Solving Hard Fourier Series: Let f = e^(cos(x^2)) for 0 < x < 2pi

[saxen]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I don't even know where to start.

All help is much appreciated!

 

[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...

 

[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.

 

[saxen]

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...

Hmm, maybe something about pointwise convergence?

 

[saxen]

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.

Can't I also think like this:

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

 

[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.

 

[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*

 

[lanedance]

so the function isn't period, but we take the periodic extension of the portion on [0,2pi)

and at a discontinuity Fourier series will converge to the average of the endpoints as you have found
http://tutorial.math.lamar.edu/Classes/DE/ConvergenceFourierSeries.aspx

note that even though the series converges to the average of the endpoints, it will still have an "overshoot" at any discontinuity regardless of how many terms you take in the series.
http://en.wikipedia.org/wiki/Gibbs_phenomenon

 

[saxen]

so the function isn't period, but we take the periodic extension of the portion on [0,2pi)

and at a discontinuity Fourier series will converge to the average of the endpoints as you have found
http://tutorial.math.lamar.edu/Classes/DE/ConvergenceFourierSeries.aspx

note that even though the series converges to the average of the endpoints, it will still have an "overshoot" at any discontinuity regardless of how many terms you take in the series.
http://en.wikipedia.org/wiki/Gibbs_phenomenon

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

 

[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:
$$\frac{f(a - 0) + f(a + 0)}{2}$$
In your case:
[tex] $$\frac{f(2\pi - 0) + f(+0)}{2}$$<br /> Calculate this limit.[/tex]