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If a function $f(x)$ is defined in $[-L,L]$, then Fourier series is defined as $$f(x)=\frac{A_0}{2}+\sum_{n=1}^{\infty}[A_n \cos (n\pi x/L)+ B_n \sin (n \pi x/L)],$$ where $$A_n=L^{-1}\int_{-L}^{L} f(x) \cos (n\pi x/L)~ dx,\quad B_n=L^{-1}\int_{-L}^{L} f(x) \sin (n\pi x/L)~ dx.$$

I would like to know why we have half of $A_0$, here,

MathDona
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    see https://math.stackexchange.com/questions/752465/why-is-the-zeroth-coefficient-in-a-fourier-series-divided-by-2 – ani Oct 03 '22 at 04:18
  • One uses orthonormal basis to form the Fourier series. The function $f_0=\mathbb{1}{[-L,L]}$ is part of the basis and has norm $\frac{1}{2L}\int^L{-L}|f_0|^2 =1$. Notice that $\langle f,f_0\rangle =\frac{1}{2L}\int^L_{-L}f=\frac{1}{2}A_0$ – Mittens Oct 03 '22 at 04:26
  • Integrate to get $\int_{-L}^L f ,dx = A_0/2 \cdot 2L$ – Mason Oct 03 '22 at 05:38

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