Timeline for Discrete Fourier transform of the Ramanujan's sums
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 9, 2017 at 10:48 | comment | added | user70925 | Because I'm interested in the eigenvalues of the circulant matrix of size $\phi(n)$ associated to the $x(l)$. So I can't freely extend the range of $l$. | |
| Sep 9, 2017 at 7:41 | comment | added | reuns | The discrete Fourier transform of $x(l)=\sum_{l \in \mathbb{Z}_n^*}^n \zeta_n^{kl},l=0 \ldots n-1$ is obvious. Why would you want to consider $l=0 \ldots \phi(n)-1$ ? | |
| Sep 8, 2017 at 18:51 | comment | added | user70925 | This is indeed, but the formulas known for Ramanujan's sums didn't helped to settle this sum. I edited the title accordingly nonetheless, it might be a bit clearer for specialists. | |
| Sep 8, 2017 at 18:40 | history | edited | user70925 | CC BY-SA 3.0 |
Added the link to Ramanujan's sums
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| S Sep 8, 2017 at 17:08 | history | suggested | Somos | CC BY-SA 3.0 |
Fixed typo. Space edits.
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| Sep 8, 2017 at 16:51 | review | Suggested edits | |||
| S Sep 8, 2017 at 17:08 | |||||
| Sep 8, 2017 at 16:46 | comment | added | Somos | The coefficient of $X^k$ is Ramanujan's sum $c_k(n)$. | |
| Sep 8, 2017 at 15:46 | comment | added | user70925 | It's corrected, thanks for noticing. | |
| Sep 8, 2017 at 15:45 | history | edited | user70925 | CC BY-SA 3.0 |
added 3 characters in body
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| Sep 8, 2017 at 15:34 | comment | added | Marcel | perhaps you mixed up $k$ and $n$? The coefficient of $X^k$ doesnt depend on $k$... | |
| Sep 8, 2017 at 15:23 | history | asked | user70925 | CC BY-SA 3.0 |