Fourier Series and the first term

[rdfloyd]

I wasn't really sure where to post this because I am covering this in 2 classes (Math and Physics). Figured this would be my best bet.

The Fourier series of some Function is $a_{0}/2+etc...$. I've looked in several textbooks but none explain why the $1/2$ is there, and not in any of the other terms of the summation.

I do have a homework problem concerning this, but my professor said it's ok to not explain this part of the Fourier series. I'm intrigued by this now, so I'd like to know.

 

[micromass]

Remember that the general term of a Fourier series is

$$<f,e_i>$$

where the $e_i$ is normalized (has norm 1)

We want to define

$$a_n=\frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos(nx)dx$$

If n>0, then this holds. The $1/\pi$ comes from normalizing the cosine. That is, the above is actually equal to

$$<f,\cos nx>$$

but cos(nx) does not have norm 1, but rather pi. So we must divide by pi to normalize.

We want the formula for a_n to hold for n=0 as well. But in this case, we have

$$<f,1>$$

and the 1 is not normalized and has norm 2pi. So in order to normalize the thing, we must divide by 2pi. Division by pi is already taken care of in the definition of a_n, so we must also divide by 2.

 

[rdfloyd]

So, why doesn't the 2 tag along with the rest of the terms? In this case, it seems like there is a piecewise function under conditions n=0 and n>0.

Sorry if I'm asking a dumb question.

 

[Office_Shredder]

Integrating from -pi to pi:
The function f=1 yields 2pi
The function f=cos²(nx) yields pi.

 

[mathman]

Integrating from -pi to pi:
The function f=1 yields 2pi
The function f=cos(nx) yields pi.

You should have cos²(nx), not cos(nx).

 

[Office_Shredder]

You should have cos²(nx), not cos(nx).

Thanks. I edited my original post to avoid confusing anybody

 

[paul2211]

I watched a video by MIT's OCW for 18.03 when I was doing this a while back, and the prof derived the Fourier series and explained the 1/2 quite well