Mathematics - 2014-03-14 (page 1 of 5)

Pedro Tamaroff

12:00 AM

STOP.

STOP. STOP.

Symbols are useless now.

usukidoll

What are you talking about?

Pedro Tamaroff

If $R$ is a relation on a set $S$, then $RR$ is defined as the set of pairs $(x,y)$ for which, for some $z\in S$; $(x,z),(z,y)\in R$.

The question is, does $(x,y)\in RR$ imply $(x,y)\in R$.

This holds when $R$ is transitive. For $(x,y)\in RR$ means exactly $(x,z),(z,y)\in R$ for some $z$.

usukidoll

my book doesn't write it as $RR$ it wrote it as a composite relation

Pedro Tamaroff

Same thing.

usukidoll

ok so how do I prove the transitive part

Alraxite

12:03 AM

$(x,y)\in R\circ R$. Then by definition $(x,v)\in R$ and $(v,y)\in R$ for some $v\in S$...

Pedro Tamaroff

Since $R$ is transitive, $(x,z),(z,y)\in R$ implies $(x,y)\in R$.

Alraxite

and we know R is transitive..

Pedro Tamaroff

This concludes the proof: $(x,y)\in RR\implies (x,y)\in R$ when $R$ is transitive.

usukidoll

so I didn't need to expand to go all the way to $(x,z)$?

Pedro Tamaroff

I don't know what you mean by that.

usukidoll

12:05 AM

Definition 6.2.3 states that $R$ is transitive if $( \forall x, y,z \in S)((x,y) \in R \land (y,z) \in R) \rightarrow (x,z) \in R)$

my previous homework was prove that R is transitive and the end result became (x,z) which I had no idea how to get there

ugh I'm making this hard for myself... should've subsituted v = z for def 6.3.9 and I could see it right there

$R_2 \circ R_1 =[(x,y) \in S \times S :(\exists z \in S)((x,z) \in R_1 \land (z,y) \in R_2$.
$R \circ R =[(x,y) \in S \times S :(\exists z \in S)((x,z) \in R \land (z,y) \in R$.

it is transitive... seen x y z on here and in the transitive def

Pedro Tamaroff

@seaturtles I guess I have an alternative to Rotman's solution.

Alraxite

@usukidoll What exactly are you trying to prove?

and what is the problem you're facing?

usukidoll

this
http://math.stackexchange.com/questions/711529/r-is-transitive-if-and-only-if-r-circ-r-subseteq-r

Alraxite

Okay. Can you prove $R$ is transitive $\implies R\circ R\subseteq R$?

usukidoll

I could try

Definition 6.2.3 states that $R$ is transitive if $( \forall x, y,z \in S)((x,y) \in R \land (y,z) \in R) \rightarrow (x,z) \in R)$

SO for $R$ being transitive implies that $ R \circ R \subseteq R$

Alraxite

12:14 AM

Usually to prove $A\subseteq B$, you would assume $x\in A$ then show $x\in B$.

usukidoll

Definition 6.3.9 states that we let $R_1$ and $R_2$ be relations on a set $S$. The composition of $R_2$ with $R_1$ is the relation $R_2 \circ R_1 =[(x,y) \in S \times S :(\exists v \in S)((x,v) \in R_1 \land (v,y) \in R_2$.

For this problem :$R \circ R =[(x,y) \in S \times S :(\exists v \in S)((x,v) \in R \land (v,y) \in R$.
Definition 6.2.9 states that we let $R$ be an equivalence relation on a set $S$. For each element $x \in S $ the set $[x]=[y \in S: (x,y) \in R$ is the equivalence class with respect to $R$.

Alraxite

In your case, let $(x,y)\in R\circ R$ and then try to prove $(x,y)\in R$.

usukidoll

x belongs to $ R \circ R$ and x belongs to $R$ by def 3.1.2
$[x:x \in R \circ R \rightarrow x \in R$

Alraxite

Well, if it is 'obvious' from the definition then sure.

usukidoll

yesss

now how to put transitivity into the picture because that's just without transitivity

maybe there exists some z for $ R \circ R$!?

$R_2 \circ R_1 =[(x,y) \in S \times S :(\exists z \in S)((x,z) \in R_1 \land (z,y) \in R_2$

Alraxite

12:17 AM

But usually, you would go like this: Let $(x,y)\in R\circ R$. Then $(x,z)\in R$ and $(z,y)\in R$ for some $z$. Since $R$ is transitive, $(x,y)\in R$.

And since $(x,y)$ was an arbitrary element of $R\circ R$, $R\circ R \subseteq R$.

usukidoll

sigh if only transitivity wasn't involved, I would've gotten it already. you made it look easy

oh now I'm seeing it... but that's for if we're starting at $R \circ R \subseteq R$ which means that there are x elements that belong to $R \circ R$ and $R$. Those sets must have some elements in common...

IvanMatala

Hello:)

Alraxite

Hi!

IvanMatala

can we simplify this even further such that n is outside? puu.sh/7upAp.png

can we factor the exponent n and make it outside

Daniel Fischer

@Prototank What is $\lvert z^6\rvert$ for $\lvert z\rvert = 2$, and what is a bound for $\lvert z^6 - p(z)\rvert$ there?

Pedro Tamaroff

12:32 AM

@DanielFischer Hilfe!

Daniel Fischer

@PedroTamaroff Who's bullying you?

Pedro Tamaroff

@DanielFischer Rotman.

He's saying crazy things.

Daniel Fischer

@PedroTamaroff Didn't I tell you not to listen to algebraists?

Pedro Tamaroff

@DanielFischer I forgot.

Damn it.

It's Little Red Riding Hood and the Big Bad Wolf all over again.

Consider the following.

Daniel Fischer

So, @Pedro, I saw above that he's saying even is odd. Anything else?

Pedro Tamaroff

12:34 AM

Nope.

But I need to find a proof of his claim.

Daniel Fischer

Vot Klaim?

Pedro Tamaroff

So, consider the case $t=-1+2^{m-1}$

Oh.

Let $G$ be a group. Let $x,y$ be elements of $G$, with $x$ of order $2^m$; $y^2=x^{2^r}$, $r<m$ (of course?) and $x^y=x^t$.

Then of course $x=x^{y^2}=x^{t^2}$.

So $t^2=1\mod 2^m$ and $t=\pm 1,\pm 1+2^{m-1}$.

Rotman claims that if $t=\pm 1+2^{m-1}$ then $G$ contains at least two involutions.

$x^{2^{m-1}}$ of course is one.

Daniel Fischer

@PedroTamaroff $x^y$ was $y^{-1}xy$?

Pedro Tamaroff

@DanielFischer Ja.

No.

$yxy^{-1}$.

Sorry.

$(x^y)^z=x^{zy}$. Tough luck. =P

Yeah.

Anthony

Hey when we say that things like the fourier basis are complete, is that the same kind of completeness as when we talk about a metric space?

Pedro Tamaroff

12:38 AM

@Anthony No.

Daniel Fischer

@PedroTamaroff Urk. I'm not particularly fond of the $x^g$ notation anyway, but $x^{yz} = (x^z)^y$ is YUCK!!!

Pedro Tamaroff

@DanielFischer Hehehe, I know.

Will not do it again in your presence.

I swear.

@Anthony An orthogonal system is complete if $\langle x,0\rangle=0$ for every $x$ in the system implies $x=0$.

For example, for continuous functions (or pushing it a little further, Lebesgue integrable functions) over $[0,2\pi]$, $\langle e^{int}:n\in\Bbb Z\rangle$ is complete,.

@DanielFischer So, the point is to look at $x^ky$ for some $k$.

And show that is an involution for an appropriate $k$.

When $t=2^{m-1}+1$, things work out nicely.

But when $t=2^{m-1}-1$; we get wrong claims by Mr. Rotman.

Daniel Fischer

@PedroTamaroff Okay, slowly. So $x^kyx^ky = x^kx^{tk}y^2$.

Anthony

$\langle x,0\rangle=0$ for every $x$?

Pedro Tamaroff

@DanielFischer Indeedz.

Anthony

12:42 AM

Shouldn't that always be $0$?

Pedro Tamaroff

@Anthony Sorry, I meant $\langle x,s\rangle =0$ for every $s$ in the system. =)

My bad.

Anthony

phew

Pedro Tamaroff

If that implies $x=0$; then $S$ is complete.

@DanielFischer That gives, $k+tk+2^r=0\mod 2^m$.

Anthony

So I mean, how do we go about making things like the fourier basis?

Daniel Fischer

And that's $x^{k(1+t)+ 2^{r}}$.

Pedro Tamaroff

12:43 AM

That's what we want to happen.

@DanielFischer Yiss.

When $t+1=2^{m-1}$, we want $k2^{m-1}+2^r=0\mod 2^m$.

Anthony

It seems like a large, perfect, package, I don't even know how to go about understanding how it came to be, or how we know that it's complete while something like the polynomials (I think?) aren't.

Daniel Fischer

@PedroTamaroff Which is fine if $r = m-1$.

If not, not.

Pedro Tamaroff

@DanielFischer Sure.

But if $r$ is smaller we done goofed.

Anthony

I guess you just take it, show they're orthogonal, then show that condition you gave... Just such a large thing.

Pedro Tamaroff

@Anthony Polynomials are not orthogonal.

Well, you need another weight.

Anthony

12:46 AM

I see.

Pedro Tamaroff

But according to Weiertrass, $\langle x^n\rangle$ is complete.

Anthony

:o

Pedro Tamaroff

For continuous functions.

And then thus for every Lebesgue integrable function by density. Over compacts subsets of $\Bbb R$.

Daniel Fischer

@PedroTamaroff So let's try $x^ky^{-1}$.

Pedro Tamaroff

@DanielFischer Let's a-see.

Anthony

12:47 AM

Will projecting onto that give you something different than a Taylor expansion?

Pedro Tamaroff

@Anthony Come again?

Oh.

You need different weights, Anthony.

To make the inner product orthogonal.

Else there is no such thing as projections and whatnot.

@DanielFischer That squared looks promising, I guess.

Daniel Fischer

$x^ky^{-1}x^ky^{-1} = x^k y^{-2}yx^ky^{-1} = x^k x^{-2^r}x^{tk}= x^{(1+t)k-2^r}$

Nope. Not that.

Pedro Tamaroff

@DanielFischer Doesn't work, right?

Still $2^r$ ruining things.

Anthony

I understand that I need different weights-but in the end, you'll have some expansion onto the basic of $\langle x^n\rangle$ right? Do those coefficients agree with Taylor expansions?

Pedro Tamaroff

You see, $t-1=2^{m-1}$ works because we can divide by $2^r$ and get an odd leading term in $ak+b=0\mod 2^{m-r}$.

usukidoll

12:56 AM

finally done with this *** ***

Anthony

@PedroTamaroff?

Pedro Tamaroff

@Anthony Recall that if $f(x)=\sum a_nx^n$ and if $x$ converges in some disk $|x|<r$ then you can show $a_n=f^{(n)}(0)/n!$.

Daniel Fischer

Hmm, $y^ax^k = x^{kt^a}y^a$, so $x^ky^ax^ky^a = x^{k(1+t^a)}y^{2a} = x^{k(1+t^a)+ 2^ra}$.

Pedro Tamaroff

@DanielFischer Right, but then $a$ cycles $y$.

So maybe we're just ending up with $x^{2^{m-1}}$ again-.

Daniel Fischer

@PedroTamaroff Yes, maybe. the question is whether we can avoid that.

If $a$ is even, then $t^a \equiv 1 \pmod{2^m}$, so we have $2k + 2^ra \equiv 1 \pmod{2^m}$ to get.

Anthony

1:01 AM

$x$ converges, or $f(x)$?

Daniel Fischer

Which looks a tad impossible.

So $a$ would have to be odd, and we're back to square one because $t^a \equiv t$ then.

Anthony

And yes! So Taylor series do represent a projection onto the polynomials then? Just maybe not for divergent functions?

Pedro Tamaroff

@DanielFischer This is doomed.

I wonder if there is an errata by Rotman.

@KarlKronenfeld

Karl Kronenfeld

@PedroTamaroff ey

Pedro Tamaroff

What @DanielFischer and I are discussing.

Karl Kronenfeld

1:05 AM

lol, a bunch of symbols?

Daniel Fischer

@PedroTamaroff I'm almost sure the singular is erratum.

Pedro Tamaroff

@DanielFischer Well, I just mean a list of corrections, which include this one. =P

Daniel Fischer

Ah, a list of errata.

Karl Kronenfeld

@DanielFischer I had to look this one up once. Erratum is the singular, yes, referring to a single error.

How do you count the number of monomials of degree $d$ in $k[x_1,\dots,x_n]$?

Pedro Tamaroff

@KarlKronenfeld By degree you mean...?

Karl Kronenfeld

1:11 AM

Usual definition, sum of powers.

Pedro Tamaroff

So $x_1^2x_2+x_1$ has degree...? =)

Karl Kronenfeld

monomials, bro

Lusinghiero

I hate algebra

:)

Pedro Tamaroff

@KarlKronenfeld Oh.

Derpacola.

Karl Kronenfeld

It's just a combinatorics problem hidden in this setting. I don't know how to translate it though.

Pedro Tamaroff

1:13 AM

@KarlKronenfeld I don't get it. You just mean elements of the form $a x_i^j$ for $a\in k$ and $j>0$?

Karl Kronenfeld

@PedroTamaroff Monomials are typically assumed to be monic.

Daniel Fischer

@PedroTamaroff $x_1^3x_2^5x_3x_4^{12}$

Pedro Tamaroff

Oh.

Sorry.

Karl Kronenfeld

I'm thinking we are counting functions from the partitions of $d$ to $[n]$.

Pedro Tamaroff

My bad.

@KarlKronenfeld I agree.

Assing to each $i=1,\ldots,n$ a number $d_i$ such that $\sum d=n$.

Karl Kronenfeld

1:14 AM

But we have to account for some symmetries

Pedro Tamaroff

Then form $x^d=\prod x_i^{d_i}$.

@KarlKronenfeld Example?

We just need to find out what to divide by.

Karl Kronenfeld

It's just that I know the answer, which isn't a power of n

Pedro Tamaroff

Oh, well.

@KarlKronenfeld What's the answer?

You could've started with that! =D

Karl Kronenfeld

@PedroTamaroff What are you saying here?

Pedro Tamaroff

@KarlKronenfeld $d_1+d_2+\cdots +d_n=d$ I meant.

Karl Kronenfeld

1:16 AM

@PedroTamaroff $\binom{n-1+d}{n-1}$

Pedro Tamaroff

That gives a monomial $x_1^{d_1}\cdots x_n^{d_n}$ of degree $d$.

@KarlKronenfeld Aha. Stars and bars! =D

@KarlKronenfeld I learnt it has balls and bins. Your bins are $x_1,\ldots,x_n$. You have $d$ balls.

When you finish, you get a monomial.

Of course the balls are indistinguishable, but the bins are.

Karl Kronenfeld

Yes, right.

Cool thanks

Pedro Tamaroff

1:31 AM

@DanielFischer If you want something a bit more itneresting than the above, that is used to show that if a $p$-group as a unique subgroup of order $p$, it is either cyclic or a generalized quaternion.

Which is quite cool.

Daniel Fischer

Aha, @Pedro, why is it cool?

Pedro Tamaroff

@DanielFischer Well, it shows say the importance of involutions? (2-groups).

The case $p$ odd is quite easy though.

The tough work is for $p=2$.

Daniel Fischer

As usual. $2$ is an odd prime.

Pedro Tamaroff

Indeed. =)

Rotman says "It is not unusual that the prime $2$ behaves differently than odd primes."

Daniel Fischer

Tru dat.

2

Julien

1:35 AM

Hi, someone has a reference for the foliowing Theorem of Chudnovsky : let $f : I \to \mathbb{R}$ a continuous function defined over a segment $I = [a;b]$ wich does not containing integers. Then there exists a sequence $(P_n)_{n \in \mathbb{N}}$ of polynomials with integer coefficients converge uniformly to $f$ over $I$.

I searched but I did not find demo on the internet

Daniel Fischer

1:45 AM

@PedroTamaroff Fancy idea. $G$ has at least two involutions if $t \equiv 1 \pmod{2^{m-1}}$, not if $t \equiv 2^{m-1}\pm 1\pmod{2^m}$.

Alexander Gruber

@DanielFischer i use that a lottttt

@PedroTamaroff my advisor in texas told me that group theory is the study of what makes the number 2 special.

@DanielFischer for example the Sylow $p$-subgroups of a Frobenius group have unique elements of order $p$

Daniel Fischer

@AlexanderGruber Good. What was a Frobenius group, again?

Alexander Gruber

@DanielFischer they're a type of finite group that come up a lot, definitions here

basically you've got a nilpotent group $K$ being acted on by a group $H$ which fixes only the identity, so that $G=K\rtimes H$

(what i just said uses a lot of results and is by no means a definition, but it's the best way to think about them)

so for example you could have $C_q\rtimes C_p$ with $p\mid q-1$, with an appropriate action that doesn't fix anything nontrivial

Daniel Fischer

Ignoring the fact that I know basically nothing about group theory, that sounds interesting. How well are Frobenius groups understood, cool open problems?

Alexander Gruber

@DanielFischer a lot is known but they come up everywhere so they're part of many open problems

skullpatrol

1:59 AM

If group theory is the study of what makes the number 2 special,
then calculus is the study of what makes the number 0 special :D

Alexander Gruber

my work involves lattices of Frobenius groups (and $2$-Frobenius groups, which are a slight variation)

Daniel Fischer

Lattices as in meet and join, or which kind of lattices?

Alexander Gruber

@DanielFischer more like formations of them

there's something called a prime graph of a finite group, you take the primes dividing its order as vertices, and put an edge between $p$ and $q$ if there's an element of order $pq$

it turns out that in solvable groups, the nonedges correspond to Frobenius (or $2$-Frobenius) Hall subgroups, so i started orienting the edges of the complement according to who's acting on what

anyway, essentially you end up with a directed graph showing how a bunch of Frobenius groups interact, that is what I mean by lattices/formations

Daniel Fischer

Interesting. Thanks. I won't pretend I understand what you do, though.

(I wish I did)

Alexander Gruber

@DanielFischer it's a little involved... the idea is actually pretty simple but there are a lot of definitions so it's kind of a long story

Daniel Fischer

2:07 AM

@AlexanderGruber And I have barely read the first lines of the introduction.

Alexander Gruber

@DanielFischer usually i can explain it pretty well with a chalkboard, like most things involving graphs it's a little obtuse if you can't draw it.

Daniel Fischer

@AlexanderGruber Everything can be explained better with chalk and blackboard.

Mike

2:26 AM

hello folks

Karl Kronenfeld

hello @Mike

Pedro Tamaroff

2:49 AM

@Alexander damn it

Im going to sleep now

Alexander Gruber

@PedroTamaroff i will not be for a long time.

@Mike yo

Mike

yo @alexander

Pedro Tamaroff

@Alex what side of the us are you Near? Ill be visiting NJ and NY on august

Mike

He's in FL

bottom right

It's here

Alexander Gruber

@PedroTamaroff i'm in the south.

i'm from Ohio, which is about 8ish hours driving from NY.

but now I live in FL, which is a long way.

Mike

2:56 AM

would you be offended if someone sawed off florida @alexander

Alexander Gruber

@Mike i would not

we could succeed as an island nation

Karl Kronenfeld

I wonder where florida is headed in the gif.

Alexander Gruber

@KarlKronenfeld we're going to war with venezuela

@Mike someone should put the locations of all the chat regulars on a map.

Karl Kronenfeld

@AlexanderGruber 21st century war: the whole population comes to you at a glacial pace. I like it.

@AlexanderGruber Mike's assassins would like that.

Mike

@AlexanderGruber I'd put mine on there if you made it interactive

I wonder if you could set it up so you could sign in with your SE account

and only you could put it in?

Alexander Gruber

3:06 AM

@Mike like on google maps or something?

Mike

yeah

@PedroTamaroff

Ted Shifrin

Howdy all ... I see @Mike made it back safely and is assiduously avoiding his grading

Mike

I'm tutoring @TedShifrin

Alexander Gruber

@Mike

Ted Shifrin

Chat-tutoring, eh? Talented :)

Hi @Alex

robjohn

3:15 AM

@TedShifrin or is he tutoring on chat?

Alexander Gruber

@TedShifrin hi there

Ted Shifrin

@robjohn, he purports to be making money whilst so engaged.

Mike

@TedShifrin I think I sort of agree with the school about the rejected candidate. Her requests showed she does not want to be a teacher. But the school she wants to work at is a teaching school.

robjohn

@TedShifrin oh, then I was wrong

Mike

I am certainly not chatting when I'm tutoring :)

Ted Shifrin

3:17 AM

It's a tough call, @Mike. Even at small teaching-oriented schools there is research pressure. I think t

Alexander Gruber

i wonder how hard it'd be to start a private college

Ted Shifrin

I thought an offer was an offer, so that bothers me. She should have done negotiating on the phone, would be general consensus. Email is not for negotiating.

Be very, very rich, @Alex

Alexander Gruber

@TedShifrin maybe i can meet a few rich people

Mike

@Ted I also feel uncomfortable about rescinding an offer, but I understand why the school did so - she was showing that she wanted to spend less time teaching and more time researching. This is a noble goal at a school that values research more than teaching, but that school is focused on teaching.

(Also, I only talk here when people aren't asking questions - when they are, they are my focus.)

@AlexanderGruber how do I add my name?

Alexander Gruber

@Mike i guess search for your location, add a marker, and change it to your name

i just mussed with it till it worked

Ted Shifrin

3:27 AM

I guess I'm bothered that the candidate didn't bring this up during her campus visit, @Mike. On the other hand, I'm skeptical about reading too much in: When I interviewed at universities, faculty were prone to take my expressed devotion to teaching as anti-research, which I certainly was not.

Mike

@TedShifrin I see what you mean, but at the same time, those are very much that (barring the first two): fewer classes, a pre-tenure sabbatical...

Those are very directly "I want to teach less and research more".

@alexander i'm in

Alexander Gruber

@Mike all riiiight, nice

Ram

Hi All. How to write a 2 complex in form of vertices. I mean like if X is a 2 complex with ordering on vertices as in page 6 of the link math.unl.edu/~mbrittenham2/classwk/872s07/lecnotes/…

then why A_i = [z, w_i, v_i] not [z, v_i, w_i]

Alexander Gruber

we got Karl in there too

Mike

he dares show his face?

i bet he's trying to trick us. i bet he's really on the opposite side of the world.

so reflect through the center of the earth...

Karl Kronenfeld

3:34 AM

I added a note.

Uncomfortable with giving exact location, so I aimed at a diner in Hell, MI. '

Mike

yeah, I put mine a bit away from where I actually am

Alexander Gruber

@Mike mine's just a nearby intersection

Mike

i was trying to find the lat/long so i could find the opposite point

Alexander Gruber

should we sidebar the map for others?

Karl Kronenfeld

Map of locations of chat regulars.

2

Alexander Gruber

3:38 AM

let's include instructions also:

MSE chat dwellers: pin your location (just for fun) [instructions]

12

Ram

Hi, could some body please answer my question :(

Alexander Gruber

@Ram sorry man, I don't know.

Ram

@AlexanderGruber Thanks

Mike

@Pedro "Punto 3", ya weenie?

Pedro Tamaroff

Yes

Im on my phone it is jot easy

Mike

3:50 AM

I'm putting your name in

Pedro Tamaroff

Im the south star.

Tehehehe

Im off to sleep now.

Sawarnik

4:14 AM

@r9m I wont talk to you ever. :Plum

Mike

@Karl we've both got cool icons.

Alexander Gruber

@Mike it's more customizable than you'd think huh

Mike

yeah

Sawarnik

@AlexanderGruber How do I add myself in the map?

Mike

Hit the little pin button below the search bar

To the right of the hand

Karl Kronenfeld

4:26 AM

@Mike Sure not a star, but the next best thing.

Alexander Gruber

@Sawarnik add a pin with your location and name it after yourself

Karl Kronenfeld

@Sawarnik You're logged in (with a google account) there, right?

Sawarnik

Yes, ok.

Look where I am now?

Mike

i'd prefer that you actually put roughly where you are

why did we lose all our icons? :(

Karl Kronenfeld

what happened to my skull and cross bones?

Alexander Gruber

4:30 AM

@Mike idk weird

Mike

i'm putting mine back in

Alexander Gruber

somebody must have been messing with the styles

Sawarnik

@Mike happy now

Mike

mine's gone again.

i'm annoyed.

Alexander Gruber

4:54 AM

Map instructions: click the pin icon under the search bar, click on your location, then name it after yourself. Alternatively, search for your location, click the pin, click add to map, then rename it.

- it doesn't have to be your exact address, just approximate location
- to change the style of your pin, hover your mouse over your name and click the paint bucket. (Do not change the layer style away from "individual styles"!)
- Please don't mess with other people's pins or add random crap to the map.

Karl Kronenfeld

5:06 AM

$$[h$]$

wow annoying

Alexander Gruber

5:25 AM

i want this book so bad

ccorn

@AlexanderGruber Yay, hardcover. I wish all the books I've bought were hardcover.

Alexander Gruber

@ccorn i wish it wasn't so expensive

Karl Kronenfeld

Jeez, down from 190

Mike

one of my friends is falling for an MLM. :/

ccorn

@AlexanderGruber: Is there also a map for chat-nonregulars? :-)

Alexander Gruber

5:39 AM

@ccorn haha, go for it

ccorn

@AlexanderGruber That seems to need a Google account. Bummer.

Karl Kronenfeld

@ccorn It's been done before. populationlabs.com/maps/World_Population_Map.png

Alexander Gruber

@KarlKronenfeld idk, i think i see me on there.

Karl Kronenfeld

:P

eXtremiity

The next book I'll buy for casual reading will be amazon.com/Perfect-Rigor-Mathematical-Breakthrough-Century/dp/…

I'm quite fascinated by Grigori Perelman.

Karl Kronenfeld

5:44 AM

Btw, a map of extraterrestrial chat-nonregulars: boallen.com/assets/images/randbitmap_true.png

Mike

I suggest this book instead

eXtremiity

I never read fictions anymore. Always non-fiction. I leave "fiction" to my maths :) .

Mike

then read borges's selected non-fiction

eXtremiity

Fantastic. I'll look into it.

Alexander Gruber

reading math all the time really does deplete one's ability to read fiction

Mike

5:48 AM

i'm going to start reading more nonfiction this summer.

one book i've been meaning to read is "wittgenstein's mistress".

Karl Kronenfeld

@Mike Way too inexpensive. Can you suggest reasonably pricy alternatives?

Mike

@Karl What's your minimum?

Alexander Gruber

it's hard to move from super dense math texts to reading linearly through a story

Mike

the solution is to read highly nonlinear postmodern books

eXtremiity

Fair point.

Karl Kronenfeld

5:52 AM

@Mike That kind of thing would actually be enjoyable to read.

Mike

@Karl I'm super into that kind of literature, I can recommend some of the best if you want.

Karl Kronenfeld

@Mike Sure

Alexander Gruber

do you like poetry?

Karl Kronenfeld

Not really, I have also not given it much of a chance though.

Mike

@karl thomas pynchon is one of the original big names. "V." is one of his best, though not his easiest, books. "the crying of lot 49" is a bit easier and gives you a taste of his style so you don't read V. and not enjoy it

one of the lighter, less difficult, but still very good and very interesting is david foster wallace's "infinite jest". this book was big for a while.

my favorite postmodern author is william gaddis. "J.R." is an awesome story told in roughly 99% dialogue

eXtremiity

5:56 AM

Added myself to the google "pin your location".

Karl Kronenfeld

@Mike Cool thanks

eXtremiity

So far away from you all. ^_^

Mike

@karl of those i suggest checking out lot 49 or infinte jest first.

Karl Kronenfeld

ok