Mathematics - 2017-09-15 (page 4 of 7)

I see the fact that if for a holomorphic function $g$, $g(p) \neq 0$ there is a neighborhood $U$ of $p$ on where there is some holomorphic $k$-th root $h$ of $g$ (aka, $g(z) = h(z)^k$ for all $z \in U$) used a lot.

I always think of this as a corollary of holomorphic map lifting lemma; If $g : U \to \Bbb C$ is the map, then $\Bbb C \to \Bbb C$ given by $z \mapsto z^k$ is a holomorphic covering map away from $0$. A lift $h : U \to \Bbb C$ of $g$ is exactly what a local $k$-th root of $g$ is.

I guess analysts would think about it like locally choosing a branch of the $k$-th root or whatever

(No idea why I'm sharing this; probably to contribute an epsilon amount of math to this room)

Leaky Nun

everyone is infected by secret :D

Leaky Nun

10:59 AM

@TobiasKildetoft lol I'm stuck at 2) of exercise 1 already

Tobias Kildetoft

@LeakyNun That was quick

Leaky Nun

@TobiasKildetoft I had been stuck for like hours before I found you

Let $|x|=hm$ and $|y|=hn$. Then sure, $|(xy)^h|=mn$ by part 1, but it isn't true that $|xy|=hmn$ (by the example above)

and I have been struggling to find a suitable power that makes it work

Tobias Kildetoft

@LeakyNun Given that $|x| = n$ and $m\mid n$ what is $|x^m|$?

Leaky Nun

I can't prove that $(xy)^h$ must have a "primitive $h$-root"

@TobiasKildetoft n/m

Tobias Kildetoft

@LeakyNun Ok, do you know how to make two numbers coprime while keeping their least common multiple?

Leaky Nun

11:02 AM

@TobiasKildetoft only arbitrarily

Tobias Kildetoft

Then consider that a bit more closely

Leaky Nun

@TobiasKildetoft I mean, $mn$ and $1$ are coprime

Consider $a=12$ and $b=18$... their gcd is $h=6$. $a/h=2$ is not coprime with $b$, and $b/h=3$ is not coprime with $a$

$a/h=2$ is coprime with $b/h=3$, but the LCM is not preserved

one could say that $4$ and $9$ are coprime and preserve LCM?

but then $4$ and $9$ are very artbirary

@TobiasKildetoft is this what you have in mind?

Tobias Kildetoft

@LeakyNun Consider prime factorizations

Leaky Nun

$a=2^2 \times 3$ and $b=2 \times 3^2$

oh, you keep each prime power with the larger exponent

Tobias Kildetoft

@LeakyNun right

Leaky Nun

11:10 AM

I'm not sure how to state it in symbols

Tobias Kildetoft

No need to state fully in symbols. The important thing is that this can be done

Leaky Nun

does this have a name?

Tobias Kildetoft

Not sure actually. Probably

Leaky Nun

How the Axiom of Choice Gives Sizeless Sets | Infinite Series

►

13 hours ago :D

wait, it's a re-upload of an earlier video

Leaky Nun

11:36 AM

If you're from India and you're one in a million, that means there's about 1400 people just like you. =D — corsiKa 15 hours ago

lolllll

user308168

12:11 PM

It seems that the MSE moderators do not want to talk to me.

Simply Beautiful Art

@WDNWBM They are simply having bad times. Give them a bit :-)

@WDNWBM Also, you can't just pop in and out and expect a conversation

._.

John Do

@GabrielRomon my question was absolutely retarded and I answered it. Turns out I was done, simply needed to wake up

@GabrielRomon at least I've proven that there ARE stupid questions ;)

Simply Beautiful Art

Lol

Alessandro Codenotti

How can I align the two "if" in LaTeX in a definition like this: $$a^n=\begin{cases} a \quad \text{if }n=1 \\ a\ast(a^{n-1})\quad \text{if }n>1 \end{cases}$$

Simply Beautiful Art

@AlessandroCodenotti Replace \quad with &

Alessandro Codenotti

12:22 PM

wow that was easier than expected

Simply Beautiful Art

You generally don't need the \text{if } IMHO, it is usually implied.

Alessandro Codenotti

thanks!

Simply Beautiful Art

np

user308168

I wanted to speak about my problem when at least one of them is here, but it seems that I should speak about it with you.

user308168

I am MathematicsAminPhysics, and I want to speak about the suspension of this user .

Simply Beautiful Art

12:26 PM

@WDNWBM :|

**sighs** That, I can't help you with

And if you are found avoiding the suspension via different accounts, they will likely suspend/block your IP address

2

user308168

I have been suspended because of posting a question on the Meta for 30 days? Do you not think this punishment is extraordinarily heavy?

Simply Beautiful Art

@WDNWBM No.

user308168

People with more than 10k rep can see that post:math.meta.stackexchange.com/questions/26903/…

Leaky Nun

@WDNWBM it's just the problem of timing.

Simply Beautiful Art

@WDNWBM Since that was closed by a moderator outside of MSE, I suspect that either the math mods weren't quite sure what to do with you or requested some outside helps.

I can't say much about it though

user308168

12:33 PM

Which was closed by a moderator outside of MSE?

Simply Beautiful Art

@WDNWBM the one you linked was deleted by shog9, the community manager of SO and Stackexchange.

2

user308168

@SimplyBeautifulArt Why?

Balarka Sen

It's a bad idea to discuss these before the suspension on your account lifts.

4

Simply Beautiful Art

@WDNWBM how should I know?

Balarka Sen

You're just going to get yourself IP banned

user308168

12:37 PM

Am I doing an illegal thing?

user308168

I have been suspended by the MSE moderators, not by the PSE ones.

Simply Beautiful Art

Can't help you with any of that stuff mate

user308168

I want only to know the reason of my suspension.

user308168

@Danu Hello. Maybe you can help me since you are a moderator.

Simply Beautiful Art

Darn, I think I caught the sniffles.

Tanner Swett

12:42 PM

Can a 5x5 square be cut into 4 pieces which can be reassembled into a 3x3 square and a 4x4 square?

Simply Beautiful Art

@TannerSwett 4 equal pieces?

Oh, never mind

Leaky Nun

Can a circle be cut into a finite number of pieces which can be reassembled into a square?

(spoiler: it can)

hola, aqui va

Simply Beautiful Art

@WDNWBM If that is what you wish, I suggest you not use accounts to contact the mods. Try email instead.

Akiva Weinberger

Ah, I've heard of this. Don't know the details, though. Are the regions disconnected dust, like in Banach-Tarski?

Leaky Nun

@AkivaWeinberger I also don't know the details

Akiva Weinberger

12:45 PM

Found a link

https://en.wikipedia.org/wiki/Tarski's_circle-squaring_problem

Leaky Nun

@AkivaWeinberger care to share?

Tanner Swett

And I mean "reasonable" pieces here, the sort of thing you could cut out with a pair of scissors.

Leaky Nun

@TannerSwett I'm just diverting :P

Akiva Weinberger

> In particular, it is impossible to dissect a circle and make a square using pieces that could be cut with scissors (that is, having Jordan curve boundary). The pieces used in Laczkovich's proof are non-measurable subsets.

user308168

I want to chat to them

Leaky Nun

12:46 PM

> Laczkovich actually proved the reassembly can be done using translations only; rotations are not required.

Simply Beautiful Art

@WDNWBM well, we warned you, didn't we?

Akiva Weinberger

@TannerSwett Yes

Leaky Nun

@AkivaWeinberger :O

user308168

@SimplyBeautifulArt warned what?

Leaky Nun

oh, it was obvious

Akiva Weinberger

12:48 PM

Lemme draw a picture

Leaky Nun

AAABB
AAABB
AAABB
CCCBB
CCCDD

Akiva Weinberger

Yeah

Tanner Swett

Tangential: reading about Osgood curves convinced me that the Jordan curve theorem really isn't obvious at all.

Leaky Nun

BBBB
BBBB
CCCD
CCCD

Akiva Weinberger

Ah, yep @LeakyNun

Leaky Nun

12:48 PM

@AkivaWeinberger eso es lo que pensabas?

Akiva Weinberger

Sí, hasta una rotación

Yeah, up to rotation

Tanner Swett

¿Por qué hablamos en español? :)

user308168

@SimplyBeautifulArt It is very strange to me that a Meta post has been closed by a community manager.

Akiva Weinberger

Por que él sabe que yo sé español

y por eso le gusta hablar conmigo en español

Leaky Nun

cosas pasan sin razon

Tanner Swett

12:51 PM

Yo no hablo español ni un poquito, entonces no entiendo nada que dicen. :D

Nah, just kidding. I do, in fact, know some Spanish.

user308168

@SimplyBeautifulArt Thanks for telling that to me.

Tanner Swett

Looks like I actually want "así que" or "por lo que" in there instead of "entonces"?

This chat room is now about Spanish.

user308168

I stop talking about this subject now. Continuing it is very dangerous for me.

Leaky Nun

1:10 PM

14

A: Fields of arbitrary cardinality

Let $F$ be a finite field, and $X$ an infinite set, let $\hat X$ be the set of all finite strings of things in $X$ where two strings are equal if they differ only by their order (e.g. $x_1x_3x_2=x_1x_2x_3$). Then let $A$ be the set of formal sums $\{\sum_{i=1}^n f_ix_i\mid n\in\Bbb N,\ f_i\in F\...

John Do

reputation limits suck... anyone wanna comment something to help someone out because I can't?

Leaky Nun

@JohnDo ?

John Do

@LeakyNun well therre's someone who's asking for a hint on someone else's answer but he's gotten no response

I want to help him but I don't have the reputation necessary to comment

Leaky Nun

cut the opening

John Do

what?

Vrouvrou

1:30 PM

hello, I have $\|u\|_{\Phi}=\inf\{\lambda>0, \Phi(\frac{|u|}{\lambda})\leq1\}$ and $\|\nabla u\|_{\Phi}+\mu\|u\|_{\Phi}=1$ how I can simplify the expression with deleting the $\inf$ ?

mixedmath

@WDNWBM I'm here now

what's up?

Anonymous I

Can someone please help me with the proof? I've answered it myself all the way down. math.stackexchange.com/questions/2425275/…

I'm almost finished

user308168

@mixedmath Can I talk to you in private?

mixedmath

@WDNWBM Is there something wrong with chatting here?

Vrouvrou

?? someone have an idea ?

user308168

1:34 PM

@mixedmath It is better to speak about in a private room if it is possible.

mixedmath

Very well. Join me in here: chat.stackexchange.com/rooms/65652/temp-chat-with-mixedmath

Semiclassical

To: Seasonal allergies

Subject: F*** off

Simply Beautiful Art

@Semiclassical :-/

SteamyRoot

I feel your pain.

Semiclassical

1:55 PM

I may have to start taking my seasonal allergy med before I go to bed rather than when I wake up

As it is, the first hour of my day on campus is just ugh

SteamyRoot

I hope they're different from my meds, then.

Mine have increased heart rate as a side effect... it basically feels like I've downed a few energy drinks if I take them before sleeping (or trying to sleep, at least)

Semiclassical

I havent used them before going to bed in a while, so I dunno

SteamyRoot

If they do, it's worth trying with a lower dose first. Worked for me

Semiclassical

Right

I use loratadine, and that's a small once-a-day pill

Easy enough to split it though

mixedmath

allergies suck. And allergy meds make me feel fuzzy

Abcd

2:32 PM

Consider$$y= \dfrac{2x}{1+x^2}$$, then the range of the expression $y^2+y-2$ is___________
My attempt:
I found out the range of $y \forall x\in R$ which is $[-1,1]$

Thus, the function is maximised at $y =1$ with value $0$ and minimised at $y= 0$ with value $-2$. However, answer given is $$[\dfrac{-9}{4},0]$$
Where have I gone wrong?

please don't answer if it wastes your time or sounds like a lazy HW problem.

Leaky Nun

@TobiasKildetoft typo on exercise 3? (x,0) is not an element of the infinite dihedral group you constructed

@Abcd why is it minimized at $y=0$?

Sha Vuklia

Guys, so my book says that an ellipse in standard form has the equation $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, with bladibla. Now I’m wondering if there is difference in meaning between saying that a set 'has an equation', or ‘satisfies an equation’?

Leaky Nun

@ShaVuklia no difference.

a set has an equation which every point satisfies.

Abcd

@LeakyNun Got my answer now :). It's minimised at $-\dfrac{\Delta}{4a}$. I realised my mistake and corrected it too. I had analysed it erroneously before.

Leaky Nun

@Abcd good for you

Sha Vuklia

2:38 PM

@LeakyNun okay thanks

Secret

I found it interesting about the conspiracy theory in this chat with nature itself:

They seemed to coordinate so well in a way such that they never allow me to get the level needed to unleash the infinity bomb

Well I guess that's a good thing

4 hours ago, by Leaky Nun

everyone is infected by secret :D

Also lol

The Weirdness spreads. But as long healthy amount of maths that makes sense is kept, I am happy

Abcd

@Secret I wanted to ask you what koolman did that annoyed you?

Sha Vuklia

2:56 PM

sry for the deletes

SteamyRoot

(removed)

Removed parties seem to have become a part of MSE chat culture :P

Secret

@Abcd Help vampire (all the way from 2 years ago to as recent as just a week ago in h bar, heather's comment can support that)

Meanwhile, My recent progress with the harmonic sum $\sum_{m=1}^n \frac{H_{m,r}}{m^s}$ is moot because it gives me a 0=0 error. Guess that might mean a direct algebraic manipulation and reduction in terms of harmonic numbers may not be possible for such general case. I will be more satisfied if the nonexistence of closed form with this property can be proved however...

[Random]

Yet another notion of an elegant closed form

BAYMAX

any help in probability related question

here

Secret

A closed form under this definition is considered elegant if given a map $f \in \text{maps}$ and some operator $O$ $O(f) \in \text{maps}$ where $\text{maps}$ is the class of all maps

For example:

$\int \sin x dx =-\cos x +C$ is elegant because cos and sin belongs to the same class of functions

How is this related to harmonic number series? :

It is found that for some simple sums of the form $\sum_{m=1}^nH_{m,r}$ the closed form is often a polynomial of generalised harmonic numbers

and we conjectured that all harmonic sums have elegant closed forms as polynomials of generalised harmonic numbers

Alessandro Codenotti

@Akiva are you here? What would you suggest as a first book in music theory?

Secret

3:12 PM

The harmonic numbers and fibonacci numbers have recursion relations in similar ways, thus another thing we conjectured is that sums of ratios of fibonacci numbers may have a elegant closed form consists of only fibonacci numbers

Leaky Nun

@AlessandroCodenotti How Equal Temperament Ruined Harmony (and Why You Should Care)

mixedmath

@LeakyNun that looks interesting

Leaky Nun

well basically if an octave is 1:2 and a fifth is 2:3, then you can't do the circle of fifth thing

Akiva Weinberger

@AlessandroCodenotti No idea, sorry

Alessandro Codenotti

What boook did you use?

Akiva Weinberger

3:18 PM

I only started learning it in school a few days ago

brb

Alessandro Codenotti

Ah, sorry, I thought you had been learning it for a couple of years, don't know why

Faust

anyone on that knows anyhting about groups?

Leaky Nun

@Faust just ask

Faust

Im trying to figure ut all the possible orders of the elements of the ax+b group

i got 2

and infinity

anything else im missing?

Leaky Nun

what is the ax+b group?

Faust

3:35 PM

i dont know how to type matrices

its the line 1

a b

0 1

that matrix

those are the elements in it

Leaky Nun

is it 2 or infinity?

Faust

i said when a=1 has order 2

and a not one it has order infinity

Leaky Nun

I don't see when you said that...

Faust

Im trying to figure ut all the possible orders of the elements of the ax+b group
i got 2
and infinity

Leaky Nun

you didn't mention a=1

Faust

3:37 PM

well thats how i get 2

Leaky Nun

alright

Faust

i think a has to be nonzero

nvm a def has to be non zero

Leaky Nun

( 1 b ) ( 1 b ) = ( 1 b+b )
( 0 1 ) ( 0 1 ) = ( 0  1  )

So I assume a and b are elements of Z/2?

Faust

no

sorry i guess a=1 and b must be non positive

nvm

i have no diea

Leaky Nun

but then the order isn't 2 as I demonstrated

Faust

3:39 PM

a=-1

Leaky Nun

( -1 b ) ( -1 b ) = ( 1 0 )
(  0 1 ) (  0 1 ) = ( 0 1 )

Faust

of i forgot the identity it has order 1

Leaky Nun

( a b ) ( a b ) = ( a^2 ab+b )
( 0 1 ) ( 0 1 ) = (  0    1  )

Faust

yeah i forgot theat -1*-1 was 1

Leaky Nun

( a b ) ( c d ) = ( ac ad+b )
( 0 1 ) ( 0 1 ) = (  0   1  )

(a,b) -> (a^2,ab+b) -> (a^3,a^2b+ab+b) -> ..

Faust

3:43 PM

order of an element is a product with itself till it reaches I though not anthore element?

Leaky Nun

@Faust yes

I'm just thinking out loud

Faust

ah

Leaky Nun

a^n-1=0

a=1 (rej) or -1

you're right, if a=-1 then order is 2; if identity then order is 1; otherwise order is infinity

Faust

ah kk thanks =)4

Semiclassical

This is a bit bizarre. I'm grading my students' first quantum hw, and one of them consistently used $\delta$ instead of $\partial$ in writing partial derivatives

Dodsy

3:52 PM

@Semiclassical so I figured out today that the prof is on page 26

Semiclassical

Lin alg?

Dodsy

just by looking at the page he was on

and finding it.

because the textbook was infront of me :)

yes.

still no required readings but I'm going to read to about page 35

and then I should be caught up.

Semiclassical

Good plan

Dodsy

Pretty crazy though...

Then in calc today I realized I made a huge mistake on my assignment

Dodsy

4:09 PM

@Semiclassical perhaps you could help me a litttleeee bit

Dodsy

4:32 PM

nvm dood

Mr. Xcoder

ಠ_ಠ My book has $\mathbb{R}\supseteq\mathbb{Q}$ on the cover ಠ_ಠ

Dodsy

:o

what does the backwards subset mean

Mr. Xcoder

@Dodsy Superset. In this case, superset or equal.

Dodsy

I see.

Mr. Xcoder

I don't understand why superset or equal

Dodsy

4:36 PM

oh the line underneat?

I don't think that means equal

Mr. Xcoder

@Dodsy That's what it means.

I know that for sure.

Dodsy

For instance for subset

there can be a subset or a subset with a line underneath

and they can usually be used interchangeably

because it "may" be equal"

Leaky Nun

@Mr.Xcoder I generally include the line unless I'm emphasizing that they aren't equal, in which case I use $\subsetneq$ :P

because a naked $\subset$ is too ambiguous

Dodsy

huh interesting.

Leaky Nun

the fact that $\Bbb R \ne \Bbb Q$ is not important in your book's context

Mr. Xcoder

4:38 PM

@Dodsy Not really. $\{1, 2, 3\} \subset \{1,2, 3\}$ is false, whilst $\{1, 2, 3\}\subseteq \{1, 2, 3\}$ is true.

Leaky Nun

@Mr.Xcoder the former is ambiguous.

Mr. Xcoder

@LeakyNun Not really.

Leaky Nun

it can be true depending on convention

@Mr.Xcoder yes, really.

Dodsy

True but you could say $ \{1,2,6\} \subseteq \{1,9,4\}$

Leaky Nun

@Dodsy eh, not really

Mr. Xcoder

4:39 PM

@LeakyNun Then why would $\subseteq$ exist at all?

Dodsy

ah sorry

Leaky Nun

@Mr.Xcoder because $\subset$ is ambiguous.

Dodsy

bad example

if we said $U=\{1,2,3,4,5\}$ and $A = \{2,4\}$ $B=\{1,3,5\}$ then we could say that $A \subseteq U$ , right?

Mr. Xcoder

@LeakyNun Well, one thing I know for sure: My maths professor always says that $\{1, 2\}\not\subset \{1, 2\}$ :P

Leaky Nun

@Mr.Xcoder what???

Mr. Xcoder

4:41 PM

@Dodsy Yes, correct.

Leaky Nun

\not\subset

I avoid using \subset in general.

Mr. Xcoder

Bad mathjax

Leaky Nun

$\subseteq$ and $\subsetneq$.

Dodsy

hm

Mr. Xcoder

Hehe

Dodsy

4:42 PM

so it can be used interchangably in one direction.

?

is that the rule?

Mr. Xcoder

For what?

Leaky Nun

@Dodsy you can say so

Dodsy

gotta run.

Leaky Nun

@Dodsy see you

Dodsy

@Mr.Xcoder so if I had a set c given above sets already given, such that the set was $\{1,2,3,4,5\}$ we can't say that set C is a subset of set U, we have to say that it's a proper subset.

bye leaks.

Mr. Xcoder

4:45 PM

> leaks

@Dodsy You can say $C\subseteq U$.

Leaky Nun

@Mr.Xcoder so what was the context of $\Bbb R \supseteq \Bbb Q$?

Mr. Xcoder

@LeakyNun It was written on the cover of my maths book... And I didn't really get why not $\Bbb{R}\supset\Bbb{Q}$

Leaky Nun

which convention does your math book use?

Mr. Xcoder