text:
\begin{document}
\thispagestyle{empty}
\begin{center}
{\Large {\bf On the universality of Somos' constant}}
\end{center}
\begin{center}J\"org Neunh\"auserer\\
Technical University of Braunschweig \\
joerg.neunhaeuserer@web.de
\end{center}
\begin{center}
\begin{abstract}
We show that Somos' constant is universal in sense that is similar to the universality of the Khinchin constant. In addition we introduce generalized Somos' constants, which are universal in a similar sense.\\
{\bf MSC 2010: 11K55, 37A25}~\\
{\bf Key-words: Somos' constant, universality, representations of real numbers, ergodic transformations}
\end{abstract}
\end{center}
\section{Introduction and main result}
Let us first recall the Khinchin constant
\[ K=\prod_{i=1}^\infty \left(1+\frac{1}{i(i+2)}\right)^{\log_2 i} = 2.6854520010\dots~~.\]
By the famous theorem of Khinchin \cite{[KI]} this constant is universal in the following sense: For almost all real numbers $x$ the geometric mean of the entries of the continued fractions of $x$ converges to $K$. We consider here Somos' constant
\[ \sigma=\prod_{i=1}^\infty \sqrt[2^i]{i}= 1.6616879496\dots~~,\]
which first appeared in \cite{[SO]} in the context of the quadratic recurrence $g_{n}=ng_{n-1}^{2}$, see also page 446 of \cite{[FI]}.
In the recent past this constant raised some attention, see for instance \cite{[GS],[M],[SL]}. We will show that the Somos' constant is universal in a sense that is similar to the universality of the Khinchin constant. In \cite{[NE]} we represent real numbers $x\in(0,1]$ in the form
\[ x=\langle n_{1},n_{2},n_{3},\dots\rangle:=\sum_{k=1}^{\infty}2^{-(n_{1}+n_{2}+\dots+n_{k})}\]
with $n_{k}\in\mathbb{N}$ and show that the representation is unique. Replacing the continued fraction representation by this representation, we obtain the universality of Somos' constant.
\begin{theorem}
For almost all $x=\langle n_{1},n_{2},n_{3},\dots\rangle\in(0,1]$ we have
\[ \lim_{i\to\infty} \sqrt[i]{n_{1}n_{2}\dots n_{i}}=\sigma.\]
\end{theorem}
We will prove this theorem in the next section. In the last section we will introduce generalized Somos' constants, which are universal with respect to a modification of the representation used here.
\section{Proof of the main result}
Consider the map $T:(0,1]\to(0,1]$, given by $T(x)=2^{i}x-1$ for $x\in (1/2^{i},1/2^{i-1}]$ and $i\in\mathbb{N}$. The relation of this transformation to the expansion of real numbers, defined in the last section is given by
\begin{lemma} Let $x=\langle n_{1},n_{2},n_{3},\dots\rangle\in(0,1]$.
For all $k\in\mathbb{N}$ we have $T^{k-1}(x)\in (1/2^{i},1/2^{i-1}]$ if and only if $n_{k}=i$.
\end{lemma}
\proof Obviously $T(\langle n_{1},n_{2},n_{3},\dots\rangle)=\langle n_{2},n_{3},n_{4},\dots\rangle$. Since $x\in (1/2^{i},1/2^{i-1}]$ if and only if $n_{1}=i$ the result follows immediately. \qed\\~\\
To apply Birkhoff's ergodic theorem we prove:
\begin{proposition}
The Lebesgue measure $\mathfrak{L}$ is ergodic with respect to $T$.
\end{proposition}
\proof
For an open interval $(a,b)\subseteq [0,1]$ we have
\[ \mathfrak{L}(T^{-1}((a,b)))=\mathfrak{L}\left(\bigcup_{i=1}^{\infty}(a/2^{i}+1/2^{i},b/2^{i}+1/2^{i})\right)\]
\[ = \sum_{i=1}^{\infty}2^{-k}\mathfrak{L}\left((a/2^{i}+1/2^{i},b/2^{i}+1/2^{i})\right)=\sum_{i=1}^{\infty}2^{-i}(b-a)=b-a=\mathfrak{L}((a,b)).\]
Hence $\mathfrak{L}(T^{-1}(B))=\mathfrak{L}(B)$ for all Borel sets $B\subseteq (0,1]$, which means that $\mathfrak{L}$ is invariant under $T$.
Let $B$ be a Borel set with $\mathfrak{L}(B)<1$, which is invariant under $T$; that is $T(B)=B$.
Note that for all $k\in\mathbb{N}$ the intervals of the form
\[ I_{m_{1},\dots,m_{k}}=\{\langle n_{1},n_{2},n_{3},\dots\rangle|n_{i}=m_{i}\mbox{ for }i=1,\dots,k\}\]
build a partition of $(0,1]$, where the length of the partition elements is bounded by $1/2^{k}$. By Lebesgue's density theorem for every $\epsilon>0$ there is an interval $I= I_{m_{1},\dots,m_{k}}$ such that $\mathfrak{L}(I\backslash B)\ge (1-\epsilon)\mathfrak{L}(I)$. Since $T^{k}(I)=(0,1]$ we have
\[ \mathfrak{L}((0,1]\backslash B)\ge \mathfrak{L}(T^{k}(I\backslash B))\ge (1-\epsilon)\mathfrak{L}(T^{k}(I))= 1-\epsilon.\]
Hence $\mathfrak{L}(B)=0$. This proves that $\mu$ is ergodic.
\qed.\\~\\
Now we are prepared to prove Theorem 1.1. Let $f(x)=\sum_{i=1}^{\infty}\log(i)\chi_{(1/2^{i},1/2^{i-1}]}(x)$, where $\chi$ is the characteristic function.
By lemma 2.1  we have $f(T^{k-1}(x))=\log(n_{k})$ for $x=\langle n_{1},n_{2},n_{3},\dots\rangle$.
Applying Birkhoff's ergodic theorem to $T$ with the $L^1$-function $f$, we obtain
\[ \lim_{i\to\infty}  \frac{1}{i}\sum_{k=1}^{i}\log(n_{k})=\lim_{i\to\infty}\frac{1}{i}\sum_{k=1}^{i}f(T^{k-1}(x))=\int_{0}^{1} f(x)dx \]
\[ = \sum_{i=1}^{\infty}\log(i)2^{-i}\]
for almost all $x=\langle n_{1},n_{2},n_{3},\dots\rangle\in(0,1]$. Taking the exponential gives the result.
\section{A generalisation} Let $b\ge 2$ be an integer.
It is easy to show that a real numbers $x\in(0,1]$ has a unique representation in the form
\[ x=\langle n_{1},n_{2},n_{3},\dots\rangle_{b}:=(b-1)\sum_{k=1}^{\infty}b^{-(n_{1}+n_{2}+\dots+n_{k})}\]
with $n_{k}\in\mathbb{N}$, the argument can be found in $\cite{[NE]}$. Now consider the map $T_{b}:(0,1]\to(0,1]$, given by $T_{b}(x)=b^{i}x-(b-1)$ for $x\in ((b-1)b^{-i},(b-1)b^{1-i}]$ and $i\in\mathbb{N}$. Using the argument in the last section with respect to $T_{b}$ instead of $T$ we obtain:
\begin{theorem}
For almost all $x=\langle n_{1},n_{2},n_{3},\dots\rangle_{b}\in(0,1]$ we have
\[ \lim_{i\to\infty} \sqrt[i]{n_{1}n_{2}\dots n_{i}}=\prod_{i=1}^\infty \sqrt[b^i]{i^{b-1}}=:\sigma_{b}.\]
\end{theorem}
We like to call $\sigma_{b}$ for $b>2$ a generalized Somos' constant. These constants are universal with respect to the base $b$ representation $\langle n_{1},n_{2},n_{3},\dots\rangle_{b}$. The generalization given here is slightly different from the generalization of Somos' constant studied in \cite{[SH]}, which is not related to universality.\footnote{They consider $\sqrt[b-1]{\sigma_{b}}$.} \\
We like to end the paper with a nice expression of generalized Somos' constants $\sigma_{b}$ using values of the generalized Euler-constant function
\[ \gamma(z)=\sum_{i=1}^{\infty}z^{i-1}\left(\frac{1}{i}-\log(\frac{i+1}{i})\right),\]
where $|z|\le 1$.
\begin{proposition} For all integers $b\ge 2$ we have
\[\sigma_{b}=\frac{b}{b-1}e^{-\gamma(1/b)/b}.\]
\end{proposition}
\proof We have
\[ \gamma(1/b)=\sum_{i=1}^{\infty}(b^{-i+1}/i-b^{-i+1}\log(i+1)+b^{-i+1}\log(i))\]
\[ =b( \sum_{i=1}^{\infty}b^{-i}/i-\sum_{i=1}^{\infty}b^{-i}\log(i+1)+\sum_{i=1}^{\infty}b^{-i}\log(i))\]
\[ =b(\log(b/(b-1))-b\sigma_{b}/(b-1)+\sigma_{b}/(b-1))=b\log(b/((b-1)\sigma_{b}))\]
using
\[ \sum_{i=1}^{\infty}b^{-i}/i=\log(b/(b-1)) \mbox{ and }\log(\sigma_{b})=\sum_{i=1}^{\infty}b^{-i}(b-1)\log(i).\]
Hence $e^{\gamma(1/b)}=(b/((b-1)\sigma_{b}))^{b}$ and $e^{\gamma(1/b)/b}=b/((b-1)\sigma_{b})$ given $\sigma_{b}=be^{-\gamma(1/b)/b}/(b-1)$.\qed~\\~\\
Estimates of $\gamma(1/b)$ can be found in \cite{[M]}.

config: arxiv


TITLE: Production function (economy)
QUESTION [0 upvotes]: A company produces one good using two factors of production factors. If $x$ and $y$ denotes the units of the factors used by the company, the technology function is given by $F(x; y) = xy^2$. In the market of the production factors the price of the factor $x$ is 5 but the price of $y$ is unknown. However we can observe that when the total production of the company is $Q = 1000$ units the company uses 10 units of $x$ and 10 units of $y$.
(a) If the company acts rationally (it wants to minimize the cost of the production) which is the price of the factor $y$?
(b) Which is the marginal increment of the cost of production if the company decides to (marginally) increase the production?

REPLY [1 votes]: First you should convince yourself that the inputs that firm chooses satisfy $xy^2=1000$. If not, i.e. if $xy^2>1000$ then the firm could reduce cost by choosing an $x$ smaller than 10, therefore coming up with an alternative plan $(\tilde{x},\tilde{y})$ that has a smaller cost than $(10,10)$ but that it still guarantees an output at least 1000.
Once we establish this we can replace $x=\frac{1000}{y^2}$. This means the firm chooses only $y$ (i.e. by choosing $y$ it implicitly chooses $x$).
Problem is $min \{p_x \frac{1000}{y^2}+p_y y\}$. Here $p_x=5$. First order conditions tell you that $p_y=\frac{10000}{y^3}$. Since $y$ is the optimized value, it is $10$, hence $p_y$is 10.
The answer to your last question is given by the Lagrange multiplier, which after you know prices, you find it as $\lambda=.05$


text:
\subsection{Double Roman domination}
\label{sec:doubleRoman}

This problem was first defined in~\cite{doubleRoman} and proved to be NP-complete for bipartite and chordal graphs in~\cite{doubleRomanNPc}.

A \emph{double Roman dominating function} on a graph $G$ is a function $f \colon V(G) \to \{0,1,2,3\}$ having the property that if $f(v) = 0$, then vertex $v$ must have at least two neighbors assigned 2 under $f$ or one neighbor $w$ with $f(w) = 3$, and if $f(v) = 1$, then vertex $v$ must have at least one neighbor $w$ with $f(w) \geq 2$. The \emph{weight} of a double Roman dominating function $f$ is the sum $f(V(G)) = \sum_{v \in V(G)} f(v)$, and the minimum weight of a double Roman dominating function on $G$ is the \emph{double Roman domination number} of $G$.

We can model the Double Roman domination problem as a {\rLCproblem{1}} in the following way:
\begin{itemize}
\item $L_{v} = \{0,1,2,3\}$;

\item $(\WeightSet, \wlesseq, \weightsSum) = (\mathbb{R} \cup \{+\infty\}, \leq, +)$;

\item $\textsc{w}_{v, i} = i$;

\item $\begin{array}[t]{rrlrl}
check(v, c)
		& =		& (c(v) = 0	& \Rightarrow	& (\existsFormula{u,w \in N_G(v)}{u \neq w \land c(u) = c(w) = 2})\\
		&		&			& \lor			& (\existsFormula{u \in N_G(v)}{c(u) = 3}))\\
		& \land	& (c(v) = 1	& \Rightarrow	& \existsFormula{u \in N_G(v)}{c(u) \geq 2}).
\end{array}$
\end{itemize}

It is easy to see that some of its variants, such as perfect~\cite{perfectDoubleRoman}, independent~\cite{independentDoubleRoman}, outer independent~\cite{outerindepDoubleRoman} and total~\cite{totalDoubleRoman}, can be modeled by making slight modifications to the previous items.

By Corollary \ref{col:mainCorollary}, the Double Roman domination problem can be solved in polynomial time for graphs of bounded treewidth. However, we show a constant partial neighborhood system and therefore obtain a linear-time algorithm:
\begin{itemize}
\item $\NeighborhoodSet{v}{i} = \{0,1,2\} \times \{0, 1\}$;

\item $n \neighborhoodSum{v}{i} n' = (\min(n_1 + n'_1, 2), \min(n_2 + n'_2, 1))$;

\item $\toNeighborhood{v}{i}(u, j) = \begin{cases}
(0, 0) & \text{if } j \leq 1 \\
(1, 0) & \text{if } j = 2 \\
(0, 1) & \text{if } j = 3;
\end{cases}$

\item $check_{v,i}(n) = (i=0 \Rightarrow n_1 \geq 2 \lor n_2 \geq 1) \land (i=1 \Rightarrow n_1 + n_2 \geq 1)$.
\end{itemize}

Notice that this partial neighborhood system simply counts (until it saturates) the number of neighbors assigned with 2 and also the ones assigned with 3. For other versions of the double Roman domination problem, we might require to also count the number of neighbors assigned with 0 or 1.


TITLE: What are the generating partitions of the odometer?
QUESTION [5 upvotes]: According to the countable generator theorem, every ergodic invertible measure-preserving transformation has a generating partition.
What are the generating partitions of the dyadic odometer ? I don't find the answer in textbooks. Is there a "canonical" one among them ?

REPLY [4 votes]: Quite a nice one is the two-set partition $A,A^c$, where $A$ is the set of points with an even number of terminal 0's:
$$
A =  \bigcup_{k \geq 0} A_k
$$
where $A_k = \bigl\{(x_1, x_2, \ldots) \mid x_1=\ldots=x_{2k}=0 \text{ and } x_{2k+1}=1 \bigr\}$.
Or, if you work with the odometer acting on the space $[0,1[$:
$$
A = \bigcup_{k \geq 0} \left[\frac{1}{2^{2k+1}}, \frac{1}{2^{2k}} \right[
$$
This partition is generating for the following reason. Code a trajectory $x, Tx, T^2x, \ldots$ by a sequence $(v_0, v_1, \ldots)$ of $a$'s and $b$'s according to whether it's in $A$ or not.
Then you can get the first digit $x_1$ of $x$ by looking at the blocks of four consecutive terms of the $v$ sequence:

if the first digit of $x$ is $x_1=1$, then the block $(v_0,v_1,v_2,v_3)$ of four consecutive terms is one of $aaab$, $abaa$ or $abab$ (that is $(v_0,v_1,v_2,v_3)$ is $a*a*$, with at least one of the stars a $b$);
if $x_1=0$, then $(v_0,v_1,v_2,v_3)$ is $*a*a$ with at least one of the stars a $b$.

In particular, the sets of possible codes are disjoint, so that this information suffices to determine the first digit of $x$.
To get the second digit of $x$, look at blocks of eight consecutive terms:

if the first two digits of $x$ are 00, then the code is $*aba*aba$ with at least one of the $*$'s an $a$;

if the first two digits of $x$ are $01$, then the code is $ba*aba*a$ with at least one of the $*$'s an $a$;

if the first two digits of $x$ are 10, then the code is $aba*aba*$ with at least one of the $*$'s an $a$;

if the first two digits of $x$ are 11, then the code is $a*aba*ab$ with at least one of the $*$'s an $a$;

Again, these sets of possibilities are disjoint, so that one can recover the first and second digits of $x$ from 8 terms of the $v$ sequence.
And so on, looking at the block of $2^{n+1}$ consecutive terms of the $v$ sequence determines the first $n$ digits of $x$.


TITLE: Are there non-continuable functions that become continuable when raised to some power?
QUESTION [5 upvotes]: Let $X$ be a complex algebraic variety (integral, separated scheme of finite type over $\mathbb C$) and $U\hookrightarrow X$ an open subvariety. I will say that $f\in\mathscr O_X(U)$ is continuable if there exists some $g\in \mathscr O_X(X)$ with $g|_U=f$. My question is simply: Can it happen that $f$ is not continuable, but there exists a $k\in\mathbb N^+$ with $f^k$ continuable?
I was debating with a friend who thinks it is possible but could not give an example, while I thought that it is impossible but wasn't able to give a satisfying proof. So, I'd be really glad if you could provide one of the two (that is, a simple example of such a phenomenon or a proof that it can not happen).

REPLY [2 votes]: This answer is a generalization of Martin's answer.
Claim: If $X$ is a normal complex algebraic variety, then any function $f\in \mathcal{O}_X(U)$ with $f^k$ continuable must be continuable too.
Proof:
Notice first that wlog we can choose $U:=\{x\in X:f\in\mathcal{O}_{X,x}\}$ to be the maximal open subset of $X$ such that $f|_U$ is regular. If $U\neq X$, by algebraic Hartog theorem $X\setminus U$ is of purely codimension $1$, which means $X\setminus U$ is a finite union of distinct prime Weil divisors $D_i$.
Since $f^k$ is regular on $X$, we have $\operatorname{ord}_{D_i}(f^k)\geq 0$, hence $\operatorname{ord}_{D_i}(f)=\operatorname{ord}_{D_i}(f^k)/k\geq 0$, which means $f\in \mathcal{O}_{X,\zeta_i}$, where $\zeta_i$ is the generic point of $D_i$. Therefore, $\zeta_i \in U\cap D_i=\emptyset$, we get a contradiction.
Edit: Actually here is a much simpler proof:
Suppose $X$ is covered by affine open subsets $U_j$, then $\mathcal{O}_X(U_j)$ is integrally closed by the normality of $X$. Since $f$ is regular at the generic point of $X$, $f$ belongs to the quotient field of $\mathcal{O}_X(U_j)$ for every $j$. Since $\mathcal{O}_X(U_j)$ is integrally closed and $f^k$ is regular on every $U_j$, so $f$ is also regular on every $U_j$, hence $f$ is regular on $X$.


TITLE: How can this vector component be negative?
QUESTION [0 upvotes]: Problem

Let vector A $ = \langle a, 2\rangle$ and vector B $ = \langle 1, 3\rangle$. For what values of $a$ is the component of A along B
  negative?

Computationally, I know how to get the answer, but intuitively, the component of A along B being negative doesn't make sense to me. If B was along either the x-axis or y-axis, then the component of A being negative would make sense but B is not along any particular axis. B is just pointing in some random direction, so how can the length of the projection of A onto this random direction be given a sign?

REPLY [0 votes]: If I understand the question with what I think is the usual, standard meaning of things, the component of $\;A\;$ along $\;B\;$ is just the length of the orthogonal projection of $\;A\;$ on the subspace $\;Span\{B\}\;$, so:
$$\text{comp}_B\,(A):=\frac{A\bullet B}{\left\|B\right\|}=\frac{(a,2)\bullet (1,3)}{\left\|(1,3)\right\|}=\frac{a+6}{\sqrt{10}}$$
End the argument now.


TITLE: proof inequality . Thomae's function
QUESTION [2 upvotes]: I was looking at this post here, I'm trying to understand but I do not understand this:
Let $x\in \mathbb{Q^+}$. How I can show $x > f(x)$.
The function $f$ is defined by $f(x)=x$ if $x$ is irrational, and $f(x)=p\sin\frac1q$ if $x=\frac{p}q$ is written in lowest terms.

REPLY [1 votes]: If $x>0$ then $\sin x<x$.  Therefore
$$
p \sin\frac 1 q < p\cdot\frac 1 q = \frac p q.
$$


TITLE: I do not know how to proof this. An undirected graph $G$ with $n$ vertexes is connected if that graph has more than $\frac{(n-1)(n-2)}{2}$ edges.
QUESTION [0 upvotes]: I do not know how to proof this.
An undirected graph $G$ with $n$ vertexes is connected if that graph has more than $\frac{(n-1)(n-2)}{2}$ edges.

REPLY [3 votes]: Let $G(V,E)$ be a graph with $|V|=n$ vertices and $|E|=m$ edges. We need to show that if $m>(n-1)(n-2)/2$, then $G$ is connected. It's sufficient to show the contrapositive, i.e., if $m\leq (n-1)(n-2)/2$, then $G$ is disconnected.
First note that a disconnected graph with the most edges is the union of a complete graph on $n-1$ vertices and an isolated vertex. In this case, the disconnected graph has $\binom{n-1}{2}=(n-1)(n-2)/2$ edges. QED.


TITLE: Surface tension molecular theory
QUESTION [3 upvotes]: How does the surface of water behave like a stretched membrane? What's the cause for that?
I know that the surface molecules experience a net force downward so there is dense layer of water molecules in the surface but how does this leads to minimizing the surface area? Please help me clear my doubts.

REPLY [1 votes]: Every one of the water molecules in the bulk liquid exerts a small amount of attractive force upon its nearest neighbors. The total energy of the ensemble of water molecules is minimized when the molecules are pulled close to one another by this force.
When a portion of liquid water is in contact with air, the molecules right at the surface have only half as many nearest neighbors with which to share that force of attraction, and as a result they attract one another more strongly at the surface than they do in the bulk. To minimize their energy, they want to minimize the distance between them, and this makes the surface of the water act like a stretched membrane which strives to minimize its area.


text:
\section{Preliminaries} \label{sec:problem}
\vspace{-3.2pt}

We consider the problem of training an NN to approximate the dynamics of an unknown discrete-time monotonically non-decreasing and stable system~\eqref{eq:system}, by training from sample system trajectories.

We first formally introduce

such systems as follows.

For a given initial state,
we refer to the corresponding solution $x(t)$ of~\eqref{eq:system}
as a \emph{trajectory} of the system.

The system~\eqref{eq:system} is a \emph{monotone} system,
if it preserves some partial order $\preceq$ on $\real^n$.

In this work, we consider a common partial order~\cite{smith_MonotoneDynamicalSystems_2008}
defined as
\begin{equation} \label{eq:preceq}
    x \preceq y \iff x_i \leq y_i \textrm{ for all } i \in [n],
\end{equation}
where
$[n] = \{1, \ldots, n\}$,
$x, y \in \real^n$, and
$x_i, y_i \in \real$ denote their $i$-th entry.

\paragraph*{Monotone systems.}
The system~\eqref{eq:system} is \emph{monotone} on domain $D \subseteq \real^n$
if for any two trajectories $x_1(t), x_2(t) \in D$,
it holds that
\begin{equation}
    x_1(0) \preceq x_2(0) \implies x_1(t) \preceq x_2(t) \textrm{ for all } t \in \nat.
\end{equation}

The system~\eqref{eq:system} is \emph{monotone}
if and only if the function $f$ is monotonically non-decreasing in the common sense -- i.e., if for any two inputs $x_1, x_2 \in \real^n$,
it holds that $x_1 \preceq x_2 \implies {f} (x_1) \preceq {f} (x_2)$.

\begin{example} \label{ex:scaler monotone system}
A scalar linear system $x(t+1) = a x(t)$ is monotone on the domain $[0, +\infty)$ for any $a \geq 0$, since for any two initial states $x_1 (0) \leq x_2 (0)$, the two corresponding trajectories satisfy $x_1(t) = a^t x_1(0) \leq a^t x_2(0) = x_2(t)$.

Therefore, the ordering between the initial states is preserved during the evolution of two trajectories for all times $t \in \nat$.
\end{example}

It is important to highlight that the monotonicity is defined for the (initial) state not for the time,
as illustrated in \cref{fig:path of monotone systems}.

From~\cref{ex:scaler monotone system}, when $a \in (0,1)$, the trajectory $x(t) = a^t x(0)$ decreases with the time $t$; yet, the system is still monotone with respect to the initial state $x(0)$.

The monotonicity of the system~\eqref{eq:system} can be equivalently characterized by the gradient of function $f(x)$, as captured in the following lemma from.

\begin{lemma}{\cite{smith_MonotoneDynamicalSystems_2008}} \label{lem:monotone system}
The system~\eqref{eq:system} is monotone
if and only if $\frac{\partial f}{\partial x_i} \geq 0$
for each entry $x_i, (i=1,...,n)$ of $x\in\mathbb{R}^n$.

Specially, if the system~\eqref{eq:system} is linear
-- i.e., $f(x) = A x$ for some $A \in \real^{n \times n}$,
then it is monotone if and only if
each entry $A_{ij}, (i,j\in[n])$ of the matrix $A$ satisfies that $A_{ij}\geq 0$.
\end{lemma}

\paragraph*{Lyapunov functions.}
The system \eqref{eq:system} is \emph{globally asymptotically stable}
(or \emph{stable} for short), if and only if it has a (discrete-time) Lyapunov function $V(x)$
\cite{khalil_NonlinearSystems_2002}.

Suppose that $x = 0$ is the stable point of the system (i.e., $f(0) = 0$)
in general, a stable point $x_0$ can be moved to $0$
by substituting $x$ with $x - x_0$ in the system \eqref{eq:system}.

Then $V(x)$ should satisfy the \emph{Lyapunov condition} that
\begin{equation} \label{eq:Lyapunov}
    V(0) = 0, \textrm{~~ and ~~} \forall x \neq 0, V(x) > 0, \textrm{~~ and ~~} V\left(f(x)\right) - V(x) < 0.
\end{equation}

The Lyapunov function can be viewed as a `potential' with zero value at the stable point and positive values elsewhere.

For stable systems,
since the \emph{discrete Lie derivative} $V\left(f(x)\right) - V(x)$ is negative,
the (positive) value of the Lyapunov function should decrease along the system path
so that it finally converges to zero at the stable point.

For monotone stable systems, the Lyapunov function
can always be written as the maximum of scalar functions~\cite{smith_MonotoneDynamicalSystems_2008} -- there exist scalar functions $V_i: \real \to \real$ such~that

\begin{equation}
\label{eq:monotone Lyapunov}
V(x) = \max_{i \in [n]} V_i (x_i), \quad \textrm{ for } \quad [n] = \{1, \ldots, n\}.
\end{equation}

\eqref{eq:monotone Lyapunov} provides a foundation of our technique to efficiently learn Lyapunov function of $\hat{f}$ by an~NN~$\hat{V}$.


TITLE: Unwanted curvature in fourier series
QUESTION [1 upvotes]: As a hobby project I'm trying to implement something with a Fourier series for an arbitrary function. I have a library that can integrate real functions numerically. I'm using the complex notation for a Fourier series, so I had to rewrite the expression for $c_n$ to:
$$
c_n = \int_0^1 r(t) \cos{\lambda_n(t)} dt + i \int_0^1 r(t) \sin{\lambda_n(t)} dt
$$
where $\lambda_n(t) \equiv \phi(t) - 2 \pi n t$ and some original function $f(t) = a(t) + i b(t) = r(t) \exp{i \phi(t)}$.
Then, I calculate some number $N$ coefficients and determine:
$$
g(t) = \sum_{n=-N}^{+N} c_n \exp{2 \pi i n t}
$$
Plotting this function $g(t)$ in white and the original function $f(t)$ in red, this results in:

As can be seen, the approximation has some weird curvature. I suspect this is caused by the boundary conditions or the fact that it is periodic and "jumps" from start to end. Any idea what is wrong here? I would like to achieve a similar result as in the last 3B1B video.
Edit: Additional discovery: for $N >> 1$ this effect blows up. See this for $N=75$:

For $N=85$:

REPLY [1 votes]: This issue was caused by using a wrong integration algorithm. Just using https://en.wikipedia.org/wiki/Trapezoidal_rule with small steps (1e-4) MASSIVELY improves the results.


TITLE: $p_0 = 1,~~p_i = c^{p_{i−1}} . $ Find the largest $c$ such that this sequence is bounded.
QUESTION [2 upvotes]: For a given $c$ define the power tower $p$ of $c$ as follows: $$p_0 = 1,~~p_i = c^{p_{i−1}} . $$ Find the largest $c$ such that this sequence is bounded.
Clearly $c=1$ works. So assumer $c>1.$
Note that for $c>1,$ $c^n$ is increasing as $n$ increases.
There are two cases:

There exists $x\ge 1$ such that $c^x\le x$
There does not exist $x\ge 1$ such that $c^x\le x$

Note that if there exists $x\ge 1$ such that $c^x\le x$ and hence we get the sequence $\{c^n\}$ to be bounded above by $x.$ Since $\{c^n\}$ is bounded above then so is $\{p_i\}.$
But I couldn't understand the following part.

After this we have to find the maximum value of $c$ such that there exist $x$ such that $c^x\le x\implies x^{1/x}\le c$ or find maximum value of $c$ which is ( after differentiating) we get $e^{1/e}.$
My doubt is, I get that $c^x-x$ is increasing but it might be bounded above so how can we say that there exists $A$ such that $c^A>A+c-1$  ( we know that $c^x>x$ for all $x>1$).
Moreover, if $c^x-x$ is increasing, then in the interval $[1,A]$ wouldn't $c-1$ be the minimum?
And lastly, I did not understand how $p_i=c^{p_{i-1}}\ge p_{i-1}+m.$ But I understood that if $p_i=c^{p_{i-1}}\ge p_{i-1}+m$, then we must have $p_i\ge 1+m\cdot i.$
Any help or other solution is appreciated.

REPLY [1 votes]: Explicitly writing few terms
$$1, c, c^c , c^{c^c}, c^{c^{c^c}}, \ldots$$
$c=1$ satisfies the boundedness condition. To find better solution let us assume $c>1$. Also, let $t_n$ is the $n$th term of the sequence.
Let the limit of the sequence is denoted by $x$, i.e. $$\lim_{n \to \infty} t_n = x$$ Therefore, $$c^x-x =0$$ because $c>0$.
We want to find the largest $c$ which satisfies this condition. From the equation above
$$ \frac{1}{c} \frac{dc}{d x} =\frac{\log(x) -1}{\log(x)^2}$$
Solving to $x=e$ and $c = e^{1/e}$
p.s.: Edited my completely wrong answer.


TITLE: How to solve $y'=\frac {e^{x-y}}{y-1}$?
QUESTION [1 upvotes]: Could you please give me some hint how to solve this problem.

Suppose that $y(x)$ solves the differential equation $y'=\dfrac {e^{x-y}}{y-1}$ and that $y(x)\to 0$ when $x\to -\infty$.
  Compute $y\left(\ln\left(2-\frac 3e\right)\right)$.

I tried to solve the DE:
$\frac{dy}{dx}=\frac {e^x} {e^y\left(y-1\right)}$
$e^y\left(y-1\right)dy=e^xd$
$\int^y\left(t-1\right)e^tdt-\int^x e^udu=C$
$\left(y-2\right)e^y-e^x=C$
$lim_{x\to-\infty}\left(y-2\right)e^y-\lim_{x\to -\infty}e^x=C \Rightarrow C=0$.
Thus we have $\left(y-2\right)e^y=e^x$,
but this is not helping to evaluate $y\left(ln\left(2-\frac 3e\right)\right)$:
$2-\frac3e=\left(y-2\right)e^y$.
What did I do wrong?

REPLY [2 votes]: One knows that $(y(x)-2)\mathrm e^{y(x)}=\mathrm e^x+C$ for every $x$, for some $C$. If $y(x)\to0$ when $x\to-\infty$, then $(y(x)-2)\mathrm e^{y(x)}\to-2$ when $\mathrm e^x\to0$ hence $C=-2$ and $(y(x)-2)\mathrm e^{y(x)}=\mathrm e^x-2$ for every $x$. If $x_0=\ln(2-3/\mathrm e)$, this yields $G(y(x_0))=-3/\mathrm e$, where $G$ is the function $$G(t)=(t-2)\mathrm e^t.
$$
Unexpectedly (to me), some more work is needed to deduce $y(x_0)$ from this.
Here is a graph of the function $G$:

Thus, the equation $G(t)=-3/\mathrm e$ has two solutions, one negative $t=-1$ and the other positive $t=2+W(-3/\mathrm e^3)\approx1.82$. Note that $y(-\infty)=0$ and that $y'(x)=(y(x)-1)\mathrm e^{y(x)-x}$, hence $y'(x)\lt0$ for every $x$, in particular the function $x\mapsto y(x)$ is decreasing and $y(x_0)\lt0$. Thus, $y(x_0)$ is the negative solution of the equation $G(t)=-3/\mathrm e$, that is, $$y(x_0)=-1.$$


TITLE: If $N \unlhd A_n$, then elements of order $2$ or $3$ in $N$ imply $N$ contains generator
QUESTION [1 upvotes]: I am not assuming we know $A_n$ is simple for this question.
Suppose that $n >4$ and that $N$ is normal in $A_n$. If $p \in N$ is a permuatation that has order either $2$ or $3$, then I want to show $N$ contains either a $3$-cycle or a product of disjoint transpositions.
In either case, we can write $p$ as the product of disjoint $2$-cycles or $3$-cycles, respectively. Then somehow we want to use that $N$ is normal by finding a suitable conjugation of $p$. But I cannot figure out how to do this.

REPLY [3 votes]: In the case of order $2$, note that single 2-cycles are not in $A_n$. Given an element like $\alpha=(1\,2)(3\,4)(\cdots)\in N$ where the dots represent a non-negative number of 2-cyles, we als have its conjugate $\beta=(1\,3)(2\,4)(\cdots)\in N$, as well as their product $\alpha\beta=(1\,4)(2\,3)$. As $n>4$, we also have $(1\,5)(2\,3)$ and the product $(1\,5\,4)$.
In the case of order $3$, suppose we have $\alpha=(1\,2\,3)(\cdots)$ where the dots represent a non-negative number of 3-cyles. Then we also have $\beta=(1\,3\,2)(\cdots)$ and also $\alpha^2\beta=(1\,2\,3)$. By conjugation, we also have $(1\,2\,4)$ and the product $(1\,3)(2\,4)$.


TITLE: Sheaf as a functor
QUESTION [7 upvotes]: Let $X$ be any topological space, $S$ - any category (e.g. of sets). Consider a new category $C$: its objects are only open subsets of $X$ and a set of morphisms from $U$ to $V$ is nonempty if and only if $U\subset V$. In this case we define the only one morphism $\phi_{UV}:U\rightarrow V$, that includes naturally $U$ into $V$. Thus we can define a presheaf of sets on $X$ as a contravariant functor $\Phi$ from $C$ to $S$ (that functor turns inclusion $U\subset V$ into restriction $res_{VU}$ - a morphism in $S$).
Perhaps, the following question is too popular, but i cannot get the answer myself.
Is it possible to define a sheaf in the same way? I mean some simple categorical terms.

REPLY [12 votes]: Yes, it is. If the target category is complete, the gluing axiom translates into $F(U) \rightarrow \prod_i F(U_i) \rightrightarrows \prod_{i,j} F(U_i \cap U_j)$ being a difference kernel diagram. This is actually a very common definition of sheafs, especially since it is no longer dependend on the starting category being the category of open sets of a topological space. On can (and does) define (pre)sheaves on any category in this way. This is especially useful when working with Grothendieck topologies etc.
Greetings.


TITLE: If an electron can be in any orbit around the nucleus in Bohr's model, then what determines the typical atomic size?
QUESTION [2 upvotes]: An electron can goto higher orbits in its excited state so what determines the typical atomic size of an atom
note that I am asking only from the viewpoint of Bohr's orbit.

REPLY [1 votes]: You may use expectation
$$\langle r \rangle_{1s}=\int_0^{\infty} r|\psi_{1s}(r)|^2 4\pi r^2 \, dr=\frac{3a_0}{2}$$
which is three halves of the Bohr radius.
In general, for hydrogen-like atom in particular state (without screening):
$$\langle n\ell |r| n\ell \rangle=
\frac{n^2 a_0}{Z} \left[ \frac{3}{2}-\frac{\ell(\ell+1)}{2n^2} \right]$$


TITLE: Some questions on the Frattini subgroup
QUESTION [3 upvotes]: According to Derek J.S Robinson's A Course in the Theory of Groups, the Frattini subgroup of a group $G$, denoted $\mathrm{Frat}G$ is defined to be the intersection of all maximal subgroups of $G$. When $G$ has no maximal subgroup, $\mathrm{Frat}G$ is set to be $G$ itself. It can be proved that $\mathrm{Frat}G$ is in fact the set of all nongenerators of $G$. (Here, a nongenerator for a group $G$ is an element $a$, such that for any subset $X$ of $G$, if $G = \langle a,X\rangle$, then $G = \langle X\rangle$.) The definition and property can be applied to find the Frattini subgroups of a given group, but I still have problems in doing this.

Find the Frattini subgroup of $D_n$, the dihedral group of order $2n$.

$D_n = \langle a,b | a^n = b^2 =1, abab =1\rangle$. The subgroup $\langle a\rangle = \{ 1, a, a^2, \cdots, a^{n-1} \}$ is maximal in $D_n$, so $\mathrm{Frat}D_n \subseteq \langle a\rangle$. If $n$ is odd, then $\langle b\rangle = \{ 1, b \}$ is also a maximal subgroup of $D_n$. $\langle a\rangle \cap \langle b\rangle = 1$, so $\mathrm{Frat}D_n =1$. But what will happen if $n$ is even? Do I have to test whether $a^k$ is a nongenerator for any integer $k$, $1<k<n-1$?

Prove that the Frattini subgroup of $S_n$ is trivial.

I think of the proof in these two directions: no nontrivial element of $S_n$ is contained in every maximal subgroup of $S_n$, and, no nontrivial element of $S_n$ is a nongenerator. As $A_n$ is a maximal subgroup, $\mathrm{Frat}S_n \subseteq A_n$. Moreover, the Frattini subgroup is normal in $S_n$, so if $n \geq 5$, this subgroup must be $A_n$ or trivial. If for any given nontrivial element of $A_n$, I could find some maximal subgroup of $S_n$ not containing it, then I could prove the result. But I don't know how to find them.

Let $G$ be a finite $p$-group. Then $\mathrm{Frat}G = G'G^p$.

In the proof, the author said:

As $G$ is nilpotent and satisfies the normalizer condition, we have $M \lhd G$ for a maximal subgroup $M$ of $G$. Moreover, $|G:M| = p$. Hence $G'G^p \leq \mathrm{Frat}G$.

I don't know why this is true. It is easy to see that $G^p \leq \mathrm{Frat}G$, because $G^p$ is contained in every maximal subgroup of $G$. But why is $G'$ contained in $ \mathrm{Frat}G$? This is not true in general, for example, let $G = S_3$, then $a \in G'$ but $a \notin \mathrm{Frat}G$. Is it true for all the $p$-groups? all the finite $p$-groups? Why is it true?
Thank you very much.

REPLY [2 votes]: If $1 \le d < n$ and $d$ is a divisor of $n$, then $\langle b, a^d \rangle$ is a proper subgroup of $D_n$.  Which of these subgroups are maximal?  What do you get when you intersect the maximal ones?


text:
\section{Appendix: Complete Proof of LS-CS Stability}
We prove Theorem \ref{stabres2} by induction. First notice that using condition \ref{initass} of the theorem, the result holds for $t=0$. This proves the base case. For the induction step, assume that the result holds at $t-1$, i.e. $\tDelta_{t-1} \subseteq \Sset_{t-1}(d_0)$, and $|\tDelta_{t-1}| \le (2d_0-2)S_a$,  $|\tDelta_{e,t-1}| =0$, $|\tT_{t-1}| \le S_0$. Use this to prove that the result holds for $t$.

Since $T_t = \tT_{t-1}$, so $|T_t| \le S_0$.

Notice that $N_t = N_{t-1} \cup \Aset_t \setminus \Rset_t$, $\Rset_t \subseteq N_{t-1}$ and $\Aset_t \subseteq N_{t-1}^c$. Since $\Delta_{e,t} = \Nhat_{t-1} \setminus N_t =  \Nhat_{t-1} \cap [(N_{t-1}^c \cap \Aset_t^c) \cup \Rset_t] \subseteq \tDelta_{e,t-1} \cup \Rset_t = \Rset_t$. The last equality follows because $|\tDelta_{e,t-1}| =0$. Thus $|\Delta_{e,t}| \le |\Rset_t| = S_a$.

Now, $\Delta_t = N_t \setminus \Nhat_{t-1}  = (N_{t-1} \cap \Nhat_{t-1}^c \cap \Rset_t^c) \cup (\Aset_t \cap \Nhat_{t-1}^c) = (\tDelta_{t-1} \cap  \Rset_t^c) \cup (\Nhat_{t-1}^c \cap \Aset_t)$. Thus, $\Delta_t$ will be largest if $\tDelta_{t-1}$ is largest and is disjoint with $\Rset_t$ (i.e. $\tDelta_{t-1} =\Sset_{t-1}(d_0)$) and $\Nhat_{t-1}$ is disjoint with $\Aset_t$.
Thus, in the worst case, $\Delta_t =(\Sset_{t-1}(d_0)  \cap  \Rset_t^c) \cup \Aset_t$. If $d_0>1$, $\Sset_{t-1}(d_0)$ is not empty, $\Rset_{t} \subseteq \Sset_{t-1}(d_0)$ and $\Aset_t \subseteq \Sset_{t-1}(d_0)^c$, and thus in the worst case, $|\Delta_t| = |\Sset_{t-1}|-|\Rset_t|+|\Aset_t| =(2d_0-2)S_a - S_a + S_a = (2d_0-2)S_a$. If $d_0=1$, $\Delta_t = \Aset_t$ and so $|\Delta_t|=S_a$.
Thus, $|\Delta_t| \le \max(S_a,(2d_0-2)S_a)=k^a$.

From the above discussion, $\Delta_t \subseteq \Sset_{t-1}(d_0) \cup \Aset_t \setminus \Rset_t$. So, in the worst case, $\Delta_t$ contains $S_a$ elements each with magnitudes $M/d, 2M/d \dots d_0M/d$ and $S_a$ elements each with magnitudes $M/d, 2M/d \dots (d_0-2)M/d$.
Consider the detection step at time $t$. Apply Lemma \ref{detectcond} with $S_T = S_0$, $S_\Delta = k^a$, $b= d_0M/d$ and $\gamma=1$. With this choice of $\gamma$, $\Delta_1 \subseteq \Iset_t(d_0)$ and so $|\Delta_1| \le S_a$. Also, $\Delta_2 \subseteq \Sset_t(d_0) \setminus \Dset_t(d_0)$. Thus $\|x_{\Delta_2}\|^2 \le e^2$.
Since conditions \ref{measmodel} and \ref{add_del} hold, all the undetected elements of $\Iset_{t}(d_0)$ will definitely get detected at time $t$. Thus $\tDelta_{\dett,t} \subseteq \Delta_t \setminus \Iset_{t}(d_0) \subseteq \Sset_{t-1}(d_0) \cup \Aset_t \setminus (\Rset_t \cup \Iset_{t}(d_0)) = \Sset_t(d_0) \setminus \Dset_{t}(d_0)$. The last equality follows from (\ref{sseteq}). Thus, $|\Delta_{\dett,t}| \le \max(0,(2d_0-3)S_a)=k^d$.

 In the detection step, there are at most $f$ false additions (from condition \ref{addthresh}) and thus $|\tDelta_{e,\dett,t}| \le |\Delta_{e,t}| + f \le S_a+f$.

Also, $|\tT_{\dett,t}| \le |N_t| + |\tDelta_{e,\dett,t}| \le S_0+S_a+f$.

Apply Lemma \ref{nofalsedelscond} with $S_T = S_0+S_a+f$, $S_\Delta = k^d$, $b_1 = d_0M/d$, and use the fact that $\tDelta_{\dett,t} \subseteq \Sset_t(d_0) \setminus \Dset_{t}(d_0)$ to bound $\|x_{\tDelta_{\dett}}\|^2$ by $\sm$. Since conditions \ref{measmodel} and \ref{add_del} hold, all elements of $\tT_\dett$ with magnitude greater than or equal to $b_1=d_0 M/d$ will definitely not get falsely deleted. But nothing can be said about the elements of $\Dset_{t}(d_0)$ (or other smaller ones which may have got added) and in the worst case $\tDelta_t$ may contain all of these elements, i.e. it may be equal to $\Sset_t(d_0)$. Thus, $\tDelta_t \subseteq \Sset_t(d_0)$ and so $|\tDelta_t| \le (2d_0-2)S_a$.

Apply  Lemma \ref{truedelscond} with $S_T = S_0+S_a+f$, $S_\Delta = k^d$ and use $\tDelta_{\dett,t} \subseteq \Sset_t(d_0) \setminus \Dset_{t}(d_0)$ to bound $\|x_{\tDelta_{\dett}}\|^2$ by $\sm$. Since condition \ref{delthresh} holds, all elements of $\tDelta_{e,\dett,t}$ will get deleted by the deletion step. Thus $|\tDelta_{e,t}|=0$.

Also, $|\tT_t| \le |N_t| + |\tDelta_{e,t}| \le S_0$.

This finishes the proof of the induction step and hence of the theorem.

For the first part of Corollary \ref{stabres2_cor}, apply Theorem \ref{stabres2} to bound $|\tDelta_{t-1}|,|\tT_{t-1}|,|\tDelta_{e,t-1}|$. The discussion above then gives the desired bounds on $|\Delta_t|,|T_t|,|\Delta_{e,t}|$ and $|\tDelta_{\dett,t}|,|\tT_{\dett,t}|,|\tDelta_{e,\dett,t}|$.
The second and third part follow using $\tDelta \subseteq \Sset_t(d_0)$ and $\Delta \subseteq \Sset_{t-1}(d_0) \cup \Aset_t \setminus \Rset_t$.