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15:00
The mathworld page for field is also an immediate Google hit as well
Also, so many people here ask simple proofs that can even be found in most textbooks. It is as if they don't read a textbook.
The phrase "due diligence" comes to mind.
A lot of questions on the main site seem lacking in that respect
In the least, they should write the question in proper English, which is also often not the case, lol.
They try to
2/3 of people who speak English are not native speakers
Ehhh, sometimes they try to
15:06
I only recently learnt that lel is a variant of lol.
Have you ever heard the difference between an acronym and an initialism?
But I can understand non-native English speakers having trouble
lol is an acronym, while lol is an initialism.
Should the second be lel ?
Anonymous
Okay. I'm trying to explain in my own words what I understood from Wikipedia's "Classical definition". A field is any set, such that on any random pair of elements if we apply the operations "addition" and "multiplication" then we get an element belonging to the same set. Furthermore, the operations are to satisfy the properties of : Associativity, Commutativity, Distributivity and also Additive/Multiplicative identity/inverses. Makes sense? @JasperLoy @Semiclassical I think I got it now
@Blue Yes, something like that. I think you got it.
Anonymous
Thanks!
@Semiclassical Should be pronounced as "ell oh ell is an acronym, while lawl is an initialism"
The point being that people pronounce it either way
15:09
I see
LASER = light amplified by stimulated emission of radiation
It doesn't really work in a text context
SCUBA = self contained underwater breathing apparatus
SCUBA = self-contained underwater breathing apparatus
DAMMIT
Some people read EULA as you-ler, for End User Licence Agreement.
15:11
Are you British, by any chance
I am not British by nationality, and I don't live in Britain.
However, I use British spelling in written English.
For a specifically American one: COBRA = Consolidated Omnibus Budget Reconciliation Act
Wow, haven't heard of COBRA, LOL.
Which is why searching for cobra insurance on google gives you different results than you'd expect
UNESCO, or you-nes-co, is United Nations Educational Scientific and Cultural Organisation
15:13
@JasperLoy I only ask because "la" and "ler" would only sound the same in a non-rhotic accent, like most British accents
Re: EULA
@AkivaWeinberger Yes, of course the American pronounce the rrrrrrrr.
We do indeed.
Plus or minus a Boston.
You can already tell from the way I spell licence.
That requires me to remember how to spell license
(thanks, autocorrect)
Or practise.
British spelling has licence/license and practice/practise, but American spelling only has license and practice.
These are two very special example words.
15:16
i don't think it's an American thing, but material/materiel
I think material has a different meaning from materiel.
Anonymous
I sometimes mix up British and American spellings and make a hodge-podge :P
Or a hocus pocus
Or a Harry Potter
Materiel = military materials :)
15:18
meanwhile there are enterprise and prize, both coming from the same root :-)
Maybe different dictionaries will also list different allowed variants, depending on the lexicographers.
I like the fact that patience and passion come from the same root word
Enterprice sounds like a failed version of eBay
Enterprise is the Star Trek ship.
Though with that one there's a modern holdover in that Christians refer to the suffering and death of Jesus as "the Passion of the Christ "
So that usage points to where the word passion comes from
15:22
Suffering can also be a form of passion. Human emotions are complicated.
3
Other than etymonline.com, several free online dictionaries also give brief etymologies.
Suffering for a purpose can be a form passion, we like to think of our emotions as complicated things, but they're just chemical interactions in our brains.
I am not sure they are just chemical interactions.
There might be a mind beyond the physical brain.
And the chemical changes might be a result and not a cause of the emotions.
@Semiclassical but "patience" is from Latin patientia and "passion" is from Latin "passio"
15:27
Just because we don't fully understand the mind (and consciousness) yet, doesn't mean we'll lean back to supernatural explanations, there are good reasons to believe that only the physical exists.
whereas "enterprise" and "prise" are both from French -prise (one with enter before it)
In Spanish, paciencia and pasión
per defintiion: $\mathrm{grad} f(x) = (\partial_1f(x), \ldots, \partial_nf(x))$. Is $\mathrm{grad} f = (\partial_1f, \ldots, \partial_nf)$ a correct notation, too?
@LeakyNun yes, a slider. I have already uninstalled Geogebra though.
@Kirill geogebra is so good
basically the only free 3D modelling tool
@LeakyNun no it is so bad
15:33
let's agree to disagree
@LeakyNun From the 'patience' entry: Late 12c., "sufferings of Christ on the Cross," from Old French passion "Christ's passion, physical suffering" (10c.), from Late Latin passionem (nominative passio) "suffering, enduring," from past participle stem of Latin pati "to endure, undergo, experience," a word of uncertain origin.
so it comes from 'pati'.
@Semiclassical which source is this?
Of course patience comes from patientia
15:35
you linked me to the 'passion' entry and claimed that it is from the 'patience' entry?
I can't find a single "patience" in your link
No. I claim that 'passion' ultimately derives from Latin 'pati'
@Semiclassical well I did not say otherwise
And that's definitely the same as what underlies the Latin patientia
15:36
yes, that is correct
So same root word.
It's not entirely surprising when one thinks about how the words are pronounced rather than written.
@Semiclassical that's just a coincidence
the present stem is pat-, while the perfect participle stem pass- is from earlier pats-
...and yet we pronounce 'patience' with the same 'sh' sound as in 'passion'
15:40
the -i- that follows both stem make the former with a "ch" sound and the latter with the "sh" sound
then the "t" in the "ch" sound got lost in French, and we are left with the same consonant
huh.
language is weird.
correction: it made a "t" into a "ts" sound, which became an "s" sound later
and then combine it with the "i' again to get the "sh" sound
patsientse -> pasiense -> pashense
hmm. so they've got the same root word but not the same root pronunciation, so to speak.
in Latin it would be as written, pa-ti-en-ti-a, and pas-si-o
@Semiclassical and then sound changes merged the pronunciation in English
@Semiclassical you might want to know that "lesson" and "lection" are from the same Latin word
I don't know what 'lection' means.
15:50
you might be more familiar with the word "lecture" but it wouldn't be from the exact same word
Noun[edit]
lection (plural lections)

(obsolete) The act of reading.
(ecclesiastical) A reading of a religious text; a lesson to be read in church etc.
ah, sure.
Is there any sense in which $\dfrac{\sin\pi x}{\pi x}$ is the most natural extension of $\Bbb N\to\Bbb N:x\mapsto\begin{cases}1,&x=0\\0,&\rm else\end{cases}$ to the reals?
Rather than, say, that plus a 1-periodic function
@AkivaWeinberger heh?
I don't understand the $\Bbb N \to \Bbb N$ part
16:06
So I've just been relaxing, enjoying my summer for the first time in 5 years.
get a call, shit has hit the fan.
@LeakyNun Domain and codomain. But I meant $\Bbb Z\to\Bbb Z$, whoops
but you said it is to the reals
Extension to the reals
$\frac{\sin\pi x}{\pi x}$ extends that function to the reals
Hi all. Does anyone know how to find rigid triples of conjugacy classes in GAP/another computer algebra system?
Doubt it ^
16:09
I'll pay someone 1 bolivar at the end of the year if they do.
@AkivaWeinberger I don't get it
@AlwaysNeedHelp googles conjugacy classes
googles rigid triples
@LeakyNun Graph $\frac{\sin{\pi x}}{\pi x}$ (or rather it's continuous extension), look at what it does to the integers
16:25
@AkivaWeinberger Almost always 0?
Except it sends $0$ to $1$
Oh yeah, if anyone wants to try and make a really large number/function, please support my proposal:
https://codegolf.meta.stackexchange.com/a/13524/58880
Does somebody know this guy: $\lim_{h \to 0, h \ne 0}\frac{r(h)}{h}$ and his brother $\lim_{h \to 0 h \ne 0}\frac{\mid r(h) \mid}{\mid h \mid}$? They come from the formula $f(x+h) = f(x) + f'(x)h + r(h)$. What are they?
@Kirill Are we assuming $r(0)=0$?
Oh, I see
@SimplyBeautifulArt I don't know. It is about the differentiablity in $\mathbb{R}^n$
16:30
Well,
$$\frac{r(h)}h=\frac{f(x+h)-f(x)-f'(x)h}h=\frac{f(x+h)-f(x)}h-f'(x)$$
Take $h\to0$ and you should get $0$ for the limit I think.
if r=f(x+h)-f(x)-f'(x)h then r(h)~f''(x)h^2 no? (also, shouldn't r depend on x too?)
What is the idea behind this? What is $r(x)$? A useless piece of function, or how?
@Kirill the remainder
because $f'(x) = \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}{h}$
in one-dimensional analysis I could imagine this approximation as the limit of the secant line going towards the tangent line. Here we have derivatives of $f$. Where do they come from?
@LeakyNun this formula is not ok for $\mathbb{R}^n$ for $n\ge2$.
never mind I'm out of this
16:36
@Kirill r stands for "remainder"
@arctictern remainder of what?
f(x+h)=f(x) is a 0th order approximation
f(x+h)=f(x)+f'(x)h is a 1st order approximation
@Kirill remainder of taking away the approximation
with division, if you want to approximate 1776 with a multiple of 10, you have 1770, and the remainder (difference between estimate and true value) is 6
f(x+h)=f(x)+f'(x)h+f''(x)h^2 would be a 2nd order approximation
@arctictern is there any verification why we can approximate the function with its derivatives?
you mean does Taylor's Theorem have a proof? sure, google away.
@arctictern shouldn't it be $2!$ in the 2. order?
16:39
err, right
@arctictern so, Taylors' theorem does it?
"does it"? the fact you're talking about is called taylor's theorem
@arctictern really? I am reading the part "Differentiable functions" and see no word about it
MVT covers it too
@SimplyBeautifulArt Middle value theorem?
or multi-variable topology? :)
16:43
mean value theorem
ok
its' version for a one-dimensional space is quite ok, but for others is quite strange. It says, "there is a value, so, that the UNequality holds". I haven't even thought about connecting these two theorems. Can one derive Taylor from MVT?
I still hope that Mr. Taylor wrote his formula not randomly and had proved it by induction, so that there is a logical argument, why derivatives can approximate a given function.
What would be a good question to ask to get points from
if you want negative points then 0.999=1
16:58
@AkivaWeinberger

A good question would be to ask whether or not anyone knows a GAP implementation to search for rationally rigid triples of conjugacy classes of certain finite simple groups.
Does Euler's Theorem mean that a the 0 node simple graph has 2 regions (faces)?
Seems unlikely, if you get (one side of) a blank sheet of paper you've drawn it and where are the two faces?
Maybe that specific example is outside of the image/domain of the equation
Hey :)
@alan2here Well, the point of Euler's theorem more generally is that it tells you about the space in which your graph lives. For instance, you could embed a planar graph on the surface of a torus; in that case, V-E+F wouldn't be 2.
But if you've got a single node then you you haven't got enough data to say anything about the space in which your graph lives.
It does produce the correct answer with one node, this could be by axident.
Just with 0 nodes, and therefore 0 edges, it doesn't.
17:06
True.
V-E+F for the 3 node cyclic graph will be (3-3+2 = 2) for both a torus and a flat plane.
hmm, yeah.
I guess you've got to have edges loop around the space for this to work, on the plane I guess this means going off to infinity then reconnecting on the other side.
that graph is effectively confined to only a patch of the torus in this case, though.
if you put the three nodes so that the edges wind around the torus, then there's only one face.
I was thinking of a sort of plus that joins up to itself, but thats an even simpler example :)
17:10
But that still doesn't quite work, since V-E+F=3-3+1=1 not 0.
There are certainly more non-working edge cases for the torus than the plane.
Yeah. Someone who knows algebraic topology could say where we're going wrong.
The faces wouldn't be homeomorphic to disks
Same reason it doesn't work for disconnected graphs in the plane
I read faces as feces, lol.
I generally think of a disconected graph as multiple different graphs so that didn't even occur to me to try.
17:13
hmm, yeah.
And by plane I mean sphere, actually
If you consider the image there as a way to produce the torus
maybe it's true to say that if you need to have enough information in the graph to understand the space it's on then you don't need nearly as much information for a plane so it's nice and simple
Then that'd really be one vertex, two edges, and one face. So V-E+F=0.
With the key being that now the one face really is homoemorphic to a disk.
How to solve the inequality: $0<= (k-3)/4(k-1)<=1$
Where have I reached? For the part <=1 ,$ (2+3k)/(k-1) >=0$ How to continue from here?
17:17
TY all
I know there's a similar picture for a genus-g surface, but I can't remember. Is it a 2n-gon?
so $0 \leq \frac{k-3}{4(k-1)} \leq 1$
@LucasHenrique Yes
as $k \ne 1$, let's make up some cases
first case: $k > 1$
Yes
17:21
@Semiclassical 4g gon
multiply everything by $4(k-1)$ (which is positive) to get $ 0 \leq k-3 \leq 4k - 4$
blah, you're right.
which, speaking of, what are we talking about? :P
I am back home, officially
Congrats, Balarka.
solve this inequality, intersect the solution with $(1, \infty)$
17:23
Hi @Ted
Well. It's got half of those edges identified, so that's E=2g. And there's still one face, so F=1. So V-E+F=V-2g+1, which should be 2-2g. So there's only one vertex?
@LucasHenrique How can 4(k-1) be only positive?
and do the analogous think with $k < 1$, but just revert the inequalities
@Semiclassic: You get a $g$-holed torus if you identify (appropriately) the edges of a $4g$-gon.
@Semiclassical Yeah, all the vertices gets identified.
17:23
Huh.
@LucasHenrique Okay, thanks.
Hi @Alessandro
@Abcd that's our supposition: $k > 1 \implies k - 1 > 0 \implies 4(k-1) > 0$
There's a specific gluing information of the edges, like Ted says, of course.
17:24
@LucasHenrique yes
I also changed $2k$ to $4g$ :P
Hi @Lucas @Abcd
I'm trying to stare at the g=2 case and convince myself that there's only one vertex.
@TedShifrin Hey!
17:24
@Ted I corrected that one before you said so, so I #snip'd you on that one :D
Oh, I missed that. :)
hey @Ted! :)
Though this image help:
I'm only here for mere instants, anyhow. Have to go to the dentist and get 3 crowns installed (fun, fun, fun).
some of those identifications are easier to see than others.
17:25
Yikes, @Ted.
for instance, the endpoints of the c-edges are obviously the same.
Better exercise, @Semiclassic, is to cut the 8-gon in half and see how that's putting two 1-holed tori (each minus a disk) together. That's the connected sum all the topology people in here talk about.
the $n$ holed torus has letters $aba^{-1}b^{-1}cdc^{-1}d^{-1}\cdots$ on the $4n$-gon
Hmm, I think I see it.
I definitely see it at the level of a 2-torus.
17:27
That is sufficient.
I always did this at the end of my diff geo class when I was discussing hyperbolic geometry.
Of course, that also fits with being able to view a genus g surface as a sphere with g handles.
@TedShifrin A guy in the workshop kept drawing a hyperbolic 9-gon instead of an octagon when discussing conformal vs. hyperbolic structures on surfaces
Oops. But mathematicians can't count past 3 anyhow.
Reminds me of the novel Watership Down, which is about rabbits. They've only got four digits on their paws, so their counting goes: one, two, three, four, many.
17:30
I remember that book.
It's a good one, though I haven't read it in years.
I lent it to a friend 30 years ago and haven't seen it since.
I have heard of it, but never read it.
I actually am not sure where my copy is, come to think of it. I imagine it's at home somewhere.
I carefully read my three religious textbooks the previous week, however.
17:33
It's funny. I gave a pretty precise hint on a multivariable analysis question and then afterwards two advanced people with lots of rep gave complete proofs by contradiction. I put on both "Why is everyone so enamored of proofs by contradiction?" ... Another occurrence of people with rep doing someone's complete question rather than leaving it to the OP to think through.
Your textbooks are religious, Balarka?
@TedShifrin So the connected sum would be: Take two surfaces, cut out disks from them, and identify the resulting boundaries?
Right. Balarka can lecture you on this. I'm off to the dentist :)
You have to identify with "matching" orientations, of course.
17:34
@Secret oh, this is mean
See y'all later.
Bye, @Ted
I can kinda see how it'd be a useful idea, since in taking the disk you modify each surface in an obvious way.
bye @Ted
Also, yeah: Eliot, "The Waste Land" is my Veda, Dostoyevsky, "Notes from The Underground" is my Bible and Borges, "Labyrinths" is my Quoran
17:36
Re: Double torus as octagon
That's what the covering space of the double torus looks like
I kinda want to see that in Beltrami-Klein now
though that would just make the central octagaon even bigger I think
@AkivaWeinberger What time does the clock say? :D
17:38
(Beltrami–Klein is the projection that has true straight lines, though circles and angles get messed up)
@Akiva The octagons would be honest to god octagon there, right?
Yeah, but at most one would be a regular octagon
I actually want to see like a GIF of a ball traveling on an octagon, like a billiard ball, but instead of bouncing off the wall it obeys the quotienting of the sides and comes out one of the other sides
but at a weird angle, determined by how it would work if the octagon were on a Beltrami-Klein model of that tiling
I want to learn hyperbolic geometry properly at some point.
17:43
What I'd like to see (as in actually handle) at some point is a crochet model of the hyperbolic plane
I think Henry Segerman has a 3D printed version on his YouTube channel
but made of lots of small triangular parts latched together so that it's not rigid
Is that like a C1 embedding of $\Bbb H^2$ in $\Bbb R^3$ or what
For the details on the hyperbolic crochet stuff, see here: math.cornell.edu/~dwh/papers/crochet/crochet.html
I thought you could always embed a compact subset of it, just not the whole thing
or something
Complete negatively curved surfaces do not embed in $\Bbb R^3$ I don't think
Not smoothly
Hey everyone!
I just realized how to solve a completely unrelated problem
@Daminark Hi, Doctor Nick!
hi chat
@Balarka that's a sweet theorem
$\hat\chi$
@EricSilva Yeah. I think I actually honestly know the proof for compact surfaces only.
what is le problem, @Akiva?
Hi @Daminark
17:52
So there's a quick proof that $(a+b\sqrt n)^k$ is never an integer when $a,b\ne0$, $k>0$, and $n$ is positive nonsquare
the proof of Hilbert's theorem is pretty nontrivial
However, this fails for negative $n$, as $(1\pm\sqrt{-3})^3=-8$.
The questions is if $1\pm\sqrt{-3}$ (and $-1\mp\sqrt{-3}$) are the only counterexamples.
@AkivaWeinberger yeah. I felt bad, hahah
And I think they are
Oh wait hold on my proof idea doesn't work
Never mind
@AkivaWeinberger :O
17:55
I think it does follow from the much more higher-powered theorem that cyclotomic polynomials are irreducible, but I don't know how that's proven
How's it going @Lucas, @Akiva, @Eric, and @Balarka?
@AkivaW That's a cute question.
I'm fine, what about you @Daminark? :)
and it seems like there should (hopefully) be an easier proof
lecture prepping @DAminark
17:55
@BalarkaSen I don't know if you saw the proof for $n$ positive?
@Eric nifty, what's it on?
@Lucas I'm aight, thanks!
regularity theory for area minimizing currrents
@AkivaWeinberger well, I can express it as $x + y\sqrt{n}$ with both $x$ and $y$ positive
Nice
The GMT stuff has an amount of PDE in it, right?
@AkivaWeinberger No but I'd try doing binomial expansion. I think there's a short proof from considering (a + bsqrt(n))^k + (a - bsqrt(n))^k or something
17:59
@BalarkaSen Yeah, it's an integer iff it equals its conjugate, so we want $(a+b\sqrt n)^k=(a-b\sqrt n)^k$, but that's impossible since $|a+b\sqrt n|=|a-b\sqrt n|$ implies either $a=0$ or $b=0$
@Daminark yeah it does, but you don't need PDE for what I'm doing
@Akiva Mhm.
it's the other way around, the stuff I will lecture on is good for PDE applications

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