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1:32 AM
If $\ast$ is a binary operation on a set $S$, we say a subset $X \subseteq S$ is closed under $\ast$ if $x, y \in X \implies x\ast y \in X$.
Is there a name for a subset $X$ for which $x\ast y \in X \implies x, y \in X$?
 
1:45 AM
Reminds me of prime ideals, but I don't know any general terminology
 
2:00 AM
yeah
maybe there's something like this in more general algebra?
 
Bob
will things like general algebra be useful outside of academia?
 
Applied category theory is an academic discipline in which methods from category theory are used to study other fields including but not limited to computer science, physics (in particular quantum mechanics), control theory, natural language processing, probability theory and causality. The application of category theory in these domains can take different forms. In some cases the formalization of the domain into the language of category theory is the goal, the idea here being that this would elucidate the important structure and properties of the domain. In other cases the formalization is used...
 
is it useful to know which regular n-gons can be constructed with ruler and compass
 
2:16 AM
Not really, but the constructibility and field extension approach is very elegant proof
This site is somethin else
Renaming BananaCats to AfterPaper, and the design is totally changing
I'm going to make a general math tool, but aimed at higher math
such as group theory
 
@shi hi
 
2:39 AM
A lamination of a surface is a partition of a closed subset of the surface into smooth curves. It may or may not be possible to fill the gaps in a lamination to make a foliation.
How does one determine whether it's possible or not to fill the gaps in a lamination to make a foliation?
 
 
12 hours later…
2:56 PM
hi, i and one of my friends willing to use these se chatrooms for studying like some people do
but he don't have an account, even he creates one now it'll take some to reach 100 points
do you know any other mathjax enabled chat systems/projects? i took a look at google for such projects but didn't succeed to find a suitable one
maybe i'll write one, eventually
 
3:11 PM
Howdy everybody
For my research I'm using an argument containing the general form of the Heaviside cover-up method. One of the guys had the following expression for it:
$$
\frac{{{\prod}_{i=1}}^A(x-a_i)}{{{\prod}_{i=1}}^B (x-b_i)} &= \delta_{A,B} + {{\sum}_{i=1}}^B \frac{1}{x-b_i} \left[{{\prod}_{j=1}}^A (b_i-a_j)\right]\left[{{\prod}_{\substack{j\neq i\\1}}}^B \frac{1}{b_i-b_j}\right].
$$
For the case $B\geq A$.
I'm interested also in the case where a part of the $x-b_i$ have power $2$
So I've been looking at that case, and it appears this just spawns more terms, and an additional summation within each term
Now I'm considering the complicated case where I admit that it would be nice to have some verification,
So I got something like this for the extra terms:
$$
\sum_{i=1}^{C} \frac1{x-c_i} \left[\prod_{j=1, j\neq i}^{C} \frac1{c_{ij}^2\right]\left[\prod_{j=1}^{B}\frac1{c_i-b_j}\right}\left[-\sum_1^B \frac1{c_i-b_j} + \sum_{1}^C \frac{2(-1)^j}{c_{ij}}\right]
$$
For the case without numerators, though this one is pretty much subject to changes of the poles
So I'm treating the denominators as the part that matters
Is there any reference for this stuff?
 
 
3 hours later…
5:56 PM
Hi all.
 
"all"
Doesn't seem very crowded tonight :P
 
I like to be optimistic sometimes.
 
fair enough
 
Oh, some life.
 
6:09 PM
hallo Leute
 
Hi, demonic @Alessandro.
And @Leaky, although I highly doubt that's proper German usage.
 
Guten Abend @Leaky
 
@TedShifrin that's what she says every episode
 
We'll ask the resident Germans.
 
Hallo Leute is fine :P
 
6:11 PM
Okey.
 
Also, hallo Leute
:D
 
I can't imagine saying Bonjour, gens in French.
 
you can't translate phrases literally, I guess
@EdwardEvans Schach?
 
Definitely not.
 
Nein danke, ich bin gerade vom Radeln heimgekommen
 
6:14 PM
Radeln?
 
I am stuck with this question and hint: math.stackexchange.com/questions/2039873/…
 
Fahrradfahren
 
ich habe jetzt zwei Woerte gelernt :P
 
I don't quite see how to compute with the residue theorem here, or how this integral relates to the original question.
 
Fahrradfahren und Radeln? Oder Fahrradfahren und heimkommen?
 
6:15 PM
@anakhro: The first comment is exactly what I would do.
 
Fahrradfahren und Radeln
 
aso haha
 
@TedShifrin How does one compute the residue $Res(f;a_k)$?
 
Radeln ist eher Dialekt
 
What are the poles and corresponding residues of $z^n/P(z)$?
 
6:16 PM
Poles are the zeros of P(z), right?
 
Yes, and they're all simple zeroes.
How do you find a residue at a simple pole?
 
The only way I recall is using the Laurent expansion.
And then finding the coefficient at -1
But finding the Laurent expansion of $z_n/P(z)$ seems to be difficult.
 
I'm a big fan of Laurent expansions, but it's worth knowing that the residue of $f/g$ at $z=a$ is $f(a)/g'(a)$ when $g$ has a simple zero at $a$. You can get this immediately out of Laurent.
 
O.
 
Oh, assuming $f(a)\ne 0$, of course.
 
6:19 PM
I will put that on my list to check.
So that explains the residue theorem + how it relates to the rest of the question.
Okay. So then the estimation theorem
 
What's that?
 
So the absolute value of this sum I get (the one I need to prove is zero) is less than or equal to $2\pi R\cdot\max_{|z|=R}\frac{z^m}{P(z)}$
But because deg P > m, then the max part is 1/R or "better".
 
Oh, I don't call that anything. $\pi R$, not $2\pi R$.
Don't we have $m\le n-2$?
 
Yeah.
 
That's crucial.
You're being sloppy.
 
6:23 PM
Yes, I definitely am. But I think the point is that the numerator there has degree less than the denominator, and this is how we get it to go to zero?
(I promise I will work out the details)
 
You need better than "less".
 
Do I need better than "2 less"? :P
 
For $R$ large enough, $|P(z)|\ge R^n/2$ on $|z|=R$ (assuming $P$ monic).
No, "2 less" is precisely what you need.
 
Hmmm.
 
6:39 PM
@TedShifrin how do you get $\pi R$ and not $2\pi R$. It is supposed to be the circumference, right?
OH
I see why we need n-2
Silly me. :P
 
 
2 hours later…
8:26 PM
 
 
1 hour later…
Bob
9:32 PM
Hi
 

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