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00:00
@TedShifrin Take those two points, and the line through them?
@TedShifrin: Can I ask you a silly geometry question?
00:13
@FernandoMartin What is the question?
@FernandoMartin (Also, we have a rule here: don't ask to ask!)
Ok
I have to calculate the curvature and torsion of some curves.
@FernandoMartin OK?
There's a standard formula for that but everything gets really messy real fast. I don't get the point of the exercise.
Heh. What are the formulae?
for instance
00:16
0
Q: Measure Theory: How to prove this statement using one of these results?

TwinkRESULT 1: If $I_1, I_2,..., I_n$ are disjoint open intervals and $E$ is a bounded set such that $E \subset \cup_{k=1}^n I_k$ then $m^*(E)=\sum_{k=1}^n m^*(E \cap I_k)$. (where $m^*(S)$ is the Lebesgue outer measure of $S$.) RESULT 2: If $\{I_k\}_{k=1}^\infty$ is a disjoint collection of open int...

help please
ayuda por favor
given a curve $f$, its curvature is $\frac{\vert f' \times f''\vert}{\vert f'\vert ^3}$
Even silly things like $(x,x^2,x^3)$ are quite annoying.
@Twink Use [text](link) please!
I don't know if I'm missing something.
@FernandoMartin It gives a number, what's the fuzz?
00:18
I don't have to calculate the curvature at a given $x$, I have to find a general expression.
For most curves, that's impossible, in the sense that the definition I mentioned above is the most you can get. So I don't know if I'm missing something important, since the exercise makes no sense.
Maybe the point of the exercise is "see? curvature is hard to calculate".
@FernandoMartin Hehe, let me try one.
can someone please help
2
Q: What is the probability that the student knew the answer to at least one of the two questions?

JeffA student takes a true-false examination containing 20 questions. On looking at the examination the student and that he knows the answer to 10 of the questions which he proceeds to answer correctly. He then randomly answers the remaining 10 questions. The instructor selects 2 of the questions at...

Exercise 7@PeterTamaroff
Stuff like (c). What's the point of it? It's just plugging in in the formula. It makes no sense.
@Jeff [text](link) please!
00:22
@Jeff: There's already three good answers there.
lol
text
lol
is the second answer to the question correct? it looks detailed but im not sure if its accurate
@Jeff: Why do you think it isn't correct?
@FernandoMartin nothing in particular, i just wasn't sure
Come on take a look at my question @PeterTamaroff please
00:33
@PeterTamaroff: I have a little silly algebra problem if you want
@FernandoMartin Go ahead.
Do you know what an integral domain is?
@FernandoMartin Yes.
Ok, maybe you know this already. Prove that any finite integral domain is a field.
lol
00:34
@FernandoMartin I know it, yes. =)
Haha, ok.
everyone wants @PeterTamaroff to help them
Now prove that a finite division ring is a field.
@FernandoMartin That one is hard! You won't trick me! Heheheh!
(just kidding, that's a somewhat well known theorem and is not easy)
STOP KNOWING SO MUCH PETER.
00:36
one potato two potato
@FernandoMartin please leave Peter alone or he won't be able to help me
:-(
@Twink I am reading your question.
:D ty
@Twink Sadly, I know little about measure theory. What is the def of outer measure, again?
the inf
00:37
@FernandoMartin Can't help it. =D
of the sums of the lenght intervals
the lenght of the intrvals
that cover the set
$$m^\ast(A)=\inf \left\{\sum \ell(I_j):A\subseteq \bigcup I_j \right\}$$
?4
yes but
l(Ij)
the lenght
Sure, yes.
OK.
and Ij are open
00:39
@Twink That is probably irrelevant.
Isn't it immediate? The sets are well separated, so you can choose disjoint coverings
not if you want to use the Heine Borel theorem
no
that's not so immediate
read my comment on one of the comments
in the question
@Twink: you can clearly get disjoint coverings by intervals, the question is how to make these finite
00:41
i gave a counterexample
if you have an infinite collection of disjoint intervals in a bounded set, the lengths of the intervals and the distances between them must get arbitrarily small
thus the separation property guarantees the covers are finite
@Twink I bet you can just make a $\leqslant,\geqslant$ argument.
I don't know anything about measure theory, but it's reasonable to think that $\leq$ holds in general.
It holds in probability spaces, which are the only ones I know, haha.
outer measure is subadditive, yes
yes but I think this is to prove additivity
when the sets are disjoint
we already have the equality in the two results
we used $\leq$ and $\geq$ proving that
00:49
Well, to prove $\geq$, just prove that, given any covering $L$, there's a finer covering $L'$ such that $(L'\cap A)\cap(L'\cap B)=\emptyset$. I think that suffices, but I may be wrong.
all you need is the fact that an open set in $\mathbb R$ is a countable disjoint union of open intervals along with the second result
don't even need compactness argument unless you want to just use the first result
hi
Rehi @Peter, @Fernando, @Anthony, @Twink ...
01:04
hihi
@TedShifrin @Fer had a question for you.
hi Ted
@Peter, did you get the last link?
@TedShifrin Oh? Let me see.
Link? What do you mean?
No, I meant the lines problem ... Last step.
01:07
@TedShifrin Oh, I said
1 hour ago, by Peter Tamaroff
@TedShifrin Take those two points, and the line through them?
@Fernando, what can I do for you?
@TedShifrin: it's a general question.
The line is your 4th line?
In most cases, there's no nice closed form for the curvature and torsion of a curve, right?
I mean, I have a formula, but once I plug in the derivatives of my curve it doesn't get any better than that (in general).
What do you mean, Fernando? Sure, there are easy formulas in terms of the param
01:09
@TedShifrin Right.
No, that's a closed form :)
@Ted: Yes, I know there are formulas. But for instance, even in a "nice" curve as $(x, x^2, x^3)$ there's no nice closed form.
@Peter: So how many lines meet all 4 of our lines?
I'm asking this because I have an exercise that consists in calculating curvature and torsion for a couple of curves, and in most cases it's hopeless.
I'm not sure what the point of the exercise is.
Hopeless? I'm confused.
01:11
@TedShifrin Infinitely many?
Hmm. For instance, the curvature for $(t,t^2,t^3)$ at $t$ is $\frac{\vert (6t^2,-6t,2-2t)\vert}{\vert (1,2t,3t^2)\vert ^2}$
Remember we built the surface out of our 3 lines. You have two families of skew lines ... The $x$ lines and the $y$ lines.
That's just plugging in the derivatives in the formula. There's not much that I can do to simplify that expression.
@TedShifrin The $x$ lines and the $y$ lines?
01:14
Hmm, my $t$s turned into $u$s
There's an error, but, yeah, you're right, and it's not an interesting exercise. Look at my notes for better curves :)
Ok, thanks for your help!
No, they're $t$s, but your exponent is wrong in the denom.
You're right, it should be 3, not 2.
Yes, @Peter, fixing $y$ gives you a line in the saddle surface, etc.
01:16
I started typing $t$s and then started copying from my notebook where I used $u$s so I got them mixed up, but then I corrected them.
@TedShifrin Oh, well.
I'm lost.
Remember we said two lines through each point? Thus, two families of lines.
Each $x$ line meets every $y$ line but no other $x$ line, and vice versa.
@TedShifrin Ah, OK.
I am confused. In $\Bbb R^3$ doesn't $z=xy$ show that we can find $3$ skew lines such that infinitely many lines pass through them simultaneously?
Yes. True. But now we add the fourth.
BTW, which family were the three lines I gave you in?
@TedShifrin The $x$ lines.
01:24
Cool :)
What I was meaning to say is this.
Take any line skew line $tv+w$. Then it always crosses $z=xy$ at two points.
Generically, yes.
@TedShifrin Hm, I think you lost me.
Sorry.
Where?
@TedShifrin I don't know how to move on. I thought my idea
2 hours ago, by Peter Tamaroff
@TedShifrin $$\left( {t{v_2} + {w_2}} \right)\left( {t{v_1} + {w_1}} \right) = t{v_3} + {w_3}$$ always has a solution in $\Bbb C$.
was going to lead to something.
01:32
Yes, it does! Two solutions. So, two points on our fourth line and on our saddle surface.
Leading question. If a line meets all three of our first lines, what can you say about it?
Hmm...
I am guessing it should be on the surface, but I don't know why.
Bingo.
But why?
01:36
Well, in how many points does that line meet our surface?
@TedShifrin Three.
Think about your algebra from two hours ago.
wonders if @Peter is OK
@TedShifrin I am.
I stopped thinking because I didn't want to get frustrated.
01:48
Awww ... Polynomial of degree $2$ with $\ge 3$ roots ...?
@TedShifrin Oh, that.
OK.
So it is in the surface.
Yuppers.
Go algebra :)
So, what now?
So how many lines meeting our 3 lines meet the 4th as well?
@TedShifrin Only two?
01:51
Huzzah!
High five!
I don't understand something.
How can we be sure three lines cannot pass through four?
Restate?
Well, "we" just proved two lines can meet four.
What about three?
(Is it a dumb question?)
01:55
Ohh ... It would mean that your fourth line meets the surface in 3 points, and hence is in the surface. The fourth line was supposed to be in general position.
@TedShifrin Ah. OK.
Different flavor math from what you're used to with analysis.
@TedShifrin Aha.
Hello, if I have $F' = (P(x)\phi(x))' = P'(x)\phi(x) + P(x)\phi(x)'$ where P(x) is a polynomial of degree n, what would be the derivative if I have to do $F' = (\frac{P(x)\phi(x)}{Q(x)})'$
I'll tell you a different topological way of looking at this question some other day when I haven't burnt you out.
01:58
@TedShifrin No, ge ahead!
@masfenix Use the quotient rule.
You'll have to use the product rule in the num. then.
That is $$d\frac{fg}h=\frac{d(fg)h+dh (fg)}{h^2}$$
It's about a 4-dim manifold called the Grassmannian of lines in 3-space.
@TedShifrin OK.
@Peter: numerator
@TedShifrin Derp.
I should stick to "up" and "down".
Didn't want @mas confused.
02:01
@TedShifrin Of course.
"De" in denominator must mean "down" :)
thanks, its actually a question in functional analysis. I am to show that $\phi(x) = \frac{-1}{1 - |x|^2}$ if $|x| < 1$, 0 if $|x| \geq 1$ is a function that belongs to the space $C_0^\infty$, ie the set of all continuous, infinitely differentiable functions SUCH THAT the support is COMPACT.
@masfenix Do you mean $\exp \phi$?
It doesn't live there ...
so i took that function and i did a few derivatives. It follows that $\phi(x)' =\frac{ P_{3n-2}(x)*\phi(x)}{Q_{2n}(x)}$
02:05
@masfenix Again, I think you mean $\exp \phi(x)$.
Yes, sorry $\phi(x) = exp(.)$ where dot is my function from before
@masfenix OK. I have answered that here already.
@TedShifrin Not going to tell me about the Grassmanian?
so sorry, now i have a polynomial of degree 3n - 2 and another polynomial Q with degree 2n so I am try to prove it is infinitely many times differentiable by induction.
@PeterTamaroff how long was it? Maybe I can go back up in chat to look at what you said
Best to do $ 1/x^2$ instead.
@ted for the denominator?
02:07
Let's save it for later, @Peter. :)
Yes. Then you're simplifying the algebra and you just compose at the end.
@masfenix Here.
@TedShifrin Better $x^{-1}$!
Right. Mea culpa :)
Mi*
@Twink Nope. Latin.
@Twink: Latin!
LOL
02:10
xD ok sorry
@PeterTamaroff so I can rewrite my eqn as $\phi(x)' = (P(x)Q(x^{-1})\phi(x))'$. what would be the corresponding product rule/chain rule
@masfenix Did you read my answer?
@anon Yelloes!
yello
hi anon
I know you hate me but say hello:-(
02:14
@PeterTamaroff I didn't really get why you did the steps you did
@masfenix Because it works =)
@Ted, @Peter: Sorry to bother, but what does that the line problem say again?
@TedShifrin do you know about measure theory?
@FernandoMartin Dude, you're not bothering.
Haha, ok, just trying to be polite.
02:18
@TedShifrin I'm afraid to enunciate the theorem wrongly. "Take four skew lines in $\Bbb C^3$. At most how many lines meet them all?"
@PeterTamaroff, Peter I would like to show that the support is compact for $\phi$. Is there a method to showing that?
@masfenix Is $[-1,1]$ compact?
yes.
OK, so what's the fuzz?
No fuzz just smacked myself in the back for not seeing that.
02:35
@PeterTamaroff I am having a hard time going through these derivatives, any suggestions?
@masfenix I don't understand what you're doing.
@Fernando: Four skew lines in general position.
Basically, math.stackexchange.com/questions/119858/… but for the function i described above.
General position means affine independence?
No, pairwise skew but generic ... When you get into the problem you find what that means.
@Twink: I have known measure theory, but it's basically my least favorite part of analysis. Lebesgue integrals are fine :) plenty of people here who know it better than I.
@TedShifrin I am still not sure why there cannot be three lines touching four lines in general position.
@PeterTamaroff basically, imgur.com/IQIPMDk are the derivatives. I can that it is in the form $D^n \phi(x) = \frac{P_{3n-2}(x) \phi(x)}{Q_{2n}(x)}$
@masfenix Yes, but you're not supposed to do that. Did you read my answer? Induct.
Yes, so now I am trying to show that for $D^{n+1}$.
02:42
@Peter: Basically, it would require the fourth line to lie on your saddle surface, which is special position, not general :)
@TedShifrin What saddle surface?
$z=xy$
The one created by the first three lines? (Should it be $z=xy$ always?)
I got that. I just didn't know why it should always be $z=xy$.
No, it won't always. That depended on the three lines I gave you. This is where projective equivalence comes in ...
@TedShifrin Oh, OK. But the point is the three lines give a surface. OK.
02:45
A special surface, yes.
@TedShifrin Yes. Still have a lot to learn about geometry. =D
You can't know everything at 20. :) you already are ahead of most undergrads and plenty of grads ...
@TedShifrin Of course. Time to time.
The "of course" is in response to "You can't know everything at 20.", of course!
"Of course" also applies to the second part as well
of course, of course, a horse is a horse
02:58
@FernandoMartin Heh, wouldn't want to go around presuming.
03:09
@Anthony, I think that ditty needs a shuffle :)
close enough :P
@TedShifrin Ah?
Yes, don't want @Peter's head too swollen. He's already gonna destroy me in tennis.
@TedShifrin Grrrrrrrrrrr
03:11
@TedShifrin I am off now, need to sleep, 8hs tomorrow!
Bedtime for me, too. Happy week!
@FernandoMartin Hope to see you guys tomorrow.
Cheers.
user61230
03:51
Hello!
user61230
Are there any math books which teach by asking the reader to derive proofs, formulas, theorems, etc.?
Peter is not present. really eerie.
@Emrakul There is one by I.M Yaglom called Convex Figures
@Emrakul However, generally books of this style are very out of fashion. My personal recommendation is when reading a book read the theorem and cover up the proof, try to prove it yourself, if you cant get it, read a line of the proof, try to prove again, etc
Speaking of books, does anyone know of a book about the more exotic number?
Such as transcendental numbers, non computable irrationals, transfinite numbers, etc.
Im looking for something that is a little bit rigorous, not a general-public level book
user61230
04:14
That's unfortunate. Thank you, though!
04:36
how can i integrate (e^ax-e^bx)dx/[(1-e^ax)(1-e^bx)] from 0 to infinity
w|a says it doesn't converge because of blowup at x=0
@anon do you go to school
yes
where
somewhere, over the rainbow
04:46
lol
is it a university?
@anon how did you select which schools you wanted to apply to?
by living in the area
what do you mean
i don't get it
Thinks of integral domain on first read
 
3 hours later…
07:55
@Kal, Hello .
I just want to let you know; I find <the chained ring> in a research a bout the complement of zero-divisor graph of commutative rings
 
4 hours later…
user87637
11:40
@Ethan You should really apply to UCLA since it is close to your home and it is a top school in mathematics. Also, look at the curriculum to see if you like it as for your other choices, and the cost of living and things like that.
user87637
I will be applying to UCLA for grad school. I hope to see Ethan and robjohn there if I make it in time, lol.
12:46
Hi @WillieWong long time no see?
:-)
 
1 hour later…
14:05
hey anyone there?
I am trying to understand the definition of c(n) in this post on mathoverflow: mathoverflow.net/questions/142777/…
I asked on the website but the explanation is not clear to me.
Should I post this as a question on mathematics stackexchange.
14:47
@Shahab Dd you read about Van der Waerden's Theorem?
The answerers in that thread seem to assume that as given.
@robjohn: of course
I am the poster of the question in the thread
I know that
I don't understand the argument for the lowerbound given as a comment by fedja
@Shahab Okay, I have almost no knowledge about Ramsey Theory, so I can't go into that :-)
okay, thanks.
can i post my doubt on mathematics stackexchange
14:52
Anybody familiar with Galois groups?
@robjohn What?
@robjohn what does that mean?
@Shahab links added
lol
@robjohn: that's funny. Maybe Frank has a question though.
15:06
@Shahab Yes, I know, but I am not an algebra person, so that is all I could add. Sorry to interfere.
@FrankScience: I don't much algebra myself, but why don't know you ask your question. Maybe someone else will answer it.
 
1 hour later…
16:19
What is conjugation by elementary matrices ?
@robjohn Do you know what is it?
@user43418: Do you know what conjugation is?
16:36
Greetings
Yes: $P^{-1}AP$ no?
@FernandoMartin
Exactly
An elementary matrix is just the identity after a single row operation.
They are used to represent row operations.
@FernandoMartin Ok so I have the following matrix: $ A=\begin{pmatrix} a & b \\ c & d \end{pmatrix} $ and I am being asked to show that using conjugation by elementary matrices, one can "eliminate" the a entry.
What does "eliminate" mean ? Equal to zero ?
I guess so.
Try to replicate the Gauss-Jordan algorithm
Does it mean that $a$ must disappear alltogether or the location 1,1 must be equl to 0
16:45
I'm not sure, try to do both
By Gauss-Jordan Algorithm you mean to try and reduce the matrix to the the identity matrix by elimentary operation right ?
@FernandoMartin
@FernandoMartin
@user43418 yes

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