Mathematics - 2017-07-26 (page 4 of 5)

Tobias Kildetoft

19:00

@Typhon So you are claiming that "greater in real magnitude" is not a total order?

Typhon

i dont know

Suppose not. Then, a >= c and b >= d. Multiplying the latter equation by x we get that a >= c and bx >= dx. Adding the two equations together we get a + bx >= c + dx. However, this contradicts our presumption that a + bx < c + dx. Therefore our assumption was false and the statement is true. []

Fawad

Do any one have general form of integral of f(f(x))dx ?

Tobias Kildetoft

@Typhon You negated the condition incorrectly

Typhon

XD

no wonder you are confused

my god I must be not on point today

it's supposed to be:

"For a quadratic real integer a + bx another element c + dx is greater in real magnitude if and only if a < c or b < d"

Tobias Kildetoft

still not correct

Typhon

19:02

...

Tobias Kildetoft

Seriously, you really need to look at a few small examples

Typhon

yeah, it is

i just proved it

why are coming down on me anyway? I was just talking to akiva.

Tobias Kildetoft

One of your conditions imply the inequality, the other is implied by it

neither is equivalent to it

Typhon

what are you talking about???

there's only one condition

a + bx < c + dx

Fawad

@Typhon a<c or b<d

Typhon

19:05

@Fawad that's what my post SAYS.

"For a quadratic real integer a + bx another element c + dx is greater in real magnitude if and only if a < c or b < d"

Fawad

Should there be and ? I think so

Typhon

no

it's not and

here's the proof

Suppose not. Then, a >= c and b >= d. Multiplying the latter equation by x we get that a >= c and bx >= dx. Adding the two equations together we get a + bx >= c + dx. However, this contradicts our presumption that a + bx < c + dx. Therefore our assumption was false and the statement is true. []

Fawad

Whatever. Hope tobias helps you

Typhon

@Fawad he's not helping me...

i was merely sharing a proof with Akiva and he started attacking it. :/

Fawad

Alright

You know calculus ?

Typhon

19:08

how is that relevant?

and yes

Fawad

7 mins ago, by Fawad

Do any one have general form of integral of f(f(x))dx ?

Astyx

What's wrong is "Suppose not"

Typhon

@Astyx how so?

proof by contradiction

Astyx

If you suppose not, you get that "$a\ge c$ or $b\ge d$"

Typhon

no

READ THE STATEMENT

"For a quadratic real integer a + bx another element c + dx is greater in real magnitude if and only if a < c OR b < d"

De Morgan's laws

In propositional logic and boolean algebra, De Morgan's laws are a pair of transformation rules that are both valid rules of inference. They are named after Augustus De Morgan, a 19th-century British mathematician. The rules allow the expression of conjunctions and disjunctions purely in terms of each other via negation. The rules can be expressed in English as: the negation of a conjunction is the disjunction of the negations; and the negation of a disjunction is the conjunction of the negations; or the complement of the union of two sets is the same as the intersection of their complements...

Fawad

19:10

I think multiplying by x he means multiplying by |x| because inequality will change for negative value multiplication

Typhon

@Fawad x cannot be negative

or at least I'm pretty sure it isn't.

either x or its conjugate is

we'll just assume x is the positive conjugate. One of them always is.

unless it's a complex number

but then ordering goes out the window

@Fawad no I definitely meant multiplying by x. It's just I'm pretty sure that one of the two solutions to x^2 + ax + b = 0 is positive. But perhaps I'm thinking of x^2 + b = 0. Now my head hurts. XD

Semiclassical

@Fawad Not in any generic way. Best case scenario is that $f(x)$ and $f(f(x))$ both have valid Taylor series in the region of integration, so that you can write out the composition of series and integrate term-by-term.

Typhon

@Fawad Here's a counter example. x^x.

Fawad

@Semiclassical oops. For indefinite inegration?

Typhon

then it becomes (x^x)^(x^x)

Semiclassical

19:15

same idea, really.

Typhon

@Fawad that's what he was saying.

Semiclassical

except even harder to control since the domain isn't specified.

Typhon

having an indefinite integral directly implies that you have a definite one

by the second fundamental theorem of calculus

@AkivaWeinberger Somehow when writing that I forgot to specify that we are discussing Z[x] where x^2 + b = 0 and b is negative. My bad.

In that case, the proof is trivial and a mirror image for when I did b = -3

Fawad

Ooh. By substituting tylor series we get polynomial integrals

Semiclassical

An example where this works fine is $f(x)=e^x$.

Typhon

19:18

except the integral of e^x is trivial

like.... as trivial as evaluating e^x.

Semiclassical

Sure, but $e^{e^x}$ isn't. He asked how to integrate $f(f(x))$.

Typhon

@Semiclassical oh.... duh

Semiclassical

Hmm. The leading terms of the Taylor series about $x=0$ for that is $e^{e^x}=e+ex+ex^2+\frac{5e}{6}x^3+\frac{5e}{8}x^4+\cdots$

Typhon

@Fawad For fractional polynomials, yes. It's called expansion, long division and partial fraction decomposition.

in other words, simplify the function first

Fawad

Yeah

Typhon

19:21

other than that...

Semiclassical

Which means that the first few derivatives of $e^{e^x-1}$, evaluated at $x=0$, are 1,1,2,5,15,52...

all integers. hmmmmmmm

Secret

$$\int f(f(x))dx = ???$$

Let $f = \sin$

$$\int \sin (\sin (x))dx=\int \sin (\frac{e^{ix}-e^{-ix}}{2})dx = \int \frac{e^{i(\frac{e^{ix}-e^{-ix}}{2})}-e^{-i(\frac{e^{ix}-e^{-ix}}{2})}}{2}dx = \frac{1}{2} \int e^{i(\frac{e^{ix}-e^{-ix}}{2})} dx - \frac{1}{2} \int e^{-i(\frac{e^{ix}-e^{-ix}}{2})} dx$$

and then have no idea, but semi has a special function for the integral of the form:

$$\int e^{e^x}dx$$

Typhon

welp... good luck

Semiclassical

It's a sign of the stuff I get interested in that I wonder what sequence those derivatives correspond to.

Typhon

@Semiclassical if I have a monic polynomial with integer coefficients and the roots are irrational real numbers, is there ever a time when both roots are negative? I cannot think of such a case, but I know it has to exist...

Semiclassical

19:23

at which point I thank goodness for OEIS

Typhon

lol

Semiclassical

"Both roots." Do you intend that to be a monic quadratic?

Secret

actually, is $$\int e^{i\sin x}dx$$ simplifiable to some known special functions?

Typhon

yeah, sorry. Must be tired today.

i keep doing that

ugh

Semiclassical

@Secret Definitely:

Jacobi–Anger expansion

In mathematics, the Jacobi–Anger expansion (or Jacobi–Anger identity) is an expansion of exponentials of trigonometric functions in the basis of their harmonics. It is useful in physics (for example, to convert between plane waves and cylindrical waves), and in signal processing (to describe FM signals). This identity is named after the 19th-century mathematicians Carl Jacobi and Carl Theodor Anger. The most general identity is given by: e i z cos ⁡ θ = ...

Fawad

19:24

wolframalpha.com/input/?i=Integral+of+sin(sin(x))dx

Secret

wolfram is relatively suck at indefinite integrals

Semiclassical

Yeah.

SteamyRoot

@Typhon Just pick roots like $-2+\sqrt{2}$ and $-2-\sqrt{2}$ ?

Secret

right, in that case, we only need to justify interchange integral with that bessel sum to get it simplier

Typhon

@SteamyRoot true but is that the solution to a quadratic?

hrmm

x + 2 = sqrt(2)

SteamyRoot

19:26

$x^2 - Sx + P$

Secret

(whcih I am poor at beause I know almost no real analysis)

Typhon

ah

Semiclassical

$(x+2-\sqrt{2})(x+2+\sqrt{2}=(x+2)^2-2$, yup.

Typhon

welp

that puts bullet holes through my whole proof

Semiclassical

@Secret I always like the Jacobi-Anger identity, because it's a Fourier series with Bessel functions as coefficients :)

Typhon

19:27

that's what I get though for trying to claim that all quadratic rings share the same nontrivial property. XD

I feel like I should know by now: They don't all have the same properties

yet I keep finding myself doing that

XD

wait a second!

if b divides a

Semiclassical

Hah, of course this is familiar: oeis.org/A000110

Typhon

then x and x' divide each other

Semiclassical

Huzzah Bell numbers

Typhon

therefore $x/x' = \pm 1$

but if it were 1

then that would be a double root

which is an integer

Semiclassical

(I'm sure I've seen $e^{e^x-1}=\sum_{n=0}^\infty B_n\frac{x^n}{n!}$, but I never remember these things off the top of my head)

Typhon

19:31

@Semiclassical is there a quadratic with a double root that is irrational?

i think not

wow though

I just figured out what case to examine

Semiclassical

if it's monic and integral, no.

Typhon

thought so

Semiclassical

the constant term would be the product of the roots, and since the root is double it'd be irrational.

Typhon

oooh

of course

|a| = |x + x'|

Secret

$$\int \sin (\sin x)dx = \int \frac{e^{i\sin x}-e^{-i\sin x}}{2} = \frac{1}{2}\int e^{i\sin x} - e^{-i\sin x}dx$$

Use Jacobi-Anger expansion

$$\frac{1}{2}\int e^{i\sin x} - e^{-i\sin x}d = \frac{1}{2} \int \sum_{n \in \Bbb{Z}} J_n(1)e^{inx} - \sum_{n \in \Bbb{Z}} J_n(-1)e^{inx}dx$$

and good luck figuring out how to show whether you can interchange integrals with sums...

Semiclassical

19:33

@Secret This is why I much prefer definite over indefinite integration, heh.

Typhon

@Semiclassical thanks

Semiclassical

np

Typhon

@AkivaWeinberger turns out my whole proof from earlier is garbage. I'm going to go and write the proof now on the server. I'll send you screenies of it when it is done. It partially relies upon out modular arithmetic thing for this case.

Secret

I like indefinite because if you can figure it out, you get all integrals of that type, and no more magic and confusion

Semiclassical

The other reason I like definite integration is because it often can be understood as contour integration

And then different solutions to the same ODE can be understood as just different integration contours of the same integrand.

Typhon

19:35

@Semiclassical technically i could handwave a bunch as it's just me writing math proofs while playing on a minecraft sever to share with a few curious individuals.... but my morality won't let me do that. XD

Semiclassical

lol

Secret

I found it odd why we cannot indefinite contour integrate, is is because we will be literally doing a path integral in the complex plane?

Typhon

it is actually fun to try and cram proofs into the little tiny books

or worse, use signs. XD

bye guys

Semiclassical

Well, what's the actual definition of an indefinite integral? It's the general solution to the ODE $y'(x)=f(x)$.

Secret

yup, and that's a family of them

Semiclassical

19:37

Right. $y(x)=F(x)+C$.

Secret

An indefinite complex integral is probably the general solution of the cauchy riemannian PDE and whatever the PDE the integrand is linked to

Semiclassical

thing is, in writing down a particular representative of that family, one inevitably has a boundary value e.g. $F(0)$

In which case we can write that $F(x)$ as $F(x)=\int_0^x f(x')\,dx'$.

In which case that's really just a particular case of the definite integral: You integrate along a path from $0$ to $x$.

In complex analysis, though, the notion of a path from $0$ to $x$ is no longer so simple. You could do the straight line path, or you could do something twisty.

However, now one has to worry about obstacles like poles, or branch cuts, or what have you.

So one usually gives up on trying to talk about the antiderivative in a generic sense, and instead focuses precisely on integrating along a particular path.

Secret

I see

Daminark

@Ted now that I think about it, I'm not sure that what I said works, about finding an analytic primitive by integrating, since we don't know that $\int_{z_0}^z f(z)dz$ exists without some kind of path invariance, which is the sort of thing you'd only know after the integral theorem, so that attempt is circular anyway

Semiclassical

@Daminark Probably a good place to start is to show that $\int_\Delta f(z)dz=0$ when $\Delta$ is some triangle in the complex plane and $f(z)$ analytic in that triangle.

Daminark

19:52

@Semi the thing I was trying to do is make the Stokes' proof of the integral theorem work for all holomorphic functions

Eric Silva

@Daminark the triangles thing @Semi suggested is how it's done without $C^{1}$ i think

Cauchy-Goursat is the name or something

Semiclassical

That immediately gives a version of path invariance, since it means that integrating from a vertex A to a vertex B is the same as integrating from A to C + C to B

Daminark

@Eric is it at least nicer than all that stuff in Titchmarsh?

Eric Silva

it's pretty nice

Daminark

I just didn't like that proof

Semiclassical

19:55

I forget how the proof works, tbh. Some combination of the estimation lemma and splitting the triangle up into four pieces.

Eric Silva

the proof is still a bunch of estimates

but like you will not be able to avoid it

i think that's a nice one though cuz you just get down to holomorphicity

but you can do so much better

Daminark

How so?

Eric Silva

you don't even need to assume the derivatives exist a.e.

and then using the real good stuff i mentioned earlier, you don't even need to assume that the derivatives classically exist at all, only weakly

Semiclassical

I mostly like it because it reminds me of finite elements stuff.

Eric Silva

@Semi what is that, I've heard of it before

Semiclassical

19:58

basically, it's a way to solve pde's

Daminark

Hmm, so wait if you know that if the weak derivative of a function exists... wherever it needs to, and weakly, then you get that it's holomorphic

Eric Silva

right right

Semiclassical

Finite element method

The finite element method (FEM) is a numerical method for solving problems of engineering and mathematical physics. It is also referred to as finite element analysis (FEA). Typical problem areas of interest include structural analysis, heat transfer, fluid flow, mass transport, and electromagnetic potential. The analytical solution of these problems generally require the solution to boundary value problems for partial differential equations. The finite element method formulation of the problem results in a system of algebraic equations. The method yields approximate values of the unknowns at discrete...

Eric Silva

@Daminark, if it solves C-R yeah

Daminark

Oh okay so weak derivatives don't actually give you C-R automatically

Eric Silva

19:59

of course not

Daminark

Wait what exactly is a weak derivative? Is it just derivative in the weak topology?

Semiclassical

Basically, you split the region of your problem up into a lot of little regions

Eric Silva

it needs to solve something

@Daminark no it's like, a derivative in the sense of integration by parts for ex

or like a distributional thing, which you would learn about if you take the analysis sequence next year

Semiclassical

You then approximate the PDE on each region and solve each of them symbolically. That gives you a huge list of algebraic equations to solve.

Which is something computers are good at.

Daminark

Lol, still gotta figure that out

Eric Silva

20:01

@Semi cooool

Semiclassical

Eric Silva

@Daminark basically elliptic operators (like C-R) are extremely nice

it was one of hilbert's problems to prove that solutions to elliptic problems were regular

Semiclassical

Ultimately it comes down to converting a PDE problem to a linear-algebra problem.

Eric Silva

as with all things differential ultimately

i usually describe linear algebra as "the only thing we really understand"

Semiclassical

And the matrices you get from those are sparse, since each finite element of your region is only adjacent to a handful of other elements.

Secret

20:04

I wonder if we will ever evolve to be able to deal with nonlinear operators directly...

Semiclassical

One bit on the FEM Wiki page which might be of interest: Prior to doing discretization, one first rephrases the relevant BVP in weak form.

See for instance the examples starting here: en.wikipedia.org/wiki/…

And, heh, the last sentence in their discussion of this step: "This solution is a-priori only a member of $H_0^1(0,1)$, but using elliptic regularity, will be smooth if $f$ is."

Eric Silva

ya good shit

this is my favorite theorem maybe

it's top 5 at least

that you can go from living in a very general sobolev space to being smooth

Daminark

Dayum

Semiclassical

A pity not all operators are elliptic :)

Eric Silva

and the proof isnt even that hard in modern language

it's like lots of energy estimates

@Semi ya but all my favs are tbh

Semiclassical

20:12

Alas, not all of mine are.

The wave equation being the most obvious one.

Eric Silva

o true the wave equation is a goodie

hyperbolic equations scare me still

Daminark

Schlag intensifies

Semiclassical

I think failures of ellipticity are a big problem in QFT? But don't quote me on that.

Eric Silva

QFT seems like a big spook

Semiclassical

QFT is scary.

Eric Silva

20:15

it's like one of those things that the snotty physics majors say they do to intimidate ppl into thinking they're hyperintelligent

at least around here that's literally the only context ive heard people talk abt it

Daminark

I mean, many physics majors talk about it because they're actually into it

Eric Silva

@Daminark ive had at least 5 in my house who did what i described

Semiclassical

The thing with QFT is that, however esoteric it may seem, it works damn well in its domain of applicability

Daminark

I know at least a few people into physics who only talk about it when they're bringing up some story from the class, or amongst each other

A friend of mine took QFTCS and actually liked it

Eric Silva

@Daminark im mostly talking about first years who use big words to sound smart, of which there are a lot

Semiclassical

20:20

The experimental success of QED is one of the singular achievements of modern physics

Eric Silva

not the kind of people who could take grad classes and do well

Daminark

Lol, most of the first years I know into science and all are actually rather timid

Semiclassical

Where things get hairier is QCD, because the problems get a lot harder

Eric Silva

@Semi what is QCD?

Avantgarde

quantum chromo dynamics

Semiclassical

20:22

Right.

Eric Silva

@Daminark idk man Im dating a physics major and she agrees that the physics major culture is pretty bad here and rife with what i described

Daminark

Like they'll talk about the continuum all day because they're chill with it, but most of them dare not say a word beyond that which they're comfortable with for fear of being stupid

Semiclassical

It's the QFT of the strong nuclear force, basically

Eric Silva

and she's involved in a lot of the undergrad clubs and stuff

@Daminark you might just know good people

Semiclassical

QED is quantum electrodynamics, so it accounts for the electromagnetic force

Eric Silva

20:23

Chromodynamics?

Daminark

Yeah actually what's the deal with the whole, strong/color stuff?

Avantgarde

yes

Daminark

What caused the other name to just step in the picture?

Semiclassical

Eh. That's used because quarks can have three possible kinds of a property

Eric Silva

the ? meant id like an explanation of what it meant

Akiva Weinberger

20:24

@Typhon Re: $ax+b>cx+d$ iff $a>c$ or $b>d$; the problem is "iff"

Semiclassical

And they decided to call those three possible properties as red, blue, green

Akiva Weinberger

You could have $ax+b<cx+d$ and still have $a>c$

Semiclassical

Hence, color

Eric Silva

ah ok

Daminark

@EricSilva Maybe, I dunno. For some reason on the whole my experiences with other students have been better than yours, maybe it's a house thing or w/e

Semiclassical

20:26

That might seem silly, but quarks aren't an object of everyday experience

So you might as well call that property 'color'

Eric Silva

@Daminark i also have more experience with more people than you

i go to parties and whatnot

Daniel Castle

Anyone know if an expression can be found for the highest power of 2 some even n is divisible by?

Akiva Weinberger

@Semiclassical Are the three colors symmetric?

Like, are the laws of physics invariant under arbitrary permutations of them

Daminark

Perhaps, though to my understanding people often drink a good bit of alcohol at parties, so they're more likely to be at their worst

Akiva Weinberger

@DanielCastle I don't think there's an actual expression for it (for reasonable definitions of "expression"), but I've seen people write $\nu_2(k)$ for the highest power of $2$ a rational $k$ is divisible by

Eric Silva

20:29

alcohol doesn't usually make you a worse person than you already are

Daminark

Like, I have many reasons for not wanting to drink, the main one being religious but my second place is basically that I'm straight up terrified that everything is gonna go to hell if I have anything less than full cognitive control

Akiva Weinberger

@DanielCastle en.wikipedia.org/wiki/P-adic_order

Daniel Castle

@AkivaWeinberger Oh ok, thanks - I've generated some graphs of it and there is an obvious pattern

Semiclassical

Well, it's worth keeping in mind that quarks aren't something you can observe directly. No one has ever detected a free quark. What we do observe are, say, protons, and those consist of two up quarks and one down quark

Daminark

Well, I think that a lot of people are iffy at controlling it, and if your judgment is impaired sufficiently, it'd make sense that the "I'm not sure if this is a good idea" filter is... not exercised to full potential

Akiva Weinberger

20:31

@DanielCastle Kind of like the markings on a ruler, yeah?

Daniel Castle

@AkivaWeinberger yup

Eric Silva

@Daminark this is sketchy, mostly what im talking about is attitudes and not actions

Semiclassical

(Up/down are other possible properties of quarks which go under the heading of 'flavor' for reasons I've never understood)

Eric Silva

alcohol might exaggerate the extent to which you express and exaggerate your attitudes, but it doesnt really change what they are

Mike Miller

@Daminark That's not an accurate description of what alcohol does to people

Semiclassical

20:33

Thing is, all three colors have to show up in a proton's quarks.

So a proton is a color-symmetric object.

Daminark

I mean, mostly I care about what people do or say rather thank think. If people have iffy attitudes but keep them contained then I'm chill. Either way, the religious reason is sufficient for me to not do so

@Mike oh huh

Akiva Weinberger

I have tasted wine, and regretted it every time

Mike Miller

It impairs your filter, for sure, but like Eric said this more broadly means you're willing to do things you wanted to but had too much of a barrier to doing before

Akiva Weinberger

("Tasted" means drops, by the way)

Mike Miller

"drunk actions sober thoughts"

Yeah I'm definitely not trying to make you drink!

Daminark

20:34

Oh that wasn't in response to you

Eric Silva

@Daminark bad attitudes affect the culture even if they're unspoken. those bad attitudes can hurt people in real ways even if they're unspoken (they frequently are spoken by sober people here anyway)

Hushus46

Hello all

Akiva Weinberger

Judaism, on the other hand, doesn't prohibit alcohol… more like the opposite, to be honest

Avantgarde

@Astyx You got into ENS?

Astyx

Nupe

Akiva Weinberger

20:35

I still doubt I'll ever drink more than a few drops at a time

Sharath Zotis

hey guys

Avantgarde

Oh, I read a starred message. Is that another institute?

Semiclassical

The more generic phrase is that all hadrons (particles made from quarks) have total color charge of zero

Avantgarde

@Hushus46 Hi :)

Eric Silva

@Akiva i used to say that but i developed a taste for it, not everybody does though

Astyx

20:36

It is @Avantgarde

Akiva Weinberger

@Semiclassical Total color charge?

Does red+green+blue=0 or something?

Avantgarde

@Astyx Congratulations! Is it within your top choices, I suppose?

Semiclassical

Yep.

Eric Silva

i mean I like hard drinks that a lot of people think tastes bad

Astyx

Yup, it's my second choice

I am really happy

Semiclassical

20:37

Basically it's like three vectors in the plane at 120 degree angles

Avantgarde

@Astyx Awesome! Good luck!!! :D

Astyx

Thanks ! :)

Akiva Weinberger

So if I somehow switched all red quarks in the universe with green quarks and vice versa, instantaneously, would anything change? @Semiclassical

Eric Silva

@Daminark so i know a few people in organizations like society for women in physics, and the kind of terrible shit that actually happens in the physics department here is like on full display. Things as they are now are actually pretty bad, there's lots of good discussions on it that they host and i recommend you go (as i would anyone).

Semiclassical

My guess is no, but I'm not a QCD guy

Avantgarde

20:39

@EricSilva What terrible things?

Alessandro Codenotti

Hi chat

Avantgarde

@AlessandroCodenotti Hey

Daminark

Huh, then this is one of the things where Islam is an outlier among the three. @Akiva

Eric Silva

lots of the broader societal problems that socialize women away from stem in the first place for example @Avantgarde, a lot of people say openly sexist things and it's not treated like a problem, etc etc

Avantgarde

@EricSilva I see.

Daminark

20:41

And merp @Eric

(Also sidenote merp @Titchmarsh because screw that book, but anyway back to the topic at hand...)

Eric Silva

there was a talk by the director of physics at argonne natl lab on the subject last year, it was pretty great

Akiva Weinberger

Bye

Semiclassical

There's a lot of -isms in academia in general tbh

Eric Silva

oh actually i think she does QCD funnily enough

phy.anl.gov/mep/staff/Hafidi_K.html

yup

Daminark

That's not good, like it'd be nice if we could just have a reset button and put people with instructions to do it right this time

Semiclassical

20:43

Lol

Daminark

Undertale intensifies

(I only realized afterwards that this was kinda Undertale-esque)

Eric Silva

@Daminark during her talk she actually said something along the lines of "I didn't believe sexism in stem was still athing, then I got involved in politics, and then I realized holy shit, this actually ruins peoples careers and not enough people are talking about it"

Semiclassical

Welcome to human society, it's kinda fucked up

Eric Silva

when you said undertale the title music took over my brain for a second

Daniel Castle

tfw the requirements for understanding a wikipedia article is five more wikipedia articles

feelsbadman

Daminark

20:46

Lol, you should check out this thing on Youtube called Glitchtale

It's like, really good

Semiclassical

What bothers me personally more than sexism (which I'm not in a position to experience myself) is the extent to which academia self-selects for certain personality types

Daminark

Is that so? Which ones?

Eric Silva

i dont follow the undertale fandom @Daminark

SteamyRoot

@AkivaWeinberger If you're going to switch two quark colours, you'd better switch their anticolours too.

Eric Silva

i dont follow any fandoms for games i like

Mike Miller

20:48

@EricSilva good

Semiclassical

Not sure how I'd describe it tbh

Eric Silva

i get what you mean @Semi

Mike Miller

"extreme neuroticism and paranoia"

SteamyRoot

Also, quarks actually swap colour with eachother through gluon emission/absorption.

Eric Silva

society kind of also engenders hatred of math in a lot of young people of color and that makes me really mad

Daniel Castle

20:51

Wait what part of the world is this occuring @EricSilva

Eric Silva

The US

gian

How do you render MathJax in chat?

Daniel Castle

oh, im in the uk

Eric Silva

idk anything abt the uk

so i cant speak for it

Daniel Castle

It's a bit different, I wasn't aware of that issue actually

SteamyRoot

20:52

@gian Check the links in the chat description, top-right.

Eric Silva

i saw it a lot growing up, it affects young black people and latino disproportionately @Daniel

Semiclassical

@EricSilva Hatred seems the wrong word. Disdain, maybe

Eric Silva

that's closer to what i meant

Daniel Castle

Really a shame @EricSilva

Luckily in the UK I haven't experienced anything like that

Eric Silva

yeah i know a lot of people who would probably be pursuing a stem type career if they were luckier

Semiclassical

20:55

"I don't understand math, but who needs to?"

Daminark

@MikeMiller meh, that's a personally shift I'd be a bit reluctant to induce

Daniel Castle

Is it true numbers interested in STEM in the US are declining?

Mike Miller

oh no, I'm just saying I started that way

Semiclassical

i doubt it's declining as a total, if only because the population keeps rising

Daniel Castle

proportionally perhaps

Semiclassical

20:57

That I could believe

Eric Silva

I don't have the numbers, it's probably not keeping pace with other parts of the world

Daminark

Ah, I see

Semiclassical

It's tricky, though, because there's another knee jerk reaction: "you're good at math/science? Ok, then you should go get a PhD"

Daniel Castle

Well we've had a bunch of people come in to schools to promote STEM degrees

yeah, I guess you don't want to push people down something they dislike also

SteamyRoot

The Belgian's government wants to promote STEM. So they made a website where they list all STEM-related professions.

Semiclassical

20:59

And that's a reflex I'm not so fond of, if only because the amount of tenure track positions is so much smaller than the supply of PhDs

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$Mathematics$ Mathematics