Mathematics - 2017-10-18 (page 1 of 2)

Simply Beautiful Art

00:02

@RGS Would you like the generalized version of my problem?

RGS

Sure

Simply Beautiful Art

Or rather, one generalization

RGS

but I am about to go off

Simply Beautiful Art

Oh, okay

Let $G_0(n)$ be how many steps it takes for $n$ to reach zero in my sequence.

Let a number be in base-b,1 SGF if it is of the form:
$G_0(x_k) + G_0(x_{k-1}) + \dots + G_0(x_0) + b + b + \dots + b + c$
where $x_k \ge x_{k-1} \ge \dots \ge x_0 \ge b$, and $x$ is in base-b,1 SGF.

For example, $8 = G_0(G_0(1)) + G_0(1)$ in base-1,1

The approach is similar. Start by writing in base 1, then go to base 2 and subtract 1, then base 3 and subtract 1, etc.

Let $G_1(n)$ be how many steps it takes for $n$ to reach zero in the above sequence...

And then it loops around to give $G_k(n)$

:-) All for the joy of creating absurdly large finite numbers.

$G_2(n)$ grows faster than the Goodstein sequence, which @AkivaWeinberger may find interesting.

RGS

way too sleepy to understand that

:P

Simply Beautiful Art

00:09

G'night!

RGS

gotta go to sleep

thanks :P

Simply Beautiful Art

Wait!

@RGS You should change the printing statement to " + "

It should have plus signs in your program.

Leaky Nun

I wonder if this probability is valid and already has a name: take sample space $\Omega = [0,1]$, and sigma algebra $\mathcal F = \mathcal P(\Omega)$. Take an enumeration of the rational numbers $\varphi: \Bbb N \to \Bbb Q$. Then, define $P : \mathcal F \to [0,1]$ as follows: $$P(E) = \sum_{n \in \Bbb N} \begin{cases} 2^{-n} & \varphi(n) \in E \\ 0 & \varphi(n) \notin E \end{cases}$$

I think this is uncountably additive

(cc @AlessandroCodenotti)

Leaky Nun

00:27

@Ted hi

Ted Shifrin

Rehi

I don't even know what uncountably additive means.

Simply Beautiful Art

@RGS In theory you could golf your code even further. Whenever you have something like ...+1 if ... you can shorten it to ...+1if .... You don't need a space between a number and an if/else statement. Also, can you use ternary expressions?

Kasmir Khaan

@TedShifrin Hello Ted :D

Leaky Nun

@TedShifrin usually probabilities are countably additive: if you have a countable set of disjoint events $E_n$, then $\displaystyle P\left(\bigcup E_n\right) = \sum P(E_n)$

this solves the usual problem of "the probability of each point is 0 so they can't add up to 1"

Meow Mix

00:44

hi @TedShifrin

user84215

03:22

@MartinSleziak Thanks for your invitation.

0celo7

04:54

Why wouldn't you ask that on physics stack exchange?

Secret

05:05

in The h Bar, 4 mins ago, by Secret

It may seemed that we are given 24 hours, but in reality, in order to have a healthy body, and combined with the transportation time, we technically only have 13-15 hours per day (including 7 hours of sleep)

There's an insane amount of time wasted on just traveling from A to B as part of the routine (We are not talking about travelling in terms of relaxing or sighseeing like those in trips)

Sure, with the advert of mobiles and other electronic devices, we can try to squeeze some of that unutilised time, but there are still a lot of jobs that only a desktop computer can do and thus all tha…

Silent

06:24

This says in the notes that topology generated by the basis $\cal C=\{[a,b): a<b; a,b\in \Bbb Q\}$ is not comparable to the countable complement topology on $\Bbb R$.

I think since any basis for the countable complement topology is $\Bbb R$ with countably many points removed, we can find $[a,b)$ with $a,b$ rationals such that $[a,b)$ is in that basis for the countable complement topology, so topology generated by $\cal C$ is finer than the countable complement topology. Am I right?

Alessandro Codenotti

Why is $[a,b)$ in the cocountable topology? Is its complement countable?

Silent

@AlessandroCodenotti I am not saying that $[a,b)$ is in the cocountable topology, I am saying that if $B$ is basis for cocountable topology and $x\in B$ then there exists $[a,b)$ such that $x\in [a,b) \subset B$.

Alessandro Codenotti

Your last sentence is very unclear, is $x$ a point or a set?

Silent

@AlessandroCodenotti Its a point.

Alessandro Codenotti

06:39

Elements of $B$ are sets though

Silent

@AlessandroCodenotti, if you have Munkres' Topology, i am using lemma 13.3 given on P81.

@AlessandroCodenotti How so? $B$ is a basis element for cocountable topology, an example of $B$ is $\Bbb R-\{0\}$, so elements of $B$ are reals.

Alessandro Codenotti

Oh, ok, you wrote $B$ is basis earlier so I thought $B$ to be a basis not an element of a basis

What you said earlier it's true then, but it's not helpful in establishing whether one of the two topologies is finer than the other

Silent

omg

Alessandro Codenotti

$\tau$ finer than $\eta$ means that all open sets in $\eta$ are open in $\tau$, right?

Silent

yes yes, that's how Munkres defines it.@AlessandroCodenotti

Shall i ask this on main site?

Alessandro Codenotti

06:49

I think you're close enough to work it out, earlier ypu concluded that the topology generated by $[a,b)$ is finer the cocountable one, if that were true all of the sets with countable complement should be open in the topology generated by $[a,b)$

Have you seen what a separable space is already?

Kirill

09:12

@KonformistLiberal hah :) I am fan of Mr. Knuth

Laska

09:56

How do I get "start Mathjax to work", please? I don't understand "drag to bookmark bar". There is nothing obvious on the chat page. Sorry novice here.

Kirill

there is something else there @Laska

save it to bookmarks

what does $m+1$ in $x_i^{(m+1)}$ mean in the Jacobi method description?

btw what is the start vector?

Laska

10:12

@Kirill: I right-clicked on link in math.ucla.edu/~robjohn/math/mathjax.html to "add it to bookmarks", but that only affected what was visible in that mathjax page. It doesn't affect what I see here. I am sure this is extremely obvious when you know what to do.

TastyRomeo

If it's on your bookmarks

Look at this page and then click on the bookmark

Laska

you mean Firefox bookmark? I click on bookmark there and a new window opens, but nothing in it.

TastyRomeo

You drag it to your bookmarks, don't click on it there.

Kirill

@Laska save what is said to be saved in your bookmarks, and click every time you enter this chat

hi @TastyRomeo! Any idea about that?

24 mins ago, by Kirill

what does $m+1$ in $x_i^{(m+1)}$ mean in the Jacobi method description?

Kirill

11:24

Some MatLab profi here?

Secret

12:10

[Random]

thatsmathematics.com/mathgen/…

Secret

13:04

Map of Computer Science

►

Leaky Nun

13:35

I went to a talk about finite fields and Weil conjectures

Balarka Sen

good shit

Leaky Nun

they spent 99% of the time on finite fields and 1% on the first conjecture

Balarka Sen

oh

anon

what was the venue / audience composition?

Leaky Nun

”undergraduate colloquium”

anon

13:37

if it's not a strong pure math undergrad community then I'd believe that

Leaky Nun

i would have learnt more by googling weil conjecture at home

@anon it’s imperial college for god sake

Balarka Sen

I have actually forgotten the main statement of Weil conjectures. I think one of them says that the zeta function of a F_p-variety is rational?

Which follows from the Lefschetz fixed point theorem of the etale or l-adic or whatever the heck cohomology theory, if memory serves

Leaky Nun

@BalarkaSen yes that one

@BalarkaSen they never managed to prove anything

Balarka Sen

I don't expect them to

Leaky Nun

they mentioned galois though

Balarka Sen

13:42

I went to a graduate level student seminar on Weil conjectures; they just explained the etale cohomology theory axiomatically and how the Weil conjectures follow from that machinery, but I suspect it's very hard to construct the etale cohomology theory.

Leaky Nun

they didn’t mention those

they literally stated the first conjecture

and then dismissed us

Balarka Sen

Well this stuff is nontrivially hard, so

BAYMAX

@Semiclassical perhaps you might be knowing FORTRAN?

Alessandro Codenotti

hi chat

Balarka Sen

Hi @Alessandro

ÍgjøgnumMeg

13:52

@Alessandro Yo

Balarka Sen

I have had too much of ODEs today

Alessandro Codenotti

Any positive amount of ODEs is too much ODEs in my opinion

ÍgjøgnumMeg

^

Balarka Sen

lol

it's good ass stuff

Alessandro Codenotti

The only thing I know about ODEs is Picard-Lindelöf

Anyway back to the only allowed topic: algebra

Balarka Sen

13:57

Picard-Lindelof is useful. It's also important that solutions of C^k ODE's depend C^(k-1)-ly on the initial conditions

Alessandro Codenotti

I want to construct an example of an irreducible but not separable polynomial, I know I need to work over an infinite field of positive characteristic

Balarka Sen

Ah I asked this to Mathei a few days ago

Alessandro Codenotti

Which means $\Bbb F_p(X)$ because that seems nicer than $\overline{\Bbb F_p}$

Really? I must have missed it

Balarka Sen

$X^p - t^p$ in $\Bbb F_p(t)[X]$ works I think

Uh, no, try $X^p - t$

That splits as $(X - \sqrt[p]{t})^p \in \overline{\Bbb F_p(t)}[X]$, so you're rekt

Alessandro Codenotti

Nice one

Bysshed

14:03

does anyone know how to render the mathjax in chat on chrome on android? I tried this approach, tinyurl.com/cfqcvpc, but with no success.

Alessandro Codenotti

This example has convinced me that in all commonly encountered cases separability is free

Balarka Sen

Like Mathei and I discussed, separability is about having Hausdorff fibers of covering spaces (tee-hee)

Which is pretty generic

Alessandro Codenotti

I know very little about covering spaces, mostly the definition, but it looks like you need to have seriously ugly spaces for that to fail

Balarka Sen

Oh I mean, fibers of a covering space are discrete spaces.

It's just than in field theory it's sort of like a generalization where the fibers could be non-Hausdorff

That's what nonseparability brings us

Alessandro Codenotti

Oh, wait, fibers over a point

I was thinking about the $p^{-1}(U)$ part being covering spaces, but of course $p^{-1}(\{x\})$ is discrete

Balarka Sen

14:08

ah yes that is what i meant

Leaky Nun

14 hours ago, by Leaky Nun

I wonder if this probability is valid and already has a name: take sample space $\Omega = [0,1]$, and sigma algebra $\mathcal F = \mathcal P(\Omega)$. Take an enumeration of the rational numbers $\varphi: \Bbb N \to \Bbb Q$. Then, define $P : \mathcal F \to [0,1]$ as follows: $$P(E) = \sum_{n \in \Bbb N} \begin{cases} 2^{-n} & \varphi(n) \in E \\ 0 & \varphi(n) \notin E \end{cases}$$

@AlessandroCodenotti did you read this?

Alessandro Codenotti

nope

Leaky Nun

could you read this?

Balarka Sen

no

Leaky Nun

ok

Balarka Sen

14:11

no

Alessandro Codenotti

hmmm it looks like a proper measure to me

Leaky Nun

@AlessandroCodenotti thanks

I wonder if it is uncountably additive

Balarka Sen

no

Alessandro Codenotti

If you pick uncountably many disjoint sets in $[0,1]$ all but countably many will have measure $0$

Leaky Nun

@AlessandroCodenotti so it reduces to countable additivity, which means the answer is yes

mercio

14:13

it looks uncountably additive to me

Leaky Nun

@mercio thanks

mercio

also it reminds me of (I think ?) the devil staircase

Leaky Nun

didn't really get inspiration from the devil's staircase

I got the inspiration from a strictly increasing function that is discontinuous at every rational number

$$f(x) = \sum_{q \in \Bbb Q \cap (-\infty,x)} 2^{-\varphi^{-1}(q)}$$

mercio

yeah

Leaky Nun

strictly

this is like impossible lol

Balarka Sen

14:17

functions can be bad

Leaky Nun

anyone who thinks they can imagine the rational points inside the real line cannot.

Balarka Sen

bad functions are not necessarily fun

Leaky Nun

but functions

Semiclassical

@BAYMAX nooooope

Balarka Sen

functors are fun

BAYMAX

14:24

ok

Any MSE site I can ask about this

unfortunately Computational science SE chat not so much active

0celo7

@BAYMAX what are you trying to do?

Secret

These two colors are different, but they feel the same

0celo7

@Secret colors in pictures are never what they seem.

Secret

Yeah, thus sometimes when I pick out colors from my photos, I tried to use the sliders to recreate the color I actually saw in the real scene as close as possible. It works ok for most colors except near the violet side of the spectrum (which is beyond the range of RGB galunets)

So, in the above pics, that pale greenish looking tile as seen from my computer screen and my eyes is the closest color I saw in the real beach

(But the tile will look very different in color on another computer due to resolutions and its color range)

It will be great if there exists some kind of linear map that maps color representations from one computer to another, so that when transferring photos from one computer to another, they will all look the same to you regardless of the computer display

Krijn

14:49

Heyo

Balarka Sen

Hi @Krijn

Krijn

What's up mathematically Balarka

And, in real life, too

Balarka Sen

Learning full blown dynamics of ODEs

Unfortunately nothing ever is up in real life, only down :)

bolbteppa

Why is $\mathrm{SL}(2,\mathbb{R})$ a torus topologically?

Balarka Sen

It is not. $\pi_1(\text{SL}(2, \Bbb R)) \cong \Bbb Z$; the torus has fundamental group $\Bbb Z^2$.

bolbteppa

15:02

Wait, "from the topological point of view it is homeomorphic to the inside of a torus"

Is that better?

Balarka Sen

Yeah that's a lot better. That should be true.

bolbteppa

haha

Balarka Sen

So, yeah, you can see this as follows

$\text{PSL}(2, \Bbb R)$ acts on the hyperbolic plane $\Bbb H^2$ by isometries

bolbteppa

So $\mathrm{PSL}(2,\mathbb{R})$ is apparently the group of matrices such that $x' = \begin{bmatrix} a & b \\ c & d \end{bmatrix}(x) = \frac{ax+b}{cx+d}$ right?

Balarka Sen

In fact this action extends to the unit tangent bundle $T_1 \Bbb H^2$, defined by $g(p, v) = (gp, dg_p v)$ where $g$ is the isometry of $\Bbb H^2$ corresponding to the matrix in PSL_2(R).

@bolbteppa Well, no. SL_2(R) is the group of such matrices with determinant 1.

anon

15:05

That action (on the complex plane) is true for all matrices in GL(2,R), but PSL(2,R) is what the action factors through

bolbteppa

Oh, $\mathrm{PSL}(2,\mathbb{R})$ is that action up to scalar multiples of the numerator and denominator (which cancel) right?

anon

PSL(2,R) is a group, not an action

It's SL(2,R) mod +/-Id

And SL(2,R) is the subgroup of GL(2,R) of matrices with det 1

(as blarka said)

I am saying blarka deliberately because it's fun

Balarka Sen

I like that

I should change my name to blarka

@anon Right, the point is, that action restricted to $\text{PSL}(2, \Bbb R)$ is actually free + transitive

(I am sure you know all this)

anon

yep

and KAN is a beautiful way to prove it

well, not free on H

it's regular on the unit tangent bundle of H

AN acts regularly on H, and K=Stab(i) acts regularly on i's unit tangent space

Balarka Sen

Of course, I agree. It's free+transitive on T_1H

@anon ah interesting

bolbteppa

15:11

Okay that's making more sense, so $\mathrm{PSL}(2,\mathbb{R})$ is a group whose action is to produce two lines $(ax+b,cx+d)$, such that $(\lambda(ax+b),\lambda (cx+d)) = (ax+b,cx+d)$,

Balarka Sen

@bolbteppa Er? It acts by fractional linear transformations on the upper half plane

bolbteppa

"to each operation in this group correspond two real linear substitutions in two variables. The homographic group ($\mathrm{PSL}(2,\mathbb{R})$), therefore, has one-valued and two-valued representations, but no multivalued representations of order higher than two. "

Krijn

@BalarkaSen :(

bolbteppa

Very confusing

Balarka Sen

@anon I guess the way I would see the action is transitive is as follows. Take $(p, v)$ and $(q, w)$ on $T_1 \Bbb H^2$, and join $p, q \in \Bbb H^2$ by a geodesic. Translation from $p$ to $q$ by this geodesic, composed with a hyperbolic rotation takes $(p, v)$ to $(q, v')$ to $(q, w)$. That establishes transitiveness.

That's the boring proof though :)

(Freeness is easy)

@bolbteppa In any case, if you're prepared to believe all this, then by orbit-stabilizer, $\text{PSL}_2(\Bbb R)$ is diffeomorphic to the unit tangent bundle $T_1 \Bbb H^2$. But that's a circle bundle on a disk, so $S^1 \times D^2$ is the topology.

$\text{SL}_2(\Bbb R)$ is just a double cover of this, so has the same topology.

bolbteppa

15:19

I wish I could understand it :p

Balarka Sen

I suspect there are simpler proofs using just group decompositions. KAN is perhaps one way

bolbteppa

Seems like, using group actions, you can say it in an easier way, but then justifying it, or just generalizing to groups, uses that kind of language

KAN is another thing I've seen yeah

Balarka Sen

Actions let you identify abstract objects like groups to topological objects on which it acts, that's all, really

anon

KAN is very simple. you use A to radiate from the origin, which gets a point to the height you want, and N translates left and right. uniqueness is obvious. you can use K to rotate a tangent vector around i to begin with.

Balarka Sen

Right, and A and N are both 1-parameter subgroups there, so you end up with R x R x S^1 as topology

bolbteppa