Welcome to our community

Be a part of something great, join today!

Problem Of The Week # 317 - Aug 01, 2018

Status
Not open for further replies.
  • Thread starter
  • Moderator
  • #1

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,925
  • Thread starter
  • Moderator
  • #2

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,925
Congratulations to Opalg for his correct solution. Honorable mention goes to greg1313 and kaliprasad for proofs that used the value of $\zeta(4)$ without proof. You can read Opalg 's solution below.


Let $f(x) = x^2$, on the interval $[-\pi,\pi]$. The Fourier coefficients of $f$ are given by \(\displaystyle \hat{\ f}(n) = \frac1{2\pi}\int_{-\pi}^\pi x^2e^{-inx}dx.\)

When $n=0$, that gives \(\displaystyle \hat{\ f}(0) = \frac1{2\pi}\int_{-\pi}^\pi x^2dx = \frac1{2\pi}\left[ \frac{x^3}3\right]_{-\pi}^\pi = \frac{\pi^2}3.\)

For $n\ne0$, integrate by parts twice to get $$\begin{aligned} \hat{\ f}(n) &= \frac1{2\pi}\int_{-\pi}^\pi x^2e^{-inx}dx \\ &= \frac1{-2in\pi}\left[ x^2e^{-inx}\right]_{-\pi}^\pi - \frac i{n\pi}\int_{-\pi}^\pi xe^{-inx}dx \\ &= 0 + \frac1{n^2\pi}\left[ xe^{-inx}\right]_{-\pi}^\pi - \frac1{n^2\pi} \int_{-\pi}^\pi e^{-inx}dx \\ &= \frac{2(-1)^n}{n^2} .\end{aligned} $$ By Parseval's theorem, \(\displaystyle \frac1{2\pi}\int_{-\pi}^\pi |f(x)|^2dx = \sum_{\Bbb N}|\hat{\ f}(n)|^2.\) Since \(\displaystyle \int_{-\pi}^\pi |f(x)|^2dx = \int_{-\pi}^\pi x^4dx = \frac{2\pi^5}5,\) it follows that $$\frac{\pi^4}5 = |\hat{\ f}(0)|^2 + 2\sum_{n=1}^\infty|\hat{\ f}(n)|^2 = \frac{\pi^4}9 + 8\sum_{n=1}^\infty \frac1{n^4},$$ so that $$\sum_{n=1}^\infty \frac1{n^4} = \frac18\left(\frac{\pi^4}5 - \frac{\pi^4}9\right) = \frac{\pi^4}{90}.$$ The sum of the even-numbered terms of that series is \(\displaystyle \sum_{n=1}^\infty \frac1{2^4n^4} = \frac1{16}\frac{\pi^4}{90}.\) So the sum of the odd-numbered terms is \(\displaystyle \frac{15}{16}\frac{\pi^4}{90}\). The alternating sum \(\displaystyle \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{n^4}\) is then the sum of the odd terms minus the sum of the even terms, namely \(\displaystyle \left(\frac{15}{16} - \frac1{16}\right)\frac{\pi^4}{90} = \frac{7\pi^4}{720}.\)
 
Status
Not open for further replies.