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Feb
22
comment Theorems Implying their Own Generalization
@M10687: I don't know if your wording "which were later found" was meant to indicate a historical advance from specific to general or not. I gave my favorite nontrivial example, but this is a common phenomenon. For example, statements about simple functions are frequently used (as lemmas) to prove statements about measurable functions in measure theory.
Feb
22
answered Theorems Implying their Own Generalization
Feb
18
awarded  Mortarboard
Feb
17
revised Finding where the tail starts for a probability distribution, from its generating function
Cleaned up the language a bit, used ceiling instead of round for the approximations to N, as suggested by Kirill.
Feb
15
comment Finding where the tail starts for a probability distribution, from its generating function
@Kirill: Indeed, I did round and agree ceiling is more appropriate. Very nice additional information. I didn't arrive at $\alpha$ through Newton's method, but I find it interesting that $\alpha$ is the first iterate if the initial guess is $z = 1$.
Feb
14
answered Finding where the tail starts for a probability distribution, from its generating function
Feb
12
comment $S_n$ as a Coxeter group? (with Matsumoto's theorem)
+1: Bjorner and Brenti's coverage of these topics is very accessible. Matsumoto's theorem is theorem 3.3.1 in that book. They attribute the theorem to Tits. Geck and Pfeiffer's Characters of Finite Coxeter Groups and Iwahori-Hecke Algebras attribute it to Matsumoto. It is Theorem 1.2.2 in that book.
Feb
11
answered free semi-group
Jan
14
comment Embeddings into symmetric structures
I don't think this is exactly what you seek, but it is interesting and related: math.stackexchange.com/questions/1701/…
Jan
11
answered If $w'(\beta)<0$ and $\ell(w)+\ell(w')=\ell(ww')$, then $ww'(\beta)<0$?
Dec
6
comment Are you allowed to substitute for imaginary numbers on a real integral?
The $i$ is from $dx = i dc$, not from $f(ci) = i f(c)$.
Nov
6
comment The longest word in Weyl group and positive roots.
It occurs to me that I was mixing orders a bit: Given any reduced decomposition for $w$, we can order the positive roots sent negative by $w$ in the order in which the reflections sent them to a negative root. The longest element maps all positive roots to negative roots. The order of the < in my comment is such that there exists a reflection ordering equal to it. (Hence the conflation of these two orders in my first comment.)
Nov
6
comment The longest word in Weyl group and positive roots.
For two vectors in $\mathbb{R}^n$ relative to a fixed ordered basis $(a_1,\ldots,a_n) \leq_{lex} (b_1,\ldots,b_n)$ if $a_1 < b_1$ or if $a_1 = b_1, \ldots a_i = b_i$, but $a_{i+1} < b_{i+1}$. Precedence of comparison goes to the initial coordinate where the coordinates differ. (It is similar to alphabetical order, hence the name.) I believe what you are referring to is "height". Among simple roots $\alpha_1,\alpha_2$, $\alpha_1 < \alpha_2$ in lex order but are incomparable in the order you mention. The $<$ I was using is decomposition-based, but can be made to match lex order on roots.
Nov
5
comment The longest word in Weyl group and positive roots.
Section 5.2 of Bjorner and Brenti's "Combinatorics of Coxeter Groups". The order is described within the proof of Proposition 5.2.1 on page 137.
Jul
5
awarded  Yearling
Mar
31
comment Lemma regarding subgroup of group of automorphisms
The automorphism $\tau$ is applied to the coefficients of $P(x)$: If $P(x) = a_n x^n + \cdots + a_0$, then $\tau P(x) = \tau(a_n) x^n + \cdots + \tau(a_0)$. Since $\tau$ extends to a ring homomorphism from $A[x]$ to $A[x]$, we can apply $\tau$ to each factor in the product representation. The point is that $\tau P(x) = P(x)$ implies $\tau(a_i) = a_i$ for any coefficient $a_i$ of $P(x)$. Since this holds for all $\tau \in G$, the coefficients of $P(x)$ are in $A^G$.
Mar
26
answered 4D Geometry Book
Mar
22
comment What happens if singleton set is not closed
The trivial topology (only $\emptyset$ and $X$ are open) is an example of a topology on $X$ such that singleton sets are not closed, assuming $|X| > 1$.
Mar
6
answered Reference request: calculus of variations
Mar
3
revised How would I put these recurrence relation terms into a summation?
Inserted \left and \right on parentheses that needed them.