Timeline for how to get the coefficient of a special term in the expansion of the graph polynomial?
Current License: CC BY-SA 4.0
27 events
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| Oct 30, 2019 at 3:28 | vote | accept | Jacob.Z.Lee | ||
| Jun 22, 2018 at 10:56 | comment | added | Jacob.Z.Lee | @AbdelmalekAbdesselam Thank you very much! yes, actually , it is a multisection of a power series, which was first discovered by Thomas Simpson. | |
| Jun 21, 2018 at 21:12 | comment | added | Abdelmalek Abdesselam | @FedorPetrov: I think the root of unity trick is called "Simpson's dissection". It is mentioned in the book by Andrews, Askey and Roy on special functions. | |
| Jun 17, 2018 at 2:08 | comment | added | Jacob.Z.Lee | I mean, did you introduce the matrix $A$ by the coefficient formula? | |
| Jun 17, 2018 at 1:38 | comment | added | Jacob.Z.Lee | @FedorPetrov Does the matrix $A$ come from the coefficient formula? | |
| Jun 15, 2018 at 8:04 | comment | added | Fedor Petrov | $tr(A^{m})=\sum A_{i_1,i_2} A_{i_2,i_3}\dots A_{i_m,i_1}$ over all $m$-tuples $(i_1,\dots,i_m)$ of the rows of $A$. In our situations the rows of $A$ are enumerated by proper 3-colorings $[a,b,c,d,e]$ of a 5-cycle, and we have to sum up exactly such a product (for $m=2n$). | |
| Jun 14, 2018 at 2:05 | comment | added | Jacob.Z.Lee | @FedorPetrov How does the matrix $A$ come from? Does it from the coefficient formula? Why the answer is the trace of $A^{2n}?$ It has taken me a few days to think about it, but I am still confused. Would you like to explain it in detail? Thanks a lot! | |
| Jun 7, 2018 at 22:25 | comment | added | Jacob.Z.Lee | @FedorPetrov Good. Yes, there is 30 possible values for $[a_i,b_i,c_i,d_i,e_i].$ I will try to find them (eigenvalues of $A$) out. | |
| Jun 7, 2018 at 6:34 | comment | added | Fedor Petrov | We have 30, I guess, possible values for $[a_i,b_i,c_i,d_i,e_i]$ (proper 3-colorings of a 5-cycle.) Denote them $\{P_i\},1\leqslant i\leqslant 30$, further, if $P_k=[v_1,v_2,v_3,v_4,v_5]$ ($v_6:=v_1$), and $P_m=[u_1,u_2,u_3,u_4,u_5]$, we denote $h_k=(-1)^{\sum_{i=1}^5 v_i}\prod_{i=1}^5 \frac{v_{i+1}-v_i}{\binom{4}{v_i}}$; $A_{k,m}=h_k \prod_{i=1}^5 (u_i-v_i)$. The answer is the trace of the matrix $A^{2n}$. It remains to find the eigenvalues of $A$. | |
| Jun 7, 2018 at 0:00 | comment | added | Jacob.Z.Lee | For each i, let $A_i=\{ 0,1,2\},i=1,2,\cdots, 2n$, $[a_i,b_i,c_i,d_i,e_i]$ is a permutation from the set $\{ 0,1,2\}$. There are many cases,not like $C_3\times C_{2n}.$ The cases about $C_5\times C_{2n}$ become more complicated. | |
| Jun 7, 2018 at 0:00 | comment | added | Jacob.Z.Lee | @FedorPetrov When we consider the case about $C_5\times C_{2n}$ and we look for the coefficient of $\prod a_i^2b_i^2c_i^2d_i^2e_i^2$ in the polynomial $F:=\prod_{i=1}^{2n} (a_{i+1}-a_i)(b_{i+1}-b_i)(c_{i+1}-c_i)(d_{i+1}-d_i)(e_{i+1}-e_i)\cdot \prod_{i=1}^{2n} (a_i-b_i)(b_i-c_i)(c_i-d_i)(d_i-e_i)(e_i-a_i)$, where indices are taken modulo $n$. Can we solve this problem through above methods? | |
| Jun 5, 2018 at 21:35 | comment | added | Jacob.Z.Lee | @FedorPetrov Great ! I have found the proof of this formula from the references and your paper with Karasev. Thanks a lot! | |
| Jun 5, 2018 at 21:07 | comment | added | Jacob.Z.Lee | @FedorPetrov Great. I got it. Thank you very much. | |
| Jun 5, 2018 at 15:53 | comment | added | Fedor Petrov | You may read the proof of the formula in any of references I give in the referred MO answer, for example, in our paper with Karasev which I may send you if you write an e-mail. | |
| Jun 5, 2018 at 15:51 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
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| Jun 5, 2018 at 15:43 | comment | added | Fedor Petrov | @Jacob.Z.Lee we fix $\pi_1$ (6 ways to do so), after that we choose $\varepsilon$'s so that their sum is divisible by 3 and sum up $\prod \varepsilon_i$. | |
| Jun 5, 2018 at 15:42 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
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| Jun 5, 2018 at 15:16 | comment | added | Jacob.Z.Lee | @FedorPetrov I calculated some of them.The value of the polynomial should be $4^{2n}\prod_{i=1}^{2n}\epsilon_i$ and the denominator in the formula is always $4^{2n}$, aren't they?. how can we get the equality of the following: $6\sum_{3|\varepsilon_1+\dots+\varepsilon_{2n}}\varepsilon_1\cdot \ldots \cdot\varepsilon_{2n}=6\sum_{3 |m-n} (-1)^m\binom{2n} {m}.$ and why the sum equals $\frac13((1-1)^{2n}+w^{-n}(1-w)^{2n}+w^{-2n}(1-w^2)^{2n})=\frac23 (-3)^n$. Would you like to explain them in detail? I will appreciate it. | |
| Jun 5, 2018 at 14:45 | comment | added | Jacob.Z.Lee | @FedorPetrov Great works! Thanks a lot! I will check it . But where can I find the proof of this formula about the coefficient of $\prod_{i=1}^{n}x_i^{d_i} $ in the expansion of the polynomial $ f(x_1, x_2,\cdots, x_{n})$? | |
| Jun 4, 2018 at 16:58 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
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| Jun 4, 2018 at 16:52 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
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| Jun 4, 2018 at 16:19 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
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| Jun 4, 2018 at 16:15 | comment | added | fedja | Ah, yes, sorry. I just misunderstood the argument entirely. No objection then. | |
| Jun 4, 2018 at 15:37 | comment | added | Fedor Petrov | @fedja I am afraid that I do not understand you. Which second and fourth powers? | |
| Jun 4, 2018 at 15:02 | comment | added | fedja | Erm... The non-zero value may also come from the product of second and fourth powers. How do you account for those? | |
| Jun 4, 2018 at 14:02 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
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| Jun 4, 2018 at 13:40 | history | answered | Fedor Petrov | CC BY-SA 4.0 |