Use the NewtonRaphson method to solve f1x1x2 x12 plus x22 m

Use the Newton-Raphson method to solve f_1(x_1,x_2)    =    x_1^2 plus x_2^2 minus 1 = 0 f_2(x_1,x_2)    =    X_1 plus X_2 = 0 with an    initial guess = x_1^0   and x_2^0. Write MATLAB code  to solve the problem. Do 5 iterations and put your results in    the    Table 1.

Solution

Table

iteration x1 x2

0 1 -1

1 0.75 -0.75

2 0.7083 -0.7083

3 0.7071 -0.7071

4 0.7071 -0.7071

5 0.7071 -0.7071

%main function here

function [x,iter] = newtonm(x0,f,J)
% Newton-Raphson method applied to a
% system of linear equations f(x) = 0,
% given the jacobian function J, with
% J = del(f1,f2,...,fn)/del(x1,x2,...,xn)
% x = [x1;x2;...;xn], f = [f1;f2;...;fn]
% x0 is an initial guess of the solution
N = 100; % define max. number of iterations
epsilon = 1e-13; % define tolerance
maxval = 10000.0; % define value for divergence
xx = x0; % load initial guess
while (N>0)
JJ = feval(J,xx);
if abs(det(JJ))<epsilon
error(\'newtonm - Jacobian is singular - try new x0\');
abort;
end;
xn = xx - inv(JJ)*feval(f,xx);
if abs(feval(f,xn))<epsilon
x=xn;
iter = 100-N;
return;
end;
if abs(feval(f,xx))>maxval
iter = 100-N;
disp([\'iterations = \',num2str(iter)]);
error(\'Solution diverges\');
abort;
end;
N = N - 1;
xn
xx = xn;
end;
error(\'No convergence after 100 iterations.\');
abort;
% end function

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%supporting function save in different .m file

function [f] = f2(x)
% f2(x) = 0, with x = [x(1);x(2)]
% represents a system of 2 non-linear equations
f1 = x(1)^2 + x(2)^2 - 1;
f2 = x(1)+x(2);
f = [f1;f2];
% end function

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

%supporting function save in different .m file

function [J] = jacob2x2(x)
% Evaluates the Jacobian of a 2x2
% system of non-linear equations
J(1,1) = 2*x(1); J(1,2) = 2*x(2);
J(2,1) = 1; J(2,2) = 1;
% end function

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

%calling function from command window

x0=[1;0] % initial guess

[x,iter] = newtonm(x0,\'f2\',\'jacob2x2\')

xn =

1
-1

xn =

0.7500
-0.7500

xn =

0.7083
-0.7083

xn =

0.7071
-0.7071

xn =

0.7071
-0.7071

x =

0.7071
-0.7071

iter =

5