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3:06 AM
Hey @anon maybe you can provide me with some motivation as why that happened
So in the notion of semi-direct product
let me first define the groups
Let H and K be groups and let $\psi$ be a homomorphism from K into Aut(H)
and we can construct a group G of ordered pair (h,k) with h $\in H$ and k $\in K$. We showed that $H \cap K = 1$ by seeing its isomorphic copies and obviously they intersect trivially.
But of course the people who did this thing they had think about it, so I was wondering can you figure out from those assumptions that $H \cap K = 1$ without resorting to the semi-direct group G ?
 
3:31 AM
maybe also @TedShifrin if you have anything to say about this
 
 
3 hours later…
6:19 AM
@KarimMansour Still here?
 
Anyone here?
 
Huy
probably
 
Yay :D
Do you happen to know about differential geometry?
 
Huy
a bit
and you?
 
None at all
 
Huy
6:28 AM
:(
 
I don't even know non-differential geometry
 
:o
I want to learn differential geometry, and I had a few questions, if you guys have the time :D
 
Huy
ask questions, don't ask to ask
 
yes @TobiasKildetoft
 
Thank you. My current goal is to learn differential geometry to begin studying general relativity. I've heard that you need to know differential equations to learn differential geometry, and I have taken a course in ODE. I recently bought a book in PDE to learn, so I can also learn quantum mechanics. I've taken up to multivariable calculus, but I want to know, what else would be required for me to learn differential geometry, other than differential equations
 
Huy
6:36 AM
I only know the very basics about differential equations and even less about PDEs and I think you can manage well without much knowledge of the two.
Both for differential geometry and for quantum mechanics.
You'd better do some functional analysis for QM.
 
and operator algebra
 
Huy
Then again, maybe we think of different things when we talk about differential equations.
Of course knowing either of the two won't hurt you and will make your knowledge of mathematics more complete, but they really aren't required if you have a good background of the basics like linear algebra and analysis, imo
 
I've heard of functional analysis, but what is that? And what about operator algebra, I've never heard of that?
 
Huy
oh, topology is most certainly useful to study differential geometry
in functional analysis you usually study vector spaces similarly to in linear algebra, but now they are often infinitely-dimensional and usually have additional structures like an inner product (which induces a norm and thus a topology)
study of functional analysis automatically leads to some very basic PDEs
 
its algebra of continous linear operator on topological vector space.
 
Huy
6:41 AM
do you know any topology? I think it could be very useful for differential geometry and functional analysis
 
C∗-algebras (pronounced "C-star") are an important area of research in functional analysis, a branch of mathematics. A C*-algebra is a complex algebra A of continuous linear operators on a complex Hilbert space with two additional properties: A is a topologically closed set in the norm topology of operators. A is closed under the operation of taking adjoints of operators. C*-algebras were first considered primarily for their use in quantum mechanics to model algebras of physical observables. This line of research began with Werner Heisenberg's matrix mechanics and in a more mathematically developed...
 
Huy
(and those two are usually the things a mathematician studies before looking at GR or QM)
 
Hmm, I've taken a course in linear and abstract algebra, but I'm pretty sure we only scratched the surface. Are those related to vector calculus? And no, I don't know any topology
 
Huy
isn't vector calculus usually covered in multi-variable calculus?
 
What I covered in multivariable calc was things like multiple integration and its applications, flux, green's theorem and whatnot. I don't think I actually learned much vector calculus
 
Huy
6:45 AM
you don't need to be an expert in vector calculus to study the other things mentioned. just solid knowledge of calculus. implicit function theorem and the like should be well known
(for differential geometry)
 
Ahh. Do you know any books that are useful for functional analysis?
 
@JonSnow books with "functional analysis" in the title are a good place to start?
 
Huy
here's a book my professor is currently working on: https://dl.dropboxusercontent.com/u/2098511/FAnotes.pdf
:D
apart from that, I didn't study any English books, so you'd have to ask other people.
 
I see, I'll take a look at that. @Ale
 
NP Glad to help.
Libraries are great.
 
6:49 AM
@AlecTeal, true, but I was wondering if anyone had a personal recommendations
 
Functional analysis is a broad topic. Some books focus on different areas.
It took me a while to find one that ... oh bugger I'm making them sound like religions.
Just go look at some books!
 
Huy
@JonSnow: Here's a list from a professor of my uni with recommendations to start functional analysis: www2.math.ethz.ch/education/bachelor/lectures/hs2015/math/fa/…
But I think you should really start with topology for both functional analysis and differential geometry.
 
Yes if you are not confident with metric spaces, MOST books will be out of reach
 
@KarimMansour I am not quite sure what you mean by your question about semidirect products
The fact that the subgroups intersect trivially is pretty much obvious
 
Huy
@JonSnow: Before you ask, the standard textbook for an intro to topology is Munkres.
 
6:52 AM
Or "Introduction to topology" by Bert Mendelson
Which is great because that book is like £3.00 (I love Dover)
maths.kisogo.com/… (Lots of links) I found this book because Proof Wiki has a lot of references citing it too.
maths.kisogo.com/… (Munkres search, but these hit other books by Munkres so may not be a fair test)
 
Haha, thanks guys. @Huy you mentioned earlier "Topology for functional analysis and differential geometry". Would it be a big difference if I get a book that doesn't particularly focus on those two topics?
Also, @AlecTeal, My PDE book is by Dover, and it cost $10. It's actually an amazing book for the price.
 
Huy
you need to study topology, then whichever of the other two you prefer. any topology book suffices
 
@JonSnow which one? I may have it! I really recommend that Bert Mendelson one, I also recommend the Munkres one (I've got second edition, VERY happy) - I did Mendelson first, which I think was right to do.
@JonSnow in this picture, blue is unread, green is read, orange is read a lot and red is read the most:
 
Huy
I'm off to some lectures, see you around
 
I have the by Farlow "Partial Differential Equations for Scientists and Engineers".
Have fun
 
6:57 AM
@JonSnow this picture is my path through the book with time (YES, got that book) same colour scheme
So as you can see I've really hammered that book.
 
@AlecTeal, what do they mean by pre, norm, and post?
 
pre is the contents pages and stuff, post is after the main part of a book (some books have funny sections like A for answers and such)
So this is Munkres' topology:
 
How do you keep track so effectively? :O
 
I write something like [59- when I start notes at page 59, then I write like [59-61] if I end at 61. I cross this out when I've added it to my software.
But you can see how the shorter and very good Mendelson book let me skip vast chunks of Munkres (which has some tough preliminaries - but they always are) and use it more as a reference. I'd say they were an effective combo.
 
It seems like Mendelson's book would be a better one for self-study in my opnion
opinion*
 
7:04 AM
Oh I don't go to lectures.
 
I'm an engineering/physics major, so I don't have time to enroll in these classes
 
I was really really pissed off by the TINY desks and trying to understand the accent and stuff, I also prefer to read and stuff. So I just purchased like 3 meters squared of desk surface area instead.
 
Do you still enroll in the classes though?
 
I do the homework and exams as it is set.
Anyway, get Mendelson's book. Do it, get Munkres - reference it. If your course isn't called "Topology" you wont need much from Mendelson.
 
My reason for getting the book isn't for a course, though, it's to gain a better understanding before learning differential geometry
Is that unnecessary?
 
7:09 AM
Depends what you want to learn TBH. If you're going for manifolds and that sort of differential geometry then HELLS YES. You need Topology
If it doesn't go further than radius of curvature and stuff (which I encountered at A-level, before university) you should be find. I've read books called Differential Geometry which do not mention the word "topology" and other ones which start with "Topological Manifolds"
 
7:39 AM
can you explain @TobiasKildetoft
 
@KarimMansour Now very well unless you explain more carefully what you are unsure of.
 
in the definition of semi-direct group
we have H and K are groups
but we don't have any information about them
only that there exist a map $\psi$ from K into Aut(H)
how do you see that $H \cap K = 1$?
without actually knowing about semi-direct product
 
@KarimMansour It makes no sense to ask whether two groups intersect trivially just like that
you need them to be subgroups of some other group for that
 
well, in the semi-direct group proof we showed that they intersect trivially
 
@KarimMansour No, you showed that the natural copies of those groups intersect trivially inside the semidirect product
 
7:43 AM
oh I see
 
this is no different from the case of direct products
(which is just a special case of a semidirect product)
 
yeah
so we can have H and K not intersecting trivially but their natural copies intersecting trivially?
 
@KarimMansour again, it makes no sense to ask whether two groups "in the wild" intersect trivially
as abstract sets, they could be equal or disjoint, and it would says (almost) nothing about them as groups
 
I mean given that there exists a map $\phi : K \rightarrow Aut(H)$
 
@KarimMansour such a map exists for any pair of groups
 
7:51 AM
I see
can you give an example
 
@KarimMansour an example of what?
 
why does any pair of group has a map from K to Aut(H)
?
 
@KarimMansour just send everything to the identity map
(this is the only one we are guaranteed to have)
 
I see
 
 
3 hours later…
10:54 AM
Water on Mars:
Leading the pack of doubters is waning US shock jock Rush Limbaugh, who was quick to denounce NASA's announcement as both unimportant and probably a conspiracy to bolster the case for climate change. He said that NASA has lied about environmental data in the past, so why should it be believed about water on Mars – if that's even its real name.

"They're lying and making up false charts and so forth, so what's to stop them from making up something that happened on Mars that will help advance their left-wing agenda on this planet?
The fuck America.
 
 
2 hours later…
12:36 PM
I have a kind of meta question
3
A: Limitation of integration by parts to compute $\int e^{-x}(\cos wx + w\sin wx)dx$

UserXHint; Separate the intgral into 2 parts like this; $$\int e^{-x}(\cos wx + w\sin wx)dx= \int e^{-x}(\cos wx) dx+\int e^{-x}(w\sin wx)dx$$ Then integrate by parts the first one 'till magic happens. Hint#2; $e^{-x}\cos wx$ can be written as $-\cos wx \times -e^{-x}$

I answered to this and another guy gave an answer that the questionee can't understand due to it requiring higher knowledge. Should the answerer elaborate to make it easier for the questionee to understand it or just leave it there for people that can appreciate it as it is?
 
Huy
it's really up to him, but if he wants his answer to be useful to the questioner, he should elaborate
there's no point forcing him to do so if he doesn't want to
many good and interesting questions can't be answered such that the answer is fully comprehendible for the questioner
 
I think you both mean questioner, not questionee (the latter would be the person being questioned)
 
1:05 PM
How does one write commutative diagrams in here? I have a question.
 
@DanielFischer Hi
 
@DanielFischer How goes it?
 
So-so.
 
Hi all
 
1:11 PM
hi
 
I earned a popularity question badge. The simpler the title the more views you get.
@skillpatrol I don't even remember asking the question.
Hi Fischer @DanielFischer
 
Hi @LucioD.
 
@DanielFischer Is Bobby Fischer the most famous of the Fischers?
 
@LucioD Among chess players certainly. For other population groups, here are a few other famous Fischers mentioned.
 
$$\require{AMScd}
\begin{CD}
\pi_{1}(X,x_0) @>(h_{x_{0}})_{\bullet}>> \pi_{1}(Y,y_0)\\
@VV\hat{\alpha}V @VV\hat{\beta}V\\
\pi_{1}(X,x_1) @>(h_{x_{1}})_{\bullet}>> \pi_{1}(Y,y_1)\\
\end{CD}$$.

I have to show that the following diagram commutes . Given that $h:X\to Y$ such that $h(x_0)=y_0, h(x_1)=y_1$.
Any hints on how should I prove this?
 
1:17 PM
@DanielFischer I see...So this conversation has been had already...
 
Nothing new under the sun ;-)
Seems that doesn't work in chat, @Rememberme.
 
@DanielFischer Any way I can write commutative diagrams here ?
 
@Rememberme I think I have seen some here, but I don't know how to do it.
 
How did that suddenly come?
 
1:24 PM
@Rememberme proper capitalisation of the package name helps ;)
 
Oh thanks !! :) @DanielFischer
@Tobias You are there?
 
@Rememberme Plug in the definitions. Take a loop with base point $x_0$, and look at what both compositions make of it. There may be a relation between $\hat{\alpha}$ and $\hat{\beta}$ required.
 
Okay let me see
 
@DanielFischer For a square $p \times p$ complex matrix the definition of positive definite matrix is usually given as $x^{*}Ax > 0$.

I was recently given a definition of a positive square matrix $A$ as being one such that $\langle A h, h \rangle \geq 0$ for $h$ in Hilbert space $\mathcal{H}$, are you familiar with how these definitions are connected?
 
@LucioD What's the standard inner product on $\mathbb{C}^n$?
 
1:35 PM
@DanielFischer I think it would be $\langle u,v \rangle = u^{\ast}v$ where $u^{\ast} = \overline{u^{T}}$.
Not sure why it's displaying like that.
 
@LucioD Because you used * instead of \ast. That confuses the parser here, for * is Markdown markup.
(And there was a pair of *s)
@LucioD Okay. (Although mathematicians usually make it linear in the first and antilinear in the second argument.) Now look and see whether you can see the similarity between $x^\ast Ax$ and $\langle x,Ax\rangle$.
 
@DanielFischer You get $\langle x, Ax \rangle = x^\ast A x$, but if we are in some $\mathcal{H}$ then why would you assume that the inner product is defined in the same way as it is for $\mathbb{C}^{n}$? Are you assuming finite dimensional $\mathcal{H}$?
@DanielFischer It has to be finite dimensional if $A$ is acting on $h$?
 
1:57 PM
@LucioD In an arbitrary Hilbert space, $x^\ast Ax$ makes no sense. The thing that makes general sense (although $A$ isn't a matrix generally but an operator) is $\langle x, Ax\rangle$. In the case of $\mathbb{C}^n$, one can write that in the form $x^\ast A x$.
 
@DanielFischer Oh so you are just saying that the idea follows from the case of $\mathbb{C}^{n}$ since this can be written in both ways, which is not necessarily true for arbitrary Hilbert spaces?
 
2:16 PM
@KarimMansour unfortunately I cannot understand what your question is
 
Hi, does anyone know any good website(s) where I can read up on the application of integration to find mean values and centroids? Please let me know if you do :)
 
@DanielFischer For the case of a matrix we can go the other way as well since $\langle x, Ax \rangle = x^{T}Ax$?
 
2:34 PM
@LucioD I don't understand. For $x\in \mathbb{C}^n$ and $A\in \mathbb{C}^{n\times n}$, we have $\langle x, Ax\rangle = x^\ast Ax$. The left hand side makes general sense, the right hand side doesn't. (Hence it is much preferable to use the left hand side, except for things that are specific to $\mathbb{C}^n$.) Now, what do you mean with "go the other way"? What goes which way?
 
@DanielFischer uhmmmm....Yeah as your wrote it as what I meant. The right hand side has the advantage of not necessarily needing an inner product.
 
@LucioD But things are more transparent if you don't hide the principles behind $\mathbb{C}^n$. Even when working in $\mathbb{C}^n$ it is often helpful to use the abstract properties of inner products.
 
@DanielFischer OKay. In the original question could I reason that $h$ is p-dimensional if $A$ is a $p \times p$ matrix?
 
2:57 PM
@LucioD If $A$ is a $p\times p$ matrix, then $Ah$ isn't defined except for $h \in \mathbb{C}^p$.
 
@DanielFischer One last thing.
If you have a $2 \times 2$ matrix $A$ where the entries are also $2 \times 2$ matrices, you have top left entry of matrix
A=
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}
top right
\begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix}
bottom left
\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}
bottom right
\begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}.

Is it clear that this matrix is not positive definite?
 
@LucioD Depends on what counts as "clear". It's a $4\times 4$ matrix with determinant $-1$, so it's not positive definite, nor is it negative definite.
 
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaahhhhhhhhhh!!!!!!!!!!!!!!!
Sorry.
@DanielFischer So positive definiteness is characterized by the value of the determinant?
Fuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuukkkkkkkkkkkkkkkkkkkkkkkkkkkkkk....Sooooooooooooo‌​ooooooooorrrrrrrrrrrrrryyyyyyyyyyyyyy.
@DanielFischer I see the determinant is always positive if it is positive definite.
 
3:14 PM
@LucioD Not quite as simple. I think it is Sylvester's criterion, but could be I got the wrong name. A hermitian matrix is positive definite if and only if the determinants of the principal minors are all positive.
In particular, the determinant must be positive, but that's not sufficient.
(For even dimensions, $A$ and $-A$ have the same determinant.)
 
@DanielFischer I was at this post.
@DanielFischer Will look up Sylvester's criterion now.
@DanielFischer Oh okay so you check the upper left corners until you check the whole thing and if they are all positive then you have a positive definite matrix.
 
Right. To find that a matrix is not positive definite, it may be quicker to use some other square submatrix, not an upper left one. In the example above, the central $2\times 2$ matrix was $\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}$, and one knows that that isn't definite.
 
@DanielFischer No what I described was a $2 \times 2$ matrix $A$, with $a_{11} = \begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}$

$a_{12} = \begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix}$

$a_{21} = \begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix} $

and

$a_{22} = \begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}. $\
 
3:31 PM
@LucioD Just look at it as a $4\times 4$ matrix with complex entries. That you write it in a more complicated way makes no difference.
Unless you want to interpret it so that "positive definite" has no clear meaning.
 
@DanielFischer Are those the only two options...
 
@LucioD Can't swear to it. But nothing much else comes to mind.
How did you want it to be interpreted?
 
@DanielFischer I want to see that positive definite has a clear meaning but I don't see how you are equivalently viewing it as a $4 \times 4$ matrix with complex entries...you are viewing it in this way so that it is a complex Hermertian matrix which is a requirement for using Sylvesters criterion?
 
@LucioD The first thing to find out is what sort of things can be positive definite or not.
 
@DanielFischer Continue...
@DanielFischer What sorts of things can be positive definite?
 
3:43 PM
Well, sesquilinear forms can be positive definite (or not). And stuff we can identify with sesquilinear forms.
 
@DanielFischer Okay I will never forget that.
@DanielFischer One thing.
@DanielFischer Was I right in stating that you wanted to change the way the matrix was viewed so that it was Hermertian matrix and that would allow the use of Sylvester criterion?
 
Can we give an infinite field a topology such that functions from the field to the field are continuous?
 
@DanielFischer
But Daniel how in Jesus's name can you just view it as

$\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}. $

simply because that makes life easier!...
 
In the context, that was, up to isomorphism, the only thing that made sense that I could think of.
Hi @Ted.
 
3:54 PM
Hi @DanielF
 
@DanielFischer It's quite good...otherwise there wasn't a clear way of interpreting the original matrix so that positive definiteness makes sense?
Hi @TedShifrin
 
Hi @Lucio
 
@TedShifrin Maths is troubling...
@TedShifrin Daniel is leading me astray. He has had enough of my nonsense...
 
So is life :)
Daniel never leads people astray.
 
@TedShifrin True...I lead myself astray to be completely honest...
@TedShifrin Life and maths...horrible stuff...
 
3:58 PM
LOL, poor baby :)
 
@TedShifrin And you want to top it off with well placed session of mocking...
 
Not mocking ...
 
@LucioD Faced with a matrix $$\begin{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} & \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \\ \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix} & \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \end{pmatrix}$$ and the question whether that is positive definite, have you any idea but interpreting it as a matrix corresponding to a linear map $(\mathbb{C}^2)^2 \to (\mathbb{C}^2)^2$?
 
Hello@Ted
 
4:01 PM
@TedShifrin It's unpleasant.
 
Hi @Remember
Lucio, relax. We're all friends.
 
@DanielFischer Yes that's an idea I agree with.
 
@Ted What do you think of this question?
http://chat.stackexchange.com/transcript/message/24543999#24543999
 
Well, and $(\mathbb{C}^2)^2$ is isomorphic to $\mathbb{C}^4$.
 
All functions? @Remember .. Just the discrete topology.
 
4:04 PM
@DanielFischer I understand thanks. I was just joking about leading astray...*cough* Ted came and started causing trouble.
 
causes more trouble
 
Anything except the discrete topology?
I mean not the trivial ones. Some nice topology
 
I don't see where you're using the algebra of having a field, @Remember.
 
I really did not understand what you meant @Ted
 
You phrased the question in terms of a field. Why do we care?
 
4:10 PM
Thats always a difficult question to answer. I really don't know. I wanted to think of such a field to see what interesting can I get out of such a field.
 
But once you consider all functions, you're ignoring the algebra and it's just about sets.
 
Okay so I guess I have to restrict the types of function I am considering to get something interesting
OKay , So what if I think of continuous functions if and only if they are polynomials. By that I mean to say that
Which infinite fields are there such that if I equip them with a certain topology the functions from the field to the field are continuous if and only if they are polynomial functions. @Ted
 
4:36 PM
Think about $\Bbb R$ or $\Bbb C$, @Remember.
 
4:48 PM
Hello @TedShifrin
I am in number theory mode today
After reading about arithmetic derivatives
And using derivation on UFDs
awesome stuff
I'm sure there is a Lie structure in their somewhere
Too bad I don't know enough to say anything smart about it
 
5:02 PM
70 colored balls are placed in an urn, 10 for each of the seven rainbow colors.

What is the expected number of distinct colors in 20 randomly picked balls?
 
this looks a nice question to answer,and cheer up my mood screwed during this fugly day.
 
@N3buchadnezzar Nice problem
I tink you can solve it with a markov chain
 
@Alizter I tried brute force first :p
 
Yes but imagine you are sitting in a room with no time
There are usually very elegant solutions to these types of problems
 
5:18 PM
yeah
Well my brute force has not yet given any satisfactory answer
 
Hmm
1/7 chance they will pick a certain colour
are they returned to the urn?
What is the probability that one colour will be picked?
That is just 1/7^20
 
Well atleast two colors will be picked
since we are picking 20 and there are only 10 of each color
 
No
Wait
Yes
Hmmm
Are we returning balls?
 
OK
 
5:25 PM
good, found the answer, time for coffeeee
2
 
Huy
sorry I didn't do the numbers but why is the problem difficult with brute force?
 
@Huy Oh nothing big, I am just churning thorugh a large number of samples to get a high accuracy so it takes some time.
I also figured it was an interesting problem =)
 
Huy
ok
 
I am getting 9.04902260000000069 which seems reasonable?
 
Huy
there are only 7 different colours no?
 
5:29 PM
Oh snap, yeah. I assumed 10
Need to do some simple changes in the code then
 
Huy
code? aren't you doing it with pen and paper? :D
 
You said brute-force :D
 
Huy
well brute force still gives you some sort of formula
and then you plug in some numbers
or isn't that what you meant by bruteforce?
ah you mean just trying everything out and counting?
 
@Huy whistles suspiciously Yeah, thats what i did. Just churned through 6 million picks
 
Huy
like run through all possibilities and have 7 counters that go up by one every time the possibility occurs
and then compute the average
 
5:32 PM
i think that would take a long time no?
 
Huy
and why not just compute the expectation value like usually? :D
well yeah, that's why I didn't even think of that
and take a long time, idk. today computers do that in probably less than half a minute
 
70C20 = 1.61884603662657876*10^17 Mmmm
 
Huy
^^
 
Had a hard time doing it on pen and paper as well
 
@Agawa001 Should it not be the other way round?
Coffee=>Answer?
 
6:15 PM
@Huy bah :p
 
6:45 PM
@N3b: This is a variant of the coupon problem, where one asks the expected number of turns to get a certain number of distinct coupons. The expected number of turns to get $10$ different colors, for example, is $1+10/9+10/8+10/7+10/6 +\dots + 10/1\approx 29.3$. And the expected number of turns to get $9$ different colors is approx. $19.3$. So this seems to agree with your numerical work.
Wait, I'm assuming an infinite supply of each color of ball. Not the same problem.
 
7:31 PM
hello, i have a small question on topology, i have a space $E$ with this topology $\theta=\{G\subseteq E, \mathrm{card}(E\setminus G)<+\infty\}\cup\{\emptyset\}$
such that $card(E)=\infty$
i want to find $\overset{\circ}{A}$where $card(A)=\infty$ and $card(E\setminus A)=\infty$
please
 
@TedShifrin finite
 
@TedShifrin can you help me please
 
7:48 PM
Hello!! Does the following stand? $$-a k_1 +b k_2 -c k_3 \leq |-a+b-c| \max \{k_1 , k_2 , k_3\}$$
@robjohn @DanielFischer do you have an idea?
 
@robjohn did you continue working on my problem?
@Hippalectryon $$\int_0^1 \int_0^1 \int_0^1\frac{dx \ dy \ dz}{(1+x) (1+y) (1+z) (1+x+y+z+x y+x z+yz+9 x y z)}$$
 
@Chris'ssistheartist Have you solved the generalization ?
 
@Hippalectryon Yeah.
 
8:05 PM
@Hippalectryon :D
 
8:16 PM
if $x$ is a $p$-dimensional standard normal vector and $A$ is some matrix, then $x^T A x$ converges to the trace of $A$. At what rate?
 
Huy
converges while what happens?
 
sorry, i mean as the number of samples and dimensionality go off to infinity, but say that $p/n$ converges to a constant
 
Salut, @Hippa. T'es méchant comme toujours?
Hi @Huy.
 
Huy
hi @TedShifrin. Off to bed soon, sorry :D
 
I'm procrastinating; I'm supposed to be writing a book review, and I disliked it and find this all very painful.
 
8:19 PM
@TedShifrin Bien sûr ! Ca va ?
 
Huy
then make sure the book review shows how much you disliked it
 
Oui, plus ou moins, et toi?
 
Oui oui ça va :)
 
So, anything interesting happening here?
 
Not really :(
 
8:22 PM
@ted il est $toujours$ méchant, même pas besoin de poser la question...
 
@Ramanewbie très méchant
 
@ted Surtout avec moi
 
Et toi, tu n'es jamais méchant, n'est-ce pas, Ramanewb?
 
@ted Jamais, je suis toujours gentil
 
Hmm, il faut que je pose la question à vos parents ...
 
8:25 PM
@ted alright
@ted @hippa How much can you get ? typing-speed-test.aoeu.eu
 
@Ramanewbie That sounds just like the title of the spam mails I get ..
"How much can you score ? Click here to find out !!"
 
@Chris'ssistheartist Actually, I have not had a chance to start, sorry.
 
@hippa lol exactly but this isn't a spam
 
hi @robjohn
 
8:30 PM
@MaryStar I would highly doubt it, simply considering $a=1,b=2,c=1$ and $k=\{0,1,0\}$
 
@Chris'ssistheartist What's the $\phi$ here ?
 
@Hippalectryon Lerch transcendent
 
@TedShifrin Hey, Ted! Our rain has gone far away.
 
@robjohn Thanks, we got it in Paris this week :(
 
@Hippalectryon It's $\Phi$.
 
8:31 PM
@Hippalectryon we need it here. Severe drought.
 
no, @Hippa, that was the east coast's and Bahamas' (?) bad storm.
 
@robjohn wher do you live (country)
 
Phi (uppercase Φ, lowercase or ; Ancient Greek: ϕεῖ, pheî, [pʰé͜e]; modern Greek: φι, fi, [fi]; English: /faɪ/) is the 21st letter of the Greek alphabet. In Ancient Greek, it represented an aspirated voiceless bilabial plosive ([pʰ]), which was the origin of its usual romanization as "ph". In modern Greek, it represents a voiceless labiodental fricative ([f]) and is correspondingly romanized as "f". Its origin is uncertain but it may be that phi originated as the letter qoppa and initially represented the sound /kʷʰ/ before shifting to Classical Greek [pʰ]. In traditional Greek numerals, phi has...
 
@Ramanewbie 345
 
@hippa CPM or WPM
 
8:32 PM
cpm
69 wpm
with two fingers
 
@AlecTeal: Does the enumeration of the permutations that I was describing make sense now?
 
@hippa Why 2 fingers ? isn't it slower ?
 
Because I always type with 2
 
@Ramanewbie USA
 
@hippa it's slower but it's better for fingers musculation
 
8:34 PM
@Ramanewbie I don't care about that -_-
 
@hippa I doubt USA's rain can reach France
@hippa about speed or about fingers musculation ?
 
@Ramanewbie both
 
@hippa You should care about both
@hippa Consider typing fast and you'll gain quite much time
 
@Chris'ssistheartist I'll look at this later :D thanks. I doubt I'll manage to do it though :c
 
@hippa And what is your fingers aren't muscled at all ? You could have a car crash because your fingers aren't strong enough
 
8:36 PM
@Hippalectryon They are related to some of my advanced research, there are poor odds to come up with some brilliant stuff, but never lose the hope. I worked much in this area. You need a chain of results to get the marvellous picture in place. ;)
 
@Ramanewbie Good, I don't drive. Now, I thought this was supposed to be a math room -_-
 
@hippa Indeed. What is the simplest algorithm to compute 4 digits numbers squares instantly ? (without a calculator)
 
'instantly'
It's not hard to do a multiplication by hand ... Even if it's an 8 digits number ...
 
@hippa Yes, because some people manage to do that. Instantly = less than 1 sec
 
@Ramanewbie Some people can do 11 digits multiplications in one second. Some people can find 5th roots in one second. That by no means implies that you can.
 
8:41 PM
My original question was not well posed. Let me ask again. For a given dimensionality $p$, let $x$ be a draw from $N(0,I_p)$ and let $A$ be an independent $p\times p$ matrix. Then as $p\to\infty$, does $x^T A x$ converge to the trace of $A$? I think it does (or at least to a multiple of it), but am not sure at what rate
 
Does the following stand? $$|-\frac{5}{24}h^2f^{(4)}(x_1)+\frac{64}{24}h^2f^{(4)}(x_2)-\frac{81}{24}h^2f^{(4)}(x_3)|\leq |-\frac{5}{24}h^2+\frac{64}{24}h^2-\frac{81}{24}h^2|\max_x |f^{(4)}(x)| \\ = |-\frac{5}{24}h^2+\frac{64}{24}h^2-\frac{81}{24}h^2| ||f^{(4)}||_{\infty}$$

where $h>0$
 
I think it is connected to the sum of the product of i.i.d. normals
 
@hippa I doubt they can, since the difficulty is exponential relatively to the number of digits you compute
 
@Ramanewbie It's really easy do do it with some training... multiplication is pretty basic
 
@hippa Really ? Even 5 digits number squared, I doubt anyone can do it within 1 sec
 
8:45 PM
Doubt again.
 
i will
 
Is anyone familiar with divided differences?
 
dumb question, but what is the distribution of the product of a standard normal and another independent standard normal?
 
@MaryStar Why do you write $-\frac5{24}h^2+\frac{64}{24}h^2-\frac{81}{24}h^2$ separately instead of $-h^2$?
 
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