Mathematics - 2015-01-14 (page 3 of 4)

N3buchadnezzar

19:03

@DanielFischer You are blue, abedi abede.

Mike Miller

a friend of mine sings "if I was green I would die" instead

Daniel Fischer

@N3buchadnezzar What does "abedi abede" mean, and where does it come from?

N3buchadnezzar

@DanielFischer Da Ba Dee

Daniel Fischer

> Unfortunately, this video is not available in your country because it could contain music, for which we could not agree on conditions of use with GEMA.

N3buchadnezzar

@DanielFischer Last attempt

Mike Miller

19:07

Ah, music is not allowed in Germany.

A very serious country indeed.

N3buchadnezzar

@MikeMiller They have march music and classical music.

Daniel Fischer

@N3buchadnezzar Aha. Not my type of music, but I've heard worse.

@N3buchadnezzar And we have/had Modern Talking and Tokyo Hotel (or was it spelled Tokio?). We can do much much worse than Marschmusik.

Hippalectryon

@MikeMiller $$\sum _{k=1}^{\floor \left(x\right)}\frac{\phi \left(\frac{x}{k^2}\right)}{k^2}$$ ? That isn't close to a staircase function..

Mike Miller

No, @Hippalectryon, I said define it piecewise.

Actually, I guess something like you're doing would work.

Actually... that looks like it works.

:P

Hippalectryon

O_o how ?

Even $$\sum _{k=1}^{\floor \left(x\right)}\frac{\phi \left(\frac{\floor \left(x\right)}{k^2}\right)}{k^2}$$ isn't monotone

Mike Miller

19:13

how is that not monotone?

iwriteonbananas

@BalarkaSen ok, balarka...just tell me why we cant contract the cone of hawaiian earring to the interesting point pls

Hippalectryon

@MikeMiller gyazo.com/1fa5a3e914f971ed229d8cf7db706ad7

Mike Miller

That's not $\phi(x)$.

Hippalectryon

@MikeMiller Oh wait q_q I forgot it had to be $0$ outside. Wouldn't that make it $\phi(x)$ ?

Mike Miller

this is key :)

Yes

Outside of $|x|<1$

Hippalectryon

19:17

How do I show that that sum is $C^1$ ?

Mike Miller

In a neighborhood of any point it's a sum of finitely many $C^\infty$ things./

Hippalectryon

Well $\phi$ is $C^\infty$, but I don't see why $$\sum _{k=1}^{\floor \left(x\right)}\frac{\phi \left(\frac{\floor \left(x\right)}{k^2}\right)}{k^2}$$ would be $C^\infty$

N3buchadnezzar

@DanielFischer Uuuuugh

Mike Miller

At any point $x$, there's a small neighborhood in which it's $$\sum _{k=1}^{m}\frac{\phi \left(\frac{\floor \left(x\right)}{k^2}\right)}{k^2}$$ for some $m$. Everything in the sum is $C^\infty$. Finite sums of smooth functions are smooth.

Daniel Fischer

@N3buchadnezzar Don't complain, I did warn you.

Hippalectryon

19:24

@MikeMiller $\phi(x)$ is smooth, but $\phi(\lfloor x\rfloor)$ isn't

Mike Miller

Wait

Where did that floor come from

Why is there a floor there

Hippalectryon

Idk just copy paste :c

Mike Miller

I didn't turn my ChatJax on so I never noticed it

Hippalectryon

Lol

Mike Miller

Get rid of the floor

Hippalectryon

19:25

Ok

Mike Miller

Then you're smooth :)

N3buchadnezzar

@DanielFischer Fine. What are you up to these days?

Hippalectryon

@MikeMiller It's smooth, but the derivative seems awful.

$$-\sum _{k=1}^{m}\frac{\phi \left(\frac{\left(x\right)}{k^2}\right)}{k^2}\dfrac{2x}{(1-x^2)^2}$$

Daniel Fischer

@N3buchadnezzar Trying to finish reading up on past events and general moderation strategies so that I'll have time to answer a question or two again in the foreseeable future.

Mike Miller

Oh, we've done something wrong.

The thing you're describing won't ever be more than 1.\

We want $$\sum _{k=1}^{\lfloor x\rfloor}\frac{\Phi '(k^2(x-\lfloor x\rfloor)}{k^2}$$

N3buchadnezzar

19:31

@DanielFischer Be moderate and modest in the moderation. Also remember to use the official title: Mod in the hood, yo. When entering chatrooms =)

Mike Miller

balls

that's what we want.

NO

DAMn

Hippalectryon

Lol

Mike Miller

I didn't want the prime

well, you can figure out what it's supposed to say.

Hippalectryon

Ok ok

$$\sum _{k=1}^{\lfloor x\rfloor}\frac{\Phi (k^2(x-\lfloor x\rfloor))}{k^2}$$

Mike Miller

Yes

now note that outside of a small region, each of those $\Phi$s is constant... hence has derivative zero.

Hippalectryon

19:34

But how do I get a derivative that I can use ? (to show it's $1$ somewhere on eveer interval of length ...)

Mike Miller

so you only need to care about one $\Phi$ at a time

Hippalectryon

Yep

Ugh wait, why is that ?

$|k^2(x-\lfloor x\rfloor))|<1$ in the general case

Mike Miller

hmm

ok let me fix

$$\sum _{k=1}^{\lfloor x\rfloor}\frac{\Phi (k^2(x-k))}{k^2}$$

finally this works

Hippalectryon

Hmm what about the derivatives this time ?

Mike Miller

now the thing I said above applies

Hippalectryon

19:40

I don't see why we can only consider one function at a time. What if $x-k$ is small enough ?

Chris's sis

Greetings

Hippalectryon

@Chris'ssis Hello !

Chris's sis

@Hippalectryon Hello

Mike Miller

@Hippalectryon If $|x-k| < 1/k^2$, is $|x-n| < 1/n^2$ for any other integer $n>1$?

Hippalectryon

Nope

:)

Mike Miller

19:43

right - triangle inequality. (I needed to demand $n,k>1$ because 1/1+1/4 > 1... so 1.8 works just fine as an x)

Hippalectryon

Hm ok so $f'(x)=\left(\frac{\Phi (k^2(x-k))}{k^2}\right)'$ for a given $k$ which depends on $x$.

Mike Miller

yep

Hippalectryon

Which is a rather ugly derivative

Mike Miller

Chain rule says that's equal to $\Phi'(k^2(x-k))$...

Hippalectryon

Well it's equal to $-\dfrac{2x\phi(k^2(x-k))}{(1-x^2)^2}$

Mike Miller

19:45

Working too hard

Phonon

hi guys, can I ask one tiny permutation/combination question?

Mike Miller

Is there a point $c$ such that $\Phi'(c) = 1$?

Hippalectryon

@Phonon Nope, it's forbidden :P

Go ahead lol

@MikeMiller I hope so q_q

Mike Miller

Actually, not even that. There's some positive $a$ and some $c$ such that $\Phi'(c) = a$. That's automatic - $\Phi$ is monotonic nonconstant.

Phonon

I have only two distinct elements, say 2 letters, i and j, in how many ways can I create distinct n element strings from such i and j, if I know that I have k times i occuring and n-k times j occuring?

Hippalectryon

19:47

@MikeMiller That wouldn't work. Take $f(x)=1-1/x$. It is monotonic onoconstant, but its derivative$\to0$

Mike Miller

What?

Phonon

(without repetition)

Mike Miller

@Hippalectryon Yes, that has nonzero derivative everywhere on its domain. I assume you want the domain to be positive reals.

No, I don't care that its derivative goes to zero. I care that it has nonzero derivative somewhere.

Hippalectryon

@MikeMiller I'm only looking at the limit in $+\infty$, so a positive domain is ok.

Mike Miller

Because $\Phi'(c) = a$, automatically $f'(k+c/k^2) = a$.

Hippalectryon

19:49

@MikeMiller But we want a function with a derivative that does not go to $0$

Mike Miller

That's a sequence of points that goes to infinity at which the derivative is a fixed positive number. So $f'$ does not have a limit as $x$ goes to $\infty$.

Hippalectryon

Uh why wouldn't that apply for $1-1/x$ then ?

Your $a$ depends of $x$

Or am I missing something ?

Mike Miller

yes.

Hippalectryon

@MikeMiller You said "There's some positive a and some c such that Φ′(c)=a.", but that is true for any function which is inreasing somewhere

Mike Miller

yes, that's correct.

Hippalectryon

19:55

So, how do we use that here ?

We want to show that there exists an $a>0$ such that for all x, there exists $x_0>x$ such that $f'(x_0)=a$

kave

why is (a+b)^p <= 2^p(a^p+b^p) ?

Pedro Tamaroff

@kave Convexity.

kave

@PedroTamaroff can you explain further?

Pedro Tamaroff

@kave Do you understand where I am getting at?

Do you know what "convexity" means?

kave

yes but dunno which function you want to have convex

x^p would be convex

but the 2^p irretates me. btw p >= 1

Hippalectryon

19:58

@MikeMiller So, when did we show that ? Reading again what we said, i can't find it.

kave

oh got it

@PedroTamaroff thanks

Pedro Tamaroff

@kave No problem.

Mike Miller

@Hippa You know what $f'$ is. You verify that $f'(k+c/k^2)=a$ for all positive integers $k>1$. You conclude.

Hippalectryon

What you call $c$ is what I called $x_0$, but where have we proved its existence ?

If we have its existence, then indeed we can conclude easily.

Mike Miller

Hippa, $\Phi$ is a nonconstant monotonic function.

wing

20:04

any other equal for this series? (except cosh(x)
(1+x^2+x^4+.... = (e^x+e^(-x))/2) )

Mike Miller

I could drop the word monotonic and it wouldn't matter.

There is some point $c \in (-1,1)$ that has nonzero derivative.

Hippalectryon

:O I just got it. My mind was really confused :c

evinda

could you take a look at this?
http://math.stackexchange.com/questions/1045173/which-is-the-greatest-integer-value-of-a-for-which-a-is-asymptotically-fas

Hippalectryon

@MikeMiller Ugh sorry, but I still have a problem with that q_q

@MikeMiller Let's call $\phi_k=\frac{\Phi (k^2(x-k))}{k^2}$

@MikeMiller Suppose that $\sup\phi_k'(x)=a_k>0$

What happens if $a_k\to0$ ?

Pedro Tamaroff

@Hippalectryon Hippa.

Stop.

Sit down in a table.

Hippalectryon

20:08

@PedroTamaroff ?

Pedro Tamaroff

Think about the problem by yourself for a while.

Hippalectryon

@PedroTamaroff I'm on a desk

Pedro Tamaroff

OK, no internet.

Try to answer the questions by yourself.

Else you're compulsively asking stuff that pops to mind without thinking about it.

Think about it.

evinda

@DanielFischer Could you maybe take a look at this? math.stackexchange.com/questions/1045173/…

Hippalectryon

@PedroTamaroff I've already thought about it. As i'm typing here, i'm also writing on paper. What I don't understand is that we have a function whose derivative is $f'(x)=\phi_{k_x}(x)$ where $k_x$ depends of $x$ and the $\phi_k$ are nonconstant functions. Mike told me (well, that's what I understood) that the $\phi_k$ are nonconstant hence their derivative is $>0$ somewhere hence $f'$ doesn't converge to $0$. I still don't see how that last point is made.

Mike Miller

20:12

Before you go, I messed up the definition of $\Phi$; the result doesn't look like a staircase like I wanted. You need to modify $\Phi$ so that it's constant outside of $(-1,1)$, smooth, and $\Phi(x)=0$ for $x<-1$, $\Phi(x)=1$ for $x>1$. Otherwise your function isn't monotonic.

Before I go, rather. I'm leaving. See ya.

Hippalectryon

Ok, thanks for everything @MikeMiller

Alizter

20:24

heeeuuulloww @Hippalectryon

Hippalectryon

@Alizter Hoi

Alizter

@Hippalectryon Wut arr yuu duin

Hippalectryon

@Alizter Maaathhhs with SANIC

Alizter

@hippal wat sanic?

Hippalectryon

SANIC, GOTTA GO FAST

Ted Shifrin

20:49

@Hippa: Je suis désolé que je sois disparu :)

Hippalectryon

@TedShifrin uh ?

Ted Shifrin

You asked me a question and I had to leave. No, $f'$ doesn't have to go to $0$ as $x\to\infty$ if $f(x)\to\infty$. Try $f(x)=e^x$.

user153330

heeellaaw

Ted Shifrin

howdy @user153330

Hippalectryon

@TedShifrin $f$ needs to have a real limit

Ted Shifrin

20:52

Oh, your statement was ambiguous. So $f$ is increasing and $f\to c$ as $x\to\infty$?

user153330

@TedShifrin how're you prof? =)

Ted Shifrin

doing ok, thanks ... :)

Mike Miller

I'm here for a moment before I go. @Ted: Don't start over, we constructed one... there's just a detail he's missing.

Hippalectryon

@TedShifrin Yep

Ted Shifrin

Oh, ok ... Never mind. You constructed one where the derivative doesn't go to 0, right?

Hippalectryon

20:54

@TedShifrin Mike showed me somthing I should be able to turn into an answer, though

Ted Shifrin

OK, good, I don't need to burn any more brain cells.

Mike Miller

Lol

Hippalectryon

:D

user153330

@Chris'ssis hi, how are you?

Chris's sis

@user153330 Hi. Not that bad, thanks. You?

user153330

20:57

@Chris'ssis tired these days

Chris's sis

@user153330 I know what you mean. I need to do something and buy a large black(white)board and work there. I'm sick and tired of sitting down.

user153330

@Chris'ssis a white board is better

Chris's sis

@user153330 Yeah, I need a large one and then continue my work there.

kave

in analysis we defined measurable function as functions so that the preimages of intervalls are measureable which is equivalent to that preimages of borel sets are measureable.
in probabilty theory we defined measureable functions as functions so that preimages of all _measureable_ sets are measureable, which leads to some (for me bigger) differences, e.g. in analysis the chaining of two measureable functions doesn't need to be measureable, in probability theory though it does.
is it normal that there are two different definitions for something so basic?

Chris's sis

Actually I walk around my table and try to solve another problem. :-)

user153330

21:02

@Chris'ssis a black board is better iff you have one of those (those are the same chalk that are used in MIT)

Chris's sis

@user153330 Those from MIT are probably pretty expensive.

Huy

@Chris'ssis: When I'm done with my studies, I'll totally need a whiteboard in my office at home. I'm looking very forward to it.

kave

@Chris'ssis buy window pencils for less then 10 bucks

that is how I do/did it

Huy

Like Nash in the movie?

kave

no black or whiteboard whatsoever needed

yeah I got the idea from the movie probably :D

but they do that in a lot of cool movies

Chris's sis

21:06

@kave Only if I had those windows from "A beautiful mind" :-)))

Huy

I'd find that a bit weird, tbh.

Chris's sis

@kave Me too. It's like the water and air to me, I need it! :-)

Huy

I'd go for a whiteboard over writing on my windows any time.

kave

@Huy but 1. whiteboards cost, 2. as a student, my room isn't that big that I like to put other big stuff into here :D

Huy

@kave: Yes, that's why I'm only getting a whiteboard after I'm done with my studies (hence working more).

@kave: I wouldn't dare to ruin my view by writing on my windows in my current student's flat. :P

kave

21:09

I can live without my view :D it isn't that great sadly

Huy

@kave: There's the difference. :D

kave

yeah that could just be it :D

Chris's sis

This is what I proposed to myself today

Prove without p & p that $$\lim_{p\to\infty} \frac{\sqrt{2p+1}}{2^{2p}}\sum_{j=0}^{p} \frac{(-1)^j}{2j+1}\binom{2p+1}{p-j}=\sqrt{\frac{\pi}{2}}$$

2

Without many boring calculations and without leaving unclear steps. All must be straightforward and the limit is dead in one line.

Things are clear for an eye with some practice, but explaining things a bit rigorously it can take some more.

Considering some first values of $j$ we see immediately why the limit is $\sqrt{\pi/2}$, and that's because we have in mind the asymptotic for the central binomial coefficient.

Hippalectryon

21:48

@Chris'ssis How would you do $\displaystyle\int_a^\infty\dfrac{\cos x}{\sqrt{x^\alpha-\sin x}}dx,\alpha>1$ (and $a>0$ such that $x^a=\sin x$) ?

Ted Shifrin

@Hippa: (1) What you have isn't right. (2) Please do not use $a$ and $\alpha$ in the same problem.

Hippalectryon

@TedShifrin What do you mean "isn't right" ?

Ted Shifrin

I believe you mean that $a^\alpha = \sin a$.

Hippalectryon

Oops bad notation i'll use \beta

Ted Shifrin

Nods

Hippalectryon

21:58

@Chris'ssis $\displaystyle\int_a^\infty\dfrac{\cos x}{\sqrt{x^b-\sin x}}dx,b>1$ and $a>0$ such that $a^b=\sin a$

Chris's sis

OK

@Hippalectryon Having given that condition, I had in mind the change of variable $\sqrt{x^b-\sin(x)}=y$

Ted Shifrin

It's not totally obvious to me that it converges when $b<2$, say.

Hippalectryon

@Chris'ssis I didn't manage to solve it, but tell me if you find something :)

@TedShifrin It ?

Ted Shifrin

But I was hesitating ...

Hippalectryon

22:16

@TedShifrin Any hint on the convergence of $\displaystyle\int_0^\infty\dfrac{x}{1+x^4|\sin x|^{3/2}}dx$ ?

Ted Shifrin

@Hippa: I have no idea, although I would vote against.

Mike Miller

@Ted It strikes me that one could do a lot of elementary topology by using handlebody decompositions; it gives nice proofs of classification of surfaces, Poincare duality... I wonder if one could make this a pedagogically reasonable approach for a first course.

Ted Shifrin

Takes a reasonable about of work (i.e., differential topology) to understand them in the first place, though.

Mike Miller

I guess one often does the same with triangulations instead.

Ted Shifrin

The differential topology course I took from Hirsch did a lot of what you say.

Mike Miller

22:23

Yes, that's fair enough. That's why I'm wondering if it's a reasonable approach instead of just saying "What an idea!"

Ted Shifrin

My view is that unless one already knows a lot of topology, it won't make much sense.

Mike Miller

Such an approach would probably take as an axiom handlebody decompositions. But one usually does that with triangulations, too...

I guess so.

Shame.

Ted Shifrin

@Hippa: I continue not to believe your integral converges when $b<2$. Here is the graph of the integral of $(\cos x)/x^{2/3}$, for example.

Hippalectryon

@TedShifrin TBH I haven't studied the convergence in detail. It might not converge everywhere.

Ted Shifrin

I think you need $b>2$ or something ...

But I'll let @Chris'ssis decide :P

@Hippa: It sorts of feels like the Dirichlet test for convergence of series. So maybe you can get it to work, and the convergence is just very, very slow.

Lord_Farin

22:34

I'm frustrated to say I'm unable to find a duplicate for this trivial question.

Ted Shifrin

I'm frustrated that people can't take the time to look up formulas on the web.

Hippalectryon

@Lord_Farin Then leave it. That way, it will be easier to mark the next one as a duplicate.

Lord_Farin

Should someone find an appropriately general treatment, consider adding it to this meta thread.

Ted Shifrin

At this point in my life, I have a lot more to do than to play policeman of the postings here.

Gato

@TedShifrin@Hippalectryon Bonsoir!

user159870

22:36

Hello everyone! I have found a german phrase but I don't know what it means. I hope someone speak german here! The phrase is : "Das Ganze ist mehr als die Summe seiner Teile."

Ted Shifrin

Bonsoir, @Gato.

Hippalectryon

@Gato Bonsoir !

@user159870 Have you tried google ? ............

Lord_Farin

@user159870 The whole is more than the sum of the parts.

Gato

@Hippalectryon What's up? La rentrée?

Hippalectryon

@Gato Surtout, les concours dans ~11 semaines

Gato

22:37

@TedShifrin how are you?

user159870

@Lord_Farin Interesting! But what does this mean?

Ted Shifrin

It's a famous philosophical saying, @user159870.

Lord_Farin

@TedShifrin A wise decision. I get annoyed sometimes and decide to hunt for the dupe instead of closing as "off-topic -> context".

Gato

@Hippalectryon Ah oui bientôt, prêt?

Hippalectryon

@Gato Pas encore lol

Gato

22:38

@Hippalectryon J'imagine...

user159870

@TedShifrin Could you explain this saying? I have never heard of it.

Ted Shifrin

@user159870: Something like this. You get a lot more out of a wonderful poem than you do out of its individual words.

Mike Miller

@Lord_Farin Tragically, you voted to reopen this question, leaving nobody to close it now that it's actually a duplicate.

user159870

@TedShifrin Thanks!

Ted Shifrin

I hardly find it tragic, @Mike.

Mike Miller

22:40

Sorry to hear that, @Ted.

Lord_Farin

@MikeMiller I couldn't care less. Why don't you just leave the post alone?

Mike Miller

Why not do the same on main?

Gato

@TedShifrin giving a sequence how can I find the $\lim \sup(u_n)$?

user159870

Also I read a joke but I don't get it.

It is: $\forall \forall \exists \exists$

Can you explain it please? @TedShifrin

Lord_Farin

@MikeMiller I think you know the difference between a congratulatory thread and a solution to a mathematical problem.

Ted Shifrin

22:42

I have no idea, @user159870

Lord_Farin

The purposes of main and meta are wholly different, and my closing habits reflect that.

Ted Shifrin

Depends on the sequence, evidently, @Gato.

Mike Miller

If it makes sense, @user159870, I assure you it's not funny.

user159870

@Lord_Farin u?

Lord_Farin

No clue.

user159870

22:43

DO YOU WANT TO SAY THAT I AM NOT FUNNY????? @MikeMiller

Ted Shifrin

correct, @user159870

Mike Miller

That's not what I said, but I can say it if you want.

user159870

No, please no @MikeMiller HHAHAHHAHHAHA

Kevin Driscoll

@TedShifrin I bet it does converge but not uniformly

Ted Shifrin

that makes no sense, @Kevin

We have a single function here.

Oh, you mean uniformly in $b$? I'm worried about $b$ fixed.

Gato

22:46

@TedShifrin Right, but I don't understand how does it works. The definition is $\lim \sup=\inf_{n\in \Bbb{N}}[\sup_{k\in \Bbb{N}}(u_{n+k})]$, for example I have $(1,1/2,2,1/3,\cdots,1/n,n,\cdots)$, I don't see how can I compute the lim sup.

Ted Shifrin

Another way of thinking of it (equivalent) is to take the sup of all subsequential limits, @Gato. But with your definition, it should be clear that if you look at all the terms except the first 100, what's the sup of what's left?

Lord_Farin

@Gato $\limsup$ is infinite for this sequence.

Ted Shifrin

@Lord: Sometimes people give answers way too quickly ... on main and in here.

Lord_Farin

@TedShifrin I guess I read "don't see how I can compute" in the context of "because it won't be finite".

Apologies for spoiling.

Ted Shifrin

LOL ... I'm not entirely sure what @Gato meant, but still ...

Kevin Driscoll

22:49

@TedShifrin Maybe I used the wrong word. Mathematica gives a finite answer for $\int_0^{\infty} \cos{x}/x^{2/3} dx$ but cannot give a finite answer for $\int_0^{\infty} \lvert \cos{x}/x^{2/3} \rvert dx$

Gato

@TedShifrin Ok I see, so yeah it's $+\infty$ we can 'extract' the sequence $(n)$. OK the word $compute$ was not appropriate.

user159870

I found this joke on a site saying that one of the signs that you are a mathematician is that you understand this joke.

We all haven't understood this joke. Result? We aren't mathematicians.

@TedShifrin @Lord_Farin @MikeMiller

HAHAHHAHAHAH

Kevin Driscoll

But I came late to the party, so it's possible I missed something important

Ted Shifrin

So what's lim sup for this sequence, @Gato: $a_n = (-1)^n(2-1/n)$.

Lord_Farin

@user159870 Perhaps the site is wrong.

Ted Shifrin

22:51

smacks @user159870

Lord_Farin

2 days ago, by Ted Shifrin

I think I've made this room too violent.

Ted Shifrin

I am cognizant, @Lord :D

Lord_Farin

@TedShifrin :P. Thanks for expanding my vocabulary.

Gato

@TedShifrin $2$? (even subsequence).

Ted Shifrin

You didn't know that word? Like cognescenti? cognition? :D

Hippalectryon

22:53

@TedShifrin Hello, cognizant ! I'm Hippalectryon. :3

Ted Shifrin

yup @Gato

Nick

@Lord_Farin Potatoes! :D

Lord_Farin

@TedShifrin I figured it'd be something like that, but no, hadn't encountered it before.

Ted Shifrin

But do you see how to get it from your definition, too, @Gato? Here's the reason for the definition. What if the first 10 terms were 1,2,3,4,5,6,7,8,9,10, and then we use my sequence?

hi @Nick

Nick

@Hippalectryon Hello Hippa, I'm the Game :D

Hippalectryon

22:54

@Nick nooo you just made me lose :/

Kevin Driscoll

Ya I definitely used the wrong phrase, but I can't remember what the right one is

Hippalectryon

Shame on you @Nick

Ted Shifrin

You mean it doesn't converge absolutely, @Kevin.

Nick

@TedShifrin Greetings, wise owl :)

Hippalectryon

Owls don't cook

Kevin Driscoll

22:55

@TedShifrin Yes, you're right

user159870

An other sign is: It takes you 20 minutes to split a bill of $84.76 three ways after adding a 15% tip.

Is this true for you? @TedShifrin @Lord_Farin @MikeMiller

Nick

@Hippalectryon Shame on me? No, shame on you! Shame, Shame, puppy shame!

Ted Shifrin

No, @user159870, and I usually get stuck figuring out bills at restaurants when I go with non-mathematicians.

Kevin Driscoll

"because you must be good with numbers" ?

Ted Shifrin

They know I'm not a number person.

user159870

22:58

So maybe the negation of all these signs shows that we are mathematicians!

AHHAHAHAHA

@TedShifrin

Nick

@KevinDriscoll "I'm well versed in the sets of all numbers, I'm great with numbers. I suck at arithmetic, though."

Ted Shifrin

ok, @user159870 ... enough.

Hippalectryon

@Nick I am Kasp Swordbeer, and I shall slay you where you stand, you night-prowling beast of shadow!

Lord_Farin

^ This.

user159870

Sorry, if I have disturbed. @TedShifrin

Gato

22:59

@TedShifrin If I think correctly if $u_n$ is majorée (bounded below?) we have a subsequence witch converges to the lim sup, and if $u_n$ is not bounded below a subsequence converges to $+\infty$. In any case we cannot 'extract' a subsequence that converges to a limit >lim sup.

Transcript for

$Mathematics$ Mathematics