bio | website | math.sunysb.edu/~azinger |
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location | ||
age | ||
visits | member for | 3 years, 1 month |
seen | Jun 24 '13 at 16:56 | |
stats | profile views | 107 |
May 31 |
awarded | Critic |
Apr 14 |
comment |
On the proof of Robert Lipshitz's formula on Maslov index.
Why not just e-mail, Robert? |
Mar 31 |
answered | Difference between parallel transport and derivative of the exponential map |
Feb 19 |
awarded | Enthusiast |
Jan 29 |
answered | Analytic curve on Riemann surface |
Jan 28 |
revised |
Tubular neighborhoods of chains
added 36 characters in body |
Jan 28 |
revised |
Tubular neighborhoods of chains
added 156 characters in body; deleted 103 characters in body |
Jan 28 |
answered | Tubular neighborhoods of chains |
Jan 24 |
awarded | Scholar |
Jan 24 |
awarded | Supporter |
Jan 24 |
awarded | Student |
Jan 21 |
awarded | Editor |
Jan 21 |
revised |
Orientations for pseudoholomorphic curves with totally real boundary condition
added 773 characters in body |
Jan 21 |
answered | Orientations for pseudoholomorphic curves with totally real boundary condition |
Oct 10 |
awarded | Teacher |
Oct 10 |
answered | Why are Gromov-Witten invariants of K3 surfaces trivial? |
Jun 7 |
comment |
number of weighted trivalent trees
Thank you very much, everyone. Paul's answer in particular completely resolves my question. Here is a direct reformulation of his argument. Let $a_n$ be the weighted number of graphs with $n+1$ marked points divided by $n!$ and $f(x)=x+\sum_{n\ge2}a_n x^n$. From the natural recursion for $a_n$, $$f(x)=x+f(x)+(f(x)-1)\sum_{k=1}^{\infty}\frac{f(x)^k}{k} \quad\longrightarrow\quad (1-f(x))\ln(1-f(x))=-x.$$ Thus, $f(x)=1-e^{W(-x)}$. Using the $r=-1$ case of the above formula then gives $a_n=(n-1)^{(n-1)}/n!$. |
Jun 7 |
accepted | number of weighted trivalent trees |
Jun 5 |
comment |
number of weighted trivalent trees
This is indeed related to moduli spaces (of stable maps). The structure coefficients in formulas for genus 0 Gromov-Witten invariants are sums over $N$ marked trivalent trees. These can be bounded by the number of weighted trees in my question. I can show this number is bounded above by $C^N\cdot N!$, which is good enough for me. However, instead of adding another half a page proving this, I thought I might be able to quote something from the literature (and then apply Stirling's formula to get $C^N\cdot N!$). Since valence - 3 is so natural for trivalent, I thought this formula were known... |
Jun 5 |
asked | number of weighted trivalent trees |