Mathematics - 2018-12-05 (page 1 of 2)

Mike Miller

00:00

I do gauge theory. I know a little symplectic geometry because those fellas hang around near me. But I don't know much

Some similar techniques

Zee

Weird , idk why I thought you did symplectic . Well I am also interested in guage theory actually

Leaky Nun

Don't you just hate it when the author adds a "hint" in square brackets right after the exercise?

Mike Miller

yeah

i think hints are ok but maybe should go in the back of the book or smth?

@TedShifrin OK, tried and true: I spent a long time tryuing to prove it's false, so time to start finding an example instead.

Rithaniel

00:17

So, I have definition for a simply connected space which states that $(X,\tau )$ is simply connected if it is path connected and, given $f:[0,1]\rightarrow X$ and $g:[0,1]\rightarrow X$ are two paths with $f(0)=g(0)$ and $f(1)=g(1)$, then there is a homotopy $H:[0,1]\times [0,1]\rightarrow X$ with $H(x,0)=f(x)$ and $H(x,1)=g(x)$.

Using this definition, I need to show that a space is simply connected iff any loop with base point $x_0$ is homotopic to the constant function $h(x)=x_0$, but this seems way too easy. I mean, $h(x)$ has endpoints both equal to $x_0$, so I feel like I can just restate the definition to answer the problem. It makes me suspicious that I'm missing something.

The reverse direction I suppose is the tricky part, though.

Mike Miller

Homotopic through maps which have $f_t(1) = x_0$?

If so, it's as easy as you think it is.

If not, it takes a little work, and the easiest argument that comes to mind uses the definition of the fundamental group and an identification of its conjugacy classes.

Alternatively it requires an explicit construction which looks kinda like the latter.

Rithaniel

A loop here is $L(0)=L(1)=x_0$

(L for "loop.")

Mike Miller

ok, I wanted to think of maps from the circle instead of the arc. In any case the answer is "yes", all of your loops always fix the bit that goes to the basepoint.

If your definition was only $L(0) = L(1)$ it would be harder.

Rithaniel

Really? I'm not immediately seeing how that complicates the issue.

Mike Miller

00:36

Write down a proof of the converse direction, checking at every step if you use that assumption.

If you write it down and don't see where it happens, post it here and I'll show you.

Rithaniel

00:46

Actually, I need the assumption that the space is path connected, right? That not assured by loops being homotopic to the constant function?

(It's not actually stated in the question)

(The fact that we can assume the space is path connected, that is. It's just $X$.)

Mike Miller

Sure.

"Paths rel endpoints are homotopic" and "Loops rel basepoint are homotopic" are equivalent independent of path connectedness. By picking a basepoint you chose a component to care about.

Rithaniel

Also, an outline of my proof in the opposite direction:
We know that functions which are homotopic to each other form a group. In other words, if a function is homotopic to a second, and that second function is homotopic to a third, you know that the first function is homotopic to the third. Next, given any path, we will build a loop which moves along the path and then retracts steps back to the start of the path. Badabing Badaboom.

Mike Miller

That's not what it means to form a group. The word you want is called an equivalence relation.

I don't understand what your argument is trying to say.

You never used the word "loop" or the definition of path and it's not clear to me what you're proving in the end

Rithaniel

01:07

Well, the thought is that we take a function $f:[0,1]\rightarrow X$ and then define the path $p(x)=f(2t)$ if $0\leq t\leq\frac{1}{2}$ and $p(x)=f(2-2t)$ if $\frac{1}{2}\leq t\leq 1$. Now we have a loop. It is homotopic to the constant function $h(x)=f(0)$. Given any path with starting point $f(0)$, we now know they're homotopic to $h(x)$, and since homotopies are transitive (also, you're right, I don't know why the word "group" was in my head), all these functions are homotopic.

In fact, I think this is stronger than simply connectedness, because I've shown that end points can vary, right?

Well, I haven't shown anything, yet. This is just outline stuff.

Mike Miller

01:22

@Rithaniel When you allow the endpoints to very everything is trivial. That's what you're observing.

Rithaniel

Alright, so is there a flaw in my proof wherein I assumed from the beginning that end points can vary? Stuff like that has slipped by me before, so I wouldn't be surprised.

Mike Miller

Of course you can collapse the interval to a point. But trying to get every loop to be homotopic without moving the endpoints - and exploiting this fact about the interval - is going to be impossible.

It might help your intuition to think of this as being maps from a circle (with a basepoint fixed). That's what the idea is supposed to capture.

You can't do any tricky stuff like that there.

Rithaniel

What I'm thinking about is that I should be able to turn a path into a loop, deform it to another loop, and then disconnect the loop. Doing this I should be able to go from any path to any other path, right?

Mike Miller

The evil part is "disconnect the loop".

Rithaniel

Ah, I see.

So, I'd need to show that any loop can be "chopped off" into a path which goes from one point on the loop to another point on the loop?

Or perhaps try a more cautious approach?

Mike Miller

01:38

You're going way off the deep end for a relatively simple proof. It should be symbol pushing.

Wait, @Rithaniel.

Can you remind me precisely what you're trying to prove, and define the terms involved? I'm a little confused about something.

Rithaniel

I need to prove that a space "$X$ is simply connected iff any loop (with base point $x_0$) is homotopic to the constant function $k(x)=x_0$ for $x\in [0,1]$." The forward direction I simply argue from definition.

"iif = if and only if" and the simply connectedness uses the definition I gave above (I could write it all out again, but, like any definition, it's a little longwinded).

Mike Miller

When you say "loop", you never mean a map from the circle, right? But rather a map from the interval with certain specified boundary values?

And when you say "homotopic", how precisely is this different from what is written in your definition of simple connectedness?

OH MAN

Rithaniel

Actually, I was not given a definition for "loop"

Oh man? Ohman? Ohmen? Omen? Is this an omen?

Mike Miller

So I now see that I have been leading you astray. The content of what you're trying to prove is very different than what I thought you were trying to prove.

I thought you were essentially trying to prove the following. "Every pair of maps $f,g: [0,1] \to X$ with $f(0) = f(1) = x_0$ are homotopic through maps with the same property" equivalent to "Every pair of maps $f, g: S^1 \to X$ with $f(1) = x_0$ are homotopic through maps with the same property."

This is immediate by folding up the interval. Those two descriptions are exactly the same thing. Not much has to be proved.

What you're worried about is trying to show that $f \sim g$ for all $f,g$ iff $f \sim 1$ for all $f$.

And your proof is fine. The fundamental group is a group. So $f \sim g$ is equivalent to $fg^{-1} \sim 1$. And if you know $h \sim 1$ for all $h$, you're done.

Rithaniel

01:56

Woot!

Mike Miller

Sorry for leading you astray. I hope at least it's clear how my mind got there.

Rithaniel

I was beginning to suspect that this was a much more difficult problem than I had initially suspected.

Though, the problem you thought it was seems like an interesting one.

Mike Miller

You should try to understand my brief comment about why those two concepts are the same. :) It involves the definition / universal property of quotient maps.

Rithaniel

and I think I can see the shadow of why not having the $x_0$ makes it more difficult.

Mike Miller

Well, regarding that, the bit in your proof hwere you said "equivalence classes form a group with concatenation", that's where you secretly used the basepoint.

To concatenate you need your intervals to start and finish at the same point.

Rithaniel

02:00

Ah, okay. I completely missed that, too.

Secret

04:01

Hmm

Is that a convoluted way of saying:

$\mathcal{S}^p = \mathcal{P}(\mathcal{S}) \cap (\mathcal{P}(A)-\{A\})$ ?

Secret

05:17

[Random]

A fresh look at topology:

Recall that given any set $A$, the powerset $\mathcal{P}(A)$ can be identified with the set of all indicator functions of $A$

A topology $\tau$ is thus a subset of $\mathcal{P}(A)$ satisfying the following properties:

Sharath Zotis

There is an empty tank that has a hole in it. Water can enter the tank at a rate of 1 gallon per second. Water leaves the tank through the hole at a rate of 1 gallon per second for each 100 gallons in the tank. How long, in seconds, will it take to fill the tank with 50 gallons of water?
Anyone have any idea how to approach this problem

Secret

1. $0_A, 1 \in \tau$

2. $\sum_{a \in \tau} 1_a \in \tau$

Sharath Zotis

@Secret any idea about the problem above

Secret

05:33

3. $\prod_{n < \omega} 1_{a_n} \in \tau$

@SharathZotis that's a standard problem in 1st year calculus. Have you wrote down the differential equation for the change in volume of water?

Sharath Zotis

Hmm Im having a little bit of trouble doing that

Secret

What have you tried so far?

Sharath Zotis

so from what I remember from Calc the $V'(t) = $ Rate at which water enters the tank minus the rate at which water exits the tank

So the rate at which water enters the tank is 1 gallon/s

and the rate at which water leaves the tank is $\frac{1}{100} \cdot V(t)$

where V(t) is the volume at the time t

But I can't have $V'(t)$ depend on $V(t)$

Secret

Right so you have Rate = rate in - rate out. Plugging the terms in, you have a linear ODE:

Rithaniel

Why not?

Secret

05:49

$$V'(t)=1-\frac{1}{100}V(t)$$

Which can be easily integrated

Sharath Zotis

lol, for some reason I kept on writing $V(t) = 1 - \frac{1}{100}V(t)$

instead of $V ^ { \prime } ( t ) = 1 - \frac { 1 } { 100 } V ( t )$

thank you for your help

wait @Secret how do you integrate $\frac{1}{100}V(t)$

Nvm thank you

CowperKettle

07:57

> Scientists at the University of Oxford may have solved one of the biggest questions in modern physics, with a new paper unifying dark matter and dark energy into a single phenomenon: a fluid which possesses 'negative mass." If you were to push a negative mass, it would accelerate towards you.

Yet another news that I'll never be able to comprehend

phys.org/news/2018-12-universe-theory-percent-cosmos.html

theantomc

08:31

Hi @Secret

I m going a little bit foreward... I understand that i need to calculate a moment, because i cant apply directly the Markov inequality (becase i have a weak assumption) and after apply the inequalty. So i understand just the flow

Silent

10:39

This is rudin's proof about Riemann Stieltjes integral works fine for addition of function.

In this proof, i can't understand the last line: 'If we replace $f_1$ and $f_2$ in (21) by $−f_1$ and $−f_2$, the inequality is reversed, and the equality is proved.'

How does this work?

Vrouvrou

12:25

Hello

please I need help here:math.stackexchange.com/questions/3026084/…

I don't find the proof

Vrouvrou

13:04

@Astyx hello

Vrouvrou

13:44

hello

someone know the essential sup

?

Liad

14:32

@AlessandroCodenotti hi

@Vrouvrou $||f||_\infty$ ?

Vrouvrou

@Liad this

Liad

i can try to help.. what's the question?

Vrouvrou

how to prove 1 ans 2

Liad

sorry not familiar with this :/

Vrouvrou

do you know property of esssup

$esssup(p(x)/[p(x)-1])$

what can be equal to ?

Semiclassical

14:41

@Silent presumably the point is that, if $f_1,f_2\in\mathscr{R}(\alpha)$, then so too are $-f_1,-f_2$

Vrouvrou

@Semiclassical can you help me?

Semiclassical

you can then run the argument of the proof again for this pair, and (apparently) you'll get the same result except with $f_1,f_2,f=f_1+f_2$ replaced by $-f_1,-f_2,-f=-f_1-f_2$

which amounts to the same result as (21) except with the inequality symbol reversed. (i.e. $-x\leq -y$ is the same as $x\geq y$)

So then you're in a scenario where $x\leq y$ and $x\geq y$ therefore $x=y$.

Vrouvrou

no one know essential sup >?\

tarit goswami

15:46

Is there any Journal in Combinatorics or discrete math which can be understood upto a good extent by an Undergrad?

Vrouvrou

Is $ess sup(1/p(x))=1/ess inf(p(x))$ ?

Secret

16:22

Essential supremum and essential infimum

In mathematics, the concepts of essential supremum and essential infimum are related to the notions of supremum and infimum, but adapted to measure theory and functional analysis, where one often deals with statements that are not valid for all elements in a set, but rather almost everywhere, i.e., except on a set of measure zero. == Definition == Let f : X → R be a real valued function defined on a set X. A real number a is called an upper bound for f if f(x) ≤ a for all x in X, i.e., if the set f − 1 ...

Too much measure theory to comprehend

Also, too many pings

Liad

16:36

@MatheinBoulomenos let $H = Z(f)$ the locus of zeros of a polynomial $f$ , i want to describe in terms of f the algebra $k[\Bbb P^n - H]$ of regular functions on $\Bbb P^n$. i want to find generators and deduce directly that $k[\Bbb P^n -H]$ is finitely generated. any chance you can help me with this question?

anakhro

Hey hot dogs and cool cats

Silent

@Semiclassical Thank you very much, semi!

Mary Star

17:16

Hello!!

Does the diophantine equation $x^2+2y^2=z^2$ have only positive solutions?

Fargle

@MaryStar Say you had a positive solution. Could you derive a negative solution from it?

Mary Star

Since we have squares the negative one will be also a solution, or not? @Fargle

Fargle

I agree with that reasoning.

Mary Star

So it is enough to calculate the positive solutions and from there we get all the solutions, right? @Fargle

Fargle

In this case, yes. Each positive solution $(x_0,y_0,z_0)$ will determine 7 others.

Semiclassical

17:27

Do diophantine equations allow any of the variables to be zero?

Rithaniel

17:39

I believe diophantine equations are just over the integers, and 0 is an integer.

Mary Star

@Fargle Ok! Thanks!!

Semiclassical

@Rithaniel makes sense. in that case you end up with annoying things like (1,0,1) determining only 3 other distinct solutions

Fargle

Yeah, I should have provided for that.

Semiclassical

it's still the case, though, that the nonnegative solutions are sufficient to give representatives of all solutions

it's just that the size of each class of solutions will differ depending on how many zeros you allow

...but, also, there's just not that much interesting about the solutions with zeros.

so focusing on positive solutions is sensible

Vrouvrou

18:07

hello

any help for this please :math.stackexchange.com/questions/3027403/…

Semiclassical

It's always fun when you come up with an integer sequence that isn't in the OEIS, lol

Ted Shifrin

Hi @Semiclassic and @Fargle

Fargle

Heya @Ted

Ted Shifrin

Well, things sure are lively around here ...

Rithaniel

As always :P

Ted Shifrin

18:22

Often they're too lively for me ...

Semiclassical

hi @ted

Rithaniel

Fair enough. I suppose I'm used to chatrooms where no one ever shuts up.

Semiclassical

I converted my polytope code into sage, and it's working pretty well

Ted Shifrin

Well, there are times where literally 5 people are asking me for help simultaneously ... that's no fun.

Aha @Semiclassic. Well, you've hit variants of Matlab, Maple, and Mathematica.

Semiclassical

yep

main problems with matlab being that it's not open source (so I can't use it on my laptop anymore---my school license for it expired) and that the code I had in matlab was doing numerical linear algebra

first of those wasn't such a big issue, since I can still stop by campus for that

Ted Shifrin

18:26

The few times I tried to learn Matlab, I went crazy. So I gave up. I did learn Maple at some point, because of our calculus labs were done in Maple and I wrote several new labs.

Semiclassical

the second, though, is a pretty big deal when dealing with many many points

and in that regard Sage is way better since I can have it work over the rationals

Ted Shifrin

Right ... Sage is very seriously used by number theory types ...

Semiclassical

yeah

Doesn't mean it's not slow when I start to crank up the complexity, of course

Ted Shifrin

I hate it when someone emails me about a (brief) answer I posted 3+ years ago ...

You getting your tickets for Italy?

Semiclassical

not immediately. that's still a few months away (end of May)

haven't really started thinking about the logistics yet

Ted Shifrin

18:28

Ah ...

Hi @Tobias

Tobias Kildetoft

@TedShifrin Hi

Semiclassical

These polytope things are in connection with that, though probably not as a starring role

(Largest case I've managed to run so far ends up as a 3D polytope with 3076 faces and 4588 vertices, lol. Have not attempted to make Sage plot that one yet.)

Ted Shifrin

Wow, you sure have gotten up there, @Semiclassic.

Semiclassical

yeah.

and I finally actually managed to get over my confusions with one of my larger-but-still-small cases

Ted Shifrin

Well, congrats. I'll stick to standard simplices :P

Semiclassical

18:33

in particular I think I have a good way of finding points in my original nD polytope which end up landing at the vertices of the final 3D polytope

Érico Melo Silva

hi chat

Ted Shifrin

Oh, hlo @Eric. I just this second answered your email.

Érico Melo Silva

oh word

my internet is weird and my inbox isn’t loading so i can’t read it lol

everything is borked so ima just sign off and go do some complex geo problems

Ted Shifrin

Oh ... only your internet is weird? :P

I am just asking if you wanna discuss some complex geo so I can write my emails ... but I said we can skip it 'cuz I doubt you need my emails anyhow. But I'm still happy to do whatever in the next few weeks.

Érico Melo Silva

ya i mean regardless i wanna talk some complex geo now that my schedule is way less awful

Ted Shifrin

18:39

OK, I'm available most of the time.

Érico Melo Silva

submitting all those apps by itself took 4 hours just going through forms and making sure everything is in order it was a true nightmare lol

Ted Shifrin

Well, I had to do it all with paper and mail, centuries ago, so you are still spoiled :P

Érico Melo Silva

@TedShifrin aight well ima go off to do complex geo probs so i will ask u if any interesting developments or impasses come about

Ted Shifrin

Okey dokey.

Érico Melo Silva

tchau tchau for now

Semiclassical

18:47

@TedShifrin Here's one part of this which is sorta strange.

One thing I know about my polytopes is that they each have to lie in the set defined by $x^2+y^2+z^2 \leq 2xyz+1$ where $(x,y,z)\in [-1,1]^3$

Now, in the small cases, the following is true: Every vertex of my 3D polytope lies on the boundary of that set.

If I enumerate my cases of interest as n=1,2,3,..., this works up until n=3.

But once I reach n=4, it seems as though I start getting vertices which lie in the interior.

Some of the vertices lie on the boundary, but not all of them.

Ted Shifrin

I know Hilbert Cohn-Vossen talk about polytopes in high dimensions and the weird things that happen. I wonder if some of this is related to the shrinking of the unit ball as dimensions go up. I've never really worked on this.

Semiclassical

i think 'concentration of measure' is another relevant phrase

I mean, "weird things happen in high dimensions" is not so strange

So I'm not shocked, but it was still not what I had expected

Vrouvrou

19:07

@TedShifrin bonsoir

Ted Shifrin

bonsoir

@Mathein @Leaky: Have a non-Noetherian example for this?

Vrouvrou

pouvez vous m'aider en théorie de la mesure ?

Ted Shifrin

Non, cette sorte de questions n'est pas du tout pour moi.

Vrouvrou

merci

loch

@TedShifrin k[t,t^{1/2}, t^{1/4}, t^{1/8},...]?

Ted Shifrin

19:18

Oh, sorry, @loch, I should have tagged you, too. Oh, yeah, that should give it easily.

Answer the question :P

Mike Miller

Non-Noetherian isn't in the title, though

Ted Shifrin

So? :)

Mike Miller

Doesn't it feel like making the ring bigger should make it harder to be a UFD?

Anyway, point made

Ted Shifrin

I hadn't thought of a Noetherian way to get infinitely many factorizations ...

But rschwieb makes an interesting point I wouldn't have thought of ...

Tobias Kildetoft

@TedShifrin But that ring is not Noetherian, since that would precisely make all elements be factorizable into irreducibles

Ted Shifrin

19:26

Which ring?

Tobias Kildetoft

The example given by rschwieb

Ted Shifrin

Right ... So I guess you're commenting that my saying non-Noetherian is a necessary requirement.

Tobias Kildetoft

Right, all Noetherian rings have factorization into irreducibles

Ted Shifrin

Seems like one of you guys should put another comment/answer there.

Tobias Kildetoft

since repeated factorization would lead to an increasing sequence of ideals

Balarka Sen

19:28

Hi everyone

Fargle

Heya @Balarka

Ted Shifrin

hi, a @Balarka

christmas_light

hi chat

Ted Shifrin

hi light

Perturbative

Hey everyone

christmas_light

19:34

consider the ideal $I=(x^{a_n+1}_n-x^{a_n+1}_1,\ldots,x^{a_2+1}_n-x^{a_2+1}_1) \subset k[x_1,\ldots,x_n]$, where $a_2\leqa_3\leq\ldots\leq a_n$. How can I prove $I$ is the ideal of $\prod_{i=1}^n a_i$ distinct points in $\mathbb{P}^{n-1}$?

Balarka Sen

Just take $C(\Bbb R)$. That's a non-UFD, non-Noetherian thing :3

Ted Shifrin

What's an element with infinitely many distinct factorizations? I suppose you just modify what I was thinking of with loch's example.

Balarka Sen

$f = (f^{1/n})^n$

For each $n$ that gives you a different factorization

Do you want infinitely many distinct factorization into irreducibles?

christmas_light

I tried some examples on macaulay2, finding the primary decomposition of $I$

but i don't know how to prove, or just to see, the general case

Daminark

Everyone is here!

Super Smash Bros. Ultimate - Everyone is here! (Nintendo Switch)

►

Ted Shifrin

19:41

hi Demonark

@christmas_light: Are you sure you have all your indices correct in that question?

As always, you should write down some simple, concrete examples, for starters.

Perturbative

So weird question, does there exist a function $f : A \to \emptyset$ where $A$ is nonempty?

Ted Shifrin

No.

Perturbative

So follow up question, is $H_n(\emptyset)$ defined to be $0$, the trivial group?

christmas_light

oh the product starts from $i=2$, sorry. it's taken from this article arxiv.org/abs/1110.0745 ...i'll try and return to the chat if i can't find a solution ;)

Leaky Nun

@Perturbative it's not really "defined" to be so so much that it happens to be so

Rando Hinn

19:48

I'm back

Leaky Nun

I'm front

Rando Hinn

this time with a guess that needs validating

lol :D

Balarka Sen

@TedShifrin So I suppose if $f : M \to \Bbb R$ is a smooth function, by Thom transversality theorem you can make $j^2 f : M \to J^2(M, \Bbb R)$ transverse to the subset $Z \subset J^2(M, \Bbb R)$ consisting of 2-jets where both the formal derivative and the formal 2nd derivative are degenerate. $J^2(M, \Bbb R)$ fibers over $M \times \Bbb R$ with fiber $\Bbb R^n \times \Bbb R^{n(n+1)/2}$ and $Z$, fiberwise, has codimension $n+1$ in there (formal derivative is degenrate iff it's 0, the 2nd formal derivative is degenerate if $\det = 0$).

Rando Hinn

Anyways, I've been given this monstrosity desmos.com/calculator/pgab0ri3al and asked to rotate it around the X axis, and find the volume of it, if the circle forms an empty part inside the body.

Am I correct in believing that I want to find piintegral((6-6*(x3))*e**(7x)) at x = 1 and subtract piintegral((6-6*(x3))*e**(7x)) at x = -20/7 ?

Ted Shifrin

@christmas_light: I'm worrying about the polynomials you have written down.

Rando Hinn

19:50

and then find the circle's volume

and subtract

Balarka Sen

That implies at every point of $M$, either $df \neq 0$ or $d^2 f \neq 0$ - i.e., $f$ is Morse.

This proves density of Morse functions in $C^\infty(M, \Bbb R)$

Ted Shifrin

Since I can prove that with simple transversality, I'd say that's overkill, a @Balarka.

Rando Hinn

Should I do anything in regards of this shape being symmetrical on both sides of the x axis?

or is it still the same integral?

Balarka Sen

@TedShifrin I think I can use this proof to prove Cerf's theorem on homotopy of Morse functions.

Roughly speaking this means $f$ generically only has fold singularities - those are the points where $df = 0$ and $d^2 f \neq 0$. In Cerf's theorem you get finitely many slices in a homotopy of Morse functions where you get cusp singularities.

Ted Shifrin

@RandoHinn I have no idea what you're talking about.

Oh, that sounds reasonable, a @Balarka.

christmas_light

19:54

unfortunately they appear without reason on the paper, the only thing is that they're contained (obviously) on the ideal $(x^{a_1+1}_1,\ldots,x^{a_n+1}_n)$, which is the perp ideal of the monomial $x^{a_1+1}_1 \ldots x^{a_n+1}_n$

Rithaniel

Trying to figure out a good proof for $\mathbb{R}^2\setminus\{ (0,0)\}$ not being simply connected. Unfortunately, all the proofs I've found use theorems I don't know yet.

Ted Shifrin

@christmas_light: You're missing my point. You have $x_n$ and $x_1$ in both (every?) binomial.

Huh? @Rithaniel

Adam

what are the acceptable words I use to refer to this? $$\lim _{n\rightarrow N}a_{{n}}b_{{n}}\neq \lim _{n\rightarrow N}
\left( a_{{n}}\right)
\left(
\lim _{n\rightarrow N}\left(b_{{n}} \right)
$$

Rando Hinn

@TedShifrin If you look at my desmos

christmas_light

oh sorry, now i see

Rando Hinn

19:56

the vertical line and the fancy function (not sure it persists colors, so won't refer to them by color) limit a shape, the volume of which I have to find

Ted Shifrin

It doesn't help me, @RandoHinn. Describe the region in words and equations for me. Forget about the circle. You can always remove that ball after you do the problem.

Rando Hinn

Yup

Ted Shifrin

@Adam: For starters, do you mean $n\to\infty$?

christmas_light

@TedShifrin it's $(x^{a_n+1}_n-x^{a_n+1}_1,\ldots,x^{a_2+1}_2-x^{a_2+1}_1)$

Perturbative

@Rithaniel $\mathbb{R}^2 \setminus \{(0, 0)\}$ is homotopy equivalent to $S^1$ and $\pi_(S^1) \cong \mathbb{Z}$ so $\pi_1\left(\mathbb{R}^2 \setminus \{(0, 0)\}\right) \cong \mathbb{Z}$

Ted Shifrin

19:57

OK, @christmas_light. Now work out a few simple examples, say with $n=2$ and $n=3$, for starters.

@Perturb: Way too fancy.

@Rithaniel: Draw a picture. Do you really think it's NOT path connected?

Perturbative

Then since $\pi_1\left(\mathbb{R}^2 \setminus \{(0, 0)\}\right)$ is non-trivial we have that $\mathbb{R}^2 \setminus \{(0, 0)\}$ is not simply connected

Ted Shifrin

Oh, you said PATH connected.

Perturbative

@TedShifrin Rithaniel wanted to show it's not simply connected

Ted Shifrin

He originally put PATH connected.

Rithaniel

Yeah, that was a typo, originally, sorry.

Ted Shifrin

19:58

That's a big deal.

So what do you know about simple connectivity?

Rithaniel

Indeed, I usually check over my posts before hitting enter, but I didn't that time.

Adam

$$\lim _{n\rightarrow N}(a_{{n}}b_{{n}})\neq \lim _{n\rightarrow N}
(a_{{n}})\lim _{n\rightarrow N}(b_{{n}}) $$ sorry fixed the latex there ok sure yes it lets suppose it is an infinite limit @TedShifrin

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$Mathematics$ Mathematics