Write 2 function files in matlab : one for fixed-point iteration for a system of equations and one
for Newton’s method for a system of equations. Determine what your inputs and outputs should
be. Test the code on the following system. the code must also take convergence into
consideration.
x2- y
x- y2
Solution
Matlab codes;
Main function;
%clearing window, variables and figures
clear all;
close all;
clf;
clc;
fixed()
newton()
%fixed approach function
function fixed()
fprintf(\'\ Fixed point approach\ \');
syms x;
f1=x; %first function
f2=sqrt(x); %second function
%for function 1;
%equation is in terms of x,x=0;
x=sym(0);
fprintf(\'\ Value of function is 0 as x is already at zero\ \');
%Checking convergence;
f1dash=1 %Finding derivative f1 w.r.t x
x=sym(0);
fprintf(\'\ The first function converge as f1dash is constant\ \');
fprintf(\'\ Value of function is 0 as x is already at zero\');
%for function 2;
%equation is in terms of x,x=0;
x=sym(0);
%Checking convergence;
f2dash=(1/2)*(1/sqrt(x)); %Finding derivative w.r.t x
fprintf(\'\ The second function converges as for negative infinity, absolute of f2dash is =0 or <1\
\');
fprintf(\'\ Value of function is 0 as x is already at zero\');
%Newton raphson method approach
function newton()
fprintf(\'\ Newton raphson approach\ \');
syms x;
f1=x; %first function
f2=sqrt(x); %second function
%for function 1;
xo=sym(1);
error=5;
while(error>.01)
f1dash=diff(f1,x);
x=sym(xo);
dx=-eval(f1)/f1dash;
x1=xo+dx;
x=sym(x1);
fc=eval(f1);
x=sym(xo);
fe=eval(f1);
error=fc-fe;
xo=x1;
end
xo
f2=sqrt(x); %second function
%for function 1;
xo=sym(1);
error=5;
while(error>.01)
f2dash=diff(f2,x);
x=sym(xo);
dx=-eval(f2)/f2dash;
x1=xo+dx;
x=sym(x1);
fc=eval(f2);
x=sym(xo);
fe=eval(f2);
error=fc-fe;
x0=x1;
end
xo
fprintf(\'\ Both the functions converges for any value of x\ \');
Result;
Value of function is 0 as x is already at zero
f1dash =
1
The first function converge as f1dash is constant
Value of function is 0 as x is already at zero
The second function converges as for negative infinity, absolute of f2dash is =0 or <1
Value of function is 0 as x is already at zero
Newton raphson approach
xo =
0
xo =
1
Both the functions converges for any value of x.
Matlab functions for fixed point iteration and Newton's method to solve a system of equations
1. Write 2 function files in matlab : one for fixed-point iteration for a system of equations and one
for Newton’s method for a system of equations. Determine what your inputs and outputs should
be. Test the code on the following system. the code must also take convergence into
consideration.
x2- y
x- y2
Solution
Matlab codes;
Main function;
%clearing window, variables and figures
clear all;
close all;
clf;
clc;
fixed()
newton()
%fixed approach function
function fixed()
fprintf(' Fixed point approach ');
syms x;
f1=x; %first function
f2=sqrt(x); %second function
%for function 1;
%equation is in terms of x,x=0;
x=sym(0);
fprintf(' Value of function is 0 as x is already at zero ');
%Checking convergence;
f1dash=1 %Finding derivative f1 w.r.t x
x=sym(0);
fprintf(' The first function converge as f1dash is constant ');
fprintf(' Value of function is 0 as x is already at zero');
%for function 2;
%equation is in terms of x,x=0;
2. x=sym(0);
%Checking convergence;
f2dash=(1/2)*(1/sqrt(x)); %Finding derivative w.r.t x
fprintf(' The second function converges as for negative infinity, absolute of f2dash is =0 or <1
');
fprintf(' Value of function is 0 as x is already at zero');
%Newton raphson method approach
function newton()
fprintf(' Newton raphson approach ');
syms x;
f1=x; %first function
f2=sqrt(x); %second function
%for function 1;
xo=sym(1);
error=5;
while(error>.01)
f1dash=diff(f1,x);
x=sym(xo);
dx=-eval(f1)/f1dash;
x1=xo+dx;
x=sym(x1);
fc=eval(f1);
x=sym(xo);
fe=eval(f1);
error=fc-fe;
xo=x1;
end
xo
f2=sqrt(x); %second function
%for function 1;
xo=sym(1);
error=5;
while(error>.01)
f2dash=diff(f2,x);
x=sym(xo);
dx=-eval(f2)/f2dash;
3. x1=xo+dx;
x=sym(x1);
fc=eval(f2);
x=sym(xo);
fe=eval(f2);
error=fc-fe;
x0=x1;
end
xo
fprintf(' Both the functions converges for any value of x ');
Result;
Value of function is 0 as x is already at zero
f1dash =
1
The first function converge as f1dash is constant
Value of function is 0 as x is already at zero
The second function converges as for negative infinity, absolute of f2dash is =0 or <1
Value of function is 0 as x is already at zero
Newton raphson approach
xo =
0
xo =
1
Both the functions converges for any value of x
