How can it be proved that for a given $f(x)$ with a period $p$, that for any integer $n=1,2,\dots$, the product $np$ is also an period of the function $f(x)$?
I know that the definition of a periodic function is that there is some non-zero value, $p$ (called the period) such that $f(x+p) = f(x)$ for all $x$.
So I'm guessing that the proof may starts off with something like
$f(nx) = f(nx +np) \dots $ ?
But I'm really no sure how to prove this one.
Best Answer
By the definition of periodic, there is a number $p$ so that for every $x$, $$ f(x+p)=f(x). $$ We can then proceed essentially by induction: we have that $1p$ is a period, by definition. Now suppose that $np$ is a period, so $ f(x+np)=f(x) $. But then $$ f(x+(n+1)p) = f((x+np)+p) = f(x+np) = f(x), $$ so if $np$ is a period, so is $(n+1)p$. Thus induction gives that $np$ is a period for every positive integer $n$. It is clear that $-p$ is a period if and only if $p$ is: if $p$ is a period, $$ f(x) = f((x-p)+p) = f(x-p), $$ so $-p$ is, and vice versa.