yc,iter,x]=bisect_ver2(f,xl,xu,es) xr =

14.9001 err =

4.8828e‐04 yc =

0.0022 iter =

13 x =

Columns 1 through 11

14.0000 15.0000 14.5000 14.7500 14.8750 14.9375 14.9063 14.8906 14.8984 14.9023 14.9004

Columns 12 through 13 14.8994 14.8999

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Advantages and Disadvantages of Bisection Method

Advantages

• Simple and easy to implement

• One function evaluation per iteration

• The size of the interval containing the zero is reduced by 50% after each iteration

• The number of iterations can be determined a priori

• No knowledge of the derivative is needed

• The function does not have to be differentiable Disadvantage

• Slow to converge

• Good intermediate approximations may be discarded

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§ 9.2 Open Methods

• There exist open methods which do not require bracketed intervals

• Newton‐Raphson method, Secant Method, Muller’s method, fixed‐point iterations

• First one to consider is the fixed‐point method

• Converges faster but not necessary converges

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• Newton and Newton-Raphson are just different names for the same method.

• King of the root‐finding methods

• Newton‐Raphson method Based on Taylor series expansion

Newton's Method

Given an initial guess of the root x

₀

, Newton‐Raphson method uses information about the function and its derivative at that point to find a better guess of the root.

Assumptions:

– f(x) is continuous and the first derivative is known – An initial guess x

₀

such that f’(x

₀

)≠0 is given

             



 





_ _



2 i 1 i i

1 i i i

1

i

x x

! 2 x f

x x f x f x

f 

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Truncate the Taylor series to get

 x _i  f   x _i f   x _i x _i x _i 

f  1     1 

At the root, f ( x _i+1 ) = 0 , so

  ^x _i ^f   ^x _i ^x _i ^x _i 

f   

 1

0  

  ⁱ _i

i

i f x

x x f

x 1   

Newton’s Method : Cont.

2020 43

root

x*

0 4 3 )

( _x _ _x 4 _ _x _ _ f

Newton’s Method : illustration

False position - secant line Newton’s method - tangent line

x _i x _i+1

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Newton’s Method

• Note that an evaluation of the derivative (slope) is required

• You may have to do this numerically

• Open Method – Convergence depends on the initial guess (not guaranteed)

• However, Newton method can converge very quickly (quadratic convergence)

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Newton’s Method : Algorithm

• Step 1 : Start at the point (x ₁ , f(x ₁ )).

• Step 2 : The intersection of the tangent of f(x) at this point and the x‐axis.

x ₂ = x ₁ ‐ f(x ₁ )/f `(x ₁ )

• Step 3: Examine if f(x ₂ ) = 0

or abs(x ₂ ‐ x ₁ ) < tolerance,

• Step 4: If yes, solution x _r = x ₂

If not, x ₁ x ₂ , repeat the iteration.

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47

Newton’s Method : Graphical Depiction ‐

• If the initial guess at the root is x _i , then a tangent to the

function of x _i that is f’(x _i ) is extrapolated down to the x‐axis to provide an

estimate of the root at x _i+1 .

• Consider the tangent line L to the curve y = f(x) at the point (x ₁ ,f(x ₁ )) and look at the x‐intercept of L, labeled x ₂ .

• Here’s the idea behind the method.

– The tangent line is close to the curve.

– So, its x‐intercept, x

₂

, is close to the x‐intercept of the curve (namely, the root r that we are seeking).

– As the tangent is a line, – we can easily

find its x‐intercept

NEWTON’S METHOD

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• To find a formula for x ₂ in terms of x ₁ , we use the fact that the slope of L is f’(x ₁ ).

• So, its equation is:

y ‐ f(x ₁ ) = f’(x ₁ )(x ‐ x ₁ )

• As the x‐intercept of L is x ₂ , we set y = 0 and obtain:

• 0 ‐ f(x ₁ ) = f’(x ₁ )(x ₂ ‐ x ₁ )

• If f’(x ₁ ) ≠ 0, we can solve this equation for x ₂ :

• We use x ₂ as a second approximation

• to r.

NEWTON’S METHOD

1 2 1

1 ( ) '( ) x x f x

  f x

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• Next, we repeat this procedure with x ₁ replaced by x ₂ , using the tangent line at (x ₂ ,f(x ₂ )).

This gives a third approximation:

• If we keep repeating this process, we obtain a sequence of approximations

x ₁ , x ₂ , x ₃ , x ₄ , . . .

3 2

2 2

( )

'( ) x x f x

  f x

THIRD APPROXIMATION

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• In general, if the nth approximation is x _n and f’(x _n ) ≠ 0, then the next approximation is given by:

• If the numbers x _n become closer and closer to r as n

becomes large, then we say that the sequence converges to r and we write:

• The sequence of successive approximations converges to the desired root for functions of the type illustrated in the previous figure. However, in certain circumstances, it may not converge.

1

( )

'( )

n

n n

n

x x f x



  f x

SUBSEQUENT APPROXIMATION

lim

_n

n

x r





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• Consider the situation shown here.

• You can see that x ₂ is a worse approximation than x ₁ . – This is likely to be the case when f’ (x

₁

) is close to 0.

It might even happen that an approximation falls outside the domain of f, such as x ₃ .

– Then, Newton’s method fails.

– In that case, a better initial approximation x

₁

should be chosen.

NON‐CONVERGENCE

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A Simple MATLAB Program

end

x f

x x f

x

n i for

x f n Assumputio

x x f x f Given

i i i

i

' ( )

) ( : 0

__

__________

0 ) ( '

), ( ' ), (

1

0 0









end

X FP X F X X i for X

PROGRAM MATLAB

) ( / ) ( 5 : 1 4

%





2020 53

• Example 1:

• find the third approximation x ₃ to the root of the equation x ³ – 2x – 5 = 0

• Starting with x ₁ = 2,

• We apply Newton’s method with

f(x) = x ³ – 2x – 5 and f’(x) = 3x ² – 2

– Newton himself used this equation to illustrate his method.

– He chose x

₁

= 2 after some experimentation

because f(1) = ‐6, f(2) = ‐1 and f(3) = 16

(28)

55

Example (Cont.)

2130 . 0742 2 . 9

0369 . 4375 2 . ) 2 ( '

) (

4375 . 16 2 3 9 ) ( '

) (

33 3 4 33 ) ( '

) ( 1

4 3 '

2 2 2

3 1 1 1

2 0 0 0

1 2

































x f

x x f

x

x f

x x f

x

x f

x x f

x x x (x) f

: 3 Iteration

: 2 Iteration

: 1 Iteration

56

Example

k(Iteration)

x

k

f(x

k

) f’(x

k

) x

k+1

|x

k+1

–x

k

|

0 4 33 33 3 1

1 3 9 16 2.4375 0.5625

2 2.4375 2.0369 9.0742 2.2130 0.2245

3 2.2130 0.2564 6.8404 2.1756 0.0384 4 2.1756 0.0065 6.4969 2.1746 0.0010

(29)

57

Disadvantages of Newton’s Method

• If the initial guess of the root is far from the root the method may not converge.

• Newton’s method converges linearly near multiple zeros { f(r) = f’(r) =0 }. In such a case, modified algorithms can be used to regain the quadratic convergence.

Example 2:

Use Newton’s method to find correct to eight decimal places.

– First, we observe that finding is equivalent to finding the positive root of the equation x

⁶

– 2 = 0

– So, we take f(x) = x

⁶

– 2 – Then, f’(x) = 6x

⁵

So, Formula of Newton’s method becomes:

Choosing x

₁

= 1 as the initial approximation, we obtain:

As x

₅

and x

₆

agree to eight decimal places, we conclude that

to eight decimal places.

6

2

6

2

6 1 5

2 6

n

n n

n

x x x

 x

  

2 3 4 5

1.16666667 1.12644368 1.12249707 1.12246205 x

x x x



6 

2  1.12246205

(30)

Example 3:

Find, correct to six decimal places, the root of the equation cos x = x.

– We rewrite the equation in standard form: cos x – x = 0 – Therefore, we let f(x) = cos x – x

– Then, f’(x) = –sinx – 1

So, Newton’s formula becomes:

1 cos

sin 1

cos

sin 1

n n

n

n n

n

x x

x

x x



  

 

  



2020 59

• To guess a suitable value for x ₁ , we sketch the graphs of y = cos x and y = x.

– It appears they intersect at a point whose x‐coordinate is somewhat less than 1.

Example 3 : Cont.

2020 60

(31)

• So, let’s take x ₁ = 1 as a convenient first approximation.

– Then, remembering to put our calculator in radian mode, we get:

– As x

₄

and x

₅

agree to six decimal places (eight, in fact), we conclude

that the root of the equation, correct to six decimal places, is 0.739085

2 3 4 5

0.75036387 0.73911289 0.73908513 0.73908513 x

x x x



Example 3 : Cont.

2020 61

Flow Chart – Newton ^Start

Input: x_o,



_s, maxi



_a=1.1i=0



_s

Stop

while

a>s&

i <maxi

i=1 or x_n=0

x₀=x_n

n o100%

a n

x x

x

 



Print: x_o, f(x_o) ,



_a, i

  ⁰

0 '

0

1

n

f x

x x

f x

i i

 

 

(32)

A MATLAB Program : Chapara, page 171

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,...):

% 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 if nargin<3

error('at least 3 input arguments required')

end

2020 63

if nargin<4|isempty(es) es=0.0001;

end

if nargin<5|isempty(maxit) maxit=50;

end iter = 0;

while (1)

xrold = xr;

xr = xr - func(xr)/dfunc(xr);

iter = iter + 1;

if xr ~= 0

ea = abs((xr - xrold)/xr) * 100;

end

if ea <= es | iter >= maxit break

end end

root = xr ;

2020 64

Lecture 9 Root Finding: