I'm hoping to use the Gauss-Newton method to solve a non-linear least squares problem where the solution $\boldsymbol x$ must be non-negative.

To do this can I instead solve for $\pm \sqrt{\boldsymbol x}$ and square it? Specifically, is the following method valid:

  1. Replace $x_i$ (the $i_{th}$ element of $\boldsymbol x$) with $(x'_i)^2 $ in the cost function (where $x'_i$ is the $i_{th}$ element of the new state vector $\boldsymbol x'$).
  2. Use Gauss-Newton to find the $\boldsymbol x'$ that minimizes the new cost function (with no non-negative constraint).
  3. Square each element of $\boldsymbol x'$ to yield the $\boldsymbol x$ which minimizes the original cost function and satisfies the non-negative constraint.

It seems valid to me but has not performed well when I have tried it.

  • $\begingroup$ Have you tried the transformation like $x_i = \exp(z_i)$ ? $\endgroup$ – cpraveen Jan 26 at 14:43
  • $\begingroup$ No, I’ll give that a shot soon, thank you for the suggestion. Is there a theoretical basis for why this option would perform better? $\endgroup$ – specarmi Jan 26 at 16:14
  • $\begingroup$ $x_{1}=exp(z_{i})$ is a one-to-one invertible transformation, while $x_{i}=z_{i}^{2}$ isn't. Your proposed problem transformation creates lots of local minima that may cause problems in the optimization. $\endgroup$ – Brian Borchers Feb 10 at 23:38

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.