# Equality of fouriers expression

1. Mar 23, 2010

### samreen

When we are resolving a periodic piecewise continuous function in a fouriers series of harmonic terms, as follows: f(x)= a0/2 +Σancos nΠx/L + Σ bnsin nΠx/L, period 2L

we are essentially expressing a sectionally discontinuous function (with finite no of such finite discontinuities, if at all) by a continuous function, arent we? As in the series is assuming the mean value of the the function at evry point, to define itself. Correct? So how does the equality hold, if you consider the definition rigorously? Im at a loss here :’(

2. Mar 23, 2010

### Eynstone

By Fejer's theorem, the series converges to (1/2)(f(x+) + f(x-)) at a discontinuity ( & not f(x)).

3. Mar 23, 2010

### samreen

exactly. thats what i mean by 'assuming mean value at every point'. so the series IS defined at discontinuous points of the function, or points of no definition. so i return to my question: how does the equality hold?

4. Mar 29, 2010

### jostpuur

Your question "how does an equation hold?" is not very clear.

It holds so that the left and right side are equal?

Anyway, if you were interested in some rigor assumptions that guarantee the convergence, here's something:

http://mathworld.wolfram.com/DirichletFourierSeriesConditions.html

When I read about convergence from one book, it used an assumption about bounded variation property, but it could be it is the same thing as this finite number of extrema property. You can formulate the assumptions in different ways.

5. Mar 29, 2010

### Mute

A sum of infinitely many analytic functions may indeed by a non-analytic function. This is the crux of how a partition function in statistical mechanics can represent more than one phase.

Of course, this doesn't necessarily resolve your question, because, for example, it is well known that at the discontinuous points the fourier series overshoots the value of one by some amount. This is known as the Gibbs phenomenon.

In any event, it may be best to consider the fourier series not as being equal to the original function, but as being a representation of it, which turns out to be perfect for continuous functions but has some imperfections for discontinuous ones.

6. Mar 29, 2010

### jbunniii

Very simple, equality does NOT hold at points of discontinuity. That is, if

$$a_n = \int f(x) e^{-inx} dx$$

it isn't necessarily true that

$$f(x) = \sum_n a_n e^{inx}$$

for all $x$.

In fact, it isn't necessarily true if $f$ is continuous. The best that can be said in general is that if $f$ is continuous (or more generally, $f \in L^p$ for some $1 < p < \infty)$, then its Fourier series converges to $f(x)$ for almost every $x$. That result was not known until the 1960s (Carleson's theorem).

Last edited: Mar 29, 2010
7. Mar 29, 2010

### jostpuur

You shouldn't state claims like this. You never know if there is still something that you don't know about yet.

There exist assumptions which will guarantee point wise convergence too. If $f$ is continuous and has the bounded variation property at some environment of $x_0$, then the Fourier inverse transform will converge towards the value $f(x_0)$.

8. Mar 29, 2010

### jbunniii

I should have worded it more carefully. Here is what I meant to say: Even a continuous function isn't guaranteed (without further assumptions) to have a pointwise convergent Fourier series, although almost-everywhere convergence is guaranteed by Carleson-Hunt.

9. Mar 30, 2010

### samreen

i looked up the bit about the overshoot (gibbs penomenon)..thanks, was interesting. and i'm absolutely sure about the answer to my question now. thanks again, everyone! the equality does NOT hold rigorously, if you want to be a purist about it. in fact, i went back and looked up quite a few books which have chosen to do away with the equality, and instead represent the series by an approximation ( as in ~ rather than =). look up Ross, if thats a help.