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If $f(x+1) + f(x-1) = \sqrt3f(x)$, then what is the period of $f(x)$?

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And what have you tried? – J. M. May 7 '11 at 19:09
Hint: Mathematica says that your $f(x)$ are of the form $f(x)=e^{-\frac{1}{6} i \pi x} \left(a+b e^{\frac{i \pi x}{3}}\right)$ for some real $a,b$ – Listing May 7 '11 at 19:53
@user3123 : Actually, I think Mathematica's answer is not exhaustive. You can take $a$ and $b$ to be any $1$-periodical function. – Joel Cohen May 7 '11 at 20:45
ok I did not check that – Listing May 7 '11 at 22:44

You can start solving the equation $(E_{\lambda})$ (where $\lambda$ is a complex number) :

$f(x+1) = \lambda \, f(x)$

If $\lambda_1$ and $\lambda_2$ are the roots of $X^2 - \sqrt{3} X + 1$, you can check solutions to $(E_{\lambda_1})$ and $(E_{\lambda_2})$ are solution to your equation. Conversely solution of your equation are linear combinations of the previous ones.

Finally (and most importantly), you can wonder how I came up with this idea and what is the general setting in which this idea could be applied.

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$$ f(x+1)+f(x-1) = \sqrt{3} f(x) $$ $$ \sqrt{3}f(x+1) + \sqrt{3}f(x-1)= 3f(x) $$ $$ f(x)+f(x+2)+f(x-2)+f(x)= 3 f(x)$$
$$ f(x+2)+f(x-2)=f(x) $$ $$ f(x+4)+f(x) = f(x+2) $$ Adding last two equations give , $$ f(x+4)+f(x-2)= 0$$ $$ f(x+10)+f(x+4)= 0 \implies f(x+10) = f(x-2) \implies f(x+12)=f(x) $$ Thus period = 12 $\Box$

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To be strickt: this only shows that the period is at most 12. – Winther Jul 9 '14 at 5:23

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