Mathematics

Indeed, looks promising.

robjohn

12:21 AM

$$
\begin{align}
\prod_{k=0}^\infty\sqrt[\Large2^k]{1+\frac{2}{2^{2^k}+2^{-2^k}}}
&=\prod_{k=0}^\infty\left(\frac{\left(2^{2^{k-1}}+2^{-2^{k-1}}\right)^2}
{2^{2^k}+2^{-2^k}}\right)^{1/2^k}\\
&=\prod_{k=0}^\infty\left(\frac{\left(1+2^{-2^k}\right)^2}{1+2^{-2^{k+1}}}\right)^{1/2^k}\\
&=\prod_{k=0}^\infty\frac{\left(1+2^{-2^k}\right)^{1/2^{k-1}}}{\left(1+2^{-2^{k+1}}\right)^{1/2^k}}\\[9pt]
&=\left(1+2^{-2^0}\right)^{1/2^{-1}}\\[24pt]
&=\frac94
\end{align}
$$

12:32 AM

@robjohn Hello!

robjohn

@Charlie hi, how are you?

@robjohn I'm fine, and you? :)

robjohn

@Charlie pretty good. Getting ready to take the dog for a walk.

@robjohn great!

btw, I like when @anon explains non-math stuff. It's funny.

@Charlie Link?

12:46 AM

@PeterTamaroff I hear a ping, but I don't see who pinged

@Charlie Should I call you sillypantalones to wear your mysterious anger away?

@PeterTamaroff I'm listening to things...

:P'

@Charlie You are a mystery.

12:55 AM

@PeterTamaroff yes, I am

@PeterTamaroff what made you think it?

@Charlie Common sense.

@PeterTamaroff hmm

1:11 AM

Aren't we all? @Charlie, you feeling better?

@TedShifrin Yes, very, Thanks for asking! and how are you?

Just fine, thanks.

@TedShifrin :)

@PeterTamaroff I couldn't find

i was looking for one specifically

After reading 40 pages of starred messages...

look what i've found, @PeterTamaroff:

Mar 16 '12 at 16:36, by anon

@Kanna: When people downvote your posts arbitrarily just imagine them in their underwear.

1:27 AM

still wonder how that user knew what bdate I have on the mse system

@anon maybe he's looking over your shoulder right now

@anon how are you

?

pretty good

@anon good to know

1:42 AM

@TedShifrin Hey are you around?

I have a question

@anon Hey

yea

I have a question

consider $G = \{\left(\begin{array}{cc} y & x \\ 0 & 1 \end{array}\right)$

with $y > 0$

as a topological space just the upper half plane in $\Bbb{R}^2$ @anon

The book I am reading says that the left invariant measure on $G$ is given by $y^{-2} dydx$

@anon what does $y^{-2}$ there mean?

I assume $dydx$ is just the ordinary Lebesgue measure

y^-2 means y^-2

so if you wanted to measure a set A, you would integrate y^-2 dydx over that region

Hmmmm

Ok if $A$ is the unit square with vertices $(0,1),(0,2),(1,1),(1,2)$

for example an invariant measure on the topological group $\Bbb R^\times$ is $dx/|x|$, so $\mu(A)=\int_A\frac{dx}{|x|}$ for any measurable $A\subseteq{\Bbb R}^\times$

you can check by substitution that satisfies $\mu(cA)=\mu(A)$ for any real $c\ne0$

1:49 AM

@anon Is that something similar to the "Haar invariant measure"?

@user38268 then that would be $\int_0^1\int_1^2y^{-2}dydx$

@PeterTamaroff same thing

Hmm this dx thing is confusing me because I thought to define the integral we have to define a measure first.

@anon Oh. =)

@user38268 we're defining one measure in terms of another

@anon You're saying from the Lebesgue measure we define the new measure above.

1:50 AM

yes

@user38268 They just defined it, didn't them?

for example one may write $d^\times x=\frac{d^+x}{|x|}$ or such things to distinguish if desired

@anon Why the absolute value? I heard people talk about how the $\Gamma$ function is $t^{x-1}e^{-t}$ w.r.t. the Haar invariant measure.

Or is it just the case $\int_0^\infty$ in $\Gamma$ leaves it $>0$?

the topological group $\Bbb R^\times$ includes negative numbers.

@anon So to see left invariance of my measure I need to know $\int_{gA} y^{-2} dxdy = \int_A y^{-2} dx dy$ for any $g \in G$

1:53 AM

correct

goodnight all

night

@PeterTamaroff so yes, if you restrict yourself to positives the |-| is superfluous

@TedShifrin Hey.

Rehey

I'm rading your book at the moment. I should get a schedule to study! =D

1:59 AM

Not to mention your actual classwork :)

@TedShifrin Well, that I work on, mostly at uni, worry about it there. It isn't too challenging, yet.

Ah ....

@TedShifrin I guess I will have to worry more when I start studying stuff I don't know about Linear Algebra, dunno.,

My students are freaking out with closed sets, interior, and frontier points.

But then again, I am finding stuff easier than it was some months ago, so maybe it will be just fine.

@TedShifrin Why? Did you show them metric spaces before topological ones?

2:02 AM

We're only in $\Bbb R^n$ in my book. Hard enough.

@TedShifrin Oh. Why are they freaking out?

Did you sort out Jean-François?

@TedShifrin Dunno, he went away.

Ah. Learning logic and language is hard for most.

@TedShifrin Sure. But maybe there is something they don't grokke or feel is unnatural?

2:08 AM

Evening!

Today a physics student was freaking out about the theorem that given a set of generators in $V$ and another in $W$ of the same size, there is one and only one linear transform with $T(v_i)=w_i;i=1,\ldots,m$.

Then I threw weirdness of limits in $\Bbb R^2$ at 'em. Start derivatives tomorrow ...

(I mean base in $V$, gens in $W$)

Hmm, two people randomly answered some old question of mine today. I wish I had that a couple of months ago.

I tried to tell him it was something "natural", made him think how base make up everything, and uniquely, thus a LT is fully determined by its values on the base, and such.

@FernandoMartin Hey.

@FernandoMartin Which one?

Q: Basic categories cheat sheet

2:10 AM

Some stuff I wanted to know about categories

12

Has anyone come across a cheat sheet containing basic properties of the most well-known categories (i.e. does it have (co)products, (co)equalizers, (co)limits, etc?)?

Well, yeah, linear maps are abstract for most physics students. Wait for quantum mechanics :)

Count me out @Fernando.

Haha, it's okay, I needed that for revising for an exam I took a while ago; it's no longer needed.

But it's weird that someone got to answer it after such a long time.

Sometimes things go unnoticed ...

@PeterTamaroff: Could you find a hostel?

@FernandoMartin Yes, Freedom.

2:15 AM

That's the same we're staying in.

Did someone cancel their reservation?

Where are you traveling?

(is that correct? is it 'their'?)

@FernandoMartin Wouldn't know. I am sharing a four person's room with a girl named Yamila.

@FernandoMartin (yes ;) )

@PeterTamaroff No clue

Well, that's good

@TedShifrin: There's some national inter-university meeting of sorts

@TedShifrin It's the Annual meeting of the Arg. Math. Association, and we're going to the "students" part of it.

(at least me)

2:18 AM

Awesome. I just was asked to give a lecture for undergrads at the southeast Math Assn of America meeting this spring. Have to come up with something ...

@PeterTamaroff I thought the courses were longer

There's a lot of free time actually

@FernandoMartin Micro-tourism!

Haha, yup. I've never been to Rosario before.

@FernandoMartin Neither have I. If the weather is still shitty like this, I have some good games, an XBOX controller and an i7 =D

Hahah

It was a total pain to go to class today.

2:21 AM

@FernandoMartin Yeah. Lights went out about 4:30 pm.

Oh. Didn't know that, I arrived at 5

Well, gotta go

See you!

@FernandoMartin Bye byes.

@MarianoSuárez-Alvarez Hey! Fernando just left.

Hi

Fernando?

Ah

@MarianoSuárez-Alvarez Did the power out affect you today?

nope

2:25 AM

@MarianoSuárez-Alvarez Oh. I thought of the CS guys.

@TedShifrin I can't find in your book where you define orientation (for Green's theorem.)

@MarianoSuárez-Alvarez I'll have to get a different hostel for Friday, since Freedom is full. =/

Counterclockwise, with region on your left.

@PeterTamaroff Ouch

when are you leaving?

@MarianoSuárez-Alvarez I don't know. Maybe I should leave Friday night? When does it end?

Can someone give me the reasoning for why the reputation incentives on this site are biased towards answers over questions?

official activities end on Fri

2:28 AM

@Alexander Biased?

there is a lunch on saturday, I guess

@MarianoSuárez-Alvarez Ah. I guess I will figure it out.

@PeterTamaroff I guess I just mean that an answer point is 2x a question point and there are many more badges for answers than questions.

@Alexander Maybe one shouldn't worry that much.

@TedShifrin "...with region on your left." What do you mean by that?

@PeterTamaroff I think you're right. It doesn't really bother me, I was just curious as to whether there was some philosophical reason.

2:31 AM

@Alexander, the idea is that what makes the site valuable is the answers

a place where all people do is ask questions is not that useful!

on MO originally votes on questions and on answers were all 10 pts, iirc

when we moved here, reputtion was recalculated according to the new rules

and I don't think it changed anything significantly

people with a significant number of good questions has an even more significant number of answers

Here, I would say that the most prolific answerers have very few questions

@MarianoSuárez-Alvarez Yes, that's true. But some people are here just to teach, say Brian M. Scott. =)

@MarianoSuárez-Alvarez That makes good sense. My original thinking was that a good question causes much more activity in the site users than a good answer, but I can see both sides of the coin.

*from the site users

@Alexander, but you want to keep those that are good at answering hooked, too

more so than those asking

because otherwise who will the askers ask?

people who can answer questions are rarer than those who can ask questions, at least at the level of M.SE

on MO it is a bit different: asking a good question at that level is also rare

@MarianoSuárez-Alvarez Yes, I can see that. Also, it appears that there are very few dedicated question askers. The answerers form the bulk of the community

I suppose I need to think of some good questions to ask. Even better if I don't know the answer. :)

2:44 AM

@TedShifrin I have a question.

LOL ... I didn't answer the last one, but you can figure it out.

We have seen that whenever $ω = df$ for some function f, it is the case the integral along any closed curve of $ω$ is $0$. So certainly we expect that the size of $d\omega$ on a region will affect the integral of $\omega$ along the boundary of that region.

@TedShifrin (The one of $\Bbb C^3$?)

No, I was meaning your question about "on the left."

@TedShifrin Oh.

@TedShifrin I don't understand what you mean by that.

Walk around with your body along +z-axis. Put your left arm out. It should point into the region. I can say this all purely mathematically. That is done in 8.6 for Stokes's Thm and boundary orientation on manifolds with boundary.

Walk around the boundary, I mean.

Ethan

2:50 AM

$$n^{x}+(d(n)-1)n^{x/2(1-\frac{1}{d(n)-1})}\leq \sigma_x(n)$$

n>1

@TedShifrin Can I get the mathematical def?

You want the outward normal to the bdry followed by tangent to the bdry to give the standard orientation on $\Bbb R^2$.

@TedShifrin Right, what I guessed. Though it isn't immediate to see, maybe, how a surface is oriented?

Green is only for regions in the plane.

You need Stokes in general.

Oh, so 8.5. I lied.

@TedShifrin Tsk tsk!

2:57 AM

Gewiß.

@AlexYoucis It was "the grave diggers" it seems.

@TedShifrin Googles.

@TedShifrin "Surely"?

is the product of two independent variables also independent

For sure :)

Certainly.

why

Sorry, @Elp, I wasn't answering you.

3:02 AM

@TedShifrin bummer

@TedShifrin well if this questions piques your interest:

If X N(0; 1), Z is independent of X with P(Z = 1) = 1 - P(Z = -1) =
0.5, and Y := XZ, then X and Y are uncorrelated but not independent.

True or false

I'm no expert on probability ...

the wikipedia article does even discuss independence of the product. en.wikipedia.org/wiki/Product_distribution

and bummer again

Seems unlikely that $XY$ and $X$ would be independent.

i agree but need to explain

I'm going go with that the joint pdf can't be broken up into discrete things

Thomas Andrews

I'm not sure I've ever voted to reopen a question in M.SE.

3:21 AM

@Thomas: What's your point?

Thomas Andrews

No point, just an observation.

Oh, ok ... I'm getting more and more annoyed by both questioners and answerers, so I might vote to close me.

cyberskull

I vote for you to stay open to suggestions :-)