6:01 PM
how do we prove gamma log convex?
is that related to eulers integral that ended up with $\frac{\pi}{\cos{\pi z}}$ or whatever it was?
hi

Have you seen [this}(http://en.wikipedia.org/wiki/Bohr%E2%80%93Mollerup_theorem)?

wow

@OldJohn what is the question?

I want $$\,\lim_{n\rightarrow\infty}\frac{n^xn!}{(x+n)(x+n-1)\ldots(x+1)x}$$ too
this result seems important for something I forgot which

6:18 PM
@JayeshBadwaik why is the gamma function log convex
Maybe need to prove that second derivative of $\log\Gamma(z)$ is always positive
first derivative is $\Gamma'(z)/\Gamma(z)$ - that seems to ring a distant bell

that's not too scary
but the second derivative is

@OldJohn okay.

But I can't stop and enjoy this one - going out for an Indian meal :)

bye! :)

6:22 PM
@OldJohn enjoy!!

Bye for now

@cassandra0 What do you "want" from that?

@PeterTamaroff Did you get that nilpotent thing without the determinant?

@JayeshBadwaik I went off to get a haircut
So no
=P

6:25 PM
I did write something though.

Okay.

FUUU
My keyboard is getting all filled up with hair

@PeterTamaroff Don't you have a always bath after haircut?

> The analog of the gamma function over a finite field or a finite ring is the Gaussian sums, a type of exponential sum.
in what sense?

6:27 PM

6:48 PM
anyone have a good PDF on the theory of the Gamma function?

@cassandra0 Artin's Gamma Function?

Eulers $$\int_0^\infty e^{-t} t^{-s-1} dt$$

@cassandra0 ?

what

I have Artin's Gamma Function pdf
you want it?

6:52 PM
oh I see
thanks
hi charlie
I am sorry this integral is due to Legendre

@Charlie Good afternoon.

Euler did the product formula

@Charlie Wassup?

6:55 PM
@Charlie Doing okay. Solving some integrals.

@Charlie I thought you were gone

@PeterTamaroff me too

@Charlie ORLY?
But why?

nothing special

7:01 PM

@Charlie Boooh!

@PeterTamaroff haha

@Charlie No, really

@PeterTamaroff how are you?

7:05 PM
@Charlie I can't complain.

@PeterTamaroff hmm

@Charlie Why are you "Hmm"ing me?

@PeterTamaroff just an expression, Pedro
when you don't have or don't wanna say anything else you say hmm, has 9000 meanings

@Charlie I see. I choose meaning 3486.
"Oh, God. Did that really happen?"

7:09 PM
@PeterTamaroff or the first: "nice"

@Charlie Oh, look at that. How is math?

I don't know, but will ask it:"hey math, how are you?"
Math says it's fine.

@Charlie Wrong. Math has tons of problems! How can it be OK? How!?!?

@PeterTamaroff everyone has to face its own problems and move on...

7:12 PM
@Charlie That is questionable.

Here we go...

@Charlie I just feel that is just a classical "cliche" thing to say.

"So close, no matter how far..couldn't come much more from the heart.."
@PeterTamaroff yes. I'm cliche. I'm full of cliche
Hey @peter!

@Charlie HAHAHA WHUUT?

7:19 PM
@PeterTamaroff what about that strike that happened there?

@Charlie Strike?

@Charlie Oh. Just a big ass strike. But our government is one big asshole, so they neglect it,.

@PeterTamaroff pfff...

7:22 PM
@Charlie And that's why we're like we're. Fucked up politicians are fucked up.

yes... motherfuckers
Jigglypuff jiggly...igllypuff

@Charlie hahahahaha wtf

You don't know jigglypuff?

@Charlie I certainly do.

Yay!

7:32 PM
Does jiggly puff have a meaning other than the lullabies?

I don't think so

hmm....

@PeterTamaroff Was the protest about the minimum wages?

7:33 PM

@Charlie yup

@PeterTamaroff hmm.

@JayeshBadwaik do you like pokémon, Jay?

@Charlie I never watched it, so I do not know. I only saw the first episode I think.

7:38 PM
oh!

My brother was a big fan though.

hahaha

hilarious!

Pikachu... this one has electrical powers right? Hmm.

7:40 PM
Pika pikaaa

Then let me ruin your childhoods for you: Brentalfloss - STDs: Gotta Catch 'Em All (moderately NSFW).

@kahen that's ridiculous

lol

@kahen and stupid

Oh yes, very. And impossible to unhear after you've heard it.

7:45 PM

I was going to say something..but i forgot

I was going to forget something..but i remembered :-D
2

@Argon It works, doesn't it?

7:50 PM

@robjohn I thought to use Fubini, we had to first prove that the integral does converges.

@robjohn Why?

And here, we did not know the value, so it was kind of difficult to apply

@aDangerousIdea You were going to forget something and failed.

do you have a recommendation for log-convexity and stirlings approximation of log Gamma?

7:53 PM
@cassandra0 I have a proof. is that what you are looking for?

yes :D

@robjohn The original integral in question converges though.
$\int_R \pi/2 - \arctan x^2 dx$

@cassandra0 look here

thank you !!

7:58 PM
how come you post so much brilliant material
3

@robjohn sorry for not writing it very well...

@JayeshBadwaik It was fine, I just didn't note the $x^2$

@robjohn Ahh. Hmm..

@cassandra0 flattery will get you everywhere :-)

you' posts are helping me a lot studying Gamma :)

8:02 PM
@cassandra0 $\Gamma$ is an important function
and $\gamma$ is an important constant :-)

I've not had $\gamma$ come up yet

@robjohn I'd love to get to know wether or not it is irrational (which I guess it is) or trascendental.

$\displaystyle\gamma=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1k-\log(n)\right)$

I don't see this: we have $\psi(x) - \psi(\frac{1}{2}x) < \frac{3}{4}x$ and then it says summing over $x,\frac{1}{2}x,\frac{1}{4}x,\frac{1}{8}x$,.. we get $\psi(x) < \frac{3}{2}x$. I got $\frac{11}{4}x$ though.
maybe it's 8/2 it's hard toread

@cassandra0 That would surely be simplyfied.,

8:07 PM
@cassandra0 Note that $(\psi(x)-\psi(x/2))+(\psi(x/2)-\psi(x/4))+(\psi(x/4)-\psi(x/8))+\dots\lt\frac34x+\frac38x+\frac3{16}x+\dots$

You get that $\psi(x)-\psi(0)\lt\frac32x$

thanks a lot

Assuming that $\psi$ is continuous at $0$

yeah it is
btw anon showed me yesterday how to derive betrands postulate* from the prime number theorem (* for sufficiently large N)

8:11 PM
@cassandra0 Is that the one with the two constants?

there's a prime between n and 2n

@cassandra0 Ah, yes

it's such a neat theorem, turns out there's $k$ primes between n and 2n if you take n big enough too

What form of the error term did you use?

from $\pi(n) \sim n/\log(n)$
which is equivalent to $\pi(n) - n/\log(n) = 1 + o(n/\log(n))$
Ramanujans paper shows how to calculate $k$ :D

8:14 PM
@cassandra0 Oh, right, you are only looking for $n$ large enough
The error term for $\frac{n}{\log(n)}$ is $\frac{n}{\log(n)^2}$

oh I don't really know what that means then

@rob: good afternoon

@cassandra0 $$\left|\pi(n)-\frac{n}{\log(n)}\right|\le\frac{n}{\log(n)^2}$$
@Ilya Hey there. I was just wondering when I would see you again. Are your holiday travel plans unaltered?

@robjohn, how do you derive this?

@rob: nope. Btw, I've tried to ping you by Skype today - but apparently you've not seen that

8:19 PM
@Ilya I have now :-)
@cassandra0 It comes from using $\mathrm{li}(x)$

seems a very good result to have
although it doesn't tend towards zero

@cassandra0: it tends as $n\downarrow 0$

lol

@cassandra0 There is no error for $\pi(x)$ that tends to 0

Hi @OldJohn how was the food?

8:30 PM
@aDangerousIdea it was excellent, thanks :)

I'm getting addicted to this chat and the site again...
I had so calm period from September till November

@Ilya I understand...

@Ilya JUst let go, man.

I finished reading the paper now :D

nice gravatar, @S.D.

8:38 PM
@Ilya: thanks :)

I had to skip the bits about Gamma approximations but I can learn this now

@Charlie have you seen this?

@S.D. Hello!

@tohecz hi!

8:46 PM
aha

I find it very curious that you evaluate Gamma'(1) by doing Gamma'(1)/Gamma(1)

There's this thing: Consider $a=a_1a_2a_3\dotsm\in\{0,2\}^\omega$ and compute $g(a)=\sum_{k\geq1} a_k 3^{-k}$

user19161
Essentially the question I answered today could have gotten me the cap of 200, so I lost over 100 points of rep. =)

user19161
Hmm, where is Aaron? I am waiting to give him a song...

8:49 PM
then the image of whole $\{0,2\}^\omega$ is the classical Cantor set as a subset of $\mathbb R$

@WillHunting what about Charlie? Where is Charlie?

user19161
@Charlie Charlie is with Chaplin.

@WillHunting And Will is hunting

user19161
@Charlie Yes, for some important things...

@WillHunting yes...

8:51 PM
and since Cantor set is closed, it means that subset of $\{0,2\}^\omega$ is closed iff its image under $g$ is closed in $\mathbb R$. Do you get it?
btw, hi @Will!

user19161
@tohecz Oh never seen you here before!

@WillHunting I'm here for the first time, because @S.D. asked me something math-related.

user19161
@tohecz Ah, I have also never seen S.D. in this chat. I will be hanging out exclusively here till the end of my life. Essentially, I have retired from TeX and Eng. =)

@WillHunting noooo!

@aDangerousIdea "You were weak when I found you"

8:54 PM
@tohecz aha, so...

@S.D. so for a set in $\{0,2\}^\omega$, you can first take a closed set in $\mathbb R$, do intersection with the Cantor set and then apply $g^{-1}$.

@S.D. and since bounded closed (=compact) sets in $\mathbb R$ are isolated points, intervals, and finite unions of these, ... you see

user19161
See guys this is the question that got me a ridiculous number of votes. math.stackexchange.com/questions/247710/…

@aDangerousIdea you know where this line come from?

user19161
8:56 PM

@tohecz I'm quite tired and sleepy already - so apparently I'll realize it tomorrow

Hiz

@Charlie Star Wars?

@tohecz: but maybe you can tell me - this is yes (only cylinders are) or no?

8:57 PM

hello

@S.D. I believe it is yes: only cylinders and isolated points.
and finite unions of these two

@WillHunting weird :)

@cassandra0 hello

user19161
@S.D. Hehe. Well there are many weird things on SE, such as the users. =)

8:58 PM
@tohecz isolated points are not clopen

@S.D. oh of course, you're right. So only finite unions of cylinders

what's up

Cantor space it very strange. And every space of infinite words over finite alphabet with at least 2 letter is the same as Cantor space

@WillHunting If f is a function, what is the difference between multiplying $f$ n times, and composition of f n times?

user19161

9:01 PM
@WillHunting That is $f(x)^n$, not $f^n$ :-)

@tohecz oh, so we can talk in terms of words. Then essentially I wonder which languages are clopen. Omega regular are not

I just showed $S=T$ by proving that for every $\epsilon>0$, $|S-T|<\epsilon$

@S.D. of course, I studied combinatorics on words.

Mother of farfetchedness?

user19161
@JayeshBadwaik Ah, OK in case that was not a joke, well, it is as written in my answer. It could mean either thing.

9:02 PM
@tohecz: also not all regular are

@WillHunting :-D
@Argon

user19161
@argon It's Friday Aaron.

@WillHunting So it is!!

My internet company promised me 3 Mbps, it gives me 4 Mbps (punches fists in air) :P

9:03 PM
hi @Argon

user19161

@WillHunting YAY!
@cassandra0 Hi again!!!

user19161
@Argon Do you know this song was flagged once in another room?

Anon gave me a realy really good answer about my prime number question

@WillHunting HAHA why?

user19161
9:05 PM
@Argon I don't know.

now I get what PNT says and why it's stronger than chebychev

@cassandra0 Why?

@S.D. regular are problematic. They don't have nice topological properties

user19161
@cassandra0 Are you specialising in analytic number theory?

Moreover, you have to say what do you mean by regular if you speak about infinite sequences.

9:05 PM
@argon

@Charlie Hi Marilia! How are you?

@Argon Fine, thanks, and you?

@Charlie Fine, as always!

@Argon Good!

Yep

9:07 PM
@tohecz all possible infintie concatenations of course - I hope there is no much ambiguity in embedding regular into infinite
@tohecz which of them have?

@S.D. so you take an $L$ regular and you do $L^\omega$?
@S.D. basically, none. Speaking about languages and about Cantor topology has one thing in common (words) but that's it.

a chebychev type bound just tells us a function has asymptotic growth that stays in a sector, it could potentially oscillate or have other perturbations though. PNT gives the exact asymptotic growth in the large scale. It's not possible to change the base of the logarithm or scale $\frac{x}{\log(x)}$ by a multiple .

I see...

user19161
@argon I love this song. I was listening to it earlier. youtube.com/watch?v=kybeq2dWBf8

@WillHunting Ok, I will listen

9:11 PM
nope, I take w\in L finite word and embed it as all a cilynder starting with w

@S.D. consider the language $a^*b$ and the set of cylinders $C=a^*b \{a,b\}^\omega$. then $C$ contains everything but a single point $a^\omega$.

user19161
@Argon Also I just returned from a walk outside, and it is going to be 1 Dec!

@WillHunting classical

@WillHunting Soon!

9:13 PM
@WillHunting YAY!

This is better, in my opinion.

You like this type of song?

user19161
@Charlie Yes, I think the girl is very beautiful.

@tohecz so what? it refers to the duality of Globally and Eventually

9:16 PM
@S.D. so, it has no nice topological properties in general, even such a simple example is "weird"

oh yeah, regular languages are just open ;)

or maybe, let me think, maybe the result will be always an open set, but I would have to think trice before approving it. And it's quite late here as well.

because they can be verified based on finite prefixes
which accounts to the existnce of an open neighborhood

@S.D. I see it another way: countable union of (cl)open is open ;)

@Argon Love is like Argon

9:20 PM
Hahaha!

@tohecz :-p

user19161
I have a feeling @argon looks like the guy in that video I just posted.

@WillHunting HAHAHAHAHA!!

user19161
@Argon Hehe, don't you like it? I think he looks rather cute.

@WillHunting It's funny that you would think that was me!

9:22 PM
@S.D. well, I hope you found this discussion valuable, I'm gonna go now I think. I had a busy week with a busy weekend to come

user19161
@Argon Erm, I have weird thoughts all the time. They just come naturally, like how I see patterns...

@tohecz yeah, thanks a lot

@S.D. you're welcome
so see ya!

see you

9:25 PM

Bye guys!

@Charlie bye!

@Charlie Bye!

@Argon Bye Aaron!

later

9:27 PM

user19161
This is a nice song in Mandarin. youtube.com/watch?v=aEeTuhvJRHs

@WillHunting Thanks to the subtitles, I can sing along with him!
=D

user19161
@Argon What? You know Mandarin? =)

9:31 PM
@WillHunting Nope :)

user19161
@Argon The title means Love is Eternal.

@Argon What did you think of the Oxygen?

user19161
@OldJohn Yes.

9:33 PM
Hi @OldJohn!

user19161
@OldJohn Wow, you know Chinese!

@Argon hi there
@WillHunting not very much - I have forgotten most that I once knew

user19161
@OldJohn Me too. =)

@WillHunting :)
fascinating language, but never got the hang of the tones

user19161
@OldJohn You are only 100 from 6k, gonna reach there soon, by the weekend I guess. =)

user19161
9:40 PM
Wow, how does BMS get OVER 9000 per month, literally? LOL

user19161
That's like OVER 300 per day!

@WillHunting I might not get there until next week - not answered anything for a couple of days ...

BMS will be 3rd overall soon...
at that rate.

user19161
I think I will be satisfied with 10k, so 3k more to go. =)

does anyone see why (1 + (a-b)/b)^b ?
I figured it out
I thought i was binomial theorem but it's much simpler

user19161
9:52 PM
Hmm, I will go eat some sandwich now...

@WillHunting Have fun

I don't understand why $e^{a-b}$ shouldn't it be $e^{b}$?
what does e even mean
oh it's eulers constant