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okay, so if you've studied PDEs you know the value of a fourier series, and the difficulty of determining a fourier coefficient. my question relates to finding this coefficient. briefly, i'll define a fourier series as [tex]f(x)=\sum_{n=0}^{\infty} A_n\cos\frac{n\pi x}{L}+B_n\sin\frac{n\pi x}{L}[/tex] when solving for [itex]A_n[/itex] one way (the only way i know) is to multiply both sides by [itex]\cos\frac{m\pi x}{L}[/itex]. now, when we integrate over [itex][-L,L][/itex] the sine term vanishes by its orthogonality with cosine and we are left with [tex]\int_{-L}^{L}f(x)\cos\frac{m\pi x}{L}dx=\sum_{n=0}^{\infty} A_n \int_{-L}^{L} \cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}dx[/tex], where the r.h.s=0 if [itex]m \neq n[/itex] thus we know [itex]m = n[/itex] (for now let's assume [itex]m=n \neq 0[/itex]). [itex]m = n \implies \int_{-L}^{L} \cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L}dx=L[/itex] thus [tex]\sum_{n=1}^{\infty}A_n=\frac{1}{L}\int_{-L}^{L}f(x)\cos\frac{n\pi x}{L}dx[/tex] but every book i see states simply [tex]A_n=\frac{1}{L}\int_{-L}^{L}f(x)\cos\frac{n\pi x}{L}dx[/tex] where did the sum go???

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# Fourier coefficients

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