Grigory M
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 Jan17 comment Closed-form solution for $f(n) = \sum_{k>0}\binom{n}{2k}x^{k}$ without $\sqrt{x}$ @1234 $\frac{P(x)+P(-x)}2$ extracts even terms from any polynomial $P$, you know... Jan17 comment Closed-form solution for $f(n) = \sum_{k>0}\binom{n}{2k}x^{k}$ without $\sqrt{x}$ Well, do you know the answer for just $\sum \binom nlx^l$? Jan17 comment Elementary proof of the fact that any orientable 3-manifold is parallelizable Do you know the standard proof («$w_1=w_2=0$ implies parallelizable by elementary obstruction theory» + «$w_1=0$ implies $w_2=0$ by Wu's formulas»)? It's not that hard — and at least the first part is, in a sense, the most straightforward approach possible (but I indeed don't know any intuitive explanation of the second part). Jan16 comment If $a,b,c$ are real numbers all less than or equal to $1$ such that $a+b+c=0$ , then is it true that $(1-a)(1-b)(1-c) \le 1$? «which is true» — why, actually? ($ab$ can be negative) Jan16 comment On ${-1 \choose 0}=1$, can I assume that $\frac{(-1)!}{(-1)!}=1$? OK, I apologise — this question was not completely clear as written — but it's a) not a duplicate of the linked question; b) can be reasonably answered. Jan16 comment On ${-1 \choose 0}=1$, can I assume that $\frac{(-1)!}{(-1)!}=1$? Jan15 comment Unimodality of q-binomial coefficients Unimodality of q-binomial coefficients is a difficult theorem, proved more than 20 years after it was conjectured. If you're really interested in a proof — it's easy to google references. Jan15 comment Finite summation with binomial coefficients, $\sum (-1)^k\binom{r}{k} \binom{k/2}{q}$ @655321 though there is one example that is explained in detail in many books (e.g. in Enumerative Combinatorics): Lagrange inversion formula Jan15 comment Finite summation with binomial coefficients, $\sum (-1)^k\binom{r}{k} \binom{k/2}{q}$ @655321 I'm afraid I don't know a good reference... Jan14 comment Embedding of two-dimensional CW complexes which induces a zero homomorphism on second homotopy groups x-posted to MO: mathoverflow.net/q/193845 Jan14 comment When $\frac{C(n, k)}{n^{k-1}} > 1$? Well, for $n=k!+t$ we need to compare $(1+t/k!)\cdot(1-1/(k!+t))(1-2/(k!+t))\ldots(1-(k-1)/(k!+t))$ withs 1. Shouldn't be hard (take logarithm, bound it...) — and at least genesis of $s_k$ is clear. Jan13 comment Summation with Binomial Coefficients, $\sum (-1)^k \binom{m_1}{k} \binom{m_2}{k}$ Jan13 comment Solve easy sums with Binomial Coefficient possible duplicate of Understanding a combinatorial relation. Jan13 comment Summation with Binomial Coefficients, $\sum (-1)^k \binom{m_1}{k} \binom{m_2}{k}$ related: $\sum(-1)^k\binom mk^2$ Jan12 comment Groups with no nontrivial topology (Re: «I am quite interested in how the problem changes if G is infinite or finite.») That's easy to answer: changes from completely trivial (finite) to very hard problem that stayed open for almost 40 years (infinite). Jan12 comment Groups with no nontrivial topology Jan10 comment Are Exponential and Trigonometric Functions the Only Non-Trivial Solutions to $F'(x)=F(x+a)$? related: math.stackexchange.com/q/199691 Jan9 comment Finite summation with binomial coefficients, $\sum (-1)^k\binom{r}{k} \binom{k/2}{q}$ off the top of my head, something like math.stackexchange.com/a/609202 should work (but finding a bijective proof would be, perhaps, more challenging) Jan9 comment Finite summation with binomial coefficients, $\sum (-1)^k\binom{r}{k} \binom{k/2}{q}$ @Marko tomorrow (or later) — maybe; if you have time now — please just go ahead Jan9 comment A three variable binomial coefficient identity Thank you! I'm awarding the bounty now — and will try to understand the proof later.