mathematics and computing- sc file

8/10/2019 mathematics and computing- SC File

1/38

www.mcdtu.wordpress.com

www.mcdtu.wordpress.com| 1

SCIENTIFIC COMPUTING

LAB FILE


2/38



ASSIGNMENT 1

BISECTION METHOD

SOURCE CODE:

% Bisection method to find the root of the function f(x)=0

% INPUT - f is the function% - a and b are the left and right endpoints% - tol is the tolerance

% OUTPUT - c is the root% - yc= f(c)% - err is the error estimate for c

% PROGRAM STARTS HERE

clear all; close all; clc;

disp('__________________________________________________________________')disp('--- Name: A T I N M I N O C H A, Roll No. 2K11/MC/017 ---')disp('--- Assignment 1, Bisection Method ---')

disp('--- Date - 27/02/2013 ---')disp('__________________________________________________________________')

% INPUTS: Enter the followingf = inline('sin(x) - tan(x^2) + exp(x)'); % Function in f(x)=0a = -1.0; % Upper initial guessb = 0.0; % Lower initial guesstol=.000001; count=0;fprintf('The initial guess a = %g \n',a);

fprintf('The initial guess b = %g \n',b);

% SOLUTION STARTS HEREya=f(a);yb=f(b);if (ya*yb > 0)

disp('No root lie in this interval');return;

endwhile( b-a > tol)

count=count+1;

c=(a+b)/2;yc=f(c);


3/38



if yc==0fprintf('The equation is sin(x) - tan(x^2) + exp(x)and the root

is x = %g \n',c);elseif yb*yc>0

b=c;

yb=yc;else

a=c;ya=yc;

endendc=(a+b)/2;err=abs(b-a);yc=f(c);fprintf('\n The equation is sin(x) - tan(x^2) + exp(x) and the root isx = %g \n',c);

fprintf('\n Number of iterations are = %g \n',count);% END OF THE PROGRAM


4/38



OUTPUT:


5/38



ASSIGNMENT 2

NEWTON RAPHSON METHOD

SOURCE CODE:

% Newton-Raphson method to find the root of the function f(x)=0

% INPUT - f is the function% - f1 is the derivative of f% - xn is previous value% - tol is the tolerance% - fxn=f(xn)

% OUTPUT - xn1 is the root

% - fxn1= f(xn1)



disp('__________________________________________________________________')disp('--- Name: A T I N M I N O C H A, Roll No. 2K11/MC/017 ---')disp('--- Assignment 2, Newton Raphson Method -

--')disp('--- Date - 27/02/2013 ---')disp('__________________________________________________________________')

% INPUTS: Enter the following% GRAPHsyms x;x= -2:0.2:2;f = inline('cos(x)-x^2-x'); % Function in f(x)=0

f1 = inline('-sin(x)-2*x-1'); % Derivative of the functionplot(x,cos(x)-x.^2 -x,'--rs') % Plotting the graph of the functiongrid on;

disp('------- ------- SOLUTION STARTS HERE ------ -------')disp('Since the graph crosses the x-axis at two points,')disp('The function has two roots. Therefore, we need to take')disp('two different approximations and solve them seperately.')

% FIRST ROOT

disp('------- ------- FIRST ROOT OF f(x) ------ -------')


6/38



xn=-1; % Initial guesstol=.000001; count=0;disp(sprintf('The initial approximation of the first root is %g',xn));fxn=f(xn);f1xn=f1(xn);

xn1=xn-fxn/f1xn;while( abs(xn1-xn) > tol)

count=count+1;if fxn==0

xn1=xndisp(sprintf('The equation is cos(x)-x^2-x and the root is x =

%g',xn1));else

xn=xn1;fxn=f(xn);f1xn=f1(xn);

xn1=xn-(fxn/f1xn);end

enddisp(sprintf('\n The equation is cos(x)-x^2-x and the root is x =%g',xn1));disp(sprintf('\n Number of iterations are = %g',count));

% SECOND ROOTdisp('------- ------- SECOND ROOT OF f(x) ------ -------')xn=0;% Initial guess

tol=.000001; count=0;disp(sprintf('\nThe initial approximation of the second root is%g',xn));fxn=f(xn);f1xn=f1(xn);xn1=xn-fxn/f1xn;while( abs(xn1-xn) > tol)


xn1=xndisp(sprintf('The equation is cos(x)-x^2-x and the root is x =

%g',xn1));else

xn=xn1;fxn=f(xn);f1xn=f1(xn);xn1=xn-(fxn/f1xn);

endenddisp(sprintf('\n The equation is cos(x)-x^2-x and the root is x =%g',xn1));disp(sprintf('\n Number of iterations are = %g',count));

% END OF THE PROGRAM


7/38



OUTPUT:


8/38



ASSIGNMENT 3

GAUSS SEIDEL METHOD

SOURCE CODE:

% Gauss-Seidel method is used to find the solution of a system of% equations!

% INPUT - x, y, z and w are the initial approximations.

% OUTPUT - x, y, z and w are the solutions to the given system of% equations.



disp('__________________________________________________________________')disp('--- Name: A T I N M I N O C H A, Roll No. 2K11/MC/017 ---')disp('--- Assignment 3, Gauss - Seidel Method ---')disp('--- Date - 20/03/2013 ---')

disp('__________________________________________________________________')

% INPUTS: Enter the following

disp(sprintf('\nSystem of equations is:\n1. 4x+y+w=2\n2. x+4y+z=-2\n3.y+4z+w=2\n4. x+z+4w=-2'));

x=0; % Initial approximationsy=0;z=0;

w=0;tol=.000001; % tol is the limit of while loop.count=1; % count is used to calculate no. of iterations.

disp(sprintf('\nInitial guess =[%g %g %g %g]',x,y,z,w));

disp('------- ------- SOLUTION STARTS HERE ------ -------')% SOLUTION starts here!

X0=[x;y;z;w];

x= 1/4*(2-y-w);y= 1/4*(-2-x-z);


9/38



z= 1/4*(2-y-w);w= 1/4*(-2-x-z);X1=[x;y;z;w];

while(max(X1-X0) > tol)

count=count+1;X0=[x;y;z;w];x= 1/4*(2-y-w);y= 1/4*(-2-x-z);z= 1/4*(2-y-w);w= 1/4*(-2-x-z);X1=[x;y;z;w];end

disp(sprintf('\nNo. of iterations are: %g',count));disp(sprintf('\nThe solution of system of system of equations =[%g %g

%g %g]',x,y,z,w));



10/38



OUTPUT:


11/38



ASSIGNMENT 4

SECANT METHOD

SOURCE CODE:

% Secant Method to find the root of the function f(x)=0

% INPUT - f is the function% - xn and xn0 are the previous values% - tol is the tolerance% - fxn=f(xn)% - fxn0=f(xn0)

% OUTPUT - xn1 is the root

% - fxn1=f(xn1)



disp('__________________________________________________________________')disp('--- Name: A T I N M I N O C H A, Roll No. 2K11/MC/017 ---')disp('--- Assignment 4, Secant Method -

--')disp('--- Date - 20/03/2013 ---')disp('__________________________________________________________________')


syms x;x= 0:0.2:1;f = inline('cos(x)-x.*exp(x)'); % Function in f(x)=0

plot(x,cos(x)-x.*exp(x),'--rs') % Plotting the graph of the functiongrid on;

disp('------- ------- SOLUTION STARTS HERE ------ -------')disp('The graph crosses the x-axis between 0 and 1.')

xn=1; xn0=0; % Initial guesstol=.000001;count=0;disp(sprintf('The initial guesses are %g and %g.',xn0,xn));

fxn=f(xn);


12/38



fxn0=f(xn0);xn1=xn-fxn/(fxn-fxn0)*(xn-xn0);while( abs(xn1-xn) > tol)


xn1=xn;xn=xn0;

disp(sprintf('The equation is cos(x)-x*(e^x) and the root is x= %g',xn1));

elsexn=xn1;fxn=f(xn);fxn0=f(xn0);xn1=xn-fxn/(fxn-fxn0)*(xn-xn0);

endend

disp(sprintf('\n The equation is cos(x)-x*(e^x) and the root is x =%g',xn1));disp(sprintf('\n Number of iterations are %g',count));



13/38



OUTPUT:


14/38



ASSIGNMENT 5

CROUTS METHOD

SOURCE CODE:

% Crout's method is used to find the solution of a system of% equations!

% INPUT - A and B are the matrices of the system of equations.% - A = LU% - X is the solution matrix of the system.% - L is the lower triangular matrix.% - U is the upper triangular matrix.% - Z = UX

% OUTPUT - x, y, z and w are the solutions to the given system of% equations.



disp('__________________________________________________________________')disp('--- Name: A T I N M I N O C H A, Roll No. 2K11/MC/017 ---')

disp('--- Assignment 5, Crouts Method ---')disp('--- Date - 03/04/2013 ---')disp('__________________________________________________________________')


disp(sprintf('\nSystem of equations is:\n1. 2x+1y+-4z+1w= 4\n2. -4x+3y+5z-2w=-10\n3. 1x-1y+1z-1w= 2\n4. 1x+3y-3z+2w=-1'));

% SOLUTION starts here!

A=[2, 1, -4, 1; -4, 3, 5, -2; 1, -1, 1, -1; 1, 3, -3, 2]B=[4 ;-10 ;2 ;-1]disp('------- ------- SOLUTION STARTS HERE ------ -------')

l11=A(1,1);l21=A(2,1);l31=A(3,1);

l41=A(4,1);u12=A(1,2)./l11;


15/38



u13=A(1,3)./l11;u14=A(1,4)./l11;l22=A(2,2)-l21.*u12;l32=A(3,2)-l31.*u12;l42=A(4,2)-l41.*u12;

u23=(A(2,3)-l21.*u13)./l22;u24=(A(2,4)-l21.*u14)./l22;l33=A(3,3)-l31*u13-l32*u23;l43=A(4,3)-l41*u13-l42*u23;u34=(A(3,4)-l31*u14-l32*u24)/l33;l44=A(4,4)-l41*u14-l42*u24-l43*u34;

disp(sprintf('\nLower triangular matrix is\n'));L=[l11 0 0 0;l21 l22 0 0;l31 l32 l33 0; l41 l42 l43 l44]

disp(sprintf('\nUpper triangular matrix is\n'));

U=[1 u12 u13 u14; 0 1 u23 u24; 0 0 1 u34; 0 0 0 1]

Z=(inv(L))*B;disp(sprintf('\nThe solution of system of system of equations is'));X=(inv(U))*ZAinv=inv(U).*inv(L)



16/38



OUTPUT:


17/38




18/38




19/38



ASSIGNMENT 6

NEWTONS FORWARD AND BACKWARD DIFFERENCE

INTERPOLATION METHOD

SOURCE CODE:

% Newton's Forward and Backward Interpolation Difference Formula!

% INPUT - x is the matrix for which value is given.% - y is the matrix giving values corresponding to x.% - x1 is the matrix whose values are to be found.



disp('__________________________________________________________________')disp('--- Name: A T I N M I N O C H A, Roll No. 2K11/MC/017 ---')disp('--- Assignment 6, Interpolation ---')disp('--- Date - 06/04/2013 ---')disp('________________________________________________________________

__')

% INPUTS:Enter the following

x= [0 0.1 0.2 0.3 0.4 0.5]y= [-1.5 -1.27 -0.98 -0.63 -0.22 0.25]x1= [0.15 0.25 0.45]

% SOLUTION starts here!

for l=1:3p=x1(l);

disp(sprintf('\n\nApplying Newton Forward Interpolation differenceformula for x = %g',x1(l)));

% Newton's Forward Interpolation Formula

n = length(x); % Calculating length of xa(1) = y(1);for k = 1 : n - 1

d(k, 1) = (y(k+1) - y(k))/(x(k+1) - x(k)); % Finite forwarddifference matrixend


20/38



for j = 2 : n - 1for k = 1 : n - j

d(k, j) = (d(k+1, j - 1) - d(k, j - 1))/(x(k+j) - x(k));end

end

for j = 2 : na(j) = d(1, j-1);

endDf(1) = 1;c(1) = a(1);for j = 2 : n

Df(j)=(p - x(j-1)) .* Df(j-1);c(j) = a(j) .* Df(j);

endfp=sum(c);disp(sprintf('\nWe get value as %g',fp)); % Finding sum

disp(sprintf('\n\nApplying Newton Backward Interpolation differenceformula for x = %g',x1(l)));

% Newton's Backward Interpolation Formulafor i=1:n

diff(i,1)=y(i); % Finite backward difference operator matrixendfor j=2:n

for i=n:-1:jdiff(i,j)=diff(i,j-1)-diff(i-1,j-1);

end

endanswer=y(n);h=x(n)-x(n-1);s=(p-x(n))/h;for i=1:n-1

term=1;for j=1:i

term=term*(s+j-1)/j;endanswer=answer+term*diff(n,i+1); % Finding sum

end

disp(sprintf('\nWe get value as %g',answer));end% END OF THE PROGRAM


21/38



OUTPUT:


22/38




23/38



ASSIGNMENT 7

TRAPEZOIDAL METHOD

SOURCE CODE:

% TRAPEZOIDAL METHOD to perform Numerical Integration!

% INPUTS - f is the fucntion e^-(x^2)=0 we need to integrate.% - x0 is the lower limit of integration.% - x1 is the upper limit of integration.% - n is the number of mesh points% i.e no of intervals between x0 and x1.% - h is the length of the interval.% Area gives the result of integration of curve f between f0 and f1.



disp('__________________________________________________________________')disp('--- Name: A T I N M I N O C H A, Roll No. 2K11/MC/017 ---')disp('--- Assignment 7, Trapezoidal Method ---')

disp('--- Date - 10/04/2013 ---')disp('__________________________________________________________________')

% INPUTS:Enter the following:

f=inline('exp(-x^2)');%inline function definitionx0=0;x1=1;n=20; % No. of intervals to be present between 0 and 1.

f0=f(x0);f1=f(x1); % Computation of function values at end points.h=(x1-x0)/n; % Value of length of the intervals.f2=0;%initialize f2 to zeroa=x0;

for i=1:n-1

x2= a+i*h;f2= f2+ f(x2);

end

disp('The answer for integration is: ');area= h*(f0+f1)/2+ h*f2 % Prints the result of integration.


24/38





25/38



OUTPUT:


26/38



ASSIGNMENT 8

SIMPSONS METHOD

SOURCE CODE:

% Simpson's Method is used for Numerical Integration!

% INPUT - f is the function% - f0 and fn are 1st and last values% - h is the difference between 2 values of x

% OUTPUT - fx is the required function% PROGRAM STARTS HERE


disp('__________________________________________________________________')disp('--- Name: A T I N M I N O C H A, Roll No. 2K11/MC/017 ---')disp('--- Assignment 8, Simpsons Method ---')disp('--- Date - 11/04/2013 ---')disp('________________________________________________________________

__')


f=inline('exp(-x^2)'); % Functionf0=f(0); % Value of the function at 0fn=f(1); % Value of the function at 1h=0.05;n=20; % No. of divisionsfx=h/3*(f0+fn);if mod(n,2)==0 % mod is used to find the remainder

for i=1:19k=i*h;if mod(i,2)==0

fx=fx+2*h/3*f(k); % Formula when i is a multiple of 2else

fx=fx+4*h/3*f(k); % Formula when i is not a multiple of 2end

endenddisp(sprintf('\nThe equation is exp(-x^2) and its integration from x

= 0 to 1 by Simpsons method is\n %g',fx));



27/38



OUTPUT:


28/38



ASSIGNMENT 9

ROMBERGS METHOD

SOURCE CODE:

% Romberg's Method is a method used for Numerical Integration

% INPUT - f is the function.% - f0 and fn are 1st and last values.% - h1,h2,h3 are the difference between 2 values of x.

% OUTPUT - Q1,Q2,Q3 are the required functions at h1,h2,h3 resp.% - R1 and R2 are the Richardson solutions.% - S is the required function value.



disp('__________________________________________________________________')disp('--- Name: A T I N M I N O C H A, Roll No. 2K11/MC/017 ---')disp('--- Assignment 9, Rombergs Method ---')

disp('--- Date - 17/04/2013 ---')disp('__________________________________________________________________')

% INPUTS: Enter the followingf=inline('(cos(x)*log(sin(x)))/(1+(sin(x)^2))'); % Given functionf0=f(pi/4); % Value of the function at 0.fn=f(pi/2); % Value of the function at 1.h1=pi/8;n1=(pi/2-pi/4)/h1; % No. of divisions

Q1=h1/2*(f0+fn); % Applying Trapezoidal Rule.for i=1:n1

k=i*h1;Q1=Q1+h1*f(pi/4 + k);

endh2=h1/2;n2=(pi/2-pi/4)/h2;Q2=h2/2*(f0+fn);for i=1:n2

k=i*h2;Q2=Q2+h2*f(pi/4 + k);

endR1=(4*Q2-Q1)/3;


29/38



h3=h2/2;n3=(pi/2-pi/4)/h3;Q3=h3/3*(f0+fn); % Applying Simpson's ruleif mod(n3,2)==0 % mod is used to find remainderfor i=1:n3

k=i*h3;if mod(i,2)==0

Q3=Q3+2*h3/3*f(pi/4 + k); % Formula for i multiple of 2else

Q3=Q3+4*h3/3*f(pi/4 + k); % Formula for not i multiple of 2end

endendR2=(16*Q3-Q2)/15;S=(16*R2-R1)/15;disp(sprintf('\nThe equation is (cos(x)*log(sin(x)))/(1+(sin(x)^2))

and its integration \nfrom x = pi/4 to pi/2 by Rombergs Method is\n%g',S));



30/38



OUTPUT:


31/38



ASSIGNMENT 10

PICARDS METHOD

SOURCE CODE:

% Picard's Method is used to find solution of a differential equations

% INPUT - f is the function% - y0 is the solution at initial value of x=0.4

% OUTPUT - Y is the solution of differential equation.% - Y1 is the solution of differential equation at x=0.8.



disp('__________________________________________________________________')disp('--- Name: A T I N M I N O C H A, Roll No. 2K11/MC/017 ---')disp('--- Assignment 10, Picards Method ---')disp('--- Date - 29/04/2013 ---')

disp('__________________________________________________________________')

% INPUTS: Enter the followingf=inline('sqrt(x+y)');fint=inline('2/3*sqrt((x+y)^3)');x0=0.4;% Initial value of xy0=0.41; % Initial value of y at x=0.4Y=y0;x1=0.6;for i=1:3

y1=Y;Y=y0+fint(x1,y1)-fint(x0,y1);

enddisp(sprintf('\nThe solution of the differential equation dy/dx =sqrt(x+y) \nby Picards method at x = 0.6 is\n%g',Y));y0=Y;x2=0.8;Y1=Y;for i=1:3

y1=Y1;Y1=y0+fint(x2,y1)-fint(x1,y1);

end


32/38



disp(sprintf('\nThe solution of the differential equation dy/dx =sqrt(x+y) \nby Picards method at x = 0.8 is \n%g\n',Y1));



33/38



OUTPUT:


34/38



ASSIGNMENT 11

EULERS METHOD

SOURCE CODE:

% Euler's method is used to find solution of a differential equation.

% INPUT - f is the function.% - y0 is the solution at initial value of x=0.4.




disp('__________________________________________________________________')disp('--- Name: A T I N M I N O C H A, Roll No. 2K11/MC/017 ---')disp('--- Assignment 11, Eulers Method ---')disp('--- Date - 29/04/2013 ---')

disp('__________________________________________________________________')


f=inline('sqrt(x+y)');x0=0.4;% Initial value of xy0=0.41;% Initial value of y at x=0.4x1=0.6;h=0.2;

% SOLUTION STARTS HERE

Y=y0+h*f(x0+h/2,y0+h/2*f(x0,y0));Y1=Y+h*f(x1+h/2,Y+h/2*f(x1,Y));disp(sprintf('\nThe solution of the differential equation dy/dx =sqrt(x+y)\nby Eulers method at x = 0.6 is \n%g\n',Y));disp(sprintf('\nThe solution of the differential equation dy/dx =sqrt(x+y)\nby Eulers method at x = 0.8 is \n%g\n',Y1));



35/38



OUTPUT:


36/38



ASSIGNMENT 12

RUNGE KUTTA METHOD

SOURCE CODE:

% Runge Kuttas 4th Order Method is used to find the solution of a% differential equation

% INPUT - f is the function.% - y0 is the solution at initial value of x=0.4.




disp('__________________________________________________________________')disp('--- Name: A T I N M I N O C H A, Roll No. 2K11/MC/017 ---')disp('--- Assignment 12, Runge Kutta Method ---')disp('--- Date - 29/04/2013 -

--')disp('__________________________________________________________________')

% INPUTS: Enter the followingsyms x;f=inline('(x+y)^(1/2)');x0=0.4;% Initial value of xy0=0.41;% Initial value of y at x=0.4Y=y0;x1=0.6;

h=0.2;k1=h*f(x0,y0);k2=h*f(x0+h/2,y0+k1/2);k3=h*f(x0+h/2,y0+k2/2);k4=h*f(x0+h,y0+k3);Y=y0+1/6*(k1+2*k2+2*k3+k4);disp(sprintf('\nThe solution of the differential equation dy/dx =sqrt(x+y)\nby Runge Kutta 4th Order Method at x = 0.6 is \n%g\n',Y));k1=h*f(x1,Y);k2=h*f(x1+h/2,Y+k1/2);k3=h*f(x1+h/2,Y+k2/2);

k4=h*f(x1+h,Y+k3);Y1=Y+1/6*(k1+2*k2+2*k3+k4);


37/38



disp(sprintf('\nThe solution of the differential equation dy/dx =sqrt(x+y)\nby Runge Kutta 4th Order Method at x = 0.8 is \n%g\n',Y1));



38/38


OUTPUT:

mathematics and computing- sc file

Documents