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I have been trying to solve the following functional equation

$$f(x+2)+af(x+1)+bf(x)=0$$ for all real values of $x$.

My guess and intuition leans towards $f(x)$ being some exponential function. I started out with $f(x)=\lambda a^x $ but ended up with a quadratic which seemed like a dead end.

Any hints will be helpful.

It does look similar to the general differential equation $${d^2y \over dx^2} + a {dy\over dx} + b=0$$ which also leads to a quadratic but I'm getting messy expressions for the functional equation.

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  • $\begingroup$ The problem is simple for integer $x$ $\endgroup$ Commented May 29, 2021 at 8:39
  • $\begingroup$ @ClaudeLeibovici I wish to solve $f(x)$ for all real values of $x$ $\endgroup$ Commented May 29, 2021 at 10:03
  • $\begingroup$ Why would the quadratic seem like a dead end? It can give you some solutions. $\endgroup$
    – Sil
    Commented May 29, 2021 at 11:37
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    $\begingroup$ Essentially, the result for the integers is all you can get. More precisely, for any $ g : [ 0 , 2 ) \to \mathbb R $, you can extend $ g $ to an $ f : \mathbb R \to \mathbb R $ satisfying the desired equation; just treat every $ x \in [ 0 , 1 ) $ separately, and find the value of $ f ( x + n ) $ for all $ n \in \mathbb Z $. Even further assumptions on $ f $ like continuity/smoothness doesn't make the solution set very small (you'd only need $ g $ to be continuous/smooth with the additional assumption that its limit or the limits of its derivatives near $ 2 $ equal to the value at $ 0 $). $\endgroup$ Commented May 29, 2021 at 22:26
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    $\begingroup$ The situation is different if you restrict the case more than that, and for example require $ f $ to be real analytic. in that case, the only solutions are those that are (linear) combinations of functions of the form you've already mentioned. $\endgroup$ Commented May 29, 2021 at 22:27

1 Answer 1

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$\mathrm{f(x) = k_1\alpha ^x + k_2\beta ^x}$ is the solution for the given functional equation, where $\mathrm{\alpha}$ and ${\beta}$ are the roots of the equation $\mathrm{x^2+ax+b=0}$.

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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Aug 5, 2022 at 16:38

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