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My professor made an analogy between Fourier series and orthogonal projections, and I was hoping someone could explain that somewhat more. Basically, as I understand it:

$$a_n = \frac{1}{L} \int_L^L f(x) \cos\left(\frac{ n\pi x}{L}\right) \ dx \longleftrightarrow c_1 = \frac{v \cdot b_i}{b_i \cdot b_i},$$

where $\dfrac{1}{L}$ can be thought of as normalizing the projection, and the integrand is the inner product (equivalent to the dot product on the right side).

Am I understanding this correctly? And can someone clarify this for me?

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  • $\begingroup$ "Paralell..." Not a pun? $\endgroup$
    – Pedro
    May 1, 2014 at 23:35

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He is referring to the inner product induced by the space of square-integrable functions $L^2 [-L, L]$.

$$a_n = \frac{ \left <f(x), \cos \left ( \frac{n \pi x}{L} \right )\right >}{ \left < \cos \left ( \frac{n \pi x}{L} \right ),\cos \left ( \frac{m \pi x}{L} \right )\right >} = \frac{ \int_{-L}^{L} f(x) \cos \left ( \frac{n \pi x}{L} \right ) dx }{ \int_{-L}^{L} \cos \left ( \frac{n \pi x}{L} \right ) \cos \left ( \frac{m \pi x}{L} \right ) dx }.$$

where $\left < \cos \left ( \frac{n \pi x}{L} \right ),\cos \left ( \frac{m \pi x}{L} \right )\right > = \left\{\begin{matrix} 1& \text{if} \;n = m \neq 0\\ 0& \text{if} \;n \neq m \end{matrix}\right. $

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You are understanding it correctly. The time domain and the frequency domain are two different bases for a vector space. Both are orthogonal and can be orthonormal if you want. The functions are vectors in that space. The FFT is the transformation between these two bases. Because the vector space is infinite dimensional, you have to worry about convergence, but if the functions are nice enough it works out.

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