Mathematics - 2020-05-07 (page 1 of 2)

Thorgott

00:40

What goes awry if one doesn't use germs in order to define tangent vectors as derivations?

Leaky Nun

@Thorgott nothing?

Thorgott

then why is it done?

Leaky Nun

I don't understand your question

Thorgott

00:56

One way of defining tangent vectors at a point is as derivations on the space of germs of smooth functions at that point. I was going to ask why we use germs and don't just look at smooth functions defined on neighborhoods of the point themselves, but I just realized that doesn't carry a ring structure anymore. And looking at those defined on a fixed neighborhood sounds very worrisome. So nevermind.

Leaky Nun

well you can fix a neighbourhood

because of those bump functions tricks

@AkivaWeinberger que tal

Thorgott

yeah, I considered that heuristically, but that's enough work for germs to be the more pleasant option

Leaky Nun

great

r9m

01:28

@robjohn what kind of virus is that blue mask supposed to keep away? XD

Knight

03:45

Can someone please recommend me books on Complex Numbers? I need an introductory book which builds up the concept from the concept $ i =\sqrt {-1}$ and moves on to the polar form of complex number and Euler form (exponential form).

Then it should teach the properties of conjugate and modulus like $$ \overline{z_1 + z_2 +... z_n} = \bar{z_1} + \bar{z_2} +... \bar{z_n}$$ and $$ \big| z_1 z_2... z_n\big|= |z_1| |z_2| ... |z_n|$$

These type of things are not usually covered in Precalculus or Calculus books.

user76284

04:12

@Knight Those things follow straightforwardly from induction. Do you mean proving $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$, etc.?

proofwiki.org/wiki/Sum_of_Complex_Conjugates

Knight

@user76284 Yes, I meant the proof. Now, I realise those things are really straightforward :-)

Thanks

Akiva Weinberger

@MikeMiller Turns out this is false

Knight

@robjohn Why you have worn a mask? I thought you were inside your home

2

Akiva Weinberger

False for line segments, anyway

If you have a(n infinite) line between two boundary points, such that the sphere only intersects that line at those two points, hopefully we should still be good

Akiva Weinberger

04:37

(In the line segment version, you can take a sphere and drill a knotted hole in it that doesn't go all the way through)

Akiva Weinberger

04:59

Hm. Maybe a starting place would be to intersect the sphere with a plane that contains the line

That should give us a bunch of disjoint loops in the plane (except in degenerate cases which we can hopefully ignore, e.g. by thinking of the sphere as piecewise linear and making the plane avoid the vertices?)

Each disjoint loop would bound a disk; by looking at the loops contained in those disks, we can give the loops a tree structure and then inductively get rid of them starting from the leaves

William Sun

Is there a way to determine whether a square matrix with entries in $Z/nZ$ is nilpotent or not?

In the case of field/field of fractions of an integral domain we just calculate the minimal polynomial but what about in general?

Akiva Weinberger

Maybe you can check it mod Z/pZ for each prime factor of n?

Dunno if that's necessary

Also not sure if that works for prime powers EDIT: yeah, probably doesn't

@WilliamSun This gives at least one direction

3

A: Showing a matrix is nilpotent if its charateristic polynomial is $t^n$ mod ${\rm nil}(R)$

Let $\chi_A(T)\in R[T]$ be the characteristic polynomial of $A\in M_n(R)$. Say $\chi_A(T)\equiv T^n\bmod{\rm nil}(R)$. By the Cayley-Hamilton theorem we know $\chi_A(A)=0$. But $\chi_A(A)=A^n+f(A)$ where $f(\cdot)$ is some polynomial with nilpotent coefficients from ${\rm nil}(R)$, by the hypoth...

You can look at its characteristic polynomial mod the nilpotent elements of Z/Zn

Dunno if that's necessary though (as opposed to sufficient)

Balarka Sen

@AkivaWeinberger The problem seems to reduce to proving that if I is a line segment between two boundary points in B^3, then an embedding f : B^3 -> R^3 such that f(bd B^3) = standard S^2 satisfies f(I) = unknotted arc in f(B^3).

But that's obvious right

f is actually a homeomorphism B^3 -> B^3, so unknotted arcs go to unknotted arcs

skullpatrol

05:17

@BalarkaSen did you prove the old adage, people who go to sleep with crushing questions wake up feeling crushed?

Balarka Sen

What?

Thanks for readability :P I didn't know that was an old adage, but I guess I always feel pretty crushed

Yeah I know I wrote that I just couldn't parse your sentence

skullpatrol

Sorry, my bad.

Balarka Sen

'saright

skullpatrol

I should've used quotation marks

Akiva Weinberger

@BalarkaSen I isn't known to be a line segment between two boundary points in the geometric $B^3$. It's known to be a line segment between two boundary points in a subset of $\mathbb R^3$ homeomorphic to $B^3$.

And then the "sphere with a knotted hole drilled not all the way" thing works

Balarka Sen

05:26

I see what you're saying.

Can you find an arc between the two endpoints of the line segment lying on the boundary of the embedded B^3 such that the union of the arc and the line segment is an unknot

Akiva Weinberger

In the stronger case where the extended line doesn't hit the sphere anywhere else (which is the only case that's true), I think you can intersect the sphere with a plane containing that line

and ignore any disconnected loops

Balarka Sen

Ah so that's what you were trying to do

Makes sense

Akiva Weinberger

By the way

Reworking the argument from the start

Say we have A#B and assuming it equals 0

Can we get rid of spheres from the argument this way?:

Draw a segment between A and B (so now you essentially have a knotted graph with two vertices of degree 3)

Homotope space so that A#B goes to 0

While you do this, ensure that the third segment remains in a single plane

(Assuming that's something we can do)

Finally, A equals the left half of A#B plus the third segment. In the final configuration, that is the unknot. Similarly for B

Hm… no, I don't think this will work

I don't think you can ensure the third segment can remain in its plane

So yeah, thinking of a sphere (and its intersection with a plane) seems like the way to go

Good night

Balarka Sen

Night

Akiva Weinberger

~~(a plane through the two points where it hits the knot)~~ (the plane containing the unknot at the end)

Balarka Sen

05:38

Seems to be the right idea

Akiva Weinberger

Hm

Wait a second

OK so start from the start

We have it in the A#B configuration

and a sphere surrounding the A bit

Since A#B=0, it bounds a disc

Intersect the disk with the sphere

Balarka Sen

That can be horrendous

Akiva Weinberger

We should get a path on the sphere between the two points where the sphere hits the knot, plus some random loops

No?

And then the A bit plus that path is (a) the unknot from the disk's perspective and (b) equivalent to A itself

Balarka Sen

Why is A bit plus the path an unknot

Is it 100% obvious that bounds a disk

Ah because it's contained in the disk

Akiva Weinberger

The path divides the disk into two pieces

Yeah

Maybe that's easiest to see if we homotope space into a position where the disk is a circle (EDIT: or look at the preimage space of the disk)

I mean, this is essentially the same proof, just trying to ignore the deformations

Balarka Sen

05:44

Sounds right

Akiva Weinberger

In any case

Before this, the only proofs I knew of A#B=0 -> A=B=0 were (a) the knot genus adds and (b) the Mazur swindle

I don't remember how to prove the knot genus of A#B is the sum of the knot genuses of A and B, and besides, it involves introducing a numerical invariant

The Mazur swindle leaves the world of tame knots so I wasn't sure it applied to a situation where all deformations had to be tame

So I'm glad there's another way to see it

Now night

Balarka Sen

Lol good night

Akiva Weinberger

Oh also you can use a plane to separate A and B (in A#B) instead of a sphere, which is more symmetric

(of course, in S^3 it doesn't matter)

So if anyone asks me for a two second explanation of why it's true, I'll say the start of an idea is to cut A and B apart by a plane, and then intersect that plane with the disk that A#B bounds

Stupid Kid

06:36

Guys I need answer fast

Prove the following series not uniformly convergent on the domain given.

$\sum(x^2+n)/(x^2+n^3)$ on domain R

R is real

my book answer is when x=n

^2

x=n^2

there is a term in the series ,after the nth,which is greater than 0.5

So altough the series is convergent for each value of x by comparison with $\sum 1/n^2$

the greatest difference between the nth partial sum (I.e.$f_n$) and the limit function is not null

Please anyone answer me

I got like 11 min

@robjohn please help me with only this

@EdwardEvans are you online?

I found the infinite sum is -1/3

I got 7 min left 😱

Well I am doomed

lol

Stupid Kid

07:00

No problem I will be back after an hour please tag me when you know how to explain

I really don't understand the process unlike That I asked yesterday.

Knight

I’m re-learning Integrals and this time from Spivak’s Calculus. In the first few pages he develops an intuitive understanding of integrals. The sum area of rectangles lying below the function is denoted by $s$ and the sum of area of rectangles stretching out of function is denoted by $S$ (well they are, of course, the sum of rectangles when the function is evaluated at its minima and maxima respectively)

Then he says that the actual area A under the function must satisfy $$ s \leq A ~\text{and} ~ A \leq S $$ And then he writes “and this should be true, no matter how the interval [a,b] is sub divided “

I want to know why he put up “no matter how the interval is subdivided” and what is the real significance of that thing? Why he felt it so important that he put it in italics?

Leaky Nun

07:25

@Knight because later (presumably) he will let the subdivision's mesh tend to 0

and $s$ and $S$ will converge to the same limit (for integrable functions, that is)

and they will sandwich the "true" value of $A$

and they will be used to define $A$

robjohn

07:38

@StupidKid What is $\lim\limits_{x\to\infty}\frac{x^2+n}{x^2+n^3}$?

Knight

@LeakyNun Thanks Leaky

Hello Robbie

Stupid Kid

07:54

@robjohn wait I am finally free from restricted area lol

Knight

Hey Leaky! I wanted to ask a question from you, if you don’t mind

Stupid Kid

@robjohn It's 1

@robjohn my main question That I didn't mentioned is why x=n^2?

Sorry I was in hurry

And I don't get is after n term it says it is greater than 1/2 while the infinite sum is -1/3

Last part is triviaial but the first and middle part of answer is something that I don't understand and seeking to understand

It is more likely to be language problem for me.

Stupid Kid

08:32

@AkivaWeinberger Thanks @AkivaWeinberger For explaining. I was confused since book didn't give examples like you XD. I am really thankful or else I was not gonna continue reading geometry book.

user736948

08:59

How does one check that a solution of the Euler-Lagrange equation is actually an extremal of the functional?

Stupid Kid

@robjohn are you there?

Knight

@user736948 You can ask Physics doubts from JR sir in here

feynhat

09:14

@BalarkaSen The geometry group at IISc is organizing an online lecture series.

Link

Balarka Sen

@feynhat Looks cool.

Stupid Kid

or anyone dare to explain me?

Haven't got any explanation yet

Sha Vuklia

09:34

I have to determine
$$
\int_{C(0,1)^+}f(z)dz,
$$
where $f(z)=\frac{e^{\sin(1/z)}}{(z-2)^2}$. However, the problem is that we have in fact a product of two power series, one with negatieve coefficients, and the other with positive ones, so in order to determine the coefficient belonging to $z^{-1}$, it seems like I would need infinitely many values. The expansion of $1/(z-2)^2$ is no problem, but I believe that for $e^{\sin(1/z)}$ I would need to give the coefficients recursively. So I'm not sure how to proceed. Or maybe there would be a better way to do this? (Btw: this is homework, so I wo…

Edward Evans

Morning

Jasper

@Knight We want the integral to represent the intuitive area under a curve. If we want a definition for the integral that makes sense, then the integral should satisfy those properties, in particular, the ones mentioned there, and in particular, the ones in italics. Mathematical definitions are made to agree with intuition, and once made, they become independent of that intuition. That is how mathematics works. We wouldn't want two different subdivisions to give different results, would we?

Alessandro Codenotti

Hi @Edward

Edward Evans

Hey @Alessandro :)

Balarka Sen

Hi

Edward Evans

09:44

Waddup @Balarka

Knight

@Jasper I really felt very nice after reading your reply. Is there any way I can store some of the replies which I will be neeeding in future ?

Jasper

@Knight If you have enough rep, you can bookmark a conversation, but that is something else. You can also star messages, but that is something else. I guess you should just copy and paste in your computer if you really want to store it. No other ways.

Knight

@Jasper Thank you, I will do the same

Stupid Kid

@Jasper can you teach real analysis?

Knight

He can’t teach you anything

He can just answer (if he wants) to your some particular question

Stupid Kid

09:56

@Knight Why?

Knight

@StupidKid Do you think it is possible to teach anybody anything over a chat?

Stupid Kid

You mean explanation? Though I don't mean permanent teaching just answer particular answer. (sorry for sloppy writing)

definition of teach:cause (someone) to learn or understand something by example or experience. BTW

Knight

Yes of course you may receive answer to some particular questions if you’re clear enough and show some effort in your solution

Stupid Kid

But what if you don't even know what the answer is saying.

I mean people like me might not understand some specific part. Thus you can't put effort. I mean you need to understand people do have low capacity to understand and experienced people can teach the weaker people since they have the experience and complete understanding of it.

Knight

You have so much capability my friend, don’t underestimate yourself. Rise up my captain :-)

Just put up your problem specifically and clearly and ask for only specific help. @TedShifrin and @robjohn are the actaul teachers here. They will surely help you.

Stupid Kid

10:08

@Knight Thanks knight for motivation. I will go home and try to make it more specific and rigorous. I was in hurry and there are some social problems I have so I couldn't explain it more clearly. :)

Knight

@StupidKid Okay pal! Relax and write out your self.

Mike Miller

10:29

@AkivaWeinberger @BalarkaSen In fact it's reasonably clear why it's false --- just take any knot, delete a small ball around a point and consider the complement as an arc embedded in the ball. All arcs are unknotted so you can isotope it to be a straight line, and then the sphere will do whatever it wants to do.

Jasper

11:24

@StupidKid I cannot teach anything now, because I am only a banana. But others may be able to teach you.

Stupid Kid

Guys I got across this question in my last chapter of real analysis book.
Prove that $\sum (x^2+n)\over(x^2+n^3)$ is not uniformly convergent on the domain $\Bbb R$.
This is Book's Answer:
When $x=n^2$, there is a term in the series, after the nth, which is greater than $1\over 2$. So altough the series is convergent for each value of x by comparison with $\sum 1/n^2$, the greatest difference between the nth partial sum (i. e.$f_n$) and limit function is not null.
The first statement in answer that book made is where I don't understand. Why is $x=n^2$ even choosed? How it came? And what doe…

@Jasper No you are not. You just lack experience XD.

@robjohn I think I wrote it more clearly now.

Jasper

@StupidKid Nice that you are studying on the bus. Just remember to get off at the right stop!

Stupid Kid

@Jasper I stare in my mobile reading PDF so much that everything not near me is blur 😂😂😂. I am already at my home now lol.

Hope I can finish the last lesson of real analysis today. Got only 7 question left out of 800. So that one day I can teach other here.

May be I will attempt to do other while someone explains me this question.

@TedShifrin Can you answer my question.

Balarka Sen

13:06

@LeakyNun Do you want to do a calculus computation

Leaky Nun

@BalarkaSen yeah?

Edward Evans

$\int x dx$ go

Balarka Sen

@LeakyNun Lol nevermind

robjohn

@StupidKid $x=n^2$ gives $\frac{x^2+n}{x^2+n^3}=\frac{1+\frac1{n^3}}{1+\frac1n}=1-\frac{1-\frac1{n^2}}{n+1}\gt\frac12$ for $n\ge1$

Stupid Kid

13:31

@robjohn understood. What about that $\sum 1/n^2$ it is kinda wierd. If you need to perform comparision test you need 0<n<1.

But $x$ is not bounded but $x=n^2$

geocalc33

How do you make $f(x)$ into a p-adic function?

Stupid Kid

now for finding if it is uniformly convergent or not I think it is same procedure we did yesterday. I mean taking max and finding the limit pointwise of the function.

geocalc33

Here's my attempt. $f(x)=x.$ $f(p^{-n})=p^{-n}$

Also I have a really endocentric question. What is a hyperbolic curve? And how does it relate to a hyperbola?

A hyperbolic curve is an algebraic curve obtained by removing r points from a smooth,
proper curve of genus g, where g and r are nonnegative integers such that 2g−2+r > 0.

robjohn

13:57

@StupidKid The question is about the uniform convergence of the series for all $x\in\mathbb{R}$. This shows that there is no $N$ so that $\sum_{n\gt N}\frac{x^2+n}{x^2+n^3}\lt\frac12$ for all $x$

Stupid Kid

@robjohn It does shows that series $\le$ 1/2. But It tells nothing about uniform convergence I think it is kinda unnecessary step.

Stupid Kid

14:18

oh I kinda get it. The limit of that $\frac{x^2+n}{x^2+n^3}$ becomes 1 thus it is greater than 0.5 and 1/n^2 is just use to become null and it shows that series is convergent and in order to get uniform convergence we do the same old trick which is distance between sequence of function and its limit pointwise function reaches 0 but it is not in this case.

But series of function is convergent only when 0<n<1

But why choose $\sum 1/n^2$?

I am kinda confused and sleepy.

robjohn

@StupidKid No, it shows that the sum does not converge uniformly. That is, we cannot choose an arbitrary $\epsilon\gt0$, say $\epsilon=\frac12$, and find an $N$ so that $\sum_{n=1}^N\frac{x^2+n}{x^2+n^3}$ is within $\epsilon$ of $\sum_{n=1}^\infty\frac{x^2+n}{x^2+n^3}$ for all $x\in\mathbb{R}$.

Abr001am

Hello, how can i find questions related to the problem $2^n=pq$ i can't find any in the main. thatks a lot.

where q, p are primes

Stupid Kid

@robjohn Oh wow it already proved it is not uniform convergent lol. This means sup(that series of function ) doesn't tend to 0

Abr001am

sorry 2^n+1

Stupid Kid

For uniform convergece of the series $\sum f_n(x)$ a necessary condition is $f_n(x) \to 0$ uniformly. This means $\sup_x |f_n(x)| $ must tend to $0$.

It didn't satisfied this statement.

thus is not uniform convergent

Tomorrow I will read the chat to make every idea airtight. Good Evening and good night.

anakhro

14:39

good morning

Stupid Kid

@robjohn Thanks for explanation!!!

geocalc33

Good Morning!!

Edward Evans

@Abr001am those might have the name "Fermat semi-primes"

geocalc33

anybody ever heard of Zimmer's conjecture?

it was proven a few years ago

Knight

@anakhro Hi! Seeing you after a long time

Hi chat! I’m unable to understand something, let’s say we have a vector field $\mathbf A$ with $$ A_x = -1/2 y k ~~~ A_y = 1/2 x k~~~ A_z=0$$ then my book writes “since the $x$- component is proportional to $-y$ and the $y-$ component is proportional to $+x$ , $\mathbf A$ must be at right angles to the vector from $z$ axis which we will call $\mathbf r’$ “

I don’t know what “vector from $z$ axis” means. Please help me in seeing what is he trying to say.

RScrlli

15:03

Hey guys I have a question; If we have an orthonormal basis $(e_i)$ on a Hilbert space , can we say $\langle f,g\rangle =\sum_{i=1}^{\infty} \langle f, e_i\rangle \langle g, e_i\rangle$? This looks like a sort of Parseval identity,

Knight

@JackOhara Hello Jackie!

geocalc33

Can I get help with, Exercise 2: Given a diffeomorphism $\rho:K \to K$ where $K=L \cup L'$, where $L=\Bbb R^{1,1}$ and $L$ and $L'$ are transversal, find the probability that $\rho$ is compact. Assume that $L$ and $L'$ are finitely, analytically laminated, and densely embedded in a compact space.

robjohn

15:39

@StupidKid that's the point of uniform convergence.

@StupidKid welcome!

feynhat

16:21

@RScrlli If the base field is $\Bbb R$, then yes.

Alessandro Codenotti

@Balarka I listened to the Elder album, pretty cool

Balarka Sen

Glad you liked it

Anonymous

Does anyone know how to write down (explicitly) a homomorphism from the binary icosahedral group $2I$ to the icosahedral group $I$, with kernel $\{\pm 1\}$? (as mentioned here)

robjohn

@RScrlli if the base field is $\mathbb{C}$, then you need to use $\overline{\langle g,e_i\rangle}$

Semiclassical

16:59

0

Q: Proof that y = 2 in the goormaghtigh conjecture

I believe I have a proof that y = 2 in the goormaghtigh conjecture. For reference:Goormaghtigh conjecture I start by proving the lemma $\frac{x^{m-1}-1}{x-1} = \frac{y}{x}\frac{y^{n-1}-1}{y-1}$ Proof: $\frac{x^{m}-1}{x-1}-1 = \frac{y^{n}-1}{y-1}-1$ $\implies \frac{x^{m}-x}{x-1}=\frac{y^{n}-y}{...

2

question begins: "I believe I have a proof..."

Tobias Kildetoft

goormaghtigh? Is that one of the elder gods?

Semiclassical

ends: "I realize that the argument becomes shoddy in the last step and would like some suggestions on how to overcome it"

Balarka Sen

Lmao

Ted Shifrin

@Knight They mean the part of the position vector lying in the $xy$-plane (ignore the $z$ coordinate).

Balarka Sen

everything about this question is great

feynhat

17:00

I like the boldness.

Ted Shifrin

howdy, @Semiclassic, @Tobias, @robjohn, and, of course, a.

Tobias Kildetoft

@TedShifrin Hi

Semiclassical

when you see that kind of boldness enough, you start to see it more as carelessness than anything else

Tobias Kildetoft

@Semiclassical Take a look at the abstract of the linked paper here math.stackexchange.com/questions/3663635/…

Thorgott

what even is the proof strategy in that question

Balarka Sen

17:01

Feit–Thompson conjecture

In mathematics, the Feit–Thompson conjecture is a conjecture in number theory, suggested by Walter Feit and John G. Thompson (1962). The conjecture states that there are no distinct prime numbers p and q such that p q − 1 p − 1 {\displaystyle {\frac {p^{q}-1}{p-1}}} divides...

Lol

Feit and Thompson, nutcases

Ted Shifrin

@Semiclassic: The proof of the lemma of course assumes what's to be proved. The lemma is not remotely true, right?

Semiclassical

ow @TobiasKildetoft

@TedShifrin I mean, I don't know, I didn't actually look at what they did. just the contrast between the intro and outro

Thorgott

omg that abstract

robjohn

@TedShifrin a? Good morning!

Tobias Kildetoft

The funny thing is, it has been well-known for a long time that ABC implies FLT

Semiclassical

17:03

but, i wouldn't count on it...

@robjohn great icon btw

but yeah, when your entire argument can be summed up as "affirming the consequent"

not a great sign

Thorgott

I quite like the sentence "Nearly, ABC conjecture is interesting."

robjohn

@BalarkaSen nutcases? I went to grad school with Feit's son.

Semiclassical

lol

Ted Shifrin

@robjohn Balarka has a new name. :P

Balarka Sen

Hi @TedShifrin. You need to ping me! "a" doesn't ping me

@robjohn Oh cool.

Ted Shifrin

17:08

Some things are more important than pings :P

Balarka Sen

@TobiasKildetoft No mochi, disappointed

Tobias Kildetoft

mochi?

Balarka Sen

I have read it - links intergalactic

robjohn

@TedShifrin I must be missing something...

Ted Shifrin

Yes. :)

Semiclassical

17:10

intergalactic planetary planetary intergalactic

robjohn

@TedShifrin i feel so incomplete :-(

skullpatrol

complete the square!

user193319

2

Q: Representations of Finite Groups and Weak Containment

I am trying to prove the following: Let $G$ be a finite group and let $\pi : G \to U(\mathcal{H})$ and $\rho : G \to U(\mathcal{K})$ be irreducible unitary representations. If $\pi$ is weakly contained in $\rho$, then $\pi$ is contained in $\rho$. I am allowed to use the following fact: ...

group-theory functional-analysis finite-groups representation-theory hilbert-spaces

Balarka Sen

imagine if mochi was a troll and just wanted to use "interuniversal" as a mathematical terminology

linda Oiladali

hello, how to solve $ arcsin(x)=arcsin(a)+arcsin(b)$ where $a,b\in \mathbb{R}$

Tobias Kildetoft

17:14

@user193319 What does weak containment mean?

@BalarkaSen Some people get away with stuff like that. I mean, someone got away with naming the maximum security clearance "cosmic top secret"

Ted Shifrin

@linda: If you write $\theta = \arcsin x$, $\alpha=\arcsin a$, etc., what equations can you write down?

Balarka Sen

There's something known as electrocution of geodesics in geometric group theory

2

I think coined by Benson

feynhat

@BalarkaSen lmao... why so violent?

linda Oiladali

@TedShifrin $arcsin(x)=\theta+\alpha $

Balarka Sen

yeah thats why people changed it and use electrification now

Thorgott

17:16

that sounds epic

Ted Shifrin

@linda: Huh?

linda Oiladali

@TedShifrin sorry, $\theta=\arcsin(x)$ means that. $x=\sin(\theta)$

?

Ted Shifrin

Yes, true, but you have more stuff to write.

Balarka Sen

link.springer.com/content/pdf/10.1007%2Fs000390050075.pdf

"The term electric space was suggested to me by Bill Thurston, since geodesics in this space behave like lightning bolts shooting between metal plates (the horospheres)"

Thorgott

((Without abstract))

Astyx

17:20

Hi chat

Ted Shifrin

Salut, @Astyx

skullpatrol

Hi pal

Balarka Sen

@Thorgott That's because it's a monograph I think

linda Oiladali

@TedShifrin I don't understand the idea

Balarka Sen

GAFA publishes a bunch of monographs

Astyx

17:20

@skullpatrol Wow, new picture

skullpatrol

Yup :-)

Ted Shifrin

You did not correctly write down the original equation in terms of the new variables. I was also leaving it to you to define $\beta$ correctly.

Balarka Sen

But also maybe because Benson doesn't give a shit lol

Geometric group theorists are nutcases

skullpatrol

Gotta get with the times @Astyx

Astyx

Indeed. Hope he wears gloves and stays at home as well !

skullpatrol

17:22

will do pal

Vinothkumar Raman

I have a small question while reading Joy of cats but it seems too small to write a question. So the question is If $A$ is a subcategory of $B$ and $X \in A$ then $id$ should be reflection whether $A$ is full or not right? But I see a proposition which claims it is a reflection only when its full? Why is it so?

Alessandro Codenotti

@BalarkaSen just another good reason to do ggt

Vinothkumar Raman

I mean $id_X$ is a reflection of $X$

Semiclassical

geometrically, the point is just that arcsines are angles

linda Oiladali

$\theta=\arcsin(x)$ means that $x=\sin(\theta)$ with $x\in [-1,1]$, $\alpha=\arcsin(a)$ then $a=\sin(\alpha)$ and $b=\sin(\beta)$, a,b\in [-1,1]?

Ted Shifrin

17:24

OK, and how are $x$, $\alpha$, and $\beta$ related?

robjohn

@BalarkaSen is this related to tringulation?

2

Ted Shifrin

I guess a is now inventing words :P

Oh, that's the missing "a" !!!!

Balarka Sen

God

How do you dig this stuff up

Ted Shifrin

That's what you were missing all along, @robjohn.

robjohn

@TedShifrin we found it!!!

@TedShifrin thanks for completing the mean square!

skullpatrol

17:26

lol

Ted Shifrin

Mind your p's and q's, @robjohn.

robjohn

@TedShifrin after completing the square, I will be more discriminating...

Balarka Sen

This is ridiculous, I am getting cyberbullied by people from before the internet era

5

Ted Shifrin

Are you calling this bullying?

robjohn

@BalarkaSen This isn't bullying...

Balarka Sen

17:28

Just joking

Semiclassical

amusingly, i know a paper titled "(Don't) mind your p's and q's", in reference to p/q in classical vs. quantum mechanics

skullpatrol

Yeah, I'm on your side pal @BalarkaSen

Ted Shifrin

Of course, I was referring to $(L^p)^* = L^q$.

Balarka Sen

On my side, or by my side?

skullpatrol

on

Balarka Sen

17:30

On my sides will be rather painful

skullpatrol

@robjohn that's discrimination

Ted Shifrin

I rule this whole conversation off-sides.

Next we'll have sticking and general fighting.

skullpatrol

Please kick me out Professor

Semiclassical

at this rate we'll start tossing around fighting words like "inelegant" and "trivial"

Ted Shifrin

Oh, that would be serious.

linda Oiladali

17:32

@TedShifrin $x=\sin(\alpha+\beta)$?

Ted Shifrin

OK.

user193319

@TobiasKildetoft See this for a definition: math.stackexchange.com/questions/3576436/…

Balarka Sen

@Semiclassical (To be read in a proper English accent) You seem to be making rather obvious observations in a, how shall I say it, particularly inarticulate manner, to demonstrate your humor

linda Oiladali

@TedShifrin what I do know?

maths student

0

Q: continuity and circle

$f(x, y)=\left\{\begin{array}{ll}x^{2}+y^{2} & \text { if } x^{2}+y^{2}<1 \\ 1 & \text { if } x^{2}+y^{2} \geq 1\end{array}\right.$ Is this function continuous ? It seems like it is circle of radius 1 which is continuous. But I am not sure about it and I have seen this type of continuity problem...

limits continuity circles

Ted Shifrin

17:38

Well, I don't know what your original question was. What are you given and what does "solve" mean?

maths student

@TedShifrin are you referring with me?

linda Oiladali

@TedShifrin "Etudier l'existence de solutions selon a et b "

Ted Shifrin

@mathsstudent: Let $f(x) = \begin{cases} x, & x< 1 \\ 1, & x\ge 1\end{cases}$. Is this a continuous function?

No, @maths, not that time, obviously.

@linda: So, part of what they want you to think about is the domain and range of the $\arcsin$ function. You're not supposed to actually solve.

Edward Evans

Evening chat

Ted Shifrin

Good morning, @Edward.

skullpatrol

17:41

Hi pal

Edward Evans

@TedShifrin Good morning, even!

I'm a time-zone-ist

maths student

Yes we have to check at boundary points for that case so function is open set in interior and so continuous at interior of circle and at boundary it is continuous using definition of function @TedShifrin

linda Oiladali

@TedShifrin $arcsin:[-1,1]\to[-\pi/2,\pi/2]$

Ted Shifrin

So doesn't that give you some conditions on $a$ and $b$?

OK, @mathsstudent. Obviously the function is continuous away from $x=1$ (in my case) or $x^2+y^2=1$ (in your case). So you have to check at those points.

linda Oiladali

a+b must be in [-1,1]

Ted Shifrin

17:45

How did you get that?

maths student

@TedShifrin so $x^2 + y^2 = 1$ imply $x^2 = 1 - y^2 $ so when x approaching from outside it is equal to 1 by definition of function and from inside it is 1 since x^2 + y^2 = 1

linda Oiladali

no, not this sorry, $x=\sin(\arcsin(a)+\arcsin(b))$ so $a\in[-1,1],b\in[-1,1]$ and $\alpha+\beta\in [-\pi/2,\pi/2]$

Ted Shifrin

I think you should think about this in terms of a single variable $u=x^2+y^2$, @mathsstudent.

maths student

@TedShifrin right thanks I am over complicating things

Ted Shifrin

OK, seems correct, @linda.

@mathsstudent: The point is that $u=g(x,y)=x^2+y^2$ is a continuous function of $(x,y)$.

maths student

17:51

@TedShifrin so which property we are using to show substitution does not affect continuity ?

Ted Shifrin

Composition of continuous functions is continuous?

linda Oiladali

@TedShifrin how to answer this" representer l'ensemble des points M(a,b) pour les quels l'equation a une solution

Ted Shifrin

I think you did that in the last thing you wrote, @linda. You have to be explicit and say it in terms of $a$ and $b$.

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$Mathematics$ Mathematics