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Is there expression for an operator that gives for any analytic periodic function its period?


In my view this probably means solving the following system of equations:


against $T$.

I just wonder whether the solution to this system can be written in a form of one expression.

P.P.S. Alternatively the equation can be written as

$$\Delta f(Tz)≡0$$

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I can't think of any operator that would be able to handle both elliptic functions and the function $x-\lfloor x\rfloor$... – J. M. May 7 '11 at 8:28
What is an expression? What is a periodic function? (For example, is the indicator function of $\mathbb{Q}$ a periodic function? What is its period?) – Qiaochu Yuan May 7 '11 at 8:28
I am interested in operator that returns (minimal) period for analytic functions. – Anixx May 7 '11 at 8:57
So... it would have to return one result for exponentials and two results for elliptic functions, then? – J. M. May 7 '11 at 9:02
If you want to focus on analytic functions, do specify that. If you have further conditions on the functions you are interested in - please specify them. If you add context to why you want this operator, other ideas that might be useful can be given instead. – Asaf Karagila May 7 '11 at 10:08

If you just want an abstract operator you could define for the function $f: A \mapsto B$ the period length $P$ as (you need some norm to define whats "small"):

$P(f)=\min(p \in \mathbb{R} | \exists z: |z|=p: \forall x \in A: f(x+z)=f(x))$ if the minimum exists. If you want some generic function (that will be more useful) to actually determinate the period length you have to make some assumptions about $f$ as there probably isn't any more generic useful formula.

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Yes. I want a constructive formula that gives an actual expression or encodes an algorithm of how to find such period. – Anixx May 7 '11 at 13:04
As I said you have to make an assumption about the functions you want to look at then. – Listing May 7 '11 at 13:09
I meant analytic functions. – Anixx May 7 '11 at 13:10
So you can try to find the smallest strictly positive root (in terms of $y$) of $f(x)-f(x+y)$ which is again a numerical problem (note that it has to hold for all $x$). – Listing May 7 '11 at 13:13

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