# Fourier method | eigenfunctions

• December 18th 2009, 08:11 PM
kingwinner
Fourier method | eigenfunctions
$sin((n+1/2)(-\pi))= -1 = - sin((n+1/2)\pi)$ and
$(n+1/2)cos((n+1/2)(-\pi))= 0 = - sin((n+1/2)\pi)$ so sin((n+1/2)x) satisfies both $u(-\pi)= -u(\pi)$ and $u_x(-\pi)= -u_x(\pi)$. Yes, of course, they are eigenfunctions.