Mathematics

A subset $S$ of the plane is called "connected" if there are two open sets $A$ and $B$ where $(A\cap S)\cup(B\cap S)=S$ and where $(A\cap S)\cap(B\cap S)=\emptyset$. (In other words, $S\subseteq A\cup B$, and $(A\cap B)\cap S=\emptyset$.)

You also want $A\cap S\ne\emptyset$ and $B\cap S\ne\emptyset$ to avoid triviality

Zach Hauk

@AkivaWeinberger connected is it's complement is like

umm what's the word

oh, nvm im thinking of simply connected

simply connected has a connected complement

@Akiva so... i think the countable case is trivial

because a disconnected subset is the union of two non disjoint nonempty open sets

so in order to have that we'd have to have something to divide it... a jordan curve

which is impossible because none self-intersect or intersect with eachother

is that correct?

Ted Shifrin

12:43 AM

@Zach: That's an old-fashioned definition that used to be used for subsets of the plane when one took the complement in the Riemann sphere.

Zach Hauk

@Ted oh :[

Zach Hauk

12:59 AM

@Ted so, do you think Artin's is good for like first year algebra? I've read a bit of this other one but it was all basic group theory.

i was asking, because i think Artin's is finally in stock

Zach Hauk

1:14 AM

oh i just realized that a 2-dimensional determinant is kind of like crossing a rational equality

zed111

If WA=AW for a skew symmetric matrix W, what can we say about A. Anyone?

Can we prove A=cI where c is a constant and I is the identity matrix

Daminark

@Zach Artin is a quality book

It gives a good treatment of algebra and linear algebra

Zach Hauk

i'm rusty on my linear algebra, if $A$ is a matrix whose eigenvectors make another invertible matrix called $X$, then $X^{-1}AX$ is diagonal, right?

zed111

Yes thats true.

Zach Hauk

alright, phew :P

working with spectral theorem right now

zed111

1:24 AM

hmm that seems a nice route

Zach Hauk

what do you mean? :P

Semiclassical

What you're after is: Which matrices commute with a skew-symmetric matrix W?

zed111

to attack the problem.

Yeah

@Semiclassical

Semiclassical

I don't think it's true that the only such matrix is the identity, alas.

zed111

Even I don't think so. Actually I had to prove that A looks like cI when W(QAQ^T)=(QAQ^T)W where Q is any orthogonal matrix

Semiclassical

1:28 AM

Actually, it's definitely false. Any skew-symmetric matrix commutes with itself.

This sounds like a Schur's lemma thing, though.

zed111

okay. That makes even the 2nd statement false since it does not hold for Q=I.

Semiclassical

Is there something particular you're trying to prove beyond this?

zed111

It is a question on proving how a tensor looks like if it is frame indifferent. Here is the full question Q2

tinypic.com/view.php?pic=1j9phh&s=9#.WK-ONhIrJsM

Semiclassical

Hmm.

Could the condition be that WA=AW for every skew-symmetric W?

zed111

Yes

Semiclassical

1:39 AM

Ah. Then A=W isn't a counterexample.

W commutes with itself, but it needn't commute with another skew-symmetric matrix.

In 2D that doesn't help since there's only one skew-symmetric matrix (up to a constant factor) but in 3D you've got 3 of them.

The condition is that A commute with all 3 of them, and I think that then A=cI is required.

zed111

Right.

Semiclassical

My brain isn't up to showing that's true, alas.

Zach Hauk

oh, i don't have any LaTeX editor on this operating system

i guess ill have to use something like sharelatex

zed111

Another way is if we take the axial vector of W as v, then the condition is v \times Ax = A v\times x for every vector x. This is true if Ax = cx. Hence every x is an eigen vector of A\implies A=I.

Semiclassical

Sounds sensible.

zed111

1:47 AM

Thanks @Sem

@Semiclassical

@ZachHauk which OS?

Zach Hauk

i'm on windows 10 right now

but i was using Linux Mint 17

which is my usual OS for STEM related tasks

@Semiclassical all origin preserving isometries of $\Bbb R^n$ are linear maps, right?

i.e. rotation is a linear map with an orthonormal basis

reflection is multiplication by negative of the identity

actually, i can pose that question to anyone in this chat

zed111

I think that is true @ZachHauk

Zach Hauk

and so, therefore, linear maps which are isometries must preserve the dot product... right?

because isometries would preserve angles too

zed111

True

Zach Hauk

what i want to show is that

there exists an invertible linear map from a basis of orthonormal vectors to the identity

which would therefore prove that all orthonormal bases are linearly independent

there's probably an easier way but whatever :P

zed111

2:03 AM

Okay how are you trying to prove existence of invertible map using the dot product preservation property?

Zach Hauk

notice that the standard basis has all dot products 0

so if all the other dot products are 0

then there must be an invertible linear map (isometry!) between them

assuming that my statement is right

actually

zed111

Isn't having zero dot product, by definition, a property of orthonormal maps?

Zach Hauk

yeah

i mean, there are many ways to go about doing this

i could prove that all orthonormal matrices have inverses

and so that inverse would map it back to the identity, which has all linearly independent basis vectors

zed111

hmm you're going the right way.

Here is what I thought: Assume e_i are orthonormal basis. Then let \sigma c_i e_i=0 with at least one c_i not zero. Just one c_i zero would give the vector itself zero. Otherwise if multiple c_i's are non-zero, then you can have an eqn like \sigma c_k e_k =0. Keep k=j on LHS and remaining on RHS. Then take dot product with e_j on both sides to show c_j=0, a contradiction.

Zach Hauk

ah, something along the lines of contradiction

that works, didn't think of that to be honest :P

zed111

2:18 AM

Or even easier just write \sigma c_i e_i = 0. Then take dot product with e_j both sides to prove c_j=0 for all j.

Zach Hauk

oh boy i gotta diagonalize 8 quadratic forms @.@

i'm bringing out the ti-84

JessyunBourne

2:35 AM

Can anybody answer the follow-up questions I asked the guy who gave the second answer to this question: math.stackexchange.com/questions/2153826/…

I'm worried he's never coming back.

arctic tern

@JessyunBourne for me it's the first answer.

and no (3) is not a ring homomorphism

for instance [1][n-1]=[0] inside Z/n but (x^1)(x^(n-1))=x^n does not equal x^0.

addition in Z/n "wraps back around," whereas addition in the exponents of x in Z[x] do not wrap back around, they just keep getting bigger and bigger

Simple

Let $M$ be a random variable that is distributed uniformly on $\{0, 1, 2, ..., n\}$
for some fixed positive integer $n$. how to find the probability generating functions of M

arctic tern

step (1): look up definition of probability generating function. have you done that?

Simple

I did, $E(s^M)=\sum_{m=0}^ns^m/n$

arctic tern

how many elements does {0,1,...,n} have?

Simple

2:48 AM

n+1

arctic tern

so it should be n+1 in the denominator, right?

Simple

that is a mistake

arctic tern

okay. so now $E(s^m)=\frac{1}{n+1}\sum_{m=0}^n s^m$. Do you know the formula for $\sum_{m=0}^n s^m$?

Simple

I don't

arctic tern

do you know what kind of a sum $1+s+s^2+\cdots+s^n$ is called?

Simple

2:51 AM

what is the value of $s$

arctic tern

this whole thing is a function of s.

Simple

my book has no information about $s$

arctic tern

yeah, that's because it's a function of s

s is an independent variable, like x in f(x)=x^2.

Simple

if $s<1$, it is a geometric sum, if $s>1$, I dont know

arctic tern

it's a geometric sum either way

$|s|<1$ is only important if you're adding infinitely many terms

(which you are not)

Simple

2:53 AM

ok, thanks

my true question is, $X$ is bin(n,U) where $U$ is uniform on (0,1). I need to show $X$ is uniformly distributed on $\{0,1,2,\dots,n\}$

$G_X(s)=E(s^X)=\int_{0}^{1}\left(\sum_{x=0}^{n}\binom{n}{x}(ps)^xq^{n-x}\right)d‌s=\int_{0}^{1}(1+p(s-1))^nds=\frac{1}{n+1}\left(\frac{1-s^{n+1}}{1-s}\right)$

arctic tern

what does bin(n,U) mean?

Simple

binomial distribution with parameter $n$ and $U$

$U$ is a uniform distribution

arctic tern

so bin(n,U) is a random-variable-valued random variable? gnarly.

Simple

something like that

if I understand correctly

arctic tern

but then I don't see how it makes sense to say X is uniformly distributed on {0,1,...,n}.

or that X takes values in {0,1,...,n} at all. looks like X equals random variables bin(n,u) where u is some value in [0,1].

Simple

3:00 AM

I agree your second part

WDUK

When there are two options to factor my answer to make it simpler to read, is there a good rule of thumb to decide which one? I can pick between an x^2 or a ln(x/2) wondering what would make it easier to read for most

arctic tern

@WDUK question seems too vague. can you be more specific.

Simple

that is the entire statement on the book

WDUK

well i have this answer x^2 +4/x + 3ln(x/2)x^2 + 4ln(x/2)

I want to factor either x^2 or ln(x/2) to simplify it a bit but wondered which would be the better choice for easier reading

arctic tern

@Simple wasn't talking to you that time

Simple

3:05 AM

oh

arctic tern

@Simple but now that you mentioned it's in a book, try reading through the material that exists before the exercises

see if there's anything relevant

@WDUK just don't factor out anything

WDUK

why not ?

Simple

@arctictern the book only gives the definition and some easy example

arctic tern

the definition? an example?

Simple

definition of the generating function and an example about how to find the generating function of a constant variable

$P(x=c)=1$, $G(s)=E(s^X)=s^c$

user84215

3:19 AM

hello

user84215

can we have an uncountable set of real numbers that none of its elements is a limit point with respect to elements of the set ?

arctic tern

3:33 AM

@aminliverpool have you tried googling?

I found http://math.stackexchange.com/questions/310113/accumulation-points-of-uncountab‌le-sets

user84215

yesyes

user84215

yes

user84215

this is not the answer to my question

arctic tern

You asked if there is an uncountable set of real numbers that does not contain any of its limit points, right?

user84215

yes

arctic tern

3:40 AM

and the answers over there say "every uncountable subset of R contains at least one of its accumulation points."

user84215

accumulation point is not different from limit point ?

Akiva Weinberger

3:53 AM

@ZachHauk The standard example of something that's connected but not path-connected is the graph of $y=\sin(1/x)$ for $x\in(0,1]$ together with the line $\{0\}\times[-1,1]$ on the $y$-axis.

en.m.wikipedia.org/wiki/Topologist's_sine_curve

@ZachHauk What do you mean?

WDUK

question, if i have a function (a+b)^c can i simply differentiate a^c and b^c seperately then add them together to get f'(x)

Daminark

@aminliverpool All accumulation points are necessarily limit points

Wait actually

Limit points and accumulation points are the same

I was thinking of condensation points lmao

user84215

4:22 AM

thank you. I notice my mistake. I got the answer

Pedro Tamaroff

@TedShifrin Hehe, welp.

Akiva Weinberger

@WDUK The derivative of $1^x$ is $0$, but the derivative of $(1+1)^x$ is $2^x\ln x$, so I'd say no.

Simple

4:38 AM

Let $X(\geq 0)$ have probability generating function $G$ and set $t(n)=P(X>n)$ for the "tail" probability of $X$. How to show the generating function of the sequence $\{t(n):n\geq 0\}$ is $T(s)=(1-G(s))/(1-s)$

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$Mathematics$ Mathematics