Mathematics - 2018-05-11 (page 3 of 5)

7:00 PM

@BalarkaSen is there an abstract argument why there is no continuous injection $T^2 \to S^2$? I have an argument, but I don't like it.

B. Mehta

ah the sum should start from $n=1$, my bad

Balarka Sen

@MatheinBoulomenos So you want something easier than yelling "invariance of domain"?

ah that works

LOL

thanks B

7:04 PM

Heh, I'm able to prove it by contradiction by examining every possible case, which will do, I suppose. Though I wonder if there's a nicer/algebraic proof.

Balarka Sen

@anonra GFaux was giving you an easier one.

Hm, or was he?

anonra

Didn't lead us anywhere unfortunately.

GFauxPas

my clue isnt a good clue

i mean it's easy to show $\min(x,y) \le a + b$ but that's not enough

B. Mehta

can't you argue $\min(x,y) \leq x \leq a$ and $\min(x,y) \leq y \leq b$, so $\min(x,y) \leq \min(a,b)$

1010011010

Hi all, any authorities in the chat on conformal stereographic projections and/or higher dimensional integration? (for the lack of a better wording)

Balarka Sen

7:07 PM

The easiest idea is to order {x, y} and {a, b} without loss of generality. :P

GFauxPas

har har

Balarka Sen

@B.Mehta Yeah that works.

GFauxPas

isn't that exactly what we're trying to prove though thonk

B. Mehta

@BalarkaSen can you argue that $T^2$ isn't simply connected while $S^2$ is, so there's no continuous injection?

Balarka Sen

@B.Mehta There is a continuous injection S^1 --> S^2.

B. Mehta

7:09 PM

since that would induce an injective group hom from $\mathbb{Z}^2 \to \{e\}$ which fails

MatheinBoulomenos

every manifolds injects into $\Bbb R^n$

GFauxPas

ooh

Balarka Sen

Simple connectedness has nothing to do with being able to embed stuff

GFauxPas

oh, right

there you go @anonra

Mehta what's the question about $T^2$ and $S^2$?

Balarka Sen

The point is injective continuous maps do not induce injective homomorphisms on fundamental groups.

B. Mehta

7:11 PM

Ah, I was just about to ask that

Right, I need to revise my algebraic topology...

anonra

Quite a clever idea. Thank you all! :)

GFauxPas

that there's no continuous injection?

B. Mehta

@GFauxPas Yup!

Although it seems like asking someone else about that might be a better idea

MatheinBoulomenos

Balarka already answered it

Balarka Sen

Unanswered it, rather :3

GFauxPas

7:15 PM

$T^2 = e_0 + e_2 + e_2 + e_4$ is that right?

maybe you can make a homology argument. but the induced maps on the homology doesnt need to be injective either...

wait no that's too big

Akiva Weinberger

I'm back

Did you solve the thing?

Balarka Sen

I gave you a different problem

Fibonacci stuff

Akiva Weinberger

35 mins ago, by Balarka Sen

@AkivaWeinberger Here's another one I couldn't figure out in a reasonable amount of time. Consider this Collatz-like algorithm: $T(n) = n/2$ if $n$ is even, $T(n) = n+1$ if $n$ is odd. (a) Prove that there is some $k$ such that $T^k(n) = 1$ for all $n$ (b) Denote $c_k = \#\{n \in \Bbb N : T^k(n) = 1\}$. Prove that $c_k$ is a Fibonacci sequence.

This?^

Looking at it now

Balarka Sen

Ye

(b) is the bit I can't do

Semiclassical

Maybe think in terms of binary sequences?

Akiva Weinberger

7:21 PM

So how many steps does 11101101 base 2 take, for example

Balarka Sen

36 mins ago, by Balarka Sen

Think about binary expansions

Semiclassical

lol

Balarka Sen

32 mins ago, by Balarka Sen

Hm, if the binary expansion of $n$ is, read from right-to-left, 0...(a_1 times)...01...(b_1 times)...10...(a_2 times)...01...(b_2 times)...1..., then after $\sum (a_k + b_k) + \ell$ moves, where $\ell$ is length of the binary expansion, I think the algorithm terminates.

rekt my dudes

Semiclassical

The fact that it's n->n+1 not n->n-1 is annoying

Akiva Weinberger

11?

Semiclassical

7:23 PM

hmm, now I'm curious what the backwards tree looks like

Balarka Sen

Oh should be $\ell - 1$.

B. Mehta

I get 11 for that too, @AkivaWeinberger

for a) you can argue the number of 1s in the binary expansion never increases

Balarka Sen

Or that length of the binary number has to decrease every 3 steps.

Only time it increases is where there is a sequence of 1's, after which it becomes 10...0

B. Mehta

oh that's nice

Balarka Sen

blah

Yes.

100 -> 10

That's also nice

11101101 -> 11101110 -> 1110111 -> 1111000 -(3)-> 111 -> 1000 -(3)-> 1

B. Mehta

7:28 PM

does $c_k$ count the $n$s which take exactly $k$ steps or just hit 1 at $k$ steps

Balarka Sen

Seems to be the latter. I was confused about that a bit.

B. Mehta

you added an extra 0 in your third step

Oh

1110111 -> 1111000

rekt me

7:29 PM

my feeling is that one should be able to map this onto the n->n-1 version of the problem

Balarka Sen

11 it is

Semiclassical

and once one does that, then the binary ops are simpler

you just remove a zero from the end or turn a one into a zero at the end

of course, what one really needs to do regardless is argue that $c_k=c_{k-1}+c_{k-2}$

Balarka Sen

Well all know you're an orthogonal personality

Akiva Weinberger

2n and 2n-1 take n's steps plus 1

Semiclassical

if by that you mean i am consistently right

then cool

Akiva Weinberger

7:32 PM

Thus Fibonacci

Balarka Sen

@AkivaWeinberger confused gestures

Semiclassical

which in turn suggests doing something like c4=c3+c2 as a simple case

Balarka Sen

Oh you're using that fuckass recurrence

Akiva Weinberger

Oh wait I meant 2n is n's steps plus 1, and 2n-1 is n's steps plus 2

Semiclassical

also

Akiva Weinberger

7:34 PM

Still: this Fibonacci

Semiclassical

0 in N ?

Akiva Weinberger

Oh here's an idea

Wait I need to go

Balarka Sen

I need to sleep

B. Mehta

suspense

Balarka Sen

@AkivaWeinberger Ah yes

Clearly

Akiva Weinberger

7:36 PM

But each number has a unique binary expansion where, instead of using the digits 0 and 1, you use the digits -1 and 1

Right?

Balarka Sen

Ye

Akiva Weinberger

Or maybe not

B. Mehta

that doesn't sound true?

0celo7

@Semiclassical did you have a followup to that PDE

Semiclassical

nah, I just wanted terminology

it's the KPZ equation. (except that I didn't include the noise term)

Balarka Sen

7:38 PM

2(1) + 1 and 2^3(1) + 2^2(-1) + (-1)

@AkivaWeinberger I guess that's rip

Ok I'm off to bed

Semiclassical

@0celo7 one detail of the presentation I'm putting together which I probably will omit

Akiva Weinberger

Conjecture: Take n+1 in binary instead of n

Each 0 is worth one point and each 1 is worth two points

Semiclassical

a big part of the story is being able to turn the problem from a stochastic PDE to a path integral formulation

but the details for that are a pain. in particular, it involves something called an Ito regularization

and there is no way in the hell I'm putting that in my presentation. i'm not sure I even want to say it in passing

Akiva Weinberger

And then subtract one point?

And that should be the number of steps?

Should work

@B.Mehta @BalarkaSen Can you confirm?

B. Mehta

hmmmm

Akiva Weinberger

7:44 PM

To find the number of steps required for n, write n+1 in binary (instead of n). Then each 0 is worth one point and each 1 is worth two points; subtract one from the total to get the number of steps.

B. Mehta

that doesn't sound right... let me try and make an example

5 takes 5 steps but gets 4 points?

101 -> 110 -> 11 -> 100 -> 10 -> 1

but a bigger issue is that 6 gets one more point and takes one fewer step

in general for n odd, n+1 gets one more point and takes one fewer step

Akiva Weinberger

@B.Mehta Subtract one at the end

B. Mehta

at the end?

for n=5, you write 110 which is 5 points and subtract one to get four

unless i'm misunderstanding

Akiva Weinberger

Oh no you're right

6 doesn't work

Oh, wait

Add one

Argh hold on

B. Mehta

you still get the issue with odds

Akiva Weinberger

7:49 PM

101 takes 5 steps: 101->110->11->100->10->1

110 takes 4 steps: 110->11->100->10->1

B. Mehta

right

Akiva Weinberger

101 gets a score of four: 101+1=110, that's 1+1+2=4

B. Mehta

but for 6 you write down 111 which gets more points

Akiva Weinberger

110 gets a score of three: 110+1=111, that's 1+1+1=3

B. Mehta

but 0s are worth one point

and 1s are worth two points

Akiva Weinberger

7:50 PM

Oh derp

B. Mehta

so 101 gets a score of 5

Ah

then that sounds believable

Akiva Weinberger

Hm… wait so what's the fix?

Make 0 two points and 1 one?

That shouldn't work

OK, revision: Take n-1, and do some shenanigans in binary

Semiclassical

My silly way of simulating this in Mathematica: Start with an array of the numbers 1 through N (say, N=100)

act f elementwise on that array, count how many elements have reached 1, then remove these elements and set that as the new array

then repeat

skull

how long is your presentation suppose to be?

Akiva Weinberger

@B.Mehta @BalarkaSen Revision: take n-1 in binary. 0s are worth two points, 1s are worth one point.

And that should be the final answer.

Semiclassical

7:58 PM

between a half an hour and an hour

Akiva Weinberger

Test: 1000 should get three and 1001 should get seven

1000->100->10->1 (three)

1001->1010->101->110->11->100->10->1 (seven)

1000 gets three points: n-1 is 111, 1+1+1=3

1001 gets seven points: n-1 is 1000, 1+2+2+2=7

Semiclassical

the defect in the algorithm is that, if two numbers have the same stopping time but one is larger than N, then only the one smaller than N will be found and counted

Akiva Weinberger

101 and 110 work correctly as well.

Semiclassical

So for any N it gives at best a lower bound on the $c_k$

Akiva Weinberger

And then this is that classical result on filling a 1xn rectangle with squares and dominoes, which is well-known to be the Fibonacci sequence

thus giving another proof that it's the Fibonacci sequence.

0celo7

8:05 PM

@Semiclassical ew

Semiclassical

yeah

"Were I here to talk about the mathematical subtleties of this path integral formulation, this would be the point where I'd put up a gory-looking slide about discretization and Ito regularization...But i'm not, so f*** it, I'm moving on."

Twink

how can we count the number of arrays of $1 \times 12$ that can be formed using squares of $1 \times 1$ and rectangles of $1 \times 2$?

Semiclassical

I would suggest you take another look at the conversation that just finished.

Jake Rose

$AA^T$ is always defined for any matrix right?

Semiclassical

yeah

Akiva Weinberger

8:18 PM

@Twink Hint: Let $f(n)$ be the number of ways of doing that for a $1\times n$ thingy. How many ways can you do it if the last one is a square? How many ways can you do it if the last one is a domino?

B. Mehta

wow that's quite a coincidence

Jake Rose

And so is $A^TA$

?

Akiva Weinberger

Yeah @JakeRose

Semiclassical

yep

Akiva Weinberger

Yes to both these things

Jake Rose

8:18 PM

Thanks guys

Semiclassical

it'd better be, since you can rename $A\to A^T$

Twink

@AkivaWeinberger $f(n-1)$ and $f(n-2)$?

Akiva Weinberger

Yeah

And so you get a cool recurrence relation

Semiclassical

Here's something sorta cute.

Twink

the fibonacci sequence?

Akiva Weinberger

8:29 PM

@B.Mehta I do appreciate

Semiclassical

I mentioned that algorithm earlier. Here's a sample output for N=300 000:

Twink

but for the fibonacci sequence $f(2)=1$

Akiva Weinberger

@Twink Right, so write out the first five or so values

Semiclassical

Akiva Weinberger

$f(1)=1$, $f(2)=2$, $f(3)=3$, $f(4)=5$, $f(5)=8$, etc., right?

Semiclassical

8:30 PM

x-axis is the integer n, blue is f(n) according to the algorithm

Akiva Weinberger

It looks like $f(n)$ is the n+1th Fibonacci number

Semiclassical

and orange is the Fibonacci sequence

Akiva Weinberger

In other words, $f(n)=F_{n+1}$

Jake Rose

How do you do matrix notatin in latex?

Semiclassical

So the algorithm reproduces the Fibonacci sequence up to about n=20, but after that it starts missing large elements with shorter stopping times

what's interesting to me is that it looks more or less like a Poisson distribution

Akiva Weinberger

8:32 PM

\begin{bmatrix}1&2\\3&4\end{bmatrix} -> $\begin{bmatrix}1&2\\3&4\end{bmatrix}$

@JakeRose

Jake Rose

Thanks

Semiclassical

use pmatrix or vmatrix for variations on that

Akiva Weinberger

bmatrix for "bracket matrix". You can also have pmatrix (parentheses) and matrix (no brackets)

Or vmatrix (vertical lines), yeah

B. Mehta

@Semiclassical Your array was length 1000000, and it goes wrong after $2^{20} \approx 1000000$ which explains why it starts getting things wrong

Semiclassical

no, it was only out to 300000

B. Mehta

8:34 PM

oh lol nvm then :p

Semiclassical

now I wonder what would happen if I used the standard collatz map

Akiva Weinberger

matrix: $\begin{matrix}1&2\\3&4\end{matrix}$. pmatrix: $\begin{pmatrix}1&2\\3&4\end{pmatrix}$

Leaky Nun

13 hours ago, by Silent

@MatheinBoulomenos, Let $G=\left\{\begin{pmatrix}a&b\\0&a^{-1}\\ \end{pmatrix}:a,b\in \Bbb R, a>0\right\}$ and $N=\left\{\begin{pmatrix}1&b\\0&1\\ \end{pmatrix}:b\in \Bbb R\right\}$, how do we know that $G/N$ is isomorphic to $\Bbb R$ under addition?

Akiva Weinberger

vmatrix: $\begin{vmatrix}1&2\\3&4\end{vmatrix}$

Leaky Nun

@Silent by constructing such an isomorphism?

Akiva Weinberger

8:36 PM

@JakeRose

Leaky Nun

you should figure out what the operation on G looks like first @Silent

and the first isomorphism theorem might help

Twink

:D thanks

Jake Rose

Is the matrix for $\sum_{j,k=1}^{3} a_{ij} b_{jk}x_k$

Akiva Weinberger

Use {}, not ()

Jake Rose

I was gonna type the full matrix out but i fear Itll take me forever

So is it a nx3 matrix of a multiplied by a 3x3 square matrix of b multiplied by a $(x y z )^T$

There should be spaces between x y and z

Twink

8:39 PM

can you guys look at this question please? math.stackexchange.com/questions/2775478/…

Semiclassical

if (x1, x2, x3)=(x,y,z) then yeah

Jake Rose

Should just be x not y and z

Yeah that

But since it didnt sum over i is it just a 1x3 matrix

??

Akiva Weinberger

$$\begin{bmatrix}a_1&a_2&a_3\end{bmatrix} \begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\\ b_{31}&b_{32}&b_{33}\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$$

That was so annoying to write on mobile

Jake Rose

But it doesnt sum over i? @AkivaWeinberger

Akiva Weinberger

Oh, $A$ is $n\times 3$?

Then $ABx$ is $n\times1$

and so it would not sum over $i$

$$\begin{bmatrix}a_{11}&a_{12}&a_{13}\\ \vdots&\vdots&\vdots\\ a_{n1}&a_{n2}&a_{n3}\end{bmatrix} \begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\\ b_{31}&b_{32}&b_{33}\end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$$

Semiclassical

8:46 PM

$(n\times 3)(3\times 3)(3\times 1)=n\times 1$

Akiva Weinberger

It is an $n\times 1$ array whosez $i$th element is $\sum_{j,k}a_{ij}b_{jk}x_k$

or $a^i_jb^j_kx^k$ if we use the Einstein summation convention

(of summing over any variable that appears once as a superscript and once as a subscript)

Jake Rose

So because we havent defined i it just takes values 1 - n

Semiclassical

it's a free index at this point, yes

Jake Rose