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?


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 $g$.

  • $\begingroup$ 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

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