prove cosine(pi*t*t/8) is not periodic

Question

anonymous · Answer

A function $f(x)$ is periodic if you can find $l$ such that $f(x)=f(x+l)$ with $l>0$.

Assume the given function is periodic, then
$$\begin{align*}
\cos\left(\frac{\pi t^2}{8}\right)&=\cos\left(\frac{\pi (t+l)^2}{8}\right)\\
\frac{\pi t^2}{8}&=\frac{\pi (t+l)^2}{8}\\
t^2&=(t+l)^2\\
t^2&=t^2+2lt+l^2
\end{align*}$$
If $l>0$, this can't be true. Hence we have a contradiction, so the given function cannot be periodic.

loser66 · Answer

@SithsAndGiggles 
We have l =0 and l = -2t
Those cases show it is periodic. Am I right?

anonymous · Answer

@Loser66 the period of a function must be constant, so it can't be $l=-2t$.

As for the case $l=0$, consider the function $f(x)=x$. This function is periodic if $f(x)=f(x+l)$, which gives
$$x=x+l~~\implies~~l=0$$
but we know $f(x)=x$ is not periodic.

As I said in my first post, the period $l$ must be non-zero.