Consider the periodicity of the following function: $y=f(x)=\cos (x^2)$?

Question

A periodic function is a function with $f(x) = f(x+T) \ (T>0; ∀x ∈ D; \ D$ is the definite set). In the problem I give, I am trying to prove that there does not exist a satisfying $T$. However when I substitute $T = 2\pi$, we get $\cos[(x+2\pi)^2] = \cos(x^2 + 4x\pi + 4 \pi^2) = \cos (x^2)$, so this is a periodic function , but the answer given is a non-periodic function and I don't know what was wrong ?

Answer 1

It is not. Both the terms $4\pi x$ and $4 \pi^2$ cannot be 'canceled', since $x$ can be anything and $4\pi^2 = 2\pi(2\pi)$, so neither of them is necessarily a integer multiple of $2\pi$ (the latter never is). Even combining then we get $2\pi(2x+2\pi)$, and whenever $2x+2\pi$ is not an integer, we do not get $\cos(x^2+2\pi(2x+2\pi))=\cos(x^2)$

For example: $$\cos(0^2) = \cos(0) = 1$$ $$\cos((2\pi)^2) = \cos(4\pi^2) \approx -0.207$$

I recommend looking the graph of the function.