Can we pull $f$ from this equation?

Question

$$\sum_{k=1}^\infty f(k)\cdot e^{-k} = c$$

Can we pull $f$ function from this equation somehow? $c$ is a real constant...

Answer 1

I don't think so. Let $c=1$. Then we can let $f(X)=\delta_k(X)e^X$ for any $k\in\mathbb N$, where $\delta_k(X)=1$ if and only if $X=k$ (and $0$ otherwise). By this same reasoning, adding continuity and smoothness won't help the situation.

Answer 2

Unfortunately, no (at least, not without more information). For example, $f$ could be the constant function $$f(x)=(e-1)c,$$ but we could easily change any two function values--say $f(1)$ and $f(2)$--to something else, but keep the desired identity true. In fact, using bump functions, we could change the values of $f$ in the interval $(1/2,5,2)$ only, say, in such a way that we kept the desired identity true, and still have an infinitely-differentiable (but non-constant) function, so even that fairly strong condition won't do the trick.

Answer 3

Yes you can consider the equation as a specific value of a Z-transformed function, as "orion" said. So, you have an unknown function, $f(k)$, and its Z-transformed value at at point.

Explicitly, let $$F(z) = \sum_{k=1}^{\infty} f(k) e^{-kz}$$ be the Z-tronsform of f(k) where k is integer and z is complex. Then the inverse Z-transform would be $$f(k) = \frac{1}{2\pi i} \oint_C F(z) e^{-kz} dz$$ where $C$ is a counter-clockwise closed path encircling the origin and the region of convergence (i.e. the region where the series $F(z)$ converges).

Here, I have written everything according to $k \in [1, \infty)$, however you would find $k \in [0, \infty)$ in the literature which gives the transformations slightly different.

In your case, you have the constraint $F(1) = c$ which means that $F(z)$ hasn't got a pole at $z=1$. This helps to figure out the region of convergence, I guess.