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7:02 PM
I'm not understanding mymath
 
user19161
There are actually a number of math people on the TeX site who are very active there but not on math.
 
@anon I'm looking into this now,
 
@AndrewSalmon My math assignment is the following:

Assume that a 3x3 matrix A, has the eigen-values \lambda_1 = 1, \lambda_2 = 0.8 and \lambda_3 = 0.6 and the eigen-vectors:

$
v_1 = \begin{pmatrix}
1 \\
0 \\
2 \\
\end{pmatrix}
$
$
v_2 = \begin{pmatrix}
2 \\
3 \\
1 \\
\end{pmatrix}
$
$
v_3 = \begin{pmatrix}
0 \\
2 \\
1 \\
\end{pmatrix}
$
Every 3-vector $v$ can be written as a linear combination $v = x_1v_1 + x_2v_2 + x_3v_3$

What happens to $A^nv$ when $n \to \infty. What does that even mean?
So LaTeX doesn't work?
 
It does.
You need ChatJAX
 
No, it works.
 
7:06 PM
A^2v=A lambda v = lambda A v = lambda^2 v
 
@TheJohlin If your matrix has those eigenvalues, you can write is as $A=BDB^{-1}$ for some diagonal $D$, can't you?
 
A^n v = lambda^n v
 
@TheJohlin As $n \to \infty$, $A^nv$ goes to $0$ everywhere except when $v$ is parallel to the eigenvector $v_1$.
 
@AndrewSalmon Is that because $v_1$s eigenvalue is 1?
 
In which case $A^nv = v$.
@TheJohlin Yes.
 
7:13 PM
Hi @peter
 
@Charlie Hello.
 
@AndrewSalmon That actually clarified a few things! Do you have any other comments?
 
@PeterTamaroff how are you?
 
@AndrewSalmon Yeah, I got the script now! :)
 
7:14 PM
@TheJohlin Actually I am way wrong. $A^nv$ will be the component of $v$ in the $v_1$ direction.
 
@AndrewSalmon Could you elaborate on that, please?
 
Because the other components will go to $0$ (by linearity and the fact that their eigenvalues are less than $1$), but that component will stay constant (eigenvalue is $1$).
 
$$A^n (x_1 v_1 + x_2 v_2 + x_3 v_3) = x_1 A^n v_1 + x_2 A^n v_2 + x_3 A^n v_3 = x_1 \lambda_1^n v_1 + x_2 \lambda_2^n v_2 + x_3 \lambda_3^n v_3$$
 
@user58512 You know, I've been looking at that formula like times now and I just now realized that what it says is $A^nv = \lambda^nv$. I feel stupid now...
 
yeah
first step linearity, then use eigenvalue property
@AndrewSalmon, are you a high school student?
 
7:39 PM
Can anyone here help with a question regarding topological groups?
 
probably not, what is it?
 
I want to show that the trivial subgroup $\{e\}$ is open.
The problem I'm having is that in a totally disconnected space, the trivial subgroup is a component, and is therefore closed.
 
the trivial subgroup is open if and only if you have the discrete topology
 
Hmm... Then that must be the case here. That is, the group I'm working with must have the discrete topology.
Well, is the following claim true: If $G$ is a compact, totally disconnected (and therefore Hausdorff) topological group, then the identity component is open?
 
totally disconnected implies the trivial subgroup (which is the identity component) is open
 
7:51 PM
That's what I thought, but I'm having a hard time proving it. Or at least understanding why that is true.
 
wait, my reasoning was wrong. a subgroup being open implies closed but not vice-versa (except if it's finite index).
 
Why does finite index matter?
 
you can take finite intersections of open sets and always get an open set, but not infinite intersections in general
 
Oh right.
 
conversely, you can take finite unions of closed sets and get a closed set, but not infinite unions
 
7:56 PM
im bashing my head against this numbe rtheory stuff but it doesn't become clear
help me Namagiri
 
@Charlie Uneasy. You?
@anon I'm trying to work this out
14 hours ago, by anon
@PeterTamaroff , it suffices to show that a doubly transitive action cannot preserve a partition into two nonempty nontrivial cells A and B. both |A|,|B|>1 (why?). then pick a g then sends a pair in A to a pair of one thing in A and one in B, contra the hypothesis that the partition is stabilized.
Got half an hour to do so.
I am thinking about the first thing you say. Let $G$ act on $S$ $2$-transitively, and $S=A\cup B$, $|B|,|A|>1$ be a nontrivial partition.
We want to show that the actions doesn't stabilize this partition, and in turn this means no non trivial partition is stabilized.
 
@PeterTamaroff good, a bit drowsy, but good, writing my notes, studying
 
Assuming the first assertion, I'm thinking about $\pi(S)$ being a nontrivial parition, and assume this is stabilized. There exists $|A|,|B|>1$ . Because of the first assertion, these cells must be moved to other cells, $A'$, $B'$, and because of $2$-trans. we must be able to move them back to $A$ and $B$ which is a contradiction, yes?
@anon
 
8:12 PM
How is moving A',B' back to A,B a contradiction?
Pick a distinct pair $(u,v)\in A\times A$ and a pair $(a,b)\in A\times B$. Let $g(u,v)=(a,b)$. Hence $gA$ has nontrivial intersection with both $A$ and $B$, impossible.
 
Suppose $G$ acts on a set $X$ which is partitioned into blocks $\{\Gamma_i\}$ such that any that meet are equal. Equivalently we have an equivalence relation $x \sim y$ iff $x,y$ lie in the same block and this relation is preserved by the group action.
If $G$ was 2-transitive then pick any $a,b \in \Gamma_i$ and map one into $\Gamma_i$ and another into $\Gamma_j$: In terms of the equivalence relation this can only happen if $\Gamma_i = \Gamma_j$ (i.e. there is only a single block) or $a$ and $b$ lie in different blocks (i.e. every block is a single element)
 
8:37 PM
hi @Ethan
 
Hi
 
how are you
im bored :[
 
I'm bored too
 
@Ethan, ill tell you a problem
 
K
 
8:49 PM
@Ethan, switch the order of summation in $$\sum_{n \le x} \sum_{d | (n,D)} \mu(d)$$
 
It's Phil's arithmetic function or some such non sense
 
(also it counts something interesting.. but I wont give it away)
 
Oops nvm
 
Phil's arithmetic function?? is that a joke? lol
 
8:52 PM
Well the second sum is zero unless gcd(n,d)=1
 
yes
so take D to be the product of all primes below sqrt(n)
 
It can be expressed in terms of eyes totient function and a small partial sum, if n is divisible by d, it can be given in terms of just the totient function
Eulers*
D*
 
oops sorry, i meant below sqrt(x)
 
If x is a perfect square it all can be evaluated in terms of eulers totient function
 
I've actually not seen any totient function in this
im sure you are right but I never did that
 
9:00 PM
Here you want an asymptotic formula
 
yeah :D
@Ethan, want answer?
 
Sense there sofar exists an asymptotic formula for primorials, and I doubt there ever will be one, it will have a primorial in it
 
wel I think the primorials is equivalent to PNT in some way (maybe?)
 
Doesn't exist an asymptotic expressin for primorils
No that's the logarithm of the primorial, the chebyshev function,
 
well there's a formula for that just take exp of it
anyway it counts the number of primes between n and sqrt(n)
given D as above
 
9:04 PM
No lol just because two things are asymtotic u can't infer there expoential s are
 
sieve of Eratosthenes
I thought you could
oh yeah you'er right the error gets bigger too
so whats new?
 
If you had an asymptotic expression for primorials you could get a logarithmic error term for the prime counting function, which is incomprehensible accuurate
Y*
 
RH gives $\psi(x) \sim x + O(\sqrt{x}(\log(x))^2)$
wiki says $ p_n\# = e^{(1 + o(1)) n \log n} $
 
Yes, who ever wrote that is incorrect
 
I think its right
why can't you just do exp
that makes sense to me
 
9:17 PM
$$\sum_{p\leq x} \ln(p)\approx \sum_{n=2}^\infty\frac{\ln(x)^n}{\zeta(n)n!}$$, should be a very good approximation
 
that's crazy dude
I've never understood any formula which sums zeta functions
hmm zeta(n) gets very small doesn't it?
I can't tell if that formula converges or not
 
It does
 
oh yeah it does
 
9:37 PM
Hey
is it concave up for the whole domain?
 
no
 
no???
oh nv
m
 
Hmm, so there is an inflection point
time to recheck my work
 
it is concave down in (-2/3,0)
 
Fouind it
just a plugging into wrong error
yup :D
 
10:15 PM
whats up
 
hi gusy
guys
 
hey mick
 
hey
 
how are you
 
aparantly tommy corrected his conjecture.
im tired , wont stay long
 
10:18 PM
bye
 
lol
 
lol
im actually going to sleep soon too
 
in think his conjecture is true now
unless prime twin gaps are known to be ln(x)sqrt(x) or such
...
 
there is a bit known about prime gaps
but it requires seriously deep results about zeta
 
i said prime twin gaps
 
10:24 PM
oh
obviously nothing is known about twin gaps
 
afak nothing known about it theoretically but maybe by computer search
anyways i like the conjecture.
im off guys
bye@user58512
 
bye
 
10:47 PM
@JayeshBadwaik Broken.
 
@Arkamis Thanks.
 
Man, this Harlem Shake thing beats the crap out of me.
 
@anon Hey
@anon Where did you learn you diff. geom from?
 
I dunno, wikipedia maybe?
 
@BenjaLim Benjamin.
@anon LAWL. You rock.
 
10:56 PM
@PeterTamaroff yea
 
I didn't really learn it. Taking the class now.
 
@anon Right. Any math.se threads I can look at? I need to learn about immersions quickly for an assignment
 
not off the top of my head
 
@anon ok
 
NOOOO
 
10:58 PM
@anon WAAAAAAAAAAT?
 
my gigantic-ass 8x8 multiplication table for ${\bf F}_8$ has an error that I just discovered while filling in the last line
 
I'm just going to pretend it's not there and hope my teacher doesn't have time to check it all
2
 
@anon Does it affect all 64 entries?
 
no, I don't immediately know how many or which are affected, probably only a few entries
 
11:01 PM
@anon This is like a doctor who realized there is a biological breach... "How many have been compromised, Doc?" "I don't know! I don't know!"
 
11:50 PM
Hello. Is anybody good with electromagnetism here?
Or took it before?
 
No clue.
 
@PeterTamaroff you never took it ?
 
@Carpediem No. I'm not interested in physics.
 
@PeterTamaroff ok
 

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