Mathematics - 2022-06-18 (page 1 of 2)

3:44 AM

@BalarkaSen I found a different solution about the trivial Weyl group problem I asked before. For any compact connected Lie group $G$, there is a finite cover $\tilde{G}$ of $G$ such that $\tilde{G}\cong T^k\times L$ where $T^k$ is a torus and $L$ is a simply connected compact Lie group.

Using the fact that Weyl group of a Lie group and its tangent space is equal, we conclude $W(G) = W(\tilde{G})$. Now assume $W(G)$ is nontrivial, then $G$ is nonabelian so $\tilde{G}$ is also nonabeilan. Hence, $L$ is nontrivial so it's a product of simply connected compact Lie groups whose Weyl group is nontrivial. Since Weyl group preserves product, this shows $W(\tilde{G})$ is nontrivial.

Koro

4:52 AM

1

Q: Divergence of a series given another divergence series

Honestly, It is a homework problem. Let $\sum_{n=1}^{\infty} x_n$ be a divergent series with positive terms. We have to examine whether the following are true or false i) $\sum_{n=1}^{\infty} \frac{x_n}{1+n^2x_n}$ is convergent ii) $\sum_{n=1}^{\infty} \frac{x_n}{1+nx_n} $ is divergent I am able...

is a duplicate of math.stackexchange.com/questions/130471/…

CroCo

5:18 AM

Good morning everyone, what are the common mathematical proof methods in linear algebra?

copper.hat

5:39 AM

asking on chat in mse is one.

your question is a bit too general to get a good answer imo.

leslie townes

a lot of elementary stuff is definition chasing. you can get a surprising lot out of that alone because the definitions have been crafted to be useful and good.

also, a lot of induction arguments of varying degrees of cleverness.

less purely calculational proving than you might think.

Koro

5:59 AM

Every continuous open mapping of R into R is monotonic.

Silent

9:23 AM

How to integrate $\int _0^{\infty \:}\:\dfrac{\arctan \:ax}{x\left(1+x^2\right)}\:$ for $a>0, a\ne1$? Please provide me some hint! I tried substituting $u=\arctan ax$, but could not go much far. Then I tried using contour integral technique as integrand is an even function, on real line, but there did not know how to prove that as radius of semicircle increases, contour integral on upper half ges to zero.

Silent

9:38 AM

ok, got it here. Still, I would love to know how can I use contour integration here, if at all.

Mad Spaces

10:05 AM

@Koro Yes. and from that you can deduce if its bijective then it is also strictly monotonic

Koro

10:16 AM

@MadSpaces it is bijective.

robjohn

10:44 AM

@Silent arctan has a problem around $i$ like log has around $0$. You need a branch cut to define arctan around $i$.

The branch cut can be a cut between $+i$ and $-i$, but it prevents one from drawing a contour along the real axis and back along a large semi-circular arc.

Koro

Every real valued function on R has at most countable simple discontinuities.

To prove this, the book gives hint that involves injecting the set of such discontinuous into $\mathbb Q^3$.

robjohn

11:07 AM

how do you define a simple discontinuity?

Koro

when left hand and right hand limits exist but 1) are not equal or 2) they are equal but not equal to value of the function at that point.

robjohn

so a jump discontinuity

or one where the limit does not match the value

Koro

yes.

Mad Spaces

I have an issue understanding this.
https://math.berkeley.edu/~vvdatar/m185f16/notes/Lecture-19_Zeroes.pdf

Proof of Theorem 0.1. he says, if f is not identically zero, there exists a point (according to the lemma) there exists a point at which the derivatives until some index equals zero.. the previous lemma only states, if a function has such point to all indexes equal to zero then the function is zero. my problem is, how does he even guarantee the existence of such point?!

"IE" that the derivatives also equal to zero.

Or is he just saying "if such exist", if not we take the index to be 1 since the condition says atleast f(a) = 0?

i think yes. okay nevermind for annoying you my ladies and gents i understood it :)

geocalc33

2:07 PM

Consider $\mathcal S^n,$ the space of simply connected, closed topological $n-$manifolds as subsets of the unit $(n+1)-$cube, which include $p=(0,0,\cdot\cdot\cdot,0)$ and $q=(1,1,\cdot\cdot\cdot,1)$ and consider the maps between $\mathcal S^n$ that fix $p$ and $q.$ Does this make sense so far?

Monty

2:26 PM

I want to use IBP to conclude that $\int_{\mathbb{R}^d}g(x)Ax\cdot \nabla f(x)dx=\int_{\mathbb{R}^d} \text{div}(g(x)Ax )f(x)dx$, here $A$ is a constant $d\times d$ real matrix, and $f,g\in L^1(\mathbb{R}^d)$, but I dont know how to argue the boundary terms are zero

oops that should of beeen $\int_{\mathbb{R}^d}-\text{div}(g(x)Ax)f(x)dx$

leslie townes

that vector calc stuff is not going to make literal sense for general L^1 functions. but often in situations like this you see people prove something literally for smooth compactly supported functions and then some kind of density argument establishes whatever more generalized thing is going on.

Calvin Khor

3:23 PM

@Monty maybe you should first argue why the integrals even make sense for f,g in mere L1. Once you fix the assumptions there you might be able to get what you want

Asinomás

3:39 PM

hello good afternoon

can someone help me with a math problem

I have a number theory question

but I will not be satisfied with any answer

I want my answer to be color coded

and to make 0 sense

red green and blue

Joe Shmo

what are you talking about

Asinomás

This is not possible?

Joe Shmo

$\color{red}{red}\ \color{green}{green}\ \color{blue}{blue}$ is possible, if that's what you're talking about

Asinomás

Yeah

But isn't there a nice repository of nonsense answers with that palette?

leslie townes

some MSE answerers use color pretty effectively. i'm not sure that i've seen people gratuitously use it in ways that make no sense. i suppose i could try.

Asinomás

3:42 PM

lodged here

?

Joe Shmo

I used it spectacularly just the other day

what are you talking about dude?

Asinomás

purple is also allowed if it helps

Joe Shmo

I mean, this is the internet. You could certainly find nonsense plenty here

Mad Spaces

Is this the same as the identity theorem?, when i google "weak principle of analytic continuation" i do not see any results. The identity theorem seems to be worded differently

Asinomás

good point nvm

I will please strive to do better

leslie townes

3:46 PM

mad: yeah, that's often what people call it.

or a variant of that with slightly different hypotheses. i think U doesn't need to be open, just have an accumulation point in Omega.

maybe that's the 'strong' version?

Joe Shmo

cool, good talk

are you translating from French?

Asinomás

Who, me?

Joe Shmo

is that what's happening here?

you

Asinomás

I am translating from the frenchest one

Mad Spaces

Thanks Leslie

Asinomás

3:48 PM

that is probably what happened

but now it's all good hopefully

Joe Shmo

I loved your poem up top, fyi

Asinomás

glad to hear it :)

Joe Shmo

My query is in the theory of numbers
but I will not be satisfied with any answers, none!
so help me color code my LaTex
so it all makes 0 sense
red green and blue

Asinomás

I would have been a poet had I not ignored my guitar teacher when I was young

hmm

that doesn't really capture what I wanted though

I know how to color code latex already unfortunately

I was just looking for an answer to a question in elementary number theory

Joe Shmo

I didn't understand anything from that paragraph. Gurmischt.

Asinomás

3:53 PM

but I wanted it to be very bad, and in red, blue, yellow, green and purple

I thought this was a good place for that

Joe Shmo

why were you looking for a bad answer, so badly?

I am offended at your last remark.

Ethakka appam with Chai

@Asinomás what's the idea behind this

Asinomás

Well, I wasn't looking for it so badly, but I knew it could be found here

Ethakka appam with Chai

why?

Asinomás

like when you want a cinammon roll

and u want it to be bad

Ethakka appam with Chai

3:54 PM

what

Joe Shmo

and you are going to double down on it. I see how it is.

Asinomás

so u go to the bus stop terminal

Ethakka appam with Chai

I have never in my life wanted a bad cinamon roll

Joe Shmo

I cannot relate.

Ethakka appam with Chai

facts

Asinomás

3:55 PM

good cinnamon rolls are not a thing

Ethakka appam with Chai

how is that possible?

Asinomás

unless you are coping to the n'th degree

Joe Shmo

what. the. f***.

Asinomás

because the recipe sucks?

Joe Shmo

are you talking about

Ethakka appam with Chai

3:55 PM

do you mean good in the moral sense or ethical sense or in the health sense?

Asinomás

in the "run it with a normal objective function for a homo sapiens"

Joe Shmo

OK now you're going to flex on us with your French pastries?

is that what this is?

Ethakka appam with Chai

what's a normal objective function

Asinomás

you pick some embedding for objective functions

and run stuff

Ethakka appam with Chai

and infact, we are not homo sapiens actually

Asinomás

3:56 PM

and hope your employer won't yell at you

Ethakka appam with Chai

we are homo sapiens sapiens

Asinomás

but he wont

Ethakka appam with Chai

fyi

Asinomás

nice one

Ethakka appam with Chai

the more you know

Asinomás

3:57 PM

the more you meme

Ethakka appam with Chai

facts

but what has motivated this craving .. or urge or need?

Asinomás

I was going to answer some weird number theory problem

which could be solved by casework and CRT

Ethakka appam with Chai

you wanted to put a shitpost as an answer?

Asinomás

but I had to go to a breakfast meeting

Ethakka appam with Chai

o_o

Asinomás

3:58 PM

nah, I was going to put a normie answer

but had no time

Ethakka appam with Chai

dayum

Asinomás

but the problem admitted a dupe closure

Ethakka appam with Chai

Bruh

Asinomás

under a very vague dupe

Ethakka appam with Chai

dupe closed back into the stone age

Asinomás

3:59 PM

so I was waiting for the usual suspect to dupe hammer it

but nothing in site

Ethakka appam with Chai

I don't get it

how is dupe closure and dupe hammer different?

Asinomás

So I was looking for someone else that was versed in RGB coded hieroglyphs to hammer it

Ethakka appam with Chai

...?

Asinomás

they aren't

Ethakka appam with Chai

if it is dupe closed, how can someone further dupe hammer it

Asinomás

4:00 PM

it hasn't been dupe hammered yet

Ethakka appam with Chai

ah

Joe Shmo

you are looking to decipher a bad answer with colors?

Ethakka appam with Chai

then write the answer, what's stopping you?

I think they want to write an answer

Asinomás

it's going to take me like 20 minutes

and then someone is going to say

Joe Shmo

there's 2 of them now?

Ethakka appam with Chai

4:01 PM

getting banned by EOQS speed run

Asinomás

"please strive to not answer dupes"

Ethakka appam with Chai

sweaty gamer speedrun meme original

►

Asinomás

wow, you picked one of the worst versions of that meme

Ethakka appam with Chai

is there better version?

Asinomás

why did you back down on ur profile ?

Ethakka appam with Chai

4:03 PM

I decided to embrace the philosophy of gandhi

satyagrah

"Mahatma Gandhi to designate a determined but nonviolent resistance to evil."

Asinomás

based and regional

what a legend

Ethakka appam with Chai

lmfao

Maybe next month I will actually change my profile and name to Gandhi himself

Asinomás

let me know when

Ethakka appam with Chai

AHAHAHHAHA xD will do

Asinomás

so I can change my name to "Rephunter Chad" at the same time

Ethakka appam with Chai

4:06 PM

LMFAO

Asinomás

Sorry, it was "Rephunter Chadwick"

it performed better in the control group

Ethakka appam with Chai

which control group? o_o

Asinomás

The one I deployed in my mastodon group

Ethakka appam with Chai

you mean the ancient elephant?

Asinomás

Mastodon (software)

Mastodon is free and open-source software for running self-hosted social networking services. It has microblogging features similar to the Twitter service, which are offered by a large number of independently run Mastodon nodes (known as "instances"), each with its own code of conduct, terms of service, privacy options, and moderation policies.Each user is a member of a specific Mastodon instance, which can interoperate as a federated social network, allowing users on different nodes to interact with each other. This is intended to give users the flexibility to select a server whose policies they...

Ethakka appam with Chai

4:09 PM

this is the first time I heard of this

Asinomás

that is because you are a major noobcake

Mad Spaces

@leslietownes
https://www.dpmms.cam.ac.uk/~agk22/id-thm.pdf
I am checking out the proof of the identtiy theorem. in the before last line. he writes So γ(t0) is a non-isolated zero of h. isnt this sufficient to show a contradiction, i thought holomorphic functions can only have isolated zero points.

Ethakka appam with Chai

;-;

what is the appeal of mastodon?

leslie townes

mad: all of this stuff really depends on how you order the material. they could be on their way to proving that.

Asinomás

no appeal

Ethakka appam with Chai

4:11 PM

then why you use it?

Mad Spaces

@leslietownes Suppose you already know that. Can you stop there?

Asinomás

it is like rebel software for rebel wannabe rebels

leslie townes

am i actually going to have to click into this pdf to find out?

Mad Spaces

hmm, i dont think you need to..

Ethakka appam with Chai

but what we rebel against

le stackexchange?

Asinomás

4:12 PM

we should not rebel

because all is well and good

Ethakka appam with Chai

then why do you use mastodon

God's in his heaven, all is good with the world

Asinomás

because I am a goth highschooler

Ethakka appam with Chai

really?

quite a big brain highschooler you are ig

SHASHAANK B.H.

@Asinomás Highschooler? are u doing ur college?

Asinomás

I am doing well thank you

Ethakka appam with Chai

4:13 PM

ahahahaha

wtf question did you answer here : math.stackexchange.com/questions/1762693/…

ahhahaha

leslie townes

mad: it looks like up until that point it's only been proved for discs, so the remaining bit is to handle the case when "D" is more general than a disc. if you already had it for connected open sets you would be done.

SHASHAANK B.H.

@EthakkaappamwithChai That question is gud btw

Asinomás

@EthakkaappamwithChai why are you going against an answer with almost 50 upvotes from our prestigious users

ur out of line nooby

Ethakka appam with Chai

;-;

nooby...? :flushed:

Asinomás

Like 80% of my rep comes from answers that thoroughly suck

SHASHAANK B.H.

4:15 PM

offcourse ain't u new here?

Mad Spaces

@leslietownes I understand, however, how does this change the fact, that "as far as my understanding" a holomorphic function can not have none isolated zeros...So, i am thinking, he could just stop the proof there? but i am not sure how to go on if i just say "contradiction"

leslie townes

as i read that page, he couldn't stop there because he hasn't proved it for a holomorphic function on a general domain, only on a disc. the biggest disc around that one non-isolated point that you know to exist by the hypothesis of the theorem might not be all of the domain.

so you need to say something about the other stuff. it isn't very much to say, but if you were teaching it you'd probably include something.

Mad Spaces

Suppose you know it can not be done, what would you say " the supremum " does not exist?

leslie townes

books vary a lot in terms of how much they work in the first instance with discs vs. more general domains. for example because series developments only exist on discs and because cauchy's integral formula is maybe easier to state for circles centered on a point than it is for arbitrary curves.

be more specific. what's the "it" that cannot "be done." if you know a holomorphic function on a domain has only isolated zeros, then you're done because basically have the result already.

you wouldn't worry about curves leaving a disc and going to some other part of the domain.

Asinomás

hello Mr Leslie

leslie townes

4:20 PM

you'd skip most of that and just apply what you already know to the difference f - g.

hi asinomas

Asinomás

Have you ever had to bother EU companies ?

Mad Spaces

@leslietownes I see. okay. so if a function has none isolated zeros and its homorphic. it must be the identical zero function?

leslie townes

from time to time. and been bothered by them.

Asinomás

Is it complicated if the EU company is small?

leslie townes

mad: the domain of the function is important. you keep leaving this out. his "domain" is a connected open set, while a disc is just one example of such a thing.

Asinomás

4:22 PM

like a retali company with under 1M retail

leslie townes

a holomorphic function on a domain (in the sense of functions) that is not a connected open set can have non isolated zeros and yet not be identically zero.

Mad Spaces

@leslietownes If the set was open and connected?

leslie townes

asinomas: i only practice in the US and most of our law is territorial, so it is unlikely that such a company would be bothering people here (or being bothered by them)

Asinomás

thank u sir

leslie townes

i think what we're learning here is that it's kind of bad to use "domain" for "connected open set" because it collides with other uses of the term "domain."

it's what i'm learning, anyway.

Asinomás

4:24 PM

here is the chat room?

Mad Spaces

no leslie i am asking generally, since
I am a bit confused by this, since we had " a none zero function on open conn set that is holomorphic, can not have none isolated zeros" i am not sure if i can then deduce that, if a function on open set conn set has none isolated zeros . then it MUST be the Zero function

In matter of fact, i am going to say something embarassing, i am not sure if the ZERO function is complex differentiable... i think yes.. but for some reason not sure

leslie townes

i'm failing to understand why you keep asking, essentially, "if you already had a proof of this result, would you still write this proof of this result." no, you'd just cite the other thing.

all that's going on at the bottom of the page is using the result for discs to get it for connected open sets.

there's so much rigidity with holomorphic functions that tons of slightly technically different sounding hypotheses end up leading to the same thing (e.g. that a function is constant). this is just one of those times.

Wolgwang

Hola!

> The function $f(x) = a(x² - 1)(ax + b)$, ($a \ne 0$) has:
(A) a point of local maxima at certain $x \in \mathbb{R}$
(B) a point of local minima at certain $x \in \mathbb{R}^-$
(C) a point of local maxima at certain $x \in \mathbb{R}^-$
(D) no point of local maxima/minima

How to approach this?

leslie townes

you're gonna wanna take a derivative at some point.

maybe even two.

Wolgwang

$$f'(x)=a\cdot\left(3ax^2+2bx-a\right)\\f''(x)=a\cdot\left(6ax+2b\right)$$

leslie townes

4:40 PM

i'll assume that's right. so f'(x) = 0 is a quadratic equation. it'll have real roots or not depending on a condition involving the coefficients. note that from this approach, it isn't immediately clear that the answer might not depend on what a and b are.

you could also avoid derivatives for a minute and just look at the roots of f. f's a cubic, it's graph has a shape you might be able to recognize after learning just a few things about the roots.

this takes advantage of the fact that they've factored it for you and you even know the sign of the leading term. i'm not sure how much of that kind of analysis is covered in the average calculus class, or when it's covered (e.g. before or after plugging and chugging with f'(x))

Asinomás

leslie are u a guitar player

leslie townes

i have a guitar in my closet that i haven't played in years. i'd say no that.

Asinomás

but u used to share some very pro guitaring

Ted Shifrin

5:01 PM

@leslietownes I think this is pretty standard in complex analysis books/courses, however.

leslie townes

yes. just not the best word for this one page of this one issue.

Ted Shifrin

I'm not following what's going on. Not worth it.

Asinomás

why not worth it

Ted Shifrin

Too busy reading about how the "religious" white supremacists want me lined up against the wall and shot.

leslie townes

no big issue. it was extending a result from discs to domains that weren't discs, and when the "domain" on which the relevant thing was assumed holomorphic kept being omitted from the discussion, things got confusing for a while.

Asinomás

5:04 PM

where do they live

the white supremacists looking to like and shoot

and not coke

Ted Shifrin

The US, of course. I realize that this is an issue in many countries, but the US is quickly going backwards about 100 years.

Mad Spaces

Leslie,
i do not know how to draw conclusion to prove the theorem based on what i know, is it not obivous?

I have no idea how to do it...

Asinomás

We sold my gram's lands

Ted Shifrin

Take a path from here to there and cover it with overlapping disks.

leslie townes

you keep distinguishing what you 'know' (from where?) from the exposition on that page. i'm wondering if that is significant, i.e. you want a route to this result with randomly grabbed notes as a guide, or if you're confused by the treatment given in these specific notes.

Asinomás

5:06 PM

but where she lived it was getting eaten up by the cartel

they are using new drones with bootrstrapped granades

Mad Spaces

i already told you what i know
A holomorphic function have no none isolated zeros.
How do i prove now the identity out of this?

leslie townes

mad you keep omitting the domain of your holomorphic function from that statement. it's part of that result and it might be part of the confusion here.

Mad Spaces

Thos is what i want to prove

leslie townes

it seems to be a one-liner if you know that (as the identity principle it's stated in those notes).

Ted Shifrin

Boy, what a weak principle.

leslie townes

5:08 PM

OK, that's a different result than what i was looking at, with a stronger hypothesis. we may have been looking at different pages.

Mad Spaces

You told me its a stronger version

Ted Shifrin

I assume Mad didn't read what I typed above.

Mad Spaces

So i went to prove the weaker version.

And i failed to do that with my knowledge

I asked you how to do it, you told me its obvious

leslie townes

mad: under those hypothesis, the function h = f - g is holomorphic on Omega and has a whole disc of non-isolated zeros, so is constant.

Mad Spaces

I am dumb to see its obvious

Ted Shifrin

5:09 PM

Have we proved the strong version before we're trying to prove the weak version? Seems silly.

leslie townes

from your principle that a holomorphic function on a domain cannot have non isolated zeros (unless it is identically zero).

Ted Shifrin

shrugs and moves on

leslie townes

we're way too blended up here. i would maybe pick up a different set of notes, or go through those various lemmas in the order in which they are written.

Mad Spaces

Okay, but then it is the zero function on $U$ but we dont know how it behaves outside

leslie townes

it might not be the best order in any objective sense, but it's probably the only order that's going to make sense if you're working out of those notes.

Ted Shifrin

5:10 PM

Mad continues to ignore my sentence telling him what to do.

Mad Spaces

Sorry let me check

Ted Shifrin

Duh.

leslie townes

ted this is done in the document. we were discussing it earlier.

Ted Shifrin

Apparently it was not yet grokked.

leslie townes

he got stuck and we have now backtracked.

Ted Shifrin

5:11 PM

That's fine. It's OK to reuse a powerful technique and learn it by using it over and over.

But I apologize for interfering.

Mad Spaces

No you are not interfering, its fine, i am just a useless idiot

@leslietownes Why is it exactly constant ?

leslie townes

if you claim that you already know that a holomorphic function on a domain Omega cannot have isolated zeros without being identically zero on all of Omega, you apply that result to the difference f - g. from the hypothesis that it has an open set worth of zeros within Omega, you deduce that it is identically zero on Omega.

the key point behind every version of this type of result is you can deduce from an equality holding on a nice enough subset of your domain, that the same equality holds everywhere on the domain.

Mad Spaces

I understand what you are saying.

I am just saying,
if $h$ could be defined as $0$ on $U$ and none zero on anything else, why is that not possible?! Is it then none holomorphic

leslie townes

right. it's just not possible for a holomorphic function to do that on a connected open set.

Mad Spaces

And is that because of the continuinity ? or some other wild result

leslie townes

5:17 PM

non holomorphic functions can do that on connected open sets, and holomorphic functions can do that on not-connected open sets. but you gotta give up something.

it's a mix of the complex differentiability condition and connectedness and everything else. it's a weird result.

Mad Spaces

How do i interpet these results?? If some area has two functions which are equal, then they must be equal everwhere? this is counter intuitive

leslie townes

holomorphic functions are just very special. it's counterintuitive to be able to prove identities like that. for example if you know sin and cos have holomorphic extensions to C, you know that stuff like sin^2(z) + cos^2(z) = 1 and sin(2z) = 2 sin(z) cos(z) hold for all complex numbers z as long as you know that they hold for x in some interval of the real line (say (0, pi/2), where you can draw pictures of triangles to prove stuff like that).

it's a goofy world.

Ted Shifrin

This is the power of elliptic operators.

leslie townes

it's definitely not the way people prove identities for more general classes of functions. :)

Mad Spaces

Alright, thank you for bearing up with me.

Ted Shifrin

5:25 PM

Have you worked out the basic case so that you understand it is coming from basic facts about convergent power series on a disk? That's where the power comes from and you need to understand that.

Holomorphic = analytic ... super powerful.

Mad Spaces

No, purely because my profs script sucks ASS

leslie townes

the power of power series.

Ted Shifrin

I haven't thought about this. Can I derive the identity principle just from the Cauchy Integral Formula?

I certainly don't see how. Oh, if I know the holomorphic function is identically zero on the curve $\gamma$, I can get it's zero inside, of course. But not the strong identity principle.

leslie townes

yeah, i can't see it.

Ted Shifrin

5:41 PM

@leslie I realize that this has elementary differential geometric content, but the issue I didn't iron out is analysis/linear algebra. Any ideas would be more than welcome.

It's a very bizarre result.

Create an orthonormal basis for $\Bbb R^3$ by integrating vector fields that lie in the normal plane along a closed curve.

leslie townes

oof.

Ted Shifrin

If only an approximate delta function were truly a delta function. But wiggling to guarantee normality isn't obvious.

It would be easy (at the second point) if I didn't have to stay orthogonal to the curve there.

The paper they got this result from just states it as if it were a fact that every kindergarten student knows. Crazy.

BBIAB

leslie townes

clear as mud to me.

this is the kind of thing where i'd wonder if the author's dissertation had the details. but, no sign that this was phd work.

oh, it's a 'special case' of the first author's dissertation under jost.

hrm.

Ted Shifrin

6:02 PM

Oh, I didn't see any reference in the version I could get to.

As someone commented, it's also very closely related to a technical lemma in Nash's work on embedding!

I think my proof is very close. Just not quite there.

@leslie Where did you find that it's a special case blah blah with a reference? Not that I have access to references, but that would be useful to the OP who posted the question. ... OK, I'm walking to the neighborhood farmers market. BBIAB

leslie townes

page 4 of the paper OP linked to. the author is on researchgate but only has that paper online. i think you can find pdfs of other papers from her but no sign of the dissertation.

Ethakka appam with Chai

what's up gamers

A relaxing break from the Hot Tub Meta - Crossmauz

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one of the most famous twitch streamer as of late

Mad Spaces

6:23 PM

let $f$ be a holomorphic function on $ U - \{z_o\} $ whereas $U$ is an open subset of the complex plane.
When $f$ is none zero, then there exists some number $m$ such that $(z-z_o)^m*f(z) := g$ is holomorphe analytic contunition with $g(z_o) \neq 0$
Why is this true?

Joe Shmo

Mad, you have a really fitting name

youre always so angry

Mad Spaces

No frkn s man, if you only knew what kind of amazing life i have, and this type of stuff i need to fight with because people in authority positions can not make a coherent lecture

Joe Shmo

if the latter is your only problem

take a deep breath. go out for a jog. your life is great

Mad Spaces

i disagree

Joe Shmo

if the latter is your only problem, I disagree with your disagreement

Mad Spaces

6:31 PM

try working four jobs, have two chronical ilnesses and doing a double bachelor. oh also living alone, not that great

Do you know the answer to my problem?

It seems that this is called the "ORDER" but i am not sure which theorem states what i wrote

Joe Shmo

smells like its true, but idk. this is right up Teds ally I think

Mad Spaces

I know that a similiar statment can be made , if f(z_o) = 0. .but i am not sure about being a singularity

Joe Shmo

whats the context?

throw terms at me

give me some of the associated jargon

Mad Spaces

Its about analytical continuinity...

Joe Shmo

thanks

geocalc33

6:35 PM

Lesliecoin

Joe Shmo

looks like Ted was trying to help up top?

leslie townes

mad: it's like the result for polynomials and even has a similar proof if you know the power series expansion

e.g. if p(x) is a polynomial and p(1) = 0 you can factor out at least one copy of x - 1 from p(x) and write p(x) = (x-1)^[something] q(x) where q(1) is nonzero

[something] being the order of the root (which has an algebraic meaning in the polynomial case)

my p(x) is not the zero polynomial there, that exception comes up too even with polynomials

Mad Spaces

I understood your example about the polynomial, however, how does this relate to the function said. what theorems or subjects i need to know to understand this statement?

leslie townes

the power series expansion of a holomorphic function around a point and enough of the analysis to justify factoring a power of (z - a) out of it

Mad Spaces

For a power series, is it not needed that the function be defined on all of the subset. Said function has a singularity ( i am guessing you mean Taylor?)

leslie townes

6:48 PM

when i say 'power series' i'm ruling that out. nonnegative powers of z-a only

holomorphic functions, not ones that might have poles or other singularities

Mad Spaces

7:03 PM

I think you are possibely mixing up a zero and a isolated singularity

And i wrote it wrong.. because i am stupid

The statements states that if yadda yadda then $(z-z_0)^m * f(z) $ has an analytical continuiation $g(z)$ and this $g(z)$ is not equal to zero at Z_o

@leslietownes thats the correct phrasing, i misunderstood it first time

Wave

Is it true that if I have two real valued random variables $X,Y$ and I take a mesurable set A in the product sigma algebra and want to compute $P((X,Y)\in A)$ then $P((X,Y)\in A)=\int_{\Bbb{R}^2} 1_{A}(x,y) P_(X,Y)(dx,dy)$ where $P_(X,Y)(dx,dy)$ denotes the density if there is one.

Ted Shifrin

@MadSpaces It’s not true. Take $e^{1/z}$.

Mad Spaces

No ted, it is true... its apparently just iterativity of the Riemann removable singularities law

Ted Shifrin

No, did you say bounded?

I gave you a function that’s a counterexample …

Your prof is assuming a pole, but the hypotheses didn’t say that.

Order of zero must be finite; order of singularity can be infinite.

Mad Spaces

Let me actually you write the whole theorem i am reading:

If $z_0$ a point in the open set $U \in \mathbb{C}$ and f is holomorph on $U - \{z_0\}$ then only one of the two cases may arise
1) for $m \in \mathbb {Z} $ the function $(z-z_0)^m * f(z) $ has in $z_0$ a removable singularity. for $m_0$ the smallest such number we say that
$\omega (f,z_0) : = -m_o$ the order of $f$ in $z_0$
2) for each nighbourhood $V$ of $z_o$ follows $f(V-\{z_0\})$ is dense in $\mathbb{C}$

Ted Shifrin

7:18 PM

Note this is quite different. Look at my function. How does it fit this?

Mad Spaces

i do not think your function has a dense image around $z_0$, so situation one must arise

Actually i am not sure...

Ted Shifrin

What is $z_0$ in my case?

Mad Spaces

0

Ted Shifrin

OK. Now think about the exponential function $e^w$ for large $w$.