Many a times it is asked to find the period of combination of two functions given the period of individual functions.

Let’s take this example:

If $f(x)= \cos ax + \sin x$ is periodic, then $a$ cannot be? $\pi$, 0.3, 0.5, 5

I can see that the period of $\cos ax$ is $\frac{2\pi}{a}$ and that of $\sin x$ is $\pi$ but how to find the period of $f(x)$?

Another example of very similar concept is

If $f(x)$ and $g(x)$ are periodic functions with period $7$ and $11$ respectively. Then the period of $F(x) = f(x) g(x/5) - g(x) f(x/3)$

I know the period of $g(x/5)$ is $11/5$ and that of $f(x/3)$ is $7/3$, but how to combine the periods?

  • $\begingroup$ The keyword is commensurate. $\endgroup$ Commented Sep 7, 2020 at 8:26
  • $\begingroup$ @GerryMyerson Can you please explain me a situation where the individual periods are incommensurate? $\endgroup$ Commented Sep 7, 2020 at 10:09
  • $\begingroup$ The period of $\cos\pi x$ is incommensurate with the period of $\sin x$. $\endgroup$ Commented Sep 7, 2020 at 12:26
  • 1
    $\begingroup$ If the period of $f(x)$ is $7$, then the period of $f(x/3)$ is not $7/3$, it's $7\times3=21$. $\endgroup$ Commented Sep 8, 2020 at 11:20

1 Answer 1


The period of a function is the smallest nonzero $t$ such that

$$f(x+t)=f(x)$$ and by induction $$f(x+kt)=f(x).$$

Now if $f$ is a combination of two functions of known and commensurate periods $t',t''$, there are two relatively prime integers such that


and we have

$$g(x+k't')=g(x),\\h(x+k''t'')=h(x)$$ which shows that $f$ is certainly periodic with a period not exceeding $t$.

On the other hand, if $t',t''$ are incommensurate, $f$ is aperiodic.

Anyway, the period can be shorter, and this depends on the particular combination of $g,h$ defining $f$. Unfortunately, I know of no way to find the shorter period when there is one, in another way than... looking for the period. E.g. $\sin 5x+\cos x$ and $\sin 5x-\cos x$ are both of period $2\pi$, but their sum $2\sin 5x$ has the period $\dfrac{2\pi}5$.

  • $\begingroup$ It's really a very nice answer, very well explained. So, for $\cos ax + \sin x$, the individual periods are $ \frac{2\pi}{a}$ and $2 \pi$, how to find $k'$ and $k''$? I think we do something like this $$ a ~ \frac{2\pi}{a} = 1\times 2\pi = 2\pi $$ but $a$ and $1$ are not co-prime. $\endgroup$ Commented Sep 7, 2020 at 10:07
  • $\begingroup$ @Knight: $k',k''$ can be any relatively prime integers and $a=\dfrac{k'}{k''}$. The excluded value is obvious. $\endgroup$
    – user65203
    Commented Sep 7, 2020 at 10:11
  • $\begingroup$ Can you please stress a little more on why the function will be aperiodic when the ration of periods of individual functions is not a rational number? $\endgroup$ Commented Sep 7, 2020 at 10:14
  • $\begingroup$ @Knight: solve $k't'=k''t''$. $\endgroup$
    – user65203
    Commented Sep 7, 2020 at 10:17
  • $\begingroup$ $$ \frac{k’}{k’’}= \frac{t’}{t’’}$$ $\endgroup$ Commented Sep 7, 2020 at 10:24

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