@MikeM: Just wanted to let you know I got a downvote (and one upvote) on that complex indices question. Of course no comment why. Take a look. Was I wrong (to your knowledge) or otherwise deserving?
@TedShifrin Hey. Also, I don't know if such a thing is easily queriable. It's a complex question whose subset substring matches likely don't lead to anything constructive.
Prove that the largest power of $2$ dividing $n(n-1)(n-2)\cdots (n-2^k+1)$ is $2^k-1$ if $n = 2^k m$ where $m > 1$ is odd.
We are basically adding up the powers of $2$ in the multiples of $2$ up to $2^{k-1}$ (and $k$). Thus, since we have $$2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32$$ we wo...
@EricStucky I am trying to solve the following: Prove that the largest power of $2$ dividing $n(n-1)(n-2)\cdots (n-2^k+1)$ is $2^k-1$ if $n = 2^k m$ where $m > 1$ is odd and $k \in \mathbb{Z}^+$.
I started to realize that they aren't just the opposites of the natural numbers, but a whole different paradigm of numeric units. Naturals correspond to numeric quantities, negatives correspond to debts or removals of quantities. It took this long to have it sink in that those are very fundamentally different types of objects.
With this new perspective, I can't think of a way to justify division by a negative without falling back to the argument "it just works".
I didn't even read what you said in time. I hope you're not taking my BS to heart.
I obviously believe that you can do these things, $\frac{10}{-5} = -2$. But I just feel like I understand that concept at too high a level to explain it formally from the original foundations.
@Ted I'm not quote sure what you mean when you say "we treat $d\bar z(\bar v)$ as the same thing". The only thing I would add is an example, eg that you would write $dz \wedge d\bar z$ as $dz^1\wedge dz^{\bar 1}$.
@Axoren When constructing it from ZFC, you just kind of define it to be like that.
But if you take anything in applied math where you divide by a negative on the way to your final answer, you can probably justify it within that application.
@TedShifrin They exam didn't have any restrictions on calculators. I really just used it for one thing, which was to find the LCM or GCD of two numbers faster than in my head and to quickly generate list of natural numbers that are less than x and coprime to x
Interestingly, the example sequential set of numbers I wanted to test for being coprime was exactly what was needed in the exam, so all I had to do was press the up arrow on the TI-89
(fyi – this is about quickly getting the fellow integer generators of some cyclic group $\langle a \rangle$. $\langle 21 \rangle \in \mathbb{Z}_{30}$ was the one in the exam, which, admittedly is quite easy, but in past exams it's been $\langle 3 \rangle = U(31)$ for example which (at least for me) is a bit trickier just to do in my head)
Following are the two theorems that Hardy and Wright state in their book
Theorem A: The number of primes not exceeding $x$ is given by $\pi(x) \sim \frac{x}{\log{x}}$.
Theorem B: The order of magnitude of $\pi(x)$ is $\pi(x) \asymp \frac{x}{\log{x}}$.
where,
$f \sim \phi$ iff $\; \f...
@Evinda It is a classification of projective functors on parabolic subcategories of the BGG category $\mathcal{O}$ in type $A$. It was what I spent most of my time on during the first 8 months or so of my current postdoc
@TobiasKildetoft Interesting. So current a potdoc, one publishes papers? How do you write papers? Do you read known ones and want to answer an unanswered question?
> Formally, if r(x) is any non-zero polynomial, it must be writable as ${\displaystyle r(x)=A(x-x_{0})(x-x_{1})\cdots (x-x_{n})}$ , for some constant A.
Integrate[(xres*((x1^(-1/2 + I t + eps)*((-(g1 - x1 + 1)) - 1)))*(1/ 2 + I t - eps)^0 + xres*((x2^(-1/2 + I t + eps)*((-(g2 - x2 + 1)) - 1)))*(1/2 + I t - eps)^0 + xres*((x3^(-1/2 + I t + eps)*((-(g3 - x3 + 1)) - 1)))*(1/2 + I t - eps)^0)/(xres*((x1^(-1/2 + I t - eps)*((-(g1 - x1 + 1)) - 1)))*(1/2 + I t + eps)^0 + xres*((x2^(-1/2 + I t - eps)*((-(g2 - x2 + 1)) - 1)))*(1/2 + I t + eps)^0 + xres*((x3^(-1/2 + I t - eps)*((-(g3 - x3 + 1)) - 1)))*(1/2 + I t + eps)^0), t]
elements of $Rm$ are of the form $rm$ for $r \in R$. You just need to show that you can add stuff and subtract stuff, and remain in $Rm$.
also that 0 is in there, but yeah
also, if you want to do the other part carefully you say $r_1 (r_2m) = (r_1r_2)m$, and $r_1r_2 \in R$, so this is in $Rm$. Do you see how the parentheses matter?
So, we have that $r_1m, r_2m\in Rm$. Then $-(r_1m)=(-r_1)m\in Rm$ and $r_1m+r_2m=(r_1+r_2)m\in Rm$. Is this correct? Yes, I see how the parentheses matter... @SamuelYusim
I have recently started reading a paper "Group actions on $\mathbb{R}$-trees", and sort of been looking around for the big picture on the subject, and how it comes about @BalarkaSen
@BalarkaSen Groups acting on trees ;). Well group acting on trees seem to come up when studing splittings of groups (bass-serre theory), and has a relation to studying 3-manifolds and sufaces (rips machine ), and somehow you can get compactifications which is an analogue compactification of Teichmuller space. I am still trying to figure out what all these things are though, so I can't be of any more help at the moment
Prove that the largest power of $2$ dividing $n(n-1)(n-2)\cdots (n-2^k+1)$ is $2^k-1$ if $n = 2^k m$ where $m > 1$ is odd.
We are basically adding up the powers of $2$ in the multiples of $2$ up to $2^{k-1}$ (and $k$). Thus, since we have $$2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32$$ we wo...
Following are the two theorems that Hardy and Wright state in their book
Theorem A: The number of primes not exceeding $x$ is given by $\pi(x) \sim \frac{x}{\log{x}}$.
Theorem B: The order of magnitude of $\pi(x)$ is $\pi(x) \asymp \frac{x}{\log{x}}$.
where,
$f \sim \phi$ iff $\; \f...
