# Let $f$ be periodic function with fundamental period $\tau$. Prove that function $f(nx)$ have fundamental period $\frac{\tau}{n}$.

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?

## 1 Answer

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 $$g$$.

• Why this period of g must therefore satisfy nT≥τ ? – josf Jan 22 '20 at 19:43
• 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 $f$. – copper.hat Jan 22 '20 at 19:45