How to obtain a periodic function from a rapidly decaying function?

Question

Suppose $f(x) = \exp(-x^2)$ with $x \in [0, 3]$. How could I periodise this function to obtain an analytical form of a continuum periodic function $x \in [0, +\infty)$ with period T = 3?

Answer 1

Given a function $f$ with rapid decay at infinity, one can obtain a "periodic" version of it by summation of translates: $$F(x) = \sum_{n = -\infty}^\infty f(x-nT)\tag1$$ For your function this results in $\sum_{n = -\infty}^\infty e^{-(x-3n)^2}$.

Note that the function shown on your graph actually has period $T=6$, not $3$. With the period $3$ we get something that looks suspiciously similar to the cosine curve:

Observe that the shape of $f$ is not exactly preserved in the sum. If you want it to be preserved, you can still use (1), but make sure that $f$ vanishes outside of an interval of length $T$. E.g., let $f(x) = e^{-x^2}\chi_{[-T/2,T/2]}$.

^{(Expanding the comment by Matt L.)}