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03:18
Hey all

I have physics background but would love to understand this approach:

https://math.stackexchange.com/a/4110007/430082
Can someone recommend me some textbooks to go through?
There must be some texts written for physicists rather than for mathematicians, but we won't be likely to know those. The math sources I know are at the graduate level.
@TedShifrin I haven't been able to find any ... I would appreciate it. If you recommend me the math books. I'll try and have a go?
Spin Geometry, Lawson and Michelsohn
@TedShifrin thanks
03:34
I am curious what this symbol means
I mean that $|_0$
i don't see it
Evaluated at $A=0$.
Oh thanks.
this is quite standard
oh, yeah, the vertical bar. evaluate at A = 0.
03:37
I didn't saw that in my multivariable calculus course.
Think also about the usual notation for the fundamental theorem of calculus.
it's really common with leibniz notation for derivatives, particularly partial derivatives, as this style of notation often suppresses any other place to indicate where you're evaluating something.
this is its blessing and its curse.
So it means first set A=0 then evaluate the partial derivative of the function.
other way around.
Ah sorry I get it fundamentals theorem of calculus.
Now I need to ask physicist for the physics context :(
03:43
everything's whizzing around and interacting with everything else. that's all i know about physics.
then there is nuclear fusion, up there with quantum computing and crypto.
@WilliamJohn The physics context of what?
j=partial derivative evualated at A=0. I don't know why set A=0?
04:10
Current density at a point is a limit per area as area goes to $0$. Think about mass density (mass per volume as volume shrinks to $0$).
 
5 hours later…
09:09
May I ask if somebody could take a look at my question?
2
Q: Computing $\displaystyle\int_S x^2ydxdydz$ given a particular change of variables.

Dolphin Let $S=\{(x,y,z)\in\Bbb R^2\mid x>0, y\ge 1,z\ge 1, xyz\le 1\}.$ Find the set $S^*$ obtained by the change of variables : $$\begin{cases}x=u\\ y=\frac{u+v}u\\z=\frac{u+v+w}{u+v}.\end{cases}$$ and compute $\displaystyle\int_S x^2ydxdydz.$ My attempt: The condition $xyz\le 1,x>0$ gives $$xyz=u\cd...

hello, please how to prove this $|\frac{x^p}{1+x^p}-\frac{y^p}{1+y^p}|\leq \frac{p}{2} |x-y|$ for $x,y>1$, p>0
i tried mean value theorem
for $f(z)=\frac{z^p}{1+z^p}$ on ]x,y[
i found
$|\frac{x^p}{1+x^p}-\frac{y^p}{1+y^p}|= p\frac{c^{p-1}}{(1+c^p)^2} |x-y| $
but why \frac{c^{p-1}}{(1+c^p)^2}<\frac12$ ?
09:54
@Vrouvrou I think you meant [x,y].
@Vrouvrou $\frac 1c<1$ (since c is in ]x,y[) and therefore $\frac{c^{p-1}}{(1+c^p)}\le 1$
from mean value theorem there is $c\in (x,y)$
but i need $\frac{c^{p-1}}{(1+c^p)^2}\le \frac12$
@Vrouvrou you need $(1+c^p)^2$ in the denominator.
Now, note that $c>1\implies c^p>1\implies 1+c^p>2\implies (1+c^p)^{-1}\lt 1/2$.
yes
ok
and $c^{p-1}$?
10:05
@Vrouvrou multiply the inequalities $\frac{c^{p-1}}{1+c^p}\le 1$ and $(1+c^p)^{-1}\lt \frac 12$.
why ?
@Vrouvrou to get this.
i don't understand
can you tell exactly which part you don't understand?
we have $(1+c^p)^{-1}\lt 1/2$
then $\frac{c^{p-1}}{1+c^p)^2}\leq \frac{c^{p-1}}{4}$
why $c^{p-1}<1$ ?
?
10:33
@Vrouvrou: $\frac{c^{p-1}}{(1+c^p)}=\frac{c^p}{(1+c^p)}\color{blue}{1/c}\le 1( \color{blue}1)=1$
10:44
where is 1/2?
@Vrouvrou please see my earlier messages.
I'm struggling to prove $X = (\vec{x_1},...,\vec{x_n})^T$, then $X^T X = \sum \vec{x_i} \vec{x_i}^T$
Any ideas?
11:26
i see it
 
3 hours later…
14:03
@Vrouvrou no it's not true.
14:56
Is there an example of a function $f:\mathbb R\to \mathbb R$
that is integrable on every bounded interval and for some $x\in \mathbb{R}$ we have that $f(x)\ne \lim_{r\to 0+}\frac{1}{2r}\int_{x-r}^{x+r}f(y)dy$?
@JaakkoSeppälä try looking for non-continuous function f.
I think step function f(x)=1 on $x\gt 0$ and $f(x)=-1$ when $x\le 0$ should work. The inequality should work then for x=0.
@Govind75 $(X^\top X)_{jk} = \sum_i (X^\top)_{ji} X_{ik} = \sum_i X_{ij} X_{ik} = \sum_i (X_i \otimes X_i)_{jk} = (\sum_i X_i \otimes X_i)_{jk}$
where $a \otimes b = a b^\top$ is the outer product.
15:11
Thanks
I had another exercise that puzzles me. Let $f:[a,b]\to \mathbb R$ be integrable. How can I find a constant $c\in\mathbb{R}$ s.t. $\int_a^b (f(x)-c)^2dxs$ achieves its minimal value?
Sorry, I meant $\int_a^b (f(x)-c)^2dx$
Yeah one can't directly conclude maximum modulus principle for unbounded domain for general analytic function. If it has compact support then essentially original mmp.
15:31
Recently developed this pet peeve about students calling things "trivial" when they are not of the form "P \implies Q" with Q always true.
@anak I have long been on the record with the opinion that words like "trivial", "clear", and "obvious" should be stricken from mathematical exposition. None of these words actually add any content to a work of written mathematics.
i think they sometimes add value, but are bad pedagogically.
@XanderHenderson Amen. Had a prof who said to "only use those words if you have not checked yourself". I then am struck with guilt whenever they appear in my writing.
which i guess does mean that in a lot of settings they are absolutely bad.
@leslietownes trivially bad
15:49
@leslietownes Obviously.
@leslietownes Example?
Personally, I believe that whenever a mathematician says that something is "obvious", or "clear", or "trivial", they really mean one of the following:
1. I know how to prove it, but I am too lazy or important to bother proving it for you,
2. I don't know how to prove it, but it seems like the kind of thing which should be true, or
3. I don't know how to prove it, because it is, in fact, false.
One of the better job-talks I've seen involved a potential professor explaining their PhD research, which, ultimately, came down to showing that an "obvious" result from a 40 year old paper was actually false. But a Big Name™ has asserted that the result was true, so the received wisdom was that it was true.
it's useful to have a verbal shorthand for a detail that everybody in the audience knows, if they're only reminded with a nudge that it is known (ideally also along with what "it" is).
this is obviously audience dependent. but it's not always used to obfuscate or as part of a power trip. your prong 1 above could be "i know how to prove it, but i'm too lazy or important to bother proving it for you, or alternatively in our limited time together i'm just going to remind us that this can be done, in a way that doesn't make it the focus of the talk."
once it becomes a rhetorical move that people use to avoid details, yeah, that's bad.
which is why it probably shouldn't be used in classroom education.
but some random seminar talk, why not. go nuts. maybe not a job talk on your thesis or a talk about supposedly important work on an outstanding open problem.
I agree that there are times when the intended audience should know the proof, but I don't think that stating that the proof is obvious or clear actually adds any value beyond "It is known that..." or "It is an elementary fact that..." or "Those familiar with the field likely know that...".
i prefer "well known" to "obvious" because "obvious" just sounds bad to me. it does sound like a power trip even if it's not being intended that way.
That being said, I am generally more forgiving of "verbal" exposition, as one typically has less time to present, and less time to carefully chose one's words. But I still think that words like "obvious" are a crutch, and that folk should try to eliminate them from their mathematical vocabulary.
you do need to be able to sometimes say stuff like "as everybody ought to know, in 1913 X proved Y and this is now known as a straightforward consequence of Z theory." without immediately being hounded out of the room, or taking people back to slides of papers from 1913.
16:00
On the other hand, I started doing Toast Masters when I was in the 3rd grade, so there are a lot of "filler" words that make me cringe. Every time I say "uh..." or "um...", I hear a little bell in my head, telling me that I have done did wrong.
0
Q: Spread the butter in Analysis

SultanDeGranadaIs it true that analysts use the sentence "spread the butter" in informal contexts whenever they have to consider an expression of the form $\varepsilon/2^n$ in a proof? If so, I would be interested in finding out who came out with this idea and to hear if you know more expressions of the kind.

my analysis 1 instructor was a fan of 'obvious' and that was definitely a misuse of the term. she almost literally taught us that the name of the game was to learn which things were obvious and which weren't so we could avoid proving them.
koro: add me to the list of people who's never heard of that usage. people do sometimes say stuff like "this is an epsilon over 2^n type argument." it wouldn't surprise me if some books, particularly in measure theory where you see these arguments all the time, have come up with more colorful names.
I recall reading in a book that said by Curry -leaves trick we have:(or something similar) $|x_m-x_n|\le |x_m-x|+|x-x_m|<\epsilon/2+\epsilon/2$.
@leslietownes I've never heard of an $\varepsilon/2^n$ argument. $\varepsilon/2$, $\varepsilon/3$, and $\varepsilon/5$, sure. I've even seen an $\varepsilon/7$ once. But over $2^n$? Preposterous!
I still don't know what curry-leaves trick is.
16:04
It is when you eat all the curry, then leave before you get caught.
@leslietownes perhaps some other (more specific) buzz word should be created to indicate what you mention here. I can see how this is helpful, but at this point in time "clearly"/"trivially"/"obviously" don't clearly/trivially/obviously convey this, due to repeated misuse throughout academia.
I would like to believe that.
But why is it used in the context of definition of limits?
i think someone cooked up that term just as 'spread the butter' term in the linked post above.
@Koro Sure. It is an "obvious" metaphor, but that don't mean it's widely used.
One of my favorite analysis profs in graduate school had a lot of wacky metaphors, many of which were, I think, subtly racist. :/
wacky and subtly racist, what a combination. they don't make many of 'em like that anymore.
But his English was so marginal that I always assumed that he didn't really know what he was saying.
"So... uh... you see! When you are standing on the boundary, it is like being on a wall. And you have... uh... one foot! in the US, and one foot in... Mexico! So... that is... uh... the boundary! of a set!"
16:09
with non native speakers it can be really hard to infer intent or awareness of what something might sound like to a native speaker. let's give the benefit of the doubt and go with 'wacky.'
@leslietownes That was always my strategy.
unless he was bringing mexico into everything. but then maybe he just liked mexico.
Trump was talking about building a wall at the time. My guess is that he was trying to be current.
@leslietownes you haven't met a complex analyst with a subtle aversion to the Polish, have you.
Ugh... I should go record another lecture, so that I can pack the truck, so that I can get on the road by sunrise tomorrow. BUT I DON'T WANNA!
16:12
anak: you have to be fairly pro-polish in any kind of analysis. or topology. without them there isn't much left.
If a space ain't Polish, it ain't worth thinkin' about.
Okay, I'm lecturing now.
@leslietownes I've met someone who thinks otherwise. :(
17:01
@TedShifrin Right, I'll have a look at this later when I've seen a bit more theory
17:20
Why is GF(16) isomorphic to $\{ax^3+bx^2+cx+d+\langle x^4+x+1\rangle|a,b,c,d\in \mathbb Z_2\}$?
GF(16) is Galois field of order $2^4$.
@ShaVuklia The intersection number of a (generic) section with the zero-section is the "usual" definition. I don't know what your definition of Euler class even is (probably abstract nonsense).
@Koro Prove that $x^4+x+1$ is irreducible/$\Bbb Z_2$.
ah, because field of order p^n is unique (upto isomorphism).
The braces part is just: $\frac{Z_2[x]}{\langle x^4+x+1\rangle}$, which is a field of order 16.
Provided you prove irreducibility, yes.
@TedShifrin I was confused about the isomorphism part. It's clear now. :)
17:44
@JaakkoSeppälä Take the derivative with respect to $c$ and set it to 0. Solve for $c$.
18:13
Mexico, then Trump's Wall, then Polish, how is one to keep up! ;D
I don't.
Reminds me of a game we played in scouting: scouts form a circle. First person reads a slip of paper with a word, without speaking. They turn to second person in the circle, and whisper the word, ......, the last person in the circle whispers the word to the original whisperer. Let's just say, it rarely ended with the words matching.
@TedShifrin Good answer!! (In FamilyFeudESE)
Oh, I forget the name for that "game." It's quite famous.
The game is called "Telephone" where I grew up.
"telephone?"
18:19
Yup.
@XanderHenderson Great minds remember well!!
Chinese whispers (Commonwealth English) or telephone (American English)[1] is an internationally popular children's game. From wiki.
These days the "Chinese" part might be racist.
Everything is racist.
:/
Well, certainly it's a good thing "Indian giver" from my childhood has left the room.
That was, in hindsight, so ironic a term.
We called it Chinese Whispers, but there was no intended racist connotation. WP lists lots of names for it. en.wikipedia.org/wiki/Chinese_whispers
18:22
@TedShifrin Indeed. I remember learning to jump rope with a long rope held by two kids on each end. One kid jumps : "Chinese Japanese ..." I can't even right it, it's pretty anti asian, particularly to asian women.
@TedShifrin It isn't abstract in fact! We used linear algebra to define it. For a matrix matrix $X=[x^i_j]$ where each $x^i_j$ is an indeterminate, such that $x^i_j=-x^j_i$ it can be shown that $\det X$ is a perfect square in $\mathbb Z[X]$, and then we choose a particular root and call that the Pfaffian. And then we just use the Chern-Weil homomorphism
Yeah, the Pfaffian is very concrete. So you defined the Euler form using Chern-Weil.
Right.
OK, so the notes you say I sent you show how using the topology of the Grassmannian you can prove what I said about the intersection number.
If I in fact didn't send them, let me know. I'm leaving again, but can send later.
I looked them up yesterday, so it's confirmed:)
See you around!
@PM2Ring Indeed. Grapevine, gossip. It is a lesson against believing that the gossip (social media) hasn't been distorted.
I have been reading today about what Vietoris topology is
18:32
@amWhy In "Telephone", even when everyone is trying hard to pass the message on in without distortion, there's usually some change. But often there's at least one person in the circle who thinks it's amusing to intentionally change the word. And in the Real World, there are always parties who can get some advantage by distorting the information they have control over.
it's topology you give to the nonempty closed subsets of topological space (kind of like Hausdorff distance for metric spaces)
@PM2Ring Exactly!
@Xander You may enjoy this song. Molly Tuttle has a new band. Here's a new song, explaining that it's ok to be different. That can be a difficult concept for some country music fans. ;) Crooked Tree. Also, Big Backyard

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