7:17 AM
@anon Probably you want to argue like this: give me a set $\{a_i\}$ of left coset representatives for $A \cap C$ in $A \cap B$. You claim that these still give distinct cosets of $C$ in $B$. Why? What does it mean for two of those guys $a_i, a_j$ to give the same coset of $C$? Well, you'd need $a_j^{-1}a_i \in C$. And $A$ is a subgroup so this product is also in $A$.
And I think that does it.

How would $a_j^{-1}a_i$ also be in $A$?
oh, $a_i\in A\cap B\implies a_i\in A$.
yeah that does it
thanks

Cool.
Commensurability comes up all over the place for arithmetic lattices. I can't remember why which is kind of sad.
I mean, it's part of certain ways of defining "arithmetic" but there has to be something beyond that.

7:38 AM
hey

yo

yo
wassup @anon

nuhmuch

8:02 AM
What's being proved?

where?

What kind of software is needed for making blog entries on wordpress

8:18 AM
an internet browser is all you need

9:05 AM
wordpress is not available.

@anon You there?

@Manishearth yes

Is there a dupe of crypto.stackexchange.com/questions/3079/… on math.SE?

@Manishearth It's possible, but I don't recall a dupe.

9:10 AM
K, np
Why is it so quiet in here?

It's 4am...
(in my timezone anyway)

Normally the math room has >>3 messages a minute
@anon Ah, right :)

lol, it's 6pm in Japan

@Frank why'd you need to delete that?

3 messages a minute, really...?

9:14 AM
@ZhenLin From what I've seen before. Whenever I've popped in, there's been a lot of activity.

@Manishearth The question seems silly.

Not if someone hasn't seen products before..

Is there any way to find the asymptotics for $\sum_{k>0}\exp(-k-k^2/2n)$
For here

looks theta-function-y

Well, $\Theta_n(t)=\sum_ke^{-(k+t)^2/n}$?
Thanks, @anon
Now the Fourier series works.
$\Theta_n(t)=\sqrt{\pi n}(1+2e^{-\pi^2 n}(\cos2\pi t)+2e^{-4\pi^2 n}(\cos4\pi t)+2e^{-9\pi^2 n}(\cos6\pi t)+\cdots)$

9:33 AM
@FrankScience WHy did you flag that?

@Manishearth The first one is wrong, so I want to delete it.

You can delete your own messages, you know
There's a little dropdown menu on the left side of the message, visible on mouseover

Timeout.

well then dealwithit.stackexchange.com

Flags should only be used for serious offenses

9:36 AM
While I was editing it. the message showed up that 10 sec left, so I immediately entered <CR> but I've found that it's wrong.
Oh, without flags, how can I delete my post more than 2 mins?

@FrankScience Oh right, that didn't occur to me :\

@N3buchadnezzar the gf or the grandmother?

@FrankScience I frankly forgot about the timeout. I think I could have undone that (not sure)

@Manishearth Eh, you're moderator?

9:38 AM
@anon The grandmother.

his name ain't blue for nothin :)

Those dem royals coming in here spawing around with their blue blood

@FrankScience Chemistry moderator
We get mod powers all over chat.SE

$CH_4$

$\ce{CH4}$
(only works if you use the chem chatjax :P)

9:41 AM
ooh, chem has their own chatjax?

@anon Yep. I basically ripped off yours, and added the mhchem extension
One sec, digging
Or, I can do this in the room and chem will work: $\require{mhchem}$

Eh? You can draw the image for $C_6H_6$ (cyclic $C\equiv C=C\equiv C=C\equiv C=$) in such TeX?

(re click chatjax now)
@FrankScience Not via mhchem
2

This is an offshoot of ChatJax, which enables MathJax along with mhchem on chat. Copy the text below: javascript:(function(){if(window.MathJax===undefined){var%20script=document.createElement("script");script.type="text/javascript";script.src="https://d3eoax9i5htok0.cloudfront.net/mathjax/lat...

nice

@FrankScience See the "basic chem" section of meta.chemistry.stackexchange.com/questions/86/… for what mhchem does

9:44 AM

@FrankScience Come over to chem.SE, it'll get improved :)
@FrankScience There are other LaTeX packages (which don't work in mathjax) that do drawing

@Manishearth How active is Chem compared to Tex, Math and physics.Exchange? =)

@FrankScience Here, a TeX package that does chem drawing ctan.org/pkg/chemfig
@N3buchadnezzar Nowhere close. Especially to math

@Manishearth Chemistry seems so complicated.

We're 60 days old, give it time :P

9:46 AM
@Manishearth Pity

Something in Chemistry is related to something in graph theory

Chemical graph theory is the topology branch of mathematical chemistry which applies graph theory to mathematical modelling of chemical phenomena. The pioneers of the chemical graph theory are Alexandru Balaban, Ivan Gutman, Haruo Hosoya, Milan RandiÄ‡ and Nenad TrinajstiÄ‡. In 1988, it was reported that several hundred researchers worked in this area producing about 500 articles annually. A number of monographs have been written in the area, including the two-volume comprehensive text by Trinajstic, Chemical Graph Theory, that summarized the field up to mid-1980s. The adherents of the the...

@Manishearth
I would tell you a joke about chemicals, but I doubt it would get a good reaction...
Good luck on growing the site =)

Use [...](URI) instead.

9:51 AM
@Frank go easy on the flagging.. Big pics are fine :)

See Chat Rules.

@FrankScience Anyway, I edited it

@Manishearth Here, the direct link should be avoided.

@FrankScience Did you hear about the seriousness of flags, if nobody is killed avoid flaging it.

@Manishearth See Chat rules

9:52 AM

the chat rules describe etiquette, they aren't something you flag over

@FrankScience I saw them, it still doesn't warrant a flag

Well, thanks.

Remember, if I act on a flag, the user gets a 30 second timeout (which can escalate)

How many offences does a user have to receive a lifeban ?

9:54 AM
aaand, I dunno

Lets find out what it takes! =D
Release the Kraken!

Actually, I dunno about the escalation

But I think 2-minute-timeout seems too short.

I think mod flags resulting in a deletion are always 30second or something

I think it is 0.30 - 2 - 5 - 30 - 60

9:56 AM
Spam/offensive flags are IIRC 30 minutes. Which may escalate

Oh, there are different kinds of flags.

Yep
The one on the right is spam/offensive
I think the mod attention one is in the left dropdown
but I'm not sure--as a mod, I don't get to use that flag
:P

9:59 AM

Well
Goodbye everybody, I've chatted too much.

@FrankScience Cya some other times =)

10:24 AM
@FrankScience : You were mentioning some theorem yesterday, I think it is by definition isn't it

@RajeshD Hey

Hi @N3bu

You know some basic probability right? I am helping some younger friends of mine.

i'll try
shoot at me

You have 3 blue, 2red and 1 green pens.

10:26 AM
ok

You draw two pens at random, what is the chance of not drawing a green pen?
Well ofcouse you can use the big cannon $5C2 \cdot 1C0 / 6C2 = 2/3$

wait a bit

Or you can think like $5/6 \cdot 4/5 = 2/3$

@RajeshD What?

But I wanted to try to solve this problem using $$\frac{\text{Wanted outcomes}}{\text{Possible outcomes}}$$

10:28 AM
it is the prob of not drawing green in first chance and multiplied by prob of not drawing green in second chance

Here I thought we had 4 wanted outcomes and thirty possible outcomes
But this is inconsistent with my answers above

(5/6)*(4/5)

eg $$\frac{4}{30} \neq \frac{2}{3}$$

@N3buchadnezzar drawing red ball #1 is not equivalent to drawing red ball #2; your count of how many outcomes are wanted undercounts.

@FrankScience : You mentioned a theorem yesterday you remember

10:29 AM
@N3buchadnezzar Use $\binom nm$ instead of $nCm$.

@FrankScience first you want me to use less space, now you want me to use more ;)

@RajeshD The dense of $\Bbb Q$ on $\Bbb R$?

no

@N3buchadnezzar $\binom nm$ is a more conventional notation.

@anon ah right, so we have 6 possible outcomes.

10:30 AM
lolno

BBBRRG

Partition the outcomes into four types: (R,R), (R,B), and (B,B). Now count how many outcomes are in each category.

Yeah, thanks for clearing my head =)

The number of no-G-2-tuple?

I suppose it's drawing two pens at once, rather than with replacement

10:32 AM

$(3\text{C}2) + (3\cdot2+2\cdot3) + (2\text{C}2)$

also, the total number of outomces = 6C2 = 15, not 30
or if you order the draws, it is 30
whatever you want

@FrankScience Here

@anon Its been to long since I have had about this, but it is starting to come back to me. Thanks anon =)

@RajeshD Can you prove it?

10:36 AM
yesterday what we did is we calculated the derivative from the first principles instead of $$\lim_{x\to x_0}f'(x)$$ @Frank

@RajeshD okay

Why to prove it

@RajeshD eh?

its not what we have used we used first principles, and BTW I used $\lim_{x\to x_0}f'(x)$ and got $\infty$ instead of $0$

First, can you prove that theorem?

10:39 AM
I am saying I disproved it yesterday

@RajeshD I think you're wrong.
$f(x)=\sin\alpha x/\sin\beta x$
$f(0)=\alpha/\beta$
$f'(x)=(\alpha\cos\alpha x\cdot\sin\beta x-\beta\sin\alpha x\cdot\cos\beta x)/\sin^2\beta x$ for $x>0$.
How did you do it next?
I'll have lunch.

@Frank : $f(x) = \frac{sin((n+\frac{1}{2})x)}{sin(\frac{x}{2})}$ when $x > 0$ and $n$ when $x=0$and $f'(x) = \frac{(2n+1)sin(\frac{x}{2})cos((n+\frac{1}{2})x) - sin((n+\frac{1}{2})x)cos(\frac{1}{2}x)}{4(sin(\frac{x}{2}))^2}$ Now If I use $\lim_{x\to x_0}f'(x)$ I get $\infty$ Thats why We derived it from first principles, Infact you only showed me how to do it using Taylor series
ok
For $f'(x)$, the numerator goes to $0$ as $O(x)$ and the denominator goes to $0$ as $O(x^2)$ and hence $\lim_{x\to x_0}f'(x) = \infty$, Thats why we should use first principles I guess
@FrankScience : Sorry I think I have found my mistake, $$\lim_{x\to x_0}f'(x) \ne \infty$$
But How to prove the theorem, Please give some hints

10:58 AM
@anon yo

hey
no, but I have it

I would like some advice on like the first chapter and second chapter
chapter 0 seems a little
not many details
and quite complicated for me
Do I just skip chapter 0 and go to the second?

Noooo I still think $\lim_{x\to x_0}f'(x) = \infty$, Darn I am confused, In any case please tell me how to prove that theorem @Frank

@ZhenLin do you know?

11:11 AM
@RajeshD Both $\alpha\cos\alpha x\cdot\sin\beta x-\beta\sin\alpha x\cdot\cos\beta x$ and $\sin^2\beta x$ approaches $0$ when $x\to0$.

yes

@RajeshD There's no evidence that $O(x)/O(x^2)\to\infty$ as $x\to0$ you know.

Oh ok

for instance, $x^2=O(x)$ and $x^2=O(x^2)$, but $x^2/x^2=1$ identically
(as $x\to0$)

@RajeshD Because when $f(x)=O(x)$, the following statement could also works: $f(x)=O(x^2)$.

11:14 AM
ok

$O$-notation doesn't indicate the exact growth level.

then the best method is to fully calculate using the Taylor series is it?
ok

fully? no. maybe two terms out.

very deceptive these things

You can deal with the rest problem with Taylor series, but it seems that it's not as convenient as to deal with the original problem.

11:16 AM
yes I mean
@FrankScience : Tell me how to prove the theorem. I am kinda scratching my head on that one

But it benefits when the Taylor series is hard to compute, especially when Taylor series doesn't converge to the original function.

ok

And you can use L'Hospital rule instead.

Boss I beg to tell me how to prove it

There's another caution. When $\lim_{x\to x_0^+}f'(x_0)$ exists, we can conclude that $f'(x_0)$ is that value (one-sided), but you cannot say the inversion.

11:21 AM
plug in $\cos u=1-u^2/2+O(u^4)$ and $\sin u=u-u^3/6+O(u^5)$ for $u=\alpha x,\beta x$...

He is eager to know how to prove $\lim_{x\to x_0}f'(x_0)=A$ implies $f'(x_0)=A$.

you mean $\lim f\,'(x)$

We only prove for the right-sided case.

when $f\,'$ is continuous it's true

But not only when, guy.

11:23 AM
@FrankScience yes I want to prove it.

@RajeshD Supposing that $x_0<x\le x_0+H$, we should compute $(f(x)-f(x_0))/(x-x_0)$.

yes
lets assume $x_o = 0$
for ease

Since $f(x)$ is continuous on $[x_0..x_0+H]$ and differentable on $(x_0..x_0+H]$, we can apply mean-value theorem.
It's $f'(y)$, where $x_0<y<x$.

yes

@anon are you here?

11:28 AM
Now let $x_1,x_2,\ldots$ is any sequence converges to $x_0$ from the right-side, and never equals to $x_0$.

ok

Then we can get the corresponding $y_1,y_2,\ldots$, where $x_0<y_k<x_k$ over all $k>0$.

yes

Thus $y_k\to x_0$ and $y_k$ never equals $x_0$.

@Ilya yes

11:30 AM
yes

if $R\subset A\times B$ is a relation, how should I call
$$\{a\in A:aRb\}$$
section? Don't ask me, why do I think you should know the answer.

Now consider the identity $(f(x_n)-f(x_0))/(x_n-x_0)=f'(y_n)$.

yes

Let $n\to\infty$

ok

11:32 AM
The right-side becomes $f'(x_0)$ because the limit $\lim_{x\to x_0}f'(x)$ exists.

render

@Ilya If it was an equivalence relation, you could say equivalence class, but I'm not sure otherwise.

So $(f(x_n)-f(x_0))/(x_n-x_0)$ converges to $\lim_{x\to x_0}f'(x)$ for arbitrary sequence $x_1,x_2,\ldots$.

@anon that I know, but it's not in general an equivalence relation in my case

in prev statement left or right?

11:33 AM
thanks

@Frank

@RajeshD Edited.
@RajeshD Oh, I was wrong but timeout, so I can't edit it.
@RajeshD The right-side becomes $\lim_{x\to x_0}f'(x)$, not $f'(x_0)$.

render
ok
then

The left-hand side is not $f'(x_0)$ while we are proving, because we have not proved the existence of $f'(x_0)$

ok

11:36 AM
Now we can conclude that $f'(x_0)$ exists and $f'(x_0)=\lim_{x\to x_0}f'(x)$, because of the limits in term of sequence, which I've mentioned yesterday.
Exercise
Try to prove the theorem by the language of $\epsilon$-$\delta$, not the limits in term of sequence.

@FrankScience : i've got it, But how can we say limit exists, we can only say if at all it exists then it is $f'(x_0)=\lim_{x\to x_0}f'(x)$, am I right/wrong

Now it should be an exercise (see following), otherwise I'll spoil your ability.
Exercise 2

where is Exercise 2?
ok

If $f(x_n)$ converges whenever $x_1,x_2,\ldots$ is an arbitrary sequence which converges to $x_0$ and never equals $x_0$, we have $\lim_{x\to x_0}f(x)$ exists and all such limits are equal.

I'll try with the $\epsilon$ and $\delta$, but later after a while.

11:43 AM
Hint

@FrankScience Thats only when $f$ is continuous
that is a criterion for continuity no?
But here we do not assume $f'$ is continuous

We sometimes can restrict $x_1,x_2,\ldots$ in a neighborhood of $x_0$, because we can delete finity terms, and we can allow $x_k=x_0$ for finity many $k$. The reason is same.
@RajeshD No, $f$ is an arbitrary function.
@RajeshD The condition $x_k\neq x_0$ is very important.

In mathematics, a continuous function is a function for which, intuitively, "small" changes in the input result in "small" changes in the output. Otherwise, a function is said to be "discontinuous". A continuous function with a continuous inverse function is called "bicontinuous". Continuity of functions is one of the core concepts of topology, which is treated in full generality below. The introductory portion of this article focuses on the special case where the inputs and outputs of functions are real numbers. In addition, this article discusses the definition for the more general cas...
click on it

Ok I got it there is a misunderstanding here

11:48 AM
@anon Is the limit of function in term of sequence taught in high-school or college?

the limit of a function is in terms of open balls in the domain, not in terms of sequences usually.. but definitions of limits of functions only come in calc or analysis classes, which most don't take in hs (though some do) in the US

@anon Is open-ball $\epsilon$--$\delta$?

yeah

But I find that it's sometimes harder to think $\epsilon$--$\delta$ than the limit in terms of sequences.
Although they're usually equivalent in the particular proof, I think it's more intuitive.

user19161
@FrankScience The epsilon delta seems more natural to me as a definition, while the sequence can be very useful for proving things.

11:58 AM
@Jasper Yes, I meant proving and thinking, not definition.

user19161
@FrankScience Sure, people think differently.

@JasperLoy Yes. My ability of reasoning is very bad, so it's complicated for me to understand something like $\forall\epsilon\exists\delta\forall$.

12:13 PM
@Frank : My question is why would the sequence $f'(y_n)$ converge in the first place, we have not assumed continuity of $f'$, we do have a case of $f'(x_o)$ not existing, right

@RajeshD $f'(y_n)$ converges to $\lim_{x\to x_0}f'(x)$.

@FrankScience yes, hence there is a case for when both the limits on either side of '=' being $\infty$ am I right?

@RajeshD Well, if $A_n=B_n$ for all positive integer $n$, and $\lim B_n$ exists, then $\lim A_n$ exists and equals to $\lim B_n$.

@FrankScience I agree but what i am saying is that there is a case when both $A_n$ and $B_n$ diverging which has to be made explicit, saying if at all $f'(x_0)$ exists the it is equal to $\lim_{x\to x_o}f'(x)$. I am agreeing with the proof, but I am saying an extra case to be added to the statement
I am saying both sides can be $\infty$
$f'(x_o) = \infty$ and $\lim_{x\to x_o}f'(x) = \infty$

@Eugene Are you around? Do you have a copy of Hatcher?
Who has a copy of Hatcher and has read chapter 0 on CW complexes and chapters 1 and 2?
Where are all the algebraic topologists?

12:24 PM
Brian Scott?
where is he now a days

yes

@FrankScience : Thank you for teaching me the proof

1 hour later…
1:28 PM
@BenjaLim it's available online man

1:46 PM
@Eugene Hey
Do I just skip chap 0?
It seems very fuzzy everything there
like as if the reader is assumed to already have some knowledge of AT
For example in chap 0 I think he assumes that readers have already some familiarity with quotient spaces and stuff like that
Wow I have exactly 7000 rep OMG OMG OMG

2:34 PM
@BenjaLim No, it is the opposite in fact.

Yeah, you have negative 7000 rep.

Chapter 0 is where you learn how to change your mindset and think geometrically, put the algebra in the background.
lol
@BenjaLim You don't need to know any Algebraic Topology already to read Chapter 0, it's there to get your intuition flowing. You may not understand everything and all the technical details inside and out when you go through it, but often with math you aren't meant to the first time anyway, just say "ok I'll try to understand what this means a bit", trust your instincts, and go for the ride. Every loose end will get tied up later.
2

2:58 PM
@RagibZaman So true :-). (for all mathematics)

@BenjaLim hatcher is a little tough. maybe just stick it out?

3:19 PM
Hatcher talks alot. If you prefer someone more straight to the point I'd suggest Vick.

@BenjaLim and honestly some authors (cough silverman cough) write chapter 0s that aren't meant to be read.

3:34 PM
good day all!
I see I missed Frank Science. I just spent way too long adding to my answer to his question

@robjohn Davide mentioned that #C00 is generally nicer than red. I use this now.

@BillDubuque I will try that.
@BillDubuque It looks a little darker. I guess it is less intrusive.

@robjohn Exactly. red looks neon on my lcd (non-calibrated wide-gamut)

@BillDubuque In my graphics I usually use #00A000 instead of full green for that same reason.

4:38 PM
Hello

5:03 PM
Ah, I see now :-)
@BenjaLim Congrats! although now it is 7010 :-)

@robjohn Sorry, they don't allow me to "chat" in rehab ;-)

5:27 PM
I am boggled. I consider myself a complete ignoramus when it comes to physics, but it appears that I have just answered someone's peculiar physics question to his complete satisfaction.
I am never quite sure what to make of it when something like this happens. Most likely, I was just the lucky person to respond first.
@BillDubuque I agree. I noticed some years ago that #F00 and \$0F0 do not look as good on my talk slides as darker colors.

@MarkDominus These are the things I don't worry about at all. If I would do that it would mean my life is going perfect and smooth as apparently I would not have more serious things to worry about! 8-).

the funny thing about serre is that when he publishes a book called "a course in arithmetic" it's a graduate text in mathematics...

user19161
@Eugene What is so funny about that?

It makes him lol, Brother. Amusing things are subjective. I might find it funny if my colleague gets wet because he fell in the water during a "raft". He might not 8-).

@JasperLoy usually arithmetic is deemed as an elementary topic

user19161
5:40 PM
@JonasTeuwen What you just said makes me lol bro.

user19161
@Eugene There is also Weil's "Basic number theory".

user19161
@robjohn I noticed long ago...

user19161
Why do authors like to add "basic" or "advanced" or "introduction" to their titles when these words are not even well-defined.

user19161
They should just help everyone save some ink and space on the paper.

Because it is just a bloody title. He might as well have named it "Unicorns! But then better, just like apples!".
There is no magic going on there, Sir.

5:45 PM
@JasperLoy serre's arithmetic is famous for being misleading however

user19161
@Eugene Yes, someone I knew told me he was reading it on the train and a woman was staring at the book because the contents and title did not match.

user19161
@JonasTeuwen Now you have caught iyengar's virus of calling everyone sir.

Did she have X-ray eyes, Sir?

user19161
No, but maybe she taught he was cute.

@JasperLoy I have learnt that when in doubt it is better to be too polite than too little! 8-).

5:48 PM

user19161
What's this thing about X-ray eyes? Is it some internet meme? Everywhere I see it it is starred.

@Eugene I am in my office as we speak, door closed. Is that okay with you Sir? Does this satisfy the requested properties?

@JonasTeuwen yes sir. thank you sir

I know nothing about X-ray eyes. I was just wondering how you can read the books title and contents at the same time! Unless...

user19161
There are now two viruses in this room. One is the deletion virus and the other is the sir virus. Actually there is a third, the flagging virus.

5:52 PM
@JasperLoy you are so astute sir

@Eugene AYOOOOOOO

user19161