non linear equation using newtons method

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Ronald Aono 2019 年 12 月 8 日
Answered: Walter Roberson 2019 年 12 月 8 日
% start iteration loop
if x2 == 1.0, x2 = 1.001; end % don't allow x2 = 3 as guess
x1 = sqrt(4-(4*x2.*x2)); % satisfies eqn. 1
xo = [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
keep getting the following error
Matrix dimensions must agree.
Error in HW_6c (line 20)
xnew = xold - J\f;

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採用された回答

Walter Roberson 2019 年 12 月 8 日
xold is not defined in your function before the first time you use it. However, your use of
xo = [x1 x2 ]';
hints that xold is likely to be 2 x 1. If so then when you do
xnew = xold - J\f;
then because J\f would be 1 x 2, you would be getting a 2 x 1 minus a 1 x 2. In R2016a or earlier that would be an error.
In R2016b or later, it is well defined, and would give you a 2 x 2 array as a result. That would probably trigger problems further down the road.

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