0
$\begingroup$

Let $f$ be periodic function with fundamental period $\tau$, prove that function $f(nx)$ have fundamental period $\frac{\tau}{n}$.
My idea:
First I need to show that $\frac{\tau}{n}$ is a period. $$f(n(x+\frac{\tau}{n}))= f(nx + \tau) = f(nx).$$ But how to show that this is the smallest such? Any idea?

$\endgroup$
1
$\begingroup$

Let $g(x) = f(nx)$. You have shown that if $f$ has a period $\tau$ then $g$ has period ${\tau \over n}$.

The same reasoning shows that if $g$ has period $T$ then $f$ has period $nT$.

If $\tau$ is a fundamental period of $f$ then any $T$ period of $g$ must therefore satisfy $nT \ge \tau$, or in other words, $T \ge { \tau \over n}$. Hence $ { \tau \over n}$ is a fundamental period of $g$.

$\endgroup$
2
  • $\begingroup$ Why this period of g must therefore satisfy nT≥τ ? $\endgroup$ – josf Jan 22 '20 at 19:43
  • $\begingroup$ If $T$ is a period of $g$ then $nT$ is a period of $f$. Since $\tau$ is the fundamental period of $f$ we must have $nT \ge \tau$ otherwise this would contradict $\tau$ being the fundamental period of $f$. $\endgroup$ – copper.hat Jan 22 '20 at 19:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.