Mathematics - 2022-07-17

shintuku

00:09

i'm still somewhere in there

i'll have about a week free to do a bit more until january heh

i absolutely need to finish the book before the 2023 semester starts

Akiva Weinberger

01:27

@BalarkaSen That's a lot

(^that's an edited version of what I actually meant to say, which was just a string of profanity)

@leslietownes lol

leslie townes

:)

Koro

01:54

@TedShifrin not allowed. It doesn’t have a finite non zero limit point.

Koro

03:32

thanks. But still, I'm stuck. I'll explain where: Suppose that Q+ is enumerated as $q_1,q_2,...,q_k,...$. Q+ has a non zero limit point (take $\pi$ for example). Let q be a non zero limit point of Q+. Then, there is a k>1 in N such that 1/k< q<k. Since q is a limit point, there is a $q_i$ in (1/k,k).
Take the q_i with minimum subscript that lies in (1/k,k) and call it $r_1$. Note that $(1/k, k)\subset (1/k^2, k^2)$. Choose the $a_i$ with min. subscript such that it lies in $(1/k^2, k^2)$ call it $r_2$.

(we are choosing $r_2$ s.t. $r_2\ne r_1$. This is possible as q is a limit point.)

nope. this is still wrong.

Koro

03:49

instead of $(1/k^2,k^2)$, I should consider $(\frac 1{k^{\sqrt 2}}, k^{\sqrt 2})$. Then, I construct a sequence $r_n$ such that $\frac {1}{k^{\sqrt n}}<r_n<k^{\sqrt n}$. Then $r_n^{\frac 1n}\to 1$.

But this creates problem when $q=\frac 1{k^{\sqrt j}}$ for some j or when $q=k^{\sqrt l}$ for some l.

:(

Even if this problem is sorted out. There is no reason to believe why the sequence $r_n$ will contain every element of Q+.

(for example: N and 2N have the same cardinality but are not the same sets.)

leslie townes

koro what happens if you toss countably many 1's into ted's example

Koro

Leslie, that's also not allowed. :(

leslie townes

i stand by my two-starred remark

2

Koro

we are given a countable infinite set of positive numbers and I think that a set by definition has no repetition allowed.

@leslietownes : )

leslie townes

ok, interleave 1 - 1/1000000^n with ted's example?

Koro

04:02

looks like a counterexample.

But the source of the problem is P. Orno, Math. Magazine, Problem 1021

by some search online, I found math.stackexchange.com/questions/2669206/…

but I don't understand the answer.

in particular the (unique) order isomorphism part

leslie townes

f sends n in N to the nth element of N \ f(N)

i haven't digested the proof but that is that part of it

Koro

leslie, why is the above not a counterexample?

leslie townes

dunno, i don't work on this stuff

facultymembers.sbu.ac.ir/shahrokhi/ProBookMathAnal2.pdf has a solution although you will be making a request to iran

i'm probably on 20,000 US watch lists now

Koro

I dropped a comment on the post.

@leslietownes that's the book. :)

I have the following problem with the solution in the book:

27 mins ago, by Koro

But this creates problem when $q=\frac 1{k^{\sqrt j}}$ for some j or when $q=k^{\sqrt l}$ for some l.

dtn

07:21

0

Q: Differential Equations with Special Functions: Are analytical solutions possible?

I want to ask a question to specialists, I am very interested in getting their advice and opinion. Let's take elliptic integrals and hypergeometric functions and write down any differential equations in which they enter. For example, these: $x'=-EllipticF(x(t)+1|-1)-EllipticF(\sin (x)|1)$ $x'=_{1...

BAYMAX

08:32

Given $|x_{i}|<1,|y_{i}|<1$. If $\Pi_{1=1}^{n} x_{i} (1+y_{i}) < 0$, then $\Pi_{1=1}^{n} y_{i} (1+x_{i})$ cannot be less than zero.

I am thinking if this si true?

As $|x_{i}|<1, |y_{i}|<1 \forall i=1,\ldots,n$

so $\Pi_{i=1}^{n} |x_{i}|<1$

doesnot look correct

one potato two potato

@BalarkaSen Oh I've heard that fibrant/cofibrant replacement term. I was about to ask any motivation or application of that replacement in topology at that time.

Calvin Khor

09:16

@Koro if you have at least one limit point say L, then you have infinitely many terms which are in the interval (L-1, L+1). Enumerate these as b_n, and the rest of the points in the sequence as c_n. So for your rearrangement, you use N-1 terms from the b sequence, where N is chosen so that $1-1/2<c_1^{1/}<1+1/2$. Then for the Nth term, you put. c_1. Now you continue using M-N-1 bs where M is chosen so that $1-1/3< c_2^{1/M} <1+1/3 $. Then for the Mth term you put c_2.

BAYMAX

Any graph come to mind which covers all the quadrants except the third quadrant

say in the x-y plane

Calvin Khor

And so on; in this way you obviously use all the bs, and we slowly pick off the cs one by one, so they're all used

@leslietownes oh no, this comment is self-referential

Hi, can you please explain why the countable set $\{e^{n^2}: n\in \mathbb N\}\cup \{1+1/n: n\in \mathbb N\}$ is not a counterexample to the statement in OP? — Koro 5 hours ago

$\{ e^{n^2} : n\ge 1\} = \{ e, e^4, e^9, e^{16},\dots \}$ Observe $e^{1/3}\in (1-1/2, 1+1/2)$. So choose the first few terms of the sequence is $1+1/1, 1+1/2, e$. The next term from the bad sequence we need to add is $e^4$. Well, $e^{4/15} \in (1-1/3, 1+1/3)$. So our sequence is now $$1+1/1, 1+1/2, e, 1+1/3, 1+1/4, \dots, 1+1/13, e^4.$$ Then we need to add $e^9$...etc.

Balarka Sen