Mathematics - 2013-11-24 (page 1 of 3)

usukidoll

00:00

for n = 2

because the base is $n!$/$(s-a)^n$

Alexander Gruber

@anon yep, absolutely, that's what i'm trying to break out of. :p

Alec Teal

/me returns!

Alexander Gruber

@KarlKronenfeld that's what people always tell me to think of, is vector spaces, but those aren't too much better than modules for me. (i mean, i get the basics, i've passed all the classes and everything. but i don't understand them, you know?)

anon

I think it is dangerous to think of modules like vector spaces

I came to module theory through representations, where a G-rep is essentially a k[G]-module

so modules are like linearized versions of group actions

usukidoll

bah I'll do it on paper and then scan here k ^^ for the induction

Don Larynx

00:11

hey anyone why is it that the sequence $\{\frac{1}{n}\}$ diverges in the metric with $d(x, y) = |x - y|$?

shouldn't it converge to $0$?

because for any $\epsilon$ you can find two numbers that satisfy $|x_1 - y_1| < \epsilon$?

anon

does your metric space happen to not include 0?

because 1/n certainly converges in R with the euclidean metric

Charlie

that was deep

anon

(or perhaps it's possible you mean to be talking about the sequence of partial sums)

Don Larynx

@anon just making sure.

Right, it's an open set.

so it doesn't contain its limit point, 0.

hi @charlie.

Charlie

@DonLarynx olá

that means hello

Don Larynx

00:22

Charlie, I'm hispanic. :)

(soy hispano)

(soy sauce hispano)

usukidoll

Trying to prove by induction...
Show by induction that $£(x^ne^{ax})$ = $n!/(s - a)^n$ for s > a, and
n = 1,2, ..... I've done n = 1 already. I just need to do know if I did n = 2 right...

https://docs.google.com/file/d/0B54O0sv0MQILUV9GclRGS0ZNQlk/edit

Don Larynx

(soy sauce his pan, ohhhh) @charlie

@usukidoll you can't do that for infinitely many $n$.

like by hand

you have to assume $P(n)$ holds

and show $P(n+1)$ is true

Charlie

@DonLarynx hmm weird

usukidoll

well that's strange because another study buddy of mine says that I could do it through integration by parts over and over for n = 1,2,3...k

Don Larynx

Then show us your work! :)

usukidoll

00:26

x.x I uploaded the one for n =2

unless the link didn't work... I changed the access on the image

Don Larynx

ah okay, didnt see that

usukidoll

docs.google.com/file/d/0B54O0sv0MQILUV9GclRGS0ZNQlk/edit

ADR

Let me ask an stupid question, what is the normal bundle of a point (as a submanifold)?

usukidoll

D: I don't know x.x

Ted Shifrin

@ADR: The tangent space of the ambient manifold.

Alexander Gruber

00:43

@anon do you have any kind of way to visualize the linearization there

i also think it's easier as group actions, but i'm not always sure how to view the decomposition into subspaces

anon

the actions can be added instead of just multiplied, which makes sense because they act on things that can be added

Karl Kronenfeld

It is pretty safe to think of vector spaces as the launchpads and targets for linear maps. The reason why you can't use the basis set alone is that you are interested in the kernel and image spaces corresponding to these maps. In this sense module theory is not too different, since you often do not even have bases work with so you want structures with scalar-multiplication-preserving maps and the substructures corresponding to the kernel and images of these maps.
A completely different thought: I found it enlightening to note that all abelian groups are $\mathbb Z$-modules and your average …

ADR

@TedShifrin thanks!

Ted Shifrin

@ADR: Most welcome.

anon

there is a functor, for example, that takes a G-set X and gives the free k-vector space generated by X as a k[G] module, with the G-action just extended from X linearly. this is a concrete kind of "linearization." in general the ring acting on a module need not "come" from a group action in this way, but this gets the idea started.

ADR

00:48

@TedShifrin one more question, do you know where I can find examples of sprays? Especially I'm interested in sprays on $S^1$

Ted Shifrin

Interesting that I've avoided these my whole career. Isn't a spray just a lift of a differential equation on $M$ to one on $TM$ (e.g., geodesic sprays are the flows on $TM$ that project to geodesics on $M$)?

usukidoll

yay got through the integral for the sinax problem...just need the induction proof part and I'm done with this thing.

ADR

@TedShifrin yes, I'm trying to find an spray on $S^1$ for wich not all maximal solution curves are defined on all of $\mathbb{R}$. (It's an exercise on Bröcker's Introduction to differential topology) but I don't have a clue on how to tackle it.

Ted Shifrin

Ah, it's just like finding a vector field on $\Bbb R$ that is not complete.

usukidoll

awww D:

Alec Teal

01:02

Evenin Ted

ADR

@TedShifrin and then lifting to $TM$ and prove that it is an spray i.e. $\xi(sv) = Ts(s\xi(v))$ for all $s\in \mathbb{R}$ and $v\in TM$, but thats when things get messy

Ted Shifrin

Since the tangent bundle of $S^1$ is trivial, can't we actually do it on $\Bbb R$ and pull back to $S^1\times\Bbb R$? I can't decipher what you just wrote :(

Hi, @Alec!

ADR

@TedShifrin ouch, its on page 112 of Brocker's books.google.com.mx/…

Alec Teal

@TedShifrin you usually spoil me with help :P I have used intuition and concluded that there's some link betwen the Schwarz (is there a T? My book says nay) inequality and the determinant of a "gram matrix" - however it is only by chance that I know of the Gram matrix. Because of this I want to do it by definitions, math.stackexchange.com/questions/578614/… but no one has replied 'cept one person.

@ADR "A first course in geometric topology and differentia geometry" ftw.

Ted Shifrin

@ADR: I would try to take a non-complete vector field on $\Bbb R$ and cook up a vector field corresponding to it in $TS^1$.

ADR

01:14

@AlecTeal on wich page I can find that?

Alec Teal

It's a book.

Ted Shifrin

No t @Alec. I'll look.

ADR

@AlecTeal yes I know, I meant on wich page I can find second order differential equations and sprays in that book

Alec Teal

@TedShifrin "t"? (to?)

Also @ADR was it you saying yo passed courses for something but didn't really understand it?

Enjoys Math

Look at this epic answer I made: math.stackexchange.com/questions/578610/…

I'm done for the night

Alec Teal

01:16

@ADR I checked the index, can't find it sorry.

Ted Shifrin

No t in Schwarz. I think Will overshot your knowledge, but his answer is the right direction. Consider the matrix $B$ whose columns are formed by your $v_i$, and consider the rank of $B$ vis-à-vis the rank of $B^\top B$, which is your Gram matrix.

Alec Teal

So it cannot be proved by manipulating definitions?

Also if you use his answer, the matix elements will take the form $\sum(<a,b><c,d>)$ (there wil be multiplications, and there will be adds)

Enjoys Math

Schwanz?

Alec Teal

I don't know how to get rid of them nicely, bounding the terms doesn't help with determinants @TedShifrin (no "Schwarz")

Ted Shifrin

No, not with any elegance. If the vectors are dependent, you should get one column of $G$ as a linear comb of the others, but the independent case isn't obvious to me.

LOL@EnjoysMath

Enjoys Math

01:23

@usukidoll whatre u working on?

Alec Teal

@TedShifrin I could perhaps default to induction, starting with n=2.

Enjoys Math

The Schwanz space is a sausage fest

Alec Teal

Not sure how I'd do that. If I try will you tell me "WTF IS THAT? BURN IT!" or "well done" @TedShifrin

Ted Shifrin

Why so stubborn, @Alec? There's beautiful geometry in the Gram determinant. See the end of section 2.1 of my diff geo notes!

Alec Teal

@TedShifrin I know, that's how I know what the Gram Matrix is.

But I stumbled across this from a totally different direction.

Ted Shifrin

01:26

I don't know where you're getting those products of dot products from Will's.

Alec Teal

Anyway, perhaps it's not worth it. I honestly thought it'd be pretty straight forward, having said that @TedShifrin I still want to try induction because the case k=2 is very special.

@TedShifrin a matrix of things of the form <a,b> timesed by a matrix of things of the same form, involves adding and multiplying things of the form <a,b>

Ted Shifrin

No, the $k=2$ case generalizes completely by extending orthonormally to a basis.

Alec Teal

/me is unfortunately overshot.

Ted Shifrin

No, the original matrix is just the vectors. You get dot products when you multiply by the transpose.

Alec Teal

It'd also work if I used the dot product, but I want to do it with <,>

I'm not sure it'd work otherwise.

Ted Shifrin

01:29

Same thing. Different notation.

Alec Teal

(Matrix multiplication, summing the <,>? )

Enjoys Math

@Charlie look got answer-robbed: math.stackexchange.com/questions/578610/…

Alec Teal

All <,> are the same is a VERY bold claim, dot product works nicely, with <,> we only know three things.

Ted Shifrin

No, $v_i^\top v_j=\langle v_i,v_j\rangle$.

Alec Teal

I'm sure you can back it up though :P How is matrix multiplication defined? If not adds and multiplications.

@EnjoysMath you at one point say 0 is an eigenvector, even I know that's wrong.

Enjoys Math

01:31

oh, doesn't matter in that post though

a detail easily forgotten

Alec Teal

I know, just saying.

Don't forget I said :P

Enjoys Math

They didn't even mention that $v \neq 0$ in the OP.

amatuers

Alec Teal

@TedShifrin is Matrix multiplication better thought of as what you just put (where vi and vj are rows/cols of a matrix?) because if I use that then the dot-product (the "adding and timesing I've been clinging to) becomes fine.

Charlie

@EnjoysMath hmm...

Alexander Gruber

@EnjoysMath that sucks. i upvoted you.

Enjoys Math

01:34

thx @AlexanderGruber

I may have done a real fly-by over his head in the answering

:D

usukidoll

@EnjoysMath prove by induction

I'm not sure if I did it right for n =2

sorry sANDwich in hand

@EnjoysMath Trying to prove by induction...
Show by induction that $\mathscr L$$(x^ne^{ax})$ = $n!/(s−a)^n$ for s > a, and
n = 1,2, ..... I've done n = 1 already. I just need to do know if I did n = 2 right... https://docs.google.com/file/d/0B54O0sv0MQILUV9GclRGS0ZNQlk/edit has n = 2

Pedro Tamaroff

@TedShifrin Helloes.

@usukidoll You want \mathscr L

usukidoll

yes because this is laplace

yay

Enjoys Math

Why not start with the base case $n = 0$, that seems the easiest

Pedro Tamaroff

@usukidoll Well, do you know that in general $\mathscr L(F(t)e^{ta})(s)=\mathscr L(F(t))(s-a)$?

usukidoll

01:45

well I started n = 1...

that's according to the Laplace table right?

or is this something else?

Enjoys Math

Now prove for $n = k$, assuming $n = k-1$ case is true.

Pedro Tamaroff

@usukidoll It is usually called a "shift property" of something similar.

usukidoll

that's like the same thing... like n = k n = k+1 n = k-1

Enjoys Math

Consider the integral $\int_0^{\infty} (x^n e^{(a-s)x})dx$

usukidoll

ok :D

Let n = k
$\int_0^{\infty} (x^k e^{(a-s)x})dx$

pourjour

01:52

supposing $\lambda \in \mathbb{R}$ and $(a,b) \in \mathbb{R}^2$ as a<b and $f[a,b] --> \mathbb{R}$ and f(a)-f(b)=0

Enjoys Math

Let $u(x) = x, dv = x^{n-1} e^{(a-s)x} dx$

usukidoll

integration by parts... super -_-

Enjoys Math

perform the $uv - \int v du $ trick or 'integration by parts'

usukidoll

u = $x^k$
du =$kx^(k-1)$

doh!

Enjoys Math

what

$u(x) = x$!!!

usukidoll

01:54

-.- my k-1 is not shifting up

oh u = x du = dx

Enjoys Math

!

:D

:'D

usukidoll

I'm new at Latex. Don't judge T_T

Enjoys Math

okay o_O

usukidoll

seriously T_T I'm learning

pourjour

how can I prove that there exists a $c \ ]a,b[ f'(c)=- \lambda\frac{f(c)}{c}$

Enjoys Math

01:55

you want to use \in for c \in (a,b)

$c \in (a,b)$

usukidoll

u = x du = dx
dv = $x^{n-1} e^{(a-s)x} dx$
v = $x^{n-1} \frac{-1}{a-s}e^{(a-s)x} dx$

Enjoys Math

yepperdy depperdy I think!

now compute the rest and show the result

see if that works

YOU GOT THIS!!!

usukidoll

alrighty integration by partsu

pourjour

@EnjoysMath yes

Enjoys Math

is $f$ continuous?

@pourjour

pourjour

02:00

@EnjoysMath it's continuous and derivable on [a,b]

Enjoys Math

cool

Well use the intermediate value theorem.

Pedro Tamaroff

@pourjour differentiable

pourjour

@PedroTamaroff yeah my english still bad

@EnjoysMath how could be that

Pedro Tamaroff

@pourjour Where are you from?

pourjour

on which function

@PedroTamaroff you forgot easily about me

usukidoll

02:02

$xe^{a-s}x$=$\frac{-x}{a-s}e^{a-s}xx^{n-1}$-$\int$ $\frac{-1}{a-s}e^{a-s}xx^{n-1}$

Enjoys Math

Mean value theorem

[[Image:Mvt2.svg|thumb|300 px|right|For any function that is continuous on [a, b] and differentiable on (a, b) there exists some c in the interval (a, b) such that the secant joining the endpoints of the interval [a, b] is parallel to the tangent at c.]] In calculus, the mean value theorem states, roughly: that given a planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints. The theorem is used to prove global statements about a function on an interval starting from local hypotheses about derivative...

usukidoll

come on x's that should be at the a-s y u no go up?

Pedro Tamaroff

@TedShifrin

You there?

Enjoys Math

@usukidoll I think you're missing some symbols on the LHS of that

usukidoll

T_T like x.x

pourjour

02:04

@EnjoysMath I know about MVT but on which function I'm going to apply it on

Ted Shifrin

Howdy @Pedro ... I wasn't here, although it apparently thought I was :)

usukidoll

$x^ne^{a-s}x$=$\frac{-x^n}{a-s}e^{a-s}xx^{n-1}$-$\int$ $\frac{-1}{a-s}e^{a-s}xx^{n-1}$

Pedro Tamaroff

@TedShifrin Ah!

pourjour

I need more clarification I spent all the day on it

usukidoll

what a mess

Pedro Tamaroff

02:06

@pourjour Ah, I haven't forgotten.

usukidoll

then supposedly I have to either take the integration by parts again on the right hand side of the equation right??? @EnjoysMath

pourjour

nice :D

@EnjoysMath concerning using MVT f(b)-f(a)=0

Ted Shifrin

@Alec: It is often useful to keep in mind two alternative definitions of matrix mult by vector: dot product of row with vector, or linear comb of columns, weighted by entries if vector.

Tough being indispensible @Pedro :)

Pedro Tamaroff

@TedShifrin Come again?

Ted Shifrin

Your needed clarification! :)

Pedro Tamaroff

02:09

@TedShifrin With what?

Ted Shifrin

Oh, never mind. I was commenting on your exchange with @pourjour.

usukidoll

?

Enjoys Math

@pourjour in the MVT set $f'(c) = \frac{f(a) - f(b)}{a-b} = \frac{f(c)}{c} \lambda$ and solve for $\lambda$. Assuming $f(c) \neq 0$ of course.

oh $f(a)$ not defined?

usukidoll

$x^ne^{ax-sx}$

Pedro Tamaroff

@TedShifrin Oh, =)

How are you doing?

You were kinda sick, right?

Enjoys Math

02:15

Define $f(a) := \lim_{x \to a^+} f(x)$, similar for $f(b)$ and perform MVT on it

Ted Shifrin

Much better, thanks. Been making my Thanksgiving menu for 12.

usukidoll

$x^ne^{ax-sx}$=$\frac{-x}{a-s}e^{ax-sx}x^{n-1}$-$\int$$\frac{-1}{a-s}e^{ax-sx}$
$x^{n-1}$

there we go...so the x..sjldkfs

Pedro Tamaroff

@TedShifrin Ah, that is nice.

usukidoll

so I have to integrate the right hand side again...the one after the integral sign right @EnjoysMath

Enjoys Math

idk, seems to have blown up

I don't think you have the far right correct

usukidoll

02:19

yeah because the code broke off

$x^ne^{ax-sx}$=$\frac{-x}{a-s}e^{ax-sx}x^{n-1}$-$\int$$\frac{-1}{a-s}e^{ax-sx}$ $x^{n-1}$

Pedro Tamaroff

@TedShifrin I have forgotten how one proved that for any base $\{v_1,\ldots,v_n\}$ of a vector space $V$; there is one and only one inner product that makes it an orthonormal basis.

usukidoll

see?

Enjoys Math

Okay, @usukidoll start over with $u(x) = x^n$

usukidoll

oh putting the space works XD

Ted Shifrin

We've not discussed this, @Pedro. But isn't it immediate?

usukidoll

02:21

oo trickster.. $u(x)$=$x^{n}$ $du$=$n$$x^{n-1}$

robjohn

@IanMateus Sorry, I was out.

Ted Shifrin

Hi, @robjohn.

Pedro Tamaroff

@TedShifrin Uniqueness is.

Enjoys Math

you tricky devir

usukidoll

udv = uv- duv

Pedro Tamaroff

02:23

@TedShifrin For existence I define $\langle v,w\rangle =(v)_B\overline{(w)}_B^t$ IIRC.

robjohn

@TedShifrin hey there

usukidoll

$x^ne^{ax-sx}$=$\frac{-x}{a-s}e^{ax-sx}x^{n-1}$-$\int$$\frac{-1}{a-s}e^{ax-sx}x^‌{n-1}$

Enjoys Math

@usukidoll should get $\frac{1}{a-s} x^n e^{(a-s)x} - \frac{n}{a-s} \int e^{(a-s)x} x^{n-1} dx$ the last term being inductively what we already know by formula.

usukidoll

$x^ne^{ax-sx}$=$\frac{-x}{a-s}e^{ax-sx}x^{n-1}$-$\int$$\frac{-1}{a-s}e^{ax-sx}$n‌$x^{n-1}$

arjkhdfklg'

Pedro Tamaroff

@anon Why did you delete your fields answer?

anon

02:24

closed as duplicate

usukidoll

$\frac{1}{a-s}$ $x^n e^{(a-s)x}$ - $\frac{n}{a-s}$ $\int$ $e^{(a-s)x$ $x^{n-1}$ dx$

Ted Shifrin

So is existence, @Pedro: $\langle \sum a_iv_i,\sum b_jv_j\rangle =\sum a_ib_i$ satisfies the definition.

anon

hahaha, you voted to undelete

Ted Shifrin

Hi @anon

Pedro Tamaroff

@TedShifrin Yes, $\langle v,w\rangle =(v)_B\overline{(w)}_B^t$.

anon

02:25

hello

Pedro Tamaroff

@anon I did.

usukidoll

ack hold up everyone lol trying to get the $ on the things

Enjoys Math

@usukidoll the seem to match

Ted Shifrin

So, @Pedro, is there an issue?

usukidoll

MATTEEE!!!!! wait in Japanese XD

Pedro Tamaroff

02:26

@TedShifrin Na.

Don Larynx

I just find it beautiful and weird that two open sets cant cover each other...

Enjoys Math

@usukidoll what are you trying to prove again?

\frac{1}{a-s} x^n e^{(a-s)x} - \frac{n}{a-s} \int e^{(a-s)x} x^{n-1} dx

Pedro Tamaroff

@DonLarynx $\Bbb R$ can certainly cover $\Bbb R$.

Ted Shifrin

They can drape nicely :P

usukidoll

prove by induction

Don Larynx

02:27

@Pedro yes, but do you see what I mean?

usukidoll

Show by induction that $£(x^{n}e^{ax})$ = $\frac{n!}{(s - a)^n}$ for s > a, and
n = 1,2, ....

Pedro Tamaroff

@DonLarynx No, I don't.

usukidoll

I know that's the wrong symbole for Laplace but yeah I'm not scrolling up XD

anon

(seconds after which, usukidoll scrolls up...)

Enjoys Math

$\frac{1}{a-s} x^n e^{(a-s)x} - \frac{n}{a-s} \int e^{(a-s)x} x^{n-1} dx \\ = \frac{1}{a-s} x^n e^{(a-s)x} - \frac{n}{a-s} [\frac{n!}{(s-a)^n}]_0^{\infty}$ Calculate the last term

@anon it's not on my screen

Ted Shifrin

02:29

British pounds sterling?

usukidoll

I give up T_T

for repeating the question in Latex it's humbug

Ted Shifrin

tries $\mathcal L$ on for size

Enjoys Math

Just copy what I had so far, hover over my post, show math as latex, copy / paste

Don Larynx

@Pedro Let $\epsilon, \delta, \tau > 0$. Now pick any point in the open set $A = N_{\epsilon}(1)$. Now pick the neighborhood that is so large that it may cover $A$ but it can't because there is some $N_{\epsilon + \delta}(1)$ that covers that neighborhood. but then that neighborhood is covered by an $N_{\epsilon + \delta + \tau}(1)$, and so on....

Pedro Tamaroff

@DonLarynx Can you state what you want to prove?

dfeuer

02:32

Hi @PedroTamaroff and @DonLarynx.

Don Larynx

That any open set covers any open set (contained in a segment) in $\Bbb{R}$.

Hi @dfeuer. I had a dream that I obtained an F on my ODE test.

Pedro Tamaroff

@DonLarynx Again?

Ted Shifrin

I'm afraid that doesn't make sense @DonL ... To me.

Don Larynx

Okay I am missing details, hold on

dfeuer

I answered my own question at math.stackexchange.com/questions/578414/… and I'd love to get a sense of whether it's correct and whether it could be improved. (The answer being based on concepts from the comments to the question).

Ted Shifrin

02:34

Any open set in $\Bbb R$ is a countable union of intervals?

Don Larynx

@Pedro @Ted let the metric we are working in be $\Bbb{R}, d$.

with d being the usual metric

Yes, I suppose @Ted

usukidoll

$\frac{1}{a-s} x^n e^{(a-s)x} - \frac{n}{a-s} \int e^{(a-s)x} x^{n-1} dx \\ = \frac{1}{a-s} x^n e^{(a-s)x} - \frac{n}{a-s} [\frac{n!}{(s-a)^n}]_0^{\infty}$
that's just evaluating from 0 to infinity on the right or one more integration by parts?

dfeuer

@DonLarynx, I would consider it a nightmare if I dreamed I were taking a differential equations class!

Don Larynx

@Ted so is my proof ok?

@dfeuer it's not as bad as it seems, you get used to de's near the end of the semester.

Ted Shifrin

Proof? What proof?

Don Larynx

02:36

by then it is usually too late for some

masfenix

Hey guys, I want to show the convergence of this integral: $$ \int_{\mathbb{R}^d} (1 + |\xi|^2)^{\frac{s}{2}} d\xi$$ I know the answer is obtained using converting this to polar/spherical

Pedro Tamaroff

@TedShifrin Suppose $B$ is the basis I want to use to define an inner product. Then $$(v)_B\overline{(w)}_B^t=C(E,B)(v)_E\overline{(w)}_E^t \overline{C(E,B)}$$ Right or left?

Don Larynx

@Ted chat.stackexchange.com/transcript/message/12323816#12323816

masfenix

but how do you turn a integral over $R^n$ into polar?

Pedro Tamaroff

@DonLarynx I cannot understand what you're writing there.

Ted Shifrin

02:36

Yikes ... I can't be 3 places at once!

Pedro Tamaroff

@masfenix Depends on the integral! =)

Don Larynx

@Pedro we are working in $\Bbb{R}, d$

masfenix

the answer is $s < -d/2 $ so Im trying to show that

usukidoll

@dfeuer I'm in a differential equations class...out of all the assignments out there this is the worst one

Ted Shifrin

With $r$ and the surface element on $S^{n-1}$, @masfenix.

usukidoll

02:38

but yay this induction thing is my last problem

nd then I'm free free

from this insanity

Pedro Tamaroff

@masfenix Ah, what you have there is $1+x^2+y^2+z^2+\dots$?

usukidoll

change it to spherical coordinates XD

masfenix

@TedShifrin thanks, i'll look into that but I am not sure. @PedroTamaroff No, its not a multi-index, @usukidoll any ideas?

Ted Shifrin

$d\xi = r^{n-1} dr d\sigma$

Pedro Tamaroff

@masfenix Isn't $\xi$ a point in space and $|\xi|^2$ its norm squared?

usukidoll

02:39

I'm not there yet T__T

I thought you were asking a calculus question

Ted Shifrin

Yes @Pedro

Pedro Tamaroff

@TedShifrin Did you see my question on the inner product thing? I am wrong, I think. The matrices should be inside.

Ted Shifrin

I'm trying to catch up. I'm slow on the iPad.

Pedro Tamaroff

@TedShifrin Nevermind, I can check my notes =)

Ted Shifrin

Your transposes are on the wrong side, too, unless you're using row vectors, @Pedro

Pedro Tamaroff

02:44

@TedShifrin I use row vectors.

Don Larynx

@Ted, @Pedro...any thoughts again? Please?

Pedro Tamaroff

$(v)_B$ is row, $(w)_B^t$ is column.

Ted Shifrin

Ah. So have I been in my grad course, but never in any other course.

Pedro Tamaroff

@DonLarynx You still haven't stated what you want to prove precisely.

usukidoll

Evaluate the last term... by having n = 0 to n = infinity?

masfenix

02:44

@TedShifrin where is $d\sigma$ coming from

usukidoll

~_~

Pedro Tamaroff

@TedShifrin Darn. I need to start using column vectors from now on.

=)

Ted Shifrin

I'm guessing he's trying to prove what I said, @Pedro, but he hasn't confirmed.

Don Larynx

@Pedro, YES i Did

@Pedro any open set is a countable union of open intervals.

this is what happens when you leave someone hanging

Pedro Tamaroff

@DonLarynx OK. So first one needs to show their are a union of open intervals.

Can you do that?

Ted Shifrin

02:46

That's the area element on the sphere, @masfenix. Try polar and spherical to see the pattern.

Don Larynx

oops

Pedro Tamaroff

@DonLarynx So, can you share your thoughts on how to prove that?

@TedShifrin OK, $$\langle v,w\rangle=\overline{(w)^t_E}A^\ast A(v)_E$$

Column vectors from now on, Ted.

$A=C(E,B)$.

Ted Shifrin

Looks good @Pedro.

Pedro Tamaroff

@TedShifrin =)

Don Larynx

Let $A$ be an open set contained in a metric $X, d$ with the usual metric. Then for all points $p_i \in A$ and $q_j \in X$ such that $d(p, q) < \mbox{max}\hspace{3 pt}d(p_i, q_j) = \delta$, the set of all $q_j$ is contained within $\bigcup_{i = 1}^kN_{\delta}(p_i)$. Thus $d(p, q_j) < \delta$.

I gave it my best shot

usukidoll

02:52

lost cause...that's what I describe my homework as...

gonna take a break and start on Calc iv

Pedro Tamaroff

@DonLarynx You're in $\Bbb R$. Work in $\Bbb R$. Pick an open set $A$ in $\Bbb R$.

Don Larynx

I was trying to generalize it now pedro.......

Ted Shifrin

Let's get the easy case right first. It's not true in an arbitrary metric space.

Don Larynx

Okay then. Let $A$ be an open set contained in $\Bbb{R}$. Then for all points $p_i \in A$ and $q_j \in \Bbb{R}$ such that $d(p, q) < \mbox{max}\hspace{3 pt}d(p_i, q_j) = \delta$, the set of all $q_j$ is contained within $\bigcup_{i = 1}^kN_{\delta}(p_i)$. Thus $d(p, q_j) < \delta$.

Pedro Tamaroff

@DonLarynx What are he $p_i$ and $q_j$?

At any rate, it is not generally true you can cover an open set $A$ by a finite number of open sets.

And you seem to be trying to do that.

I don't really understand what you're doing, though.

Don Larynx

02:58

Ah, not finite

hmm...

Ted Shifrin

He's trying to take balls of maximum radius around each point of $A$?

shuts up now

Pedro Tamaroff

@DonLarynx If you want to, I can tell you what to do.

Ted Shifrin

Try my hint?

masfenix

@TedShifrin do you have a PDF or a wiki link on how to do polar in n dimensions? I've actually never done one

Don Larynx

basically what ted said

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$Mathematics$ Mathematics