Mathematics - 2020-05-01 (page 2 of 2)

Mats Granvik

15:19

@Knight Hi.

Knight

@MatsGranvik Hello!

Mats Granvik

I don't know the answer to your question. I can try it in Mathematica after I finish eating this sandwich.

Knight

@MatsGranvik Okay, please do that.

How was the sandwich :-) ?

Mats Granvik

It is a baguette.

Knight

Wow!

Mats Granvik

15:33

Mathematica says that it can't do the limit, but it leaves it unevaluated so the limit should exist:
Clear[n, x, k];
x = 2;
Limit[x/n*Sum[Log[1 + k^2*x^2/n^2], {k, 1, n}], n -> Infinity]

The limit gives me for x=2 and n=10^6:
1.4331748698945606`

While the integral gives me for N[Integrate[Log[x], {x, 1, 2 + 1}], 12]
1.29583686600

Knight

@MatsGranvik I’m sorry but I really don’t know about Mathematica

What are your conclusions ?

Mats Granvik

@Knight Can't say for sure, but they seem to be not equal. Also why do you have x+1 in the upper integration limit? Did you mean to write some other number there equal to x in the limit like: $\int_{1}^{X+1} ln(x) dx$

Or: $\int_{1}^{x_1+1} ln(x) dx$

Knight

Because that limit of sum was equal to $f(x)$

Can we convert that limit into an integral in any other way?

Mats Granvik

I can't say, I don't know.

Knight

Thanks for your time.

Did you take coffee or tea with that sandwich ?

Mats Granvik

15:46

@Knight Neither, I quit coffee and tea about 5 or 6 years ago.

Knight

@MatsGranvik WHY?

Mats Granvik

Because my boss does not drink coffee or tea, and I thought it was a good idea to follow his example.

$$\int_1^{x+1} \log (x) \, dx = ConditionalExpression[(x+1) (log(x+1)-1)+1,Re(x)>=-1[Or]x[NotElement][DoubleStruckCapitalR]]$$

This is what Mathematica says when you have x+1 in your upper integration limit.

Knight

What does it mean?

I have never used Mathematica, just heard about it

Mats Granvik

ConditionalExpression[1 + (1 + x) (-1 + Log[1 + x]), Re[x] >= -1 || x \[NotElement] Reals]

|| is the word Or

The latex got messed up, but a conditional answer to the integral anyways.

Re[x]= Real part of x

x does not belong to the reals.

Knight

Okay

feynhat

15:52

@Knight Write $1 + \frac{k^2}{n^2}$ as $\left( 1 + \frac{k}{n} \right)^2 - \frac{2k}{n}$.

Knight

@feynhat I tried but it didn’t help

feynhat

ok

user434058

Do you guys add "A Geometric Approach" after everything? : A Geometric Approach

user434058

I think I should've written "A Recursive Approach"...

feynhat

16:16

@LeakyNun How do we know that any discrete subgroup of $\Bbb R$ is isomorphic to $\Bbb Z$?

Leaky Nun

@feynhat the first positive element is a generator

feynhat

first?

How do we know there is a least positive element?

Can they come very close to 0, like 1/n.

They can't because discrete.

But still is there a way to argue that there is a least positive element?

@LeakyNun

Thorgott

can't happen, because they contain $0$

so if they get arbitrarily close to $0$, they contain a limit point

by translation you can obtain that any such non-discrete subgroup is in fact dense

feynhat

@Thorgott Neat.

nbro

16:37

@Thorgott Are you familiar with stochastic approximation theory?

robjohn

@Knight $\int_0^x\log\left(1+t^2\right)\mathrm{d}t$

user736948

What is in general meant by a generator of a free group? For example here: "Given a set S, there is a function i: S → F(S), known as the canonical inclusion, sending each
element of S to the corresponding generator of F(S)." .

Tobias Kildetoft

@user736948 Are you familiar with how the free group is defined?

user736948

Set of equivalence classes of words with letters in S. Group operation being concatenation.

Tobias Kildetoft

Right, and those letters are the generators

abhas_RewCie

16:44

Hei, does anyone here knows Solomonoff Induction?

Solomonoff Induction - Why that is even called Induction?

user736948

Oh!

abhas_RewCie

Mathematically speaking, Induction is defined as a method which proves a statement is true for all Natural Numbers.

But, Solomonoff thing does completely different...

Tobias Kildetoft

@abhas_RewCie No, induction does not need to be about natural numbers

user736948

So the canonical inclusion mentioned in my previous post just sends each letter to it's corresponding equivalence class

abhas_RewCie

@TobiasKildetoft Why?

user736948

16:47

To be precise... the generators are the equivalence classes of each letter, no? @TobiasKildetoft

Thorgott

no, I'm not

abhas_RewCie

@TobiasKildetoft tutorialspoint.com/discrete_mathematics/…

...?

Tobias Kildetoft

@user736948 Right, though it is usually not very frtuiful to think in terms of equivalence classes for free groups

@abhas_RewCie Just because some tutorial site claims it does not make it true

induction is a technique that applies to any well-ordered set

user736948

@TobiasKildetoft Thanks! That clarifies things for me.

abhas_RewCie

@TobiasKildetoft means?

Tobias Kildetoft

16:49

@user736948 It is easier for most purposes to think in terms of reduced words

@abhas_RewCie Not sure what you mean.

Since each equivalence class contains a unique reduced word

abhas_RewCie

@TobiasKildetoft You told that induction isn't for Natural Numbers, it can be for other sets too? Is that what you told na?

Tobias Kildetoft

Yes, it can be used for any set with a suitable relation

(a bit of extra care needs to be taken though)

abhas_RewCie

@TobiasKildetoft Okay... So, Induction is a technique for proving statements? right.?

Tobias Kildetoft

right, usually

abhas_RewCie

@TobiasKildetoft usually? means? I thought it's always...

Tobias Kildetoft

16:52

Well, as a mathematician I am loath to make absolute statements

Thorgott

I mean, there is no categorical definition for "induction", is there?

Tobias Kildetoft

Anyway, I wrote a blog post about the general version of induction here math.blogoverflow.com/2015/03/10/when-can-we-do-induction back when MSE had a blog

abhas_RewCie

@TobiasKildetoft Then, what Solomonoff Induction even does? It just assigns probability distribution to functions, it isn't any close to induction.

Tobias Kildetoft

@Thorgott Not one that I know at least (and if there were, then probably it would be the other sort of induction anyway)

@abhas_RewCie Ahh, that is probably a case of the other sort of induction then, where the word derives from "inducing"

so you induce from something to something else

usually it means to expand the range of applicability of something to something larger, but often not in a way where you can just restrict to get back the original

abhas_RewCie

@TobiasKildetoft Totally didn't get it. Can you please explain with an example or references please?

Tobias Kildetoft

16:55

I have no idea what the specific example you mention is, so unfortunately not

abhas_RewCie

@TobiasKildetoft references?

Tobias Kildetoft

not for your example, no. And other references might be too far from the topic to be useful

abhas_RewCie

Hmm... Okay. Thanks :-)

Ah... some info here - Algorithmic Probability

abhas_RewCie

17:21

Elon Musk is gone crazy on Twitter.

Oh My god. What has happened to him...

Mats Granvik

17:38

@TedShifrin Did you see my latest question?

abhas_RewCie

17:56

@abhas_RewCie Bye

Thorgott

If $U\subseteq S^1$ is open and $U=-U$, then $U^2$ is open in $S^1$

Leaky Nun

@Thorgott so you used both additive and multiplicative notation

Thorgott

each to be interpreted pointwise

geocalc33

18:23

anyone ever come across the hadamard product of matrices?

Hadamard product (matrices)

In mathematics, the Hadamard product (also known as the element-wise, entrywise or Schur product) is a binary operation that takes two matrices of the same dimensions and produces another matrix of the same dimension as the operands where each element i, j is the product of elements i, j of the original two matrices. It should not be confused with the more common matrix product. It is attributed to, and named after, either French mathematician Jacques Hadamard or German mathematician Issai Schur. The Hadamard product is associative and distributive. Unlike the matrix product, it is also commutative...

geocalc33

18:40

anyone know how to fibrate something?

2

feynhat

@Thorgott $z \mapsto z^n$ is an open map on $S^1$, no?

Thorgott

no, that's the same as saying $z\mapsto\sqrt[n]{z}$ is continuous, which fails if you move around the circle once

or wait, it's not the same as saying that

feynhat

what does $\sqrt[n]{z}$ even mean?

Thorgott

say, $e^{i\theta}\mapsto e^{i\theta/n}$

geocalc33

who is writing nLab? they are so advanced. (nvm it's an organization so multiple folks are probably writing for it)

Thorgott

18:54

but if you look at where a small open strip of the circle around $1$ is mapped, the image isn't open

feynhat

@Thorgott I can't see this. If I take an open arc about 1 and square it, would I get another longer open arc?

Ted Shifrin

@feynhat: And fatter.

geocalc33

does anyone want to do the pre-calculus AND can you even do precalc on manifolds?

Ted Shifrin

Yes, $z\mapsto z^n$ is an open map. It's a local diffeomorphism.

feynhat

@TedShifrin fatter? We are in S^1. There is no width.

Thorgott

19:02

oh, I'm being stupid

Ted Shifrin

Thorgott was talking about strips around the unit circle. Sorry.

Astyx

hi

Ted Shifrin

I shouldn't jump into middles of conversations.

Salut @Astyx

geocalc33

how do you fatten something?

like that's exactly what I need for the fibrations.

hi astyx.

Thorgott

not that that matters to it being an open map $S^1\rightarrow S^1$, but is it a local diffeomorphism around $0$ too? @TedShifrin

Ted Shifrin

19:05

No, of course not.

geocalc33

hi thorgott

Ted Shifrin

It is if you remove $0$, but not in an actual neighborhood of $0$.

$0$ is a branch point, as the $n$ sheets all meet there.

feynhat

If we think of $S^1$ as $\Bbb R / \Bbb Z$. The map $t \mapsto nt$ induces $z \mapsto z^n$. I start with an open set in $S^1$ and lift it to $\Bbb R$ (it is open because quotient), $t \mapsto nt$ is open. Then I project back to $S^1$ (this map is open, right?).

geocalc33

@EnjoysMath what?

I have a prophecy that you will solve the twin primes conjecture in less than 5 years

feynhat

* Maybe we need $t \mapsto 2\pi nt$ on $\Bbb R$.

EnjoysMath

19:12

@geocalc33 I don't think so. It's probably impossible to solve :)

geocalc33

yeah I know. I dropped the glass orb afterwards and it caused you not to :(

Balarka Sen

Covering maps are open maps, @feynhat

feynhat

Oh. Right.

So, we're done.

geocalc33

foliation on the punctured plane $X=\Bbb R/\{0\}.$ S Say you fatten each $y_k=k/x$ for $k \in \Bbb R$. You fatten the $y_k.$ To fatten them you have to embedd into $\Bbb R^3$ correct?

so you could define a radius now on the $y_k.$ in the following manner. $y_k=x$

so slice the hyperbola with $f(x)=x$

and then revolve the $y_k$ around the fixed points i suppose.

then you get these tori

that are infinitely long basically.

Alessandro Codenotti

19:34

Hi @Balarka

I listened to the Ulver album, but it didn't fully convince me

Astyx

Hi @ShaVuklia

feynhat

Night @BalarkaSen @TedShifrin @Thorgott @AlessandroCodenotti

Edward Evans

@Alessandro not even Eos?

Alessandro Codenotti

Hi Feynhat

Sha Vuklia

olas @Astyx

Astyx

19:35

How are you ?

Balarka Sen

@AlessandroCodenotti Aw

Night @feynhat

Thorgott

night

Sha Vuklia

@Astyx I'm doing fine. Had a stressful week in fact due to deadlines 'n stuff, but today is pretty relaxed. You?

Astyx

Cool ! Pretty much the same for me, my last exam was today

Alessandro Codenotti

@EdwardEvans That's good, the whole album is really, it just didn't convince completely for some reason

Edward Evans

19:37

Fair one lol

Alessandro Codenotti

Might just be because I'm enjoying heavier stuff lately

Sha Vuklia

O, nice

Balarka Sen

I discovered some very heavy Opeth

Alessandro Codenotti

Opeth are great

Balarka Sen

it's an old album called "My Arms, Your Hearse"

Alessandro Codenotti

19:38

Do you guys know Rolo Tomassi?

Balarka Sen

who's he

Alessandro Codenotti

They're a british band. They make a very heavy and experimental mix of post hardcore and mathcore

Balarka Sen

ah ok

Edward Evans

Don't know them lol

Astyx

I've been listening to polyphia recently

Edward Evans

19:41

sounds like a parody of animals as leaders

Alessandro Codenotti

Try this

Edward Evans

listening

Astyx

Polyphia's very similar to animals as leaders

Edward Evans

yeah I'm not a big fan of Polyphia

Balarka Sen

I never listened to AAL

Alessandro Codenotti

19:42

@EdwardEvans Nah I would say that AAL make "modern" instrumental progressive metal

They're pretty good

Edward Evans

@Alessandro I was just making a dumb joke about the name Rolo Tomassi

Alessandro Codenotti

I heard them live once

Edward Evans

They were e x c e l l e n t live

Alessandro Codenotti

Oh ok

Yeah they're incredible musicians

Edward Evans

I spent about 7 years trying to play the start of Tempting Time

Alessandro Codenotti

19:43

And they had Plini and Intervals opening who were both incredible as well

Edward Evans

yeah we must've seen them on the same tour

Balarka Sen

Plini is great

Astyx

Tosin Abasi - Rolo Tomassi

Edward Evans

aye hahaha

Alessandro Codenotti

@EdwardEvans Oh the first song or two might trick you

Edward Evans

19:45

hahahaha I just hit Aftermath

and I'm like

iS tHiS AlEkSaNdRa DjElMaSh

Balarka Sen

i have been listening to classic death metal lmao

shade empire and stuff

Astyx

You have 9 string guitar ?

Balarka Sen

very classic corn

Edward Evans

I have an 8 string lol

Alessandro Codenotti

You have to wait until she starts screaming

@EdwardEvans A true djentleman

Edward Evans

19:46

death metal aint kvlt envvgh

Balarka Sen

hot take

Alessandro Codenotti

I only have a 5 strings bass :C

It's a fretless one tho

Edward Evans

bassist who does analysis, anything else you wanna do to destroy yourself?

2

Balarka Sen

lmao

Edward Evans

hahaha jkjk

Alessandro Codenotti

19:47

lol

@EdwardEvans which one?

Edward Evans

wait which what

which song am I on?

Alessandro Codenotti

which guitar

Like which model

Edward Evans

ohhh

Alessandro Codenotti

I can't English

Edward Evans

I have a Schecter Omen 8

fairly entry level 8 string but I'm a bedroom guitarist

Balarka Sen

19:49

@EdwardEvans i like how these old power-influenced death metal songs all start with some epic keyboard intro, and ends with some techno shit happenins

its the same structure lolol

Alessandro Codenotti

Schecter basses are pretty good so I assume their guitars are too

@EdwardEvans Anyway you might notice a slight change in Rituals (the third song)

Edward Evans

Yeah this is fokken cool

Alessandro Codenotti

(also note that they only have one singer, she is doing both the clean and the scream)

Edward Evans

@Balarka I don't really know much death metal, I just hate on it because Varg killed Jeremy Epstein

Yeah it's great

Balarka Sen

lmfao

"Death" is a good death metal band (loool)

honest

Edward Evans

19:51

ohh yeah I know Death

and Morbid Angel?

Balarka Sen

havent listened to much by them

Edward Evans

@Alessandro really loving this album man

Alessandro Codenotti

I'm glad, it's great

Mike Miller

@BalarkaSen mathoverflow.net/a/56458/40804

Balarka Sen

Yeah I have seen this argument before

Mike Miller

20:05

New to me is all

Wondering if there's a geometric argument that $[\Bbb{RP}^n, \Bbb{RP}^n] = \Bbb Z/\chi(\Bbb{RP}^n)$, determined by degree

geocalc33

20:18

I answered a question on math stack exchange!

My day is done

but I don't understand the question very well so maybe I should not have answered it

$\displaystyle \inf_{f \in \bar{C}} \sup_{x\in [0,1]} \frac{x f(x)}{\int_0^1 f(t) dt}$

what does that exactly mean

you want the max f for the max x right?

infimum (supremum)

Balarka Sen

@MikeMiller Is that what you meant? If $n = 2$ the right hand side of your equation is $\Bbb Z/1$, the trivial group, which is not right.

Given a map $f : \Bbb{RP}^n \to \Bbb{RP}^n$, restrict to $\Bbb{RP}^{n-1}$. We know $[\Bbb{RP}^{n-1}, \Bbb{RP}^n] = [\Bbb{RP}^{n-1}, \Bbb{RP}^\infty] = H^1(\Bbb{RP}^{n-1}, \Bbb Z_2) = \Bbb Z_2$, so there are exactly two such maps upto homotopy, standard inclusion or collapse. By homotopy extension property we can homotopy $f$ so that $f|\Bbb{RP}^{n-1}$ is either standard inclusion or collapse as well.

If $f|\Bbb{RP}^{n-1}$ is a collapse, then $f$ gives rise to a class in $[S^n, \Bbb{RP}^n] = \pi_n \Bbb{RP}^n = \Bbb Z$.

If $f|\Bbb{RP}^{n-1}$ is the standard inclusion, then $f$ is a map of pairs $(\Bbb{RP}^n, \Bbb{RP}^{n-1}) \to (\Bbb{RP}^n, \Bbb{RP}^{n-1})$ which is identity on $\Bbb{RP}^{n-1}$, so lifts to a map $(D^n, \partial D^n) \to (D^n, \partial D^n)$ which is identity on $\partial D^n$. Such a map has to be relatively homotopic to the identity, right? By coning it off.

Mike Miller

20:37

I meant Z/2 or Z, lol. My bad

@BalarkaSen In the first case you have to show which of those maps are homotopic as maps on RP^n. I'm a little confused about the second case though.

Balarka Sen

Right, so I think all that needs to be understood is the kernel of $[S^n, \Bbb{RP}^n] \to [\Bbb{RP}^n, \Bbb{RP}^n]$ given by precomposing with $\Bbb{RP}^n \to \Bbb{RP}^n/\Bbb{RP}^{n-1}$

Unsure how to go about doing that.

Mike Miller

I am really confused about the second case though! For n odd there should be Z many

These should be the maps of odd degree

Balarka Sen

I think the kernel of the map above is trivial or $2\Bbb Z$ depending on dimension? The second case doesn't matter it seems, all it contains is identity - if that's the case it doesn't contradict what you said.

Mike Miller

What you get seems to be that it lifts to a map which is +-1 on the boundary, but whatever, not a big difference

Balarka Sen

Yeah, I meant to correct myself on that

Should i use the Puppe sequence to compute the kernel of $[\Bbb{RP}^n/\Bbb{RP}^{n-1}, \Bbb{RP}^n] \to [\Bbb{RP}^n, \Bbb{RP}^n]$? Guess so

Mike Miller

20:51

No I think there's something wrong with the second case.

The however many fold suspension of z^k induces a map of S^n of degree k, which descends to a selfmap of RP^n of degree k

Balarka Sen

What is degree if $n$ is even

mod 2 degree I guess

Victor

Hi, Any person can give any suggestion on this one?math.stackexchange.com/questions/3654195/…

Simple

21:14

in Calculus and analysis, yesterday, by Simple

Let $\Omega$ be an open subset of $\mathbb{C}$ and let $T \subset\Omega$ be a triangle whose interior is also contained in $\Omega$. Suppose that $f$ is a function holomorphic in $\Omega$ except possibly at a point $w$ inside $T$. Prove that if $f$ is bounded near $w$, then $\int_{T}f(z)dz=0$

Let $D_{\epsilon}(w)$ be an open disk of radius $\epsilon$ centered at $w$. There exists a $\epsilon_0>\epsilon>0$ such that the closure of $D_{\epsilon}(w)$ of $\mathbb{C}$ in $\Omega$. For any triangle $T$ in $\Omega$, let $T_{\epsilon}$ be the contour which is the union of $T\setminus\,D_{\epsilon}(w)$ and the boundary of $D_{\epsilon}(w)$ that is inside in the triangle $T$ is enclosed by $T$. Since $f$ is holomorphic in $T\setminus\,D_{\epsilon}(w)$, by Cauchy's Theorem, we have
$$\int_{T_{\epsilon}}f(z)dz=0$$

any comments are appreciated

Alessandro Codenotti

Still here? @Balarka

Balarka Sen

ye

Alessandro Codenotti

Do you want to listen to some GGT stuff and tell me if it sounds sensible? I'm getting very confused by some stuff and I think explaining it to someone would be helpful

Rubber duck debugging

Balarka Sen

Feel free to rant on garbology, I'll listen in from time to time. No guarantees I can help though

Mike Miller

@BalarkaSen Sorry, when n is even I bet I agree there's only the identity in the second case but I'm not sure how to prove it

But when n is odd, the odd degree maps descend to odd degree maps of RP^n, which must induce 1 on pi_1

Balarka Sen

21:19

Oh OK

Whereas my first class of maps all induce 0 on pi_1

So they can't come from the first class

Victor

Hi, Anyone can offer any help on this? (1/t^3)'=100t'', May you find all t constant that satisfy this equation?

Mike Miller

Yeah

Balarka Sen

Annoying. How do you compute these things generally?

Mike Miller

I don't really know. I was hoping that in this simple case we could work just a little harder than Hopf's theorem.

Ted Shifrin

@Victor What variable are you taking derivatives with respect to? And what do you mean by $t$ constant?

Mike Miller

21:26

@BalarkaSen Oh, here's the point. You can get a map $(D^n, \partial D^n) \to \Bbb{RP}^n$ which is the double cover of the usual inclusion on the boundary. But why should you be able to lift it up to the disc as codomain?

Balarka Sen

Oh ok no

Let's see. A map $\Bbb{RP}^n \to \Bbb{RP}^n$ which is identity on $\Bbb{RP}^{n-1}$ lifts to a map $(S^n, S^{n-1}) \to (S^n, S^{n-1})$ by map lifting lemma

On the $S^{n-1}$ factor it definitely has to be identity on antipodal though

vesii

21:48

Hello, I can't think of an example of a singular matrix $M$ that a unique has LU factorization. can someone suggest an example $3\times 3$?

Mike Miller

@BalarkaSen Yeah. And it's not hard to imagine those of any degree n

Balarka Sen

yeah it doesnt give me anything

how did you compute the Z/2, Z thing

it annoys me that i cannot do basic computations haha

Mike Miller

It's stated somewhere

I don't have any clean argument but there should be one

My idea is that if n is odd you can distinguish between all the different possible maps using the degree, while if n is even there should be some trick to cancel out the maps of even degree

Balarka Sen

it does seem that it's determined by degree if n is odd and degree mod 2 if n is even yeah

Mike Miller

Like pushing a point around an orientation reversing loop to change its sign and then cancelling them

Balarka Sen

22:03

Maybe try $n = 2$ like this? If $f : \Bbb{RP}^2 \to \Bbb{RP}^2$ is a smooth map, make $f$ transverse to $\Bbb{RP}^1$. The preimage is a bunch of closed curves in $\Bbb{RP}^2$ with Mobius normal bundle. These kind of guys upto cobordism in $\Bbb{RP}^2$ maybe determines the homotopy classes?

If you have a single Moebius loop collapsing the exterior gives a map to RP^2 for sure

Coalescing multiple Moebius loops might give trivial loops, in which case I expect the map to be trivial in the first place

This sounds finicky. I don't like it

Meh

BAYMAX

Hi

Am revisiting abstract algebra after a bit of time

any idea to this *) Let R be a ring and A,B are ideals of R such that AIB = (0)

I think I is also an ideal

Balarka Sen

@MikeMiller I think I see why, for example, the degree 2 map $S^n \to S^n$, when descended to $S^n \to \Bbb{RP}^n$ and precomposed by $\Bbb{RP}^n \to \Bbb{RP}^n/\Bbb{RP}^{n-1} = S^n$ to get $\Bbb{RP}^n \to \Bbb{RP}^n$, is always nullhomotopic.

It's because if you lift it up, you get $S^n \to S^n$, the map which folds the two hemispheres of the domain sphere to the upper hemisphere.

So I think the first case always gives me $\Bbb Z_2$'s worth of maps

Mike Miller

22:23

No, the first case includes all even degree maps in the case that n is odd

It'll be either Z in each case or 1 in each case

Balarka Sen

Oh, right.

I give up

Mike Miller

I'm really sure this is Hopfy

Balarka Sen

I have run out of ideas to calculate. Tell me if you find a way

Mike Miller

We agree that

user714630

23:19

@BAYMAX A and B can have nontrivially intersecting annihilators and you choose a nonzero element from there, call the singleton set I. Are you given additional hypotheses?

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