Mark Wildon
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Registered User
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Apr 10 |
answered | Partitions-Sum of divisors identity |
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Apr 3 |
comment |
Identity involving Fresnel integrals That's interesting. I stopped at the final integral because it can be done in a routine way with a reduction formula: write $(1-u^2)^{2n} = (1-u^2)^{2n-1} - 2u(1-u^2) u/2$ and use integration by parts on the second term. I don't know of a combinatorial proof. |
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Apr 3 |
accepted | Identity involving Fresnel integrals |
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Apr 2 |
revised |
Identity involving Fresnel integrals edited tags |
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Apr 2 |
answered | Identity involving Fresnel integrals |
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Apr 1 |
answered | Motivation and Intuition for Sprague-Grundy Theorem |
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Apr 1 |
comment |
Motivation and Intuition for Sprague-Grundy Theorem Just to expand on the final line: in the game $G + \star n$, if the first player plays to $\star m$ in $G$ where $m < n$, then the second player can play in $\star n$ to give $\star m + \star m = 0$; if the first player plays to $\star m'$ in $\star n$ where $m' < n$ then the second player can play in $G$ to $\star m'$ to give $\star m' + \star m' = 0$. (We know that $\star m'$ is an option of $G$ because $n$ is the minimum excluded option.) Hence $G + \star n = 0$ and $G = \star n$. |
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Feb 21 |
revised |
Box removing operators on partitions Many corrections. |
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Feb 20 |
revised |
Box removing operators on partitions Corrected LaTeX. |
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Feb 20 |
revised |
Box removing operators on partitions Repaired argument so it works for the question as posed. |
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Feb 20 |
revised |
Box removing operators on partitions deleted 2505 characters in body |
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Feb 20 |
revised |
Box removing operators on partitions added 25 characters in body |
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Feb 20 |
answered | Box removing operators on partitions |
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Feb 16 |
accepted | Known decomposition of $\bigwedge^k Sym^d \mathbb C^n$ in special cases? |
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Feb 15 |
comment |
Known decomposition of $\bigwedge^k Sym^d \mathbb C^n$ in special cases? Thank you for the reference. As you say in your paper, it's a remarkable isomorphism. |
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Feb 14 |
revised |
Known decomposition of $\bigwedge^k Sym^d \mathbb C^n$ in special cases? Forgot to say that \lambda has distinct parts |
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Feb 14 |
comment |
Known decomposition of $\bigwedge^k Sym^d \mathbb C^n$ in special cases? When $n=2$ the plethysm ${\rm Sym^k}{\rm Sym^d}V$ is given by the Cayley-Sylvester formula. I can't see why this is equivalent to knowing your plethysm. But I think it should be possible to work out the constituents of $\bigwedge^k ({\rm Sym}^d V)$ using similar arguments with formal characters of ${\rm SL_2}(\mathbb{C})$. |
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Feb 14 |
answered | Known decomposition of $\bigwedge^k Sym^d \mathbb C^n$ in special cases? |
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Jan 12 |
comment |
An inequality for the ratio of standard Young tableau with {1,2,…,k} in the first row I agree with darij grinberg's comment. Do you know if there is a refinement to a chain of results: $\frac{\dim \lambda / (k)}{\dim \beta / (k)} \ge \frac{\dim \lambda / (k-1)}{\dim \beta / (k-1)}$ for $k \ge 1$? |
