Mathematics - 2013-02-26 (page 2 of 3)

Andrew

04:00

@user43758 I went and read about determinants and some stuff about matri...matricies?....apparently it's -1 because the matrix will always be right or left handed?

Yea that.

or rather

lefthanded

Peter Tamaroff

@user43758 That is not right.

Andrew

and not righthanded

Peter Tamaroff

@user43758 Where can I see the original exercise?

user43758

Wait

@PeterTamaroff

Peter Tamaroff

@user43758 Oh, OK.

Andrew

04:04

How can i be equal to j and also not equal to j...JW

in option 3..

Peter Tamaroff

@user43758 Yes, there is a big typo there.

Andrew

probably means S

Peter Tamaroff

I think they mean $s\neq j$

Andrew

yea

makes a lot more sense that way

is this for algebra?

user43758

yes

Andrew

04:06

Yea..the letters gave it away..

user43758

So it is s \neq j

So let me try again

Andrew

what are r and s

jw

user43758

variables between 1 and n

Andrew

yea, i figure matrices don't have negative indices, but do they increment with rows or columns or what?

Peter Tamaroff

@user43758 As far as I can see, it is a permutation of the identity matrix.

Andrew

04:10

well that can't have a determinant of -1...

Peter Tamaroff

@Andrew Sure it does.

Any permutation of the identity matrix has determinant $1$ or $-1$

Andrew

Oh, right

oh, yea ok

I just realized why that is the case.

Peter Tamaroff

@user43758 It $E^{r,s}$ is the identity matrix with the $r$th and $s$th row changed if $r\neq s$.

Andrew

so it's a left handed identity matrix?

Peter Tamaroff

Since it is a $1$ (odd) step permutation it has det $=-1$

@Andrew Are you joking now?

Andrew

04:13

I mean, if it's a permutation of an identity matrix, well I can only think of two of those, and one of them is \ and the other is /

Peter Tamaroff

@Andrew There are $n!$ permutations of the identity matrix of in $n\times n$.

Andrew

So unfortunately I'm afraid I'm not joking

user43758

@PeterTamaroff So in the case of 4x4 it's

1010
0100
0010
0001

Peter Tamaroff

@user43758 What is that?

$E$ what?

user43758

take (r,s)=(1,3)

Andrew

04:15

well then, there are a lot more identity matrices than purplemath.com/modules/mtrxmult3.htm told me about

Peter Tamaroff

@Andrew there is only one for each dimension.

Andrew

are there n dimensions?

user43758

I what I wrote correct or not ?

What is E then ?

Peter Tamaroff

@user43758 No. $a_{11}$ should be zero because $1=1$ but at the same they share the same first coord with $1,3$

Also $a_{31}$ should be $1$, not zero!

And $a_{33}$ should be zero, not $1$.

Just follow the rules.

Andrew

@PeterTamaroff Hmm, yea, I need to grade papers - as interesting as this is, I don't have the mathematical background, or the plugin that makes the stuff between dollar signs look like something other than PHP variables

@user43758 @PeterTamaroff Have a good night =D

Peter Tamaroff

04:19

@Andrew You can MathJAX by following the link to the right that say LaTeX support for chat, in the starred msgs.

user43758

So It's

0010
0100
1000
0001

@petertamaroff

Peter Tamaroff

@user43758 Yes.

And that is a permutation of the id. matrix.

Just rows $1$ and $3$ are interchanged.

Andrew

@PeterTamaroff Hey thanks now everything you type looks cool and mathy instead of claustrophobic and over-punctuated

@PeterTamaroff Oh when you said permutation you meant rows...or columns....or sets of rows or columns..that makes a lot more sense

user43758

But how can I show that the determinant of this equals -1 ?

Peter Tamaroff

@user43758 You know how column and row operations affect determinants?

Changing a row with another changes the sign.

user43758

04:24

Yes but you think that it is enough of an argument

?

Andrew

@user43758 induction

Peter Tamaroff

@Andrew Nah.

It isn't necessary.

Andrew

@PeterTamaroff awww ok =(

Peter Tamaroff

@user43758 I think so. What you ought to show is that your matrix is indeed what we say it is.

user43758

How do I "show" that

Gustavo Bandeira

04:25

habitrpg.com

user43758

I was planning on just drawing it

Peter Tamaroff

Let $C^{r,s}$ be the canonical matrix: zeroes everywhere but $a_{rs}$ where you have a $1$.

Andrew

Ok guys, it actually was really fun =D ttys

user43758

@PeterTamaroff yes..

Peter Tamaroff

You want to show $E^{r,s}=\operatorname{Id}-C^{r,r}-C^{s,s}+C^{r,s}+C^{s,r}$

Note $r\neq s$

user43758

04:30

What is the canonical matrix ?

Is it the matrix with all the terms equal to zero except the one situated in r s ? (which equals 1)

Peter Tamaroff

Yes, I just told you =)

5 mins ago, by Peter Tamaroff

Let $C^{r,s}$ be the canonical matrix: zeroes everywhere but $a_{rs}$ where you have a $1$.

user43758

Sorry I am very tired

Peter Tamaroff

@user43758 You need to read, that's all, no biggie.,

@user43758 What time is it? Also, wouldn't you like to pick a username?

user43758

11:30

I generally don't stay late

I am more of a morning person

And the determinant would be what ?

I am mean using the expression you gave me

Peter Tamaroff

$-1$, because we swapped two rows.

awllower

04:57

Hello everyone!

I wonder:
Whether it is true or not: the subgroups of proucts of cyclic groups are products of subgroups?

Maybe I have given a wrong answer...

user43758

@PeterTamaroff One more question and I finished my Problem set

Peter Tamaroff

@user43758 OK.

user43758

It is the one I posted online. Nobody knew how to prove it without using $rank(t_A)=rank(A)$

Peter Tamaroff

@user43758 What is $t_A$? Also, what about your username?

user43758

transpose of A

How do you change the username

I changed it online

but it doesn't seem to change on the chat

Peter Tamaroff

05:08

@user43758 Oh, it takes a while. Try refreshing.

user43758

No I did it about an hour ago

Peter Tamaroff

@user43758 How do you define the rank of a linear transformation? As the rank of the matrix of $f$?

user43758

2

Q: Matrix, Ranks and Rows

Let $f:V \rightarrow W$ be a linear transformation. Given bases $\{v_i\}_{1\leq i \leq n}$ and $\{w_j\}_{1\leq j \leq m}$ of V and W, respectively, $f$ has an associated $m \times n$ matrix $A$. I am having trouble showing the the dimension of the row space equals the rank of $f$. I can't use th...

Peter Tamaroff

"The column rank of a matrix A is the maximum number of linearly independent column vectors of A. The row rank of a matrix A is the maximum number of linearly independent row vectors of A. Equivalently, the column rank of A is the dimension of the column space of A, while the row rank of A is the dimension of the row space of A."

I can't really help now, sorry. It is 2 am here, got to go.

user43758

ok good night

awllower

05:11

Good night!

Peter Tamaroff

Bye byes, peoples of the math.

user19161

05:31

Just got the linear algebra badge.

awllower

@JacobBlack Congrats!

Now you are a linear algebraic master!

:)

user19161

@awllower Even though I cannot row reduce!

awllower

You can reduce Bananas!

2

in your stomach

2

user19161

Hahaha.

awllower

:-)

Andrew

05:43

Ok guys serious math question here

Who's up for the challenge?

awllower

06:06

What challenge?
I am not so good, so forgive me if I fail.

Hm
I have to depart now.
Later then. :)

skullpatrol

06:47

later

robjohn

Somebody upvoted 8 of my posts today. 7 of those were removed.

skullpatrol

+1

net

gain

>8(

user19161

@robjohn Script runs at GMT 0300.

robjohn

@JacobBlack That would be during the time I was gone for park and dinner.

user19161

I shall not go into any more details, in case people try to game the system.

Ben W.

06:59

How do you know that?

user19161

I have psychic powers. I am not of this world.

Ben W.

I believe it.

user19161

I saw a user profile that said she was battling cancer. I wish her well...

Ben W.

I remember seeing that to. Props to her for continuing to study math.

user19161

We might have seen different people, does it start with a J?

Ben W.

07:04

Can't remember. She was asking a question from Lang. That's all I recall.

I was interested in the question, so I clicked on her profile.

user19161

Sounds like it then.

user19161

Hmm, perhaps I will drop her a comment...

The Substitute

07:22

Hello. What does (a_j ^X) mean? Is this standard notation to say that (a_j) is a countable sequence?

user58512

I over ate now I feel sick :(

skullpatrol

07:53

Hi @Sanchez what's happening?

Sanchez

hi @skullpatrol, what's up?

skullpatrol

@Sanchez Chillin'

Sanchez

lol, same

skullpatrol

cool

user58512

ugh math.stackexchange.com/users/14615/sylvain-julien

user19161

08:36

@TheSubstitute You must tell us where you saw it. Usually $(a_n)$ refers to the sequence $a_1,a_2,a_3,\ldots$.

user19161

@user58512 What is it about this user? Why the ugh?

pourjour

09:29

0

Q: equation of $(E):z^2-2mz+1=0$ in $\mathbb{C}$

Suppose the equation $(E):z^2-2mz+1=0 \quad / m\in \mathbb{C}\quad z\in\mathbb{C}$ and we suppose $z_{1}$ and $z_{2}$ are the two solution of this equation. How can I prove that $|z_1|+|z_2| = |m-1|+|m+1|$?

complex-analysis complex-numbers

help

Gustavo Bandeira

09:48

Would it make sense to say that $\mathbb{C}$ is closed under the operation of square root?

Orange Harvester

@GustavoBandeira yes.

Gustavo Bandeira

@OrangeHarvester Thanks.

pourjour

10:34

1

Q: Solution of $(E):z^2-2mz+1=0$ in $\mathbb{C}$

Suppose the equation $(E):z^2-2mz+1=0 \quad / m\in \mathbb{C}\quad z\in\mathbb{C}$ and we suppose $z_{1}$ and $z_{2}$ are the two solution of this equation. How can I prove that $|z_1|+|z_2| = |m-1|+|m+1|$?

complex-analysis complex-numbers

any help please

Chris

Good morning

Ethereal

@Chris Are you the brother of Chris's Sister?

Chris

Not that I know :)

Ethereal

Guys, how do you generate a polynomial given a sequence?

Chris

Sorry but I've never done any statistics in high school, and this question is probably way too simple to make a post out of:

If each year my factory has a 1/10 chance to break down, what are the chances of it being broken down after four years?

I thought 4/10 initially, but I'm confused about how the chance remains independent for each year...

Sure this is trivial for you, but if you could give me a tip that'd be great.

Orange Harvester

10:41

@Ethereal langrange's interpolation polynomial

Ethereal

@OrangeHarvester Oh.

user19161

11:03

I have moved from google to yahoo to bing search, I am still deciding between the three.

Orange Harvester

there is probably no difference between yahoo and bing

they both use the same backend

pourjour

is there anyone here with some knowledge at complex number

s

awllower

11:33

Well, at least I know what the minimal polynomial of $i$ over $R$ is.

It is $x^2+1$

:)

skullpatrol

@Chris 4/10 sounds right to me.... as far as the chance remaining independent for each year that is a given for the problem.

user58512

12:34

@awllower, hello

awllower

Hello

user58512

why are all the zeros of zeta on the line 1/2?

awllower

Had I known, I would be awarded Fields...

user58512

is there no idea known?

awllower

Colorful picture.

Prove it by using Birch Swinnerton Dyer?

user58512

12:36

ok

user58512

12:53

"Indeed, it's a nice exercise in a first-year algebraic geometry course to compute the genus of X^n + Y^m = 1 (by hand!). The sequence of blow-ups needed to resolve the singularity mimics the Euclidean algorithm used to compute gcd(m,n)." - interesting

Kasper

13:12

Does anybody here understand why this doesn't work at mathjax in stackexchange?
http://html5mathml.googlecode.com/svn/trunk/longdiv-mj-mml2.html

$\longdiv{534}{17}$

or at least not with chatjax

user58512

not a real command

Kasper

but at this website: http://html5mathml.googlecode.com/svn/trunk/longdiv-mj-mml2.html
they also use mathjax, and there it does work

$$
\longdiv{12345678912}{325}
$$

user58512

no

view the html source you can see them specifically adding the custom longdiv command

hey pourjour

Orange Harvester

That is one stupid long longdiv command.

pourjour

@user58512 hi

user58512

13:19

whats up

pourjour

@user58512 fine just preparing to math exam for tomorrow

Kasper

@user58512 ah, I see..

user19161

15:06

I am now the grey square.

awllower

Hello grey banana

Some fungus have grown?

Haha :)

skullpatrol

We're almost the same color :-D

awllower

:-D

Orange Harvester

Jacob is skull, skull is jacob.

skullpatrol

How symmetric.

awllower

15:15

Jacob Skull est, et vice versa.
Hm, I thought that Latin was very symmetric, but it appears that English is more so?

pourjour

any geek at complex numbers?

Orange Harvester

@awllower yes, simpler grammar, simpler construction! more symmetry, follows principle of least action. :P

awllower

:P

user58512

how do you express that a is a power of 2 in the language of rings

I got $$\forall x,y \exists z, a = xy \to (x = 2z \wedge x=1)$$

but can we do it with all the same type of quantifier?

Tobias Kildetoft

@user58512 why would you use such formal language for speaking about rings?

user58512

15:28

I don't know, just thinking about this

his formula must be wrong though or at least convoluted

user58512

15:44

similar here

robjohn

@pourjour why not just ask the question?

pourjour

@robjohn ok

actually I have many question for example we suppose $z=1+e^{i2\theta} / \theta \in (0;\pi)$

how can I prove that the set of points $M(z) \in (C)$ as (C) is a circle

I tried this $|z-e^{i2\theta}|=1$

I deduced that M(z) is in the circle with center $e^{i2\theta} $ and a ray r=1

is it correct

z is the affix of M

user58512

16:04

I don't understand your notation about M an C ,you didn't define them

how can a set of points be a member of the circle

\subseteq rather than \in ?

robjohn

@pourjour How about $|z-1|=\left|e^{i2\theta}\right|=1$?

user58512

16:33

@robjohn, good day

robjohn

@user58512 how goes?

user58512

pretty good but I need to get on with some work soon, not sure what

I am being lazy

robjohn

@user58512 welcome to the time sink :-)

user58512

I asked my teachera question about something and he just said it was hard to explain,.. lol

well he said it was easy, but hard to explain

robjohn

@user58512 did you ask here?

user58512

16:36

no

robjohn

@user58512 ah

user58512

I dont know anyone from that class so I cant talk to them about it

Orange Harvester

use this as an excuse to get to know people.

Matt N.

Hi everyone.

user58512

hi

Orange Harvester

16:41

@MattN. Hello.

Matt N.

Is this site on the way downhill or is it just me?

Not that it is a matter of great importance.

user58512

@MattN., in what respects

Matt N.

People are assholes in the sense that there is tons of downvoting and not as much upvoting.

user58512

yeah, I guess that is true a bit

Matt N.

I meant compared to two years ago. Never mind.

But I fear I might be one of those who lament that the "times" are not as good as they used to be and if these people were right the world would have to have ended already since then the "times" would have been getting worse for 2000 years.

user58512

16:45

every website gets worse, doesn't it?

Matt N.

Don't worry I don't mind.

Oh, wow.

And ever since you had a boost in productivity? : )

@user58512 Everything always gets worse, doesn't it?

Orange Harvester

Yeah, I am actually happy.
No, not a boost in productivity, but have not missed anything actually.

user58512

not everything

Matt N.

@user58512 Including McDonald's chocolate sundae.

Orange Harvester

@MattN. I feel every internet site reaches a phase where it enters "Eternal September" and then reputation and rankings and related votes cease to matter.

I feel it happens not only with internet, but with everything.

Matt N.

16:48

@OrangeHarvester I think they still matter tons. But they just aren't there. Even though needed badly.

The more educated people are craving rep but they don't get as much as they used to because the user base has changed. So they contribute less.

Or stop contributing.

Orange Harvester

@MattN. Sorry, I phrased it wrongly, what I meant was, they cease to reflect the the the actual quality demographics.

Matt N.

Alright.

Orange Harvester

How is your set theory coming off?

Matt N.

Slowly. But thanks.

@OrangeHarvester How are your studies coming along?

Orange Harvester

@MattN. They are coming off well. Right now concentrating on some measure theory and group theory.

Matt N.

16:58

Good.

Charlie

Hi people

Matt N.

Hi there.

Peter Tamaroff

Has anyone watched the Hunger Games?

Matt N.

Yep, it was ok. Much better than expected.

Peter Tamaroff

Good, good.

@MattN. What are you up to?

Matt N.

17:07

@PeterTamaroff Thinking about making something to eat. And yourself?

Peter Tamaroff

@MattN. I had breakfast a while ago, now I'm watching that movie. I have yet again been defeated by an exercise in Spivak.

It is one of the few that keep beating me.

Orange Harvester

Which one?

Peter Tamaroff

Let $a_n$ be $$\frac{1}2,\frac 13,\frac 23,\frac 14,\frac 24,\frac 34,\frac 15,\frac 25,\frac 35,\frac 45,\frac 16,\dots$$

Prove the sequence is uniformly distributed over $[0,1]$

Orange Harvester

these are just the rationals

Ahaa, I see.

Uniformly distributed $\equiv$ dense?

Matt N.

Heh, is this just an enumeration of the rationals?

Orange Harvester

17:14

yes

Matt N.

I guess one would have to prove it. But I believe you : )

It certainly looks that way.

Peter Tamaroff

@OrangeHarvester Nope.

@MattN. And it is not an enumeration of the rationals. Well, it is an enumeration of the rationals in $[0,1]$, but it repeats the elements.

Matt N.

Ah. Yes.

Peter Tamaroff

Let $a<b\in [0,1]$. Define $$N(a,b:n)=\{j\leq n:a_j\in[a,b]\}$$. We say that $a_n$ is UD in $[0,1]$ if for each $0\leq a<b\leq 1$

$$N(a,b;n)/n\to b-a$$

Orange Harvester

@PeterTamaroff I see. I just looked up the definition in spivak.

amWhy

17:17

Yikes, talk about an attitude! Read the comments here

@Orange I know I've been remiss in getting out an email to you! Nothing personal...just bogged down a bit.

user58512

@amWhy, yeah that sucks

amWhy

@user58512 OP must be having a bad day!

Peter Tamaroff

The interesting thing is that if $(a_n)$ is UD over $[0,1]$ and $f$ is integrable over $[0,1]$ $$\int_0^1 f=\lim\frac{f(a_1)+\dots+f(a_n)}n$$

user58512

how are you doing

Matt N.

I will refrain from commenting. I read the thread about plagiarism committed by you (amWhy) on meta.

But that answer does indeed not seem to be helpful to OP.

amWhy

17:21

@user58512 Hanging in there...rough going, but I I can only move forward...

Matt N.

Now before I get dragged into pointless discussions about pointless virtual ongoings I will be off. I'll see you all later! Have a nice day!

Orange Harvester

@MattN. Bye!

@PeterTamaroff Do you want an idea or do you want to struggle some more?

Peter Tamaroff

@OrangeHarvester For what?

Orange Harvester

@PeterTamaroff for the above problem.

user58512

I'm just worried I run out of teabags

Peter Tamaroff

17:24

@OrangeHarvester Oh. The one aboutthe sequence right? Not about the integral.

Orange Harvester

@PeterTamaroff Yes, about the sequence.

Peter Tamaroff

@OrangeHarvester Oh. What is the idea behind the hint? What are you looking at?

user58512

is this even true? math.stackexchange.com/questions/315050/…

Orange Harvester

@PeterTamaroff idea behind the hint! that is going to be difficult to say.. but I can probably say that first prove for a specialized subset of values of $a,b$ and then extend to all others! (you said idea behind the hint, so I am intentionally cryptic, if it is too cryptic, I will have to give the hint directly)

@user58512 What is $[2n]$?

Peter Tamaroff

@OrangeHarvester Well, I am convinced it doesn't hurt to look at just rational values of $a$ and $b$.

@OrangeHarvester I guess it is $\{i:1\leq i\leq 2n\}$?

Orange Harvester

17:30

@PeterTamaroff okay, the list notation, hmm.

@PeterTamaroff Yes. It is easy to prove for all rational $a,b$. Then, I guess, we can extend to reals by some sort of convergence.

Peter Tamaroff

@OrangeHarvester "Easy to prove". Thanks, bro!

Orange Harvester

@PeterTamaroff Hehe. Sarcasm? I feel the extending part is the more involved one, hence my idiom.

user58512

uck, the way you get it is that it has to contain a double of some number

@amWhy, my problem is I have too much to do, so I end up doing nothing :(

amWhy

@user58512 Ohhhh, I TOTALLY know what you mean!

Peter Tamaroff

If you display the number in a triangular manner. Like this

$\frac 1 2$
$\frac 1 3,\frac 2 3$
$\frac 1 4,\frac 2 4,\frac 3 4$

You can find out when $\frac k n$ will repeat.

It is exactly $n$ rows down and $k$ columns right.

And only then.

user58512

17:38

@amWhy, did oyu see my geometry question: I was tossing and turning in bed twisting spheres around but came up with nothing math.stackexchange.com/questions/314131/picture-of-a-4d-knot

Peter Tamaroff

Take $1/2$. It repeats in $2/4$ which is $2$ rows and $1$ col right.

user58512

it's fun to think about

amWhy

+1 good question...I spend many sleepless nights contemplating open problems/unsolved questions !

user58512

:D

amWhy

@user58512 Awesome!!!

Orange Harvester

17:43

@PeterTamaroff Spivak has given the repeated sequence, so the solution might lie in evaluating the number of such numbers between $[a,b]$ using some facts from the sequence itself rather than trying to operate solely on the number line and then adding up the repeated numbers.

user58512

the next problem would be prove it's knotted.. there doesn't seem to be any invariant like 3 coloring..

@amWhy, I am thinking of dropping number theory but I dont know if I should :/

amWhy

@user58512 Is that one of your current classes?

user58512

yes

Peter Tamaroff

@OrangeHarvester OK; just for the sake of it, how would you prove that for $a=1/2$, $b=3/4$ the limit is indeed $1/4$? Wouldn't it follow from some bounding or squeezing?

amWhy

@user58512 Is it a time-constraint matter? Personally, I think, when possible, do fewer things very deeply, vs. too many things if it means only surface learnlng.

user58512

17:52

yeah it feels like a waste of time

idk

(because I'm not taking the exam for it)

I'd be better off learning that stuff slower and deeper

like you say

Orange Harvester

Possibly there is a method that way. What I was thinking is like follows:
A number $a_s = k/j$ is between $p/q$ and $r/q$ if $pj<kq<rj$. So, from here you get a bound on possible values of $k$. And given any $n$ you can determine where $s < n$ by the bounds determined by $k+j$.

Transcript for

$Mathematics$ Mathematics