7

Parseval's Identity: For continuous $f: [- \pi , \pi] \to \mathbb{R}$

$$ \sum_{n=- \infty}^{+ \infty} |c_n|^2 = \frac{1}{2 \pi} \int_{ - \pi}^{ \pi} |f(x)|^2dx, \text{ where } c_n = \frac{1}{2 \pi} \int_{- \pi}^{ \pi} f(x)e^{-inx}dx$$

Problem: Using Parseval's Identity, consider the even function $f:= \frac{1}{2} - \frac{\pi}{4} \sin \left( \frac{x}{2} \right), \ x \in [0, \pi] $ to compute the sum: $$\sum_{n =1}^{+ \infty} \frac{1}{(4n^2-1)^2} $$

This is my first attempt to work with Parseval's Identity and Fourier coefficients. I did read this similar question beforehand, but it didn't help me to complete this task Use Fourier series for computing $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$.

My approach: I will merely write down the results, I did all the calculations by hand on paper and then double-checked by using Mathematica. Writing out the entire process would only make this post blow up in length further.

To get the Fourier Coefficients I did integrate using only the definition as above, I first did look at the case for $n=0$. I got $$ c_0 = \frac{1}{2}$$ for the case $ n \neq 0$ I applied integration by parts two times and after a pretty tedious process I did end up with $$c_n = \frac{ in \cos (n \pi)}{4n^2-1}= \frac{in (-1)^n}{4n^2-1} $$ Which was a very satisfying result, because now I have $$|c_n|^2 = \frac{n^2}{(4n^2-1)^2}$$which somewhat looks very close to what I am supposed to get, except for that annoying $n^2$ in the numerator.

Finally evaluating $$ \frac{1}{2 \pi} \int_{- \pi}^\pi |f(x)|^2dx = \frac{1}{32}(8+ \pi^2)$$

Doubts and Questions:

  • I have a feeling that my method of integrating was wrong, because I did not make use of $f$ being even which would mean that $f(-x)=f(x)$ and therefore integration from $- \pi, \pi$ would result to be twice the original integral with bounds from 0 to $2 \pi$
    • I did not make use of the statement $f$ is even, because when I plot $f$ it does not look even at all to me, so I wonder about this statement in general. I suppose it has something to do with the definitions of Fourier Coefficients and their application to $A$-periodical Functions.

  • If the above procedure would happen to be right, how could I get the final statement? More precisely formulated, how would I possible get from $$ \sum_{- \infty}^{+ \infty} \frac{n^2}{(4n^2-1)^2}$$ to the Sum that starts from $1$ and ends at $\infty$ ?

Community
  • 1
Spaced
  • 3,499
  • Are you sure of your calculations? – Mhenni Benghorbal Mar 19 '14 at 22:12
  • @MhenniBenghorbal, unfortunately I have to answer with no, you will find it under my "Doubts and Questions" section that I am not 100% positive about my calculations. My main issue is the statement that $f$ is even which, after plotting the graph of $f$, makes no geometric sense to me whatsoever. I could not teach Mathematica to consider $f$ being even, but all the above Calculations were done by hand and then checked in Mathematica CAS. – Spaced Mar 19 '14 at 22:19
  • 1
    To use Parseval's Identity you need to have a function defined on a symmetric interval. In your question you just gave $f$ defined on $[0,\pi]$. How are you extending it to $[-\pi, \pi]$? – Git Gud Mar 19 '14 at 22:20
  • @GitGud, I did not extend it in my question if you wanted to know that. If you wanted to hear about a possible approach to extend the above function to the interval $[- \pi, \pi]$ I have to confess that I do not know, I could take twice the values maybe or use periodicity of $\sin$ – Spaced Mar 19 '14 at 22:27
  • @Spaced You need to extend it, I think. It's hard to believe that they want you to take $f$ defined by the same expression in $[-\pi, \pi]$. It's possible, but unlikely, in my opinion. have you tried to put the sum in mathematica just to check? – Git Gud Mar 19 '14 at 22:37
  • Thanks for your response @GitGud, I guess with the sum you meant the one I have given in my last row, by Mathematica, this sum would converge, which is indeed no good sign. So far I haven't heard about "extending definitions" or ever used it, nor seen it in class. Although it sounds to me like something that can easily be done. I am sorry if the problem is unclear in that manner, but its a 1:1 copy from my Problem Set (optional - not accredited problem), I will try to consult my tutor. Thanks for your insight. – Spaced Mar 19 '14 at 22:43
  • Is your function an even function? – Mhenni Benghorbal Mar 19 '14 at 23:36
  • @MhenniBenghorbal, yes the function is supposed to be even as defined on $[0, \pi]$, I did highlight that again in my question. – Spaced Mar 20 '14 at 08:42
  • @Spaced I was gonna answer your question, but before doing that I checked if the sum was right. It isn't as you can see here, but it's close. Can you check if you can correct the error somewhere? If you can't, ping me and I'll help you. – Git Gud Mar 20 '14 at 23:18

1 Answers1

2

Making use of the fact that $f$ is an even function, we extend $f$ as follows: \begin{align}f:& \begin{cases} [0, \pi] &\longrightarrow \mathbb{R} \\x & \longmapsto \displaystyle \frac{1}{2} - \frac{\pi}{4} \sin \left( \frac{x}{2}\right) \end{cases} \tag{given} \\\\\\ f:& \begin{cases} [ - \pi, \pi] & \longrightarrow \mathbb{R} \\ x & \longmapsto \begin{cases} f(x), & 0 \leq x \leq \pi \\ f(-x), & - \pi \leq x \leq 0 \end{cases} \end{cases} \tag{extended} \end{align} With this definition of $f$ we have the property that for $ n \neq 0$ (for $n = 0$ analogous) \begin{align}c_n = \frac{1}{2 \pi} \int_{- \pi}^{ \pi} f(x) e^{-inx}dx = \frac{1}{2 \pi} \int_{- \pi}^0 \underbrace{f(x)}_{=f(-x)} e^{-inx}dx + \int_0^\pi f(x)e^{inx}dx \end{align} A simple substitution $u=-x$ for the first integral leads to: \begin{align}c_n = \frac{1}{2 \pi} \int_0^\pi f(x) \underbrace{\left( e^{inx}+ e^{-inx} \right)}_{= 2 \cos (nx)}dx \end{align} which would finally lead to the correct, symmetric Fourier coefficients \begin{align} c_n = \frac{1}{8n^2-2}= \frac{1}{2} \cdot \frac{1}{4n^2-1} \text{ with } c_{-n}=c_n\end{align} Computing in a likewise manner $$ \frac{1}{ 2 \pi} \int_{- \pi}^\pi |f(x)|^2dx = \frac{\pi ^2-8}{32} $$ leads to the final result: $$ \sum_{n=1}^\infty \frac{1}{4n^2-1}= \frac{\pi^2-8}{16} $$

Spaced
  • 3,499