Mathematics - 2013-11-10 (page 1 of 2)

Pedro Tamaroff

00:00

@FernandoMartin $\sup$?

Kevin Driscoll

00:17

@PedroTamaroff $\inf$

Pedro Tamaroff

00:28

@KevinDriscoll Dude

I cannot kill Queelag.

I started over.

Killed Asylum Demon, Havel, Capra, Taurus, Butterfly, Big Ass Rat, some Knights.

Cannot kill Queelag.

Kevin Driscoll

@PedroTamaroff That's sad. Queelag is usually the first boss that REALLY sucks. Like Taurus and Capra are hard, but Queelag is just TERRIBLE. Starting over may help considering how scattered your levels were, but Queelag is always hard

Pedro Tamaroff

@KevinDriscoll Well, now I focused on Strength, Endurance, Vitality.

I guess my "build" is better now.

Kevin Driscoll

Ya thats totally reasonable

Need to find a reasonable weapon

Pedro Tamaroff

Gargoyle Halberd is doing quite good.

I made it divine... don't know if I should be regretting it.

Kevin Driscoll

Its ok to start out with, but weapons that are easier to upgrade and scale quickly outpace it

Oh, Halberd does scale. Its a D tho which isnt so good

Pedro Tamaroff

00:31

Gargoyle Halberd,.

It's a little more badass.

Kevin Driscoll

Divine is ok. Not as much raw damage but it gets some magic damage

Ya I meant Gargoyle Halberd

Pedro Tamaroff

@KevinDriscoll Right, it gets me 117 raw plus 147 magic.

But I don't see it doing 260 dmg.

Kevin Driscoll

I wont because split damage is bad

enemimes have a defense against eacg kind of damage

but instead of doing something smart and averaging their D against your attack

it actually applies each defense separately to each type of damage

so split damage is usually worse than 1 type of damage cuz the enemy gets 2 kinds of defense against it

@Pedro

Pedro Tamaroff

FU

@KevinDriscoll Well, they say divine wpns are good against the undead, so maybe I have a good time down at the Catacombs. =O

I killed Lautrec, too.

Got two Embers.

I am basically where I left, except for the Gaping Dragon. shivers

@KevinDriscoll You know what's the saddest thing of it all?

Kevin Driscoll

00:58

Oh, you ALREADY killed Lautrec? Cheater.

Gotta wait and do it in Anor Londo

@PedroTamaroff Whats the saddest thing?

Pedro Tamaroff

@KevinDriscoll MUA-HA-HA.

@KevinDriscoll I get to see some mofo kill Queelag with 2398398 less lvls than me.

Kevin Driscoll

lol ya

people ahve done lvl 1 runs

Pedro Tamaroff

@KevinDriscoll Seen that. 30 min run.

Crazy.

Impossible.

Kevin Driscoll

I think some ppl have enen done lvl 1 broken sword only runs

Fernando Martin

hey there @Pedro

Kevin Driscoll

01:02

Gwyn must take 4ever

Pedro Tamaroff

@KevinDriscoll WAHAHHATHAHAHA

@FernandoMartin Heya.

Charlie

Why matrices product is the way it is?

Pedro Tamaroff

@Charlie Compose linear transformations-.

Charlie

@PedroTamaroff I want details

Pedro Tamaroff

@Charlie OK.

Charlie

01:08

I want deep explanations

Pedro Tamaroff

Take a vector space $V$ of dimension $n$.

Take a basis $\{v_1,\ldots,v_n\}$.

agent154

Oooh... I think I understand this... :P @PedroTamaroff

At least so far

Pedro Tamaroff

Write $f(v_j)=\sum_{i=1}^n a_{ij}v_i$

Then your matrix is $[a_{ij}]$

Charlie

@PedroTamaroff but the concept of matrix product came after the topic of vector spaces? :/

Pedro Tamaroff

And if you write any vector $v=\sum_{k=1}^n\beta_kv_k$ $$f(v)=\sum_{i=1}^n\sum_{k=1}^n\beta_k a_{ik}v_k$$

That is the usual matrix times vector thingy.

@Charlie Ah?

Fernando Martin

01:13

@Pedro: conocés alguna biblioteca grande de propósito general que haga préstamos a domicilio?

Pedro Tamaroff

@FernandoMartin Nah.

Fernando Martin

Ok, gracias igual

Amzoti

What is the protocol to report someone who is posting a large number of bogus answers?

Charlie

@PedroTamaroff didn't make sense to me

agent154

@Amzoti Don't they just get downvoted to oblivion?

Pedro Tamaroff

01:15

@Amzoti Where?

@Charlie What doesn't makes sense?

Charlie

@PedroTamaroff explaining something a bit simpler with a more complex concept

Amzoti

This particular users post end up on cues for low quality and in the Tools review area. There are LOTS of them

on the MSE main site

Fernando Martin

@Charlie: I'd say explaining simple things via complex concepts is a general, recurrent theme on mathematics

agent154

groan

Fernando Martin

hahahaha

Pedro Tamaroff

01:20

@Amzoti Cannot you just say which user it is?

Charlie

@FernandoMartin but where it came from? I mean, what does it mean? What it represents?

Fernando Martin

It represents exactly what Pedro said

leo

@agent154 ha ha ha ha

Amzoti

khosrotash

Charlie

@FernandoMartin I'm not content with that explanation

Fernando Martin

01:24

why not?

Argon

@FernandoMartin It's not intuitive. Clearly someone came up with the concept for a reason from a priori reasons.

Charlie

@FernandoMartin it seems more like one way of seeing it than what it is In fact

Fernando Martin

well

check Arturo's answer here

It's basically what @Pedro said though

Pedro Tamaroff

Arturo used to give incredibly long answers.

Fernando Martin

I'll be back in a while

The Chaz 2.0

01:33

Would anyone with some sweet graphing software be willing to show me the graph of $e^{1/z}$ near the singularity $z=0$? I'm just learning about Casorati-Weierstrass and ... well, it's somewhat mind-blowing!

agent154

@TheChaz2.0 wolfram doesn't work?

The Chaz 2.0

@agent154 it works; it's just somewhat ... opaque. Though maybe I don't know how to adjust the scale.

Also, they keep wanting me to upgrade or something.

leo

01:47

@TheChaz2.0 look at

\documentclass{standalone}

\usepackage{tikz}
\usetikzlibrary{arrows}

\newcommand{\tipx}[3][below]{\draw (#2,2pt) -- (#2,-2pt) node[#1]{#3}}
\newcommand{\tipy}[3][right]{\draw (-2pt,#2) -- (2pt,#2) node[#1]{#3}}
\newcommand{\tx}[1][]{\tipx{#1}{$#1$}}
\newcommand{\ty}[1][]{\tipy{#1}{$#1$}}

\begin{document}
\begin{tikzpicture}[>=stealth,font=\small]
	\draw[<->] (0,-10)--(0,10) node[above]{$y$};
	\draw[<->] (-10,0)--(10,0) node[right]{$x$};
	\clip (-10,-10) rectangle (10,10);
	\draw[samples=300,domain=-10:-0.01] plot(\x,{exp(1/\x)});

or wolframalpha.com/input/?i=plot+exp%281%2Fx%29

The Chaz 2.0

@leo What should I be looking at? Sorry to be so dense... is this for a particular software?

leo

@TheChaz2.0 LaTeX

The Chaz 2.0

Hold on... haven't got the plugin on this browser

leo

@TheChaz2.0 it would not work. You have to copy and paste it somewhere and then compile it in your local machine

agent154

@TheChaz2.0 You can try it at www.writelatex.com I think

The Chaz 2.0

01:52

Oh. Ok thank you. Also, I didn't realize that "plot" was an option in W|A.

@agent154 Cheers

agent154

Save that, if you have windows, look for miktex

leo

@TheChaz2.0 W|A?

agent154

Wolfram Alpha I think

The Chaz 2.0

Wolfram|Alpha

leo

ah, yes :)

The Chaz 2.0

01:55

Thanks, everyone! I'm off to ponder :)

leo

@TheChaz2.0 have fun

robjohn

02:37

@TheChaz2.0 does one get wet when one ponders?

@AlexanderGruber: hey there

Alexander Gruber

@robjohn hiya

Enjoys Math

pink floyd's on spotify now

it wasn't always

Alexander Gruber

@EnjoysMath whoa- awesome. i was just looking for them earlier.

Enjoys Math

02:52

How do you take the Fourier transform of $1$, I'm getting weirdness at the end when inserting $\infty$'s

Oh, it doesn't have a converging one, lol

its not rapidly decreasing or anything like that

robjohn

@EnjoysMath It is the dirac delta distribution

@EnjoysMath convolution of functions corresponds to multiplication of the Fourier transforms and vice versa. think about that and you'll see :-)

Enjoys Math

03:21

What math do you study?

Zibadawa

Is that your new pickup line?

leo

pickup line

Fernando Martin

03:59

Hey, does anyone here know about basic algebraic number theory?

I'm trying to calculate the ring of integers of $\mathbb{Q}(\sqrt{2}, \sqrt{3})$

I've calculated the discriminant, which is 96^2

leo

@FernandoMartin How calculate? I gues @AlexanderGruber should know

Alexander Gruber

well

should be stable under the Galois group right?

what's that leave you with for options?

Vrouvrou

04:50

hello

can someone help me please

I was wondering if this is singular homology or anything else. But if you know already that $H_0(X)\isom\mathbb Z$, then you seem to know all elements of $H_0(X)$. What are they (@John has answered this already)? Are they in the image of $H_0(Y)$? — Carsten Schultz 6 hours ago

Sanchez

what is $X$ and $Y$?

Vrouvrou

$X$is a path connected space and $Y\subset X$

Sanchez

I'm not sure what is the precise question you have in mind, but the elements of $H_0(X)$ are really, maps from a single point to $X$ modulo connected component

Vrouvrou

my question is hy $H_0(X,Y)\simeq 0$ using the fact that $H_0(Y)\rightarrow H_0(X)$ is surjective

Sanchez

Do you know how $H_0(X,Y)$ is defined?

Vrouvrou

05:00

$H_0(X,Y)=C_0(X,Y)/B_0(X,Y)$

$B_0(X,Y)=im (\partial : C_1(X)/C_1(Y)\rightarrow C_0(X)/C_0(Y)$

@Sanchez are you ok ?

Sanchez

yea, I was trying to see if I can do this by diagram chasing

never mind, maybe I should directly deal with the definition

Vrouvrou

ok

Sanchez

never mind, diagram chasing is better

Just draw this chain complex

$0 \to C_0(Y) \to C_0(X) \to C_0(X,Y) \to 0$

and the one for $C_1$

Of course we also have the maps from $C_1(Y)$ to $C_0(Y)$ and so forth

can you see what the cokernels are, for $C_1(Y)$ to $C_0(Y)$ and $C_1(X)$ to $C_0(X)$ and $C_1(X,Y)$ to $C_0(X,Y)$?

@Vrouvrou, I guess you are not here. Anyway, you can then finish off by snake lemma.

Vrouvrou

05:16

no i'm here

fppf

@Sanchez Hey

Sanchez

@fppf, hey

I see that you are changing names forever

:P

fppf

@Sanchez I have a question about Borel measurability which i'm confused about

Sanchez

go on

fppf

@Sanchez I want to prove that if $E,F$ are Borel measurable subsets of R^n and R^m

then $E \times F$ is Borel measurable as a subset of $R^{n+m}$

@Sanchez Now if $E,F$ are open cubes this is immediate

Sanchez

05:17

then?

fppf

Ok but now I think I should be using something about the Borel $\sigma$-algebra being the smallest one containing all open sets

@Sanchez I don't know what to do next.

Vrouvrou

@Sanchez sorry i don't uderstand "and the one for $C_1$"

sorry @fppf

fppf

@Sanchez I don't understand a lot of the measure theory stuff.

Sanchez

@Vrouvrou, the exact sequence $0 \to C_1(Y) \to C_1(X) \to C_1(X,Y) \to 0$

Vrouvrou

ah ok

so we see the exactenes?

?

Sanchez

05:27

@Vrouvrou, what exactness?

@fppf, it would be good to assume $F$ is an open cube first

So fix $F$ to be a fixed open cube, then look at the algebra on $R^n$, collecting all the sets $S$, such that $S \times F$ is Borel measurable

Vrouvrou

exactness of the diagram

Sanchez

check that this is a sigma algebra that clearly contains all open sets of $R^n$, so it contains the Borel sigma algebra.

In particular for any borel measurable $E$, $E \times F$ is borel measurable. Now fix $E$, and do the same thing for $F$.

@Vrouvrou, the two rows are exact by definition

Vrouvrou

so what we do this diagram ?

Sanchez

Do you know snake lemma?

it tells you that the cokernels are exact on the right

In particular, $H_0(Y) \to H_0(X) \to H_0(X,Y) \to 0$ is exact

Vrouvrou

ok

so

if $H_0(Y) \to H_0(X) \to H_0(X,Y) \to 0$ is exact

$H_0(X)\simeq \mathbb{Z}$ and $H_0(Y)\simeq \mathbb{Z}$

right

so $H_0(Y)\rightarrow H_0(X)$ is surjective

Sanchez

05:35

$H_0(Y)$ may not be $\mathbb{Z}$, if it's not path connected

In any case, the reason why $H_0(Y) \to H_0(X)$ is onto is that a generator of $H_0(X)$ would be the map from a point to any point in $Y$

which clearly lies in the image of $H_0(Y)$

Vrouvrou

oooo ok

why we need the surjectivity ?

please

Sanchez

You want to show $H_0(X,Y)$ = 0

Vrouvrou

YES

Sanchez

The exactness of that thing does not give you that, unless $H_0(Y) \to H_0(X)$ is onto

Vrouvrou

i don't see this

Sanchez

05:40

so from the exactness of that thing, can you see what $H_0(X,Y)$ is isomorphic to?

Vrouvrou

i just see that $j:H_0(X)\rightarrow H_0(X,Y)$ is surjective

Sanchez

what is its kernel?

fppf

@Sanchez Right, I will do that. Just before I leave, do you have any recommendations for measure theory exercises to do (borel measures, etc)?

Vrouvrou

and that if $i:H_0(Y)\rightarrow H_0(X)$, ker j=im i=H_0(X)

fppf

@Sanchez I will leave you and @Vrouvrou to your conversation.

Sanchez

05:44

@fppf, not really. Don't waste time on boring stuff :P

What book are you learning from though?

fppf

@Sanchez I have an exam unfortunately. Stein and Shakarchi volume 3

Sanchez

I see. I never read that.

fppf

chapter 6 on abstract measures doesn't really have any good exercises.

Sanchez

Cohn's measure theory book has many exercises

fppf

@Sanchez paul cohn?

Sanchez

05:45

Not sure. But if you google cohn measure theory there should be only one book

fppf

right. Seems a little too much though

Sanchez

@Vrouvrou, ker j = im i, but it may not be the same as $H_0(X)$ (I'm only looking at the exact sequence now)

Wow, the course is so packed

usukidoll

Hi everyone.

fppf

@Sanchez I know there is shit ton of material in that course

Vrouvrou

@Sanchez why it's not the same ? what can be im i ??

@usukidoll hi

fppf

05:48

@Sanchez We haven't really had time to grasp anything

Sanchez

@fppf, anyway I'm not sure; since I'm not familiar with Stein's book. You can probably do some drilling from Cohn, but I don't have recommendations about it since I didn't do it too

That's normal. You are packing functional analysis and measure theory in one quarter

usukidoll

I'm finally able to chat yay :D. So, how is everyone?

Sanchez

@Vrouvrou, okay. If you are using the condition $H_0(Y) \to H_0(X)$ is onto then yes im i = $H_0(X)$. Otherwise why should they be equal at all?

@fppf, doesn't matter. But to continue along your path, pay some attention to functional analysis

measure theory is meh, you don't need to care too much. All you will use is probably this and that Haar measure anyway

fppf

@Sanchez If I want to communicate personally with you, how can I do that?

Sanchez

I think you sent me an email once?

fppf

05:50

ah ok.

@Sanchez Do you have any recommendations for exercises on fourier transforms?

Sanchez

Stein 1 should have some, no?

Vrouvrou

from the exactness of $H_0(Y)\rightarrow H_0(X)\rightarrow H_0(X,Y)\rightarrow 0$

fppf

yea but that is retarded in some sense because it doesn't assume any measure theory.

Sanchez

What's different?

fppf

well ok.

I will try that.

Sanchez

05:52

What extra input does measure theory have?

Make sure you know Fourier transform stuff well by the way

fppf

well you get stuff like the fourier transform is a bijective map from $L^2$ to itself.

@Sanchez why?

Sanchez

That's just language, not really content

fppf

right.

Sanchez

You are interested in number theory/representation theory/AG for now, right?

fppf

yes.

Sanchez

05:53

Say Tate's thesis = Fourier analysis on adeles

fppf

@Sanchez I'd say leaning towards rep theory/AG.

@Sanchez I've looked at borel-weil

Sanchez

If you read automorphic forms you will also see some of those ideas

I see

fppf

@Sanchez But I'm open to any of them.

Sanchez

In any case I just wanted to say that generalizations of fourier analysis occur in number theory

fppf

Yeah certainly.

usukidoll

05:54

What is the best way to succeed in writing a proof paper?

Sanchez

@Vrouvrou, that does not tell you that $H_0(Y) \to H_0(X)$ is onto at all

fppf

@Sanchez How did you guess I was interested in those above?

Vrouvrou

no we have that $i$ is onto because the generator of $H_0(X)$ would be the map from a point to any point in $Y$

Sanchez

@fppf, your profile a while ago lol

fppf

ah ok.

Sanchez

05:57

@Vrouvrou, so you are using the fact that $H_0(Y) \to H_0(X)$ is onto here. This does not come from the exact sequence

fppf

@Sanchez see you. I'm off to study more.

Sanchez

Bye @fppf

Vrouvrou

yes but keri-im i from the exact sequence

Sanchez

Yes

Vrouvrou

ker j=im i

Sanchez

05:58

So the exact sequence says $H_0(X,Y) = H_0(X)/im i$

and onto says RHS is 0

Vrouvrou

why it say's that $H_0(X,Y) = H_0(X)/im i$

Sanchez

It's from the exact sequence

$H_0(X) \to H_0(X,Y)$ is onto

So some isomorphism theorem says $H_0(X,Y)$ is isomorphic to $H_0(X)/ker j$

Jaycob Coleman

I would appreciate comments or critiques anyone has of this question

Sanchez

Exactness tells you $ker j = im i$, like you said

Vrouvrou

yes

Sanchez

06:01

Does it make sense now?

Vrouvrou

but i never see that exactness sys that $H_0(X,Y)$ is isomorphic to $H_0(X)/ker j$

Sanchez

First isomorphism theorem, or whatever it's called

When you have a surjective map from an object $A$ to another object $B$

Vrouvrou

ooo ok

Sanchez

then $B$ is isomorphic to $A/kernel of the map$

Vrouvrou

thank you

@Sanchez thank you

Vrouvrou

06:21

@Sanchez if you have a time can you see this : math.stackexchange.com/questions/551575/… ,the question 2)

i'm ok with @Christoph Pegel in the Hint 1) but 2) i don't understand it

user2493976

06:33

How to evaluate the integral $\int_{0}^{100}e^{x-[x]}$

Zibadawa

Compare $x-[x]$ on the interval $[0,1]$ and on the interval $[1,2]$.

Sanchez

@Vrouvrou, we already know that $H_0(X,Y)$ is isomorphic to $H_0(X)/im i$

The rest comes from explicitly writing down the map $i$ and the generators of $H_0(X)$ and $H_0(Y)$

You can verify this on your own:
$H_0(X)$ has a $\mathbb{Z}$-basis, each element corresponds to a connected component of $X$: for a connected component $C$, the corresponding element is the map from a point to $X$ with image being any point $C$.

the image of $H_0(Y)$ then corresponds to: map from a point to connected component $C$ of $X$ that intersects $Y$

the quotient is then: map from a poitn to connected component $C$ of $X$ not intersecting $Y$

This is what the question is asking for. Also, by "connected" I meant path-connected in my last messages

Vrouvrou

06:52

ok thank you

Faraad

Does anyone want to talk real analysis?

Sanchez

07:04

@Faraad, what do you want to talk about?

eXtremiity

07:48

Need a bit of help here.

Anyone available?

J. W. Perry

@eXtremiity Assume that I am not here, but feel free to speak your level headed mind at any time on topics of general discussion and math questions alike. People come out of the cracks :).

eXtremiity

Hahaha :D, it's alright - I've decided to ask it on the forum. Thanks though:) .

J. W. Perry

cool :))

Chris's sis

08:04

Greetings the great ones!

(I'm preparing for some work --- many unsolved questions have been waiting me)

Chris's sis

09:09

@robjohn are you around?

Jack M

09:22

Are there subfields strictly between C and R?

I thought there were none until I found this mathoverflow.net/questions/51085/…

anon

no, there are no subfields between R and C

Sanchez

Your link does not contradict it.

Jack M

OH

thanks @Sanchez

Jack M

10:35

here's a question that's been bothering me for a while

how do we really know that $y=ax+b$ is the equation for a line?

do we simply have to define a line as a set of points verifying that?

Alizter

10:46

Where would be a good place to start solving this?

How many polynomials $P(x)$ are there such that the coefficients of $P(x)$ are integers from $0$ to $24$ (inclusive) and $P(5)$=$1200$?

Realz Slaw

$5^k=1200$ << this is your max exponent I think

Alizter

@RealzSlaw I think 4 is my max

Realz Slaw

then you can simply count :P

$25^4$ iterations later you'll have an answer

Alizter

@RealzSlaw I am trying to find an elegant way of doing this :P

Realz Slaw

you can eliminate about half of them by being smarter :P

lol

well

if it is max 4-degree polynomial, then it is max 4 answers to quartic equation

solve for quartic equation?

Alizter

10:51

@RealzSlaw You make it sound easy

Realz Slaw

Alizter

yeaa

Realz Slaw

that'll give you up to 4 answers

Alizter

I have it as a poster in my room :P

Realz Slaw

lool

@Alizter you can write it as an SAT problem

or SMT

Alizter

10:54

What do those mean

Realz Slaw

mmm I am a CS guy

basically it means brute force

but more elegant

Alizter

Cs?

Realz Slaw

computer science

Alizter

ah

I am writing a brute force in C# now

Realz Slaw

you rewrite it as a circuit

and then you send it to an sat solver and ask what inputs satisfy the output

pretty powerful stuff

but intractable

so still a cheatful answer :P

Alizter

10:58

Brute force gave me 241

Realz Slaw

mmm cool

Alizter

static void Main()
{
int count = 0;

for (int i = 0; i < 2; i++)
{
for (int k = 0; k < 10; k++)
{
for (int j = 0; j < 25; j++)
{
for (int l = 0; l < 25; l++)
{
for (int f = 0; f < 25; f++)
{
if (5 * 5 * 5 * 5 * i + 5 * 5 * 5 * k + 5 * 5 * j + 5 * l + f == 1200)
{
count++;
}
}
}
}
}
}

Console.WriteLine(count);
Console.ReadLine();

}

Now I want to do it the number theory way

Realz Slaw

/me points <-- that way -->

oh wait

mmm I just thought of another way

Alizter

@RealzSlaw As a cs what is your language of choice?

Realz Slaw

I currently use python

mmm the way I am thinking of is basically integer-linear-programming :(

you can possibly do something with a system-of-linear-equations here, but it won't restrict the sizes

so I am not sure how to count the solutions

yeah it would prolly just give you this formula as the output, basically useless mm

I can think of ways of optimizing your program though

but it still isn't number theory :P

robjohn

12:26

@Chris'ssis I am now :-)

Chris's sis

@robjohn I sent something to you (that you might like).

:-)

robjohn

@Chris'ssis Got it ... I will look at it when I finish fixing an answer I just gave.

Chris's sis

@robjohn ok

Mark S.

12:50

If I just posted an answer that I didn't mean to make community wiki, when is it worth flagging to have it undone?

And no, I was not posting on a CW question, I clicked CW thinking of posting something that was just in the comments and then changed my mind without unclicking it

Alizter

@Chris'ssis Did I get the answer right?

Chris's sis

@Alizter which answer do you refer at? That one with $\zeta(3/2)$?

Alizter

Yes

Chris's sis

@Alizter Yes, this is the correct answer.

Alizter

Dances around

Chris's sis

12:55

@Alizter :-)

Mark S.

@Robojohn, you're a mod, so if you have any input on my question above, it would be most appreciated (even if that input is just "post your question on meta")

@robjohn, you are not "robojohn". I never realized...

Alizter

Why do 5 people like $\pi^2$ is the area of a unit circle?

Mark S.

They think the error is cute and/or are poking fun at Jack M, presumably.

Transcript for

$Mathematics$ Mathematics