text:
\subsection{Proof of Lemma \ref{lemma:bound-h0t}}
\label{sec:proof-lemma-bound-h0t}

Recalling that $\bm{0} \leq \sum_{i=1}^t \prod_{j=i+1}^{t} ( \bm{I} -\bLam_j)\bLam_i \one \leq \one$ (cf.~\eqref{eqn:cat}), we obtain

\begin{align}
	\| \bm{h}_{0,t}\|_{\infty}
	&\leq \gamma \Big\|\sum_{i=1}^{t}\prod_{j=i+1}^{t}\big(\bm{I}-\bm{\Lambda}_{j}\big)\bm{\Lambda}_{i}\Big\|_{1}
	 \big\|(\widetilde{\bm{P}}-\bm{P}\big)\overline{\bm{V}}\big\|_{\infty}

	\leq \gamma \big\|(\widetilde{\bm{P}}-\bm{P}\big)\overline{\bm{V}}\big\|_{\infty}.
	\label{eq:H0t-simple-bound}

\end{align}

As a result, it remains  to upper bound $\big\|(\widetilde{\bm{P}}-\bm{P}\big)\overline{\bm{V}}\big\|_{\infty}$.

Suppose that $\widetilde{\bm{P}}$ is constructed
using $N$ consecutive sample transitions. Without loss of generality,
assume that these $N$ sample transitions are the transitions between
the following $N+1$ samples
\[
(s_{0},a_{0}),(s_{1},a_{1}),(s_{2},a_{2}),\cdots,(s_{N},a_{N}).
\]
Then the $(s,a)$-th row of $\widetilde{\bm{P}}$ --- denoted by
$\widetilde{\bm{P}}(s,a)$ --- is given by

\begin{align}
	\widetilde{\bm{P}}(s,a)=\frac{1}{K_{N}(s,a)}\sum_{i=0}^{N-1} \bm{P}_{i+1}(s,a) \overline{\bm{V}}\ind\{(s_{i},a_{i})=(s,a)\}=\frac{1}{K_{N}(s,a)}\sum_{i=1}^{K_{N}(s,a)} \bm{P}_{t_{i}+1}(s,a) \overline{\bm{V}}	,
\end{align}

where $\bm{P}_i$ is defined in \eqref{eq:defn-Pt}, and $\bm{P}_i(s,a)$ denotes its $(s,a)$-th row. Here, $K_N(s,a)$ denotes the total number of visits to $(s,a)$ during the first $N$ time instances (cf. \eqref{eq:defn-Kt}), and
$t_k:= t_k(s,a)$ denotes the time stamp when the trajectory visits ($s,a$) for the $k$-th time (cf \eqref{eq:defn-tk-sa}).

In view of our derivation for \eqref{eq:independence-Markov-chain}, the state transitions happening at time $t_1,t_2,\cdots,t_k$ are independent for any given integer $k>0$. This together with the Hoeffding inequality implies that

\begin{equation}
	\mathbb{P}\left\{ \frac{1}{k} \left|\sum_{i=1}^{k} \big(\bm{P}_{t_{i}+1}(s,a)-\bm{P}(s,a) \big)\overline{\bm{V}}\right|\geq\tau \right\} \leq2\exp\left\{ -\frac{k\tau^{2}}{2\|\overline{\bm{V}}\|_{\infty}^{2}}\right\} .
	\label{eq:Hoeffding-Ptilde}
\end{equation}

Consequently, with probability at least $1-\frac{\delta}{|\mathcal{S}||\mathcal{A}|}$
one has
\[
	\left| \frac{1}{k} \sum_{i=1}^{k} \big(\bm{P}_{t_{i}+1}(s,a)-\bm{P}(s,a)\big)\overline{\bm{V}}\right|\leq\sqrt{\frac{2\log\big( \frac{2N|\mathcal{S}||\mathcal{A}|}{\delta} \big)}{k}}\big\|\overline{\bm{V}}\big\|_{\infty},\qquad1\leq k\leq N.
\]

Recognizing the simple bound $K_{N}(s,a) \leq N$, the above inequality holds for each state-action pair $(s,a)$ when $k$ is replaced by $K_{N}(s,a)$. Conditioning on these $K_{N}(s,a)$, applying the union bound over all $(s,a)\in \cS\times \cA$, we obtain

\begin{align}
	\big\|(\widetilde{\bm{P}}-\bm{P})\overline{\bm{V}}\big\|_{\infty}
	\leq
	\max_{(s,a)\in \cS\times \cA} \sqrt{\frac{2\log\big( \frac{2N|\mathcal{S}||\mathcal{A}|}{\delta} \big)}{K_N(s,a)}}\big\|\overline{\bm{V}}\big\|_{\infty}
	\label{eq:Hoeffding-Ptilde2}
\end{align}

with probability at least $1-\delta$.

	In addition, for any $N \geq \tcov$, Lemma~\ref{lemma:Bernstein-state-occupancy} guarantees that with probability $1-2\delta$, each state-action pair $(s,a)$ is visited at least $N\mumin/2$ times, namely, $K_N(s,a) \geq \frac{1}{2} N \mumin $ for all $(s,a)$. This combined with \eqref{eq:Hoeffding-Ptilde2} yields

\begin{align}
	\big\|(\widetilde{\bm{P}}-\bm{P})\overline{\bm{V}}\big\|_{\infty}
	&\leq
	\sqrt{\frac{4\log\big( \frac{2N|\mathcal{S}||\mathcal{A}|}{\delta} \big)}{N\mumin}}\big\|\overline{\bm{V}}\big\|_{\infty} \nonumber \\
	& \leq \sqrt{\frac{4\log\big( \frac{2N|\mathcal{S}||\mathcal{A}|}{\delta} \big)}{N\mumin}}\left( \big\|\overline{\bm{V}} - \bm{V}^{\star} \big\|_{\infty}  + \big\|  \bm{V}^{\star} \big\|_{\infty}  \right) \nonumber \\
& \leq \sqrt{\frac{4\log\big( \frac{2N|\mathcal{S}||\mathcal{A}|}{\delta} \big)}{N\mumin}}  \big\|\overline{\bm{V}} - \bm{V}^{\star} \big\|_{\infty}  + \frac{1}{1-\gamma}\sqrt{\frac{4\log\big( \frac{2N|\mathcal{S}||\mathcal{A}|}{\delta} \big)}{N\mumin}}\label{eq:Hoeffding-Ptilde2}
\end{align}

with probability at least $1-3\delta$, where the second inequality follows from the triangle inequality, and the last inequality follows from $\big\|  \bm{V}^{\star} \big\|_{\infty}\leq \frac{1}{1-\gamma}$. Putting this together with \eqref{eq:H0t-simple-bound} concludes the proof.


TITLE: Equality Constraint Optimizations
QUESTION [1 upvotes]: I have an optimization problem of this sort
$min_{(x_1, x_2) } 4x^2_1 + x^2_2$
subject to  $x_1 -2x_2 + 5 = 0$
I am trying to solve for the optimal points using the first and second derivative tests by making $x_1$ function of $x_2$.
I have that $x_1 = 2x_2 +5$ and the minimization problem is of the form $17x_2^2 - 80x_2 +100$
From this point I'm guessing that first derivative of $17x_2^2 - 80x_2 +100$ =0. Hence $x_2 = 80/34$ I'm not sure if this is correct or how to do the second derivative test.

REPLY [1 votes]: Another way, without derivative -
$x_1 = 2x_2 - 5$
So, $4x_1^2+x_2^2 = 4(4x_2^2-20x_2+25)+x_2^2 = 17x_2^2-80x_2+100$
$ = 17(x_2^2 - \frac{80}{17}x_2+\frac{100}{17}) = 17[(x_2-\frac{40}{17})^2 + \frac{100}{17^2}$.
Min of square will be zero so min value of $4x_1^2+x_2^2$ is $\frac{100}{17}$ when $x_2 = \frac{40}{17}$. Can also get the value of $x_1$.


TITLE: Show that $\mathfrak{o}(2l+1)$ coincides with its derived algebra
QUESTION [1 upvotes]: Let $\mathfrak{o}(2l+1)$ the (odd) orthogonal Lie algebra defined as the Lie algebra of matrices $x$ such that $sx+x's=0$ ($x'$ is the transpose), where
$s=\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & I_l\\
0 & I_l & 0
\end{pmatrix}.
$
I want to show that $\mathfrak{o}(2l+1)\subseteq [\mathfrak{o}(2l+1),\mathfrak{o}(2l+1)]$.
From the definition it is not hard to find a basis of $\mathfrak{o}(2l+1)$ as vector space, so one strategy is to compute the brackets of the elements of the basis in order to find that all of them can be expressed in terms of brackets of the basis. This is not conceptually hard but it's certainly tedious, so my question is the following:

Is there a more conceptual way to show $\mathfrak{o}(2l+1)\subseteq [\mathfrak{o}(2l+1),\mathfrak{o}(2l+1)]$ using the implicit definition of the algebra?

REPLY [2 votes]: It is sufficient to notice that this algebra is semisimple, which may be verified by checking that its Killing form is nondegenerate. Of course this requires some computation also. Another route is to find a Cartan subalgebra and the corresponding root system. This is done in many textbooks in Lie theory. I would be very happy to learn about a way to get the desired result with no computations at all, but I'm affraid this may be impossible.
Another way to show that the algebra is semisimple would be to construct any nondegenerate invariant bilinear form - and it is obvious that the Hilbert-Schmidt form does the job. Then you know that it is reductive, so it remains to show that the center is trivial.


TITLE: What is the third (matroid) flat axiom?
QUESTION [2 upvotes]: Given: a finite set $E$ and a set family $\mathscr{F} \subseteq 2^E = \mathscr{P}(E)$.
Every source known to me gives these as the first two  axioms for $\mathscr{F}$ to be a family of matroid flats:

$E \in \mathscr{F}$.
$F_1, F_2 \in \mathscr{F} \implies F_1 \cap F_2 \in \mathscr{F}$.

Question: (a) What is the correct form for the third axiom? Can it be stated precisely?
(b) What is the conclusion of the third axiom when applied to $E \in \mathscr{F}$?

E.g. how should one understand "the empty set partitions the empty set"? Isn't the empty union equal to $\emptyset$, not to $E$? And if the third axiom makes no sense when applied to $E$, and $E \in \mathscr{F}$ is only included so that $\mathscr{F}$ isn't empty, then why not say $E \not\in \mathscr{F}, \mathscr{F} \not= \emptyset \subseteq 2^E$?

Wikipedia says: 3. If $F \in \mathscr{F}$, then the set of all $G \in \mathscr{F}$ which cover $F$ (meaning that $G \supsetneq F$ but there is no $H \in \mathscr{F}$ between $F$ and $G$) partition the elements of $E \setminus F$.
Encyclopedia of Mathematics gives two supposedly equivalent forms:
3.(a) If $G_1, \dots, G_k \in \mathscr{F}$ is the family of flats which cover $F \in \mathscr{F}$ (i.e. each $G_i$ contains $F$ properly with no $H \in \mathscr{F}$ in between), then $G_1 \setminus F, \dots, G_k \setminus F$ partition $E \setminus F$.
3.(b) For all $F \in \mathscr{F}$, $\bigcup \{ F' \in \mathscr{F}: F' \succ F  \} = E$.
Oxley says: If $F \in \mathscr{F}$ and $\{ G_1, G_2, \dots, G_k\}$ is the set of minimal members of $\mathscr{F}$ that properly contain $F$, then the sets $G_1 \setminus F, G_2 \setminus F, \dots, G_k \setminus F$ partition $E \setminus F$. These notes parrot this.
Gordon, McNulty say the same thing as the first axiom given by the Encyclopedia of Mathematics, except that they don't even bother to try to explain what "cover" means.

Attempt: Are either of these definitions of the "flat cover" correct?
(a) $cover: \mathscr{F} \to 2^{2^E}$, $cover: F \mapsto \{G \in \mathscr{F}: G \supsetneq F, \forall H \in \mathscr{F}, F \subseteq H \subseteq G \implies H = F \lor H = G  \}$
(b) $cover: \mathscr{F} \to 2^{2^E}$, $cover: F \mapsto \{G \in \mathscr{F}: G \supsetneq F, \forall X \in 2^E, F \subsetneq X \subsetneq G \implies X \not\in \mathscr{F} \}$
EDIT: What I had here before was totally wrong (among other things, it should be $G \setminus F$, not $E \setminus G$). Also the definitions of cover incorrectly said $G \supseteq F$ instead of $G \supsetneq F$.
Then define $[cover(F) \setminus F] \in 2^{2^E}$ by $[cover(F) \setminus F] := \{ G \setminus F: G \in cover(F) \}$.
My understanding of "partition" is the following:
$[cover(F) \setminus F]$ partitions $E \setminus F$ $\iff$ $\bigsqcup\limits_{G \in cover(F)} G \setminus F = E \setminus F$, where $\bigsqcup$ denotes disjoint union.
EDIT: What follows is also totally incorrect and based on a false premise. We're interested in $G \setminus E$ (which does not exist because $cover(E) = \emptyset$), not in $E \setminus E$, which does exist but is the empty set. I.e. using the right definition $cover(F) \setminus F$ it is an empty union, using the wrong definition $E \setminus cover(F)$ it is a non-empty union of empty sets.
When $F = E$, we should have that $E \setminus F = \emptyset$, so $$ \bigsqcup\limits_{X \in cover(F)} E \setminus X = \emptyset \,,$$ thus this is supposed to be either an empty union, or a non-empty union of empty set(s).
If it is supposed to be an empty union, then we should have that $cover(E) = \emptyset$. If it is supposed to be a non-empty union of empty set(s), we should have that $cover(E) = \{ E \}$.
In particular, should we have that $cover(E) = \{ E \}$ or $cover(E) = \emptyset$?
Using definition (a), $cover(E) = \{E\}$, (b) is supposed to be equivalent to (a), so to get $cover(E) = \emptyset$ one needs to change $H = F \lor H = G$ in the definition of (a) to $H = G$ or something else entirely.

REPLY [1 votes]: Neither of your definitions of "cover" is correct, because you seem to have forgotten about the requirement that $G$ strictly contains $F$.  If $F,G\in\mathscr{F}$, then $G$ covers $F$ if $F\subsetneq G$ and there does not exist any $H\in\mathscr{F}$ such that $F\subsetneq H\subsetneq G$.
The third axiom can then be stated as follows.  For any $F\in\mathscr{F}$ and any $x\in E\setminus F$, there exists a unique $G\in\mathscr{F}$ such that $G$ covers $F$ and $x\in G$.  In other words, every element of $E\setminus F$ is in $G\setminus F$ for a unique $G$ which covers $F$, so the collection of such sets $G\setminus F$ is a partition of $E\setminus F$.
In the case $F=E$, there is nothing special going on.  There do not exist any $x\in E\setminus F$, so the requirement is vacuous.  There are also does not exist any $G\in\mathscr{F}$ which covers $F$ (since there is no element of $\mathscr{F}$ which properly contains $F$), but that is fine, since the empty set is a partition (the only partition!) of the empty set.

Note that actually the uniqueness requirement here can be omitted.  If $G$ and $H$ both cover $F$ and $x\in G,H$, then $G\cap H\in\mathscr{F}$ by the second axiom.  We have $F\subsetneq G\cap H \subset G$ (the first inclusion is strict since $x\in G\cap H$), so $G\cap H=G$ since $G$ covers $F$.  Similarly, $G\cap H=H$, so $G=H$.
So it suffices to assume the union of all the elements of $\mathscr{F}$ which cover $F$ contains all of $E\setminus F$.  This is almost the same as requiring that the union of all the elements of $\mathscr{F}$ which cover $F$ is $E$ itself, which is what the second definition in the Encyclopedia of Mathematics says.  Indeed, as long as there exists some element which covers $F$, then the union always contains $F$, so all that matters is whether it contains $E\setminus F$.  However, when $F=E$, there is nothing that covers $F$.  So the second definition in the Encyclopedia of Mathematics is incorrect and should be modified to only apply when $F\neq E$.


TITLE: Choosing a surface that makes the flux of F maximal,
QUESTION [2 upvotes]: For a closed surface S in $R^3$, consider the flux of F, given by the usual flux integral.  For what choice of S will the flux be maximal?
So, I want to apply the divergence theorem and instead look at the triple integral of the divergence of the given vector field F (over a solid that is enclosed by S.)
Since the flux integral = the divergence integral, I can aim to maximize the divergence integral.
My computation of the divergence gives me -3($x^2 + y^2 + z^2 - \frac{5}{3}$), so this would be the integrand in the triple integral.
It sounds reasonable that in order to maximize this triple integral, I should maximize the integrand.  How do I do that?
My attempt was to make $x^2 + y^2 + z^2$  - $\frac{5}{3}$ < 0, since there's a factor of -3 to consider.  negative * negative will give me a positive integrand - which would be good for maximizing the flux of F.
Then this tells me that I should choose a sphere of radius $\sqrt{(\frac{5}{3})}$.
I carried out my work and it looks like I got the correct answer.
But I feel like I chose my sphere ...by luck.
How do I know for sure that I've maximized the integrand, simply by making $x^2 + y^2 + z^2$  - $\frac{5}{3}$ < 0?  Could I have done even better, achieving a better maximum?  I simply knew that this factor had to be < 0, since it was being multiplied by -3.
Thanks,

REPLY [3 votes]: There is a very simple underlying principle. You choose the surface $S$ so that the region $V$ it bounds is precisely all the points with $\text{div}\,\vec F \ge 0$. Doing so will give you the greatest possible value for the integral $\iiint_V \text{div}\,\vec F\,dV$. (If you do not enclose all of those points, you get a smaller integral, and if you enclose a region with points where $\text{div}\,\vec F<0$, this makes the integral smaller as well.)
This problem has a solution precisely when the region $\{(x,y,z): \text{div}\,\vec F \ge 0\}$ is a (closed) bounded region. If the region is unbounded, there is no optimal surface $S$.


TITLE: Verifying the Markov property
QUESTION [3 upvotes]: We throw a dice infinitely often. Define $U_n$ to be the maximal number shown up to time $n$. How can I verify that
    $$
\mathbb{P}(U_{n+1}=u_{n+1}|U_n=u_n,\ldots,U_1=u_1)=\mathbb{P}(U_{n+1}=u_{n+1}|U_n=u_n),
$$
    i.e. that the Markov property is fullfilled?

I have no idea, I only can reproduce the definition:
$$
\mathbb{P}(U_{n+1}=u_{n+1}|U_n=u_n,\ldots,U_1=u_1)=\frac{\mathbb{P}(U_{n+1}=u_{n+1},U_n=u_n,...,U_1=u_1)}{\mathbb{P}(U_n=u_n,...,U_1=u_1)}.
$$
But I do not know how to show the Markov property now.-
Hope you can help me.
Greetings

REPLY [2 votes]: This is essentially asking "Prove that the probability of the $n+1^{th}$ event depends only on the outcome of the $n^{th}$ event, and not on the ones prior to that." In this context, it is clear that the probability of $U_{n+1}$ being $u_{n+1}$ definitely depends on what $U_n$ was - clearly, $U_{n+1}\geq U_n$, however, knowing what $U_{n-1}$ was provides no new information (or knowing $U_{i}$ for any $i<n$, to be more general).
To prove this, you can simply write out the distribution of $U_{n+1}$ given the past history - splitting into three cases. For $u_{n+1}<u_n$, which can never happen since the maximum roll always increases, we get,
$$P(U_{n+1}=u_{n+1}|U_n=u_n,\ldots,U_1=u_1)=0$$
Then, for $u_{n+1}=u_n$, where the new die roll is equal to or less than some previous one, and hence could be any value less than or equal to $u_n$, we get
$$P(U_{n+1}=u_{n+1}|U_n=u_n,\ldots,U_1=u_1)=\frac{u_n}{6}$$
And for $u_{n+1}>u_n$, where the new die roll is the new record, we have simply the probability of that roll occurring
$$P(U_{n+1}=u_{n+1}|U_n=u_n,\ldots,U_1=u_1)=\frac{1}{6}.$$
However, notice that the left side of these equations only uses the value of $u_n$ - thus, the probability of $P(U_{n+1}=u_{n+1}|U_n=u_n)$ would be equal to the above, since it would give the exact same expression; we gained no new information by knowing the entire history of $U_i$, and hence the Markov property holds.


TITLE: Find the differential (if exist) of the function $h(\vec{x}) = \frac{f^3(\vec{x})+f(\vec{x})g^2(\vec{x})}{f^2(\vec{x})+g(\vec{x})}.$
QUESTION [0 upvotes]: Let $A\subset\mathbb{R}^n$  a noempty open set and $\vec{x}_0\in A$. Let $f,g:A\to\mathbb{R}$ two differentiable function in $\vec{x}_0$ so that, $g(\vec{x})>0$, $\forall \vec{x}\in A$. Consider the function
$$h:A\to\mathbb{R}\ ,\ \vec{x}\mapsto h(\vec{x}) = \frac{f^3(\vec{x})+f(\vec{x})g^2(\vec{x})}{f^2(\vec{x})+g(\vec{x})}.$$
Show that $h$ is differentiable in $\vec{x}_0$ and calcule   $Df(\vec{x}_0)$.

REPLY [0 votes]: We can write that $h\left(\vec{x}\right)=\psi\circ\phi\left(\vec{x}\right)$ with
$$\begin{array}{ccccc}
\phi & : & \mathbb{R}^n & \longrightarrow & \mathbb{R}^2\\
& & \vec{x} & \longmapsto & \left(f(\vec{x})^3+f(\vec{x})g(\vec{x})^2,f(\vec{x})^2+g(\vec{x})\right)
\end{array}$$
and
$$\begin{array}{ccccc}
\psi & : & \mathbb{R}^2\setminus\left(\mathbb{R}\times\{0\}\right) & \longrightarrow & \mathbb{R}\\
& & \left(u,v\right) & \longmapsto & \frac{u}{v}
\end{array}.$$
This is consistent because $f(\vec{x})^2+g(\vec{x})>0$. Now we have
$$\mathrm{d}\phi\left(\vec{x}_0\right)[\vec{y}]=\left(3f\left(\vec{x}_0\right)^2\mathrm{d}f\left(\vec{x}_0\right)[\vec{y}]+\mathrm{d}f\left(\vec{x}_0\right)[\vec{y}]g\left(\vec{x}_0\right)^2+2f\left(\vec{x}_0\right)g\left(\vec{x}_0\right)\mathrm{d}g\left(\vec{x}_0\right)[\vec{y}],2f\left(\vec{x}_0\right)\mathrm{d}f\left(\vec{x}_0\right)[\vec{y}]+\mathrm{d}g\left(\vec{x}_0\right)[\vec{y}]\right)\quad\quad\quad\forall\vec{y}\in\mathbb{R}^n\quad\quad\quad(1)$$
and
$$\mathrm{d}\psi\left(u_0,v_0\right)[h,k]=\frac{h}{v_0}-\frac{u_0k}{v_0^2}\quad\quad\quad\forall\left(h,k\right)\in\mathbb{R}^2\quad\quad\quad(2).$$
Hence, $h$ is differentiable at the point $\vec{x}_0$ and its differential is given by the well known formula
$$\mathrm{d}h\left(\vec{x}_0\right)[\vec{y}]=\mathrm{d}\psi\left(\phi\left(\vec{x}_0\right)\right)\circ\mathrm{d}\phi\left(\vec{x}_0\right)[\vec{y}]\quad\quad\quad\forall\vec{y}\in\mathbb{R}^n$$
that is, $\left(u_0,v_0\right)$ in $(2)$ are replaced by $\left(f(\vec{x}_0)^3+f(\vec{x}_0)g(\vec{x}_0)^2,f(\vec{x}_0)^2+g(\vec{x}_0)\right)$ and $\left(h,k\right)$ in the same formula $(2)$ becomes the (long) vector of $\mathbb{R}^2$ in $(1)$.


TITLE: How to give a combinatorial proof for this forumula
QUESTION [2 upvotes]: I need to give a combinatorial argument that $$S(n,m) = \sum_{i =0} ^{n-1} {n -1 \choose i} S(i,m-1)$$
Where $S(n,m)$ is the Stirling numbers of the second kind.
Here is my attempt.
Well first thing to know here is that $S(n,m)$ is the number of ways to distribute $n$ distinct among $m$ non empty identical containers.
So If by somehow I managed to show that the right hand side of the above equation mean the same thing then I am done right !
Now for $ 0 \leq i \leq n-1$, let $i$ count the number of distinct objects that we actually distribute (The number of objects that got distributed from $n-1$ objects). This number of distinct objects can be chosen in ${n-1 \choose i}$ ways, Now the number of ways that we can distribute those $i$ objects among $m-1$ non empty identical container is ${n-1 \choose i} S(i,m-1)$.
But if we kept doing that we will get $S(n-1,m-1)$ and not $S(n,m)$.
I feel like I am missing something here, and something is wrong with my reasoning. How do I proceed with this proof or re correct it.

REPLY [3 votes]: Say that the objects being distributed are the integers $1,2,\ldots,n$. Let $i$ be the number of objects that are not in the same container as $n$. Then $0\le i\le n-1$, there are $\binom{n-1}i$ ways to choose those $i$ objects, and there are $S(i,m-1)$ ways to distribute them amongst the remaining $m-1$ containers.


text:
\section{Indexed Summation of Sum of Mappings}
Tags: Summations

\begin{theorem}
Let $\mathbb A$ be one of the [[Definition:Standard Number System|standard number systems]] $\N, \Z, \Q, \R, \C$.
Let $a, b$ be [[Definition:Integer|integers]].
Let $\closedint a b$ denote the [[Definition:Integer Interval|integer interval]] between $a$ and $b$.
Let $f, g: \closedint a b \to \mathbb A$ be [[Definition:Mapping|mappings]].
Let $h = f + g$ be their [[Definition:Pointwise Addition|pointwise sum]].
Then we have the equality of [[Definition:Indexed Summation|indexed summations]]:
:$\displaystyle \sum_{i \mathop = a}^b \map h i = \sum_{i \mathop = a}^b \map f i + \sum_{i \mathop = a}^b \map g i$
\end{theorem}

\begin{proof}
The proof proceeds by [[Principle of Mathematical Induction|induction]] on $b$.
For all $b \in \Z_{\ge 0}$, let $\map P b$ be the [[Definition:Proposition|proposition]]:
:$\displaystyle \sum_{i \mathop = a}^b \map h i = \sum_{i \mathop = a}^b \map f i + \sum_{i \mathop = a}^b \map g i$
=== Basis for the Induction ===
Let $b < a$.
Then all [[Definition:Indexed Summation|indexed summations]] are [[Definition:Zero (Number)|zero]].
Because $0 = 0 + 0$, the result follows.
=== Basis for the Induction ===
Let $b = a$.
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = a}^b \map h i
      | r = \sum_{i \mathop = a}^a \map h i
      | c =
}}
{{eqn | r = \map h a
      | c =
}}
{{eqn | r = \map f a + \map g a
      | c = {{Defof|Pointwise Addition}}
}}
{{eqn | r = \sum_{i \mathop = a}^a \map f i + \sum_{i \mathop = a}^a \map g i
      | c =
}}
{{eqn | r = \sum_{i \mathop = a}^b \map f i + \sum_{i \mathop = a}^b \map g i
      | c = as $a = b$
}}
{{end-eqn}}
Thus $\map P a$ is seen to hold.
This is our [[Definition:Basis for the Induction|basis for the induction]].
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge a$, then it logically follows that $\map P {k + 1}$ is true.
So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
:$\displaystyle \sum_{i \mathop = a}^k \map h i = \sum_{i \mathop = a}^k \map f i + \sum_{i \mathop = a}^k \map g i$
from which it is to be shown that:
:$\displaystyle \sum_{i \mathop = a}^{k + 1} \map h i = \sum_{i \mathop = a}^{k + 1} \map f i + \sum_{i \mathop = a}^{k + 1} \map g i$
=== Induction Step ===
We have:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = a}^{k + 1} \map h i
      | r = \sum_{i \mathop = a}^k \map h i + \map h {k + 1}
      | c = {{Defof|Indexed Summation}}
}}
{{eqn | r = \sum_{i \mathop = a}^k \map h i + \map f {k + 1} + \map g {k + 1}
      | c = {{Defof|Pointwise Addition}}
}}
{{eqn | r = \sum_{i \mathop = a}^k \map f i + \sum_{i \mathop = a}^k \map g i + \map f {k + 1} + \map g {k + 1}
      | c = [[Indexed Summation of Sum of Mappings#Induction Hypothesis|Induction Hypothesis]]
}}
{{eqn | r = \sum_{i \mathop = a}^k \map f i + \map f {k + 1} + \sum_{i \mathop = a}^k \map g i + \map g {k + 1}
      | c = [[Commutative Law of Addition]]
}}
{{eqn | r = \sum_{i \mathop = a}^{k + 1} \map f i + \sum_{i \mathop = a}^{k + 1} \map g i
      | c = {{Defof|Indexed Summation}}
}}
{{end-eqn}}
So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
{{qed}}
\end{proof}

config: wiki


TITLE: Problems with any non-Kolmogorovian (frequentistic, subjective etc.) approaches to probability
QUESTION [4 upvotes]: When introducing Kolmogorov's axiomatic approach probability, it is often claimed that this is are a way out of the problems associated with the following two interpretations of probability:
• frequentistic approach: probability is the limit of relative frequence of an event occuring in long strings of repetitions of an experiment
• subjective approach: the (hypothetical) monetary value that I would put on a bet that an event occurs
My question is: What exactly are the problems, associated to these
interpretations?
[Careful! Long sentence ahead!] Are they merely philosophical (e.g. in case of the
frequentistic approach: we can't associated a probability to all events,
that we would like to associated probabilities to, since not all events
can formulated within repeatable experiments; e.g. to estimate probability
of a politician being elected, we can hold 1000 times an election,
to approximate, as the fraction of those 1000 times in which he was
elected, the probability of election) or are there ``hard'' mathematical
obstacles that arise when one formalizes these other approaches and
states them axiomatically (e.g. in case of the frequentistic approach:
we assume that $(\Omega,\mathcal{F})$ is a measurable space, let
$A\in\mathcal{F}$ be arbitrary and consider for each $k\in\mathbb{N}$
a finite sequence $(B_{i}^{k})_{i\leq k}\in\mathcal{F}^{k}$ such
that $\lim_{k\rightarrow\infty}\frac{N_{k}((B_{i}^{k})_{i\leq k})}{k}$
exist (which then necessarily lies within $[0,1]$), where $N_{k}$$((B_{i}^{k})_{i\leq k}):=|\{i:B_{i}^{k}=A\}|$;
the value of this limit is then called probability of $A$. For the
subjective approach I don't know how to formalize this)?

REPLY [3 votes]: You can easily dispose of random variables for elementary probability (on finite sample spaces), just look at Feller’s volume 1 which does not formally introduce them until over halfway through the book. However, in the theory of stochastic processes, especially in continuous time, the concept of random variable is indispensable.
In short the first half of this post addresses the fact that the elementary approach fails (somewhat dramatically!) when combinatorics no longer applies. The second half recounts a passage of Cramer essentially commenting that while other axiomatic approaches to probability theory are certainly valuable, none seem to be as amenable to actually doing mathematics as Kolmogorov's.

Allow me to try to digest the introduction I referenced as well as provide some quoted passages from H. Cramer's book on distributions:
First on the introductory chapter of Probability with Martingales. As often in probability models, we have some event $A$ we are interested in [in the chapter its the event of extinction of the Branching process], and a sequence of events $A_n$ that are associated with $A$ somehow. When, how, and why can we say $\mathbb{P}(A)=\lim_n \mathbb{P}(A_n)$? In the specific case $A_n=\{Z_n =0\}$, the event of extinction in the $n$-th generation and $A=\{Z_m = 0 \text{ for some } m\}$ the event of extinction, ever, and $\pi_n := \mathbb{P}(Z_n=0)$ and $\pi:=\mathbb{P}(Z_m = 0 \text{ for some } m)$. It is intuitive to posit that this limit holds in this case, but how do we prove it? And even if we do prove it, what if I have a different model now, or some slight variation?
In the elementary theory we can identify the sample space with the set of all outcomes without issue and combinatorics greatly aids in our work. Now recall for this branching process, for $n\in \mathbb{Z^+}$ and $r\in \mathbb{N}$, the RV $X_r^{(n+1)}$ represent the number of children in the $(n+1)$-st generation of the $r$-th animal from the $n$-th generation (if there is one). Then the total number in the $(n+1)$-th generation is given by
$$Z_{n+1}=X_1^{(n+1)}+\dotsc +X_{Z_n}^{(n+1)}$$
The $X$ are non-negative integer valued RVs and the doubly-infinite sequence $X_r^{(s)}$ for $r,s\in \mathbb{N}$ is assumed to be IID. For a single copy of $X$ we might just as well say $\Omega = \{0,1,2,\dotsc\}$. But for this experiment, as long as we want to compute $\pi :=\mathbb{P}(Z_m = 0 \text{ for some } m)$, we need to know the results of the entire sequence $(X_s^{(r)}: r,s \in \mathbb{N})$ as a single outcome of our experiment. We aren't just interested in one generation or one animal, so our sample space needs to reflect that. Thus following the author now,

We could follow elementary theory in taking $\Omega$ to be the set of all outcomes, in other words, taking $\Omega$ to be the Cartesian product
$$ \Omega = \prod_{r,s \in \mathbb{N}} \mathbb{Z}^+,$$
the typical element of $\Omega$ being
$$\omega = (\omega_s^{(r)} : r,s \in \mathbb{N}),$$
[a doubly-indexed sequence of non-negative integers]
and then setting $X_s^{(r)}(\omega)=\omega_s^{(r)}$. Now $\Omega$ is an uncountable set, so that we are outside the "combinatorial" context which makes sense of $\pi_n$, in the elementary theory. Moreover, if one assumes the Axiom of Choice, one can prove that it is impossible to assign to all subsets of $\Omega$ a probability satisfying the "intuitively obvious" axioms and making the $X$'s IIDRVs with the correct common distribution. So, we have to know that the set of $\omega$ corresponding to the event "extinction occurs" is one to which one can uniquely assign a probability (which will then provide a definition of $\pi$). Even then, we have to prove $\pi=\lim \pi_n$ [emphasis mine].

Kolmogorov's measure-theoretic treatement resolves this (or settles on a compromise, depending on your preferences). We know exactly which sets can be measured (those in $\Sigma$ or $\mathcal{F}$), and we have useful limiting properties like left-continuity of measures and $\sigma$-additivity for computations. Further this answer discusses why finite additivity is not enough, even for countable sample spaces. Finally, you might be interested in example 3.7 Coin Tossing on page 32, Chapter 3 of the same book. It seems similar/related to the example at the end of the OP.
Let me also add that it is hard to formulate and state results about existence of continuous modifications of stochastic processes, let alone prove them (or any other sample-path property for that matter) without random variables, for example Kolmogorov’s continuous modification theorem: en.m.wikipedia.org/wiki/Kolmogorov_continuity_theorem.
Note also the followingm with regards to the book Rogers and Williams Volume 1, p123: The canonical sample space is nice (then the sample-path is the outcome of the experiment, as it is in elementary theory), but "probability theory gets most of its depth from being able to construct (certainly non-canonical!) processes from other processes by time transformations, or as solutions of SDEs."
(This book, volume 1, opens up with a survey of results on Brownian motion, exemplifying nearly every property/concept in the general theory (martingale, gaussian process, markov process, infinitesimal generator etc). The next chapter reviews measure theoretic probability, stochastic processes (Daniel-Kolmogorov theorem), and discrete time and continuous time martingales. After is the general theory of Markov processes.)

Now I leave you with a moderately lengthy passage from Cramer's little book on the difficulties with other axiomatic approaches proposed then (first printed 1937, 2nd edition 1962):

The axiomatic basis of a theory may, of course, always be constructed in many different ways, and it is well known that, with respect to the foundations of the Theory of Probability, there has been a great diversity of opinions.
The type of statistical regularity indicated above [previous page contained a discussion on frequentist's perspective] was first observed in connection with ordinary games of chance with cards, dice, etc., and this gave occasion to the origin and early development of the theory. [...] This led to the famous principle of equally possible cases which, after having been more or less tacitly assumed by earlier writers, was explicitly framed by Laplace, as the fundamental principle of the whole theory. Throughout the whole century following the publication of Laplace's classical treatise, a large amount of work has been spent on the discussion of this principle.
During the course of this discussion, it has been maintained by various authors that the validity of the principle of equally possible cases is necessarily restricted to the field of games of chance, so that it is wholly incapable of serving as the basic principle of the theory. Attempts have been made$^1$ to establish the theory on an essentially different basis, the probabilities being directly defined as ideal values of statistical frequencies. The most successful attempt on this line is due to v. Mises who endeavors to reach in this way an axiomatic foundation of the theory in the modern sense.
The fundamental conception of the v. Mises theory is that of a "Kollektiv", by which is meant an unlimited sequence $K$ of similar observations, each furnishing a definite point belonging to an a priori given space $R$ of a finite number of dimensions. The first axiom of v. Mises then postulates the existence of the limit
$$\lim v/n = P(S),$$
for every simple sub-set $S\subset R$ while the second axiom requires that the analogous limit should still exist and have the same value $P(S)$ for every sub-sequence $K'$ that can be formed from $K$ according to a rule such that it can always be decided whether the $n$-th observation of $K$ should belong to $K'$ or not, without knowing the result of this particular observation$^2$. [$v$ is the number of observations of the event $S$ out of $n$ trials, described on an earlier page]. It does, however, seem difficult to give a precise mathematical meaning to the condition printed in italics, and the attempts to express the second axiom in a more rigorous way do not, so far, seem to have reached satisfactory and easily applicable results [emphasis mine]. Though fully recognizing the value of a system of axioms based on the properties of statistical frequencies, I think that these difficulties must be considered sufficiently grave to justify, at least for the time being, the choice of a fundamentally different system.
The underlying idea of the system that will be adopted here may be roughly described in the following simple way: The probability of an event is a definite number associated with that event; and our axioms have to express the fundamental rules for operations with such numbers.
1 For history of these attempts, cf. Keynes, chaps VII-VIII
2 The second axiom as given by v. Mises is somewhat more complicated. It can, however, be shown that this is equivalent to the simpler statement given above.

from Random Variables and Probability Distributions 2nd edition by H. Cramer, p3-5.


TITLE: How does the identity follow?
QUESTION [2 upvotes]: I can show the simple algebra for the first part, but then having issues showing, how $|z_1||z_2| = |z_1z_2|$ follows from it?

REPLY [1 votes]: $$(|z_1||z_2|)^2=z_1\,\overline{z_1}\,z_2\overline{z_2}=z_1\,z_2\,\overline{z_1}\,\overline{z_2}=(z_1\,z_2)\overline{z_1\,z_2}=|z_1\,z_2|^2$$


TITLE: Feasible subgraphs containing at most $d-2$ nonsimple vertices
QUESTION [0 upvotes]: In a paper concerning the reconstruction of polytopes from their graphs I have the following lemma

Lemma 4.4:
Let $P$ be a $d$-polytope , and let $H$ be a feasible subgraph of $G(P)$ containing at most $d-2$ nonsimple vertices. If $G(F)$ of some facet $F$ is contained in $H$, then $H\cong G(F)$

The proof starts by stating that it is true if $H$ and $F$ have the same vertices. Then we suppose that the vertex of $F$ are strictly included in those of $H$. Then it says:

in which case any path from a vertex in $H\backslash G(F)$ to a vertex in $G(F)$ must pass through a non-simple vertex, since simple vertices have the same degree in both $H$ and $G(F)$

I don't get this. To start with there doesn't even necessarily exist any non simple vertices.
PS: A feasible sub-graph of $G(P)$ is a $(d-1)$-connected sub-graph of $G(P)$ whose simple vertices (i.e. simple in $P$) each have degree $d-1$ and each non-simple vertex has degree $\geqslant d-1$

REPLY [1 votes]: Since $F$ is a facet of $P$, any simple vertices of $P$ which are in $F$ have degree $d-1$ in $F$.
Suppose $v$ is a vertex in $H$ but not in $G(F)$, and consider a path from $v$ to some vertex in $G(F)$. Suppose $w$ is the first vertex of $G(F)$ along this path. If $w$ is simple in $P$, then it has degree $d-1$ in both $G(F)$ and in $H$; so all the edges incident to $w$ in $H$ are also in $G(F)$, and so all its neighbors are in $G(F)$. Thus $w$ could not be the first vertex on the path which is in $G(F)$, contradiction.
So, the first vertex of $G(F)$ along the path must be a non-simple vertex.
(As you say, there aren't necessarily any non-simple vertices, which is fine. That would just mean that $G(F)$ and $H$ must coincide, which is what we're trying to prove.)


TITLE: Two applications of limit theorems - which one is wrong?
QUESTION [0 upvotes]: Let $X_1,X_2,\ldots,X_n,\ldots$ be IID random variables, each one of the exponential distribution, with the mean 1.
In order to calculate $\lim_{n\rightarrow\infty} P(\frac{1}{n}\sum_{i=1}^{n}X_i \leq 1)$, we may use CLT, and obtain: $$\lim_{n\rightarrow\infty} P(\frac{1}{n}\sum_{i=1}^{n}(X_i-1) \leq 0) =\lim_{n\rightarrow\infty} P(\frac{\frac{1}{n}\sum_{i=1}^{n} X_i-1 }{\frac{1}{\sqrt{n}}} \leq 0) = P(Z \leq 0)=\frac{1}{2},$$
where $Z$ is normally distributed. However I see another solution. By the law of large numbers, $\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^{n} X_i = 1$ in probability and therefore also in distribution. Therefore $\lim_{n\rightarrow\infty} P(\frac{1}{n} \sum_{i=1}^{n} X_i \leq 1) = F_{\delta_1} (1) = 1_{[1,\infty)} (1) = 1$, where $F_{\delta_1}$ is the distribution function of the distribution concentrated at 1.
Could you please help me and explain, which solution is wrong and why?

REPLY [3 votes]: Convergence in distribution means that the cdfs. converge to the limiting cdf. at continuity points of the limiting cdf.
The point $1$ is not a continuity point of $1_{[1,\infty)}$.
It does imply that, for any $\delta>0$,
$\lim_{n\rightarrow\infty} P(\frac{1}{n} \sum_{i=1}^{n} X_i \leq 1-\delta) = 0 $
and $\lim_{n\rightarrow\infty} P(\frac{1}{n} \sum_{i=1}^{n} X_i \leq 1+\delta) = 1 $, so the results are consistent. It says nothing about
$\lim_{n\rightarrow\infty} P(\frac{1}{n} \sum_{i=1}^{n} X_i \leq 1)$.


TITLE: Extensions obtained adding torsion points of an elliptic curve
QUESTION [11 upvotes]: When adding to the rational the $p$-torsion points $E[p]$ of an elliptic curve we obtain an extension containing the $p$-th roots of the unity, and whose Galois group can be embedded in $GL(2, \mathbb{F}_p)$. To what extent are such extensions coming from elliptic curves?
I mean, assume $K/\mathbb{Q}$ to be an extension whose Galois group can be embedded in $GL(2, \mathbb{F}_p)$ and containing the $p$-th roots of the unity (which is required to expect a positive answer), is $K$ obtained adding to $\mathbb{Q}$ the torsion points of some elliptic curve defined over the rationals?
Note that I'm not considering a particular Galois representation, but just Galois groups that can be embedded in some way into $GL(2, \mathbb{F}_p)$. Thanks!

REPLY [14 votes]: This question is discussed very carefully in Section 3 of the paper Mod $p$ representations
on elliptic curves, by Frank Calegari (available here).
In particular, after noting that the answer is positive when $p \leq 5$ (as was already
observed in the comments above), he proves (in Theorems 3.3 and 3.4) that if
$p \geq 7$ then there exists a continuous representation $\rho: Gal(\overline{\mathbb Q}/
\mathbb Q) \to GL_2(\mathbb F_p)$ with cyclotomic determinant which does not come from
the $p$-torsion of an elliptic curve.  (He also notes that the same result was established
by Dieulefait, in the paper Existence of non-elliptic mod $\ell$ Galois representations for every  $\ell > 5$.)


TITLE: Standardization of coset tables in GAP
QUESTION [5 upvotes]: I struggle to understand the notion of a standard coset table. In particular, I do not understand what the GAP function 'StandardizeTable' really does. My guess is that it relabels the cosets such that the resulting coset table is standard. My question is: Can it happen that, during this standardization procedure, the coset $1$ (which should correspond to the given subgroup) is relabeled to something else? Or is the coset $1$ always fixed?

REPLY [4 votes]: Coset 1 is special, as it is the only coset that forms a subgroup. It therefore is fixed, when standardizing. The numbering of the other cosets is somewhat arbitrary (and will depend on how the coset table was obtained). Standardization removes this arbitrariness.
Standardization assumes that the cosets (different from 1) are re-numbered according to the order in which they first arise in the coset table. (This will still depend on the generating set used to represent the group.) That is, the first rows of the coset table will look rather like 2,3,4,5,... rather than randomly chosen.
The main reason for doing this (apart from tidyness) is that it leaves a unique coset table and thus allows for testing equality of subgroups through the associated standardized coset tables.
If you only work with one subgroup (or table) there is basically no need for standardization.


TITLE: Rational functions on algebraic curves
QUESTION [4 upvotes]: This might be super elementary but, after having read abstract concepts of algebraic curves, I have trouble dealing with actual examples. For instance, why is the $\phi=\dfrac{y}{x}$ a rational function on the curve $F=y^2+y+x^2$? I know that any rational function on this curve should be of the form $\{\phi=\dfrac{f}{g}:f,g\in K[x,y]/(F), g\neq 0\}$, but what do I need to actually check to show that this is a rational function on $F$?
Also, if anybody can give more examples of this kind I will be very grateful.
Thanks for your help in advance.

REPLY [2 votes]: Your definition of a rational function is just fine for where you're at and your function exactly fits it. To see this, note that you need to verify that $f=y$ and $g=x$ are elements of $K[x,y]/(F)$ and that $g\neq 0$ (as a function). But this is clear.


text:
\subsubsection{Molecular Modelling}

At IFP Energies Nouvelles, the researchers of the Department of Thermodynamics and Molecular Modelling elaborate models for the structure of molecules such as hydrocarbons and alcohols.
In the following numerical study we consider a system of \emph{pentane} molecules at a \emph{reduced temperature} of $0.75$.
Pentane is an \emph{alkane} with five carbon atoms ($\mathrm{C}_5\mathrm{H}_{12}$, \textit{cf.} Figure \ref{figure:GIBBS:pentane}.)
\begin{figure}
	\center
	\includegraphics[scale = 0.1, bb=0 0 1557 603]{pentane.png}
	\caption{The structural formula of pentane. (Wikipedia)}
	\label{figure:GIBBS:pentane}
\end{figure}
Here we use the Anisotropic United Atoms model (AUA) \cite{Ungerer00} to represent the molecular structure of pentane.
The carbon ($\mathrm{C}$) and hydrogen ($\mathrm{H}$) atoms in each of the two terminal \emph{methyls} ($\mathrm{CH}_3$) and each of the three \emph{methylene} bridges ($\mathrm{CH}_2$) are treated as a single \emph{interaction center}.
To each of these two types of interaction centers correspond three parameters of the model:
an energy parameter ($\varepsilon_{\mathrm{CH}_3}$ and $\varepsilon_{\mathrm{CH}_2}$), a size parameter ($\sigma_{\mathrm{CH}_3}$ and $\sigma_{\mathrm{CH}_2}$), and a displacement parameter ($\delta_{\mathrm{CH}_3}$ and $\delta_{\mathrm{CH}_2}$). These six parameters will be the components of the set of input parameters of the numerical codes under consideration:
\begin{equation*}
	\xx=\left(\varepsilon_{\mathrm{CH}_3},\varepsilon_{\mathrm{CH}_2},\sigma_{\mathrm{CH}_3},\sigma_{\mathrm{CH}_2},\delta_{\mathrm{CH}_3},\delta_{\mathrm{CH}_2}\right)\in\xR^6.
\end{equation*}\indent

Here the matter of interest is the prediction of two macroscopic properties of a pentane system in \emph{chemical equilibrium}: the \emph{volumetric mass density} $\rho_\mathrm{liq}^\mathrm{eq}$ of the liquid phase and the \emph{vapor pressure} $P_\mathrm{gas}^\mathrm{eq}$ (pressure of the gas phase).
The algorithm carried out to numerically predict $\rho_\mathrm{liq}^\mathrm{eq}$ and $P_\mathrm{gas}^\mathrm{eq}$ -- for a given set of parameters $\xx$ -- relies on the \emph{fundamental postulate of statistical mechanics}: an isolated system in equilibrium is found with equal probability in each of its accessible \emph{microstates}.
Thus, $\rho_\mathrm{liq}^\mathrm{eq}$ (respectively $P_\mathrm{gas}^\mathrm{eq}$) can be approximated by the average of the values of the volumetric mass density of the liquid phase $\rho_\mathrm{liq}$ (respectively the vapor pressure $P_\mathrm{gas}$) of a large number of accessible microstates of the system.
Therefore the algorithm is essentially a loop, each step consisting of the following two instructions:
\begin{enumerate}[(1)]
	\item randomly generate an accessible microstate of the system (characterized by the positions of the molecules and the volume of the system),
	\item compute the value of $\rho_\mathrm{liq}$ (respectively $P_\mathrm{gas}$) given by the model for this particular microstate.
\end{enumerate}
Let us denote $G_{\rho_\mathrm{liq}}(\xx)$ (respectively $G_{P_\mathrm{gas}}(\xx)$) the probability density function of the random value $\rho_\mathrm{liq}$ (respectively $P_\mathrm{gas}$) computed at each loop step.
The two instructions above are iterated a hundred million times.
After the execution of the loop, the arithmetical mean of these millions of values is computed -- it is a Monte Carlo approximation of the integral of $G_{\rho_\mathrm{liq}}(\xx)$ (respectively $G_{P_\mathrm{gas}}(\xx)$) -- and is returned by the algorithm as the prediction of $\rho_\mathrm{liq}$ (respectively $P_\mathrm{gas}$.)
The whole algorithm is carried out in about twenty-four hours on a supercomputer (for each set $\xx$ of input parameters.)\newline\indent

In order to circumvent this time-consuming process, we want to build a metamodel of the numerical codes mapping the input parameters $\xx$ to the probability density functions $G_{P_\mathrm{gas}}(\xx)$ and $G_{\rho_\mathrm{liq}}(\xx)$ of the random outputs. We had at our disposal a sample of size $N=50$. Figure \ref{figure:GIBBS:density:pdf} and Figure \ref{figure:GIBBS:pressure:pdf} show the corresponding $50$ probability functions for volumetric mass density and vapor pressure respectively.

\begin{figure}
\center
\includegraphics[scale=0.4, bb=0 0 1080 432]{density_massvol.pdf}
\caption{GIBBS test case: distributions of the volumetric mass density for different sets of input parameters.}
\label{figure:GIBBS:density:pdf}
\end{figure}

\begin{figure}
\center
\includegraphics[scale=0.4, bb=0 0 1080 432]{density_pressure.pdf}
\caption{GIBBS test case: distributions of the vapor pressure for different sets of input parameters.}
\label{figure:GIBBS:pressure:pdf}
\end{figure}

\subsubsection{Kernel regression}

We applied the kernel regression method on the numerical codes providing the volumetric mass density of the liquid phase and the vapor pressure.
The results provided by the $\xLn{2}$ estimator with isotropic bandwidth are provided in Table \ref{table:GIBBS:KR:massvol:err} for the volumetric mass density and in Table \ref{table:GIBBS:KR:pressure:err} for the vapor pressure.
The errors have been computed by the Leave-One-Out methodology as in Section \ref{castem:kr}.
We can see that most of the results are disappointing: all relative errors are higher than $30\%$, excepted -- surprisingly -- the relative Hellinger error of the approximation.
The latter amounts to nearly $7\%$ for the volumetric mass density, and only nearly $0.05\%$ for the vapor pressure.
The errors for quantile 1\% and quantile 25\% are particularly high.
The Hellinger estimator give almost the same results: the difference between the error for the $\xLn{2}$ estimator and the error for the Hellinger estimator is lower than $10^{-5}$ in every case.
That is why they are not charted here.

\begin{table}
\center
\begin{tabular}{ccc}
\hline
L1  & L2 & Hellinger\\
151.1\% & 375.9\% & 7.089\% \\
\hline \\
\hline \\
Mean & Variance & Quantile 1\%\\
50.77\% & 98.07\% & 513.8\% \\
\hline \\
\hline \\
Quantile 99\% & Quantile 25\% & Quantile 75\%\\
30.44\% & 198.4\% & 41.52\% \\
\hline
\end{tabular}
\caption{GIBBS test case: kernel regression relative errors of the $\xLn{2}$ estimator for the volumetric mass density.}
\label{table:GIBBS:KR:massvol:err}
\end{table}

\begin{table}
\center
\begin{tabular}{ccc}
\hline
L1  & L2 & Hellinger\\
123.7\% & 254.1\% & 0.04878\% \\
\hline \\
\hline \\
Mean & Variance & Quantile 1\%\\
69.86\% & 80.72\% & 667.63\% \\
\hline \\
\hline \\
Quantile 99\% & Quantile 25\% & Quantile 75\%\\
74.82\% & 156.1\% & 73.5\%\\
\hline
\end{tabular}
\caption{GIBBS test case: KR relative errors of the $\xLn{2}$ estimator for the vapor pressure.}
\label{table:GIBBS:KR:pressure:err}
\end{table}

In Figure \ref{figure:GIBBS:KR:massvol:plot} and Figure \ref{figure:GIBBS:KR:pressure:plot} we plot the probability density functions of the volumetric mass density and the vapor pressure with their kernel regression approximations, for arbitrarily chosen sets of input parameters.
We can see how poorly the densities are approximated.

\begin{figure}
\center
\includegraphics[scale = 0.4, bb=0 0 720 576]{erreur_KR_massvol.pdf}
\caption{GIBBS test case: distribution of the volumetric mass density and its KR approximation for the sets of input parameters 1, 2, 3 and 4.}
\label{figure:GIBBS:KR:massvol:plot}
\includegraphics[scale = 0.4, bb=0 0 720 576]{erreur_KR_pressure.pdf}
\caption{GIBBS test case: distribution of the vapor pressure and its KR approximation for the sets of input parameters 31, 32, 33 and 34.}
\label{figure:GIBBS:KR:pressure:plot}
\end{figure}

\subsubsection{Functional decomposition methods}

In this section, the four functional decompositions MMP with $\xLn{2}$ and Hellinger distances, AQM and CPCA are compared. Figure \ref{figure:massvol:decompositions} represents, in logarithmic scale, the relative errors on the different norms, modes and quantiles versus the basis size (from 1 to 20) for the volumetric mass density of the liquid phase.
The error curves are globally decreasing excepted the one corresponding to the variance relative error for AQM.
Moreover, the relative error on variance is high even for basis with 20 functions and especially for AQM.
The behavior of the AQM approximation is more erratic for the 99\% quantile.
Both MMP methods perform worse on the 25\% and 75\% quantiles than other methods.
Other relative errors are low.
CPCA gives better results for $\xLn{1}$, $\xLn{2}$, Hellinger distances and the 25\% quantile. For 25\% and 75\% quantiles, errors on both MMP bases are higher than for others methods.

\begin{figure}
\center
		\includegraphics[scale=0.35, bb=0 0 1296 864]{fig_massvol_decompositions.pdf}
	\caption{GIBBS test case: comparison of the relative errors on different quantities for the volumetric mass density in function of the size of the basis $q$. Blue: MMP decomposition given by $\xLn{2}$ norm. Red: MMP decomposition given by the Hellinger distance. Black: AQM decomposition. Green: CPCA decomposition.}
\label{figure:massvol:decompositions}
\end{figure}

The results for the vapor pressure are presented in Figure \ref{figure:pressure:decompositions}. The errors on all quantities of interest are more steady than for the volumetric mass. In this case, CPCA and AQM have very different behavior compared to both MMP methods. For all quantities of interest except $\xLn{1}$, CPCA and AQM errors have a plateau for higher basis sizes. For higher basis sizes, MMP method performs better than AQM and CPCA for all studied quantities, except the 25\% quantile. Thus, contrary to what happens in the other presented numerical examples, CPCA method does not perform better than AQM and MMP for distances.
For MMP decompositions, the errors on the mean and quantiles display a jump for basis sizes of 12 or 13.

\begin{figure}
\center
		\includegraphics[scale=0.35, bb=0 0 1296 864]{fig_pressure_decompositions.pdf}
	\caption{GIBBS test case: comparison of the relative errors on different quantities for the vapor pressure in function of the size of the basis $q$. Blue: MMP decomposition given by $\xLn{2}$ norm. Red: MMP decomposition given by the Hellinger distance. Black: AQM decomposition. Green: CPCA decomposition.}
\label{figure:pressure:decompositions}
\end{figure}

config: arxiv


TITLE: Question based on finding the dividend
QUESTION [3 upvotes]: The least number which on division by $35$ leaves the remainder 25 and on division by 45 leaves the remainder $35$ and on division by $55$ leaves the remainder $45$ is _______

I understood the difference between divisor and remainder is 10 in all the three cases.
I want to know why the dividend will be LCM of (35, 45 and 55)-10
This is a gmat exam question.

REPLY [1 votes]: Note:
$$X=35m+25=45n+35=55p+45\Rightarrow$$
$$X+10=35(m+1)=45(n+1)=55(p+1)\Rightarrow$$
$$min(m,n,p)\iff LCM\{35,45,55\} \Rightarrow$$
$$X=35(m+1)-10=45(n+1)-10=55(p+1)-10.$$


TITLE: Two vector spaces with same dimension and same basis, are identical?
QUESTION [7 upvotes]: Let $V$ subspace of $W$ and both have same dimension and same basis. Then can we safely say that $V= W$ ?
I believe yes. For example there may be an element $x \in V$ written as a linear combination of the basis elements. This linear combination is unique.
Now let's take another vector $y \in W$. The $y$ is written as a linear combination of the basis vectors.
We equate the two linear combinations, and since the basis elements are linearly independent, we get that $x=y$.
Thus $V=W$.
Do you agree?

REPLY [2 votes]: Here is another proof. We know that $V$ is a subspace of $W$ and $\dim(V) = \dim(W)$. Now suppose that V is a proper subset of W. We know that V must have a complement that is a subspace of $W$. Let that complement be $U$. We have that:
$$U + V = W\\U\cap V = \varnothing$$
By Greenman's theorem, we have:
$$\dim(U+V) = \dim(U) + \dim(V) - \dim(U\cap V)$$ which, using the relations above simplifies to:
$$\dim(W) = \dim(U) + \dim(V) $$
Now, it is obvious that $\dim(U) \ge 1$: Suppose $\dim(V) = 0$, then $V = \varnothing$. But this is impossible since $U + \varnothing = U $, which contradicts the fact that $U$ is a complement of $V$ in $W$.
It follows immediately that $\dim(V) < \dim(W)$. This is a contradiction. Therefore $V$ is not a proper subset of $W$, in other words:
$$V = W$$


TITLE: XP Decay mode of radioactive nucleus
QUESTION [2 upvotes]: The decay mode of Carbon-8 is listed as 'XP' in this table.  None of the references I looked at listed XP as a decay mode.  What is it?

REPLY [3 votes]: According to http://en.wikipedia.org/wiki/Isotopes_of_carbon#Table it emits two protons


text:
\subsection{Problem Statement}
\label{sec:gen_prob}
In this work, we develop a class of algorithms to solve unconstrained minimization problems

\begin{align}
	\min_{x\in\mathbb{R}^n} J(x)
	\label{eq:opt_prob}
\end{align}

for which a closed form expression of $J:\mathbb{R}^n\rightarrow \mathbb{R}$ may be lacking, and only zero-order information in terms of function evaluations are available to find a local minimizer $x^*\in\mathbb{R}^n$ of $J$. The algorithms we propose are of the form

\begin{align}
	x_{k+1} = \mapA{M}{k}{x_k,J(x_k)},\quad k\ge 0,
	\label{eq:algo}
\end{align}

where we call $\map{M}{k}:\mathbb{R}^n \times \mathbb{R} \rightarrow \mathbb{R}^n$
 the \textit{transition map}  and $h \in \mathbb{R}_{>0}$ is the step size.

The main idea is to design the transition maps in such a way that for every $k\in\mathbb{N}$, the $m$-fold composition of these maps, i.e.,

\begin{align}
	x_{k+m} = \left( \map{M}{k+m-1} \circ \cdots \circ \map{M}{k}  \right) (x_k,J(x_k))
	\label{eq:composition}
\end{align}

approximates a gradient descent step, i.e.,

\begin{align}
	x_{k+m} = x_k - h\nabla J(x_k) + \mathcal{O}(h^{3/2})
	\label{eq:grad_descent}
\end{align}

as visualized in \Cref{fig1:macu_motivation}.

\input{Fig/macu_motivation1}

Hereby, we impose the following  structure for the transition maps

\begin{align}
\begin{split}
&\mapA{M}{k}{x_k,J(x_k)} = x_k + \sqrt{h}\alpha_1 s_k\Big(J(x_k)\Big)
\\
&~~~~~~~~~~~~~~~~\,+ \sqrt{h}\alpha_2 s_k\Big(J\big(x_k+\sqrt{h}s_k(J(x_k))\big)\Big) \\
&s_\ell (J(x_k)) = f(J(x_k))u_\ell+g(J(x_k))v_\ell
\end{split}
\label{eq:TM}
\end{align}

with parameters $\alpha_1,\alpha_2 \in \mathbb{R}$
where $\alpha_1 + \alpha_2 \not = 0$.
We call $s_\ell:\mathbb{R}\rightarrow \mathbb{R}^n$ the \textit{evaluation map}, $f,g:\mathbb{R}\rightarrow \mathbb{R}$ the \textit{generating functions} and $u_\ell,v_\ell\in\mathbb{R}^n$ the $m$-periodic \textit{exploration sequences}. Note that for $\alpha_2 \neq 0$, only two evaluations of $J$ per iterations are necessary. We elaborate on the choice of this structure for the algorithm in the next section.

The main goal of this work is to characterize and design
\begin{enumerate}[leftmargin=13pt]
	\item $m$-periodic exploration sequences $u_\ell$, $v_\ell$, and
	\item gradient generating functions $f$ and $g$,
\end{enumerate}
 such that \eqref{eq:algo} with transition map \eqref{eq:TM} yields \eqref{eq:grad_descent}.

We will at various points make the use of one or both of the following assumptions.
\begin{enumerate}[leftmargin=2.5em, label=(A\arabic*)]

	\item The functions $f,g$ are of class $C^2(\mathbb{R},\mathbb{R})$ and the objective function $J$ is of class $C^2(\mathbb{R}^n,\mathbb{R})$.
	\label{assump:smooth_fg}

	\item The objective function $J$ is radially unbounded and there exists an $x^* \in \mathbb{R}^n$ such that $\nabla J (x)^\top (x-x^*) > 0$ for all $x\in \mathbb{R}^n \backslash \{x^*\}$.
	\label{assump:smooth_J}

\end{enumerate}
We note that \ref{assump:smooth_J} will not be required for the design of the algorithms, but only when we analyze their performance.

The implementation of the algorithms, however,  is not limited to the class of objective functions satisfying \ref{assump:smooth_fg}.

\subsection{Related Results}
\label{sec:existing_results}
The structure of the transition map in \eqref{eq:TM}  was introduced by the authors of the present work in \cite{jfDerFree} for one dimensional problems. Therein,  two cases were considered, specified by the parameter setting $[\alpha_1\ \alpha_2] = [1 \ 0]$, as so-called \textit{single-point} map

\begin{align}
	\!\mapA{M}{k}{x_k,J(x_k)}\! = \mapA{E}{k}{x_k,J(x_k)}\! := x_k + \sqrt{h}s_k(J(x_k))
	\label{eq:map_E}
\end{align}

and by $[\alpha_1 \ \alpha_2] = [\nicefrac{1}{2}\ \nicefrac{1}{2}]$, as so-called \textit{two-point} map

\begin{align}
\mapA{M}{k}{x_k,J(x_k)} &= \mapA{H}{k}{x_k,J(x_k)} := x_k + \frac{\sqrt{h}}{2}\left[s_k\Big(J(x_k)\Big) \right. \nonumber \\
 &\left.~+s_k\Big( J(x_k+\sqrt{h}s_k\big(J(x_k)\big) \Big)\right].
 \label{eq:map_H}
\end{align}

The algorithms relying on transition maps \eqref{eq:map_E} and \eqref{eq:map_H} are called single and two-point algorithm, respectively, reflecting that  the number of function evaluations of $J$ at each iteration is one and two. This type of map structure was inspired by the well-known Euler and Heun (trapezoidal) numerical integration methods respectively (thus the naming of the maps $E$ and $H$), i.e., executing a single integration step with step size $\sqrt{h}$ of
the differential equation

\begin{align}
	\dot{x}(t) = s\big(J(x(t))\big) = f\big(J(x(t))\big)u(t) + g\big(J(x)\big)v(t)
	\label{eq:ode}
\end{align}

with piece-wise constant $m\sqrt{h}$-periodic inputs $u(t),v(t)\in\mathbb{R}^n$ for $t\in[\ell\sqrt{h},(\ell+1)\sqrt{h}]$ with $\ell \in \mathbb{N}$, yields \eqref{eq:map_E} and \eqref{eq:map_H}, respectively.

Note that \eqref{eq:ode} is well known
as an approximate gradient descent flow in the context of extremum seeking control (cf.~\cite{durr2013lie}).
For a detailed explanation of the proposed class of algorithms and the continuous-time algorithm \eqref{eq:ode} plus how non-commutativity comes into play, we refer to \cite{jfDerFree,jfESNCM}.

For the coordinate-wise descent case (see Lemma~1 and Lemma~2 in \cite{jfDerFree})  the choice of exploration sequences

\begin{align}
\label{eq:input_coord}
\begin{split}
	&u_\ell = \bar{u}_\ell e_i,\ \ v_\ell = \bar{v}_\ell e_i \ \ \text{with} \ \
	i = \corrected{\lfloor\ell/4\rfloor\,\text{mod}(n)+1}
	\\
	&\bar{u}_\ell = \begin{cases}
	~~~1 &\ell=0 \\
	~~~0 &\ell=1 \\
	-1 &\ell=2 \\
	~~~0 &\ell=3 \\
	\bar{u}_{\ell-4} & \text{else}
	\end{cases}, \quad
	\bar{v}_\ell = \begin{cases}
	~~~0 &\ell=0 \\
	~~~1 &\ell=1 \\
	~~~0 &\ell=2 \\
	-1 &\ell=3 \\
	\bar{v}_{\ell-4} & \text{else}
	\end{cases}
	\end{split}
\end{align}

with $m=4n$ leads to the evolution of $x_k$ with $[\alpha_1\ \alpha_2]~=~[1\ 0]$ such that

\begin{align}
	x_{k+m} &= x_k + h\Big\lbrace([f,g](J(x_k))  \nonumber \\
	 & - \frac{1}{2}\frac{\partial (f^2+g^2)}{\partial J}(J(x_k)) \Big\rbrace \nabla J(x_k) + \mathcal{O}(h^{3/2})
	 \label{eq:taylor_E}
\end{align}

and with $[\alpha_1\ \alpha_2] = [\nicefrac{1}{2}\ \nicefrac{1}{2}]$ such that

\begin{align}
x_{k+m} &= x_k + h\Big\lbrace([f,g](J(x_k)) \Big\rbrace \nabla J(x_k) + \mathcal{O}(h^{3/2}),
\label{eq:taylor_H}
\end{align}

where $[f,g] := \frac{\partial g}{\partial J}f - \frac{\partial f}{\partial J}g$ is the \textit{Lie bracket} of $f$ and $g$. A simple calculation shows that the term in brackets in \eqref{eq:taylor_E} and \eqref{eq:taylor_H} is identical to $-1$ for $f(J(x)) = \sin(J(x))$ and $g(J(x)) = \cos(J(x))$, hence \eqref{eq:grad_descent} is recovered.

The  exploration sequence above is constructed in such a way that components of the gradient are approximated sequentially for the multi-dimensional setting, hence, coordinate-wise.

In \Cref{fig:ncm}, the exploration sequence and the gradient approximation is visualized
for the scalar case $x_k \in \mathbb{R}$ ($n=1, m=4$).

In summary, the existing procedure is limited and mimics a coordinate-wise descent algorithm.
Further, only a single exploration sequence was presented as well as a single pair of generating functions.

There are, however, many ways to construct exploration sequences and generating functions, especially in the multi-variable case.

Since different exploration sequences and generating functions lead to different
properties of the algorithm, it is the goal of this work to provide solutions for a flexible design and constructions of exploration sequences in the multi-variable setting and to characterize a large class of generating functions.

\input{./Fig/commutativ_maps}


TITLE: Sum of vector spaces
QUESTION [0 upvotes]: Let $ V $ be a vector space, and $ W_1,\dots,W_r \subseteq V$ be subspaces of $V$. Now I'm trying to prove $$W_1 + \cdots + W_r = \operatorname{span}(W_1 \cup \cdots \cup W_r).$$
I start with "$ \subseteq $".
Let $ x \in W_1 + \cdots + W_r$ be arbitrary. So we have $x = w_1 + \cdots + w_r$ such that $w_i \in W_i$ for all $i=1,\dots,r$.
From here I have no idea how to proceed.

REPLY [2 votes]: $\subset$ :
Since each of your $w_i \in W_i$, $x$ is a linear combination of elements of $(W_1 \cup ... \cup W_r)$, and therefore of span$(W_1 \cup ... \cup W_r)$.

$\supset$ :
Let $x \in$ span$(W_1 \cup ... \cup W_r)$. Then $x = \sum_i \lambda_i y_i$, where $y_i \in W_1 \cup ... \cup W_r$, so that for each $y_i$ there exists $j$ such that $y_i \in W_j$. Since $W_j$ are subspace of V, then $\lambda_i y_i \in W_j$, and so $\sum_i \lambda_i y_i \in  W_1 + ... + W_r$.


TITLE: A version of Brower's fixed point theorem for contractible sets?
QUESTION [4 upvotes]: Brouwer's fixed point theorem states that a continuous map $f:B^n\to B^n$ ($B^n\subset\Bbb R^n$ being the $n$-dimensional ball) has a fixed point. It is clear that we can replace $B^n$ with a space $X$ homeomorphic to the ball.

Question: Is there a version of this fixed point theorem that only requires $X$ to be compact and contractible? Do we need more, e.g. locally contractible or something like this? I am happy to assume that $X\subset \Bbb R^n$ for some finite $n\ge 1$.

This question has a note about a contractible space for which the fixed point theorem does not hold, but it is unclear to me whether the counterexample stands if $X$ is required to be a subset of a finite-dimensional Euclidean space.

REPLY [7 votes]: There are contractible compacta without the fixed point property. In fact,

Theorem (Kinoshita): There is a contractible compact subset of $\mathbb{R}^3$ which does not posses the fixed-point property. $\square$

In a positive direction, here is a result which generalises John Palmieri's answer and follows directly from the Brouwer fixed point theorem. The idea hinges on the following observation.

If $Y$ is a space with the fixed-point property, and $X\subseteq Y$ is a retract of $Y$, then $X$ has the fixed-point property.

To leverage this we have:

Any compact CW complex $X$ embeds in $D^n$ for some $n\in\mathbb{N}$. If $X$ is contractible, then it is a retract of $D^n$.

In fact every locally-finite CW complex of dimension $n$ embeds in $\mathbb{R}^{2n+1}$ as a neighbourhood retract (see Fritsch, Piccinini, Cellular Structures in Topology Th.1.5.15). Of course every compact CW complex is finite-dimensional, and has bounded image under the previous embedding. That a finite complex should be a retract of any disc in which it embeds is a consequence of Paul's answer here.
Now, in light of the previous two observations we use Brouwer's fixed point theorem to conclude the following.

Every compact CW complex (triangulable or not) which is contractible has the fixed-point propety. $\square$

Of course the full machinery is more powerful, and has application outside the realm of CW complexes.

Every finite-dimensional compact metric space which is both contractible and locally contractible has the fixed-point property.

The argument is as before. Show that the spaces in question embed in some disc as a retract and quote the Brouwer fixed point theorem.


TITLE: $T:V \to W$ and $S: W \to V$ such that $T\circ S = I_w$ but $S\circ T \neq I_V$
QUESTION [0 upvotes]: Working through some questions and I'm stuck on the following:
Find an example of vector spaces $V$ and $W$, and linear transformations $T:V \to W$
and $S: W \to V$, such  that $T\circ S$ = $I_w$ but $S\circ T \neq I_V.$

REPLY [0 votes]: Well, first, if $T\circ S = I_W$, then we know that $S$ must be injective, since it has a left-inverse.  We also see that $T$ is surjective for similar reasons.  This helps us think of some examples.  The easiest one I can think of is setting $W = \mathbb{R}$ and $V = \mathbb{R}^2$.  We then let $S:V\to W$ be the inclusion map $S:x\mapsto (x, 0)$, and $T:X\to V$ be the projection map $T:(x, y)\mapsto x$.  It is pretty easy to verify that $T\circ S$ is the identity on $\mathbb{R}$, but $S\circ T$ is not the identity on $\mathbb{R}^2$, since, for example, $(S\circ T)(1, 1) = (1, 0)$.


TITLE: Deriving conserved charges from the equations of motion
QUESTION [3 upvotes]: It is very well established how to derive conserved charges associated to the symmetries of Lagrangian using the Noether's theorem. Also in the Hamiltonian formulation, we know how to derive the charges associated to symmetries of the Hamiltonian action, using the symplectic structure etc.
My question: Suppose I have a set of equations without knowing the explicit action or Hamiltonian. I realize that the equations of motion are invariant or covariant under some transformation. Is there a way to associate conserved charges (or better conserved currents) associated to this symmetry transformation without using Lagrangian or Hamiltonian?
This question was answered here for the very special case of a system of particles where the equations of motion take a "natural form" $m_i\ddot{{\vec x}}_i=-{\vec\nabla} V(x^j)$. However, this is very limited and does not apply to a field theory or a gauge theory in general (e.g. consider general relativity without any special gauge condition).
Application A good example of this issue is the Electric-magnetic duality symmetry. Remind that Maxwell equations are invariant under the electric-magnetic duality $(E,B)\to (B,-E)$. With some effort, this symmetry can be also realized at the level of the action as shown originally by Deser and Teitelboim (here). Then using the Noether procedure, it turns out that the Noether charge corresponding to duality is helicity which is a physical observable.
In linearized GR same thing happens. The eom is duality symmetric. One can also find a duality invariant action (here)
However, the problem becomes more interesting in full GR. In this case, the Einstein equations still have an electric-magnetic duality under rotating the electric and magnetic parts of the Weyl tensor into each other (see e.g. here or section 7.2 of the book by Christodoulou and Klainerman). However, this symmetry cannot be realized at the level of the action. Indeed there is a no-go theorem by Deser et al. (here) showing that the duality symmetry cannot be realized at the level of the action. So the question is whether there is a conserved quantity that correspond to the duality symmetry in GR or not?

REPLY [2 votes]: A simple comment. Consider a modification of the equation you quoted by adding friction:
\begin{equation}
\ddot{\vec x} = -k \dot{\vec x}-\vec\nabla V(x).
\end{equation}
Clearly, there is time translation invariance, however, there is no energy conservation. Similarly, you can drop $V(x)$, in this case there is also space translation that is a symmetry of EOM, but again momentum is not conserved. The bottom line is that invariance of EOM does not guarantee existence of a conserved quantity.


TITLE: Verifying simplification of simple logic expressions
QUESTION [1 upvotes]: I am trying to simplifying these 3 expressions, and I have actually done them, but can someone check if it's correct? Thank you.

$x \vee y \land y \vee \neg x$
$x \land y \vee y \land \neg x$
$\neg(\neg(x \land y) \vee x)\vee y$

The first one:
$x \vee y \land y \vee \neg x = x\vee(y)\vee\neg x = (x\vee\neg x)\vee y=1\vee y=1$
Second one:
$x\land y \vee y \land \neg x = (x \land y) \vee (y \land \neg x) = (y \land x) \vee (y \land \neg x) = y \land (x \vee \neg x) = y \land 1 = y$
Third one:
$\neg(\neg(x\land y)\vee x)\vee y = \neg(\neg x \vee \neg y \vee x) \vee y = \neg(1\vee\neg y)\vee y=\neg(1)\vee y=0\vee y=y$
Thank you for helping.

REPLY [1 votes]: Yes, those are all correct.   Also, your steps are clear, so it was easy to follow your reasoning.


TITLE: Combining the duality principle and the graphical method
QUESTION [0 upvotes]: I am trying to minimize this linear program by combining the duality principle and the graphical method:

I can't seem to find an example of how to approach this, can anyone show me how I would go about this?

REPLY [0 votes]: The dual problem is:
objective function
$\text{max} \ -y_1+y_2$
constraints
$2y_1+y_2 \leq 2$
$y_1+3y_2 \leq 3$
$-2y_1+y_2 \leq 3$
$3y_1+2y_2 \leq 6$
$-2y_1+y_2 \leq 4$
$y_1,y_2$ are unrestricted.
You have to solve every constraint for $y_2$. Then you can draw the constraints in a $y_1-y_2$-coordinate system. If you consider the inequality-signs you will get a solution space.
Now you set the objective function equal to zero and solve the equation for $y_2$. After you have drawn it in the coordinate system you have to move it parallel till you just touch the solution space.


TITLE: Question on current flux
QUESTION [0 upvotes]: I found this in Jack Vanderlinde's Classical Electromagnetic Theory, Problem 1-13:
"A fine needle emits electrons isotropically at a steady rate. Find the divergence of the current density and the resulting current flux at distance r from the point in the steady state."
Now, for the divergence of the current density, I figured it's about continuity equation, and the value is $-\frac{\partial \rho}{\partial t}$ at any distance from the needle. But i don't understand about current flux; what did the question ask, or what current flux even mean.

REPLY [0 votes]: current density flux is zero in this case as it is a steady current. current density is a VECTOR FIELD  and like any other vector field it can have a divergence. this intuatively mean , given you have a pillbox drawn ( gaussian surface) around some current density. its surface integral must be zero... aka, the same amount of charge remains inside this box at all times
and yes, it directly relates to the continuity equation. in integral form the continuity equation is :
the closed surface flux  integral  of J = -d/dt of the volume integral of charge density .
where the  closed surface  encloses the volume


TITLE: Where are we using $C^1$ in this proof?
QUESTION [2 upvotes]: This must be something really easy/standard, but for some reason I can't see what's wrong with this argument.
Let $f : \mathbb{C} \rightarrow \mathbb{C}$ be any complex function, and write $f = u + iv$, with $u, v : \mathbb{C} \rightarrow \mathbb{R}$.
I know that if $f$ is $C^1$ in the real sense and $f$ satisfies the Cauchy-Riemann equations, then $f$ is holomorphic. My definition of holomorphic here is that the limit
$$\lim_{\zeta \rightarrow 0} \frac{f(z + \zeta) - f(z)}{\zeta}$$
exists and is a complex number. This is not hard to prove, as I outline below, but I'm having trouble seeing where in this proof I'm actually using the continuity of the derivative of $f$. Here's the argument:
Proof: Let $z_0 = x_0 + iy_0 \in \mathbb{C}$, and let $A = u_x(x_0, y_0)$, $B = u_y(x_0, y_0)$, $C = v_x(x_0, y_0)$ and $D = v_y(x_0, y_0)$.
Since $u$ is differentiable at $(x_0, y_0)$, we can write
$$\epsilon_1(h, k) = u(x_0 + h, y_0 + k) - u(x_0, y_0) - Ah - Bk,$$
where $\epsilon_1(h, k)/\|(h, k)\|$ goes to zero as $(h, k) \rightarrow 0$. Now, for a moment there I thought this was the part where you need continuity of the partials $u_x$ and $u_y$, but note that they're being computed at a fixed point $(x_0, y_0)$, and the fact that $\epsilon_1$ goes to zero above is just the definition of differentiability, isn't it?
If we accept this and do the same expansion for $v$, then it's easy to see that the aforementioned limit will exist and will be a complex number, because of the Cauchy-Riemann equations.
So, again, my question is:
Where in the above proof are we using that $f$ is $C^1$, and not just differentiable?
Thanks!

REPLY [2 votes]: EDIT:I misunderstood your definition.
I'ld like to point out When is a function satisfying the Cauchy-Riemann equations holomorphic? instead.

Example Consider the function$$f(z)=e^{-\frac{1}{z^4}} \ \ \forall z \neq 0$$
$$f(0)=0$$
This function satisfies cauchy Riemann equation but is not holomorphic in any open neighbourhood of $0$.


TITLE: Can every odd integer greater than $1$ be written as a product of fractions $\frac{4m+1}{2m+1}$?
QUESTION [10 upvotes]: Can every odd integer greater than $1$ be written as a product of fractions of the form $\frac{4m+1}{2m+1}$, $m$ a positive integer?
Here is a proof if the fractions were instead $\frac{4m-1}{2m+1}$. I tried to mimic it, but it doesn't work.

Since the fractions available have odd numerator and denominator, the same will be true of any product, so even integers aren't reachable. Odd integers can be reached, however.
Let $f(q)$ be the fraction $\frac{4q-1}{2q+1}$. Then $f(1)=1$, $f(4)f(7)=3$ and $3 f(4)=5$. So $1,3,5$ are obtainable.
Suppose now that $m\ge 2$ and we've obtained all odd numbers up to and including $4m-3$. Since $2m+1\le4m-3$ we can use $2m+1$ to obtain
\begin{gather*}
4m-1 = (2m+1) f(m),\\
4m+1=(2m+1) f(3m+1).
\end{gather*}
Hence $4m-1$ and $4m+1$ are obtainable as well, so having obtained $5$ we can then reach all odd positive integers.

Any help appreciated!

REPLY [2 votes]: For integer $a ≥ 0$ define $f(a) = (4a + 1) / (2a + 1)$.
Let $g_k(a) = f(a) \cdot f(2^1a) \cdot f(2^2a) \cdot … \cdot f(2^{k-1}a)$. $g_k(a)$ is a telescopic product, and $g_k(a) = (2^{k+1}a + 1) / (2a+1)$.
We have $g_k((b-1) / 2) = (2^k b - 2^k + 1) / b$.
We have $g_k((cy - 1) / 2) = (2^k cy - (2^k - 1)) / cy = 2^k - (2^k - 1) / cy$.
Pick an odd x > 1. Find the largest k such that x+1 is divisible by $2^k$, so $k ≥ 1$. Let $y = (x + 1) / 2^k$, then y is odd, $y < x$, and $2^k y = x + 1$.
Let $c = 2^k - 1$, then $y \cdot g_k((cy - 1) / 2) = 2^k y - 1 = (x + 1) - 1 = x$, so x is formed by multiplying y by k values of the form $f(a)$.
Together with 1 = f(0) this shows by induction that every odd number is a product of numbers of the form f(a). I think the number of values f(a) used according to this proof is k whenever $2^{k-1}<x<2^k$. Not sure if more or fewer values could be used.


text:
\section{Introduction}

Much progress has been achieved recently in Bayesian inference of high dimensional problems. It has been possible to develop
rigorous methods in order to derive exact ``single letter'' variational formulas for the mutual information
in the asymptotic limit of the number of variables tending to infinity, when the prior and all hyperparameters of the problem are assumed to be known (this is referred to as the Bayes-optimal setting).
Such formulas have often been first conjectured on the basis of the replica and cavity methods of statistical mechanics of disordered spin systems
and are also known as ``replica symmetric'' formulas \cite{MezardParisi87b,MezardMontanari09}. Examples where {\it full} proofs have been achieved are
random linear estimation and compressed sensing \cite{barbier_allerton_RLE,barbier_ieee_replicaCS,ReevesP16}, learning for single layer networks
\cite{barbier2017phase}, generalized estimation in multi-layer settings \cite{2018arXiv180509785G,2018arXiv180605451A}, or low-rank matrix and tensor estimation \cite{XXT,2017arXiv170200473M,2017arXiv170108010L,2016arXiv161103888L,2017arXiv170910368B}. Invariably, the Guerra-Toninelli interpolation method \cite{guerra2002thermodynamic} has been used to derive one-sided bounds (we note that \cite{ReevesP16} is exceptional and does not seem
to rely directly on the same interpolation). For the converse bounds typically other ideas have usually been
necessary, such as
spatial coupling \cite{XXT,barbier_allerton_RLE,barbier_ieee_replicaCS} or the rigorous version of the cavity method  \cite{2017arXiv170200473M,2017arXiv170108010L,2016arXiv161103888L}. Recently two of us introduced
a new interpolation scheme, called {\it adaptive path interpolation method}, that allows to derive the replica symmetric formulas in a more straightforward and unified manner
\cite{BarbierM17a}, \cite{BarbierMacrisCW2019}. The new method is quite generic once the mean field solution has been identified and
is directly applicable when the concentration
of the ``overlap'' can be proved. Roughly speaking in Bayesian inference the ``overlap'' is the inner product between the random vector to be inferred and the ground truth vector.
For models on dense graphs this concentration follows from variants of Ghirlanda-Guerra identities \cite{GG} adapted to Bayesian inference combined with the so-called Nishimori identities.

The successes of the adaptive interpolation method have so far been limited to inference models with a dense underlying factor graph.
It is therefore desirable to see to what extent the method can be developed when the factor graph is instead sparse.
Typical examples of such systems are Low-Density Parity-Check codes, Low-Density Generator-Matrix (LDGM) codes, or the Stochastic and Censored Block Models.
It is fair to say that the replica symmetric formulas for the mutual information is much more complicated in such models.
Indeed, besides the measurements (or channel outputs): (i) the sparse graph is also random; (ii) the single letter variational problem involves a functional
over a set of probability distributions (instead of scalars as in the dense graph case). Existing rigorous
derivations of the replica formulas have so far been achieved using a combination of the interpolation method
(first developed by \cite{FranzLeone} for sparse models) and spatial coupling \cite{MacrisGiurgiuUrbanke2016} or the rigorous
version of the cavity method \cite{aizenman2003extended,Coja-Oghlan:2017}.

In this work we
consider a simple version of the Censored Block Model \cite{6949658,abbe2013conditional,DBLP:journals/corr/ChinRV15,7282642}, for which we
fully develop the adaptive interpolation method. We believe that this constitutes a first step towards an
analysis of more complicated models via this method. A summary of the present analysis was presented in \cite{2018arXiv180605121B}.

In the Censored Block Model one has a set of $n$ hidden binary variables. One observes $\alpha n$ products of random $K$-tuples, where
$\alpha > 0$ is called the fraction of measurements, through a noisy channel. The goal is to reconstruct an estimate of the hidden variables from the noisy observations.
There are other interpretations of this model. For example, it can be interpreted as a Low-Density Generator-Matrix code ensemble, with design communication rate $1/\alpha$, on a
factor graph with degree $K$ factor nodes and variable nodes with Poisson degrees (when $n\to +\infty$). Another possible interpretation is,
as a model of statistical mechanics, namely an Ising model on a sparse random graph with $K$-spin interactions.
The Censored Block Model has been discussed when the measurement channel is a Binary Symmetric Channel in \cite{6949658} and the
replica formula proven in this case \cite{Coja-Oghlan:2017}. Here, we consider a simpler situation where the measurement channel is the Binary Erasure
Channel (BEC) for which the adaptive interpolation method can be completely developed. As we will see this method requires concentration results for a whole set of suitable ``overlaps''
and requires new ideas in the case of sparse graphs.
Here, we solve this issue for the BEC, and it is currently the only aspect of the method that is missing for extending our analysis to other channels.

The paper is organized as follows. In Section \ref{sec:2} we give a precise formulation of the model and state the main result of this paper (Theorem \ref{RSconj}).
In Section \ref{prelimtool} we review
two important tools used throughout our analysis, namely the Nishimori identities and the Griffiths-Kelly-Sherman inequalities. The adaptive interpolation method for the sparse graph models
is formulated in Section \ref{interp} and the core of the proof of Theorem \ref{RSconj} is also developed. This section contains the main new technical ideas of this paper. Overlap concentration
is proved in Section \ref{sec:overlap} and \ref{sec:thermal-and-quenched}. A series of more technical results are found in Section \ref{sec:proof} and in the appendices.


TITLE: About associated Legendre equation.
QUESTION [1 upvotes]: Is there someone that maybe can explain where come from the substitution $y(z) = (1-z^2)^{m/2}\cdot u(z)$ in the associated legendre equation? It come from series power? I can't manage the complete solution of this equation. Actually i need only to understand why we use this substitution.  Sorry my rusty english.

REPLY [1 votes]: I'm not sure what you would like to know, but I will try to
give you some explanantions regarding the relation between Legendre equation and associated Legendre equation.
As you know, Legendre differential equation is,
\begin{equation}
\begin{aligned}
(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+n(n+1)y=0
\end{aligned}
\end{equation}
If you differentiate the above equation once, you obtain,
\begin{equation}
\begin{aligned}
(1-x^2)\frac{d^3y}{dx^3}-4x\frac{d^2y}{dx^2}+[n(n+1)-2]\frac{dy}{dx}=0
\end{aligned}
\end{equation}
Differentiating twice, you obtain,
\begin{equation}
\begin{aligned}
(1-x^2)\frac{d^4y}{dx^4}-6x\frac{d^3y}{dx^3}+[n(n+1)-6]\frac{d^2y}{dx^2}=0
\end{aligned}
\end{equation}
and so on.
Differentiating $m$ times $(n>m)$, you will find that the general equation can be obtained as,
\begin{equation}
\begin{aligned}
(1-x^2)\frac{d^{m+2}y}{dx^{m+2}}-2 (m+1)x\frac{d^{m+1}y}{dx^{m+1}}+[n(n+1)-m(m+1)]\frac{d^my}{dx^m}=0
\end{aligned}
\end{equation}
You can prove the above relation by the induction method.
Now, if you transform, $y=(1-x^{2})^{m/2}z$ and put it into the differential equation, known as the associated Legendre differential equation below,
\begin{equation}
\begin{aligned}
(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+\left[n(n+1)-\frac{m^2}{1-x^2}\right]y=0
\end{aligned}
\end{equation}
you will see that the equation becomes,
\begin{equation}
\begin{aligned}
(1-x^2)^{m/2}
\left
[(1-x^2)\frac{d^{2}z}{dx^{2}}-2 (m+1)x\frac{dz}{dx}+[n(n+1)-m(m+1)]z\right]=0
\end{aligned}
\end{equation}
So this leads us to the relation,
\begin{equation}
\begin{aligned}
z=\frac{d^my}{dx^m}
\end{aligned}
\end{equation}
which indicates that, if y is the solution of Legendre differential equation $P_n(x)$,
\begin{equation}
\begin{aligned}
P_n^m(x)=(1-x^{2})^{m/2}\frac{d^my}{dx^m}=(1-x^{2})^{m/2}\frac{d^mP_n(x)}{dx^m}
\end{aligned}
\end{equation}
will be that of the associated Legendre differential equation.
In this note, please be reminded that I have omitted the discussion of the function of second kind, $Q_n(x)$ and $Q_n^m(x)$, for simplicity. But the story I believe is basically the same.
Hopefully, my explanation serves your question.
*P.S. I have added some replies in response to your comment.
I don't inderstand what is meant by "artificial". The function that satisfies the differential equation written above, i.e.,
\begin{equation}
\begin{aligned}
(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+\left[n(n+1)-\frac{m^2}{1-x^2}\right]y=0
\end{aligned}
\end{equation}
which is a more general form of Legendre equation, is the associated Legendre function, $P_n^m(x)$ expressed here in the note.
Of course, since the Legendre function is polynomial and can be expressed as, for example,
\begin{equation}
\begin{aligned}
P_n(x)=\sum^{[n/2]}_r(-1)^r\frac{(2n-2r)!}{2^n r!(n-r)!(n-2r)!}x^{n-2r}
=\frac{1}{2^n n!}\frac{d^n}{dx^n}(x^2-1)
\end{aligned}
\end{equation}
(ther are several other ways to express the summation form) you can easily see that the associated Legendre function expressed as $(1-x^2)^{m/2}P_n(x)$ is also a polynomial function. For more detail about the form of the polynomial, see for example, "Handbook of Mathematical Functions", by M. Abramowitz and I. A. Stegun. If you prefer to use the polynomial form, you can do that, but it's just doing the same thing as I have explained.
In the equation,
\begin{equation}
\begin{aligned}
(1-x^2)\frac{d^{m+2}y}{dx^{m+2}}-2 (m+1)x\frac{d^{m+1}y}{dx^{m+1}}+[n(n+1)-m(m+1)]\frac{d^my}{dx^m}=0
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
\frac{d^{m+2}y}{dx^{m+2}}=\frac{d^2}{dx^2}\left(\frac{d^{m}y}{dx^{m}}\right)
=\frac{d^2}{dx^2}z
\end{aligned}
\end{equation}
and
\begin{equation}
\begin{aligned}
\frac{d^{m+1}y}{dx^{m+1}}=\frac{d}{dx}\left(\frac{d^{m}y}{dx^{m}}\right)
=\frac{d}{dx}z
\end{aligned}
\end{equation}
and so on, which thereby indicates that the function $(1-x^2)^{m/2}z$ satisfies the associated Legendre differential equation and as a whole the above equation becomes,
\begin{equation}
\begin{aligned}
(1-x^2)^{m/2}
\left
[(1-x^2)\frac{d^{2}z}{dx^{2}}-2 (m+1)x\frac{dz}{dx}+[n(n+1)-m(m+1)]z\right]=0
\end{aligned}
\end{equation}
as I have explained in the note.


TITLE: How to prove simple function is measurable
QUESTION [7 upvotes]: I'm  aware of the definition of the measurable function. But I was wondering how to prove simple function is measurable? It would be better have some detailed proof.

REPLY [10 votes]: Consider a simple function
$$
f = \sum_{i=1}^na_i1_{E_i},
$$
where $E_1,\ldots,E_n$ are disjoint measurable sets. To see that $f$ is measurable, it suffices to show that
$$
f^{-1}(-\infty,a]
$$
is measurable for every $a\in\mathbb{R}$.
Given $a\in\mathbb{R}$, note that
$$
f^{-1}(-\infty,a] = \bigcup\{E_i \mid i\in\{1,\ldots,n\}\ \text{such that}\ a_i\le a\}
$$
is a union of measureable sets. Therefore $f$ is measurable.


TITLE: Why is there a sudden drop-off in half-life of isotopes at around 130 neutrons? Is there a name for this?
QUESTION [13 upvotes]: Pertaining to the chart of nuclides, there is a region above Bismuth, in which the relatively continuous trend of stability is interrupted by a batch of isotopes all with extremely short half-lives. I know that "magic numbers" contribute to the stability of nuclei via shell-filing, is there such a thing as an "anti-magic" number which leads to this "bay of instability" style feature? I assume this is a well studied phenomenon, I just know not what keywords to look for in order to look into it further.

I find it particularly fascinating as it is so close to the last few stable elements, the drop-off in half-life is precipitous and it affects a lot of isotopes in this area. Is it something to do with how alpha decay operates? Every isotope in this loosely-defined zone seems to undergo alpha decay. I also note that generally, above Bismuth the 128-neutron line is extremely unstable, with half-lives much more gradually increasing as $N$ increases.

REPLY [7 votes]: The instability is actually centred on N=128 neutrons, not N=130, and is caused by N=126 being a neutron magic number.  Isotopes with magic numbers of neutrons have greater binding energies than those that don't, so it is energetically favourable for isotopes with N=128 to decay to an N=126 isotope by alpha decay. This causes a peak in alpha decay energy ($Q_\alpha$) at 128 neutrons and corresponding dip in half-life ($t_{1/2}$), as shown in Figure 3 from this paper.  According to the Geiger-Nuttall Law, the half-life for alpha decays decreases exponentially with increasing $\sqrt{Q_\alpha}$.
I don't know if the dip has a name.


TITLE: Why don't we take this term $D_{\mu}D_{\nu}F^{\mu\nu}$ in Lagrangians?
QUESTION [2 upvotes]: Why don't we take
$$D_{\mu}D_{\nu}F^{\mu\nu}$$
in Lagrangians?

REPLY [10 votes]: We do if there is a color trace. The term $D_{\mu}D_{\nu}F^{\mu\nu}=\frac{1}{2}[D_{\mu},D_{\nu}]F^{\mu\nu}$ is proportional to $F_{\mu\nu} F^{\mu\nu}$.

REPLY [1 votes]: As the more upvoted answer said, if there are color indices then the covariant derivative doesn't commute with itself and the expression you wrote makes sense.  If not, symmetry arguments about the symmetric nature of the derivative and the connection and the anti-symmetric nature of the curvature tensor are enough to reason like I did below that the argument is zero.  The two derivatives are symmetric, and the curvature is anti-symmetric, and so you are contracting a symmetric matrix with an anti-symmetric one, which is zero.
$$
D_{\mu}D_{\nu}F^{\mu \nu} = -D_{\mu}D_{\nu}F^{\nu\mu} = -D_{\nu}D_{\mu}F^{\nu\mu} =-D_{\mu}D_{\nu}F^{\mu \nu} \implies D_{\mu}D_{\nu}F^{\mu \nu} = 0
$$