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I'm talking about the incredible answer from Mark Eichenlaub.

Fourier transform for dummies

He said :

"we must allow the circles to have complex radii. It's the same thing as saying the circles have real radii, but they do not all have to start at the same place."

and

"If your path closes on itself, the Fourier transform turns out to simplify to a Fourier series. Most frequencies are no longer necessary"

could one explain them to me please?

I still have troubles to understand the importance of the imaginary part in fourier coefficient (even though they give the sinus part, since you can see it by the fact that if a fonction is even then the fourier transform is in R). I don't understand if they have any meaning like kinda the phase or anything else.

Moreover, i don't understand the link between Fourier series and the Fourier transform.

On the one side, if we assume strong enough hypothesis on the function $f$ : $$ f(x) = \int_{\mathbb R} \mathcal F (\alpha) e^{ 2 \pi \alpha x } d \alpha $$ on the other : $$ f(x) = \sum_n \frac 1 T \int_{0}^T f(y) e^{ \frac {2 i \pi (x-y) } T } dy$$ which I think is equal to :

$$ f(x) = \sum_n \frac 1 T \int_{ \mathbb R^+ } f(y) dy$$

when you take the limit of T going to infinity, the usual case of the fourier transform ; where $T$ is the period of the function $f$

Marine Galantin
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1 Answers1

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The polar representation of a complex number is $r e^{i\theta}$ where $r \ge 0$ is the distance from the origin and $\theta$ is the argument (the angle from the positive $x$ axis).

A circle in the complex plane, centred at $0$, with (positive) radius $r$, consists of all points at distance $r$ from the origin, and thus can be written as $z = r e^{i\theta}$. With $r$ replaced by $r e^{i\alpha}$ , it becomes $z = r e^{i\alpha} e^{i\theta} = r e^{i(\theta + \alpha)}$. This is still the circle of radius $r$, but $\theta$ has been shifted.

Robert Israel
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