Prove that the function $f(x)=\cos x^2$ is not a periodic one .

Question

I was going through a chapter about functions from " Problems in Calculus of One Variable " by I.A Maron . There was an example given in the book :

Prove that the function $f(x)=\cos x^2$ is not a periodic one .

The solution given in the book is as follows:

Let us prove the contrary . Suppose the function has a period $T$ ; then the identity $\cos(x+T)^2\equiv\cos x^2$ is valid . By the condition of equality of cosines for a certain integer $k$ , we have , $x^2+2Tx+T^2\pm x^2\equiv 2\pi k .$ But this identity is impossible, since k may attain only integral values , and the left member contains a linear or quadratic function of the continuous argument $x$.

However, I am not getting how are they concluding the identity "$x^2+2Tx+T^2\pm x^2\equiv 2\pi k $" is impossible? Now, $k$ attains integral values but $\pi$ is irrational. How to make the contradiction seem more obvious ? Or rather, how is it so much immediate?I am not quite getting it...There might be similar questions related to the same topic but I can't seem to find it either...

Answer 1

Suppose there exists $T>0$ such that $$\cos(x^2)=\cos\left({(x+T)}^2\right),\quad x\in\Bbb R.\tag{1}$$ Define $k:=\lfloor\frac{T^2}\pi\rfloor$, which means that $k\pi\le T^2<(k+1)\pi$. Then $$I:=\left(0,\sqrt{1+(k+1)\pi-T^2}-1\right)$$ is a non-empty interval of $[0,\sqrt\pi]$ such that $k\pi<{(x+T)}^2<(k+1)\pi$ for all $x\in I$. Now, observe that all three numbers $x^2+k\pi$, ${(x+T)}^2$ and $(2k+1)\pi-{(x+T)}^2$ all belong to the interval $[k\pi,(k+1)\pi]$ (which has length $\pi$). Thus:

1. If $k$ is even, then $\cos(x^2+k\pi)=\cos(x^2)=\cos\left({(x+T)}^2\right)$ by (1), implying $x^2+k\pi={(x+T)}^2$. Hence $k\pi=2x+T^2$ for all $x\in I$, which is absurd (you may differentiate this equality to get the contradiction $0=2$). “A linear function cannot be constant on a non-empty interval.”
2. If $k$ is odd, then $\cos\left((2k+1)\pi-{(x+T)}^2\right)=-{\cos\left({(x+T)}^2\right)}=\cos(x^2+k\pi)$ by (1), implying $(2k+1)\pi={(x+T)}^2+x^2+k\pi$ for all $x\in I$. Again, this is absurd (you get $0=2$ again by differentiating twice this equality). “A quadratic function cannot be constant on a non-empty interval.”

This proves that $T$ cannot exist.

Answer 2

Note $f(x)=\cos(x^2)$ and assume that $f(x+T)=f(x)$ for all $x$.

Then $f(T)=f(0)=1$ so $\cos(T^2)=1$ and $\sin(T^2)=0$. Besides $$f(x+T)=\cos(x^2+2xT+T^2)=\cos[(x^2+2xT)+T^2]=$$ $$=\cos(x^2+2xT)\cos(T^2)-\sin(x^2+2xT)\sin(T^2)=$$ so we have for all $x$ $$f(x+T)=\cos(x^2+2xT)=f(x)=\cos(x^2)$$ so $$\cos(x^2+2xT)-\cos(x^2)=0$$ in other words, we have for all $x$ that $$-2\sin(x^2+xT)\sin(xT)=0$$ which is absurd because none of the three factors in the $LHS$ is null.We are done.