# 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. :)

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.

• That little formula has been hard at work for you. +1 Sep 27, 2015 at 18:45

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 $\sqrt270-\sqrt90$

• the first zero point has to be $1.25$, how can I calculate it? Thank you for your help :) Sep 27, 2015 at 17:50
• Your approximation is in radians, I used degrees. The exact answer of your 1.25 is the square root of $\pi/2$ Sep 27, 2015 at 17:52
• Hmmm that's strange. Somebody gave me a down vote, but there is no comment or explanation...Peculiar to say the least. Sep 29, 2015 at 13:46
• Umm, what if the period might be greater than say $\sqrt{450}$? Apr 19, 2017 at 12:22
• @Sawarnik. Ok, what is the next zero after $\sqrt{450}$? Apr 19, 2017 at 13:51

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.

• Can you maybe show me how can I prove the uniformly continuity of this function? This is my 4 week at the Calculus I. course and I've never heard about it before. Sep 27, 2015 at 17:56
• Sep 27, 2015 at 18:22
• math.stackexchange.com/questions/775045/… Sep 27, 2015 at 18:23

$f(x)$ 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 $0$).

• Your more "extensive" argument doesn't seem very valid to me. For example, if we consider the function $\sin(\frac1{x})$ for x>0, in intervals of amplitude equal to the largest zero of the function, we will obtain a periodic function, but however, the distance between its zeros converges also to 0. Oct 30, 2022 at 8:03