« first day (1510 days earlier)      last day (3510 days later) » 

5:02 PM
Bye Dr. @Nick!
 
wb
 
@Nick You like being high in the morning ? Uh ...
2
 
That sounds wonderful actually
I've got no deadlines for a week :)
 
I've got deadlines every day :/
 
5:17 PM
Every day?
Sounds fun!
 
busy me :( ... hi and bye :(
 
5:33 PM
Why Cleo never answer my questions? Maybe they are too easy?
 
@Chris'ssis Too hard :P :P
@Chris'ssis Maybe because they get answered too fast / have too few upvotes
 
did i miss anything?
 
@BalarkaSen The bus. RUN
 
@Hippalectryon aha
 
throws a table at @Hippa
 
5:35 PM
How many tables do you just have ?
 
uncountably many oh whoops i mean countably many. i don't want the axiom of choice to interfere in the choice of my tables.
 
I have $\huge\aleph_{\omega_\pi}$ tables :D
 
You must have been quite a horrible case for your interior designers and landscapers
 
5:37 PM
@BalarkaSen I've already starred it
 
@Hippa Excellent. The deletion time is over. Now I have evidence of your betrayal to show to Ted. evil twisted laugh
 
I have already seen it.
 
I mean, he already knows :)
 
Ah.
Right.
@PedroTamaroff
 
5:45 PM
@Chris'ssis I have derived the following approximation through magic: $\frac{1}{2} \pi (-n) n \log (n)$
 
of what, @Mick?
 
@MickLH What is that?
 
an approximation to: $$\sum _{i=1}^n \left(\sum _{j=1}^n \left(\frac{1}{i i+j j}-\frac{1}{2} \pi \log (n)\right)\right)$$
 
what is $\pi(-n)$?
 
oh, oops
 
5:47 PM
@BalarkaSen $\pi\cdot (-n)$ ?
 
@Hippalectryon that is unusual.
 
Yes it's that, sorry for the bad LaTeX
 
as $\pi(x)$ represents the prime counting function usually
 
I let a computer program generate the latex for me, it did a silly job
 
I don't believe that's true @Mick
the error is bound to be $o(n)$ which it's apparently not.
if you mean upper bound, then it's a pretty naive one
 
5:49 PM
I didn't claim that it's a good approximation by any means
I just found it interesting that something I was working on was so similar visually
 
i am claiming that it's a bad approximation
both the error and the relative error varies hugely as $n$ grows. in fact, they tend to $\infty$
 
Hey everyone
 
Hello @Kaj
 
hi
 
Having a better semester @BalarkaSen? Or more grade school BS?
 
6:01 PM
School :>
 
what semester, @Kaj?
 
@BalarkaSen The third one :P
 
Well, I mean how's school going for you?
 
not bad
but particularly boring
 
Yeah, I remember how that goes
 
6:03 PM
i am fiddling with commutative rings. too hard.
 
@KajHansen Hey handsome!
 
@WillHunting, I see you've taken my potential username
 
@KajHansen Yes, as far as I know, I am the only user who ever used this username.
 
@robjohn I think we won't see soon an answer to that question. :-)
 
@Chris'ssis I think it may be very hard.
 
6:11 PM
@robjohn After a while, I'll probably offer a bounty for it, like 500 points.
@robjohn Yeah, it is.
 
6:27 PM
@Hippalectryon you might ask your professors to compute that limit I posted. :-)
 
@Chris'ssis They won't do free work :)
They already have lots of things to do
 
I will think out load about this @robjohn @Chris'ssis. $r_2(n)$ in number theory is known as the number of ways $n$ can be expressed as sum of two squares. The usual Euler-type approach to count these partition thingys are to use generating functions, so that gives you $\left ( \sum_{n = -\infty}^\infty x^{n^2} \right)^2$ which is nothing but $\vartheta^2(x)$, $\vartheta$ being Jacobi theta function. In fact, one can product an explicit closed form for $r_2(n)$ using some theta-sum identities.
Now here we want $r_2(n, \ell)$ which counts the number of $(1, 1) < (a, b) \leq (\ell, \ell)$ such that $a^2 + b^2 = n$. So the relevant generating series must be $\left ( \sum_{k \leq \ell} x^{k^2} \right)^2$. I am not familiar with any closed form for sections of thetanulls but probably a good asymptotic exists. But in any case, it'd be only the first step.
That is the number theoretic approach I have in mind.
@Chris'ssis Hey, don't forget who posted the problem in chat ;)
 
@BalarkaSen I'd like you to know I already searched on google for some "theta function asymptotics" :-))))))
 
@Chris'ssis Many are known. I am not looking for theta function asymptotics.
I am looking for asymptotics for theta function partial sums.
i.e., "sections" of theta functions.
 
:17798320 Well, yeah, theta function partial sums ... :-)
 
6:32 PM
Ah, you did? Got anything relevant?
 
@BalarkaSen There is some paper on Jstor I cannot read. :-(
 
Link me.
 
Greetings everyone!
 
that's not relevant, @Chris'ssis. "partial theta function" is something vastly different
well, so it seems from looking at the first page
but thanks for the effort!
 
6:37 PM
@BalarkaSen I don't find anything else :-( I would have liked to compute that limit, but it seems too crazy ...
@rehband Hi
@BalarkaSen I might read some of Ramanujan work on this subject ...
 
@Chris'ssis I saw today that my uni offers a course which uses Furdui's book! I was amazed :D
 
@rehband great ;)
 
@Chris'ssis yep it is (but don't forget who posed the problem! =) )
@Chris'ssis yes, they are great
i can't fully understand them though
they are too hard for me
 
@BalarkaSen I might have thought of it some time ago when I met problem $1.28$ $b)$ in Ovidiu's book.
 
yes, you should have, but you didn't =D
 
6:42 PM
@BalarkaSen actually that version crossed my mind, really, but nothing more
 
@Chris'ssis What might you have thought of?
 
yeah i recall that you worked on different kind of generalizations @Chris'ssis
like n-dimensional sums and whatnots
 
@BalarkaSen this is that problem $$\lim_{n\to\infty} \left(\sum_{i=1}^{n} \sum_{j=1}^{n}\frac{1}{i+j}-(2n+1)\log(2)+\log(n)\right)=-1/2-\gamma$$
 
mmhmm, i solved that a few weeks ago didn't I?
 
@rehband I'm referring to this limit proposed by @robjohn while we were in a discussion today
11
Q: Computing $\lim\limits_{n\to\infty} \Big(\sum\limits_{i = 1}^n \sum\limits_{j = 1}^n \frac1{i^2+j^2}-\frac{\pi}{2} \log(n)\Big)$.

Chris's sisIn the chatroom we discussed about the asymptotic of $\displaystyle \sum_{i = 1}^n \sum_{j = 1}^n \frac1{i^2+j^2}$, and if we think of the inverse tangent integral, it's easy to see that $\displaystyle \sum_{i = 1}^n \sum_{j = 1}^n \frac1{i^2+j^2}\approx \operatorname{Ti}_2(n)\approx \frac{\pi}{2...

 
6:46 PM
@Chris'ssis Thanks :)
 
it's a good question
 
@BalarkaSen Yeap
 
@Chris'ssis Any idea on $\displaystyle\sum_{i,j\in[|0,n|]}\dfrac{\max(i,j)}{\min(i,j)}$ ?
 
well, have fun with that. i'll think about it in my way whenever i get time @Chris'ssis and let you know of any progress
 
@BalarkaSen Perhaps the key is to get the proper asymptotic. I'm very curious about the value of the real limit.
 
6:48 PM
mmhmm
 
btw, upvoted and starred.
bitcoin haha
 
Btw I just changed my name :D
 
Your avatar changed too
 
Hi Will Hunting
 
6:50 PM
@KajHansen That was a while ago
 
bleh @Hippa
bad name
 
Hehe
Best name evar
Now you're gonna lose every day
 
(╯°□°)╯︵ ┻━┻
 
@Hippalectryon check some small cases, it should be pretty easy.
 
6:52 PM
@Chris'ssis $\displaystyle\sum_{i,j\in[|0,n|]}\dfrac{\min(i,j)}{\max(i,j)}$ was easy, but I can't manage to do this one
 
it's patently obvious.
=P
i kid
whoops i have to go bye.
 
@Hippalectryon Don't give up, work hard. :-)))
 
@Chris'ssis Give me a hint :/ I've been trying for some time, and I'm currently trying to solve another exercise math.stackexchange.com/questions/941837/…
 
6:55 PM
See you man
 
@KajHansen click the link though
the ping is a link
 
Link?
Ah
 
@Hippalectryon check some small cases ... (as previously said)
 
OMG
 
The ping is a link The cake is a lie
 
6:56 PM
LAUGHING SO HARD
 
heh
bu-byes
 
@KajHansen Is "The blue book of grammar and punctuation" very popular in the US?
 
Never heard of it
 
@Hippalectryon What is this weird set GL(E) ?
 
@rehband automorphisms in E
 
6:57 PM
@Hippalectryon I see
@Hippalectryon What's the dimension of L(E)?
 
$n^2$
 
@Hippalectryon Right
 
@rehband Left
 
@Hippalectryon Up
 
@rehband Down
 
7:02 PM
(highfive)
 
(brofist)
 
lol
 
@rehband General linear group.
 
@BalarkaSen Yep :)
 
So @Hippa
It's the group of nXn invertible matrices over the field E
 
7:12 PM
@BalarkaSen Not exactly. It's isomorphic to it.
 
@TheGame Nitpick :P
 
Ahh, I see what you mean @TheGame
But it's just abstractly isomorphic to Aut(E).
 
Hehe
 
The best interpretation of GL(E) is as the matrix group.
 
Why would it be the 'best' ?
@BalarkaSen $GL(E)$ is the set of automorphisms
 
7:14 PM
throws a table you don't have to be so nitpicky, @The
 
That's its only definition
lelele
@rehband You're gonna loose every day nao :P
 
@TheGame What am I gonna lose?
 
GL(E) is known as the matrix group all over the world, @The
It's universally accepted so.
 
@rehband Don't you know what the game is ? en.wikipedia.org/wiki/The_Game_(mind_game)
 
(+1) agree with Barlarka. That's how I learned it
 
7:15 PM
@BalarkaSen Not in France xD
we French are Martians
 
Oh everything is the wrong way in France
 
@TheGame Lol first time I see that
 
At least we drive on the good side :D
 
French people are lunatics
 
LOL, @rehband
 
7:17 PM
Hehe
 
@rehband My. Name. Is. NOT. Barlarka.
Stop Germanifying names.
 
Proud to be french
 
Fine! I'll just call you "Sen" from now on!
 
Find the error ^
 
LOL, @rehband, that's much too formal.
 
7:19 PM
@TheGame I also need a hint for $$\lim_{n\to\infty} \frac{2^n}{n!}\int_1^{\infty}\int_1^{\infty} \cdots \int_1^{\infty} \frac{1}{x_1\cdot x_2 \cdots x_n (\operatorname{max}(x_1,...,x_n))^2}\ dx_1 \ dx_2 \cdots dx_n$$
 
@BalarkaSen Is Barlarka Sen your real name?
 
@Chris'ssis I'm not good enough :) I can't help you on those problems. Maybe in other fields.
 
@rehband No.
 
@BalarkaSen How about Balarka Sen?
 
Yes.
 
7:20 PM
@BalarkaSen are you a muslim? I always felt that.
 
@BalarkaSen Yo Balark :P
 
@Chris'ssis No.
 
@BalarkaSen OK
 
@rehband Balarka is common.
@Chris'ssis Sens are never muslims.
AFAIK
 
7:21 PM
@rehband 'i am a bad boy' Hmm I see
 
@BalarkaSen It's a pretty cool name
@TheGame Lol
 
@BalarkaSen OK
 
kicks @Hippa
@Chris'ssis For example you could search for Amartya Sen. Best known Sen I could think of.
 
@BalarkaSen I see.
 
Balarka had clever ancestors!
 
7:28 PM
throws table at @Khallil
throws tables at everyone
 
@BalarkaSen $\in$ everyone
 
Was totally worth loading ChatJax for that one
 
For what @Mick?
 
the joke with the $\in$ symbol
 
Could do way better
 
7:30 PM
eh
a table hits himself
 
@BalarkaSen $\subsetneq$ @TheGame
 
@TheGame $\cong$ @Hippalectryon $\subset$ $\emptyset$
 
Therefore I do not exist
Therefore I can talk bad about you as much as I want :D since no one will be talking
 
My set inclusions are not strict, @TheGame
You are set-isomorphic to the null set.
 
Someone should create the account @YouLost :D
@BalarkaSen You are isomorphic to chemistry dat insult
 
7:36 PM
I am only one in my isomorphism class.
Unique.
 
@BalarkaSen You need to stop throwing tables.
Also, Ted needs to stop smacking people.
 
Who starred that?
You fool.
 
Math clearly induces physical violence
 
In a few hours, my first psychotherapy session in my life will begin.
I hope it helps me.
 
@WillHunting Hope it helps you
:D
I read you mind
 
7:39 PM
@TheGame Are you Hippa?
 
I am.
@WillHunting And you just lost. :D
 
@WillHunting Are you Jasper?
 
@TheGame Is Hippa your real name?
 
@WillHunting No it's short for Hippalectryon
 
@BalarkaSen Yes, the one and only Will Hunting, lol.
 
7:40 PM
Haha, that's a good one @TheGame
 
Too few people know what Hippalectryons are :/
 
@TheGame What are they?
 
Even google autocorrect doesn't put it in its list
A hippalectryon (or hippalektryon, from Greek ἱππαλεκτρυών) is a type of fantastic hybrid creature of Ancient Greek folklore, half-horse and half-rooster, with yellow feathers. The front half is that of a horse, the rear half a rooster's wings, tail and legs. The oldest representation currently known dates back to the 9th century BCE, and the motif grows most common in the 6th century, notably in vase painting and sometimes as statues, often shown with a rider. It is also featured on some pieces of currency. A few literary works of the 5th century mention the beast, though no myths related to it...
 
Half-horse half-rooster?
It makes sense why that does math!
 
:D
 
7:43 PM
I think I will switch to AmE from now. So I will spell color instead of colour, lol.
 
AmE ?
 
American English
BrE is British English
 
8:23 PM
HI I want to check somethin in algebra
 
what is it?
 
I've got a really nice sum for which I've found a solution by differentiating geometric series. If anyone can offer an alternate solution, by all means. $$ \sum_{n=2}^{\infty} (n^2-1)z^{n} = \dfrac{z^2(3-z)}{(1-z)^3} $$
^_^
 
@TheGame why the silly name?
 
@Alizter Because you just lost :D
@Khallil $- \begin{cases} \frac{z^{2}}{- z + 1} & \text{for}\: \left\lvert{z}\right\rvert < 1 \\\sum_{n=2}^{\infty} z^{n} & \text{otherwise} \end{cases} + \begin{cases} \frac{4 z^{2} \left(z^{2} - 3 z + 4\right)}{\left(- z + 1\right) \left(4 z^{2} - 8 z + 4\right)} & \text{for}\: \left\lvert{z}\right\rvert < 1 \\\sum_{n=2}^{\infty} n^{2} z^{n} & \text{otherwise} \end{cases}$
 
Oops. I forgot the specify the condition that $|z| < 1$, @Hippa.
 
8:29 PM
If $G/N \cong H$ and $H\cap N=1$ and $N\unlhd G$, then $G\cong N\rtimes H$. Since $NH$ is a group we have the following $$NH/N\cong H/N\cap H=H\cong G/N $$ Therefore $G\cong NH\cong N\rtimes H$
 
@Khallil Too late :D
 
@TheGame is that above
 
@JoseAntonio Uh sorry i'm not good enough in that field.
 
Ok don't worry thanks anyway :)
Someone else?
 
@DanielFischer
 
8:32 PM
@TheGame Quoi?
 
Thanks
 
@DanielFischer Maybe you can help @JoseAntonio
It's not my field.
 
@JoseAntonio What is the question?
 
I think is the easiest way to show that $G$ can be broken into s $N\rtimes H$ But since is is my first course in algebra I don't completely sure. First of all @DanielFischer Hi
The problem is the following If $G/N \cong H$ and $H\cap N=1$ and $N\unlhd G$, then $G\cong N\rtimes H$.
My answer is so simple: Since $NH$ is a group we have the following $$NH/N\cong H/N\cap H=H\cong G/N $$ Therefore $G\cong NH\cong N\rtimes H$
But I'm not completely sure if the above argument is correct. I think is the easiest way. Only using the isomorphism thms.
and sorry for the multiple questions in this days, hahaha.
 
why write NH/N ~ H/N^H = H ~ G/N when we already know G/N = G/N? what are you trying to get at?
oh, you're trying to show G=NH I take it?
 
8:40 PM
Yes
 
I'll let @anon handle this, he knows more algebra than I.
 
to do that, pick g in G, let h be its image under G->G/N~H, then since g is in the coset hN of the kernel N you know that g=hn for some n in N. this proves G=HN, or G=NH if you use right cosets.
then you want to actually exhibit an isomorphism $N\rtimes H\to NH$
which will just be (n,h)->nh
then argue it's a bijective group homomorphism
 
the first point is not completely clear is clear that g goes to some h under the composition. But the point g is in the coset hN
is not completely clear for me.
 
hN is the unique coset sent to h, and g is sent to h, so g must be in that coset
anyway bbl
 
but why not $hN$ can be to some other $h*$
sorry if is a stupid question
 
8:48 PM
ah I suppose the assumed isomorphism G/N~H is not necessarily the obvious one
so that is a good question
 
how do we know that the map is the identity under the cosets $hN$ and not ¨permutes¨ the h in the coset to some other h
 
since the isomorphism G/N~H could be an exotic one you can use your reasoning (mention the lattice correspondence theorem you used) to show G=NH
 
yes but it's never specified in the problem. FOr that reason I try to use the isomorphism thms. instead.
so in the general case it´s possible to use the correspondence thm, right?
 
@JoseAntonio you're right, we don't. if Aut(H) is nontrivial there will be exotic isos.
@JoseAntonio yes, even if the relevant groups are infinite (so cardinality arguments don't work) the lattice correspondence theorem is valid
 
Thanks
@anon thanks for your help. I will continue studying
 
8:52 PM
np
 
I've to go I hope to find someone if I have other question, hahaha by the way the videos of the extension school in harvard are really a gem bye bye
 

« first day (1510 days earlier)      last day (3510 days later) »