Timeline for The $2\pi$ in the definition of the Fourier transform
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 26, 2019 at 17:16 | comment | added | Charles Hudgins | Fascinating. I think this is the first time I've really appreciated how remarkable it is that $2\pi$ appears in that equation. | |
| Jun 26, 2019 at 2:01 | comment | added | WhatsUp | Yes. It all comes to the identity $e^{2\pi i} = 1$. | |
| Jun 25, 2019 at 1:28 | comment | added | Charles Hudgins | So is the idea roughly that we're implicitly using an isomorphism between $S^1$ and $\mathbb{R}/\mathbb{Z}$ that scales volumes by $\frac{1}{2\pi}$? | |
| Jun 24, 2019 at 11:25 | comment | added | WhatsUp | I don't have much detail here, but the deeper reason should be the fact that $x \mapsto e^{2\pi ix}$ has kernel $\mathbb{Z}$ and that $\mathbb{R}/\mathbb{Z}$ has total volume $1$ under Lebesgue measure. | |
| Jun 23, 2019 at 12:59 | comment | added | Charles Hudgins | I realize this post is from over 2 years ago, but I've been wondering why the constant gets pinned to $2\pi$. One can show that it must be $2\pi$ by considering test functions, but I wonder if there isn't something deeper going on. Is there a more abstract way to see why it should be $2\pi$? | |
| Mar 23, 2017 at 14:27 | history | answered | WhatsUp | CC BY-SA 3.0 |