Mathematics - 2013-06-09

Gustavo Bandeira

00:36

Peter Tan

Peter Tamaroff

01:46

@Ethan I almost derped.

Gustavo Bandeira

03:25

@Ethan You reminded me of this.

amWhy

03:52

Slowwwwwwww night..... yawn :-O

Bandeira Gustavo

05:40

EPIC RAP BATTLES: Qiaochu Vs. Bill Dubuque!

@skullpatrol "define define" was one of the best things I read in life.

Sometimes I read stuff in death, necromancer powers and stuff.

Little Child

06:03

^_^ good morning !

Bandeira Gustavo

06:21

Hello.

Little Child

Whats up ?

Bandeira Gustavo

Not much. just reading Derbyshire's unknown quantity.

Little Child

Cool !

What time of the day is it in your country ?

Bandeira Gustavo

03:24p.m.

Little Child

cool !@

Bandeira Gustavo

06:25

Why?

Little Child

Just asked :)

Bandeira Gustavo

Erotic.

Little Child

WHAT ? :-D

Bandeira Gustavo

Erotic.

Little Child

Why ? :p

Bandeira Gustavo

06:32

Dunno. I day that when there s nothing to say.

Little Child

LOL.. to creep people out ? :-D

Bandeira Gustavo

My reasons are a matter of secrecy.

Little Child

ok

Bandeira Gustavo

I'm going, cya.

Little Child

same here, cya

Andrew

06:38

Hello

IS anyone one?

user412173

08:42

hi

The Substitute

09:39

Is there a way to show the Hausdorff Dimension of the Cantor set is log2 / log 3 other than using the definition of the Hausdorff Dimension twice? I think I remember seeing someone obtain the answer without using the definition directly but using some measure theory.

Chris's wise sister

09:57

Greetings

Chris's wise sister

10:40

@robjohn are you around?

robjohn

@Chris'swisesister yes

sorry, I was writing a reply to a comment.

Dominic Michaelis

10:58

Does someone have the mac lanes categories for working mathematicans in his near ?

ah nevermind i just switched the notation in my proof wrong :D

Chris's wise sister

11:26

@robjohn sorry for delay (I didn't notice you answered me back - my sound was on mute). Are you now?

@robjohn could you take a look on it? imagebin.org/260789

robjohn

@Chris'swisesister okay

Chris's wise sister

@robjohn thanks!

@robjohn I think something is wrong there.

@robjohn my notations failed in some places.

At the beginning it's $$\lim_{n\to\infty} \int_a^{\infty} f_n= \int_a^{\infty} \lim_{n\to\infty}f_n= \int_a^{\infty}f$$

robjohn

11:57

@Chris'swisesister It looks fine.

Chris's wise sister

@robjohn thank you! :-)

Peter Tamaroff

13:14

@TheSubstitute There is a über informal proof.

Chris's wise sister

I ask myself if there is a nice, simple way to prove the convergence of $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{k+n}}{k+n}$ and then extend the solution to the multiseries. (I'm not interested in computing its exact value)

Peter Tamaroff

@Chris'swisesister Well, the inner series converges by Leibnitz.

And it is a cut off sine.

Chris's wise sister

@PeterTamaroff There must be some work on Leibniz's multiseries test but I didn't find it on google.

Peter Tamaroff

Rememeber $$\sum_{k\in\Bbb Z}\frac{(-1)^k}{a+k}=\frac{\pi}{\sin\pi a}$$

Chris's wise sister

hmmm

Peter Tamaroff

13:21

Also, $$\sum_{n=1}^\infty \frac{(-1)^n}{x+n}$$ converges uniformly over $\Bbb R_{\geq 0}$

Chris's wise sister

@PeterTamaroff I've just found this: link.springer.com/article/10.1007%2FBF01086823#page-1

BenjaLim

13:37

@robjohn Are you here?

robjohn

@BenjaLim yes

BenjaLim

13:52

@robjohn I have posted a question on main.

Peter Tamaroff

@BenjaLim I have been told you have to be careful about what measure you use on $E$ and which on $f^{-1}(E)$.

The condition is that for every Borel measurable $E$, $f^{-1}(E)$ is Lebesgue measurable.

BenjaLim

I am only talking of the Lebesgue measure

Peter Tamaroff

@BenjaLim Well, but the definition is that a function $f:\Bbb R\to\Bbb R$ is Lebesgue measurable if the preimage of a Borel measurable set is Lebsgue measurable. Ask Mariano if you'd like.

BenjaLim

I don't know what you're talking about

Peter Tamaroff

See the second paragraph here

@BenjaLim In your post you have a typo "$f:\Bbb R^d\to f$."

Dominic Michaelis

14:00

@PeterTamaroff well that is only half true. That is the usual definition of measurable on $\mathbb{R}^n$, or at least equivalent to the usual definitions. If you talk about arbitrary $\sigma$ algebras the definition that a function is measurable whenever the preimage of a measurable set is measurable

Peter Tamaroff

@DominicMichaelis I am talking about $\Bbb R^n$ in particular, that is my point.

Dominic Michaelis

In my lecture we defined that a function is measurable if for all $\alpha \in \mathbb{R}$ the set $$ \{ x \in \mathbb{R}^n : f(x)< \alpha \}$$ is measurable. This one is obviously equivalent to yours

Peter Tamaroff

14:54

No upvotes? :snif:

Argon

@PeterTamaroff There there Pedro

Mats Granvik

15:27

@Chris'swisesister I posted a question at main about the Euler-Mascheroni constant:
$\gamma = \int_0^n \left(\int_k^{\infty } e^{a b} \, da\right) \, db-\int_k^{\infty } \left(\int_0^n e^{a b} \, da\right) \, db$
http://math.stackexchange.com/questions/415491/is-this-euler-mascheroni-constant-calculation-from-double-integrals-a-true-ident

Chris's wise sister

@MatsGranvik awesome question (+1) Where did you meet it?

Mats Granvik

@Chris'swisesister thanks

@Chris'swisesister It started with your suggestion to evaluate a similar triple integral at x=1/Exp[1], here in this chat room. I was away from my computer for two weeks though.

Chris's wise sister

@MatsGranvik ah, I've just remembered it !

Chris's wise sister

15:53

@MatsGranvik I like your researcher attitude of discovering math things, and in a way I do the same thing on my own (as an amateur), but you know, sometimes life gets in the way and I cannot attend long enough what I like. At any rate, my mind is full of ideas and I could write down things all day. :-)

Blackoffe

Hello :)

I'm new here, could I ask math question in this room?

Peter Tamaroff

@Blackoffe Sure,.

Mats Granvik

@Chris'swisesister I do mathematics as an amateur as well. My day job is working in a chemical laboratory.

Blackoffe

Well, I've been struggling with proving $V(0) = Spec(R)$, could somebody help me?

Peter Tamaroff

@Blackoffe Ugh, I don't know if I can help you there. Could you give me some background?

Chris's wise sister

16:08

@MatsGranvik nice

Blackoffe

Umm yeah, given V(0)=\{P \in Spec(R)| 0 \in P\} and $Spec(R)=\{I| I prime ideal of R\}$

Peter Tamaroff

@Blackoffe $V(0)=\{P\in{\operatorname{Spec}}(R)\; :\;0\in P\}$ and ${\operatorname{Spec}}(R)=\{ I\; :\; I\text{ is a prime ideal of } R\}$

That in ideal is prime means $ab\in I\implies a\in I \wedge b\in I$, yes?

Blackoffe

yes

Peter Tamaroff

Ok. So, suppose $I$ is a (prime) ideal. Is $0\in P$?

Blackoffe

yes,,I can hardly prove if 0 is in any set in Spec(R)

Peter Tamaroff

16:12

You're defining an ideal by an additive subgroup such that $a\in R,b\in I\implies ab,ba\in I$, yes?

If it is an additive subgroup, it contains $0$.

Blackoffe

yes,given $R$ is a ring

Peter Tamaroff

Or in a more silly fashion, $0\in R$,$a\in I$ so $a0=0\in I$.

=)

Blackoffe

oh i see..

Peter Tamaroff

Are you sure this is the problem? Because it seems completely irrelevant that the ideals are prime. That's why I think I am missing something.

Blackoffe

yeah i'm pretty confused too, but there's something that makes me think over my proof because of your statement just now :)

@PeterTamaroff thanks a lot, I found the proof (^_^)

JohnMerlino

16:38

is it true all perfect square trinomials will only intercept the x axis once? I just tried it with 3 and all of them only intercepted x axis once

ok nevermind ignore above]

Peter Tamaroff

16:59

@Blackoffe How does it go?

anon

17:59

@PeterTamaroff The $V(S)$ thing is more general though: it is defined as $V(S):=\{P:P\supseteq S\wedge P~\rm prime\}$. These are precisely the closed sets of ${\rm Spec}$.

(i.e. the Zariski topology). indeed computing V(0) is trivial though.

Peter Tamaroff

@anon Oh, OK...

The Substitute

19:02

@PeterTamaroff @PeterTamaroff, what is it? Is there a link?

Peter Tamaroff

@TheSubstitute Well, look at it this way.

Take a square. If you triple it's side length, you get 9 copies, so its dimension is 2, since 3^2=9.

If you get a line and triple it, you get 3 copies, so it's dimension is 1, since 3^1=3.

If you get a cube and triple it's side length, you get 27 copies of itself.

So its dimension is 3, because 3^3=27.

Take now the Cantor set.

Say, over $[0,1]$. If you triple it, you get $[0,3]$. But by subtracting the middle third $[2,3]$ you now see there are two copies of the Cantor set.

So if $x$ is its dimension, $3^x=2$

Thus, $x=\log 2/\log 3$.

The Substitute

@PeterTamaroff, thank you! Is there a name for the type of dimension you are using?(box dimension?) When does that dimension agree with Hausdorff dimension?

Peter Tamaroff

19:24

@TheSubstitute Hmm... no idea!

Basically we kinda have a function $A$ that assigns a ¿size? $A(S)$ to a set $S$ and we we're looking at $A(\lambda S)=\lambda^{D(S)}A(S)$, right?

And we wanna call this $D(S)$ the dimension of $S$?

It is all too fishy to tell.

The Substitute

@PeterTamaroff, got it. Very helpful!

Vrouvrou

19:50

hello

can someone help me please

{math.stackexchange.com/questions/415649/…}

exitingcorpse

20:57

what exactly does the number next to the triangle (near reputation) mean?

Julian Kuelshammer

@exitingcorpse number of bronze badges

exitingcorpse

i just got one for reading the entire about page in search of what it meant

it's called "informed"

woooo irony

Julian Kuelshammer

21:12

:-)

JohnMerlino

22:46

if x is equal to √-16 , where exactly is that on the cartesian plane?

When I run that into calculator it gives back: 4 i

Peter Tamaroff

@JohnMerlino Ugh... better put it as $4i$.

This is the complex number $(0,4)$.

user1

23:26

@JohnMerlino Since $(4i)^2=4i4i=16i^2=-16$, the calculator is right that one of the square roots of $-16$ is $4i$.

Peter Tamaroff

@user1 What's up, nerdizzle?

user1

@PeterTamaroff You know, all sorts of things are up, and I am sure someone helped you determine those in the past.

Peter Tamaroff

@user1 Those?

user1

@PeterTamaroff I suspect both the sky and a roof are over your head.

Peter Tamaroff

@user1 That is true.

They are complex structures.

Takes time to understand them.

user1

23:32

@PeterTamaroff I guess it would not be worthwhile to get into an argument with KeyIdeas already here.

Peter Tamaroff

@user1 Not personal experience, but according to others, my only recommendation should be: RUN AS FAST AS YOU CAN, AND DON'T LOOK BACK!

user1

@PeterTamaroff :)

Peter Tamaroff

@user1 I'm reading a proof of Cantor Berstein for the first time.

user1

@PeterTamaroff I barely remember any proofs of it. Is it by contradiction?

Peter Tamaroff

@user1 Nope, it is an explicit construction of a bijection given two injections! =)

Not funny! Not funny!

user1

23:42

@PeterTamaroff The proof that I am looking at now is not exactly an explicit construction, since it has one of those "without loss of generality" phrases.

Peter Tamaroff

@user1 Really?

WLOG is just fine. It is part of the proof.

If you prove it rigorously, I'm OK with it,.

user1

@PeterTamaroff It does not explicitly construct it for all $A$ and $B$, though.

I am fine with the proof btw; I was still wrong with guessing that it is by contradiction.

Voyage

When you have a common distribution function of two random variables, how can one immediately check if the random variables are independent or not?

user1

@PeterTamaroff Actually, if $f:A\to B$ and $g:B\to A$ are injective and you construct $b:A\to g[B]$ bijective, then $g^{-1}\circ b:A\to B$ is bijective and explicitly constructed.

You are right. :)

Peter Tamaroff

@user1 Aha.

I think! =)

user1

23:58

@PeterTamaroff Therefore, you are.

(not that I was worried about you being some bot)

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$Mathematics$ Mathematics