Is there a function $f$ such that $f(-x)=\sum_{k=0}^{+\infty}{\frac{f(k)}{k!} x^k}$?

Question

You can rewrite this as the requirement that $(-1)^k f^{(k)}(0)=f(k)$ but I do not feel this helps much. I also saw some similarities with Ramanujan's master theorem/interpolation formula but I also got stuck there. The closest function I found was $\cos(\pi x)$ but this function still has an additional $\pi^k$, which I feel is not easily fixable unless I am missing something. Is there a such function $f$ that is not exactly the zero-function?

Answer 1

If a function $f$ solves the delay differential equation

$$ f'(x) = -f(x+1), \tag{1} $$

then $f^{(k)}(x) = (-1)^k f(x+k)$ and hence $(-1)^k f^{(k)}(0) = f(k)$ will be satisfied. Then, assuming $f(x)$ takes the form $f(x) = e^{rx}$ and plugging this to $\text{(1)}$, we find that $f$ is indeed a solution of $\text{(1)}$ if

$$ r + e^r = 0. \tag{2} $$

This equation has infinitely many complex roots with values approximately

\begin{gather} -0.567143, \\ 1.53391 \pm 4.37519 i, \\ 2.40159 \pm 10.7763 i, \\ 2.85358 \pm 17.1135 i, \\ 3.16295 \pm 23.4277 i, \\ 3.39869 \pm 29.7313 i, \\ \vdots \end{gather}

In fact, any linear combination $f(x)=\sum_{n=1}^{N} c_n e^{r_n x}$ with each $r = r_n$ satisfying $\text{(2)}$ also solves $\text{(1)}$ by linearity.

Remark. The zeros of $\text{(2)}$ are of the form $-W_k(1)$, $k\in\mathbb{Z}$, where each $W_k$ is a branch of the Lambert W-function. In particular, $-W_0(1) \approx -0.567143$ corresponds to the negative of the Omega constant.