Lec 8 Secant

7/27/2019 Lec 8 Secant

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Roots of Nonlinear Equations

Topic: Secant Method

Dr. Nasir M Mirza

Numerical Methods

Email: [email protected]


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Secant Method

)(xf

)f(x-= xx

i

iii

1

f(x)

f(xi)

f(xi-1)

xi+2 xi+1 xiX

ii xfx ,

1

1 )()()(

ii

ii

i

xx

xfxfxf

)()(

))((

1

11

ii

iiiii

xfxf

xxxfxx

Newton-Raphson Method

Approximate the derivative


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Secant Method

)()(

))((

1

11

ii

iiiii

xfxf

xxxfxx

Geometric Similar Trianglesf(x)

f(xi)

f(xi-1)

xi+1 xi-1 xiX

B

C

E D A

11

1

1

)()(

ii

i

ii

i

xx

xf

xx

xf

DE

DC

AE

AB


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Algorithm for Secant MethodTo find a solution to f(x) = 0 given initial guess as x0 and x1.

INPUT: initial guess x0 nd x1, max. number of iterations (N) andtolerance (TOL).

OUTPUT: approximate solution

Step 1: Set i = 1;

y0 = f(x0); y1 = f(x1);Step 2: while i N do step 3 to 6

Step 3: Set xi+1 = xiyi(xixi-1)/(yiyi-1)

Step 4: If | xi+1xi | < TOL then

OUTPUT ( xi+1 ); procedure completed successfully

STOP

Step 5: Set i = i+1

Step 6: Set xi = xi+1 ; yi = f(xi);

Step 7: OUTPUT: Method failed after N iterations

STOP


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Step 1

010x

1

1

x

- xx=

i

iia

Calculate the next estimate of the root from two initialguesses

Find the absolute relative approximate error

)()(

))((

1

11

ii

iiiii

xfxf

xxxfxx


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Step 2

Find if the absolute relative approximate error is greaterthan the pre-specified relative error tolerance.

If so, go back to step 1, else stop the algorithm.

Also check if the number of iterations has exceeded themaximum number of iterations.


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Example

You are working for DOWN THE TOILET COMPANY thatmakes floats for ABC commodes. The ball has aspecific gravity of 0.6 and has a radius of 5.5 cm. Youare asked to find the distance to which the ball will getsubmerged when floating in water.


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Solution

The equation that gives the depth x to which the ball issubmerged under water is given by

Use the Secant method offinding roots of equations tofind the depth x to which theball is submerged under water.

Conduct three iterations toestimate the root of the aboveequation.

423 10x99331650 -.+x.-xxf


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Graph of function f(x)

-0.02 0.00 0.02 0.04 0.06 0.08 0.10 0.12

-0.0004

-0.0003

-0.0002

-0.0001

0.0000

0.0001

0.0002

0.0003

0.0004

0.0005

f(x)

x

4

23

10x9933

1650

-.+

x.-xxf


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Iteration #1

%61.22

0.06461

05.0,02.0

1

10

100

01

01

a

x

xfxf

xxxfxx

xx


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Iteration #2

%525.3

06241.0

x

06461.0,05.0

2

01

011

12

10

a

x

xff

xxxfxx

xx


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Secant Method algorithm

Do while |x2 - x1| >= tolerance value 1

or |f(x3)|>= tolerance value 2

Set x3 = x2 - f(x2)*(x2 - x1)/(f(x2)-f(x1))

Set x1 = x2;

Set x2= x

3;

END loop


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Example 2:

Let us solve thefollowing problem:

x4 - x10 = 0

Then f(x) = x4 - x10

df/dx = 4x3 - 1

Let us first draw a

graph and see where are

the roots;

The roots are between (-2, -1), & (1, 3)

-3 -2 -1 0 1 2 3

-20

0

20

40

60

80

f(x)

x


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Matlab pogram for graph

% --------------- graph of a function

clear all

fun = @(x) x^4. - x - 10.0 ;

xmin = -3.0; xmax = 3.0; max_points = 100;

del = (xmax - xmin)/max_points;x(1)= xmin ; y(1)=0.0;

for j=2: 1 : max_points

x(j) = x(j-1) + del ; end

for i = 1: max_points

f(i)= fun(x(i)) ; y(i)=0.0 ; endaxis( [xmin xmax -20. 80.0] )

hold on

plot(x, f, x, y, 'LineWidth',2)


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Solution

4375.30.10)50.1()50.1()(50.1

0.80.8

.)2(0.1)0.8(0.1

0.80.10.)1()1()(,0.1

0.80.10.)2()2()(,0.2

4

11

10

10001

4

00

411

xfx

xfxf

xxxfxx

xfx

xfxIteration 1:

2812.4)(,8767.1

)0.8(4375.3

)0.1(5.1)4375.3(5.1

4375.30.10)50.1()50.1()(,50.1

0.80.10.)1()1()(,0.1

22

01

01112

4

11

4

00

xfx

xfxf

xxxfxx

xfx

xfx

Iteration 2:


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Solution

5951.0)(6678.1

)4375.3(2812.4

)5.1(8767.1)2812.4(8767.1

2812.40.10)8767.1()8767.1()(,8767.1

4375.30.10)50.1()50.1()(50.1

33

12

12223

4

22

411

xfx

xfxf

xxxfxx

xfx

xfxIteration 3:

Iteration 4:

0857.0)(6933.1

)2812.4(5951.0

)8767.1(6678.1)5951.0(6678.1

5951.00.10)6678.1()6678.1()(,6678.1

2812.40.10)8767.1()8767.1()(,8767.1

44

23

23334

4

23

4

22

xfx

xfxf

xxxfxx

xfx

xfx


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Solution

0022.0)(6976.1

6976.1

0857.00.10)6933.1()6933.1()(,6933.1

5951.00.10)6678.1()6678.1()(,6678.1

45

34

34445

4

44

433

xfx

xfxf

xxxfxx

xfx

xfxIteration 5:

Iteration 6:

0001.0)(6975.1

6975.1

0022.00.10)6976.1()6976.1()(,6976.1

0857.00.10)6933.1()6933.1()(,6933.1

46

45

45556

4

45

4

44

xfx

xfxf

xxxfxx

xfx

xfx

The root seems to be -1.6975 with deviation of 0.0001.


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Computer Program in MATLAB

% Use of secant method to find the roots of F(x)=0% Input: xleft,xright = left and right brackets of the root

% n = (optional) number of iterations; default: n =6

% Output: x = estimate of the root

n=8; xleft = -2.0; xright = -1.0;

a = xleft; b =xright; % Copy original bracket to local

variablesfa = a^4 - a - 10.0; % Initial values

fb = b^4 - b - 10.0;

fprintf(' k a b xm f(xm)\n');

for k=1:n

xm = b - fb*((b-a)/(fb-fa)); % Minimize roundoff

fm = xm^4 - xm - 10.0; % f(x) at xm

fprintf('%3d %12.8f %12.8f %12.8f

%12.3e\n',k,a,b,xm,fm);

a = b; fa = fb; b = xm; fb = fm;

end


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Results

>>

k a b xm f(xm)

1 -2.00000000 -1.00000000 -1.50000000 -3.438e+000

2 -1.00000000 -1.50000000 -1.87671233 4.282e+000

3 -1.50000000 -1.87671233 -1.66776026 -5.959e-001

4 -1.87671233 -1.66776026 -1.69328962 -8.570e-002

5 -1.66776026 -1.69328962 -1.69757795 2.181e-003

6 -1.69328962 -1.69757795 -1.69747151 -7.683e-006

7 -1.69757795 -1.69747151 -1.69747188 -6.851e-0108 -1.69747151 -1.69747188 -1.69747188 0.000e+000

>>

Result of the computer program: xleft = -2.0 , xright = -1.0and iterations n = 8.

The root value is x = -1.69747188


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Advantages

Converges fast, if it converges; Requires two guesses that do not need to bracket the

root.

It is used to refine answers obtained by other

methods such as bisection. Since this method requires a good first approximation

therefore it gives rapid convergence.


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Problems with the Secant Method

This method can be

catastrophic if thepoints are near apoint where the firstderivative is zero.


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Drawbacks

Division by zero

10 5 0 5 10

2

1

0

1

2

f(x)

prev. guess

new guess

2

2

0

f x( )

f x( )

f x( )

1010 x x guess1 x guess2

0 xSinxf


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Drawbacks (continued)

Root Jumping

10 5 0 5 102

1

0

1

2

f(x)x'1, (first guess)x0, (previous guess)Secant linex1, (new guess)

2

2

0

f x( )

f x( )

f x( )

secant x( )

f x( )

1010 x x 0 x 1' x x 1

0 Sinxxf

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Lec 8 Secant