Mathematics - 2017-09-27 (page 2 of 5)

Leaky Nun

10:01 AM

@Secret there is no contradiction

Star Wars: I Feel a Conflict

►

0:37 there is no conflict

@AkivaWeinberger hi

Secret

Uh, you cannot have an odd number equal to an even number?

Leaky Nun

@Secret the un-trigonometric steps are wrong

$\cos(\pi-be)=\cos(a)$ implies $\pi-be=a+2n\pi$ or $\pi-be=-a+2n\pi$

$\sin(-be)=\sin(a)$ implies $-be=a+2m\pi$ or $-be=-a+(2m+1)\pi$

actually, you can directly from $e^{bi(\pi+e)}=e^{ia}$ read out $b(\pi+e) = a+2n\pi$

which to my mind is rather pointless

Secret

10:40 AM

Yeah I just rechecked, the "or" cases will allow the attempted contradiction to fail and remain consistent.

Anyway, the reason that I suddenly want to check the irrationality of $\pi + e$ is because if this is true, then what I previously said about the cosets of $\Bbb{R}/\Bbb{Q}$ not overlapping for any two irrational generating element will become false

Leaky Nun

@Secret it is an open problem

(yes, we still haven't determined if $\pi+e$ is rational)

@Secret well one only has to look for $\sqrt2+\Bbb Q$ and $(\sqrt2+1)+\Bbb Q$.

Secret

That example of $\sqrt{2}$ and $\sqrt{2}+1$ should have been ruled out in my previosu description, though I don't know how to describe that

Leaky Nun

you can't

Secret

perhaps, all irrationals that don't differ by an integer or rational "in some obvious manner"

but then $\pi + e$ pose difficulties

ACuriousMind

If $a$ and $b$ are irrational, then either $a+b$ or $a-b$ is irrational. That's all you can say.

Leaky Nun

10:44 AM

@ACuriousMind right

@Secret remember: there are uncountably many irrationals, but only countably many which you can describe

Secret

yeah, we don't have the luxury of a computer with countably infinite strings

Leaky Nun

@AkivaWeinberger hmm?

you mean algebraic certainly

Akiva Weinberger

mlerrp

Oh wait

Jasper

@LeakyNun a and b are letters of the alphabet, lol.

Akiva Weinberger

Uh, derp

Leaky Nun

10:47 AM

@Jasper go back to your iceberg

Akiva Weinberger

Yeah, sorry

Jasper

@LeakyNun Hey, you flying off this week right?

Akiva Weinberger

If $a$ and $b$ are both transcendental, then $a+b$ and $ab$ can't both be algebraic

Leaky Nun

@Jasper tomorrow

Akiva Weinberger

is what I meant

Leaky Nun

10:48 AM

@AkivaWeinberger right

Akiva Weinberger

It is too early in the morning for me, bye for a bit

Jasper

What? So many (removed)

Akiva Weinberger

I made a stupid mistake

So I cleansed it with fire

Secret

$\pi e =\frac{a}{b} \implies \ln \pi = \ln a - \ln b$ gets nowhere

Leaky Nun

@Secret it's an open problem

if it can be solved by you (or me or anyone else in this room), it would have been solved a decade ago

3

Secret

10:49 AM

I know it is open, but I don't know why (in terms of complexity or other notions) it remains open

In fact, is it possible that the solution to it is fundementally undecidable?

Leaky Nun

@Secret we've discussed this before

Sep 16 at 18:20, by Leaky Nun

plot twist in mathematics: what if it turns out that $e+\pi \in \Bbb Q$ is independent of ZFC?

Read from here

Akiva Weinberger

I think the statement "$e+\pi$ algebraic" is equivalent to some statement of the form $\exists n,\phi(n)$ in PA, where each $\phi(n)$ is decidable (aka either true or false)

Secret

Sep 16 at 18:29, by Daminark

There's nothing else to say. If it is independent, well it's true and its negation is false. So not independent

Akiva Weinberger

I don't know enough logic to say if that implies the whole thing is decidable

Hm, I guess it isn't necessarily

Secret

Sep 16 at 18:22, by Semiclassical

I find it hard to be anything but childish when it comes to questions about $e+\pi$ tbh

Akiva Weinberger

10:53 AM

But it does mean it's definitely either true or false regardless of whether or not ZFC can prove anything about it

since PA has a canonical model

Secret

Actually, given the actual thing that forced me to revisit that topic today, knowing the irrationality or not of $\pi + e$ has an importance outside of number theory: It gives us information on the structure of the cosets in $\Bbb{R}/\Bbb{Q}$

Akiva Weinberger

You could definitely ask on Math SE if it's possible for it to be independent. I feel like there might be some metamathematical theorems that are relevent.

Leaky Nun

good luck constructing $e$ in PA @AkivaWeinberger

Akiva Weinberger

You can't construct $e$ itself but you can construct relevant stuff

Leaky Nun

@AkivaWeinberger well you can

Akiva Weinberger

10:55 AM

Like, you can define a function in PA equivalent to "$a<eb$"

the same way you can define a function in PA equivalent to $a<\sqrt2b$

(it's simply $a^2<2b^2$)

(but clearly for $e$ it would be more complicated)

Leaky Nun

5

Q: Constructing the reals from the integers

A map $f\colon\mathbb{Z}\longrightarrow\mathbb Z$ is called a quasi-homomorphism if the set$$\{f(m+n)-f(m)-f(n)\,|\,m,n\in\mathbb{Z}\}$$is bounded. Let $R$ be the set of these functions. Let's consider the binary relation $\sim$ in $R$ defined by$$f_1\sim f_2\iff\{f_1(n)-f_2(n)\,|\,n\in\mathbb{Z}...

Relevant

Akiva Weinberger

Oh, cool

That's a cool way to go about it

Instead of reals ($r$), you think about linear maps $x\mapsto xr$

but since those can't exist exactly in $\Bbb N\to\Bbb N$, we think about things that are approximately linear maps

That's a really cool idea

Secret

List of formulae involving π

The following is a list of significant formulae involving the mathematical constant π. The list contains only formulae whose significance is established either in the article on the formula itself, the article Pi, or the article Approximations of π. == Euclidean geometry == π = C / d {\displaystyle \pi =C/d} where C is the circumference of a circle, d is the diameter. A = π r 2 {...

Akiva Weinberger

In any case, by asking if $e+\pi$ is algebraic, you're asking if a certain function is in a certain countable collection of functions

Secret

Is now tempted to use the Wallis product, and some product representation of $e$ and see if the resulting closed form of $\pi e$ can be computed exactly

Akiva Weinberger

11:00 AM

I heard somewhere, incidentally, that ZF-C and ZFC agree on all arithmetic statements

That is, the presence or absence of choice can't affect any arithmetic statements

(That includes statements about Turing machines, since those can be framed in PA, which is also interesting)

Leaky Nun

@AkivaWeinberger yes

Secret

https://en.wikipedia.org/wiki/List_of_representations_of_e
and.... most of them are so complicated

Leaky Nun

wait a minute

@AkivaWeinberger $x \mapsto \lfloor xr \rfloor$

Akiva Weinberger

Right

Those are approximately linear maps, and would serve as canonical representatives of each class

FuzzyPixelz

Okay, so if a function like $f\left(x\right)=\frac{1}{x}\left(\frac{\sqrt{1+x}-1}{x}-\frac{1}{2}\right)$ is undefined at $x=0$. I can simply rewrite it as $f_0(x)=\frac{-1}{2\left(1+\sqrt{1+x}\right)^2}$. Literally $f_0=f$ so $f$ is both defined and undefined at $x=0$ depending on who you ask?

Akiva Weinberger

11:05 AM

@FuzzyPixelz They don't have the same domain, so they're not the same function

Leaky Nun

"For even if AC is as a matter of fact false, it cannot lead to false elementary assertions about the integers, unless ZF already does." — Kenny Lau Sep 23 at 14:14

"There is a metatheorem that the Axiom of Choice is not necessary to prove any statements of Arithmetic, any proof of a statement about integers that uses AC can be constructively transformed into a proof which does not use that axiom." — Kenny Lau Sep 23 at 14:15

@AkivaWeinberger found it

Akiva Weinberger

$f$ has domain $[-1,0)\cup(0,\infty)$ and $f_0$ has domain $[-1,\infty)$ (I think)

FuzzyPixelz

Haha that was both fast and convincing, thanks @AkivaWeinberger

Akiva Weinberger

However, yes, where they're both defined, they're equal

If you restrict $f_0$ onto $f$'s domain (i.e. delete $0$ from the domain), then you get $f$.

FuzzyPixelz

Yes, I forgot that the domain takes part in the definition of a function, silly me.

Leaky Nun

11:11 AM

@FuzzyPixelz there's a difference between high school and uni

in high school, you do everything informally, and the domain comes from the function definition (and every function is a real function)

in uni, you define the domain first, and functions can be between any sets

Secret

The following is expected to lead to nowhere:
$\pi + e = \sum_{k=0}^{\infty}\frac{2^{k+1}k!^2}{(2k+1)!} + \sum_{k=0}^{\infty}\frac{1}{k!} = \sum_{k=0}^{\infty}\frac{2^{k+1}k!^2}{(2k+1)!}+\frac{1}{(2k+1)!} - \sum_{k=0}^{\infty}\frac{1}{(2k)!}$

bleh, back to set theory

FuzzyPixelz

We have seen that, I know that we can define a function as a transformation or mapping between two sets. However all that knowledge was abolished quickly once we started Calculus (Limits and continuity now), so yes you're right.

@LeakyNun

Akiva Weinberger

Also, I haven't double-checked your algebra, so I'm just trusting you when you say that they're equal (where both defined)

FuzzyPixelz

I double-checked using Desmos @AkivaWeinberger thanks again !

Leaky Nun

@FuzzyPixelz it is a transformation between two sets

Akiva Weinberger

11:17 AM

@FuzzyPixelz Yeah people tend to focus on real functions for calc and other physics-like stuffs

This distinction comes up once again if you ever deal with complex functions

as $\Bbb R\to\Bbb R:x\mapsto x^2$ and $\Bbb C\to\Bbb C:x\mapsto x^2$ are fairly different things

FuzzyPixelz

@LeakyNun shrug Of course, why wouldn't high school math be as formal as it should, I mean it's not asking so much.

Leaky Nun

@FuzzyPixelz let’s learn ZF in high school :p

Balarka Sen

what are we doing

Leaky Nun

learning ZF in high school

Balarka Sen

keep up the good work

FuzzyPixelz

11:21 AM

@LeakyNun I'm all in, replace Calculus with ZF and you get a whole generation of theoretical scientists and a much larger majority of people dropping out of school for mysterious reasons, I wouldn't mind that though.

Balarka Sen

No, you won't get that

Akiva Weinberger

Wait quick question @BalarkaSen

FuzzyPixelz

@BalarkaSen An overthrowed government ?

Balarka Sen

Calculus is a fundamental stepping stone of mathematics (and science in general). Axiomatic set theory is a specialized branch of mathematics that is absolutely not an absolute essential.

Akiva Weinberger

Zed-Ef or Zee-Ef?

Leaky Nun

11:23 AM

I would rather replace pi with 3.2

Balarka Sen

@Akiva I spell it as Zed-Ef-Cee

Well, if you have ZFC

Leaky Nun

@AkivaWeinberger same here

Akiva Weinberger

Heh

Balarka Sen

I have a great meme idea now.

Akiva Weinberger

I'm American, so I've always been reading your comments as "Zee-Ef-Cee"

I've also been reading them with an American accent 'cause I guess my "internal reading voice" has an American accent if that makes any sense

FuzzyPixelz

11:25 AM

@BalarkaSen Of course, a whole generation of people producing useless theories.

Akiva Weinberger

In any case, it's weird that the alphabet song doesn't rhyme for you guys

Leaky Nun

@BalarkaSen tell us

@AkivaWeinberger it does

Balarka Sen

@LeakyNun You know how Jake Paul and Team 10 wears those "OFC" shirts? Ohio Fried Chicken?

Akiva Weinberger

"A B C D E F Gee / H I J K LMNO Pee / Q R S, T U Vee / W X, Y and Zed"

Balarka Sen

To be continued

Akiva Weinberger

11:26 AM

That final rhyme fails

Leaky Nun

I say zee there

Secret

$\int_{\Bbb{R}}e^{-x^2}dx = \sqrt{\pi}$
$e^{\frac{1}{2}}\int_{\Bbb{R}}e^{-x^2}dx = \int_{\Bbb{R}}e^{-x^2+\frac{1}{2}}dx$ computing...

nope, it just returns what we put it, bleh

Leaky Nun

@BalarkaSen and then?

FuzzyPixelz

If $f$ is a continuous function on an interval $[a,b]$ such that $f(a)f(b)\lt 0$, then there is at least a $c\in (a,b)$ such that $f(c)=0$

Leaky Nun

IVT

FuzzyPixelz

11:35 AM

That follows from the intermediate value theorem

Leaky Nun

sniped!

FuzzyPixelz

I'm wondering if there are any alternative ways of writing that,

Why not this:

Leaky Nun

there isn’t

ivt is the path

ivt is the light

FuzzyPixelz

If $f$ is a continuous function on an interval $[a,b]$ such that $f(a)f(b)\le 0$, then there is at least a $c\in [a,b]$ such that $f(c)=0$

Leaky Nun

ivt is the way

FuzzyPixelz

11:37 AM

I'm mean like.. Why can't we have $f(a)=0$ or $f(b)=0$ @LeakyNun

Leaky Nun

or else fa fb would equal 0

Secret

Btw, I found it amazing that google search give no examples of improper integrals that evaluates to e

and there are tons of integrals that give $\pi$ containing expressions

FuzzyPixelz

Why not ! It would give a solution to $f(x)=0$ anyway.

Leaky Nun

well I didn’t see that you turned < into <=

but right it would just be a or b

Akiva Weinberger

11:55 AM

@FuzzyPixelz This is also true

and equivalent to IVT

FuzzyPixelz

The theorem statement is ambiguous, we can't know if $0\in [f(a),f(b)]$ or $0\in (f(a),f(b))$ @AkivaWeinberger

Corrected, my bad

Akiva Weinberger

If $f(a)f(b)\ne0$ then clearly $c\ne a,b$ so $c\in(a,b)$, and otherwise if $f(a)f(b)=0$ then one of $c=a$ or $c=b$ is true

Oh, you edited it to something slightly different than what I thought you were gonna edit it to

Still, just break it into cases on whether or not $f(a)f(b)=0$

In the end, what matters for the IVT is that continuous functions can't go from negative to positive (or vice versa) without crossing 0

FuzzyPixelz

Thanks, my teacher didn't approve and basically told me to accept the statement as we wrote it in class... ._.

Leaky Nun

@FuzzyPixelz state the theorem

FuzzyPixelz

The problem isn't the theorem, because it allows for both of the statements I mentioned earlier @LeakyNun

Leaky Nun

12:04 PM

@FuzzyPixelz I want to see how the theorem is stated

@FuzzyPixelz @_@

FuzzyPixelz

For every $\lambda$ between $f(a)$ and $f(b)$, there is at least a number $c$ between $a$ and $b$ such that $f(c)=\lambda$ @LeakyNun

Leaky Nun

I kindly refer you to $f(x)=1-x^2$ where $f(-1)=f(1)=0$ yet $f(x) \ne 0$ for $-1<x<1$ :P

FuzzyPixelz

Of course we assume that $f$ is continuous on $[a,b]$

Secret

Unrelated mumble: Using (forgot theorem name, but likely related to lindlemann weistrass) given $a$ a transcendental and $n$ an integer $> 0$, then $\{a^n\}_{n\in\Bbb{N}}$ is algebraically independent.

This means given any polynomial $P(x)=\sum_{k=0}^m b_kx^k$, with algebraic coefficients, since each $b_kx^k$ must be transcendental, the consequence of these summands being algebraically independent means that $P(a)$ is not only nonzero, but always transcendental.

(Need to figure out how to prove this properly). This means, $P(\Bbb{R})$ actually partition the reals nicely into transcendenta…

Leaky Nun

@Secret it certainly isn't

FuzzyPixelz

12:11 PM

@LeakyNun I kindly accept your reference sir, and I apologize for my terrible comprehension and my unfortunate tendency to ask silly questions.

Secret

O wait, forgot that algebraic times transendental will give another transendental very different from the one starting with, thus we are bumping into that $\pi e$ issue again, my bad.

FuzzyPixelz

But I can't help myself shrug literally.

Leaky Nun

@Secret I was referring to the first paragraph

Secret

But what about rational coefficients polynomials, that should close right since e.g. $pe^2+qe$ for $p,q$ rational can never yield an algebraic number?

Leaky Nun

@Secret that's linearly independent

Secret

12:14 PM

So for $P$ with rational coefficients. If $a$ is transcendental, then $P(a)$ must be transendental since the powers of $a$ are (forgot) linearly independent?

Leaky Nun

@Secret if $P(a)$ were algebraic, then $a$ would be algebraic

let $f(P(a))=0$ where $f$ is a polynomial

Then $f \circ P$ is a polynomial where $a$ is a root

i.e. $(f \circ P)(a)=0$

so $a$ is algebraic, contradiction

FuzzyPixelz

Don't you think that referring to "functions", one of the most important concepts in all of maths as $f$, $g$ and $h$ is a lack of creativity ?

Perhaps a little boring over time ?

Leaky Nun

@FuzzyPixelz what, do you want me to write $\Bbb F$?

$\varphi$?

$\eta$

FuzzyPixelz

Yes sir ! By the way, my message wasn't intended to be a reply to yours.

Secret

Hmm, so that would mean if $a$ is transcendental and $P$ has rational coefficients, then $P(a)$ has to be transcendental. So that means the algrebraic and transendental numbers are separatedlly closed under the action of the set of all rational polynomials.

That is if $P(a)$ is algebraic, then $a$ is algebraic
and if $P(a)$ is transcendental, then $a$ must be transendental.
I am not sure if this is interesting

But as you said, my claim about $Q(a)$ transcendental -> $a$ transcendental would not hold for $Q$ with algebraic coefficients

Leaky Nun

12:28 PM

for all non-zero polynomials $P$ with algebraic coefficients, $a$ is algebraic iff $P(a)$ is algebraic.

when did I say that it would not hold?

Secret

17 mins ago, by Leaky Nun

@Secret I was referring to the first paragraph

Işık Kaplan

Hey guys

Secret

meh nvm, read that wrong

Işık Kaplan

Can anyone help me out this one?

Leaky Nun

@IşıkKaplan <BDC = 45, <ADC = 135, use law of sines to find CD, then law of cosines to find BC

Secret

12:29 PM

But anyway, at least transcendentals under polynomial maps can never jump to algebraic numbers and vise versa

Işık Kaplan

I usually don't ask this kind of things but this, I'm apparently missing something, it has been one hour

DC is 8sqrt3-8

Secret

So given the reals, the action of the polynomials over the reals will give two cosets, the algebraic numbers and the transcendentals

Leaky Nun

@IşıkKaplan then use law of cosines on BCD

Işık Kaplan

But either I am completely wrong or the choices are wrong

Secret

(actually, we might not call these cosets as polynomials are not always invertible)

Leaky Nun

12:31 PM

@Secret they do have additive inverses

polynomials over a field form a ring

Işık Kaplan

If it is not a problem for you can you try to solve it for couple minutes?

Secret

so I guess the algebraic and transcendentals in the polynomial ring form two ideals that partition the reals

Leaky Nun

@IşıkKaplan DC is 8sqrt2

Işık Kaplan

As I said I've been trying to solve this for an hour and I'm not exaggerating.

Leaky Nun

how did you get 8sqrt3-8?

Işık Kaplan

12:33 PM

Wait let me draw it

I know this is completely wrong but I can not see it, is it possible I've become retarded in the last couple hours?

The last line is not y^2 but y.

Leaky Nun

you most certainly didn't calculate CD.

I already gave you a solution @IşıkKaplan

you're complicating everything

Işık Kaplan

12:49 PM

I know and thanks again for the solution but I still don't understand if I'm making a mathematical error or if my steps are faulty.

Leaky Nun

@IşıkKaplan may I repeat, that you didn't find CD

that you were wrong to claim that you found DC=8sqrt3-8

because x isn't DC

y^2 = (2x)^2 + (2x)^2 - 2(2x)(2x)cos(150)
y^2 = 2(16sqrt3-16)^2 [1-cos(150)]
y^2 = 256(sqrt3-1)^2 [2+sqrt3]
y^2 = 128(sqrt3-1)^2 [4+2sqrt3]
y^2 = 128(sqrt3-1)^2 [1+sqrt3]^2
y^2 = 128(3-1)^2
y^2 = 512
y = 8sqrt2

I have no idea what you're doing in your second paragraph

@IşıkKaplan

Işık Kaplan

1:07 PM

Thank you so much for the step by step solutions, I am apparently having a brain fart, I'm going to try to solve it again this night.

But again, thank you so much!

Akiva Weinberger

@FuzzyPixelz Did the teacher actually write it with $“f(a)f(b)<0”$?

That's weird. It's more common to see the (slightly less general but more intuitive) formulation $“f(a)>0$ and $f(b)<0”$.

Lembik

hi all.. here is a puzzle....if Alice and Bob each have a bit string of length n, how much data (in bits) does Alice have to send Bob so that Bob can compute the Hamming distance between the two strings mod 4 ?

Leaky Nun

@Lembik does Alice know Bob's string?

Lembik

1:22 PM

@LeakyNun no

otherwise it's rather easy :)

Mary Star

1:39 PM

Could someone of you take a look at my question about the exterior product: math.stackexchange.com/questions/2446994/… ?

Jester Tran

@Lembik I know how to compute the Hamming distance between two strings but to compute it "mod 4"?? What does that mean?

I know what "mod 4" means but don't know what it means with this usage

Akiva Weinberger

A number is "0 mod 4" if it's a multiple of 4, it's "1 mod 4" if it's 1 more than a multiple of 4, etc

Bob just needs to know how far above the closest multiple of four it is

Jester Tran

OH, hamming distance mod 4, just read it wrong

Balarka Sen

I have to decide what math I want to do now

Alessandro Codenotti

What are the options?

TastyRomeo

1:53 PM

study inter-universal Teichmüller theory

And verify the proof of the abc-conjecture, thanks :)

Balarka Sen

no pls stop

just because i am a memer, you don't have to meme at me

i am on a srs crisis now

Alessandro Codenotti

Learn measure theory then help me with my doubt

Balarka Sen

@Alessandro First options are Riemann surfaces, ODEs, then Riemannian geometry perhaps, or revisiting covering space theory

Jester Tran

@AkivaWeinberger @Lembik I can't see it. Currently stuck at sending n bits... at least

TastyRomeo

Riemann surfaces are pretty cool.

Or do Riemannian geometry like it was taught at my uni: no vector bundles and differential forms :P

Lembik

1:57 PM

@JesterTran it just means x mod 4 if x is the Hamming distance

@JesterTran so Bob has to output 0,1,2 or 3

Jester Tran

@Lembik yeah, my lower bound is n bits. I'm stuck

Lembik

@JesterTran if you can prove a lower bound of n that is interesting

Leaky Nun

@AlessandroCodenotti what doubt?

Alessandro Codenotti

The one from yesterday

Balarka Sen

@Tasty I am currently going through a block about a section on Forster

Alessandro Codenotti

1:59 PM

Actually I think I know how to solve it, but I need to check the details

Jester Tran

@Lembik I must be misinterpreting the question then. If Alice sends n bits of information (which is her bit string), then Bob knows both strings and easily finds the Hamming D

Balarka Sen

I guess I could do Riemannian

yeah I am not doing bundle geometry or moving frames

Leaky Nun

17 hours ago, by Alessandro Codenotti

So, suppose $f:\Omega\to\Bbb R$ is a function, with $\Omega\subseteq\Bbb R^n$ open and connected (we can also take $\Omega=\Bbb R^n$ if it's easier) and that $\int_{B_r(x)}f(y)\mathrm{d}y=0$ for all balls $B_r(x)$ contained in $\Omega$

17 hours ago, by Alessandro Codenotti

Without assuming continuity must it be $0$ almost everywhere?

@AlessandroCodenotti this one?

Alessandro Codenotti

precisely

Lembik

@JesterTran right..that's an upper bound

@JesterTran we want a lower bound

Jester Tran

2:04 PM

@Lembik so both upper and lower bound, so has to be $n$ lol. At least I know that I understand the question now

NV-US

hello

Lembik

@JesterTran :)

NV-US

If H has index 2 in G, then H is normal in G. Is the converse also true?

if not, any counterexamples?

Leaky Nun

@NV-US of course not

there are many normal subgroups with index way bigger than 2

just pick any abelian group

Alessandro Codenotti

@NV-US Are all subgroups of $\Bbb Z$ all of index $2$? Surely they're all normal so...

NV-US

2:07 PM

every subgp of abelian is normal, hence follows, ty

what about non abelian?

@LeakyNun

Jester Tran

@LeakyNun I swear you're a freshman/sophomore in pure mathematics. I met you in some programming chat and also did not see you this active in the mathematics chat. You previously had some dark coloured display picture

Leaky Nun

@AlessandroCodenotti how would you describe the elements in $\varepsilon_0$ to someone without any knowledge of ordinal?

@JesterTran hmm?

Alessandro Codenotti

"very big"?

Leaky Nun

@AlessandroCodenotti I mean the elements inside

e.g. $\omega^\omega$ contains the polynomials with natural number coefficients

NV-US

any example in a non abelian group??

Leaky Nun

2:09 PM

@NV-US just pick a group large enough

Alessandro Codenotti

It's probably easier to just explain what ordinals are

Leaky Nun

@NV-US {1,-1} is normal in Q8

the quaternion group of order 8

@AlessandroCodenotti hmm

NV-US

ty

Jester Tran

@LeakyNun I certainly remember having a few chats with you here or in codegolf. Trying to recall what you helped me with

Leaky Nun

@JesterTran I do am active on both sites

and the fact that I'm a freshman in pure mathematics is clearly written in my bio :P

however I never had some dark coloured display picture

Faust

2:17 PM

chat all buggy

Jester Tran

I didn't check as to test my memory. I wish I had zoomed in on your old display picture :(

Leaky Nun

@JesterTran oh, that one

it's dark red

brown

Jester Tran

ahh ok haha

Faust

i am trying to prove the following corollary let u and v be distinct vertices in a non separable graph G. if H obtained by adding a new vertex w and jioning w to u and v then H is nonseprable.

i meant i know there are two ID paths to u and v from any vertex.

i then wanted to follow across the edge from u to w then the from w to v

and claim it was a cycle

but i cant

because i dont know that the the thwo paths are internally disjiont

GFauxPas

3:22 PM

So for my complex analysis homework, I was tasked with defining a function holomorphic except at 6 points. I used $\dfrac 1 {1-z^6}$ because there was no requirement that the function be everywhere continuous. But if I wanted to force continuity everywhere, how would I go about the problem?

Alessandro Codenotti

3:40 PM

Try doing it with 1 point

GFauxPas

Hm

Balarka Sen

Wait, you want a holomorphic function which is continuous everywhere but holomorphic everywhere except one point? That can't happen?

GFauxPas

why not?

Balarka Sen

That point is a removable singularity, isn't it?

GFauxPas

what about holomorphic everywhere except several points?

maybe?

Semiclassical

3:50 PM

holomorphic except at finitely many points is easy enough.

the issue is the continuity bit

Balarka Sen

It can't happen

TastyRomeo

You'd need something that is continuous everywhere, but not differentiable (or such that Cauchy-Riemann don't hold)

Balarka Sen

Riemann's removable singularity theorem extends it over those points holomorphically

TastyRomeo

But that would again require discontinuous derivatives and such

GFauxPas

hmm okay

Semiclassical

3:51 PM

I must be tired. I'm forgetting where $f(z)=\sqrt{z}$ would land in this classification.

since $f(z)\to 0$ as $z\to 0$, though $f'(z)$ isn't well-defined as $z\to 0$.

Balarka Sen

square root of z is not even a well defined function on C

Semiclassical

point.

GFauxPas

just choose a branch?

Semiclassical

if you choose a branch, it's not continuous along the branch

Balarka Sen

^

GFauxPas

3:52 PM

ah

Semiclassical

it's only continuous across the branch if you're working on the branched cover

Balarka Sen

choosing a branch means you choose a well-defined lift over C minus an axis

minus half an axis

Semiclassical

in which case it's not really a function on C anymore

Balarka Sen

yup

GFauxPas

it is if you make the domain closed on one side and open on the other?

Semiclassical

3:53 PM

how does that help?

with continuity, I mean.

GFauxPas

I guess it doesnt

Alessandro Codenotti

@BalarkaSen I was trying to get him to work it out by himself!

Semiclassical

lol

Balarka Sen

@Alessandro Oops!

GFauxPas

well I didn't learn Riemann's removable singularity theorem yet :P

Semiclassical

3:54 PM

hmm

GFauxPas

complex analysis is so interesting, I just need to make sure I know my foundations well enough

last class we did series and power series, but we didn't yet prove that:

Semiclassical

At the same time, how does one interpret that for $\sqrt{z}$ if you think of that as a function on the branched cover?

GFauxPas

$f$ holormophic iff $f$ = its power series

Semiclassical

^ in a given nbhd

GFauxPas

which i read online

ah

Balarka Sen

3:56 PM

@Semiclassical Interpret which?

Semiclassical

i'm not awake, so I'm probably not thinking clearly. But:

Balarka Sen

I think over the branched cover the function is just $z$.

The lift of the function along $\Bbb C \to \Bbb C$, $z\mapsto z^2$, I mean

Semiclassical

hmm

in which case you'd not even have a singularity.

Balarka Sen

Right.

Semiclassical

but that seems odd, since while $\sqrt{z}\to 0$ was well-defined the first derivative wasn't continuous at zero.

GFauxPas

3:59 PM

oh, this is interesting

Balarka Sen

I'd be careful about taking anything at zero because the branched cover is, by definition, not biholomorphic near that.

Semiclassical

hmm

Balarka Sen

The thing is if you baleet 0

so you get

GFauxPas

I asked if $\lim$ can be replaced with $\lim \sup$ in the ratio and root test?

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$Mathematics$ Mathematics