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3:00 AM
And in colors like hot pink, electric yellow and lime green
do you prefer the red design or blue
The 3Blue1Brown YouTube channel does black backgrounds really well
ccool @MikeMiller thanks
or this red theme
which one?
@meow-mix Make the lines thicker, it's really hard to see them
And brighter
brighter?! they're pure white :P
Not the red one
@meow-mix add some element of green; green is naturally brighter to the eye
@meow-mix But you're right, I think it's not as important as just making them thicker
3:05 AM
@DHMO I'm too tired to do proofs idk :(
so which would you guys like more
red, green, or blue?
@GFauxPas but you can make my proof tidier
green is brightest to the eye
because of evolutionary reasons
Maybe tomorrow
doesn't mean its a better color :P
Did you guys read about how people might not have perceived blue the way we do as recent as several centuries ago?
It's so weird
3:07 AM
Yeah, lots of people don't believe that
Did they reference the "winedark sea" thing?
So many languages didn't have a word for "blue"
@GFauxPas I assume that this used the fact that the Greeks called the sky "bronze"
Didn't? Don't
The main argument was the lack of the word for it
I think this isn't a problem of perception, but a problem of necessity.
3:08 AM
@GFauxPas But they've tested monolingual speakers who don't have words for certain colors. They see them just the same way we do
How many things in nature, especially in certain regions, are blue besides the sky?
They cited a study with results to the contrary
There are some languages that just have languages for black and white (or, rather, "light" colors and "dark" colors) and can't distinguish beyond that
@GFauxPas That study is in the minority
did you know that brightness is logarithmic
Most things are
Or, rather, our perception of them is
3:10 AM
and that the color gray on your computer screen is only 25% the brightness of white
The concept of linguistic relativity concerns the relationship between language and thought, specifically whether language influences thought, and, if so, how. This question has led to research in multiple disciplines—especially anthropology, cognitive science, linguistics, and philosophy. Among the most popular and controversial theories in this area of scholarly work is the theory of linguistic relativity (also known as the Sapir–Whorf hypothesis). An often-cited "strong version" of the claim, first given by Lenneberg in 1953, proposes that language structure determines how we perceive the world...
It's also worth it to note that perception changes like that are highly unlikely to crop up as a result of evolutionary changes, especially considering the timescale, and how many different cultures from parts of the world with very little direct contact with each other eventually gave blue its own word.
The Greeks called the sky "bronze" not because it didn't look blue to them, but because they didn't use "bronze" so much as a color. There was something more to it--likely its brightness (akin to a gleaming bronze shield, for example).
Also, as near as we can tell, the makeup of our cones hasn't changed in a long, long time--we share our trichromatic nature with many animals. However, some people are born as tetrachromats.
And wine, despite its redness, has a certain way of reflecting light that is similar to the dark waters of the ocean
Hence "winedark" — not the color, but the way it shines when lit
Or something similar
@AkivaWeinberger The weirdest thing to me about all of this is how convincing the Berlin-Kay hypothesis is.
I'm imagining you'd need to look at a barrel of wine to see it
Why can't you get a badge for accepting your own answer? It seems to me you should get something for it...
3:19 AM
@DHMO I just had a thought
maybe you can use the other form of induction
That is, I wonder why it would be that red is one of the first colors distinguished. Possibly blood?
@Fargle Is that the one that says that it doesn't much depend on the language?
"let $x$ be the smallest number that is not equal to $a$ and is neither $a < x$ nor $x < a$
or culture?
@GFauxPas "smallest" is circular
3:20 AM
isn't it equivalent to induction
@AkivaWeinberger Yeah. That there's generally a hierarchy--B/W are named first, then red, then yellow/green, then blue, then brown, then others
I guess it is
Pas pas fizz fizz
Oh what a relief it is
what's the name of that form anyway
it has a name
It's interesting how English distinguishes between bright and dark red (that is, "red" and "pink") but not bright and dark blue (though Russian does — it doesn't have one word that means "blue," but rather two words that mean "light blue" and "dark blue")
3:22 AM
well-ordering principle
that was it
Also, Vietnamese doesn't distinguish between green and blue.
(It's been translated as "grue.")
@AkivaWeinberger Does Russian distinguish between light and dark red?
So the sky and the grass are the same color in Vietnam.
That doesn't mean one can't distinguish between them in Vietnamese, but you use adjectives to do it: "leaf grue" and "ocean grue"
@JessyCat Only in the same way that a light blue object and a dark blue object are "the same color"
Okay, so that means that Vietnamese kids don't draw extremely boring pictures.
3:24 AM
Different hex values, say, but the same basic color category in the language
@Fargle I don't know
The thing that makes this so hard to study is how easy it is to conflate a name for a color with the perception of it.
There's a definite cognitive bias of "well surely I give different names to different colors, so they must not see them as different" but the light/dark blue example is a good way to dismantle said bias.
If you give someone a bunch of weird shades of colors and tell them to group them by similarity, it makes sense for their decisions to be influenced by language, since "these are called by the same word" is a similarity
for example, "light red" is called "pink"
@AkivaWeinberger also, language pretty much determines how the brain sees things
language and culture are intertwined
And it's not as though any grouping is preferred over another.
In a manner of speaking, green and blue really are quite similar.
@DHMO That's essentially the Sapir–Wharf hypothesis, the strong version of which has been rejected
3:26 AM
For example, copper (II) ions in aqueous solutions are blue/green
and various weaker versions are doubtful at best
@DogAteMy thanks for the information
This feels off ^
lol @dhmo the potw has been there since may
let me change that
But continuing what I was saying before, what you'd want in that experiment is them to group them only by how similar they look. But that's harder to test.
3:29 AM
@GFauxPas nice
The thing that bugs me most about Sapir-Whorf is how it's used on the Internet to decry people who speak different dialects
There's a sub on Reddit for that, by the way. /r/badlinguistics
@Fargle for example?
@AkivaWeinberger Hehe, I'm already there.
Ah. May I ask what your Reddit username is?
3:31 AM
@DHMO It's going to disgust me to type this, but stuff like "people who speak in ebonics are literally dumbing themselves down"
@DHMO AAVE (African American Vernacular English), also known as "ebonics", is widely considered by linguists to not be "English with mistakes"
@Akiva "Wellthenpeople"
I don't post much. Mostly stuff about math, or social justice issues, or music.
but people tend to just use it as evidence of how black people are stupid
or stuff like that
Maybe that's too strong.
It also pops up in a different form whenever there's a new slang term and people get all prescriptivist over it.
@DHMO hi!
3:34 AM
"Oh my goodness, 'bae' isn't even a word!" Or, "Why does everyone know what 'fleek' means but can't tell the difference between 'your' and 'you're'?"
A quick joke: At Yale, a visitor asks a student, "Do you know where the library is at?" "I do," the student replies, "but at Yale, we don't end sentences with prepositions." The visitor replies, "OK, do you know where the library is at, asshole?"
@Ramanujan hi
@GFauxPas nice
@AkivaWeinberger Yessssssssssssssssssssssss
My favorite joke about prescriptive grammar is:
It is wrong to ever split an infinitive.
@GFauxPas I don't understand why $(1)$ must be true
for a total-ordering on $\Bbb C$
3:39 AM
@DHMO "Fourty-one: Be consistent."
there's a result about inverting switching inequalities
Or, rather, "41) Be consistent" probably works better
> Place pronouns as close as possible, especially in long sentences, as of 10 or more words, to their antecedents.
@AkivaWeinberger i don't get this
The word "their" refers to the word "pronouns"
but they are very far apart
I mean I don't get "41) Be consistent"
3:42 AM
Oh, sorry. It's the difference between "41.", which follows the style that all of the other points used, and "41)"
oh DMHO I know why
because otherwise you could add the equalities together to get $0 \prec 0$
All of the others are "number, dot", so don't suddenly use "number, parenthesis"
@GFauxPas you know, the link to total ordering given in the theorem does not exclude this...
@AkivaWeinberger ok
it's in the "compatible with" part
you can impose a total ordering on $\mathbb C$, but addition and multiplication will ruin it
maybe it hsould be more explicit
you can edit it :)
@GFauxPas I mean, that does not rule out 0<0
the total ordering is not strict
3:45 AM
ordering has to be antisymmetric
but it is not a strict ordering
hmm, so, put it on the discussion page
if it's proof of the week, you better make sure it's correct
or rather
maybe it should be called "complex numbers cannot be strictly ordered"
does anyone here know if theres a weak ordering on $\mathbb C$ compatible with addition and multiplication?
What's a weak ordering?
Oh. Well if you have a weak ordering, then you get a strict ordering by saying "$a\prec b$ iff $a\preceq b$ and $a\ne b$"
So shouldn't they be equivalent?
Like, if no strict order is compatible, shouldn't that mean no weak order is compatible either?
3:49 AM
DHMO and I are trying to figure out what precisely is the problem that would arise if we had $0 \prec 0$
which is the contradiction this proof rests on
what would be the problem is saying $0 \prec z \land 0 \prec -z$
@GFauxPas then you would have $z \prec 0$ (compatibility with addition)
but then an ordering has to be antisymmetric
oh by adding $z$ to both sides
so there's your answer
3:53 AM
so edit the proof making explicit what the problem would be :)
but you might need to add it to the proof
you have editing rights
fist bump
as cool as $\mathbb C$ is, the total ordering of $\mathbb R$ will always ensure it has a place in my heart
What about $\Bbb H$
3:59 AM
I never played with those, though I read about them
I recall reading that sim cards use quaternion algebra
ok heres another animation i just made
Yeah, I think it's 'cause they're good at doing 3D rotations
it shows how projective points are affected by projective transformations
that's cool, I dont really understand it though
@GFauxPas I've made the proof very messy lol
4:14 AM
I think it's clearer than before though
following typical italic conventions, $i$ should really be non-italic, but when I see it that way it looks weird, I'm so used to it italic
same with $e$
$i$ should be non-italic???
Usually variables are italic and constants are upright
but $\mathrm i$ looks weird
I dunno, I've always used italics for $i$ and $e$ and stuff like that
Right, I'm saying, people typically italicise $i$ and $e$
4:19 AM
Hello, someone have an idea about this : math.stackexchange.com/questions/2052878/…
hmm now I cant find the source that constants are not italicized
in most places I've seen its italicized
but I recall reading somewhere that constants should be upright and variables italicized
im trying to find a source
hmm maybe it was a dream
@GFauxPas how many hours is it until tomorrow?
In my time zone, 27 min
4:33 AM
@GFauxPas it's just tiring to tex upright e and i
but @G have you heard that it's better to?
This is my time zone. I own it.
okay good night my internet mathematician comrades
You also own many wine burgers
How original.
4:34 AM
@AkivaWeinberger Yeah, this is a little test of rendering rotations using quaternions in javascript I did jsfiddle.net/xmd9suLp/1
@GFauxPas I think it really is up to you... You see both in papers
@AkivaWeinberger your first name is Jewish and your last name is German; what are you?
Jewish. As for nationality-wise, my father's father was Polish, my father's mother is American, my mother's mother is American, and my mother's father is Polish but he moved to Argentina when he was four
But both parents are American
@DHMO Also, Jewish and German aren't mutually exclusive
@AkivaWeinberger we call those yiddish
Yiddish is a language.
do you speak Yiddish?
4:41 AM
are you monolingual?
English, a bit of Hebrew, a bit of Spanish
so you're an American
@DHMO ? Where did that come from?
@G.Bergeron As a language, Yiddish is essentially German mixed with some Hebrew
And used to be spoken widely among European Jews
primarily in Germany and Poland, I think
4:43 AM
@DHMO Yeah, sorry, forgot to say.
@AkivaWeinberger So it's late for you too...
Watching a video about boredom. "...[Drug] addicts' reported levels of boredom are the only reliable indicator of whether or not they will stay clean." Interesting
it was a joke @DogAteMy
... trying to convince myself to work on a paper...
4:46 AM
I think in terms of my identity I would say I'm first a New Yorker, second a Jew, third an American, and last a human (though the last two might be switched around, I'm not sure)
Were I more perfect I'd put the "human" thing first
Are you frum?
Just curious
@AkivaWeinberger Not far from me... but it seems you're giving a fair bit if thoughts about this identity thing...
Ortgodox jewish
4:48 AM
I'm "Modern Orthodox"
"Frum" is Yiddish
Indeed 'tis
I consider myself modern frum :)
@DogAteMy do you have Jewish religious affiliations?
4:49 AM
Who is this @dogatemy , he's not showing for me
It's Ted's nickname for me
apparently somebody changed name
No, Ted gave it to me
4:49 AM
I didn't choose this
Is this guy like the resident professor here?
Who, Ted? Yeah, essentially
Well if that's your name, that's what we should call you
My old username was columbus8myhw ("Columbus ate my homework")
so he decided to call me DogAteMy
(Columbus isn't actually my dog's name)
I've never been to New York even though it's so close
4:51 AM
Come to Brooklyn!
Actually, no, stay home, it's too cold out
Colder here
Where are you? CT?
Up north
I don't live in Brooklyn but I go to college there
4:52 AM
Ahh. So not as close as I was thinking
I've never been to Montreal
@AkivaWeinberger Yeah, at some point, but I have some friend to visit in Boston first
Yes! We beat you by -7 over here :D
hi chat i need help
Apparently it's like $60 to get a train ticket to up there
at the lowest
Yeah, most people just drive, it's really not that far
Too young so far
4:55 AM
I can walk that far
But then again I'd be too young to go to Montreal by myself by train
Meh, I don't have a car...
I can walk that far will you help me now ?
@AkivaWeinberger I don't think there's a minimum age
I'd want to ask my parents first
4:56 AM
@KasmirKhaan about what?
@AkivaWeinberger Understandable, anyway you probably need their signature to get a passport
I already have one
why is absolute value of the integral allways less than or equal the integral of the absolute value of function ?
is this triangle inequality ?
Sum of abs < abs of sum
Think about the negative terms
Same with integral, it's just an infinite sum
does it work in complex as well ?
4:58 AM
thanks bergeron :)
Suppose that the integral is positive.
I should have written <=
Then $\int f(x)dx\le\int|f(x)|dx$ since $f(x)\le|f(x)|$
@DHMO I'm on my cell phone so I can't do it myself, but would change the proof a little
$|\int f(x)dx|\le\int|f(x)|dx$
Similarly for if the integral is negative
5:00 AM
Invoking proof by cases makes it seem like you are going to keep the result, but you're going to throw it out
For complex functions it's a lot harder to prove
Say "by hypothesis, $\preceq$ is connected. This means that either ...
i just wanted some intuition on why it is the case
not really harder... the abs is always $\geq 0$
It's essentially the triangle inequality
5:01 AM
is there an easy exemple ?
IMHO that would be a better presentation
@KasmirKhaan How about $f(x)$ is equal to $a$ for $0\le x<1$ and $b$ for $1\le x\le2$.
It's gonna be true in any measurable metric space
Then the formula implies that $|a+b|\le|a|+|b|$
because that's $|\int_0^2 f(x)dx|$ and $\int_0^2|f(x)|dx$
5:02 AM
thanks alot akiva and bergeron :)
gFauxpas i cant open that link
@GFauxPas just edit it yourself
@dhmo I'm on my phone
It's basically using the mapping $z=r e^{i\theta}$ and the phase part has Abs=1
5:06 AM
But your proof is fine the way it is l, really, I shouldn't fuss
Which proof
Hi guys
Anyway, i'll go work on that paper
6:02 AM
Q: Proving that the following bilinear map is well defined.

AdeekLet M be a compact differentiable manifold of dimension n. Show that the bilinear map $H_{dR}^s(M,\mathbb{R}) \times H_{dR}^{n - s}(X,\mathbb,R) \rightarrow \mathbb{R}$ defined by $([v],[w]) \mapsto \int_{M} v \wedge w$ is well defined. I was thinking maybe I could use stokes theorem but I am no...

Q: Line integral ( Stokes' Theorem)

Kasmir Khaan$\gamma$ : intesection of $z=x^{2}-y^{2}$ with $3x^{2}+4y^{2}=1$ Am supposed to calculate the line integral over the intersection. $\int \bar{F}dr = \int \int Curl\bar{F}dS $ CurlF =$ (7y^{6},5z^{4},3x^{2})$ How to parametrize the surface here? Thanks in advance

@adeek hi
hi @KasmirKhaan
6:27 AM
what does this symbol $\sqcup$ mean?
i would guess UNION
square union? I want to know if it is the same as $\cup$
disjoint union
hi @Sophie @Kasmir
hi, and thanks
@Kasmir: It is entirely non-obvious what surface would bound that curve.
I guess all you could use is the portion of the saddle surface $z=x^2-y^2$ lying over that ellipse. So you'd parametrize by that elliptical region and use $(s,t,s^2-t^2)$. Go for it.
6:48 AM
hi @TedShifrin been waiting for you to come :D
This seems to be an ugly integral.
it is a cylinder intersection hyperbola of one sheet
A hyperbolic paraboloid — a saddle surface.
hi @TedShifrin
I think we suppose to use strokes but cant get the picture working
yes but why did you chose (s,t,s^2-t^2 ) ?
6:49 AM
The curve is sort of like the boundary of a potato chip, and you want to integrate over the potato chip.
Because that's the way you always parametrize a graph.
no i mean , where did you get s^2-t^2?
The equation of the surface.
please tell me if am wrong
what i see is
Once you use symmetry considerations, the integral won't be that bad.
like a curvy plane intersected with elliptical cylinder
it should be some ellips there
but not perfect ellips
why is this argument wrong ?
6:53 AM
I'll post the picture for you.
thanks :D
very nice thanks! what program did you use?
@TedShifrin I really don't understand this De rham cohomology stuff
Hello @Ted.
6:54 AM
Mathematica, @Kasmir.
Heya @Fargle
the way it is written in the notes isn't good at all
How goes it?
Did you learn differential forms well, Karim?
I should install that asap , have absolut no feeling for these 3d objects
It is expensive, @Kasmir, but there are some free things that work, I think.
6:55 AM
I see why now it should be (s,t,s^2-t^2 ) =P
I ll figure it out =p
no @TedShifrin I just know how to solve the problems prof is so bad..
Think carefully about symmetry and why things should integrate to 0 before you do horrendous integrals, @Kasmir. This problem is uglier than ones I assigned.
this is a lesson for me next semester to enroll in a class by making sure prof are good.
I answered your question, Karim.
yeah I saw that makes sense
6:56 AM
You mean if its odd and symmetric intervall = 0 yeah ?
thanks @TedShifrin
The double integral version of that, @Kasmir.
@TedShifrin thanks alot again Ted ! you are the best! :)
am working on it now
You're welcome.
Let $V \subset X$ be inclusion of real oriented compact manifold. Assume the non-degeneracy show that via integration form over V that V determines a cohomology class $[V] \in H_{DR}^r(X,\mathbb{R})$
what does this mean lol

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