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

In summary, the Fourier series at [itex[x = 4 \pi[/itex] is the same as the value at 2 \pi as well as 0.
  • #1
saxen
44
0

Homework Statement



Let f be the 2pi periodic function defined by f(x)=e[itex]^{cos(x^{2})}[/itex]
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!
 
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  • #2
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
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
lanedance said:
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?
 
  • #5
Harrisonized said:
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)
 
  • #6
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
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
  • #9
lanedance said:
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 alot, I understand now! Its very logical when you think of it. You can claim your reward anytime.
 
  • #10
The value of the Fourier series at [itex[x = 4 \pi[/itex] is the same as the value at [itex]2 \pi[/itex] as well as [itex]0[/itex]. If we perform the Fourier series expansion of a function that has a finite jump at some point [itex]x = a[/itex], then it will have the value:
[tex]
\frac{f(a - 0) + f(a + 0)}{2}
[/tex]
In your case:
[tex]
[tex]\frac{f(2\pi - 0) + f(+0)}{2}
[/tex]
Calculate this limit.
 

1. What is a Fourier series?

A Fourier series is a mathematical tool used to represent a periodic function as a sum of sine and cosine waves of different frequencies. It is named after the French mathematician Joseph Fourier.

2. What is the purpose of solving a Fourier series?

The purpose of solving a Fourier series is to approximate a periodic function with a simpler combination of sine and cosine functions. This can be useful in applications such as signal processing and image compression.

3. How do you find the coefficients of a Fourier series?

The coefficients of a Fourier series can be found using the Fourier series formula, which involves integrating the function over its period and multiplying by certain trigonometric functions. In some cases, the coefficients can also be found using symmetry or other properties of the function.

4. Why is e^(cos(x^2)) a particularly hard function to solve for a Fourier series?

e^(cos(x^2)) is a particularly hard function to solve for a Fourier series because it is not a simple combination of sine and cosine functions. It is also an exponential function, which can be difficult to manipulate and integrate.

5. What is the domain and range of the function f = e^(cos(x^2)) for 0 < x < 2pi?

The domain of f = e^(cos(x^2)) for 0 < x < 2pi is 0 < x < 2pi, as specified in the given problem. The range of the function is all real numbers greater than or equal to 1, since e^(cos(x^2)) is always positive and has a minimum value of 1.

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