Show that $f(x):=\cos(x^2)$ is not periodic. How can I proof that the following function  $f(x):=\cos(x^2)$ is not periodic? I think that I should find the zero points of the function but I don't know how to calculate it.
Thank you very much for your help. :)
 A: Result: If a function $f$ is continuous and periodic on $\mathbb{R}$ then it is uniformly continuous on $\mathbb{R}.$
Again you can easily see that the function $\cos(x^2)$ is continuous but not uniformly continuous on $\mathbb{R}.$ Hence it is not periodic.
A: Ok, find zeros: $cosx^2=0$ means $x^2=90 $ plus its "multiples".
First zero is at $\sqrt90$, second zero is at $\sqrt270$, third zero is at $\sqrt450$. Now $\sqrt450 - \sqrt270$ is not equal to $\sqrt270-\sqrt90$
A: Suppose $f(x)$ has a period $L>0$. Then by definition,
$$ 0=f(x+L)-f(x) = \cos{((x+L)^2)}-\cos{(x^2)} $$
for all $x$, and applying the prosthaphaeresis formula
$$ \cos{A}-\cos{B} = -2\sin{\left( \frac{A+B}{2} \right)}\sin{\left( \frac{A-B}{2} \right)} $$
(I need these relations on speed-dial...) gives
$$ 0 = -2\sin{\left( \frac{(x+L)^2+x^2}{2} \right)}\sin{\left( \frac{(x+L)^2-x^2}{2} \right)} = -2\sin{\left( \frac{2x^2+2xL+L^2}{2} \right)}\sin{\left( \frac{2xL+L^2}{2} \right)} , $$
which is only true for all $x$ if $L=0$ (one of the factors would have to be identically equal to zero, but this is obviously false for both). Therefore we have a contradiction, and there can be no such $L$, so $f(x)$ is not periodic.
A: $f(x)$ equals zero at successive (positive) points $x_1=\sqrt {\frac{\pi}{2}};  x_2 =\sqrt {\frac{3\pi}{2}}; x_3=\sqrt {\frac{5\pi}{2}}$ ; besides  $x_2-x_1=(\sqrt{\frac32}-\sqrt{\frac 12})\sqrt{\pi}=0,517638\sqrt{\pi}$ and $x_3-x_2==(\sqrt{\frac52}-\sqrt{\frac 32})\sqrt{\pi}=0,356393 \sqrt{\pi}\lt0,517638\sqrt{\pi}$. This is enough to prove  $f$ is not periodic.
(Another more extensive form is consider the function $g(x)=(\sqrt{2x+1}-\sqrt{2x-1})\sqrt{\frac{\pi}{2}}$ and verify its derivative $g’(x)= (\frac{1}{\sqrt{2x+1}}-\frac{1}{\sqrt{2x-1}})\sqrt{\frac{\pi}{2}}$ is clearly negative so $g$ is decreasing. Actually the distance between two consecutive zeros of $f$ converges quickly enough to $0$).
