How to find $f(x)+f(x+1) = e^x$?

Question

Let $x,a,b$ be real numbers and $f(x)$ a (nongiven) real-analytic function.

How to find $f(x)$ such that for all $x$ we have $f(x)+af(x+1)=b^x$ ?

In particular I wonder most about the case $a=1$ and $b=e$. (I already know the trivial cases $a=-1$ and $a=0$)

I know how to express $f(x+1)$ into a taylor series once I have the taylor series for $f(x)$ and I assume this is related ? But does it help to find a closed form solution here ? If there is a closed form solution ? Am I on right track here or do we need to use something completely different or more general ? Does this relate to the fibonacci sequence ?

Answer 1

Pick $ f(x) = \lambda e^{kx} $. Then $$ f(x) + f(x + 1) = \lambda \left[ e^{kx} + e^{k(x + 1)} \right] = \lambda (1 + e^{k}) e^{kx}. $$ By choosing $ k = 1 $ and $ \lambda = \dfrac{1}{1 + e} $, you get a solution.

Answer 2

Note that you can write $Lf = e^x$ where $(L f)(x)=f(x)+f(x+1)$ is a linear operation.

Therefore, it's enough to find a single function $f$ such that $Lf = e^x$ and all solutions will be of the form $f+g$ where $L g=0$. You can discover as in Haskell Curry's answer that $f$ can be taken to be $\frac{1}{1+e} e^x$.

Functions satisfying $L g = 0$ are called antiperiodic, one example is $\exp(\pi i x)$ since $\exp(\pi i (x+1))=-\exp(\pi i)$.