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00:00 - 14:0014:00 - 00:00

00:01
I once dreamed the first 37 million decimal places of pi but when I woke up, I forgot them.
Hahaha
Now I'm pieless
Funny
Project Gutenberg has a book which gives the first 5 million decimal places for $\sqrt{2}$ and ...
I've always been obsessed with phi
00:05
1 million places for $\sqrt{3}$ and ...
Even my phone number has phi in it
1 million places for $\sqrt{4}$ :)
HAHAHHAAHAH
00:09
@Charlie just made your day better, you're welcome
@ian hahahahah
 
1 hour later…
01:18
Linear algebra is so dry
01:34
Just back from UCLA. Now it's time to walk the dog. BBL
oh @robjohn pls do u have time for one question
after u back with the dog
01:49
if u have time can u have a look at this math.stackexchange.com/questions/659092/…
but my problem is about the upper limit in the second integral
02:08
@Danny what's up?
@robjohn its ok iam a bit frustrated here
@robjohn how are u btw
did u make some exam?
did u see the link
@Danny I commented.
@robjohn can we get it from the start, first he makes the substitution
$z =\frac{t-\mu}{\sigma}$
the u get the second integral
i still dont get how the upper limit $x$ changes to $\frac{t-\mu}{\sigma}$
02:32
@Danny If $t=x$ then what is $z=\frac{t-\mu}{\sigma}$?
that is the upper limit
@robjohn u mean that when $t$ is at is upper limit
..which is $x$, then then if we make the substitution $z =\frac{t-\mu}{\sigma}$
then $z $ will have upper limit $\frac{x - \mu}{\sigma}$
anyhow thanks @robjohn i think i got it...i need to go to sleep its 3:45 am here
02:56
@Danny good night
03:35
:(
04:21
bangs head
 
3 hours later…
06:53
okay
alright.
who knows about algebraic geometry.
@AlexanderGruber A little of this, a little of that, but probably I would be no use of you.
@BalarkaSen actually what i'm looking for is a book recommendation
@AlexanderGruber Ah, I see.
Shafarevic might help.
i'm doing all this megarigorous commutative algebra in a class and i want to see what it's building up to, but i don't just want to read hartshorne/a megarigorous algebraic geometry textbook, i'm looking for basically the "softest" book on A.G. there is so I can read it concurrently for motivation.
I know, Hartshorne is way to rigorous. Shafarevic is just the thing for you, I think.
07:00
okay, cool. thanks.
@AlexanderGruber Are you onto NT much?
@BalarkaSen i took a basic course (and then some crypto/finite field theory) but i haven't seen the arithmetic geometry side of it
i hear it's supposed to be good. i don't understand it yet.
@AlexanderGruber Whats that? NT?
07:03
@BalarkaSen yes
i even liked the bit of analytic number theory that i saw, went through chebyshev's theorem proof in detail for a course project. (which says a lot, i generally can't stand analysis.)
@AlexanderGruber I am (or I claim I am) an analytic NTst.
Although my interest much lies on TNT.
(Transcendental NT)
=D
I need a hug
@BalarkaSen i'm not sure what that is.
Transcendence theory.
A study of transcendental numbers
07:08
i didn't realize there was a whole discipline of that, besides the theory of trascendental extensions in commutative algebra.
There is a whole branch on that, yes. and that's where my interest on theory of equations came into play.
(Which I thought had no application to NT, silly me)
Funny how 'apparent' unrelated interests might be related in the most unexpected way.
@Alexander Are you interested in theory of equations?
@BalarkaSen i'm not sure what you mean
wouldn't that just be, yknow, math?
Theory of equation consists of galois theory, solving lower degree equations on $\mathbb{ \bar Q}$ and solving higher degree equations (quintics, sextics, etc) in larger extensions of operations.
@BalarkaSen so you mean that it is galois theory?
maybe what we have here is just a difference in terminology.
in any case, i actually do not have too much interest in galois theory, unfortunately. some of it is interesting, but i am more interested in other areas of algebra.
@AlexanderGruber Not really, galois theory is just a part of it. For example, 'how to solve quintics" raise a pretty good question on this branch.
07:21
@BalarkaSen but galois theory addresses how to solve quintics. maybe in my neck of the woods the phrase galois theory is used more broadly than in yours.
And my own-made schematic partially answers "how many ways there are to solve a quintic"
@AlexanderGruber Galois theory says that quintics cannot be solved using elementary field operators. It does not answer how to solve, does it?
image not found
great way to solve something eh
@BalarkaSen well, yes, especially if you take a constructivist viewpoint. cardano's method, for example, is also galois theory.
@AlexanderGruber Perhaps. But I like to keep it theory of equations, since sometimes much more than that of galois theory is needed to answer some questions.
Hello, Shobhit.
I mean, @IntegralsandSeries
 
1 hour later…
08:55
@Sush $\{X\ge x\}$
@Sush Since $y_n\lt x$, $x$ is in the intersection
09:08
@robjohn, thank you so much!
09:30
@robjohn, how to show that $$F(x)=\begin{cases}0 & \text{ for $x<0$}\\\frac{x}{1+x} & \text{ for $x\geq0$}\end{cases}$$ is not differentiable at $0$?
09:51
@Sush what are the left and right derivatives at $0$?
Left is 0, Right is 1/(1+x)^2 and at 0 this equals 1.
@robjohn
Thank you.
@Sush :-)
10:10
I am stuck.
Hello, @Sush
$\mathop {\lim}_\limits{x \to \infty} {{x^2 + 2x + 3} \over {2x^2 + x + 5}}^{{3x - 2} \over {3x + 2}}$
dat broken latex
yeah...
sorry about that.
here it is
$$\lim\limits_{x \to \infty} {\frac{x^2 + 2x + 3}{2x^2 + x + 5}}^{\LARGE{\frac{3x - 2}{3x + 2}}}$$
I think it is $e^{1/2}$
W|A says it is 1/2
:(
10:24
$\lim\limits_{x \to \infty} {\frac{x^2 + 2x + 3}{2x^2 + x + 5}} = \frac12$
and since the limit of the exponent is 1...
@MartinSleziak oh
why didn't that occur to me?
10:39
@PedroTamaroff There are answers in the back?
> the UMMO star is a good star,but if there exists any earth's scientist can solve the UFO without my theory? i will not migrate there!

-- i am alien
@KarlKronenfeld Apparently?
Also... hai.
Whuzz poppin'?
nothin', nothin'
arghhhhhhhhhhh! bangs head
10:54
$\lim\limits_{x \to a}f(x)^{\Large{g(x)}} = (\lim\limits_{x \to a} f(x))^{\Large{(\lim\limits_{x \to a}g(x))}}$ ?
right?
@ShuklaSannidhya Sometimes.
And not necessarily always.
So when it is not?
@BalarkaSen ?
Leave me out of these, I am not an analyst. Pedro can answer this well, though!
11:09
@DanielFischer: I apologize for not @-pinging on my edit. The whole intention of the thank you was to get your attention, but I accidentally left out the @-ping.
@Pedro Is it true that $\sin$ of $\mathbb{Q(\pi)}$-rational arguments doesn't take all of $\Bbb{\bar Q}$?
@robjohn The trouble was that you commented and after that temporarily deleted. Then when you undeleted after correcting, I got no notification.
You mean whether $\sin(\Bbb Q(\pi))\supseteq \overline{\Bbb Q}$?
Not really.
(And although it showed in my inbox, the notification for your comment didn't stay when you deleted the answer.)
11:12
I mean $\sin(\Bbb Q(\pi)) \subset \bar{\Bbb Q}$
Note that I used $\subset$ not $\subseteq$
@DanielFischer sorry about that.
Right? @Pedro?
@robjohn No problem. Most of all, the point of my second comment was the pun about letting the moderators do the work.
I am too sick to think about it clearly.
@BalarkaSen Sick? What do you have?
11:14
Severe cold.
What does $\bar{\mathbb Q}$ mean?
Well, I have no idea. You could note that using the addition formulas, you're kinda looking at terms of the form $$\sum q_i \sin(r_i \pi^k)$$
Or something like that.
@KarlKronenfeld Alg. closure.
@KarlKronenfeld Algebraics.
Well, some more with cosines too.
Anyway, no idea.
I'd highly doubt you can write $e$ in terms of cosines and sines of rational multiples of powers of $\pi$.
11:16
Ok. I conjecture that $\sin\left(\mathbb{Q}(\pi)\right)$ does not contain $\frac{1}{3}$
@PedroTamaroff I said $\Bbb{\bar Q}$
$e$ is not an algebraic.
@BalarkaSen I know.
Just thinking out loud.
@PedroTamaroff Provably true. Sine of rational multiples are always algebraic.
Hey fellas
Hey @Mike
@Mike Can you have a look at my question?
I mean, prove that $\text{arcsin}(1/3)$ is $\Bbb{\bar Q}(\pi)$-transcendental?
11:20
@Balarka The 1/3 thing?
Oh
Okay, I've had al ook :P
@BalarkaSen Did you see the solution to the $\sum \varepsilon\exp(-n^2\varepsilon^2)$ thingy?
@DanielFischer Morning?
@PedroTamaroff What solution? $\sqrt{\pi}/2$ thingy?
Yep, I saw it.
@PedroTamaroff About mid-day.
Ya.
@KarlKronenfeld I was wondering about something.
@PedroTamaroff sure
11:31
@DanielFischer Dunno if you want to comment on this, @Mike?
@Pedro I rarely want to comment on anything
@KarlKronenfeld An abelian group $A$ is divisible iff it is $p$-divisible for any prime $p$, meaning any elt as a $p$th root.
Right or left?
@PedroTamaroff What should I want to comment on or not?
@PedroTamaroff Yeah, kind of.
Now, the cyclic group $C_p$ is $q$ divisible for any prime $\neq p$.
11:33
Yep
In fact, divisible for any $(k,p)=1$
Thus, to make it divisible, it suffices to add the $p^n$-th roots. This is precisely accomplished by adding $C_{p^k}$ for $k>1$.
I'd agree with that rationale.
And seems to be the minimal construction that does so?
I wanted to ask @anon =D
It is provably so.
@KarlKronenfeld Oh!
11:35
@PedroTamaroff Recall the structure theorem of divisible groups. There is just no other construction to try.
@KarlKronenfeld Ah, you're highfalutin. Still a bit over my head.
You know, I failed to prove $\bf P$ is $p^k$ divisible algebraically.
@Karl He's just learning about divisibles, I don't think he has the structure theorem :p
@Mike I linked him to it the other day.
Oh.
No excuses @Pedro!
I used its elts are of the form $\exp(2\pi ikp^{-m})$, but I was thinking I should be able to prove it without using the explicit form of the elts.
11:38
deargod
what is P even
Prüfer group.
So, under what condition $\lim\limits_{x \to a}f(x)^{\Large{g(x)}} = (\lim\limits_{x \to a} f(x))^{\Large{(\lim\limits_{x \to a}g(x))}}$ ?
Prufer group.
Darn.
11:38
Take le logs.
Hi! Someone have an idea for my exercise please, it will be very helpful math.stackexchange.com/questions/659023/…
@PedroTamaroff I'm seeing if an answer of mine is relavent
That seems like a silly way to descripe it
@PedroTamaroff Ah... I would have said $\overline{\mathbb{Q}}=\mathbb{R}$, but I think too analytically rather than algebraically.
11:40
@robjohn =)
$\Bbb Z[1/p]/\Bbb Z$
@Mike Why?
@Mike YOUR NOTATION SUCKS!
@BalarkaSen Whoa, what?
@Pedro If I write it like that we're basically done.
11:41
That's standard notation...
@Mike I don't understand that one.
He's saying I'm saying that to you :p
@Pedro You don't understand the notation?
@PedroTamaroff I am talking Lang
(Nevermind)
@Mike I do. I don't understand why that is $\bf P$.
@Mike that looks like the base-$p$ representations of $\mathbb{Z}^{\ge0}$
Also, why don't you like ${\bf P}=\bigcup C(p^k)$?
11:42
@Pedro $\bf P \subset S^1 \simeq \Bbb R/\Bbb Z$
@Mike Yes.
in particular $\bf P \subset \Bbb Q/\Bbb Z$
since the rationals you may have are specifically those that are a negative power of $p$ times an integer...
$\bf P = \Bbb Z[1/p]/\Bbb Z$
I don't get it.
Where'd I lose you
@Mike Clever.
11:44
@Pedro I'm using your explicit description of elements
@Mike For example, what is the class of $1+p^{-2}+p^{-4}$?
Is that an element of the Prufer group...?
Oh
It is an element of $\Bbb Z[p^{-1}]$.
You mean what happens when you mod out by Z
11:46
Yeah, that's $p^{-2}+p^{-4}+\Bbb Z$
It works the same way $\Bbb R/\Bbb Z$ does, it's just that our starting ring is smaller.
And that to the power of $p^4$ is zero.
For example.
Yah
But when you notate this way it's obvious that your group is p-divisible :)
@PedroTamaroff btw, I won't go into detail about my purely algebraic approach... :)
@Mike Why?
@Pedro Just explicitly divide by $p$!
11:49
Well, yes... but that's not what I was after.
That's... pretty much what everything is gonna take the form of. Maybe in disguise, but it's all essentially the same.
So elements of $\bf P$ are integer polynomials in $p^{-1}$ with no independent term and whose independent term gets killed when it appears.
yop
In fact all coefficients can be made coprime with $p$.
right, you can essentially take coefficients mod p
11:52
So, an iso with $\bigcup {\bf G}_{p^k}$ would be...
Where ${\bf G}_n$ are the $n$th roots of unity.
That one should be pretty easy
You can arbitrarily pick generators.
Oh, maybe you have to be a little more careful than that.
@Mike Minimal polynomial of $\sin$ of $\Bbb{Q}(\pi)$-rational arguments have cyclic Galois group, right?
@Balarka I don't know.
11:56
Darn.
But I believe I am more or less correct.
Just hoping someone would check that.
(I am too ill)
proof revisions and added proof assignments are crazy ba ram ba ba bahhh
@Mike $\Bbb Q/\Bbb Z$ is iso to $\bigcup_{n\geqslant 1} C(n)$?
@robjohn Why are you so mean?
@Pedro Right, as long as the union isn't taken disjointly, but you wouldn't do that
@Mike Sure.
12:03
It starts to become unclear what the union means at this point. This kinda motivates the definition of direct limit.
@robjohn You answered the darn thing here and you can't even answer my question?
thanks @Karl
@BalarkaSen Maybe he's busy.
Don't expect express service here.
Possible.
OK.
@robjohn (I upvoted it as an apology for calling you mean, by the way)
But then you are mean squared, no?
12:07
@Pedro Actually, no, I don't believe that's an isomorphism (after translating it appropriately to inverse limits like Karl suggests)
@Mike The books says $\Bbb Q/\Bbb Z\sim\{z:z\text{ is a root of unity }\}$
@Mike You do want a colimit (aka direct limit), I think.
@Karl I don't see what that limit would be.
Lemme think on it.
You take the directed set given by divisibility of integers.
No, that won't work. If you have an element of $C(n)$ and one of $C(m)$, pick some $p$ coprime to both of those. Then they both map to the same element of $C(p)$
So that the direct limit of $C(n)$ will be trivial.
12:11
Why do they have to map to some element of $C(p)$?
What are you talking about...? =D
Ah, point taken, I was just taking the limit over $\Bbb N$
Which is not the right system here.
So $\sin(x)$ is algebraic if and only if $x$ is in the extension $\Bbb Q(\text{arcsin}(\alpha))$ where $\alpha \in \Bbb Q$?
Is that right? @Pedro @Mike
Why is it important that the stab of $x_1x_2+x_3x_4$ is iso to $D_8$? @Mike
@Pedro Depends on your definition of important.
12:14
And I mean the stab under the action of $S_4$.
The book says it will appear latter on, prolly Galois stuff?
@PedroTamaroff Yes.
@BalarkaSen I have no idea.
I guess you might use that for galois stuff, or something.
@BalarkaSen Ask me analysis stuff, not that. =)
Don't ask me anything, I'm lazy.
12:15
@Mike is mean.
@Karl Onto transcendental NT?
I am generally too lazy to think off and on about the hard questions here.
@BalarkaSen God no
Calling a whambulance. Please hold.
@Karl I have a question for you.
@Mike Is it hard?
Here it goes.
12:17
@Karl It's the Hodge conjecture.
@Karl I'd also settle for proofs of the standard conjectures.
oh, now we're talking
12:20
@BalarkaSen which question?
@BalarkaSen You asked that an hour ago and I was not pinged.
How would I know which one are TNTst and which are not?
=D
@BalarkaSen Sorry, but I don't have the chat window up all the time nor am I at my computer all the time. I'm sorry if you feel slighted.
12:25
@KarlKronenfeld Every subgroup of $Q_8$ is normal, right...?
@Pedro Correct. List them first.
@robjohn Just confirm whether I can ping you on TNT Qs or not.
@BalarkaSen what are TNT Qs?
Transcendental number theory questions.
=D
12:27
@BalarkaSen You would probably have to ping me in chat unless I have posted an answer or a comment to the question.
@robjohn trasc numb thry questions
@Pedro Isn't every subgroup of $Q_n$ normal?
@robjohn Ok, alright.
@Mike $Q_n$?
Ugh I hate that notation
Don't say it, @BalarkaSen
@Mike By $Q_8$ I mean quaternions, I take by $Q_n$ you mean the generalized quaternions?
12:28
Oh.
Ha, notations.
Nevermind, now I hate your notation ten times more.
@Mike =D
I thought you were writing $Q_{2n} = D_n$
Dihedral group with $2n$ elements.
Haha.
I never used that.
12:29
This crap isn't standardized well enough.
@Mike No, but $QD$ is the quasidihedral group.
$D_{2n}$ is standard, AFAIK
No, $D_n$ is standard.
@PedroTamaroff D&F says different
I will fight you.
12:30
@Mike Come at me bro.
@Mike $\text{Dih}_{2n}$ is sometimes used too.
Differential equation gets to the bottom of this notational mess.
@Mike Heh.
I send you a dove.
There's no need to tell me these things, I'm very prompt. Check back again. :P
A.B.O.T. by U.D.
12:38
@Mike LOL dude.
I was asking Karl, not you.
You were asking Karl about sending doves?
I'm so confused.
OH
OH GOD
HAHAHAHAHAHAHA
@KarlKronenfeld Mike likes you.
That was unprofessional.
I like you too, @KarlKronenfeld.
This is so off limits.
12:40
this is far too awk for me to handle
and it's way past my bedtime
so adieu
@Mike Don't worry.
See, you guys drove Karl off the chat.
That was mean.
Okay, I have to got to go. Bye
13:43
@PaulEpstein Morning.
@PedroTamaroff what's goin' on?
Drawing lattices of subgroups.
with Karl?
Just alone.
13:56
@AntonioVargas
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