Timeline for Need explicit formula for certain "$q$-numbers" involving gcd's
Current License: CC BY-SA 3.0
26 events
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Dec 7, 2015 at 15:39 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
corrected alignment in the "o" list
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| Jul 2, 2015 at 14:50 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
added 102 characters in body
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| Jul 2, 2015 at 2:49 | comment | added | მამუკა ჯიბლაძე | @OfirGorodetsky Thanks for the link! Reading that - seems an interesting idea to give it a combinatorial interpretation in terms of the Möbius function of some poset. In fact this might be important for the original problem which is of combinatorial nature. | |
| Jul 2, 2015 at 2:45 | comment | added | მამუკა ჯიბლაძე | @alpoge But this is great! I was sticking to $e_0$ so much it never occurred to me that dropping it gives such a nice orthogonal basis! It is true that I still need $e_0$ essentially (it is the unit of certain algebra), but your version does almost all of it. Could you please post this as an answer? It's worth it in any case, and if nobody will find a way to incorporate $e_0$ I will go for it | |
| Jul 1, 2015 at 21:55 | comment | added | Ofir Gorodetsky | Just a remark\restatement, nothing fancy: Let $A_{i,j} = q^{(i,j)}$ be a $(n+1) \times(n+1)$ matrix. Decompose $A$ as $B^T B$ (Cholesky-Decomposition; $B$ can be upper-triangular). The condition that $o_i = \sum_{j} o_{i,j} e_j$ are orthonormal w.r.t to the given inner product is the same as requiring that $B o_i$ are orthonormal w.r.t to the ordinary inner product, so we can take $o_i = B^{-1} e_i$. Hence the problem reduces to Cholesky-decomposing $A$, and inverting. Bruce Sagan's talk here about GCD-matrices might provide useful input: users.math.msu.edu/users/sagan/Slides/mfp5.pdf | |
| Jul 1, 2015 at 21:06 | comment | added | alpoge | Formally doesn't $o_n := \sum_{d\vert n} \mu(d) e_{n/d}$ for $n > 0$ and $o_0 := e_0 - \sum_{n\geq 1} o_n$ work? (Not that that helps much, since the expression for $o_0$ is nonsense.) | |
| Jul 1, 2015 at 18:04 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
Added the q=-1 case
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| Jul 1, 2015 at 17:59 | comment | added | მამუკა ჯიბლაძე | @Wolfgang Well $q=-1$ looks somehow less hopeless but still I have no clue :) I'll add it too | |
| Jul 1, 2015 at 7:23 | comment | added | Wolfgang | and I suppose the same with $q=-1$ doesn't reveal more either? | |
| Jul 1, 2015 at 7:18 | comment | added | Wolfgang | I had skipped $n=6 $, so that was a wrong conclusion of mine! :(. | |
| Jul 1, 2015 at 6:23 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
Alternative expressions for e_n seem to be more readable
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| Jul 1, 2015 at 6:17 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
Alternative expressions for e_n seem to be more readable
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| Jul 1, 2015 at 5:37 | comment | added | მამუკა ჯიბლაძე | @PeterMueller I've added that | |
| Jul 1, 2015 at 5:34 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
Explained a question raised in a comment
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| Jul 1, 2015 at 5:31 | comment | added | მამუკა ჯიბლაძე | @PeterMueller I use $\gcd(0,0)=0$, I think it is not senseless in view of $\gcd(0,n)=\gcd(n,n)=n$ for all $n$. | |
| Jun 30, 2015 at 21:52 | comment | added | Peter Mueller | @OP: What is $\langle e_0,e_0\rangle$? It seems to me that the problem is not well formulated, as $gcd(0,0)=\infty$. In particular, I don't know how to interpret your assertion $0=\langle o_0,o_1\rangle=\langle e_0,e_1-qe_0\rangle=q-q\cdot q^{gcd(0,0)}$. | |
| Jun 30, 2015 at 20:48 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
Added a table according to the request from a comment
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| Jun 30, 2015 at 20:06 | comment | added | მამუკა ჯიბლაძე | @Wolfgang I've looked at it, what happens that seeminglu each of these polynomials (well, starting from 3) is divisible by $(x-1)^2(x+1)$, but the quotient seems impenetrable to me. I'll add it to the question though | |
| Jun 30, 2015 at 13:01 | comment | added | მამუკა ჯიბლაძე | @Wolfgang Many thanks for the suggestion, sounds promising! I'll definitely try it | |
| Jun 30, 2015 at 13:00 | comment | added | Wolfgang | Have you tried putting $q=1$ and replacing $e_i$ by $x^i$ in the $o_j$ expressions? The resulting polynomials seem to split into linear and quadratic factors, and maybe you'll find some patterns in those factors, which might give some ideas to start with. | |
| Jun 30, 2015 at 8:45 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
deleted 39 characters in body
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| Jun 30, 2015 at 8:34 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
added 25 characters in body
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| Jun 30, 2015 at 8:28 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
Added the inverse transformation
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| Jun 30, 2015 at 8:11 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
Just noticed the fact...
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| Jun 30, 2015 at 7:37 | history | asked | მამუკა ჯიბლაძე | CC BY-SA 3.0 |