Mathematics - 2013-11-06 (page 1 of 2)

Enjoys Math

00:00

Charlie Sheen made the best alien movie ever

leo

Or perhaps calculating some Galois groups

Charlie

Charlie Sheen is funny

Enjoys Math

"would you like to see the ruins, my friend." - The Arrival

Charlie

@EnjoysMath i like top gang

Enjoys Math

top gun?

Charlie

00:06

@EnjoysMath top gang

user71494

Hello, can anyone help me? When I integrate floor(x) it gives me Floor[x]^2 + Floor[x])/2 + (Floor[x]+1)*(x - Floor[x] but when I finde the derivative of that it gives me 1 - Floor[x] (-1 + Floor'[x]) + (-1/2 + x) Floor'[x]. Why?

user71494

Also, the plot of it looks similar to the floor(x) plot: goo.gl/hXplm2

user71494

...is anyone there?

Enjoys Math

00:22

noone's here

lolcats

agent154

@EnjoysMath I'm here now, but not that that matters... :P

Enjoys Math

@agent154 humble as always

user71494

@agent154 @EnjoysMath Could any of you help me?

Enjoys Math

First you need to $\LaTeX$ify everything

then we can read it

we're mathematicians not computers

Although it was a human computer's birthday yesterday

user71494

But it doesn't work...

agent154

00:27

I don't think I could be of help with that... Not sure how to integrate and differentiate the floor function.

@Runemoro math.ucla.edu/~robjohn/math/mathjax.html

That's how you get latex to work in chat

Enjoys Math

Put shit in enclosing dollar signs: $

user71494

Ok

user71494

I found that the integral of $floor(x)$ is $\frac{floor(x)^2 + floor(x)}{2} + (floor(x)+1)(x-floor(x))$

user71494

But when I find the derivative of that it gives me something different

user71494

than floor(x)

agent154

00:38

Maybe somebody will come along who can help... or maybe you can post it as a question on math.stackexchange.com... Sadly I am of no help

Sanchez

@Runemoro, why is the integral differentiable?

You are also differentiating the floor function. Which is not diff at integers, and is 0 elsewhere

When you sub. $floor'(x)=0$ back in you will probably recover floor(x) again, at non-integer points

user71494

@Sanchez No, I don't

user71494

@Sanchez Oh wait, I do. Sorry

Kevin Driscoll

00:56

You have to record Floor as a distribution and find its distributional derivative

similar to finding the derivative fo the heaviside function

Alexander Gruber

i have analysis. where is @PedroTamaroff.

user71494

Yes, I know, I was having problems with it because wolfram alpha's Floor' function is weird

user71494

wolframalpha.com/input/?i=plot+Floor%27%28x%29+from+0+to+1

pourjour

hi folks

Karl Kronenfeld

@AlexanderGruber If it was sisylana, I would be able to help you.

leo

01:04

@AlexanderGruber cool! He is probably sleeping

Alexander Gruber

@KarlKronenfeld sounds like the name of a succubus

@leo is it night where he lives?

agent154

@AlexanderGruber Sadly I have analysis questions too and nobody I know to ask.

Alexander Gruber

@agent154 i'm awful but you can try me

agent154

0

Q: How to conclusively prove the interior of a set

I'm fairly confident I understand the meaning of the 'interior' of a set, but I can't figure out how to prove conclusively what said interior is for a given set. Consider this definition: Let $S$ be a subset of $\mathbb{R}$. A point $x\in\mathbb{R}$ is an interior point of $S$ if there exists...

leo

@AlexanderGruber it is, not too late but yes. He said he was tired and sleepy

Alexander Gruber

01:06

i guess i'll have to ask it on main

leo

@agent154 do you mean prove that something is the interior of a set?

agent154

@leo I suppose... I'm asked to find the interior of a set, and I want to be able to show work as to why it is what I say it is without just writing an answer

This is part of a homework assignment

Enjoys Math

@agent154 You know that $Int(A\cup B) \supset Int(A) \cup Int(B)$

agent154

@EnjoysMath That was never brought up insofar as I know. My chapter on this subject never says that

Enjoys Math

First, before going on in your book

Karl Kronenfeld

01:09

@agent154 In your examples, you have to show that the interior of a given set is empty. In other words you have to show that any point $S$ is not in the interior of $S$. So for your first set, take any $\frac 1n$ and find the nearest points of $S$ to $\frac 1n$.

Enjoys Math

prove that the definition of interior is equivalent to $Int(A) = \bigcup_{U \subset A, U \ open } U$

agent154

@KarlKronenfeld The nearest point to $\frac{1}{n}$ is $\frac{1}{n+1}$

Karl Kronenfeld

@agent154 Ok, and on the other side?

agent154

$\frac{1}{n+1}<\frac{1}{n}<\frac{1}{n-1}$ Am I understanding your question?

Karl Kronenfeld

@agent154 Right (assume $n>1$ btw), now any neighborhood $U$ of $\frac 1n$ intersects the open set $I=(\frac1{n+1},\frac1{n-1})$ in an open set. But what is $U\cap I\cap S$?

[I am using the convention that all neighborhoods are open, the argument can be easily adapted to the other convention.]

leo

01:15

@agent154 your intuition is fine. And the how to prove it is fine as well. Indeed, if you want to prove that the interior of a set $S$ is empty, you have to prove that for any $x$ there are no nbghd containing $x$ which are contained in $S$. Or equivalently, you have to prove that any nbghd intersects the complement of $S$

pourjour

I was wondering how to conclude that $e_r = \cos \theta e_x + \sin \theta e_y$

agent154

@KarlKronenfeld Well, $I\cap S=\{\frac{1}{n+1},\frac{1}{n-1}\}$ and any neighborhood $U$ of $\frac{1}{n}$ with sufficiently small radius will give $U\cap\{\frac{1}{n+1},\frac{1}{n-1}\}=\emptyset$... or am I off base?

Wait -- $I\cap S$ isn't even that, since the endpoints are not in $I$.

$I\cap S=\{\frac{1}{n}\}$

Karl Kronenfeld

@agent154 Yep, what can you conclude about $U\cap I\cap S$, then?

agent154

Well, any $U$ will have $\frac{1}{n}$ in it right? Then shouldn't $U\cap I\cap S=\{\frac{1}{n}\}$?

Karl Kronenfeld

Yes.

However, if $U\subseteq S$, then $U\cap S\cap I$ would be open.

agent154

01:22

So that shows that $\frac{1}{n}$ is isolated?

pourjour

any help it's about cylindric ref

Karl Kronenfeld

@agent154 Indeed, there is no neighborhood of $\frac 1n$ contained in $S$ because if we assume to the contrary, we get a contradiction.

leo

@AlexanderGruber what is it about?

agent154

@KarlKronenfeld Here's another related question -- is the given definition of $S$ open, closed, neither, or both? The definition I have for an open set is one where $S=S^0$

Kevin Driscoll

@pourjour what do you want to start with?

Karl Kronenfeld

01:25

@agent154 I am not sure what you mean, are you asking whether $\{\frac1n:n\in\mathbb N\}$ open?

pourjour

@KevinDriscoll from the lhs

agent154

@KarlKronenfeld yes

I guess you'd need to know the interior first before making that conclusion?

Karl Kronenfeld

@agent154 Yes.

Kevin Driscoll

@pourjour I mean what do you want to take as already known about cylindrical coordinates?

agent154

@KarlKronenfeld I'm not sure I follow this reasoning... Maybe I'm missing some minor definition or detail in my already weak understanding.

pourjour

01:29

@KevinDriscoll take what ever you need I just need a basic proof

Karl Kronenfeld

@agent154 Yeah, I see what you mean. How do you define neighborhood?

There is an easy way out of this tangle using most definitions of neighborhoods.

agent154

@KarlKronenfeld The $N_{\epsilon}(x)$ neighborhood is $(x-\epsilon,x+\epsilon)$

Or let me copy from my text

Dave

hey

agent154

Let $x\in\mathbb{R}$ and let $\epsilon>0$. A neighborhood of $x$ is a set of the form $$N(x;\epsilon)=\{y\in\mathbb{R}:|x-y|<\epsilon\}.$$

Dave

anyone can check my linear algebra question ? trying to close that homework but not sure about that one math.stackexchange.com/questions/538807/…

Karl Kronenfeld

01:33

@agent154 Ok, so $U$ takes the form $(\frac1n-\varepsilon,\frac1n+\varepsilon)$. If $U\subseteq S$, then $U\cap I\subseteq S$. In other words, $U\cap I\cap S=U\cap I$. Now, just note that $U\cap I$ contains a neighborhood of $\frac 1n$ as well.

agent154

@Dave That's above my knowledge of Linear Algebra, sorry...

Dave

:\

Kevin Driscoll

@pourjour If the position in Cartesian coordinates is given by $\vec{r} = x \hat{x} + y \hat{y} + z\hat{z}$ then a vector which points in the x-direction is $\frac{\partial}{\partial x} \vec{r}$

@pourjour Which in this case is just $\hat{x}$

@pourjour you can generalize this to other coordinate systems in cylindrical $\vec{r} = r \cos{\theta} \hat{x} + r \sin{\theta} \hat{y}$ so a vector that points in the $r$ direction is given by $\cos{\theta} \hat{x} + \sin{\theta} \hat{y}$

@pourjour and we got lucky, its already normalized so it is in fac t aunit vector

agent154

@KarlKronenfeld I'm really not following, sorry. Let me see if I can explain what I don't get

So for $\frac{1}{n}$ to be in the interior of $S$, I just need to find a single neighborhood of $\frac{1}{n}$ that is a subset of $S$ right?

pourjour

@KevinDriscoll thanks :D

agent154

01:42

And $\frac{1}{n}$ itself is not a neighborhood, because we need a positive non-zero radius.

Karl Kronenfeld

@agent154 Right.

@agent154 You suppose for the sake a contradiction that $U$ is such a neighborhood.

Dave

@robjohn thanks !!

agent154

$U$ is what, exactly? Just any neighborhood such that the radius is positive?

Karl Kronenfeld

@agent154 It is a neighborhood of $\frac 1n$ contained in $S$.

We are showing that mere existence of $U$ leads to a contradiction.

agent154

@KarlKronenfeld OK, so I'm assuming that $\frac{1}{n}$ is in the interior

Karl Kronenfeld

01:45

@agent154 Yes, sorry I wasn't clear about that initially.

agent154

OK, so we have $U=(\frac{1}{n}-\epsilon,\frac{1}{n}+\epsilon)\subset S$.

I guess where I'm lost is why this implies that $U\cap I\subset S$

Karl Kronenfeld

Well, any point of the intersection of two sets is a point of both sets. So if $x\in U\cap I$, then $x\in U$. Moreover, $x\in S$ since $U\subseteq S$.

agent154

$(\frac{1}{n}-\epsilon,\frac{1}{n}+\epsilon)\cap(\frac{1}{n+1},\frac{1}{n-1})$ depends on $\epsilon$, no?

Karl Kronenfeld

@agent154 That's right.

agent154

OK, so then.. if $x\in U\cap I$, then $x\in U$ and $x\in I$...

robjohn

01:49

@Dave I made an additional suggestion regarding organization

agent154

OK... I'm following a bit now.. (My brain isn't good at skipping some steps)

Dave

@robjohn yeah I saw that. I also realised that I typed 2 times my first vector,

@robjohn what do you think about the part where I state how to get the under face of the cube, is that right or wrong, why do we need that if we have the other equations ?

agent154

So..

\begin{align}x\in U\cap I&\Rightarrow x\in U\\&\Rightarrow x\in S&&(\text{by assumption that $U\subset S$)}\end{align}

Karl Kronenfeld

Yep

agent154

OK, but we're not done right?

robjohn

01:52

@Dave isn't that covered in your other faces? I think he was just writing what you did with other vector notation

Karl Kronenfeld

Therefore $U\cap I\subset S$.

agent154

Wait - my set theory is a bit shakey... is that a bidirectional implication? $x\in U\cap I\iff x\in U$?

Dave

@robjohn ok, thanks a lot.

:)

Karl Kronenfeld

@agent154 No, the set of all points for which the backward implication does not hold is called the set difference $U\setminus I$.

robjohn

@Dave any time

agent154

01:55

OK, so $U\cap I\subset S$... So how do we get from here to the fact that $U\cap I\cap S$ is open?

Karl Kronenfeld

@agent154 Okay, we have to show $U\cap I$ is open.

Or rather, a neighborhood $\frac 1n$.

agent154

Well, it is open if and only if all of its contents are interior points right?

Is that what we're going to use?

Karl Kronenfeld

@agent154 Right, let's not do that though.

Actually, ignore both things I said there. lol

Okay, we have to show that $U\cap I$ contains a neighborhood of $\frac 1n$.

agent154

We assumed as much earlier on didn't we?

We assumed that $U\subset S$

Karl Kronenfeld

@agent154 The only relevant facts we know right now are that $U=(\frac 1n-\varepsilon,\frac1n+\varepsilon)$ for some $\varepsilon$ and the definition of $I$.

agent154

02:01

OK

Karl Kronenfeld

The way to show what we want to show is to first find an interval $V=(\frac1n-d,\frac1n+d)$ within $I$.

agent154

So let $U'=(\frac{1}{n}-\delta,\frac{1}{n}+\delta)$ for some $\delta\neq\epsilon$...

Karl Kronenfeld

@agent154 why?

pourjour

@KevinDriscoll thanks :D

The Substitute

Off topic: I don't think this question is serious enough to ask on the main site. If $A$ is a subset of a ring $R$ closed under addition and multiplication with elements of $R$ and $f(x)=A+x, x\in R$ is a homomorphism, why would the kernel of $f$ be $A$ instead of 0 or the empty set? Is this a convention when the output of the function is a set?

agent154

02:03

Well, I'm thinking if we need to show that $U\cap I$ contains a neighborhood of $\frac{1}{n}$, we can't assume our previous neighborhood is contained in $U\cap I$ can we?

That would be the case only if $U\subset I$

Karl Kronenfeld

Right. When $\varepsilon$ is small, just use $U$. When $\varepsilon$ is large, we need to construct a neighborhood contained in $I$ in the first place.

agent154

BRB, gonna get coffee

@KarlKronenfeld ok I'm back

Karl Kronenfeld

@agent154 There is a cute little corollary to what I said.

agent154

to which comment?

Karl Kronenfeld

We only need to resolve how small $\varepsilon$ needs to be for $U$ to be the wanted neighborhood.

(I was referring to my most recent comment by the way)

agent154

02:14

OK, so let $\varepsilon=\min(?)$

Karl Kronenfeld

Call it $\delta$, but the idea is that if $\varepsilon\le\delta$, then $U\subseteq I$.

We just have to determine a good value for $\delta$.

@agent154 So $\delta$ will be the minimum of two values, namely the distances from the endpoints of $I$.

agent154

So $I=(\frac{1}{n+1},\frac{1}{n-1})$

$\frac{1}{n}$ is definitely in the middle somewhere, but not right in the center. I'm trying to visualize geometrically where $1/n$ is relative to the endpoints of $I$

It's closer to $\frac{1}{n+1}$ than to the other side

Karl Kronenfeld

@agent154 That's true.

agent154

So let $\delta=|\frac{1}{n+1}-\frac{1}{n}|$?

Karl Kronenfeld

Yes.

To use another letter. Let $V$ be $N_\delta(\frac1n)$.

agent154

02:20

OK, so $N_{\delta}(\frac{1}{n})\subseteq I$ given the construction of $\delta$

OK, now where are we... scrolls up

So now we have $N_{\delta}(\frac{1}{n})\subseteq U\cap I$

Karl Kronenfeld

Not exactly. :) There are two cases, depending on the relationship between $\varepsilon$ and $\delta$.

agent154

I ought to write this down on paper... I keep forgetting details and getting lost in the mess of messages

Thanks for the help and patience by the way. Most people who try to help don't go into so much detail with me and get frustrated when I don't get their hints right away.

Karl Kronenfeld

@agent154 I am glad to help. It is absolutely necessary that you understand how to show the little steps. And at this stage, it is as important as those strategic guidelines that comprise hints.

agent154

I need to ask again - where did $I$ come from again?

Alexander Gruber

@leo it's asking for an alternative, direct proof of a relatively basic topology/analysis result that i have only seen done by contradiction

agent154

02:27

$(\frac{1}{n+1},\frac{1}{n-1})$

Alexander Gruber

it's probably not hard, i'm just not very strong in analysis and i can't think of a way to do it.

Karl Kronenfeld

@agent154 That's an interval such that $I\cap S=\{\frac1n\}$.

Enjoys Math

What's the question?

Alexander Gruber

mine?

Karl Kronenfeld

@agent154 The explicit form is helpful when constructing $\delta$, so we chose the largest possible $I$.

Enjoys Math

02:29

yes

yours, my friend

Alexander Gruber

theorem is from baby rudin:

a compact set E with infinitely many points contains a limit point.

Enjoys Math

What's the question?

lolcats, prove that?

Karl Kronenfeld

@AlexanderGruber How is compactness defined?

Alexander Gruber

@KarlKronenfeld every open cover has a finite subcover.

Enjoys Math

You can prove that using the finite intersection property that certain compact sets satisfy.

Let me go to my notes goes to notes

Fernando Martin

02:32

All compact sets satisfy the FIP

agent154

@KarlKronenfeld I need to go back and ask why we needed $I$ in the first place.

Fernando Martin

That's equivalent to being compact

Enjoys Math

There's a non-empty intersection involved is my point

Alexander Gruber

so the proof, generally, is, take a small enough open neighborhood around each of the points, so there's a finite subcover, so the set isn't infinite, contradiction.

i'm wondering if there's a direct/constructive proof

Karl Kronenfeld

@agent154 The key property is that $I\cap S$ does not contain a neighborhood.

agent154

02:33

OK... I'll just work until the end and work backwards, maybe I can fill in the missing pieces. I feel like I could go back and ask "why" about everything we've done so far so I'll hold off

Fernando Martin

@AlexanderGruber: Have you proved that compact sets have the Bolzano-Weierstrass property?

Karl Kronenfeld

@agent154 Do you remember/understand my "split into cases" message?

Fernando Martin

If you have, you can pick an infinite countable subset from your compact set

Alexander Gruber

@FernandoMartin well, the space isn't necessarily real

Fernando Martin

BW holds for compact metric spaces

Alexander Gruber

02:35

it's just a topological space

agent154

Let me try to summarize what we have from the start to see what I'm missing..

Fernando Martin

Does Baby Rudin work with topological spaces?

Enjoys Math

$X$ is compact iff for every coll $\mathscr{C}$ of closed sets in $X$ such that any finite intersection in the set is non-empty, we have that the intersection over the whole collection is non-empty.

Alexander Gruber

@FernandoMartin it's not just for euclidean spaces?

Enjoys Math

@Alexander use that

Fernando Martin

02:38

Bolzano-Weierstrass holds for compact metric spaces

it's actually equivalent to compactness

Alexander Gruber

oh i see

agent154

$S=\{\frac{1}{n}:n\in\mathbb{N}\}$. $N_{\epsilon}(\frac{1}{n})=U=(\frac{1}{n}-\epsilon,\frac{1}{n}+\epsilon)$. $I=(\frac{1}{n+1},\frac{1}{n-1})$. $U\cap I=\{\frac{1}{n}\}$. We assumed that $U\subset S$, so therefore $(U\cap I)\subset S$. We then constructed $V=N_{\delta}(\frac{1}{n})$ using $\delta=|\frac{1}{n+1}-\frac{1}{n}|$.

Fernando Martin

@AlexanderGruber: having the BW property is that every sequence has a convergent subsequence

That's sometimes called sequential compactness

agent154

@KarlKronenfeld Isn't it trivial that $(U\cap I)\subset S$ though? After all, $\{\frac{1}{n}\}\subset\{\frac{1}{n}:n\in\mathbb{N}\}$

Enjoys Math

@Alexander, are 1-pt sets closed in your hypothesis?

Karl Kronenfeld

02:40

@agent154 $I\cap S=\{\frac 1n\}$, not $U\cap I$.

Fernando Martin

In any case, if you already have that result proved, you can construct a limit point: just pick a sequence with non-repeating terms, since it has a convergent subsequence, there's a limit point

agent154

@KarlKronenfeld Well, $(\frac{1}{n}-\epsilon,\frac{1}{n}+\epsilon)\cap(\frac{1}{n+1},\frac{1}{n-1})=\{\frac{1}{n}\}$

Karl Kronenfeld

@agent154 no

agent154

Are we not using the fact that $\frac{1}{n}\in\mathbb{Q}$?

Karl Kronenfeld

@agent154 That is irrelevant.

agent154

02:44

OK, I just scrolled back and refreshed my memory... so $U\cap I\cap S=\{\frac{1}{n}\}$

Karl Kronenfeld

@agent154 That's the one.

@agent154 The next big claim is that either $V$ or $U$ is contained in $U\cap I$.

agent154

Well, isn't it sufficient to use just $V$ since it's the smallest?

There doesn't appear to be a need for two cases

If $U\subset(U\cap I)$, then definitely $V\subset(U\cap I)$

Karl Kronenfeld

$V$ is a subset of $I$ and not necessarily $U$.

This is based on the fact that $\delta$ was constructed independently of $\varepsilon$.

Argon

@robjohn Are you around?

agent154

OK

Karl Kronenfeld

02:50

@agent154 Now, recall that $U\cap I\cap S=U\cap I$.

And recall $U\cap I\cap S=\{\frac 1n\}$.

agent154

So $U\cap I=I\cap S=U\cap I\cap S=\{\frac{1}{n}\}$

Karl Kronenfeld

@agent154 I don't see the second equality.

Oh, of course.

agent154

Well, if $I\cap S=\{\frac{1}{n}\}$, and $U\cap I\cap S=\{\frac{1}{n}\}$,

Karl Kronenfeld

Yes, it's true. :)

agent154

OK, so onto the two cases then?

Karl Kronenfeld

02:54

What we have is that either $U$ or $V$ is contained in $U\cap I=U\cap I\cap S$, which contradicts the hypothesis that $U$ is a neighborhood and the fact that $V$ is a neighborhood.

agent154

So neither can actually be a neighborhood

Karl Kronenfeld

Well, we have our contradiction. That's the end of the proof.

agent154

Because $U\cap I\cap S=\{\frac{1}{n}\}$, and that is a singleton, so there is no subset other than itself and $\emptyset$?

Karl Kronenfeld

Right.

agent154

OK... this is indeed a lot to tie together. I'm gonna review it all and see if I can understand it.

Karl Kronenfeld

02:59

@agent154 I'd normally say to draw a picture, but this proof by contradiction can't all be pictured at once because the initial construction is impossible. However, there are two threads that can maybe be drawn in two separate pictures.

@agent154 I have to go. The other problem calculating the interior of the set of rationals has to be treated in a completely different (and in my opinion a less intuitive) way.

agent154

OK, thanks

Enjoys Math

03:15

$Int(A) = \bigcup_{O \ open, \ \ O\subset A} O$

$Cl(A) = \bigcap_{C \ closed, \ \ C \supset A} C$

Enjoys Math

03:28

Flight w Denzel Washington is on NF, ladies & gentlemen

leo

@AlexanderGruber Did you asked it on main?

robjohn

@Argon I am back

agent154

@EnjoysMath I saw that movie... heard it was an excellent film, though I wasn't as impressed as I expected to be. The ending sorta surprised me though.

Alexander Gruber

03:44

@leo not yet i'm workin on it myself first

leo

03:56

@AlexanderGruber did you mind if I ask you which result it is?

3

A: How to find $f$ if $f(f(x))=\frac{x+1}{x+2}$

If one is looking only for solutions that are fractional-linear, $f(z)=(az+b)/(cz+d)$, then the two that @sos440 has found are the only ones, as suggested by the interesting response of @i707107. Here’s a somewhat more elaborate response than that of @MatthewConroy, his answer really disposes of...

cool

agent154

Am I correct in calculating that $\{x:(x-3)^{2}\geq 1\}=(-\infty,2]\cup[4,\infty)$?

Alexander Gruber

04:43

@leo yeah, i mentioned it above. i'm looking for a direct proof that if $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.

in the book it's proved by contradiction, and the proof is clear and pretty easy, but i want to try a constructive-ish approach and see if i can make it work

robjohn

05:13

@agent154 yes

Zibadawa

05:26

@AlexanderGruber $E$ is not empty, so let $x_1\in E$. $E\setminus \{x_1\}$ is non-empty since $E$ is infinite, so let $x_2\in E$. You can thus iteratively define a (non-constant) sequence $\{x_n\}$ in $E$. Since this is also a sequence in $K$ and $K$ is compact, the sequence must have a limit point (it need not converge, but it has a limit point). That uses axiom of choice (the existence of each $x_i$ doesn't, but I'm pretty sure actually selecting it to build up the sequence does).

Well, I should say "never-repeating" or something instead of "non-constant". And compactness of $K$ forces the limit point to be in $K$, if that wasn't clear. The basic idea is that every infinite set contains a countably infinite subset. You can view any such set as a sequence through a choice of enumeration.

Mariano Suárez-Alvarez

05:44

Everyone should read Bass' Algebraic K-theory...

robjohn

@MarianoSuárez-Alvarez Do we need to understand it?

Alex Mardikian

Revised my question here:
http://math.stackexchange.com/questions/550691/asymptotic-solution-for-tn-6tn-4-n-lg-n
Would anyone mind taking a look at it? Its about giving a bound on a recurrence relation for a finite number of iterations.

Mariano Suárez-Alvarez

the first few 500 pages are amazingly accesible

2

it gets hard after that :-)

the other volumes I have not tried, tbh

robjohn

@MarianoSuárez-Alvarez The first few 500 pages? there are multiples of 500 pages?

Mariano Suárez-Alvarez

the first volume is almost 800

robjohn

05:47

Ah, that's better...

Mariano Suárez-Alvarez

it's typeset using typewriters, so there's lots of spacing, though

that anyone had the energy to typewrite 800 pages is amazing

the prime character is notoriously ugly

Vrouvrou

@Zibadawa hello

Zibadawa

hi

Vrouvrou

how are you ?

Zibadawa

@AlexMardikian I commented on the page as well, but could you explain why the Master Theorem doesn't seem to apply?

@Vrouvrou It's complicated. You?

Mariano Suárez-Alvarez

05:58

there are few things more annoying that findign that there are tons and tons of papers on a subject and no nontrivial examples

Zibadawa

Possibly because there are none! It's happened before.

Alex Mardikian

@Zibadawa I commented as to why it cannot be case 2 of the Master Theorem.

Mariano Suárez-Alvarez

@Zibadawa there are many, but no one cares to exhibit them

Alex Mardikian

Give me a minute for determine the other 2 cases

Zibadawa

I don't suppose your professor proved an even more masterful theorem?

Alex Mardikian

06:05

Can't be case 1 or 3 by definition, $f(n) = n \lg n$ cannot be $\Theta(n^c)$ for any $c \in \mathbb{R}$.

No, the master theorem in lecture was the same as the Wikipedia, which is why I am pretty confused about him telling me it is applicable to this $T(n)$.

Zibadawa

Well, then, it doesn't really sound like you can use it. The example wikipedia has for Case 2 suggests you can sometimes solve the recurrence relation directly.

That or maybe the professor had a typo in the problem, and he was so convinced he didn't have one that he didn't even really look at your solution and notice that the coefficients/powers weren't what he wanted.

leo

06:26

@AlexanderGruber Oh, I see. Tell me if you find something nice :-)

it's interesting to see what people can come with

user51547

07:54

anyone here?

robjohn

08:56

@user51547 yes ;-)

123kid

09:55

Off topic brain fart: I know that any polynomial of the form $x^n-1 $ can be factored as $(x-1)(x^{n-1}+...+1)$, yet what is the context in which it matters if n is prime?

Anthony Carapetis

@123kid: something to do with cyclotomic polynomials maybe?

123kid

10:08

@AnthonyCarapetis is it that $x^{n-1}+...+1$ will be irreducible iff n is prime. I think that's it.

Anthony Carapetis

@123kid: sounds right.

123kid

thanks for the guidance

robjohn

10:33

@123kid Yes. If $n$ is not prime $\frac{x^n-1}{x-1}$ is reducible, and I believe that Eisenstein shows that $\frac{(x+1)^n-1}{x}$ is irreducible if $n$ is prime

Gustavo Bandeira

@MarianoSuárez-Alvarez This one?

Anz Joy

11:08

Any body good at permutations??

Nick

12:50

Guys, I have a question. Does $x\in(0,1]$ mean that $x=1$ ?

and also what does the expression in the above image mean?

Antonio Vargas

13:26

@Nick, $x \in (0,1]$ means that $0 < x \leq 1$.

robjohn

@Nick No, it means that $0\lt x\le 1$

@AntonioVargas Good answer :-)

Antonio Vargas

@robjohn jinx

@Nick, the $\wedge$ symbol ("wedge") means "and".

Nick

Whole numbers

Antonio Vargas

What's W?

@Nick, Yeah, if $x$ is a whole number and $x \in (0,1]$ then it must be $1$.

Since $1$ is the only whole number in the interval $(0,1]$.

Nick

@Antonio: I thought the inverted cup from set theory (which stood for intersection) meant $and$

Antonio Vargas

13:33

They're related, but $\wedge$ is used in logical statements and $\cap$ is only used between two sets.

So you could say $x \in (0,1] \wedge x \in W \Rightarrow x = 1$, for instance.

Or you could say $(0,1] \cap W = \{1\}$

Nick

:D Wow, I feel like I'm learning the syntax of Math.

Antonio Vargas

heh yup

ok well I gotta go, cya

Nick

@AntonioVargas: Thank you.

Marienplatz

13:55

Hi folks!

I have a question about probability. What I don't understand is what the correlation between each person whose a probability of getting cholera is .15 and the other people in a population. The complete question is as follows. There is a research in one flooded region, that the probability of a man attached by cholera is .15. If the total citizen in that region is 640 persons, what is the number of person estimated attached by cholera?

user96977

how do i prove that a group homomorphism satisfies: H(a^-1) = H(a)^-1

user96977

i have already prove that H(1) = 1 because H(a)H(1) = H(a*1) = H(a)

Alex Mardikian

For Sylow p-subgroups

If $p^a m$ such that $p$ does not divide $m$ satisfies that $a = 1$, there are (p-1)n_p elements of order $p$ exactly right?

Is this not true for $a > 1$ generally?

user96977

@AlexMardikian hi, i notice you are doing group theory. can you please take a look at my question. i'm sorry that i cannot help with yours.

Alex Mardikian

If you let any element $g \in G$ for your group, you know that $g$ has and inverse $g^{-1}$

lets say your maps is $\phi : G \rightarrow H$

user96977

14:09

yes

Alex Mardikian

and the identity of $G$ is $1_G$ and similarly the identity of $H$ is $1_H$

You have that $\phi(1_G) = 1_H$

user96977

yes, but this is a result of the group homomorphism preserving multiplication...

user96977

it's not an axiom as such

Alex Mardikian

So $\phi(1_G) = \phi(gg^{-1}) = \phi(g)\phi(g^{-1}) = 1_H$

user96977

yes, ok

Alex Mardikian

14:11

as $\phi(g)\phi(g^{-1}) = 1_H$, it must be that \phi(g) and \phi(g^{-1}) are inverses of each other in $H$

user96977

ok

Alex Mardikian

Multiply both sides of $\phi(g)\phi(g^{-1}) = 1_H$ by $\phi(g)^{-1}$ on the left and your done

to cancel out the $\phi(g)$

Ignore when I said "as $\phi(g)\phi(g^{-1}) = 1_H$, it must be that \phi(g) and \phi(g^{-1}) are inverses of each other in $H$". Too late to delete.

user96977

hmm ok

user96977

i still think it should be possible without this sort of inference

user96977

thank you

Alex Mardikian

14:19

This just uses the fact that you already have shown that $\phi(1_G) = 1_H$ and that you have left cancellation in $H$.

user96977

so you say f(g)^-1 * f(g) * f(g^-1) = f(g^-1)

user96977

i'm still uncertain about how f(g)^-1 * f(g) = 1

Alex Mardikian

$f(g)$ is an element in $H$

and $f(g)^{-1}$ is its inverse in $H$.

just think about how $f(g) = h$ for some $h$ in $H$.

Since its an element in $H$, which is a group, it must have an inverse. So you just take $h^{-1}$ in $H$

user96977

ok

Alex Mardikian

Since $h = f(g)$, $h^{-1} = f(g)^{-1}$ and you have what you need.

Vrouvrou

16:24

hello

can someone help me

0

Q: Question on relative homology

I have this question and I'd like an idea to solve it: If $Z_p(X,Y)=\lbrace \sigma\in C_p(X), \partial\sigma\in C_{p-1}(Y)\rbrace$, $1)$prove that $H_p(X,Y)$ is isomorphic to $Z_p(X,Y)/(B_p(X)+C_p(Y))$, $2)$ deduce that $H_0(X,Y)$ is the free module generated by the path connected components o...

algebraic-topology homology

need help to prove the isomorphism

hhh

17:09

Do you know whether there is some general theory about ordering a multilinear function into a hypercube like the below?

Or theory about ordering a power-set into a hypercube?

user96977

17:30

if bab=a^-1, does that imply that (ba)^2 = 1

Zibadawa

Multiply all sides of the first identity by a on the right

or left, either way

Nimza

17:44

Hi, is there a name for transformations of differential operators of the type $D \mapsto gDg^{-1}$, where $g$ is a smooth nonvanishing function?

pourjour

18:15

how can I write $e_r$ using $e_x and e_y$

Charlie

18:27

@pourjour hi

pourjour

@Charlie hi mari how are u doing?

Charlie

@pourjour I'm fine, and you???

pourjour

@Charlie fine too thanks

Charlie

@pourjour are you working hard?

Nimza

@Charlie bonsoir!)

Charlie

18:34

@Nimza olá!

@Nimza how are you?

Nimza

@Charlie norm, I'm at work now, and you?

Charlie

@Nimza I'm fine

@Nimza I don't know if I should study algebra or try to program :(

Nimza

@Charlie do you have a choice?

Charlie

@Nimza I can choose

Nimza

@Charlie I would choose algebra, of course

Charlie

18:41

@Nimza I'm tempeted to choose it too :/

Sohaib I

18:57

Can we find the inverse of non square matrices

?

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$Mathematics$ Mathematics