Mathematics

3:03 AM

I am told that $\lim_{x \to 6} \frac{f(x)-15}{x-4} = 8$ and $\lim_{x \to 4}\frac{f(x)-15}{x-4} = 3$. I've easily determined that $\lim_{x \to 6} f(x) = 31$. How do I find the $\lim_{x \to 4} f(x)$?

cubesnyc

i am looking over a research paper on poker suit isomorphism, and one of hte introductory paragraphs is rather cryptic to me

the paragraph reads as follows:

Our first building block will be a procedure for indexing
M-rank sets, which are sets of M cards of the same suit.
With out loss of generality, let us consider the ranks to be in
{0, 1, . . . , N} and consider a set in decreasing order.
The colex function (Bollobas 1986), provides precisely
this indexing. As much of the indexing scheme is based off
the same principles, we describe the construction here.
Without loss of generality, we order the sets lexographically. For example if M = 2, h4, 1i > h3, 0i > h2, 1i. Given

the actual paper is here

cs.cmu.edu/~kwaugh/publications/isomorphism13.pdf

i am having a hard time following his illustration that <4,1> > <3,0> > <2,1> or what that even means since it seems like it wasnt explained anywhere?

Jeff

4:16 AM

@Jeff Is it as simple as the same method for the first one, but realizing that since $\lim\limits_{x \to 4}\frac{-15}{x-4}$ DNE then the whole limit can't exist? Also, what happened to the "Reply" feature in this chat?

Ted Shifrin

5:36 AM

@Jeff The reply did just disappear. Dunno! If the limit the quotient exists and the denominator goes to $0$, what must the numerator go to?

Astyx

10:25 AM

Why does the Dolbeault cohomology use the $\bar \partial$ operator and not the $\partial$ one ?

Mike Miller

10:47 AM

@Astyx You want the 0th cohomology to be holomorphic forms

Astyx

Oh that makes sense

What happens if you take $\partial$ ? I expect you get something isomorphic ?

Mike Miller

I don't really remember! I think that might interchange (p,q) and (q,p) cohomology

Yeah seems right

Balarka Sen

Scary

Astyx

I think the isomorphism is just conjugation actually

Mike Miller

Yeah

Knight

12:41 PM

Can someone tell me where is Ron Maimon these days?

Mike Miller

Hasn't he been banned for years

Knight

I thought someone might now where is he now

Mike Miller

At home or work, most likely

Googling his name in quotes quickly shows he doesn't seem to be active much anywhere on the internet under his own name

Knight

Okay

adamaero

2:06 PM

What's the name for the curve plotting points produces? It's used in controls engineering...

tip of my tongue..

With Root Locus...

A term to describe the plotting of a shape?

Or the path..

Lol, I think it's just the locus.

Leaky Nun

zero-locus?

Astyx

2:29 PM

Is the point of the Dolbeault cohomology that it's a finer one than de Rham cohomology because it imposes holomorphic forms ?

But it's still the same as any complex cohomology no ?

I've heard all cohomology theories were "isomorphic"

Mike Miller

2:42 PM

@Astyx I don't understand this statement

@Astyx Any collection of groups $E^\bullet(X,A)$ whose input is a pair of a topological space $X$ and a subspace $A$, and which satisfies the Eilenberg-Steenrod axioms, is canonically isomorphic to singular cohomology $H^*\bullet(X,A;G)$ with coefficients in some Abelian group $G$

The input for Dolbeault cohomology is a complex manifold, which is more structure than a topological space and for which that structure exists only on very few topological spaces

It's completely incomparable, at least from abstract principles like you're talking about

If $M$ happens to be a closed Kahler manifold it follows from Hodge theory that $H^n_{dR}(M; \Bbb C) = \bigoplus_{p+q=n} H^{p,q}_{Dolbeault}(M)$

But this requires work and is again nothing to do with general principles

Astyx

@MikeMiller Oh ok, I didn't know the details of this

Hence my confusion

Mike Miller

Nobody should have told you that without telling you what it means :(

Astyx

Many people have haha

Mike Miller

There are far more things that people call cohomology than things satisfying the Eilenberg-Steenrod axioms

Astyx

So homology of complex manifolds is "richer" than homology of topological manifolds

Mike Miller

2:52 PM

It's unrelated

They measure different things

That there is a relationship at all is miraculous and what gives rise to the modern subject of Hodge theory

Astyx

Cool, thanks !

Balarka Sen

3:29 PM

The Dolbeaut double complex filters the de Rham complex, it's a structure coming out of the complex structure - this is the best you can say

The double complex cohomology of del + delbar is de Rham, but how is it related to del and delbar cohomology? I guess is unclear

Mike's statement is saying if M is Kahler then the spectral sequence degenerates at E^2

Astyx

Do you know JP Demailly's Complex Analytic and Differential Geometry ? Do you recommend it ? @MikeMiller

Mike Miller

I know nothing, especially about complex geometry books

Astyx

www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf

Mike Miller

Seems very specialized to me. I dunno what your goal is

I've personally never heard Lelong numbers much less read 70 pages about them

Astyx

I'm having a course that covers roughly the first chapter this week

Balarka Sen

3:36 PM

Yikes

Very scary

Mike Miller

Sounds like an SCV course

Balarka Sen

Whoa wait

Pseudoconvexity

Great stuff!!

I will read

Mike Miller

Tfw

Astyx

It looks cool if you ask me

Alessandro Codenotti

That's a scary looking book

Adeek

3:48 PM

hi @BalarkaSen @MikeMiller

Astyx

hi

Adeek

Hi @Astyx

I want to discuss a theory I am developing with someone interested in algebraic geometry.

anyone wants to discuss

Mike Miller

Hi; I've got nothing to contribute about that, sorry

Adeek

it is a theory that would connect geometry and algebra

@MikeMiller fair.

@Astyx would you like to discuss ?

does Ted still come here ?

Astyx

I'm not very knowledgeable in algebraic geometry

I doubt I could contribute anything useful

Adeek

3:54 PM

I see.

@TedShifrin can you check my email in regards to a theory that in development by me.

@TedShifrin Thank you so much.

@MikeMiller how are you ? Are you a postdoc now ?

Tobias Kildetoft

Isn't algebraic geometry precisely a theory that connects geometry and algebra?

Adeek

Sure I mean I guess you could call Cohomological analysis

my theory

It would connect a lot of areas not just algebra and geometry.

but also analysis as well.

@TobiasKildetoft

Balarka Sen

Seems useful

Adeek

@BalarkaSen very useful yeah.

I emailed Ted the main idea just want some critique towards it.

I am planning to publish it during my PhD.

Balarka Sen

Do you use higher category theory

Adeek

4:02 PM

yes

@BalarkaSen can I email you the idea

Balarka Sen

I wouldn't understand it

Adeek

I put it in layman's term

The main idea for people that don't know any higher categories

Balarka Sen

Oh but I'm just not familiar with much of algebraic geometry, or algebra or geometry for that matter. I think Ted will have useful feedback

Just cool to hear you used higher category theory in algebraic geometry to connected algebra and geometry

Adeek

yeah and I will generalize K-theory and regulators and other stuff as a way to link different cohomologies generated by my theory and provide a way to measure perspective.

I call it cohomology of perspective

my theory.

Mike Miller

I'm technically a postdoc, yeah. I'm not paying much attention now because I'm preparing some calculus lectures

Adeek

4:07 PM

oh I see @MikeMiller very cool.

brb

Adeek

4:48 PM

Hi Ted. "Sorry left my computer unattended"

Knight

4:58 PM

@MikeMiller What level?

Adeek

Ted if you could reply back from email that would be nice :)

Ted Shifrin

I just got it. Honestly, I know nothing about this stuff, so I cannot say anything useful.

Adeek

@TedShifrin I see no problem :). I hope everything is ok with you.

Ted Shifrin

Pandemics, fires, dangerous incompetent president ... yup, just fine.

Adeek

yes that is why I asked :S

I hope things goes towards the better soon

Ted Shifrin

5:10 PM

We all do ... :)

Silent

Mike Miller

@Knight multivariable integration

Silent

Since above $y$ is transcendental over $F$, we have that $F(y)$ is isomorphic to $F(x)$. But, then again, in the above result we claim that $[F(x) : F(y)]$ may be greater than one! I am not able to digest this.

Leaky Nun

$[F(x) : F(x^2)] = 2$

Knight

If you don’t mind terribly I would like to have your lecture (whether it is video or PDF).

Silent

5:15 PM

Sure!

Leaky Nun

put $y = x^2$ in your query

user193319

Given a ring $(R,+,\cdot)$, let $(R^{op},+,\star)$ be the opposite ring formed from it. I am trying to show that $Hom_{R}(R,R) \cong R^{op}$ as rings. What are the ring operations on $Hom_{R}(R,R)$? Usual function addition and composition?

Leaky Nun

yes

note that the Hom is R-module Hom, i.e. R-linear maps

you can't add two ring homs together

user193319

You can't add two ring homomorphisms? Why can't I define $f + g$ as $(f+g)(r) := f(r)+g(r)$? Why doesn't that work?

Thorgott

of course you can add them, but the resulting map is usually not a ring hom anymore

user193319

5:25 PM

Oh, I am confused...so then what are the ring operations on $Hom_{R}(R,R)$?

Mike Miller

I'd rather not share lectures with people not taking the class, sorry.

I'm certain that there are very polished lectures on the same material you can find elsewhere

Knight

@MikeMiller I understand the reason, there are chances of stealing, eh?

Silent

@LeakyNun Thanks :)

Thorgott

pointwise addition as addition and composition as multiplication @user193319

Knight

I shouldn’t have asked for it in first place, I never thought how corrupt someone might estimate me.

Ted Shifrin

5:31 PM

It's not a matter of corruption, @Knight.

Knight

Is it about fees?

Ted Shifrin

But there's an implicit contract with the students that this is a course for them, not to have their names and faces made public. When my lectures were videoed and put on YouTube, my students agreed to it.

Mike Miller

There's nothing interesting enough to steal from a calculus lecture, I assure you

Ted Shifrin

I've had sleepless nights wondering if people might steal my computation of gas mileage as an application of linear approximation for functions of two variables. :P

Mike Miller

I would just want them to be much more polished were they to be for "public consumption". Part of Ted's implicit contract (to me!) is that I am in communication with students afterwards --- if something is confusing, I can clarify more or less immediately

Making something public extends that privilege to others; but I only get paid for answering some of those questions ;)

Ted Shifrin

5:39 PM

You mean I get paid for answering questions? Who knew!

Oh, you do :D

Mike Miller

For now, at least

Balarka Sen

academia destroyed not anymore mwahaha

Ted Shifrin

Howdy, a @Balarka.

Balarka Sen

all the math you have learnt will be forgotten

4

im not shocked

@TedShifrin Hi!

Ted Shifrin

This is sounding like Hippa's meme.

Balarka Sen

5:41 PM

:D

Ramanewbie

@TedShifrin kind of

Ted Shifrin

You were here years ago for Hippa's meme?

Damn, that was something like 4 years ago.

Ramanewbie

Sure
I can't get if off my mind

Balarka Sen

Has to be at least 6 years

Ramanewbie

You don't remember me @TedShifrin :o

Ted Shifrin

5:43 PM

Well, those videos were made spring 2014, so I suppose 6 years is barely possible.

Balarka Sen

Oh ok

I have been a member of MSE for 6 years and I feel like I have seen that meme at the very beginning

Time flies

DarkRunner

6:03 PM

Hi all, I'm confused about the partial derivative of a cross product

I understand that $\frac{d}{d t}(\mathbf{u}(t) \times \mathbf{v}(t))=\mathbf{u}^{\prime}(t) \times \mathbf{v}(t)+\mathbf{u}(t) \times \mathbf{v}^{\prime}(t)$

I'm trying to use that to prove $v \times E = r \times \frac{\partial E}{\partial t}$

If anyone can help, that would be great

Mike Miller

That's not a partial derivative --- partial derivatives are things you take of multivariable functions, and this is a 1-variable function (of t!)

You seem to be using facts from physics here that you're not telling us

What are r, v, E?

DarkRunner

@MikeMiller Sorry, let me clarify: $E$ is the electric field of a charge. $v$ is the velocity of that charge, and $r$ is the position function

Balarka Sen

$E$ is always in the direction of $r$, so $r \times E = 0$

Use that plus the cross product rule you just wrote.

Knight

@BalarkaSen $r$ is the position of source particle

Mike Miller

There's a sign error then

DarkRunner

6:11 PM

@BalarkaSen OK, so here's what I have:

I know by the Cross Product rule that $\frac{\partial}{\partial t}{r \times E} = \frac{\partial r}{\partial t} \times E + \frac{\partial E}{\partial t} \times r$

Balarka Sen

@Knight The position of the source particle is the origin. $r$ is the position vector from the origin to the test charge, or the other way around - I don't keep track of sign.

Knight

@BalarkaSen particle is moving with $v$

Balarka Sen

What

DarkRunner

@Knight That's right

Balarka Sen

How does that contradict what I said

DarkRunner

6:13 PM

Simplifying, I have $\frac{\partial}{\partial t}{r \times E} = v \times E + \frac{\partial E}{\partial t} \times r$

Knight

@BalarkaSen You said particle is at the origin

Mike Miller

@DarkRunner You swapped the order here on that second term... Should be r x E_t

ABHIJIT BHATTACHARJEE

Hello everyone i want to ask a question about composition series of finite groups. We usually construct a composition in a finite group by a sequence of maximal normal subgroups. Can we construct a composition series in a finite group by a sequence of maximal normal subgroups ?

DarkRunner

@MikeMiller OK, got it

I think I messed up my $LaTeX$ by mistake

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$Mathematics$ Mathematics