Hello everyone :)

This room is so silent today :o

I CAN'T HEAR YOU @Jasper

:D

Sexy eyes, right?

So they are more sexy than yours?

Ohh. So you're comparing yourself to a yellow cartoon boy with sexy eyes?

Nice. I normally ask questions.

xD

My mathematical maturity is not high enough for me to answer questions yet. Maybe in some months.

We haven't. First time I see you here

I'm usually here but not everyday.

Ohh

I've always visited SE as guest. But then I decided to make an account. Totally worth it.

So many brilliant people in here to talk to :D

Heya @Sab

Speaking of brilliant people... Hi @TedShifrin :)

Optimization problems in math is a pain.

One of the most important reasons for math, both pure and applied, @Sab

:O

So it's normal I find it hard? @Ted ?

@Jasper I won't be sleeping a lot for the next 2 weeks. Exams coming up ;-;

It begins with physics and I got 13 chapters to revise/learn :S

@Jasper Nop, first year college.

I'm at a good place but the pace is extremely fast.

@Chris'ssis I am far from done. It is like a massive brick wall. I am very interested to see this solution.

@Alizter It can be finished in many ways. First, try to create your own solution. (don't give up)

I have numerically had a go and I get Pi/2 as an approximation

My maths exams are in 3 weeks though. So I guess I will have ample time to practice questions at 3-4 hours a day

I need to make sure I get my proofs perfect

@Jasper: You like my low hanging snark comment?

@Ted Are you familiar with Apostol Calculus?

Yes, @Sab, somewhat.

There's one in the library but it's the first edition, will that be good to learn from?

I guess there are some stuff which are new

@JasperLoy yeah. It's like extremely expensive. But we have it in our library and noone borrows these books.

@Sab: If you can't have Spivak, Apostol is next best.

@TedShifrin Is the first edition okay? In terms of content?

When I wanted to learn squeeze theorem, it wasn't in the first edition. So I'm assuming it doesn't have some new stuff in it

I don't know what changed between first and second. What's very unusual about Apostol is that he starts with integrals!

Exactly lol

Oh, the squeeze theorem is there. I guarantee.

That's why I didn't use Apostol. It started with integrals. But I guess now I can since that's what we are doing.

@Ted in 1st edition? I think the squeeze theorem is in a question but it's not explained.

@Sabಠ_ಠ Hey.

The good thing about these books is noone borrows them, so basically I take it and renew it every week. That way it remains on my shelf lol

Hey @Sawarnik

Has the ping sound changed?

Not really @Sawarnik

Ooh, back to normal.

He actually has what he calls the Squeezing Principle in the second edition, in the chapter on continuity. You sure?

In the second edition he has it. But in the first it's in a question if i'm not mistaken.

It's fundamental for analysis, @Sab, so that shocks me.

I found it weird too @Ted and I started to think that the squeeze theorem was something introduced after the book was released.

Or maybe he explains it in his volume 2 of first edition?

has math.se lost all the great probabilists?

I know we lost Douglas Zare

and cardinal seems to have gone

Neither @Sab, I assure you :)

Ah

I hope he explains optimization in the first edition.

I'll go borrow the book Monday.

Bien sûr!

This year I'm quite moving with the lectures and sometimes late. I'll try to learn second year stuff during my holidays

@Alizter Be unstoppable - youtube.com/… :-). This is the proper state of mind: put down all the problems you meet, give them no chance to escape alive.

@JasperLoy Well he is clearly not wrong!

@JasperLoy lol

You figure it out, its been in chat a few times

@JasperLoy ...

@JasperLoy Thank you. It's my favourite colour, and it seems evident that it's yours just as well.

@JasperLoy we need to raise money to get more probabilists! :)

@JasperLoy That's the default name.

@Chris'ssis I don't know what to do with the functions.

A small mistake to fix. It's $$\sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{\Gamma(k+1)\Gamma(n+1)}{\Gamma(k+n+2)}\left(\frac{1}{n+1} + \frac{1}{ n +2 } + \cdots + \frac{1}{ n + k + 1} \right) = \zeta(2)$$

@Chris'ssis Is it a property of the functions or the structure of the limit I should look at?

So technically, the smallest size of a rectangle is actually a square?

Most people would say a square is a special type of rectangle, no?

@Sabಠ_ಠ With a fixed perimeter, yes.

@skullpatrol yes. and that is correct.

@Sabಠ_ಠ Oh no, not the smallest, the largest no?

@Alizter Have you thought for a second to squeeze it?

@ParthKohli Hi :)

The rude boy won't reply, I know. Isn't it, @Parth ?

@Sawarnik This has nothing to do with the length of the perimeter.

@Chris'ssis I have heard of mean value and squeeze however I have never studied them in depth or looked at their properties. Maybe I should?

I was actually doing an optimization problem and it seems like every time I try to find the smallest possible rectangle it turns out to be a square

@Sabಠ_ಠ Do you have a fixed perimeter?

@robjohn I posted above a very nice double series. (maybe it's too easy for you)

Fixed area @Sawarnik

@Sabಠ_ಠ Uh oh. Ok.

If i find the dimensions for smallest possible perimeter the l and w will always be equal from what I did

I'm not sure if it's true though or it was just a coincidence that 3 different trials gave the same result/

@Sabಠ_ಠ what are you optimizing?

@skullpatrol That I was wondering too.

:-)

@skullpatrol some basic stuff, trying to learn the chapter. Basically, find the dimensions of a rectangle of 850 cm^2 whose perimeter is as small as possible.

so l = w

@ParthKohli Besides, if you don't know, since probably I am a room owner, I saw your three deleted abusive messages long ago.

So based on that, I deduced that the smallest possible rectangle would be a square if the area is fixed, no matter what the area is

The smallest possible perimeter would be a square.

yup. :)

I never knew that and then I did an optimization question :D

@Sabಠ_ಠ Its simply AM-GM.

AM-GM?

@Sabಠ_ಠ What about encircling the area with a circle?

@Sabಠ_ಠ Yes, by AM-GM $l+w \ge 2\sqrt{lw}$. So the smallest perimeter is to be $l+w=2\sqrt{lw}$. Square to get $(l-w)^2=0$, hence $l=w$.

@skullpatrol how about encircling the are with an ellipse?

I guess if we encircle it with an ellipse it will turn out to be a circle>?

@Sabಠ_ಠ Don't you know about this?

@Sawarnik That's what I call as Optimization. lol

That's exactly what I did to reach my conclusion @Sawarnik

@Sabಠ_ಠ I think the circle will be the smallest possible circumference.

@Sabಠ_ಠ Good work. Though I expect the exercise was wanting calculus.

Its easy through that too.

Well I used a bit of caculus

not really :O

Thinking about it, I think I used just algebra

xy = 1000

2x+2y

make y s.o.f

subs. in 2x+2y

oh ya I differentiated

that was the calculus part

@Alizter Trying to squeeze it might be a good idea. Or how about using Taylor series?

Hi all! A quick question just to make things clear: Let X be an $n$-dimensional random variable following some distribution (say the normal one) with pdf $f$. I would like to determine the locus that encloses the a% of the population of X. Is that correct to demand $f(x) \leq a f_{max}$? Thanks!

So far I have intuitionally decided that the Cos can just be ignored and it is an average of Arcsecs over N

@Chris'ssis I give up I have just made a mess. This one is too difficult.

@Alizter Never give up.

@Balarka Sawarnik has just kicked me out of the room by taking me off the list. Please tell him he is now on my ignore list, thank you :-)

@skullpatrol I can read here. I had briefly kicked you in, but to return the deeds you are out again.

@Sawarnik You should stop this childishness of yours.

@BalarkaSen You are the room admin. Do you what you think is best. I have no objections.

@Sawarnik My say : You will be admin as before, and let skull write.

@skullpatrol That leaves you with few supporters anyways pal :-)

In case you were you serious.

@nullgeppetto: That doesn't sound right at all. First of all, there will generally be infinitely many choices. And the answer has to involve an integral, of course.

Hello Professor @TedShifrin

hi mr @skull

@TedShifrin I'm back.

@TedShifrin, yes that seems right.. I am a bit confused.. I mean that if we would like to find the iso-density locus where the density is equal to some constant, then we would set f(x)=c, an the locus would be an ellipsoid (in the Gaussian distro case). Then, I think that if we would demand that f(x) = (1-a)*f_{peak}, where a=0.997, we would have the locus that encloses the 99.7% of the total population. Does that seem rational to you?

@Sab @skull: The general curve question is a hard theorem in differential geometry, called the isoperimetric problem ... But yes, circles are optimal.

Me too @Pedro

@TedShifrin, no that's not correct..

No, @nullgeppetto, if I want $\int_{-b}^b f(x)dx = a$, I need to know what the pdf function $f$ is ...

Were you tennising, @Pedro?

@TedShifrin Yas.

@TedShifrin yes, I see..

I play early in the morning, @Pedro. National collegiate competition going on now at campus. Actually, UGA women are playing now, but I'm not there.

Men lost, sadly, last night. But I'm going to try to go most of the remaining matches.

@TedShifrin That's nice. I miss the tournament atmosphere.

I have hopes of making it to Wimbledon and the Australian and US Opens before I die, @Pedro :P

@TedShifrin, I think I need to reformulate the "99.7% of the total population". Probably I am talking about the population that has densities lower than the max density... f(x)<c

@TedShifrin Well, you can certainly go to the US Open!

Don't you mean for the integral to be a certain fraction of $1$, @nullgeppetto?

Australian Open, on the other hand.

That's Hell on Earth, right there.

@PedroTamaroff Have you ever thought seriously about competing?

Once I've retired, I can go to the US Open ... it won't be during classes.

@skullpatrol When I was a teen I did.

But never in major leaguges. =)

@TedShifrin, no, now that I am thinking about, no. I need it simple

@Pedro will probably have a better life competing in mathematics :P

Harder on the brain, but not harder on the body. :D

@TedShifrin, I want to have as an answer the interior region of an ellipsoid, so, I need to ask different. How about the above reformulation?

Let's make it a one-dimensional problem and solve that first, @nullgeppetto.

@TedShifrin I read something the other day.

Pretty nice read.

You want the interval centered at $0$ (symmetry will make a unique answer) with what I said, don't you?

@Pedro, I figure you read something more or less every day :D

@ParthKohli "Stop disturbing me or I will report you to the moderators again." You can. Let me report your deleted messages.

www-math.mit.edu/~rstan/papers/ubc.pdf

Ah, Richard Stanley. Brilliant man.

@TedShifrin Hehe, I do.

@TedShifrin, hmm, let's forget about the fraction of population...

Stanley did some beautiful work in the 70's tying combinatorics into fundamental algebraic geometry.

Well, @nullgeppetto, what do I remember, then? :)

@PedroTamaroff Thanks for sharing pal

@TedShifrin That interests me a lot.

@TedShifrin, to put it differently, what does the form f(x)<c actually mean, where f is the n-dimensional Gaussian distro pdf?

I will try to read some of the references in that link.

The work I'm referring to was Hard Lefschetz Theorem + combinatorics.

Not so much, really, @nullgeppetto.

I come up with a new result ...

$$\sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{\Gamma(k+1)\Gamma(n^2+1)}{\Gamma(k+n^2+2)}\left(\frac{1}{n^2+1} + \frac{1}{ n^2 +2 } + \cdots + \frac{1}{ n^2 + k + 1} \right) = \zeta(4)$$ and hence I make lots of generalizations.

If I have a discrete random variable, I suppose I'm asking for what events have probability $<c$, but in the continuous case it's not analogous.

@Chris'ssis That looks waaaaay too made up. Just saying. =P

@TedShifrin, yes.. Unfortunately I am confused now... I'll take a look again and I'll be back if necessary.. Thanks anyway!

Sorry to have confuzled you, @nullgeppetto.

@TedShifrin, :) no it's not your fault! Here is late and I am tired!

Well, maybe Pinnocchio can help :D

@robjohn You here?

@PedroTamaroff what is the meaning of "made up" there? Like "crazy"? :-)

:D

maybe "contrived" @Chris'ssis

@Chris'ssis It looks like the sum is produced to give such result.

Doesn't look like something that'd pop up naturally.

@Sawarnik yes

@PedroTamaroff Right, I got that. :-)

@TedShifrin OK :-)

a lot of math is "contrived" :-)

@robjohn Can you come to the Room of General Conversation for a bit :)

I mean it has a VERY NICE FLOW! You would love my proof there! ;)

@Pedro: I learned something in elementary linear algebra thinking about and answering this a few minutes ago.

Not the truly interesting stuff, @skull.

Homework problems, often, yes.

@TedShifrin agreed.

@TedShifrin Cool. I learned something concerned with matrices two days ago: one can compute the invariant factors of a matrix over a PID by taking GCDs of minors.

$$\sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{\Gamma(k+1)\Gamma(n^p+1)}{\Gamma(k+n^p+2)}\left(\frac{1}{n^p+1} + \frac{1}{ n^p +2 } + \cdots + \frac{1}{ n^p + k + 1} \right) = \zeta(2 p), \space p \in \mathbb{N}$$

Yes, @Pedro, that's not usually enunciated, but it should be clear in hindsight.

@TedShifrin Yeah, sure. Still, it's something any Linear Algebra student can grasp.

I think that the Smith Normal form is very enlightening and useful.

Yes, although most people don't learn it 'til grad school.

Maybe more enlightening than the alternative enunciation of the "fundamental structure theorem"?

@TedShifrin Weird.

It's the algorithm for obtaining the latter.

It is ridiculously simple. I mean, I took some time to grasp the algorithm. But the idea is simple.

@TedShifrin I know.

One can't teach everything in a first course. I don't have time to do Jordan form in my wonderful multivariable math class, although I allude to it briefly.

@TedShifrin Sure. But that's multivariable, not Linear Algebra say.

no, no, my course has more linear algebra than our usual linear algebra course.

remember, it's an integrated year-long course in both linear algebra and multivariable calculus/analysis ...

@TedShifrin Year-long? OUCH =D

You seriously didn't think I covered that whole book in one term, did you?

With a class full of you I could, but I don't have that many at that level :P

@robjohn Now he has removed my access to chat.

@TedShifrin Hehe no, but neither I thought you taught it all!

Pretty much, plus a few things not in there.

I talked about complex eigenvalues and 45 minutes of intro to complex analysis at the last lecture this past year.

That's nice. The guys surely need to have certain brain stamina though.

Don't be sexist :P

Ah?

Yes, it takes determination and dedication to survive me :

@TedShifrin I'll probably take complex analysis next semester.

But the course isn't deep or whatever, they say.

@TedShifrin I thought "guys" was generic. Like girls say "Hey you guys!"

What level course? Here we teach a mickey-mouse course for undergrads, but I have the strong students (who know analysis) take the graduate course.

OK, as long as you meant it generically :D

@TedShifrin I admit that when I think of a student, I think of a male.

But I don't know if that's me being sexist or just me being male.

Who knows.

@TedShifrin Well, I'll surely read Rundin's Analysis for that one, decide how far I want to get.

Interesting datum for you: In my diff geo class, of the top 5 students, 4 were women. Of the bottom 4 students, all were women. :( Then again, the male in the top 5 is gay ... so ... :P

Rudin's Analysis for what @Pedro?

@TedShifrin "Real & Complex Analysis", I mean.

Ah, not my favorite book for complex variables.

@TedShifrin Alforhs is another pick right?

I don't know where the "h" goes.

Yes, I prefer that, although both of them are dated.

=D

Ahlfors

OK, time to go cook. Bubye.

@TedShifrin Cheers. I had some good pasta today =D

Well, I can extend it to the real numbers with some restrictions ...

$$\sum_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{\Gamma(k+1)\Gamma(n^{3/2}+1)}{\Gamma(k+n^{3/2}+2)}\left(\frac{1}{n^{3/2}+1‌​} + \frac{1}{ n^{3/2} +2 } + \cdots + \frac{1}{ n^{3/2} + k + 1} \right) = \zeta(3)$$

There is a great thing here though ...

If I compute this in 2 different ways I get things like the ones of Ramanujan ...

@skullpatrol Why do you think its because of oyu?

Hi pal @ParthKohli talk to you here in 24 hours :(

@Chris'ssis Does it behave with rationals?

@Alizter Do you refer to that limit?

@Chris'ssis No, to the zeta series.

@Alizter Yeah, sure. ($p>1/2$)

@Chris'ssis Then what real restrictions does it have?

TFW there are 2 good answers with different methods and you like them both equally but don't know which one to accept as the answer

@Chris'ssis Why not make the second summation start from 1?

then you have a cleaner summand

@Alizter I wanna keep the nice result in zeta in the right side.

@r9m you have a nice series above. You might like to try it. :-)

(newly created)

I need some sleep though.

@Chris'ssis What about $$\sum^\infty_{n=1} \sum^\infty_{k=1} \sum^k_{j=1} \frac{\Gamma(k)\Gamma(n^p+1)}{\Gamma(n^p+k+1)(n^p+j)}=\zeta(2p)$$

@Alizter Ah, I see. Well, I like the way I put all there. :-) (aesthetically I mean)

Wait no I messed up big time on that index

@Chris'ssis thats one hell of a dragon :P .. okay I will try :) .. although I am more likely to be roasted by dragon fire :P

@r9m :-)))

Alright no too much damage

Even if I write that out like yours it still will look good :)

Just gets rid of k+1 with k

$$\sum^\infty_{n=1} \sum^\infty_{k=1} \frac{\Gamma(k)\Gamma(n^p+1)}{\Gamma(k+n^p+1)}\left(\frac1{n^p+1}+\frac1{n^p+2}+‌​\cdots+\frac1{n^p+k}\right)=\zeta(2p)$$

@Alizter Yeah, but to tell the truth that is also a hint ... (that small detail)

Oh whoops

What can I say! I just can't help solving things :) bad poker face

Well, yes, it looks nicer this way. I'm out for some sleep now.

Good night!

Oh interesting @Seaturtles. He edited his question, apperently it was the square root in the original question. Which makes their answer.. odd.

le sigh

Well, I am off back to probability and graphs.. good luck y'all!

Does anyone know where I can find the more recent Crux Mathematicorum publications ? (I am looking for vol. 36)

@r9m Other than buying them?

@mixedmath yes :D

I don't think there's much of a business in pirating math journals that become free after a little bit