Using a Fourier Cosine Series to evaluate a sum

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richyw
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Homework Statement



a) Show that the Fourier Cosine Series of [itex]f(x)=x,\quad 0\leq x<L[/itex] is
[tex]x ~ \frac{L}{2}-\frac{4 L}{\pi ^2}\left[\left(\frac{\pi x}{L}\right)+ \frac{\cos\left(\frac{3\pi x}{L}\right)}{3^2}+\frac{\cos\left(\frac{5 \pi x}{L}\right)}{5^2}+\dots\right][/tex]

b) use the above series to evaluate the sum[tex]1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...[/tex]

Homework Equations



Fourier Cosine Series General form

The Attempt at a Solution



So I have done part a, but I am lost on how to do part b. I don't understand how to get the even terms? Perhaps I need to differentiate the series, which would pull an n out, making the n's even.

To differentiate term by term a cosine series I need f'(x) to be piecewise continuous. Which I think it is.
 
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Fixed LaTeX:

richyw said:

Homework Statement



a) Show that the Fourier Cosine Series of [itex]f(x)=x,\quad 0\leq x<L[/itex] is
[tex]x \sim \frac{L}{2}-\frac{4 L}{\pi ^2}\left[ \cos\left(\frac{\pi x}{L}\right)+ \frac{\cos\left(\frac{3\pi x}{L}\right)}{3^2}+\frac{\cos\left(\frac{5\pi x}{L}\right)}{5^2}+ ...\right][/tex]

b) use the above series to evaluate the sum[tex]1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...[/tex]

Homework Equations



Fourier Cosine Series General form

The Attempt at a Solution



So I have done part a, but I am lost on how to do part b. I don't understand how to get the even terms? Perhaps I need to differentiate the series, which would pull an n out, making the n's even.

To differentiate term by term a cosine series I need f'(x) to be piecewise continuous. Which I think it is.
 
micromass said:
Fixed LaTeX:

thanks. I did too. I had to quickly reinstall LaTeXiT from macports :smile:
 
micromass said:
Anyway, what you need to do is choose ##x## (and perhaps ##L##) suitably such that you can extract the series ##\sum \frac{1}{n^2}## out of there.

but my Fourier series only has the odd-n terms?
 
I'll assume the Fourier expansion is correct.

What you need to do is to choose ##x## (and perhaps ##L##) wisely in order to extract an interesting series.

Now, also note the following:

Take

[tex]S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + ...[/tex]

then

[tex]S = \left(1 + \frac{1}{3^2} + \frac{1}{5^2} + ... \right) + \left(\frac{1}{2^2} + \frac{1}{4^2} + ...\right)[/tex]

But

[tex]\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + ...= \frac{1}{4}\left(1 + \frac{1}{2^2}+ \frac{1}{3^2} + ...\right)= \frac{S}{4}[/tex]

So differentiating is unnecessary.
 
ok my initial guess was wrong obviously. because it would not make the n's even, it would make them not squared...
 
ok. I don't really have time to show my work, but I ended up with π^2/6. Is this the correct answer?

the trick I didn't get was that part with the s/4
 
micromass said:
Fixed LaTeX:

Actually, this is the Fourier series of the even function ##f(x) = |x|, -L \leq x \leq L##, extended to the whole real line as a periodic function of period ##2L##. Basically, the person posing the question needs to specify ##f(x)## outside the desired interval ##[0,L]##. If that is not done correctly the Fourier series won't be pure 'cosine' and/or might not be continuous at ##\pm \, L##, leading to a mis-match between f and the sum of the series at those points.