ex∼A0+∞∑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=2L∫L0excosnπxdx
But this exercise asks for a different solution.
We can apply term by term differentiation to yield:
ex∼−∞∑n=1nπLAnsinnπxL
Setting the two definitions of ex equal yields the following which is equality for [0,L]:
A0+∞∑n=1AncosnπxL∼−∞∑n=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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