Toronto Math Forum

APM346-2016F => APM346--Lectures => Chapter 4 => Topic started by: Shaghayegh A on November 14, 2016, 02:18:44 PM

Title: HA 6, sections 4.3-4.5, problem 6
Post by: Shaghayegh A on November 14, 2016, 02:18:44 PM

http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter4/S4.5.P.html

For problem 6f, I get $b_n=0$, which I know is wrong.
I have $$b_n=\frac{2}{\pi} \int_0 ^{\pi} \sin((m-1/2)x) \sin((n+1/2)x) dx \\
=1/pi \int_0 ^{\pi} \cos((m-n-1)x) -\cos((m+n)x) dx =0$$
Separately for $n \neq m-1$ and $n=m-1$. Why am I getting b_n=0?

Title: Re: HA 6, sections 4.3-4.5, problem 6
Post by: Victor Ivrii on November 15, 2016, 07:03:03 AM

The last line of calculations is wrong. We already counted
$\in_0^\pi \cos (kx)\,dx$