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Fourier series and even extension of function

  1. Jan 20, 2015 #1
    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!
     
  2. jcsd
  3. Jan 21, 2015 #2

    vela

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    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: Jan 21, 2015
  4. Jan 25, 2015 #3
    you're right.... :/ many thanks!
     
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