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2:02 PM
What's $z$ ?
 
$z$ is the variable of the generating function
 
What does the generating function represent ?
Ok i think I get it
 
It represents the number of binary strings of length m and the hamming weight r which do not contain k consecutive zeros
 
It's $\sum_{n=0}^{\infty}a_nz^n$
 
yes
 
2:10 PM
If I were you I'd ask with a comment on the answer
I doubt I can help you
 
Okay, thanks. I will try it!
 
in The Periodic Table, 1 min ago, by Secret
> The hessian is singular
 
2:28 PM
Not even power series allow you to integrate through singularities, right?
 
Hi
I am at a loss here -- I am stuck (and have been) on a particular question, and have n clue about what to do to get some help about it
https://math.stackexchange.com/questions/2286468/k-functional-between-ell-1-and-ell-2-for-a-specific-sequence
Basically, I have already put two bounties on it to "draw attention" (the second expires tomorrow), but in vain.
Is it just that no one knows, or is it that bounties are inherently useless given the number of questions appearing on Math.SE?
 
For me it's definitely in the "I don't know" category.
 
(I have considered Mathoverflow, but I am not sure it fits the bill there -- even though is is a research question. Plus, I don't have much reputation there)
 
Yeah, I'm never sure when to do MSE vs. MO.
 
@Semiclassical That makes two of us :)
 
2:32 PM
In favor of MO for your problem is that 1) you've already had it on MSE for quite some time, to no effect, and 2) you've got references which you cite.
Against it is the fact that I dunno what the threshold really is.
 
Yup. Most of the MO posts are a variation of "I am not sure I can parse the title, and when by chance I can it gets answered in a fraction it takes me to read the question"...
 
I remember someone here plays bass?
 
Oh. Here's a probably elementary question.
 
@Semiclassical I read that as 'Here's a probability question' and got all excited :(
 
2:40 PM
Suppose I have two matrices $ab^T-ba^T,cd^T-dc^T$ where $a,b,c,d$ are are all column vectors of the same size.
These are both manifestly antisymmetric, and any linear combination of them is also antisymmetric.
 
@SoumyoB One day, maybe
 
I'll have a probability exam in July, just wait :P
 
Antisymmetric matrices form a vector subspace don't they ?
 
Sure.
 
don't you guys ever get bored of doing just pure math all the time?
 
2:41 PM
No
 
analysis and analysis, and number theory and stuff
 
But what's not obvious to me: Given a particular linear combination of them, can I always find vectors $u,v$ such that said combination equals $uv^T-vu^T$?
 
I get bored of doing analysis, but geometry and algebra are another matter
 
Probably not @Semi
 
@AlessandroCodenotti I suppose you haven't solved many problems on sigma algebras and filtrations?
 
2:43 PM
Good morrow, chat.
 
In favor of it being true is the following counting: The antisymmetric matrices of dimension $n$ comprise a $n(n-1)/2$-dimensional subspace.
 
happy birthday @Fargle
 
@SoumyoB wait huh what
 
You can probably find $a,b,c,d$ such that the sum of you matrices is of rank $4$
 
2:44 PM
Oh, wait.
 
therefore there could not be such $u,v$
 
@fargle it's a trick I use, I win with a probability of $\frac{1}{365}$ and if I do I'm like "Yeah I'm a psychic blah blah blah"
 
To do this, just pick the canonical basis in $\Bbb R^4$
 
@SoumyoB Ah, fair enough.
Like that one XKCD comic.
 
Hmm. For $n$ sufficiently large this definitely doesn't work, since $n(n-1)/2$ grows faster than the $2n$-dimensional space of matrices of the form $uv^T-vu^T$.
I should admit at this point that what I specifically have in mind is the case of $n=3$.
 
2:46 PM
@SoumyoB And if you don't don't you look like a fool. So you look like a fool $364\over365$ of the time. Not sure it's worth it :p
 
In which case the respective dimensions are $3(2)/2=3$ and $2(3)=6$.
 
@Semi $n=4$ makes it impossible. I doubt $n=3$ works
 
Right.
 
Pick the canonical base $e_1,e_2,e_3$ of $\Bbb R^3$
 
So while it's definitely false once you get to $n=4$, it's not at all obvious to me whether it's true in $n=3$.
 
2:47 PM
Let $a=e_1, b=e_2, c=e_2, d= e_3$
 
@Astyx it's worth it, I'm a probabilist, I like gambling
 
Then if you let $A$ be the first matrix, $B$ the second
the image of $A$ is $span(e_1, e_2)$, the one of $B$ is $span(e_2, e_3)$
And the image of $A+2B$ is $\Bbb R^3$ unless I'm mistaken
It feels like I'm mistaken though
 
mmkay, so the matrices of interest are $A=e_1 e_2^T-e_2 e_1^T$ and $A=e_2 e_3^T-e_3 e_2^T$.
 
No $(A+2B)e_1 =-e_2, (A+2B)e_2 = e_1-2e_3$ and $(A+2B)e_3 = 2e_2$
 
erk, should've been B= for the second matrix I said. wooops
 
2:53 PM
So my argument fails for this example
 
@Semiclassical $A$ is for "$A$buse of notation"
 
snerk.
Looks like one can write $A+\lambda B=uv^T-vu^T$ with $u=e_1-\lambda e_3,v=e_2$.
So it looks like it may be true for $n=3$.
 
We can restate this as "the set of matrices $uv^T -vu^T$ is a vector space" right ?
 
I think so?
 
@SoumyoB depends on what "bored" means
@mathvc_ M_f/S^n is S^n with an (n+1)-disk attached to it by a degree m map $\partial$ D^(n+1) --> S^n
Now do it by cellular homology
 
3:10 PM
@BalarkaSen In fiber bundles, horizontal lifts of paths provide diffeomorphisms between fibers. Is it right to say that this is essentially due to existence, uniqueness and smooth dependence of solutions to ODEs?
 
Yeah exactly
 
I should really learn how to prove such theorems :P
(I mean the ODE results)
 
how can i get a pattern avatar?
 
After I discussed with my supervisors tmr, I might consider asking on the main CSE site about how to handle optimisation where the hessian end up singular
 
Wait, are the antisymmetric matrices of order 3 all singular ?
 
3:14 PM
Yep.
 
molecular hessians are symmetric
mine is a 66x66 matrix
 
Different context, I think. @Secret
 
So me trying to find a rank 3 matrix was just me being silly
 
@Danu They're basically Banach fixed point theorem
on function spaces
look up "Picard-Lindelof"
 
Oh and in general antisymmetric matrices of odd order are singular
 
3:18 PM
@BalarkaSen Hmm
 
Is that so? I'm forgetting why.
 
Because all characteristic values are imaginary and their sum must be real
 
A^T = -A implies det(A) = -det(A) no? so det(A) = 0
I don't get it
 
Well, their sum more specifically must be zero since antisymmetric matrices are traceless.
Right, that does it.
 
In odd dimension that is
 
3:20 PM
Right.
 
@BalarkaSen that was one of those black magic proofs the first time I saw it
 
So antisymmetric matrices of rank 3 have caracterisitc polynomial $X(X^2+a^2)$ for some real $a$
This means they are diagonalisable
at least for $a\ne0$
And unless I'm saying nonsense it should also be true for $a=0$
 
More precisely, if your antisymmetric matrix is of the form $a_1 E_1+a_2 E_2+a_3 E_3$, where $E_1=e_1 e_2^T-e_2 e_1^T$ and so forth cyclically, then $a^2=a_1^2+a_2^2+a_3^2$.
(At least that's what I seem to be getting from Mathematica)
Which dovetails nicely with the point re: all eigenvalues being imaginary.
 
So yeah, if $a=0$ the matrix is 0
 
Right. And the 0 matrix is already diagonal.
 
3:25 PM
Why is my brain so slow today ><
What's more they are of rank either 0 or 2
 
Is 0 the only rank 0 matrix?
 
Yes
Since its image is $\{0\}$
 
So every nonzero antisymmetric matrix of order 3 is rank 2.
 
Which means we can write those $ab^T + cd^T$ where a,b are linearily independant and same for b and d
 
@AlessandroCodenotti yeah it's a cool proof
 
3:29 PM
Right. But we need that to be antisymmetric still.
Can we get from there to it being of the form $ab^T-ba^T$?
 
So we get $ab^T + cd^T = -ba^T - dc^T$
 
So you'd have $ab^T+ba^T=-c d^T -d c^T$.
which...???
 
I guess you can impose $b$ and $d$ to be orthogonal ?
 
Eh, that doesn't seem right.
 
$b$, not $c$
 
3:31 PM
hmm. i dunno.
 
@Semiclassical were you the dude here that plays bass?
 
Yes, take an orthogonal basis of the preimage
 
Nope. @SoumyoB
 
Then you have $b$ and $d$ orthogonal
@SoumyoB it was Alessandro, as he claimed earlier
 
@SoumyoB I am, I told you earlier as well!
 
3:35 PM
@AlessandroCodenotti nice to meet you again
You might like this song
there's a lot of bass in it
 
I'm trying to think of my favorite bass lines in songs.
And the one which is coming to mind, for whatever reason, is this: youtu.be/yX6FsTIq6ls
 
Hi guys, can the abel summation formula be somehow modified to make 0 <= n <= x instead of 1<=n<=x?
 
https://soundcloud.com/soumyo-biswas/hot-alkali-metal-beginning-only
@Semiclassical
^my work
 
There's this link
In mathematics, Abel's summation formula, introduced by Niels Henrik Abel, is intensively used in number theory to compute series. == Identity == Let a n {\displaystyle a_{n}\,} be a sequence of real or complex numbers and Ï• ( x ) {\displaystyle \phi (x)\,} a function of class C 1 ...
where there's a generalization
y < n <= x
so I assume it can work for y = 0
however it want to be sure
 
3:42 PM
Well, it demands that $\phi(x)$ be a C^1 function
 
so?
 
Which presumably means you'd need it to be continuous at x=0 for that to be applicable.
 
sure, of course
but assuming that would the formula work?
 
Probably? I can't really say from that article.
 
Can't listen to it now, I'll check it out later
 
3:45 PM
any reference about that?
 
Well, they cite Apostal on that Wiki page.
 
I just realized I can produce fantastic pieces of stream-of-consciousness from normally written passages by translating between various languages in Google Translate
I am going to use and abuse this
 
lol
 
Zee
^bad at keeping secrets
 
secrets? this is the brave new cut-up technique man
this needs to be publicized more
 
3:55 PM
@BalarkaSen I had wonderful works under a variety of languages, Google Translate can create a sense of a known flow path that included translation and writing
In many languages, including translation, I have a wonderful book, gives a sense of flow to Google
And one last one : Some are language and translation, and with the arrival of Google
I'd say it's pretty efficient
 
After a few iterations it gets pretty good
@BalarkaSen After some English->Bengali->Spanish->English cycles, this becomes "Secret? This brave new strategy to spend the people. This should be more"
spending people is a very good literary idea
 
Hey @BalarkaSen wanna help me understand something hopefully not-too-difficult
 
I can try.
 
So take a fiber bundle $\pi:M\to B$ with fibers $F$. I pick a horizontal complement $H$ to the vertical subbundle and call a vector field $X\in \Gamma(TM)$ basic if it is horizontal and $\pi$-related to a vector field on the base.
If I consider a path in the base I have horizontal lifts $\tilde\gamma_p$ for any $p\in F_{\gamma(0)}$
They define the diffeomorphism $\tau_\gamma:F_{\gamma(0)}\cong F_{\gamma(t)}$ via $\tau_\gamma(p)=\tilde\gamma_p(t)$.
This is a one-parameter family of diffeomorphisms I guess
I want to understand the following claim: "$\tau_\gamma$ is part of a (local) one-parameter group of diffeomorphisms associated with a basic vector field"
The obvious basic vectors to look at are the derivatives of the $\tilde\gamma_p$, but those don't give a global vector field in $M$ in any obvious way to me. Just as the vector field $\dot\gamma$ is only defined along $\gamma$, I only see how to define an associated basic vector fields on the fibers over the image of $\gamma$. Is this somehow enough?
 
4:14 PM
No, I mean, it's a local 1-parameter family of diffeomorphisms of F, right?
 
You mean $F$ viewed not as embedded inside $M$?
Ah
 
F is just the fiber. Every fiber is canonically F
 
(the terminology "basic" is sort of out of place in this context but I guess that's not a big problem)
 
I imagine being associated to a basic vector field means exactly what you think it does; it's associated to $\gamma'$. Sure it extends to all of $M$; bump it up outside a nbhd of $\gamma$ in $B$
 
Eh, no that doesn't work
For one, obviously $\gamma$ might self-intersect.
In general you can't just push forward vector fields to vector fields
But I can probably still make sense out of this working just with $F$
 
4:19 PM
No, self-intersection is a dumb issue. But I see; I was thinking of $\pi$ being the tangent bundle; what does it mean to compute "rate" in an arbitrary vector bundle? $\gamma'$ is not the right thing to look at
$\gamma'$ lives in $TB$, not $M$. I suspect you do something with the connection?
 
$\gamma'$ is what $\tilde\gamma'$ pushes down to because of the equation $\pi\circ \tilde\gamma=\gamma$
 
Then why does that not work? You get $\gamma'$ below is a section of $TM$ over $\gamma$, right? Extend that to a section around a tubular nbhd $U$ of $\gamma$/
and 0 outside $U$
I don't get it
 
I have a sequence of derivatives of injective analytic functions on a compact set that converges uniformly to the analytic function $g$. Is $g$ a derivative of an injective function?
 
@BalarkaSen $\gamma'$ is a section of the pullback bundle $\gamma^*TB$, of course. It simply does not yield a section of $TB$.
 
I am confused. I said that $\gamma'$ lives in $TB$, not $TM$. Do you agree with that or do you not?
If you do, this whole approach is garbage, because we want a section of $TM$
 
4:30 PM
Similarly, I get a family of sections $\tilde\gamma_p'\in\tilde\gamma_p^*TM$ for each $p\in F_{\gamma(0)}$ so basically "a vector field on the fibers above the image of $\gamma$"
But that doesn't extend to a vector field on all of $M$
@BalarkaSen $\gamma'$ does not live in $TB$ naturally. It lives in $\gamma^*TB$. But you can think of it as living of the part of $TB$ that lies over the image of $\gamma$, sure.
 
I am too tired to do this. I give up. I suspect thinking about it all carefully should easily tell you why a path in $B$ gives a horizontal section of $TM$
 
The fact that $\gamma$ can self-intersect already shows that it never can give a global section of $TM$.
 
Why do you insist on self-intersecting paths? Why not embedded paths?
 
It's just the most obvious case where it doesn't work
It also doesn't work in general for an embedded path
 
Yeah but it's a bad example
 
4:35 PM
but I don't know any obvious examples
Pushing forwards vector fields is almost never possible
 
what are we proving and should i just go back to slep
 
@MikeMiller Hi Mike!
 
go back to slep
 
@Danu For reference, this is where the question is stated @MikeMiller
 
I'm writing a bit about Riemannian submersions and I'm trying to understand some detail.
Here is the setup:
Oh, @Semi gave it already
 
4:36 PM
:)
 
On phone no arrows
 
Ah
 
Do you know if a path in $B$ gives a section of $TM$ over that path? if you do it's all trivial. if you don't i dunno
 
27 mins ago, by Danu
So take a fiber bundle $\pi:M\to B$ with fibers $F$. I pick a horizontal complement $H$ to the vertical subbundle and call a vector field $X\in \Gamma(TM)$ basic if it is horizontal and $\pi$-related to a vector field on the base.
 
sure
basic here means "like the base", as opposed to simple
 
4:38 PM
@BalarkaSen Over the path, of course
 
might be worth sahing
 
Sure, haha
"easy" vector field :D
 
now I want to see an 'acidic' vector field :P
 
@Danu Then why can you not use a partition of unity to get a global section? if $X$ is a section of $E/B$ along a path $\gamma$, then you can always extend that to a section of $E$. That is what I want to understand
am I understanding the picture incorrectly?
 
4:40 PM
If you have a vector field along a path it does not extend to a v.f. on a neighborhood of the path in general
 
For embedded paths, of course it does?
 
are you just trying to understand what the derivative is...?
covariant
 
No, I'm trying to parse that one sentence that I give in my setup
 
Is there an obvious example of how an embedded path can fail to extend to a v.f.?
 
The reason why is because I'm looking at these Riemannian submersions and I want to show that $\tau_\gamma$ is an isometry between the fibers.
 
4:42 PM
That might settle things.
 
@BalarkaSen I'm not sure.
 
No, a vector field along an embedded path always extends to a vector field on a tubular neighborhood of the path
Look at a union of vector bundle charts along the path
and extend it identically on the fibers nearby
This is just a get-your-hands-dirty argument
 
Actually, the reason why I was looking at that turns out to be non-existent
I actually need only "horizontal vectors", not a "basic vector field", so never mind this.
 
user image
3
but yeah, whew
I was probably not going anywhere with that
 
Sloppy proof by Besse
Sorry for bothering you
I love captain Haddock though.
 
4:52 PM
I wasn't bothered :D I am glad to not waste your time too much
Tintin is my favorite comic book series all time
 
Mine too, probably.
 
I read it a bunch as a kid since they had it at our local library.
Haven't looked at in a long while unfortunately.
 
mrkrpxzkrmtfrz
 
Which is too bad, because it's meant that the main Tintin-related thing in my headspace now is this Cracked article: cracked.com/blog/…
(not a list, amazingly, but an actual bit of fiction.)
which is hilarious in a disturbed sort of way.
 
so I was thinking about en.wikipedia.org/wiki/Cayley–Bacharach_theorem
damnit
 
5:00 PM
cool theorem
 
I hate when that happens.
 
it's a theorem that says that given 8 points in the plane there is a 9th point such that every cubic going through the first 8 also go to the 9th
 
weeeeiiird
 
and obviously, this 9th point's coordinates should be expressible as rational fractions in the coordinates of the first 8
and I " have " one of degree 71 in the numerator and 70 in the denominator if i'm not mistaken
 
So there's no family of cubics with exactly 8 common intersections.
 
5:02 PM
that's bezout's theorem
 
well two cubics have to intersect in 9 points, couting multiplicities
 
n-degree curve and m-degree curve intersects in nm many points modulo multiplicities
 
"According to Bézout's theorem, two different cubic curves over an algebraically closed field which have no common irreducible component meet in exactly nine points (counted with multiplicity). The Cayley–Bacharach theorem thus asserts that the last point of intersection of any two members in the family of curves does not move if eight intersection points (without seven co-conic ones) are already prescribed."
They mention that it serves to unite Pappus's hexagon theorem, Pascal's theorem, and the associativity of the group of elliptic curves. That's pretty neat.
 
yup
 
i think 6 points actually determine a conic?
 
5:10 PM
i think it's 5
for conics
 
yeah 5
dimension of moduli space of conics is binomial(2+2,2) - 1 = 5
 
I hate when 10 different people ask the very same question 10 times in 5 days
 
5:15 PM
Whenever I see crap like this I think of the opening to the Beatles' "Paperback Writer"
 
and eg to Semi
 
_Dear Sir or Madam, will you read my book?_
_It took me years to write, will you take a look?_
bah
@BalarkaSen Yeah. At least that person edited their post to include more info. (A pity that info doesn't include their attempt to actually put it together.)
 
True enough.
 
The first one, by contrast...yeah, they can [insert colorful invective here]
 
It annoys me that there's literally nothing except the questions and the relevant definitions though, with a "Prove it"
 
5:19 PM
Yeah.
The info they added was relevant, sure, but it doesn't actually tell us what they tried to do with it.
I forget. Is there a name for the set of elements in a group commuting with a particular element?
 
Centralizer, I think
 
Yeah, that'll do it.
 
if you pick three complex numbers $z_1,z_2,z_3$ and compute $z_1z_2\overline{z_3}+z_1\overline{z_2}z_3 + \overline{z_1}z_2z_3$ you get a complex number whose argument tells you the direction of the euler line of a triangle whose sides have the directions of $z_1,z_2,z_3$
 
That sounds like a translation of a barycentric-coordinate statement into complex number form
Maybe it should be trilinear coordinates instead. I dunno.
 
Hi. Suppose that $B_1$ is a basis for $V$. $B_2$ is also a set of independent vectors, but it's not a subset of $V$, and $V \subset span(B_2)$. I suppose that there is no name for $B_2$ in such a case.. Anyway, I am looking for some analogies of a "minimal" option and "bigger" ones.. For example, the coarsest topology and the finest topology, trivial and discrete $\sigma$-algebras.. what else? Preferably in statistical/probabilistic context
 
5:26 PM
and iirc, if you resize them so that $z_1+z_2+z_3=0$, the magnitude of the thing is proportional to the area of the triangle times GO (or whatever other distance you measure on the euler line)
wait no
it shuold be area squared
sweats uncomfortably as something is not homogeneous
area squared divided by the distance GO ?
 
"Thou shalt have sensible units."
Kinda funny how much of error-checking in physics or geometry problems can be boiled down "do the units work?"
 
yup
 
Though what's even funnier is that if you go into QFT stuff, you end up always doing units where $\hbar = c = 1$.
 
Hey guys
what does the little arc mean between two listed points? Like (-1,2)arc(1,4) ?
 
that's how i got atleast 5 extra points in my multiple choice physics test
 
5:30 PM
In which case: unit of length = unit of time = unit of 1/energy = unit of 1/momentum
 
I forgot everything about physics ;w;
 
you mean like $\arc{ab}$?
aww, I was hoping that'd work
 
Yeah.
 
$\overarc{ab}$
I give up
 
(\arc)
Wait that was dumb
$\arc$
 
5:31 PM
There doesn't seem to be an obvious Latex symbol for it, oddly.
 
weird...
$\overparen{\rm CH}$
that's certainly not it
 
I think that symbol usually denotes a circular arc
 
those are not points
 
oh $\cap$
 
and it's an intersection
of intervals
 
5:33 PM
Yeah, that's the intersection between two intervals
 
Oh, I see.
 
$[a,b]$ is the set of points $x$ such that $a\leq x\leq b$.
If one of the sides is a paren, then the endpoint isn't included.
 
Oh that makes sense.
 
So it's asking about the set of points such that $-1< x\leq 2$ and $1\leq x<4$.
If you draw these as intervals on the real line, their intersection should be evident.
 
It's not looking good for me, for getting accepted into school, unfortunately.
I'll just be a janitor I suppose.
 
5:38 PM
That sucks. You could apply again next year?
 
Yeah, I've just been waiting so long. 4 years to get into school. I used to work full-time nights at a hotel, and all night I'd do physics homework just to get it done. I know I'm sort of acting entitled to enrollment...
 
And, I was so terrible at physics that each question would often take me hours, though I found it rewarding when I did figure it out.
 
Yeah, trying to be in college is plenty of work on its own
 
Yeah really though haha.
I envy people who had their stuff together in high school.
All of my friends have graduated by now.
My cousin who is 3 years younger than me is in second year of university.
 
5:42 PM
That's how I feel w/r/t the job market, tbh
 
Oh yeah, that's tough too.
 
Blah, I hate that time of day when my brain starts doing this:
 
What's it doin'.
 
lmfao.
 
5:48 PM
lolyes
I love the Kareem Abdul Jabar bit
but then it's hard for me to find a part in that movie I don't love
 
The one where the kid recognizes him?
 
Right.
 
That's a good bit.
 
"Listen kid. I've been hearing that crap ever since I was at UCLA. I'm out there busting my buns every night. Tell your old man to drag Walton and Lanier up and down the court for 48 minutes!"
 

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