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Convergence to pi^2/6 using Fourier Series and f(x) = x^2 
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#1
Nov2209, 07:42 PM

P: 176

1. The problem statement, all variables and given/known data
Using the Fourier trigonometrical series for [tex]f(x) = {x^2},{\rm{ }}0 \le x < 2\pi [/tex], prove that [tex]\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} = \frac{{{\pi ^2}}}{6}[/tex] 3. The attempt at a solution This is more of a "what am I doing wrong question". First, because I'm not in the period defined by the trigonometrical coefficients, I have to change the limits using that: [tex]0 \le x < 2\pi [/tex] [tex]  \pi \le x  \pi < \pi [/tex] [tex]\begin{array}{l}  \pi \le t < \pi , \\ t = x  \pi , \\ f(t) = f(x  \pi ) \\ \end{array}[/tex] Then I find the Fourier coefficients with this in hand: [tex]{a_0} = \frac{1}{\pi }\int\limits_{  \pi }^\pi {f(t).dt = } \frac{1}{\pi }\int\limits_0^{2\pi } {f(x).dx = } \frac{1}{\pi }\int\limits_0^{2\pi } {{x^2}.dx = } \frac{8}{3}{\pi ^2}[/tex] [tex]{a_n} = \frac{1}{\pi }\int\limits_{  \pi }^\pi {f(t).\cos (nt)dt = } \frac{1}{\pi }\int\limits_0^{2\pi } {f(x).\cos (n(x  \pi ))dx = } \frac{1}{\pi }\int\limits_0^{2\pi } {{x^2}.\left[ {\cos (nx)\cos (n\pi )  \sin (nx)\sin (n\pi )} \right].dx = } [/tex] [tex]{a_n} = \frac{1}{\pi }\int\limits_0^{2\pi } {{x^2}.\cos (nx).{{(  1)}^n}dx = } \frac{{{{(  1)}^n}}}{\pi }\left( {\left. {\frac{{2x.\cos (nx)}}{{{n^2}}}} \right_0^{2\pi }} \right) = \frac{{2{{(  1)}^n}}}{{{n^2}}}[/tex] [tex]{b_n} = \frac{1}{\pi }\int\limits_{  \pi }^\pi {f(t).\sin (nt).dx = } \frac{1}{\pi }\int\limits_0^{2\pi } {f(x).\sin (n(x  \pi )).dx = } \frac{1}{\pi }\int\limits_0^{2\pi } {{x^2}.\left[ {\sin (nx)\cos (n\pi ) + \sin (n\pi )\cos (nx)} \right].dx = } [/tex] [tex]{b_n} = \frac{{{{\left( {  1} \right)}^n}}}{\pi }\int\limits_0^{2\pi } {{x^2}.\sin (nx).dx = } \frac{{{{\left( {  1} \right)}^n}}}{\pi }\left( {\left. {(\frac{2}{{{n^3}}}  \frac{{{x^2}}}{n})\cos (nx)} \right_0^{2\pi }} \right) = \frac{{4{{\left( {  1} \right)}^{n + 1}}}}{n}[/tex] Then, using Parseval's identity: [tex]{\left\ {f(x)} \right\^2} = \frac{{{a_0}^2}}{2} + \sum\limits_{n = 1}^\infty {{a_n}^2 + {b_n}^2} [/tex] [tex]\frac{1}{\pi }\int\limits_0^{2\pi } {{{({x^2})}^2}.dx} = \frac{{{{\left( {\frac{8}{3}{\pi ^2}} \right)}^2}}}{2} + \sum\limits_{n = 1}^\infty {{{\left( {\frac{{2{{(  1)}^n}}}{{{n^2}}}} \right)}^2} + {{\left( {\frac{{4{{(  1)}^{n + 1}}}}{n}} \right)}^2}} [/tex] [tex]\frac{{32}}{5}{\pi ^4} = \frac{{32{\pi ^4}}}{9} + \sum\limits_{n = 1}^\infty {\frac{4}{{{n^4}}} + \frac{{16}}{{{n^2}}}} [/tex] [tex]\frac{{112}}{9}{\pi ^4} = \sum\limits_{n = 1}^\infty {\frac{{4 + 16{n^2}}}{{{n^4}}}} [/tex] And here I reach a deadpoint. What do I do? 


#2
Nov2209, 07:55 PM

HW Helper
Thanks
PF Gold
P: 7,659

You are extending f(x) periodically with period [itex]2\pi[/itex]. You get the same thing when you integrate a periodic function over any period. That is, if g(x) has period P, then
[tex]\int_a^{a+P} g(x)\, dx = \int_b^{b+P} g(x)\, dx [/tex] That means you don't have to use [itex]\pi[/itex] to [itex]\pi[/itex] in your formulas for the coefficients; you can use 0 to [itex]2\pi[/itex], so you don't have to fiddle with changing the x^{2}. The other thing you are missing is using the Dirichlet theorem which tells you what the series will converge to at 0 and [itex]2\pi[/itex]. Notice that your periodic extension of your function has a jump discontinuity at those points. 


#3
Nov2209, 08:27 PM

HW Helper
Thanks
PF Gold
P: 7,659

Also, one other question. Are you sure your function x^{2} wasn't given on [itex](\pi,\pi)[/itex] in the first place?



#4
Nov2309, 09:27 AM

P: 176

Convergence to pi^2/6 using Fourier Series and f(x) = x^2
And what do you want me to do with the borders? I know that they will converge to the lateral limits average but I don't know what you mean. And yes, x^2 was defined in (0, 2Pi). 


#5
Nov2309, 11:18 AM

HW Helper
Thanks
PF Gold
P: 7,659




#6
Nov2309, 11:25 AM

P: 176




#7
Nov2309, 12:01 PM

HW Helper
Thanks
PF Gold
P: 7,659

No. You don't have to change anything but the limits. If f(x) has period [itex]2\pi[/itex] then
[tex]\frac 1 {\pi}\int_{\pi}^{\pi} f(x)\cos{nx}\, dx = \frac 1 {\pi}\int_{0}^{2\pi} f(x)\cos{nx}\, dx[/tex] and ditto for the sine. In the current problem the latter is much preferred because it is the only one in which you can use [itex]f(x) = x^2[/itex]. And looking ahead, I'm guessing you might want to use the even periodic extension of x^{2}, which would be [itex]4\pi[/itex] periodic. As I mentioned before, the usual series for this is the periodic extension of x^{2} on [itex](\pi,\pi)[/itex], but that is apparently not what you have been asked to do. 


#8
Nov2309, 12:14 PM

P: 176

[tex]{a_0} = \frac{1}{{2\pi }}\int\limits_{  2\pi }^{2\pi } {{x^2}dx} [/tex] [tex]{a_n} = \frac{1}{{2\pi }}\int\limits_{  2\pi }^{2\pi } {{x^2}\cos (\frac{1}{{2\pi }}nx)dx} [/tex] [tex]{b_n} = \frac{1}{{2\pi }}\int\limits_{  2\pi }^{2\pi } {{x^2}\sin (\frac{1}{{2\pi }}nx)dx} [/tex] Is that what you're saying? 


#9
Nov2309, 12:43 PM

HW Helper
Thanks
PF Gold
P: 7,659

There shouldn't be any [itex]\pi[/itex] in your cosines and sines. [itex]2p=4\pi[/itex] so
[tex]\sin{\frac{n\pi x}{p}} = sin{\frac n 2 x}[/tex] 


#10
Nov2309, 12:46 PM

P: 176

But that's the way?



#11
Nov2309, 12:47 PM

HW Helper
Thanks
PF Gold
P: 7,659




#12
Nov2309, 12:51 PM

P: 176

OK, thanks. One question, though, when do I know when to use the periodic extensions (even an odd extensions)?



#13
Nov2309, 01:01 PM

HW Helper
Thanks
PF Gold
P: 7,659




#14
Nov2309, 06:29 PM

P: 176

Thank you.



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