can anyone verify by b_n for these series...thx!
sin(2n-1)(pi) / 2 b_n = pi/2
cosn(pi) b_n = (pi)
cos(n[pi]) / sqrt(n) b_n = 1 / sqrt(n)
cosn(pi) / (n+1) b_n = (pi) / (n+1)
nosn(pi) / nē b_n = 1/nē
Also, it would help if you would at least make your equations readable...
The first one for example, I can't tell if it's $\displaystyle \displaystyle \frac{\sin^{2n-1}\pi}{2}b_n$, $\displaystyle \displaystyle \frac{\sin{[(2n-1)\pi]}}{2}b_n$, $\displaystyle \displaystyle \frac{\sin^{2n-1}{\pi}}{2b_n}$ or $\displaystyle \displaystyle \frac{\sin{[2(n-1)\pi]}}{2b_n}$...