Computing a Jacobian Numerically using 5pt stencil approximation

2017 10 27
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更新时间：2017 10 28
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Hi guys! I am trying to compute a jacobian numerically using a 5-pt stencil approximation. my array F contains two functions:
x = [x1;x2];
f = @(x1,x2) func1(x);
g = @(x1,x2) func2(x);
%Assigning functions to an array
F(1,1) = {f};
F(2,1) = {g};
Then, I am trying to compute my jacobian but am not getting proper results.
function [J,h] = jacob(F,x)
[n,m] = size(x);
h = zeros(n,1);
%Initialize Jacobian
J = zeros(n,n);
%Numerical computation of Jacobian using 5-pt stencil approximation
for i = 1:n
for j = 1:m
%If i == j, h takes the value of the step size
if i == j
h(i) = 1e-3;
end
J(i,j) = (F{i,1}(x(j)+2*h(j)) + 8*F{i,1}(x(j)+h(j)) - 8*F{i,1}(x(j)-h(j)) + F{i,1}(x(j)-2*h(j)))/(12*h(j));
h(j) = 0;
end
end
end

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Matt J
Matt J 2017-10-27
I am trying to compute my jacobian but am not getting proper results.
as demonstrated by...?
Walter Roberson
Walter Roberson 2017-10-27
The question is how you know you are getting results that are not proper. What should we be looking at? If we were to make a change to your code in hopes of fixing the problem, then how would we know if we had succeeded ?
Cassandra Athans
Cassandra Athans 2017-10-27
Basically, the problem is that when using the function handle, the value of x = [x1,x2] is automatically fixed. In F = [f;g] where f = f(x1,x2) and g = f(x1,x2).
When I try evaluating F at x+h or x-h, the value of x will not change. I am therefore having trouble evaluating F at different values of x, or manipulating x itself.

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 采纳的回答

Walter Roberson
Walter Roberson 2017-10-27

0 个投票

f = @(x1,x2) func1([x1,x2]);
g = @(x1,x2) func2([x1,x2]);
However, in your code you invoke
F{i,1}(x(j)+2*h(j))
so you carefully defined a function handle to take two values, but you are passing in only one value.

2 个评论
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Cassandra Athans
Cassandra Athans 2017-10-27
when invoking
F{i,1}(x(j)+2*h(j))
should I then pass in 2 values? It seems as if when I change what is in (...), the value of F{i,1} does not change. It remains constant @x = [x1;x2]
Walter Roberson
Walter Roberson 2017-10-28
func1 = @(xy) xy(1).^2 - sin(xy(2));
func2 = @(xy) 2*xy(1) + cos(xy(2));
f = @(x1,x2) func1([x1,x2]);
g = @(x1,x2) func2([x1,x2]);
F(1,1) = {f};
F(2,1) = {g};
F{1,1}(7,1/2)
F{2,1}(7,1/2)
syms x y
F{1,1}(x, y)
F{2,1}(x, y)
Remember, when you just look at F{1,1} you are just looking at the function handle, without having evaluated it. You need to pass arguments to get evaluation.

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