Timeline for The discrete Fourier transform's Gaussian-like eigenvector
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Jul 24, 2022 at 2:29 | vote | accept | bobuhito | ||
| Jul 18, 2022 at 13:43 | comment | added | Fred Hucht | The first order correction $c_1$ can be obtained by first order perturbation theory. For $N\to\infty$, the leading eigenvector $|v_0^{(1)}\rangle$ is Gaussian, with width $\sigma^2=N/(2\pi)$, while in 0th order $\lambda_0^{(0)}=4$ and $|v_0^{(0)}\rangle=\{1,0,0,\ldots\}$. Therefore, we have $$ S |v_0^{(1)}\rangle = \lambda_0^{(1)} |v_0^{(1)}\rangle + \mathcal{O}(N^{-2}).$$ This gives $\lambda_0^{(1)}=4-2\pi/N+\pi^2/N^2+\ldots$, which is correct to first order. Higher orders are more involved. | |
| Jul 17, 2022 at 21:36 | comment | added | bobuhito | Any simple explanation for why $c_1$ appears to be exactly 2? or, if that is easy, why $c_2$ appears to be exactly 1/2? I think that these two terms should just come from the first and second derivatives (with respect to $\omega$) of your best numerical $\lambda$ table. | |
| Jul 17, 2022 at 19:15 | history | edited | Fred Hucht | CC BY-SA 4.0 |
Typos fixed
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| Jul 17, 2022 at 13:02 | history | edited | Fred Hucht | CC BY-SA 4.0 |
Added the comment below to the answer.
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| Jul 16, 2022 at 22:55 | comment | added | Fred Hucht | With 100 digits precision, I got some more terms, $$ \ldots +\frac{32519 \pi ^7}{5160960 N^7} + \frac{135001 \pi ^8}{20643840 N^8} + \frac{45750727 \pi ^9}{5945425920 N^9} + \frac{1198585643 \pi ^{10}}{118908518400 N^{10}} + \ldots . $$ | |
| Jul 16, 2022 at 18:19 | comment | added | Fred Hucht | The fit gives the exact expansion (at least up to $c_4$) of $\lambda_0(N)$ around $N=\infty$. This expansion obviously has simple rational coefficients in $\pi/N$. One could also rewrite the expansion in terms of $\omega=2\pi/N$, and get a rational expansion in $\omega$ around $\omega=0$. | |
| Jul 16, 2022 at 17:32 | comment | added | bobuhito | I am kind of surprised that the low-order c coefficients do not find a more-precise fit. It is as if they were forced to be of the rational form 2/A (where A is an integer). Anyway, as a more thorough check, I wonder how nice the coefficients would look if you expand using a 10th order polynomial in e/N or sqrt(2)/N (or any other famous irrational constant). | |
| Jul 16, 2022 at 16:44 | comment | added | Fred Hucht | No, I still allow for higher order terms, see my updated answer. | |
| Jul 16, 2022 at 16:38 | history | edited | Fred Hucht | CC BY-SA 4.0 |
Added more material as denoted
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| Jul 16, 2022 at 13:52 | comment | added | bobuhito | Wow! But, my instinct is that $c_5$ should be rational and the "..." part would then have bigger (also rational) coefficients to account for the difference. Aren't you simply forcing the "..." to zero for your "high-precision results"? | |
| Jul 16, 2022 at 9:11 | history | edited | Fred Hucht | CC BY-SA 4.0 |
deleted 79 characters in body
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| Jul 16, 2022 at 8:38 | history | edited | Fred Hucht | CC BY-SA 4.0 |
added 34 characters in body
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| Jul 16, 2022 at 8:29 | history | edited | Fred Hucht | CC BY-SA 4.0 |
Fixed error in $c_5$.
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| Jul 16, 2022 at 8:15 | history | edited | Fred Hucht | CC BY-SA 4.0 |
Added table and $c_5$.
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| Jul 16, 2022 at 8:00 | history | answered | Fred Hucht | CC BY-SA 4.0 |