Non-trigonometric Continuous Periodic Functions I've seen lots of examples of periodic functions, but they all have one thing in common: They all involve at least one trigonometric term (e.g. $\sin\theta$, $\cos\theta$, etc.). My question is simple: are there any continuous, differentiable periodic functions that do not involve trigonometric terms? If not, why?
 A: The simplest infinitely differentiable non-trigonometric* function I can think of is $$f(x)=\sum_{n\in\mathbb Z} e^{-(x-n)^2}\tag{1}$$
Periodicity is clear; differentiability follows from the fact that every derivative of $e^{-x^2}$ is of the form $p(x)e^{-x^2}$ for some polynomial $p$, and   the series $$\sum_{n\in\mathbb Z} |p(x-n)| e^{-(x-n)^2}$$ converges uniformly on every bounded interval. 
The function (1) is sometimes called the periodized Gaussian, although it seems that the same term is used for the nondifferentiable functions obtained by taking a central piece of Gaussian curve and repeating it. 
(*) Not-explicitly-trigonometric. As others said, there is always a trigonometric series lurking in background.
A: One insight of Fourier was that more-or-less every periodic function should be expressible as a Fourier series $a_o+\sum_{n\ge 1} (a_n \sin nx+b_n\cos nx)$. The quality of the convergence depends on the smoothness of the function in question, unsurprisingly. So, apart from technicalities, every periodic function is so-expressible.
At the same time, it is very interesting to "make" periodic functions by taking something like a Gaussian $e^{-\pi x^2}$ (as in @NormalHuman's answer) and "wind up" by summing translates, to force periodicity. With a reasonable function $f$ (for example, in the Schwartz class: infinitely differentiable and it and all derivatives are rapidly decreasing), there is "Poisson summation"
$\sum_{n\in \mathbb Z} f(n) = \sum_{n\in \mathbb Z} \widehat{f}(n)$, where $\widehat{f}$ is Fourier transform. This well-known and easily Google-able relation follows from the general expressibility of periodic functions by Fourier series. Among other things, this gives an expression for Fourier coefficients of by-force-periodic functions: by basic change-of-variable properties of Fourier transform,
$$
\sum_{n\in \mathbb Z} f(n+x) \;=\; \sum_{n\in\mathbb Z} \widehat{f}(n)\,e^{2\pi inx}
$$
A charming identity, in my opinion. :)
A: Take
$$f(\theta)=\theta^4-2\pi^2\theta^2$$
for $-\pi\le x\le\pi$, and extend periodically.  This is not explicitly a trig function, though as pointed out in a comment, by using Fourier series it can be written as an infinite sum of trig functions.
Also notice that while the derivative of $f$ exists, its third derivative does not.  If you want an example which is arbitrarily often differentiable you will need something more intricate.
A: Well first of all, I should point out that a function does not need to be defined algebraically in order to be considered a valid function; it only needs to be defined in a logically-consistent way such that every value in its domain maps to a single range value. So I could say something like:

Let f(x) be the function of x which is 0 at even integers, 1 at odd integers, and for non-integer values of x, interpolates linearly between the two surrounding integral points to give a value between 0 and 1.

I don't think that's what you meant by "function", however.
According to Wikipedia (https://en.wikipedia.org/wiki/Periodic_function#Definition):

A function f is said to be periodic if, for some nonzero constant P,
  it is the case that
$$f(x+P)=f(x)$$
for all values of x in the domain.

The function f(x) = 0 fits that definition, and it is clearly continuous as well.
However, you probably meant "non-constant" as well. If you add that requirement, you can still use Euler's formula:
$$e^{ix}=\cos x+i\sin x$$
Which allows you to rewrite the sine and cosine functions in non-trigonometric terms:
$$f(x)=\Re(e^{ix})=\cos x$$
$$f(x)=\Im(e^{ix})=\sin x$$
But if you consider that to be cheating, there's one more solution I can think of: using the modulo operation, you can make anything periodic. So all you need to do is find a function g, having two points a and b in its domain where a≠b, but g(a) = g(b). Then the following will be a continuous, periodic, possibly non-trigonometric function:
$$f(x)=g([x\bmod(b-a)]+a)$$
If g is an even function, finding a and b is trivial: simply pick any nonzero real number for b, and its negative for a. For instance:
$$g(x)=x^2$$
$$(a,b)=(1,-1)$$
$$f(x)=[(x\bmod2)-1]^2$$
There are probably other answers as well, but that's what I can think of.
