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12:01 AM
@Astyx Well this is just lovely I must say. Quadratic instead of exponential decay since the partial sum is $$(i,j)\mapsto \sum_{k=1}^{i+j-1}k + j = \frac{1}{2}(i+j-1)(i+j) + (i+j-1)j$$ if I did everything correctly, right?
12:26 AM
Define "infinite binary tails" to be the equivalence classes of infinite binary sequences where two sequences are equivalent if they disagree at finitely many points
Then without the axiom of choice, you cannot prove that the set of infinite binary tails can be given a total order
(note that I said total order, not well order!!)
Disagree at finitely many points?
Why do you call them "tails" if they are equivalence classes?
akiva: "cannot" prove? what if you just assume that as an axiom? is it equivalent to some weak version of AC?
I wouldn't call f(x) mod k a "tail" given f(x) is translated down to zero by an interval of k in the domain.
Well, at least not from the perspective of modular reduction.
1 hour later…
1:38 AM
Hi. I am solving an exercise that's telling me to estimate the value of $\sqrt{4.1}\cdot \sin\left ( 0.8 \right )$ using the partial derivatives of $f\left ( x,y \right )=\sqrt{x}\sin y$, but I'm unsure what value to use as an approximation for $y$, or rather what value the exercise expects me to use.
Choosing $y=\frac{\pi }{4}$ gives a very good estimate, but the result (assuming my solution is correct) contains both $\pi$ and $\sqrt{2}$, both of which are irrational. Choosing $y=1$ instead yields a rational number, but the error is substantial (almost 40% of the true value).
1:55 AM
Who knows what trig functions of 1 are? That’s ridiculous.
You're right. What was I thinking ...
Certainly irrational ….
I don't know how mistakes like this slip by me. My brain just took a mental shortcut and decided that sin(1) = 1 and cos(1) = 0.
2:27 AM
@AMDG 'Cause it's all the information that's contained at the "end" of the sequence
@leslietownes Cannot prove from ZF, I mean. (Not sure if it's equivalent or if it's weaker. I think weaker.)
@AMDG Sorry, what is k - the equivalence relation? What is an "interval of k"?
3:08 AM
@AkivaWeinberger nice :). How did you know this? Did you discover it on your own?
@Cosinux No wonder the error was out of bounds :)
@Koro It was a few years ago so I forget but I'll put 70% odds on "yes"
3 hours later…
6:28 AM
@XanderHenderson I remember seeing the Weaire-Phelan structure years ago, but that question has inspired me to work on an interactive 3D diagram. So far, I've only done the pyritohedron.
You can parametrize pyritohedra with a parameter $0<h<1$. h=0 gives a cube, h=1 gives a rhombic dodecahedron. h=1/2 gives the pyritohedron used in the Weaire-Phelan structure, and h=1/phi gives a regular dodecahedron. Here's an anim from Wikipedia which illustrates the family.
Here's my Sage / Python script which lets you view any member of the family in 3D.
6:51 AM
@TedShifrin Fair point, although you can approximate sin & cos by hand to a couple of digits using Taylor series. You get better approximations using Padé. Robjohn posted some a few months ago. chat.stackexchange.com/transcript/message/58919555#58919555 i.stack.imgur.com/FBaeo.png
7:27 AM
looks like a square with indigestion.
:) My program lets you set h to values outside [0, 1]. That gives you weird non-convex things.
2 hours later…
9:41 AM
A: John Lee Exercise 5.44

one potato two potatoThe following is a sketch of the proof and notations are abused frequently. Let $p\in\partial M$ and choose a smooth boundary chart $(U,\varphi)$ with local coordinate $(x^i)$. Since $f$ is a smooth function, $f:U\to [0,\infty)$ is a smooth function. For $v\in T_pM$, write $v = v^i{\partial\over\...

I answered my own question. Can someone please check if this argument is correct?
seems good to me skimming over it
1 hour later…
10:54 AM
Have anyone heard of supporting manifolds before
I have a related question.
Does there always exist a closed hyperplane in a locally convex space not reducing to a point.
The whole concept of codimension is very confusing to me
11:41 AM
So I've proved that in locally convex spaces those are precisely kernels of linear functionals
So existence of closed hyperplane reduces to existence of continuous linear functionals. And they always do in a non-trivial locally convex space
11:54 AM
Turns out I didn't need anything like hyperplanes for the argument they gave me.
4 hours later…
3:34 PM
@PM2Ring But this was the original question :)
4:29 PM
Hi! I wanted to know the mistake in my solution:
for m; i selected $4$ girls and let them be a group, they can be arranged in $4!$ ways; so total ways= $5_{C_4}\times4!\times7!$ but this would include the cases in which 5 girls are arranged together as well so $m=5_{C_4}\times4!\times7!-5!\times6!$ but in the soln they have subtracted $2\times5!\times6!$
why an extra factor of 2?
I understand other solutions but wanted to know my mistake in mine
4:58 PM
Because XY (X one girl, Y four girls) and YX can both occur and you've counted it twice.
5:50 PM
@AkivaWeinberger :)
@Jakobian my question was wrong.
I thought that R is well ordered, which is not true.
6:03 PM
@Koro I'm somewhat curious: is your nick an anagraph from Cyrillic or is it just me?
6:36 PM
@TedShifrin If say $Y$ is $G_1G_2G_3G_4$ and $X$ is $G_5$ then $XY$ will be $G_5G_1G_2G_3G_4$ and $YX$ will be $G_1G_2G_3G_4G_5$ won't both these cases be already included in $6!\times 5!$?
6:50 PM
I didn’t want to type out all the details. The point is that $1-2345$ and $1234-5$ both appear and must be removed when counting $m$.
3 hours later…
9:54 PM
Is one of these variants of the change of variables formulas a generalisation of the other? Or are they equivalent?
I have to say that the second one (Thm. 14.6) technically uses the Lebesgue integral, while the first (Thm 15.2.2) uses the Riemann integral
I don't know if this matters
Since we're considering continuous functions (on a bounded+closed set) I believe both notions (Riemann vs. Lebesgue) coincide
yes, that seems right. what is a 'figure'? the difference, if any, might be in that.
A (bounded) set whose boundary has measure zero
ok, so in the first one the mapping function g is C^1 on the closure of a 'figure', and in the second the mapping function (now called f) is only continuous? is that right? i guess 14.6 is more general.
oh, there might also be a gap between measure zero and 'content zero' but maybe not. i think under the usual definition, not.
in both cases the integrand is assumed to be continuous
i've clarified my comment above. i was referring to the map that changes variables, not the integrand.
10:01 PM
in both cases the transformation is C^1, no?
in the second statement f is assumed to be C^1
oh, it is.
i suffer from f-g-K blindness.
yea, it doesn't help that f plays the role of the integrand in one statement, and the role of the transformation mapping in the other
i think it's the same theorem, up to whether it's clear that 'measure zero' and 'content zero' are the same thing. they probably are, but they might not be defined that way.
@Koro In fact every well-order of the reals must be rather complicated (in a technical sense)
yea I'm willing to assume that content zero = measure zero (I can check later)
under that assumption I'm interested how to transform one statement in the other
hm, I think 15.2.2 => 14.6 is clear to me
assuming that the open set $G$ containing the figure has measure zero
which I'm willing to assume
10:09 PM
Measure zero and Jordan content zero aren’t the same.
I'm not dealing with measure zero as far as I am aware
in both cases it's Jordan content zero
OK … so who cares what the difference is?
I see I should have said Jordan content zero instead of measure zero
what difference?
I'm not dealing with the measure zero notion
between the two versions I’m too lazy to think about
I want to know if they are equivalent
10:12 PM
why? What is the issue?
and my guess is yes, minus perhaps that we should technically consider opens whose boundary has Jordan content zero
there is no issue, I just want to know if they are equivalent
I've already read the proof of both statements
but I wondered if I had to do that
if they're equivalent, I didn't have to
the proofs are quite different in taste
Both are weaker than needed. You don’t need the one-to-one and invertible derivative everywhere. The second one seems weaker as it requires those conditions on a larger set than that on which you will integrate.
So not logically equivalent at all.
yea that's what I'm thinking too
wait, the second one weaker?
Think about polar coordinates and the origin.
I thought the first one is weaker
10:17 PM
Yes, that stupid figure crap.
ah wait I see what you're saying
I thought I had a (hand-wavy) argument for 1=>2 (assuming additionally that in 2 it is assumed that the open containing the figure has boundary with content zero)
The first one (Munkres, I assume) is integrating over an open set, the second over a “figure.” But you can take the interior of the figure.
oh right, fair enough
In 2 you need that larger set $G$.
I have to think for a sec, but thanks for helping me through this
10:20 PM
I personally never taught my students to integrate over open sets.
I'd say one drawback of dealing with opens is that you have to mention that things can be extended continuously to the closure for example
to get compactness (having assumed boundedness already)
Since Riemann is based on rectangles (closed), it’s not worth it to me.
fair enough
I spent plenty of time thinking about the pedagogy and exposition….
I can imagine that
I'm just trying to organise my thoughts about multivariate analysis by going through books for the first time
(which is something I hadn't done during my bachelors, partly because we weren't taught from books)
10:37 PM
And you’re not looking at mine … Shame on you!
well I have
thing is though
I started this because I wanted to find a proof of
Green's theorem
(not invoking Stokes)
and your book only proves it for a special case
so once I had invested my time in a book that did precisely what I was looking for
I was already "too much in it" anyways
so I continue using that book for other multivariate-analysis stuff too
ted's like that. you have to pay extra for the complete theorems. you need the 'deluxe edition' of the textbook.
Is a Borel probability measure on a compact Hasdorff space a Radon measure?
10:54 PM
In my book it's said that the set of all probability measures on a compact Hausdorff space $X$ is identified, using Riesz representation theorem, with a $w^*$-compact set $B_{C(X)^*}$.
It's a bit much to digest for me in one go
A: Is a probability measure on a compact space regular?

user940Here's a famous example of the counter-intuitive behaviour that you want. The Dieudonné measure $\mu$ is a Borel probability measure on the compact space $X=[0,\omega_1]$, where $\omega_1$ is the first uncountable ordinal. It has the property that $\mu(K)=0$ for every compact subset of $X\setm...

I'm really confused now
11:09 PM
They later cite "Lectures on Choquet's theorem" by Phelps who defines probability measures to be Borel and regular
They assumed it's a part of the definition
Apostol in his Advanced Calculus book has a very general proof of Green's Theorem. I don't see the point of it.
I have never needed to apply it more generally than what I proved in my book (any region that can be covered by a finite union of parametrizations by rectangles). Of course, we could prove it with triangles near the boundary, with the edges of the triangles converging to the boundary curve, but ... meh.
I'm actually AGAINST proving things in their greatest possible generality. Pedagogy is more important than that unless you're a pedantic graduate student.

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