Mathematics - 2013-03-30 (page 1 of 4)

Hi. Off topic question: Is $(a_1,a_2, \ldots ,a_k)$ countable if $a_i \in \mathbb{Z}$ AND $ k\in \mathbb{Z}$ ? On one hand, I think the answer is no since it seems to imply that the reals are countable. On the other hand, a countable union of countable sets is countable.

Peter Tamaroff

@GitGud You use idioms I don't know, that's the thing.

amWhy

@JasperLoy oh, it was 2 unupvotes on questions...

Peter Tamaroff

@TheSubstitute It is countable.

user19161

@amWhy No, that is not what your rep history says Amy.

Peter Tamaroff

@TheSubstitute Why would it imply the reals are countable?

Git Gud

12:18 AM

@PeterTamaroff I meant everything literally, no idoms used.

Peter Tamaroff

@GitGud You mean South Korean girls will be able to escape?

Git Gud

No, I mean they will be trying to escape.

Peter Tamaroff

@TheSubstitute That is $\Bbb Z^k$. $\Bbb R$ is something totally different.

@GitGud Isn't NK the shitty one?

user19161

Anyway, my rep does not matter. I am deleting my account soon.

amWhy

@JasperLoy two unupvotes from questions, I see my answer which (last night) was upvoted, then unupvoted, then reupvoted!

Git Gud

12:20 AM

@PeterTamaroff Yes. Why am I not making sense?I'll try to explain:

amWhy

I might too...

user19161

@amWhy No, that is not what your rep history says.

Git Gud

@PeterTamaroff NK and SK got at it, SK people (including girls) run away from there and go to other countries. It doesn't matter who the bad guys are.

The Substitute

@PeterTamaroff, got it.

Peter Tamaroff

@GitGud Oh, well. Buy why only girls?

@TheSubstitute Please, what made you think that implies $\Bbb R$ is countable?

Git Gud

12:22 AM

@PeterTamaroff Not only girls, of course. But those are the ones I'm interested in.

Peter Tamaroff

@GitGud You like SK girls?

Git Gud

@PeterTamaroff Of course!

Peter Tamaroff

They aren't pretty in general I think...

Git Gud

Enjoy: youtube.com/watch?v=yGO84ZoYXts

amWhy

@JasperLoy Well, I saw it with mine own eyes! Maybe the second question was (downvoted/un-upvoted yesterday...On two of the unupvoted questions, I then received upvotes, even though I didn't edit.

user19161

12:24 AM

@amWhy I think you need to read carefully, that is not what you described.

amWhy

@Jasper, I also had two downvoted answers, but made up for the lost rep with upvotes.

@JasperLoy Whatever, Jasper, I'm just saying something weird was/is happening, but I'm not going to let it get to me.

Peter Tamaroff

@GitGud Doesn't make the cut.

Git Gud

@PeterTamaroff More left for me! :P

user19161

Hahahaha, this is so weird...

Peter Tamaroff

@GitGud Come to Argentina

user19161

12:27 AM

Someone is playing with my rep...

user19161

You know, the people on this site keep amazing me...

Git Gud

@PeterTamaroff That's not good for my health.

amWhy

Some of us play, in a good way ;-)

user19161

No, this is not good.

user19161

I am not playing games with these people.

user19161

12:29 AM

But they might be playing with me.

amWhy

@JasperLoy No I'm not saying loss of rep is a good thing...especially when targeted.

user19161

Anyway, whoever playing these games should stop.

user19161

Not that I mind, but I don't wanna see them get a suspension...

amWhy

I couldn't agree more.

Git Gud

Don't admins have access to info regarding who's upvoting and downvoting?

user19161

12:33 AM

Anyway, I don't care about this, as I am deleting my account soon...

amWhy

When I said playing in a good way, it had nothing to do with playing around with downvotes...I take downvotes seriously.

My downvoting, I take seriously.

When I get downvotes, I often consider them as legit, except when they seem from out of nowhere.

@GitGud I was thinking of checking that out once, but it's usually serial, rapid, in succession downvotes that draw attention.

Jonas Teuwen

That was good.

I found a jam session.

There was beer, but also a jam session.

user19161

@JonasTeuwen Where?

Jonas Teuwen

If you jammed, you got a free drink.

Just, in the neighborhood.

Epsilon ~2km.

2

user19161

@JonasTeuwen Hahahaha, anyway you should spell "neighbourhood" since you follow BrE?

amWhy

12:42 AM

@JonasTeuwen So did you jam?

Jonas Teuwen

Bloody monkey, no BrE when too much beer.

I can hardly resist doing that, yes, when there is such a thing.

The Substitute

@PeterTamaroff, since the sequences grow arbitrarily large, I somehow assumed we could get any decimal expansion. However, each given sequence in my set is finite. So we cannot get $\pi$ for example. Silly mistake. :)

Peter Tamaroff

@TheSubstitute Yes, good.

In particular $\Bbb N^{ \Bbb N}\simeq 2^{\Bbb N}\simeq \Bbb R$

user19161

Goodnight @peter @jonas @anon @ethan I will see you in my dreams...

Peter Tamaroff

@JasperLoy LOL, OK.

Jonas Teuwen

12:49 AM

Good night.

Holy cow, my beer urge is not saturated.

Need to go back for some 'am.

amWhy

Good night @Jasper...(I know you left me out of your farewell, but I thought I'd say it anyway)

Potato

We can solve the diophantine equation x+y+z=100 in nonnegative integers using the "bars and boxes" combinatorial method. You have 100 dots, and you want to place 2 partition markers among them, so you get 102C2. Is there a way to generalize this (by a change of variable, perhaps) to equations like x+2y+5z=100? I vaguely recall this being possible.

Git Gud

@Potato Hi. Did you read that article?

Potato

@GitGud Yes. In fact I read it quite long ago too, and it's where I found that anecdote about Tao only playing computer games and doing math.

Peter Tamaroff

@Potato Cannot you make it a special case of $x+y+y_1+z_1+z_2+z_3+z_4+z_5=100$?

Git Gud

1:04 AM

oh :( Sorry I wouldn't help.

Potato

@GitGud It was nice to see it again! I had forgotten about it.

@PeterTamaroff How so?

Do 3x+5y=100 or something if that's simpler to explain.

Peter Tamaroff

@Potato Well, you restrict that $y=y_1$ and $z_1=\dots =z_5$.

I am thinking on the run though.

Potato

@PeterTamaroff But then the bars and boxes method doesn't work.

Peter Tamaroff

Might not work.

@Potato I mean: First use the bars and boxes-

Then, divide by the fact you want to cross out some repetitions.

Maybe I am talking nonsense. I know zero combinatorics.

Potato

It's not going to be as simple as a division.

I guess you could consider $z$ to be each nonnegative integer in turn, solve for $x$ and $y$ in this case to get a binomial coefficient, then sum them using some identities.

Well hmm

nerdy

1:14 AM

Consider system Ax=b with matrix A 3x3 and vector x=[a1 a2 a3]^T. If the system has a two-dimensional NullSpace, then we are going to choose 2 free variables to represent it.From all the three variables ( a1,a2 and a3) only some can be treated as free variable, the other(s) must be treated as a leading variable.

I wanted to ask whats the physical meaning ( regarding to the graph of the equations, their intersection etc ) of the variables that can be treated as free variables, and the variables that must be treated as leading variables.

Peter Tamaroff

@nerdy The physical meaning?

nerdy

yea

geometrically

you know, a system Ax=b represnts a solution set that is the intersection of the graph of all linear equations

the system Ax=b represents a solution set ( Nullspace ) that is the intersection of the graph of all linear equations shifted to the origin

but i wanna ask about this free variables, whats special about them that we can verify in this intersection graph

ops the system Ax=0*

in the third sentence

My problem is every book or lecture you find on linear algebra just states this

take a system

put it's matrix into RREF

the columns witouth pivots represents the free variables

the pivot columns represents the leading variables

thats all they say

nothing else, they don't explain that we actualy can swap some columns to interchange free variables,

and they dont explain what the free variables have that can be verified in the intersection graph

thats what im trying to discover

Peter Tamaroff

@nerdy "what the free variables have that can be verified in the itnersection graph" I don't understand.

nerdy

lol me neither... i wanna discover why for a system 3x3 for example, why in some cases the y is the free variable

and some cases the x is the free variable

how do these 2 systems differ

Peter Tamaroff

No, I mean I don't understand what you're saying.

nerdy

1:22 AM

why one system has y as free variable while the other has x as free variable

Peter Tamaroff

@nerdy Well, that depends on your choice. It is the same.

The free variables just give you the dimension of the solution set.

nerdy

given any system with n variables

and suppose the system is implemented by a matrix with nullspace in 2 dimensions

can i choose ANY 2 of the n variables

Peter Tamaroff

If you have one free variable, say in $K^{3\times 3}$, you get a line. If you have two free variables, you get a plane.

nerdy

to be the free variables ?

i know.

George V. Williams

@Potato, AFAIK the way used to find the number of solutions to those Diophantine Equations is a generating function. I don't think there's any other way using bars and boxes, but I would love to be proven wrong.

nerdy

1:23 AM

given any system with n variables
and suppose the system is implemented by a matrix with nullspace in 2 dimensions
can i choose ANY 2 of the n variables to be the free variables ?

ANY ?

Peter Tamaroff

@nerdy Try it with $n=3$.

nerdy

even those variables associated with the columns ( of the matrix in RREF ) with a pivot

ok

so the 3x3 system Ax=b

Peter Tamaroff

Say $$2x+3y+z=1\\ -x+2y-z=3$$

I didn't think it through, though.

nerdy

i know

Peter Tamaroff

You should get 1 free variable probably.

nerdy

1:24 AM

1 free variable /

ok

can i choose x y or z to be the free variable ? -

Peter Tamaroff

Anyways, the answer is yes.

nerdy

can i choose any1 ?

Peter Tamaroff

Yes.

nerdy

oO

how

okay i understand

pelase

Peter Tamaroff

@nerdy ?

nerdy

1:26 AM

look at this case,please : postimg.org/image/qo5z1ins1

Sal ( from khan academy ) says that the free variables are x2 and x4

My mind instantaneously goes ... BUT WHYYYYYYYYY ?

what if i want to choose x1 and x2 as the free variables

am i not allowed ? :( -

Peter Tamaroff

@nerdy Let me see.

nerdy

thanks man, i appreciate, i need to get rid of this doubt before i sleep today im with this confusion since yesterday and i can only move on if solve any doubt i have

Peter Tamaroff

OK.

nerdy

move on learning linear algebra

Peter Tamaroff

You have $$x+2y+3w=2\\z-2w=5$$

Say we want to put it all in terms of $x,w$.

Then $z=2w+5$, so we're done with $z$.

Then $y=\frac 1 2 (2-3w-x)$, and we're done with $y$.

That's it.

You get a space of the form $\left(x,1-\frac 3 2 w-\frac 1 2 x,w,2w+5\right)$

nerdy

1:30 AM

i cant see the code you are typing

can we choose x and y to be the free variables ?

or the elading variables ?

Peter Tamaroff

@nerdy Sure.

Here for LaTeX in Chat.

nerdy

Peter, we have x + 2y + 3w = 2 and z - 2w = 5

how can we choose x and y to be the free variabels

Peter Tamaroff

It is an hyperplane $\Bbb H=x(1,-1/2,0,0)+w(0,-3/2,1,2)+(0,1,0,5)$

nerdy

?

$\left(x,1-\frac 3 2 w-\frac 1 2 x,w,2w+5\right)$

ops

Peter Tamaroff

@nerdy Put $w$ in terms of $x,y$, then $z$ in terms of $w$, which you already put in terms of $x,y$.

nerdy

1:34 AM

can we put x and y in terms of z and w ?

can we choose x and y to be the leading variables ?

Peter Tamaroff

@nerdy Yes.

nerdy

well

hmhmhm

w = 2/3 + x/3 + 2y/3

then z - 2(2/3 + x/3 + 2y/3 ) = 5

Peter Tamaroff

@nerdy Yep.

nerdy

then we have f(x,y) = z + c1 and f2(x,y) = w + c1

so we now solve for x ?

and then find y ?

Peter Tamaroff

@nerdy What do you mean by "solve for $x$"?

nerdy

1:39 AM

we ended up with two equations in the form of d1.x + d2.y = w + c1 and e1.x + e2.y = z + c2

we still dont have the leading variables equation

that should be x = f(w,z ) and y=f(w,z)

ok, Peter, just found what

we can choose any 2 to be the free variables

as i choose 2 different free variables am i changing the solution set?

Peter Tamaroff

@nerdy No, not at all.

nerdy

what is different in each solution set ( originated by choosing each pair of free variables ) ?

lets suppose solution set 1 has x and y free varibles, solution set 2 has z and y as free variables

Peter Tamaroff

@nerdy Nothing. The coefficients are just scalar multiples.

nerdy

whats the difference ebtween solution set 1 and solution set 2 ?

Peter Tamaroff

@nerdy The idea is this:

nerdy

1:50 AM

;)

Peter Tamaroff

Take our equation $$x+2y+3w=2\\z-2w=5$$

We can argue the following:

First look at the homogeneous system:

$$x+2y+3w=0 \\ z-2w=0$$

It's solution space has dimension 2.

Indeed, we can write it out as $(-2y-3w,y,2w,w)$, for example.

This is the span of two vectors:

$\langle (-2,1,0,0),(-3,0,2,1)\rangle$

Note that they are just two linearly independent vectors that satisfy the equations. We needn't talk about "free variables" and similar stuff. That is just auxiliary language.

nerdy

damn can you write witouth latex ?

i couldnt make the chat work

latex chat

;(

Peter Tamaroff

@nerdy Didn't you get the ChatJAX?

nerdy

couldnt make it work. let me see again

Peter Tamaroff

Just paste the script to your bookmark.

nerdy

1:54 AM

what script ?

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

i have this page

Peter Tamaroff

"^^drag this^^ to your bookmark bar or right click on it to add it as a bookmark. "

nerdy

what should i drag ?

Peter Tamaroff

"start ChatJax"

nerdy

the adress under "Start ChatJax" link

Peter Tamaroff

That blue thingy.

nerdy

1:56 AM

the adress under "Start ChatJax" link or "Start ChatJax"

Peter Tamaroff

Not the address. The link itself.

Click on "start chatjax" and you'll see what happens in that page

Then just bookmark that link, and it will work on any page.

nerdy

damn ;( have big trouble with it. When i click in start chatjax, nothing happens

lactually when i cliuck on it

it transformar the latex code

it transforms

what is on the page

that

Peter Tamaroff

@nerdy Yes.

@nerdy OK, so can you see the solution set is <(-2,1,0,0),(-3,0,2,1)>?

Then we will obtain the full solution set of the non homogeneous equation by "shifting" the solution subspace of the homogeneous eqn by x=(a,b,c,e) a particular solution of the non homogeneous system.

That is, our solution subspace is $S+\bf x$

These are called linear varieties.

(well, at least that is the literal spanish to english translation I can give you)

nerdy

2:25 AM

i see Peter, but this particular solution you mentioned differs

with the free variables we choose

Peter Tamaroff

2:38 AM

@robjohn

@anon

anon

2:49 AM

yes

Peter Tamaroff

@anon What would you recommend: Rudin or Apostol's Mathematical Analysis? I vouch for Apostol.

In fact I am thinking of buying a used copy next week.

anon

haven't looked at either

Peter Tamaroff

@anon I feared so. That's why I pinged @rob

robjohn

3:24 AM

@PeterTamaroff yes?

Peter Tamaroff

@robjohn Rudin or Apostol?

robjohn

@PeterTamaroff for what?

Peter Tamaroff

@robjohn "Mathematical Analysis"

That is, Rudin's "Analysis" or Apostol's?

robjohn

@PeterTamaroff You mean baby Rudin? I don't think I've used Apostol's Analysis book. I used his calculus book in high school

Peter Tamaroff

@robjohn Right.

Peter Tamaroff

4:14 AM

@robjohn You there?

Ethan

@PeterTamaroff know any good movies

Peter Tamaroff

@Ethan The Tree of Life.

Peter Tamaroff

4:54 AM

@ZevChonoles Hey.

Zev Chonoles

@PeterTamaroff: Hey, what's up?

Peter Tamaroff

@ZevChonoles Little analysis problem: Let $x\in \Bbb R$, $n\in\Bbb N$.

I want to prove there exists $h,k$ with $0<h\leq n$ for which $|hx-k|<1/n$

Now, I have a hint.

Jorge Fernández

Why is zev blue?

in the letters

Peter Tamaroff

@JorgeFernández He's a moderator.

Jorge Fernández

nice

Peter Tamaroff

4:56 AM

I'm told to consider for $t=0,1,\dots,n$ the $n+1$ numbers $\{tx\}$ (i.e. the fractional part of $tx$)

Then I'm told to prove some pair differs at most by $1/n$.

@ZevChonoles

Zev Chonoles

A classic case of the pigeonhole principle :)

Peter Tamaroff

I did think the $n+1$ was stressing that.

Zev Chonoles

Break up the interval [0,1) into the $n$ pieces $[0,\frac{1}{n}),[\frac{1}{n},\frac{2}{n}),\ldots,[\frac{n-1}{n},1)$.

Ethan

math.stackexchange.com/questions/346204/asymptotic-expansion/…

Peter Tamaroff

@ZevChonoles Oh, OK.

Ethan

4:58 AM

read the first answer lol

Peter Tamaroff

Click!

@ZevChonoles Beautiful proof. Simple and efficient!

Zev Chonoles

@PeterTamaroff: Glad I could help! If I remember correctly, Dirichlet came up with this argument, and this is why some people call the pigeonhole principle the "Dirichlet box principle" or something like that.

Peter Tamaroff

@ZevChonoles Yeah. It has a crazy German name, also!

It is literally translated as the "drawer principle".

@ZevChonoles I was asking people: Apostol or Rudin?

(their "Mathematical Analysis" books, I mean)

Zev Chonoles

I used Apostol in my very first real analysis class, and I think it was decent; only oddity I remember (a minor one at that) was that he defines measurable function as the basic notion, and then defines a measurable set to be one whose indicator function is measurable.

Peter Tamaroff

@ZevChonoles Pardon the ignorance: why is that odd?

Zev Chonoles

5:05 AM

The standard system (or at least, how I've always seen it) is to define measurable set by introducing the Borel sigma-algebra, applying Caratheodory's extension theorem to get the Lebesgue measure, and then defining measurable function as one which pulls back measurable sets to measurable sets.

Peter Tamaroff

@ZevChonoles OK. Whatever that means =) I'll get to that, eventually. Does Rudin follow that programme?

What is not that nice about Rudin is he gives sometimes too dry of an exposition.

Zev Chonoles

I've only ever glanced at Rudin, unfortunately, so I'm not sure off the top of my head. And flipping through it now, looking for a mention of Caratheodory, I'm reminded of one of the criticisms I've heard leveled at Rudin: he barely names any theorems. Until someone pointed it out, I never really appreciated how much having a name associated to a theorem helps you store it mentally.

Peter Tamaroff

@ZevChonoles Hmm, that is pretty interesting.

Peter Tamaroff

5:57 AM

Yay! I learned yet another proof of the Heine Borel theorem in $\Bbb R^n$.

user19161

6:33 AM

I see the other blue square has come to this chat, the great Zev, no less.

κρανίοπεριπολία

Have you ever considered becomig a moderator? @JasperLoy

user19161

@κρανίοπεριπολία Not really. And I am not suitable for the position, considering that I can get too emotional at times.

Matt N.

7:13 AM

@JonasTeuwen I have no idea! I don't speek Greek except for "Feta", "calimera" and "calispera" (also cannot write it as you can see).

κρανίοπεριπολία

7:52 AM

Where would math be if Fermat had Post-it notes?

@robjohn great works of art should be enjoyed by all :-D

user19161

@κρανίοπεριπολία Ah, I just saw it in his profile too!

user19161

8:13 AM

Ah, the serial downvotes on my posts yesterday have been reversed!

user19161

Just did my 5 edits today. One of the things is to add mathbb.

Jonas Teuwen

I am only editing phd-thesis.tex and its children today.

Enjoy.

user19161

@JonasTeuwen Good good. You will be Professor in a few years!

Jonas Teuwen

I hope so... 8-).

user19161

Don't forget me when you win the Fields medal!

Jonas Teuwen

8:20 AM

How can I ever forget you?

In any case, I'm going out now: see you!

user19161

Enjoy!

nerdy

10:09 AM

can anyone help me here with a RREF ?

i have a 4 variable system and 2 degree of freedoms.I want to express all the possible pairs as free variables but i can't seen to express x1 and x2 as free variables

postimg.org/image/qo5z1ins1.

Vrouvrou

10:28 AM

hi

h is an isomorphisme if h is one to one and preserves the structure,olso for h−1 right ?

pen

amWhy is 40k... what the hell

Nimza

Good day everybody!

What may mean notation $u(x) \in C^{1,1}[0,1]$?

and even $C^{1,\frac{1}{2}}[0,1]$

pen

10:59 AM

Can I ask questions related to programming?

On main...

Nimza

@pen I think it depends on kind of question. If it is about computational algorithms, for example, I would ask on scicomp.stackexchange.com

pen

@Nimza It's about finding the number of geometric progressions given some constraints.

Nimza

@pen I think it is ok then. As for scicomp too

I think that $C^{m,n}$ may stay for $C^m$ equipped with $H^{m,n}$ norm

Euclidean

11:50 AM

In euclidean ring we associate a d value to every element expect 0 which is non-negative. But then how could a polynomial ring with it's degree as d value could be euclidean, as the constants in the ring have degree 0, which is not non-negative?

Dominic Michaelis

huhu

Euclidean

Hmm

ok, so 0 is non-negative :|

κρανίοπεριπολία

0 is neither positive nor negative.

Dominic Michaelis

0 is like the swiss

neutral

Euclidean

natural number starts from 0 or 1. which is more accepted?

Transcript for

$Mathematics$ Mathematics