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From equation (II) in Newman's A simple proof of the partition formula: $$f(z) = \sqrt{\frac{2 \pi}{1-z}} \exp\left(-\frac{\pi^2}{6(1-z)}+\frac{\pi^2}{12} \right) (1+O(1-z))$$ Note that Newman's $f …

There should be a relationship like this, because the heat function proof of Hirzebruch-Riemmann-Roch uses a much more refined equality between hilbert functions and Laplacian spectra. Let me see if I …

Meanwhile, I'll answer the other question about random walks: Yes, it is quite easy to prove random walks in $\geq 3$ dimensions are transient without Stirling's approximation. Indeed, this is the sta …

Your sequence is bounded by $(125+\epsilon)^n$. Obviously, this isn't close to a good bound, but it answers the question. We start by bounding a different question: Let $\Gamma_n$ be the convex hull …

I think I might see what was confusing me. This is really a comment, but it's too long for the comment thread. As my example, let's take $$A(t) = \frac{1}{1+t^2} \begin{pmatrix} 2 & t \\ -t & -2 \end{ …

$, and asymptotics are given by Stirling's formula. …

I'll lay out the starting steps; I hope that after that it won't be much work for you to fill in on your own. To be clear, the process is that there are a succession of dances. At the start, $n$ coup …

No. Take $f(x) = 1 + y^2 + (xy-1)^2$. Let $$D = \{ (x,y) : 0 \leq x,\ 0 \leq y \leq 2,\ xy \leq 2 \}.$$ Then $f(x,y) \leq 3$ on $D$, so the integral of $1/f$ is bounded below by $(1/3) \mathrm{Area}( …

I've played around with this a bit. I have a slick lower bound, but not a slick upper bound. We start with the $\Gamma$-integral: $$n! = \int_{x=0}^\infty x^n e^{-x} dx = \int_{y=-n}^\infty (n+y)^n e^ …

This is an attempt to summarize some work related to question 2 which I do not fully understand myself. I am summarizing Sections 1.1 and 1.2 of Dudko's thesis, whose exposition is excellent (includin …

Back on Scott Aaronson's blog, I gave an argument that $e^z+z-1$ should have an analytic compositional square root. The important difference between this function and $e^z-1$ was that the fixed point …

Let me see if I can summarize the conversation so far. If we want $f(f(z)) = e^z+z-1$, then there will be a solution, analytic in a neighborhood of the real axis. See either fedja's Banach space argu …