m.mehtab

7/28/2019 M.MEHTAB

1/9

ES-341

NUMERICAL ANALYSIS

MATLAB ASSIGNMENT

NAME: MUHAMMAD MEHTAB

REG NO.: 2009174

FACULTY: ME

DATE: 17-OCT-2011

Submitted to : Dr. Siraj Ul Haq

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General Code For Newton Methodfunction [ output_args ] = Untitled7( input_args )

%UNTITLED7 Summary of this function goes here

% Detailed explanation goes here

fprintf('Solution to Question 3b using newton Method\n\n');x101=input('enter lower limit:\n');

x202=input('enter upper limit:\n');

tolerence=input('enter tolerance:\n');

p0=input('enter first guess:\n');

n=2;u=0;pn=1000;

disp('n p'); disp('----------------------');

fprintf('%.0f %9.3f \n',u,p0);

u=u+1;q=p0;

while (abs(pn-q)>tol)

a=funcsecant(p0) b=func2(p0); q=p0;

pn=p0-(a./b);

fprintf('%.0f %9.6f \n',n,pn);p0=pn; n=n+1;

end

fprintf('solution is %9.6f\n\n',pn);

end

General Code For Secant Method

x101=input('enter lower limit:\n');


tollerence=input('enter tolerance:\n');


p1=input('enter second guess:\n');

n=2; u=0; pn=1000;disp('n p'); disp('----------------------');

fprintf('%.0f %9.3f \n',u,p0);u=u+1; fprintf('%.0f %9.3f \n',u,p1);

q=p0;


a=funcsecant(p0); b=funcsecant(p1); q=p1;pn=p1-(b.*(p1-p0))./(b-a);

fprintf('%.0f %9.6f \n',n,pn);

p0=p1 p1=pn; n=n+1;

end


General Code for Newton Modified Method

x1=input('enter lower limit:\n');


tol=input('enter tolerance:\n');


%%p1=input('enter second guess:\n');

n=2; u=0; pn=1000;

disp('n p'); disp('----------------------');

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u=u+1; q=p0;


a=funcsecant(p0);

b=func2(p0);

c=func3(p0);q=p0;

pn=p0-((a.*b)/((b.^2)-(a.*c)));fprintf('%.0f %9.6f \n',n,pn); p0=pn; n=n+1;

end


Q1)y=(x.^x)-(3.*x)+(sin(3.*x))

function [ y ] = funcsecant( x )



y=(x.^x)-(3.*x)+(sin(3.*x));

end

function [ z ] = func2( x )



z=(x.^x)*(1+log(x))-3+3.*cos(3.*x);end

function [ y ] = func3( x )



y=((x.^x).*(1+log(x)).^2)-(9.*sin(3.*x));

end

RESULT

Question 1 :

y=ln(x-1)+cos(x-1)-sin(x.12);


y=ln(x-1)+cos(x-1);

end

function [ z ] = func2( x )z=(1/(x-1)-sin(x-1));

end

function [ y ] = func3( x )

y=(-1/((x-1).^2)-cos(x-1));

end

Result:

Newton Method

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enter lower limit:

1.3

enter upper limit:

1.5enter tolerance:

10^-5enter first guess:

1.4

n(no of iterations) p(Values)

0 1.400

2 -0.089115

3 0.036332

4 -0.362741

5 -0.249266

6 -0.154254

7 -0.138628

8 -0.139177

9 -0.139176

Solution= -0.139176

NEWTON MODIFIED

enter lower limit:

1.3

enter upper limit:

1.5

enter tolerance:10^-5

enter first guess:

1.4

n(no of iterations) p(Values)

0 1.400

2 -0.089115

3 0.036332

4 -0.362741

5 -0.249266

6 -0.154254

7 -0.138628

8 -0.1391779 -0.139176

10 -0.139176

Solution= -0.139176

Secant Method

enter lower limit:

1.3

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enter upper limit:

1.5

enter tolerance:

10^-5

enter first guess:1.4

enter second guess:1.5

n p

----------------------

0 1.400

1 1.500

2 -0.634765

3 -0.259479

4 -0.515309

5 -0.209034

6 -0.136615

7 -0.133114

8 -0.1396579 -0.139174

10 -0.139176

solution is -0.139176

QUSTION 2

y=(x.^2)-(2.*x.*exp(-x))+exp(-2.*x) /34;




y=(x.^2)-(2.*x.*exp(-x))+exp(-2.*x);end




z=(2.*x)-(2.*(exp(-x)-(x.*exp(-x))-(2.*exp(-2.*x))));

end

NEWTON METHOD

enter lower limit:0

enter upper limit:

1

enter tolerance:

10^-5

enter first guess:

.5

enter second guess:

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.4

n p

----------------------

0 0.500

1 0.4002 0.518386

3 0.5455324 0.558665

5 0.561870

6 0.563888

7 0.565898

8 0.566374

9 0.566849

10 0.567031

11 0.567100

12 0.567127

13 0.567133

solution is 0.567133

QUESTION 3

y=cos(x+(2).^(.5))+(x.*((x/2)+((2).^(.5))));




y=cos(x+(2).^(.5))+(x.*((x/2)+((2).^(.5))));

end




z=-sin(x+(2).^(.5))+x+(2).^(.5);

end

Newton Method

enter lower limit:

-2enter upper limit:

-1

enter tolerance:

10^-5

enter first guess:

-1.5

n p

----------------------

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0 -1.500

2 -1.478551

3 -1.462465

4 -1.450402

5 -1.4413556 -1.434569

7 -1.4294808 -1.425664

9 -1.422801

10 -1.420654

11 -1.419044

12 -1.417836

13 -1.416931

14 -1.416251

15 -1.415742

16 -1.415360

17 -1.415074

18 -1.414860

19 -1.41470320 -1.414589

21 -1.414501

22 -1.414473

23 -1.414473solution is -1.414473

===========================================================================

==========

QUESTION 4



% Detailed explanation goes herey=(x.^3)-(3.*(x.^2)).*(2.^(-x))+(3.*x).*(4.^(-x))-(8.^(-x));

end




z=(3.*(x.^2))-3.*((2.*x).*(2.*(-x))-(x.^2).*(2.^(-x)).*(0.6931471806))+3.*(4.^(-x)-x.*(4.^(-

x)).*(1.386294361))+(8.^(-x).*(2.079441542));end

Newton Method

enter lower limit:

0

enter upper limit:

1

enter tolerance:

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10^-5

enter first guess:

.5

enter second guess:

.4n p

----------------------0 0.500

1 0.400

2 0.524045

3 0.568238

4 0.584777

5 0.598859

6 0.609079

7 0.616948

8 0.622863

9 0.630733

10 0.635225

11 0.63668512 0.637788

13 0.639249

14 0.640352

15 0.64055716 0.640711

17 0.640827

18 0.640981

19 0.641032

20 0.641069

21 0.641148

22 0.641157

solution is 0.641157

===========================================================================

======

QUESTION 5




y=exp(6.*x)+3.*((ln(2)).^2).*(exp(2.*x))-(ln(8)).*(exp(4.*x))-(ln(2)).^3;

end

function [ z ] = func2( x )%UNTITLED4 Summary of this function goes here


z=6.*(exp(6.*x))+6.*((ln(2)).^2).*(exp(2.*x))-4.*(ln(8)).*(exp(4.*x));

end

Newton Method

enter lower limit:

-1

enter upper limit:

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0

enter tolerance:

10^-5

enter first guess:

-.5n p

----------------------0 -0.500

2 -0.352638

3 -0.285436

4 -0.247646

5 -0.224740

6 -0.210322

7 -0.201052

8 -0.195013

9 -0.191048

10 -0.188430

11 -0.186696

12 -0.18554513 -0.184778

14 -0.184265

15 -0.183916

16 -0.18366917 -0.183460

18 -0.183103

19 -0.182643

20 -0.182815

21 -0.182901

22 -0.182924

23 -0.182926

solution is -0.182926

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