I Sturm-Liouville Eigenfunctions

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Hi PF!

Given ##y''+\lambda^2y=0## and BCs ##y'(0)=y'(1) = 0## we know eigenfunctions are ##y=\cos (n\pi x)##, and for ##n=1## this implies there is one zero on the interval ##x\in(0,1)##. However, I read that for SL problems, the ##jth## eigenfunction has exactly ##j-1## zeros on ##x\in(0,1)##, implying there should be no zeros for ##n=1##, but there is. Can someone reconcile this?
 

Orodruin

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The first eigenfunction is ##n=0##, not ##n=1##.
 

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