Mathematics - 2020-10-07 (page 1 of 2)

Joe Shmo

00:01

@Sophie, sorry for the late response, just got around to a piece of paper, and looks like you might have gotten what you were looking for? Anyway, this is too fun to pass up, and perhaps you should try to play with it yourself to get some more intuition.. the efficient way is to correspond your group to $S_3$ and see that the relationship indeed hold in $S_3$, but perhaps it's instructive (and certainly fun) to play with what you already have. So, you have
$a^2 = b^3 = (ab)^2 = e \iff a = a^{-1};\ b^2 = b^{-1};\ ab = (ab)^{-1} = b^{-1}a^{-1}$. Then I claim that the generated group is the fo…

well, iterated $ab^j \ldots ,\ b^ja \ldots$, and composed arbitrarily with existing strings.. but we never have new elements that aren't already in the set I described because of the reduction rules we deduced..

anakhro

@BalarkaSen what did you not like about it?

user2103480

00:25

Do cocycles in dynamical systems have anything to do with those in algebraic topology?

Billy Rubina

Given two groups with the same cardinality, is it true that this condition alone guarantees that there is an isomorphism between them?

Joe Shmo

@Billy Rubina, certainly not

consider $\mathbb{Z}_4$ and $\mathbb{Z}_2 \times \mathbb{Z}_2$

anakhro

@user2103480 terrytao.wordpress.com/2008/12/21/…

user2103480

Ah yeah found it too, so it's group cohomology without spelling it out

Billy Rubina

@JoeShmo Thanks!

Joe Shmo

00:36

Ah, @Sophie, missed one - $(aba)b = (ab)^2 = e \iff aba = b^{-1} = b^2$. @BillyRubina, sure thing.

Sophie

@JoeShmo Interesting

Joe Shmo

Also, $ab^2a = ba^2 = b$, I think that's all of them..

Charlie

00:52

@MikeMiller I was asking in what area of mathematics the idea of a clifford algebra is first introduced, since they come up in quantum field theory rather out of the blue, I was curious what kind of mathematics texts would talk about them in a non-physics related context.

Mike Miller

01:16

@Charlie right but you have a hidden assumption that they are usually introduced at all!!

They show up in spin geometry. Up to you whether you already want to call that physics.

Then when you study index theory of operators you find a great many of those indices are dirac operators. Index theory and spin geometry inform each other, traditionally

anakhro

01:46

@MikeMiller any idea of what "contractible" might mean in a context where there is no obvious topology? Like something like "the space of assignment of signs to the vertices of a simplex is contractible"?

I should say "space of such assignments" as if it is a proper subset

I am thinking it might be something like they are all in bijection?

Mike Miller

02:03

@anakhro Can you send me the precise sentence?

W/ context

anakhro

It was during a lecture, so it's kind of vague as it is, sadly. I will get more information hopefully and report back when I do.

skullpatrol

03:02

RIP

Rithaniel

04:08

Anyone here ever try to compose music? I know there is a degree of mathematical theory to music, but I've never known if the connection is deep to enough to allow skills to directly translate from math to music.

Ted Shifrin

04:32

@Rithaniel: If you decide to compose twelve-tone music, then you find the Klein four-group everywhere. :P

Billy Rubina

@Rithaniel I'm currently reading about: en.wikipedia.org/wiki/Schillinger_System

Franz Absil has some stuff on it: youtu.be/EvTXx7Z29cU

I bought his book in the Schillinger Method.

And I kinda went to math because of: springer.com/gp/book/9783764357313

Edward Evans

06:47

Morning/afternoon/evening

Suppose I have a group hom $f : G \to H$ for which $\ker f$ is open and let $U \subseteq H$ be an open neighbourhood of $1$. Then $\ker f \subseteq f^{-1}(U)$, so can I conclude that $f^{-1}(U)$ is open by writing it as a union of translates of $\ker f$?

That seems non-dumb to me

Sayan Chattopadhyay

07:02

@Rithaniel I have been trying to learn jazz for sometime now. What I have seen and heard mostly is that the mathematical aspects come when you deal with various tunings mostly. So if you go to things like microtonality then you can do some cool things.

Leaky Nun

@EdwardEvans in general a group that contains a nonempty open set is open

Edward Evans

@Leaky simply because translation is a homeomorphism right? I mean in my case that $f^{-1}(U) = \bigcup_{g \in f^{-1}(U)} g\ker f$

Leaky Nun

wait, $U$ isn't a group

I guess it still works

Edward Evans

Aight ty

@Sayan @Rithaniel Check out Adam Neely if you're interested in this kinda thing :D Sample video: youtube.com/watch?v=ghUs-84NAAU

Sayan Chattopadhyay

For instance things like Shepard's tones and stuff happens because of superimposing sine waves. But such ideas like Shepard's tone also happen for tuning systems as well. Look at the work of Benedetti

@EdwardEvans lol that's where I took my inspiration from

Edward Evans

07:13

hahah, I love his videos

once I saw the word mIcRoToNaLiTy it instantly made me post Adam Neely

I'm like an Adam Neely sleeper agent

Sayan Chattopadhyay

Yup he's an awesome dude.

Leaky Nun

what about Jacob Collier

Edward Evans

yeah that guy is cool too, but Adam Neely is more likeable (imo obv)

still, Jacob Collier is crazy

Sayan Chattopadhyay

Jacob Collier is not human, hence it's better for our souls to prefer Adam Neely

Edward Evans

hahaha

Sayan Chattopadhyay

07:18

For all the djent heads here, listen to Tigaran Hamasayan

Leaky Nun

@EdwardEvans so, what are you studying

@Astyx salut !

Edward Evans

I'm just studying the first chapter of a book called "The local Langlands correspondence for $\operatorname{GL}_2$"

and this chapter is about locally profinite groups and smooth representations of them

but I haven't yet gotten to smooth reps

Astyx

Salut

Leaky Nun

@EdwardEvans so your map in fact factors through a finite group?

Edward Evans

err maybe, the group is only locally profinite though, not profinite, so it's not a limit of finite groups

but every open nbhd of 1 contains a profinite subgroup

Leaky Nun

07:27

aha, so like Qp

Edward Evans

sure, in fact the additive and multiplicative groups of any non-archimedean local field are locally profinite

Leaky Nun

sure

who needs Qp if you have $\Omega_p$

Edward Evans

the point of introducing them is to do fourier analysis on them I guess

but I'm not there yet

lol

Leaky Nun

(I dont think enough people know about $\Omega_p$)

Edward Evans

don't people usually write $\Bbb C_p$

or is that a different thing

Leaky Nun

07:32

$\Bbb C_p$ is not complete enough

Edward Evans

wot

I thought $\Bbb C_p$ was the completion of $\overline{\Bbb Q_p}$

Leaky Nun

Spherically complete field

In mathematics, a field K with an absolute value is called spherically complete if the intersection of every decreasing sequence of balls (in the sense of the metric induced by the absolute value) is nonempty: B 1 ⊇ B 2 ⊇ ⋯ ⇒ ⋂ n ∈ N B n...

Edward Evans

Interesting, I didn't know that :)

Leaky Nun

I dont think it's particularly useful lmao

Edward Evans

well it mentions that it's useful in non-archimedean functional analysis, which is relevant to the Iwasawa theory of p-adic lie groups (according to some research statement I read on some prof's homepage)

Leaky Nun

07:38

fair enough

Edward Evans

that's pretty niche though hahaha

Leaky Nun

I only know of it because of Scholze's paper

Edward Evans

Ah that might be where I've seen the symbol haha

Astyx

10:10

Thoughts on Vakil's The Rising sea ?

Edward Evans

@Alessandro is a fanboy

Leaky Nun

@Astyx c'est plein d'examples utils

Astyx

(exemples utiles)

Leaky Nun

10:28

merci

Rithaniel

11:42

@EdwardEvans that was an interesting video. Gave me something good to start my day with, too.

Edward Evans

@Rithaniel nice :)

Alessandro Codenotti

12:42

a fanboy might be too strong

But his notes were the best source I found when I took AG

Astyx

I'm following them as I progress through the AG course I'm taking

I find my prof hard to follow

Alessandro Codenotti

I think that AG is inherently hard to follow

Astyx

I finally got a grasp of the goal of it, or so I think

Edward Evans

the goal of AG is to be so difficult and so ubiquitous within pure mathematics that the bar for entry is high enough that nobody can finish a PhD in pure mathematics

it's a conspiracy from the top profs in the world to keep their positions

this shit goes right to the top man

Astyx

Does anyone else find it odd that no promine....

Edward Evans

12:54

rofl

Alessandro Codenotti

Grothendieck actually faked his death but he his still pulling the strings from the shadows

Edward Evans

Scholze is just Grothendieck in a wig

Astyx

Grothendieck is actually Michael jackson

Edward Evans

confirmed

woah

I wish Lukas were still here and could just teach me number theory

Astyx

I'm taking number theory this year

Edward Evans

12:58

The dream

Alessandro Codenotti

@Astyx what kind

Edward Evans

@Alessandro I just looked at his WhatsApp to see if he had changed his status or smth and it said "Online"

Astyx

Algebraic

Edward Evans

yaaas

Alessandro Codenotti

@EdwardEvans You could also write him you know :P

@Astyx Nice

Edward Evans

13:01

Yeah I already did and he just read it

I don't like to pry

Astyx

I think the prof intends to do a quick overview of analytical methods as well

Edward Evans

some class number formula or smth

Charlie

13:14

@MikeMiller ah I see haha, ty

Astyx

IIRC the reason Clifford Algebra come up in quantum field theory is because they are essential in representation of symmetry groups used in particle physics

So there's a particular interest for physicists to study them, and that involves maths

jeea

Does anyone know of method of finding Jordan form Jordan chains without backtracking

I mean a simple way to find the generalised eigenspace

Astyx

Physicists, not physicians

Charlie

yes they are used to construct representations of the Lorentz group, but as with most things in physics the mathematics is introduced without much justification. I was just curious where they arise in mathematics

And the wikipedia page was beyond me

Astyx

wikipedia pages often are

Mike Miller

13:25

They give you representations of the Lorenz group yeah but also of the Spin groups

Astyx

In an additive category, why is the morphism $A\to 0\to B$ necessarilly the zero morphism of $Hom(A,B)$ ?

Sayan Chattopadhyay

I remember seeing this neat proof of the dirac equation using clifford algebras in Folland's book

Mike Miller

An additive category is one that has all biproducts, right?

Astyx

yup

Mike Miller

Meh I don't want to think about this, sorry

Astyx

13:29

I feel this should be very elementary but I don't see it

Mike Miller

I tried to add it to an arbitrary element and I would have to draw a diagram

Forget that

Oh

@Astyx If $f: A \to B$ is a morphism then $f_*: \text{Hom}(X, A) \to \text{Hom}(X, B)$ probably has $f_*(0) = 0$, is that right?

Astyx

It looks right to me

Mike Miller

Then here you have the composite $\text{Hom}(A, A) \to \text{Hom}(A, 0) \to \text{Hom}(A, B)$

The second group is zero

And that composite is in the image of the map $\text{Hom}(A, 0) \to \text{Hom}(A, B)$

Astyx

Oh ok

Thank you

Mike Miller

To prove the claim about $f_*$ I assume you have to work

Astyx

13:33

It's because $f_*$ is a group morphism

Mike Miller

Still seems like you have to work to show that

If you're chasing it down to first principles

Maybe it's straightforward though

Astyx

My definition of additive category assumes group operation in $Hom(A,B)$ is distributive over composition

So $f\circ (a+b) = f\circ a + f\circ b$ ie $f_*(a+b) = f_*(a)+f_*(b)$

Mike Miller

I'm sure it follows from the existence of all bipiroducts

MathStudent

13:52

Hi there, does anyone know if $|f|$ is generally integrable for an integrable function $f$ or if that's only true for Riemann-integrability?

Also, how well known is this (in general or just for Riemann-integrability), is this well known enough that I as a math student at the end of my bachelor's degree should know this? (I think I've got some pretty major knowledge gaps in my basic math knowledge so I'm not sure)

Astyx

Doesn't integrable precisely mean that $|f|$ is integrable ?

Balarka Sen

What

is new

My diffgeo assignment wanted us to prove something about osculating circles I just defined them as having centers that are focal points for the Gauss map

Morse theory indomitable

Astyx

14:08

What's blowing up in the context of AG ?

Balarka Sen

You take a point

and then boom it

Alessandro Codenotti

and then you put some projective nonsense where that point used to be

Mike Miller

You turn x into CP^{dim - 1}

Alessandro Codenotti

And magically singularities are gone

Mike Miller

It's like taking a 2D diagram of a knot and pushing the crosisng strands off each other

Balarka Sen

14:10

you look at a scheme X and a sheaf of ideals I and take proj of I^0 + I^1 + ...

no u

Astyx

No, keep going, that resembles what my prof said

Balarka Sen

AHAHA

Alessandro Codenotti

a sheaf of ideals? Don't you blow up along closed subschemes usually? Or is that the same thing

Astyx

So you get a graded ring and take the proj space ?

Mike Miller

Awful people

Balarka Sen

14:13

@Astyx Yes I have no intuition why this is the right notion @Alessandro You have a sheaf of ideals corresponding to a closed subscheme, given by the ideal of functions vanishing on it

But this is more general because you can blowup along singular nonsense like (x^2)

The resulting objects are Not Nice

Alessandro Codenotti

Ah ok

Balarka Sen

instead of getting rid of singularities you get singularities. I think if you blow C^2 at (x^2, xy, y^2) you get Whitney umbrella

Mike Miller

@BalarkaSen Of course not, you blew them up

Balarka Sen

lmfao

@Astyx Why do you want to know this

Aren't you thinking about like complex manifolds

Astyx

To understand the course I'm following

Balarka Sen

14:16

what is the name of the course

Astyx

Algebraic geometry: schemes and cohomology

Balarka Sen

You're doomed my man

Astyx

Aren't we all ?

Balarka Sen

mf DOOMed

Alessandro Codenotti

I suppose that means sheaf cohomology

Astyx

14:17

I just accepted my demise

MathStudent

@Astyx Hm not sure if that's an equivalent definition to the definition I know which is that $f$ is $\mu$-integrable if $\int f^+ d\mu < \infty$ and $\int f^- d\mu < \infty$ for $f^+ = max(f,0)$ and $f^- = max(-f,0)$ (it then holds that $\int f \dmu = \int f^+ d\mu - \int f^- d\mu$).

Astyx

Yes, but $\int |f| = \int f^+ + \int f^-$

($|f| = f^+ + f^-$)

Balarka Sen

What to study

Maybe some

SINGULARITY THEORY

MathStudent

@Astyx Ah yes. Now I got it. Thank you!

Astyx

You're welcome

Alessandro Codenotti

14:28

@BalarkaSen Something written by Gromov, of course

Balarka Sen

close, Arnold

Alessandro Codenotti

I don't think I've ever read anything written by him

Astyx

Schwarzenegger ?

2

Alessandro Codenotti

I'm reading a paper my supervisor suggested to me now, and I only care about metrizable stuff, but the author does everything very generally. Long story short uniformities are a pain

Astyx

"I'm a stud. I'm ballsy. I don't take no shit from anyone. I compute my cohomology sequences anywhere I want. I don't have to find a hideout place, like you."

MathStudent

14:37

Pretty embarrassed about how long that took me to get. Anyone got any tips about how to be smarter and/or think faster? lol

Balarka Sen

@Astyx "Arnold" "Compute my cohomology sequences" BRUH

Astyx

Couldn't think of anything better

MathStudent

More sleep and caffeine I guess lol. Read. Meditate. Eat healthy (especially fish and green leaves). Anything else? Although I guess that's already a lot

Astyx

Practice and go to the bottom of things

That's my advice in maths in general

But everyone has a different experience of course

Balarka Sen

Evidently spectral sequences were not for
Arnold. Nonetheless, there is such a thing as
Arnold’s spectral sequence [9], a humble object
in the world of his discoveries, resembling the
asteroid Vladarnolda in the solar system (the
stability of which he proved approximately at the
time of our conversation in the canteen), named
after him.

MathStudent

14:46

@Astyx Thanks, that's certainly good advice for math.

ShankRam

For the extended hamming code, van lint does this calculation for expected number of errors

ibb.co/17hdc1s

I'm not able to understand why the summation has an (i+1)(n choose i).. instead of i*(n choose i)

user10478

15:26

Heya, what does it mean to say that a mathematical fact is true only "formally"?

Balarka Sen

its not true, informally

user10478

:P

I have glanced at the Wikipedia article relating mathematical formalism to the Platonism question, but I don't think this is the same distinction.

Balarka Sen

What is the specific context?

user10478

My only guess is that it might be intended to mean that the statement is mathematically true, but has no other application in math.

1

A: A Few Conceptual Questions About Laplace Transforms and Moment Generating Functions

The MGF is $\mathbb{E}(\exp(tX))$ and, for a discrete random variable, the probability generating function (whose coefficients are the PMF) is $\mathbb{E}(t^X)$, so passing from one to the other amounts at least formally to a substitution $t \mapsto \log t$. If $X$ is a continuous random variabl...

It is used several times in this answer and its comments.

Balarka Sen

Here they mean in the sense of formal power series

user10478

15:30

i.e., "at least formally you get the "moments" of the original function $f$ back"

Balarka Sen

You forget about convergence and play around with symbols

See here

user10478

Okay, so for an analogy, would we say, "formally, $1$ divided by anything is its multiplicative inverse," where "formally" implies not worrying about the case of $0$?

Balarka Sen

You can say $1/x \cdot x = 1$, formally, yes. I wouldn't write it as you did.

A lot of mathematics is about making things as "free" as possible, and this is the context in which I understand the term "formally". With the least possible structure and with the least possible relations between the symbols involved, the identity is true.

It is not used frequently, so I wouldn't recommend using it frequently. It's usage will become clear as you understand the context its used in (eg I have seen it used very little outside of people talking about formal power series, category theory, or mathematical logic)

user10478

Okay, I think it's counterintuitive relative to the English meaning of the word; I would expect this to be called "casual," and "formal" to reference the more verbose, technically accurate description of the fact. But I get it now, tyvm.

Balarka Sen

Haha, that's a good point.

Sayan Chattopadhyay

15:38

Casual power series lol

user10478

Nonchalant power series

Balarka Sen

Flippant power series

Sloppy series bruh

Astyx

Spooky power series for october

user10478

Bruh do you even power series?

Sayan Chattopadhyay

Nah I laurent

Balarka Sen

15:43

@SayanChattopadhyay "laurent expand" used to be a popular joke in the hostel corridors in here

(if you dont speak hindi you're not supposed to get it)

Sayan Chattopadhyay

I do and I don't get it, wth

Balarka Sen

lmao

F

you have to pronounce laurent correctly

Sayan Chattopadhyay

lmfaoooo

I was pronouncing laurent the wrong way this whole damn time

Balarka Sen

typical

@feynhat

Astyx

Anyone got a link to a proof that division algebra on finite fields are commutative ?

Balarka Sen

15:52

its called Wedderburn's little theorem if that helps

the proof is basically class equation

Astyx

Cheers

Akiva Weinberger

17:36

0

Q: Is this knotted graph knotted?

Above, I've drawn a knotted 3-valent graph. I suspect that it's not isotopic to the "unknotted" version below, but I'm not sure. Is it? I know about fundamental groups of knot complements, but it seems like that's probably more work than necessary. The easiest way would be to show that it's the ...

?

Hawk

18:08

in the gcd(a,b) = d, there exists ax + by = d what are the (x,y) called here?

maths student

19:17

Compute $\mathrm{x}=8623^{18691^{*} 72}$ mod 1997 using the smallest number of operations (multiplications).

How to do this ?

Any kind of hint is ok ?

user193319

19:29

Am I right in thinking that $\displaystyle \int_{y_0}^{y} \frac{2x}{x^2 + t^2} dt = 2x(\tan^{-1}(\frac{y}{x}) - \tan^{-1}(\frac{y_0}{x}))$?

athos

19:46

Hi i wonder, if 1/n is a repeating decimal, in which case will the repetend has odd digits?

e.g. 1/7 = 0.(142857) the repetend has 6 (even) digits.

porridgemathematics

20:11

athos all n such that there exists some odd number o for which 10^o = 1 (mod n)

user193319

Is it true that if $D \subseteq$ is a domain(?), $\gamma_1$ and $\gamma_2$ are two closed paths in $D$ that are homotopic, and $f$ is analytic on $D$, the $\int_{\gamma_1} f(z) dz = \int_{\gamma_2} f(z) dz$?

porridgemathematics

user193319 yes it is

user193319

Hmm...something seems off though...shouldn't those integrals both be zero?

porridgemathematics

no, they need not be closed paths

oh whoops, by closed paths you meant a loop ?

yes then they should both be zero

user193319

Yeah. E.g., a circle.

This is confusing...I am trying to integrate $\int_{C} \frac{Log(z+2}{(z+1)^2 z}$ over the curve $C = \{z \in \Bbb{C} \mid |z| = \frac{3}{2} \}$. I want to isolate the points $0$ and $-1$with smaller circles and split the integral into two pieces...but something seems off.

How does one do this?

porridgemathematics

20:20

so integrate over the circle of radius 3/2?

user193319

The circles would be $C_1 = \{z \in \Bbb{C} \mid |z+1| = \epsilon\}$ and $C_2 = \{z \in \Bbb{C} | |z| = \epsilon \}$, where $\epsilon$ is small enough so that the circles don't overlap and they stay inside $C$.

I don't think I can do that directly because we have singularity (poles?) at $0$ and $1$, which are interior to $C$.

porridgemathematics

aren't Log(z+2)'s poles on {(x,0): x<=-2}?

i mean its branch

user193319

No, you have to look at the denominator to.

porridgemathematics

oh sorry i misread

user193319

In that form you can't simply apply Cauchy's integral formula.

porridgemathematics

20:22

thought that was a product

okay, so can you not just apply residue theorem?

user193319

No, we haven't learned that yet.

porridgemathematics

or is that not allowed or something

well, the proof of residue theorem will then provide for a means to solve this problem

the same steps of the proof that is, will be applicable here

(just substitute in the special case of your function and follow the steps)

user193319

I don't understand.

The problem should be simpler than that.

I don't have time to work through that proof...My homework is due shortly.

Astyx

What is $Log(z+2)$ ?

porridgemathematics

they mean the principal branch

i.e. Log(z) having branch cut on the negative real axis

Astyx

20:28

But the principal isn't analytical on C is it ?

porridgemathematics

yes, but Log(z+2) is analytic on the domain they have provided

the issues are the poles that arise due to the denominator

Astyx

Oh yeah my bad

user193319

I know that I am supposed to break $\displaystyle \int_{C} \frac{Log(z+2)}{(z+1)^2 z}$ into $\displaystyle \int_{C_1} \frac{\frac{Log(z+2)}{z}}{(z+1)^2} dz$ and $\displaystyle \int_{C_2} \frac{\frac{Log(z+2)}{(z+1)^2}}{z} dz$, but that doesn't seem right because those should be $0$.

porridgemathematics

why should those be zero?

there are singularities inside those discs

maybe you should just parameterize the circle and do the integral by hand

(if you don't want to use residue theorem)

user193319

But $C_1$ doesn't contain the singularity of $\frac{Log(z+2)}{z}$, and it is analytic on the interior of $C_1$.

Same goes for $C_2$.

It's not that I don't want to use the Residue theorem, I am not allowed. And my professor wants me to do it this way, not explicitly parametrize the circle.

He wants me to use Cauchy's integral formula

porridgemathematics

20:32

ah okay, yes I see what you are trying to do

wait but why are we talking about Log(z+2)/z

what did you do to the (z+1)^2

Astyx

Why would they be zero ?

user193319

@Astyx Because, e.g., $\frac{Log(z+2)}{z}$ is analytic on and in $C_1$ and $C_1$ is a closed curve, so the integral is $0$....But maybe I am wrong.

Astyx

But you're not taking the integral of that, you're taking the integral of that multiplied with $1/z$ or $1/(z+1)^2$

So the result is given by the Cauchy integral formula, as you stated above

Balarka Sen

Im back

Are we dividing by zeroes

user193319

@Astyx Oh, so are you saying that $\displaystyle \int_{C} \frac{Log(z+2)}{(z+1)^2 z}$ literally equals $\displaystyle \int_{C_1} \frac{\frac{Log(z+2)}{z}}{(z+1)^2} dz + \int_{C_2} \frac{\frac{Log(z+2)}{(z+1)^2}}{z} dz$?

Astyx

20:41

Yes, because the path $C$ is homotopic to a path that starts with $C_1$, goes to $C_2$ in a straight line, does $C_2$ and comes back to $C_1$ taking the same path in the other direction

Something like $C\sim C_1 + l + C_2 -l$

Where $l$ is a path linking $C_1$ and $C_2$

porridgemathematics

but how does cauchys integral formula work here, since C_1,C_2 still have singularities inside?

Astyx

The integrands are of the form $f(z)/(z-\zeta)^n$ with f analytical, which is exactly what the Cauchy integral formula lets you compute

porridgemathematics

oh sorry! im getting that confused with cauchy goursat lol

its been a while since i've done complex analysis

you're completely right

yup that should definitely work now

deformation theorem + cauchy integral formula

Joe Shmo

@AkivaWeinberger, I don't understand the last picture. What do you mean?

Astyx

I thought he was only drawing a stick man

Everstudent

20:47

Hello everyone, I was wondering about this thing: If we take the left and right cosets of a subgroup H of G, can I prove that for some Ha, there isn't a b such that Ha=bH, ie. ${Ha: a \in G} \neq {bH: b \in G}$. I am thinking I can do it by counter-example, ofc I will need a non-abelian G, since for an abelian G, H would be normal, so the left cosets would be the same as the right ones.

Balarka Sen

He's asking if those collapse moves preserve knottedness

Akiva Weinberger

@JoeShmo Click on the question for more words

Joe Shmo

hrmp hrmp because the two aren't homotopically equivalent

I'll click for more words

Akiva Weinberger

If you take a Hopf link and connect the components by an arc in a simple way, the resulting shape is called Hopf handcuffs

Balarka Sen

Hopfcuff lol

Akiva Weinberger

20:49

If you do the transformation in the second image to the Hopf handcuffs, you get my knotted graph from the first image.

I don't claim that the result of the transformation is isotopic to the original.

Astyx

@BalarkaSen That sounds like a Hogwarts house

Balarka Sen

Hopflecuff

Astyx

You're a knot theorist, Harry

Joe Shmo

gotcha

Akiva Weinberger

I found a reference that answers my question (see comments)

but I still think a better proof is out there

Everstudent

20:51

So now that I am thinking about it, is there a connection between the right and left cosets of a group that's not normal? I mean both types of cosets obey Lagrange, but I am looking if there is another connection.

Akiva Weinberger

especially since the question of iterating the transformation is still open

Like, you can do that transformation again and again, and intuitively the result is still linked

and I don't have a proof

Balarka Sen

harry potter hyper distorted | sound effect

►

Astyx

I have transcended my material existence after hearing this

user193319

Btw, am I right in thinking that $\displaystyle \int_{y_0}^{y} \frac{2x}{x^2 + t^2} dt = 2x(\tan^{-1}(\frac{y}{x}) - \tan^{-1}(\frac{y_0}{x}))$?

Akiva Weinberger

@BalarkaSen Harry did you put your name into the speakers of fire

Balarka Sen

20:53

LOL

Astyx

said Dumbledoor calmly

Akiva Weinberger

I AM CALM

Balarka Sen

Astyx your youtube recommendation is my youtube recommendation

Alessandro Codenotti

@Astyx DIDDAYAPUTYANAMEINDAGOBLETOFFAIA

Akiva Weinberger

"Calmer than you are" - Walter from The Big Lebowski

Astyx

20:55

@user193319 almost, there's no 2x, just 2

Joe Shmo

Hopflecuff.. I chuckled

Balarka Sen

Chuckle... I am in chuckles.

Akiva Weinberger

By the way

Joe Shmo

whats up with all the harry potter references, anyway?

Akiva Weinberger

I only found like one reference that called it Hopf handcuffs

but it makes sense to me so I'm going with it

@JoeShmo Astyx started it

Astyx

20:59

Sorry

Balarka Sen

Ahem, I'd like to buy a pair, ahem.

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$Mathematics$ Mathematics