Hard fourier series

1. Sep 15, 2011

saxen

1. The problem statement, all variables and given/known data

Let f be the 2pi periodic function defined by f(x)=e$^{cos(x^{2})}$
for 0 < x < 2pi. What is the value of the Fourier series at x=4pi

2. Relevant equations

3. The attempt at a solution

I don't even know where to start.

All help is much appreciated!

2. Sep 15, 2011

lanedance

well it should be equivalent to the value of the function (can you think of any useful convergence theorems?), that is unless the function is misbehaving at that point...

3. Sep 15, 2011

Harrisonized

I'm pretty sure that f(x)=e^cos(x2) isn't a periodic function, but let's pretend that it is. If you want to generate the Fourier series for the interval 0<x<2π, then the series is going to loop itself every 2π. In other words, for the series, f(x)=f(x+2π). Therefore, f(0)=f(4π) (for the series only!). That means the only value there is to check is f(0), which equals e.

4. Sep 15, 2011

saxen

Hmm, maybe something about pointwise convergence?

5. Sep 15, 2011

saxen

Can't I also think like this:

f(4pi)=f(2pi + 2pi)=f(2pi)

6. Sep 15, 2011

saxen

Ok, the answer to this problem is

(1/2)*(e^cos(4pi^2)+e)

I am so confused. I offer beer and hugs in return for explinations.

7. Sep 15, 2011

Harrisonized

I'm confused too, but no matter. From the looks of it, the way they wanted you to make the Fourier series depended on both endpoints. Therefore, the "answer" (if it can be called that) is the average of f(0) and f(2π). *shrugs*

8. Sep 15, 2011

lanedance

9. Sep 15, 2011

saxen

Thanks alot, I understand now! Its very logical when you think of it. You can claim your reward anytime.

10. Sep 15, 2011

Dickfore

The value of the Fourier series at [itex[x = 4 \pi[/itex] is the same as the value at $2 \pi$ as well as $0$. If we perform the Fourier series expansion of a function that has a finite jump at some point $x = a$, then it will have the value:
$$\frac{f(a - 0) + f(a + 0)}{2}$$
$$[tex]\frac{f(2\pi - 0) + f(+0)}{2}$$