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As part of my education, I took upon myself to understand where the Fourier series functions come from, I did some digging, and found out that the vector space accommodating the function is a Hilbert space, the $L^2$-space. It provides this nifty formula for series expansion of a vector (function) given an orthonormal system of vectors spanning $[1, +\infty]$ in said Hilbert space.

Nifty Theorem

I thus proved that $e^{nwti}$ in interval $(t,t+T)$ is orthogonal using inner-product, and thus I ended up with,

$$x = \sum_{n=1}^{\infty} \langle f(x), e^{nwti}\rangle e^{nwti}$$

But that sadly doesn't match with the equations of the exponential Fourier series,

enter image description here

Especially, the $c_k$ with its $1/T_0$, where did that come from? I would like to know if my deduction is valid and why is it different from the exponential series I found many times online.

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"given an orthonormal system of vectors" the nifty formula does not follow! The formula is true for a complete orthonormal system.

Say $e_n\in L^2([0,2\pi])$ is defined by $$e_n(t)=e^{int}.$$

You're more or less looking at $\{e_n:n>0\}$. That's an orthonormal system, but it's not complete. Otoh $\{e_n:n\in\Bbb Z\}$ is a complete orthonormal system.

Edit: Now the OP asks "why do we divide the inner-product by the period?". It seems I misunderstood the question.

We don't so much divide the inner product by the period as divide the integral by the period to define the inner product. When I said that the $e_n$ above were an orthonormal set in $L^2([0,2\pi])$ I intended the inner product $$<f,g>=\frac1{2\pi }\int_0^{2\pi}f(t)\overline{g(t)}\,dt.$$If we use $\int_0^{2\pi}f\overline g$ as the inner product instead then $||e_n||=\sqrt{2\pi}$, so $(e_n)$ is not quite orthonormal.

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  • $\begingroup$ Yes, that does make sense, but if the equation works for the set you gave me, then why is it still different than the one provided in the Fourier series, in other words, why do we divide the inner-product by the period? $\endgroup$ – Whiteclaws Mar 10 at 18:25
  • $\begingroup$ Finally clicked for me, thank you :) $\endgroup$ – Whiteclaws Mar 10 at 23:52
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The point is how you define that inner product. Therefore, if you define $$\langle x, y\rangle := \dfrac{1}{T_0}\int_{t_1}^{t_1+T_0}xy^*\mathrm{dt}$$ for any $x,y$ being signals with period $T_0$ and $y^*$ is the complex conjugate, you'll have that formula. Since $(e^{nwti})^*=e^{-nwti}$, then for each coefficient $c_k =\langle f(t), e^{kwti}\rangle$.

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  • $\begingroup$ But the inner product of an L2-space is without the 1\T, why is the inner product defined as such? $\endgroup$ – Whiteclaws Mar 10 at 20:12
  • $\begingroup$ It is not without the factor. $L^2$ spaces are defined with respect to a certain measure, often taken to be the Lebesgue, but the measure can vary wildly. $\endgroup$ – Cameron Williams Mar 10 at 22:39

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