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comment Simply Transitive Group Actions and Functions
A counterexample to your guess: If $G = X = S_3$, and $G$ acts by left multiplication, then left multiplication by $(1 2)$ is a bijection $f:X \rightarrow X$, yet $(1 2) gx = g (1 2)x$ is not true for all $x$ and $g$ in $S_3$. (Any nonabelian group causes problems for the guess.)
Aug
15
comment Is det(A) maximal, if det(A+E) is maximal?
Thank you - sorry the answer (a partial one at that!) is long. The short version is that there are formulas for $\det(A+E)$ involving the matrix $A^c$ obtained by switching 1's and 0's. These can be applied to show that the conjecture is true if $M > 3m - 3$, where $M$ and $m$ are the maximal 1-2 and 0-1 determinants, respectively. When that bound doesn't hold, we've seen that $A^c$ can take on different values, so the possibility is there for the conjecture to fail. I thought I should give at least some justification for my claims, though!
Aug
14
answered Is det(A) maximal, if det(A+E) is maximal?
Jul
5
awarded  Yearling
Jun
24
comment bijective function $h:\mathbb{N}\rightarrow A\cup B$ from bijective functions $f:\mathbb{N}\rightarrow A$, $g:\mathbb{N}\rightarrow B$
The answer in the question you linked to is correct up to a minor detail/possible typo.
Jun
24
comment Prove that a function is bijective
You mean $h(n+1) = g(k)$, where $k$ is the smallest natural number such that $g(k) \not\in \{h(1),\ldots,h(n)\}$ (if $h(n) = f(m)$ for some $m$)? After all, the sets $A$ and $B$ need not be well-ordered themselves.
Jun
4
comment Relations counting in two sets
Your 2^15 should be 2^{15}. (This was my first attempted outside edit - I guess the site requires more "substantial" edits!)
Jun
3
comment angles between simple roots are obtuse, problem with proof
The negation of $(\beta,\alpha) \leq 0$ is that $(\beta, \alpha) > 0$, so you should change those weak inequalities to strict ones in your argument. The argument works nonetheless.
Jun
3
revised angles between simple roots are obtuse, problem with proof
Minor edits
Jun
3
answered angles between simple roots are obtuse, problem with proof
Jun
3
answered Mathematical concept for formal languages
May
27
comment Divergence of the sequence $\sin(n!)$
Same question for $\cos(n!)$, with some interesting commentary and answers: math.stackexchange.com/questions/8690/is-there-a-limit-of-cos-n
May
18
comment Problem with commutator relations
No problem: Actually, just to backtrack a bit on that previous thought - how do we know $H'''(0) = 0$? Here is where it helps to rearrange $H''(\lambda)$ so that you have $H''(\lambda) = 0$, regardless of $\lambda$, not just $H''(0) = 0$: If $H''(\lambda) = 0$, then $H'''(\lambda)$ and all higher derivatives equal. So there is a point to rewriting it in a nice form.
May
18
comment Problem with commutator relations
So, it all falls out nicely, once you calculate $H''(0)$ using the commutator hypotheses for $C$ and $D$.
May
18
comment Problem with commutator relations
Good point. I suppose we don't need the commuting operator. We just need to evaluate at $\lambda = 0$ to see it. I saw it easier after rearranging and applying commutativity where I could.
May
18
comment Problem with commutator relations
It's not that $H'(\lambda) = -DC + CD$; that's not true according to your first calculation. It's that $H'(0) = -DC + CD$.
May
18
comment Problem with commutator relations
No, $H'(0)$ isn't $0$, since $e^0$ is the identity operator. By commuting, I mean that $D e^{-\lambda D} = e^{-\lambda D} D$. If you are wondering what I mean by expansion, I mean this: $De^{-\lambda D} = D - \lambda D^2 + \lambda^2 / 2 D^3 +...$ You can factor this as $D(I - \lambda D + ...)$ or as $(I - \lambda D + ...) D$. This is why the two operators commute.
May
18
comment Problem with commutator relations
Also, the expansion in $\lambda$ consists of evaluating $H(0)$, $H'(0)$, etc, and writing $H(\lambda) = \sum_{n=0}^\infty \frac{H^{(n)}(0)}{n!} \lambda^n$. Since the desired answer stops at the $\lambda$ term, we expect (or at least hope!) the second derivative (and hence all remaining derivatives) to be 0. So, you likely need to use $[C, [C,D]] = 0$ or $[D,[C,D]] = 0$ to show that.
May
18
comment Problem with commutator relations
The following may help: $D$ and $e^{-\lambda D}$ commute. This is because the terms in the expansion of $D e^{-d\lambda D}$ consists of a scalar times a power of $D$, so you can factor $D$ on either the right or the left.
May
8
revised Asymmetric roles that are symmetric in every instance
deleted 3811 characters in body