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So by graphing the function it is clearly non-periodic, but I would like to know how to solve it in a more mathematical way. Is there is a way to expand this function somehow that I forget?

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3 Answers 3

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The function has the value $1$ at $0$ and it approaches $0$ as $ t \to \infty$. So it cannot be periodic.

If it has period $p$ then $f(np)=f(0)=1$ for all $n$ so $f(np)$ does not tend to $0$.

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The function is continuous, but not bounded. Every continuous periodic function is bounded.

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  • $\begingroup$ This is "funny" to see that other answers focused on the $\infty$ part while the problem was much easier the other way. $\endgroup$
    – nicomezi
    Jan 9, 2020 at 8:59
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If the function was periodic with period $T$, then the sequence $\left(e^{-2kT}\cos(2\pi kT)\right)_{k\in\mathbb Z_+}$ would be constant. Since$$\lim_{k\to\infty}e^{-2kT}\cos(2\pi kT)=0,$$we would then have$$(\forall k\in\mathbb Z_+):e^{-2kT}\cos(2\pi kT)=0.$$But $e^{-2\times0\times T}\cos(2\times\pi\times0\times T)=1$.

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