12:00 AM
Thank you for your time @robjohn, @tb and @Srivatsan.

It is today, once again.

It was a new day yesterday, but it's an old day now.
2

12:12 AM
I wonder if in this question \lambda(dx) means d\lambda(x) or does it mean something else?

That seems to confirm that they mean the same. Too many notations!
Salieri said something like that :-)
It was Emperor Joseph II

1:07 AM
BBL

See you later, @robjohn

1:33 AM
@robjohn Sometimes Bunyakovski, sometimes Bunyakowsky (the transliteration I'm familiar with), and sometimes other spellings. At least his situation's not as complicated as our friend Pafnuti...

2:19 AM
hey, @J.M., I am looking for old questions asking for books/references on basic combinatorics and counting. How do I search for questions fitting a particular tag?
In this case, [reference-request]?
Oh, of course. :) I found it in the faq.
Thanks!

Oh. Glad you found it. You know you can combine tag searches, right?

Yes, I am doing it.

(I was busy editing, so I didn't see you until the "Thanks!")

3:06 AM
Yes, thanks.

3:50 AM
Hi everybody! Can anybody help me with a cryptography related question that I have?

Hi. It might interest you to know that there's a crypto.SE; you might be able to get more expedient help there, as the people who are good with crypto aren't in this room at the moment...

3 hours later…
6:27 AM
It seems difficult to be confused about the meaning of 1_{[0,x]}
at least in the given context

Does anyone have their Ahlfors in front of them? I have a quick question.

6:51 AM
I can get it if needed.

I think I have it figured out.
He just implicitly asserts but does not prove that the criteria for an infinite product to be analytic is the same as an infinite sum of functions: we have to have uniform convergence on compact subsets, right?

@Potato where are you looking?

In the third edition, page 194, the paragraph after equation (17)

Morning! Just a quick hello before I do some work.

@Matt hello!

6:58 AM
@Potato The next paragraph after that paragraph gives the conditions for absolute convergence.
He doesn't prove it, but it is a pretty standard result.

@robjohn Ok, thanks. I was just looking for clarification there.

@Potato This is the Weierstrass factorization that was being discussed here yesterday

@robjohn Yes, that's on the next page.
@robjohn Ahlfors's proof of Stirling's formula still makes me cry. I'm going to have to write it up myself with all the details to really understand it, one of these days.

@Potato I don't know if my derivation is any better, but it might be worth a look.
Beware the ASCII art.

@robjohn I'll take a look. Thanks.
So we skipped harmonic functions in class. Could you explain why log |f(z)| is harmonic when f(z) is not 0? Does the absolute value sign screw up taking the necessary derivatives?

7:22 AM
No, I assume that you are taking f to be holomorphic.

@robjohn Yes, we assume f is holomorphic. Specifically, why is f holomorphic and why does (43) on page 205 hold?

Then, log|f(z)| = Re(log(f(z))) and the real part of a holomorphic function is harmonic

Oh yes, indeed.

and log(f(z)) is well-defined if f is not 0 in the domain
and holomorphic

Ok, so I understand now why it's harmonic. I'm still a bit confused about (43) though.

7:27 AM
I have (43) on page 207 the mean value theorem?

Thanks for the reference on that, I'll see if I can untangle it myself.

yes
I have it figured out now, I've just never seen the mean value theorem before.

@robjohn: how do you display tex in chat?
oh
no
it's just a gif

http://latex.codecogs.com/gif.latex?\log|f(0)|=\frac{1}{2\pi}\int_0^{2\pi}\log|‌​f(re^{i\theta})|\;\mathrm{d}\theta

7:36 AM
@robjohn I'm wondering if you can get that specific formula just from Cauchy's integral formula. We didn't use the mean value theorem in class and I can't decipher my notes.

@Potato You did harmonic functions without doing the mean value theorem?

We never did harmonic functions.

Ah
No, pass the URL to the dialog for upload...

Yay!

There :-)

7:37 AM
so Cauchy's gives us the formula but without the absolute values

Cool, thanks, @robjohn!

@Matt Note that the URL is pinned to the chatroom in the quotes on the right of the page.
It is pinned at the top (the hollow star)

Yours is, because it got starred, but mine isn't
or not yet maybe

Mine is pinned so that people can use the codecogs URL

@robjohn so using cauchy with a circle around the origin gives us the same formula but without absolute values...

7:40 AM
Asaf, who owns this room, pinned it.
@Potato yes
So take the real part of both sides....
and what do you get?

@robjohn Ah, ok
many thanks!

@robjohn: I see. I bookmarked it to the Chrome tab while you wrote that.
Ok, I'll see you later folks!

You have to remember that if your LaTeX has spaces, escape them as %20
Oh, well...
@JM!
@JM Good afternoon

Good afternoon.

Afternoon?
It's 2 am here.

7:49 AM
It's still Monday night here :-)
for another 10 minutes

I need to finish reviewing my notes on Jensen's formula and get to bed so I don't miss lecture tomorrow.
:(

@Potato: you are on the east coast?
and Jensen's is very important.

@Potato Ah, well, it's 16:00 where I am...

@JM 15:50

Okay, almost... :D
Ten minutes is short...

7:52 AM
@robjohn Midwest
@robjohn if I was a good student, I would also read ahead and look at the proof of Hadamard's factorization theorem before lecture tomorrow so I don't get lost, but alas it is late

@Potato Ah, yes. I thought, it was 11 here, and you said 2 there, but it is almost 12 here. My son is in Chicago, so he is in the same timezone

Coincidentally, I almost went to UChicago...

My son is at Columbia College Chicago

Very spiffy. I'm not up to speed on my art colleges but I do know that one.

He is majoring in game design

7:56 AM
@robjohn What does that mean? Is it oriented towards arts or science?

@robjohn how do you feel about that?

@Srivatsan It is an arts oriented program, he wants to write backstories or articles about games.

I was brought up to only consider a liberal arts education. No vocational training for me.

@Potato It is a new field since I went to college, but it seems to be a good one, especially since he is not into the art or the programming, he may find a niche.

@robjohn Ok, thanks.

7:58 AM
@Srivatsan He is a writer, and he does well at it.

aw well I need to get going to bed
night everybody
thanks for the help

@robjohn Um, nice.
@Potato 'night.

@Potato sleep well

@Srivatsan Ouch.
It is hard to tell if the OP is confused or just can't express what (s)he means.

8:03 AM
@robjohn =)
By the way, it seems that sqrt(3)+sqrt(2) seems to be a pretty good approximation for pi. :) I am currently breaking my head about it.
In contrast, the weaker upper bound of 2 sqrt(3) is much easier.

@JM The point of the "magic typing" comment?

Yes.

@JM :)

I think it is because of the immense amount of typing that that seems to have taken.

Or may be, magic typesetting?

8:22 AM
I scripted it... :) I'm lazy.

The entry of J^{-1} looks impressive.

(generating the matrices, that is)

Ugh...
sorry, wrong link. // now corrected. :)

@Srivatsan Hmm. Leslie Lamport wrote most of the early documentation for TeX and LaTeX

@robjohn True, but why is that relevant here?

8:30 AM
Well, what part of Lamport's stuff is making you go "ugh"?

@Srivatsan does it need to be relevant? I was just recognizing the name. It might be relevant in that he must have thought about writing up proofs while thinking about LaTeX.

@JM The proofs seem to be too low-level for me. As it is, I am struggling to connect the concepts. If every paper is going to be such an overload of routine information, I am not sure I can cope up... =)
@robjohn Hmm, you might be right.

Well, who decides what's routine? :)
(easy: the audience)

@JM The author, keeping in mind the target audience. =)

@Srivatsan He may be thinking of writing up proofs that a computer could parse and digest. Many of the leaps we make in human proofs are too big for a computer to grok.
The audience is then computer proving algorithms.

8:35 AM
That might be a good idea. But that seemed to advocate writing structured proofs in textbooks and papers.
If it's my Mac who is going to read the paper, then why do I care about the overload? =)

@Srivatsan yes, but his motivation for the level of the proof seems to be oriented so that the steps are so small that anyone could follow them, even if they are baby steps.
@Srivatsan precisely :-)

// But then you will also give me a version I can read, right?
@robjohn No, I don't get this. Assuming that I follow each individual step, should I follow the whole proof?

@Srivatsan and that there might be something completely different.

@Srivatsan when starting out in proofs, students are lucky simply to follow each step; forget the big picture aspect of the proof.
Later will come the understanding of the proof.

8:39 AM
@QED Yes, it's magic typing after all. =)

Post this "Durand–Kerner method" thing to reddit to get lots of upvotes

@robjohn For me, it's usually the opposite. I start checking the local steps more and more while the full picture slowly fades out of sight.

Can you prove the FTA with it?

@Srivatsan But you are not a neophyte proof reader. You are reading a proof for the big picture.
Different audience.

@QED I don't get it. Also, I'm not too fond of reddit... :)

8:42 AM
Don't get what?

So I tend to spend some considerable time understanding the structure of the proof. I guess that corresponds to the "high level proof" idea from Lamport's article...

@QED IIRC that's how Weierstrass proceeded. That's why the method is sometimes called Weierstrass-Durand-Kerner.
@QED why I should give it to reddit...

@Srivatsan In the beginning of proofs, readers start with the microscope and then back off. More seasoned readers start at the other side of the room and then step up to the microscope.

oh just because I don't have an account but I know this is the sort of thing that r/math would upvote a lot since it's very nifty and you can understand the idea of it in a short time

@robjohn So I am more a beginner than a seasoned reader. =)
I guess it's just easier to be local than think global.

8:44 AM
Great! "Combining this with ideas of Weierstrass, we obtain a completely effective proof of the fundamental theorem of algebra"

@Srivatsan No, I thought you were saying you started with the main ideas and then checked the steps as the full picture faded away.

@QED Oh. The method is simple enough that I know some people use this algorithm on their programmable calculators.
@QED Where did you read this? :)

@robjohn I used to start from the details. But then to compensate, I am learning to explicitly think about the structure of the proof. // Now I am better than before, but not fully seasoned, you could say. =)

I'm googling for a write-up of an FTA proof using this: came across that math.rutgers.edu/seminars/…

@Srivatsan You used to. Everyone used to.
3

8:47 AM
Oh well. :)

@Srivatsan So you are not a beginner. :-p

Cool, that's encouraging. Thanks :=)

@Srivatsan I hope that I will always be learning, thus always between beginner and expert. It would be terrible to be an expert and not have anything to learn :-)
4

@QED Nice. Sadly there doesn't seem to be slides of that talk anywhere...

I'll link them if I find them
(still searching)

8:51 AM
One more thing. Not only is it hard to write a fully structured proof, it is also hard to write the high level proof intuition lucidly. Don't know what's Lamport's take on that.

I like this because it'll be a hands-on proof of the FTA, unlike the ones I know where you sort of cast a bunch of spells any prove it by complete accident

It seems we really need two proofs: one for the man and the other for his machine.

high level proof intuition = big picture

@Srivatsan You hit the nail on the head.

@Srivatsan audience tuning
A function in L^p is not defined pointwise...
They are defined almost everywhere, they are not guaranteed to be defined at any given point.

8:59 AM
May be, a.e., f_n(x) converges to f(x)?

@QED If you want to try looking for the original paper: "Neuer Beweis des Satzes, dass jede ganze rationale Function einer Veränderlichen dargestellt werden kann als ein Product aus linearen Functionen derselben Veränderlichen"
(yes, long title. :) )

I would need a translation :(

@Srivatsan That must be it, but then L^p doesn't have much to do with it other than each function is in L^p

@QED Unfortunately, I don't know if anybody's written up a version of the proof in English...

@robjohn Yes, something seems off.

9:06 AM
The algorithm seems to have two steps: Weyl-quadtree which subdivides the plane to get you some approximate points, then Newton iteration to increase their accuracy.

That's the fancy version. In practice, most implementations just start with complex points spaced uniformly around a circle of preset radius.
and that radius is calculated from the coefficients.

and do they all converge to distinct roots?
or do you just knock at least one at a time out, then repeat?

Well, the convergence slows down when the polynomial has repeated roots (as is usual with Newton-Raphson).
but yeah, the idea is to start with n approximations, apply the WDK iteration on each, lather, rinse, repeat until convergence.
which makes for O(n^2) effort per iteration
(assuming you're using Horner for the polynomial evaluations)

this is really cool!

Ah, here is the original Weierstrass paper...

9:31 AM

Howdy folks.

hello

Morning, Asaf.

@QED: Your new question is kinda vague. What sort of infinitary logic? Is the axiom of determinacy assumed (allowing interpretation of ill-founded chains of quantifiers)?

9:48 AM
I don't know
I've never head of this axiom before

Hey Asaf.

@QED: It is a technical axiom that appears naturally in descriptive set theory (hence useful for measure theory, and thus to probability).
There is a question about it and the axiom of choice. I wrote this huge answer and tried to explain it a little bit.
I have class. See you later.

10:11 AM
Hello, I don't understand the answer .. =\
7

My Question is: Examine the continuity of $$f(x) = \lim_{n \to \infty} \frac{x}{(2\sin{x})^{2n}+1} \qquad \text{for} \ x \in \mathbb{R}$$ How can I do this? Honestly, speaking I have $\text{no idea}$ of doing, this as this seems to be a tough limit. What I only know is that as $n \to \infty$ ...

probably just ignore it
looks like a really weird textbook problem

I have to ignore my question then
2

Possible Duplicate: Testing continuity of the function $f(x) = \lim\limits_{n \to \infty} \frac{x}{(2\sin{x})^{2n}+1} \ \text{for} \ x \in \mathbb{R}$ I cannot figure out, at which points the function is discontinuous, the only thing came to my mind is solving $1+(2\sin x)^{2n} = 0$ w...

I tried to solve it by methods of testing continuity, but the question which a duplicate made it more complicated ..
Anyone can help me

I don't understand you, @Gigili. How did the answers make it more complicated?
And no, setting the denominator to 0 and solving it will not help.

you have to say what you want help with

@Srivatsan I mean, what would I do when I see a question asking about set of discontinuous points is things like setting the denominator to 0

10:21 AM
Well, that would work for "nice" functions that do not involve limits in their definition.
Here the limit n -> infty makes things more complicated. In this case, we should first unravel the limit to find the function explicitly.
(If you want a recipe for finding the points of continuity of a function like this, I think that would be hard to come by.)
The idea is this: First fix some x. Let us bother about the limit as n -> infty.
As a function of n, the numerator is a constant and the denominator is 1 + (some number a independent of n)^{2n}.

@Srivatsan CAn you explain a question for me??

Now, the limit of the numerator is easy: just x. The limit of the denominator would depend on the number a.
@RamanaVenkata Shoot your question but I might not be in a position to explain it. (Haven't slept in a while...)

Right, and the discontinuity will happen at 1/2?

@Gigili When |a| = 1/2. (Not when x = 1/2.) Here a is (2 sin x).

@Srivatsan Got it, thank you so much. Hope you sleep tight tonight.

10:31 AM
@Gigili I don't expect to sleep any time soon =) . But may be I should. Thanks....

Let F:X x Y-> Z. We say that is continuous in each variable separately if for each y_0 in Y the may h: X-> Z defined by h(p)=F(p,y_0) is continuous and for each x_0 in X the map k:Y->Z defined by k(q)=F(x_0,q).Show that if F is continuous then F is continuous in each variable./
I couldn't understand the continuous in each variable part what does that mean??

Let's take continuity in x.
Now consider restricting the function to x by setting the y-value to some fixed quantity: namely, y_0. The resulting function is a function of just x; a usual single-variable function. That is called h in your definition.
Then f is continuous in x, if for all y_0, this single variable function (in x) is continuous.
Does this make sense?

Just wait a min
so the map changes every time we change y_0??

Yes, that's correct.
It might help to draw a picture. Suppose you have a function f(x,y).
Now, consider a line y=y_0 and restrict the function to this line. Obviously the result is a single variable function and it depends on the set (line) to which you are restricting to...

10:46 AM
Simply to say y_0 is a just a parameter

That's a good way to see it.

11:06 AM
@Gortaur ? Can I ask you a follow-up question on conditional expectation?

12:06 PM
I'm back baby!

12:28 PM
@Sunrise yep. only I am busy now and will run away soon

1:08 PM
'Morning, Asaf & Gortaur.

@Srivatsan In the local uniformity there are some continous remaining :)

Done. =)
Thanks..

1 hour later…
2:40 PM
You hear that, guys? Those are crickets...

=P)
I miss them here =)

The city is not friendly to crickets...

@JM Yes, we used to have a bunch of them chirping all day long after the rains.
I don't remember if I was bothered by them though

For me, it depends.
Well, for crickets. Frogs are an entirely different matter...

Oh boy. Frogs are terrible.
Actually, this cricket thing was only brief for me. Then we moved out to some other place that only had frogs around.

2:51 PM
Ugh.

Ugh indeed. =)

Assume you have already solved the problem; then you can use this complicated integral to see that the answer is 2...

I think he's putting the cart before the horse...

3:32 PM

=)
That question does not seem too difficult for me. Why is it acceptable at MO? // [I don't know the answer to the question, but I am still judging =)]

I don't understand either.

It's more a , that's why I removed the previous tag...
Still, it may be better to ask Lenstra himself...

3:51 PM
I don't know whether this is appropriate but can anyone help me out here (see my latest comment)?

4:22 PM

@tb :) making a small edit.
Hoping it isn't too detailed.

No, no, don't worry. Let's hope he can cope with the sum of squares, at least :)

@Srivatsan Hi )

hi Gortaur =)

@tb hi to you as well )

4:24 PM

hi Gortaur :)

@Srivatsan sorry, I was busy and couldn't answer your 'hi' though I've seen it

I note that Sasha gave the name, while Sri gave the formula...

@Srivatsan Well, if a computer algebra system is used this way I'm pretty sure it will start breeding a worm for lack of challenge.

@JM Vieta's formula is a wee amount of detour, I think. =) After all, given the roots -a_i, you construct the polynomial (x+a_i), expand and extract coefficients from it...

4:29 PM
@JM hi )

I am using Vieta's formula to mean the relationship between coefficients and roots, but of course, that seems a little narrow use of that term.

...it ain't a detour. That's what's it's called. :) Or if we should be strict, one of the Vieta formulae...
hey Gortaur.

In order to complete the greeting round: hi all!

:D hi t.b.

hi J.M., tb
We ran out of people to say hi to.
hi Tim =)

4:32 PM
On the other hand, I'd use a computing environment for the last sum I gave here, as I distinctly recall going through two sheets of paper for the last time I tried it out by hand...

My rep just overshot the last 4 digits of my phone number.

Yes. It was numb for a few minutes...

Try guessing what the OP meant: i doesn't lie in the disco, so =)

But Sri, where else would these complex animals party?

Not sure why the OP objects, but I don't mind them partying in the disco at all. Those beauties have some nice curves.
Ok, off to donutland.

4:39 PM
Mmm... donuts...

Ah, you also stopped at the exponential integral... I couldn't go past it either.

Hi Srivatsan! Hi, all!

@Srivatsan, just an integral that came up in group theory. thanks, your answer is good enough for me, but i asked a few other people in person, and they might want to know how to get the exact value

@JackSchmidt Ok, I will mull over it, but I don't think I have any ideas at all. Nice question by the way... =)

4:44 PM
Sometimes I enjoy watching you all talking here. Call me a stalker. Haha

@JackSchmidt It looks to me that you'll need results from here to proceed from the integral Srivatsan obtained. I'm a bit too knackered to manipulate symbols today, however...

@JM Thanks for the link. Let's hope that says something about the existence/value of the limit.

5:05 PM
I'm off. Later.

5:33 PM
Hello @JackSchmidt. May I ask you a question? I browsed your webpage and found your slide ms.uky.edu/~jack/2008-12-02-Defense.pdf interesting. Was it written in LaTex?
If yes, how is its code/template like?

@Tim do you know beamer)?

I do. But I don't know how to make slides as fancy as Jack's.

@Tim: yes, it was written in LaTeX using the "beamer" package. There are many tutorials on the web about beamer. The diagrams were written in a latex package called tikz using its "tree" things
The preamble is mostly:
\documentclass[gray,bigger]{beamer}
\mode<presentation>
{
\useoutertheme[subsection=false,footline=authortitle]{miniframes}
\useoutertheme[subsection=false]{smoothbars}
\usecolortheme{seagull}
\usecolortheme{rose}
\usecolortheme{seahorse}
\setbeamercovered{transparent}
\usefonttheme{structurebold}
}
use \section and \subsection to alter the top few lines, and I used a simple macro for making power-point-esque slides:
\newcommand{\stdframe}[2]{\begin{frame}\frametitle{#1}\let\OldItem=\item\renewco‌​mmand{\item}{\ifx\VSkipItem\Undefined\def\VSkipItem{1}\vfill\else\vfill\fi\OldIte‌​m}\leftmargini=-0.5em\vspace{-1.5em}\begin{itemize}#2\end{itemize}\end{frame}}
Then each slide is just \stdframe{Hi I am the title}{\item My first point, \item My second point, \item A hat to fit them both }

hi all

5:50 PM
@Alex hi

@Jack: Thanks! I tried based on your code but it doesn't compile. Anything missing here:
\documentclass[gray,bigger]{beamer}
\mode<presentation>
{
\useoutertheme[subsection=false,footline=authortitle]{miniframes}
\useoutertheme[subsection=false]{smoothbars}
\usecolortheme{seagull}
\usecolortheme{rose}
\usecolortheme{seahorse}
\setbeamercovered{transparent}
\usefonttheme{structurebold}
}

\newcommand{\stdframe}[2]
{\begin{frame}\frametitle{#1}\let\OldItem=\item\renewcommand{\item}{\ifx\VSkipItem\Undefined\def\VSkipItem{1}\vfill\else\vfill\fi\OldIte‌m}\leftmargini=-0.5em\vspace{-1.5em}\begin{itemize}#2\
Maybe, did you also learn from somewhere?

Spivak makes a nice observation in his lectures on physics: dim O(n) = n iff n = 3, that's why there's cross product related to rotations in R^3 ^_^

@JonasTeuwen: I noticed that you had been drinking 3 days in a row, possibly 4 if you did on Friday.

@Matt Are you his mom?

Don't think so.

5:59 PM
Then why do you care that he drinks himself into pity or self misery every single day?

@AsafKaragila: ha ha