Mathematics - 2014-09-25 (page 1 of 5)

Anyway, there was a question saying "Show that $\omega$ a 1-form is locally of the form $\lambda df$ for some smooth maps to $\Bbb R$ $\lambda, f$ iff $\omega \wedge d\omega = 0$."

Ted Shifrin

That's an exercise next spring in my course.

Mike Miller

It's a nice one.

Ted Shifrin

It is just a line of high school algebra :)

Mike Miller

1:18 AM

Mm? Is it easier than talking about distributions?

Ted Shifrin

Yessssss!

Mike Miller

oh

well, what I said was still valid... so.

The iff is that easy? Not just one direction?

Ted Shifrin

Ah, I only assign only if ... If takes Frobenius in some form.

Mike Miller

Oh, well yes, that part was easy.

Ted Shifrin

Yeah, you be right.

Mike Miller

1:20 AM

Then it sounds like we have the same argument in mind

Ted Shifrin

I told you I was forgetful; I'm also senile.

Mike Miller

Ah, we're the same.

Balarka Sen

1:52 AM

@MikeMiller You're too young to be ~~dead~~ senile. No. No. Please no. snif

Alexander Gruber

2:18 AM

@anon you can take a person to be something with an enterance and exit time

to a particular room

those would be randomly distributed (for now, no distribution is assumed)

anon

is there only one person in this model? if not, which person is w(t) talking about?

Balarka Sen

Hello @Alexander @anon

anon

hi

Alexander Gruber

@anon w(t) would be talking about the average of the time that each person so far has spent in the room

Ice Boy

Hi @robjohn

Alexander Gruber

2:20 AM

so if i come in at t = 0 and leave at t = 2 and you come in at t = 1 and leave at t = 2 then we've got an average of 1.5 for w(t)

anon

okay, so that's what you mean

Balarka Sen

anyone got a cool algebra problem?

Alexander Gruber

i think that we can say that that is $\frac{1}{t}\int_{0}^t N_i(s)ds$ with $N_i(t)$ being the number of people in room $i$ at time $t$

Mike Miller

@BalarkaSen sure

Balarka Sen

go on

Mike Miller

2:23 AM

you never said you wanted it

Alexander Gruber

($N(t)$ would be integer valued, and not continuous, but still Riemann integrable I'm pretty sure.)

Balarka Sen

bangs head

I already said that I know the solution

And in any case it's not algebra

@MikeMiller =( no problem for me? sirius?

robjohn

@IceBoy hey there

Mike Miller

@BalarkaSen Let $R$ be a UFD such that all chains of prime ideals $\mathfrak p_0 \subsetneq \mathfrak p_1 \subsetneq \dots \subsetneq \mathfrak p_n$ have $n \leq 2$ (i.e., $R$ has dimension 2). then for any non-unit $\pi \in R$, $R[1/\pi]$ is a PID.

Balarka Sen

Oho.

Ice Boy

2:27 AM

@robjohn how are your bunnies doing?

Balarka Sen

@MikeMiller Is that supposed to be hard?

Mike Miller

I make no presumptions about its difficulty

Balarka Sen

You mean you haven't tried it?

Mike Miller

I've done the problem.

Balarka Sen

Ah, OK.

Well, I'll do it later =P Noted down.

Thinking would continue in January 2016.

Mike Miller

2:31 AM

bah

if you're not gonna think I'm not gonna provide

Balarka Sen

I will think.

Mike Miller

okie

Balarka Sen

I was just asking about difficulty because the last problem you gave took me about half a week

Mike Miller

:P

robjohn

@IceBoy very well. We just found out they are both male. That is good; we won't end up with a bunny farm ;-)

Balarka Sen

2:33 AM

better than funny barm

wait that went as a failed pun, didn't it?

Ice Boy

@robjohn good, I was wondering about that to.

Alexander Gruber

@TedShifrin hi!

@BalarkaSen hi

Mike Miller

hello

Balarka Sen

@Alexander Pedro-style : HERRO

Alexander Gruber

How are you two

Balarka Sen

2:47 AM

Fine.

Alexander Gruber

(alex gruber style: unpunctuated, with no pings)

Mike Miller

i am ok

Balarka Sen

Twiddling with commutative algebra.

Alexander Gruber

(also known as "inept")

Balarka Sen

haha @Alexander

Alexander Gruber

2:48 AM

@MikeMiller how is california this time of year

Mike Miller

quite nice

a few weeks ago it was grody

Balarka Sen

and we are dying with the frequent heavy rains

water water everywhere

Alexander Gruber

@MikeMiller I bought a new pair of sunglasses the other day

Balarka Sen

Oh, @AlexanderGruber, have you seen this?

Alexander Gruber

with which i have achieved victory in my war with the sky

Mike Miller

2:50 AM

did you just link a chat message with a link to another chat message

@AlexanderGruber ah; my solution is to close my eyes

Balarka Sen

moi? nope.

Alexander Gruber

@BalarkaSen Hahaha well, that's rather consistent

@MikeMiller he sure did

Balarka Sen

or right, yeah. i thought i linked to the search result directly

Mike Miller

I've mastered the art of walking with my eyes closed in places I know the topology of

Balarka Sen

@MikeMiller Topology of places?

Alexander Gruber

2:51 AM

@MikeMiller facebook.com/photo.php?fbid=10101961609016815&l=e4a8a78ebb

Balarka Sen

Everything is flat then!

Mike Miller

those are awesome

@BalarkaSen if you think that you've never seen a building

Balarka Sen

well, topologically speaking everything contracts to a point

Alexander Gruber

@BalarkaSen Not me.

Mike Miller

no it certainly does not

Balarka Sen

2:53 AM

@AlexanderGruber You torus?

Mike Miller

plus, homeomorphism is not homotopy equivalence

get outta here

Balarka Sen

droops ears leaves chat

Mike Miller

@BalarkaSen you can stay if you can solve a galois theory problem for me

Balarka Sen

OK

Mike Miller

I posed it to somebody on facebook, so if you solve it before he does you win

let $L/\Bbb Q$ be Galois, and $K \subset L$ be the subfield generated by all roots of unity in $L$. Suppose that $L = \Bbb Q[a]$, where $a^n \in \Bbb Q$ for some positive integer $n$. Show that $\text{Gal}(L/K)$ is cyclic

Balarka Sen

2:57 AM

Hmm. Thinking out loud : 1 --> Gal(K/Q) --> Gal(L/Q) --> Gal(L/K) --> 1 is short exact, Gal(L/Q) is cyclic.

heh?

Mike Miller

I've deleted anything that might be seen as a hint

Balarka Sen

oh right cyclic group quotiented are cyclic

Mike Miller

also, your arrows are pointing in the wrong direction

Balarka Sen

heh?

Mike Miller

in your sequence

Balarka Sen

3:00 AM

oh yes right

well, i am thinking in the right direction

so it doesn't matter

1 --> Gal(L/K) --> Gal(L/Q) --> Gal(K/Q) --> 1

Mike Miller

aye

Balarka Sen

You mean $L=\Bbb Q(a)$, surely?

both are the same in any case but standard notations ... =)

Mike Miller

think about that

Balarka Sen

grabs a cup of ~~ffee~~ coffee

@MikeMiller what d'you mean? $a$ is an algebraic so surely $\Bbb Q [a] \cong \Bbb Q(a)$?

Mike Miller

right

that's why it doesn't matter how I wrote $L$

Balarka Sen

3:06 AM

i said standard notations sigh

you are a professional expert in wasting times

Mike Miller

I'm not the one nitpicking identical notation

Weierstraß Ramirez

Hello, I have a bold question: when a function is said to be Ck, it means that its first k partial derivatives exists and are continuous, does this means that even the crossed term partial derivatives are continuous? Or does the concept is only valid for derivatives of the same variable?

Pedro Tamaroff

@WeierstraßRamirez "A function" is very broad.

Ice Boy

and "Pedro" is very bold :D

Pedro Tamaroff

Actually I have very thick blonde hair.

Weierstraß Ramirez

3:13 AM

Just had the doubt... I know its completely dull, but if you guys have answer to it I'd be great.

Pedro Tamaroff

Again, do you mean $f:\Bbb R^n\to\Bbb R^m$?

Weierstraß Ramirez

Yes.

Ted Shifrin

Hi @Pedro

All mixed partials of order up to $k$, @WeierstraßRamirez

Pedro Tamaroff

@TedShifrin How's it going?

I wonder what's the weather now around there.

Balarka Sen

Let me make sure if what I'm think is right or not

Ted Shifrin

3:17 AM

Super depressed ar how my probability exam is turning out ... Retirement looks certain.

Balarka Sen

@MikeMiller What d'you mean by "$K \subset L$ is generated by all roots of unity in $L$"?

Pedro Tamaroff

@TedShifrin =/ Did it go really bad?

Mike Miller

@BalarkaSen I mean precisely that.

Balarka Sen

Do you mean $K$ is generated over $\Bbb Q$ by all roots of unity that is in $L$?

Ted Shifrin

Getting chillier, at last. I don't want to get sick.

Pedro Tamaroff

3:18 AM

@BalarkaSen He means $K\subset L$ is generated by all roots of unity in $L$.

Mike Miller

Yes.

Balarka Sen

Cool.

@Pedro I figured that.

Ted Shifrin

Not done grading yet, @Pedro, but yes. A few did fine.

Balarka Sen

Well, then $Gal(L/K)$ fixes all the roots of unity in $L$. And the $\alpha$ by which $L$ is generated over $K$ is moved up to $\zeta^k \alpha$ by elts of $Gal(L/K)$

Pedro Tamaroff

@TedShifrin Probability is very counterintuitive!

@FernandoMartin Harro fmartin

Fernando Martin

3:19 AM

hey hey hey

Mike Miller

hello fernando

Pedro Tamaroff

Awaits for deep question on Galois stuff

Ice Boy

hi pal

Mike Miller

@FernandoMartin tell me how to prove Artin's lemma

fast

Ted Shifrin

Some of them have no idea what's going on with basic arguments that I've taught numerous times. :(

Pedro Tamaroff

3:20 AM

OH.

Fernando Martin

@MikeMiller: grab a copy of Lang

Balarka Sen

@MikeMiller what lemma?

Fernando Martin

that should do the trick

Mike Miller

boo

Balarka Sen

what does it state?

Ted Shifrin

3:20 AM

hi @Fernando

Pedro Tamaroff

@FernandoMartin supahhotfire.gif

Fernando Martin

Hi @Ted

Mike Miller

@BalarkaSen one problem at a time, and you've got two

Balarka Sen

OK, OK

Fernando Martin

long time no see

Balarka Sen

3:22 AM

So the action of $Gal(L/K)$ moves $\alpha$ to $\zeta \alpha, \zeta^2 \alpha, \zeta^3 \alpha$ and whatnots. Hmm.

$\prod_k \left ( x - \zeta^k \alpha \right )$ is left invariant under the action.

\O_o/ I am just messing about, aren't I?

Mike Miller

well, yes, but how else does one solve problems

Balarka Sen

let $\zeta$ be the dratted $n$-th root of unity. then the invariant polynomial is $x^n - \alpha^n$.

erm but that doesn't help.

meh i think i'll give up this approach

Wohahwhawhawha wait

@MikeMiller Let $n$ be the integer for which $\alpha^n \in \Bbb Q$. Then $x^n - \alpha^n \in \Bbb Q[x]$. And the splitting field is precisely $L/\Bbb Q$.

Mike Miller

right

Balarka Sen

Then $Gal(L/\Bbb Q)$ is cyclic, right?

Mike Miller

no

Balarka Sen

3:34 AM

say what

Mike Miller

probably not

if it is, you'll have to prove it

I don't have a counterexample in my hat but there's surely one

Balarka Sen

i dun believe you.

Mike Miller

if you believe it's true, prove it

Balarka Sen

oh noes it's false.

take $x^4 - 2$ over $\Bbb Q$. Galois group is dihedral.

in fact take $x^3 - 2$ =P

Mike Miller

woo

Balarka Sen

3:40 AM

it startes to look hopeless to moi. i'd better get some sleep.

Ice Boy

later pal

Balarka Sen

i'm still right here

Ice Boy

don't you want to get some sleep?

Balarka Sen

i do, but i am not going to

Balarka Sen

3:54 AM

@MikeMiller $Gal(L/K)$ just acts on $\alpha$ by moving it to $\zeta^k \alpha$, whereas all $\zeta$s are left fixed. so elts $a + b\alpha$ of $K(\alpha)$ are moved to $a + b \zeta^k \alpha$. $x \mapsto x\zeta$ is a generator for these, right?

Mike Miller

what's x

Balarka Sen

i meant $\alpha$

Mike Miller

no reason that map in general should actually define an automorphism

or that it should fix $\zeta$s

Balarka Sen

hmm right

humph

Balarka Sen

4:27 AM

@MikeMiller Are you there?

Mike Miller

sorta

Balarka Sen

Well as $L/\Bbb Q$ is galois, and as $\alpha \in L$, the $n$-th root of unity is in $L$ automatically, right?

So $L \supseteq \Bbb Q(\zeta_n)$.

you following, @MikeM?

Mike Miller

I can't right now, I'm sorry

I shoulda said no :P

Talk later

Balarka Sen

Byes.

sigh

Martin Sleziak

6:24 AM

There was a recent discussion on meta: Are congratulary post off-topic on meta? Or only some of them?. As an experiment, I have created a separate chatroom which could (possibly) serve as a place for such posts; perhaps more suitable than meta.

$Mathematics$ In praise of Math.SE site and its users

achievements, milestones, interesting statistics

Dan Brumleve

6:35 AM

thanks martin

anyone know about this? math.stackexchange.com/questions/945311/…

Martin Sleziak

Maybe you could add Wikipedia link or some short explanation what smooth number is to your question.

Admittedly, people who do not know this notion (like me) will probably not be able to answer your question. But the question will be more self contained and you will help a few people to learn something new.

Chris's sis

Greetings

@robjohn have you seen Kirill's answer? Crazy awesome ...

Martin Sleziak

@DanBrumleve Maybe you can also mention your question in number theory chatroom. There are not many people who visit it, but those who do might be interested in the question.

Committing to a name

@Chris'ssis Did you have any progress on the Cleo Fey situation?

robjohn

@Chris'ssis To what? I am just about to finish my asymptotic expansion.

Chris's sis

6:44 AM

2

A: Computing $\lim\limits_{n\to\infty} \Big(\sum\limits_{i = 1}^n \sum\limits_{j = 1}^n \frac1{i^2+j^2}-\frac{\pi}{2} \log(n)\Big)$.

The limit in question is equal to $$\def\tfrac#1#2{{\textstyle\frac{#1}{#2}}} \tfrac14\pi\log\left(\frac{16\pi^3e^{2\gamma}}{\Gamma(\frac14)^4}\right) -G-\zeta(2) \\ = -0.82586\ 11759\ 78831\ 08201\ 02008\ 35613\ 80953\ 63017\ 94512\ 34066\ 96955\ 08772 \ldots $$ in terms of Euler's $\gamma$ and ...

Anthony

Yo.

Chris's sis

@Committingtoaname These days I'm busy with some other kind of stuff ... no time for that ...

Dan Brumleve

added wiki link for smooth

i will check out number theory room

also i had this math.stackexchange.com/questions/943964/… which is sort of the motivation for the other question

robjohn

@Chris'ssis Having that formula with the Sierpinski constant at hand makes it pretty easy. Showing that would be interesting, but I guess we can take it for granted. However, I like to understand all steps. I will finish my answer with the asymptotic expansion anyway.

Chris's sis

@robjohn I wasn't aware of that formula :-(

robjohn

6:52 AM

@Chris'ssis I wasn't either. I would like to see a derivation of it.

Chris's sis

@robjohn Me too!

robjohn

@Chris'ssis when I finish the asymptotic expansion (which is interesting in the fact that I use some divergent series), I can verify that numerical answer :-)

Chris's sis

OK :-)

robjohn

I use the divergent series $$\sum_{j=0}^\infty(-1)^j\binom{2j}{n} =\mathrm{Re}\left[\frac{i^n}{(1-i)^{n+1}}\right]$$

which can be written as a finite sum of binomial coefficients

Chris's sis

hmmm, interesting

robjohn

6:58 AM

You multiply each term by $\alpha^{2j}$, so that the series converges, then let $\alpha\to1$

This allows us to get an asymptotic series for $$\sum_{k=1}^n\frac1{n^2+\alpha^2j^2}$$

then let $\alpha\to1$

Anthony

Why do we require the base elements for a product space to consist of all but finitely copies of the entire spaces?

@MikeMiller Oh hey.

Mike Miller

hello

Chris's sis

7:14 AM

@robjohn Then it seems that if we have a triple sum, we should consider a sphere sum, shouldn't we? (referring to the Kirill's answer)

robjohn

@Chris'ssis I am not sure I follow you...

Chris's sis

(this is only if out there is an asymptotic for such a sum)

@robjohn In 2 dimensions we have square and circle for the distribution of the numbers , right? In 3 dimensions we have the cube and the sphere. (at least I mean, but this might be convenient for us)

robjohn

@Chris'ssis Oh, this is just an extension of the problem, not dealing at all with my asymptotic expansion?

Chris's sis

@robjohn Exactly. :-)

robjohn

@Chris'ssis Oh, sorry... I am concentrating on finishing my answer, so I wasn't going that way

Chris's sis

7:26 AM

@robjohn OK. I'm very curious about what you finally get. Don't hurry. :-)

Dan Brumleve

7:37 AM

math.stackexchange.com/questions/945378/what-is-an-axiom-schema other answerer deserves an upvote i think but i am conflicted about it

Chris's sis

8:40 AM

@robjohn another solution in place

0

A: Computing $\lim\limits_{n\to\infty} \Big(\sum\limits_{i = 1}^n \sum\limits_{j = 1}^n \frac1{i^2+j^2}-\frac{\pi}{2} \log(n)\Big)$.

I happened to struggle on the same problem 3 years ago. Here's another approach. Start from $$\sum_{n=1}^\infty\frac1{x^2+n^2}=-\frac1{2x^2}+\frac{\pi}{2x}+\frac{\pi}{x(e^{2\pi x}-1)}$$ From this we have \begin{array} 1\sum_{m\le N}\sum_{n\le N}\frac{1}{m^2+n^2}&=\sum_{m\le N}\left(-\frac{1}{2m^...

I'm amazed of what happens there. Niceeeeeeeeeeeeeeeeeeeeeeee!!!

Chris's sis

9:05 AM

@robjohn The last answer sent me to the recovery room, I need some time to recover myself ... that one was a bomb of beauty ...

robjohn

9:42 AM

@Chris'ssis I'm going to have to look at those. I am still transferring my series to LaTeX. I should be finished in a bit and verify the answers numerically.

Chris's sis

@robjohn OK

bbl

Anastasiya-Romanova

10:05 AM

Does anyone here know how to use Mathematica?

robjohn

@Chris'ssis I am now computing the sum for $n=10000$ to 80 places so that I can use it to compute the constant to 72 places

Anastasiya-Romanova

Hello...

Knock, knock...

Committing to a name

Not well @Anastasiya

Anastasiya-Romanova

10:21 AM

@Committingtoaname Do you know how to use Mathematica?

Committing to a name

Very barely

@Anastasiya-Romanova I could try to help you

I can do generating functions, plotting, sums, simplification, integrals and some other stuff

Anastasiya-Romanova

Does Mathematica always work online? I mean we must connect to internet to be able to run it

Committing to a name

I have never tried to be honest, but I doubt it. You would lose Wolframalpha utility, but otherwise it should work

Anastasiya-Romanova

When I turn off my internet connection, I can't run Mathematica.

No, it doesn't work

Committing to a name

I'll turn my internet off to try it and come back in a min

Anastasiya-Romanova

10:27 AM

OK

Committing to a name

It gave me a warning that no connection was detected, and I clicked Create new 'notebook', and everything seemed fine

Anastasiya-Romanova

@Committingtoaname It still doesn't work. Here is the output when I work offline: FetchURL::offline: Mathematica is currently configured not to use the Internet. To allow Internet use, check the "Allow Mathematica to use the Internet" box in the Help [FilledRightTriangle] Internet Connectivity dialog.

I want to use Mathematica in offline mode if it's possible

Huy

@Anastasiya-Romanova: You bought Mathematica without knowing whether it's possible to use it offline in advance?

Or maybe the trial version is limited?

Anastasiya-Romanova

No, I don't know that

Huy

It works perfectly offline for me.

Anastasiya-Romanova

10:35 AM

Could you please tell me how to do that?

Huy

I just started it, no magic involved.

Anastasiya-Romanova

@Huy Try click Help -> Internet Connectivity and un-check Allow Mathematica to access the internet

Does it still work?

Huy

I just shut down my laptop and have to do some other stuff before I'm leaving for university, sorry. If you're here in the evening and still have the problem I can try to help.

Gustavo Montano

@PedroTamaroff. Your stared comment made me LOL ! ^_^ !!

Committing to a name

You have a legal copy @Anastasiya-Romanova?

Anastasiya-Romanova

10:40 AM

@Committingtoaname Yep!

@Huy OK, I'll wait

Committing to a name

What version and type of license?

Gustavo Montano

ooooooooo

Anastasiya-Romanova

@Committingtoaname 9.0.0.0

Huy

@Anastasiya-Romanova: Don't you have access to the newest?

Anastasiya-Romanova

@Committingtoaname How can I see the type of license?

@Huy I don't get it your question

Committing to a name

10:43 AM

and you have it open, press file and click new notebook and then?

I meant student edition, professional etc

Anastasiya-Romanova

@Committingtoaname It perfectly works if I work in online mode

Huy

@Anastasiya-Romanova: IIRC, the current version is 10.0.1 and I'm wondering why you're using 9.0.0.0 instead.

Anastasiya-Romanova

Oh, I dunno that. Maybe student

Committing to a name

I have version 9.0 personally, it is student edition

Huy

@Committingtoaname: I have a student's version as well, but always the most recent one, 10.0.1 atm.

Anastasiya-Romanova

10:45 AM

@Huy That's I get from my dad

Huy

Okay.

Committing to a name

I can't actually find where to update it

Gustavo Montano

Hmmm, what's a good program to draw 3D surfaces?

Anastasiya-Romanova

@Huy Should I buy the version 10 now?

Gustavo Montano

Preinstalled "Grapher" on Mac OS is pretty good.

Huy

10:47 AM

@Committingtoaname: user.wolfram.com/portal I can download the most recent version there.

@Anastasiya-Romanova: I don't think it would be worth it but I was just wondering why you didn't use the most recent one.

Anastasiya-Romanova

@Huy I just asked from my dad I wanted Mathematica & he gave me this version

Committing to a name

I am only allowed 9.0.1, it says that is my product version, oh well haha

robjohn

@Chris'ssis I've posted my better asymptotic expansion and using it and computing the double sum for $n=10000$, I get the constant to 70 places.

Anastasiya-Romanova

@Huy So if I want an iPhone, it means I should buy the latest one?

Huy

@Anastasiya-Romanova: Depends on your needs. My new iPhone 6 will arrive tomorrow.

Gustavo Montano

10:50 AM

Have you seen the bending?

It's terrible!

Apple have not responded yet...I'm not quite sure what they're going to do about it.

user1326876

Probably not the right room, but, is it against the rules to ask for coding examples on math.se?

I can't seem to find them :/

Anastasiya-Romanova

@Huy Now you're becoming annoying

Huy

@GustavoMontano: I bought iPhone 4 despite the antennagate and never experienced comparable problems.

@Anastasiya-Romanova: Or you're just oversensitive.

Analysis

I can swear that @Huy is a nice guy

5

Gustavo Montano

Hopefully it isn't a global problem.

Anastasiya-Romanova

10:53 AM

@Huy: What do you mean by this one: "My new iPhone 6 will arrive tomorrow."? Are you trying to show off here or what?

Chris's sis

@robjohn That job is crazy impressive. I would have never ever thought of that. :-)

Anastasiya-Romanova

No one asks your new iPhone 6

Huy

@GustavoMontano: Also, for now, Apple are actually replacing bent iPhones until they have figured out whether it is actually a problem on their side and what they will do about it.

@Anastasiya-Romanova: You brought up the iPhone, not me.

Gustavo Montano

I would buy the iPhone 6. But a jailbroken iPhone is far superior to my needs.

Anastasiya-Romanova

But I didn't ask about your damn iPhone 6

Gustavo Montano

10:54 AM

@Anastasiya-Romanova is mad.

When a girl is mad, you say sorry.

Huy

@GustavoMontano: You can't expect any company to just blindly announce "we will ship out new unbendable iPhones for everyone next week" - they have to inspect the problem closely.

Gustavo Montano

Obviously xD.

robjohn

@Chris'ssis I have computed the cosh sum in karvens answer... I need to find it.

Huy

@GustavoMontano: Similar problems also occured with different phones but the press never cared because it wasn't Apple. For example, check out this thread: forum.xda-developers.com/showthread.php?t=2287882&page=1

Chris's sis

@robjohn Yeah, I remember that! I wanted to tell you that. Just let me know if you find it.

robjohn

10:56 AM

@Chris'ssis before I read the rest of his answer, I want to see if I can compute the limit using that sum...

Gustavo Montano

Hmmm, that is true. It is Apple. They have quite the reputation.

Huy

@Anastasiya-Romanova: Believe me, if I wanted to show off I wouldn't do it in an online chatroom where nobody will ever meet me anyways. No point in doing that. Just calm down. I didn't mean to offend you.

@Anastasiya-Romanova: Me bringing up my iPhone was meant in a humourous way but unfortunately you didn't take it that way.

I need to leave for lunch now. Have a nice day everyone!

Analysis

Cya @Huy!

Chris's sis

@robjohn I got 25 votes on that question, it seems people are in love with that. Indeed, it's an exceptionally nice question.

and a "Good question" badge

@robjohn you should pose questions more often :-)

robjohn

@Chris'ssis I was going to ask that one since I did say "the interesting question is..." but you beat me to it.

Chris's sis

11:08 AM

:D

Random Variable

@robjohn When you have the time, could you look at the proof of theorem (1) in the following answer? math.stackexchange.com/questions/884021/… I'm pretty sure it's not correct because it returns the wrong value for $s=1$. Did things go awry when he changed the order of integration?