Convergence to pi^2/6 using Fourier Series and f(x) = x^2

  1. 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]
    - \pi \le t < \pi , \\
    t = x - \pi , \\
    f(t) = f(x - \pi ) \\

    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]{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. jcsd
  3. LCKurtz

    LCKurtz 8,448
    Homework Helper
    Gold Member

    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 x2.

    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.
  4. LCKurtz

    LCKurtz 8,448
    Homework Helper
    Gold Member

    Also, one other question. Are you sure your function x2 wasn't given on [itex](-\pi,\pi)[/itex] in the first place?
  5. Let me see if I get it: you're saying that x^2.cos(nx) and x^2.sin(nx) are 2Pi periodic, so that change of limits that produced the (-1)^n was wrong? I didn't have to do anything, just integrate them between 0 and 2Pi?

    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).
  6. LCKurtz

    LCKurtz 8,448
    Homework Helper
    Gold Member

    No, I am not saying that x2cos(nx) and x2sin(nx) are periodic. I am saying that the periodic extensions of them are periodic, and that is what you are expanding in a FS. I didn't go through your steps to see if you had any mistakes an your calculation of an and bn. I am just saying you can use the usual formulas on any period you wish, and in the present case, it is much easier to use [itex](0,2\pi)[/itex].
    The reason I asked this is because, if I recall correctly, evaluating the even extension of x2 on [itex](-\pi,\pi)[/itex] at 0 is the way I have seen the problem done.
  7. But still, the expressions for an and bn, etc, are made for (-[tex]\pi[/tex], [tex]\pi[/tex]), I still have to make some correction when I calculate them for (0, 2[tex]\pi[/tex]), right?
  8. LCKurtz

    LCKurtz 8,448
    Homework Helper
    Gold Member

    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 x2, which would be [itex]4\pi[/itex] periodic. As I mentioned before, the usual series for this is the periodic extension of x2 on [itex](-\pi,\pi)[/itex], but that is apparently not what you have been asked to do.
  9. So I use the even periodic extension for f(x): that is, f(x) = x2 for -2[tex]\pi[/tex]<x<2[tex]\pi[/tex], and then I make the following calculations?:

    [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?
  10. LCKurtz

    LCKurtz 8,448
    Homework Helper
    Gold Member

    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]
  11. But that's the way?
  12. LCKurtz

    LCKurtz 8,448
    Homework Helper
    Gold Member

    And, once more, I will add that I suspect you are doing the wrong problem. Maybe there is a typo somewhere. Because if you take the periodic extension of x2 on [itex](-\pi,\pi)[/itex] and evaluate its series at [itex]x = \pi[/itex] you will get your desired answer.
  13. OK, thanks. One question, though, when do I know when to use the periodic extensions (even an odd extensions)?
  14. LCKurtz

    LCKurtz 8,448
    Homework Helper
    Gold Member

    That's a good question. It depends on what you intend to do with the series, but generally, the smoother the periodic function is, the faster the series will converge. In this case, if you take the even extension of x2 you get a continuous function whereas if you take the odd extension you get jump discontinuities at the ends. And the even extension has no sine terms so that's half the work. Another hint in this problem would be the series you are asked about has terms like 1/n2 in it which suggests you need a continuous periodic function.
  15. Thank you.
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?