TITLE: Complex n-th root question
QUESTION [0 upvotes]: Let $m$ and $n\neq0$ be any two integers.Show that $z^{m/n}=\left(z^{1/n}\right)^m$ has $n/(n,m)$ distinct values, where $(n,m)$ is the greatest common divisor of $n$ and $m$. Prove that the sets of values of $(z^{1/n})^m$ and $(z^{m})^{1/n}$ coincide if and only if $n$ and $m$ are relatively prime.
For the first part as much  as I know the $n$th root of $z^{m}$ would have n distinct values,which differ by $\frac{2km\pi}{n}$ in the argument where $k=0,1,...,n-1$. So supposing that $(n,m)=a$ I would get  $$\frac{2k\frac{m}{a}\pi}{\frac{n}{a}},$$ where $k=0,1,2...,\frac{n}{a}-1$, but I am not sure how to finish this. I mean the second part.

REPLY [3 votes]: Without loss of generality we take $|z| = 1$. Write $z$ in polar form as $z=e^{i\theta} = e^{i\theta + 2\pi ki}$. Then
$$z^{\frac{1}{n}} = \exp\left(\frac{i\theta}{n} + \frac{2\pi ki}{n}\right)$$
$$z^{\frac m n} = \exp\left(\frac{im\theta}{n} + \frac{2m\pi ki}{n}\right)$$
as $k$ ranges through the integers, the value repeats itself if and only if $\frac{2m\pi k}{n}$ becomes a multiple of $2\pi$. This happens if and only if $\frac{mk}{n}$ is integral. If we let $d=\gcd(m,\ n)$ and write $m = m'd,\ n=n'd$ then
$$\frac{mk}{n} = \frac{m'k}{n'}$$
so that the exponents complete a full cycle of values whenever
$$k\equiv 0\pmod{n'}$$
so that there are precisely $n' = \frac n d$ distinct values of $z^{\frac m n}$
For the second part, consider the difference in the two expressions
$$\left(z^{\frac 1 n}\right)^m = \exp\left(\frac{im\theta}{n} + \frac{2m\pi ki}{n}\right)$$
and on the other hand
$$\left(z^{m}\right)^{\frac 1 n} = \left[\exp\left(im\theta + 2km\pi i\right)\right]^{\frac 1 n} = \left[\exp\left(im\theta + 2k\pi i\right)\right]^{\frac 1 n} = \exp\left(\frac{im\theta}{n} + \frac{2k\pi i}{n}\right)$$
Notice that the second expression takes on different values as $k$ ranges through
$$k \equiv 0, 1,\ 2, \cdots, n-1 \pmod n$$
If the first set takes on corresponding values, that means $mk\pmod n$ must be a rearrangement of the residues $0,\ 1,\ cdots,\ n-1$. In particular, this means that the linear congruence
$$mk \equiv 1 \pmod{n}$$
has a solution for $k$ so that $\gcd(m, n)$ must divide $1$ by the linear congruence theorem. Conversely, if $m$ and $n$ are coprime, then
$$mk \equiv mk'\pmod{n} \implies k\equiv k'\pmod{n}$$
so that the former ranges through the entire residue classes of the latter and hence takes on the same set of $n$ values.


TITLE: Energy extraction from the layers of sea water?
QUESTION [1 upvotes]: Could it be possible/reasonable to extract Peltier thermoelectric energy by using a long rod of suitable material that connects the cold bottom of the ocean and the warm surface layer?

REPLY [3 votes]: Not really. If you use highly thermally conductive rods (like copper or aluminium), the depth of the ocean (let's say is about the order of 5 km) is so much greater than the radius of the metallic rod that radial heat conduction will dominate over the longitudinal one. It means that you won't be able to carry the "coldness" (or "hotteness" for that matter) of deep underwater temperatures to the surface. What will happen instead is that the temperature of the rod will be roughly the same as the temperature of the water around it. So, at the surface where your Peltier element (or better to say, thermoelectric generator) there is virtually no temperature difference due to the rod (or heat exchanger).  The rod will be essentially entirely useless. And this remains true even if you somehow encapsulate the rod in a vacuum cylinder so that the rod makes no contact with water. In that case the thermal resistance of the rod (which scales linearly with its length) is so big that the difference in temperature between the top of the rod and the bottom of the rod will be vanishingly small compared to the temperature difference between the surface of the ocean and its bottom.
In other words, no matter what you do with the rod, using a rod between the surface of the ocean and its bottom in order to establish a temperature difference across a thermoelectric generator will not work as you expect. The rod will not be able to transfer the heat from the bottom to the top, or viceversa. An analogy with an electrical circuit would be that you have in mind that using a copper wire between two elements will be fine because it conducts electricity well. However the reality is that the wire is such that its resistance is insanely high despite that its conductivity is also very high. For example if the wire has a diameter of a few micrometers and a length of millions of kilometers, you end up having a resistance of mega ohms despite the fact that you used copper. That's roughly equivalent to the case you have in mind.


TITLE: Recursive sequence, $x_{1} \geq 0, x_{n+1}=\sqrt{x_{n}+2}$
QUESTION [5 upvotes]: Recursive sequence, $x_{1} \geq 0, x_{n+1}=\sqrt{x_{n}+2}$ and it is requested to prove that $\lim_{n \to \infty} x_n=2$.
This is a common problem, but I found it quite more difficult when the value of $x_1$ is not established, as we cannot say it is a monotonous sequence, and as a part of the exercise we are given a step and a suggestion, which I am pretty sure they can be done by induction. Nevertheless, I am still unable to see the path to prove it.

$|x_{n+1}-2|\leq\frac12|x_n-2|, \forall n \in \mathbb{N}$, and as a suggestion for the problem, note that
$|x_n-2|\leq \frac1{2^{n-1}}|x_1-2|, \forall n \in \mathbb{N}.$

REPLY [7 votes]: $$x_{n+1}=\sqrt{x_n+2}$$
$$x_{n+1}^2=x_n+2$$
$$x_{n+1}^2-4=x_n-2$$
$$(x_{n+1}-2)(x_{n+1}+2)=x_n-2$$
We have $x_{n+1}+2 \ge 2$ since $x_{n+1} \ge 0$.
$$x_{n+1}-2 = \frac{x_n-2}{x_{n+1}+2}$$
$$|x_{n+1}-2| \le \frac{|x_n-2|}{2}$$
Hence we have $$|x_{n+1}-2 | \le \frac{|x_1-2|}{2^n}$$
Hence as $n \to \infty$, we have $x_n \to 2$.


TITLE: How to not use stars and bars to solve unordered sampling with replacement?
QUESTION [2 upvotes]: Say I want to find the number of different ways of unordered sampling k elements from n distinguishable elements with replacement, and a common way is to use the method of stars and bars. Although this is artful and easy to understand, I want to know whether there are other methods to solve this problem, and if there are, what are those methods? And the more direct, the rawer, the better!
BTW, I have searched in a few textbooks on this problem, and it seems all of them are using the above method.

REPLY [4 votes]: If you multiply out $(1+x+x^2+\ldots)^n$ distributively then you get a term $x^{k_1}x^{k_2}\>\cdots\> x^{k_n}$ for each independent sampling of a term  in each of the factors. Collecting terms according to increasing powers of $x$ we obtain a formal series $\sum_{k\geq0} a_k\>x^k$, whereby $a_k$ counts the number of samplings $(k_1,k_2,\ldots,k_n)$ with $k_1+k_2+\ldots+k_n=k$.
Now
$$(1+x+x^2+\ldots)^n={1\over (1-x)^n}=\sum_{k=0}^\infty{-n\choose k} (-x)^k$$
(binomial series). It follows that
$$a_k=(-1)^k{-n\choose k}={n+k-1\choose k}\ .$$


TITLE: Is anything lost by simply saying that a charge is nothing but the divergence of a field?
QUESTION [4 upvotes]: If we propose that 1) there are fields and 2) that some fields can have places where they converge (that is to say, places where del dot the field is not zero) then we have the definition of such a region, and we might as well call it a charge. In other words, instead of proposing that a a divergence is non-zero due to a charge, why should we not simply define a charge as a divergence / convergence of a field? Is there any aspect or property of a charge (of any kind, not just electric) that is not adequately described by this definition?

REPLY [1 votes]: Basically, electromagnetism requires (a) electric and magnetic fields and (b) sources for those fields, i.e. charge density and current. The point of Maxwell's equations then is to relate the fields to the charges so that you can determine $\vec E$ and $\vec B$ from $\rho$ and $\vec j$ (up to vacuum solutions). In other words, the equation $\vec\nabla \cdot \vec E=\rho$ gives a connection between two quantities, and you know one of those, you can deduce something about the other one. Furthermore, you can deduce charge conservation as a nontrivial condition on $\rho$ and $\vec j$, and that conservation really motivates the interpretation of current as flow of charge.
On the other hand, if you simply define $\tilde\rho:=\vec\nabla \cdot \vec E$ (and presumably analogously for the current $\vec{\tilde{j}}$), Maxwell's equations are empty: You have fields $\vec E$ and $\vec B$, and the divergence of $\vec E$ is an arbitrary quantity you have just decided to call $\tilde \rho$. (Also, charge conservation becomes an empty identity: $\partial_t \tilde\rho =\vec\nabla \cdot \vec{\tilde j} $, but that is just shorthand for $\partial_t \vec{\nabla} \cdot \vec{E} =\vec\nabla\cdot \left(\vec\nabla\times\vec B +\partial_t \vec E\right)$, i.e. an identity and not a relation between physical quantities.
The situation is maybe somwhat similar to the equation of motion inclassical mechanics: in $m\vec a=\vec F$, you need an independent expression for $\vec F$ to solve for the trajectory of the particle.


text:
{\bf Problem.} How many distinct three-digit positive integers have only odd digits?
{\bf Level.} Level 2
{\bf Type.} Counting & Probability
{\bf Solution.} There are 5 odd digits, and thus, each digit has 5 different possibilities, making for $5^3 = \boxed{125}$ possibilities.

set_name: MATH


text:
\begin{document}

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\numberwithin{equation}{section}

\title{Quasiregular mappings of polynomial type in $\R^{2}$}

\author{Alastair Fletcher and Dan Goodman \footnote{The first author is supported by EPSRC grant EP/G050120/1.} }

\maketitle

\begin{abstract}

Complex dynamics deals with the iteration of holomorphic functions. As is well-known, the first functions to be studied which gave non-trivial dynamics were quadratic polynomials,
which produced beautiful computer generated pictures of Julia sets and the Mandelbrot set. In the same spirit, this article aims to study the dynamics of the simplest non-trivial quasiregular mappings. These are mappings in $\R^{2}$ which are a composition of a quadratic polynomial and an affine stretch.

MSC2010: 30C65 (Primary), 30D05, 37F10, 37F45 (Secondary)

\end{abstract}

\begin{figure}
  \caption{The set of non-escaping points for $f(z) = (h_{0.8,0}(z))^{2}-0.21-0.78i$, a Douady dragon.}\label{0.8K_-0.21-0.78icfig}
\begin{center}
    \includegraphics{0.8K_-0.21-0.78ic.eps}
\end{center}
\end{figure}

\section{Introduction}

The field of complex dynamics, the study of iteration of analytic functions in the plane, goes back nearly a century to Fatou and Julia. However, there has been a surge of recent interest in the field, following on from computer generated pictures of Julia sets and the Mandelbrot set and led by the work of Douady and Hubbard, e.g. \cite{DH}. This illustrated how an iterative system with a very simple description, namely a quadratic polynomial, could have very complicated behaviour. There are several excellent introductions to the theory, for example \cite{Beardon, Milnor, MNTU}.

More recently, the iteration of quasiregular mappings in $\R^{n}$ has been studied, motivated by the fact that several key tools in complex dynamics have analogues for quasiregular mappings, for example Rickman's theorem generalizing Picard's theorem, and Montel's theorem. In fact, direct analogues of the Fatou and Julia sets can be defined for a special class of quasiregular mappings, all of whose iterates have distortion bounded by some fixed number. These are the so-called uniformly quasiregular mappings, introduced in \cite{IM} and studied in a number of papers. We restrict ourselves to mentioning \cite{HMM} and the interested reader can find further references contained therein.

For general quasiregular mappings, it is no longer possible to define the Fatou set for the simple reason that the iterates will have no common bound on the distortion (see the standard references \cite{Rickman,IM2} for the theory of quasiregular mappings). It is however possible to define the escaping set $I(f)$, the set of those points which iterate to infinity, which is a key object in complex dynamics. It is well known that for an analytic function, the boundary of $I(f)$ coincides with the Julia set. It is therefore natural to consider $\partial I(f)$ for quasiregular mappings and see to what extent it can be considered an analogue of the Julia set. Recent papers in this direction include \cite{Bergweiler,BE,BFLM}.

Of particular relevance here is \cite{FN}, where it was shown that for quasiregular mappings of polynomial type, as long as the degree of the mapping is larger than the distortion, $\partial I(f)$ is an infinite, completely invariant perfect set. Further, the family of iterates of $f$ is not equicontinuous at any point of $\partial I(f)$.

In this paper, in the same spirit as the study of iteration of quadratic polynomials, we aim to analyze the boundary of the escaping set for the simplest quasiregular mappings with non-trivial dynamics; namely the composition of quadratic polynomials and quasiconformal mappings with constant complex dilatation.

The outline of the paper is as follows. In section 2, we will cover some preliminary definitions and results. In section 3, a canonical form for the type of functions we are interested in will be derived.
In section 4, generalizations of results in \cite{FN}, on the boundary of the escaping set, to our functions will be obtained.
Section 5 deals with the connectedness or not of $\partial I(f)$, and section 6 introduces a generalization of the Mandelbrot set and covers some of its properties.

The images in this paper were computed using the Python programming language and the
NumPy (Numerical Python) extension package. The code is available on the
second authors' website
\url{http://thesamovar.net/mathematics/qrdynamics}.

The authors would like to thank Dan Nicks for interesting and stimulating discussions.

\section{Preliminaries}

We first collect some definitions and results that we will use. A quasiregular mapping $f:G \rightarrow \R^{n}$ from a domain $G \subseteq \R^{n}$ is called quasiregular if $f$ belongs to the Sobolev space $W^{1}_{n, loc}(G)$ and there exists $K \in [1, \infty)$ such that
\begin{equation}
\label{eq2.1}
\av f'(x) \av ^{n} \leq K J_{f}(x)
\end{equation}
almost everywhere in $G$. Here $J_{f}(x)$ denotes the Jacobian determinant of $f$ at $x \in G$. The smallest constant $K \geq 1$ for which (\ref{eq2.1}) holds is called the outer distortion $K_{O}(f)$ of $f$. If $f$ is quasiregular, then we also have
\begin{equation}
\label{eq2.2}
J_{f}(x) \leq K' \inf _{\av h \av =1} \av f'(x) h \av ^{n}
\end{equation}
almost everywhere in $G$ for some $K' \in[1, \infty)$. The smallest constant $K' \geq 1$ for which (\ref{eq2.2}) holds is called the inner distortion $K_{I}(f)$ of $f$. The maximal distortion $K=K(f)$ of $f$ is the larger of $K_{O}(f)$ and $K_{I}(f)$. In dimension $2$, we have $K_{O}(f) = K_{I}(f)$.

The degree of a mapping is the maximal number of pre-images and is in direct analogue with the degree of a polynomial. A quasiregular mapping is said to be of polynomial type if its degree is uniformly bounded at every point, or equivalently, if $\av f(x) \av \rightarrow \infty$ as $\av x \av \rightarrow \infty$.

\begin{theorem}[\cite{FN}]
\label{FN}
Let $n \geq 2$ and $f:\R^{n} \rightarrow \R^{n}$ be $K$-quasiregular of polynomial type. If the degree of $f$ is greater than the inner distortion $K_{I}$, then $I(f)$ is a non-empty open set and $\partial I(f)$ is perfect.
\end{theorem}

Denote by $B(f)$ the branch set of $f$, that is, the set where $f$ is not locally injective.
A quasiconformal mapping is an injective quasiregular mapping. The following result says that in dimension $2$, a quasiregular mapping can be factorized into two mappings, one of which deals with the distortion and one which deals with the branch points.

\begin{theorem}[Stoilow factorization, see for example \cite{IM2} p.254]
\label{Stoilow}
Let $f:\C \rightarrow \C$ be a quasiregular mapping. Then there exists an analytic function $g$ and a quasiconformal mapping $h$ such that $f = g \circ h$.
\end{theorem}

Stoilow factorization tells us what the branch set of a quasiregular mapping in $\C$ can be.

\begin{corollary}
Let $f:\C \rightarrow \C$ be quasiregular. Then $B(f)$ is a discrete set of points. In particular, if $f$ is quasiregular of polynomial type,
then $B(f)$ is a finite set of points.
\end{corollary}

\section{Canonical form}

It is well-known that every quadratic polynomial is linearly conjugate to $z^{2}+c$ for an appropriate $c \in \C$. We will find an analogous canonical form for compositions of quadratic polynomials and affine stretches.

Consider a quasiconformal mapping $h_{K,\theta}:\C \rightarrow \C$ which stretches by a factor $K >0$ in the direction $e^{i\theta}$. If $\theta =0$,
then $h_{K,0}(x+iy) = Kx+iy$. For general $\theta$, pre-compose $h_{K,0}$ by a rotation of $-\theta$ and post-compose by a rotation of $\theta$ to give the expression
\begin{equation*}
h_{K,\theta}(x+iy) = x(K\cos ^{2} \theta + \sin ^{2} \theta) + y(K-1)\sin \theta \cos \theta
\end{equation*}
\begin{equation}
\label{eq3.4}
+ i \left [ x(K-1)\cos \theta \sin \theta + y (K \sin ^{2} \theta + \cos ^{2} \theta ) \right ]
\end{equation}
or
\begin{equation}
\label{eq3.4a}
h_{K,\theta}(z) = \left ( \frac{ K+1}{2} \right ) z + e^{2i \theta}\left ( \frac{ K-1}{2} \right )\zbar.
\end{equation}
Using the formula for complex dilatation (see \cite{FM}), we see that
\begin{equation}
\label{eq3.3}
\mu _{h_{K,\theta}} = e^{2i\theta} \frac{K-1}{K+1},
\end{equation}
and so $\av \av \mu_{h_{K,\theta}} \av \av _{\infty} <1$ which means that $h_{K,\theta}$ is quasiconformal with constant complex dilatation.
Solving the Beltrami equation $f_{\zbar} = \mu f_{z}$ (see \cite{FM}) with the complex dilatation of (\ref{eq3.3}) gives a quasiconformal map which is unique if we require the solution the fix three points.
Therefore given $K>0$ and $\theta \in[-\pi,\pi]$, the unique solution of the Beltrami equation with dilatation (\ref{eq3.3}) which fixes $0, \infty$ and $e^{i(\theta + \pi/2)}$ is the stretch given by (\ref{eq3.4}).

Given a stretch $h_{K,\theta}$, we will represent it as a point in $\C \setminus \{ 0 \}$, given by the point $Ke^{i\theta}$.
Note that a stretch of factor $1$ in any direction is just the identity.

\begin{proposition}
\label{canonicalform}
Let $f:\C \rightarrow \C $ be a composition of a quadratic polynomial and an affine stretch of the form (\ref{eq3.4}). Then $f$ is linearly conjugate to
\begin{equation}
\label{eq3.8}
f_{K,\theta,C} := (h_{K,\theta})^{2}+C
\end{equation}
for some $Ke^{i\theta} \in \C \setminus \{ 0 \}$ and $C \in \C$.
Moreover, if we insist that $Ke^{i\theta} \in U$, where
\begin{equation}
\label{eq3.20}
U = \{ 0<\av z \av < 1, -\pi / 4 < \arg (z) \leq \pi / 4\} \cup \{ \av z \av > 1, -\pi / 4 < \arg (z) \leq \pi / 4\} \cup \{ 1 \},
\end{equation}
then such a representation is unique.
\end{proposition}

\begin{proof}
Let $g(z) = az^{2}+bz+c$ be a quadratic polynomial, where $a,b,c \in \C$ with $a \neq 0$,
let $h=h_{M,\phi}$ for $M>0$ and $\phi \in [-\pi,\pi ]$, and write $f(z) = g(h(z))$.
We need to know how $h$ behaves under pre-composition by translations and dilations.
Let $\alpha( z) = Az$ for $A \in \C \setminus \{ 0 \}$.
Then using (\ref{eq3.4a}),
\begin{equation*}
h(\alpha(z)) = h(Az) = \left ( \frac{ M+1}{2} \right ) Az + e^{2i \theta}\left ( \frac{ M-1}{2} \right )\overline{A}\zbar
\end{equation*}
\begin{equation}
\label{eq3.6}
= A \left ( \left ( \frac{ M+1}{2} \right ) z + e^{2i (\theta- arg(A))}\left ( \frac{ M-1}{2} \right )\zbar \right ) = Ah_{M,\theta - arg(A)}(z).
\end{equation}
Let $\beta(z) = z + B$ for $B \in \C$.
Again using (\ref{eq3.4a}), and noting that $h$ is $\R$-linear,
\begin{equation}
\label{eq3.7}
h(\beta(z)) = h(z) + h(B).
\end{equation}
Using (\ref{eq3.6}) with $A=1/a$,
\begin{equation*}
\alpha^{-1} \circ f \circ\alpha(z) = a \left ( a h_{M,\phi}(z/a)^{2} + b h_{M,\phi}(z/a) +c \right )
\end{equation*}
\begin{equation*}
= h_{M,\phi + arg(a)}(z)^{2} + b h_{M,\phi + arg(a)}(z) +ac
= \left ( h_{M,\phi + arg(a)}(z) + \frac{b}{2} \right )^{2} +ac - \frac{b^{2}}{4}.
\end{equation*}
Applying (\ref{eq3.7}) with $B = h^{-1}_{M,\phi + \arg(a)}(-b/2)$,
\begin{equation*}
\beta^{-1} \circ \alpha^{-1} \circ f \circ \alpha \circ \beta (z)
= h_{M,\phi + arg(a)}^{2}(z) + ac - \frac{b^{2}}{4} - h^{-1}_{M,\phi + arg(a)}(-b/2).
\end{equation*}
Therefore $f$ is linearly conjugate to (\ref{eq3.8}) with $K=M$, $\theta = \phi + \arg(a)$ and $C = ac - b^{2}/4 - h^{-1}_{K,\phi + arg(a)}(-b/2)$.

For the uniqueness,
we note that the choice of $K>0$ and $\theta \in [-\pi ,\pi ]$ for a given constant complex dilatation (\ref{eq3.3}) is not unique. There are the obvious symmetries
$(\theta \mapsto \theta + \pi)$ and $(K \mapsto 1/K, \theta \mapsto \theta + \pi /2)$. These correspond to the facts that a stretch in a direction $\theta$ is the same as stretching in the direction $-\theta$, and that stretching a factor $K$ in the direction $\theta$ is the same, up to post-composing by a conformal dilation, as stretching by a factor $1/K$ in the perpendicular direction. That is, $h_{K,\theta} = h_{K,-\theta}$
and $h_{K,\theta} = Kh_{1/K,\theta + \pi /2}$. There are no other such symmetries.

We see that $f_{K,\theta,C}$ is linearly conjugate to $f_{1/K,\theta +\pi/2, CK^{2}}$ via the conjugation $L(z) = z/K^{2}$. So if $Me^{i(\phi + \arg(a))} \notin U$, then we can apply such a conjugation (and take $-\phi - \arg(a)$ if necessary) so that $1/M e^{i(\phi +\arg (a) +\pi /2)} \in U$.

Recalling that all stretches with $K=1$ are identical, we see that if we require $Ke^{i\theta} \in U$, then the canonical form for $f_{K,\theta,C}$ is unique.
\end{proof}

For brevity, we will define the set
\begin{equation*}
QA = \{ f:\C \rightarrow \C : f=g \circ h, g(z)=z^{2}+c, h=h_{K,\theta}, c\in \C, Ke^{i\theta} \in U \}.
\end{equation*}
We will also mention the set
\begin{equation*}
PA = \{ f:\C \rightarrow \C : f=g \circ h, g \text{ is a polynomial of degree at least }2, h =h_{K,\theta} \},
\end{equation*}
and so $QA \subset PA$.

We observe that the dynamics of $f$ in Proposition \ref{canonicalform} correspond to those of $\widetilde{f}$, in particular, $z \in I(f)$ if and only if $z \in I(\widetilde{f})$.
Therefore, every composition of a quadratic polynomial and an affine stretch is linearly conjugate to some element of $QA$, and so we just need to study the dynamics of mappings in $QA$.

\section{The boundary of the escaping set}

The escaping set is defined as
\begin{equation*}
I(f) = \{ z \in \C : f^{n}(z) \rightarrow \infty \}.
\end{equation*}
Recalling Theorem \ref{FN}, the requirement that the distortion is smaller than the degree is a necessary one as the following example shows.

\begin{ex}
\label{ex1}
Consider the winding map $f:(r,\theta) \mapsto (r, 2\theta)$ in polar coordinates.
This map decomposes as $f= g \circ h$ where $g(z)=z^{2}$ and $h(r,\theta ) = (r^{1/2},\theta)$.
The partial derivatives of $h$ are
\begin{equation*}
h_{r} = \frac{e^{i\theta }}{2r^{1/2}}, \:\:\:\:\: h_{\theta} = ir^{1/2}e^{i\theta}
\end{equation*}
and so the complex dilatation is
\begin{equation*}
\mu _{h} = e^{2i\theta} \frac{ h_{r} + \frac{i}{r}h_{\theta}}{h_{r} - \frac{i}{r}h_{\theta}} = \frac{-e^{2i\theta}}{3}.
\end{equation*}
We conclude that $\av \av \mu _{h} \av \av _{\infty} = 1/3$ and the distortion of $h$ is $2$. Therefore the distortion of $f$ is $2$, and clearly the degree of $f$ is $2$, but $I(f)$ is empty, since $\av f(z)\av =\av z\av$ for every $z \in \C$.
\end{ex}

Therefore, Theorem \ref{FN} only applies to those $f \in QA$ with $\av (K-1)/(K+1) \av <1/3$ by (\ref{eq3.3}). However, in our special situation, we can actually deduce the results of Theorem \ref{FN} by modifying the proof.
For $f \in QA$, write $f=g \circ h$, where $g(z)=z^{2}+c$ and $h=h_{k,\theta}$ for $K e^{i\theta} \in U$ and $c \in \C$. We first estimate the growth of $h$.

\begin{lemma}
\label{hlemma}
Let $h$ be as above and set $L = \max \{ K, 1/K \}$.
Then for any $z \in \C$,
\begin{equation*}
\frac{\av z \av}{L} \leq \av h(z) \av \leq L \av z \av.
\end{equation*}
Further, $h$ is $L$-bi-Lipschitz, that is
\begin{equation*}
\frac{\av z - w\av}{L}   \leq \av h(z) - h(w) \av \leq L \av z - w \av
\end{equation*}
for all $z,w \in \C$.
\end{lemma}

\begin{proof}
This follows immediately from the definition of $h$, since
\begin{equation*}
L_{1}\av z \av \leq \av z \av \leq L_{2} \av z \av
\end{equation*}
where $L_{1} = \min \{K,1\}$ and $L_{2} = \max \{K,1\}$.
The fact that $h$ is bi-Lipschitz follows from the $\R$-linearity of $h$.
\end{proof}

We extend Theorem \ref{FN} in dimension $2$ as follows.

\begin{theorem}
\label{FNext}
Let $g$ be a polynomial of degree $d \geq 2$ and let $h$ be $L$-bi-Lipschitz. Let $f=g \circ h$, then $I(f)$ is a non-empty open set and $\partial I(f)$ is a perfect set.
\end{theorem}

\begin{proof}
A bi-Lipschitz mapping is quasiconformal, and so $f$ is quasiregular.
The first step is to show that $I(f)$ is non-empty.
Since $h$ is $L$-bi-Lipschitz and $g(z) = z^{d}(1+o(1))$ for large $\av z \av$, we have
\begin{equation*}
\av f(z) \av = \av g(h(z)) \av  = \av h(z) \av ^{d} (1+o(h(z))) \geq \frac{\av z \av ^{d}}{L^{d}} (1+o(z)).
\end{equation*}
Therefore there exists $R>0$ such that
\begin{equation*}
\av f(z) \av >2 \av z \av,
\end{equation*}
for $\av z \av >R$, and we can conclude that this neighbourhood of infinity is contained in $I(f)$.
The openness of $I(f)$ follows from the continuity of quasiregular mappings and the fact that $I(f)$ contains a neighbourhood of infinity. Clearly $I(f)$ is completely invariant, and therefore $\partial I(f)$ is completely invariant.

The fact the $I(f)$ is open means that $I(f)$ has no isolated points. To show that $I(f)$ is perfect, we therefore have to show that $I(f)^{c}$ has no isolated points. This is the part of the proof that requires modification when compared to Theorem \ref{FN}. Exactly as in the proof of that theorem
(we omit the details here, see \cite{FN}), we assume for contradiction that if $z_{0}$ is an isolated point of $I(f)$, and see that then $I(f) = \C \setminus \{ z_{0} \}$.

Since $f(z_{0}) = z_{0}$, we must have $i(z_{0},f) = d$, where $i$ is the local index. Therefore
\begin{equation*}
f(z) = g(h(z)) = (h(z)-h(z_{0}))^{d} + z_{0}.
\end{equation*}
Using again the fact that $h$ is bi-Lipschitz,
\begin{equation*}
\av f(z) - z_{0} \av = \av h(z) - h(z_{0})\av ^{d} \leq L^{d}\av z - z_{0} \av ^{d} < \frac{\av z- z_{0} \av}{2}
\end{equation*}
for $\av z-z_{0} \av <2^{1/(1-d)}L^{d/(1-d)}$.
nd so there is a neighbourhood of $z_{0}$ which is not in $I(f)$, giving a contradiction.
\end{proof}

We remark that in the Stoilow decomposition of the function in Example \ref{ex1}, it is easy to see that the quasiconformal mapping is not bi-Lipschitz.

\begin{corollary}
\label{QAescset}
Let $f \in QA$. Then $I(f)$ is a non-empty open set, and $\partial I(f)$ is a perfect set.
\end{corollary}

\begin{proof}
This is an immediate consequence of Lemma \ref{hlemma} and Theorem \ref{FNext}.
\end{proof}

The following theorem is proved in \cite{FN}, and the proof goes through exactly as there and so is omitted.

\begin{theorem}
\label{list}
Let $f$ be as in Theorem \ref{FNext}. Then for any $k \geq 2$, $I(f^{k}) = I(f)$. The family of iterates $\{f^{k}:k \in \N \}$ is equicontinuous on $I(f)$ and not equicontinuous at any point of $\partial I(f)$. The set $\partial I(f)$ is infinite. The sets $I(f)$, $\partial I(f)$ and $\overline{I(f)} ^{c}$ are all completely invariant. The escaping set is a connected neighbourhood of infinity.
\end{theorem}

In particular, we have the conclusions of Theorem \ref{list} for $f \in QA$. One may be tempted to ask whether the proof that $\partial I(f)$ is perfect goes through as soon as $I(f)$ contains a neighbourhood of infinity.
A modification of the winding map shows that this is not the case.

\begin{ex}
For $\lambda >1$ and $k \in \N$ with $k \geq 2$, define $f(re^{i \theta}) = \lambda r e^{i k\theta}$. This mapping decomposes as $f=g \circ h$ where $g(z)= z^{k}$ and
$h(z) = (\lambda r )^{1/k} e^{i \theta}$. We can calculate that the distortion of this mapping is $k$, that every point except $0$ escapes, and so $\partial I(f) = \{ 0 \}$. Clearly $\partial I(f)$ is not a perfect set. We note that $h$ is not bi-Lipschitz.
\end{ex}

\section{Connectedness of $\partial I(f)$}

Let $f$ be quasiregular of polynomial type and either the distortion of $f$ is smaller than the degree or $f \in QA$. Then we define the set of points whose orbits remain bounded
\begin{equation*}
N(f) = \{ z \in \C : \av f^{n} (z) \av <T , \text{ for some } T < \infty, \forall n \in \N \}.
\end{equation*}
Clearly $N(f) = I(f)^{c}$ and $N(f)$ is completely invariant by Theorem \ref{list}. This set is the direct analogue of the filled-in Julia set $K_{f}$ for polynomials, but here we are reserving the use of the symbol $K$ for distortion.

Recall the branch set $B(f)$ is the set where $f$ is not locally injective. In this section, we will give proofs for $f \in QA$, but the proofs will work equally well for $f \in PA$ or when the degree of $f$ is larger than the distortion.
We first need a quasiregular version of the Riemann-Hurwitz formula.

\begin{theorem}[Riemann-Hurwitz formula]
Let $D_{1},D_{2}$ be domains in $\overline{\C}$ whose boundaries consist of a finite number of simple closed curves. Let $f(z)$ be a proper holomorphic map of $D_{1}$ onto $D_{2}$ with $L$ branch points including multiplicity. Then every $z \in D_{2}$ has the same number $d$ of pre-images including multiplicity and
\begin{equation}
\label{rhformula}
(2-d_{1}) = d (2-d_{2}) - L,
\end{equation}
where $d_{j}$ is the number of boundary components of $D_{j}$.
\end{theorem}

\begin{corollary}
\label{rhcor}
The Riemann-Hurwitz formula (\ref{rhformula}) holds when $f:D_{1} \rightarrow D_{2}$ is a quasiregular mapping of degree $d$ in $D_{1}$.
\end{corollary}

\begin{proof}
By Theorem \ref{Stoilow}, we can write $f=g \circ h$, where $g$ is holomorphic and $h$ is quasiconformal. Since $h$ is quasiconformal, it does not alter the number of boundary components of $D_{1}$, $h$ has no branch points in $D_{1}$ and every $z \in h(D_{1})$ has exactly one pre-image under $h$ in $D_{1}$. Therefore every $z \in D_{2}$ has $d$ pre-images under $g$ in $h(D_{1})$ including multiplicity, $g$ has $L$ branch points including multiplicity in $h(D_{1})$ and so we can apply (\ref{rhformula}) to $g:h(D_{1}) \rightarrow D_{2}$.
\end{proof}

We now move on to our first connectedness result.

\begin{figure}
  \caption{The set of non-escaping points for $f(z) = (h_{0.8,0}(z))^{2}-1.25$, a pinched basilica.}\label{0.8K_-1.25cfig}
\begin{center}
    \includegraphics{0.8K_-1.25c.eps}
\end{center}
\end{figure}

\begin{figure}
  \caption{The set of non-escaping points for $f(z) = (h_{0.7,\pi /12}(z))^{2}-2.297+0.295i$, a banking airplane.}\label{0.7_pi-12K-2.297+0.295icfig}
\begin{center}
    \includegraphics{0.7_pi-12K-2.297+0.295ic.eps}
\end{center}
\end{figure}

\begin{theorem}
\label{QAconn}
Let $f \in QA$. Then $N(f)$ is connected if and only if $I(f) \cap B(f) = \emptyset$.
\end{theorem}

\begin{proof}
Recall from the proof of Theorem \ref{QAescset} the existence of a neighbourhood of infinity $U=\{ \av z \av >R \}$ contained in $I(f)$.

First assume that $I(f) \cap B(f) = \emptyset$. Then $f^{j}$ gives a quasiregular $2^{j}$-sheeted covering of $U_{j} = f^{-j}(U)$ onto $U$ with no branching,
since $f^{j}$ has degree $2^{j}$. Since $U \cup \{ \infty \}$ is simply connected, so is every $U_{j} \cup \{ \infty \}$. Since $I(f) = \cup _{j=0}^{\infty} f^{-j}(U)$, then $I(f)$ must be simply connected and hence $N(f)$ is connected.

Conversely, suppose that $I(f) \cap B(f)$ is not empty. Let $M$ be the smallest value of $j$ such that $U_{j} \cap B(f) $ is not empty.
Applying Corollary \ref{rhcor} to the quasiregular map
\begin{equation*}
f: (U_{M} \setminus \overline{U_{M-1}}) \rightarrow (U_{M-1} \setminus \overline{U_{M-2}})
\end{equation*}
with $d_{2} = 2$, $d=2$ and $L \geq 1$, we see that $d_{1} \geq 3$. This implies that $U_{M}$ is not simply connected and hence $N(f)$ is not connected.
\end{proof}

Since $0$ is the only branch point of $f \in QA$, this theorem is only interested in whether $0$ escapes or not. However, we have formulated it this way, because the proof is equally valid for $f \in PA$ or when the degree of $f$ is larger than the distortion, both cases where $f$ may have more than one branch point.
Examples of connected $N(f)$ are the Douady dragon of Figure \ref{0.8K_-0.21-0.78icfig}, the pinched basilica of Figure \ref{0.8K_-1.25cfig} and the banking airplane of Figure \ref{0.7_pi-12K-2.297+0.295icfig}.

Moving back to the case $f \in QA$, if $N(f)$ is not connected, then the following result tells us that $N(f)$ is actually infinitely connected.

\begin{theorem}
\label{infconn}
Let $f \in QA$. If $I(f)$ contains $B(f)$, then $N(f)$ is infinitely connected.
\end{theorem}

\begin{proof}
Since $B(f) \subset I(f)$, the forward orbit of $B(f)$,
\begin{equation*}
B^{+}(f) = \bigcup _{j=1}^{\infty} f^{j}(B(f))
\end{equation*}
accumulates only at infinity. Hence there exists a simple closed curve $C$ such that the interior $\Omega _{1}$ of $C$ contains $N(f)$ and the exterior $\Omega _{2}$ contains $B^{+}(f)$.
Since $\Omega _{2}$ is a neighbourhood of infinity and contained in $I(f)$,
\begin{equation*}
I(f)= \bigcup _{j=1}^{\infty} f^{-n} ( \Omega _{2})
\end{equation*}
and hence there exists $M$ such that $f^{M}(B(f) \cup \Omega_{2}) \subset \Omega _{2}$.
This implies that $f^{-M}$ has $2^{M}$ single valued locally injective quasiregular branches on $\Omega _{1}$, denoted by $h_{j} $ for $j=1,...,2^{M}$.

Then the sets $\{ U_{j} = h_{j}(\Omega _{1})\}$ are pairwise disjoint, and we call this collection of sets $X_{1}$. Each $U_{j}$ is compactly contained in $\Omega_{1}$ and contains $2^{M}$ images
$\{U_{j,k} = h_{j}(U_{k}) \}$ for $k=1,...,2^{M}$, and we denote the collection of such sets over all $j,k$ by $X_{2}$.
One can inductively define the collection $X_{n}$ for every $n \in \N$, and by construction,
\begin{equation*}
N(f) = \bigcap _{n=1}^{\infty} X_{n}.
\end{equation*}
Since $X_{n+1} \subset X_{n}$, we see that $N(f)$ is infinitely connected.
\end{proof}

For quadratic polynomials, if $I(f)$ contains $B(f)$, then $\partial I(f)$ is totally disconnected, i.e. every connected component is a point. This is not necessarily the case for $f \in QA$, as suggested by Figure \ref{0.8K_-1.1,0.003icfig} for $K=0.8$ and $c=-1.1+0.003i$, where each connected component appears to be a continuum. One can check that $0 \in I(f)$ by using the condition in Theorem \ref{thm6.3} below, and so $\partial I(f)$ really is disconnected, by Theorem \ref{infconn}, as claimed.

\begin{figure}
  \caption{The boundary of $I(f)$ for $K=0.8$, $\theta = 0$ and $c=-1.1+0.003i$}
  \label{0.8K_-1.1,0.003icfig}
  \begin{center}
    \includegraphics{0.8K_-1.1,0.003ic.eps}
  \end{center}
\end{figure}

\section{Parameter space}

For each pair $(K,\theta)$ such that $Ke^{i\theta} \in U$, we can consider the parameter space obtained by varying $c$ over functions in $QA$. We are led to the following definition.

\begin{definition}
Let $Ke^{i\theta} \in U$, and $f_{c} \in QA$ with decomposition $f_{c}=g_{c} \circ h$ where
$h= h_{K,\theta}$ and $g_{c}(z) = z^{2}+c$ for $c \in \C$.
The $(K,\theta)$-Mandelbrot set $\mathcal{M}_{K,\theta}$ is defined be the set of those $c \in \C$ for which $0 \notin I(f_{c})$.
\end{definition}

\begin{theorem}
We have the following characterisation of $\mathcal{M}_{K,\theta}$:
\begin{equation*}
\mathcal{M}_{K,\theta} = \{ c \in \C : \partial I(f_{c}) \text{ is connected} \}.
\end{equation*}
\end{theorem}

\begin{proof}
By the canonical form for functions in $QA$, the only branch point of $f_{c}$ is $0$. By Theorem \ref{QAconn}, $0 \notin I(f_{c})$ if and only if $I(f_{c}) \cap B(f_{c})= \emptyset$ if and only if $N(f_{c})$ is connected if and only if $\partial I(f_{c})$ is connected. The last equivalence follows from Theorem \ref{list} and the fact that $I(f_{c})$ is a connected neighbourhood of infinity.
\end{proof}

\begin{figure}
  \caption{$(K,0)$-Mandelbrot sets for $K=0.7$ to $K=1.2$ in $0.1$ increments.}\label{kmandelbrotfig}
  \begin{center}
    \includegraphics{kmandelbrot.eps}
  \end{center}
\end{figure}

\begin{figure}
\caption{$\mathcal{M}_{0.7,\pi/12}$}\label{M0.7_pi-12fig}
\begin{center}
\includegraphics{M0.7_pi-12.eps}
\end{center}
\end{figure}

As is well-known, the Mandelbrot set $\mathcal{M}_{1,0}$ is contained in the disk $\{ \av c \av \leq 2\}$. We will show that the $K$-Mandelbrot set is also a bounded set.

\begin{theorem}
\label{thm6.3}
Let $Ke^{i\theta} \in U$. Then
\begin{equation*}
\mathcal{M}_{K,\theta} \subset \{ c \in \C : \av c \av \leq 2 L_{1}^{-2} \},
\end{equation*}
where $L_{1} = \min \{K, 1 \}$.
Further, $\mathcal{M}_{K,\theta}$ is compact and can be characterised as the set of $c \in \C$ for which $f_{c}^{n}(0)
\leq 2 L_{1}^{-2}$ for all $n \in \N$.
\end{theorem}

\begin{proof}
Fix $(K,\theta)$ such that $Ke^{i\theta} \in U$, and recall from Lemma \ref{hlemma} that $h = h_{K,\theta}$ satisfies $\av h(z) \av \geq L_{1} \av z \av$ for all $z \in \C$, where $L_{1} = \min \{K,1\}$.
Assume that $L_{1} ^{2}\av c \av  > 2$. Then $\av f(0) \av = \av c \av$ and
\begin{equation*}
\av f^{2}(0) \av = \av h(c)^{2} + c \av \geq \av c \av (L_{1} ^{2} \av c \av - 1).
\end{equation*}
For $n \geq 2$, assume that
\begin{equation*}
\av f^{n}(0) \av \geq \av c \av (L_{1}^{2} \av c \av - 1)^{2^{n-2}}.
\end{equation*}
Then
\begin{equation*}
\av f^{n+1}(0) \av \geq \av f ( f^{n}(0)) \av \geq \av h( f^{n}(0))^{2} + c \av
\end{equation*}
\begin{equation*}
\geq L_{1} ^{2}( \av c \av (L_{1} ^{2} \av c \av - 1)^{2^{n-2}})^{2} - \av c \av \geq \av c \av ( L_{1} ^{2} \av c \av - 1)^{2^{n-1}}
\end{equation*}
when $L_{1} ^{2} \av c \av > 2$. Therefore, by induction, $\av f^{n}(0) \av \rightarrow \infty$ and $c$ is not in the $(K,\theta)$-Mandelbrot set.

For the second part of the theorem, assume that
\begin{equation*}
\av f_{c}^{m}(0) \av > 2L_{1}^{-2} + \epsilon > 2L_{1}^{-2}
\end{equation*}
for some $m \in \N$ and $\epsilon >0$.
First note if $\av c \av = \av f_{c}(0) \av > 2L_{1}^{-2}$ then $c \notin \mathcal{M}_{K,\theta}$.
So if $\av c \av \leq 2L_{1}^{-2}$, then
\begin{equation*}
\av f_{c}^{m+1}(0) \av = \av (f^{m}_{c}(0))^{2} +c \av \geq  \left ( 2L_{1}^{-2} + \epsilon \right ) ^{2} - 2L_{1}^{-2}
\end{equation*}
\begin{equation}
\label{eq6.1}
\geq 2L_{1}^{-2} \left ( 2L_{1}^{-2} - 1 \right ) +4L_{1}^{-2} \epsilon.
\end{equation}
If $K \geq 1$, then $L_{1} = 1$ and (\ref{eq6.1}) implies that $\av f_{c}^{m+1}(0) \av  \geq 2+4\epsilon$. Therefore by induction,
\begin{equation*}
\av f_{c}^{m+k}(0) \av \geq 2 + 4^{k}\epsilon
\end{equation*}
and we can conclude that $0 \in I(f_{c})$ and so $c \notin \mathcal{M}_{K,\theta}$.
On the other hand, if $K <1$, then $L_{1}^{-2} >1$ and (\ref{eq6.1}) implies that
$\av f_{c}^{m+1}(0) \av \geq 2L_{1}^{-2} + ( 2L_{1}^{-2})^{2}\epsilon$ and so by induction,
\begin{equation*}
\av f_{c}^{m+k}(0) \av \geq 2L_{1}^{-2} + (2L_{1}^{-1})^{2k} \epsilon
\end{equation*}
and again $c \notin \mathcal{M}_{K,\theta}$.
This argument shows that the complement of $\mathcal{M}_{K,\theta}$ is open and so $\mathcal{M}_{K,\theta}$ itself is a compact set.
\end{proof}

See Figure \ref{kmandelbrotfig} for various Mandelbrot sets with $\theta = 0$. It may appear that a bulb is detached from the main cardioid for $K<1$, but the following theorem shows that, among other things, these two components of $\mathcal{M}_{K}$ are attached by a segment contained in $\R$. See also Figure \ref{M0.7_pi-12fig} for this effect.

\begin{theorem}
If $\theta = 0$, then
\begin{equation*}
\mathcal{M}_{K,0} \cap \R = \left [-\frac{2}{K^{2}},\frac{1}{4K^{2}} \right ].
\end{equation*}
If $\theta \neq 0$, then there exists an angle $\phi_{0} \in [0,2\pi]$ and a real number $\eta$ such that the line segment
\begin{equation*}
te^{i\phi_{0}} \subset \mathcal{M}_{K,\theta},
\end{equation*}
for
\begin{equation*}
t \in \left[ -\frac{2}{\eta},\frac{1}{4\eta} \right ].
\end{equation*}
\end{theorem}

\begin{proof}

If $K \in  \R^{+}$, $\theta=0$ and $c \in \R$, then $f_{c}^{n}(0) \in \R$ for all $n \in \N$. It is well-known that $\mathcal{M}_{1} \cap \R = [-2, 1/4]$, i.e. $0$ does not escape under iteration of $x^{2}+c$ only when $c \in [-2,1/4]$. Since $f_{c}(x) = K^{2}x^{2}+c$ for $x \in \R$, then this observation implies that
\begin{equation*}
\mathcal{M}_{K,0} \cap \R = \left [-\frac{2}{K^{2}},\frac{1}{4K^{2}} \right ].
\end{equation*}

For the second part, assume that $\theta \neq 0$, $K>0$ and set
$h=h_{K,\theta}$ and $f=h^{2}$. Clearly $h$ maps rays emanating from $0$ to other such rays and has obvious fixed rays. We first show that $f$, which also maps rays to rays, has at least one fixed ray too.

One can calculate that $h$ is given, in polar coordinates, by
\begin{equation}
\label{eq6.4}
h(re^{i\varphi}) =
r\left ( 1+  (K^{2}-1)\cos ^{2}(\varphi - \theta ) \right )^{1/2} \exp \left [ i \left (\theta + \tan ^{-1} \left ( \frac{\tan (\varphi - \theta )}{K} \right ) \right ) \right ].
\end{equation}
For $\varphi \in [0,2\pi]$, write $L_{\varphi}$ for the ray
$\{ te^{i\varphi}: t\geq 0 \}$. Then it is easy to see that $f$ maps the ray $L_{\phi}$ onto $L_{T(\phi)}$, where
\begin{equation}
\label{eq6.2}
T(\phi ) = 2\theta +2 \tan ^{-1} \left ( \frac{\tan (\phi - \theta )}{K} \right ).
\end{equation}
To show that $f$ has a fixed ray, we therefore need to find a solution $\phi _{0}$ to $T(\phi) = \phi$. Rearranging (\ref{eq6.2}), this means we need to solve
\begin{equation}
\label{eq6.3}
\tan \left ( \frac{\phi}{2} - \theta \right) = \frac{\tan (\phi - \theta )}{K}.
\end{equation}
Let $t=\tan [(\phi - \theta)/2]$. Then (\ref{eq6.3}) and the addition formula for the tangent function yield
\begin{equation*}
\frac{t - \tan (\theta /2)}{1-t \tan (\theta /2)} = \frac{2t}{K(1-t^{2})}.
\end{equation*}
Rearranging this equation, we need to find a zero of the cubic polynomial
\begin{equation*}
P(t):= Kt^{3} - (2+K)\tan (\theta /2 ) t^{2} +(2-K)t + K\tan (\theta / 2).
\end{equation*}
Every cubic with real coefficients has a real zero, and we claim that $P(t)$ has a zero $t_{0}$ in $[-1,1]$. To see this, note that $P(1) = 2(1-\tan (\theta/2))$ and $P(-1) = 2(-1-\tan(\theta/2))$. Since
$\theta \in [-\pi /4, \pi /4]$, we have $\tan (\theta/2) \in [-\sqrt{2}+1,\sqrt{2}-1]$ and the claim follows.
Therefore, with
\begin{equation*}
\phi _{0} = \theta + 2 \tan ^{-1} t_{0},
\end{equation*}
$f$ fixes $L_{\phi_{0}}$.
Using (\ref{eq6.4}), we see that $f$ acts on the fixed ray $L_{\phi_{0}}$ by
\begin{equation*}
f(re^{i\phi_{0}}) =
r^{2}\left ( 1+  (K^{2}-1)\cos ^{2}(\phi_{0} - \theta ) \right ) e^{i\phi_{0}}.
\end{equation*}
The final claim of the theorem follows with
\begin{equation*}
\eta = \left ( 1+  (K^{2}-1)\cos ^{2}(\phi_{0} - \theta ) \right )
\end{equation*}
by using the same argument from the first part of the theorem.
We remark that if $\theta = 0$, then $\phi_{0}=0$ and $\eta = K^{2}$, agreeing with the first part of the theorem.
\end{proof}

We remark that the parameter space for mappings in $QA$ is actually two complex dimensional, since we have
a different mapping for each pair $(Ke^{i\theta},c)\in U \times \C$. Denote by $\Omega \subset U \times \C$
the set of $(Ke^{i\theta},c)$ such that $f_{K,\theta,c}^{n}(0)$ is bounded.
Each $(K,\theta)$-Mandelbrot set
is a one dimensional slice of this larger parameter space $\Omega$. Similarly, one can consider one dimensional slices of $\Omega$ where $c$ is fixed, and $Ke^{i\theta}$ varies.
As an example, Figure \ref{c_-1.5fig} is the slice of $\Omega$ through $c=-1.5$. Note the expected rotational symmetry. Further, the slice $c=0$ is the whole of $U$. A natural question to ask is whether the set $\Omega$ is connected in $\C^{2}$, just as the Mandelbrot set is connected in the parameter space of quadratic polynomials?

\begin{figure}
\caption{$c=-1.5$ parameter slice}\label{c_-1.5fig}
\begin{center}
\includegraphics{c_-1.5.eps}
\end{center}
\end{figure}

config: arxiv


TITLE: Energy power relation
QUESTION [1 upvotes]: If we have an energy equation like
$E=A\sin(wt-kx)$
If we differentiate it w.r.t. time, we will get power.
If we again differentiate the last equation, it will be termed as power per unit time.
But what it will represent and what is it's significance?

REPLY [0 votes]: But what it will represent and what is it's significance?

It will represent how quickly your power is changing. A positive value indicates that the power is increasing and a negative value indicates that the power is decreasing. As far as I know it has no particular physical significance.


TITLE: L'Hospital's rule with the indeterminate form of infinity minus infinity
QUESTION [1 upvotes]: I am having trouble finding a way to take $\lim\limits_{x \to \infty} 2e^{2x} - e^{4x}$ with L'Hospital's rule.

REPLY [0 votes]: You don’t need l’Hopital, simply observe that
$$2e^{2x} - e^{4x}=e^{2x}(2 - e^{2x})\to -\infty$$
which is obvious since $-e^{4x}$ dominates as $x$ increases.


TITLE: Mental multiplication of two digit numbers
QUESTION [4 upvotes]: There is a common way of mentally doing squares of two-digit numbers all the way up to 19*. This is how it works (example is for calculating $13^2 = 13*13 = xyz$):

The ones digit is the ones digit of the two numbers (3 for both) multiplied -> $13^2 =xy9$
The tens digit is the ones digit added together (3+3 for the example) -> $13^2 = x69$
The hundreds digit is the tens digit multiplied (1*1) -> $13^2 = 169$

Now, there can be some confusion with carrying over. So, I am doing another example of $15^2 = 15*15=xyz$:

$z = 5*5 = 25$ but the z can only be one digit long so it is added to the calculated y (just like how we do in regular multiplication). So $z=5$
$y = 5+5 = 10+2 = 12$ (the 2 comes from the previous step). Again, y can only be one digit long so the 1 is carried over and $y=2$.
$x=1*1=1+1 = 2$.

$15^2 = 15*15 = xyz = 225$
So, I thought, why don't I use this method to also do multiplication of all two digit numbers. I tried with numbers less than 20 and it worked. Here is another worked example of $13*15 = xyz$

$z = 3*5 = 15$ but the 1 is carried over so $z=5$
$y = 3+5 = 8+1 = 9$
$x=1*1 = 1$

$15*13 = xyz = 195$
Now, I then tried with one number less than 20 and one in between 20 and 30. Here's a worked example (I am probably overwhelming you guys with examples but this example is important!) of $25*13 = xyz$:

$z=5*3 = 15$ but the 1 is carried over so $z=5$
$y=5+3 = 8+1 = 9$
$x = 2*1 = 2$

So, $25*13 = xyz = 295$. But, the actual result is 325! It is off by 30. What is going on? After doing this several times, I noticed that my calculation is always off by 10 times the ones digit of the less than 20 number. So, you need to add that amount to the result. So, since the ones digit of the less than 20 number is 3, you multiply by 10, which gives you 30 and add to the result, which was 295. This gives you the correct answer of 325.
I tried to use this method with one number less than 40 but greater than 30 and another one less than 30 and greater than 20 (for example, doing $35*21$) and it did not work. I could not figure out by what factor it was off.
Then I tried with one number less than 40 but greater than 30 one less than 20 and greater than 10 and the result was off by 20 times the ones digit of the number less than 30 and greater than 20. For example, for $35*14$, I get 410. but the actual result is $410+(20*4)=490$ which is the correct answer.
Therefore, I have created a formula for multiplying $ab*1d$

$z=b*d$
$y=b+d$
$x=a*1=a$
$ab*1d=xyz+(d*10*(a-1))$

Question
So my main question is why does this method work and how? Why do I need to add the factor of $(d*10*(a-1))$? How can I extend this method for it to work for all two-digit numbers?
*I actually independently discovered this method (although obviously not being the first to do so) in bed when trying to find all numbers that are decimal concatenation of squares

REPLY [6 votes]: Everything you wrote comes from the identity
$$(10 a + b)(10 c + d) = 100 ac + 10 (bc + ad) + bd .$$
If $a=c=1$, which is the same as asking that both factors have two digits and do not exceed 20, we have many semplifications:
$$ (10 + b)(10 + d) = 100 + 10 (b+d) + bd. $$
Therefore we have $x=1, y=b+d$ and $z=bd$, as you said (remembering to carry over if necessary).
If $c=1$, which is the case $ab \cdot 1d$ in your notation, we have
$$ (10 a + b)(10 +d) = 100a + 10 (b + ad) +bd. $$
Now, your algorithm says
$$ y = b+d + d(a-1) $$
(since this is the meaning of adding $10 \cdot d(a-1)$), and this actually turns out to be $b + ad$, as required.


TITLE: Existence of only finitely many solutions
QUESTION [0 upvotes]: Let $r$ be a rational number greater than $1$. Prove that there are only finitely many natural numbers $x,y,z$ such that $$(x+1)(y+1)(z+1)=rxyz.$$
Progress: For $r=8$, the only solutions are $x=y=z=1$. If $r>8$, then clearly there are no solutions. I'm having trouble showing the same for $r\in(1,8)$. Any hints or solutions are welcome.

REPLY [0 votes]: x,y,z positive integers.
(1 + 1/x)(1 + 1/y)(1 + 1/z) = r.
Often there will be no solution.  r = 7 for example.
Assume that x,y and z's of the solutions diverge.
That forces r = 1, contradicting the assumption of solutions.
Wlog assume the z's are bounded and the x,y's of solutions diverge.
For each solution, there's another solution with larger x,y.
Thus (1 + 1/x)(1 + 1/y) is smaller, 1 + 1/z is larger and z is
smaller.  But z can be smaller only finite many times.
Wlog, assume y,z's of the solutions are bounded and x's diverge.
Each time x gets larger, 1 + 1/x gets smaller and (1 + 1/y)(1 + 1/z)
gets larger but only finite many times because each time x,y change
and there's just finite many ways they can change.
Thus the x,y,z's of solutions are all bounded
resulting in only finite many possible solutions.


TITLE: Poles of a function defined in terms of an integral
QUESTION [1 upvotes]: Suppose $\rho: [0,1] \rightarrow [0,\infty)$ with the following two
properties:
$$\int_0^1  \rho(x) dx = 1$$
and
$$\int_0^1 \rho(x) x dx =\frac{1}{2} $$
Now let
$$w(s) \equiv \int_0^1 \rho(x) x^s dx$$
for $s\in \mathbb{C}$.
And define $F(s) = \frac{1}{\frac{1}{2} - w(s)}$. What can we say about the poles of $F$? Clearly $s=1$ is a pole because of $w(1) = 1/2$. I have read, without proof, that

The $s=1$ pole is the one with largest real-part
All other poles come in pairs of complex conjugates, that is if $s$ is a pole then also $\overline{s}$ (the complex conjugate)

I would like to know why this is the case, i.e. see a proof. But unfortunately I am not able to do so. The only thing I got was an inequality for the integral summand in the denumerator of $F$, i.e. when $s=\alpha+i\beta$
$$\begin{eqnarray}
\vert w(s) \vert &\equiv& \vert \int_0^1 \rho(x) x^s dx\vert\\
&\leq&   \int_0^1 \vert  \rho(x) x^s \vert dx \\
&\leq& \int_0^1 r^{\alpha} \rho(x)dx \\
&\leq& \xi^{\alpha} \int_0^1 \rho(x)dx\\
&\leq& \xi^{\alpha}
\end{eqnarray}
 $$
In the second-last step I used the first mean value theorem for integration and therefore $\xi\in[0,1]$. Now any $s$ that is a pole of $F(s)$ has $w(s)=\frac{1}{2}$ and hence
$$\frac{1}{2} \leq \xi^{\alpha}$$
In case of the $s=1$ pole the equality is given, because for the $s=1$ case we have $\xi=\frac{1}{2}$.
But I am not sure if that is of use in general to show that any other $\alpha<1$. It would only help if I could show that $\xi>\frac{1}{2}$ for any $s\neq 1$ and hence one gets a contraction if $\alpha>1$:
$$\frac{1}{2} \leq \xi^{\alpha} < \xi < \frac{1}{2}$$
But I am not sure if that works.
Any help is appreciated. Many thaks!
EDIT
I understand the second point, that is all poles have to come in complex conjugate pairs. Suppose $s$ is a pole then $w(s) = \frac{1}{2} \in \mathbb{R}$ and hence $\overline{w(s)} = \frac{1}{2}$. And
$$\overline{w(s)} = \overline{\int_0^1 \rho(x) x^s dx} = \int_0^1 \rho(x) \overline{x^s} dx = w(\overline{s})$$
That is if $s$ is a pole then also $\overline{s}$.

REPLY [1 votes]: To show that the real of a pole is not more than $1$:
If the real part $\alpha$  of $s = \alpha + i \beta$ is more than one, then for $0<x<1$ you get $|x^s| = x^\alpha  < x$, so
$$
\Bigg |\int _0 ^1 p(x) x^s dx \Bigg|
\le \int _0 ^1 p(x)|x^s| dx
= \int _0 ^1 p(x)x^{\alpha} dx < \int _0 ^1 p(x)x dx = 1/2
$$
so $|w(s)| < 1/2$, therefore the denominator of $F(s)$ does not vanish and $s$ is not a pole.
To show that $s = 1$ is the only pole whose real part is $1$:
Say $s = 1 + i \beta$ is a pole, so $w(s) = 1/2$. Then
$$
\int _0 ^1 p(x) x e^{i \beta \log x} dx
= \int _0 ^1 p(x) x \cos(\beta \log x) dx + i \int _0 ^1 p(x) x \sin(\beta \log x) dx
= 1/2
$$
The imaginary part has to vanish, so you're left with
$$
1/2
= \int _0 ^1 p(x) x \cos(\beta \log x) dx
$$
If $\beta \ne 0$, the for some value of $x$ in $0 < x < 1$, we have $|\cos(\beta \log x)| < 1$ and you get
$$
1/2
= \Bigg| \int _0 ^1 p(x) x \cos(\beta \log x) dx \Bigg |
< \Bigg| \int _0 ^1 p(x) x dx \Bigg |
= 1/2
$$
So there can't be a pole when the real part is one and the imaginary part is non-zero.


text:
\section{Experiments}

\label{sec:experiments}

 \vspace{-2pt}

In this section, we illustrate our methods. Since computing the
persistent homology of the Vietoris-Rips (VR) filtrations built on top
of the large samples is infeasible, we resort to the subsampling
strategy described in Section \ref{sec:subsampling}. More formally,
let $X_N=\{x_1, \dots, x_N \}$ and $Y_N=\{y_1, \dots, y_N \}$ be two
large point clouds. We draw $n$ subsamples
each of size $m \ll N$ points from
$\mu$ and $\nu$, the discrete uniform measures on $X_N$ and $Y_N$, and
we compare the corresponding average landscapes and closest subsample
landscapes, induced by the persistent homology of the VR filtrations
built on top of the subsamples.  We apply this technique to two
examples.

\begin{figure}[!ht]
\centering
\includegraphics[width=5.3in]{figs/Data}
\caption{Left: 3D shapes of the first experiment.
Middle and Left: 500 random points from the magnetometer data of the second
experiment.}
\label{fig:data}
\end{figure}

{\bf 3D Shapes.}  We use the publicly available database of
triangulated shapes \citep{sumner2004deformation}. We select a
single {\it pose} (\#2) of 4 different classes: {\it camel, elephant,
flamingo, lion}. The 4 shapes are represented in Figure
\ref{fig:data}. In practice, each shape consists of a 3D point
cloud embedded in the Euclidean space, with a number of vertices that
ranges from 7K to 40K.  The data are normalized, so that the diameter
of each shape is 1.  For $n=100$ times we subsample $m=300$ points
from each shape; then we select the closest subsample to the
corresponding original point cloud and compute $4\times n$ persistence
diagrams (dimension 1), one for each subsample. See Figure \ref{fig:animals}:
the
plot on the left shows the landscapes corresponding to the closest
subsamples of $m$ points among the $n$ different subsamples from each
shape; the plot in the middle shows the empirical average landscapes
within each class, computed as the pointwise average of $n$
landscapes, with a 95\% uniform confidence band for the true average
landscape, constructed using the method described in
\cite{chazal2013stochastic}; the dissimilarity matrix on the right
shows the pairwise $\ell_\infty$ distances between the average
landscapes (scale from yellow to red), which, according to Theorem
\ref{thm:stabW}, represent a lower bound for the pairwise Wasserstein
distances of the discrete uniform measures on the 4 different shapes.

\begin{figure}[!ht]
\centering
\includegraphics[width=5.4in]{figs/animals4}
\caption{Subsampling methods applied to 3D shapes. For $n=100$ subsamples of
size $m=300$, for each shape, we constructed the landscapes of the closest
subsample (left), the average landscape with 95\% confidence band (middle) and
the dissimilarity matrix of the pairwise $\ell_\infty$ distance between average
landscapes.}
\label{fig:animals}
\end{figure}

\begin{figure}[!ht]
\centering
\includegraphics[width=5.4in]{figs/activities4}
\caption{Subsampling methods applied to magnetometer data.  For $n=80$
subsamples of size $m=200$, for each activity, we constructed the landscapes of
the closest subsample (left), the average landscape with 95\% confidence band
(middle) and the dissimilarity matrix of the pairwise $\ell_\infty$ distance
between average landscapes.}
\label{fig:activities}

\end{figure}
{\bf Magnetometer Data.}
For the second example, we consider the problem of distinguishing  human
activities performed while wearing inertial and magnetic sensor units. The
dataset is publicly available at the UCI Machine Learning
Repository\footnote{
http://archive.ics.uci.edu/ml/datasets/Daily+and+Sports+Activities} and is
described in \cite{barshan2013recognizing}, where it is used to classify 19
activities performed by 8 people that wear sensor units on the chest, arms and
legs. For ease of illustration, we report here the results on 4 activities
(walking, stepper, cross trainer, jumping) performed by a single person (\#1).
We use the data from the magnetomer of a single sensor (left leg), which
measures the direction of the magnetic field in the space at a frequency of
25Hz. For each activity there are 7,500 consecutive measurements that we treat
as a 3D point cloud in the Euclidean space. As an example, Figure \ref{fig:data}
shows 500 points at random for 2 activities (walking and using a cross trainer).
As in the previous example, for $n=80$ times, we subsample $m=200$ points  from
the point cloud of each activity, construct the landscapes of the closest
subsamples, the average landscapes (dim 1), and the dissimilarity matrix based
on the $\ell_\infty$ distances of the average landscapes. See Figure
\ref{fig:activities}.
To the best of our knowledge persistent homology has never been used to
study data from  accelerometers or magnetometers before. A remarkable advantage
is that the methods of persistent homology are insensitive to the orientation of
the input data, as opposed to other methods that require the exact calibration
of the sensor units; see, for example, \cite{altun2010comparative} and
\cite{barshan2013recognizing}.


text:
\section{Concentration of the free entropy}\label{app:concentration_free_entropy}
Consider the inference problem~\eqref{interpolation_model_R}.
Once both observations $\bY^{(t)}$ and $\widetilde{\bY}^{(t,R)}$ have been replaced by their definitions, the associated Hamiltonian reads:
\begin{multline}
\cH_{t,R}(\bs ; \bS, \bZ, \widetilde{\bZ}, \bW)
\triangleq
\sum_{j=1}^{n} \frac{\lambda R}{4} x_j^2 - \frac{\lambda R}{2}\,X_j x_j - \sqrt{\frac{\lambda R}{2}}\,\widetilde{Z}_j x_j \\
+\sum_{\bi \in \cI}
\frac{\lambda(1-t)}{2n^2} x_{i_1}^2 x_{i_2}^2 x_{i_3}^2
- \frac{\lambda(1-t)}{n^2}\, X_{i_1} X_{i_2} X_{i_3} x_{i_1} x_{i_2} x_{i_3}
- \frac{\sqrt{\lambda(1-t)}}{n}\, Z_{\bi} x_{i_1} x_{i_2} x_{i_3} \;.
\end{multline}
In this section, we show that the free entropy
\begin{equation}
\frac{1}{n} \ln \cZ_{t,R}\big(\bY^{(t)},\widetilde{\bY}^{(t,R)}, \bW\big)
= \frac{1}{n} \ln\int dP_s(\bs) \, e^{-\cH_{t,R}(\bs ; \bS, \bZ, \widetilde{\bZ}, \bW)} \;.
\end{equation}
concentrates around its expectation.
We will sometimes write $\frac{1}{n} \ln \cZ_{t,R}$, omitting the arguments, to shorten notations.

\begin{proposition}[Concentration of the free entropy]\label{prop:concentration_free_entropy}
Suppose that \ref{hyp:S_bounded_support}, \ref{hyp:varphi} hold.
There exists a polynomial $C(\Vert \varphi \Vert_{\infty}, \Vert \varphi' \Vert_{\infty}, \Vert \varphi'' \Vert_{\infty}, M_S, \lambda, R)$ with positive coefficients such that $\forall t \in [0,1]$:
\begin{equation}\label{bound_variance_free_entropy}
\E \Bigg[\Bigg(\frac{ \ln \cZ_{t,R}}{n}
	- \E\bigg[\frac{\ln \cZ_{t,R}}{n} \bigg]
	\Bigg)^{\!\! 2}\:\Bigg]
	\leq \frac{C(\Vert \varphi \Vert_{\infty}, \Vert \varphi' \Vert_{\infty}, \Vert \varphi'' \Vert_{\infty}, M_S, \lambda, R)}{n} \;.
\end{equation}
\end{proposition}
\begin{IEEEproof}
First, we show that the free entropy concentrates on its conditional expectation given $(\bW, \bS)$.
Thus, $\nicefrac{\ln  \cZ_{t,R}}{n}$ is seen as a function of the Gaussian random variables $\bZ$, $\widetilde{\bZ}$ and we work conditionally to $(\bW, \bS)$:
$g(\bZ, \widetilde{\bZ}) \equiv \nicefrac{\ln  \cZ_{t,R}}{n}$.
By the Gaussian-Poincar\'{e} inequality (see \cite[Theorem 3.20]{boucheron_concentration}), we have:
\begin{equation*}
\E \Bigg[\Bigg(\frac{ \ln \cZ_{t,R}}{n} - \E\bigg[\frac{\ln  \cZ_{t,R}}{n} \bigg\vert \bS, \bW \bigg]\Bigg)^{\!\! 2}\,\Bigg]
\leq \E\big[\big\Vert \nabla g(\bZ,\widetilde{\bZ}) \big\Vert^2 \,\big] \;.
\end{equation*}
The squared norm of the gradient of $g$ reads
$\Vert \nabla g\Vert^2 =
\sum_{\bi \in \cI} \vert\nicefrac{\partial g}{\partial Z_{\bi}}\vert^2
+ \sum_{j} \vert\nicefrac{\partial g}{\partial \widetilde{Z}_{j}}\vert^2$.
Each of these partial derivatives takes the form
$\nicefrac{\partial g}{\partial x} = -n^{-1} \big\langle \nicefrac{\partial \mathcal{H}_{t,R}}{\partial x} \big\rangle$. More precisely:
\begin{equation*}
\frac{\partial g}{\partial Z_{\bi}}
= n^{-1}  \frac{\sqrt{\lambda (1-t)}}{n} \, \langle  x_{i_1} x_{i_2} x_{i_3} \rangle_{t,R} \quad ; \quad
\frac{\partial g}{\partial \widetilde{Z}_{j}}
= n^{-1} \sqrt{\frac{\lambda R}{2}} \, \langle x_j \rangle_{t,R} \;.
\end{equation*}
We see that $\vert \nicefrac{\partial g}{\partial Z_{\bi}} \vert \leq \frac{\sqrt{\lambda}}{n^2} \Vert \varphi \Vert_{\infty}^3$
and $\vert \frac{\partial g}{\partial \widetilde{Z}_{j}} \vert \leq n^{-1}  \sqrt{\frac{\lambda R}{2}} \Vert \varphi \Vert_{\infty}$.
Therefore:
$$
\Vert \nabla g(\bZ,\widetilde{\bZ}) \Vert^2 \leq \frac{\lambda^{\nicefrac{3}{2}}}{6n} \Vert \varphi \Vert_{\infty}^6 + \frac{\lambda R}{2n}\Vert \varphi \Vert_{\infty}^2 + \cO(n^{-2})\;.
$$
Making use of the Gaussian-Poincar\'{e} inequality, we obtain (the term $\cO(n^{-2})$ is neglected):
\begin{equation}\label{bound_variance_GP_1}
\E \Bigg[\Bigg(\frac{ \ln \cZ_{t,R}}{n} - \E\bigg[\frac{\ln  \cZ_{t,R}}{n} \bigg\vert \bS, \bW \bigg]\Bigg)^{\!\! 2}\,\Bigg]
\leq \frac{\lambda \Vert \varphi \Vert_{\infty}^4}{6n} (\sqrt{\lambda} \Vert \varphi \Vert_{\infty}^2 +  3 R) \;.
\end{equation}
Next we show that $\E[\nicefrac{\ln  \cZ_{t,R}}{n} \vert \bS, \bW]$ concentrates on its conditional expectation given $\bS$.
Thus, $\nicefrac{\ln  \cZ_{t,R}}{n}$ is seen as a function of the Gaussian random variables $\bW$ and we work conditionally to $\bS$:
$g(\bW) \equiv \E[\nicefrac{\ln  \cZ_{t,R}}{n} \vert \bW, \bS]$.
We will again invoke the Gaussian-Poincar\'{e} inequality (see \cite[Theorem 3.20]{boucheron_concentration}).
To lighten the equations we drop the subscripts in the Gibbs bracket $\langle - \rangle_{t,R}$, we introduce the notation $\widetilde{\E} \triangleq \E[\cdot \vert \bS,\bW]$ and we define the following quantities:
\begin{align*}
\bX' = \varphi'\bigg(\frac{\bW \bS}{\sqrt{p}}\bigg) \quad ; \quad \bx' = \varphi'\bigg(\frac{\bW \bS}{\sqrt{p}}\bigg) \;.
\end{align*}
The squared norm of the gradient of $g$ reads
$\Vert \nabla g\Vert^2 = \sum_{i, j} \vert\nicefrac{\partial g}{\partial W_{ij}}\vert^2$ where $\forall (i,j) \in \{1,\dots,n\} \times \{1,\dots, p\}$:
\begin{align*}
\frac{\partial g}{\partial W_{ij}}
&= \cO(n^{-\nicefrac{5}{2}})+
\frac{1}{2n}
\sum_{\substack{k = 1\\ k \neq i}}^n \sum_{\substack{\ell = 1\\ \ell \neq k,i}}^{n}
\bigg(-\frac{\lambda(1-t)}{n^2\sqrt{p}}  \widetilde{\E} \langle x_{i} x'_i s_j x_{k}^2 x_{\ell}^2 \rangle
+\frac{\lambda(1-t)}{n^2\sqrt{p}}\, S_j X'_{i} X_{k} X_{\ell} \widetilde{\E}\langle x_{i} x_{k} x_{\ell}\rangle\\
&\qquad\qquad\qquad\qquad\quad\;\:
+\frac{\lambda(1-t)}{n^2\sqrt{p}}\, X_{i} X_{k} X_{\ell}\,\widetilde{\E} \langle s_j x'_{i} x_{k} x_{\ell} \rangle
+\frac{\sqrt{\lambda(1-t)}}{n\sqrt{p}}\, \widetilde{\E} Z_{i k \ell} \langle s_j x'_{i} x_{k} x_{\ell} \rangle\bigg)\\
&\quad
+\frac{1}{n} \sqrt{\frac{\lambda R}{2p}}
\bigg(\!-\sqrt{\frac{\lambda R}{2}} \widetilde{\E}\langle s_j x'_i x_i\rangle +  \sqrt{\frac{\lambda R}{2}}\,S_j X'_i \,\widetilde{\E}\langle x_i \rangle
+  \sqrt{\frac{\lambda R}{2}}\, X_i \,\widetilde{\E}\langle s_j x'_i\rangle
+ \widetilde{\E}\widetilde{Z}_i \langle s_j x'_i \rangle\bigg).
\end{align*}
The term $\cO(n^{-\nicefrac{5}{2}})$ comes from those triplets $\bi$ in $\cI \triangleq \{(i,j,k) \in [n]^3: i \leq j \leq k \}$ whose elements are non unique.
In order to further simplify these partial derivatives, we do an integration by parts with respect to the Gaussian noises $\bZ$ and $\widetilde{\bZ}$. It yields:
\begin{align*}
\widetilde{\E} Z_{i k \ell} \langle s_j x'_{i} x_{k} x_{\ell} \rangle
&= \frac{\sqrt{\lambda (1-t)}}{n}\widetilde{\E}\langle s_j x'_{i} x_i x_{k}^2 x_{\ell}^2 \rangle
-  \frac{\sqrt{\lambda (1-t)}}{n}\widetilde{\E} \langle s_j x'_{i} x_{k} x_{\ell} \rangle  \langle x_{i} x_{k} x_{\ell} \rangle \;;\\
\widetilde{\E}\widetilde{Z}_i \langle s_j x'_i \rangle
&= \sqrt{\frac{\lambda R}{2}}\widetilde{\E}\langle s_j x'_i x_i\rangle - \sqrt{\frac{\lambda R}{2}}\widetilde{\E}\langle s_j x'_i\rangle \langle x_i\rangle\;.
\end{align*}
Therefore, $\forall (i,j) \in \{1,\dots,n\} \times \{1,\dots, p\}$:
\begin{multline*}
\frac{\partial g}{\partial W_{ij}}
= \cO(n^{-\nicefrac{5}{2}})
+\frac{\lambda R}{2 n\sqrt{p}}
\bigg(S_j X'_i \,\widetilde{\E}\langle x_i \rangle
+  X_i \,\widetilde{\E}\langle s_j x'_i\rangle
- \widetilde{\E}\langle s_j x'_i\rangle \langle x_i\rangle\bigg)\\
+\frac{\lambda(1-t)}{2n^3\sqrt{p}}
\sum_{\substack{k = 1\\ k \neq i}}^n \sum_{\substack{\ell = 1\\ \ell \neq k,i}}^{n}
\bigg(\!
S_j X'_{i} X_{k} X_{\ell} \widetilde{\E}\langle x_{i} x_{k} x_{\ell}\rangle
+ X_{i} X_{k} X_{\ell}\,\widetilde{\E} \langle s_j x'_{i} x_{k} x_{\ell} \rangle
-\widetilde{\E} \langle s_j x'_{i} x_{k} x_{\ell} \rangle  \langle x_{i} x_{k} x_{\ell} \rangle \!\bigg)\,.
\end{multline*}
Making use of the boundedness assumptions, we obtain the following uniform bound on the partial derivatives:
\begin{equation*}
\bigg\vert \frac{\partial g}{\partial W_{ij}} \bigg\vert
\leq \cO(n^{-\nicefrac{5}{2}}) +
\frac{3 \lambda M_S}{2n\sqrt{p}}\Vert \varphi \Vert_{\infty} \Vert \varphi' \Vert_{\infty}\big(\Vert \varphi \Vert_{\infty}^4 + R \big)\;.
\end{equation*}
Therefore, $\Vert \nabla g(\bW) \Vert^2 \leq \frac{9 \lambda^2 M_S^2}{4n}\Vert \varphi \Vert_{\infty}^2 \Vert \varphi' \Vert_{\infty}^2\big(\Vert \varphi \Vert_{\infty}^4 + R \big)^2 + \cO(n^{-3})$ and the Gaussian-Poincar\'{e} inequality yields (the term $\cO(n^{-3})$ is neglected):
\begin{equation}\label{bound_variance_GP_2}
\E \Bigg[\Bigg(\E\bigg[\frac{\ln  \cZ_{t,R}}{n} \bigg\vert \bS, \bW \bigg] - \E\bigg[\frac{\ln  \cZ_{t,R}}{n} \bigg\vert \bS\bigg]\Bigg)^{\!\! 2}\,\Bigg]
\leq
\frac{9 \lambda^2 M_S^2}{4n}\Vert \varphi \Vert_{\infty}^2 \Vert \varphi' \Vert_{\infty}^2\big(\Vert \varphi \Vert_{\infty}^4 + R \big)^2 \;.
\end{equation}
Finally, it remains to show that $\E[\nicefrac{\ln  \cZ_{t,R}}{n} \vert \bS]$ concentrates on its expectation.
We will show that the function
$$
g: \mathtt{S} \in [-M_S,M_S]^p \mapsto \E[\nicefrac{\ln  \cZ_{t,R}}{n} \vert \bS = \mathtt{S} ]
$$
has bounded differences. To do so, we will show that the partial derivatives of $g$ are uniformly bounded.
Then we will apply the bounded differences inequality, also called McDiarmid's inequality, to get the concentration result (see \cite{McDiarmid}, \cite[Corollary 3.2]{boucheron_concentration}).
Similarly to what has be done with the random vector $\bS$, we define $\mathtt{X} = \varphi\big(\frac{\bW \ttS}{\sqrt{p}}\big)$, $\mathtt{X}' = \varphi'\big(\frac{\bW \ttS}{\sqrt{p}}\big)$ and $\mathtt{X}'' = \varphi''\big(\frac{\bW \ttS}{\sqrt{p}}\big)$.
For $\ell \in \{1,\dots,p\}$, the partial derivative of $g$ with respect to its $\ell^{\mathrm{th}}$ coordinate:
\begin{align}
\frac{\partial g}{\partial \ttS_\ell}
&= \frac{\lambda(1-t)}{n^3\sqrt{p}}
\sum_{\bi \in \cI}
\E\big[(W_{i_1 \ell} \ttX'_{i_1} \ttX_{i_2} \ttX_{i_3} + W_{i_2 \ell} \ttX_{i_1} \ttX'_{i_2} \ttX_{i_3} + W_{i_3 \ell} \ttX_{i_1} \ttX_{i_2} \ttX'_{i_3} )
\langle x_{i_1} x_{i_2} x_{i_3}\rangle\big\vert \bS = \ttS \big]\nonumber\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad
+\frac{\lambda R}{2n\sqrt{p}}
\sum_{i=1}^{n} \E\big[W_{i\ell}\ttX'_i \langle x_i \rangle\big\vert \bS = \ttS \big]\nonumber\\
&= \cO(n^{-\nicefrac{3}{2}}) + \frac{\lambda(1-t)}{2n^3\sqrt{p}}
\sum_{i=1}^{n}\sum_{\substack{j = 1\\ j \neq i}}^n \sum_{\substack{k = 1\\ k \neq i,j}}^{n}
\E\big[W_{i \ell} \ttX'_{i} \ttX_{j} \ttX_{k}
\langle x_{i} x_{j} x_{k}\rangle\big\vert \bS = \ttS \big]\nonumber\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad
+\frac{\lambda R}{2n\sqrt{p}}
\sum_{i=1}^{n} \E\big[W_{i\ell}\ttX'_i \langle x_i \rangle\big\vert \bS = \ttS \big]\label{formula_dg/dS_l}\;.
\end{align}
Once again the triplets $\bi \in \cI$ whose elements are non unique are accounted for with the term $\cO(n^{-\nicefrac{3}{2}})$, which is negligible compared to the others.
An integration by parts with respect to $\bW$ gives for all $(i,j,k,\ell) \in \{1,\dots,n\}^3 \times \{1,\dots,p\}$ such that $j \neq i$ and $k \neq i,j$:
\begin{align}
&\E\big[W_{i \ell} \ttX'_{i} \ttX_{j} \ttX_{k}
\langle x_{i} x_{j} x_{k}\rangle\big\vert \bS = \ttS \big]\nonumber\\
&\qquad= \frac{1}{\sqrt{p}} \E\big[ \ttS_\ell \ttX''_{i} \ttX_{j} \ttX_{k}
\langle x_{i} x_{j} x_{k}\rangle\big\vert \bS = \ttS \big]
+\frac{1}{\sqrt{p}} \E\big[ \ttX'_{i} \ttX_{j} \ttX_{k}
\langle s_\ell x'_{i} x_{j} x_{k}\rangle\big\vert \bS = \ttS \big]\nonumber\\
&\qquad\qquad
-\E\bigg[\ttX'_{i} \ttX_{j} \ttX_{k} \bigg\langle \!\! x_{i} x_{j} x_{k} \frac{\partial \cH_{t,R}}{\partial W_{i\ell}} \bigg\rangle\bigg\vert \bS = \ttS \bigg]
+\E\bigg[\ttX'_{i} \ttX_{j} \ttX_{k} \langle x_{i} x_{j} x_{k} \rangle \bigg\langle\frac{\partial \cH_{t,R}}{\partial W_{i\ell}} \bigg\rangle\bigg\vert \bS = \ttS \bigg]\;;
\label{1st_expectation_dg/dS_l}
\end{align}
\begin{align}
\E\big[W_{i\ell}\ttX'_i \langle x_i \rangle\big\vert \bS = \ttS \big]
&= \frac{1}{\sqrt{p}} \E\big[\ttS_\ell \ttX''_i \langle x_i \rangle\big\vert \bS = \ttS \big]
+ \frac{1}{\sqrt{p}} \E\big[\ttX'_i \langle s_\ell x'_i \rangle\big\vert \bS = \ttS \big]\nonumber\\
&\qquad\quad
-\E\bigg[\ttX'_i \bigg\langle\! x_i \frac{\partial \cH_{t,R}}{\partial W_{i\ell}} \bigg\rangle\bigg\vert \bS = \ttS \bigg]
+\E\bigg[\ttX'_i \langle x_i \rangle \bigg\langle\frac{\partial \cH_{t,R}}{\partial W_{i\ell}} \bigg\rangle\bigg\vert \bS = \ttS \bigg]\;.
\label{2nd_expectation_dg/dS_l}
\end{align}
Here $\cH_{t,R} \equiv \cH_{t,R}(\bs; \ttS, \bZ, \widetilde{\bZ}, \bW)$.
In order to prove the concentration result that we aim for, we need to make sure that both expectations\eqref{1st_expectation_dg/dS_l} and \eqref{2nd_expectation_dg/dS_l} are $\cO(n^{-\nicefrac{1}{2}})$.
The main difficulty resides in managing the terms where partial derivatives $\nicefrac{\partial \cH_{t,R}}{\partial W_{i \ell}}$ appear.
We have already dealt with these partial derivatives when proving the concentration with respect to $\bW$, and we found:
\begin{align*}
\frac{\partial \cH_{t,R}}{\partial W_{i \ell}}
&= \cO(n^{-\nicefrac{3}{2}})+
\frac{1}{2}
\sum_{\substack{j = 1\\ j \neq i}}^n \sum_{\substack{k = 1\\ k \neq i,j}}^{n}
\bigg(-\frac{\lambda(1-t)}{n^2\sqrt{p}} x_{i} x'_i s_\ell x_{j}^2 x_{k}^2
+\frac{\lambda(1-t)}{n^2\sqrt{p}}\, \ttS_\ell \ttX'_{i} \ttX_{j} \ttX_{k} x_{i} x_{j} x_{k}\\
&\qquad\qquad\qquad\qquad\qquad\quad
+\frac{\lambda(1-t)}{n^2\sqrt{p}}\, \ttX_{i} \ttX_{j} \ttX_{k} s_\ell x'_{i} x_{j} x_{k}
+\frac{\sqrt{\lambda(1-t)}}{n\sqrt{p}}\, Z_{i j k}  s_\ell x'_{i} x_{j} x_{k}\bigg)\\
&\qquad\qquad\quad
+ \sqrt{\frac{\lambda R}{2p}}
\bigg(-  \sqrt{\frac{\lambda R}{2}} s_\ell x'_i x_i +  \sqrt{\frac{\lambda R}{2}}\,\ttS_\ell \ttX'_i x_i
+  \sqrt{\frac{\lambda R}{2}}\, \ttX_i \, s_\ell x'_i
+ \widetilde{Z}_i s_\ell x'_i \bigg)\;.
\end{align*}
For $(i,\ell) \in \{1,\dots,n\} \times \{1,\dots,p\}$ define
\begin{multline}
\mathcal{A}_{i\ell}
\triangleq \frac{\lambda(1-t)}{2n^2\sqrt{p}}
\sum_{\substack{j = 1\\ j \neq i}}^n \sum_{\substack{k = 1\\ k \neq i,j}}^{n}
\big(-x_{i} x'_i s_\ell x_{j}^2 x_{k}^2
+\ttS_\ell \ttX'_{i} \ttX_{j} \ttX_{k} x_{i} x_{j} x_{k}
+ \ttX_{i} \ttX_{j} \ttX_{k} s_\ell x'_{i} x_{j} x_{k} \big)\\
+ \frac{\lambda R}{2\sqrt{p}}
\big(- s_\ell x'_i x_i + \ttS_\ell \ttX'_i x_i
+ \ttX_i \, s_\ell x'_i \big)\;.
\end{multline}
Note these two simple facts about $\mathcal{A}_{i\ell}$:
\begin{align}
\frac{\partial \cH_{t,R}}{\partial W_{i \ell}}
&= \cO(n^{-\nicefrac{3}{2}}) + \mathcal{A}_{i\ell}
+
\frac{\sqrt{\lambda(1-t)}}{2n\sqrt{p}}
\sum_{\substack{j = 1\\ j \neq i}}^n \sum_{\substack{k = 1\\ k \neq i,j}}^{n} Z_{i j k}  s_\ell x'_{i} x_{j} x_{k}
+ \sqrt{\frac{\lambda R}{2p}} \widetilde{Z}_i s_\ell x'_i \;;\label{link_A_dG/dW}\\
\vert \mathcal{A}_{i\ell} \vert
&\leq \frac{3\lambda}{2\sqrt{p}} \, M_S \Vert \varphi \Vert_{\infty} \Vert \varphi' \Vert_{\infty}\big(\Vert \varphi \Vert_{\infty}^4 + R \big)\;.\label{upperbound_A}
\end{align}
Plugging the identity \eqref{link_A_dG/dW} in both \eqref{1st_expectation_dg/dS_l} and \eqref{2nd_expectation_dg/dS_l} and making use of the upper bound \eqref{upperbound_A}, we obtain:
\begin{align*}
&\big\vert \E\big[W_{i \ell} \ttX'_{i} \ttX_{j} \ttX_{k}
\langle x_{i} x_{j} x_{k}\rangle\big\vert \bS = \ttS \big] \big\vert\\
&\;\leq \cO(n^{-\nicefrac{3}{2}}) + \frac{M_S \Vert \varphi \Vert_{\infty}^4 }{\sqrt{p}}
\Big(\Vert \varphi \Vert_{\infty} \Vert \varphi'' \Vert_{\infty}
+ \Vert \varphi \Vert_{\infty} \Vert \varphi' \Vert_{\infty}^2
+ 3\lambda\,\Vert \varphi \Vert_{\infty}^6\Vert \varphi' \Vert_{\infty}^2
+  3\lambda\, \Vert \varphi \Vert_{\infty}^2\Vert \varphi' \Vert_{\infty}^2 R\Big)\\
&\quad\;
+\frac{\sqrt{\lambda(1-t)}}{2n\sqrt{p}}
\sum_{\substack{j' = 1\\ j' \neq i}}^n \sum_{\substack{k' = 1\\ k \neq i',j'}}^{n}\!
\big\vert \E\big[\ttX'_{i} \ttX_{j} \ttX_{k} Z_{i j' k'} \big(\langle x_{i} x_{j} x_{k} s_\ell x'_{i} x_{j'} x_{k'} \rangle
-  \langle x_{i} x_{j} x_{k} \rangle \langle  s_\ell x'_{i} x_{j'} x_{k'} \rangle\big)
\big\vert \bS = \ttS \big]\big\vert\\
&\quad\;
+\sqrt{\frac{\lambda R}{2p}} \big\vert\E\big[\ttX'_{i} \ttX_{j} \ttX_{k} \widetilde{Z}_i \big(\langle x_{i} x_{j} x_{k}  s_\ell x'_i \rangle
-\langle x_{i} x_{j} x_{k} \rangle \langle  s_\ell x'_i \rangle \big)\big\vert \bS = \ttS \big]\big\vert\;;
\end{align*}
\begin{align*}
&\big\vert \E\big[W_{i\ell}\ttX'_i \langle x_i \rangle\big\vert \bS = \ttS \big] \big\vert\\
&\qquad\leq \cO(n^{-\nicefrac{3}{2}}) + \frac{M_S}{\sqrt{p}}
\Big(\Vert \varphi \Vert_{\infty}\Vert \varphi'' \Vert_{\infty}
+ \Vert \varphi' \Vert_{\infty}^2
+ 3\lambda\,\Vert \varphi \Vert_{\infty}^6\Vert \varphi' \Vert_{\infty}^2
+  3\lambda\, \Vert \varphi \Vert_{\infty}^2\Vert \varphi' \Vert_{\infty}^2 R\Big)\\
&\qquad\qquad\qquad
+\frac{\sqrt{\lambda(1-t)}}{2n\sqrt{p}}
\sum_{\substack{j' = 1\\ j' \neq i}}^n \sum_{\substack{k' = 1\\ k \neq i',j'}}^{n}
\big\vert \E\big[\ttX'_{i} Z_{i j' k'} \big(\langle x_{i} s_\ell x'_{i} x_{j'} x_{k'} \rangle
-  \langle x_{i} \rangle \langle  s_\ell x'_{i} x_{j'} x_{k'} \rangle\big)
\big\vert \bS = \ttS \big]\big\vert\\
&\qquad\qquad\qquad
+\sqrt{\frac{\lambda R}{2p}} \big\vert\E\big[\ttX'_i \widetilde{Z}_i \big(\langle x_i  s_\ell x'_i \rangle
-\langle x_i \rangle \langle  s_\ell x'_i \rangle \big)\big\vert \bS = \ttS \big]\big\vert\;.
\end{align*}
By integrating by parts with respect to $\bZ$ or $\widetilde{\bZ}$, we can now show that both upper bounds are $\cO(p^{-\nicefrac{1}{2}})$. This is because $Z_{ij'k'}$ and $\widetilde{Z}_i$ appear in the Hamiltonian $\cH_{t,R}$ via the terms $\frac{\sqrt{\lambda (1-t)}}{n}x_i x_{j'} x_{k'} Z_{ij'k'}$ and $\sqrt{\frac{\lambda R}{2}} x_i \widetilde{Z}_i$, respectively.
In the end, for all $(i,j,k,\ell) \in \{1,\dots,n\}^3 \times \{1,\dots,p\}$ such that $j \neq i$ and $k \neq i,j$:
\begin{align*}
\big\vert \E\big[W_{i \ell} \ttX'_{i} \ttX_{j} \ttX_{k}
\langle x_{i} x_{j} x_{k}\rangle\big\vert \bS = \ttS \big] \big\vert
&\leq \frac{M_S \Vert \varphi \Vert_{\infty}^4 }{\sqrt{p}}C_1\;;\\
\big\vert \E\big[W_{i\ell}\ttX'_i \langle x_i \rangle\big\vert \bS = \ttS \big] \big\vert
&\leq \frac{M_S}{\sqrt{p}} C_1\;;
\end{align*}
where
$C_1 \triangleq \Vert \varphi \Vert_{\infty}\Vert \varphi'' \Vert_{\infty}
+ \Vert \varphi' \Vert_{\infty}^2
+ 6\lambda\Vert \varphi \Vert_{\infty}^6\Vert \varphi' \Vert_{\infty}^2
+  6\lambda \Vert \varphi \Vert_{\infty}^2\Vert \varphi' \Vert_{\infty}^2 R$.
Going back to the identity \eqref{formula_dg/dS_l}, these upper bounds yield
$\big\vert \frac{\partial g}{\partial \ttS_\ell} \big\vert \leq \frac{\lambda M_S}{2 p} (\Vert \varphi \Vert_{\infty}^4 + R)\,
C_1$
uniformly in $\ttS \in [-M_S,M_S]^p$ and $\ell \in \{1\dots,p\}$ (we neglect the term $\cO(n^{-\nicefrac{3}{2}})$ that should appear in the upper bound). Hence $g$ has bounded differences (this is a simple application of the mean value theorem):
\begin{equation*}
\forall \ell \in \{1,\dots,p\}:
\sup_{-M_S \leq \ttS_1,\dots,\ttS_p, \ttS'_\ell \leq M_S}
\big\vert g(\ttS) - g(\ttS_1,\dots,\ttS_{\ell-1},\ttS'_\ell,\ttS_{\ell+1},\dots,\ttS_p)\big\vert
\leq \frac{C_2}{p}\,;
\end{equation*}
where
$C_2 \triangleq \lambda M_S^2 (\Vert \varphi \Vert_{\infty}^4 + R)\,C_1$.
By McDiarmid's inequality:
\begin{equation}\label{bound_variance_McDiarmid}
\E \Bigg[\Bigg(\E\bigg[\frac{\ln  \cZ_{t,R}}{n} \bigg\vert \bS \bigg] - \E\bigg[\frac{\ln  \cZ_{t,R}}{n}\bigg]\Bigg)^{\!\! 2}\,\Bigg]
\leq
\frac{C_2^2}{4p}\;.
\end{equation}
Combining the inequalities \eqref{bound_variance_GP_1},  \eqref{bound_variance_GP_2} and \eqref{bound_variance_McDiarmid} yields the final result.
\end{IEEEproof}


TITLE: Is the space of random continuous functions separable?
QUESTION [3 upvotes]: Let $\Omega$ and $X$ be compact metric spaces with $P$ a probability measure on $\Omega$.
Then we get the following definition of random continuous functions from "Random Probability Measures on Polish Spaces, H. Crauel page 21": let $f : X \times \Omega \to \mathbb{R}$ be a function such that:

for all $x \in X$ the map $\omega \mapsto f(x, \omega)$ is measurable,
for all $\omega \in \Omega$ the map $x \mapsto f(x, \omega)$ is continuous and bounded,
$\omega \mapsto \sup\{|f (x, ω)| : x \in X\}$ is integrable with respect to P (it is
measurable by separability of X).

If $f$ and $g$ are both functions satisfying (1)-(3) then identify f and g if $P(\{\omega : f(\cdot, \omega) \not= g(\cdot, \omega) \}$.
Finally, a random continuous function is (the equivalence class of)
a function $f : X \times \Omega \to \mathbb{R}$ satisfying (1)–(3) above. The set of all random continuous functions is a linear space, denoted by $C_\Omega(X)$ and a norm can be defined by
$$
|f|_\infty = \int \sup_{x \in X}|f(x, \omega)|dP(\omega).
$$
My question: is $C_\Omega(X)$ separable?

REPLY [1 votes]: Yes, this space is separable! Let us first show that $f$ is measurable according to the product $\sigma$-algebra.
Proof: $\Omega$ is a separable space. Thus, let $(x_n)_{n \in \mathbb{N}}$ be some dense set. Let be $B_{1/n}(x_i)$, $i=1,\ldots,i_n$ be a finite cover of $X$. For minimal $i=1,\ldots,i_n$ with $x \in B_{1/n}(x_i)$ we define $f_n(x,w) = f(x_i,w)$. In other words
$$f_n(x,w) = \sum_{i=1}^{i_n} 1_{A_{i,n}}(x) f(x_i,w),$$
where $A_{i,n} = B_{1/n}(x_i) \setminus (B_{1/n}(x_1) \cup \ldots B_{1/n}(x_{i-1}))$. This is measurable. And if $x \in A_{i(x),n}$, then $|x-x_{j(x,n)}| < 1/n$. And thus $f_n(x,w) \rightarrow f(x,w)$, because of the continuity. So, $f$ is measurable.

This proof gives already rise for a proof of the separability!

Note that, since $\sup_{x \in X} |f_n(x,w)|  \leq \sup_{x \in X} |f(x,w)|$, we get by the dominated convergence theorem also $|f_n-f|_\infty \rightarrow 0$. So it remains to approximate $f_n$ by a fixed set of countable functions, which satify (1) and (2). One problem is that $f_n$ does not satisfy (2).
So, we need to modifty our argument: Since $(B_{1/n}(x_i))$ is an open cover and $X$ compact, we find a finite number of partitions of unity, i.e. $h_{i_1,n}, \ldots h_{i_n,n}$ non-negative continuous functions such that

$\sum_{j=1}^{i_n} h_{i_j,n} = 1$
$\{ x \in X : h_{i_j,n} >0\}$ is contained in $B_{1/n}(x_{i_j})$.

Define now
$$f_n(x,w) = \sum_{j=1}^{i_n} h_{i_j,n}(x) f(x_{i_j},w).$$
This functions are according to the product $\sigma$-algebra. Moreover, for fixed $(x,w) \in X \times \Omega$, we have
$$|f_n(x,w)-f(x,w)| \leq \sum_{j=1}^{i_n} h_{i_j,n}(x) |f(x_i,w) - f(x,w)| \leq \sup_{i,|x-x_i| \leq 1/n} |f(x_i,w)-f(x,w)|.$$
By continuity, the last term converges to zero. The same argument as before shows also that $|f_n - f|_\infty \rightarrow 0$.

We have showed that functions of the form $\sum_{i=1}^n f_i(x) g_i(w)$ with $f_i \in C(X)$ and $g_i \in L^1(\Omega)$ are dense.

Let $(u_i)_{i \in \mathbb{N}}$ be a dense set in $C(X)$ and $(v_i)_{i \in \mathbb{N}}$ a dense set in $L^1(\Omega)$, then finite sums of products $u_i v_j$ are dense in $C_\Omega(X)$.

Fix $n \in \mathbb{N}$ with $\|f -f_n\| \leq \varepsilon$.
Take $u_{a_1}, \ldots, u_{a_n}$ with $\|h_{i_j,n} - u_{a_j}\| < \varepsilon/ i_n$.
Also take $v_{b_1}, \ldots, v_{b_n}$ with $\|f(x_{i_j},\cdot) - v_{b_j}\| < \varepsilon/ i_n$. (Note that $|v_{b_j}|_\infty \leq \varepsilon + |f|_\infty$.)

Define
$$g = \sum_{j=1}^{i_n} u_{a_j} v_{b_j}.$$
Then
\begin{align}
|f_n-g|_\infty &\leq \sum_{j=1}^{i_n} \int \sup_{x \in X} |u_{a_j}(x) - h_{i_j,n}(x)| |v_{b_j}(w)| \mathop{d \mathbb{P}(w)} + \sum_{j=1}^{i_n} \int |v_{b_j}(w) -f(x_{i_j},w)| \mathop{d \mathbb{P}(w)} \\
& \leq \varepsilon (1+ |f|_\infty) + \varepsilon.
\end{align}


TITLE: Is there a formula for solving the congruence equation $ax^2 + bx + c=0$?
QUESTION [3 upvotes]: Using the quadratic formula, we have either 0, 1, or 2 solutions. I wonder if we could use it this formula for congruence? Or is there a formula to solve quadratic equation for congruence?
Edit
Assume that $ax^2 + bx + c \equiv 0 \pmod{p}$, where $p$ is prime with $(a, p) = 1$, then is there a formula for this equation?
Thanks,

REPLY [1 votes]: Here (link) is a thorough discussion of the steps in reducing general moduli quadratic equation problems to those of prime moduli, including the case $p=2$.


TITLE: Proof that $\sum_{i=0}^n i^k = \sum_{i=0}^n (n-i)((i+1)^k-i^k)$
QUESTION [2 upvotes]: This is something I stumbled upon a few years back, and has a geometrically intuitive explanation, as well as being useful in summing powers of integers. However, with some occasional attempts at an inductive proof, nothing's quite worked. This came from looking at the staircase pattern the sum of powers makes visually. For example, when $k = 2$, $\sum_{i=0}^n i^2$ directly sums the square layers making up a quarter pyramid, while $\sum_{i=0}^n (n-i)((i+1)^2-i^2) = \sum_{i=0}^n (n-i)(2i-1)$ breaks up each layer into nested pairs of edges, with $2i+1$ being the volume of the edges, and $n-i$ being the number of edges of that size. You can use this to calculate, for example, $\sum_{i=0}^n i^2 = \sum_{i=0}^n (n-i)(2i+1) = (2n-1)\sum_{i=0}^n i - 2\sum_{i=0}^n i^2 + n\sum_{i=0}^n 1$, which then rearranges to the well-known $\sum_{i=0}^n i^2 = \frac{(2n+1)n(n+1)}{6}$, substituting $\sum_{i=0}^n 1 = n+1$ and $\sum_{i=0}^n i = \frac{n(n+1)}{2}$. Can anyone see a way to prove the formula, or know how to, as I'd be rather surprised if this was a brand-new result, given how visually straight forward it is?

REPLY [2 votes]: More generally, it's true that $\sum_{i=0}^n f(i) = \sum_{i=0}^n (n-i)\bigl(f(i+1)-f(i)\bigr)$ for any function $f$ with $f(0)=0$. This is just a summation by parts formula:
\begin{align*}
\sum_{i=0}^n (n-i)\bigl(f(i+1)-f(i)\bigr) &= \sum_{i=0}^n (n-i)f(i+1) - \sum_{i=0}^n (n-i)f(i) \\
&= \sum_{i=1}^{n+1} (n-(i-1))f(i) - \sum_{i=0}^n (n-i)f(i) \\
&= (n-((n+1)-1)f(n+1) \\
&\qquad{}+ \sum_{i=1}^n \bigl( (n-(i-1)) - (n-i) \bigr) f(i) - (n-0)f(0) \\
&= 0f(n+1) + \sum_{i=1}^n f(i) - nf(0) = \sum_{i=1}^n f(i).
\end{align*}


TITLE: Linearisation of a group
QUESTION [5 upvotes]: If $G$ is a connected  Lie group  acting on a vector $\mathbb{C}$-space $V$  then it is well known that the algebra of invariants  $\mathbb{C}[V]^G$ coincides with  the algebra of invariants $\mathbb{C}[V]^L$ of  corresponding Lie algebra $L.$
Question. Let now $G$ be a finite group, $V$ be its representation. Is there exists a Lie algebra $L$ and its representation on $V$ such that $\mathbb{C}[V]^G=\mathbb{C}[V]^L$?
It is easy to see  that it impossible  for symmetric group $S_n$. But maybe there are classes of finite groups for which can be found the positive answer?
Edit. Let $V=< v_1,v_2,\ldots,v_n >$  be standard representation of the symmetric group $S_n.$ Suppose that there exist a derivation $D=f_1 \frac{\partial}{\partial x_1}+\cdots+ f_n \frac{\partial}{\partial x_n}, f_i \in S(V)$  of the symmetric algebra $S(V)$ such that $D(I)=0$ for all $I \in \mathbb{C}[V]^{S_n}$.
Let $I_1,I_2,\ldots, I_n \in S(V)$ be a minimal generating set for $\mathbb{C}[V]^{S_n}.$  Since $D(I_k)=0, \forall k,$  we get  the system of polynomial equations on $f_1,f_2,\ldots,f_n:$
$$
f_1 \frac{\partial I_1}{\partial x_1}+\cdots+ f_n \frac{\partial I_1}{\partial x_n}=0,\\
f_1 \frac{\partial I_2}{\partial x_1}+\cdots+ f_n \frac{\partial I_2}{\partial x_n}=0,\\
\ldots \\
f_1 \frac{\partial I_n}{\partial x_1}+\cdots+ f_n \frac{\partial I_n}{\partial x_n}=0.
$$
The Jacobian of system of invariants $I_1,I_2,\ldots, I_n$ is not zero. It follows   the  system has only trivial  solution $f_i=0.$

REPLY [3 votes]: You asked for it! Here's the Hilbert series argument. I need to assume that the derivation $D$ is homogeneous, or equivalently comes from extending a linear map $V \to V$. This covers the case of a Lie algebra action on $V$.
First, recall that $\dim S^n(V^{\ast}) = {\dim V + n - 1 \choose \dim V - 1}$ has asymptotic growth rate $\Theta(n^{\dim V - 1})$ for $\dim V$ fixed and $n \to \infty$.
Let $g_n = \dim S^n(V^{\ast})^G$ denote the dimension of the $n^{th}$ graded component of $\mathbb{C}[V]^G$. An explicit expression for the Hilbert series $\sum g_n t^n$ can be obtained from Molien's theorem which shows in particular that the leading term in the asymptotic expansion of $g_n$ is at least $\frac{1}{|G|} n^{\dim V - 1}$ (hence, as above, that $\mathbb{C}[V]^G$ has Krull dimension $\dim V$). There should be a more elementary way to obtain this estimate.
On the other hand, let $D$ be a derivation induced from a linear map $V \to V$. Let $e_1, ..., e_n$ be a Jordan basis of $D$ with generalized eigenvalues $\lambda_1, ..., \lambda_n$. Then the monomial $e_1^{k_1} ... e_n^{k_n}$ in $\mathbb{C}[V]$ is a generalized eigenvector of $D$ with generalized eigenvalue $\sum k_i \lambda_i$. In particular, the $n^{th}$ graded component of $\mathbb{C}[V]^D$ has dimension at most the number of monomials of degree $n$ of generalized eigenvalue $0$. The exponents of these monomials are determined by the kernel of the map
$$\mathbb{Z}^n \ni (k_1, ..., k_n) \mapsto \sum_{i=1}^n k_i \lambda_i \in \mathbb{C}.$$
If at least one of the $\lambda_i$ is nonzero, the kernel of the above map is a submodule of rank at most $n-1$, from which it follows that the dimension $d_n$ of the $n^{th}$ graded component of $\mathbb{C}[V]^D$ has asymptotic growth rate at most as fast as $O(n^{\dim V - 2})$.
If all of the $\lambda_i$ are equal to zero, then $D$ is nilpotent. In this case we need to inspect the action of $D$ on monomials more carefully. Permute the Jordan basis so that $D(e_i)$ is either $e_{i-1}$ or $0$ and so that $D(e_2) = e_1$, and consider the lexicographic order on $\mathbb{C}[V]$ with respect to the $e_i$. Then $D(e_1^{m_1} ... e_k^{m_k})$ has leading term $m_2 e_1^{m_1+1} e_2^{m_2-1} ... e_k^{m_k}$, from which it follows that the image of $D$ acting on $S^n(V^{\ast})$ has dimension at least the number of monomials with $m_1 \ge 1$, or the dimension of $S^{n-1}(V^{\ast})$. The two dimensions are polynomials in $n$ with the same leading term and $d_n$ is their difference, so we again find that the asymptotic growth rate of $d_n$ is at most $O(n^{\dim V - 2})$.