# How to check if $e^{-2t}cos(2 \pi t)$ is periodic/non-periodic?

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?

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 $$0$$.

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

• This is "funny" to see that other answers focused on the $\infty$ part while the problem was much easier the other way. Jan 9, 2020 at 8:59

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 $$e^{-2\times0\times T}\cos(2\times\pi\times0\times T)=1$$.