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[note: also under discussion in sos math cyberboard]

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- Dec 18th 2009, 08:11 PMkingwinnerFourier method | eigenfunctions
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[note: also under discussion in sos math cyberboard] - Dec 19th 2009, 04:42 AMHallsofIvy
$\displaystyle sin((n+1/2)(-\pi))= -1 = - sin((n+1/2)\pi)$ and

$\displaystyle (n+1/2)cos((n+1/2)(-\pi))= 0 = - sin((n+1/2)\pi)$ so sin((n+1/2)x) satisfies both $\displaystyle u(-\pi)= -u(\pi)$ and $\displaystyle u_x(-\pi)= -u_x(\pi)$. Yes, of course, they are eigenfunctions.