• Support PF! Buy your school textbooks, materials and every day products Here!

Fourier series and even extension of function

  • #1
184
0
I'm trying to solve this exercise but I have some problems, because I haven't seen an exercise of this type before.

"[itex]f(x)= \pi -x[/itex] in [itex][0, \pi][/itex]

Let's consider the even extension of f(x) in [itex][-\pi, \pi][/itex]
and write the Fourier Series using this set ( [itex]\frac{1}{\sqrt{2 \pi}}, \frac{1}{\sqrt {\pi}} \cos nx[/itex] )
and use the Parseval identity to prove:
[itex]\sum \frac{1}{(2k+1)^2}= \frac{\pi ^2}{8}[/itex]"

My attempt:

[itex]a_0= \frac{1}{\sqrt {2 \pi}} \int_{-\pi}^{\pi} f(x)=\frac{1}{\sqrt {2 \pi}} \int_{-\pi}^{\pi} (\pi -x) dx= \frac{2 \pi^2}{\sqrt {2\pi}}[/itex]

[itex]a_n= \frac{2}{\sqrt \pi}\int_{0}^{\pi} f(x)\cos nx dx=\frac{2}{\sqrt \pi} \frac{1-cos \pi n}{n^2}= \frac{2[1+(-1)^n]}{ \sqrt {\pi n^2}}[/itex]

So,

[itex]f(x)=\frac{2 \pi^2}{\sqrt {2\pi}}+\sum_{n=1}^{\infty}\frac{2[1-(-1)^n]}{ \sqrt {\pi n^2}} \frac {\cos nx}{\sqrt \pi}[/itex]

Could you tell me if these steps are correct? Are the formulas for Fourier coefficients correct?


For the second part of the exercise, I have and enormous doubt:
It was told me that in similar cases, Parseval Identity is:

[itex]\int_{-\pi}^{\pi} |f(x)|^2 dx= a_o^2+ \sum |a_n|^2[/itex]

I don't know how to do [itex]|a_n|[/itex], because there is the term [itex](-1)^n[/itex]..
And I have a lot of doubts to obtain the result [itex]{\pi}^2/8[/itex] using the results that I have obtained...

Many thanks for your precious help!
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,584
1,206
No, your results aren't correct. The ##n^2## shouldn't be inside the square root in ##a_n##. It also doesn't look like you integrated correctly. In your expression for ##a_0##, you're integrating over the entire interval, but you didn't use the even extension for f(x). Finally, the constant term in the series isn't correct. It should be of the form ##a_0/\sqrt{2\pi}## where ##1/\sqrt{2\pi}## is there because it's the basis function.

You can simplify the expression for the series a bit by considering what happens when ##n## is even and when ##n## is odd.
 
Last edited:
  • #3
184
0
you're right.... :/ many thanks!
 

Related Threads on Fourier series and even extension of function

  • Last Post
Replies
2
Views
7K
  • Last Post
Replies
4
Views
2K
Replies
15
Views
2K
Replies
1
Views
498
  • Last Post
Replies
1
Views
2K
Replies
10
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
466
  • Last Post
Replies
1
Views
2K
Replies
12
Views
18K
Top