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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.

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  • $\begingroup$ 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

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