- #1
ognik
- 626
- 2
Hi - frustratingly I get some problems right 1st time, others just defy me (Headbang)
$f(x) = -x, [-\pi,0]; = x, [0,\pi]$
I get $a_0 = \pi$ and $a_n = \frac{-4}{\pi \left(2n-1\right)^2}$ which agrees with the book - but I thought I'd check $b_n$ for practice, it should = 0 according to the book, but I got:
$ \frac{1}{\pi} \left[ \int_{-\pi}^{0}-x Sinnx \,dx + \int_{0}^{\pi}x Sin nx \,dx \right] $
$= \frac{2}{\pi}\int_{0}^{\pi}x Sin nx \,dx $
$ = \frac{2}{\pi}\left[ \frac{x}{n}\left(-Cosnx\right) \right]^{\pi}_0 + 0 \ne 0$?
$f(x) = -x, [-\pi,0]; = x, [0,\pi]$
I get $a_0 = \pi$ and $a_n = \frac{-4}{\pi \left(2n-1\right)^2}$ which agrees with the book - but I thought I'd check $b_n$ for practice, it should = 0 according to the book, but I got:
$ \frac{1}{\pi} \left[ \int_{-\pi}^{0}-x Sinnx \,dx + \int_{0}^{\pi}x Sin nx \,dx \right] $
$= \frac{2}{\pi}\int_{0}^{\pi}x Sin nx \,dx $
$ = \frac{2}{\pi}\left[ \frac{x}{n}\left(-Cosnx\right) \right]^{\pi}_0 + 0 \ne 0$?
