I'm a PhD student in Astronomy currently and a little rusty in mathematics after my project has become very coding-intensive.

I was wondering if there is an alternate method to solve for a specific function, i.e.

$$f(x) = \int^\infty_0 g(y)\exp(-\frac{[x-k(y)]^2}{2}) dy$$

I know the exact curve for $f(x)$ and $g(y)$. And also have information on the shape of k(y), namely that it is monotonic in y, ($dk/dy > 0$).

The technique I used to approach this was to discretise this into N number of $x_i$ and $y_j$ parts, effectively creating a system of $i$ equations with $j$ unknowns (in the form of $k(y_j)$). I would then use some iterative method to solve for all the $k(y_j)$'s simultaneously and do a non-parametric curve fit to it.

I am running into huge issues trying to solve the problem in this way. My Jacobian is almost singular/ill-conditioned and the usual method of gradient descent is not so effective with large number of parameters (my N is set to 50). It's also not so easy to implement the monotonicity condition for $k(y)$ by this method - and without enforcing this condition, I feel like the solution is degenerate.

Is this the smartest method to go around solving this? Am I overlooking a simpler solution?

I should add that I'm not looking to guessing an analytical form in $k(y)$ with less parameters (although the fit is convincing, but I like to go the full mile with the most accurate representation of $k(y)$).

EDIT: I've convinced myself that there's no exact solution. Found a minimizer that gave out more or less an OK solution.

EDIT again: Maybe not trivial to solve, but you can take FT over x both sides, make a substitution and reduce the problem to a convolution where you obtain $g(y) dy/dk$.

  • $\begingroup$ Is $k(\infty)=\infty$? $\endgroup$ – Szeto Jun 14 '18 at 22:39
  • $\begingroup$ Yes, k is strictly monotonic. $\endgroup$ – keven ren Jun 14 '18 at 22:51
  • $\begingroup$ It is not clear what function is known and what function in unknown. In other words, what are you looking for? In terms of what ? $\endgroup$ – JJacquelin Jul 13 '18 at 8:53
  • $\begingroup$ Sorry, I'm not quite following your question. I was looking to solve for the function, k(y). I have the functions, f(x) and g(y). $\endgroup$ – keven ren Jul 31 '18 at 3:34

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