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7:00 PM
@DanielFischer Hello Daniel.
 
@PedroTamaroff Hi Pedro, it's too late to say hi to me, lol.
 
hello again
 
@MikeMiller Your lack of rep makes you miss some belated posts.
 
@Pedro Are any of them good?
 
@MikeMiller Nope.
One was wrong.
 
7:13 PM
Salut drowned @GTR, hi mr @Pedro @Mike @Jasper @DanielF
 
@cc I thought about it, but it seems too good to be true.
 
oh, heya @Studentmath
 
@cc doing it all the way through, I got to.. well, certainly not $Var(X)$. Hey Prof. @Ted! How are you?
 
@Pedro "good" doesn't mean "right". it means "funny".
 
Salut @Ted, comment vas tu?
 
7:14 PM
>$i$ is defined as $\sqrt{-1}.$ And we know that the rule $$\sqrt{z\,}^2=z$$ holds for all $z$ in $\mathbb R$. Thus $i^2=\sqrt{-1}^2=-1.$
 
@DanielFischer thanks. I do see why it's differentiable, but has unbounded derivative, hence not lipschitz. But uniform continuity? Not trivial to prove
 
ça va bien, merci, @DanielF, und Dir? :P
 
@G.T.R Continuous, compact interval.
 
@Studentmath: Since you're a probability expert, you might enjoy this post.
 
@G.T.R It is. Cantor.
 
7:15 PM
@TedShifrin Gut, danke der Nachfrage.
 
@DanielFischer sure, shame on me
 
@Pedro Could hve just said it wasn't funny.
 
@Ted far from it, but I start to like it. Checking!
 
@Pedro Heine
 
Heinrich Heine?
 
7:16 PM
@G.T.R Yas.
Heine-Cantor-.
 
Eduard, iirc, @Ted.
 
Diesen Heine kenn'ich nicht. :)
 
@ted salut
@DanielFischer I remember I learnt a poem that had "bescheide Blumen" in it, but I forgot the author and the title
 
too flowery for you, @Gabriel?
 
That's rather interesting, I think I have a proof for it I done as homework somewhere, will try to find..
 
7:20 PM
German can do flowery?
 
@Ted I forgot too much about German
 
Eine kleine Nachtmusik.
 
Cool, @Studentmath. He wanted to know what was wrong. I didn't think about finding a correct solution :)
 
Hello @TedShifrin @PedroTamaroff
 
Actually, @Mike, yes.
hi @Balarka
 
7:22 PM
Yes, that's true.. Are you currently catching up with probability before you start teaching it?
 
@BalarkaSen I have a problem for you. Show that $\Bbb Q^\times$ has subgroups of finite index for any integer.
Don't google it.
 
@PedroTamaroff OK.
I don't understand the question.
"for any integer"?
 
Do you mean there's always a $G$ for a given $n$ with $[\Bbb Q^\times : G]$ finite and $n = |G|$?
 
Still, @Ted, I prefer the sound of French.
 
7:26 PM
No, that is impossible.
 
@PedroTamaroff That I have to prove too.
 
I mean that for any $n>0$ there is $G\leqslant \Bbb Q^\times$ with $|\Bbb Q:G|=n$.
@BalarkaSen $\Bbb Q$ is infinite, so there are no finite subgroups of finite index!
 
@PedroTamaroff Oh, ok.
 
I value the music of the words more than I do the words themselves, usually.
 
@PedroTamaroff Makes sense.
Let me work my head round it.
 
7:28 PM
Well, since I majored in French, I concur, @Mike, but I like German too.
 
@Vrouvrou did you get some help?
 
@Studentmath: "Catching up"? Sort of ... Never have learned any of it systematically — just bits and pieces.
Hi @robjohn
 
@BalarkaSen We always have $|G|=|H||G:H|$.
 
@PedroTamaroff Yeah, Lagrange.
 
@TedShifrin hey
 
7:29 PM
@Ted Have you read any Joyce?
 
I just can't think too fast.
 
hmmm, probably not, @Mike.
 
@Pedro How about infinite cyclic groups?
$\{p, p^2, p^3, \cdots\}$
Prime $p$.
 
How is that a group? @Balarka
 
@TedShifrin Snap.
 
7:31 PM
@Ted He's precisely what I was thinking of when I said the music of the words. One needn't understand a single thing he's saying to appreciate it. (I sure as hell didn't try to analyze Finnegans' Wake.)
 
@BalarkaSen how is that cyclic? it never cycles...
 
Right, just introduce rationals.
 
@BalarkaSen That doesn't have finite index either.
 
$\{\cdots, p^{-1}, 1, p, \cdots\}$
 
Don't forget $1$, @Balarka ...
 
7:32 PM
@TedShifrin My thinking is too semigroupoidal.
 
That's just $\Bbb Z$ in disguise. What did I stumble into the middle of?
 
@TedShifrin Yup.
So do we have infinite copies of $\Bbb Z$?
 
@BalarkaSen $\Bbb Q^\times$ is isomorphic to infinities copies of $\Bbb Z$ and something more.
You can give an isomorphism.
 
You can look at $\oplus \Bbb Z$ or $\prod \Bbb Z$ if you want ...
 
@PedroTamaroff Ah. $\{1, -1\}$ is a finite cyclic among these infinite beasts.
$\Bbb Q^\times \cong \Bbb Z/2\Bbb Z \times \bigoplus \Bbb Z$
 
7:35 PM
\bigoplus
 
Thanks.
 
Oh wow @Ted, so you are learning it all now?
 
Nah, I'll be learning most of it as I go :P
 
I really like it, it can pose some really great question. Guessed so :P
Two pages ahead of the class
 
When I was young and energetic, I taught a year-long applied math course where I learned 70% of the course as I taught it. That was fun.
More than two pages, @Studentmath :)
I always think through lectures at least a week ahead so that I can write problem sets.
 
7:37 PM
@PedroTamaroff I came up with an interesting problem in galois theory this month.
 
@BalarkaSen But finish solving this one!
 
Stern taskmaster @Pedro :D
 
@PedroTamaroff OK, OK.
 
@Ted do you make mistakes by purpose at times?
 
Of what are you accusing me, @Studentmath?
Salut @Sab
 
7:38 PM
Hello everyone :)
Salut @Ted :)
 
All done with exams?
 
Nop Maths on Friday
 
Ah, exciting.
 
Hi there do anyone know I to develop $(x_1+2x_2+3x_3)^4$?
I know the multinomial theorem and how many coefficients are there.
but unlike binomial theorem how do I find each coefficient?
 
I'm revising right now actually and that's why I'm here :P
 
7:39 PM
That's what the multinomial theorem tells you, @gbox!!
 
I'm revising functions and I realize that I'm not very good with functions
 
Calculus is all functions, @Sab. :D
 
I know :P
I can use them but deep down I don't know about them enough
 
Heya @Kevin
 
I know (f+g)(x) = f(x) + g(x) but why does it work
?
 
7:40 PM
Good afternoon folks. Howdy @Ted
 
@TedShifrin what will be the first coefficient? $ \frac{4!}{4!0!0!} $?
 
If you mean the coefficient of $x_1^4 x_2^0 x_3^0$, yes @gbox.
 
Have a nomenclature question: Does the name O(1,1) make sense for a group? If so whats its English name?
 
It's definition of $f+g$, @Sab.
 
@Sabಠ_ಠ That's the definition of addition of functions.
 
7:41 PM
Ya. it's a definition. But I don't see it very well
 
It's the Lorentz group in dimension $2$, @Kevin.
 
@TedShifrin and what about the 4th one? it can be $ \frac{4!}{3!1!0!} $ or $ \frac{4!}{2!2!0!} $ ?
 
I feel I'm using functions without a thorough knowledge of them :/
 
@Ted Ah I suppose that makes sense since you can think of this thing as having its motion confined to a hyperbola
 
Don't say the *blah*th one, @gbox. Write out the monomial explicitly.
Huh? @Kevin
 
7:42 PM
I am thinking something in the lines of the fact that are only countably many copies of $\Bbb Z$ in the direct product, @Pedro
 
Maybe people refer to it as the orthogonal group with signature $(1,1)$, @Kevin. There are $O(p,q)$ for all $p,q$.
 
@TedShifrin is it 6 or 4?
 
@TedShifrin Is first year stats hard?
 
Functions are rules that assign to each $x$ in the domain a real number, @Sab. And we're using the algebraic structure in the range to add, multiply, etc., functions.
I don't know what your course is, @Sab, and I really don't know much stats at all.
 
Is there always a surjective homo from any pair of groups of countable basis?
 
7:44 PM
@TedShifrin I know the binomial it starts with 0 and goes up to some n.
 
you have a triple sum, @gbox, so there's no standard order here.
 
There are stuff like probability, poisson, hypothesis testing, chi-square(w/e)
 
You have a sum of $x_1^a x_2^b x_3^c$ over all sorts of $a,b,c$. Decide what $a,b,c$ are and then you know the coefficient.
 
It's only 1 semester and I want to do it but I'm not sure if stats is worth it
 
Well, probably half is probability and half is application to interpreting statistics meaningfully, which actually is worth understanding, @Sab, even if you want to be "pure."
But I never took such a course. Now I'm more interested than I was at your age.
 
7:46 PM
So it's worth doing it then
 
@TedShifrin and there is no formula to find all of the 15? like (4,0,0) (1,2,1) and so on?
 
@Daniel do you remember my question from yesterday? Well the function $\frac{1}{1+x^3\sin(x)}$ Does the trick
 
I'm actually interested in stats because it is needed in artificial intelligence(which I really want to do later on)
But my workload will be huge next semester and I have to deal with it.
 
Sure, @gbox. The formula is the multinomial coefficient $\dfrac{n!}{a!b!c!}$ for the coefficient of $x_1^ax_2^bx_3^c$ with $a+b+c=n$.
 
How about this : $\Bbb Z/2\Bbb Z$ multiplied by $\bigoplus \Bbb Z$ is just $\{\cdots, p^{-2}, 1, p^2, \cdots\}$. Does it make sense to think of $\{\cdots, p^{-n}, 1, p^n, \cdots\}$?
 
7:47 PM
@BalarkaSen Forget about $\Bbb Q$. Use the isomorphism, and give me a subgroup of finite index of $\bigoplus \Bbb Z$.
 
Stats will be — for most people — easier than the ideas (proofs) in calculus, @Sab.
 
@PedroTamaroff Of course. $\bigoplus n\Bbb Z$.
I was being silly.
 
That's good then. I'll be doing it for sure then :)
 
@BalarkaSen No, that has infinite index.
The quotient is still infinite!
You want a finite quotient.
 
Right.
 
7:49 PM
@robjohn i dont understand where we use the continuity of $u\mapsto u^{±}$ ?
 
I have to think.
 
I actually find the proofs in calculus easy so far, but I've not really applied them per say, I just know them. For example I know Rolle's theorem but I'm not sure how I can apply it in a question or what kind of question can be asked around it.
 
@G.T.R Shouldn't that be $\sin^2 x$? With $\sin x$, you have poles.
 
@TedShifrin but I know that after I open the expression
 
@Daniel oops forgot the square. $\frac{1}{1+x^3\sin^2(x)}$ It works actually if you replace $3$ by anything $>2$
 
7:50 PM
Good.
 
The interesting sort of question, @Sab, of which you can find one in one of my exams, is to apply Rolle's Theorem and the Intermediate Value Theorem to show a certain $f(x)$ has precisely $3$ roots or $4$ roots, etc.
 
Ohh
We got this as a multiple choice question last time
and I got it wrong
 
@Ted My hyperbola comment comes from the fact that this group is apparently a symmetry of the classical version of the problem I'm working on. And I know that the motion of the object in that case is confined to a hyperbola in the phase space. So maybe not surprising that the Lorentz group shows up.
 
Well.
 
You have to apply Rolle's Theorem a few times ... @Sab.
 
7:51 PM
just because I knew the roots lied b/w something using IVT but I couldn't find how many roots exactly were there
 
@gbox ... I'm not sure what your problem is here.
 
@G.T.R t'as fait $A$ anneau, $x\in A, x^2=x\Rightarrow A$ commutatif ?
 
So basically if I really knew my proof I would have found it
 
@Sab: If you had ($\ge$) 3 roots, then $f'$ would have to have at least how many roots? $f''$? $f'''$?
@Hippa: et $x^3=x$? :) (Ugh.)
Et bonsoir @Hippa
 
@PedroTamaroff $p_1^{\alpha_1} \cdot p_2^{\alpha_2} \cdots p_n^{\alpha_n} = n$.
 
7:53 PM
@TedShifrin how do I calculate each of the coefficient
 
@TedShifrin Non, $x^4=x$ c'est là que je suis bloqué - et bonsoir :D
 
@Sab: It's not memorizing the proof of Rolle's Theorem ... it's knowing interesting ways to apply the theorem itself.
 
@BalarkaSen Yeah, though a simpler thing would be to quotient by $n\Bbb Z\oplus \Bbb Z\oplus \cdots$. =)
I was being silly.
 
@PedroTamaroff Meh. The last step was easy.
 
I guess f''' is the answer @Ted
 
7:53 PM
I am sillier than you.
 
@TedShifrin I got the first 3, but not the 4th and so on
 
Je crois qu'il y a un morceau de Jacobson là-dessus en général, @Hippa.
 
f*** is the answer.
Of course it is!
 
@TedShifrin jamais entendu parler :/
 
@Hippalectryon done it, yes
 
7:54 PM
ou peut-être Kaplansky ... J'ai oublié.
 
haha, @Pedro, the answers are obvious but thinking is nontrivial
 
@G.T.R et avec $x^4=x$ ?
 
@Hippalectryon too
 
Si $x^{\phi(n)} = x$ pour une function quelconque $\phi$, je crois que ...
 
Now, let's get back to business. I have a galois theory problem.
 
7:54 PM
@G.T.R tu fais comment ? (pour $x^4=x$)
 
@Ted Jacobson rings iirc
 
@gbox: Seriously I don't understand. Pick any monomial (pick $a$, $b$, $c$) and then figure out its coefficient. There are a lot of 'em to do. In most applications, you care about one particular one.
 
how would you prove that f(x) = 2+x+arctan(x) has an inverse?
 
I thought so, @GTR. This is the part of algebra I abhor the most :D
 
@TedShifrin thanks
 
7:56 PM
I finally got 1300.
 
I know that f^(-1)(f(x)) = x can prove it, but I don't know how to proceed
 
$L/\Bbb C(z)$ and $L'/\Bbb C(z)$ be galois extensions, $\text{Gal}(L/\Bbb C(z)) \cong \text{Gal}(L'/\Bbb C(z))$. Is $L \cong L'$? I have no counterexample neither any proof using classical galois theory. @PedroTamaroff
 
You can't find $f^{-1}$ explicitly, @Sab. What theoretical tool will tell you there exists an inverse function?
 
@Hippalectryon try to figure out identities
 
Let me think
 
7:57 PM
You're welcome, @gbox. I'm sorry if I didn't say exactly what you wanted.
 
@BalarkaSen I don't know what a Galois extension is. =)
 
@Ted what about Wedderburn theorem?
 
@PedroTamaroff It's a normal and separable extension, lol.
 
I think I saw that for the last lecture of my undergraduate algebra course, @GTR. What is it?
 
@PedroTamaroff Eh, just an extension with galois group nontrivial. Like, for example, $\Bbb Q(\sqrt[3]{2})/\Bbb Q$ is not.
 
7:58 PM
@JasperLoy No shits.
 
@JasperLoy That wouldn't make sense to anybody.
 
@Pedro's algebra education hasn't included Galois theory yet, as far as I know.
It makes sense to someone who already knows what it is, @Balarka :D
 
@Ted I can't figure it out
I'm lost.
 
The function has to be one-to-one in order to have an inverse, @Sab.
 
Yeah
ono-to-one and onto
 
7:59 PM
@TedShifrin I can't make sense of separable extensions. Of course, I nearly always work over $\Bbb C$, haha
 
So, is that one? @Sab
 

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