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newtons non linear method

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Ronald Aono
Ronald Aono 2019-12-10
评论： darova 2019-12-10
% 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
  3 个评论
显示 1更早的评论
Ronald Aono
Ronald Aono 2019-12-10
first i had it has a 1x1 but the answer is a 2x1
darova
darova 2019-12-10
WHat we gonna do about it?

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