anyone here?

yes

Hi @QED

hey

do you know how I can ask a moderator to move my question to stats.SE? It was suggested in one of the comments, and I agree with the suggestion...

just don't know how to find the moderator... :)

I think you can flag the question

hmmm i don't see the option for that...

ahh

i see it now

Flagged it for moderator's attention.

Thanks @QED -- appreciate your help!

@Bullmoose you mean this one?

yes

though someone upvoted it recently

there is some overlap of the user bases, and this question can get good answers on both sites, I think. However the chances of getting an answer soon are better on stats.SE, I believe. Therefore I voted to migrate it over there.

Ok, thanks!

Yeah, I think that the question fits on either site. It seems to me that there is less traffic on stats.SE

Certainly, yes. It shouldn't be too long before it's moved. Good luck with your question!

thanks!

this is more of a curiosity/self-learning question

the method analyses so beautifully, so I thought it might be "special" in some way

Hey, I just want to confirm: there exists no meromorphic function with poles at $1/n$, $n\in N$, right?

@Potato that is true.

Hi rob. About your Euler-Mascheroni computation: did you happen to remember the method you used? Brent-Macmillan or something else?

@JM I think there is a link to the page that has the algorithm. I'll check.

@JM Here

The C(m+j,j) is a binomial coefficient, right?

@JM I believe that my algorithm is similar to what you get by applying Euler Series Transformation to alternating harmonic series.

@JM yes

@robjohn Interesting. :) The only series I've tried the Euler transformation on for $\gamma$ is Vacca's (formula 24 here).

And formula 16 here.

@JM `"?Global\`*"` would have done the trick here, I think

To morrow top, you not of the whats.

@tb Weirdly enough, the last time I tried out the backslash in between backticks, it failed spectacularly... thanks!

@AsafKaragila The what morrow, and not you top of to.

What's up?

My thumbs. :D

The Fonz, is that you???

No way, no sharks to jump on...

@JM they did some tweaking of the code, but escaping is still very confusing. Hendrik gave some instructions in his answer and the comments here

Did you guys vote your jury duty on the meta thread?

@tb Must've missed that... :)

@AsafKaragila I don't think there's any permutation left for me, hence a simple-minded: morning, Asaf

@AsafKaragila I was the first juror...

:-)

@AsafKaragila you mean the guy who had an upvote and an accept and wanted to have a badge for that? Yes, I think I was the last jury member to vote

So we need to find an executioner.

Also, this requires two more diligent hitmen.

Oh yeah, note that I can now delete answers too!

On the delete page I have exercised this privilege several times last night after gaining the trust of the system.

@AsafKaragila Can you try your newly acquired abilities on this?

Done.

I am going now. See you guys later.

See you!

later

Morning hit squad.

This looks familiar, doesn't it?

Does it still count as duplicate if it's beta squared?

@Matt given that "another" hadisanji asked the same question earlier, I think I'm going to vote for closure.

Morning, by the way

Hi, all!

@Daniil Hi there. What's up?

nm, woke up 3 hours ago, but still haven't started studying :[

Have prob theory exam on saturday.

What about you, Matt?

Is the internet keeping you from it?

I've woken up 42 min ago, made some coffee, spotted a dupe and done some laundry. As of now I'm waiting for someone to hand me over their laptop because mine is broken and this ipad is.... a pain.

The internet is a distraction, sure, but the main problem is my laziness, I guess.

Does this chat work well in iPad?

Not at all :(

@tb The hit squad must be asleep, it still has only 2 votes.

@Matt apparently :) There was some hitting earlier today, so we're all exhausted...

Come to think of hitting: I'll be hit on again tomorrow. Now I'm thinking of not going to that lecture but I think I should.

@tb hitting on rajesh? : )

@Matt Nah, I've given up on that dream :)

: )

I need more questions asked so I can answer some.

Why does $\int^{+\infty}_{-\infty} t e^{-0.5 t^2} dt = 0$?

@Daniil Your integrand's odd. ;)

The integrand is odd $f(-x) = -f(x)$

ah

Ffffff

why didn't I see that

Thanks guys.

This is unacceptable.

I hope my snow leopard comes back soon.

Ouch.

Could you also edit the title, please?

Looks very ugly indeed.

@tb Done.

@Matt All you have to do is install the new STIX fonts.

@Daniil In view of your exam you may feel free to answer this question.

@ZhenLin This isn't my laptop so I'd rather not install anything.

@Matt I don't really understand the question.

What is X_3 there? And where did the numbers came from?

That's unfortunate. Perhaps use Firefox instead, if that's available.

@Daniil I don't know either but I suspect X_3 really is meant to stand for 3X. We'll find out soon.

3X does not make any sense either. If it's a fair die then P(X = 1) = ... = P(X = 6) = 1/6, which means that E(X) = 1/6 + 2/6 + ... + 6/6 = 3.5

and E is linear, so E(3X) = 3E(X)

Indeed. But somehow they got these numbers. Maybe it's not fair dice they're using.

@ZhenLin I'm amazed. While Safari looks the same as Chrome, FF looks normal. I'm puzzled. Thanks.

@Matt Safari and Chrome are essentially the same browser: they both use the WebKit rendering engine. Firefox is different from the ground up.

StackExchange keeps telling me I have notifications but I just keep seeing the same old ones...

@JM Nice new outfit! :)

I'd like to second that.

@tb I think I figured out why continuous and compactly supported implies uniformly continuous. For an epsilon and a point f(x) in the range there is a delta ball in the domain such that its image is contained in the epsilon ball. The collection of all these delta balls covers the domain and that's compact so has a finite subcover. Then choosing the smallest delta of all these gives the delta that makes it uniformly continuous.

I guess I could've looked this up.

@tb I drew a blank on tori, so I went for Barnsley's fern instead...

@JM Her Majesty will like it! Fern by a British mathematician makes it twice as appealing!

@Matt Yes, that's the basic idea. However, there's some fleshing out to do: why exactly is it that any two points that aren't more than delta apart are contained in the same ball? (that's actually not true, so you should phrase this a bit more carefully)

Maybe it works better by contradiction.

Anyway, I shouldn't be posting attempts of proof in here as this chat is unforgiving. I was going to delete it but then it was too late already.

The idea is fine, you just need to do a little bit more work: here's a hint

@JM I refreshed because this comment hinted that maybe you had changed your avatar to a fractal.

Oh, now I see I was right.

None of my torus experiments the past few days came out the way I wanted 'em (but maybe I'll use them in the future), so I got lazy and used one of my old Christmas avatar standbys...

@JM I guess the candy-cane torus did not meet your standards :-)

@robjohn Well, you did it already, and I was looking to make a slightly different version... :)

Maybe I'll try again when I'm not burned out on geometric transformations...

Have you been working with a lot of geometric transformations recently?

Yes, in playing around with tori. :)

BTW, do you render your tori in Mathematica?

@robjohn All of 'em. :)

@JM good, then I am not that far off the beaten track :-)

The "chocolate" one, the one that looked like a cage... those were all Mathematica -generated.

I thought that perhaps you were using some faster package.

@robjohn As it happens, Mathematica is now actually faster than POV-Ray... :D

@JM Are you using Mma 8?

Yes, I am.

I guess that is still the latest version.

I've had it for a while, so I wasn't sure if there was a newer version.

I think so; I'd think there'd have been buzz if 9 was coming out.

That's true. I am glad they came out with the Home Edition. I used to get a similarly priced educational edition from Academic Superstore, but getting it directly from Wolfram is good (in case I am without university discounts at some point).

@Daniil It was supposed to be x cubed.

Still missing a vote.

@Matt thwak!

@robjohn Heh. Someone deleted my comment. What a funny thing to do.

@Matt on that question?

@Matt ah, now it makes sense

@robjohn Yes! As you can see the comment that responded to it is still there.

@Matt I don't think that anyone but a mod can remove someone else's comment, unless that is a 20K ability.

@Matt Was your comment linked to the dupe?

@Matt you've become a victim of this feature

@Matt btw. did you see my hint to this?

Ah, t.b. has linked to the explanation.

@tb I was just typing a reply when I was flooded by messages.

@Matt or a "feature" :-)

@robjohn that was the intention :) thanks for providing it

@robjohn Of course it's a feature and not a bug. : )

@robjohn With how SE works sometimes, the line between "bug" and "feature" gets blurry...

@JM Yes. : )

@JM it's been $\color{red}{\rm (status{-}bydesign)}$ed

@tb space before the ed after the $

@robjohn really? it works for me

@tb Oops, I had refreshed my page since I had run the bookmark :-)

@tb Ah, yes, to get JM's new avatar. Now I remember.

@tb Oh, I see. The only thing that I missed was that my delta balls weren't necessarily subsets of the delta_i balls in the finite subcover and therefore the f values not necessarily contained in the epsilon balls. Nice.

@Matt If you are talking about the Lebesgue number problem, I think I remember proving something with the maximum of the set of distance functions to the complements of each open set in the cover. Note that the maximum of a finite collection of continuous functions is continuous. Find the minimum of this maximum and note that it is >0.

@robjohn Sorry, was afk. No, I was talking about my failed attempt to prove that continuous and compactly supported implies uniformly continuous.

Still quite bothersome. Thanks.

I've come to understand why people do speedballs.

I found how awesome rapid intake of caffeine and beer can get.

@Matt Choose $\epsilon>0$. Since the function is continuous, at each point $x$, there is a ball $B_x$ of radius $r_x$ centered at $x$ so that $|f(y)-f(x)|<\epsilon$ for all $y\in B_x$.

@AsafKaragila BOINGGG!!

@robjohn Yes.

What about Boeing industries?

@robjohn unbelievable, I also have it for the last 3 days

@Matt So you can cover the compact support with finitely many of these balls.

@Ilya rather annoying :-(

@robjohn I did that. Although now that I come to think of it I'm not sure why I need the finite subcover. I can just pick the smallest delta in the original cover.

@Matt uniformly continuous where?

@Matt the smallest $\delta$? there are infinitely many, the infimum might be 0.

@Ilya On $X$ if $f \in C_c ( X)$.

@Matt at some point you will have to use that your function has compact support.

@robjohn Thank you. Nice : )

@Matt then you wouldn't need compactness, which is crucial - see Rob's comment

(now we're three saying the same :))

@Matt but then you must resort to the Lebesgue number idea...

so that if two points are close enough they are in the same Ball.

@tb ))

X isn't compact but I'm doing the argument on the support of f.

@tb Three is an infinite integer.

@robjohn I thought I had figured that out: Pick the smallest delta in the finite subcover and then the delta balls are in the delta_i balls of the finite subcover so all the f values are epsilon close together.

Maybe I still don't understand it.

@Matt: Regarding your dating advice from the other day, I woke up this morning and remembered that between my first and second year as an undergrad I hooked up with this girl from my classes (though she was in comp. sci.) and later I found out that she was dating someone and were in an open relationship at the time.

@AsafKaragila Sounds awful. How did that make you feel? (I'm going to be hit on again tomorrow, I wonder what I've done wrong to deserve this : ( )

@Matt Well, I was hoping to hook up with her again but by the time I found out they already closed the relationship and outed it (it was a bit secret at first...)

@Matt another way to work it is to cover the compact set with balls of half the radius needed to insure that $f$ in the ball differs less than $\epsilon/2$ from $f$ at the center.

@Matt then let $\delta$ be the smallest of the radii of the finite subcover chosen.

@robjohn I thought that that's what I was doing, except for the delta half bit in it.

@Matt pick any $|x-y|<\delta$ and then both are within $2\delta$ of any ball containing $x$ thus, $|f(x)-f(y)|<|f(x)-f(\text{center})|+|f(y)-f(\text{center})|<\epsilon$

@Matt choosing the balls to be half the radius required for $\epsilon/2$ is important to this method.

THE END IS NIGH!! My LaTeX editor is open and I am writing something.

@AsafKaragila what are you going to write?

@AsafKaragila I wonder if you have a shorthand for \emptyset

@robjohn My idea for a very useful lemma, which will allow me to avoid complicated forcing arguments.

@robjohn His memoirs.

@Matt What memoirs?

@Ilya No, I also use \varnothing.

@AsafKaragila about girlfriends from CS

@AsafKaragila "Memoirs about my mother"

@AsafKaragila matter of taste?

@Matt "Memoirs on set theory and me momma."

@Ilya I never had girlfriends from CS. I only hooked up with two chicks from there... the others were either humanities or biology students.

: D

@Matt Your mother.

@JM The set of your mother.

:-D

@AsafKaragila Hook up = euphemism for to have sex with, I take it.

@Matt Obviously.

: ) Never heard that before.

@Matt You have to admit "hook up" is a bit more dignified than the alternatives...

@JM What's wrong with sleep with? That's what it is and I see nothing undignified about it.

@JM Porked, chafed, played pelvic football with, farked, forked, violated, etc etc

@JM It doesn't seem too much more dignified once you immediately associate it with its meaning.

@robjohn Also true. : D

@Matt That sounds more euphemistic to me than "hook up", but that's just me.

@JM Also me.

@JM hook up sounds miserable

@Matt I saw "hook up" and immediately thought of carnal meaning.

@AsafKaragila "played pelvic football" - eww, dude.

@JM I can easily make it waaaaaaay worse ;-)

Certainly, but let's run the hell away from that path...

@JM Second that. But I hear bad sex is common. And that sounds like an instance of it.

@Matt I have to admit that only one girl I had bad sex with. I've had my share of girls too, so I know what I am talking about.

On that note... "um-friend" is probably the only thing that sounds more euphemistic to me than "sleep with"...

@JM FWB (friend with benefits) (not fried with benefits)

@Matt Actually, when I think back on which sex I can call "pelvic football", it was one of the wilder nights of my life and a pretty great sex too.

@robjohn The "B" portion's pretty "nudge-nudge-wink-wink" already to me. :) I'm weird that way. (And that second bit is sorta kinda perturbing...)

But I digress. I am writing about symmetric extensions and permutations of the extension itself.

@robjohn Fries With Burger?

And I hadn't finished with fixing my flawed proof.

@Matt I thought I fixed it...

I didn't have time to read it. I'm making coffee.

@Matt Just assume $0=1$.

A nice side effect of being forced to use FF is that I get to try the MathJax bookmark. Nice.

@Matt you never got it to work with Chrome?

I had thought you did.

@Asaf: you didn't have to... :)

@JM Huh. I did not know that :-)

I have a real analysis question

@JM, @rob: are you here?

@Ilya yes

could I ask you to take a look on my proof (it's short) and my intuition?

@Ilya which proof?

simple problem in real analysis

I'll put it here then

@Matt The Lebesgue number should fix the simpler argument, but halving the distances doesn't require the Lebesgue number.

@Rob Let $\phi:\mathbb R\times\mathbb R^n\to\mathbb R^n$ be $C^1$-map and $h:\mathbb R^n\to\mathbb R$ be $C^1$-function. Define $S = \{x:h(x) = 0\}$ and pick up $t(x):\mathbb R^n\to\mathbb R$ such that $$ \phi(t(x),x)\in S$$ Suppose that for some ball $x\in U_0$ the function $t(x)$ is uniquely defined and differentiable. What is the rank of the map $g(x):=\phi(t(x),x)$

@robjohn But in my original broken proof the thing I needed to fix is fixed by using tb's comment about the Lebesgue number : )

@Matt yes. I think I was agreeing with that :-)

Namely, the question is if the rank of Jacobian matrix $Dg(x)$ can be $n$ for some $x\in U_0$

Now I'm quite glad I posted it here. Thank you, robjohn and tb!

I argue that if $|h(x)|>0$ on $S$ then it's not true: we have $$\nabla h(g(x)) = 0 $$ since $g(x)\in S$ for all $x$ and hence $$ \nabla h(g(x)) = (\nabla h)|_{g(x)} Dg(x) = 0$$

I tired myself out from writing.

Also the caffeine is wearing off.

@Ilya yes, I was just about to say that $h$ would need to be $0$.

@AsafKaragila You don't last long, do you?

@Matt Because writing details is very hard for me. I hate it.

@Rob since the vector $\nabla h$ is non-zero, $0$ is a eigenvalue of $Dg(x)$ hence determinant is zero and hence rank is at most $n-1$. Am I right?

@Ilya yes.

I am going for lunch. Bye

and my intuition was that since $S$ is $n-1$-dim and $g:\mathbb R^n\to S$ then it cannot have a full rank

@AsafKaragila have fun

@robjohn thanks a lot

@AsafKaragila Bon apétit!

See you later Asaf!

@robjohn Is that like a ritual of yours? You always spell it wrongly the first time and then correct it.

@Ilya You would need to have some region with codimension $0$ in $S$

@robjohn that's the point. if $S$ is codim-1 then rank is at most $n-1$. I've never worked properly with these statements but they seem very intuitive for me

@Matt If I miss it the first time, then I will correct it.

@Ilya yep

@Ilya So unless $h=0$ on some "thick" region, you have dim $n-1$

@robjohn aha, that's given by the non-degenerating gradient which is normal to $S$

I still remember this from my Math Analysis course 7 years ago ;)

@Ilya :-)

@AsafKaragila It would be interesting to hear the other parties' views on that. I'm not suggesting anything, I'm just saying.

@Matt I first read his message as '[...] only one girl I had bed sex with'

since I wasn't here for a while - I thought you've already started to discuss places

: )

@Matt I don't mind. I can give you names and facebook pages of all except one girl that I don't know what was her last name.

@AsafKaragila : D That would be too much information. I was just teasing you anyway.

@Matt I know you did. I just don't mind sharing this sort of information. Be forewarned, though, about half of those girls hate me for breaking their hearts eventually.

@AsafKaragila you were dating only with girls who're registered in facebook?

Oh, the professor is confused again...

@AsafKaragila That's nothing to be proud of. And I think sharing someone's name on the internet is not something you should do.

@AsafKaragila you bastard, real gentleman breaks girl's leg at most

No, I just assume that they all have facebook accounts... I have no actual knowledge about those things.

@Matt Obviously I won't do it here...

@tb weird

@tb I'm considering editing my comment into "...Prof. Chun-Yue...", for the lolz.

@Matt that would be nice ))

@Ilya Yes but also spiteful. I'm not sure I like that. I wouldn't like spiteful comments to my questions either.

@Matt I feel that half the comments I receive (here and on MO) are spiteful.

@Matt sometimes I do things that I don't like to people that I don't like. Not here at MSE anyway

@Ilya I caught myself doing that too a couple of times in real life. But here, where communication is asynchronous, it's easy to avoid.

@Matt :-p

I attribute his new acceptance rate to my comment.

Hello, can somebody tell me how I find Google's cash in hand from their balance sheet here: google.com/finance?q=NASDAQ:GOOG&fstype=ii

is it "Cash & Equivalents"??

@its_meα if only you knew how wrong place you've chosen to ask it :D

@Matt still, calling himself professor...

I know :( But I couldn't find ppl on appropriate chat rooms

personal finance or maybe quant.stackexhange

There's none on quant

& only 1 inactive mod on personal fin.

Does the site appear funny for any of you? The logo is missing for me, the voting buttons are all gone...

@Srivatsan Looks as always for me.

@Srivatsan looks normal to me.

I am worried now. It's fine for me in my Firefox...

Thanks, Matt and robjohn.

@its_meα is that a dorsal fin?

@robjohn lol, I meant finance

@its_meα I know :-)

:2798356 Thanks.

And I was worried that animated gifs would be annoying in answers.

:)

Just came across it, and I thought it was awesome jump!

@its_meα If you want to, just put some text before the link.

@robjohn, Your answer is very nice. Thanks.

However, I wonder what's up with the low vote count for all the answers in that thread...

@Srivatsan ah, you've had a chance to look at it :-) I am still working on simplifying the proof of monotonicity.

@Srivatsan The question itself has gotten a lot of votes though

@robjohn Oh, I was away the whole of yesterday.

@Srivatsan Yes, I noticed that you were gone for almost 24 hours.

@Srivatsan I've seen that a few times already: question's interesting, answers are (very) good, but they require you to actually think! :o

@robjohn You know, I wouldn't say no to a lot more votes though =)

@Srivatsan say "no" to a lot more votes?

@robjohn Yikes, that word went away when I was editing my comment here...

Thanks.

@Srivatsan I might draw up graphs for several $n$.

@tb Yes, it was a bit hard even for me.

I have only read through them cursorily. I would like to spend some time playing with them.

Ideally I would like to reconstruct the answers after getting only a gist of the idea.

@robjohn Yes, even I was going to suggest that myself. David had a number of graphs but he removed them.

@Srivatsan Mathematica verifies that the derivative is positive, but the formula is pretty unweildy.

@robjohn Like I said, I have only looked through your answer. I haven't worked on it, so I will take your word for it. =)

Good, that has become my new top-voted question now. =)