When a periodic function is squared (or cubed, and so on...) does it always lose its periodicity?

Question

For instance $$\sin^{2}\left(-\frac{\pi}{6}\right) = \sin^{2}\left(\frac{\pi}{6}\right)$$

i.e., $\sin^2 (x)$ is an even function and loses the $2\pi$-periodicity of $\sin x $.

Is this true in general?

Does $\tan^2 x$ lose the $\pi$-periodicity of $\tan x$? The $\tan^2 x$ function still blows up wherever $\cos^2 x$ is equal to zero - I am currently studying $\tan^2 x$'s singularities to try to understand a solution to a problem I've been working on.

Answer 1

If $f : \mathbb R \to \mathbb R$ is periodic with period $T$, i.e., $f(t + T) = f(t)$ for all $t$, then certainly $f^n$ is also periodic of period $T$ as $$f^n(t) = (f(x))^n = (f(t + T))^n = f^n(t+T)$$

However $f^n$ can also pick up other shorter periods, as your example of $f = \sin$ illustrates: $\sin$ has period $2\pi$ as does $\sin^2$; but $\sin^2$ also has period $\pi$. The minimum period of $\sin^2$ of $\pi$ is therefore less than the minimum period of $2\pi$ for $\sin$.

$\tan$ has period $\pi$, as does $\tan^2$. What $\tan^2$ does not have is a minimum period less the minimum period of $\tan$.

Answer 2

It's still true that $$ \sin^2(x+2\pi)=\sin^2x $$ so certainly $\sin^2$ is periodic with period $2\pi$. It has a different minimum period, since, as you observe, $\sin(-x)=-\sin x$, so also $$ \sin^2(x+\pi)=\sin^2 x $$

Answer 3

What do you think of the periodic function having $1$ for period and defined by $f(x)= 0$ for $ x \in [0,1/2)$ and by $f(x)=1$ for $ x \in [1/2,1)$?

Answer 4

Let $f$ be a periodic function of (minimum) period $T$ and let $g(x)=\left(f(x)\right)^2$.

$\textbf{Proposition:}$ $g$ have a (minimum) period of $\frac{T}{2}$ iff $f(x+\frac{T}{2})=\pm f(x)$

$\textbf{Proof:}$ Left to the reader.

Answer 5

How about the Dirichlet's function? $$\mathbf 1_\Bbb Q: \Bbb R\ni x \mapsto \begin{cases} 1 & \mbox{for } x \in \mathbb Q, \\ 0, & \mbox{for } x \notin \mathbb Q. \end{cases}$$ It is periodic with each rational number being its period, and is retains all those periods when raised to a natural power $$\forall_{n\in\Bbb N^+} (\mathbf 1_\Bbb Q(x))^n = \mathbf 1_\Bbb Q(x)$$

Same applies to any periodic function $\Bbb R\to \{0, 1\}$.