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I am wondering whether there exists a class of oscillating functions that are distinct from trigonometric functions. The only oscillating functions I could think of are of the $e^{ix}$ and $(-1)^x$ varieties, but these are easily expressed as trigonometric functions (or sums thereof). I'm looking for functions that cannot be expressed as finite sums of trigonometric functions (but functions that are not themselves finite sums either).

I'm not sure how best to phrase this, but I'm looking for non-trivial answers to this. Its trivial to find a function that has the same value at two distinct x-values and tile it infinitely. I'm looking for something that can be expressed in a simpler way than an infinite piece-wise function.

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Do you allow infinite sums of trigonometric functions? Are you familiar with Fourier series? What about a square wave, or sawtooth wave? –  Potato Oct 4 '12 at 23:29
What do you mean by "oscillating"? Fourier analysis shows that every "nice" periodic function can be written as an (infinite) sum of trigonometric functions. –  Henning Makholm Oct 4 '12 at 23:30
@Potato Yes, I am familiar. I should've specified; something that cannot be written as a finite sum of trigonometric functions. Updating the question.. –  Aaron Dufour Oct 4 '12 at 23:32
The Stone-Weierstrass theorem ensures that the set of trigonometric polynomials are dense in the set of continuous functions in the uniform norm, which basically means that for every continuous function and a chosen very small error, you can find a sum of trigonometric terms which approximates that function with that error everywhere. –  Ragib Zaman Oct 4 '12 at 23:32
@AaronDufour Functions that can be written as finite sums of trigonometric functions have a finite Fourier series - finding the fourier series "exposes" if it can be written as a finite sum of trig functions. So a sawtooth wave will never be able to be written as a finite sum of trig functions. Another way to see that is to note that any finite sum of trig functions is smooth while a sawtooth has non-differentiable points. –  Ragib Zaman Oct 4 '12 at 23:34
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6 Answers

up vote 1 down vote accepted

The graph of $f(x) = x \pmod n$ for any integer $n$ is periodic. In case you are not familiar with modular arithmetic, $f(x)$ is the remainder of $x$ after division by $n$. As an example, here is $f(x) = x \pmod 5$, courtesy of WolframAlpha:

x mod 5

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This is what I was looking for! Something that I'm familiar with but completely blanked on. Thanks –  Aaron Dufour Oct 5 '12 at 6:19
This has an infinite Fourier expansion due to having only one discontinuity per cycle. –  Ross Millikan Jun 11 '13 at 18:15
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What about a "saw" function, with graph something like:

   .     .     .
  . .   . .   . .
 .   . .   . .
.     .     .

and then extended in the obvious way? But of course this can be represented via infinite sums of trigonometric functions ... that is called Fourier theory

A formula for the above saw function might be $$ f(x) = \begin{cases} x, & \text{if $0\le x \le 1$} \\ 2-x, & \text{if $1 \le x \le 2$} \end{cases} $$ and then extended periodically.

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How can this be expressed algebraically? –  Aaron Dufour Oct 4 '12 at 23:38
updated the answer to answer that. –  kjetil b halvorsen Oct 4 '12 at 23:45
You may want to adjust the intervals so that $f(1)$ results in 1 value only. –  Emmad Kareem Oct 4 '12 at 23:56
Like in the other answer, I'm looking for a non-trivial expression. In looking for something periodic, its trivial to tile a function over a short range infinitely. Can this be expressed more simply? (Something to do with floor/ceiling functions, perhaps?) –  Aaron Dufour Oct 4 '12 at 23:58
@AaronDufour Periodic means tile a function infinitely. Trigonometric functions don't appear to be defined in this way, but the fact that they discard multiples of $2\pi$ is equivalent to just defining it on $[0, 2\pi]$ and then tiling. –  Austin Mohr Oct 5 '12 at 2:59
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Let $f(x) = e^{-1/x(x-1)}$ for $0<x<1$ and $f(0)=f(1)=0$. Extend periodically to all of $\mathbb R$.

The result is $\mathcal C^\infty$ and (by construction) periodic, but it is not a finite sum of trigonometric functions. Such a sum must be analytic but our $f$ is not analytic at the integers.

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Can this be expressed algebraically? –  Aaron Dufour Oct 4 '12 at 23:38
I'm not sure what you mean by that. You can certainly frame the result with words that belong to algebra (the set of horizontally scaled trigonometric functions does not algebraically span the vector space of periodic functions), but particularly enlightening it ain't. –  Henning Makholm Oct 4 '12 at 23:39
The obvious way to express what you've written is as an infinite piecewise function. Can it be expressed more simply than that? –  Aaron Dufour Oct 4 '12 at 23:41
My point is that you can do this for any function that has the same value at two different x values, by tiling them infinitely. Can this one be represented in a simpler way than that? –  Aaron Dufour Oct 4 '12 at 23:56
Generally periodic extension for "any function that has the same value at two different $x$ values" will not produce a $\mathcal C^\infty$ result. –  Henning Makholm Oct 4 '12 at 23:59
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Mostly what comes to mind is BESSEL FUNCTIONS, of which there are many kinds. Note that $$ f(x) = \frac{\sin x}{x} $$ extends with $f(0) = 1$ to an function that oscillates but cannot be written as a sum of sines and cosines.

If you are insisting on genuinely periodic functions, you are out of luck, as such will have Fourier series if piecewise continuous.

On the other hand, here is one that has roots at constant intervals and constant amplitude, $$ g(x) = \left( 2 + \cos \left( \frac{8 x^2}{\pi} \right) \; \right) \; \sin x $$ but has no Fourier series. Would I lie?

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To extend Austin's answer: as I have noted here, one could use the sawtooth function $x\bmod 1$ to represent any periodic function on the real line, if you know the "repeating unit". To summarize the point of that answer: if you want your function to be a repeated version of the function $f(x)$ over the interval $[a,b]$, and $f(a)=f(b)$, then $f\left(x\bmod(b-a)\right)$ is the function you want. One can then choose an $f$ that involves no trigonometric functions whatsoever.

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If you extend the (-1)^x , where x is an unknown function,to coin-flips, or random {-1,1}.The series, S=∑{-1,1} can be a non-symmetric random walk with zeros proportional to the geometric mean of #+1's and #-1's. There are 2^n of these "functions". It may be impossible to express them algebraically or as trigonometric functions. They could be a "function" if stored on a computer.( I use 1e6 digits of pi as a source.)

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