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Given f(x) which is periodic on 0<x<L, so period is L, then Fourier series isf(x)∼1L∑n=−∞∞cnein2πLx Where cn=⟨n|f⟩=1L∫0Lf(x)e−in2πLxdx
The basis are |n⟩=1Le−in2πLx and L is the period.
Fourier transform for non periodic f(x) is (sum above becomes integral)f(x)=12π∫−∞∞ckeikxdkck=∫−∞∞f(x)e−ikxdx
This gives rise toδ(x−x′)=12π∫−∞∞eik(x−x′)dk
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