« first day (2448 days earlier)      last day (2870 days later) » 

17:00
that is interesting
generate a non cyclic proper subgroup if I'm not mistaken
how do you prove it?
Which part ?
that it is a proper subgroup
Show it does not contain $a$ for instance
17:03
how?
In fact $ab$ and $ba$ might suffice
not really
$(ba)(ab)(ba) = baabba = bbba = a$
Indeed
You can show the extremities always are $ab$ or $ba$ by induction
And that's sufficient
consider $(baba)' = abbabb$
which contradicts your theorem
Oh yeah I frogot about inverse :p
But you can go around that
Either the two extremities are inverses, in which case you take the inverse of your element
Or one isn't and you're done
For this, let $e = abab, f = baba$, and compute $e^2, f^2, ef, fe, ef', e'f$
And then write any element of the span $e^{n_1}f^{m_1} \dots e^{n_k}f^{m_k}$
Where the $n_i, m_i\in \Bbb Z^*$
17:15
$\begin{array}{rcl}
e^2 &=& abababab \\
f^2 &=& babababa \\
ef &=& ababbaba \\
fe &=& babbab \\
ef' &=& abababbabb \\
e'f &=& bbabbababa
\end{array}$
Except $n_1$ and $m_k$ which can be 0
Hellô, if $y\inR $ does $(y)^(1÷3)\in R $?
@Vrouvrou yes
$f:x \mapsto x^3$ is bijective
But I need ^(1÷3)
Yes
$f^{-1} : x \mapsto \sqrt[3]x$
17:25
Good day!
Suppose $f$ and $g$ are analytic on a domain $D \subseteq \mathbb {C}$ and that $fg \equiv 0$ on $D$. Show that either $f \equiv 0$ or $g \equiv 0$ on $D$.
@Harry: Remember that domains are, by definition, connected.
@TedShifrin hi
1
A: Non- linear ODE's with no independent variable

DHMOThe first one: $$\begin{array}{rcl} y &=& y' \ln y' \\ W(y) &=& \ln y' \\ y' &=& e^{W(y)} \\ \dfrac{\mathrm dy}{\mathrm dx} &=& e^{W(y)} \\ \dfrac{\mathrm dx}{\mathrm dy} &=& e^{-W(y)} \\ x &=& \displaystyle \int e^{-W(y)} \ \mathrm dy \\ &=& \displaystyle \int e^{-u} \ \mathrm d(ue^u) \\ &=& \di...

Hey @TedShifrin, I was thinking of a complex analysis approach
17:40
OK, @DHMO.
Well, you need to use a fundamental result from complex analysis, yes, @Harry.
I guess I can figure that one out, thanks, @TedShifrin. I do have another question, though
Is this true? If $H$ is a subgroup of $G$, and both are finite groups, then $\#H \mid \#G$
Yes, @DHMO, that's one of the fundamental theorems from the beginning of group theory.
@TedShifrin that's interesting
That is Lagrange's theorem!
17:42
Salut Ted
Salut, Astyx. Guten Tag, @DanielF.
I'm asked to use the Cauchy Integral Formula on $\int_0^\pi \dfrac {d \theta} {(a + cos \theta)^2}$ where $a > 1$. I cannot seem to find a way to convert this integral to a form that can use the Cauchy Integral Formula
@Harry: Hint: Can you make up a problem $\int_0^{2\pi} ...$ that will solve this one?
@DHMO, yes, it is indeed Lagrange's Theorem. Orders of subgroups divide the order of the group
@HarryEvans thanks
17:44
anybody got an answer to my earlier message on hilbert spaces? or care to have a look?
@DHMO: The key idea is cosets.
Bon jour @Ted.
@TedShifrin, I know that the integral I wrote is one half of the integral when taken from 0 to $2\pi$. But the denominator is not of a form $(z - a)^{n+1}$
@Harry: What if you write $z=e^{i\theta}$?
I miss seeing you around, @DanielF.
Too little time to hang around in the chat much these days.
17:49
Well, I hope you are enjoying yourself some, @DanielF.
How to prove that $|G:H \cap K| \le |G:H||G:K|$ where $H$ and $K$ are subgroups of $G$?
hi @Alessandro
@TedShifrin I do. Not all day, but every day at least a little.
Is it true that $|G:H \cap K| = \operatorname{lcm}(|G:H|,|G:K|)$?
17:51
I'm glad, @DanielF, seriously.
I doubt it, @DHMO.
@DHMO One method is to find an injection $G/(H\cap K) \hookrightarrow (G/H) \times (G/K)$.
I doubt that, too, @DanielF. :)
Hi, Demonark.
How's it going @Ted?
(Also hello everyone!)
Still sick (growl), but otherwise just fine. How you?
@TedShifrin then how would I approach it?
17:55
What's $|G:H|$ for groups?
@AlessandroCodenotti number of cosets of $H$ in $G$
@DHMO: If you didn't know Lagrange's Theorem, you probably don't have much experience with groups. But I would suggest thinking about a reasonably small non-abelian group where you can see everything that's going on.
alternatively, $|G/H|$
Well, today we did the three types of integration
At any rate, I'm confident it's false.
17:56
So that was pretty cool
the three types, Demonark?
Well, "the" is not the right way to say it
But Riemann, Lebesgue, and Henstock
@TedShifrin are all $\Bbb Z_n$ groups abelian?
Yes, @DHMO.
I suppose so
17:57
What's the simplest interesting symmetry group?
$S_3$?
I don't recall Henstock, Demonark.
What's the Henstock integral?
OK, which is also $D_3$, the group of symmetries of an equilateral triangle. Either way, play around with that, @DHMO.
@TedShifrin heh
well I can find 2 non-trivial subgroups
namely, $C_3$ and $\sigma$
17:58
You can find lots more than that.
@TedShifrin I thought there are only $6$ elements
How many subgroups are there?
At least five.
including the two trivial subgroups?
Then six.
5+2=6
17:59
I was counting one of the trivial ones the first time; thank you very much.
I think all subgroups are cyclic
They are.
so i can just $C_3 \sigma$
The idea is that if you have a tagged partition, and a gauge $\delta: [a,b]\to (0,\infty)$, you say that the partition is $\delta$-fine if $[x_{i-1},x_i] \subset (t_i - \delta(t_i),t_i+\delta(t_i))$
18:00
or $\sigma C_3$
@DHMO. Stop talking and sit down and think.
@TedShifrin salut
@AlessandroCodenotti arrivederci
And then you say that the number $I$ is the Henstock integral of $f$ on $[a,b]$ if for any $\epsilon$, there exists a gauge $\delta$ and a $\delta$-fine partition $P$ such that $|\sum_P f - I| < \epsilon$
It's not much of a modification to the definition of a Riemann integral (special case where the gauge is constant) but at least on $\mathbb{R}^n$ it's more general than even the Lebesgue integral
Does it incorporate Riemann-Lebesgue $\int f\,d\alpha$ somehow for a discontinuous $\alpha$?
I believe so, though it was created more to solve the problem of $\infty - \infty$
Like when you split up a function to positive and negative parts
18:05
So Riemann-Lebesgue will give you as a special case a sum of values of $f$ at particular points. Can you get that with Henstock?
I'm not totally sure, we hadn't talked about it much, it was a side note in passing. We mentioned that Riemann or Lebesgue integrable function is Henstock integrable, and that you can always integrate a derivative and get the original function
Lebesgue integrable with respect to the standard Lebesgue measure on $\Bbb R^n$ ... but presumably not weird measures.
We mentioned that it doesn't get you stuff like probability
I'm guessing my hunch is right about Riemann-Stieltjes and counting measures.
So for that reason the Lebesgue integral still has a reason for existing
:P
18:11
@TedShifrin hi
@Daminark hi
Hi, Karim.
Hey @Adeek
@TedShifrin have you heard of Qing liu algebraic geometry ?
No, but I'm not current ...
it is really nice intro to algebraic geometry
once exams are done I will start reading it
18:12
But yeah, also I'm not sure if the set Henstock integrable functions gives you a nice space like the $L^p$ spaces
You mean a Banach space, Demonark?
Yeah
shrug
We'll be using only the Lebesgue integral from now on anyway so yeah
I'm still gonna give you some 2- and 3-dimensional integrals to evaluate :P
18:16
th... three? That's... a lot of dimensions...
are you there @DHMO?
shivers They won't be too hard, right? Like linear functions over rectangles, right?
My students had to do some in 4, Demonark, and integrals over some 3-dimensional manifolds in 4 dimensions ... :P
Sid
Sid
jee advaced anyone?
Demonark: You think you're so masterful living in infinitely many dimensions ...
18:18
That sounds really tough
becomes an algebraist
Lol jk
I dunno, I like infinite dimensions though, they're fun
I wonder, if there are convergent examples of such expressions:
$$\lim_{R\to \infty}\int^{(R)}f(x)d^Rx$$
Well, $f(x)=e^x$ is easy, since it is just an eigenfunction of the integral operator, but are there others
But yeah in total fairness I'm probably losing/going to lose some computational skill this quarter in particular
First quarter had some long ones, as well as last year, but starting from second quarter it was mostly just a couple small ones here and there, and this quarter likely none
So it probably would be worthwhile to practice at least some
Elaboration: $\int^{(R)}=\underbrace{\int \cdots \int}_{\textrm{R times}}$
@Secret I was very confused for a second there
yeah, there seemed to be no nice notations for integrals beyond quadruple integrals
Therefore if $f(x)=e^x$ (and let initial condition be 0, thus the constant term will always be zero) we get the following countable sequence:

$$\begin{matrix}e^x & e^x & e^x & e^x & \cdots\end{matrix}$$

which is a "constant" sequence that obviously "converges" to $e^x$
18:29
Hi again
Hey @Astyx!
How are you ?
Random variables $X$ and $Y$ have joint density function
$$
f(x,y)=\begin{cases}
c(x^2+\frac{1}{2}xy)&\text{if }0<x<1,0<y<2,\\
0&\text{otherwise.}
\end{cases}
$$
I have to find the value of the constant $c$. I started as follows:
$$
\int_{x=0}^1\int_{y=0}^2c(x^2+\frac{1}{2}xy)dydx=\int_{x=0}^12cx^2+x\,dx=\frac{2}{3}c+\frac{1}{2}.
$$
Since this should equal 1, we have: $c=\frac{1}{2}\cdot\frac{3}{2}=\frac{3}{4}$. However, my solution set says that $c=\frac{6}{7}$. So where did I make a mistake?
What is a primitive of $x\mapsto x$ ?
$1/2x^2$?
18:32
I'm alright, how about you?
@ShaVuklia Re read yourself :)
I'm fine thanks :)
which part?@Astyx
How's the exam situation?
I've reread it countable times :P
Second line of equations
18:34
You made the standard mistake for which I always yelled at my students.
You did not integrate $x$ with $x^2/2$
Do not distribute $c$. Factor it out once and for all.
ohhhh
right
haha :P thanks
Break these bad habits!
@Daminark Exams start in 36 hours or so
18:35
but, I thought I made things easier by distributing?
why is it better not to distribute then?
And here I am sitting on my chair solving rubik's cubes
NOOOO ... you made it easier to screw up !! Always happens.
@Astyx: Bonne chance :)
I'm not even sure if I can do the integral without distribution:P I'm gonna see if I know how it works
Merci :)
He means get it out of the integral @Sha
OHhhh!
righttt
that is indeed better
18:37
Funny how people never trust me ... :D
Poor Ted :p
Exactement.
hahaha :P no, I thought you meant I had to integrate $c(x^2+\frac{1}{2}xy)$. But it is true that there have been instances when I didn't show trust where due :P
Good luck @Astyx!
Funny story also. For Centrale-Supelec I mistakenly registered myself to take the exam in Poitiers instead of Paris
So I'll be doing them in Poitiers
18:42
How far an excursion is that?
300 km or so
But there is a direct TGV connection so it's okay
Oh damn
Convenient
Is TGV an international slang btw ?
No, I think it's specific to France
It's probably quicker for me to get to Poitiers than to get to the place the exams would have been in Paris
18:44
Or at least to French-speaking countries
Je m'en doute.
But would a random person still understand what it is ?
Nope
But, as I've taken it several times and will be taking it again in June, I'm fully aware :P
Pretty crazy error, Astyx.
18:46
Yeah, I'm not 100% sure it's my fault really
My favorite word for this is ... ditzy.
ditzy ?
Oh, how is it not your fault? Sort of the way I ended up with a train ticket in Italy where we switch trains and end up in second class, even though I never saw any way to avoid that? The "help" people were worse than unhelpful.
I'm pretty sure I remember taking Poitiers and not having any other choice on the registration website. This being said, my memory might be failing me
ditzy = out of it, dumb ... :P
18:48
In fact taking it in Poitier might even be an improvement after all, since it's very calm and quiet there (compared to Paris)
Italy is conservative, no?
But you have to allow all sorts of time for travel and getting lost.
I'll be staying there for 4 nights, so I have a whole day to find my way, and once I'm there I've been told it's only 5 minutes away, so that's settled
Oh, not to mention hotel :P
But it should be a fun adventure.
Centrale is not really a school I want in any case
Maybe I'll go to the Futuroscope instead ...?
:p
18:51
@TedShifrin I want to make sure I understand something correctly
here it should be finitely generated as a ring over $\phi(A_0)$ right ?
Karim: I have told you multiple times not to ask me algebra stuff. I haven't thought about this in 40 years.
oh okay
Seriously. I'm getting annoyed.
ok sorry
@Astyx Come to Chicago so you never have to compute things ever again!
(@Ted Must think I'm a great influence right about now)
18:57
Do people in Chicago never compute anything ?
I mean computation isn't like, nonexistent
Reverso translates "ditzy" as "écervelé", didn't realise it was that bad :p
But say, as opposed to the typical track of students doing mostly computational calculus, linear algebra, and ODEs before reaching proofs
Here it's more, Spivak first year, which is proofs plus a couple examples here and there to make sure you don't totally forget how it works
Random variables $X$ and $Y$ have joint density function
$$
f(x,y)=\begin{cases}
e^{-x-y}&\text{if }x,y>0,\\
0&\text{otherwise.}
\end{cases}
$$
Find $\mathbb P(X+Y\leq 1)$. I started as follows:
$$
\mathbb P(X+Y\leq 1)=\int_{x=0}^1\int_{y=1}^{1-x}e^{-x-y}dydx=\int_{x=0}^1-e^{-1}+e^{-x-1}dx=-e^{-1}-e^{-2}+e^{-1}=-e^{-2}.
$$
This can't be correct. The answer should be $1-2e^{-1}$. Could someone tell me where I made the mistake?
Then second year, you do one of the 3 analysis classes. First is more calculation-based, has the feel of rigorous multi plus a side of metric spaces. Second is Rudin throughout the year, third kinda fast tracks through Rudin in a couple weeks and focuses on functional analysis, measure theory, and whatever the professor feels like doing
19:01
@Sha $y=0$ as a bound, not $1$
@Daminark I really don't understand how you can do maths without doing proofs ...
@Astyx oh yea, thanks
@Astyx A lot of schools in America have very small math departments and get a good amount of funding from having service courses to engineering departments and all that, and don't have the resources to have a side track that just jumps into proofs for actual majors
A lot of schools are somewhere in the middle
UGA, for example, has an honors multivariable calc class that Ted used to teach, which is taken right after first year calculus and is proof-based (though computation does actually exist in a non-trivial way)
From what I've gathered, the main places where undergrad math looks like how it's done in France would be Harvard, Princeton, Chicago, and so forth
Those math departments have more room to have math classes outside the standard funnel of "math for science/engineering"
@Sha it's not $-x-1$ but $-x$ once you've integrated once
@Daminark I believe that works better for producing mathematicians, but I guess engineers don't care
Yeah, that's why I really cannot appreciate more having come here
I'm now in the fast track class and we're covering a lot of very fun content
The speed makes it such that you might get through material only vaguely, or at a cost of some computational fluency
But it's a whole lot of fun
Are you doing only maths ? Or do you also take physics, chemistry etc ?
19:13
So, we have a core curriculum
So everyone does some amount of humanities, social science, civilization studies, physical science, biology, and math
We're doing approximately the same for preparation of the exams, I can't wait to be in school so I can learn more deeply about stuff
Oh all that
That takes about a third of the time
Then a third is dedicated to major requirements (give or take)
Then a third is just whatever you want to do
Second major, minor, going deeper into your field to prepare for grad school, random classes for fun, whatever you feel like
I got through most of them, I just have to finish one more quarter in biology along with 2 in civilization
Is civilization like history ?
And what are humanities and social science ?
Yeah, and humanities is stuff like philosophy, art, linguistics, that kind of thing
Social science usually entails stuff like psychology, politics, economics, and so forth
The core classes have mostly the feel of a discussion-based class on political philosophy
(core in social science)
Sounds cool
19:19
Yeah, most of the core classes have been very enjoyable
The math department requires you to take a few classes outside math as part of the major, in either computational neuroscience, physics, chemistry, statistics, or computer science
(Which is not a decision I agree with, like they should decrease this requirement and add point-set topology and complex analysis at least...)
I'll be satisfying those more by theoretical compsci classes, like algorithms and combinatorics
Yeah I also have computer science, and I enjoy it tbh
I'll possibly take, like, general relativity or something, plus some philosophy classes and one on Italian Renaissance
But from here on it'll mostly be math and theoretical compsci
Yeah same
Though I'm more into the theory side, along with functional programming
C just annoyed me to no end
Oh we mostly do theorical stuff, with graphs, trees, languages
And the language we use is CamL, so lots of functionnal programming :)
Nice
The first class I took in compsci used Typed Racket
I'm also hoping to get into Haskell
I've barely ever heard of those
19:24
Alright, well, I'm gonna have to head out in order to finish my sosc stuff now, I've got class in about half an hour
Oh they're pretty fun, I probably would say Haskell is better
See ya
any way i can see the chat when i last logged on
19:43
hello please if $y$ is real then $(y)^(1/3)$ is real ?
@Vrouvrou Usually yes
Use y^{1/3}
why ? please because i found this and i don't undertand :quora.com/What-is-value-of-1-1-3
That's why I said usually
so not alwayse ?
So here's the thing. $x^2=1$ does not mean $x=\sqrt1$, but rather $x=\pm\sqrt1$.
19:45
it can be real it cn be not
?
Well note the above
@SimplyBeautifulArt Depends if you mean principal root or not
There are two different solutions to $x^2=1$
@MeowMix Yes, I was getting there :P
@Vrouvrou Likewise, there are three solutions (not necessarily unique) to any third degree polynomial.
Also, there is always a real root of $x^3-y = 0$
(thus implying that $y^{1/3}$ always has a real solution)
One of the solutions to $x^3=1$ is $x=1$. But the other two solutions are not real.
19:48
@Vrouvrou The function y=x^(1/3) is defined for whole real number line, No matter what is x if it is a real number the output will be real too.
It's also not differentiable on its domain :]
@Vrouvrou So to sum it up, our point is that the equations $x^2=1$, $x^3=1$, and $x^3=-1$ have multiple solutions. But one of them is always a real solution.
my problem is that i want to prove that $(R,d)$ where $d(x,y)=|x^3-y^3|$ is complete, and i found that any Cauchy sequence converge to $(l)^{1\3}$ where $l$ is real to say that $(R,d)$ is complete i need to have that $(l)^{1\3}$ is real
If $X$ and $Y$ have joint density function
$$
f(x,y)=\begin{cases}
\frac{1}{2}(x+y)e^{-x-y}&\text{if }x,y>0,\\
0&\text{otherwise,}
\end{cases}
$$
find the density function of $X$ and $Y$. I was considering two things:
1) I could check whether $X$ and $Y$ are independent, and if so, use the convolution formula $f_Z(z)=\int_{-\infty}^\infty f_X(x)f_Y(z-x)dx$,
2) I could differentiate $\mathbb P(X+Z\leq z)=\int_{-\infty}^\infty\int_{-\infty}^{z-x}\frac{1}{2}(x+y)e^{-x-y}dydx$.
I'm not sure if 2) is reasonably doable. Are these my two only options? Or is there a better way?
@ShaVuklia yes, and it's over
19:56
What do you mean, @LeGrandDODOM ?
@ShaVuklia there's nothing more to write
ohh, you're referring to that other question
yea, that's solved, thank you
for your last question, you should have seen in class that you get marginal densities by integrating the joint density with respect to one variable, letting the other one be fixed
If you haven't, it's a consequence of Fubini's theorem
yes I've seen that
but that's only useful if they are independent, right?

« first day (2448 days earlier)      last day (2870 days later) »