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I know that we can represent some function $f$ in this way:

$$f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n\cos\left(\frac{n\pi t}{L}\right) + \sum_{n=1}^\infty b_n\sin\left(\frac{n\pi t}{L}\right)$$

Because we can suppose $f$ is written in this form, and then multiply if by $\cos(x)$ and also by $\sin(x)$, integrate in its period and then use some properties of orthogonal integration to zero out every coeficiente except the one you need. Then you isolate $a_n$ and $b_n$ to get:

$$\begin{align}a_0 &= \frac{1}{L}\int_{-L}^{L}f(t)dt \\a_n &= \frac{1}{L}\int_{-L}^{L}f(t)\cos\left(\frac{n\pi t}{L}\right) dt \\b_n &= \frac{1}{L}\int_{-L}^{L}f(t)\sin\left(\frac{n\pi t}{L}\right) dt\end{align} $$

This is nice, I understand how and why we can write the coefficients in this way. I've learned all this in a book, but it didn't explain me where the $a_0$ comes from.

However, my real question is: how we go from this, to the Fourier transform? What's so good about Fourier Series that able me to create a transformation? What's the intuition? What is the reason someone, looking for these series, imagine a transformation?

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  • $\begingroup$ Why can't you see how they find $a_0$? $\endgroup$ – Mhenni Benghorbal Sep 14 '14 at 5:38
  • $\begingroup$ See this where they go from Fourier series to Fourier transform, A technique which is used in quantum mechanics. $\endgroup$ – Mhenni Benghorbal Sep 14 '14 at 5:54
  • $\begingroup$ @Guerlando OCs: $a_0$ and $b_0$ comes from setting $n=0$ in the formulas for $a_n$ and $b_n$. Not just "some function" can be written this way, but any continuous function on a limited interval... $\endgroup$ – Lehs Sep 14 '14 at 6:01
  • $\begingroup$ @Lehs so why they make people remember the formula for $a_0$? And why they don't make one for $b_0$? $\endgroup$ – Guerlando OCs Sep 14 '14 at 21:48
  • $\begingroup$ @Guerlando OCs: I don't know, but $b_0=0$. Now, hope someone can explain the transform for us. $\endgroup$ – Lehs Sep 14 '14 at 22:13
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There is a quite nice way of explaining the connection of Fourier transforms and Fourier series using distributions. Let us first write the function $f$ as a complex Fourier series $$ f(x)=\sum_{n=-\infty}^\infty c_n e^{in\pi x/L}. $$

The next step is to compute the distributional Fourier transform of a pure exponential of the form $g_n(x)=\exp(in\pi x/L)$, which is given by $\mathcal F(g)=\delta_{n/L}$. The delta distribution $\delta_x$ is explained here. It is loosely speaking just a spike located at $x$ and $0$ elsewhere, I don't want to go into detail on distributional Fourier transforms. Since the Fourier transform is linear we can transform $f$ to find $$ \hat f(\xi)= \sum_{n=-\infty}^\infty c_n \delta_{n/L}, $$ so $\mathcal F(f)$ is a train of spikes. The spacing between these spikes is $1/L$, which corresponds to the periodicity of the function $f$. Now if we want to Fourier transform a general square integrable function we can not assume any periodicity so we let $L \rightarrow \infty$. By doing so the spacing between the spikes becomes smaller and we get closer and closer to cover the whole real line. I understand that this argument is not air tight but it should give you an intuition of what the connection between Fourier series and Fourier transforms is.

In general the standard Fourier transform is defined for square integrable functions, so they need some decay. Fourier series are intended to be computed for functions that are periodic, so they have no decay. The distributional Fourier transform I used above is defined for both classes so it can be used to develop an intuition for the connection. I can also recommend reading Stephane Mallat's bood A wavelet tour of signal processing, page 50.

Hope this helps...

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