1
$\begingroup$

When solving a system of equations, it is sometimes faster to directly pass the Jacobian to FindRoot (https://reference.wolfram.com/language/tutorial/UnconstrainedOptimizationSpecifyingDerivatives.html).

However, in my (huge) fixed point problem, it seems that mathematica spends most time on computing the inverse of the jacobian. I can compute this inverse faster, as I have information about the structure of this function (i.e., I am explicitly structuring the inverse rather than use the Inverse command). Is there a way to pass this "jacobian inverse" directly into the FindRoot function?

$\endgroup$
  • 1
    $\begingroup$ I expect FindRoot does not invert the Jacobian, but uses the stabler LinearSolve[]. Can you just implement Newton-Raphson manually? $\endgroup$ – Michael E2 Sep 23 '16 at 12:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.