Why must the angle contain $\pi$ for $\cos$ be periodic?

Question

The $\cos(\frac{n}{6})$ is aperiodic, and $\cos(\frac{n\pi}{6})$ is periodic, why? What role does the $\pi$ play? By the way, $n$ must be integer

I have seen this question,Why is cos(n/6) aperiodic? but the answer just show that it is not of the form $2\pi(mN)$,but it did't explain why should that $\pi$ exist,and the comment below said that we can't find the next larger (integer) value of n such that $\cos(\frac{n}{6})=1$ after $n=0$, however, yes we can, the $n=360 \times 6$ can let $\cos$ value back to the $0$ again. So, I want to ask why must the $\pi$ exist to let $\cos$ be a periodic?

Answer 1

I will reuse a former answer of mine too: Proof of complex conjugate symmetry property of DFT. It relates to how sines and cosines can be defined. One answer is: from the exponential, and thus derives $\pi$. There exist other constructions, this one is (imho) elegant.

The periodicity of the cosine comes from the fact it is defined as the real part of the cisoid, or complex exponential, $e^{ix}$. In W. Rudin's Real and Complex analysis (very first pages, 1 to 3 of the prologue), $e$ comes first, and $\pi$ appears subsequently.

One first defines for any complex $z$:

$$e^z=\sum_{n=0}^{+\infty} \frac{z^n}{n!}$$

which is an absolutely convergent series. It is its own derivative. And then you get some other results, like $e^z$ is never equal to zero. But the two most striking ones are:

There exists a positive number $\pi$ such that $e^{\pi j/2} = j$ and such that $e^z = 1$ if and only if $z/(2\pi j )$ is an integer.

<p><span class="math-container">$e^z$</span> is <span class="math-container">$2\pi j$</span> periodic.</p>

From this, you define the sine and the cosine as the imaginary and real parts. The proof is quite interesting. It for instance remarks that $\cos 2 <0$, and since $\cos 0 =1$, by continuity, there should exists a constant at which the cosine vanishes:

This should answer to Why does $\pi$ exist, with its role unveiled, and why the cosine is periodic.

The answer to the "next integer" is given by Matt's comment.

Answer 2

The confusion comes from the missing details that authors usually skip or falsely assumed about the mathematical background of their students.

Let us see why the first signal is aperiodic whereas the second one is periodic. The periodicity of a signal holds if we can show $x(n)=x(n+N)$, otherwise, the signal is nonperiodic. Let's see that by starting with $$ \begin{align} x(n+N) &= \cos( \frac{n}{6} + \frac{N}{6}) \\ &= \cos(\frac{n}{6})\cos(\frac{N}{6}) - \sin(\frac{n}{6})\sin(\frac{N}{6}) \end{align} $$ In order for $x(n)=x(n+N)$ to hold, $\cos(\frac{N}{6})=1$ and $\sin(\frac{N}{6})=0$. We search for the smallest value of $N$. This is true if $\frac{N}{6}=2\pi \implies N = 12\pi$. Indeed, if $N=12\pi$, we get $$ \begin{align} x(n+N) &= \cos( \frac{n}{6} + \frac{N}{6}) \\ &= \cos(\frac{n}{6})(1) - \sin(\frac{n}{6})(0) \\ &= \cos(\frac{n}{6}) \\ &= x(n) \end{align} $$ But $N$ must be a positive integer, therefore, the signal is nonperiodic. In other words, there is no such a positive integer so that $x(n)=x(n+N)$. Now let's prove the periodicity of the second signal. Simply start with $$ \begin{align} x(n+N) &= \cos( \frac{\pi n}{6} + \frac{\pi N}{6}) \\ &= \cos(\frac{\pi n}{6})\cos(\frac{\pi N}{6}) - \sin(\frac{\pi n}{6})\sin(\frac{\pi N}{6}) \end{align} $$ In order for $x(n)=x(n+N)$ to hold, $\cos(\frac{\pi N}{6})=1$ and $\sin(\frac{\pi N}{6})=0$. We search for the smallest value of $N$. This is true if $\frac{\pi N}{6}=2\pi \implies N = 12$. Indeed, if $N=12$, we get $$ \begin{align} x(n+N) &= \cos( \frac{\pi n}{6} + \frac{\pi N}{6}) \\ &= \cos(\frac{\pi n}{6})\cos(\frac{\pi N}{6}) - \sin(\frac{\pi n}{6})\sin(\frac{\pi N}{6}) \\ &= \cos(\frac{\pi n}{6})(1) - \sin(\frac{\pi n}{6})(0) \\ &= \cos(\frac{\pi n}{6}) \\ &= x(n) \tag{1} \end{align} $$ Eq.(1) proves the periodicity of $\cos(\frac{\pi n}{6})$.

As you can see, using trigonometric identities allow us to see clearly why the signal is periodic. It has nothing to do with $\pi$ rather we could find the smallest value of $N$ that allows us to reach $x(n)$ from $x(n+N)$. Again, these kinds of details are crucial but authors tend to ignore them.

Answer 3

Cuz today is Pi Day. Happy Pi Day everyone!

Seriously though, to answer this belated question in somewhat more of a simple intuitive way.

The period of the trig functions is $2\pi$. In order for a trig signal to be periodic (hitting the same domain values, repeatedly), it's RadiansPerSample factor has to be a rational (as in not irrational) factor of $2\pi$. Since $\pi$ is quite irrational, this means the argument has to be $ r \pi $ where $r$ is a rational number (ratio of integers, say $a/b$).

$\frac{1}{6}$ is not a rational multiple of $\pi$, but $\frac{\pi}{6}$ is.

Nuttin' to it.