Periodicity of the triginometric function $g(x) = \sin\Big(2 \pi b (x-a) \cos\big(2 \pi c (x - a) \big)\Big) + d$

Question

I have a trigonometric function and I'm interested to know whether or not it has a period. At this stage I'm fairly certain that it is not periodic. However, I don't know how to prove it. Can anyone help?

This is the function:

$$g(x) = \sin\Big(2 \pi b (x-a) \cos\big(2 \pi c (x - a) \big)\Big) + d$$

Thanks

Answer 1

Substitute $\xi=2\pi(x-a)$ and set $d=0$ w.l.o.g, then your period $T$ should satisfy

$$\sin[(\xi+T)b\cos((\xi+t)c)]=\sin[\xi b \cos(\xi c)]$$ i.e. for $bc\neq 0$ the arguments must differ by an integer multiple of $2\pi$ for all $x$:

$$(\xi+T)b\cos(\xi x+Tc)=\xi b \cos(\xi c) + 2\pi k(x)$$

with some function $k(x)\in\mathbb{Z}$. As this function has to be smooth and is restricted to $\mathbb{Z}$, it must be constant and you get

$$\xi \cos(\xi c + Tc)+T\cos(\xi c +Tc)=\xi\cos(\xi c)+2\pi k,\qquad \forall \xi.$$

This is obviously not possible for $T\neq 0$.

Edit: Of course as mentioned, the specific case $bc=0$ has to be excluded.

Answer 2

In some cases the function is periodic. Specifically, if $c=0$ and $b\neq 0$, the function has period $\frac1b$.