## Sunday, November 04, 2007

### Damn that a_0!

I had to release my neck a minute ago, before I actually did become a figment of everyone's imagination! The redness is still there, and my heart rate has only just calmed down.

It is unfortunate and necessary, that we be examined. In an ideal world this wouldn't happen, however in a sense this examination is a good thing. I mean would I ever intentionally try to make an effort to understand Fourier Series if I wasn't going to be examined on them?

The answer to that question (before you guess incorrectly!) is yes I would want to understand them! I don't deny disliking the course at times, but never have I denied its right to be called interesting. And it is due to this interestingness of it, that I would actually want to understand it.

Anyway, I will now ramble incoherently for a few minutes about Fourier Series. Any periodic function (regardless of its behaviour) can be represented by a Fourier series. A Fourier series is basically a trigonometric series of the form:

$f(t) = \displaystyle\frac{a_0}{2} + \displaystyle\sum_{n=1}^{\infty} (a_n \cos (nt) +b_n \sin (nt))$,
where the function has period 2\pi.

The coefficients a_n (n=0,1,...) and b_n (n=1,2,...) are known as Fourier coefficients, and the aim of the game is calculate them. They are found by the following integrals:

$a_0 =\displaystyle \frac{1}{\pi} \int^{2\pi}_{0} f(t) dt$

$a_n =\displaystyle \frac{1}{\pi} \int^{2\pi}_{0} f(t) \cos (nt) dt$

$b_n =\displaystyle \frac{1}{\pi} \int^{2\pi}_{0} f(t) \sin(nt) dt$.

We can easily verify the above. For a_0 just integrate the series from the 0 to 2\pi. For a_n, first multiply both sides by cos(mt) and then once again integrate everything from 0 to 2\pi. (Note the orthogonality relations which I sort of mentioned here, need to be used. Consider the case when m=n \neq 0) For b_n multiply the series by sin(mt) and proceed in a similar way.

It is potentially a tedious job to do, and that is why we are called students and asked to do so!

The one thing that had me scratching my head and maybe strangling myself (on this occasion!) was the a_0 term. Why was it there I asked myself? I am pretty sure that the lecturer did mention this, but I am also equally sure that I wasn't listening at that time! Today the message hit home. Sin and cosines are sort of(!) symmetric - well if you were take the average value of a function with sines and cosines, it would be zero. However, if our function (which we wanted to find a Fourier series for) was non-negative so had a positive average value, we would need that constant!

But there is another big BUT. I understand that, and after doing all the hard work i.e. horrible integrals (read evil evil twice by parts integrals!) I make a silly mistake. If a_0 = 1, I write f(t)= 1+ \sum....

Has that induced some banging? I forget to divide the darn thing by 2! It's a_0/2 - not a_0. GAH. Consequently, when the question asks about convergence, I am a dead duck. I am hoping that the redness around my neck sticks around tomorrow, so that I don't make this dodo mistake again. I am sure that the redness can be reintroduced (if worse comes to worse).

By the way, please excuse any mathematical mumbo jumbo in this post (corrections welcome!) I am only half way through this night - got lots left I am afraid. I also think I should mention that it has been a whole year since I have been camping on the floor! It was reading week last year when I decided that I would no longer move things from the bed to the floor, from the floor to the bed, from the bed to the floor (...), and instead would sleep on the floor. Little did I know that a year on I would still be on the floor. Now to half-series and other coordinate series. What joy!

Беанс said...

BTW - I haven't really been strangling myself! (Just a lot of wall-banging).

Jake said...

"Anyway, I will now ramble incoherently for a few minutes about Fourier Series. Any periodic function (regardless of its behaviour) can be represented by a Fourier series."

But don't forget that the Fourier Series won't neccesarily converge to the function although I guess Dirichlet's Theorem gives a heck of a lot of functions to work with (and information about the value (of the series) at finite discontinuities)

Беанс said...

But doesn't the question of convergence only arise once we have a function represented with a Fourier series?

Once we have a series it can do three things - we only "care" for when it converges, hence Dirichlet's conditions.

(His names pops up in analysis too (duh), but I am sure of reading it in my notes!)