Under what condition on $a$, the function $\cos ax \cos (ax+\pi)$ will be periodic over $x>0$?

Question

I have this function $\cos ax \cos (ax+\pi)$ for $x>0$, and I want to know under what condition on $a$, it will be periodic?

Answer 1

This function is periodic for every $a$. You can try to find out what is his littlest period.

You can transform the function to $\cos(ax)\cos(ax+\pi)=-\cos^2(ax)$ because $\cos(x+\pi)=-\cos(x)$.

The littelest period of $\cos^2(x)$ is $\pi$.

Hence, the littelest period of the function is $\frac{\pi}{|a|}$ for $a\not =0$.

For $a=0$, the function is constant equals to $-1$, so we can't define a littlest period (the function is periodic for every real).

Indeed, let $T$ the littelest period of our function.

We have : $-\cos^2(ax)=-\cos^2(a(x+T))=-\cos^2(ax+aT)$.

$|aT|$ must be equal to $\pi$.

EDIT :

The op. actually wanted to find for which $a$, $\cos(ax)\cos(a+\pi)x$ is periodic.

We can use this simple argument :

If $\cos(a+\pi)\not = 0$, then $\cos(ax)\cos(a+\pi)x$ is not bounded.

Indeed, $|\cos(a\frac{n\pi}{a})\cos(a+\pi)\frac{n\pi}{a}|\rightarrow +\infty$ if $a\not =0$.

If $a=0$, it's a linear function which is not bounded.

The function can't be periodic with this condition because a periodic function is continuous in a segment $[0,T]$ and then bounded in this segment. By periodicity, it's bounded everywhere.

If $-\cos(a)=\cos(a+\pi)=0$, the function is constant and then obviously periodic.

So the function is periodic if and only if there exists $k\in \mathbb{Z}$ such as $a=\frac{\pi}{2}+k\pi$.