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Find the Fourier series of the function$\ f(x) = x^2; 0 < x < 2\pi $ of period $\ 2\pi $ and find the sum of the infinite series $$\ (i) $$ $$\sum_{n=1}^\infty \frac{1}{n^2}$$ $$\ (ii) $$ $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2}$$

I have tried to solve the second part of the math using fourier cosine series.My point of view was that the function is even and it is defined over $\ [0,2\pi]$ as like the half range fourier function defined over [0,L].

Cosine Series: $\ f(x) =\frac{a_0}{2} +\sum_{n=1}^\infty a_n\cos \frac{n\pi x}{L} $

$\ a_0=\frac{2}{\pi} \int_0^\pi x^2 dx =\frac{2\pi^2}{3}$

$\ a_n = \frac{2}{\pi} \int_0^\pi x^2 \cos nx dx =\frac{2}{\pi}[\frac{x^2}{n}\sin nx +\frac{2x\ cosnx}{n^2} -\frac{2}{n^3}\ sinnx]_0^\pi $

$\ =\frac{2}{\pi} \frac{2\pi}{n^2} (-1)^n $ $\ =\frac{4}{n^2} (-1)^n $

$\ f(0)=\frac{f_L(0)+f_R(0)}{2}=0 $ [I am not sure about this line]

$\ f(x) =\frac{\pi^2}{3} +\sum_{n=1}^\infty \frac{4}{n^2} (-1)^n \ cos nx $

$\ f(0)=\frac{\pi^2}{3}+4\sum_{n=1}^\infty \frac{1}{n^2} (-1)^n $

$\ 0= \frac{\pi^2}{3} + 4 \sum_{n=1}^\infty \frac{1}{n^2} (-1)^n $

$\ \sum_{n=1}^\infty \frac{1}{n^2} (-1)^n = -\frac{\pi^2}{12} $

$\ \sum_{n=1}^\infty \frac{1}{n^2} (-1)^{n+1} = \frac{\pi^2}{12} $

Asm Arman
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  • https://math.stackexchange.com/a/1824700/254075 – sharding4 Jun 03 '17 at 19:08
  • @sharding4 the problem you have mentioned here has a range of [-pi,pi].But the problem i have mentioned here has a range of [0,2pi].So,how could I calculate the discontinuity of the function?I have just started the fourier series.So,please pardon me if i have said something wrong. – Asm Arman Jun 03 '17 at 20:22
  • That is a bit harder, but I don't think it changes things too much. You can't exploit that the function is even. And you'll need to evaluate $f(\pi)$ in order to find the sums you're after. Have you learned Fourier Series in term of the $\sin$ and $\cos$ functions or $e^{i\theta}$? – sharding4 Jun 03 '17 at 20:34
  • I think you should get something like $a_0=4\pi^2/3$ and $a_n=4/n^2$ and $b_n=-4\pi/n$. Be careful with your signs and you should be able to deduce the sum of the series you're supposed to find. – sharding4 Jun 03 '17 at 21:10
  • @sharding4 it's a even function.so,$\ b_n$ should be zero,shouldn't it?And why I have to evaluate $\ f(\pi)$ ?It doesn't seem to be my point of discontinuity.I have learned in terms of sin and cos but not in $\ e^{i\theta}$ form – Asm Arman Jun 03 '17 at 21:26
  • No when defined on $(0,2\pi)$ and extended using periodicity $x^2$ is not even. You'll evaluate $f(\pi)$ in order to get one of the series you're looking for. The $\sin$ terms will all drop out. – sharding4 Jun 03 '17 at 21:33
  • What do you get for the Fourier coefficients of $f(x)$ ? – reuns Jun 03 '17 at 21:35
  • @sharding4,can you provide me with an answer?.I am in confusion. – Asm Arman Jun 03 '17 at 21:42
  • @user1952009,I have tried it with normal fourier series and half range cosine fourier series.So,the coefficients are different for two different series. – Asm Arman Jun 03 '17 at 21:44
  • The question is about the Fourier series $$g(x) = \sum_{n=-\infty}^\infty c_n e^{i n x}, \qquad c_n = \frac{1}{2\pi} \int_0^{2\pi} f(x) e^{-inx}dx$$ what do you get for $c_n$ ? Also does the Fourier series converges, for what $x$, to what function ? – reuns Jun 03 '17 at 21:47
  • @user1952009,pardon me if I am saying something silly.But I am not acquainted with the 'converges' thing.I am guessing that you are asking about the continuity of the function.if I am wrong,would you please briefly explain the the 'converges' thing. – Asm Arman Jun 03 '17 at 22:06
  • Come on, there are several questions. The first is : what do you get for $c_n$ ? – reuns Jun 03 '17 at 22:22
  • $\ c_n =\frac{4}{n^2}$ – Asm Arman Jun 03 '17 at 22:31
  • This is not correct. – reuns Jun 03 '17 at 22:33
  • Then,I don't know what is wrong.could you please solve it for me? – Asm Arman Jun 03 '17 at 22:40

0 Answers0