Then only one I really cared about on SO I just got.
Unsung Hero.
Those were some yummy enchiladas.
It's like a spicy meat pie.
Dunno didn't get the badge, there should be a qualifier on the description that says it's also the highest score or top score per question, or something like that.
I tried sharing my answer on Reddit, but I think most of them came over and downvoted me for it.
@AaronHall I don't think so. From what I gathered on MSO, it works as described. The accepted answer must have a score greater than 10, and yours must be the highest voted, and have a score more than twice that of the accepted answer. There have been several questions on MSO where people wondered why they didn't get it when one (or both) was equal.
I have a solid foundation, I like to think about the interplay of different parts of math, and I'm insistent on doing/understanding examples whenever possible.
@Ted The category of C* algebras is isomorphic to the opposite category of compact Hausdorff spaces. Apparently this is the motivation for noncommutative geometry.
I fear I'm falling into the induction trap here. Given a connected graph G with n vertices, I am constructing G' where every vertex in G' represents a single spanning tree in G, and every pair of vertexes in G' are adjacent iff the corresponding spanning trees in G have n(G)-2 common edges. I am trying to prove that G' is connected and to find its diameter
I am thinking of proving by induction that if d(x,y)=a (where x and y are vertexes in G') than the corresponding trees in G have n(G)-2*a common edges..
Am I possibly falling into the induction trap in the n+1 step?
@PedroTamaroff I think you guys were too hard on this question. Obviously the OP needs to be reminded what a limit is and how it differs from a regularized divergent sum. However, there are some manipulations of sums that apply to regularizations (c.f. some arguments of Euler) whereas there are others that don't.
This leads into the meaningful and fruitful discussion of why some manipulations work and some don't. The answer, IIRC, has to do with a "graded" aspect to divergent series (in the words of Matt E) which may or may not be perturbed by certain manipulations.
@PedroTamaroff apparently the intermediate rings between a dedekind domain and its fraction field are precisely the localizations if and only if its class group is torsion
@anon Anyways, today I proved that localization preserves UFDness and PIDness. Also, that if A is a UFD and p is a prime, A_(p) is a PID (I observed the ideals are (p)^n), and Karl more generally taught me that if A is local and the powers of the maximal ideal intersect at the zero ideal, every ideal is a power of it.
I don't get this, I keep getting a false result. If d(x,y)=a-->n(G)-2a, and since $n(G)-2a\ge 0$ must uphold I get that $n(G)/2 \ge d(x,y)$, hence $n(G)/2 \ge diam(G')$, but that's not the result I should get...
Given a connected graph G with n vertices, I construct G' so that every vertex in G' corresponds to a single spanning tree of G, and two are adjacent iff the corresponding spanning trees have n(G)-2 common edges. Trying to prove G' is connected and to find diam(G')
For any positive integer $k$, let $S_k$ denote the sum of the infinite geomet- ric progression whose first term is $\dfrac{k-1}{k!}$ and and common ratio is $\dfrac1k$. The value of the expression $S_1+S_2+\dots$ is ?
I think it is $e$
But am not sure as $S_1=0$ and $S_2=1$, $S_3=1/2!$, etc. So, it could be $e-1$.
Let's say you've got a compact topological space $X$. Then the space of continuous functions to $\Bbb C$, $C(X)$, is a $C^*$ algebra (with pointwise addition, pointwise multiplication, $^*$ the adjoint operator, and $||\cdot||$ the $\sup$ norm.
Oh, so even though our group is finite and its image in the GL(V) is finite, we might not be able to find a smaller subgroup of GL(V) that contains this image?
My question is: If it is so easy to construct an infinite dimensional representation, why are they making such a big deal out of constructing an infinite dimensional representation of the monster? Any math major can do it
well a) who's "they" and b) just because you have an infinite dimension rep - which might be the easiest way to define it - doesn't mean it's obvious how to show that this is just an "extension" of a finite dimensional rep the way you describe
whoever you're reading might be mentioning it's infinite dimensional just because the easiest way to work with your thing is with this infinite dimensional rep
i'd also like to point out that groups are for nerds
How about $V$ is a finite dimensional vector space and $\rho:G \to V$. Then, add basis vectors to get an infinite dimensional space. Then $GL(V)$ is a subgroup of $GL(W)$( Its actually the subgroup that fixes $V$). Will this work?
This was probably not important enough for flagging or posting on meta, but maybe some of the mods will notice this here in chat.
There is a question on meta which is identical copy of the question on the main. This question was closed.
Wouldn't a better solution be delete the meta question or migrate it to main and close as a duplicate?
I am not sure whether Community user bumps closed questions, if he does this question will reappear on front page of meta periodically. (I have asked about this on meta.SO.)
Huh. Someone just went through and upvoted eight answers of mine in the space of a few minutes -- each one with +9 and checkmark before the mass upvote.
@DanielFischer Heya what countour would you use to solve $\int_{-\infty}^\infty e^{-ix}/(1+x^2)\mathrm{d}x$? Semi-circle in the lower or upper half plane ?
