yes

thank you

Good. Could you show me?

yes wait..

OK.

@PeterTamaroff Pedro, have you ever read "Contributions to the creation of transfinite set theory" ?

@Charlie No.

@PeterTamaroff ah, okay

@Charlie Why?

@PeterTamaroff I don't have time to write it now. But I got it

I have to finish my problem set first

@user43758 Let me make it quick then.

@PeterTamaroff just asking, I'm taking a look at it

@user43758 Let $q\in S+p$, say $q=s+p$. Then $Aq=As+Ap=0+B=B$, so every element of this set is a solution, that is $q\in S+p\implies Aq=B$.

On the other hand, let $Az=B$. Then $z-p\in S$; since $A(z-p)=Az-Ap=B-B=0$

This means that $(z-p)+p=z\in S+p$. That is $Az=B\implies z\in S+p$. Done. @user43758 It isn't that long! =)

I wrote it down as S={x_0+\lambda_i x_i; \lambda_i \in R}

such that x_0 is a solution of the homogenous system

$$S=\{x_0+\lambda_i x_i; \lambda_i \in R\}$$

@user43758 Enclose that in dollar signs

and \forall i \in {1,...,n} , x_i the solution of the non-homogenous system

@PeterTamaroff Anyway you can't see the LaTeX expressions

@user43758 Simply write $S=\{{\bf x}:A{\bf x}=0\}$

@user43758 My eyes don't parse TeX code, sadly.

@PeterTamaroff =)

@user43758 Do you follow the proof?

Wait

I have another question which has noting to do with this

$$\begin{cases} 2x+5y-3z+u=-1\\ x-y+z=2\end{cases}$$

That's it?

They ask me to solve this using Gauss elimination process

@user43758 You do realize the solution space will be two dimensional?

That is, you have two free variables.

Yes this is what I wanted to make sure

You can express the solution as a "line".

@PeterTamaroff What I did is that I substracted the first line with 2 times the second line

@PeterTamaroff I get x-y+z=2

this is the only "nice" thing I get

@PeterTamaroff Are you here?

wohooo exactly 1000 points :)

@jacob hi, did you see that i got suspended earlier?

@JacobBlack for saying that someone should have intercourse with himself

@PeterTamaroff Thanks. Can you help me with a final one? Just to make sure

@JacobBlack i got irritated and used, no big deal

it's not that bad

@JacobBlack the weird is I called him fag 3 times and no one said anything

I added the first and the third

@JacobBlack gitgud

@PeterTamaroff Then I added the second with the resulting equation

and obtained 4x+5y=4. Do I stop here or ?

@JacobBlack I'm always too moderated, I want to freak out

@JacobBlack I can't promise anything

@JacobBlack oh yeah ;)

@user43758 Oh, but then you have more equations!

@user43758 Yes, you might start and say $z=2+y-x$. Now write $u$ in terms of $y$ and $x$

@JacobBlack He says he's not interrested, sorry. =/

Somebody should post an answer to my lame question: math.stackexchange.com/questions/313562/…

@JacobBlack But he says he might be if you get into some maths.

@Mikhail But you got an answer.

@JacobBlack Oh. But it's a plushie, it's asexual.

@PeterTamaroff I want to accept an answer

@Mikhail Life isn't all rainbows and roses, you know.

@JacobBlack Oh, boy.

Can you guess :-)

Derivatives of multiple orders?

Hey

Hey @Link

I have to find the inflection points of x(x-4)^3

and I get x = 4 as one of them

but why is the other x = 2?

that is f(x) = x(x-4)^3

@Link Take the second derivate, find the roots, profit.

I did

@Link HEYA

roots for second derv are 0 and 4

$f'=(x-4)^3+3x(x-4)^2$

$f''=3(x-4)^2+3(x-4)^2+6x(x-4)$

One root is clearly $x=4$.

Now factor $x-4$ out.

Hmm

3(x-4) + 3(x-4) + 6x?

oh

And that boils down to...

(x-4) * (3(x-4) + 3(x-4) + 6x)

$12x-24=0$

That is $x=2$.

interesting

@user43758 $\TeX$!!!

Sorry

thanks

@Link Orderly. Do things in an orderly manner.

I get it now

@PeterTamaroff, I just found it, and then simplified it a way that didn't give me 12x-24

@user43758 But yeah, that's great.

I checked on my calculator

If you take z=t

the expression looks ugly as well

Things are ugly.

In general.

I changed my avatar

@Charlie YAY!

@Argon YYAAY!

@anon

yes?

I am depressed, since I left my laptop's power cord at school. Only a matter of time till I have to hit the hay!

I might be having a stroke right now but, does it make sense to take $[a,b]\times \Bbb N$?

@anon Do not listen to me, sorry. I was thinking about another operation!

Let me explain my problem to you.

@anon

@PeterTamaroff Are you math major? What year

@user43758 Math major?

What does that mean?

@PeterTamaroff North American thing

(I'm not from the US)

oh sorry. Where are you from ?

@user43758 Argentina.

What year are you in university?

@user43758 I start first year in a few days.

So you are still in high school ?

Ah no I understood now

My bad

@user43758 I finished high school in 2011

I finished in 2009

2014 boyz!

@Argon YAY

Then I took a one year long set of courses in 2012 to be able to get to the uni. @user43758

@Argon OH MY GOD:2+1+4=7

Hahahaha

@Argon this is amazing!

@Charlie And BANANA has 3 ocurrences of A!

@PeterTamaroff it's an internal joke between Aaron and I :P

@Charlie Damn you, internalists.

@PeterTamaroff MWAHAHAHA

How can Chinese students take the AIME?

@MaoYiyi Konnichiwa!

@MaoYiyi What is that?

@Peter: Yes, but probably not for long.

@PeterTamaroff amc.maa.org/e-exams/e7-aime/aime.shtml

@BrianM.Scott Praise the heavens!

OK, this is the thing.

@PeterTamaroff Konnichiwa is Jappanese.

Let $(a_n)_{n\in \Bbb N}$ be a sequence.

@MaoYiyi That's the best I can do, sorry.

For the sake of ease consdier the unit interval.

Just wondering because at my international school the Chinese student sit and take the AIME exam, which I thought were for USA students

Then define $N(a,b;k)$ to be the number of integers $j \leq k$ such that $a_j\in [a,b]$ for $a<b\in [0,1]$

@BrianM.Scott

@PeterTamaroff Okay so far.

We say that $(a_n)$ is uniformly distributed on $[0,1]$ if, for any choice of $a<b$ we have that $$\lim\frac{N(a,b;k)}k=b-a$$

Right.

Now, I wanted to express $N(a,b;k)$ as

$$|(a_k)_{j\leq k}\cap [a,b]|$$

But this is wrong because it misses repetitions in the sequence.

I tried to fix that and said, well, let's work with pairs $(a_k,k)$ to differentiate the same elements.

But I cannot do that in $[a,b]$ because it is not countable.

@BrianM.Scott See what's my problem?

Let me think about it for a minute.

@BrianM.Scott Thanks.

Why don’t you want to work with the original definition? What do you want to do that makes it inconvenient?

Well, it is wordish. I was wondering how we would put it in terms of simple usual symbols.

Not that I have anything against it.

What does this matrix look like exactly..

@PeterTamaroff $$N(a,b:k)=\frac1k|\{i\le k:a_i\in[a,b]\}|$$

is it $$diag(1,...,1,\lambda)$$ ?

@BrianM.Scott Right. I had written that! =P

@user43758 Looks like the canonical matrix with a $\lambda$ in it, and a diagonal.

:8266963 Not necessarily.

@user43758 It has ones on the main diagonal and zeroes everywhere else except in the $(r,s)$ entry, which is $\lambda$.

It is $\lambda E^{r,s}+ \text{diag}(1,1,1,1,\dots,1,1)$

Yes so it's of the form diag(1,...,\lambda,...,1) such that the lambda is on the r,s entry

@BrianM.Scott How does one go about proving uniform distribution? I'm a little clueless.

@user43758 That notation is off. It is not what you mean it is.

I am thinking about it though, with some examples.

Am I reading this wrong ? They are asking me to show that $$\det E^{r,s}(\lambda)$$=1 ... Shouldn't it be \lambda ?

I mean shouldn't the determinant be equal to lambda ?

@user43758 If $\lambda$ isn't in a diagonal, it is $1$. Otherwise it should be $1+\lambda$, as far as I see. I mean, from what you described that matrix is.

Ah sorry, r and s are DIFFERENT integers

@PeterTamaroff Notice that your sequence comes in blocks, one for each denominator. For any $[a,b]$, once the denominator is large enough, about the right fraction of each block will be in $[a,b]$, and the approximation will improve as the denominator increases. Moreover, when $k$ is large the later blocks will dominate the earlier ones. That’s all impressionistic handwaving, but it’s the direction in which I’d look to put together a proof.

It may end up something like what I did in this answer.

so lambda is not on the diagonal

@BrianM.Scott I see. I'll be working on it. Maybe tomorrow I can give you good news. I should then prove that if $(a_n)$ is UD on the unit interval and $s$ is a step function, $$\int_0^1 s =\lim \frac{s(a_1)+\dots+s(a_n)}n$$

And use that to prove the claim for integrable $f$.

It is quite nice. I diverted from algebra for a moment (I'm reading Basic Algebra I) because this problem always beats me!

@PeterTamaroff It looks non-trivial but do-able. I’m going to have to run for now, though: the person for whom I wrote that answer has a bunch of questions.

@PeterTamaroff How do I show that the determinant equals 1 ?

@BrianM.Scott I'll help as far as I can.

@user43758 First, do you understand what that matrix is?

Well it's basically the identity matrix with a lambda at the r s position

@petertamaroff

@user43758 OK, good.

Oh, no you aren't.

why ?

No no, you're good.

So, what happens if $\lambda$ doesn't fall in the diagonal?

Then if it falls up and right the diagonal, you can just calculate the determinant as $1\times 1\times\dots\times 1=1$

If it falls under the diagonal, on the left, you just take the transpose and go back to what we just said, it is $=1$.

If it falls on the diagonal, it is $\lambda$

Why is it $1\times1\times\dotes\times1$ in the first case ?

Do you know that $\det\text{diag}(a_1,\dots,a_n)=\prod a_k$?

yes

But I have a lambda in a certain location on the matrix

OK; nevertheless, this happens for any matrix in an upper trianglar form.

it is no longer disgonal

diagonal*

AH!!

Expand it throughout the diagonal and that is it.

right

Can we calculate the determinant without using the formula for a triangular matrix?

@PeterTamaroff (because I don't think I can use it)

Well, how do you define the determinant? Because that formula is just something we obtain directly by the column-row expansion of the determinant, which we usually obtain using the defintion.

@PeterTamaroff I didn't get why this matrix is upper triangular

@charlie Because all therms under are zero

and over the diagonal there is the lambda

I mean because for i>j, we had the terms all equal to zero

But I still don't get why the determinant equals 1. I understand using the formula for a triangular matrix

but how can we calculate it using another method ?

do we calculate it according to the r-th line ?

@user43758 Ah, okay, so is zero if i>j

@user43758 'cause it seems that, to me, that the matrix would be 1 at the diagonal and lambda everywhere else

because it says the we have $\lambda$ if $(i,j)=(r,s)$, and r, s are different integers

@Charlie But look

I have place lambda on the first row

The determinant I get

If I develop according to the first row is

1\times the determinant of the identity matrix

+ (-1)^r+s\times \lambda \times determinant of the identity matrix

okay, okay

..

@PeterTamaroff

@Charlie What do you think ?

@user43758 I still don't get why do we have the zero entries

@Charlie A gente devia ganhar dinheiro vendendo identidades falsas tuas.

@GustavoBandeira C'mon.....

@GustavoBandeira Que falsas tuas ?

@user43758 Stupid question: is it a nXn matrix?

no

lol

ok

@user43758 is it like this?

@user43758 ??

Guys, I have to go, bye bye! Good night!

I tried to calculate the determinant according to the first line

So I get: 1+ (-1)^{r+s} \lambda

@MarianoSuárez-Alvarez ?

@user43758 I thought we had settled this =|

But what am I doing wrong..

@PeterTamaroff

Why am I not getting 1

@user43758 Tell me again what the defintion of your matrix is

Code that properly pleaaaaaaaaaaaase.

OK.

Then it is the unit matrix plus a canonical matrix with entries not in the diagonal.

so..

Well, do you see that a matrix in upper triangular form has what we said as a determinant?

This is my problem how do you express the determinant ?

I get 1+(-1)^{r+s} \lambda

@user43758 thats called a Kronecker delta

@user43758 written as $\delta_{i,j}$

@user43758 Try to see if in your book, you have a formula for the determinant of an uper triangular matrix.

The idea is as follows:

$$\delta_{i,j}$$= 1 if i=j and 0 if i \neq j

@user43758 yes

Develop the matrix throughout the first colum. Then it is $(-1)^{1+1}a_{11} \det(M)$ where $M$ is the minor of $1,1$, and all the zeroes under $a_{11}$ (i.e $a_{j1}=0$ ) kill the subsequent determinants (of the other minors)

Then you do the very same with $\det(M)$ until you get to $a_{nn}$ so you get the determinant is $\prod a_{kk}$

If the matrix is "lower" triangular, just use $\det(A)=\det(A^T)$

Why? We can simply develop according to the r-th row

we have to terms

a determinant multiplying 1 and another one multiplying \lambda

I want to understand why am I not getting 1..

The $r$th row? OK, what is the determinant of say

Note that the determinant of the minor of the $\lambda$ term will be zero.

Yes

ok I understood now

OK, can you see it is $1$?

the determinant multiplying the lambda equals 0

that's why

thank you!

Yep.

Sorry for being kinda slow

@user43758 Meh. Not a problem at all.

How long are you going to still be online

just in case..

I guess I will be at least 3 more hs.

I am trying to solve something.

ok

@PeterTamaroff I wrote $det(E)=1+(-1)^{r+s}\lambda\delta_{r,s}= 1$ such that $\delta_{r,s}$ is the (r,s)th minor

@user43758 Huh?

Didn't we agree it is $1$?

Yes we did

Oh, sorry.

I misread it for the Kroenecker delta.

OK, that is fine.

Try not to mix up common notation.

I know.

That's why I specified it was the minor

@PeterTamaroff What are you working on ?

@user43758 Uniformly distributed sequences. I can't even get started.

I realized how to prove something, but can't prove another more specific thing.

Anybody wanna take a crack at this? math.stackexchange.com/q/264403/12952

@AlexanderGruber Sorry I don't know what adjacent matrix is

oh, you just number the vertices on your graph and put a 1 in the $i,j$ position if vertex $i$ and vertex $j$ are connected, and a $0$ otherwise

@PeterTamaroff Are you online ?

@user43758 Yes.

Can I ask you another question ?

Yes.

Hah...

So, hashtags in resumes...yes or no?

I need to show that the determinant equals -1

@user43758 What is a determinant?

@Andrew Just google it. It is the unique multilinear alternating function such that $f(I_n)=1$.

@user43758 Try a few examples, see if you can work something out.

I did a 4x4 representation of this just to see what it looks like and it seems to me that all the terms are 1 except the one situated in the rth row

Christopher Robin just got kidnapped, sorry.

@Peter Oh, I've stumbled into math by accident.

@Andrew How?

@PeterTamaroff

@Peter I think I intended to enter a different chat

@user43758 OK, tell me what $E^{1,3}$ is over $4\times 4$.

@Andrew I see.

@user43758 You sure?

no it is a and