# Are sin and cos the only continuous and infinitely differentiable periodic functions we have?

Sin and cos are everywhere continuous and infinitely differentiable. Those are nice properties to have. They come from the unit circle.

It seems there's no other periodic function that is also smooth and continuous. The only other even periodic functions (not smooth or continuous) I have seen are:

• Square wave
• Triangle wave
• Sawtooth wave

Are there any other well-known periodic functions?

-
There are plenty. Just construct a nice $f$ on $[a,b]$ to have the same values and same derivatives at $x=a$ and $x=b$. Then extend'' $f$ to all of $\Bbb R$. –  David Mitra Nov 12 '11 at 19:10
Note that smoothness implies continuity. How are your three examples smooth? Or do you mean piecewise smooth? –  wildildildlife Nov 12 '11 at 19:11
@wildildildlife Sorry, I meant square, tri, sawtooth are the only other common periodic functions I have seen. –  bobobobo Nov 12 '11 at 19:15
$5\sin(x) + 7\cos(2x)$ is neither $\sin$ nor $\cos$. –  GEdgar Nov 12 '11 at 19:16
or something like $\frac{\sin(x)}{1+\cos^2(x)}$... –  N. S. Nov 12 '11 at 19:41

"Are there any other well-known periodic functions?"

In one sense, the answer is "no". Every reasonable periodic complex-valued function $f$ of a real variable can be represented as an infinite linear combination of sines and cosines with periods equal the period $\tau$ of $f$, or equal to $\tau/2$ or to $\tau/3$, etc. See Fourier series.

There are also doubly periodic functions of a complex variable, called elliptic functions. If one restricts one of these to the real axis, one can find a Fourier series, but one doesn't do such restrictions, as far as I know, in studying these functions. See Weierstrass's elliptic functions and Jacobi elliptic functions.

-
....but I haven't said what "reasonable" means nor mentioned the various senses in which a Fourier series might or might not converge. –  Michael Hardy Nov 12 '11 at 19:16
On the other hand, the elliptic functions are (necessarily!) meromorphic, while the exponential and trigonometric functions are periodic and holomorphic. One could think of the usual singly-periodic functions as "degenerate" doubly-periodic functions with one infinite period... –  Ｊ. Ｍ. Nov 12 '11 at 22:58

There are some unreasonable ones that are periodic.

$$f(x) =\sum_{n=-\infty}^\infty (-1)^{n}e^{-(x-n)^2}$$

is periodic with period 2. Looks like cosine but is not.

-
I'm not sure this one is "unreasonable". Doesn't it have a convergent Fourier series? –  Michael Hardy Nov 13 '11 at 4:32

This is an extension of David Mitra answer.

Pick any functions which is smooth on an open interval containing $[0,1]$.

Define $g: [0,1] \rightarrow R$ by $g(x)=x^2(1-x)^2f(x)$. Then $g(0)=g'(0)=g(1)=g'(1)=0$.

You can now prove that $h: R \rightarrow R$ defined by

$$h(x) = g( \{ x \}) \,,$$

where $\{ x \}$ represents the fractional part of $x$ is smooth periodic with period 1...

-
That's not smooth! For $f(x)=1$, the third derivative jumps at every integer. However, you could make $g(x)=e^{\left(\frac{-1}{x(1-x)}\right)}f(x)$. –  Henning Makholm Nov 13 '11 at 3:35

Of course not.

For example, $\sin_{[n]}(x)$ as shown in http://en.wikipedia.org/wiki/Functional_square_root is in fact a smooth periodic function of period $2\pi$ $\forall n\in\mathbb{R}^+$ .

-