Function with extrema 1, 1/4, 1/9, ...

Question

I am looking for a function $F(x)$ which has an infinite number of extrema, and whose values at those extrema are of the form $1/n^2$ for $n \in \mathbb{Z}$.

To clarify, I don't want a function whose extrema are at $x=1,1/4,1/9,\ldots$. Instead, I want the function itself to take these values at the extrema.

Edit: To clarify, there shouldn't be multiple extrema with value F(x)=0!

Thanks for your help!

Answer 1

Does this fit the bill?

$f(x) = \frac 1{x^2}|\cos \pi x|$

It will also have extrema equal to $0$ in between. But you can't have a chain of extrema that is monotonically decreasing. Each minimum must be less than the maxima on either side.

Answer 2

A simple one, aside from the problem near $x=0,$ is $F(x)=(\frac 1n \cos x\pi)^2$ where $n$ is the closest integer to $x$. The cosine term takes values $\pm 1$ at all the integers.