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7:00 PM
@Typhon So you are claiming that "greater in real magnitude" is not a total order?
 
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. []
 
Do any one have general form of integral of f(f(x))dx ?
 
@Typhon You negated the condition incorrectly
 
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"
 
still not correct
 
7:02 PM
...
 
Seriously, you really need to look at a few small examples
 
yeah, it is
i just proved it
why are coming down on me anyway? I was just talking to akiva.
 
One of your conditions imply the inequality, the other is implied by it
neither is equivalent to it
 
what are you talking about???
there's only one condition
a + bx < c + dx
 
@Typhon a<c or b<d
 
7:05 PM
@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"
 
Should there be and ? I think so
 
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. []
 
Whatever. Hope tobias helps you
 
@Fawad he's not helping me...
i was merely sharing a proof with Akiva and he started attacking it. :/
 
Alright
You know calculus ?
 
7:08 PM
how is that relevant?
and yes
 
7 mins ago, by Fawad
Do any one have general form of integral of f(f(x))dx ?
 
What's wrong is "Suppose not"
 
@Astyx how so?
proof by contradiction
 
If you suppose not, you get that "$a\ge c$ or $b\ge d$"
 
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"
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...
 
7:10 PM
I think multiplying by x he means multiplying by |x| because inequality will change for negative value multiplication
 
@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
 
@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.
 
@Fawad Here's a counter example. x^x.
 
@Semiclassical oops. For indefinite inegration?
 
then it becomes (x^x)^(x^x)
 
7:15 PM
same idea, really.
 
@Fawad that's what he was saying.
 
except even harder to control since the domain isn't specified.
 
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
 
Ooh. By substituting tylor series we get polynomial integrals
 
An example where this works fine is $f(x)=e^x$.
 
7:18 PM
except the integral of e^x is trivial
like.... as trivial as evaluating e^x.
 
Sure, but $e^{e^x}$ isn't. He asked how to integrate $f(f(x))$.
 
@Semiclassical oh.... duh
 
Hmm. The leading terms of the Taylor series about $x=0$ for that is $e^{e^x}=e+ex+ex^2+\frac{5e}{6}x^3+\frac{5e}{8}x^4+\cdots$
 
@Fawad For fractional polynomials, yes. It's called expansion, long division and partial fraction decomposition.
in other words, simplify the function first
 
Yeah
 
7:21 PM
other than that...
 
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
 
$$\int f(f(x))dx = ???$$

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:

$$\int e^{e^x}dx$$
 
welp... good luck
 
It's a sign of the stuff I get interested in that I wonder what sequence those derivatives correspond to.
 
@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...
 
7:23 PM
at which point I thank goodness for OEIS
 
lol
 
"Both roots." Do you intend that to be a monic quadratic?
 
actually, is $$\int e^{i\sin x}dx$$ simplifiable to some known special functions?
 
yeah, sorry. Must be tired today.
i keep doing that
ugh
 
@Secret Definitely:
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 = ...
 
wolfram is relatively suck at indefinite integrals
 
Yeah.
 
@Typhon Just pick roots like $-2+\sqrt{2}$ and $-2-\sqrt{2}$ ?
 
right, in that case, we only need to justify interchange integral with that bessel sum to get it simplier
 
@SteamyRoot true but is that the solution to a quadratic?
hrmm
x + 2 = sqrt(2)
 
7:26 PM
$x^2 - Sx + P$
 
(whcih I am poor at beause I know almost no real analysis)
 
ah
 
$(x+2-\sqrt{2})(x+2+\sqrt{2}=(x+2)^2-2$, yup.
 
welp
that puts bullet holes through my whole proof
 
@Secret I always like the Jacobi-Anger identity, because it's a Fourier series with Bessel functions as coefficients :)
 
7:27 PM
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
 
Hah, of course this is familiar: oeis.org/A000110
 
then x and x' divide each other
 
Huzzah Bell numbers
 
therefore $x/x' = \pm 1$
but if it were 1
then that would be a double root
which is an integer
 
(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)
 
7:31 PM
@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
 
if it's monic and integral, no.
 
thought so
 
the constant term would be the product of the roots, and since the root is double it'd be irrational.
 
oooh
of course
|a| = |x + x'|
 
$$\int \sin (\sin x)dx = \int \frac{e^{i\sin x}-e^{-i\sin x}}{2} = \frac{1}{2}\int e^{i\sin x} - e^{-i\sin x}dx$$

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...
 
7:33 PM
@Secret This is why I much prefer definite over indefinite integration, heh.
 
@Semiclassical thanks
 
@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.
 
I like indefinite because if you can figure it out, you get all integrals of that type, and no more magic and confusion
 
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.
 
7:35 PM
@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
 
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?
 
it is actually fun to try and cram proofs into the little tiny books
or worse, use signs. XD
bye guys
 
Well, what's the actual definition of an indefinite integral? It's the general solution to the ODE $y'(x)=f(x)$.
 
yup, and that's a family of them
 
7:37 PM
Right. $y(x)=F(x)+C$.
 
An indefinite complex integral is probably the general solution of the cauchy riemannian PDE and whatever the PDE the integrand is linked to
 
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.
 
I see
 
@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
 
@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.
 
7:52 PM
@Semi the thing I was trying to do is make the Stokes' proof of the integral theorem work for all holomorphic functions
 
@Daminark the triangles thing @Semi suggested is how it's done without $C^{1}$ i think
Cauchy-Goursat is the name or something
 
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
 
@Eric is it at least nicer than all that stuff in Titchmarsh?
 
it's pretty nice
 
I just didn't like that proof
 
7:55 PM
I forget how the proof works, tbh. Some combination of the estimation lemma and splitting the triangle up into four pieces.
 
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
 
How so?
 
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
 
I mostly like it because it reminds me of finite elements stuff.
 
@Semi what is that, I've heard of it before
 
7:58 PM
basically, it's a way to solve pde's
 
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
 
right right
 
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...
 
@Daminark, if it solves C-R yeah
 
Oh okay so weak derivatives don't actually give you C-R automatically
 
7:59 PM
of course not
 
Wait what exactly is a weak derivative? Is it just derivative in the weak topology?
 
Basically, you split the region of your problem up into a lot of little regions
 
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
 
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.
 
Lol, still gotta figure that out
 
8:01 PM
@Semi cooool
 
@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
 
Ultimately it comes down to converting a PDE problem to a linear-algebra problem.
 
as with all things differential ultimately
i usually describe linear algebra as "the only thing we really understand"
 
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.
 
8:04 PM
I wonder if we will ever evolve to be able to deal with nonlinear operators directly...
 
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."
 
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
 
Dayum
 
A pity not all operators are elliptic :)
 
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
 
8:12 PM
Alas, not all of mine are.
The wave equation being the most obvious one.
 
o true the wave equation is a goodie
hyperbolic equations scare me still
 
Schlag intensifies
 
I think failures of ellipticity are a big problem in QFT? But don't quote me on that.
 
QFT seems like a big spook
 
QFT is scary.
 
8:15 PM
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
 
I mean, many physics majors talk about it because they're actually into it
 
@Daminark ive had at least 5 in my house who did what i described
 
The thing with QFT is that, however esoteric it may seem, it works damn well in its domain of applicability
 
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
 
@Daminark im mostly talking about first years who use big words to sound smart, of which there are a lot
 
8:20 PM
The experimental success of QED is one of the singular achievements of modern physics
 
not the kind of people who could take grad classes and do well
 
Lol, most of the first years I know into science and all are actually rather timid
 
Where things get hairier is QCD, because the problems get a lot harder
 
@Semi what is QCD?
 
quantum chromo dynamics
 
8:22 PM
Right.
 
@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
 
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
 
It's the QFT of the strong nuclear force, basically
 
and she's involved in a lot of the undergrad clubs and stuff
@Daminark you might just know good people
 
QED is quantum electrodynamics, so it accounts for the electromagnetic force
 
8:23 PM
Chromodynamics?
 
Yeah actually what's the deal with the whole, strong/color stuff?
 
yes
 
What caused the other name to just step in the picture?
 
Eh. That's used because quarks can have three possible kinds of a property
 
the ? meant id like an explanation of what it meant
 
8:24 PM
@Typhon Re: $ax+b>cx+d$ iff $a>c$ or $b>d$; the problem is "iff"
 
And they decided to call those three possible properties as red, blue, green
 
You could have $ax+b<cx+d$ and still have $a>c$
 
Hence, color
 
ah ok
 
@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
 
8:26 PM
That might seem silly, but quarks aren't an object of everyday experience
So you might as well call that property 'color'
 
@Daminark i also have more experience with more people than you
i go to parties and whatnot
 
Anyone know if an expression can be found for the highest power of 2 some even n is divisible by?
 
@Semiclassical Are the three colors symmetric?
Like, are the laws of physics invariant under arbitrary permutations of them
 
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
 
@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
 
8:29 PM
alcohol doesn't usually make you a worse person than you already are
 
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
 
@AkivaWeinberger Oh ok, thanks - I've generated some graphs of it and there is an obvious pattern
 
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
 
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
 
8:31 PM
@DanielCastle Kind of like the markings on a ruler, yeah?
 
@AkivaWeinberger yup
 
@Daminark this is sketchy, mostly what im talking about is attitudes and not actions
 
(Up/down are other possible properties of quarks which go under the heading of 'flavor' for reasons I've never understood)
 
alcohol might exaggerate the extent to which you express and exaggerate your attitudes, but it doesnt really change what they are
 
@Daminark That's not an accurate description of what alcohol does to people
 
8:33 PM
Thing is, all three colors have to show up in a proton's quarks.
So a proton is a color-symmetric object.
 
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
 
I have tasted wine, and regretted it every time
 
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
 
("Tasted" means drops, by the way)
 
"drunk actions sober thoughts"
Yeah I'm definitely not trying to make you drink!
 
8:34 PM
Oh that wasn't in response to you
 
@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)
 
Hello all
 
Judaism, on the other hand, doesn't prohibit alcohol… more like the opposite, to be honest
 
@Astyx You got into ENS?
 
Nupe
 
8:35 PM
I still doubt I'll ever drink more than a few drops at a time
 
hey guys
 
Oh, I read a starred message. Is that another institute?
 
The more generic phrase is that all hadrons (particles made from quarks) have total color charge of zero
 
@Hushus46 Hi :)
 
@Akiva i used to say that but i developed a taste for it, not everybody does though
 
8:36 PM
It is @Avantgarde
 
@Semiclassical Total color charge?
Does red+green+blue=0 or something?
 
@Astyx Congratulations! Is it within your top choices, I suppose?
 
i mean I like hard drinks that a lot of people think tastes bad
 
Yup, it's my second choice
I am really happy
 
8:37 PM
Basically it's like three vectors in the plane at 120 degree angles
 
@Astyx Awesome! Good luck!!! :D
 
Thanks ! :)
 
So if I somehow switched all red quarks in the universe with green quarks and vice versa, instantaneously, would anything change? @Semiclassical
 
@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).
 
My guess is no, but I'm not a QCD guy
 
8:39 PM
@EricSilva What terrible things?
 
@AlessandroCodenotti Hey
 
Huh, then this is one of the things where Islam is an outlier among the three. @Akiva
 
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
 
@EricSilva I see.
 
8:41 PM
And merp @Eric
(Also sidenote merp @Titchmarsh because screw that book, but anyway back to the topic at hand...)
 
there was a talk by the director of physics at argonne natl lab on the subject last year, it was pretty great
 
There's a lot of -isms in academia in general tbh
 
oh actually i think she does QCD funnily enough
yup
 
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
 
8:43 PM
Lol
 
Undertale intensifies
(I only realized afterwards that this was kinda Undertale-esque)
 
@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"
 
Welcome to human society, it's kinda fucked up
 
when you said undertale the title music took over my brain for a second
 
tfw the requirements for understanding a wikipedia article is five more wikipedia articles
feelsbadman
 
8:46 PM
Lol, you should check out this thing on Youtube called Glitchtale
It's like, really good
 
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
 
Is that so? Which ones?
 
i dont follow the undertale fandom @Daminark
 
@AkivaWeinberger If you're going to switch two quark colours, you'd better switch their anticolours too.
 
i dont follow any fandoms for games i like
 
8:48 PM
@EricSilva good
 
Not sure how I'd describe it tbh
 
i get what you mean @Semi
 
"extreme neuroticism and paranoia"
 
Also, quarks actually swap colour with eachother through gluon emission/absorption.
 
society kind of also engenders hatred of math in a lot of young people of color and that makes me really mad
 
8:51 PM
Wait what part of the world is this occuring @EricSilva
 
The US
 
How do you render MathJax in chat?
 
oh, im in the uk
 
idk anything abt the uk
so i cant speak for it
 
It's a bit different, I wasn't aware of that issue actually
 
8:52 PM
@gian Check the links in the chat description, top-right.
 
i saw it a lot growing up, it affects young black people and latino disproportionately @Daniel
 
@EricSilva Hatred seems the wrong word. Disdain, maybe
 
that's closer to what i meant
 
Really a shame @EricSilva
Luckily in the UK I haven't experienced anything like that
 
yeah i know a lot of people who would probably be pursuing a stem type career if they were luckier
 
8:55 PM
"I don't understand math, but who needs to?"
 
@MikeMiller meh, that's a personally shift I'd be a bit reluctant to induce
 
Is it true numbers interested in STEM in the US are declining?
 
oh no, I'm just saying I started that way
 
i doubt it's declining as a total, if only because the population keeps rising
 
proportionally perhaps
 
8:57 PM
That I could believe
 
I don't have the numbers, it's probably not keeping pace with other parts of the world
 
Ah, I see
 
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"
 
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
 
The Belgian's government wants to promote STEM. So they made a website where they list all STEM-related professions.
 
8:59 PM
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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