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Jonas Teuwen
Jonas Teuwen
4:00 PM
@J.M. I remember the stuff which name I was forgotten! It was nystatin I believe.
 
Ilya
Ilya
$x$
 
J. M.
J. M.
@JonasTeuwen That's the candida specialist, actually. :)
 
Ilya
Ilya
hooray! it works
 
Jonas Teuwen
Jonas Teuwen
@J.M. Nice.
@J.M. Amphotericin sounds quite nasty...
 
Ilya
Ilya
say "NO" to ugly bookmarks
 
Jonas Teuwen
Jonas Teuwen
4:01 PM
@Ilya Bookmarks? :D.
 
J. M.
J. M.
@JonasTeuwen It's the nastiest of them all...
 
Ilya
Ilya
yes
 
Jonas Teuwen
Jonas Teuwen
@Ilya Elaborate please!
 
Ilya
Ilya
guys, from category-theory or a set theory perspective
 
Jonas Teuwen
Jonas Teuwen
@J.M. Mm, I'll say no for the moment to that then. Doesn't sound like a toy.
 
Ilya
Ilya
4:02 PM
if I want to consider a function $f:X\times Y\to [0,1]$
 
Jonas Teuwen
Jonas Teuwen
@Ilya Hmm... What?
 
Ilya
Ilya
I usually have to write that consider a triple $(X,Y,f)$
 
J. M.
J. M.
@JonasTeuwen It's basically the nuclear weapon of the mushroom-killers...
 
David Wheeler
David Wheeler
@FrankScience perhaps this is helpful? thezensite.com/ZenTeachings/Miscellaneous/…
 
Jonas Teuwen
Jonas Teuwen
@J.M. I like good stuff, but that is maybe a bit too much 8-).
 
Ilya
Ilya
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 $f$?
 
Jonas Teuwen
Jonas Teuwen
@Ilya Other than domain and codomain?
Oh... wait. Had to run the ChatJax.
 
Ilya
Ilya
I know that it may work like this: we consider $g:A\to B$ then $A = \text{sor}(g)$ and $B = \text{tar}(g)$
 
Jonas Teuwen
Jonas Teuwen
@Ilya Did you now get into categories?
 
Ilya
Ilya
@Jonas: no :)
just for the notational stuff
 
Jonas Teuwen
Jonas Teuwen
@Ilya Hm... What is wrong with just calling it $X$ and $Y$?
The $\text{sor}(g)$ looks a bit... fuzzy.
 
Ed Gorcenski
Ed Gorcenski
4:05 PM
We need a 2012 tag, so that I can more easily identify which questions to ignore.
 
Ilya
Ilya
@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 $(P,\mathsf L)$
 
Jonas Teuwen
Jonas Teuwen
Mmm... finite alphabet and stochastics. Sounds kickass.
What the heck are you working on? 8-).
 
David Wheeler
David Wheeler
so L labels events?
 
Frank Science
Frank Science
@DavidWheeler Maybe this helps. I do not want to argue more.
@DavidWheeler Good night.
 
Ilya
Ilya
@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 :)
 
Jonas Teuwen
Jonas Teuwen
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.
 
David Wheeler
David Wheeler
why do you need to write the 5-tuple?
 
Ilya
Ilya
@DavidWheeler well, because I consider another tuple $(E,\tilde{\mathscr E}, \tilde P, \Sigma,\mathsf L)$
 
Jonas Teuwen
Jonas Teuwen
@Ilya Send it soon! I'd like to read some esoteric stuff.
 
David Wheeler
David Wheeler
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
 
Ilya
Ilya
@David and I just would prefer to write it as $(\tilde P,\mathsf L)$
 
David Wheeler
David Wheeler
4:14 PM
for example, the formula f(x) = 2x does not define a function
 
Ilya
Ilya
@DavidWheeler I know
 
David Wheeler
David Wheeler
we have to say what "x" and "2x" are, for f to even have a meaning
 
Ilya
Ilya
@Jonas: I can send a neat 2-page part of it now
want it?
 
Jonas Teuwen
Jonas Teuwen
@Ilya Please! Are you still going on a holiday?
 
Ilya
Ilya
?
oh, yeah. 1-15 Aug
 
Jonas Teuwen
Jonas Teuwen
4:15 PM
Cool, where are you going?
 
David Wheeler
David Wheeler
that is, P itself is actually a triple, yes?
(domain, co-domain, rule)
 
Ilya
Ilya
@JonasTeuwen Sent it to you
@DavidWheeler yes
 
David Wheeler
David Wheeler
i'm just saying, if you make the definitions longer, the references get shorter
 
Jonas Teuwen
Jonas Teuwen
@Ilya The accompanying text is already nice 8-).
 
Ilya
Ilya
I thought I have to justify myself
 
Jonas Teuwen
Jonas Teuwen
4:19 PM
@Ilya I am often also surprised by such things.
 
Ilya
Ilya
@Jonas why I. G.?
 
Jonas Teuwen
Jonas Teuwen
@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-).
 
Ilya
Ilya
oh, indeed
@JonasTeuwen nice to know
oops the system!
 
robjohn
robjohn
@Ilya: good day. anything exciting going on today?
 
Ilya
Ilya
yeah, I was swearing, sorry :(
slowly walking to his corner
@Jonas: please, tell me your feedback
 
Jonas Teuwen
Jonas Teuwen
4:24 PM
@Ilya I will tell you, but first I am still typing something. I'll read it tonight, okay? :-).
 
Ilya
Ilya
whenever you have time
 
Jonas Teuwen
Jonas Teuwen
Sure.
 
user19161
4:44 PM
@Ilya Swearing at who?
 
user19161
@Ilya One usually gives feedback and not tells feedback.
 
Zhen Lin
Zhen Lin
5:32 PM
Blah, there's a user posting textbook questions and answering them himself...
 
Henry T. Horton
Henry T. Horton
@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
 
Zhen Lin
Zhen Lin
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.
 
Henry T. Horton
Henry T. Horton
Posets without the Axiom of Choice
 
Zhen Lin
Zhen Lin
5:44 PM
mmm
 
user19161
Also I saw MK's latest meta question. Not much drama, but there was a little there.
 
Zhen Lin
Zhen Lin
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...
 
Henry T. Horton
Henry T. Horton
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 ;-) – 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.
 
Henry T. Horton
Henry T. Horton
9001
 
Ilya
Ilya
5:47 PM
.
 
user19161
@Ilya What?
 
robjohn
robjohn
@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.
 
Zhen Lin
Zhen Lin
To be fair, a question like that would attract a -1 even if he didn't answer it himself.
 
Peter Tamaroff
Peter Tamaroff
5:50 PM
@HenryT.Horton Wait, that is OVER 9000.
@ZhenLin Link?
 
Zhen Lin
Zhen Lin
See the latest category-theory questions.
 
Henry T. Horton
Henry T. Horton
@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.
 
Henry T. Horton
Henry T. Horton
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.
 
Henry T. Horton
Henry T. Horton
Because he's not even on earth.
 
user19161
You are so funny, Henry!
 
Henry T. Horton
Henry T. Horton
No one's laughing. Especially not me.
 
user19161
I am sorry to laugh then.
 
Henry T. Horton
Henry T. Horton
A $2012$ question! Quick, no one answer it!
1
Q: Functional Equation with Value

AndyIf $f$ 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 :(

homework functional-equations
 
user19161
5:58 PM
@HenryT.Horton I thought it would be nuked from orbit by now, given that it was posted 2 hours ago.
 
Matt N.
Matt N.
Hi bros.
 
Henry T. Horton
Henry T. Horton
Hiho
 
Old John
Old John
Hi all
 
user19161
@MattN. Hello Matt.
 
Matt N.
Matt N.
@OldJohn Hi bro.
: )
 
user19161
6:08 PM
@OldJohn Hello John.
 
Henry T. Horton
Henry T. Horton
That's Brofessor to you
6
 
Matt N.
Matt N.
: D
 
user19161
How many hellos are we going to say?
 
Old John
Old John
Enough!
 
Henry T. Horton
Henry T. Horton
Hi
 
Peter Tamaroff
Peter Tamaroff
6:09 PM
@OldJohn Go Leibniz!
 
Old John
Old John
@PeterTamaroff Leibniz? - which?
 
Peter Tamaroff
Peter Tamaroff
Gottfried Leibniz!
 
Henry T. Horton
Henry T. Horton
Leibniz rules!
 
user19161
Newton rules.
 
Peter Tamaroff
Peter Tamaroff
@JasperLoy We shall duel to death.
 
Old John
Old John
6:11 PM
@PeterTamaroff Yes - I assumed that one - just wondered why you are excited about him
 
user19161
@PeterTamaroff Yeah, like Galois...
 
Peter Tamaroff
Peter Tamaroff
@OldJohn You seem to be proud of your Newton fractals.
 
J. M.
J. M.
@Ilya, they seem to have not heard the thing about Galois...
 
Peter Tamaroff
Peter Tamaroff
@JasperLoy Well, no. If I was the genius he was, I would not be stupid enough to duel.
 
Old John
Old John
@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...
 
J. M.
J. M.
I'll let him tell the tale...
 
robjohn
robjohn
@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. :-)
 
Peter Tamaroff
Peter Tamaroff
6:15 PM
 
Matt N.
Matt N.
@robjohn Yes.
 
robjohn
robjohn
@JasperLoy I might >8(
 
Peter Tamaroff
Peter Tamaroff
@robjohn I have something to show you.
 
robjohn
robjohn
@PeterTamaroff Show me!
 
Henry T. Horton
Henry T. Horton
Can I see too?
 
user19161
6:19 PM
@J.M. I see you have transformed again!
 
Peter Tamaroff
Peter Tamaroff
@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.
 
J. M.
J. M.
@JasperLoy Congratulations.
 
user19161
Is it that only those who believe can see?
 
J. M.
J. M.
@JasperLoy I did say I was due for a molt...
 
robjohn
robjohn
6:21 PM
@PeterTamaroff nice collection. I think I have one similar to that.
 
J. M.
J. M.
@robjohn Sure. ;P
 
user19161
@robjohn Do you know why I see nothing there?
 
robjohn
robjohn
@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.
 
Gigili
Gigili
@JasperLoy The most number of smiles? What's that supposed to mean?
 
robjohn
robjohn
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.
 
Peter Tamaroff
Peter Tamaroff
Question for the peoples of the maths:
 
Gigili
Gigili
As a relevant note, it is compulsory for all motorcyclists to wear helmets.
 
Peter Tamaroff
Peter Tamaroff
@JasperLoy Good
 
user19161
@Gigili Smiles = :-)
 
robjohn
robjohn
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
 
Peter Tamaroff
Peter Tamaroff
@Gigili People would die!
You know what an angry biker can do, don't you?
 
robjohn
robjohn
@Gigili I believe in California, it is law that you need to wear a helmet while riding a motorcycle.
 
Gigili
Gigili
@JasperLoy Really? I thought :( was smile.
 
user19161
@peter What was that thing about? You made it yourself?
 
user19161
@Gigili Well, that too, but not usually the case.
 
Gigili
Gigili
6:26 PM
@PeterTamaroff Can kill us all with one bullet?
 
Peter Tamaroff
Peter Tamaroff
@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.
 
Gigili
Gigili
@robjohn I think it's law everywhere in the world, isn't it?
 
Peter Tamaroff
Peter Tamaroff
Define a coultrafunctofilter of a pocoset as the counion of invariant shifts....
 
robjohn
robjohn
@Gigili I don't know about that.
 
user19161
6:27 PM
@Gigili Do you know that there are over 9000 countries.
 
robjohn
robjohn
@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.
 
robjohn
robjohn
barbless barb-wire
 
Gigili
Gigili
@robjohn Well, at least it's law in half of the world.
 
anon
anon
@PeterTamaroff clearly that's just the Fourier transform
 
Gigili
Gigili
6:28 PM
@JasperLoy It was seven countries last time I checked.
 
robjohn
robjohn
@Gigili That could be.
 
J. M.
J. M.
@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.
 
Henry T. Horton
Henry T. Horton
So under 9000
 
user19161
@HenryT.Horton Yes, I was using what is known as a hyperbole there.
 
anon
anon
6:30 PM
 
Henry T. Horton
Henry T. Horton
Oh... hi Jasper, I didn't see you there
 
robjohn
robjohn
@J.M. It is much more recognizable as you in the 32x32 form
 
Peter Tamaroff
Peter Tamaroff
@anon Of course. That is trivial.
 
Old John
Old John
@anon He has a different test for convergence, I believe
 
Peter Tamaroff
Peter Tamaroff
Define convergence as.....
 
Gigili
Gigili
6:31 PM
@HenryT.Horton He's tiny like that.
 
Peter Tamaroff
Peter Tamaroff
@Gigili BUUUUUUUUUUUUUURN
 
user19161
@anon It converges in the extended reals to minus infinity. That guy is a genius bro!
 
Gigili
Gigili
Are you alright @Peter?
 
robjohn
robjohn
Another voteless acceptance :-) and the OP is not a newbie.
 
Henry T. Horton
Henry T. Horton
@Gigili So that's why I want to punch him in the face... he reminds me of myself
 
Peter Tamaroff
Peter Tamaroff
6:32 PM
@Gigili Why wouldn't I be alright?
 
Gigili
Gigili
Right, what about alleft?
 
robjohn
robjohn
@HenryT.Horton you have inward directed anger?
 
Gigili
Gigili
@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.
 
Peter Tamaroff
Peter Tamaroff
Gusfrava, people. Gusfrava
 
robjohn
robjohn
6:34 PM
@Gigili You want all of Jasper but his nose?
 
Old John
Old John
Having recently discovered p-adic convergence, I am a little hesitant to shout at newbies when they post questions about series that look divergent
 
robjohn
robjohn
@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...
 
Henry T. Horton
Henry T. Horton
@robjohn $-\frac{1}{12}$
 
robjohn
robjohn
@HenryT.Horton :-)
 
Gigili
Gigili
6:36 PM
@robjohn I'm not sure HHW can do more for us.
 
Old John
Old John
@JasperLoy I suspect many of us thought that - before we were "educated"
 
J. M.
J. M.
@OldJohn I start with the assumption that everything is formal, and worry about convergence later...
 
robjohn
robjohn
@HenryT.Horton I proved that ($\zeta(-1)=-\frac1{12}$) somewhere using the Euler-Maclaurin Sum formula.
 
Old John
Old John
@J.M. That is probably a good idea - I think Euler would have approved
 
user19161
@Gigili Who is HHW? Isn't Henry HTH?
 
Old John
Old John
6:37 PM
@robjohn Divergent by almost all definitions of divergence?
 
J. M.
J. M.
@JasperLoy Henry the Horton?
 
user19161
@J.M. I doubt it, since he capitalizes the T.
 
robjohn
robjohn
@OldJohn yes, except as the analytic continuation of a function of the same form.
 
Gigili
Gigili
Horton Hears a Who.
 
anon
anon
"regularization"
 
Henry T. Horton
Henry T. Horton
6:38 PM
@robjohn Yes I was referring to $\zeta(-1)$ :)
 
user19161
@Gigili Where did that come from?
 
Old John
Old John
@robjohn Ah yes - zeta
 
robjohn
robjohn
@HenryT.Horton I figured so, therefore, my comment
 
J. M.
J. M.
Abel and Borel are quite useful when you need them...
 
Old John
Old John
@J.M. Abel summability?
 
Gigili
Gigili
6:39 PM
@JasperLoy Ask @HHW.
 
J. M.
J. M.
@OldJohn Yes, that.
 
Old John
Old John
@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.
 
J. M.
J. M.
A nice classic.
 
user19161
One treatise on infinite series is the one by Knopp: Theory and applications of infinite series
 
Old John
Old John
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.
 
J. M.
J. M.
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.
 
Old John
Old John
@J.M. I always shudder when people mention addition and primes in the same sentence :)
 
robjohn
robjohn
@J.M. why is it that you can't sum the primes?
 
Peter Tamaroff
Peter Tamaroff
@J.M. But why?!
You don't have a pattern to follow?
Like asking to sum to digits of $\pi$, say?
 
J. M.
J. M.
6:44 PM
@robjohn I talked about that here.
 
Peter Tamaroff
Peter Tamaroff
I mean, the sum of $\zeta(-1)$ depends on knowing "what comes after".
 
robjohn
robjohn
Does $\sum\limits_{p\text{ odd prime}}\frac{(-1)^{\frac{p-1}{2}}}{p}$ converge?
 
anon
anon
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.
 
robjohn
robjohn
@J.M. Cool! I have never seen that before.
 
Old John
Old John
@robjohn wouldn't that depend on results about asymptotic densities of primes of form $4n+1$ and $4n+3$?
 
anon
anon
6:49 PM
@OldJohn they have the same asymptotic density (quick addendum to Dirichlet's theorem), but they have different lower-order growth asymptotics
 
Peter Tamaroff
Peter Tamaroff
 
Old John
Old John
@anon Ah - yes - and my comment about Skewes was rubbish too - different result :(
(memory not as sharp as it used to be)
 
Peter Tamaroff
Peter Tamaroff
I know the modern symbol for $\operatorname{Bdry}A$ is $\partial A$. What would we use for $\operatorname{Int}A$?
 
Old John
Old John
I used $A$ with a little o superscript
 
anon
anon
prime number races is relevant
 
Peter Tamaroff
Peter Tamaroff
6:51 PM
$A^o$?
 
Old John
Old John
@PeterTamaroff That's what I used, yes
 
Peter Tamaroff
Peter Tamaroff
BTW why on Earth do we use $\partial$ for the boundary?
 
Henry T. Horton
Henry T. Horton
@OldJohn He asked for MODERN ;)
 
Peter Tamaroff
Peter Tamaroff
@HenryT.Horton Mean!
 
Henry T. Horton
Henry T. Horton
It says "old" right there in his name!
 
robjohn
robjohn
6:53 PM
We leave for Mammoth tomorrow for a short vacation. We'll be on the road for about 5 hours or so.
 
Old John
Old John
@HenryT.Horton ;)
@anon Thanks - I am picking up some great reading material from this room :) (mostly from you, I think)
 
Henry T. Horton
Henry T. Horton
Well I'm off to Analfest 2012
To work on my analysis skillz
 
Old John
Old John
@HenryT.Horton Don't go! - It has 2012 in the title!
 
Ed Gorcenski
Ed Gorcenski
Good afternoon!
 
Peter Tamaroff
Peter Tamaroff
@JasperLoy This song is awesome
 
J. M.
J. M.
7:01 PM
Hello @Ed.
Well, gotta go, but before that...
 
robjohn
robjohn
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 $1+2+3+4+5+6+\dots=-\frac1{12}$.
 
J. M.
J. M.
...and with that: later, y'all.
 
robjohn
robjohn
@J.M. Are they just asking about $\frac{\pi^2}{6}$ or the EMSF in general?
 
Old John
Old John
@J.M. Bye
 
robjohn
robjohn
@J.M. See you later. I hope before we leave for Mammoth.
 
J. M.
J. M.
7:06 PM
@robjohn Apparently, you'll have to address both; he doesn't get what an asymptotic series is.
Now, later, for real.
 
robjohn
robjohn
@J.M. That's what I feared
 
Ed Gorcenski
Ed Gorcenski
Shocked at how many comments regarding lack of understanding of "total derivative". Thought this was a fairly common, if seldom-used, thing.
 
Gigili
Gigili
@EdGorcenski Whatcha talk about? A Q on main or what?
And 'Ello.
 
Ed Gorcenski
Ed Gorcenski
 
Gigili
Gigili
@EdGorcenski Nice one.
 
anon
anon
7:19 PM
I asked this before, but anybody know the algorithm that determines the order tags are displayed on questions?
 
Ed Gorcenski
Ed Gorcenski
Is it not the order in which the author chooses?
 
anon
anon
no
 
Ed Gorcenski
Ed Gorcenski
Oh, thanks @Gigili, now my rep is 666.
;)
 
Gigili
Gigili
@EdGorcenski Probably from the one with most questions being tagged as?
 
Ed Gorcenski
Ed Gorcenski
That would be my second guess
 
anon
anon
7:24 PM
 
Peter Tamaroff
Peter Tamaroff
@anon HWhy?
 
anon
anon
I made the call for dupe votes before Qiaochu closed it.
 
Peter Tamaroff
Peter Tamaroff
Oh, OK.
 
anon
anon
Wait, did you vote for reopening?
 
Peter Tamaroff
Peter Tamaroff
I thought you wanted to reopen it!
 
anon
anon
7:28 PM
You minion you.
 
Gigili
Gigili
@EdGorcenski I don't know what's the point of it. I mean I don't care in whatever order they are displayed.
 
Ed Gorcenski
Ed Gorcenski
I don't know, @Gigili, I was just conjecturing the answer to @anon's question
 
Gigili
Gigili
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.
 
Ed Gorcenski
Ed Gorcenski
@Gigili do you mean in my "total derivative" answer?
 
Gigili
Gigili
7:49 PM
@EdGorcenski Uhum.
 
Peter Tamaroff
Peter Tamaroff
How is the Cantor set defined inductively?
i.e if $C=\bigcap_{n=0}^\infty C_n$, what is $C_n;n=1,2,\dots$
 
Old John
Old John
@PeterTamaroff Isn't it defined by chopping out the middle third of each interval in the preceding $C_n$?
 
Ed Gorcenski
Ed Gorcenski
@Gi
@Gigili I don't understand, either. That's what surprised me, too.
 
Peter Tamaroff
Peter Tamaroff
@OldJohn Yes.
But I'd like an inductive definition
 
Old John
Old John
@PeterTamaroff In what way inductive?
 
Peter Tamaroff
Peter Tamaroff
7:55 PM
@OldJohn $c_n$ in terms of $c_{n-1},c_{n-2},\dots$
 
Jonas Teuwen
Jonas Teuwen
Hi.
 
Old John
Old John
@PeterTamaroff $C_n$ can easily be expressed in terms of $C_{n-1}$
Wikipedia pages does it
@JonasTeuwen Hi Jonas
 
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