newtons non linear method

% start iteration loop

clear all

x2 = [1 0.5]';

x1 = sqrt(4-(4*x2.*x2)); % satisfies eqn. 1

xold = [x1 x2 ]';

itmax = 30; it = 0; tol = 1e-5; emax=1; n = length(xold);

fprintf(1,'\n Intermediate edit for NLDemo2 \n');

while emax > tol && it <= itmax

it = it+1;

x= xold;

%

% compute function vector using xold

f = [(x(1)*x(1))-(2*x(1))-x(2)+0.5;

(-x(1)*x(1))+(4*(x(2)*x(2)))-4];

%

% compute Jacobian matrix evaluated at xold

J = [2*x(1)-2 -1;

2*x(1) -8*x(2)];

%

% compute xnew

xnew = xold - J\f;

%

% calc & edit error (intermediate results)

emax = max(abs((xnew-xold)./xnew));

fprintf(1,' it = %3d max error = %8.3e \n',it,emax);

fprintf(1,' xnew xold \n');

for j = 1:n

fprintf(1,' %10.5f %10.5f \n',xnew(j),xold(j));

end

%

xold = xnew; % use current estimate as guess for next iteration

end

%

% print final max relative error and iteration count

fprintf(1,'\n Number of iterations to convergence = %3d\n',it);

fprintf(1,' Max relative error at convergence = %8.3e\n',emax);

if it >= itmax

fprintf(1,' ***** WARNING -- Hit max number of iterations!!! *****\n');

not able to make it past the first iteration, ang i getthe error below

>> HW6B_2c

Intermediate edit for NLDemo2

it = 1 max error = 1.000e+00

it = 1.686852e-01 max error = xnew xold

-0.25000 0.00000

1.00000 1.00000

Operands to the || and && operators must be convertible to logical scalar values.

Error in HW6B_2c (line 8)

while emax > tol && it <= itmax

[darova]

Can you tell the size of emax?

[Ronald Aono]

first i had it has a 1x1 but the answer is a 2x1

[darova]

WHat we gonna do about it?