Period T of function $y = \cos(nx)$

Question

There are two functions:

1) $f(x) = \cos(nx)$

2) $f(x) = \cos(x)$

$T=2 \pi$ is the fundamental period of $(2)$ function.

$T_1$ is the fundamental period of $(1)$ function.

How to prove that $T_1=\frac{2\pi}{n}$?

Answer 1

Use the definition of periodic functions, let $T_1$(the smallest non negative real) be the period of the first function so:

$$ f(x) = f(x+T_1)\quad (T_1\neq0)\\ \cos{nx} = \cos{n(x+T_1)}\\ \cos{nx} = \cos{(nx+nT_1)} = \cos{(nx+2\pi)} (I\quad used \quad the \quad periodicity\quad of\quad the \quad second\quad function)\\ nx+2\pi = nx+nT_1\\ T_1=\frac{2\pi}{n} $$

Answer 2

By definition, a function $f(.)$ is periodic with period $p$ if f(x+p) = f(x) and $p$ is the smallest positive number that satisfies the above relation.

Hence $\cos(n(x+p)) = \cos(nx +np) = cos(nx) \iff np = 2*\pi$.

Hence the period $p = \frac{2\pi}{n}$

Answer 3

The period of function (1) is the smallest positive $T$ such that $f(x+T)=f(x)$ for all $x\in \mathbf R$.

This means that $\cos(n(x+T))=\cos(nx +nT)=\cos nx$. Thus $nT\equiv 0\mod 2\pi$, or $T\equiv 0\mod\dfrac{2\pi}n$. The smallest positive value is $\;\dfrac{2\pi}n$

Answer 4

$%\ f(x/n+T_1/n)=cos(x+nT_1)=cos(x)=f(x/n)$ Take $g(x)=f(x/n)$.So $f(x)=g(nx)$

so $g(x)=cos(x)$.

$g(x)$ has period $2 \pi$.

i.e $f(x+2 \pi /n)=g(nx+2 \pi)=g(nx)=f(x)$ and $\nexists \hspace{0.3cm} \delta< 2\pi/n (i.e \hspace{.3cm} n\delta<2\pi) \hspace{0.3cm} s.t \hspace{0.3cm} f(x+ \delta)=g(nx+ n\delta)=g(nx)=f(x) $

so $f(x)$ has period $2\pi /n$.