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:

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



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.

  • $\begingroup$ :) ${}{}{}{}{}{}{}{}$ $\endgroup$ – MPW May 28 '14 at 13:40
  • $\begingroup$ Thanks! My maths skills are questionable to say the least, but I have two questions: 1) is it not true that if the period is $T$, then $g(x) = g(x + T) \forall x$? And, if so, I'm wondering if the substitution is correct? 2) should $\forall \xi$ be $\forall k$? $\endgroup$ – SnagaDuath May 28 '14 at 20:08
  • $\begingroup$ @SnagaDuath (improved comment) (1) This point I did a bit fast. If the function is invariant under replacing $x\mapsto x+T_x$ with some period $T_x\neq 0$ it is also invariant under replacing $\xi\mapsto \xi+T_\xi$ with some period $T_x\neq 0$. You have $T_\xi=2\pi T_x$ In my answer i discussed $T_\xi$. (2) No it is $\forall \xi$ meaning for all function arguments like the $\forall x$ in your comment. $\endgroup$ – flonk May 29 '14 at 18:02

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


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