Mathematics - 2014-12-13 (page 2 of 5)

Daniel Fischer

11:00 AM

@SwapnilTripathi Yes, except in extraordinary circumstances.

r9m

@DanielFischer same problem here ... no apparent elementary ways of approximating $\frac{1}{x_{2015}}$ :-) ..

Daniel Fischer

:-/

Swapnil Tripathi

@DanielFischer: Don't you plan to nominate yourself? :D (Evidently, it seems you don't, but still)

r9m

@DanielFischer but I checked a few smaller initial values in wolfram .. it shows a linearish plot of the points .. the problem is determining where the duplicate values appear ! :(

Daniel Fischer

@SwapnilTripathi I'm still hoping that Jyrki steps up to save my freedom. If he doesn't, I'll have to.

r9m

11:04 AM

like here 3 appears twice instead of a nice linear plot ! :-(

Swapnil Tripathi

@DanielFischer Good to know. :)

r9m

@Daniel san .. do you know the final result ? :-) (w|a doesn't allow me to go that far)

Daniel Fischer

I haven't verified that the floating-point errors don't accumulate enough to change the result, but I think it's decently reliable: `Prelude> iterate (\x -> x*(1+x)) (1/2014) !! 2014
0.1731222461174841`

r9m

@DanielFischer wo ! thanks ! :-)

Daniel Fischer

@r9m So, knowing what you have to prove, try proving it ;)

(And ping me when you have found a nice proof.)

r9m

11:18 AM

@DanielFischer it just shattered what I was imagining it would be :P irks ! :-)

Daniel Fischer

Yeah, one would think it grows faster, wouldn't one?

r9m

ya .. bloody deceptive sequence ... it deserves death by induction ! =( growl ..

Sush

This question is from "The Shape of Space" by Weeks. I can't understand if A Square traveled straight in his first trip, then how could he go on circular path? @DanielFischer

Daniel Fischer

@Sush "Straight" means it travels on a geodesic, I'd think. Just like when you walk straight on the surface of the earth, you also go round in (great) circles.

(Approximately, the earth's surface isn't really a sphere.)

Sush

@DanielFischer, Thanks.

r9m

11:55 AM

@Anastasiya-Romanova秀 hey ! 'sup ! ^^

Anastasiya-Romanova 秀

@r9m Hi. 'sup?

r9m

@Anastasiya-Romanova秀 I see you are running for mod election ! best of luck ! :-)

Anastasiya-Romanova 秀

@r9m Thanks... $\ddot\smile$

@r9m I got this problem from my aptitude and ability tests:

What is the next number in the sequence

$$16,\,06,\,68,\,88,\cdots$$

Can you answer it?

r9m

@Anastasiya-Romanova秀 I don't know .. point reflection of 87 maybe ? =P

Anastasiya-Romanova 秀

@r9m I answered 87 too, but idk the correct answer

Daniel Fischer

12:07 PM

@Anastasiya-Romanova秀 You forgot to rotate, $L8$.

2

r9m

@DanielFischer that's not a number ! :o

Anastasiya-Romanova 秀

@DanielFischer That's funny one :D

Daniel Fischer

@r9m Base $36$?

r9m

@DanielFischer yes ! yes ! just realized and was about to edit it :D

@Anastasiya-Romanova秀 how about $$52, 63, 94, ... ?$$ :-)

@Khallil ^ the next number in the sequence ? :)

Khallil Benyattou

Afternoon, @r9m!

r9m

12:15 PM

@KhallilBenyattou stu ! :)

Anastasiya-Romanova 秀

64?

Khallil Benyattou

I'm really bad with these things!

r9m

@Anastasiya-Romanova秀 explain please :)

Khallil Benyattou

Hey, @Anastasiya. Where'd you get 64 from?

Anastasiya-Romanova 秀

\begin{align}52&\implies25=5^2\\63&\implies36=6^2\\94&\implies49=7^2\end{align}

r9m

12:18 PM

@Anastasiya-Romanova秀 HAI !! :D

Anastasiya-Romanova 秀

My brother once told me a riddle when I was 7 that took me a year to solve it

r9m

@Anastasiya-Romanova秀 wow ! what was it ? :-)

Anastasiya-Romanova 秀

Wait, mom's calling

r9m

'kay

Khallil Benyattou

By your logic, wouldn't the next number be 46, @Anastasiya?

$46 \implies 64 = 8^2$

r9m

12:23 PM

@KhallilBenyattou yes ..that's the right answer :)

Khallil Benyattou

That's a cool one! It really bends your way of thinking, @r9m. ^_^

r9m

@KhallilBenyattou ya ! ;)

Mary Star

@DanielFischer Yes, I meant $\mathbb{F}_{p^k}$.

Ok!! I see... Thank you very much!!!! :-)

Daniel Rust

Hi all

Khallil Benyattou

Hey, @DanielRust!

How are you doing today?

Daniel Rust

12:35 PM

I'm doing very well thanks

How are you, and everyone else in chat?

Typical that an election starts at my busiest time of year...

user4215

hi

can I ask you a very obvious question?

Daniel Rust

ask away

Hi @TedShifrin!

evinda

Hello @TedShifrin!!! :) Could I ask you something?

Daniel Rust

how can answering/asking maths questions be used against you?

Balarka Sen

Hi @DanielRust

Daniel Rust

12:45 PM

@BalarkaSen hi

Balarka Sen

can you help me with something?

Khallil Benyattou

I'm doing good, thank you. Struggling to wake up before midday, but I'm working on it, @DanielRust!

Balarka Sen

I want to find out the Cech fundamental group of the standard $p$-adic solenoid.

Daniel Rust

I can try

Is the Cech fundamental group something like a direct limit of fundamental groups?

Balarka Sen

Yeah, @DanielRust. The inverse limit of fundamental groups of the nerves.

Daniel Rust

12:47 PM

oh, the inverse limit ok

Anastasiya-Romanova 秀

@KhallilBenyattou That was what I meant. I forgot to flip the number.

Balarka Sen

I trust that 1-skeleton of the nerve of a sufficiently nice open cover of the solenoid is inverse limit of Cayley graphs of Z/p^iZ.

Khallil Benyattou

Ah, cool! I know what you meant, @Anastasiya. =P

Daniel Rust

well the p-adic solenoid can be written as an inverse limit of multiplcation maps on the circle, so I guess you're looking as the inverse limit of the $\times p$ map on $\mathbb{Z}$.

Anastasiya-Romanova 秀

@r9m Wait, I'll try to translate the problem into English

Balarka Sen

12:49 PM

@DanielRust i don't think so. Cech fundamental groups preserve inverse limits?

Ted Shifrin

Hi @DanielRust @evinda

Daniel Rust

I'm not sure @BalarkaSen. I've not come across it before. I'm more used to Cech cohomology which is contravariant and continuous (so it maps limits to colimits)

Balarka Sen

ah

evinda

@TedShifrin I am looking again at the exercise at which I have to find the inflection points of the curve $x^3+y^3+z^3=0$ in $\mathbb{P}^2(\mathbb{C})$.

Balarka Sen

actually my goal is to define some good notion of fundamental groups so the \pi_1 of the solenoid is $\mathbf{Z}_p$ @DanielRust

evinda

12:51 PM

@TedShifrin The equation $a^3+1=0, a\in \{X,Y,Z\} $ has also the solution $e^{\frac{i \pi (2m+1)}{3}}, m=0,1,2$, right?

So does this mean that the inflection points of $ \mathbb{P}^2(\mathbb{C})$ are the following?

$ [0,1,-1], [0,-1,1], [1,0,-1], [-1,0,1], [1,-1,0], [-1,1,0], [e^{\frac{i \pi (2m+1)}{3}},1,0], [e^{\frac{i \pi (2m+1)}{3}},0,1], [0, e^{\frac{i \pi (2m+1)}{3}},1], [1, e^{\frac{i \pi (2m+1)}{3}},0], [0, 1, e^{\frac{i \pi (2m+1)}{3}}], [1, 0, e^{\frac{i \pi (2m+1)}{3}}], m=0,1,2$ ? Or am I wrong?

Daniel Rust

and here $\mathbf{Z}_p$ is the group of $p$-adic integers?

Balarka Sen

this is motivated from the fact that Z_p acts on the solenoid properly discontinuously and freely, so if it was path connected and locally path connected there would have been a SES 1 --> \pi_1(X) --> \pi_1(X/Z_p) --> Z_p --> 1 which would let us realize Z_p geometrically

@DanielRust yeah

Anastasiya-Romanova 秀

@r9m If I'm not mistaken, the problem is like this:

"A man with a big ego lives in an old apartment at floor 10 where people who live there are very selfish and don't care anyone else. Every morning when he goes to the office, he uses the elevator to go down to basement, but when the afternoon after coming from the office he only uses the elevator to the 2nd floor and the rest of the way to his room is reached on foot by passing through the stairs. Why so? What is the most logical explanation for his action?"

Balarka Sen

i'm hoping to do this in general for profinite groups. particularly the absolute galois group over Q :P

yes, i read about the Esquisse d'un Programme lately.

:P

Ted Shifrin

@evinda: I can't read all that. $-1$ should be one of those cube roots. All,in all, you should have 9 inflection points.

Daniel Rust

12:53 PM

I think what you'll want to ask then is, does the fundamental group provide us with a cocontinuous functor?

Balarka Sen

@DanielRust actually someone on the homotopy chat room yesterday said that he suspects the cech homotopy group is trivial, so solenoid is more like universal cover of a space with \pi_1 isomorphic to Z_p than a space with \pi_1 isomorphic to Z_p

@DanielRust uh?

oh you mean the Cech fundamental group

i doubt.

evinda

@TedShifrin I find more.... :(
Is it right that the equation $a^3+1=0, a\in \{X,Y,Z\} $ has also the solution $e^{\frac{i \pi (2m+1)}{3}}, m=0,1,2$ ?

Daniel Rust

yes sorry @BalarkaSen

Ted Shifrin

Yes. You should not have more. You're double-listing.

Balarka Sen

@DanielRust I've actually got a weird kind of topology on the solenoid (not the usualy profinite one) that could make the cech \pi_1 to be Z_p.

Daniel Rust

12:55 PM

@BalarkaSen do you have a reference for the definition of the Cech fundamental group?

Balarka Sen

@DanielRust it's in nLab i guess

Daniel Rust

nLab suggests a link with shape theory, which is interesting.

but I can't see a definition

Balarka Sen

the topology i am thinking of consists of open sets (A_0, A_1, A_2, ...) \in \prod S^1 such that f_n(A_n) = A_{n-1}, where A_i are open in the i-th copy of S^1 in the inverse system

Daniel Rust

is that different to the subspace topology?

Balarka Sen

you mean the subspace topology inherited from the infinite torus \prod S^1?

evinda

1:00 PM

@TedShifrin So, don't we have to count all the points that contain a $e^{\frac{i \pi(2m+1)}{3}}, m=0,1,2$ ? :/

Daniel Rust

yes, the usual way one defines the inverse limit

Balarka Sen

hmm

maybe not.

but then the fundamental group is bound to be Z_p

@DanielRust can you visualize the nerve of the open cover of the solenoid coming from the open cover of the base S^1?

i'm having a hard time doing that

can't even see the 1-simplicias

r9m

@Anastasiya-Romanova秀 oo ! :O I have to think about it !! beats me ! :O

Daniel Rust

Haha no... although I'm pretty sure that at 'good' parts of the refinements of covers, you get a space which is homotopic to the circle

Balarka Sen

@DanielRust no wonder. the solenoid is the fiber product of S^1 and the cantor set so locally at the leafs the space would look like an S^1

but globally...

Daniel Rust

1:03 PM

so that if you restrict your inverse system to those refinements and their maps, they are homotopic to the sequence $S^1\stackrel{\times p}{\leftarrow}S^1\stackrel{\times p}{\leftarrow}\cdots$

Anastasiya-Romanova 秀

@r9m Take your time. My brother patiently waits almost a year for my answer, so do I :D

r9m

@Anastasiya-Romanova秀 okay ^^'

Lembik

hi

any probability people in? Any interest in math.stackexchange.com/questions/1059379/… ?

Balarka Sen

@DanielRust so you expect the fundamental group to be the inverse limit of Z <-- pZ <-- p^2Z <-- ... ?

evinda

$[1,0,e^{\frac{i \pi(2 \cdot 1+1)}{3}}]=[1,0,-1]$, so have to write only one of these, right?

Daniel Rust

1:06 PM

isomorphic to that I guess, rather just $\times p$ at each stage.

Balarka Sen

weird. dunno what that is.

Daniel Rust

they're the same thing, you've just composed a few of the maps that appear in my inverse system

Balarka Sen

yeah, i mean i dunno what the inverse limit is :P

Daniel Rust

oh

Integrator

@r9m @Anastasiya-Romanova秀 That man is short! And cannot reach for the lift buttons

Daniel Rust

1:08 PM

I'm more of an expert on direct limits of groups if I'm honest (that's why the contrvariance of Cech cohomology is more familiar to me)

Balarka Sen

haha i see

Daniel Rust

actually... I think that inverse limit is trivial

Balarka Sen

you do?

hmm

Daniel Rust

no element except for the identity has an infinite preimage

Balarka Sen

makes sense

Daniel Rust

1:10 PM

I think in those cases, it's worth looking at the derived inverse limit

Balarka Sen

@Daniel you mean \lim^1?

Alyosha

How do you prove that the solenoid is the fibre product of the circle and the Cantor set?

Daniel Rust

yeah

Balarka Sen

i believe that is Z_p/Z

@Alyosha there are canonical maps from Z_p and R

Daniel Rust

I have to admit I know little about $\lim^1$ though

Balarka Sen

1:11 PM

you can use those to pullback and get that it's topologically R \times Z_p/Z

evinda

@TedShifrin So, now these points remain:

Daniel Rust

My supervisor John Hunton wrote a nice paper about the $\lim^1$ invariant of tiling spaces which solved this problem of trivial inverse limits of fundamental groups

evinda

Daniel Rust

But I've not read it in a while

Alyosha

What is R?

Balarka Sen

1:11 PM

@DanielRust Oh?

interest

@Alyosha the reals :P

Anastasiya-Romanova 秀

@Integrator @r9m We have a winner here! \ $(\ddot\smile)/$

Alyosha

Oh right. I thought you meant Cantor or something.

Balarka Sen

@Alyosha think abou the solenoid as .... --> R/p^2Z --> R/pZ --> R/Z

Z/p^iZ is sitting at each stage and there is also a map from R to R/p^iZs.

evinda

@TedShifrin I uploaded a picture above.. Which other points are the same? :/

Balarka Sen

so if your solenoid is X, there are maps R \to X and Z_p \to X by uniqueness of inverse limit.

Daniel Rust

1:13 PM

@BalarkaSen it may be a tough read for someone who's not familar with shape theory/tiling spaces arxiv.org/pdf/1105.0835.pdf

Alyosha

OK. I will think about this in a while, thank you!

Balarka Sen

@Alyosha now pullback Z_p --> X <-- R

just write down the set theoretic pull back, you'll see what it is soon enough

Alex Clark, I have heard about that guy @DanielRust

wrote a lot of stuff about the solenoid embedded in R^3

Daniel Rust

Alex Clark is my other supervisor :P

Balarka Sen

fun stuff, the geometric topology involved there in computing solenoid complements in R^3.

@DanielRust oh cool

shape theory deals with totally messed up spaces :P

Daniel Rust

Haha, I have a hard time understanding shape theory sometimes

it's very unintuitive

But it seems to be the right setting to answer some of these questions

Integrator

1:18 PM

@r9m @Anastasiya-Romanova秀 and how exactly is this princess going to reward me?

Balarka Sen

@DanielRust well, i've made a deal with pathological spaces until i settle these quests with profinite groups.

:P

Daniel Rust

Haha, I'm currently studying problems in the dual theory. Groups like $\mathbf{Z}[1/p]$ and how they appear as invariants.

Khallil Benyattou

Balarka Sen

@DanielRust i've been thinking whether it'd be helpful to think about it conformally/complex analytically. that guy solenoid is essentially of the same homotopy type as the solenoid riemann surface obtained by taking inverse limit of riemann surface of w^{p^n} = z over C.

bonding maps are natural (w, z) \to (w, z^p)

Khallil Benyattou

As a quick question, is it enough to solve the two equations simultaneously and find that $(q_1 - q_2)m + (r_1 - r_2) = 0$ and deduce that both $q_1 = q_2$ and $r_1 = r_2$ for the equality to be true?

Daniel Rust

1:21 PM

If you're only looking at the fundamental group (or questions surrounding it) I'm not sure what you gain by thinking about it in any way finer than homotopically.

Balarka Sen

@DanielRust if you're gonna talk more about dual theory, i am listening.

Ted Shifrin

@evinda: You still don't understand what projective space is. Note that $[1,-1,0] = [-1,1,0]$. Also, what is $e^{\pi(2\cdot 1+1)i/3}$?

Balarka Sen

@DanielRust well just that we can think about riemann surfaces for more general problems than realizing Z_p geometrically. i think there is hope that way to think about the absolute galois group over C(z) using solenoid riemann surfaces.

Anastasiya-Romanova 秀

@Integrator I just did

Daniel Rust

@BalarkaSen Well as I already said, the Cech cohomology is a contravariant, continuous functor, so you get $\check{H}^k(X)=\lim_{\rightarrow}(H^k(X_i),f^*_i)$ where $X=\lim_{\leftarrow}(X_i,f_i)$ is some inverse limit representation.

Balarka Sen

1:23 PM

mmhmm

Integrator

@Anastasiya-Romanova秀 Okay! After election please put your photos back ;)

Daniel Rust

so for instance the first Cech cohomology of the $p$-adic solenoid is just $\mathbf{Z}[1/p]$.

Balarka Sen

yeah.

yeah it is.

Daniel Rust

So anyway, the spaces I study are these 'tiling spaces', and the question I'm currently studying is what kinds of groups can be realised as the Cech cohomology of such a space.

evinda

@TedShifrin I see... $e^{\pi(2\cdot 1+1)i/3}=-1$.
So, now the following points remain:

Balarka Sen

1:27 PM

@DanielRust i see.

Daniel Rust

I already have a partial answer of just how varied these groups can be (see here - arxiv.org/abs/1411.4991v1)

evinda

@TedShifrin Do I have to write now the points for all $m$, to see which are the same?

Balarka Sen

that's quite recent @DanielRust

Daniel Rust

@BalarkaSen It's my first paper :P

Balarka Sen

oh

well congratulations, although it's a bit late

Daniel Rust

1:29 PM

late?

Ted Shifrin

@evinda: You haven't understood my first comment. You have doubly listed points because you're not understanding what homogeneous coordinates mean. Remember that $[a,b,c]=[ta,tb,tc]$ for any nonzero $t\in\Bbb C$.

Balarka Sen

a month i would consider late enough for congratulations, @DanielRust

:P

Daniel Rust

Oh haha, well thankyou @BalarkaSen

Ted Shifrin

Congrats, @DanielRust. Even if you do math I cannot begin to fathom :)

Daniel Rust

Thanks @Ted

Khallil Benyattou

1:30 PM

You wrote your first paper? Congrats, @DanielRust!

Daniel Rust

haha, thankyou @KhallilBenyattou

Ted Shifrin

I wonder when I get congratulations for having written my last many years ago :P

Daniel Rust

Don't worry @Ted, I find your stuff unfathomable too! :D

Balarka Sen

~~differential geometry~~

ducks

Ted Shifrin

That isn't always a symmetric relation, @DanielRust, but it often is. :)

evinda

1:35 PM

@TedShifrin I multiplied this pont: $[1, 0, e^{\frac{i \pi (2m+1)}{3}}]$ by $e^{\frac{-i \pi(2m+1)}{3}}$ and I got: $[e^{\frac{-i \pi(2m+1)}{3}},0,1]$. Is this point equal to $[e^{\frac{i \pi (2m+1)}{3}},0,1]$ for some $m$ ? :/

Daniel Rust

So has anyone been keeping up with the election? I haven't had time this week to see how things are progressing

I noticed Peter was nominated

Anastasiya-Romanova 秀

@Integrator No. I'm going to be a college student, won't do silly things again.

evinda

@TedShifrin Or should I do something else? :/

Balarka Sen

There's still not much people I'd like to vote for @DanielRust

Pedro, for one. Behaviour (Care Bear/ Thursday / 900 sit ups a day / [ad infinitum...]), maybe.

Daniel Rust

I'm surprised at the number of poor nominations tbh

some people without any reviews/flags

Balarka Sen

1:40 PM

Indeed. The nominatory thread has become hilarious.

theage

Nominating yourself is as easy as clicking a link - considering how many users there are it's kind of inevitable honestly.

Khallil Benyattou

Really? Not even flags?

Very fair point, @theage!

Balarka Sen

There'd be less moderator than 3 this time if @DanielFischer doesn't step up. The positions won't even fill.

Speaking of it, what did you decide @DanielFischer?

Daniel Fischer

@BalarkaSen Still waiting for Jyrki Lahtonen. If he steps up, Jyrki, Pedro and Thomas are truly good choices, Arkamis to be considered.

Balarka Sen

Dunno about Thomas, but Arkamis looks good.

evinda

1:47 PM

@TedShifrin The points that we get for $m=2$ are the same as these that we get for $m=0$ in the projective plane right?

Balarka Sen

He seems much of a politician, looking at his comments below the nominatory post :P

off-topic : this is interesting.

Daniel Rust

A very short paper

although there's also no content apart from conjectures :P

Balarka Sen

but what that guys says/conjectures looks good

Daniel Rust

I don't know any Galois theory beyond the basics unfortunately :(

Balarka Sen

oh me neither. i know a bit about infinite galois theory, just as much to appreciate the complexity of Gal(\bar Q/Q)

i'm learning algebraic topology to understand that group a la Grothendieck

that guy in the paper says Gal(\bar Q/Q) surjects into the automorphism group of the infinite binary tree.

Integrator

1:54 PM

@Anastasiya-Romanova秀 :(

Balarka Sen

@DanielRust not sure if you know of the analogy already, but you can think of deck transformation groups of galois covering maps as galois groups

Daniel Rust

Yes, I know of the Galois correspondence (for field extensions and covering maps)

Balarka Sen

ah cool. but not only there is a galois correspondence but also a galois action

Daniel Rust

apparently there is a strong underlying theory, which I guess will bring in the Grothendieck stuff you mentioned

Balarka Sen

yeah

Daniel Rust

1:57 PM

Grothendieck scares me :P

Balarka Sen

of course, i know absolutely nothing about it

@DanielRust :P but it looks motivating stuff

Daniel Rust

Oh sure, I'm in awe of people that can understand this stuff.

Balarka Sen

his abstractness scares me too, but he did some practical nonabstract stuff later

you know about dessins d'nfants?

Daniel Rust

nope

Balarka Sen

yeah well those were the theories he developed to realize Gal(\bar Q/Q) geometrically. Beyli proved that any algebraic curve over \bar Q can be realized as a riemann surface, branched over $\Bbb P^1$ only at $0, 1$ and $\infty$

so Grothendieck constructed little graphs correspinding algebraic curves over \bar Q which is made out of taking the preimage of the interval [0, 1] in the riemann sphere on the corresponding riemann surface

the white nodes are preimages of 0, the black ones are preimages of 1, and the edges are preimages of [0, 1]

these are called dessins. he proved that Gal(\bar Q/Q) acts on the set of all the dessins.

Daniel Rust

2:02 PM

Ah nice

Balarka Sen

amazing, isn't it?

Daniel Rust

I'm always in favour of turning algebraic problems into geometric ones

Balarka Sen

there is a recent development based on this by Drinfeld which shows that Gal(\bar Q/Q) injects onto some braid group (look up grothendieck teichmuller group) and the conjecture is that Gal(\bar Q/Q) is precisely that braid group!

@DanielRust yeah realizing Gal(\bar Q/Q) gemetrically is a pretty open-ended problem the last time i looked

Alyosha

For what $n$ is it possible to glue $n-1$-dimensional faces of $\Delta^n$ together to get a triangulation of $S^n$?

Daniel Rust

ooo that does sound interesting. braid theory is one my 'pet' subjects that I learn for fun

Balarka Sen

2:04 PM

@DanielRust haha cool

maybe you should look at some of that. it's called grothendieck teichmuller theory

i know nothing about it, and i'm dying to learn that stuff

Daniel Rust

@Alyosha, can't every sphere be triangulated?

Alyosha

Maybe. It's not obvious to me that $S^2$ can be formed from $\Delta^2$ just by gluing edges together. Obviously possible to do it if you allow collapsing of edges to a point.

teadawg1337

Morning, guys :3

Daniel Rust

well, first split the sphere into two discs, and then triangulate each disc.

Alyosha

But this does not use a single $\Delta^2$, surely?

Daniel Rust

2:09 PM

Oh, you only want to use a single simplex?

Alyosha

Yes.

Sorry, I wasn't clear.

Daniel Rust

I think that's only possible for $n=1$

Alyosha

A Hatcher exercise is to show that it's possible for $n=3$.

Daniel Rust

You can probably prove that quite quickly using the Euler characteristic

Alyosha

Ex. 2.1.7

I posit that it's possible iff $n$ is odd.

Balarka Sen

2:16 PM

OK I guess I should go do some algebraic topology right now.

Bu-bye.

Khallil Benyattou

Morning, @teadawg1337!

(Well, afternoon for me, but it's still morning somewhere!)

How've you been?

teadawg1337

@Khallil Hey! Do you like number theory?

Khallil Benyattou

I really haven't done enough of it to say I like or dislike it! I mean, it does seem very useful, but I haven't seen anything that's wowed me yet, @teadawg1337. I'm guessing you're a big fan? ^_^

teadawg1337

I'm feeling better than I did yesterday, btw

Khallil Benyattou

Ah, good to hear. Me too!

teadawg1337

2:21 PM

@Khallil I'm a fan of most fields in mathematics :P

I'm working on a question related to number theory, I think it's super interesting (though my opinion is somewhat biased)

I'm almost done, I think I'll post it within the next few hours

evinda

Hey @DanielFischer !!! :) I am looking again at the following exercise:
Show that we can sort $k$ integer numbers with values between $0$ and $k^2-1$ in time $O(k)$.

Did you mean this algorithm:

1) Subtract all numbers by 1.
2) Since the range is now 0 to n2, do counting sort twice as done in the above implementation.
3) After the elements are sorted, add 1 to all numbers to obtain the original numbers.

Studentmath

How is substracting all numbers by 1 makes the range between 0 and $k^2$?

teadawg1337

2:37 PM

@Hippalectryon I just realized that $k!+m!=n^3$ is a Diophantine equation (for the purposes of the question I'm about to post on main)! It's a linear Diophantine equation since the solutions must be in integer form: $k!=k\left((k-1)!\right)$, $m!=m\left((m-1)!\right)$, and $n^3=n\times{n}\times{n}$ with $n\in\mathbb{N}$

I think it's a linear Diophantine equation, but I'm not entirely sure...

Daniel Fischer

@evinda What's with the subtracting and adding $1$? In both situations, you have the constraint $0 \leqslant k \leqslant n^2-1$. You would need to subtract and afterwards add if you started with numbers $1\leqslant k \leqslant n^2$.

evinda

@DanielFischer So don't we need to subtract all numbers by 1?

teadawg1337

@DanielFischer Could you verify my statement above?? I'm not entirely sure if I'm correct

Daniel Fischer

@evinda No. Only the radix sort.

evinda

@DanielFischer So, do we just have to apply the radix sort? :/

Daniel Fischer

2:42 PM

@teadawg1337 About the Diophantine equations? What's your definition of Diophantine equation and in particula linear Diophantine equation?

@evinda Yes.

teadawg1337

@DanielFischer A Diophantine equation is an equation in which only integer solutions are allowed, and it's to my understanding that they must also be in polynomial form. A linear Diophantine equation is a Diophantine equation with terms of degree 1(?)

Daniel Fischer

@teadawg1337 In that case: $m!$ is not a polynomial.

teadawg1337

@DanielFischer Alright, I was just making sure. Thanks for the help! :D

Studentmath

2:59 PM

Woha, this is surprising. $f:Q \to R$ where $f(x)=0$ if $x\le \sqrt2$, $f(x)=1$ if $x\ge \sqrt2$ is continous

I would expect otherwise as $Q$ is dense in $R$

Daniel Fischer

@Studentmath $g(x) = \frac{1}{x}$ is continuous on $\mathbb{R}\setminus \{0\}$, which is also dense in $\mathbb{R}$.

Studentmath

Is it continuous from $R-\{0\}$ to $R$, though?

yeah of course it is..

Daniel Fischer

The point is that the "bad point" does not belong to the domain of the function.

Studentmath

Indeed, that makes it clear - thanks @Daniel

I was trying to prove it wasn't conitnuous when I realised it was, kind of hit me

Daniel Rust

@Studentmath This is partially also because $\mathbb{Q}$ is a disconnected space, so its image doesn't have to be connected under a continuous map, unlike a connected space like $\mathbb{R}$.

Studentmath

3:09 PM

Just like $R-\{0\}$

Daniel Rust

yep

Studentmath

Or any disjoint union of subsets of $R$

Daniel Rust

well, the disjoint union of $[0,1]$ and $(1,2]$ is a connected subspace

Studentmath

Ach so, right

Huy

@DanielFischer: I want to construct a norm on $\mathbb{R}^2$ such that the unit circle w.r.t. the norm is a regular hexagon. How would you approach this task?

evinda

3:21 PM

@DanielFischer So, do we have to apply the following algorithm?

Daniel Fischer

@Huy Use the Minkowski functional (also known as gauge functional) of the hexagon. If you want an explicit expression, write the hexagon as the intersection of three rectangles and the norm as the maximum of the three norms corresponding to the rectangles.

evinda

Huy

@DanielFischer: I know two explicit forms but I have no idea how I would get there.

@DanielFischer: I know $z \mapsto \sum_{i=1}^3 |\operatorname{Re}(\zeta^i z)|$ and similarly the maximum norm for $\zeta = e^{2 \pi i/3}$ are such norms.

Daniel Fischer

@Huy If $B_p$ denotes the unit ball of the seminorm $p$, then $$\bigcap_{k=1}^n B_{p_k} = B_{\max \{p_1,\dotsc,p_n\}}.$$

@evinda If that is the algorithm of the radix sort, then yes.

evinda

@DanielFischer I used twice the variable k.. One of them should be n...
But isn't the time complexity of this algorithm $O(k^2)$? Or am I wrong? :/

Daniel Fischer

3:35 PM

@evinda It's $O(k\cdot n)$ if you correct it. But $k = 2$, so it's $O(n)$.

evinda

@DanielFischer Why is it $k=2$?

Daniel Fischer

@evinda Because all numbers are $< n^2$, and we use radix $n$, so the numbers have two digits.

evinda

@DanielFischer I haven't understood it.. From which point do we conclude that the numbers have two digits? :/

Chris's sis

@robjohn Why the message I posted it here was deleted?

12

Q: A couple of definite integrals related to Stieltjes constants

In a (great) paper "A theorem for the closed-form evaluation of the first generalized Stieltjes constant at rational arguments and some related summations" by Iaroslav V. Blagouchine, the following integral representation of the first Stieltjes constant $\gamma_1$ is given (on page 539): $$\gamma...

user111187

@Anastasiya-Romanova秀 You wanted to talk.

Daniel Fischer

3:49 PM

@evinda Two digits in base $n$. In base $b$, what is the largest number you can write with at most $d$ digits?

Anastasiya-Romanova 秀

@user111187 Ya, Hi Ruben. I wanna say I'm reluctant to answer Problem 42. I know I violate the rules but I think it's about time to end it. How do you think?

@user111187 I'm very busy these days

My focus is only here right now

user111187

@Anastasiya-Romanova秀 I would certainly like to see a proof or proof sketch of 42. You are under no obligation to participate in the contest after that, although we would miss you.

Anastasiya-Romanova 秀

@user111187 How about if I give you a hint here?

user111187

@Anastasiya-Romanova秀 Would appreciate that. I wrote the integral as a sum from 0 to $\infty$, what to do next?

Anastasiya-Romanova 秀

@user111187 Wait a sec

Khallil Benyattou

3:57 PM

3 hours ago, by Khallil Benyattou

As a quick question, is it enough to solve the two equations simultaneously and find that $(q_1 - q_2)m + (r_1 - r_2) = 0$ and deduce that both $q_1 = q_2$ and $r_1 = r_2$ for the equality to be true?

Anybody? ^_^

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$Mathematics$ Mathematics