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12:11 AM
@robjohn: Here?
 
@AsafKaragila OMG! Nice to see you!
2
 
@AsafKaragila HUGS
 
@AsafKaragila I've removed the profanity
 
@rob: Oh yeah, I left another two open flags about that user before things become... so repetitive in the comments.
 
12:14 AM
Hello @AsafKaragila have you finished your PhD?
 
If you can attend to them as well, it would be nice.
So long.
 
@AsafKaragila I will look at them tonight.
@AsafKaragila So long and thanks for all the fish
 
@robjohn I want fish too.
 
@JasperLoy Have you read HG2TG (Hitchhikers Guide to the Galaxy)?
 
@robjohn No.
 
12:17 AM
@JasperLoy Very good set of books. The movie is pretty funny, too.
 
Asaf did not talk to me, sad panda...
 
@JasperLoy I think his anger at someone's actions was just enough to overcome his self-ban from chat. Now that he has said his piece, I don't think he will back for a while.
 
@robjohn Whereas it is well known that I will create a new account after deletion, lol
 
@JasperLoy Sigh...
 
The great anon is here
 
 
2 hours later…
1:57 AM
Hi!
Anyone can help me? I want to know what's the formula use in this answer : math.stackexchange.com/a/870044/160964.
 
2:18 AM
@mick well, Ramanujan sums are related to the theta functions and theta functions are the great voodoo hauptmoduls of the modforms so no wonder there is an arithmetic connection.
I think what is more interesting in the sense of analytic number theory are Kloosterman sums, which is a generalization of Ramanujan sums.
The connection with the sum of square function with Ramanujan sums are probably related to the so-called identity $$\sum_{n \geq 0} r_2(n) \exp(i2\pi nz) = \theta^2(z)$$, $\theta$ being one of the thetas.
So it's since recently that there have been quite a hubbub about Fourier series of modular forms.
@blue
 
2:55 AM
@BalarkaSen ?
 
3:13 AM
@DanielFischer hello i edited my message
 
3:46 AM
@BalarkaSen Can you teach me analytic continuation, I don't know really where to start.
 
4:20 AM
Just continue to be analytic and you get analytic continuation.
 
r9m
4:57 AM
AMM problem 4305 (April 1950) by H. F. Sandham $$\sum_{n=1}^{\infty} \dfrac{H_n^2}{n^2} = \dfrac{17\pi^2}{360}$$
@robjohn @Chris'ssis ^ looks like it was known before 1993 :)
 
@r9m It has probably been known for quite a while.
 
r9m
@robjohn but 1950 is the oldest reference I have seen .. maybe it was known b4 that too :-)
 
@r9m My guess is that it was.
However, these things are hard to search for even in electronically accessible sources. Before 1950, things are harder to find.
 
r9m
@robjohn I haven't seen the 1950 volume directly .. it was mentioned in 1957: list of 400 best problems from (1918-1950)
sorry typo $\pi^4$ there .. not $\pi^2$
@robjohn should it mean .. it shouldn't be called Au-Yeung Series if it was known before he found it ? :P
 
5:50 AM
@r9m Names are like that. They don't always go to the discoverer, but to the person who popularized something, or got a nice proof.
 
r9m
@robjohn like the Pell equation had little to do with Pell ? :P
 
@r9m I guess... I don't really know, perhaps this was discovered in 1950, but Euler Sums have been studied for a long time.
 
They should change the name of Fermat's last theorem :-)
 
@skullpatrol to what?
 
Huy
6:06 AM
Good morning everyone.
Anyone want to help me decide where to put my PS4? :D
 
6:20 AM
in the mail, addressed to moi
 
6:40 AM
Hello?
 
7:06 AM
@robjohn The Wiles theorem.
Or the Fermat/Wiles theorem.
:-)
 
 
1 hour later…
8:42 AM
@Huy what are your first impressions about it ?
 
Huy
@G.T.R I had one before but sold it to my brother and now I bought another one because it was on sale.
@G.T.R I really like it. But I don't like the cooling system integrated.
 
9:00 AM
@robjohn hello , can you help me on uniform convex space ? please
 
9:20 AM
@Jasper I'm sure if you asked nicely enough... ;)
 
9:48 AM
@skullpatrol Wiles found a proof, but Fermat is the one who made it popular. Popular consensus may change the name to Fermat-Wiles, but that will be seen.
 
10:01 AM
hello @DanielFischer
 
@blue LEL
C'mon. $\Bbb {F}_1$? Seriously?
 
mmhmm
 
Abstraction has a limit, man.
 
But, the imagination does not.
 
Mine does.
 
10:05 AM
Why?
 
@Vrouvrou what is the question?
 
@MatsGranvik I can get you through the ideas. Are you familiar with complex analysis? If not, get yourself a book.
 
@skullpatrol dunno. i guess i need a bit of motivation for everything.
 
True, the path one travels should be carefully chosen...
 
10:08 AM
And I am not used to take risks.
F1 seems. Well. So. Well.
...
@blue What the devil is that thing? Riemann hypothesis? F1 and RH?
What the?
 
@Vrouvrou Hi. I just posted an answer to your uniform convexity question.
 
@BalarkaSen imagine, if you will, that there are various families of constructions that attach themselves to finite fields. in case after case, there is a "degenerate object" that would be associated with F_1, except if one actually plugs the trivial ring into the construction one doesn't get anything meaningful. so this degenerate-object-F1 correspondence exists on some higher level, beyond our current definitions. hence the heightened abstraction necessary to make sense of this phenomena...
 
I don't get it. What's the construction? What's the object? How are we associating?
 
I said "various families of constructions" so there are many out there
see the "properties" section on wikipedia
and especially the q-binomial subspace-counting thing, which motivates sets being F_1 spaces
 
@blue Let me look.
 
10:18 AM
@DanielFischer thank you, have you seen my comments ?
 
Is $\sigma$ not always the identity function on $F$?
Given that it's an $F$-monomorphism.
 
the number of m dimensional subspaces of $\Bbb F_q^n$ is the q-binomial $\binom{n}{m}_q$. if one naively plugs in $q=1$, this formula counts the $m$-subsets of an $n$-element set, so that's what a space over F_1 should be
 
@blue hmmm
So we can have a sensible look at the n dimensional space over F_1, if not F_1.
is that right?
but it still doesn't convince me.
 
sure, one example shouldn't be too convincing (although that's a good one because it comes with a formula and plugging in q=1). but when it's example after example and expert intuition points the way...
that's a sign there is something going on behind the scenes
 
what's the most convincing sign of existence of F_1?
 
10:23 AM
I think it was more of a cumulative thing. I'm not aware of the current state of research, but they've no doubt covered some ground in actually characterizing a definition or a set-up. search for it on arxiv.
there was one really cool determinant formula for L functions I saw on there once
 
@blue L functions over what?
 
aha! "Mapping F_1-Land"! the very first equation. (actually it's just the completed zeta function, memory was foggy)
 
Let's see.
 
Gah, I misread it.
 
@Alyosha wat
@blue hmm. so they expect $\mathbf{Spec} \, \Bbb Z$ to be a curve over some space.
Wait I was under the impression that Spec Z can be realized as some manifold with a metric, given that it has the Zariski topology.
 
10:35 AM
heya guys
 
Hi @cirpis
 
Seems that JasperLoy will create a new account.
 
11:05 AM
@anon Yo.
 
we were just talking like two seconds ago
 
Well, I have to make the impression as if @blue and @anon are two different persons, no?
Superhero secrecy.
 
as you were then
 
Why don't you change your username to analytic-non and blue to algebraic-non? That'd make much more sense about having two accounts.
 
>2, more planned.
I will farm an army of 10k+ers and take over as supreme leader.
6
 
11:09 AM
ohhemgee.
yes, it seems like your blue account will quickly reach 10k.
@anon the SE managers would have to think about merging account feature to deal with you then.
are the other masks 10k+ too?
 
I have only just begun. patience is a virtue.
 
@anon Hmm anon. I was looking at this
I am not very interested in what connection there might be with Ford circles and the upper half plane under the action of J anyway, but the tiling with Farey sequence is something my professor told me about.
If I recall correctly, the tiling of the hyperbolic plane with Farey sequence is a powerful tool in diophantine approximations.
It gives you some kind of continued fraction (what it is, I forgot) that is used to derive some lower bounds.
I can't find the reference he sent me.
Oh well. I quickly jump from one thing to another. I should really go back concentrating about my commutative algebra.
Hey @Alizter
 
11:24 AM
hey @BalarkaSen
I am confused to why I have so many stars on the board
there is a serial starer on the loose
 
yes. i noticed that.
 
weird
 
@Alizter so, any fun math you have right now?
 
I was thinking about rubiks cube
 
meh.
 
11:32 AM
least possible amount of moves such that each square is adjacent to a distinct colour
I got 6
I don't think there are any more
 
not interested.
however, this just reminded me of the four-color problem.
 
See if I can mess around with differential fields
 
@Alizter this may interest you
 
I did some crazy isomorphisms a while back
I cant find it though :(
 
@Alizter what isomorphism?
 
11:34 AM
Glois groups of some d fields to some group
memory is fuzzy
 
Galois group of fields are nonsense. You mean galois group of an extension.
What is it?
 
you know what I mean
 
yeah, yeah, but what is the problem?
 
@BalarkaSen I don't have it with me
I will need to find it
 
oh well.
 
11:36 AM
All my notebooks are red and unmarked
doesn't help when there are 10+
I have another one of those function questions
 
all my notebooks are biologically locomotive and always tend to hide behind bookshelfs.
 
$f(x^2+yf(z))=xf(x)+zf(y)$
 
@Alizter No motivation.
Not interested.
 
Same thing, find all $f:\Bbb R \to \Bbb R$ and $x,y,z\in\Bbb R$
 
Set $(y, z) = (0, 0)$.
$f(x^2) = xf(x)$
Good luck figuring out the solutions to that.
 
11:42 AM
@Alizter $f(x^2)=x\,f(x)\implies f(x)=\tfrac{cx}{e^2}$
 
@Hakim knows the solution to every functional equation. =)
 
@BalarkaSen Nope, pure luck ;)
 
Set $x = y = z$. $2f(x^2) = xf(x) + xf(x) = f(x^2 + xf(x)) = f(x^2 + f(x^2))$. Set $x^2 = t$.
Just sprouting ideas.
$f(t+f(t)) = 2f(t)$
@Alizter
 
r9m
oh! Balarka taking interest in olympiad questions ?! :)
 
setting $x=0,y=1$ we get $f(f(z))=zf(1)$, thus the function is injective, right?
 
11:51 AM
@r9m No. I am killing time.
@cirpis Not only that but $f(f(z))$ is linear. Good point.
 
hi
 
also since $f(x^2)=xf(x)=xf(x)+zf(0)$ this means that $f(0)=0$
hello
 
@Hippalectryon You are just bathing in rep from chris'sis problem ;)
 
@Alizter Indeed :D
 
Set $x = 0$ and $z = 1$. $f(yf(1)) = f(y)$. So $f(f(f(t))) = f(tf(1)) = f(t)$.
 
11:55 AM
They are like blood diamonds except for the killing part.
Who stared that?
 
Not me.
 
@Hippalectryon are you troll staring?
 
@Alizter It's not me >:O
 
You can just delete the message you don't want to be starred.
 
@BalarkaSen I don't mind but it's just annoying.
The fact that it says Alizter for more than 50% of the board
They are obviously one person because there is single star
oh come on
wow
 
11:58 AM
I do mind starring arbitrary messages.
 
Would be good if we could see who stars ...
 
So stop.
 
they just suddenly went
 
Hello
 
@Alizter Shut that mouth of yours.
 
11:59 AM
Hello
 
Now seriously there are 2 persons doing this.
 
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