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6:38 PM
hello
I need some help with mathjax
I basically have a complicated equation for a question but I can't type it in mathjax
is there any way one of you nice people would be willing to take my equation and put it in mathjax?
@EricSilva could you help?
halp
 
ok i just went through this yesterday
 
top right of the screen click the link where it says latex in chat
 
what happened
 
6:47 PM
ohh lol different link
i can view mathjax but I can't write it well
 
in terms of you knowing the language or your computer not rendering it?
 
knowing the language
 
oh ok
 
2nd time i needed it
 
that I can't help you with other than to give you a link
 
6:48 PM
@Christopher2EZ4RTZ: You should recognize the first integral as $-b/a$ times the area of a semicircle of radius $a$. The second term integrates to $2ab$, of course.
 
@TedShifrin I just need the latex to use in my question :P
 
ok gotcha
 
But I answered the question :) You want \int_{-a}^a \big(-\frac{b}{a}\sqrt{a^2-x^2}+b\big)dx ... That's the hard part.
 
hold up
 
@DavidReed i got it (martin gave it)
 
6:56 PM
Howdy, Demonark.
 
How's it going?
 
Going fine, and you?
Done with finals yet?
 
hi @TedShifrin, @Daminark.
 
hi @Antonios.
 
or should I edit it?
 
7:00 PM
Just did both my finals this morning
 
You should have some parentheses inside the integral for clarity, as I put them. Do you want an answer to your question? I already gave you most of the answer.
 
Hey Antonios!
 
Wait, two finals already? I guess they're not 3-hour finals.
 
Hi everyone! What abot link text?
 
7:01 PM
2 hour finals
 
Aha.
I was used to 3-hour, both as a student and as a prof.
 
@TedShifrin I need to post it to main (added some edits btw) so I can show my old math teacher the proof
 
Hmm, OK.
 
I see. I think I prefer 2 hours since a day like today would've been draining if I had to test for 6 hours almost straight
 
@TedShifrin does the newest version look fine?
 
7:03 PM
I prefer 3 hours to give people time to think and for the exam to be somewhat comprehensive.
 
Yeah that's fair
 
You don't need the big parentheses, unless you remove the fraction and just write 1/2 times the whole thing.
 
One of my professors gave us about 4 and a half hours for the test.
 
anything else?
 
7:05 PM
Last year she gave them unlimited time and a harder test but someone took 8-9 hours, which she felt was cruel to the TA
 
Well, Christopher, it's good enough.
 
@TedShifrin what should i title it?
 
I don't know. It's basically "What is an explanation for this equality?" You do not want to do the integral by some horrendous means. As I said, you want to observe how it's related to the area of a semicircle. And then it's immediate.
 
One of my friends has extra time for his exams (well – did in undergrad) and they were 6 hours long.
 
7:09 PM
0
Q: Proof request for finding the area of a rectangle using an ellipse and an integral.

Christopher 2EZ 4RTZ$$\frac{\int_{-a}^a (-\frac ba \sqrt{a^2-x^2} + b\;)\mathrm{d}x + \frac{\pi ab}2}2=ab$$ The above equation finds the area of a rectangle given A and B (both > 0). I wrote this myself as a challenge. Using desmos I verified that it was true (the desmos). I was wondering if someone could write a pr...

 
hey @Ted!
 
hi @Lucas.
How's life with linear algebra? :)
 
@Antonios-AlexandrosRobotis oh wow that's a lot
But yeah these two were tests that I kind of enjoyed
How's it going with you Ted and Antonios?
 
7:44 PM
not bad. Finally beginning to understand what this algebraic NT stuff is about.
 
7:54 PM
ramification = branch points of covering maps
algebraic NT = geometry done over bad fields
algebra is just geometry made complicated
 
@BalarkaSen algebraic geometry = geometry over the best fields?
 
(I am joking; don't let the pretentiousness get you)
 
Ugh geometry not again
 
@Alessandro complex geometry = geometry over the best fields
the best fields are obviously 1) C 2) C 3) C
 
Of course
Well you need an algebraically closed field and C is the only one, right? :P
 
7:57 PM
Gotta write a geometry diss track
 
@Alessandro Correct
 
"It's char p bro with the finish field flow"
 
Dabs at @Daminark's general direction
 
Hmm does this question make sense to you guys? Apparently the answer is b.
https://i.gyazo.com/60c38343bddc64b20675c8a01021196e.png
 
Hello, someone know the Lagrange differetial equation ?
 
8:11 PM
@Dragneel nope
it doesn't make sense
 
I think n%16 = 1 means n = 1 mod 16
 
@TedShifrin I took a break to study for MathCamp's quiz.
 
I have never seen that notation used anywhere except PARI/GP
 
hello @AkivaWeinberger
 
8:14 PM
@Dragneel Well, look at the prime factors. How many in the list 0,1,\dots,1000 are divisible by 2?
 
so you just need to find the proportion of order dividing 4 elements in $\Bbb Z/16$
 
sleepily Hi
 
I think that should be plenty accurate.
 
I guess I finally understand the deformation retraction of Bing's house
 
push holes into a pancake
 
8:16 PM
It seems all of the odd elements have order dividing 4
 
No, not the deformation retraction from the *ball into the house
The deformation retraction from the house to a point
The function from ${\rm BH}\times I$ to ${\rm BH}$
The main idea is this image
 
oic
@AkivaWeinberger you mean a ball btw
 
I don't think spheres def retract to points.
 
They don't
Balls do
 
8:21 PM
@PVAL-inactive Ah that's interesting. And that's why b is the correct answer, since 50% of the numbers are odd.
 
@PVAL-inactive That's equivalent to the fact that there's no retraction from $D^3$ to $\partial D^3=S^2$, isn't it
You can't deformation retract a ball onto its boundary
 
@Akiva to me that's a different statement.
 
who know how we can solve this ODE $$y(t)=t f(y'(t))+g(y'(t))$$
 
I don't see why they are the same either
 
They can both be proven using the same machinery.
 
8:23 PM
@PVAL-inactive The connection is that $D^3$ is the cone on $S^2$
You know how you can take $S^2\subseteq D^3$ and shrink it uniformly to the center?
Suppose there were a retraction $f$ onto its boundary, and put that shrinking through $f$
You'd get a way to shrink $S^2$ to a point inside $S^2$
i.e. a deformation retract of a sphere to a point
 
sure I guess that works.
 
Oh, the homotopy S^2 x I --> S^2 to the constant map gives you a map D^3 --> S^2
which is id on the boundary
I see
 
@BalarkaSen 'Cause you can factor it through S^2 x I --> D^3 --> S^2
sinxe S^2 x {1} should map to a point
 
Sure, universal property of quotient maps
 
@Dragneel I think that $\Bbb Z/n$ has non-cyclic units group unless n is prime which should imply every element has order dividing 4 immediately without checking element by element.
 
8:26 PM
If that's what it's called, so be it @BalarkaSen
Ah, universal property
I see
 
I fucked up the name
 
@PVAL-inactive Here's how I thought about it. For any integer $n$, $n*16$ seems to result in an integer than ends with either $0,2,4,6,8$, that is, even digits. And since we're looking for $n\equiv 1 (mod 16)$, only odd $n$ integers will satisfy us because of a remainder of $1$.
 
@PVAL-inactive No, $\mathbb{Z}/n\mathbb{Z}$ can have cyclic unit group without $p$ being a prime. It holds precisely for $n = p^r$ or $n = 2p^r$ for odd prime $p$ (well, plus $n=4$).
 
I knew it was something like that.
@Dragneel All odd integers though don't have remainder 1 when divided by 16, but it turns out the 4th powers do.
 
True, good observation.
Anyway, I have to go write an exam, see you later, and thanks for the help guys :)
 
9:15 PM
A cool generalization to all the "exponentio-polynomial" divisibility problems: if $m$ divides $a + d$, $(b - 1)c$ and $ab - a + c$ then $m$ divides $ab^n + cn + d$ for all natural numbers $n$.
There's a short induction proof. Anyone knows a more natural way to think about this?
Induction is basically bashing all the algebra
 
3Blue1Brown has a new video out on the following problem: Choose four points on the sphere uniformly at random. What are the odds that the tetrahedron determined by those points contains the center?
Also: Bye, see you all in ~25 hours
 
Hi
I have a question. The points of $RP^2$ are the lines of $R^3$ which go through $0$. What are the lines of $RP^2$ then? According to my resources it's the planes containing $0$ of $R^3$ but how can I see that? How is the term "line" even defined for an arbitrary space?
 
@brot Cool question. So for projective spaces, lines inside $\Bbb{RP}^n$ basically means 1-dimensional projective subspaces. So, the "linearly embedded $\Bbb{RP}^1$'s".
Since $\Bbb{RP}^1$ is the space of 1-dimensional vector subspaces of $\Bbb R^2$, the lines in $\Bbb{RP}^2$ are precisely the subspace of 1-dimensional vector subspaces of a 2-dimensional subspace of $\Bbb R^3$
That's a bit of a world-clutter there, but I hope that makes sense
 
The term "line" cannot really be defined for an arbitrary space.
 
The point is if you take a plane $S$ in $\Bbb R^3$ going through the origin, and think of lines lying on $S$ that passes through the origin, those give rise to points on $\Bbb{RP}^3$ which lie on a line.
@PVAL Well we can tell him a bit about geodesics.
 
9:26 PM
If you have the structure of a Riemannian metric (i,e, lengths of paths) you can talk about geodesics.
In this case there is a natural Riemannian structure on S^2
and the covering map to $\Bbb RP^2$ gives a "spherical" Riemannian structure on $\Bbb RP^2$.
Geodesics are locally length minimizing paths.
 
The point being that $\Bbb{RP}^n$ has a natural Riemannian metric on it, and under that the geodesics are exactly the $\Bbb{RP}^1$'s coming from the way I described, or rather, the "projective lines", as PVAL says
 
The geodesics of S^2 are just great circles and the geodesics in RP^2 are just the images of great circles under this projection.
 
@BalarkaSen I meant $\Bbb{RP}^2$ here, by the way. Unfortunate typo.
 
I think I get the idea
But what is a projective subspace? Does $\mathbb R P^n$ have vector space structure?
 
No, it has a projective space structure :)
It's really nothing complicated. A projective subspace of $\Bbb{RP}^n$ is a subspace that comes from a vector subspace in $\Bbb R^{n+1} \setminus \{0\}$ under the quotient map
You can take that to be the definition
 
9:43 PM
Ok, I see :) and then I can directly observe that in $\mathbb R P^2$ projective lines always intersect at one projective point? that's cool
 
Quite.
There's another picture I like to think about. Do you know homogeneous coordinates?
 
You mean like $(x_1:x_2:x_3)$?
 
yeah
So that's how you denote a point in $\Bbb{RP}^2$, right?
 
yup
 
So now consider the set of all points $(x_1 : x_2 : x_3)$ in $\Bbb{RP}^2$ such that $x_3 \neq 0$.
You can scale the coordinates to get $(x_1/x_3 : x_2/x_3 : 1)$
So such points all look like $(x : y : 1)$ in homogeneous coordinates
But there's a bijection between the set of such points and $\Bbb R^2$, right? $(x : y : 1) \mapsto (x, y)$.
On the other hand, consider the complement of the set, which are set of all points $(x_1 : x_2 : x_3)$ such that $x_3 = 0$. Or, well, set of points of the form $(x : y : 0)$. We can ignore the last zero coordinate and just say that this set is bijective to $\Bbb{RP}^1$, bijection being $(x : y : 0) \mapsto (x : y)$
So we have decomposed our projective space as follows: $\Bbb{RP}^2 = \Bbb R^2 \cup \Bbb{RP}^1$
Is this okay?
 
9:55 PM
hi there, can someone help me with getting the gcd of two polynomials over the field of rationals?

what is the gcd of $x^3 + 2x^2 + 2x + 1$ and $x^2 + x + 2$
 
@vanaghka euclidean algorithm
 
Yes I can follow
 
over $\mathbb{Q}[x]$
 
@vanaghka doesn't change
 
am i right with 2?
 
9:57 PM
Nice. So the way you can think about it is, $\Bbb R^2$ is not compact. But you add a "line at infinity" to it to get a compact object $\Bbb {RP}^2 = \Bbb R^2 \cup \Bbb{RP}^1$. So it's a compactification in a sense, like one point compactification, but with adding something more than a point at infinity @brot
If you think like this, then the lines $\ell$ of $\Bbb R^2$ (not necessarily passing through the origin; just a random straight line) become precisely the lines $\ell'$ of $\Bbb{RP}^2$ after the compactification
So it's just your usual concept of a line, souped up somewhat
 
@vanaghka $\gcd(x^3+2x^2+2x+1,x^2+x+2) = \gcd(x^2+1,x^2+x+2) = \gcd(x^2+1,x+1) = \gcd (2,x+1) = ???$
is $\Bbb Q[x]$ a euclidean domain?
 
How do you know that $\mathbb R P^2$ is compact?
 
Good question.
Do you know how $\Bbb{RP}^2$ can be defined as a quotient of $S^2$?
 
@LeakyNun yes. can i show you what ive worked out so far?
 
Yes, relating opposing points
 
10:03 PM
@vanaghka sure
 
Ok, and then it's compact. I see :)
 
right
 
@brot Precisely
 
it feels kind of weird
because 2 doesn't exactly divide $x+1$
 
10:11 PM
I don't quite understand the lines part though. So if I have a line in $R^2$ and then compactify it, what happens with the line on $\mathbb RP^1$?
 
The $\Bbb{RP}^1_\infty$ (line at infinity) is the extra bit you add to compactify $\Bbb R^2$. If $\ell$ was the line in $\Bbb R^2$, in the compactification $\Bbb R^2 \cup \Bbb{RP}^1_\infty$ $\ell$ intersects $\Bbb{RP}^1_\infty$ at one point.
So the line $\ell$ itself gets "compactified" to an $\Bbb{RP}^1$
It's a confusing picture but gets clearer if you think about it
 
the line at infinity has a point at infinity
double infinity
but not
 
it's turtles all the way down
worse is if you have RPinfty lmao
 
i guess $\mathbf{R}P^{\infty}$ is turtles all the way up
 
these are the kind of bullshit cell decompositions you get when your objects are algebraic though so
 
10:17 PM
i think the cell decomp of the projective spaces is p good
it's like a 7/10 for general constructions, but like a 9/10 cell complex
 
eh, it's kinda bad if you want to think about the projective space as an algebraic geometric object
like projective varieties get these kind of structures
 
i wouldnt know about that
 
where you have a massive zariski open set as your cell
top cell
and you throw it out
you get a variety again
and take a massive zariski open in it
rinse lather repeat
 
I should learn some algebraic geo
 
i agree
 
10:20 PM
maybe this winter break
 
i guess you'd immediately go to the complex geometric story
given your background
 
yeah
I have a copy of griffiths book on complex algebraic curves
 
coolio
 
so i was gonna bring that when i go home for break and give it a read
 
i never progressed much on the Forster thing i had
 
10:22 PM
@LeakyNun so it doesn't. 2 is the remainder yes. whats the next step after what i've done (if im right??)
 
rip
I tried to read this book at some point but didnt get very far cause I was doing way too much analysis
 
i want to restart it again but nobody wants to read it with me
 
and now im kind of burnt out on analysis
 
If I think about $\mathbb R P^1$ as the line at infinity, it's intuitive that $\ell$ intersects it at one point. But why can I do so?
 
well, it's the thing you add to $\Bbb R^2$ to make it compact
like you add a point to $\Bbb R^2$ to make $S^2$
and call that point the "point at infinity"
in this case you add a full projective line, so you call it the line at infinity
 
10:26 PM
And then parallel lines in $\mathbb R^2$ would intersect $\mathbb R P^1$ at the same point?
 
Yep
That's why we say the parallel lines "intersect each other at infinity"
the point of intersection lies on the line at infinity
 
How could I show that?
 
Phew, the shortest substitution I could write up from $(ab^{-1})(cd^{-1})$ to $(ac)(bd)^{-1}$ with field axioms uses a lot of associative transformations:
$$\begin{split}
&(ab^{-1})(cd^{-1})\\
=&((ab^{-1})c)d^{-1}\\
=&(c(ab^{-1}))d^{-1}\\
=&((ca)b^{-1})d^{-1}\\
=&((ac)b^{-1})d^{-1}\\
=&(ac)(b^{-1}d^{-1})\\
=&((ac)1)(b^{-1}d^{-1})\\
=&((ac)((bd)(bd)^{-1}))(b^{-1}d^{-1})\\
=&((ac)((db)(bd)^{-1}))(b^{-1}d^{-1})\\
=&((ac)((bd)^{-1}(db)))(b^{-1}d^{-1})\\
=&(((ac)(bd)^{-1})(db))(b^{-1}d^{-1})\\
=&((ac)(bd)^{-1})((db)(b^{-1}d^{-1}))\\
=&((ac)(bd)^{-1})(d(b(b^{-1}d^{-1})))\\
=&((ac)(bd)^{-1})(d((bb^{-1})d^{-1}))\\
=&((ac)(bd)^{-1})(d(1d^{-1}))\\
=&((ac)(bd)^{-1})(d(d^{-1}1))\\
 
@vanaghka oh wait, we're in $\Bbb Q[X]$
$\gcd(2,x+1) = \gcd(2,x+1-2(\frac12x)) = \gcd(2,1) = 1$
 
11:00 PM
@LeakyNun okay how was that step allowed
 
Question: given $y = e^{x + y} - e^{-x-y}$, can you get all the y's to one side?
 
11:21 PM
@user8663905 I don't believe so, no
Actually, WA gives the real solution $x=\ln\left(\frac12\left(\sqrt{y^2+4}+y\right)\right)-y$
 
what'd you enter into Wolfram to get that? I'm always struggling to get WA to do what I want it to do.
 
"solve y=e^(x+y)-e^(-x-y) in reals"
 
Alright, I wonder what the algebra is to get there. Danke for the info
 
Also, a question for anyone up to it: Given any equation between two expressions consisting of constants, the six basic trigonometric functions of $x$ ($\sin x,\cos x,\tan x,\cot x,\sec x,\csc x$), and elementary operations (addition, subtraction, multiplication, division), what would be a minimal set of trigonometric identities (e.g., $\sin^2x+\cos^2x=1$) required to verify via substitution a general equation in this form?
$x$ is assumed here to be a real number such that all expressions are defined
 
11:46 PM
2
Q: Growth rate of the nth natural number not constructable with n steps of addition and multiplication

Simply Beautiful ArtWhile messing around with the idea of ordinal collapsing functions, I stumbled upon an interesting simple function: $$C(0)=\{0,1\}\\C(n+1)=C(n)\cup\{\gamma+\delta:\gamma,\delta\in C(n)\}\\\psi(n)=\min\{k\notin C(n),k>0\}$$ The explanation is simple. We start with $\{0,1\}$ and repeatedly add it...

^ Seems to be a fun question. I myself am having fun over it lol
 
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