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Questions For any positive integer $r$, compute $$(\frac{d}{dY})^r e^{(Y^2)}| _{Y=0}.$$ The answer should directly relates to a counting problem about Feynman diagrams. Is there a tutorial for how F …

This question is motivated by Why do combinatorial abstractions of geometric objects behave so well? The algebraic geometry of Kazhdan-Lusztig-Stanley polynomials Kazhdan-Lusztig-Stanley polynomial …

$\DeclareMathOperator\Rep{Rep}$It is well-known that the Macdonald "$P$" polynomials deform the Jack "$J$" polynomials [1]. The latter have profound relations with representation theory. For example, …

In Chung and Yau's paper: "A combinatorial trace formula" (MSN), they proved a combinatorial version of Selberg's trace formula for lattice graphs. I learned also in the setup that it makes sense to d …

The first 24 lectures of Jacob Lurie on Geometric Topology [1] gave a concise introduction to the comparison of smooth manifolds and piecewise-linear manifold. In the first five lectures, it is shown …

Motivation: Euler-Maclaurin formula uses calculus to estimate discrete sums. I wonder what one can do by reverse engineering. A concrete problem I ran into is the following. Question: Let $\chi: \math …

Let $F$ be a generating function $F(x) = \sum_{i=0}^\infty f_i x^i$, and suppose that we can do operations formally without worrying about convergence issues. Define the coefficients \begin{gather*} a …

Let $L$ be a link in $\mathbb{R}^3$, with $D, D'$ be diagrams in $\mathbb{R}^2$ representing $L$. Khovanov constructed two graded chain complexes $$C_{D} = (Ch_{D}, d_{D}) \quad C_{D'}=(Ch_{D'}, d_{D' …

Both cohomology and homotopy groups capture global topological information of a manifold $X$. It is interesting to ask if they can be computed from local data. A triangulation $T$ is a natural present …

I was recently introduced to analytic combinatorics, and found the method of removing poles astonishing. More precisely, I was reading the last chapter of the popular "generatingfunctionology", in whi …

An interesting combinatorial identity is the Vandermonde convolution identity: $$ \sum_k {n\choose k}{m\choose s-k} = {n+m \choose s},$$ which can be proved by considering the coefficients in $(x+1)^{ …