@MattN and now my strange neighbor just rang on my door....

@MattN How about kittens?

@tb Wha..?

Forgot his key, apparently...

@tb Was he drunk and without pants?

And you let them in?

I would've pretended to not be home.

@MattN well, it's very hard to pretend not to be home when he glances through your kitchen window...

@AsafKaragila fortunately, not :) Slightly drunk maybe.

@tb You can always shout through the door "I'm not here, you drunken jackass!!" (in German, though)

Or just look away.

: D

People how spend too much time on the internet pretend they are not at home a lot

@AsafKaragila bi nid da du bsoffne whatever, you mean?

Ah but he only rang your doorbell because he saw you were still awake.

So if he hadn't been sure he'd not done it.

@tb Sure. Why not.

So he's not a complete idiot.

@MattN well, not seeing any light didn't inhibit his ringing the door bell in the past...

@tb Ha. You're funny.

I would've quickly made him learn that doing this is a very bad thing.

@MattN By stabbing him in the liver?

No by not letting him in.

Well, if you ever stab someone in the liver, watch out for Hep A/B/C.

I need to go to sleep.

Good night, Matt!

I'll go to sleep too...

Night, teddy.

I'll keep my wise advice to myself today.

Living in different time zones pretty much sucks.

@MattN And you won't let the next door weirdo in tonight?

@MattN Well, I'm trying not to stay up for too long tonight.

There are no next door weirdos here.

@MattN I have an indoor weirdo... but that's just me.

@MattN So I heard.

@robjohn wow!

But why divide by 4?

@tb: there is a nice smooth example built from the ideas of my piecewise linear one.

@tb Just realised that I won't be seeing much teddy bear once we're in different time zones either.

Are you moving, @tb?

No, I am.

@tb I wanted the slope at 0 to vary between 2 and 1/2 like my piecewise linear one

@robjohn I see! Very nice. The plot looks nice, too.

: )

See you tomorrow folks.

This is actually the picture I had in mind earlier, but could not write it down :-)

Bye too.

@AsafKaragila see you

Perhaps. Iran might let out with the nukes while I'm asleep...

Bye Asaf

@AsafKaragila that would probably wake one up.

@robjohn Only if you're riding the bomb...

@AsafKaragila Yee-haw!

:)

And people think that DSL is a kind of high speed internet!

Those who know, know it means Dr Strangelove!

Flouride in the water is a communist plot

@robjohn It is safe to upvote you now, isn't it?

@tb It is a new day :-)

@robjohn there!

@tb Thanks for waiting :-) I did send 45 points to the cap bin yesterday.

@robjohn how annoying...

@tb it's just part of getting too many points

For me it was just 20 but that's because I deleted two answers...

I wonder how many points Arturo would have if not for the cap.

@robjohn if this query is trustworthy, he's lost about 30-35k

(the databases are old)

(his userid is 742)

but it is accurate for a time in the past. That should be a lower bound.

well, it's about the third or fourth query I tried, the other ones just gave some nonsense.

31152

Can someone here give me a hint as to how I can prove that the product topology on $X\times Y$ is the coarsest topology for which the projection functions $p_{X}:X\times Y\to X$ and $p_{Y}:X\times Y\to Y$ are continuous?

@robjohn That's what I got, too. So he's losing about 1.5-2k per month. The databases are about two months old, so...

you've lost 830 according to that

@DavidK how do you define the product topology?

@robjohn this sounds about right.

I haven't capped badly very often.

I've lost 248

@tb Evidently nor have I :-)

@tb The product topology on $X\times Y$ is the topology generated by the basis $\mathcal{B}=\{U\times V:U\text{ is open in }X\text{ and }V\text{ is open in }Y\}$.

@DavidK Agreed. 1. are the projections continuous with respect to that topology? 2. What can you say about the sets that have to be open if the projections are to be continuous?

@tb 1. Yes, the projections are continuous with respect to that topology. 2. $\ldots$

@DavidK well, if $U$ is open in $X$ then $p_{X}^{-1}(U)$ is open in $X \times Y$. If $V$ is open in $Y$ then $p_{Y}^{-1}(V)$ is open in $X \times Y$. Now what can you say about $p_{X}^{-1}(U) \cap p_{Y}^{-1}(V)$?

@tb 2. $p_{X}^{-1}(U) \cap p_{Y}^{-1}(V)=((U\times Y)\cap(X\times V))=(U\times V)$.

and?

@tb and $p_{X}^{-1}(U) \cap p_{Y}^{-1}(V)$ is open in the product topology, since it's the intersection of two open sets?

@DavidK if $p_X$ and $p_Y$ are continuous then this set must be open.

1. says: the product topology is coarser than the topology generated by $\mathcal{B}$

(coarser = less open sets)

and 2. says: the product topology must contain the sets of $\mathcal{B}$, so it is finer.

math.stackexchange.com/q/117723/21436 needs close votes.

Taking those two together you see that the product topology as you described it is exactly the topology making the two projections continuous.

@tb Wait, the product topology is generated by $\mathcal{B}$.

@DavidK Sorry, read: the coarsest topology making the two projections continuous whenever I said product topology.

(it's getting late here)

@tb @tb I'd like that you review that question in link before you go to bed. Please votee to close as it is exact duplicate of...

@KannappanSampath I just did.

@tb Thank you so much. Have a good night's sleep. : )

@tb so "1. says: the coarsest topology making the two projections continuous is coarser than the topology generated by $\mathcal{B}$ and 2. says: the coarsest topology making the two projections continuous must contain the sets of $\mathcal{B}$, so it is finer"?

Exactly.

@tb finer than what?

@DavidK once again: let $\mathcal{T}$ be the topology generated by $\mathcal{B}$ (what you call the product topology) and let $\tau$ be the coarsest topology making $p_X$ and $p_Y$ continuous. 1. Says that $\tau \subset \mathcal{T}$, so $\tau$ is coarser than $\mathcal{T}$. 2. says that $\mathcal{T} \subset \tau$, so $\tau$ is finer than $\mathcal{T}$. Taking these together you get $\mathcal{T} = \tau$, hence the product topology $\mathcal{T}$ is equal to $\tau$,

i.e. the coarsest topology making the projections continuous

@KannappanSampath thanks, I'll be there, soon :)

@tb Oh! I see what you were saying now. Sorry for having to spell it out for me like that. Thanks!

@tb Looks like you'll need to sleep soon. : )

@DavidK oh, no problem, I made the confusing move to call both $\tau$ and $\mathcal{T}$ product topology, out of habit... So, is this settled now?

@tb Yes I think so. Some details to be worked out, but I think I get it now. Thanks again!

@tb: I added the smooth example, but its graph disappears into smallness before an entire period, so I didn't add it.

@DavidK Okay, very good then. I'm up for way too long now, so if anything remains unclear, just ping me, but you'll have to wait for a few hours before I'll answer. But there are others here who could answer that as well... Sorry for the confusion, and see you around!

@tb Thanks. Cheers!

@robjohn Wait, I read: you added it, ..., so you didn't add it? :) I'll check it out tomorrow. Good night!

@tb Good night. I added the example, but not the graph

@KannappanSampath Yes I do and I fear Matt's reaction if I don't go right now... See you tomorrow!

@robjohn I see.

Good night y'all

@robjohn Need a suggestion of notation.

Can you help me with it, please?

@KannappanSampath notation about what?

So, we have $k$ random Bernoulli Random variables, the $i$th random variable takes value $1$ with probability $p_i$ and 0 with probability $1-p_i$.

okay

Now, I would like to know $P(\sum X_i=j)$ for $j \in \{0,1,2, \cdots , k\}$. This is cyclic sum of product of $j$ $p_i$'s and $k-j$ $(1-p_i)$'s, but I dread writing it.

@Srivatsan: good morning!

Morning, Rob.

I suppose it is evening for you.

@Srivatsan Good morning. : )

@robjohn So, please tell me what do you think would be ideal way of writing that?

Good morning, Kannappan.

@KannappanSampath I'm working on it... just a minute

@KannappanSampath look at the coefficient of $x^j$ in $\prod\limits_{i=1}^k(1-p_i)+p_ix$

@KannappanSampath Is that better?

@robjohn Oh, yeah. Great. Thank you. This is a great idea. But, I wonder is it going to be any easier to write the distribution. But, never mind about this distribution.

@robjohn Sure it is!

@KannappanSampath You are doing the same calculations, but a CAS can do it in this form with a simple formula :-)

@robjohn CAS??

@KannappanSampath But I don't think there is a really simple way to compute the convolution of those distributions.

@KannappanSampath Computer Algebra System (Mathematica, etc)

@robjohn Thank you.

@Rob: Did you happen to see this question? :)

@Srivatsan Nope.

You asked it 21 minutes ago?

Yes, recently. I deleted it thinking I will give myself one last chance. If I can't solve it by the end of the day, I will undelete the question...

@Srivatsan ah :-)

Ok, something came up, got to go. I will see you guys later.

@Srivatsan see you later !

@Srivatsan see you later.

@robjohn Can I ask for more help there?

@KannappanSampath okay, but I have to leave in a short bit.

@robjohn This bit and there is another, as long as you stay, may be...

I am trying to figure out what the expectation would be.

But, I cannot succeed in getting with those simplifications.

Using $f(x)=\prod\limits_{i=1}^k(1-p_i)+p_ix$ evaluate $f'(1)$.

That is the expectation.

Ah, I have the Probability generating function, right?

yes

For higher order moments, I differentiate that many times, evaluate at $1$, is that right? @robjohn

@KannappanSampath you need to multiply by $x$ and differentiate

but essentially, yes

@robjohn for second order moment?

I have to go. Contact me later about the other thing.

@KannappanSampath don't you want to essentially find the expectation of $x^2$?

@robjohn Sure, Thank you. Have a nice time.

@robjohn No, I am playing around with the fact that sum of bernoulli is binomial in this manner. So, more I know, the better. : )

but the second moment is the expectation of $x^2$ or not?

@robjohn Yes.

I think I have had the brilliant flash.

I have PGF. So, I know MGF, differentiate that several times to know.

so $f(x)=\sum_j P(S=j)x^{j}$

Yes, but with $S$ instead of $X_i$.

then $f'(x)=\sum_j P(S=j)\;jx^{j-1}$

and $f''(x)=\sum_j P(S=j)\;j(j-1)x^{j-2}$

@robjohn Yes, right. : )

@robjohn True as well! : )

So you could take $f'(1)+f''(1)$

for the second moment

@robjohn Yes. Cool. So, this is pretty.

ok. Now I need to go. See you later

Thank you @robjohn Have a nice time. Thanks for the help. : )

Hmm. Nothing happened since I left.

Heeeeeeeeeeeeeeeey!

@JasperLoy why would you do that?

Just said hello.

@JasperLoy You're watching Will Hunting?

@JasperLoy I never knew: What mathematics does he work on in the movie?

@JasperLoy Do you recognize what is the importance of the paper he burns?

I mean, IRL, does it relate to any open problem?

@JasperLoy So, are you looking at any questions in math.SE?

@JasperLoy What's ELU?

@JasperLoy Oh. I see. And how does that relate to math? I'm interested.

@robjohn Would you like to take a look at this?

@PeterToff I have to head out to get dinner now. I will look when I get back.

@robjohn Oh, its 00:05 here, so I guess I won't be around then. Well, maybe I will, I don't know.

@PeterToff What you showed using the OP's assumption looks pretty reasonable. What's wrong?

Reminds me of the Cauchy estimates.

@Dylan Well, that you still need $|x-a| < R$

@DylanMoreland I proposed $f^{(n)}(x)\leq R^n$. At least, that gives a definitive result. Maybe I'm missing something.

$f$ is only defined on an interval to begin with, so that doesn't seem so bad.

@DylanMoreland Yes, it doesn't. Is the condition $|x-a|<R$ implicit somewhere, and I'm missing it?

Maybe the statement is just imprecise. From the inequality, which he seems pretty confident in, I think what you've done is all there is to say.

@DylanMoreland Hm. Ok.

@DylanMoreland Are you browsing any interesting questions?

I could be so wrong, though.

Not really.

@DylanMoreland Check this one out

Plz suggest me a book with a good illustrative and high emphasis of explaining concepts slowly...for the subject of derivatives of functions of the form $f : \mathbb{R}^n \to \mathbb{R}^m$..i need full mastery of the concepts....as i want to venture into learning differential geometry and so that it will be easy there

@Rajesh Why don't you ask in math.SE or look for a similar question already asked?

i am searching....just in case someone knows readily something...thanks anyway

Hi, is it possible to download 'Convex Analysis and Minimization Algorithms ' from the internet

Hello, hello, is there anybody out there?

Does anybody really care ...

Hey

Hi.

Hello!

Whatz up?

Happy International Women's Day: un.org/ecosocdev/geninfo/women/womday97.htm

Can someone here help me with a topology problem? I think I know how to solve it, but I've been awake for a long time now, and I'm not sure that I'm making any sense.

Teddy has left.

Yes, but David has arrived.

: D

Hello David!

Did you want help with topology?

@MattN Yes. But this isn't a very topological question.

Then I'm not sure I can help you but how about you just ask it?

Without having to prove a bunch more stuff, I'm trying to use some lemmas that I already know.

Namely that $\sigma:\mathbb{R}^2\to\mathbb{R}$ defined by $\sigma(x,y)=x+y$ is continuous, and the analogous statement for the multiplication map.

Good. Then how about you go by induction?

Concatenation of two continuous functions is continuous.

Here's my idea. Define $h_{1}(x)=f_{1}(x)$ and for $2\leq n\leq m$, $h_{n}(x)=(h_{n-1}(x),f_{n}(x))$.

I thought.

Morning teddy, did you sleep well?

@MattN very well :) morning

Then $S(x)=\sum_{n=2}^{m}(\sigma\circ h_{n})(x)$

Does this work as you say under the "concatenation" argument?

Oh wait, that makes me "over-count" doesn't it...

Well $\sigma$ is supposed to do the addition.

But I'm still stuck seeing what you get when you concatenate $\sigma$ with $((((\dots)\cdot) \cdot) \cdot)$

I would prove first that $f: \mathbb{R} \to \mathbb{R}^m$ given by $x \mapsto (f_1(x),\ldots,f_m(x))$ is continuous. Then use that $\sigma: \mathbb{R}^m \to \mathbb{R}$ and $\mu: \mathbb{R}^m \to \mathbb{R}$ given by $\sigma(x_1,\ldots,x_m) = x_1 + \cdots+x_m$ and $\mu(x_1,\ldots,x_m) = x_1 \cdots x_m$ are continuous and observe that $\sigma \circ f$ is your sum and $\mu \circ f$ is your product.

Right, I want to do an iteration I think...

That's what I meant by induction. Now I'm not so sure anymore what I said is actually right.

Sigh.

But now teddy gave you the answer, so: problem solved : )

But of course the direct approach by induction works, too.

I already have that $x\mapsto(f_{1}(x),\ldots,f_{m}(x))$ is continuous. For the other two I only have that $\sigma$ and $\mu$ are continuous when $n=2$. So I'd have to prove by induction that $n>2$ is continuous.

@tb Good : )

I should add, that I'm lazy.

Maybe I caught it off you.

Is it contagious?

Yes.

@DavidK use induction for the continuity of $\sigma$ and $\mu$.

@tb Okay.

Thanks guys.

For $\sigma$ it's basically the triangle inequality... For $\mu$ something along the lines that Matt suggested.

You need to know that $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^{n-1} \times \mathbb{R}$...

@tb We haven't developed the concept of homeomorphisms yet.

Then observe that $\mu_{n}(x_1,\ldots,x_n) = \mu_2(\mu_{n-1}(x_1,\ldots,x_{n-1}),x_n)$, similarly for $\sigma$.

So you're looking at a concatenation $$\mathbb{R}^{n} \to \mathbb{R}^{n-1} \times \mathbb{R} \xrightarrow{(\mu_{n-1},\operatorname{id}_{\mathbb{R}})} \mathbb{R} \times \mathbb{R} \xrightarrow{\mu_2} \mathbb{R}$$ of continuous functions.

Can this be tagged pre-calculus?

Maybe it's too advanced for that.

Not sure. I leave it to others to tag. Galois theory, polynomials or something like that.

This guy has one hell of an obsession with the quintic, lately :)

@tb What is $\text{id}_{\mathbb{R}}$?

The identity map $\mathbb{R} \to \mathbb{R}, x \mapsto x$.

@tb Ok. That's what I thought.

Here is a bit of Physics meets Math fun: E = mc^2 and a^2 + b^2 = c^2, therefore E = m( a^2 + b^2).

: )

What is a symmetric balanced incomplete block design and why should I bother to find out?

I can't think of a good answer to this.^ : )

Maybe it is not a good question.

Well, if somebody throws a bunch of symbols at me and thinks everybody should know what they're supposed to mean then I want to know why I would want to know. I didn't say it was a good question :)

In fact "I know what it means to say that an equation can't be satisfied, but I don't know what you mean when you say that expression can't be satisfied." – Gerry Myerson. Illustrates this perfectly. Many fools can ask questions that wise men can not answer ;-)

@DavidK so, have you figured it out?

@tb I'm working on an inductive argument now.

Well, I've given you the inductive step :)

@tb Yes you have. Thanks! As I suspected, my original idea didn't make much sense.

What are closed form sums?

$1+2+\cdots+n = \frac{1}{2}n(n+1)$.

for example

Ahh. yes.

What teddy said. My typing isn't fast enough.

@MattN. What about transfinite induction?

Hullo Brian

@tb That's why I deleted my comment. It was not even true.

I see :)

Hi Brian.

@teddy I'm still puzzled about yesterday. Can you tell me why you were surprised by the non-pun-carrot if you understood it in a completely innocent way?

No.

: D

It's too embarrassing :)

It's starting to be more and more puzzling. I have to give up, I suppose.

Hullo, folks. What is a non-pun-carrot?!

: D

Impossible to find out unless you have moderator powers and bother to dig through a whole spate of $\color{silver}{\rm (removed)}$ posts here in chat :)

I see (said the blind man).

It's basically a misunderstanding.

<whistles innocently> Work? What’s that?

: D

I want to be retired, too!

What is this? Yes, by an infinite measure I mean a measure that assigns the whole of $\mathbb{R}^n$ infinite measure. Try saying that three times fast.

@BrianMScott Hi Brian, would you like to discuss the criterion that a good question should satisfy?

@Skullpatrol I honestly don’t see much point. In the MSE context I’m more interested in what makes a question bad, and that can be any number of things.

I heard a talk once that said: a good question should be: 1. relevant 2. revealing and 3. accessible. Do you think a lack of any of those things makes a question bad?

Isn't this discussion bound to be somewhat too generic to be of any value?

We may find an application for MSE

@tb Yep: context matters enormously.

Hm. I don't like this kind of answers. Spam?

Hi Kannappan.

@MattN Hi. : )

@tb math.umt.edu/tmme/vol5no2and3/…

@tb This one seems on topic though.

Does all advertising count as spam?

@KannappanSampath doesn't seem to be the same paper. The authors don't match...

@tb Yes and I did not read the question. So, I don't know the relevance of the answer to the question and also journals don't match too. : )

@MattN I guess it depends on what they are advertising.

Advertising porn is definitely spam, in my opinion.

@tb Probably not spam, given the title, but probably not terribly useful $-$ especially since the journal in question is more than a little obscure.

The paper that Kannappan found is definitely relevant: the conclusion is that of the $19683$ distinct binary operations on a three-element set, only $113$ are associative.

I'm just wondering how anything addressing the case $n=3$ can be useful in the context of a question asking for asymptotics of $n \to \infty$.

That compares with $729$ that are commutative, $243$ with an identity, and $243$ with a zero.

@MattN Do you think we should count advertising Open Courseware as spam?

@Skullpatrol No.

@tb Not useful, but not completely off-topic. And it is kind of interesting that the number is already such a small fraction for $n=3$.

Agree completely

"Usefulness" would fall under the category of revealing an application, while on or off topic would be under relevance. Provided they are both accessible ;-)

Can someone help me reason out a step?

I have $Y \lt t$.

shoot

Now, I want to show that, in a sequence monotonically increasing and converging to $t$, there is a $t_n$ such that $Y \lt t_n$.

Proof by contradiction?

Do it directly :)

Fine, back from my class.

By definition: for every $\varepsilon \gt 0$ there is $N$ such that ...

$|t_n -t| \lt \varepsilon$ for all $n \ge N$.

Now choose $\varepsilon$ appropriately (monotonicity of the sequence doesn't really matter in the argument)

@JonasTeuwen hi, Jonas

Choose $\varepsilon=t-Y$?

by definition of a limit all $t_n$ are strictly within $t-Y$ of $t$ at some point

@KannappanSampath exactly

Now spell the rest out :)

Aha, yes. I think I have it now.

So, $t-t_n$ is positive as the sequence increases to $t$.

you don't need that fact

there is an $N$ such that $t-t_n \lt t-Y$ for all $n \ge N$

:3741124 Sorry, I fixed that.

If x is strictly within t-Y of t, then what is the infimum value x can take on the real line?

So, I have found infinitely many $t$ such that $Y \lt t_n$.

Is this OK?

@KannappanSampath infinitely many $n$. Yes.

@tb I should sleep sufficiently longer. I am making a dozen of mistakes. : /

But notice that, as anon and I pointed out several times already, you don't need monotonicity of the sequence $(t_n)$. If $|t-t_n| \lt t-Y$ then $t_n \gt Y$...

I am not using monotonocity, I am using the fact that sequence increases to $t$, as in all terms are less than $t$. Is that what the objection is to?

@tb This pointing out could get monotonous ...

that's also not needed, Kanna

@KannappanSampath monotone = increasing for me (here). I'm sloppy

if x>t then x>Y because t>Y

You just need that $t_n$ converges to $t$.

@BrianMScott I just did a quick search for the number of times the words: "relevant," "revealing," and "accessible" have been mentioned in the chat room; and I came up with 461 times in total, with "relevant" leading the way with 140 times. I would conclude that these words are generally important to question asking and answering at MSE.

Let me think a bit. I am messing things up. Hold on a bit...

You're not messing things up, you're using a hypothesis that isn't necessary

What you say is perfectly fine.

^^nice

@anon Well, how do you know $Y$ lies below $t$?

@Skullpatrol That inference seems at least a bit shaky: I can easily imagine any of those words being used here in contexts that have nothing to do with questions, or even with answers.

@BrianMScott True