I'm trying to write a function that accepts a function vector of any size and return the numerical Jacobian matrix.

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bhador
bhador 2018-3-6
编辑： Torsten 2018-3-7
The answer should be x=1.2812, y=2.5349 and z=-0.5022 but once I put the loop, I'm getting these values x=0.9794, y=2.9794 and z=-0.0206
This is my function script:
function J = jacobian_numeric(f,x,h)
[m,n] = size(x);
for ii = 1:n
hv = zeros(1,n);
hv(1,ii) = h;
J(:,ii) = (f(x+hv) - f(x-hv))./(2*h);
end
end
And this is the m-file that calls the function:
clear; clc;
f = @(x) [(x(1).^2 - x(2) - x(1).*x(3) + 0.25) ; (x(1).*x(2) + x(3).^2 - 3.5) ; (sin(pi.*x(1)) - x(2).*x(3) - 0.5)];
x0 = [1 ; 3 ; 0];
h = 0.00001;
tol = 1e-5;
err = 1 + tol;
iter = 0;
while err > tol
J = jacobian_numeric(f,x0,h);
xold = x0;
x0 = x0 - J \ f(x0);
iter = iter + 1;
err = max(abs((x0 - xold)./(x0)));
end
x = x0(1,1)
y = x0(2,1)
z = x0(3,1)
I tried many things but nothing worked, so I am hoping on getting help from someone here. Thank you in advance.
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Torsten
Torsten 2018-3-7
编辑：Torsten 2018-3-7
function J = jacobian_numeric(f,x,h)
n = numel(x);
J = zeros(n,n);
for ii = 1:n
hv = zeros(n,1);
hv(ii,1) = h;
J(:,ii) = (f(x+hv) - f(x-hv))./(2*h);
end
end
Wouldn't have happened without implicit expansion ...
Best wishes
Torsten.

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