A community for students.
Here's the question you clicked on:
 0 viewing

This Question is Closed

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.38(1/1^2 + 1/3^2+1/5^2 +....)

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.3use this expansion http://mathworld.wolfram.com/FourierSeriesTriangleWave.html

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.3or f(x) = x for 0<x<= pi/2 = pix for pi/2 <= x < pi find its half range cosine series, it comes out to be f(x) = pi/4  (2/pi)[cos 2x/1^2 + cos6x/3^2 + cos 10x/5^2 +...] thn put x=pi/2

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.3or simply f(x) = [(pix)/2 ]^2 in [0,2pi] find F.S expansion

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.32nd one is simpler i guess.

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.3Define a function T(x) this way T(x) = arctan(2/pi)x for pi/2<=x<pi/2 T(x) = arctan(2/pi)x for pi/2<=x<3pi/2 \[ T (x) = {8 \over\pi^2 \ }\sum_{n=1,3,5 ..}{(1)^{n1 \over 2} \over n^2} \sin (n x) \\ T(\pi/2) = 1 \\ {8 \over \pi^2 } {(1)^{n1} \over (2n1)^2} \sin((2n1)\pi/2) = 1 \\ {8 \over \pi^2} ({1^2 + 1/3^2 + 1/5^2 + .. .})=1\]

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.3was the the Q realting Fourier Transform ?? i thot it was....

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0i can do the fourier cosines series method i was wondering if there was another way ,

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.3yeah, as i said the 2nd way i shown is simpler...u want the formulas ? or know it ?

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0i have worked it out

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0\[\large{\color{grey}{π^2}}\]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.3Plot[8/Pi^2 Sum[(1)^(n  1)/(2 n  1)^2 Sin[(2 n  1) x], {n, 1, 20}], {x, 0, 2 Pi}]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.3\[ T(x) = {2 \pi} \arcsin(\sin(x))\] http://www.wolframalpha.com/input/?i=Plot+2%2Fpi+sin^1%28sin%28x%29%29

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.0just for fun\[\zeta(2)=\frac{\pi^2}{6}\]\[\zeta(2)=\sum_{n=1}^{\infty} \frac{1}{n^2}=\sum_{n=1}^{\infty} \frac{1}{(2n)^2}+\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}=\frac{1}{4}\zeta(2)+A\]where\[A=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}\]and \[A=\frac{3}{4}\zeta(2)=\frac{\pi^2}{8}\]so ur expression becomes\[8A=\pi^2\]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.3yep!! i realized that :)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.