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

P.S.

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

$f^{(n)}(0)=f^{(n)}(T)$

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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    If you wa$n$t 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.2011-05-07

1 Answers 1

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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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    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 $x$).2011-05-07