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Find the fundamental period of $x[n] = \cos(\frac{\pi}{8}n^2)$ where $x[n]$ is defined over the integers.

I start by trying to show that $x[n] = x[n+N]$, where $N$ is the period, and replace $x[n] = \cos(\frac{\pi}{8}n^2)$ with $x[n] = e^{\frac{j\pi}{8}n^2}$ and after substitution I find that for the equality to be true: $16 = 2nN + N^2$m but from here I can't see how I would find the fundamental period.

Anyone have any ideas?

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    This function is not periodic. To have period $N\neq 0$ a function must satisfy $f(x) = f(x+N)$ for all $x$. – 2017-01-17
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    I suspect $x[n]$ is only defined on the integers? Then the function is periodic with period $8$. – 2017-01-17
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    @Winther How did you find that the function has period $8$? – 2017-01-17
  • 2
    We need $\frac{\pi}{8}(n + T)^2 \equiv \frac{\pi}{8}n^2 \mod 2\pi$ (i.e. the difference between these two expressions must be an integer multiply of the period of $\cos(x)$ which is $2\pi$). If $T = 8$ then this is true so ther period is $\leq 8$. Then you need to rule out smaller (integer) periods (none of which works). – 2017-01-17
  • 0
    This exact question has been [answered](http://dsp.stackexchange.com/q/22421/4298) over at DSP.SE. – 2017-01-17

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