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22:00
@Lord_Farin There is a missing + at the end befors dots
Ok, I think I got it.
Yes.
@Lord_Farin small question: do you know a version of this lemma where $\alpha$ is not a constant and depends on some variable $m$ ?
let $\alpha \in \mathbb{R}^+$ and $p_1,p_2,\dots, q_1, q_2, \ldots \in \mathbb{N}$ such that $\left|\alpha q_n - p_n \right| \neq 0$ for all $n \in \mathbb{N}$ and $ \displaystyle \lim_{n \rightarrow \infty} p_n = \lim_{n \rightarrow \infty} q_n = \infty\,$ if $\displaystyle\lim_{n \rightarrow \infty} \left| \alpha q_n - p_n \right| = 0.\,$ Then $α$ is irrational.
@Chris'ssis Hm, it seems quite hard. (Read: I haven't solved it after one minute of thinking :P.)
@Lord_Farin :D. No worry, I only showed it to you for its beauty. :-)
@Chris'ssis If $x$ can actually be found, that's a pretty awesome formula :).
And I trust it can be if you're putting it in your book :).
22:05
@Lord_Farin Sure, we can, the problems is created by me. :-)
@user153330 Can you elaborate? In what way is $\alpha$ nonconstant?
@Lord_Farin yes, but the problem is in choosing some $q_n$ and $p_n$ that will work for all $n$
@user153330 Can $m,n$ vary independently?
@Lord_Farin yep
Because if so, then there is no material difference between $\alpha$ constant and $\alpha$ depending on $m$.
We can simply prove $\alpha_m$ is irrational for every $m$ using that same lemma.
You will have to adapt your proof that $\alpha_m$ satisfies the conditions to work for every $m$, though.
22:09
@Lord_Farin yes, but the problem is in choosing some $q_n$ and $p_n$ that will work for all $m$, that's why i ask you if there's some variation of that lemma that will save me some work
:19718468 you what?
@user153330: you got my reply earlier?
@TedShifrin yup : P
dokey okey.
@user153330 Typed here by mistake.
@PedroTamaroff dokey okey.
22:11
don't misaddress me by mistake, @Pedro
@user153330 If I read it correctly, then $p_n, q_n$ ought to be the convergents for $\alpha$. Hence they are required to be different for different $\alpha$.
@user153330: I don't think your avatar is very fair to Charlie Hebdo.
@TedShifrin it's in relation with chat.stackexchange.com/transcript/message/19718080#19718080 , i hope i'm not offensing anyone, if i'm then please say it
@Lord: Glad to see everyone's putting you to work after your holiday :)
@TedShifrin Yes :). I find pleasure in it :).
22:12
We're glad to have you back.
@TedShifrin Thanks. I'm planning on staying, I've learned to balance things a bit more.
You can cover for me when I've disappeared, @Lord :)
@TedShifrin Different socks, same foot, right? :P
I'll have to parse that one.
I won't spoil the fun until you ask me to.
22:15
@TedShifrin do you have any idea on this question chat.stackexchange.com/transcript/message/19718424#19718424 ?
@TedShifrin!!!!!!!!!
How have you beeeeeeeeeeeeeen?!
@Anthony :) Nice to see you.
Doing well, thanks, trying to get ready to make the move to CA this summer.
@Ted Apparently everyone's so glad to see me that one of the first meta threads I created had to be locked because of the large number of comments :).
You're moving?!
:O
You troublemaker, @Lord!
22:17
Hey @crl !!! Are you familiar with 2-3 trees?
That's the plan, @Anthony.
Where in CA?
can someone help to find this sum
@TedShifrin Who, me? Never. (A)
@user153330: That's Liouville's Theorem. No, I don't know what you'd hope to say if you had a sequence of $\alpha$'s.
San Diego, @Anthony, but I plan to visit LA and the Bay Area reasonably frequently.
22:18
! I'm from San Diego!
Of course, you should be gone soon, right?
Oh, cool, well perhaps you won't avoid me forever, @anthony :D
I'll be a senior next year, if that's what you me. I'm bound to see you some day.
(Plus I'm going to apply back here for grad school, so who knows)
oh, I thought you were a senior this year.
@pourjour $r$ is in ... ?
22:19
Nope! Hoping to go to school for something in Quantum Information... but we'll see.
@pourjour: What about $\sum r^n(e^{ix})^n$?
Oh, interesting.
as $r\in ]-1,1[$
So, @pourjour, réponds-moi :P
Where in SD, @Anthony?
Carlsbad/Encinitas?
Ah, cool ...
22:20
Where are you looking to move?
@TedShifrin je sais pas mais le problème c'est dans n-1
Oui, je comprends. Mais le mien est facile à résoudre et puis ...
I have friends in Kensington, @Anthony, but hoping for an upscale apartment in Hillcrest, Bankers Hill, or perhaps even Kensington for a year or two.
Exciting!
@TedShifrin je pense en pososns la raison tel que $q=re^{ix}$ c'est ça?
Yeah, scary to try to get my house ready to sell and my office emptied ... But somehow it'll happen, @Anthony.
22:23
I have problem rendering mathjax?
Evidemment, @pourjour.
What are you going to do for work?
using robjohn script
See the LaTeX in chat link on the right, @pourjour.
What work, @Anthony? I'm gonna be a bum :P
22:24
Maybe I'll tutor underprivileged kids ... maybe I'll teach a class or two as an adjunct. We'll see.
I see, you can't stop teaching!
D:
@TedShifrin you'll get the status of professor emeritus right? so you can still teach if you want ?
We'll see if I miss it. I might do some distance-learning stuff with high school kids via Stanford, @Anthony. One of my old friends mentioned that.
@TedShifrin I thought you were younger, haha.
That's good.
LOL @user153330: Emeritus status doesn't mean anything, other than having access to the library back at UGA :P
22:25
@Chris'ssis just back from the store. I will look
No, @Anthony, I'm antique :D
Hello @robjohn.
How are you doing?
rehi @robjohn
@robjohn what did you buy ? : P
@robjohn OK. I also prepared another amazing question. :-)
22:26
@TedShifrin even so I couldn't render latex on this page
You have to import the link to your browser toolbar and then click on it once you're in here, @pourjour.
@TedShifrin that's what I've did exactly
@pourjour then click on it
@Chris'ssis Is there a typo? The exponents are powers of 2 except the last one says $\phi^{2^n-2}$
@robjohn Right, no typo.
22:29
@user153330 when I click nothing is changed
Does @Chris'ssis ever make typos?
@TedShifrin Yes!
@Chris'ssis Then I don't get what the general term is.
@pourjour that's strange, what browser are you using?
is there any updates concerning the scripts
@user153330 Google chrome
22:30
I'm using it on Chrome right now.
@robjohn you mean the question is posed wrongly? Look at my answers.
I uploaded it a year ago, though.
@pourjour Are you using extensions like NoScript, HTTP Switchboard, etc.?
Those might pose some trouble.
<--- ignoramus — has never even heard of those.
@TedShifrin So you started teaching right out of grad school?
22:31
@Lord_Farin no I checked if the javascript is allowed and I found it activated
@TedShifrin Basic internet safety nowadays, Ted. I hope you do use AdBlock and/or Ghostery?
Hell, @Anthony, I started teaching when I was still in high school and in college. And I taught officially in grad school for two years. But yes.
I will try to clean the cache
Jeeeeeesus.
Champion.
22:32
@robjohn It's just unusual, but it's correct I think.
@Anthony It would be appreciated if you minded your language.
If I hadn't gotten cancer 3+ years ago, @Anthony, I probably would not be retiring for a few more years. But I figure I should make some changes ... and I'm finding more things about students these days frustrate/anger me.
@Lord_Farin You mean saying Jesus?
I consider that question rhetorical.
Apparently @Lord is heretically religious.
22:33
@Chris'ssis it is not clear what the general term is supposed to be... Do you mean $$\dots-\sqrt{\phi^{2^{n-1}}-\sqrt{\phi^{2^n-2}}}$$?
@TedShifrin I didn't you had cancer. :( But it's good that you can relax a bit then, I suppose.
@robjohn Yes. Maybe I should add that term.
@Chris'ssis Then you should put a couple of terms extra before the end
Ugh at the square root.
@robjohn Right! That's about clarity! Thank you! :-)
22:35
clarity :)
@TedShifrin Yeap! Star to you!
I figure I'd offer you some free editing for your book, @Chris'ssis :P
@TedShifrin Apparently a call for polite conversation must have a deeper cause.
@Lord: A certain amount of profanity occurs here without remark. I'm just surprised at your reaction.
@TedShifrin :-))) It's much work to do, any help is very welcome! ;)
22:38
I've never actually met anyone that was outwardly offended at someone says Jesus- Although I went to Christian school for a couple years and would nag on my mom whenever she would say it. Haha.
@TedShifrin It's not that you were wrong -- I am religious. I'm just surprised to note that so many have trouble expressing themselves without resorting to swearing.
Well, I live in the Bible belt, so I've tried to restrain myself in class from using certain epithets. Sometimes I screw up. Shrug
To us, @Lord, that was not swearing at all.
You are, in fact, over-sensitive. But I accept it.
I didn't know your appellation needed to be taken quite so seriously :D
@robjohn clarity is a very important point. The things that to me seem so obvious might not be the same for the rest.
@TedShifrin No worries.
@TedShifrin
Can you help me with something?
22:40
oh, I thought you were going to cuss me out, @Pedro.
Sure :P
Consider the ring $\mathcal O(\Bbb C)$.
That's a boring ring.
Unless you mean the sheaf.
Oh, never mind. I was being silly. Go on.
In this ring, consider the ideal generated by the various $\sin \pi z\cdot \prod_{|\nu|\leqslant n}(z-\nu)^{-1}$.
@Chris'ssis I haven't looked at your solution, but I would factor out a $\phi$ which simplifies things considerably
@robjohn Yeap. I did that too (there I tried to show 2 different ways).
22:42
$\nu\in\Bbb Z$? @Pedro
@TedShifrin Yes, but $|\nu|\leqslant n$.
I want to determine if this ideal is finitely generated.
Got it. Interesting.
Nah, it can't be.
I have no proof. Just intuition.
@robjohn How does the question itself seem to you when properly arranged as discussed?
@Chris'ssis So, without actually writing anything down, I think I get $1$
Do you agree, @Pedro?
22:43
@robjohn Yeap. Right! :-)
@PedroTamaroff You mean the ideal generated by $\{ f_n : n \in \mathbb{N}\}$, where $f_n$ is the function you wrote down?
@robjohn: You're as bad as @Chris'ssis. Doing everything with no paper/pencil/pen. :D
Yes, @DanielF, surely that's what he means.
Yes, Daniel.
@Chris'ssis It is actually independent of $n$
You nitpicker.
22:44
@robjohn True :-)
@TedShifrin lol :-)
So suppose you had a finite generating set. Then anything in the ideal would have to be nonzero at infinitely many integers, methinks.
@Chris'ssis It seems pretty straight forward, once you get the idea of factoring out the $\phi$, and noticing that $1-\frac1\phi=\frac1{\phi^2}$
@robjohn It's meant for easy ones group.
@PedroTamaroff Okay, that's the ideal of functions $g\in \mathcal{O}(\mathbb{C})$ such that there is an $n(g)$ such that $g(k) = 0$ for all $k\in \mathbb{Z}$ with $k \geqslant n(g)$.
@TedShifrin This comes just after the following observation. Suppose $G$ is a region and $A$ is an infinite locally finite subset. Then the ideal of functions vanishing almost everywhere on $A$ cannot be finitely generated.
22:46
@robjohn Indeed.
If you have a finite subset $\mathscr{G}$, then $N_{\mathscr{G}} := \sup \{ n(g) : g \in \mathscr{G}\} < \infty$.
Seems to be a special case of that, @Pedro, yes.
This is because for any $a\in A$ we can pick $f\in\mathfrak a$ that doesn't vanish on $a$.
The above follows from Weiertrass' theorem.
@pourjour: Je reviens à la suggestion que je t'ai déjà donnée.
22:47
@DanielFischer Oh, noes. My $G$ and your $G$ collided.
@TedShifrin well, I wanted to have an excuse in case my answer was not right ;-)
@Ted Where did you learn French?
High school and university, @Lord. I essentially did a French major in university.
any idea
@pourjour: If you're going to ignore my suggestions, I won't make any more of them.
heya @usukidoll
22:48
@PedroTamaroff Resolved.
anyone know mathematical biology? I know my answer is correct but my question is why the heck do I have extra stuff and the book doesn't huh?!
oh hi @TedShifrin
Oh, sneaky. You mods can edit whenever you feel like it. grumph
@DanielFischer Hehe. Though I can multiply by arbitrary functions to make $n(fg)$ as large as I want for $g$ in the set of generators.
@TedShifrin Interesting. So you combined it with maths?
@TedShifrin not I'm not ignoring them, but does it work with my context?
22:49
@PedroTamaroff Then $f_{N_{\mathscr{G}}+1} \notin \langle \mathscr{G}\rangle$.
Sure. @pourjour: One step of algebra relates the answer to mine to the answer to yours, @pourjour.
Is someone familar with 2-3 trees?
growls @DanielF for double-subscripts in here
@DanielFischer What is $N_{{\mathscr g}+1}$?
I mean we have to pick a value that we can use to cancel as much as we can, so the only way I can get the 1 to appear is if r* = a which is r* = constant so a/a = 1 because the units match and stuff oh yeah should I post my work on the chat? Or is it going to clog everything
22:50
Now Daniel is confusling me.
you mean confuzling :P
$\mathscr G$.
@PedroTamaroff A misplaced }.
@DanielFischer I suppose $N_{\mathscr G}$ is the largest $n(g)$?
@PedroTamaroff Aye.
22:51
Wait. Do you mean $N_{\mathscr G+1}$ or $N_{\mathscr G}+1$?
@TedShifrin je pense que je dois faire sortir $e^{ix}$
@TedShifrin The latter.
@pourjour: Nous sommes convaincus qu'il faut regarder $(re^{ix})^n$, n'est-ce pas?
You have a point with the double subscripts, @Ted ;)
@TedShifrin oui
22:52
bows @DanielF
OK, @pourjour, et nous savons la réponse en ce cas-là
I wish I had a business card that was this cool: kurims.kyoto-u.ac.jp/~motizuki/top-english-photo.gif
@TedShifrin oui
@Alex :)
@TedShifrin do you know mathematical biology?!
Hello there @Ted. :)
22:53
OK, @pourjour. Laquelle est la relation entre $r^{n-1}$ et $r^n$?
Nope,. @usukidoll.
awwwwwwwue
@usukidoll: There's lots, lots, lots that I know nothing about.
@ModdedBear Could I ask you something about 2-3 trees?
I know my answer is right......it matches the back of the book...I just have extra stuff attached to it and the book is wrong if it thinks that a is going to survive
you could, but I may not be able to answer
22:55
$r*r^{n-1} = r^n$
Is envious of the French talkers.
Et alors, @pourjour?
@PedroTamaroff that sounds weird, I think french speakers sounds more natural.
NO you're not, @Pedro. Besides, you need to teach me Spanish. I need to be able to communicate with all the cross-the-border Mexicans in San Diego ...
@TedShifrin je dois multiplier par r
22:56
@TedShifrin Tsk tsk tsk!
yes, in English we say **-speaker
ou diviser, @pourquoi?
morning
G'night, @Mike
@MikeMiller Morning? Hardly!
@TedShifrin I am.
Green with envy.
22:57
But you need to teach me Spanish, @Pedro, seriously :)
LOL @Alex
@TedShifrin par $e^{ix}$
Non, non. Par $r$.
It's morning somewhere, @Pedro
Oh, OK, si tu veux, mais ce n'est pas la somme exacte qu'on a discutée.
@TedShifrin You will suffer. You're aware of this, yes?
22:59
Suffer?
@robjohn I think my first way occupies too much space, and this will also be a problem for the book. I only let the part you also mentioned.
Hello @Mike.
hahahaha
@TedShifrin Yes.

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