4:00 PM
@J.M. I remember the stuff which name I was forgotten! It was nystatin I believe.

$x$

@JonasTeuwen That's the candida specialist, actually. :)

hooray! it works

@J.M. Nice.
@J.M. Amphotericin sounds quite nasty...

say "NO" to ugly bookmarks

4:01 PM
@Ilya Bookmarks? :D.

@JonasTeuwen It's the nastiest of them all...

yes

guys, from category-theory or a set theory perspective

@J.M. Mm, I'll say no for the moment to that then. Doesn't sound like a toy.

4:02 PM
if I want to consider a function $f:X\times Y\to [0,1]$

@Ilya Hmm... What?

I usually have to write that consider a triple $(X,Y,f)$

@JonasTeuwen It's basically the nuclear weapon of the mushroom-killers...

@FrankScience perhaps this is helpful? thezensite.com/ZenTeachings/Miscellaneous/…

@J.M. I like good stuff, but that is maybe a bit too much 8-).

4:03 PM
but I wonder: if $f:X\times Y \to[0,1]$ - is there a nice way to call $X$ and $Y$ from $f$?

@Ilya Other than domain and codomain?
Oh... wait. Had to run the ChatJax.

I know that it may work like this: we consider $g:A\to B$ then $A = \text{sor}(g)$ and $B = \text{tar}(g)$

@Ilya Did you now get into categories?

@Jonas: no :)
just for the notational stuff

@Ilya Hm... What is wrong with just calling it $X$ and $Y$?
The $\text{sor}(g)$ looks a bit... fuzzy.

4:05 PM
We need a 2012 tag, so that I can more easily identify which questions to ignore.

@Jonas look: I have a measurable space $(E,\mathscr E)$ and a stochastic kernel there $P:E\times \mathscr E\to [0,1].$ I also have a finite alphabet $\Sigma$ and a labeling map $\mathsf L:(E,\times \mathscr E)\to (\Sigma,2^\Sigma)$. So I consider such tuples - and each time I have to write $(E,\mathscr E,P,\Sigma,\mathsf L)$ instead of just $(P,\mathsf L)$

Mmm... finite alphabet and stochastics. Sounds kickass.
What the heck are you working on? 8-).

so L labels events?

@DavidWheeler Maybe this helps. I do not want to argue more.
@DavidWheeler Good night.

@Jonas because $E,\mathscr E$ are encoded in $P$ and $\Sigma$ is encoded in $\mathsf L$
states. But also in naturally extends to events: to each trajectory of states $x_0,x_1,\dots$ there corresponds a word
$${\sf L}(x_0){\sf L}(x_1)\dots$$
@JonasTeuwen I'll send you later the draft. There are some measurable languages - I've introduced them for my own needs - and then found that someone used it for totally different purposes. And he is the only one who used that notion, almost :)

4:10 PM
@Ilya Thanks! I would love that. Are you going on a holiday?
@Ilya Oh cool! I had that a couple of times too with different things. I love that. Well, the confirmation that it is not retarded at least, not being scooped.

why do you need to write the 5-tuple?

@DavidWheeler well, because I consider another tuple $(E,\tilde{\mathscr E}, \tilde P, \Sigma,\mathsf L)$

@Ilya Send it soon! I'd like to read some esoteric stuff.

ok, but if it's a different "P", for it to even be meaningful at all, you must have the source and targets set already

@David and I just would prefer to write it as $(\tilde P,\mathsf L)$

4:14 PM
for example, the formula f(x) = 2x does not define a function

@DavidWheeler I know

we have to say what "x" and "2x" are, for f to even have a meaning

@Jonas: I can send a neat 2-page part of it now
want it?

@Ilya Please! Are you still going on a holiday?

?
oh, yeah. 1-15 Aug

4:15 PM
Cool, where are you going?

that is, P itself is actually a triple, yes?
(domain, co-domain, rule)

@JonasTeuwen Sent it to you
@DavidWheeler yes

i'm just saying, if you make the definitions longer, the references get shorter

@Ilya The accompanying text is already nice 8-).

I thought I have to justify myself

4:19 PM
@Ilya I am often also surprised by such things.

@Jonas why I. G.?

@Ilya I also found some way to do the same estimates much easier (or actually sometimes they are even way better).
@Ilya Because he also did these things. Maybe a bit higher level, but oh well 8-).

oh, indeed
@JonasTeuwen nice to know
oops the system!

@Ilya: good day. anything exciting going on today?

yeah, I was swearing, sorry :(
slowly walking to his corner

4:24 PM
@Ilya I will tell you, but first I am still typing something. I'll read it tonight, okay? :-).

whenever you have time

Sure.

user19161
4:44 PM
@Ilya Swearing at who?

user19161
@Ilya One usually gives feedback and not tells feedback.

5:32 PM
Blah, there's a user posting textbook questions and answering them himself...

@ZhenLin Well the bad part is that they are very basic things and I don't think it helps anyone out to have them archived here

They are indeed very basic.

user19161
Yes, I saw them. I hope he does not post 9000 questions of this type here.

user19161
I also hope he does not turn out to be another MK.

Posets without the Axiom of Choice

5:44 PM
mmm

user19161
Also I saw MK's latest meta question. Not much drama, but there was a little there.

I did a little bit on constructive order theory. It turns out, for example, that one needs the axiom of choice to show that every preorder is equivalent to a poset...

Thanks for the feedback. I want to have a record of my work. So far I have only worked out one and a half exercises from this text, so it shouldn't overwhelm the site ;-) â€“ Code-Guru 7 mins ago
Sounds like this guy IS a MK

user19161
@HenryT.Horton I should ask: how many exercises are in the book? LOL.

9001

5:47 PM
.

user19161
@Ilya What?

@JasperLoy It was a short sentence. About as short as they come.

user19161
Who downvoted his two questions? Confess!

user19161
I think a downvote is too harsh really.

To be fair, a question like that would attract a -1 even if he didn't answer it himself.

5:50 PM
@HenryT.Horton Wait, that is OVER 9000.

See the latest category-theory questions.

@JasperLoy How do we know YOU didn't downvote it!?

user19161
@PeterTamaroff See code-guru's latest questions.

user19161
@HenryT.Horton Well I say now that I did not, and everyone knows I do not tell lies.

I don't know that. Who exactly IS this Jasper Loy character? What do we really know about him? Can he be trusted?

user19161
5:53 PM
@HenryT.Horton He is the last person on earth to backstab you.

Because he's not even on earth.

user19161
You are so funny, Henry!

No one's laughing. Especially not me.

user19161
I am sorry to laugh then.

A $2012$ question! Quick, no one answer it!
1

If $f$ is a strictly increasing function from the naturals to the naturals, and $f(f(x))=3x$, what are all values of $f(2012)$? I have only proven that $f(3x)=3f(x)$ but that get's nowhere :(

user19161
5:58 PM
@HenryT.Horton I thought it would be nuked from orbit by now, given that it was posted 2 hours ago.

Hi bros.

Hiho

Hi all

user19161
@MattN. Hello Matt.

@OldJohn Hi bro.
: )

user19161
6:08 PM
@OldJohn Hello John.

That's Brofessor to you
6

: D

user19161
How many hellos are we going to say?

Enough!

6:09 PM
@OldJohn Go Leibniz!

@PeterTamaroff Leibniz? - which?

Gottfried Leibniz!

Leibniz rules!

user19161
Newton rules.

@JasperLoy We shall duel to death.

6:11 PM
@PeterTamaroff Yes - I assumed that one - just wondered why you are excited about him

user19161
@PeterTamaroff Yeah, like Galois...

@OldJohn You seem to be proud of your Newton fractals.

@Ilya, they seem to have not heard the thing about Galois...

@JasperLoy Well, no. If I was the genius he was, I would not be stupid enough to duel.

@PeterTamaroff Ah yes - I spent some days investigating solving cubics using Newton's method about 35 years ago - before I had heard about fractals

user19161
6:13 PM
@PeterTamaroff You are better than him!

user19161
@J.M. I wonder what this has got to do with Ilya...

I'll let him tell the tale...

@J.M. can I still blame you for bringing it up? :-)

user19161
@J.M. I think I have been infected with your ellipsis virus...

user19161
@robjohn Also, I think you have the most number of smiles in this room. :-)

6:15 PM

@robjohn Yes.

@JasperLoy I might >8(

@robjohn I have something to show you.

@PeterTamaroff Show me!

Can I see too?

user19161
6:19 PM
@J.M. I see you have transformed again!

@J.M. I will ask you for some fancy graphs when I get to the last pages of my book topology

user19161
@PeterTamaroff I see nothing there.

@JasperLoy Congratulations.

user19161
Is it that only those who believe can see?

@JasperLoy I did say I was due for a molt...

6:21 PM
@PeterTamaroff nice collection. I think I have one similar to that.

@robjohn Sure. ;P

user19161
@robjohn Do you know why I see nothing there?

@JasperLoy can your browser display PDFs?

user19161
@robjohn It can't if there is no PDF plugin right? Well, my FF has no PDF plugin now.

@JasperLoy The most number of smiles? What's that supposed to mean?

6:23 PM
@JasperLoy Then that might be the reason.
@JasperLoy Except I can see it and I don't have one either.

user19161
@robjohn Oh I can just download it using the icon on the top right @peter, so I see it now externally.

Question for the peoples of the maths:

As a relevant note, it is compulsory for all motorcyclists to wear helmets.

@JasperLoy Good

user19161
@Gigili Smiles = :-)

6:24 PM
@JasperLoy Whatever the software they use to convert it to some thing I can see is pretty sad. Perhaps they don't have one for your browser/OS combination

@Gigili People would die!
You know what an angry biker can do, don't you?

@Gigili I believe in California, it is law that you need to wear a helmet while riding a motorcycle.

@JasperLoy Really? I thought :( was smile.

user19161

user19161
@Gigili Well, that too, but not usually the case.

6:26 PM
@PeterTamaroff Can kill us all with one bullet?

@JasperLoy No but it is a cool collection to use.
Specially for cranks that wanna make new math

user19161
@PeterTamaroff If you are talking about symbols, a lot of LaTeX guides have a list of that.

@robjohn I think it's law everywhere in the world, isn't it?

Define a coultrafunctofilter of a pocoset as the counion of invariant shifts....

@Gigili I don't know about that.

user19161
6:27 PM
@Gigili Do you know that there are over 9000 countries.

@J.M.: your gravatar looks like a bike lock for someone without a bike.

user19161
In some places I guess you need it for a bike as well.

barbless barb-wire

@robjohn Well, at least it's law in half of the world.

@PeterTamaroff clearly that's just the Fourier transform

6:28 PM
@JasperLoy It was seven countries last time I checked.

@Gigili That could be.

@robjohn It's in the same style as the Ouroboros torus I did weeks ago...

user19161
@Gigili Well, the actual number is probably in between 7 and 9000.

So under 9000

user19161
@HenryT.Horton Yes, I was using what is known as a hyperbole there.

Oh... hi Jasper, I didn't see you there

@J.M. It is much more recognizable as you in the 32x32 form

@anon Of course. That is trivial.

@anon He has a different test for convergence, I believe

Define convergence as.....

6:31 PM
@HenryT.Horton He's tiny like that.

@Gigili BUUUUUUUUUUUUUURN

user19161
@anon It converges in the extended reals to minus infinity. That guy is a genius bro!

Are you alright @Peter?

Another voteless acceptance :-) and the OP is not a newbie.

@Gigili So that's why I want to punch him in the face... he reminds me of myself

6:32 PM
@Gigili Why wouldn't I be alright?

@HenryT.Horton you have inward directed anger?

@HenryT.Horton Just punch him on the nose and leave the rest to me.

user19161
@Gigili You can't be sure I have a nose, but I do have one now.

Gusfrava, people. Gusfrava

6:34 PM
@Gigili You want all of Jasper but his nose?

Having recently discovered p-adic convergence, I am a little hesitant to shout at newbies when they post questions about series that look divergent

@OldJohn What is $1+2+3+4+5+\dots$?

user19161
@OldJohn When I was in high school, I thought that 1+1/2+1/3 would converge...

@robjohn $-\frac{1}{12}$

@HenryT.Horton :-)

6:36 PM
@robjohn I'm not sure HHW can do more for us.

@JasperLoy I suspect many of us thought that - before we were "educated"

@HenryT.Horton I proved that ($\zeta(-1)=-\frac1{12}$) somewhere using the Euler-Maclaurin Sum formula.

@J.M. That is probably a good idea - I think Euler would have approved

user19161
@Gigili Who is HHW? Isn't Henry HTH?

6:37 PM
@robjohn Divergent by almost all definitions of divergence?

@JasperLoy Henry the Horton?

user19161
@J.M. I doubt it, since he capitalizes the T.

@OldJohn yes, except as the analytic continuation of a function of the same form.

Horton Hears a Who.

"regularization"

6:38 PM
@robjohn Yes I was referring to $\zeta(-1)$ :)

user19161
@Gigili Where did that come from?

@robjohn Ah yes - zeta

@HenryT.Horton I figured so, therefore, my comment

Abel and Borel are quite useful when you need them...

@J.M. Abel summability?

6:39 PM

@OldJohn Yes, that.

@J.M. I once borrowed Hardy's book "Divergent Series" from a library, but can't remember much of it

user19161
@OldJohn I think he has a Fourier Series as well.

A nice classic.

user19161
One treatise on infinite series is the one by Knopp: Theory and applications of infinite series

6:41 PM
@JasperLoy He was a pretty good writer - I have "Theory of Numbers" and delve into it quite often
@JasperLoy Very thorough, but a bit boring, I found

user19161
@OldJohn Though his style is archaic now.

It's funny, though. You can "sum" the natural numbers, but you can't "sum" the primes. It's not immediately obvious why this is so.

@J.M. I always shudder when people mention addition and primes in the same sentence :)

@J.M. why is it that you can't sum the primes?

@J.M. But why?!
You don't have a pattern to follow?
Like asking to sum to digits of $\pi$, say?

6:44 PM
@robjohn I talked about that here.

I mean, the sum of $\zeta(-1)$ depends on knowing "what comes after".

Does $\sum\limits_{p\text{ odd prime}}\frac{(-1)^{\frac{p-1}{2}}}{p}$ converge?

I don't think the L-function associated to nontrivial Dirichlet character mod 4 has a pole at s=1, so I expect it should.

@J.M. Cool! I have never seen that before.

@robjohn wouldn't that depend on results about asymptotic densities of primes of form $4n+1$ and $4n+3$?

6:49 PM
@OldJohn they have the same asymptotic density (quick addendum to Dirichlet's theorem), but they have different lower-order growth asymptotics

@anon Ah - yes - and my comment about Skewes was rubbish too - different result :(
(memory not as sharp as it used to be)

I know the modern symbol for $\operatorname{Bdry}A$ is $\partial A$. What would we use for $\operatorname{Int}A$?

I used $A$ with a little o superscript

prime number races is relevant

6:51 PM
$A^o$?

@PeterTamaroff That's what I used, yes

BTW why on Earth do we use $\partial$ for the boundary?

@OldJohn He asked for MODERN ;)

@HenryT.Horton Mean!

It says "old" right there in his name!

6:53 PM
We leave for Mammoth tomorrow for a short vacation. We'll be on the road for about 5 hours or so.

@HenryT.Horton ;)
@anon Thanks - I am picking up some great reading material from this room :) (mostly from you, I think)

Well I'm off to Analfest 2012
To work on my analysis skillz

@HenryT.Horton Don't go! - It has 2012 in the title!

Good afternoon!

@JasperLoy This song is awesome

7:01 PM
Hello @Ed.
Well, gotta go, but before that...

You can sum the integers with $\zeta(-1)$ or you can use the relation used with
the $\zeta$ function that
\begin{align} &1-2+3-4+5-6+\dots\\ &=(1+2+3+4+5+6+\dots)-2(2+4+6+\dots)\\ &=-3(1+2+3+4+5+6+\dots)\tag{1} \end{align}
and then note that
$$\sum_{k=0}^\infty(-1)^k(k+1)x^k=\frac1{(1+x)^2}\tag{2}$$
Evaluating $(2)$ at $x=1$ and using $(1)$ gives you $1+2+3+4+5+6+\dots=-\frac1{12}$.

...and with that: later, y'all.

@J.M. Are they just asking about $\frac{\pi^2}{6}$ or the EMSF in general?

@J.M. Bye

@J.M. See you later. I hope before we leave for Mammoth.

7:06 PM
@robjohn Apparently, you'll have to address both; he doesn't get what an asymptotic series is.
Now, later, for real.

@J.M. That's what I feared

Shocked at how many comments regarding lack of understanding of "total derivative". Thought this was a fairly common, if seldom-used, thing.

@EdGorcenski Whatcha talk about? A Q on main or what?
And 'Ello.

@EdGorcenski Nice one.

7:19 PM
I asked this before, but anybody know the algorithm that determines the order tags are displayed on questions?

Is it not the order in which the author chooses?

no

Oh, thanks @Gigili, now my rep is 666.
;)

@EdGorcenski Probably from the one with most questions being tagged as?

That would be my second guess

7:24 PM

@anon HWhy?

Oh, OK.

Wait, did you vote for reopening?

I thought you wanted to reopen it!

7:28 PM
You minion you.

@EdGorcenski I don't know what's the point of it. I mean I don't care in whatever order they are displayed.

I don't know, @Gigili, I was just conjecturing the answer to @anon's question

I'm wondering why there are so many why this, I am wondering and it's so confusing comments when your answer is quite straight forward.

@Gigili do you mean in my "total derivative" answer?

7:49 PM
@EdGorcenski Uhum.

How is the Cantor set defined inductively?
i.e if $C=\bigcap_{n=0}^\infty C_n$, what is $C_n;n=1,2,\dots$

@PeterTamaroff Isn't it defined by chopping out the middle third of each interval in the preceding $C_n$?

@Gi
@Gigili I don't understand, either. That's what surprised me, too.

@OldJohn Yes.
But I'd like an inductive definition

@PeterTamaroff In what way inductive?

7:55 PM
@OldJohn $c_n$ in terms of $c_{n-1},c_{n-2},\dots$

Hi.

@PeterTamaroff $C_n$ can easily be expressed in terms of $C_{n-1}$