Numerical Jacobian in Matlab

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Daniel Wells
Daniel Wells 2012 年 2 月 5 日
コメント済み: Hannah Trachtman 2019 年 6 月 19 日
I am trying to estimate the Jacobian of a multivariable function in Matlab. I came across a method here:
that recommends using the function lsqnonlin. As an example, I define the function
f = @(x)[x(1)^2 + x(2)^2, x(1)^3.*x(2)^3];
and the point
A = [1 1];
The Jacobian of f at this point should pretty clearly be [2, 2;3, 3];
However, when I use the method described in the above link, I find that
[x, ~, ~, ~, ~, ~, jac] = lsqnonlin(f, A,[],[], optimset('MaxIter', 0))
x = [.7308 .7308]
and jac = [1.4615, 1.4615; .6252, .6252];
Clearly, something is a foot. Any ideas what is going on? Any ideas of a better way, using a native MATLAB function, to evaluate a jacobian numerically? (I don't want to do it symbolically.)

採用された回答

Anton Semechko
Anton Semechko 2012 年 2 月 5 日
lsqnonlin is not designed for the purpose of evaluating Jacobians. I thought your main concern was that it does not evaluate the Jacobian accurately. But given the above example, it does.
  2 件のコメント
Anton Semechko
Anton Semechko 2012 年 2 月 5 日
but if you insist on using lsqnonlin for the sole purpose of evaluating Jacobians then instead of MaxIter=0 you have to set MaxFunEvals=0 (i.e. optimset('MaxFunEvals', 0) )
Daniel Wells
Daniel Wells 2012 年 2 月 5 日
I do not necessarily insist on using lsqnonlin to find a jacobian, I am only trying to find a (different, see above) way to do so in Matlab. I was referred to this method from the mathworks support site, and it seems that whoever answered that was wrong. You are right that 'MaxFunEvals' needs to be 0 as well, which she did not mention. I hadn't realized this, and it is indeed the answer to what I was trying to figure out.
Thanks.

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その他の回答 (4 件)

Matthieu Heitz
Matthieu Heitz 2017 年 2 月 2 日
Hi !
If I may provide another answer that uses finite differences.
% Jacobian functor
J = @(x,h,F)(F(repmat(x,size(x'))+diag(h))-F(repmat(x,size(x'))))./h';
% Your function
f = @(x)[x(1)^2 + x(2)^2; x(1)^3.*x(2)^3];
% Point at which to estimate it
x = [1;1];
% Step to take on each dimension (has to be small enough for precision)
h = 1e-5*ones(size(x));
% Compute the jacobian
J(x,h,f)
This J function works with any function f. Note that x and the result of f must be in a vertical vector for it to work.

John D'Errico
John D'Errico 2012 年 2 月 5 日
Whats the problem?
f = @(x)[x(1)^2 + x(2)^2, x(1)^3.*x(2)^3];
jacobianest(f,[1 1])
ans =
2 2
3 3
It is on the file exchange.
  1 件のコメント
Daniel Wells
Daniel Wells 2012 年 2 月 5 日
I really do like your jacobianest a lot, by the way. I use it often, and have never had a problem with it, and don't expect the problem to be there in this case. One just has to be careful with these things.

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Daniel Wells
Daniel Wells 2012 年 2 月 5 日
I am using jacobianest already, and was looking for a second opinion on the estimation for the jacobian of a particularly nasty function that I am using, since my code is behaving poorly, which makes me have to question every function I am using already.
  2 件のコメント
John D'Errico
John D'Errico 2012 年 2 月 5 日
You can always use derivest to estimate individual components of the Jacobian. It will allow you much more control over the estimation procedure, with many more options. You can vary the order of the approximation, apply one sided or two sided estimators, change limits on the steps taken, etc. If these estimates are consistent with the Jacobian, taking into account the error estimates it provides, then you can assume it is doing its job reasonably well. Jacobianest is not as sophisticated as is derivest, with far fewer options.
Of course, one can always come up with a nasty enough function that any adaptive scheme will go wrong. For proof of that, just look at the number of people who post questions asking why quad/quadgk etc. has failed on their function. And those people are not even actively trying to make those tools fail.
Hannah Trachtman
Hannah Trachtman 2019 年 6 月 19 日
Hi John,
I realize this is many years later, but I am trying to do exactly this and having trouble. I'm working from your example code below:
fun = @(x,y) x.^2 + y.^2;
xy = [2 3];
gradvec = [derivest(@(x) fun(x,xy(2)),xy(1),'d',1), ...
derivest(@(y) fun(xy(1),y),xy(2),'d',1)]
But my function is a large vector-valued function stored in an m-file (since I was originally using jacobianest). It looks something like this:
function [calcMoments] = calculator(parameters)
alpha = parameters(1)
beta = parameters(2)
calcMoments = alpha^2 + beta^2
end
How do I modify your code to apply to this kind of function? I'm a relatively inexperienced Matlab user so I might be missing something trivial. Can you point me in the right direction?
Thank you!
Hannah

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Anton Semechko
Anton Semechko 2012 年 2 月 5 日
Actually there is non problem what so ever. The Jacobian of your function is:
J=@(x) [2*x(1) 2*x(2);3*x(2).*(x(1).*x(2)).^2 3*x(1).*(x(1).*x(2)).^2];
Evaluating J at x = [.7308 .7308]:
jac=J([.7308 .7308])
jac =
1.4616 1.4616
0.6253 0.6253
you get the same answer as that returned by lsqnonlin
  1 件のコメント
Daniel Wells
Daniel Wells 2012 年 2 月 5 日
There is something wrong, because I do not want to Jacobian evaluated at [.7308, .7308], but rather at [1, 1]. It is giving me the Jacobian at an x not of my choosing.

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