Recent content by RUrubee2

My brain works in VERY mysterious ways, but thank you. That does clear things up. I am working hard on trying to figure out where my mistake is. I get confused when sometimes the answer is (-1)^(n+1) and in other cases like: b_n=(2/pi)∫(0 to pi)sin(nx)dx I get (2/npi)(1-cos(npi))and...

@ LCKurtz The answer in the book is 3/8 for a_0/2 And the -2 to 0 disappears 'cause the function is 0 on that interval P.S. Thank you for trying to help me with the real question

The reason I "imagine" that is because I had some sort of logic in my head that said that if sin(npi/2)=something, when n is an odd number than n+1 was going to give me an odd number... I see as I try to explain that it doesn't always work. But I had "sin(npi)=(-1)^(n+1) in my class notes which...

As I said, I "know cos(npi)=(-1)^n and I was told sin(npi)=(-1)^(n+1)" so I believe you that it is wrong. I imagine that that someone wanted to say sin(npi/2)=(-1)^(n+1)? As for the "equalities", I was wondering, not stating, seeing as the equations "looked" similar. Therefore, if all this is...

Homework Statement To find the Fourier series for f(x)= 0, -2<x<0 x, 0≤x<1 1, 1≤x<2 Homework Equations f(x)=a_0/2+Ʃ(from n=0 to ∞) (a_n*cos(npix/p) + b_n*sin(npix/p)) The Attempt at a Solution So p=2, interval=[-2,2] a_0=3/4, a_n=(2/n^2pi^2)*(cos(npi/2)-1), Here is my problem: did...