# How can a Fourier expansion contain all the same info as original f'n?

1. Jul 31, 2014

### thecommexokid

We know that a function f(x) over an interval [a, b] can be written as an infinite weighted sum over some set of basis functions for that interval, e.g. sines and cosines:
$$f(x) = \alpha_0 + \sum_{k=1}^\infty \alpha_k\cos kx + \beta_k\sin kx.$$
Hence, I could provide you either with the function f(x) or with the complete set of Fourier coefficients {αk, βk}, and you would have all the same information over the interval [a, b].

But a function over an interval is an uncountably infinite amount of information: you need the value of f(x) for every real number x ∈ [a, b]. Whereas the set of Fourier coefficients is a countably infinite amount of information: you need only the value of αk and βk for every natural number k.

So how can a countable set of Fourier coefficients encode all the same information as a function over an uncountable range of real numbers?

2. Jul 31, 2014

### olivermsun

Well, step back a moment and ask how the real numbers can be constructed from the (countable) subset of rational numbers.

3. Jul 31, 2014

### thecommexokid

I don't know.

(Googling "construct real numbers from rational numbers" brings up a lot of sources, but I've never had a proper course in real analysis so I don't really understand most of the explanations.)

Separately, I fail to see the mechanism by which this would help me to understand my original question.

4. Jul 31, 2014

### micromass

Good question.

There are two issues here that explain what happens. The thing is that giving a function is not the same as giving the Fourier coefficients. Indeed, the first reason why not is that two distinct functions might have the same Fourier coefficients. Second (and most important), not every function has a collection of Fourier coefficients to begin with and even if they do, the convergence might not be nice.

So it turns out that only a special kind of function can be "decomposed" in its Fourier coefficients. And this resolves your paradox, since these special kind of functions can be described perfectly by a countable number of information. Totally arbitrary functions indeed have an uncountable number of information, but one would expect for such function that the values at distinct points are totally independent. This is thus clearly not so for the very special functions which have Fourier coefficients, otherwise we would need uncountably much information.

5. Jul 31, 2014

### olivermsun

Re: Separately

If you can construct a complete set of real numbers on some interval [a, b] from only the (countably infinite) rational numbers, then isn't there an analogy with (re)constructing a continuous real function over [a, b] using only a (countably infinite) set of real coefficients?

6. Jul 31, 2014

### olivermsun

Well, I think it's fair to start with the assumption that the function above has a Fourier series, or else the equality could not hold. I agree that various ways in which the Fourier series can be said to converge (related to how literally you want to take the equality shown above) are an interesting question.

If it exists, the Fourier series can fail to converge on a set of measure zero, but I'm not sure how that helps answer the OP's question (but I'd be interested to hear any ideas on this).

7. Jul 31, 2014

### thecommexokid

Thanks for the reply; that helps. It had occurred to me just after posting that you obviously wouldn't be able to construct a Fourier expansion for a function where f(x) is selected randomly for every x. On the other hand, the condition for Fourier expansion clearly is not as strict as demanding f(x) be continuous, since step functions and such things are one of the first examples of Fourier expansions that students ever calculate.

My guess for the condition on what you call the "very special functions" such that they can be expressed as a Fourier expansion is that they have a countable number of discontinuities on [a, b]. Is that right? Or right-ish? If not, what is the condition for a function to be "very special"?

8. Jul 31, 2014

### olivermsun

There's a whole article about it at Wikipedia: Convergence of Fourier series.

The funny thing is that the Fourier series for continuous functions do not necessarily converge everywhere (but they do "almost everywhere," which allows you to exclude, e.g., a countable set of points).

However, a function of bounded variation is sufficient for convergence everywhere.

This has a parallel to the case of Fourier transforms of discrete (sampled) functions, in which case the Fourier transform is actually unique for band-limited functions.

9. Jul 31, 2014

### micromass

So I have deliberately left out the exact condition. It basically comes down to demanding that the integrals

$$\int_{-\pi}^\pi f(x)\cos(nx)dx~\text{and}~\int_{-\pi}^\pi f(x)\sin(nx)dx$$

make sense. This is not so intuitive as demanding things like continuity.

But even for those functions convergence issues are still very subtle and problematic.

10. Jul 31, 2014

### thecommexokid

I've been off reading the article. Obviously a lot of the details depend on knowledge of real analysis I just don't have. So I won't try to tackle the issue in full generality, but instead consider a relatively simpler case. In the "Pointwise convergence" section, the article says

I take from this statement that any function f(x) that is differentiable at every point in [a, b] except for a finite number of jump discontinuities should have a unique Fourier expansion. (Obviously that is not a necessary condition on f(x), but if Wikipedia is to be believed, it is a sufficient condition.) Is this correct?

If so, then let $\mathscr{F}$ be the above set of all piecewise-differentiable functions over the interval [a, b]. Given the above, I ought to be able to pick out a single f(x) ∈ $\mathscr{F}$ with a countable amount of information. Is there a straightforward way for me to see that this should be possible?

Last edited by a moderator: May 6, 2017
11. Jul 31, 2014

### olivermsun

I think if functions $f, g \in \mathscr{F}$ have the same Fourier series, and the Fourier series converges point wise, then the two functions have to be point wise equal.

12. Jul 31, 2014

### thecommexokid

I agree. I was just wondering if there's a simple way for me to understand how a piecewise-differentiable function can be represented with a countable amount of information, other than by reference to their Fourier series. Perhaps not, but I thought I'd ask.

13. Jul 31, 2014

### mathman

A function is equal to its Fourier expansion only at all points of continuity of the function. At simple discontinuities (jumps) it takes on the average value.
To address your concern, the definition of a continuous function requires it be given at a countable (dense set) number of points - the remaining values of the function are then determined by continuity.
To illustrate: Let f(x) = 1 for irrational x, f(x) =0 for rational x. The Fourier series will end up as the series for the constant g(x) = 1.

14. Jul 31, 2014

### micromass

Things are a bit more subtle than this. A fourier series does not need to converge at a point of continuity (although it converges at most points of continuity: http://en.wikipedia.org/wiki/Carleson's_theorem). Differentiability however does work.

15. Jul 31, 2014

### jbunniii

That's true, but it is worth noting that at every point of continuity, the function value can be recovered from the Fourier coefficients using Cesaro summation (Fejer's theorem).

Therefore for any continuous function, the Fourier coefficients do indeed tell us everything we need to know to reconstruct the function.

This should not come as any surprise: we can also reconstruct a continuous function if we only know $f(x)$ for rational values of $x$.

For suitably nice functions, we we can reconstruct the entire function even if we only know its values in a tiny neighborhood of some point, or if we know its (countably many) derivatives at a single point: e.g. Taylor series.

16. Aug 1, 2014

### WWGD

The infinitely-many values of a linear transformation from, say, R^n to R^k are determined by nk values in a representing matrix. The output is infinite , but the basis/generating set is/are finite. And a continuous function on the Reals is uniquely-determined by its value on the Rationals. So the issue is, I believe, that an uncountable set can have a countable (chowder) basis. Besides, think of the map f(x)=x (a subcase of example with linear maps).

Last edited: Aug 1, 2014