Mathematics

2:06 AM

is there a symbol like $\rightarrow$ which means something like a continuous injection?

even better if it can require limits at +-infinity to be identical. for example i want to be able to say "let K:R->R^3 be a knot"

Peter Tamaroff

@DanBrumleve Just say it with words...

in a notation similar to defining the domain and range of a function, but with added constraints implied by the arrow symbol

Peter Tamaroff

I think things like that are better worded. Else it gets too crumpled.

well it can get convoluted, i could say that an unknotting is a continuous injection from the reals to the set of all continuous injections from the reals to the set of all functions from 3 to the reals....

ot else simply that an unknotting is K:R->(R->R^3)

for a suitable type of arrow

curious if anyone knows of any similar notational conveniences that can be used in the context of a definition

of a function

one example is a bidirectional arrow that can be used to denote a bijection like F:A<->B

i guess an injection could be an arrow with a bar and a continous function could be a squiggly arrow and a continuous injection could be a squiggly arrow with a bar. wonder if there is anything more standard

on that topic, is there a way to have a really long squigglyarrow in LaTeX?

2:27 AM

problem that just occurred to me... let F:A->B and G:B->A be continuous injections. does there exist a continuous bijection H:A<->B?

essentially cantor-bernstein with continuity constraint

and restricted to subsets of R

Victor

Hi, is anyone here familiar with stochastic process

@Ilya - May you help me with one question?

user19161

2:50 AM

@Khromonkey I have a feeling it is true but can't prove it either. It seems it can be done if we take the biggest divisor not greater than it, then add the next biggest divisor so that the sum is not greater than it, and so on... I tried to think of induction and pigeonhole but can't come up with anything yet.

David K.

Can someone here give me a sanity check?

Suppose I have a list of primes $p_{k}=6k+5$. Is the integer $N=6p_{1}p_{2}\cdots p_{n}-1$ of the form $6\ell+5$?

peoplepower

@DavidK. Well, I am insane myself...

David K.

@peoplepower Well I guess we can go down this road together ;)

peoplepower

@DavidK. Of course.

It doesn't matter what form the primes take.

$6(p_1p_2\dots p_n-1)+5=N$.

David K.

@peoplepower Right. Ok, I was crazy. Think I'm mostly sane again.

I'm certainly a bit dumb, but sane again at least.

Thanks!

peoplepower

2:59 AM

Happy to cure insanity.

Eric Gregor

if X,Y,Z are independent variables, can we say that $P(X,Y\mid Z)=P(X\mid Y)P(Y\mid Z)?$

@Eric yes assuming that by X,Y you mean X & Y

Eric Gregor

yes, i do

it is because P(X & Y) = P(X) P(Y). and since they are (pairwise) independent P(X | Y) = P(X) and P(Y | Z) = P(Y)

actually, i'm not sure if it's necessarily true that P(X & Y | Z) = P(X & Y).

counterexample: suppose Z = X xor Y. they are still pairwise independent but P(X,Y | Z) = 0, while it is still true that P(X,Y) = P(X | Y) P(Y | Z) = P(X) P(Y)

3:50 AM

does anybody know why Grothendieck became a recluse?

Does anyone know why anyone becomes a recluse?

Just kidding ;-)

Why?

To get away from people?

Obviously, but the question is why.

3:55 AM

@MJD Why do you want to know it?

Because they want to be alone.

Thank you, Dr. Tautology.

2

Next patient please :-D

Perelman sort of gave an explanation

Please share it...

3:58 AM

He was not happy with the ethics of the mathematical society

But I guess this speaks only for Perelmann, I guess MJD is searching for a more generalist opinion.

I'm just not sure if he/she wants to find a real answer to the question.

Gustavo: I was not searching. chat.stackexchange.com/transcript/message/8182511#8182511

@Khromonkey mathematical society specifically?

@GustavoBandeira why should every ethical question come to perelman? grothendieck had ethical issues too.

admit it you where a bit curious too. (joking)

4:02 AM

@MJD Oh, didn't see that.

Is it even known that Grothendieck is still alive? If he had died, would it be news?

If he is still alive, he turns 85 years old next month.

If I go to the Alps i'm going to look for him and deliver a pizza

He sent a letter in 2010.

we are all experts by reading his wikipedia page

@Khromonkey He lives in sourthern france/andorra

4:03 AM

Even the Nobel prize has been declined...

Yes (about the expert thing) .

the Abel prize hasn't been rejected yet has it?

@skullpatrol who declined the nobel?

I rejected it last month.

@Sanchez hi

4:05 AM

Good for you.

isn't that bad?

@Khromonkey, hi

@MJD XD

@thanks for taking your time

@OrangeHarvester Sartre declined the Literature prize; Le Duc Tho declined the Peace Prize.

I would decline the Peace Prize too if I had to share it with Henry Kissinger.

4:06 AM

Hahaha.

Hitler forced alot of scientist during the war too

@MJD ahh, okay. It seems only two people rejected on their own.

4 were forced to reject. 3 by hitler, one by soviets.

@OrangeHarvester In Feynman's memoirs he claims to have contemplated declining the Prize because he didn't want to suffer all the fuss, but he decided that to decline it would make an even bigger fuss, so he accepted.

@anon apparently he wasn't very happy at Huai Dong Cao and ZU Chiping

@MJD yes, I read that.

4:09 AM

@Khromonkey Is there a user called @thanks?

@whydoyouask?

@you'rewelcome

@Sanchez thanks for taking your time

@Khromonkey, oh you're welcome :)

I'll reject my fields medal in the future, only for fun.

4:11 AM

the fields doesnt have that much cash

@Sanchez are you from mexico?

No.

The name doesn't mean anything :)

oh

its a last name usually

I'll reject everything. I'll reject my sallary and feed from light particles.

I don't have to worry about rejecting anything

@Khromonkey Possibly interesting: ams.org/notices/200410/fea-grothendieck-part2.pdf

3

4:17 AM

I've heard there are profinite fibonacci numbers. This should mean $F:{\bf Z}\to{\bf Z}$ is extended to $\widehat{\bf Z}\to\widehat{\bf Z}$. Further it should mean that for any prime $p$ and sequence $(x_n)$ of integers, we have $$v_p(x_{n+1}-x_n)\to\infty\implies v_p(F_{x_{n+1}}-F_{x_n})\to\infty,$$ where $v_p$ is the $p$-adic valuation. Anybody see a way to develop this, either through elementary or high-powered math?

@MJD thanks it was worth reading

Anyways, time to call it a night. goodbye guys see you tomorrow.

@Khromonkey math.jussieu.fr/~leila/grothendieckcircle/biographic.php

@anon

?

Use the explicit formula for Fibonacci numbers, and consider Euler's totient function theorem in $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$

4:27 AM

binet's formula is $\displaystyle\frac{\varphi^n-\overline{\varphi}^n}{\varphi-\overline{\varphi}}$. what's Euler's theorem for a ring of integers?

I suppose it might be showing $?\to1$ in ${\cal O}/p^k{\cal O}$ with ${\cal O}={\bf Z}[\varphi]$ for every $k$.

@anon, hm it seems that there's a hole of what I said now I think about it. I'm a bit tired now so I would write an outline of a possible approach, that I am not entirely certain of.

let $\phi = \frac{1+\sqrt{5}}{2}$. You want to show something like if $m \equiv n \mod p^k$, then for $F_n = \frac{1}{\sqrt{5}} (\phi^n - \bar{\phi}^n)$ , we have $F_m \equiv F_n \mod p^{sth}$, where this sth should go to infinity as $k \to \infty$

One approach to do this is to write $\phi^m = \phi^{m-n} \phi^n$. Note that $m-n$ is quite divisible by $p$. If we can show something like $\phi^{m-n} \equiv 1 (mod p^{sth})$, or simply $\phi^{p^{\text{some power}}} \equiv 1 (mod p^{sth})$ we would be happy.

To do this, one should study $\left(\mathbb{Z}[\phi]/p^{sth}\right)^{\times}$. In particular, $\phi^{\text{order of this unit group}} \equiv 1 (mod p^{sth})$

The last line is what I mean by Euler's theorem.

right

It seems like we are missing something though. The order of the unit group doesn't look like a power of $p$ :/

Matt

5:25 AM

I have a quick question. Lets say we have a set between the integers and the rationals, X. Then Z < X < Q. If you want to show that X is countably infinite, then you can easily find a bijection to Z. But if we have some arbitrary X, then is there a theorem to use?

@Matt enumerate the rationals. then go through the list one at a time and delete entries not in X. this determines an enumeration of X.

Actually I am not so sure I understand your question. In your statement, If you want to show that X is countably infinite, then you can easily find a bijection to Z, I assume X is arbitrary (subject to Z<X<Q), yet your next sentence seems to indicate otherwise (since it begins with "But").

Matt

I understand thats the bijection, but if simply know Z < X < Q, nothing more, is there some result that says that X has to be countably infinite if its between two countably infinite sets?