12:00 AM
@N3buchadnezzar Since both integrals converge absolutely
@N3buchadnezzar but the last integral is wonky

@robjohn wonky?

@N3buchadnezzar Oh, wait you fixed it.

Sorry to interject, but if m is odd, is there an easy way to determine the smallest n for which 2^n = 1 mod m? Is this computationally as difficult as a general discrete logarithm?

@robjohn OK, Master. Let me tell then. Apostol wants to prove that if $\lim f=A$ and $\lim g=B$ then $\lim (fg)=AB$. He argues that, since $$f(x)g(x)-AB=f(x)(g(x)-B)+B(f(x)-A)$$ it suffices to prove the particular case when either $A=0$ or $B=0$, since we have alrady proven that $\lim(f+g)=\lim f+\lim g$
Gotta go to eat. BRB!!!

12:02 AM
i.e., the order of 2 in Z/m for m odd...

@Bean Either convert to binary or divide repeatedly by 2, otherwise, I don't think so.

Okay, so there's no obvious closed form for that.

@PeterTamaroff this is just simple pointwise limits; we are not looking at any special type of convergence?
@Bean Oh, I was looking at that wrong. You know by Fermat's Little Theorem that $2^{\phi(m)}=1\pmod{m}$
so $n\vert\phi(m)$

@robjohn pointwise

m is not prime

12:09 AM
@Bean if it were then I would have used $m-1$ instead of $\phi(m)$
@PeterTamaroff have you shown this for a constant times a function?

Good point. But this is not necessarily the smallest such m, is it? We only know that the order would divide \phi(m)

@Bean no, all I said was that $n\vert\phi(m)$

Right okay. The LaTeX isn't actually rendering on my machine, so it's a bit difficult to read

@Bean have you installed the bookmark?

No. I don't often use the chat...

12:14 AM
@PeterTamaroff Oh, wait; I see what he is doing...

But its easy to install, Mr. Bean.

@Bean it will make your time here much more valuable when you do come here.
@Bean you can get it here

@robjohn I've done a small game design course too.

$$\lim(f(x)g(x)-AB) = \lim f(x)\color{#C00000}{(g(x)-B)} +\lim B\color{#C00000}{(f(x)-A)}$$

@robjohn YEAH

12:20 AM
each of the things in red have a limit of $0$

@robjohn Right.

@robjohn Some threshold I haven't got to yet. I'm missing something. He argues proving the theorem when $B=0$ or $A=0$ suffices

@PeterTamaroff You want to show that $\lim f(x)g(x)=\lim f(x) \lim g(x)$ right?

@robjohn Yes. Wait.
Say $B=0$. THen
the formula above becomes
$$\lim \left( {f\left( x \right)g\left( x \right)} \right) = \lim f(x)\left( {g\left( x \right)} \right)$$???

12:25 AM
no...

@robjohn no?

What he means is that you can prove that $\lim f(x)g(x)=\lim f(x) \lim g(x)$ by assuming that $\lim g(x)=0$
That is what he means by $B=0$

@robjohn OK, and how do we use this particular case for the general one?

and in each summand, the term in red has a limit of $0$
8 mins ago, by robjohn
$$\lim(f(x)g(x)-AB) = \lim f(x)\color{#C00000}{(g(x)-B)} +\lim B\color{#C00000}{(f(x)-A)}$$
So the special case of one of them being $0$ gives the general case
It is a confusing way of saying it, but it makes sense after a bit

@robjohn Because one works on $\lim f(g-B)$ and $\lim B (f-A)$ separately?

12:30 AM
@PeterTamaroff yes

@robjohn Hmm. It is a little convoluted.

I would have written it as $\lim f(x)g(x)=\lim f(x)(g(x)-B)+\lim B(f(x)-A)+\lim AB$

@robjohn I got the point, yes.

But I think that proving it without that assumption is just as easy.

@robjohn Yes. Spivak does it in general, and it is just a matter of picking good constants to append to $\epsilon$

12:34 AM
@PeterTamaroff authors get weird ideas of what will be simpler for those they are teaching.

@robjohn Hehehe yes

@PeterTamaroff That looks interesting for someone who already understands the material, but I would not present it that way.

@robjohn It is a nice way of proving it. Very little effort indeed.

@PeterTamaroff Big Rudin is like that. The proofs are slick and simple to execute, but sometimes you really have to think to understand what is going on.
@PeterTamaroff You can easily see why the steps follow from one to the next, but the big picture requires a lot of thought
@PeterTamaroff Once you understand it, it is beautiful. Trying to understand it is not so easy.

@robjohn Well, one shouldn't try... one should go for it, if you know what I mean.

12:41 AM
@PeterTamaroff There is no "try", do or do not! -Yoda

hello

user19161
@robjohn Yes Yoda!

@leo Howdy!

@WillHunting You mean, "Yoda, yes."

12:42 AM
@PeterTamaroff He was agreeing with my attribution

@robjohn Thanks, Captain!

user19161
@PeterTamaroff No, I was calling him Yoda and saying yes to him and also choosing to omit the comma there.

@WillHunting You're a piñata, aren't you?

user19161
@PeterTamaroff What's that?

user19161
Is it like Pinocchio with the long nose?

12:45 AM
@leo I don't remember. Why?

user19161
Pinata, Pinocchio, Pedro...

A piñata () is a container made often of papier-mâché, pottery, or cloth and decorated, filled with toys and/or candy, and then broken as part of a ceremony or celebration. Piñatas are most commonly associated with Mexico, but its origins are considered to be in China. The idea of breaking a container filled with treats came to Europe in the 14th century, where the name, from the Italian “pignatta” was introduced. The Spanish brought the European tradition to Mexico, although there were similar traditions in Mesoamerica. The Aztecs had a similar tradition to honor the birthday of the god ...

@WillHunting Nay. It is a cartboard empty figure, adorned with birght coloured paper, filled with treats and surprises. You have to either pop it or tare it apart to get them.
I say pop it because here in Argentina we sometimes use overly sized baloons.

user19161
@PeterTamaroff I think you mean tear and not tare.

@WillHunting I probably do, don't I?

user19161
12:47 AM
Tare is grass and also the weight you have to subtract from the total to get the weight of the goods themselves without the container.

@PeterTamaroff I have only seen paper piñatas, never a balloon one.

@robjohn I'm pretty sure there is a duplicate in somewhere

@robjohn Because you haven't been to Argentina.

user19161
@robjohn I happen to know that word because it seems not listed in many learner's dictionaries so I had to look it up elsewhere.

user19161
@peter I saw your tennis pics online, hehe.

12:48 AM
@leo Oh, you mean the main site when you say "here"

@WillHunting Damn you! Where?

@robjohn oh,... sorry, yes :-)

user19161
@PeterTamaroff Online, the tennis blog.

@WillHunting Huh?

user19161
@PeterTamaroff Well, you know how to do a google image search don't you, just use your Spanish name...

12:51 AM
@leo Try searching. I don't know offhand since I don't usually answer measure theory questions. Someone else might be able to help. Find someone who has answered a lot of measure theory questions

@WillHunting There is only one picture of me, silly, and it is not tennis.

user19161
@PeterTamaroff OK, but that was a tennis blog. Wait I think I got the wrong guy...

@leo OMG... I am on that list! Ha!

@robjohn I was about to point that
@robjohn me too?

user19161
OK @peter I got the wrong guy, but I saw the real one on another site, hehe. I think you look like the wrong guy.

12:55 AM
@leo an all time asker and last month answerer (higher than me)

@robjohn I see. I think that if a user has at least one answer or question in the tag it appears in that list. I appear at the bottom of top answerers of my list. But you indeed appear in my list

@leo No, it is an ordered list and in the last 30 days, 1 answer was enough to get me on the list.
@leo and my original answer was much more measure theoretic than the final answer using Fatou

0

I'm not a professional mathematician nor a skilled amateur mathematician, but I have interest in knowing what is the purpose of some mathematical subjects in a simple and clear way, accessible to a layman. (eg: Clifford A. Pickover's books). Can I ask such a thing on MSE?

1:18 AM
@robjohn found!

@leo Oops, you need 3000 rep to vote to close :-(

yep

@leo ask some others to vote :-) (I'd rather not clobber with a mod vote and let non-mods do it)

@robjohn however I think that the title of the older quesion is pretty bad

@leo when you get the rep, change the title to something better. :-)
@leo suggest a better title and I will change it

1:24 AM
@robjohn by the way, can we close a old question as a duplicate of a new question (if the new one deserves his right to exist)?

@leo you just try to edit the title and it will be forwarded to a mod and you will get 2 points for that

I alreay can change the title, I'm about to do so .-)

@leo ah, okay :-)

@robjohn but, is that possible?
I'm curious

@leo is what possible?

1:27 AM
2 mins ago, by leo
@robjohn by the way, can we close a old question as a duplicate of a new question (if the new one deserves his right to exist)?

@leo I have never tried it, but I don't see why it wouldn't work.
@leo If $E$ has measure zero, then does $E^2$ have measure zero?
If you change it soon, it will be one edit.

@robjohn "measure zero" or "zero measure", which is more appropiate?

@leo that part is fine, I was looking at the verb tense.
@leo "has" and "does" and comma and question mark

@robjohn done!

@leo Very nice! and in one edit. That helps since too many edits automatically change a question to CW

1:36 AM
@robjohn Thanks :-). I'll flag the other one
nice!

How do you guys like the idea that if an answer gets an up-vote, then the question should automatically get one too?

That's a silly idea.

the flag menu have now a "It doesn't belong here or is a duplicate" option that allow us to say which is the duplicate

@JohnJunior Boooooooo!

1:41 AM
@JohnJunior there pretty bad questions with pretty good answers. It's not fair

@anon All functions that are Lipschitz continuous are continuous, right?

don't know that off the top of my head

All I'm saying is that we should try to come up with an idea to counter balance the serial down voters.

Say $|f(x)-f(y)|\leq \lambda |x-y|$ for some $\lambda$, and any $x,y\in[a,b]$.

@PeterTamaroff what's "Lipschitz continuous "?
and: como andás?

1:43 AM
In mathematical analysis, Lipschitz continuity, named after Rudolf Lipschitz, is a strong form of uniform continuity for functions. Intuitively, a Lipschitz continuous function is limited in how fast it can change: for every pair of points on the graph of this function, the absolute value of the slope of the line connecting them is no greater than a definite real number; this bound is called the function's "Lipschitz constant" (or "modulus of uniform continuity"). In the theory of differential equations, Lipschitz continuity is the central condition of the Picard–Lindelöf theorem which gua...

according to Wikipedia Lip cont functions are not only continuous but differentiable almost everywhere
which I find surprising

@PeterTamaroff then yes

@anon Sure? It says it is a strong form of UNIFORM CONTINUITY.
I was about to comment "I think that if a functions is Lipschitz (like that above) then it is uniformly continuous, right?"
That above is
2 mins ago, by Peter Tamaroff
Say $|f(x)-f(y)|\leq \lambda |x-y|$ for some $\lambda$, and any $x,y\in[a,b]$.

@anon that condition implies absolutely continuity

absoposilutelytively.
2

1:45 AM
@anon LAWL

:-O

@PeterTamaroff tapado de cosas como dicen ustedes :-)
@PeterTamaroff then the function is indeed uniformly continuous in that interval

@PeterTamaroff con mucho que hacer, digamos
pero bien :-)

1:49 AM
@leo good =)
@leo Question on analysis.
What is more practical to use in convexity $$f(x\lambda+(1-\lambda)y)\geq \lambda f(x)+(1-\lambda)f(y)$$ for any $x,y\in [a,b]$ and $0<\lambda <1$ or$$\frac{f(x)-f(a)}{x-a}\geq \frac{f(b)-f(a)}{b-a}$$ for any $x\in[a,b]$?

@PeterTamaroff I like the first one, but if you ask me why, I don't know. Perhaps because I have used the former most times

@leo The first one seems more free.

@PeterTamaroff why you say so? You have some freedom in the $x$ in the second one

@leo Yes, sure, but the other one has $x,y,\lambda$

@PeterTamaroff oh yes
you right

1:56 AM
gotto go now
bye byes

@PeterTamaroff have fun! :-)

@leo I'll surely will! Modern Combat 3: Fallen Nation it will be!

2:07 AM
@PeterTamaroff you would like mwf3

2:19 AM
g nite all

2 hours later…
3:57 AM
@MeAndMath Cheguei!

@GustavoBandeira dude, shit happened here last night. As usual, I missed it.

brb

@GustavoBandeira Everyone was in rare form here yesterday. Rob, you, Trollaroff. :P

4:22 AM
Sorry, had to take a shower
I'm back

@GustavoBandeira Okay. I wll be going for breakfast in some time. Till then, have to complete some beautiful soup hacks.
See you later.

4:37 AM
hi @DavidZaslavsky
bye David

@JohnJunior Where?

@GustavoBandeira He flew in for about a second...

You frightened him.

I think so :(
We had a chat in the physics room, I think he's one of the owners there.

Yep.
He's the mod there, isn't him?
Yep.

5:00 AM
Would I be right to assume the free Lie algebra $\cal L$ over $K$ generated on a set $X$ is constructed via the formal $K$-linear combinations of elements of $X$, the commutators of those, the commutators of those (elements as well as commutators), and so forth in the limit?

5:59 AM
huh? what is going on here with the votes?
0

Recently I'm learning logic. Here is the definition from the book "Logic For Computer Science": A countable set PS of proposition symbols: P0,P1,P2... The set PROP of propositions is the smallest set satisfy: Every proposition symbol Pi is in PROP. Whenever A is in PROP, ~A is also in PROP...

6:43 AM
@anon @robjohn can we pull another close and reopen action? Here.
Morning Will.
You need another 900 rep.

user19161
@Matt Hi! Yeah, never mind about the rep. Your food pic yesterday looks yummy.

user19161
My friend just treated me to a sashimi lunch.

@WillHunting Yes, it was. Jamie Oliver's recipes are great : ) (If you had 3k then you could vote)
Nice : )

hi @BrianM.Scott

Hullo.

6:46 AM
Hi Brian! Long time no see!

Looooooong time...

@Matt What happened to the bright red?

@BrianM.Scott I decided that I wanted a change.

@Matt You seem to have made us iconic cousins, after a fashion.

: )
I wonder when it'll change, it's one of the random ones.

6:50 AM
Does that mean that it can change spontaneously?

Yes.

That’s carrying anonymity to extremes! :-)

: )

@Matt How long will you be gone?

In a Buddhistic way.

6:52 AM
Hmph.
Any idea what happened to Arturo?

I have a question I'm quite sure you may be able to answer:
@BrianM.Scott No, but I suspect he wanted to use some time to do research. Maybe also got a girlfriend (or boyfriend), or who knows what else he could do IRL : )
I get products on both sides there.

@Matt I always did wonder how he managed to find the time to be so productive here, when he's an active member of the academic mathematical community.

@BrianM.Scott I thought since he joined SE he wasn't so active anymore IRL.
(I looked at his list of publications...)

@Matt I don’t know: he was pretty active in a couple of math help groups on Usenet before SE existed.

Oh.
I should be working.

6:57 AM
That’s where I first ran into him, in fact.
@Matt On something in particular?

@BrianM.Scott Do these math help groups on Usenet still exist?

@JohnJunior Yes, but they’re pretty dead; SE is a better bet.

@BrianM.Scott Yes. I'm trying to write a BSc's thesis.

Thanks :)

7:01 AM
@Matt How’s it going?

@BrianM.Scott Not good. I have to do it in 3 weeks total time and I have 1 week left. And only written slightly less than half of it.
Gonna be a close shave.
Ok, see you all later!

@Matt The first inequality isn’t bad: $$c_1\dots c_n=\prod_{k=1}^n\frac{(k+1)^k}{k^k}=\frac{(n+1)^n}{n!}\cdot\frac{\prod_{k=1}^{n-1}(k+1)^k}{\prod_{k=1}^nk^{k-1}}\;.$$ That last factor is $$\frac{\prod_{k=1}^{n-1}(k+1)^k}{\prod_{k=2}^nk^{k-1}}=\prod_{k=1}^{n-1}\left(\frac{k+1}k\right)^k>1\;.$$

Yes, they look pretty dead :(
hi @JonasTeuwen
look who's here :-D

7:20 AM
@JohnJunior And one of them is infested by a jackass who thinks that he’s proved the Goldbach conjecture, even though he needs to ask for basic help in the use of mathematical notation.

@BrianM.Scott Personally I think the reputation system of StackExchange is a really good thing. It is one of the reasons why there is so less spam on this site.

@JayeshBadwaik That, and the fact that when spam does appear, it’s usually flagged as such almost immediately.

7:50 AM
Hi @Mariano.
Hmm, Brian is around. Long time...
And, hello folks.

@KannappanSampath Yes, it’s been a while. I’ve been either answering questions or out riding my bicycle.

@BrianM.Scott It was summer, yeah.

@KannappanSampath I hope that it still is for a little bit!

@KannappanSampath Did you get that problem of partitions of a matrix?

7:53 AM
@JayeshBadwaik Well, I am not sure, I have clean and nice counting without enumerating. I'll think about it later after my exams.

@KannappanSampath Oh yeah, exams.
Just one more question. Is your code similar to what I told or is it some new/better method?

@JayeshBadwaik I would not say it's better. But, it is new.

@KannappanSampath Ohh. Any chance you can describe it in a short here?

@JohnJunior I’d take a look, but on dial-up it just takes too long.

7:56 AM
From the reading of the transcript, I gather @Jonas is ill. Jonas: How is your health now?

@BrianM.Scott Cycling? Cross-Country?
@BrianM.Scott Cheers!! One more dial-up user who understands the pain of it!

@JayeshBadwaik Road. This summer I’ve been lazy and have only about 2100 miles so far; I’ve been averaging 4200 miles a summer for the last ten years.

@JayeshBadwaik Hmm, may be later. I am just too tired.

Good morning

hi @JohnSenior

7:59 AM
Good Morning @JohnSenior

@KannappanSampath No problem! I am interested though, so may be sometime later.
@BrianM.Scott Wow, that is one hell of a long trip every summer. Awesome. I think I should do that myself too. Seems a good target to set for yourself.

@JayeshBadwaik Well, it’s not one trip; it’s an average of between $40$ and $50$ miles per ride.

Is any of that up hill?

@BrianM.Scott Hmm. Still, impressive it is.

8:01 AM
My other annual cycling goal is to complete at least one ride whose length in miles is at least my age in years. It gets a little harder every year, but I’m still managing: I got in a $67$-mile ride this summer, which covered it with three miles to spare.
3

@BrianM.Scott well done!

@BrianM.Scott Impressive.

@BrianM.Scott Impressive!

@JohnJunior The terrain here is rolling, so there’s definitely some up and down. The worst hills are no more than a mile long, and the very worst are a lot shorter than that.

We now have a lot of New Chat Regulars. I only hope we have not lost the old ones.

8:05 AM
I think we are going to lose Matt soon :(

New blood is good, but it’s always disconcerting to lose familiar ‘voices’.

@JohnJunior Why?

@KannappanSampath Something Matt said here earlier.

@BrianM.Scott it seems interesting that the total number of recent visitors to the room seems to stay fairly constant

@KannappanSampath Because t.b. is gone.

8:07 AM
@BrianM.Scott I am not aware, perhaps a link, if none wants to elaborate?

@JohnSenior I’ve not paid much attention recently, but that fits with my recollection.

@JohnJunior Why? Why? Why?
2

@KannappanSampath It was just a brief comment, which Matt erased very quickly.

I'll miss t.b. :(
@BrianM.Scott :(((

@KannappanSampath "Why" is a question that can be asked forever...

8:08 AM
@KannappanSampath He’s still quite active on the main site.

$\text{\color{grey}{removed}}$
$\text{\color{grey}removed}$

@KannappanSampath I was going to say the same thing which @BrianM.Scott said.

I am feeling newb now.
I seem to have forgotten the syntax.
@JayeshBadwaik I was trying an old trick. Not much luck yet. :((((((((

$\color{grey}{\text{removed}}$

$\color{grey}{\text{(removed)}}$
\color{grey}{\text{(removed)}}

8:12 AM
Thank you @Brian.

Any ideas what this might be about?

It does not show greyed to me. :-(

@JohnSenior No, and I’m inclined to agree with Fortuon.

@JohnSenior That means I have to flag the post, get it sent to heaven.

$\color{grey}{\text{(removed)}}$

8:14 AM
@JohnSenior The guy is treating this as some kind of puzzle thing? My teacher used to do that. He would have cash prizes for problems he would post for us to solve! In high school, any money was good money. :P
But I agree, not relevant here.

We do seem to get a fair number of odd questions (and questioners?) on MSE

@JayeshBadwaik I see it that way.

$\Huge\color{grey}{\text{(removed)}}$
Matt is right this is the black hole of productivity.

8:29 AM
@KannappanSampath you have wentaway as your nick in main?

$\color{red}{\text{(not removed)}}$

No moar teddybear love in chat ? =(
5

8:45 AM
$\color{red}{\text{(not}}\color{blue}{\text{ removed)}}$
$\color{red}{\Huge\text{(not}}\color{blue}{\text{ removed)}}$

Are you having fun? :-)

;-)
Learning is fun.

@JohnJunior Often, anyway.

@BrianM.Scott Agreed.

Urk. It’s coming up on 5:00; time to be thinking about bed.

8:53 AM
Learning should be fun.
@BrianM.Scott Thanks for stopping by.

@JohnJunior My pleasure. I’ll be dropping in occasionally, I expect.

@JohnJunior Thanks! I will.

1 hour later…
10:01 AM
@N3buchadnezzar he is active on the main site. Perhaps he is too busy to chat these days. It does sink a lot of time :-)

@robjohn I know he is busy chasing golden haired girls who sleeps in his bed, but even so he needs to take time to relax.
Why not spend that time here? ;)

10:23 AM
@N3buchadnezzar and who wouldn't find hanging out on a math chat alluring?
2

10:51 AM
Whee! With my latest downvote, my rep is back to $0\bmod{5}$ -_-
:6063813 open windows can disturb neighbors :-)
2

@robjohn And an open tab can run up a large bill ;-)

@JohnJunior Indeed!

For the last 10 years!

@JohnJunior 67 miles?

@robjohn Yup :0
>8(

11:05 AM
@robjohn hey
what happened to the mobius mean squares?

@Eugene I didn't feel mean enough. >8(

@robjohn ah i see
@robjohn did you see clint eastwood's dialogue with a chair?

@Eugene nope

@robjohn awww

11:11 AM
@robjohn i'm not sure. maybe. it's not actually funny but seeing him discussing with a chair is kind of hilarious

@JohnJunior: I made that picture from combining some stuff from here: tango.freedesktop.org/Tango_Icon_Library

@wj32 Very nice :)

user19161
11:52 AM
It was raining cats and dogs.

user19161
@JohnJunior I am still here bro.

@WillHunting What's up?