2:40 AM
Hi

2:59 AM
Hi @TedShifrin

Pretty quiet in here.

yeah

3:17 AM
yup
over 9,000 hours later...

2 hours later…
5:06 AM
Lets say these are the two curves.They have drawn common tangent connecting three points.Firstly are they really the minimia individually?I can find minimum values individually for quite somemore points.Also what does a single point in the common tangent represent?A detailed answer to this from a real anaysis point of view is certainly welcome.

5:40 AM
@user586228 No, not minima. The tangent line needs to be horizontal.

6:05 AM
@user586228 ooo, Gibbs free energy
Tho I can’t say I remember more than that
Something something phase coexistence
Is the meaning behind that common tangent

@user586228 not to mention I see 3 curves; black, blue, and red.
or is the black line a common tangent to the blue and red curves?

BothI think. But that seems pretty unusual physically
Some good discussion of the thermodynamics here: chemistry.stackexchange.com/questions/75035/…
@user586228 what system is this physically?
Boy eutectics are weird
@robjohn this context does confirm your query btw: the red and blue curves are in just the right positions that the black line has three tangent points

3 hours later…
9:01 AM
If $f:X\to Y$ is a homotopy equivalence and $A\subset X$ be a subspace such that $f|_A A\to Y$ is surjective, then $f|_A$ is a homotopy equivalence?

9:42 AM
@love_sodam $X = [0,1]$, $Y = \{\ast\}$, $A = \{0, 1\}$

10:02 AM
likely a stupid question but, if I have $v_0,...,v_m$ vectors such that there are $c_0,c_1$ nonzero with $c_0v_0 + c_1v_1 + c_2v_2 + ... + c_mv_m = 0$, why can I be sure that they do not span a subspace of dimension $\geq m$?
sorry, actually it still seems possible, the original question I wanted to ask was why is it impossible for there to be $c_0,...,c_m \geq 0, \sum c_i = 1$ and $c_0',...,c_m' \geq 0 \sum c_i' = 1$ such that $\sum c_iv_i = \sum c_i' v_i$ if we know $\{v_0,...,v_m \} \subset \mathbb{R}^n$ do not lie in an affine subspace of dimension $< m$?

3 hours later…
12:54 PM
Is $\mathbb{R}^\omega$ normal in the box topology?

I think that's still open
It's known to not be hereditarily normal and it is known to be normal under CH iirc

I saw this in Munkres once
@LeakyNun ok

Yeah I think Munkres mentions that it is open as a remark somewhere

Okay
It was published in '80s probably
Is it still open?

sciencedirect.com/science/article/pii/0016660X72900220 this is the reference for the CH result

1:03 PM
Thanks I would see it when I get time

Looks like a very short and readable paper, maybe I should read it too
matwbn.icm.edu.pl/ksiazki/fm/fm88/fm88113.pdf this is the reference for the result that it is not hereditarily normal (he builds a sequence $X_n$ of subspaces of $\Bbb R$ such that $\prod X_n$ is not normal in the box topology, and this embeds as a subspace of $\Bbb R^\omega$)

@Alessandro I have a curiosity related to set theory and GGT, do you have some time to listen? It might be nothing though.

Sure but there's no guarantee I'll have a good answer

I think you will. Anyway, a friend sent me the following axiom of choice puzzle: Suppose there are 100 rooms with a countable family of boxes indexed by $\Bbb N$ in each room, and each box contains a real number. You know that the $n$-th box in each room contains the same real number, regardless of the room.
After some strategizing, 100 mathematicians each go in the 100 rooms and do not communicate once they go in. Each of them are allowed to open countably many boxes in their room to see what real number it contains, but not all of them. Then they are asked to pick an unopened box and guess what real number it contains.

I think I've seen some variation of this before

1:15 PM
If 99 of them have the right guess, they win. Find a strategy to win.
I have a very simple reason that this is possible. It is because geodesic coalescence in trees.

unopened by any of the mathematicians ?
(Do they all have to make a guess for the same box)

Each mathematician goes in a unique room
They all operate in their own respective rooms.
Uninfluenced by the rest

Is the convolution of two normalized distributions (not normal, i.e Gaussians) also normalized?

@BalarkaSen what do you mean?

Here is my reformulation: Take an unrooted $3$-regular tree $T$ and consider it's set of ends $\partial T$, a Cantor set. Two (not 100) mathematicians are trying to guess a proper (not necessarily non-backtracking) path $\gamma$ in the tree, which is unknown to them. They are allowed to sample a proper subset of the vertices of $\gamma$. They take their separate samples, unbeknownst to each other, and then guess a vertex they have not sampled. If one of them succeeds, they win.

1:24 PM
So if $r_{n,i}$ is the number in the n-th box of the i-th room, then we have $r_{n,i} = r_{n,j}=r_n$ ? And at the end we ask each mathematician for $r_{n_i}$ for some $n_i$. Is $n_i$ the same for all mathematicians ?

According to that website, the convolution of a Cauchy-Lorentzian and a Gaussian is. But does it hold in general.

@Astyx Yes to the first question. At the end we ask the mathematician to guess $r_{m_i}$ for some $m_i$ such that $m_i$-th box in the $i$-th room is unopened by the $i$-th mathematician.

@BalarkaSen hmm I'm not convinced this is the same problem

Maybe not, but I think they're the same in spirit.

I assume they hear each other's answers?

1:27 PM
No.
No interaction. An independent 101-th person collects everyone's answers and then tells them if 99/100 of them guessed the right thing or not.

@BalarkaSen maybe, so how do you do your version?

@Alessandro By axiom of choice, we can choose a ray $\gamma(e)$ realizing the end $e$ for every end $e \in \partial T$.
Now partition the set of vertices of the unknown path $\gamma$ into two disjoint infinite subsets $A, B$. $V(\gamma) = A \sqcup B$.
$A$ has a limit end, let's call it $e_A$. Similarly, $B$ has a limit end, let's call it $e_B$. The only information the two mathematicians know is that $e_A = e_B$, some common end $e$.

@schn what does normalized mean

Consider the three sets of vertices $A, B, \gamma(e)$. Then by geometry of the tree, $\gamma(e)$ and $A$ must coalesce after some time $t_A$ and $\gamma(e)$ and $B$ must coalesce after some time $t_B$.

coalesce?

1:32 PM
If you have two paths in the tree giving rise to the same end, then they merge after some finite time.

So the strategy is as follows. The first mathematician samples $A$, second samples $B$. The first computes $t_A$, the second computes $t_B$, because $\gamma(e)$ is known to both of them.

Wasn't $\gamma$ to be guessed? How do they know $A$ and $B$?

Because they are allowed to sample a proper countable subset of vertices from $\gamma$.

1:35 PM
This is similar to allowing the mathematicians in the first problem to open countably many but not all the boxes

I thought they were sampling from the tree

Ah ok sorry for being confusing. I mean they can do so from the path.

ok but how do they guess $\gamma$ before the smallest of $t_A$ and $t_B$?

The point is, either $t_A \leq t_B$ or $t_B \leq t_A$. So if the first mathematician picks any vertex from $\gamma(e)$ beyond time $t_A$, and the second mathematician picks any vertex from $\gamma(e)$ beyond time $t_B$, one of them have to be right that the pick is indeed a vertex from $\gamma$
Because once coalescence happens, $\gamma$ and $\gamma(e)$ are really the same path.

Ah right they guess a single vertex

1:41 PM
Does that make sense?
Yeah

Hmm it's not clear to me why $\gamma=\gamma(e)$ after a while if backtracking is allowed. Can't one keep going back and forth?

Sorry, yes, I mean, $V(\gamma) \supseteq V(\gamma(e))$ after coalescence still right?

yeah
Ok I agree now

So my question is to what extent can you push this? The collection of paths representing the ends is sorting of like a geodesic combing but towards the ends? Also, can you say something with delta-hyperbolic background instead of a 0-hyperbolic background?
It doesn't feel like the axiom of choice aspect is interesting
There is geometry happening
This "coalescence time" is the Gromov inner product I think

Isn't there a unique path between two vertices? meaning you can find two distinct elements and claim one in between is in $\gamma$ ?

1:47 PM
Hm I'm not sure, this feels very choichey to me. This problems usually work that you use AC to pick representatives for some equivalence relations, and then the prisoners are allowed to do some stuff that let's them figure out in which equivalence class they are, and the relation is nice enough that someone has to guess correcrtly if they all guess the representative
Which is kind of what is happening here as well

Hm I see

I'm not seeing how do 99 correct guesses in the puzzle above though

In terms of big O notation is $\sum_{i=0}^{\log n}i=O(\log n)$?

@Astyx You're right, so maybe I don't want this exact setup.
One second

I thought you'd want the mathematicians to also guess the index of the vertex, but I'm not sure your solution works then
I'm working it out rn
Ah yeah that seems to work

1:55 PM
Oh, ok, the time parameter in the domain of $\gamma$, you mean

Yes

Yeah OK
Thanks haha

But even then, if you look at $\gamma_1$ and $\gamma_3$, you know $\gamma_2$

Yeah there's something off about the setup, I'll think about it later
@AlessandroCodenotti It doesn't seem like a problem to me because you have 100 paths with the same end instead of two, and you have the canonical one as well, and you measure the last coalescence time
Anyway, the setup needs to be fixed I'll try it later

@Eminem No, it is $O\!\left(\log(n)^2\right)$

2:11 PM
Let $X_n$ be a sequence of random variables. I've seen sources online use what I'm interpreting as the following inequality (letting $\epsilon > 0$):
$$\mathbb{P}\left(\limsup_{n \to \infty}\left\{X_n \geq \epsilon\right\}\right) \leq \mathbb{P}\left(\limsup_{n \to \infty}\left\{X_n\right\} \geq \epsilon\right)$$
Is this true in general, and also true if I change the inequality? I'm struggling to see *why* this is true if at all.

@BalarkaSen If you take $\gamma$ to be a sequence of vertices that stabilizes into a ray I think your solution works and mine doesn't

Hm yeah maybe that's what I want, a sequence converging to an end
Quasigeodesic or whatever they call them
@Clarinetist LHS is probability that $X_n \geq \varepsilon$ for all but finitely many $n$. But that definitely implies $\limsup_n X_n \geq \varepsilon$, yes?
So one event seems to be contained in the other

Great, thanks, and so that also holds if I change the inequality

As in? With $\leq \varepsilon$?

Yeah

2:21 PM
Seems right.

Well, actually, I'll have to think about that
Yeah
It does seem right

That's the morally correct inequality I think. For above, you actually get $X_n \geq \varepsilon$ for all but finitely many $n$ implies $\liminf_n X_n \geq \varepsilon$, which is stronger.

You need to stabilize into a ray to get $V(\gamma(e))\subset V(\gamma)$
Wait no now I'm convinced your argument doesn't work
Because $t_A$ or $t_B$ can be infinite

That is not true in any reasonable setup
Coalescence always happens in finite time in tree

Take $\gamma = \gamma(e)$, but every so often you make a small detour to a neighborhood

2:31 PM
OK
Not sure what your point is

Then A or B is going to capture those detours
So $t_A$ or $t_B$ is going to be infinite

What is your A and B?
I do not understand

Same as yours, a partition of $V(\gamma) = A\cup B$

@Clarinetist One tipp to handle these, since you had problems before:

what do the paths in your tree represent, and how does it match up with the boxes?

2:34 PM
Oh, I see your point, but I think this is not an issue upto "quasi-ness"

We're trying to figure that out

convert, at the least, the set lim sup to FOL

@BalarkaSen Right, i think you want to stabilize into a ray
(like a proper ray, without backtracking)

is a ray like a path that is infinite on one side?

Hm, probably, yeah. But there should be a nice and natural way to phrase that.

2:35 PM
@BigSocks yes, and that never backtracks

what is meant by backtracking?

You can't have $v_1\to v_2\to v_1$ as a subpath

oh ok like well founded
to have that it could not be a tree though, bc that's a cycle, right?
so no rays on trees could ever backtrack

By definition of a ray

@BalarkaSen LHS should be "Probability of the set of omega such that for all n there is an m>=n with X_m(omega) >= eps", so this should mean X_n>= eps infinitely often. IIRC lim inf is what you mean

2:40 PM
liminf{A_n} is "all but finitely many A_n's happen" yeah, maybe I messed it up above

I literally have to google every time which is which

@user2103480 What do you mean by FOL?

The inequality is still correct, fortunately.

first-order logic

ok so if I understand correctly the end $e$ is the number the mathematicians want to guess and the two disjoint sets of vertices $A$ and $B$ along the ray to the end $e$, $\gamma(e)$, are the other boxes with real numbers they can freely look at?

2:47 PM
they know e, they want to guess vertices of $\gamma$

@Clarinetist the probability statement is basically just a logic statement. If infinitely many X_n are >= eps, then the lim sup must be >= eps

If you only deal in rays, that's like asking for decimals of a real number you know

Oh, I see. Thanks

which is not interesting

ok then my understanding is almost completely off. the vertices of a given ray $\gamma$ give you a real number? if I were to guess how, I would think they'd be each successive digit in a representation of a real number, but you say this is boring, so how do we interpret the vertices in this ray?

2:56 PM
Hi all.

and if they both know the end $e$, what could the analogue of this be for the boxes problem? maybe I am still connecting it too much with the first problem, but I thought they were meant to be analogous. I understand if maybe the attention is no longer on that problem so much and that this one is maybe more interesting,

@Astyx Not quite because there's no preferred starting point of the ray
Any basepoint on the tree gives an isomorphism of $\partial T$ with a Cantor set
But this is non-canonical
Depends on the basepoint.
But yeah

@BalarkaSen Let $C_1, C_2, \cdots$ be distinct closed subsets of $[0,1]$ that cover $[0,1]$. Show that all except one is empty.

This is going to be some Baire thing isn't it

@LeakyNun I'm confused. $C_1=[0,1/2],C_2=[1/2,1]$?

3:07 PM
@AlessandroCodenotti sorry, disjoint not distinct
@BalarkaSen who knows :)

Then it's a theorem by Sierpinski

aha, why didn't i know that
what does it say?

That if you cover a continuum with countably many disjoint closed sets at most one is not empty
Usually people stumble into this trying to decide whether a countable set with the cofinite topology is path connected
Or whether path connected implies uncountable more generally

Just delete the interiors
Its a direct Baire corollary

Maybe you could take the first vertex of $\gamma$ as the basepoint to work around that ?

3:13 PM
also in the original problem idk how there is any way to make a consistently planned informed choice if all the boxes could be potentially random reals? so why would we think that looking at any of the vertices would help us guess any other vertex?

Oh, no, ok I was thinking of the case where $C_i$ are intervals. In that case you can throw away all the endpoints, then you have $[0, 1]$ minus a countable set as union of countably many opens

Can I cover it with disjoint $F_\sigma$ sets instead? What's the least $\xi$ such that it can be covered by countably many disjoint $\mathbf{\Pi}^0_\xi$ sets?

Is there a fixed-point-free involution of the Möbius band?
(continuous)

Can you not just do half a circle translation ?

Doing that twice flips it along the other axis

3:24 PM
Ah yeah, $F_\sigma$ sets is trivial

@LeakyNun The collection of all endpoints is a perfect set
But there are very few things such as a countable perfect set
This probably goes through for general closed sets but I haven't been thinking carefully

mathoverflow.net/a/48991/49381 yes it can work in general

ah yah this is the other variant of baire that i never remember

It's a variant that is often used
I think the usual BCT arguments in functional analysis use it for example

3:45 PM
0

Is there a fixed-point free continuous involution of the Möbius strip? The Lefschetz fixed-point theorem is useless here because the Möbius band is homotopic to a circle, which does have fixed-point free involutions. Beyond that, I don't know what tools to use. (You can't just rotate it 180 degre...

This should be low-hanging fruit
'cause I feel like there should be a simple answer but I don't see it
so (probably) easy points^

If there was, you'd get an involution of RP^2 with exactly one fixed point, that lifts to a Z/2-equivariant fixed point on S^2 with exactly two fixed points, but then you're screwed because all those are actually the linear ones
Which gives identity on the quotient
@AkivaWeinberger

@BalarkaSen when you say the boundary map is literally the boundary, is that just for manifolds or is there a more general statement.

I'm can only think of the following theorem I'm given: Say $M$ is smooth compact orientable $n$-dimensional, so it's boundaries are closed smooth compact orientable n-1-dimensional. Then under the boundary map in the LES, the fundamental class of $H_n(M, \partial M)$ gets mapped to the sum of the fundamental classes of the boundary components

(or, equivalently, gets mapped to a vector with each entry one fundamental class)

$\partial[\xi] = [\partial \xi]$
That is how the snake map behaves if you go back to the proof of the long exact sequence

I don't understand why that screws me
What do you mean the linear ones?

@AkivaWeinberger He means that every action of a finite group on $S^2$ is conjugate to an action by a finite subgroup of $O(3)$.

3:54 PM
An involution of S^2 with exactly two fixed points has to be rotation by $\pi$ about an axis, is what I mean

This is not obvious, easy, or trivial, but it is true

Interesting. How do you show that?
I see how that gives us the result

Bah

Orbifold shit man
Geometrization for 2-orbifolds

First show that every continuous group action is smoothable (this is not too different from the proof that every topological surface has a unique compatible smooth structure up to diffeo)

3:56 PM
Mike gonna do equivariant tOpOlGy now
Im gonna leave

Then as Balarka says generalize the proof of the uniformization theorem to orbifolds. The uniformization theorem says that every surface can be given constant curvature; the uniformization theorem for orbifolds says that this may be done G-invariantly when G is a finite group acting on the surface. In particular for M = S^2/G this justifies that the "universal covering action" is G acting on S^2 by isometries w/r/t the round metric, which is an action by elements of O(3)

Oh thank god

This is what I mean, it's too high-level. It's a good tool for big hard questions but this doesn't need that much machinery.

Oh no

how does this relate to RP^2 at all?

3:57 PM
Equivariant topology incoming
@LeakyNun Cap off the boundary circle

done, now what

Or quotient up the boundary

oh, does that give you RP^2?

Algebra brain
Learning RP^2 = Mobius/bd for the first time

really???

3:58 PM
Knows proof of FLT

@LeakyNun The Möbius band is a cylinder mod antipodes
aka sphere minus poles mod antipodes

oh wow that's nice

(or sphere minus open neighborhood of poles, depends if you want boundary)

@Mike when did you start your phd at UCLA?

@AkivaWeinberger Surfaces with Euler characteristic zero are torus, Klein bottle, cylinder, Mobius strip. Quotient of mobius strip by your antipodal action gives degree 2 cover of Mobius strip to a surface of Euler characteristic zero. This surface must again be a Mobius strip (non-orientable, noncompact and/or has nonempty boundary depending on your defn of mobius strip). But classification of covering spaces says there is one connected degree 2 cover of Mobius strip: the cylinder. Contradiction

4:00 PM
@BalarkaSen fckn hell this is so easy and I swear I followed this 5 times in the past without realizing that this is immediate. relative chain in C(X,A) = quotient of (singular simplex) such that quotient of (boundary of singular simplex) is 0 <=> boundary of singular simplex is singular chain in A, and unique up to homology

@AlessandroCodenotti 2014?

@user2103480 yup!

I see, you might have met my advisor then, she was a postdoc there in 2013/14 iirc (Aleksandra Kwiatkowska)

nah

@MikeMiller Oh yeah that makes sense
If you want points you can copy/paste that into the answer field

4:03 PM
I don't have an MSE account
You can write that up if you want

The answer is of course not
It should not be written

What about the Möbius band minus an even number of points
I think you can do the Möbius band minus two points
No never mind

You know all involutions of RP^2 by my argument

Right OK yeah

@BalarkaSen These don't need to compactify to involutions of RP^2.

4:08 PM
Does the end compactification not do that?

Surely you can do some equivariant crap
At the ends

Meh

Lol
It is your time to shine

covering spaces of surfaces with boundary...
These are $N_{1,k+1}$

lol i guess

4:12 PM
Sorry, what's $N_{a,b}$? $a$ RP2s minus $b$ points?

yee

In any case, the answer to "can I fix it by deleting some points" is "no"

Z/2-actions definitely compactify cuz you can just take the homeomorphism given by multiplication by 1 which has a compactification
@AkivaWeinberger A space is k (mod n) if it has a free Z/n action after deleting k points?
You're insane
But number of fixed points of a group action is an interesting question which was studied by Smith and only @MikeMiller among us knows Smith theory

If you weaken "continuous" to "tame" then I'm pretty sure that notion is exactly the combinatorial Euler characteristic
meaning the Euler characteristic of the cohomology with compact supports

How? You only get $\chi(X) - k$ is divisible by $n$
Well, for surfaces

4:23 PM
The only number such that $\chi(X)=k\pmod n$ for all $n$ is $k$

Oh, you're demanding for all n? I agree then

I'm fine with saying "there's a vague sense in which an open interval has $-1$ points" because that's what $\chi_c$ says anyway

Ok, got it
Studying Z/n actions for all n is also not far from studying an S^1 action
But I don't know if that can be said in a precise way
The number of fixed points under an S^1 action is the Euler characteristic

How does that work for $\Bbb R$

Not a clue. Smooth action of R on a smooth manifold is pretty much a vector field, or at least you can linearize it to be a vector field
Noncompact everything breaks down, but compact you have Poincare-Hopf

4:29 PM
No I mean how does this compute $\chi(\Bbb R)$

Oh. Erm
There is not really any good S^1-action on R lol

Oh I see it only works in the one direction

good S^1 action = finite number of fixed points
Yeah

Also surely "the number of fixed points" can never be negative while $\chi$ can be

Poincare-Hopf fixes this by computing the indices of the zeroes of the vector field

4:46 PM
@Thorgott with normalized I meant that the integrals of the PDFs equal one

huh? that's part of the definition of a PDF

You’re right, that was redundant. So the question was, if one takes the convolution of two PDFs is this then also a pdf (in the case of the Voigt profile it is)?

yes
all you have to check is that the total integral of the convolution is 1

5:16 PM
Is this stated in a theorem or is it just very obvious?

@BalarkaSen algebra time
direct limit of Z/n is Q/Z, which is dense in S1 = R/Z

lol

it's a one-line calculation

It's a 0-line calculation. Convolution of f and g is the pdf of X + Y where X and Y are RVs with pdf f and g.

that's 1 line

5:22 PM
It's 1 line but not 1 line of calculation
Mass at x times Mass at y sum over all pairs x + y = z is new Mass at z
Convolutions

quick mass
2

continuous, not discrete
what you're saying is still correct, but it does take 1 line of calculation to justify

Mass is mass
Just discretize
Take epsilon covering and collapse these to midpoints and all of mass over the interval to that point
Take limit
Woosh

5:39 PM
Thanks for the replies!

5:50 PM
@Thorgott $(\mu \ast \nu)(\Bbb R) = \int_{\Bbb R} \underbrace{\mu(\Bbb R - x)}_{=1} \nu(\mathrm{d}x) = 1$
whats a probability density

woke

category theorist brain

abstract away all the difficulties and any possibility to do this practically then be smug
gÖdEls IncOMplETeneEeez iS jUsT a SpeCIaL cEsÉ oF CuRtegery ThearRy

Im a lyrical miracle categorical individual
Skibididbidbididbid dap

LMAO

6:05 PM
not mine, german techno band said it first

its fire

@AkivaWeinberger Indeed, and the only $S^1$-surfaces have nonnegative Euler char

the lap lacian is a differential operator invented by lap lace

6:28 PM
you know that little "1" that's on the browser tab when 1 message has been sent
not showing up for me anymore
Can someone help me prove or disprove the existence of roots to this function? $\Phi(s)=\Gamma\left(1+\frac1s\right)+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\zeta(-ns).$
$\zeta$ is the analytically continued Riemann Zeta function
$\Gamma$ is the gamma function
$\Phi(s)$ is the analytic extension of the series $\sum_{n\ge 1} e^{-n^s}$ to the complex plane
for $\Re(s)<1$ and complex $s\ne 0$

3 hours later…
9:18 PM
3

Given that $y(x,{\bf w}) = w_0 + w_1x + w_2x^2 + \ldots + w_mx^m = \sum_{j=0}^{m} w_jx^j$ and there exists an error function defined as $E({\bf w})=\frac{1}{2} \sum_{n=1}^{N} \{y(x_n, w)-t_n\}^2$ (where $t_n$ represents the target value). I'm having trouble making sense of a passage in my textbo...

I dont understand this guy's answer at all, where do I start
I know derivatives and chain rule , But I dont know how did he come up with y(x,w+μw′) or E(w+h). I am having hard time understanding the bloody math behind Least squares in Bishop Pattern recognition book

9:38 PM
@fido9dido are you familiar with multivariable calculus?

9:53 PM
@Nope maybe I took it under different name long time ago
well I see it contain vector and matrix math which I know

then you might want to review that first

but essentially the answer consists of two parts:

1) computing the first derivative of the function w.r.t. all the w_i's, and set them all the zero, and the answer argues that the zero is unique

2) check that this really does define a minimum, which is the part about E(w+h) (which amounts to checking that the hessian matrix is positive definite)

this should remind you of single variable calculus, where to find the minimum of a function you first 1) compute the first derivative and set it to zero, 2) check that it genuinely defines a (local) minim

Ah Ok Thanks i will start there

10:45 PM
@Semiclassic: I realize from this post that I have no understanding whatsoever of Euler angles.

ugh, euler angles

I thought it was all you physicists who adore them.

mechanical engineers, maybe

Now it makes absolutely no sense to me. Seems very non-unique, to boot.

that's very likely
it's even a headache in actual applications. c.f. gimbal lock:

10:49 PM
hello

@TedShifrin In formal language, gimbal lock occurs because the map from Euler angles to rotations (topologically, from the 3-torus T^3 to the real projective space RP^3 which is the same as the space of 3d rotations SO(3)) is not a local homeomorphism at every point, and thus at some points the rank (degrees of freedom) must drop below 3, at which point gimbal lock occurs.

Euler angles provide a means for giving a numerical description of any rotation in three-dimensional space using three numbers, but not only is this description not unique, but there are some points where not every chang

can any of you guys endorse for combinatorics in arxiv?

Yes, I was basically observing some of that ad hoc when I tried to understand the guy's question.

from the Wikipedia article on gimbal lock

Love the topology thrown in :)

10:51 PM
ya
i really wish I could vote "????" on some questions

Mostly these days I just want to close down "here's my exam question" posts.