Mathematics - 2014-11-25 (page 4 of 5)

Hence winky face.

@Pedro: Have you solved your problem? Or do I need to cogitate?

Maybe my best bet is to pick stuff up as I go... I just know that geometry of connections is key for the stuff I want to look at, so I wanted to learn it, at least the basics.

@UserX: Mike Spivak has a question for you. :)

DIdn't I send you my geometry notes, @Mike? Geez ... :D

9:12 PM

@TedShifrin Not yet, I am trying to solve it using an answer.

If you have an answer, why aren't you done? :D

I am not quick with this things.

I am trying.

It's a nontraditional problem (hence my not knowing it instantaneously). But it's built on symmetry in the circle, obviously. I'll ponder in a moment.

He's being mean, @Pedro. He's doing it to me too. Don't let him get to you!

@TedShifrin Okay. What question?

9:14 PM

He wants to know how he can buy a copy of the book you have, @UserX :)

Jasper Loy

What book?

No idea. I borrowed it from my school about a year or two ago. I can send him photos of the book or something.

Cover your eyes and ears, @Jasper.

@JasperLoy it's a military secret

@TedShifrin Are these notes precisely what you write on the board, out of curiosity?

9:16 PM

@TedShifrin how do I contact Mr Spivak?

I just emailed him, @UserX, to ask if he minds my giving you his email? I guess that's silly, since his email is on the Publish or Perish website.

I mean his super secret behind 100 proxies email that he uses for these kind of jobs winks @JasperLoy

Jasper Loy

@TedShifrin Yes, very silly.

@TedShifrin what do I write? Hey I'm the Greek guy that has that stolen version of your never-translated-in-Greek-hardcover hardcover Greek book.

@PedroTamaroff A hint: Moving the fixed point to $\infty$ (and if there are two, the other to $0$) is usually a non-stupid move.

9:18 PM

Yeah, precisely that.

@DanielFischer I am following the advice given by Joonas.

Jasper Loy

@UserX You are 17 already, you should write what you want to say.

You can say you're the person I told him about :) @UserX

At least I can do what he asks me to do. =D

@MikeMiller I think it's repeating Greek a lot.

9:19 PM

Make sure you send him the phone number of your local police department as well, @UserX.

@PedroTamaroff What was that?

But yea that's the only problem

OK, back to the kitchen with me.

@TedShifrin what's his email?

@DanielFischer Well, I have $|z|=2$ and $0$. I want to find a transformation that sends the circle to $\Bbb R$ and $0$ to $i$. This is achieved by $i\dfrac{2+z}{2-z}$.

9:20 PM

@UserX: Since I emailed him to ask if it's ok, let me let him answer :)

Then I have $|z+1|=1$ and $i$. I want a transformation that sends this to $\Bbb R$ and $i$. If I take $S_1(z)=z+1,S_2(z)=\frac{z-i-1}{1-(1-i)z},S_3(z)=i\frac{1+z}{1-z}$, $S_3S_2S_1$ does the job.

@PedroTamaroff Oh, I thought you had $\lvert z+1\rvert = 1$ and wanted $0$ fixed.

@TedShifrin Okay then... I'll be gone for an hour. If he answers make sure you ping me so I see it.

He's giving us different problems, @DanielF.

OK, @UserX. Thanks.

It seems you still have?

9:22 PM

@DanielFischer Actually, I have $|z+1|=1$; and I want $0$ fixed and $-1$ to $i$.

Ah, that's the one you gave me, @Pedro.

@DanielFischer I see.

Yes, @Pedro, moving the fixed point to $\infty$ isn't stupid.

@DanielFischer OK.

@UserX: He says it's fine. Email me at shifrin@math.uga.edu and I'll send it to you. (Yes, I know this is silly.)

9:35 PM

@Ted He ends his definitions with a flower. Adorable.

Who? oh, Sharpe?

Yeah.

9:51 PM

@Chris'ssis Are you here ?

@Hippalectryon Not really.

:/

@Hippalectryon I was working on some integrals. Now I'm here.

@Chris'ssis I'm troubled on my result for $\displaystyle S=\sum_\mathbb{N}\dfrac{(a+k)!!}{a!!k!!}$, with the double factorial (usual notation)

@Chris'ssis I get $\displaystyle S=\sqrt{\dfrac{\pi}{2}}\dfrac{a}2\sum_\mathbb{N}B(a/2,k/2+1)$

@Hippalectryon Something is wrong there. Set $a=1$

9:56 PM

@Chris'ssis I can compute the Beta sum, but

I don't see how the $\pi$ diseappears

@Hippalectryon Does it seem to me or your sum diverges? What is $a$?

I used the formula $z!!=\sqrt{\dfrac{2^{z+1}}{\pi}}\Gamma(z/2+1)$

$a\in ]1,\infty]$

@Chris'ssis i'll write a quick latex to show you how I got that

@Hippalectryon it diverges ...

@Chris'ssis Olol

I did the inverse of what I wanted

I wanted $\displaystyle S=\sum_\mathbb{N}\dfrac{a!!k!!}{(a+k)!!}$ shame on me

OK :-)

9:59 PM

Let me correct that, i'll probably ask again smthing because I guess the $\pi$ will remain

@Hippalectryon this is what Mathematica says

(2^(-(1/2) - (1/2)*Cos[aPi])*((-2^(1/2 + (1/2)*Cos[aPi]))*Gamma[(3 + a)/2] + 2^(1/2 + (1/2)*Cos[aPi])*Gamma[(3 + a)/2]*Hypergeometric2F1[1, 1, 1 + a/2, 1] + Pi^((1/2)*Cos[aPi])*Gamma[(2 + a)/2]*Hypergeometric2F1[1, 3/2, 3/2 + a/2, 1]))/Gamma[(3 + a)/2]

@Chris'ssis 2 secs let me write it down

@Hippalectryon I'll leave soon. Tomorrow morning I have an interview ...

@Chris'ssis It will take 5 mins

OK

10:16 PM

@Chris'ssis texpaste.com/n/ix5o67yp

I can't manage to find my error

@Hippalectryon I'm not able to see what you posted there.

@Chris'ssis Grr that site errors i'll post it here

$$\displaystyle a\in]1,\infty[$$
$$\displaystyle S_n=\sum_{k=0}^n \dfrac{a!!k!!}{(a+k)!!}, z!!=\sqrt{\dfrac{2^{z+1}}{\pi}}\Gamma(\frac{z}2+1)$$
Hence
$$\displaystyle S_n=\sqrt{\dfrac{2}\pi}\sum_{k=0}^n\dfrac{\Gamma(a/2+1)\Gamma(k/2+1)}{\Gamma(a/2+k/2+1)}$$
$$=\sqrt{\dfrac{2}\pi}\frac{a}2\sum_{k=0}^n B(a/2,k/2+1)$$

We know that $\displaystyle\sum_\mathbb{N} B(a/2,k/2+1)=\dfrac{a}{a-1}$

Hence $$S_\infty=\dfrac{a^2}{a-1}\sqrt{\dfrac{1}{2\pi}}$$

@Chris'ssis Do you see the error ?

@Hippalectryon there are more mistakes. I'll look at that tomorrow and fix it.

Oh ok thanks

Tu fais trop de fautes, M @Hippa.

10:30 PM

@TedShifrin Montre les moi :-)

C'est pas à moi de le faire; c'est à Chris'ssis.

@TedShifrin Elle est parti/pas dispo

Elle est partie? :)

Pas dispo plutôt

J'ai corrigé une faute. Ça suffit. :)

10:32 PM

@Hippalectryon For instance, how did you get that $$\sum_\mathbb{N} B(a/2,k/2+1)=\dfrac{a}{a-1}$$ ?

@Ted What did you mean by "love your real names"

LOL, one's a cat typing on a keyboard, the other is "X"

I thought my name appeared as orpheous or something, does it appear like fdhsh fhsh?

yeah, like the latter

That's my 100th email(not joking), I got bored typing names

10:33 PM

you need 100 different email accounts?

I forget passwords easily. I tried 10 different passwords to open this one too

My name's John :P

Imagine how it'll be when you're my age.

John doesn't sound very Greek.

Γιάννης

It's greek for John...

I get the Johnny ... although I don't know about the s at the end :P

I wonder if Spivak will assume it's spam because of the name lol. Most male greek names have s at the end, but when you call someone by name you skip it.

Those two sentences are not connected.

10:36 PM

@MikeMiller I just noticed that the reputations of Mike Miller and Mike Miller differ by a $0$ ;)

@Chris'ssis $$\displaystyle \sum_{k=0}^n B(a,k+1)=\sum_{k=0}^n\int_0^1t^{a-1}(1-t)^k\mathrm{d}t=\int_0^1t^{a-1}\sum_{k=0}‌^n(1-t)^k\mathrm{d}t$$ $$=\int_0^1t^{a-1}\cdot\dfrac{(1-t)^{n+1}-1}{(1-t)-1}\mathrm{d}t=\left[\int_0^1t‌^{a-2}\mathrm{d}t-\int_0^{1}t^{a-2}(1-t)^{n+1}\mathrm{d}t\right]=\frac{1}{a-1}-\c‌dot B(a-1,n+2)=_\infty\frac{1}{a-1}$$

@Chris'ssis I multiplied by $a$ once by error. Still, that doesn't give the right result

Correcting that gives us $S_\infty=\dfrac{a}{a-1}\sqrt{\dfrac{1}{2\pi}}$

@DanielFischer In the same sense that the reputations of Daniel Fischer and Daniel Fischer do?

There you go with the disconnectivity again.

@MikeMiller No, $6012$ vs. $612$.

@DanielFischer I am failing.

10:37 PM

Oh, I see.

I'm not used to thinking about digits, you see.

@PedroTamaroff How far have you got?

I want a transformation that fixes $|z+1|=1$ and $0$ and sends $-1$ to $i$.

Seems like a pretty crummy metric in general.

You're not metrizable, @Mike.

@DanielFischer Well, first I used $1/z$ to send $0$ to infinity as you said.

10:38 PM

There's two of me, @Ted

My circle gets mapped to $\Re z=-1/2$.

And $-1$ is fixed.

One of you is more than sufficient. Of course, @Hippa created a second me.

Ah... this one's the real deal.

@PedroTamaroff Right. And $i$?

Or, some version of a real deal. In contrast with Hippa's, I mean.

10:39 PM

@DanielFischer $i$ is sent to $-i$.

@Hippalectryon you said that $$\sum_\mathbb{N} B(a/2,k/2+1)=\dfrac{a}{a-1}$$

@Mike: Y'all look rather different :P

@PedroTamaroff So, in that configuration, what do you want your Möbius transformation to do?

@DanielF: The cranberry/apricot hazelnut Linzertorte is next.

@DanielFischer I am not sure.

Hm.

Fix the line.

And infinity.

10:41 PM

@TedShifrin Sounds interesting.

well, it does have apricot jam as a glaze :)

And, OK and swap $-1,-i$.

So far, so good, @Pedro. And?

@Chris'ssis :O sorry, that's because on my paper sheet I used different notations. Let me correct that.

$$\displaystyle a\in]1,\infty[$$
$$\displaystyle S_n=\sum_{k=0}^n \dfrac{a!!k!!}{(a+k)!!}, z!!=\sqrt{\dfrac{2^{z+1}}{\pi}}\Gamma(\frac{z}2+1)$$
Hence
$$\displaystyle S_n=\sqrt{\dfrac{2}\pi}\sum_{k=0}^n\dfrac{\Gamma(a/2+1)\Gamma(k/2+1)}{\Gamma(a/2+k/2+1)}$$
$$=\sqrt{\dfrac{2}\pi}\frac{a}2\sum_{k=0}^n B(a/2,k/2+1)$$

We know that $\displaystyle\sum_\mathbb{N} B(a/2,k/2+1)=\dfrac{1}{a/2-1}$

Hence $$S_\infty=\dfrac{a}{a/2-1}\sqrt{\dfrac{1}{2\pi}}$$

@Chris'ssis Oh no you're right

@PedroTamaroff So, a) what Möbius transformations fix $\infty$?

10:42 PM

@Chriss'ssis: I thought it was past your bedtime.

Since it's k/2 I can't use that formula @Chris'ssis

@DanielFischer Linear ones.

Careful there.

linear in the sense of high school @Mike

Boo.

10:43 PM

@Hippalectryon It's too much mess in your proof there.

@PedroTamaroff Okay. And which of those map $\operatorname{Re} z = -\frac{1}{2}$ to itself?

LOL ... applause for @Chris'ssis.

@DanielFischer I'm writing it down.

@DanielFischer It's not so obvious to me which ones do that.

@PedroTamaroff Break it into parts, the ones keeping the left half-plane and one swapping the half-planes.

Aw come on @Pedro :)

10:45 PM

@Hippalectryon $$\sum_\mathbb{N} B(a/2,k/2+1)= \frac{2}{ a-2}+ \frac{ \sqrt{ \pi} }{2}\frac{\Gamma\left(\frac{a}{2}-1\right)}{ \Gamma \left( \frac{1+a}{2} \right)}$$

Tsk tsk @Ted

@DanielFischer Well, I think I got it.

I need two reflections.

I told you it's time to retire, @Mike. :)

First, I reflect along the line I got.

This sends $-1$ to $0$ and $-i$ to $-1-i$.

@Ted Not until I can tsk before you can delete... then it's time to retire.

10:46 PM

Then I reflect along $\Im z=-1/2$.

It'll be a matter of days, @Mike.

This gets me to where I want, @DanielFischer.

But then who will I whine to?

@PedroTamaroff And two reflections composed are?

@DanielFischer Basically I have to reflect along the line $y=x$, I think.

10:47 PM

then you and Pedro will whine at each other.

I'm out. Tomorrow I have an interview ... (early sleep is needed)

So this is all that symmetry and double ratio baloney, but with lines which makes things easier.

@Chris'ssis I was hoping to have a cancellation on the denominator, but $\sum_{k=0}^n B(A,k/2+1)=\int_0^1 t^{A-1}\dfrac{1}{\sqrt{1-t}-1}$

@Chris'ssis Ok, see you :-) thanks for the help

glares @Pedro for his baloney

Sounds like a nightmare, @Ted

10:48 PM

precisely @Mike

but you're the two competitive young guns

@DanielFischer I'm trying to write the transformation down.

How do I determine the reflection along this line?

I think my problem set has it.

You don't reflect along a line.

@PedroTamaroff Möbius transformations are orientation-preserving. You don't want a reflection. (Two composed is okay again.)

You rotate so that the two half-planes are swapped.

@DanielFischer I want to reflect along the line $\Re z=-1/2$.

10:52 PM

reminds @Pedro that $1/z$ sends upper half-plane to lower ...

@TedShifrin so does multiplication by $-1$...

Apparently, I need to find a transformation $T$ that sends this to $\Bbb R$ and then take $T\overline{T^{-1}}$.

@Pedro understands inversion in a circle, perhaps not inversion in a line :)

Your problems are having you use non-conformal stuff all of a sudden? That's bonkers.

@MikeMiller No, where did you get that?

10:53 PM

You kept saying reflection, @Pedro

@TedShifrin Well, symmetry, reflection.

@PedroTamaroff You cannot reflect across a line using a conformal transformation. In addition, you wrote $\overline{T^{-1}}$, which usually denotes complex conjugate...

@MikeMiller Yes.

symmetry!=reflection

That's not conformal.

10:54 PM

I don't know what to do then.

Conformal transformations, by definition (whatever your favorite definition is), are orientation-preserving. Complex conjugation is not.

What's the actual question?

@PedroTamaroff How would you swap the upper and lower half-plane?

I have to find a transformation that fixes the freaking line and sends $-1$ to $-i$.

@DanielFischer Inverting.

@PedroTamaroff Keeping $\infty$ fixed.

@DanielFischer A rotation.

10:56 PM

@PedroTamaroff Yes. And how about trying a rotation here?

But that doesn't fix my line.

Vincenzo Oliva

@DanielFischer

How did the answer to [this](http://math.stackexchange.com/questions/1037753/is-frac1e-gamma-log-x-prod-limits-p-x-p-textprime-fracpp-1/1038058#1038058) get the last inequality? Not only am I gaga, pretty much I haven't really dealt with infinite products. :/

@PedroTamaroff If you rotate about a point on the line, it does.

Vincenzo Oliva

(wondering why I can't make the linky thing here... or maybe I can't see iy but everyone else can)

@DanielFischer I want to rotate about $(-1/2,-1/2)$.