Saturday, 16 January 2016

fourier analysis - Find flaw in solving for coefficients of ex=A0+sumlimitsin=1nftyAncosfracnpixL

exA0+n=1AncosnπxL



The Fourier cosine series on the right will be even extended periodized version of ex.



Now, I could solve for these coefficients directly,




An=2LL0excosnπxdx



But this exercise asks for a different solution.



We can apply term by term differentiation to yield:



exn=1nπLAnsinnπxL



Setting the two definitions of ex equal yields the following which is equality for [0,L]:



A0+n=1AncosnπxLn=1nπLAnsinnπxL



From there, how can I derive An?




EDIT: I found my original flaw, that last equation is only equality for [0,L], not for [L,L]. However, I'm still not sure how to calculate An via this technique.

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