Obtaining coefficients of a sum of exponential decays Let $f:\mathbb R^+\to\mathbb R^+$ be such that
\begin{equation}
f\left(x\right)=\sum_{n=0}^\infty a_ne^{-4\pi nx}
\end{equation}
where $a_n$ are nonnegative integers. Is there some way of recovering $a_n$ from $f$? I have been trying to get them like coefficients of a Fourier series, and have tried taking a Laplace transform. Maybe something else works.
 A: You actually can do the same thing you'd do for a Fourier series. The general theory is that if we have a Hilbert space with the inner product $\langle\cdot\vert\cdot\rangle$ and an orthonormal basis $\{b_i\}_{i\in\mathbb N}$, then for any element
$$v=\sum_{k=0}^\infty a_k b_k$$
of the Hilbert space we can recover the coefficients $a_n$ by calculating the inner product of $v$ and $b_n$, since we have
$$\langle v\vert b_n\rangle=\left\langle \sum_{k=0}^\infty a_kb_k\middle\vert b_n\right\rangle=\sum_{k=0}^\infty a_k\langle b_k\vert b_n\rangle=a_n,$$
where the second equality comes from the continuity and linearity of the inner product, the third equality comes from orthonormality of the base.
This is what we do when calculating Fourier coefficients. The Hilbert space is that of functions with a given period (let's say $1$). The orthonormal basis is the set of functions $\exp\left(2\pi\mathrm inx\right)$, and the inner product wrt which it is orthonormal is
$$\langle f\vert g\rangle:=\int_0^1 f(x)\cdot g^\ast(x)\mathrm dx,$$
where $X$ is the set on which the functions are defined. Calculating the Fourier coefficients is basically calculating $\langle f\vert\exp(2\pi\mathrm inx)\rangle$. We can do the same with functions of the form
$$\sum_{k=0}^\infty a_n\mathrm e^{-4\pi nx}.$$
We have to find an inner product on this space with respect to which the given basis functions $\mathrm e^{-4\pi nx}$ are an orthonormal system. I give you this one:
$$\langle f\vert g\rangle:=\int_0^1 f\left(\frac{\mathrm ix}{2}\right)g^\ast\left(\frac{\mathrm ix}{2}\right)\mathrm dx.$$
You can show that it has all the necessary features of an inner product, and that $\langle\exp(4\pi nx)\vert\exp(4\pi kx)\rangle$ with respect to this inner product is the same as $\langle\exp(2\pi\mathrm i nx)\vert\exp(2\pi\mathrm ikx)\rangle$ with respect to the one used for calculating Fourier coefficients. Calculating these "Laplace coefficients" (I don't think that's an official term) works the same as calculating Fourier coefficients. Just that with the given functions, you only have to index them with non-negative integers.
A: Here's a way to recover the coefficients. Note that we can reduce $f$ to be a power series by "warping" its domain.
$$f\left(-\frac {\ln t} {4\pi}\right) = \sum_{n\geq 0}a_n t^n$$
Thus $$a_n=\frac 1 {n!} \left.\frac{d^n}{dt^n}f\left(-\frac {\ln t} {4\pi}\right)\right|_{t=0}\\$$
Looks great on paper. Numerically, not awesome though. Numerical differentiation is tricky.
