Parseval Identity and Fourier Series Question on function $f(x)=|x|$

Question

Trying to compute the fourier series for $f(x)=|x|$ for $f$ on $[- \pi, \pi]$ using the trig method. I have a question as to the absolute value function.

I'm using the definition of absolute value where $|x|=x$ if $x \ge 0$, and $|x|=-x$ if $x <0 $. Therefore, $$a_0=\int_{-\pi}^{\pi}f(t) dt=\int_{-\pi}^0-t dt+\int_0^{\pi}tdt$$

I want to ask if this is the correct approach because I'm not getting the numerical series $\frac{\pi^4}{96}$ when I use the parseval identity so either I'm doing an arithmetic mistake with this tideous problem or my limits of integration are wrong therefore my $a_n$ and $b_n$ are wrong. Any help is greatly appreciated. Thank you.

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Yes, that's the correct approach. Perhaps there's a mistake somewhere in your development. — DonAntonio, May 23 '14 at 23:37
@DanielFischer i have uploaded a pic from my exam. Why does it say $\pi^4/96$? — User69127, Jun 02 '14 at 06:48
It says $\frac{\pi^4}{96}$ because it's only the sum over the odd numbers, not all. Without seeing your work, I made a guess. I guessed wrong. — Daniel Fischer, Jun 02 '14 at 12:22

DonAntonio · Accepted Answer · 2014-05-30T21:51:05.107

6

You should have:

$$\begin{align*}a_0&=\frac1\pi\int\limits_{-\pi}^\pi|x|dx=\frac2\pi\int\limits_0^\pi xdx=\pi\\{}\\ a_n&=\frac1\pi\int\limits_{-\pi}^\pi |x|\cos nx\,dx=\frac2\pi\int\limits_0^\pi x\cos nx\,dx=\\{}\\&=\overbrace{\left.\frac2{n\pi}x\sin nx\right|_0^\pi}^{=0}-\frac2{n\pi}\int\limits_0^\pi\sin nx\,dx=\frac2{n^2\pi}\left((-1)^n-1\right)=\begin{cases}\;\;\;\;\;\,0&,\;\;n\;\text{is even}\\{}\\-\frac4{n^2\pi}&,\;\;n\;\text{is odd}\end{cases}\end{align*}$$

So for example, for any $\;x\;$ not at the extreme points of the periodicity intervals, we have

$$f(x)=|x|=\frac\pi2-\sum_{n=0}^\infty\frac4{(2n+1)^2\pi}\cos nx$$

In particular, for $\;x=0\;$ we get

$$\frac\pi2=\frac4{\pi}\sum_{n=0}^\infty\frac1{(2n+1)^2}\implies\sum_{n=0}^\infty\frac1{(2n+1)^2}=\frac{\pi^2}8$$

which is equivalent with

$$\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$$

Now, using Parseval's Identity, we get:

$$\frac{2\pi^2}3=\frac1\pi\int\limits_{-\pi}^\pi \overbrace{x^2}^{=f(x)^2}dx=\frac12\pi^2+\sum_{n=0}^\infty\frac{16}{n^4\pi^2}\implies\frac{\pi^2}6=\frac{16}{\pi^2}\sum_{n=0}^\infty\frac1{(2n+1)^4}$$

which is equivalent with

$$\sum_{n=1}^\infty\frac1{n^4}=\frac{\pi^4}{90}$$

edited May 30 '14 at 21:51

answered May 24 '14 at 00:06

DonAntonio

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It actually is $\pi^4/96$ not 90. – User69127 May 30 '14 at 21:38
Rats, I'm being confused: no, @User69127, it is $;90;$ in the denominator! – DonAntonio May 30 '14 at 21:49
It is $;96;$ in the denominator of the odd numbers...! – DonAntonio May 30 '14 at 21:50
But that's what we want, right? Since the even n's vanish? I get 96 and my exam question also had me proved it to be $\pi^4/96$ – User69127 Jun 02 '14 at 06:45
Please look at the pic screenshot from my exam above – User69127 Jun 02 '14 at 06:47
Well, that's exactly what we got there @User69127: the sum of the inverse of the square odd numbers is $;\frac{\pi^2}{96};$ indeed, which implies my very last, and much more well-known, equality. You seem to have confused these two infinite series, even in your reply to Daniel's comment. – DonAntonio Jun 02 '14 at 07:55
I just don't see how you get 90 though? Where is it coming from? – User69127 Jun 02 '14 at 08:26
1

Hint: $$\sum_{n=1}^\infty\frac1{n^4}=\sum_{n=1}^\infty\frac1{(2n)^4}+\sum_{n=1}^\infty\frac1{(2n-1)^2};\ldots$$ – DonAntonio Jun 02 '14 at 08:30
@DonAntonio; Could you elaborate on the step where you apply Parseval's identity? I really don't see how you get the line following that sentence. – Mathbeginner Oct 19 '18 at 15:54

Parseval Identity and Fourier Series Question on function $f(x)=|x|$

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