Periodic functions proof

988 Views Asked by At

I need some help here.

Let $f$ be a $2\pi$-periodic function, and define for an arbitrary $k\in\mathbb N$ a function $g(x) = f(kx)$. Show that $g$ is also $2\pi$-periodic.

What I've done: $$ g(x) = f(kx) $$ $$ g(x+T) = f((k+T)), \text{where } T = 2\pi. $$


But i'm kinda stuck from here. Could anybody help me?

1

There are 1 best solutions below

2
On

$$g(x+2\pi)=f(k(x+2\pi))=f(kx+2k\pi)=f(kx)=g(x)$$