Mathematics - 2013-11-18 (page 3 of 3)

Charlie

20:01

@Chris'ssis it's normal

Chris's sis

@Charlie blood sugar issues?

Charlie

@Chris'ssis nope

Chris's sis

@Charlie ok

Charlie

:/

Chris's sis

I know many young people with high blood pressure, but they are careless about that. Dizziness is often met in this case.

Charlie

20:08

@Chris'ssis I don't know :/

Jebediah Kerman

I need some help

I have to find the derivative of x-4ln(3x-9)

so thats equal to d/dx of x - d/dx of 4ln(3x-9)

d/dx of x is 1

d/dx of 4ln(3x-9) simplifies into 4/(x-3)

so that would mean that the d/dx of x-4ln(3x-9) is

1 - (4/(x-3))

but I checked the answers and thats wrong

where did I make a mistake?

hello?

Asinus

20:30

Hi,

Jebediah Kerman

hi

Asinus

I find the same answer as you

what's the answer you are supposed to find ?

Jebediah Kerman

It turns out my answer is the correct one

but I need more help

I have to find the critical points/numbers

Asinus

what do you mean by critical points ?

Jebediah Kerman

you know for finding the intervals on which the slope is increasing/decreasing

Asinus

20:33

ok

one second

Jebediah Kerman

are you aware of this type of problem?

how is $1-(4/(x-3)$ equalt to $((x-7)/(x-3))$

I don't get it

??

helloo

Asinus

20:49

sorry

you can plot the derivative

if the derivative is positive, the function is increasing

if the derivative is negative, the function is decreasing

Jebediah Kerman

I am not to plot

how do I get the critical numbers from $1-(4/x-3)$

Asinus

now the derivative is positive on ]-infinity,3[ and on ]7, +infinity[

Jebediah Kerman

I do not understand where you get 7 from

Asinus

wait a minute, i'll explain after

and the derivative is negative on ]3,7[

and it's =0 if x=7

from this you can get the variation tableau

is it ok ?

now i explain where i get 3 and 7

first thing, you notice that neither the derivative nor the function are defined for x=3

because if x=3, then x-3=0 ; but you can't divide by zero, right ?

now for the 7 :

you want to know when the derivative is equal to 0

so you have 1-(4/(x-3)) = 0

which is the same as : 4/(x-3)=1, right ?

Jebediah Kerman

yes.

Asinus

20:58

but 4/(x-3)=1 means that 4=x-3

which means that x=7

so the derivative is zero when x=7

Jebediah Kerman

thanks

that helped a lot

Asinus

you're welcome

Jebediah Kerman

bye

Asinus

bye !

Dave

21:16

hey

let's say i have 3 variables, a b and c, how do i elengantly verify that these 3 are consecutives ?

anon

presumably you mean integers instead of variables

Dave

i'll use that in a programming language

so yeah i meant integers

Enjoys Math

choose min {a,b,c}

Dave

i figured that if i sum them all, and divide by 3, i get the middle integer

Link

Hey how would I convert this into cartesian coordinates? I got this far:

$$r=2csc(\theta) = \frac{2}{sin(\theta)} --> r^2 = \frac{2r}{sin(\theta)}$$

What do I do after this?

Enjoys Math

21:20

Just use a bunch of ifs then adjust your definition of elegant, that's the elegant solution

anon

I think |(a-b)(b-c)(c-a)|=2 iff {a,b,c}={n,n+1,n+2} for some integer n. This is mathematically elegant, but probably not so in a programming context.

@Link do you mean convert to cartesian coordinates? it's already in polar.

Link

@anon, yes, my mistake.

anon

$r\sin\theta=2\iff y=2$

Link

How did you do that?

anon

$x=r\cos\theta$ and $y=r\sin\theta$, remember those

Link

21:24

Yes, I understand that.

But how can you make the rsin(theta)?

is equal to 2?

anon

$r=2/\sin\theta\iff r\sin\theta=2$

you just multiply by $\sin\theta$

Link

Ah...

I get it!

Thanks!

But.....

Doesn't that mean I just end up with x^2 + 4 = 2r/sin(theta)?

I might be missing something here?

anon

what in the world?

oh, you plugged y=2 into x^2+y^2= 2r/sin(theta). while yes, the latter equation is true, I have no idea what you want to do with it.

Link

Me neither

I'm not sure what to do to be honest.

I understand how you got the y=2=rsin(theta)

anon

are you unsatisfied with the equation y=2 for some reason? what's wrong with it?

Link

21:31

Wait....

That's the finaly equation?

anon

yes

Link

Ah!

anon

and you hurt its feelings

by thinking it wasn't good enough!

Link

Thanks...

Oh..

What about the x^2?

anon

what about it?

Link

21:32

What happens to it?

Or why isn't it needed?

anon

why would it be needed? again, what. is. wrong. with. y=2?

Link

Nothing..

anon

what "happens" to it is that x^2 only exists in really weird and overcomplicated versions of the equation y=2

there is absolutely no reason to want an x^2 term in the equation of a line

Link

Okay then.

anon

so don't bother making an x^2 term where you don't have to have one

Link

21:33

Got it.

Charlie

22:02

@anon hi, ani

Mike

22:16

I posted a question on MSE, but later on MO (as I felt it was a potentially appropriate question there); I got an answer on MO. Is it acceptable to answer my MSE question with a link to that answer to get it off the unanswered list?

Enjoys Math

@anon

Are there other constructions that are quotients of free groups that you know of other than tensor product?

Mike

every group is the quotient of a free group

Enjoys Math

Oh

anon

@Charlie hello

Enjoys Math

Let $G = {a,b,c,\dots}$, then form the free group $F(G)$

anon

22:27

@Mike yes, generally CW'd. if it were me I'd put the MO answer in my own words and add commentary to it

Mike

Yeah, I CW'ed it. (The MO question was just a link to a source; I'll link the source in the question)

Enjoys Math

Define the map $F(G) \to G$, as naturally as possible.

Then take the quotient by $\ker$

I see that

but I'm looking for exotic constructions that start with setting some expressions = 0 by saying they generate a subgroup and taking the quotient

anon

direct products, semidirect products, wreath products are all quotients of free products. every group is a quotient of the free group on its underlying set by its multiplication table. the tensor product is a quotient of the free abelian group on the underlying set of two abelian groups' direct product.

Chris's sis

@robjohn are you around?

Enjoys Math

I thought it's on their cartesian product with no meaning defined algebraically

@anon

$F(M\times N)$ for modules $M, N$.

anon

22:32

I clarified that in an edit before you even asked :)

> underlying set

Enjoys Math

oh, sneaky mofo

robjohn

@Chris'ssis yes, just posted some results about the cylinder-sphere stuff

Enjoys Math

What's a wreath product?

Mike

Wreath product

In mathematics, the wreath product of group theory is a specialized product of two groups, based on a semidirect product. Wreath products are an important tool in the classification of permutation groups and also provide a way of constructing interesting examples of groups. Given two groups A and H there exist two variations of the wreath product: the unrestricted wreath product A Wr H (also written A≀H) and the restricted wreath product A wr H. Given a set Ω with an H-action there exists a generalisation of the wreath product which is denoted by A WrΩ H or A wrΩ...

Chris's sis

@robjohn have you ever computed this without integrals? $$\lim_{n\to \infty} \sum_{k=1}^{n}\frac{k}{k^2+n^2}$$

robjohn

22:36

@Chris'ssis You mean without using Riemann Sums?

Chris's sis

@robjohn no, without integrals (here we also include the one that comes from Riemann sums)

anon

mfw toc

Enjoys Math

I would maybe subtract $\sum_{k = 0}^{-n}\frac{k}{k^2 + n^2}$ and say you're calculating $1/2$ the limit, this makes things more symmetric

@Chris'ssis

Charlie

@anon what about it?

robjohn

@Chris'ssis I think I can do it using series, but the details might be less than rigorous...

Chris's sis

22:45

@robjohn ok. I also thought of this variant.

@EnjoysMath what do you mean?

robjohn

@Chris'ssis Is there another way?

Chris's sis

@robjohn There is one, but I need to find it.

robjohn

@Chris'ssis how do you know if you haven't found it yet?

Chris's sis

@robjohn Intuition (it never lies)

robjohn

@Chris'ssis okay

Enjoys Math

22:49

It's $\lim_{n\to \infty} \sum_{k=1}^n Re(\frac{k + ni}{|k + ni|})$

Charlie

@Chris'ssis :D

robjohn

@Chris'ssis: I am currently looking to see if the result I got regarding equi-angular loci on the sphere was known.

Chris's sis

@robjohn ok. Some new discovery?

Enjoys Math

@robjohn take the wreath product of the spheres, lol j/k

robjohn

@Chris'ssis That is what I am trying to find out.

@Chris'ssis The thing about elliptic cylinders and the sphere

Chris's sis

22:53

@robjohn it would be great to come up with something new! Maybe you should publish the paper.

Enjoys Math

$\lim_{n\to \infty} \sum_{k=1}^n Re(\frac{k + ni}{|k + ni|})$. Extend this to $\lim_{n \to \infty} \sum_{k =1}^n \frac{k + ni}{|k + ni|}$. Your sum is the real part of that

Chris's sis

@EnjoysMath yes

Enjoys Math

So your some is of unit vectors

it's sum in a finite group

TEH GROUP!! D:

So whatever it approaches, it approaches some subset of that group

robjohn

@EnjoysMath Is that some holiday item where we sum the logs?

Enjoys Math

Wut?

anon

22:58

opens the pill bottle for enjoys

robjohn

@EnjoysMath wreath product sums the logs... holiday humor.

Enjoys Math

i was serious about the last bit, she's taking the real part of a sum of unit vects

Chris's sis

Or doing this

$$ \lim_{n\to \infty}\sum_{k=1}^{n}\left(\frac{k}{k^2+n^2}-\frac{1}{2(n+k)}\right)=0$$

robjohn

@Chris'ssis that looks non-trivial

Andrew D

Hey people, would someone please be able to give me a quick hint on this linear algebra question - if we have a bilinear form on a finite dimensional vector space is non-singular when restricted to the orthogonal component of some subspace, then the form is non-singular on the whole space?

Chris's sis

23:04

@robjohn this idea just came to mind.

robjohn

@AndrewD what about $x^2$ on $\mathbb{R}^2$?

restricted to the $x$-axis it is non-singular

Pedro Tamaroff

@robjohn He said bilinear form?

robjohn

@PedroTamaroff $$\begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix}1&0\\0&0\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}$$

polarize it to get the bilinear form

anon

so, that's the quadratic form

the bilinear form being $((x_1,y_1),(x_2,y_2))\mapsto x_1x_2$

Mike

@anon "mfw toc"?

anon

23:19

http://www.urbandictionary.com/define.php?term=MFW
http://www.internetslang.com/TOC-meaning-definition.asp

Mike

Oh, table of contents

Charlie

Ah @pedro !

Mike

Well, it is a book of lectures!

Charlie

@anon anon...

Andrew D

I have a very hackish idea which I'm not convinced by; suppose we have that $\phi$ is the bilinear form on $V$, we have that $U,W \leq V$ and $U$ is the right kernel of $W$ (as I don't know how to do an orthogonal symbol). Then, as $U,W \leq V$, orthog($V$) $\leq$ orthog($U$), orthog($W$) = $U$. Then as $phi$ restricted to U is non-singular, the intersection of $U$ and orthog($U$) is the zero vector, so orthog($V$) = { zero vector }, so $phi$ is non-singular

anon

23:29

:)

Andrew D

Does my argument work (it could probably do with some polishing)?

Charlie

@anon ..the things you say

Pedro Tamaroff

@Charlie Yiss?

Charlie

@PedroTamaroff a little something, not very important, but surprised me

Pedro Tamaroff

@Charlie What is that?

Charlie

23:38

@PedroTamaroff this

I can't believe it

Pedro Tamaroff

@Charlie Ah, yes. I usually avoid that page mainly because of that =)

Charlie

@PedroTamaroff it's probably a very biased sample

Pedro Tamaroff

@Charlie Yep. And people are generally on the soft side.

Andrew D

Is "ratemyprofessors" an American thing only?

Charlie

@PedroTamaroff yes...

@AndrewD "colleges and universities across the United States, Canada and the United Kingdom"

Pedro Tamaroff

23:42

Does anyone have a TeX template with titles, sections and subsections?

Mike

writelatex.com has one

Pedro Tamaroff

@Mike Ah, didn't know that one! =D

Charlie

it's a very cute one

@PedroTamaroff btw, how are you?

Pedro Tamaroff

@Charlie Pretty good, lost my voice with the tennis classes though, so no singing lessons tomorrow =/

Chris's sis

@robjohn another way is to express all in terms of digamma function and then use the asymptotic expansion of digamma to get the desired limit. (it works)

Charlie

23:45

@PedroTamaroff oh no!

@PedroTamaroff will you show us your vocal talents? ;)

Pedro Tamaroff

@Charlie Heh, not now.

Still learning.

Charlie

XD

Alizter

Looks good?

Pedro Tamaroff

@Alizter Yep =)

Andrew D

^^

Pedro Tamaroff

23:51

You proved the contrapositive, though.

Which is fine.

But there is no contradiction.

Alizter

Terminology?

Pedro Tamaroff

@Alizter You want to prove $a\implies b$ under certain hypotheses.

You proved $\neg b\implies \neg a$ under those hypothesis.

Alizter

Negate both sides ;)

Pedro Tamaroff

@Alizter I'm just saying that is not a "pure" contradiction, so to speak.

Charlie

:D:D:D

Ted Shifrin

23:53

Howdy, hoarse @Pedro

Pedro Tamaroff

It is rather a proof by contraposition.

@TedShifrin Hello!

Alizter

@PedroTamaroff Yes I know. Rigor was not my goal in this proof. It is elementary and educational rather than erm obsessive ;)

Charlie

hi @leo !!!

Ted Shifrin

Hi @Charlie

Charlie

@TedShifrin hi

Pedro Tamaroff

23:56

@Alizter I just think people need to tell apart proof methods, that's all.

Mike

Hi @Ted

Ted Shifrin

Hi @Mike

Alizter

@PedroTamaroff TBH proof methods are not analysed in enough detail at an elementary level. Contradiction - Contraption - Contrabass - They all sound similar therefore educational logic means they must be the same.

Pedro Tamaroff

@Alizter That makes no sense.

Alizter

Contraposition - Contraception

Ted Shifrin

23:58

Huh?@Charlie

Charlie

@TedShifrin yes.

Alizter

So far the word induction has 5 different meanings

leo

I've seen a definition of separable element which says: Let $F\subseteq E$ fields, $a \in E$ is separable if it is transcendental over $F$ or if its minimal polynomial over $F$ is separable. I other two sources there is no transcendence part, and many things are easily proved if we assume that separables are among algebraic. Should I point it as a mistake?

@Charlie hola

Charlie

@TedShifrin better not saying anything else, last time I discovered multiple indentities i got an advertence

Alizter

@PedroTamaroff Hows your education?

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$Mathematics$ Mathematics