Problem#1

Answer
Function file
function [root,iter]=bisection(func,xl,xu,es,maxit,varargin)

% bisect: root location zeroes

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

%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)

% output:

%root = real root

%fx = function value at root

%ea = approximate relative error (%)

%iter = number of iterations

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;

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{:});

Script file

clear;

clc;

f = @(x) (x<=0).(x.^3 + 3*x +1) + (x>0).(1 + sin(x));

xL = -2;

xR = 0.1;

[root, iter] = bisection(f, xL, xR);

fprintf('The value of the root x*: %.4f\n', root)

fprintf('The value of f(x*): %.10f\n', f(root))

fprintf('The number of iterations performed: %d\n', iter)

Output
The value of the root x*: -0.3222
The value of f(x*): 0.0000001355

The number of iterations performed: 23

Problem#2
Answer
Script file

clear all

f=@(x)x.^3+3*x+1

df=@(x)3*x.^2+3

x0=input('enter initial guess')

while abs(f(x0))>0.00001

x1=x0-(f(x0)/df(x0))

x0=x1

end

Output file
f =

@(x) x .^ 3 + 3 * x + 1

df =
@(x) 3 * x .^ 2 + 3

enter initial guess> -2

f(-0.3222)= 1.3752 x 10^(-5)

Problem#3
Answer
Bisection Method
Script File
clc;

clear all;

close all;

err=10^(-5); %%% tolerence

N=20000; %%% max iteration

fx=@(x) 1/2+x^2/4-x*sin(x)-cos(2*x)/2;

a=0;

b=pi;

for n=1:N

c=(a+b)/2;

f1=feval(fx,a);
 f2=feval(fx,b);

f3=feval(fx,c);

if ( f2 > f1 && f2 > f3 )

b=c;

else

a=c;

end

if (f3 < err )

break;

end

end

fprintf(' roots using bisection method is = %f with iteration %d \n',c,n);

Output File
roots using bisection method is = 1.570796 with iteration 20000

f(1.570796)=0.0460540184

Newton Method
Script File
clc;

clear all;

close all;

err=10^(-5); %%% tolerence

N=20000; %%% max iteration

% % % % using newton raphson method

fx=@(x) 1/2+x^2/4-x*sin(x)-cos(2*x)/2;

fdx=@(x) x/2-sin(x)+x*cos(x)+sin(2*x);

a(1,1)=pi/2;

a(2,1)=5*pi;

a(3,1)=10*pi;
%%

for m=1:3

for n=1:N

a(m,n+1)=a(m,n)-feval(fx,a(m,n))/feval(fdx,a(m,n));

k=abs(feval(fx,a(m,n+1)));

if (k<10^(-5))

break;

end

end

fprintf('With initial guess x0=%f roots is %f with iteration %d \n',a(m,1),a(m,n), n);

end

Output File
With initial guess x0=1.570796 roots is 1.891623 with iteration 440

With initial guess x0=15.707963 roots is 1.891623 with iteration 445

With initial guess x0=31.415927 roots is -0.006724 with iteration 55

F(1.891623)=1.000804x10^(-5)

F(1.891623)=1.000804x10^(-5)

F(-0.006724)=1.13027033x10^(-5)

Problem#4
Answer
Function File
function[r,itrr]=Bisectionf(f,xL,xR,tol)

% f=@(x)(x^2-7); %given

% a=input('\n Enter initial value of t i.e. a: ');

% b=input('\n Enter initial value of t i.e. b: ');

% fprintf('given value of a and b is not suitable for this method enter value of a and b such that
f(a)*f(b)<0');

% a=input('\n Enter initial value of t i.e. a: ');

% b=input('\n Enter initial value of t i.e. b: ');

% end

% tol=input('\n Enter tolerence allowed: ');

a=xL;

b=xR;

i=1;

x1=a;x=b; % just for going inside while loop for one time

itter_mat=[];% for ploting the ittreation value

ERR_mat=[];% FOR PLOTING ERROR AT EACH ITTERATION

while(abs(x1-x)>tol)

fprintf('iteration=%2d\n',i);

i=i+1;

x=(a+b)/2;

itter_mat=[itter_mat,x];

ERR_mat=[ERR_mat,abs(x-sqrt(2))];

fprintf('x=%2d\n',x)

if(f(x)*f(a)<0)

b=x;

else

a=x;

end

fprintf('iteration=%2d\n',i);

i=i+1;

x1=(a+b)/2;

itter_mat=[itter_mat,x1];
ERR_mat=[ERR_mat,abs(x1-sqrt(2))];

fprintf('x=%2d\n',x1)

if(f(x1)*f(a)<0)

b=x1;

else

a=x1;

end

end

r=x1;

itrr=length(itter_mat);

itter_mat

ERR_mat

plot(itter_mat)

hold on

plot(ERR_mat)

Script File
clc

% Setting x as symbolic variable

syms x;

% Input Section

y = input('Enter non-linear equations: ');

a = input('Enter first guess: ');

b = input('Enter second guess: ');

e = input('Tolerable error: ');

% Finding Functional Value

fb = eval(subs(y,x,b));

% Implementing Bisection Method

if fa*fb > 0

disp('Given initial values do not bracket the root.');

else

c = (a+b)/2;

fc = eval(subs(y,x,c));

fprintf('\n\na\t\t\tb\t\t\tc\t\t\tf(c)\n');

while abs(fc)>e

fprintf('%f\t%f\t%f\t%f\n',a,b,c,fc);

if fa*fc< 0

b =c;

else

a =c;

end

c = (a+b)/2;

fc = eval(subs(y,x,c));

end

fprintf('\nRoot is: %f\n', c);

end

Output File
Enter non-linear equations: > x^2-7

Enter first guess: > -1

Enter second guess: > 9

Tolerable error: > 0.001

-1.000000 4.000000 1.500000 -4.750000

1.500000 4.000000 2.750000 0.562500

1.500000 2.750000 2.125000 -2.484375

2.125000 2.750000 2.437500 -1.058594

2.437500 2.750000 2.593750 -0.272461

2.593750 2.750000 2.671875 0.138916

2.593750 2.671875 2.632812 -0.068298

2.632812 2.671875 2.652344 0.034927

2.632812 2.652344 2.642578 -0.016781

2.642578 2.652344 2.647461 0.009049

2.642578 2.647461 2.645020 -0.003878

2.645020 2.647461 2.646240 0.002584

Root is: 2.645630

Problem#5
Answer