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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 ?

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One solution to title question is $\alpha e^x$ if $\alpha (1+e)=1$. I suppose uniqueness is main issue. –  Maesumi Jan 29 '13 at 23:00
    
@Maesumi Your answer seems the only analytic solution to me. Thanks. –  mick Jan 29 '13 at 23:03

2 Answers 2

up vote 5 down vote accepted

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.

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Hmmm... Can that be done? We only have $ e^{x} $ on the right-hand side of the functional equation. –  Haskell Curry Jan 29 '13 at 23:06
    
Of course, you are correct. –  Michael Albanese Jan 30 '13 at 2:48

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 $\exp(\pi i (x+1))=-\exp(\pi i)$.

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+1 thanks for extra info ! –  mick Jan 29 '13 at 23:12

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