Mathematics

robjohn

12:00 AM

@N3buchadnezzar Since both integrals converge absolutely

@N3buchadnezzar but the last integral is wonky

N3buchadnezzar

@robjohn wonky?

robjohn

@N3buchadnezzar Oh, wait you fixed it.

N3buchadnezzar

=D

Bean

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?

Peter Tamaroff

@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!!!

Bean

12:02 AM

i.e., the order of 2 in Z/m for m odd...

robjohn

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

Bean

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

robjohn

@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 $n\vert\phi(m)$

Peter Tamaroff

@robjohn pointwise

Bean

m is not prime

robjohn

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?

Bean

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

robjohn

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

Bean

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

robjohn

@Bean have you installed the bookmark?

Bean

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

robjohn

12:14 AM

@PeterTamaroff Oh, wait; I see what he is doing...

Gustavo Bandeira

But its easy to install, Mr. Bean.

robjohn

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

@Bean you can get it here

Gustavo Bandeira

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

robjohn

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

Peter Tamaroff

@robjohn YEAH

robjohn

12:20 AM

each of the things in red have a limit of $0$

Peter Tamaroff

@robjohn Right.

robjohn

@PeterTamaroff Does that answer what you were asking?

Peter Tamaroff

@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

robjohn

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

Peter Tamaroff

@robjohn Yes. Wait.

Say $B=0$. THen

the formula above becomes

$$\lim \left( {f\left( x \right)g\left( x \right)} \right) = \lim f(x)\left( {g\left( x \right)} \right)$$???

robjohn

12:25 AM

no...

Peter Tamaroff

@robjohn no?

robjohn

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 $B=0$

Peter Tamaroff

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

robjohn

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

Peter Tamaroff

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

robjohn

12:30 AM

@PeterTamaroff yes

Peter Tamaroff

@robjohn Hmm. It is a little convoluted.

robjohn

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

Peter Tamaroff

@robjohn I got the point, yes.

robjohn

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

Peter Tamaroff

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

robjohn

12:34 AM

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

Peter Tamaroff

@robjohn Hehehe yes

robjohn

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

Peter Tamaroff

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

robjohn

@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.

Peter Tamaroff

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

robjohn

12:41 AM

@PeterTamaroff There is no "try", do or do not! -Yoda

leo

hello

user19161

@robjohn Yes Yoda!

robjohn

@leo Howdy!

Peter Tamaroff

@WillHunting You mean, "Yoda, yes."

leo

I think this was already asked here

robjohn

12:42 AM

@PeterTamaroff He was agreeing with my attribution

Peter Tamaroff

@robjohn Thanks, Captain!

user19161

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

Peter Tamaroff

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

user19161

@PeterTamaroff What's that?

user19161

Is it like Pinocchio with the long nose?

robjohn

12:45 AM

@leo I don't remember. Why?

user19161

Pinata, Pinocchio, Pedro...

robjohn

Piñata

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 ...

Peter Tamaroff

@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.

Peter Tamaroff

@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.

robjohn

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

leo

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

Peter Tamaroff

@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.

robjohn

12:48 AM

@leo Oh, you mean the main site when you say "here"

Peter Tamaroff

@WillHunting Damn you! Where?

leo

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

user19161

@PeterTamaroff Online, the tennis blog.

Peter Tamaroff

@WillHunting Huh?

user19161

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

robjohn

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

Peter Tamaroff

@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...

robjohn

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

leo

@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.

robjohn

12:55 AM

@leo an all time asker and last month answerer (higher than me)

leo

@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

robjohn

@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

leo

I was talking about the "all time" answerers list

Gustavo Bandeira

0

Q: Is this an adequate place to ask for simplifications on mathematical concepts?

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?

leo

1:18 AM

@robjohn found!

robjohn

@leo Great! post a comment and vote to close as duplicate :-)

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

leo

yep

robjohn

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

leo

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

robjohn

@leo when you get the rep, change the title to something better. :-)

@leo suggest a better title and I will change it

leo

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)?

robjohn

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

leo

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

robjohn

@leo ah, okay :-)

leo

@robjohn but, is that possible?

I'm curious

robjohn

@leo is what possible?

leo

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)?

robjohn

@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.

leo

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

robjohn

@leo that part is fine, I was looking at the verb tense.

@leo "has" and "does" and comma and question mark

leo

@robjohn done!

robjohn

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

leo

1:36 AM

@robjohn Thanks :-). I'll flag the other one

nice!

John Junior

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

anon

That's a silly idea.

John Junior

:(

leo

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

Peter Tamaroff

@JohnJunior Boooooooo!

leo

1:41 AM

@JohnJunior there pretty bad questions with pretty good answers. It's not fair

Peter Tamaroff

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

anon

don't know that off the top of my head

John Junior

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

Peter Tamaroff

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

leo

@PeterTamaroff what's "Lipschitz continuous "?

and: como andás?

Peter Tamaroff

1:43 AM

Lipschitz continuity

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...

anon

according to Wikipedia Lip cont functions are not only continuous but differentiable almost everywhere

which I find surprising

leo

@PeterTamaroff then yes

Peter Tamaroff

@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 $x,y\in[a,b]$.

leo

@anon that condition implies absolutely continuity

anon

absoposilutelytively.

2

Peter Tamaroff

1:45 AM

@anon LAWL

leo

:-O

Peter Tamaroff

@leo nada terrible, tú?

leo

@PeterTamaroff tapado de cosas como dicen ustedes :-)

@PeterTamaroff then the function is indeed uniformly continuous in that interval

Peter Tamaroff

@leo Hehhee, tapado bien o tapado mal?

leo

@PeterTamaroff con mucho que hacer, digamos

pero bien :-)

Peter Tamaroff

1:49 AM

@leo good =)

@leo Question on analysis.

I was asking this today.

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 $x\in[a,b]$?

leo

@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

Peter Tamaroff

@leo The first one seems more free.

leo

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

Peter Tamaroff

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

leo

@PeterTamaroff oh yes

you right

Peter Tamaroff

1:56 AM

gotto go now

bye byes

leo

@PeterTamaroff have fun! :-)

Peter Tamaroff

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

leo

someone with powers please help to close this

leo

2:07 AM

@PeterTamaroff you would like mwf3

leo

2:19 AM

g nite all

Gustavo Bandeira

3:57 AM

@MeAndMath Cheguei!

Jayesh Badwaik

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

Gustavo Bandeira

@JayeshBadwaik What? 0.o

brb

Jayesh Badwaik

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

Gustavo Bandeira

4:22 AM

@JayeshBadwaik lol

Sorry, had to take a shower

I'm back

Jayesh Badwaik

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

See you later.

Gustavo Bandeira

cya

John Junior

4:37 AM

hi @DavidZaslavsky

bye David

Gustavo Bandeira

@JohnJunior Where?

John Junior

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

Gustavo Bandeira

You frightened him.

John Junior

I think so :(

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

Gustavo Bandeira

Yep.

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

Yep.

anon

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?

N3buchadnezzar

5:59 AM

This integral seems tough math.stackexchange.com/questions/192106/…

huh? what is going on here with the votes?

0

Q: Is the propositional set infinitely countable

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...