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10:00 PM
Or simply the unit circle $U = \{z\in\Bbb C, |z|=1\}$
 
@Ted are you still here?
 
@Socrates: Just use the domain $[0,2\pi]$ or $[0,360]$, and say $f$ is the same at the endpoints.
No, @Alessandro.
 
ah, I'll ask you another time then :P
 
What will you ask me?
 
what is the relation between the orthocenter and the centroid of a triangle?
 
10:03 PM
oh, @heather is here ... did you work out that algebra yet?
 
Hi @heather
 
You have an exercise asking whether $\{(x,y)\in\Bbb R^2|y^2=x^3\}$ is a submanifold of $\Bbb R^2$
 
@TedShifrin, getting closer, my addition skills are no longer terrible (i hope) so that's good.
@Astyx, hello
 
I should perhaps go to bed
 
They're different except for the equilateral triangle, I think, @heather. It turns out that there's a cool result about orthocenter, centroid, and incenter. They're all collinear (and the one in the middle divides the segment in a 2:1 ratio)!!!
Yes, @Alessandro, in chapter 6?
 
10:04 PM
the 2:1 thing would be centroid, I believe.
 
No, no, there's another 2:1 :)
 
seriously?
ack.
 
I only figured out how to prove this using vector methods, though. I haven't done it "classically."
Yes, seriously :)
 
but is there a way, given the centroid, to find the orthocenter?
 
it's among your exercises for chapter 1 of Guillemin and Pollack
 
10:05 PM
Good night all
Or good day depending on where you are
 
good night
 
good night @Astyx
 
@heather: No. But the theorem is that the centroid is 2/3 the way from the orthocenter to the incenter.
Bonne nuit, @Astyx.
 
so given the centroid and the incenter you could find the orthocenter?
 
Oh, cool. @Alessandro. Just wanted to know the context. OK. It's easy if you've done a particular G&P exercise (which is covered in the my text in a theorem).
Yes, @heather ... given any 2 you can find the third.
 
10:07 PM
hmm, let me see if my textbook shows how to find incenter, and see how difficult it is.
 
@heather: The incenter should be equidistant to all three sides. Can you figure out how to get that?
@Alessandro: Have you done exercises on pp. 18-19 yet?
 
equidistant to the sides...
 
One of those will be highly relevant, @Alessandro :P
 
maybe...um, i don't know, find the equations of the lines for the 3 sides and set them equal...?
and solve for x and y...?
 
10:11 PM
Nah ...
 
that seems very wrong.
 
@Astyx IVT is the answer to both of the excercises, as I found out
 
yeah, no.
 
Correct (for the one I know about) @Socrates.
@heather: What is the set of points equidistant from two intersecting lines?
Draw pictures. (And how do you measure distance from a point to a line, by the way?)
 
I've done most of the exercises on page 11-13 so I'll start doing those now. I was thinking about arguing that this set is not a submanifold of $\Bbb R^2$ using the implicit function theorem but let's see if I find something useful in this exercises
 
10:13 PM
@Alessandro: Unfortunately, the implicit function theorem is not a biconditional.
 
@TedShifrin no, i don't
sorry, i misread: i don't know
 
For example, I can look at $x^2=0$ in $\Bbb R^2$ and that's still a submanifold.
@heather: You drop a perpendicular from the point to the line and measure. Does that make sense?
 
hmm...so solve for the intersection between the perpendicular line and the original line and then find the distance between that and the point?
 
Right. We're just talking pictures here at the moment, not actually doing algebra.
So now can you tell me the locus of points equidistant?
 
hmm, one moment
a line running straight through the intersection?
bisecting the angle created by the intersection?
 
10:17 PM
Which line? :)
Bingo. Awesome job!
Now go back to your triangle.
 
=D
okay, going back to the triangle
 
Heya @Brody
 
so...find the intersection point of each of the bisecting lines?
*all
 
Right.
 
huh.
 
10:18 PM
Hey everyone
Hi @Ted. How ya doin?
 
Doing well, thanks, and you?
 
Sick, but doing alright. Thanks
 
what do you mean with $x^2=0$?
 
so, the reason i asked was this: finding the orthocenter is a bit of a pain, so i wanted to know if finding the incenter was easier, so that you could find the incenter and centroid (which is really easy to find, by finding the average of the vectors) and then use the relation between the three to find the orthocenter.
but, i actually don't know how to find the equation of a line that bisects an angle.
(just given that information, i mean.)
 
Actually, writing down perpendicular bisectors is easier. But there are easier ways to find the orthocenter (e.g., using a little bit of linear algebra). This stuff is more conceptual and not necessarily meant for calculating. That said, you can do both incenter and orthocenter quite easily with straightedge and compass.
@Alessandro: I meant that equation to describe the $y$-axis. It is a manifold. But the implicit function theorem fails.
$f(x,y)=x^2$
Sorry you're sick, @Brody. I escaped ATL 3 days early for fear of icemageddon, thereby enraging my friends in Athens whom I never went to see. As you well know, there was nary a flake and only a little ice.
 
10:23 PM
@Astyx let $f$ be the the distance function for Alice from the point Y with {0,x} as the domain. let $g$ be the function that is the distance for Bob from Y. f(1)=0, $0\leq g(1)<x$. By IVT $f(d)=g(d)$ for some $d\in${0,x}.
 
AH, of course, that makes sense
 
meh, not complete
 
@Alessandro: So failure of the hypotheses of the Implicit Function Theorem doesn't tell you that it cannot be a manifold.
 
the endpoints are obv f(2)=x, and $0\leq g(2)<x$
 
10:29 PM
@TedShifrin Oh, that sucks... or maybe not so much for you? I don't know how it was down in ATL but in upper Cobb it snowed just a bit Friday night and the following morning. That's all, but the constant refreezing ice still spun commuters off roads this morning.
 
Well, I was remembering a few winters ago when I was stuck in my house in AThens for over 5 days. I didn't want a repeat of that. I'm never visiting ATL again other than in spring or fall. :) Get better soon!
 
Several days? Damn, didn't know it was that bad for some people. And thanks, I'm hoping to beat this crap ASAP with the semester having begun ^.^
 
how are the sides of a "circle" called that is up and down, if you would tinker a circle out of paper? and how is the side called that points towards the middlepoint of a circle and how is the side called that points in the opposite direction?
 
Is that @Null? :)
 
@Brody yes
hi
 
10:49 PM
Hi @Socrates
@Socrates I can't understand this, sorry
 
11:40 PM
Ülü
I mean, hola
 
Hey guys :) I was wondering if someone could look at my proof for a matrix calculus problem and see where I did wrong.
The problem itself is simple: gradient of $x^TAx$
I got $(A^T + A)x$, while the text says $2Ax$
Not sure if that's a trivial difference but it's enough to confuse me.
 
11:58 PM
the pdf of the work is here(at the very bottom of it): docdro.id/HrgSYDb
 

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