Write 2 function files in matlab one for fixedpoint iterati

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