I would like to find the function $e(x)$ which solves $\sum\limits_{k=1}^n e(x-x_k) = h(x)$, where $x_k$ and $h(x)$ are given. There are no restrictions on any of the $x_k$ or $h(x)$ except that $h(x)$ is an analytic function. An analytic solution for $e(x)$ would be ideal, but is not necessary - any computationally feasible expression would be helpful.
This question is a repost of the one linked here with some additional information and possible methods.
One approach is to try to find a power series for $e(x)$. Taylor expanding $e(x) = \sum\limits_{j=0}^\infty b_j x^j$, $$\sum_{k=1}^n e(x-x_k) = \sum_{j=0}^\infty b_j \left(\sum_{k=1}^n (x-x_k)^j\right)$$ $$\sum_{k=1}^n e(x-x_k) = \sum_{m=0}^\infty \left( \sum_{k=1}^n \sum_{j=0}^\infty b_j {j+m \choose m} (-x_k)^j\right)x^m$$ The first of these isolates the coefficients of $b_j$ and the second isolates powers of $x$. Because the coefficients of $x^m$ in involve all of the infinitely many $b_j$, I doubt matching coefficients to the power series of $h(x)$ would help. On the other hand, I don't know how I could pick out $b_j$ from the first equation either.
A second approach is to use a Fourier transform, which changes shifts in the argument of $e(x-x_k)$ into multiplicative phase factors, so that $$h(q)=e(q)\sum\limits_{k=1}^ne^{-2\pi iqx_k}$$ Dividing, inverse Fourier transforming, and writing out the Fourier transform of $h$, $$e(x)=\int_{-\infty}^\infty h(x')\left( \int_{-\infty}^\infty e^{2\pi iq(x'-x)}\frac{1}{\sum\limits_{k=1}^ne^{-2\pi iqx_k}} dq \right)dx'$$ $$=\int_{-\infty}^\infty h(x')\left( \int_{-\infty}^\infty \frac{1}{\sum\limits_{k=1}^ne^{-2\pi iq(x_k+x'-x)}} dq \right)dx'$$ I do not know, however, how to compute this integral, and I don't recognize it as the Fourier transform of a known function.
How could I find an expression for $e(x)$, whether an analytic expression, a power series, or some other sort of expansion?