The period of the given function?

Question

I have this function for $\{\alpha,\beta,x\}>0$ and reals. $$ \frac{\cos \left(\alpha +\frac{\beta }{2} \right)x+ \Big(5 \cos (\alpha -\beta )x+7 \cos \beta x-1\Big)\cos \frac{\beta}{2}x }{\cos \frac{\alpha}{2} x\; \cos \frac{\beta -\alpha}{2} x } $$ How to determine the period of this function?

Answer 1

You really wouldn't want to ignore the parts where it's not defined. :))) Anyhow, doing just that, note that it would be enough to find a $T > 0$ such that $\cos(\frac{\alpha}{2} x) = \cos(\frac{\alpha}{2} (x + T)),$ $\cos(\frac{\beta - \alpha}{2} x) = \cos(\frac{\beta - \alpha}{2} (x + T)),$ $\cos((\beta - \alpha) x) = \cos((\beta - \alpha) (x + T)),$ $\cos(\beta x) = \cos(\beta (x + T)),$ $\cos(\frac{\beta}{2} x) = \cos(\frac{\beta}{2} (x + T))$ and $\cos((\alpha + \frac{\beta}{2}) x) = \cos((\alpha + \frac{\beta}{2}) (x + T)).$ Note that some of these equations are redundant. Taking into consideration the periodicity of $\cos,$ we deduce that $$\begin{cases} \frac{\alpha}{2} T = 2 k \pi,\\ \frac{\beta}{2} T = 2 l \pi, \end{cases}$$ for some non-zero $k, l \in \mathbb{Z}.$ Note how the fact that all the other arguments become multiples of $2\pi$ follows immediately if only these two conditions are satisfied. Thus, we get that $T = \frac{4 k \pi}{\alpha} = \frac{4 l \pi}{\beta}$ for some non-zero $k, l \in \mathbb{Z}.$

We remark that a necessary condition for this to be possible is that the fraction $\frac{\beta}{\alpha} = \frac{l}{k}$ is rational. This is also sufficient, since if $\beta = \frac{l}{k} \alpha,$ then $T := \frac{4 k \pi}{\alpha}$ is a period for the function in question. To get the smallest period provided by this line of reasoning, you compute $\frac{\beta}{\alpha}$ and simplify it as much as you can, getting a rational $\frac{l}{k}$ with $(l, k) = 1$ and then simply define $T:= \frac{4 k \pi}{\alpha}.$

Is this THE period of the function though? It may not be, the expression of the function is pretty gnarly, but maybe it can in fact be simplified to something simple, a function whose period is easily computed. Anyhow, this is more of a brute force algorithm. Nevertheless, a period does indeed appear.