# Discrete time domain, Is cos(2n) periodic?

I have just started out learning the basics of discrete time. I would like some help to understand if cos2n is periodic or not. I know the discrete time formula for periodicity: x[n] = x[n+N]. I Also know that ω=2πf, ω=2π/Ν.

I searched the web and i found a solution that goes like so: x [ n + N ] = C o s ( 2 n + 2 N ) therefore 2 N = Ω = m 2π, m ∈ Z, (at quora.com). I do not really understand it, i think that m integer does not fit in the standard formula i know of: ω=2πf. Can you solve it using a step by step guide for 'dummies'?

Consider a discrete-time sequence x[n] based on a sinusoid with angular frequency $\omega _0$: x[n]=cos[$\omega _0$n].
If this sequence is periodic with a period of N samples, then the following must be true: cos[$\omega _0$ (n + N)] = cos [$\omega _0$n] (Eq. 1). However, the left hand side can be expressed as cos [$\omega _0$(n + N)] = cos [$\omega _0$n + $\omega _0$N]. The cosine function is periodic with a period of 2$\pi$ and therefore cos [$\omega _0$n] = cos [$\omega _0$n + 2$\pi$r] (Eq.2), for integer values of r. Comparing Eq.1 and Eq.2 we get:
$$\omega _0 N =2\pi ; \ \text{(for r=1)}$$ $$2\pi f _0 N =2\pi$$ $$f _0 = \frac{1}{N}$$
Since N is integer the signal will be periodic if $f_0$ is a rational number. Otherwise, it will be non-periodic.
In your example, $\omega _0$ = 2 so $f_0$ = $\frac{2}{2\pi}$. But you cannot find an integer division ($\frac{1}{N}$) that results in an irrational number ($f_0$). So no N can be found to meet the criteria for this signal to be periodic.