Proving that a complicated complex expotential is periodic or not.

Question

I'm trying to prove that $x[n]=\exp \left(j\left(\frac{\pi}{24} n^2+\frac{\pi}{36} n^3\right)\right), n \in \mathbb{Z}$ is not periodic. I thought of proving that there is no $N \in \mathbb{Z}$ such that $x[n] = x[n + N]$. I've reached this point:

\begin{align*} \exp \left(i\left(\frac{\pi}{24} n^2+\frac{\pi}{36} n^3\right)\right) &= \exp \left(i\left(\frac{\pi}{24} (n + N)^2+\frac{\pi}{36} (n + N)^3\right)\right) \\ \frac{\pi}{24} n^2+\frac{\pi}{36} n^3 &= \frac{\pi}{24} (n + N)^2+\frac{\pi}{36} (n + N)^3 + 2\pi m \tag{ $e^{ia} = e^{ib} \Leftrightarrow a - b = 2\pi m, m \in \mathbb{Z}$} \\[15pt] -2\pi m &= \frac{\pi\left(2 N^2+6 N^2 n+3 N^2+6 N n^2+6 N n\right)}{72} \\ -2\pi m &= \frac{\pi N^2(6 n+5)+\pi N\left(6 n^2+6 n\right)}{72} \end{align*}

This seems very complicated to me and as it is part of a question of a 2nd-year level assignment, I hardly believe that this is the way to go. Any help would be greatly appreciated.

UPDATE: Moreover, I tried plugging the equation into $\texttt{SymPy}$ and solving as to N, what I got was:

\begin{equation} N= \pm \frac{3\left(-n(n+1)+\sqrt{-96 m n-80 m+n^4+2 n^3+n^2}\right)}{6 n+5} \end{equation} Not sure if this helps in any way.

Intuitively I'd expect $N$ not to be a function of $n$, but rather a constant. Maybe I'm wrong in thinking of $m$ also as a constant and not a function of $n$, because that way I could argue that the RHS of N is always a function of $n$ and therefore $x(n)$ is not periodic since $N$ is not fixed.

EDIT: Turns out the function is periodic. Huge thanks to @Conrad for pointing it out.

Answer 1

But $x(n)$ is periodic...

Just note that $$x(n+144)=\exp\left[i\pi\left(\frac{(144+n)^2}{24}+\frac{(144+n)^3}{36}\right)\right]$$

$$=\exp\left[i\pi\left(\frac{n^2}{24}+12n+864+\frac{n^3}{36}+12n^2+1728n+82944\right)\right]$$

$$=\exp\left[i\pi\left(\frac{n^2}{24}+\frac{n^3}{36}\right)\right]\exp\left[2\pi i\left(6n+432+6n^2+864n+41472\right)\right]$$

$$=\exp\left[i\pi\left(\frac{n^2}{24}+\frac{n^3}{36}\right)\right]=x(n)$$

Answer 2

Reduce your formula into its real/imaginary parts with Euler's formula, $e^{iz}=\cos z+i\sin z$. And recall that complex numbers are equal iff (if and only if) their real and imaginary parts are also equal. For $e^{iz}=e^{iw}$, we must therefore have $\cos z=\cos w$ and $\sin z=\sin w$. (In a geometric sense, if two numbers are equal in the complex plane, they must have the same horizontal/vertical coordinate, which for $e^{iz}$ is its cosine/sine component.)

Hence, if $e^{iz}$ is periodic with some period $k$, such that $e^{i(z+k)}=e^{iz}$, we must also have $\cos(z+k)-\cos z=0$ and $\sin(z+k)-\sin z=0$. We can then use the sum-to-product identities from trigonometry, $\cos A - \cos B=-2\sin\frac{A-B}2\sin\frac{A+B}2$ and $\sin A-\sin B = 2\sin \frac{A-B}{2}\cos\frac{A+B}{2}$. This facilitates finding zeroes since a product is zero iff at least one of its terms is zero. (It would therefore be convenient if the term corresponding to $\sin\frac{A-B}2$ were zero since this would eliminate all cases. And indeed we will see that to be the case.)

Writing your expression as $e^{\pi i f(n)}$, for its real terms we have

$$e^{i\pi f(n+k)}=e^{i\pi f(n)} \\\cos \left(f(n+k)\pi\right)-\cos \left(f(n)\pi\right)=0 \\-2\sin\left(\frac{f(n+k)-f(n)}{2}\pi\right)\sin\left(\frac{f(n+k)+f(n)}{2}\pi\right)=0 \\\sin\left(\frac{f(n+k)-f(n)}{2}\pi\right)=0\quad\text{or}\quad\sin\left(\frac{f(n+k)+f(n)}{2}\pi\right)=0 $$

But observe also (by the horizontal scaling of a sine wave) that $\sin\frac{a}{b}\pi=0$ iff $\frac ab$ is an integer. So we see that at least one of $\dfrac{f(n+k)-f(n)}{2}$ and $\dfrac{f(n+k)+f(n)}{2}$ is an integer. By a similar argument with the imaginary terms, we would arrive at two fractions for which at least one must be an integer (one being $\dfrac{f(n+k)-f(n)}{2}$ repeated). This has reduced the problem to elementary number theory, as we are now solely concerned with divisibility of integer valued functions.

Substituting your $f$ and simplifying, we see that $\dfrac{f(n+k)-f(n)}{2}=\dfrac{6kn+3k^{2}+6kn^{2}+6k^{2}n+2k^{3}}{144}$. As $k$ is present in all terms of the numerator, setting $k=144$ guarantees the fraction to reduce to an integer. Therefore, this zeroes $\sin \frac{A-B}{2}$ in both the real and imaginary parts, so the original equation for periodicity must be satisfied. Hence, $k=144$ is a period. To find the fundamental period, we need only check casewise through its divisors, and we would find $k=12$.