TITLE: Left inverse of matrix exists if $A\mathbf x=\mathbf b$ has a unique solution?
QUESTION [3 upvotes]: If the equation $A\mathbf{x} = \mathbf{b}$ has a unique solution for some $\mathbf{b}$ is it true that $A$ has a left inverse? $A$ is an $m\times n$ matrix.

REPLY [1 votes]: You already reasoned (in a comment) that the null space of $A$ must be $\{0\}$, which means that the equation $Ax=b'$ has a unique solution $x$ for every $b'$ for which it has a solution at all (having at least two different solutions would by subtraction give a nonzero element of the null space, which does not exist). [Stated differently $x\mapsto Ax$ is an injective map.]
The set of such $b'$ is precisely $\def\R{\Bbb R}C=\{\,Ax\mid x\in\R^n\,\}$, which is a subspace of $\R^m$ (the column space of $A$). The above means there is a well-defined map $L:C\to\R^n$ that sends $b'\in C$ to the corresponding unique solution$~x\in\R^n$. In formula, for every $x\in\R^n$, putting $b=Ax$, one has $L(b)=x$, that is $L(Ax)=x$ for all $x\in\R$. Now show that

$L$ is a linear map
$L$ can be extended (non uniquely) to a linear map $L':\R^m\to\R^n$
The matrix of any such $L'$ is a left inverse of $A$.


TITLE: Black hole's time effect and matter (or energy) position
QUESTION [0 upvotes]: Ok, so after I've read a few (around 50 :P ) articles on subject of black holes. There are a few gaps in my understanding of it. Please follow my reasoning, I bet there is a mistake in there, and I need to know where is it.
So,
First thing, black hole has an event horizon, and a singularity which is supposed to be something really small at the center of it. But an event horizon has a certain measurable radius, and singularity is practically a dot. So there is some space between it.
Second thing, time inside a black hole goes slower. Lets say a guy Vic (victim) is falling into a black hole (traveling towards event horizon) and a guy named Obs (observer) is... well observing.
Here is the question finally:
When Vic passes the event horizon, he gets stretched, starts decomposing ('n such things), but for Obs, he does so very slowly. And that slowliness increases. Now, after Vic crosses the event horizon, he still has some space to travel between event horizon and singularity (because e. horizon has radius, and I bet my mistake is somewhere around here). Since time is slowing down from Obs point of view, Vic will not reach singularity for Obs lifetime, and possibly many Obs lifetimes. He will just keep slowly moving towards it. At least that is my understanding. This means that most matter that fell into a black hole (in recent times), still hasn't reached the center of it. So is this true?
Add rotation of a black hole to a whole thing (from our perspective of passage of time), and I'm dumbfounded.
P.S.
for better illustration, I've made a small picture of what I mean.

REPLY [1 votes]: Obs will not even see Vic going below the horizon- that would mean light is coming out of the black hole and we know that doesn't happen (putting aside Beckenstein-Hawking radiation for a different discussion).
What will actually (theoretically) happen, is that Obs will see his friend Vic ever slowing his fall but never passing the horizon, eventually freezing on the horizon and than red-shifted until Vic will not be visible (in the visible light) for Obs anymore. if Obs had a very good camera that can see infinitely large wave-lengths, he will see Vic ever slowing his endless fall towards the horizon, without ever crossing it.
as far as Vic is concerned though, he fell into the black hole in a finite time.
I would highly recommend reading "Exploring Black Holes: Introduction to General Relativity" by E.F. Taylor and J.A. Wheeler, it delves deep into space-times and black holes without heavy-duty mathematics being used (no tensors and the like)-if you understand special relativity and know how to differentiate you should understand most of it. it deals with these exact problems and analyzes them mathematically and physically in a way that gives intuition to the subject- I really enjoyed it and I think if you are into black holes, you might too.


TITLE: Shifting a function is continuous
QUESTION [4 upvotes]: I'm slightly puzzled by the following: if $g(t)$ is a function in $L^q(X)$ then we can show that $g(t-x)$ is continuous function of $t$, i.e. for $\varepsilon > 0$ we can find $\delta$ such that $d(x,y)<\delta$ implies $\|g(t-x) - g(t-y)\|_q < \varepsilon$.
But $g$ is not necessarily continuous. Is this result stating something like $g$ is continuous with respect to norm $\|\cdot\|_q$? Because if $\tau_x$ is translation by $x$ then $g(t-x) = g \circ \tau_x$ is not necessarily continuous. When continuous I mean in topology on $X$ (e.g. $X \to \mathbb R$). Does this sort of continuity have a name?

REPLY [6 votes]: Let me write $G$ instead of $X$ since we're talking about compact groups and my groups refuse to be called $X$...
Silliness aside, the point here is that for each $x \in G$ the map $\tau_x\colon L^q(G) \to L^q(G)$ is an invertible (in fact isometric) linear map. You get a group homomorphism $\tau\colon G \to {\rm GL}(L^q(G))$ and as such $\tau$ is strongly continuous (provided $1 \leq q \lt \infty$). What this means is that for each $f \in L^q(G)$ the function $x \mapsto \tau_x f$ is continuous as a function $(G,d) \to (L^q(G),\lVert\cdot\rVert_q)$, which is precisely the continuity Thomas mentions in his answer.
The reason this is the case is that the space of continuous functions $C(G)$ is dense in $L^q(G)$ and that continuous functions on compact spaces are uniformly continuous. For (uniformly) continuous functions $f\colon G \to \mathbb{C}$ we have that $x \mapsto \lVert \tau_x f - f\rVert_\infty$ is continuous and since $\mu(G) = 1$ we have that $\lVert f \rVert_q \leq \lVert f\rVert_\infty$, so for continuous functions $f$ the map $x \mapsto \tau_x f$ is continuous with respect to all $L^q$-norms.
Now, for continuous functions it is clear that $\lVert \tau_x f\rVert_q = \lVert f\rVert_q$, so $\tau_x\colon (C(G),\lVert\cdot\rVert_q) \to (C(G),\lVert \cdot \rVert_q)$ is an isometry and thus it extends uniquely to an isometry of the completion $L^q(G)$ of $(C(G),\lVert \cdot \rVert_q)$. Since $\tau_{-x} \tau_x = {\rm id}_{C(G)} = \tau_{x} \tau_{-x}$ we see that each of those extensions is invertible. To see that the resulting map $\tau\colon G \to {\rm GL}(L^q(G))$ is strongly continuous, let $h \in L^q(G)$ and $\varepsilon \gt 0$. Then there is $f \in C(G)$ such that $\lVert f - h\rVert_q \lt \varepsilon$ and by continuity of $x \mapsto \lVert\tau_x f - f\rVert_\infty$ there is $\delta$ such that $d(x,e) \lt \delta$ implies that $\lVert\tau_x f - f\rVert_\infty \lt \varepsilon$. But then
$$
\begin{align*}
\lVert \tau_x h - h\rVert_q & \leq  \lVert \tau_x h - \tau_x f\rVert_q + \lVert \tau_x f - f\rVert_q + \lVert f - h\rVert_q \\
&\leq \lVert \tau_x h - \tau_x f\rVert_q + \lVert \tau_x f - f\rVert_\infty + \lVert f - h\rVert_q \\
& \lt 3 \varepsilon
\end{align*}
$$
where we have used that $\lVert \tau_x h - \tau_x f\rVert_q = \lVert h - f\rVert_q$ by translation invariance of Haar measure.
To see that $q \lt \infty$ is essential, consider the characteristic function of a non-trivial segment $[0,\alpha]$ of the circle group $S^1$.

Added:
All I said here extends (with small modifications) to non-commutative locally compact groups, see the threads

A net version of dominated convergence? for an outline of the necessary modifications
Properties of Haar measure for a collection of references on abstract Harmonic analysis and in particular Haar measure on locally compact groups.


TITLE: Normal Mode in a Vibration?
QUESTION [1 upvotes]: What exactly is the Normal Mode?  According to me, it forms a basis functions (where Max transverse amplitude is fixed wrt position), and all arbitrary vibration in the string can be written in the form of linear form of this basis function.
Please tell me where I am not grasping the concept?

REPLY [1 votes]: Yes, the trigonometric functions of the Fourier series form an orthogonal basis for the solution.
The differential equation for transversal waves in a string can be derived by applying the second Newton's law in an elementary length, which results in$$\frac{\partial^2 y}{\partial t^2} = \frac{T}{\rho}\frac{\partial^2 y}{\partial x^2}$$Without additional information about boundary conditions, any function $y(x,t)$ that can be made a sum of the form:$y(x,t) = f(x-vt) + g(x+vt)$, where $v = \sqrt{\frac{T}{\rho}}$, and $f$ and $g$ are arbitrary functions, is a solution.
When both ends of the string are supposed fixed, $y(x,t)$ can be imagined as a periodic function on $(-\infty,+\infty)$, where the length of the string is a period. Any function like that can be expressed as a Fourier series with infinite terms. For $t = 0$, $$y = \sum_{n=-\infty}^{+\infty}c_n\exp{\left(ik_n x\right)}$$ where $k_n = \frac{2\pi n}{L}$, $L$ is the length and $c_n$ are complex coeficients (so that the terms are real).
For $t\neq 0$, we can make the coeficients $c_n$ functions of the time: $c_n = A_n \exp({ik_nvt})$ or $c_n = A_n \exp({-ik_nvt})$. That means, for each $t$ there is a different function, expressed by a different Fourier series. This procedure works because, if we join the exponetials together:
$$y(x,t) = \sum_{n=-\infty}^{+\infty}A_n(\exp{ik_n(x+vt)} + \exp{ik_n(x-vt)})$$
that is a sum of functions of $x-vt$ and $x+vt$ as required for the solution.
Each $k_n$ defines a normal mode with an angular frequency $$\omega = k_n v = \frac{2\pi n}{L}\sqrt{\frac{T}{\rho}}$$


TITLE: Example of $\overline{A}\cap\overline{D} \neq \overline{A\cap D}$
QUESTION [0 upvotes]: I want to find sets $A, D \subset \mathbb{R}$ (with the topology induced by the usual distance in $\mathbb{R}$) so that the following sets are different:
$$\overline{A}\cap D \quad ; \quad A \cap \overline{D} \quad; \quad \overline{A}\cap\overline{D} \quad ; \quad \overline{A\cap D}$$
I have tried several sets and I am unable to find a pair of sets that do right by  $\overline{A}\cap\overline{D} \neq \overline{A\cap D}$. Can someone help me with this?

REPLY [1 votes]: When you are doing these kind of closure, intersection questions, you can try to play around with accumulation points. Consider the sets:
$$ A = (0, 1) \cup \{ 2 \} $$
$$ B = \{ 1 \} \cup (2, 4) $$
We can check that
$$ \bar{A} \cap B = 1, \quad
A \cap \bar {B} = 2, \quad
\bar{A} \cap \bar{B} = 1, 2, \quad
\overline{A \cap B} = \emptyset $$
Some intuition: First, lets try $A$ and $B$ disjoint. Then $A \cap B = \emptyset$, and we focus on ensuring the 3 other sets are non empty.
Next, taking closure allows you to add an extra point. So you want the closure of $A$ to gain you an extra point, say $1$. Similarly, you want the closure of $B$ to gain you an extra point, say $2$. Then a bit of trial and error gives you a valid example.


text:
\begin{document}
\baselineskip=17pt

\title[Existence of Large Independent-Like Sets]{Existence of Large Independent-Like Sets}

\author[Robert (Xu) Yang]{Robert (Xu) Yang}
\address{Department of Pure Mathematics University of Waterloo Waterloo ON N2L 3G1, CANADA}
\email{yangxu\_robert@hotmail.com}

\date{}

\begin{abstract}
Let $G$ be a compact abelian group and $\Ga$ be its discrete dual group. For $N \in \N$, we define a class of independent-like sets, $N$-PR sets, as a set in $\Ga$ such that every $\Z_N$-valued function defined on the set can be interpolated by a character in $G$.

These sets are examples of $\e$-Kronecker sets and Sidon sets. In this paper we study various properties of $N$-PR sets. We give a characterization of $N$-PR sets, describe their structures and prove the existence of large $N$-PR sets.
\end{abstract}

\subjclass[2010]{Primary 43A25; Secondary 43A46}

\keywords{Sidon Sets, Independent Sets, Kronecker Sets}

\thanks{Research was partially supported by NSERC grant 2016-03719}

\maketitle

\section{Introduction}

Independence is a property prevalent throughout mathematics. In harmonic analysis, independence has been used to produce curious examples. For example, a perfect independent set in $\R$ was first constructed by Von Neumann \cite{Von}. Independent Cantor sets in non-discrete locally compact abelian groups were constructed by Hewitt and Kakutani \cite{HandK}, in part for showing that $M(G)$ is asymmetric \cite{H}. In these examples, by `independent' we mean algebraically independent: a set of (non-trivial) characters $E$ is independent if whenever $\ga_1,...,\ga_N \in E$, $m_i \in \Z$, and $\prod_{i=1}^N \ga_i^{m_i} = 1$, then $\ga_i^{m_i} = 1$ for all $i$.

The Rademacher functions in the dual of the infinite direct product of infinitely many copies of $\Z_2$ is a set of characters which is both algebraically and probabilistically independent. These functions have proven to be very useful in harmonic analysis. In particular, they have the property that every $\pm 1$-valued function defined on the set is evaluation at some $x$ in the group. A similar interpolation property holds for all algebraically independent sets.

A weaker notion than independence is an $\e$-Kronecker set.

\begin{defn}
Let $\e > 0$. The set $E$ is said to be (weak) $\e$-Kronecker if for every $\varphi: E \ra \BT$ there exists $x \in G$ such that $|\varphi(\ga) - \ga(x)| < \e$ for all $\ga \in E$ (resp. $|\varphi(\ga) - \ga(x)| \leq \e$ for all $\ga \in E$).
\end{defn}

This notion was inspired in part by the classical approximation theorem of Kronecker, with early work done by by Hewitt and Kakutani \cite{HandK} and Rudin \cite{Rudin2}. The terminology was introduced by Varapolous in \cite{bigV}.

The set of Rademacher functions is clearly an example of a weak $\sqrt{2}$-Kronecker set. Kronecker-like sets have been studied intensively, and are known to have many interesting properties. For example, Hare and Ramsey \cite{HareR} proved that every $\e$-Kronecker set with $\e < 2$ is a Sidon set. Graham and Hare in \cite{2012paper} introduced the weaker notion of pseudo-Rademacher sets, sets of characters where every $\pm1$-valued function is point-wise evaluation, in order to study the problem of the existence of Kronecker-like sets. Galindo and Hernandez in \cite{GaandH} and Graham and Lau in \cite{GandL} both consider interpolation sets of characters of finite order. For other references and further background information we refer the reader to \cite{Harebook}.

In this paper, we generalize this notion to $N$-pseudo-Rademacher sets (or $N$-PR sets for short), sets of characters with the property that every $\Z_N$-valued function on the set is point-evaluation. Of course, a pseudo-Rademacher set is a $2$-PR set and we will see that $N$-PR sets are $\e$-Kronecker sets for suitable $\e = \e(N)$, which tends to $0$ as $N \ra \infty$. We give an algebraic characterization of $N$-PR sets, compare them with $\e$-Kronecker sets, describe their structures and prove existence theorems (\cref{thm2.8}) of large $N$-PR sets. \cref{thm2.8} gives a new proof that any uncountable subset in $\Ga$ contains a large $\e$-Kronecker set.

\section{Characterization of $N$-PR sets}
Throughout this paper $G$ is a compact abelian group and $\Ga$ is its discrete dual group.

\begin{defn}
Let $E \subset \Ga$ be a subset and $N \in \N$. We define $E$ to be an  ``$N$-pseudo-Rademacher" set (or $N$-PR set) if for every $\varphi: E \ra \Z_N \subset \BT$, there exists $x \in G$ such that $\varphi(\ga) = \ga(x)$ for all $\ga \in E$.
\end{defn}

In this section we give an algebraic characterization of $N$-PR sets. We first establish some useful lemmas.

\begin{lem}[Quotient and Subgroup Lemma] \label{lem2.1}
Let $E \subset \Ga$, $n \in \N$, $\ga \in \Ga$ and $\Lambda \subset \Ga$ be a subgroup.

(1) Let $q: \Ga \ra \Ga/\Lambda$ be the quotient map. If $q$ is one-to-one on $E$ and $q(E)$ is $n$-PR, then $E$ is $n$-PR.

(2) Suppose $E \subset \Lambda$. Then $E$ is $n$-PR as a subset of $\Ga$ if and only if $E$ is $n$-PR as a subset of $\Lambda$.
\end{lem}

\begin{proof}
(1) Suppose $q: \Gamma \ra \Ga / \Lambda$ is one-to-one on $E$ and $q(E)$ is $n$-PR. We will show that $E$ is $n$-PR. Let $\varphi: E \ra \Z_n$ be a function. Because $q$ is one-to-one on $E$, for each $\ga, \beta \in E$, if $\beta \neq \ga$, then $\ga - \beta \notin \Lambda$. Thus, we can define $\varphi': q(E) \ra \Z_n$ via $\varphi'(\ga + \Lambda) = \varphi(\ga)$ for $\ga \in E$. Since $q(E)$ is $n$-PR, there exists $x \in \Lambda^{\perp} = \Hat{\Gamma / \Lambda}$ such that $\varphi'(\ga + \Lambda) = x(\ga + \Lambda)$ for all $\ga \in E$. As $x \in \Lambda^\perp$, $\varphi(\ga) = \ga(x)$ for all $\ga \in E$. This means $E$ is $n$-PR.

(2)  We first suppose $E$ is an $n$-PR subset of $\Ga$. Let $\varphi: E \ra \Z_n$ be a function. There exists $x \in G$ such that $\varphi(\ga) = \ga(x)$ for all $\ga \in E$. Let $x+ \Lambda^\perp \in G/\Lambda^\perp = \Hat{\Lambda}$. Since $E \subset \Lambda$, for all $\ga \in E$ we have
\ali{\varphi(\ga) =  \ga(x) = (x+ \Lambda^\perp)(\ga).}
This means $E$ is $n$-PR as a subset of $\Lambda$. The proof of the converse part of (2) is similar.
\end{proof}

The following proposition is a stronger version of Lemma 3.2 in \cite{KandR}.

\begin{prop}\label{prop2.2}
Let $E \subset \Ga$. The following are equivalent:

(1)  For all $\varphi: E \ra \BT$ with $\varphi(\ga) \in \range(\ga)$ there exists $x \in G$ such that $\varphi(\ga) = \ga(x)$ for $\ga \in E$.

(2)  $E$ is independent.
\end{prop}

\begin{proof}
Assume $E$ is independent and that $\varphi: E \ra \BT$ satisfies $\varphi(\ga) \in \range(\ga)$ for $\ga \in E$. By similar arguments to the proof of \cref{lem2.1}, if we can find $x \in \Hat{\ag{E}}$ such that $\varphi(\ga) = \ga(x)$ for all $\ga \in E$, then there exists $x' \in G$ such that $\varphi(\ga) = \ga(x')$. Thus we may assume $\Ga = \ag{E} = \bigoplus_{\ga \in E} \ag{\ga}$. For each $\ga \in E$, there exists $x_\ga \in \Hat{\ag{\ga}}$ such that $\ga(x_\ga) = \varphi(\ga)$. If $E$ is finite, we let $x = \prod_{\ga \in E} x_\ga$ and $x$ can interpolate $\varphi$ exactly. For the case that $E$ is infinite, since $G$ is compact, we let $x \in G$ be a cluster point of the following set
\ali{ \set{\prod_{\ga \in F} x_\ga : F \subset E, |F| < \infty},}
and such an $x$ can interpolate $\varphi$ exactly.

Conversely, if $E$ is not independent, then there exist $\ga_1,...,\ga_k \in E$ and $m_1,...,m_k \in \Z$ such that $\ga_1^{m_1}...\ga_k^{m_k} = 1$ but $\ga_i^{m_i} \neq 1$ for all $1 \leq i \leq k$. Consider the function $\varphi: E \ra \BT$ such that $\varphi(\ga_i) = 1$ for all $i > 1$, $\varphi(\ga_1)^{m_1} \neq 1$ and $\varphi(\ga) \in \range(\ga)$ for all $\ga \in E$. Notice that such $\varphi$ exists, because $\ga_1^{m_1} \neq 1$ means $\range(\ga_1^{m_1}) \neq \set{1}$ and hence there exists $x \in G$ such that $\ga_1(x)^{m_1}  \neq 1$. Let $\varphi(\ga_1) = \ga_1(x)$.
This function $\varphi$ cannot be interpolated by any $x \in G$.
\end{proof}

\begin{cor}
Let $N \in \N$ and $E \subset \Ga$. If $E$ is independent and $\Z_N \subset \range(\ga)$ for all $\ga \in E$, then $E$ is $N$-PR.
\end{cor}
\begin{proof}
This follows directly from \cref{prop2.2}.
\end{proof}

\begin{thm}\label{thm2.3}
The following are equivalent:

(1) $E \subset \Ga$ is $N$-PR.

(2) If $\ga_i \in E$ for $1 \leq i \leq n$ are distinct and $\prod_{i = 1}^n \ga_i^{m_i} = 1$ for some $m_i \in \Z$, then $N$ divides $m_i$ for all $i$.
\end{thm}

\begin{proof}
Suppose (2) fails. Then there exist distinct $\ga_i \in E$ and integers $m_i \in \Z$ with $\prod_{i =1}^n \ga_i^{m_i} = 1$, while $N$ does not divide $m_1$. Let $f: E \ra \Z_N$ be given by $f(\ga_1) = e^{2\pi i /N}$ and $f(\ga) = 1$ for all $\ga \neq \ga_1$. For any $x \in G$, $1 = \prod_{i =1}^n \ga_i(x)^{m_i}$, while $\prod_{i=1}^n f(\ga_i)^{m_i} = f(\ga_1)^{m_1} \neq 1$. Thus, this function $f$ cannot be interpolated by any $x \in G$ and therefore (1) fails.

Conversely, suppose (2) holds. Let $E_N = \ag{\ga^N : \ga \in E}$, the subgroup generated by $\set{\ga^N: \ga \in E}$, and $\pi: \Ga \ra \Ga / E_N$ be the quotient map. Elements in $E_N$ have the form $\ga_1^{k_1N}...\ga_m^{k_mN}$ for $\ga_1,...\ga_m \in E$ and $k_1,...,k_m \in \Z$. We claim that $\pi(E) \subset \pi(\Ga)$ is independent. Indeed, suppose $\ga_i \in E$ with distinct $\pi(\ga_i)$ and $m_i \in \Z$ such that $\prod_{i =1}^n \ga_i^{m_i} \in \ker(\pi)$. Then there exists $k_j \in \Z$, $1 \leq j \leq s$ and $s \geq n$  such that $\prod_{i =1}^n \ga_i^{m_i} = \prod_{j =1}^s \ga_j^{k_j N}$ with distinct $\ga_j \in E$. Hence, $\prod_{i = 1}^n \ga_i^{m_i - k_iN} \prod_{j = n+1}^s \ga_j^{-k_jN} = 1 \in \Ga$. By (2), $N \ | \ m_i - k_iN$ for all $1 \leq i \leq n$ and hence $N \ | \ m_i$ for all $1 \leq i \leq n$. This means each $\ga_i^{m_i} \in \ker(\pi)$ and therefore the set $\pi(E)$ is independent.

Moreover, we claim that $\pi$ is one-to-one on $E$. Suppose, otherwise, that there are $\ga_1 \neq \ga_2 \in E$ and $\ga_1 \ga_2^{-1} \in \ker(\pi)$. Then there exists $k_j \in \Z$, $1 \leq j \leq s$ and $s \geq 2$  such that $\ga_1 \ga_2^{-1} = \prod_{j =1}^s \ga_j^{k_j N}$ with distinct $\ga_j \in E$. We have $\ga_1^{1 - k_1N} \ga_2^{-1-k_2N} \prod_{j = 3}^s \ga_j^{-k_j N} = 1$. Again, (2) gives $N \ | \ 1 - k_1N$ and $N \ | \ - 1 - k_2N$, which are not possible.

Similar arguments show that elements of $\pi(E)$ have order $N$. Thus $\pi(E)$ is $N$-PR from \cref{prop2.2}. Because $\pi$ is also one-to-one on $E$, \cref{lem2.1} gives that $E$ is $N$-PR, proving (1).
\end{proof}

\begin{cor}\label{cor2.4}
Let $a,b \in \N$ be co-prime. The subset $E \subset \Ga$ is $(ab)$-PR if and only if $E$ is both $a$-PR and $b$-PR.
\end{cor}

\begin{proof}
Since $\Z_a, \Z_b \subset \Z_{ab}$, if $E$ is $(ab)$-PR, $E$ is both $a$-PR and $b$-PR. To see the converse, we assume $E$ is both $a$-PR and $b$-PR. Consider $\ga_1,...,\ga_n \in E$ and $m_1,...,m_n \in \Z$ such that $\prod_{i=1}^n \ga_i^{m_i} = 1$. Since $E$ is both $a$-PR and $b$-PR, $a | m_i$ and $b | m_i$ for all $1 \leq i \leq n$. Since $a$ and $b$ are co-prime, $ab | m_i$ for all $1\leq i \leq n$. Hence, by \cref{thm2.3}, $E$ is $(ab)$-PR.
\end{proof}

\begin{cor}\label{cor2.5}
Suppose $E \subset \Ga$ is $N$-PR and $\ag{E} \cap \ag{\ga} = \set{1}$. Then $\ga E$ is also $N$-PR.
\end{cor}

\begin{proof}
Let $\ga_1,...\ga_n \in E$ and $m_1,...,m_i \in \Z$ be such that $\prod_{i=1}^n (\ga\ga_i)^{m_i} = 1$.  Since $\ag{E} \cap \ag{\ga} = \set{1}$, we have $\prod_{i=1}^n \ga_i^{m_i} = 1$ and therefore $N | m_i$ for all $1 \leq i \leq n$. This shows $\ga E$ is $N$-PR by \cref{thm2.3}.
\end{proof}

Next we compare $N$-PR sets with $\e$-Kronecker sets. We first note that $N$-PR sets are weak $\e$-Kronecker sets for appropriate $\e$ depending on $N$.

\begin{prop}
$N$-PR sets are weak $\eKG$ for $\e = |1 - e^{\frac{\pi i}{N}}|$.
\end{prop}
\begin{proof}
Assume $E \subset \Ga$ is $N$-PR. Let $\varphi: E \ra \BT$. Define $\varphi_N: E \ra \Z_N$ by $\varphi_N(\ga) = t$, where $t \in \Z_N$ satisfies $|t - \varphi(\ga)| \leq |1 - e^{\frac{\pi i}{N}}|$. Let $x \in G$ be such that $\ga(x) = \varphi_N(\ga)$ for $\ga \in E$. We have $|\ga(x) - \varphi(\ga)| \leq |1 - e^{\frac{\pi i}{N}}|$.
\end{proof}

Of course, not every $\e$-Kronecker set in a torsion group is $N$-PR. Here is an example.

\begin{exmp}
For every $\e > 0$ and $N \in \N$, there exists an infinite $\eKG$ set $E$ with every $\ga \in E$ having finite order a multiple of $N$, but $E$ is not $N$-PR.

Let $N \in \N$ and $\e > 0$. Let $(p_i)_{i = 1}^\infty$ be an increasing sequence of primes coprime to $N$. Let $\Ga = \Z_N \oplus \bigoplus_{i \geq 1} \Z_{p_i}$. Let
\ali{S := \set{(1,0,0,...), (1, 1, 0,0,...), (1,1,1,0,0,...),...}.}
Each element in $S$ has order a multiple of $N$. It is not hard to see that if we exclude finitely many elements of small orders, we have a co-finite $\eKG$ set $E \subset S$. From \cref{thm2.3} no subsets of $S$ other than singletons are $N$-PR and therefore $E$ is not $N$-PR.
\end{exmp}

\begin{rem}
Recall that the Bohr compactification of $\Ga$, denoted by $\Bar{\Ga}$, is defined as the dual group of $G_d$, where $G_d$ is the group $G$ equipped with the discrete topology.

Let $\Bar{E}$ be the closure of $E$ in $\Bar{\Ga}$. In  \cite{Harebook}, Ex. 1.4.3, it is shown that there is an $\eKG$ set $E$ and an integer $M \geq 1$ such that $\Bar{(E \cup E^{-1})^M}$ has Haar measure 1. That is not the case for $N$-PR sets. Indeed, if $E$ is any $N$-PR set with $N \geq 3$, then the Haar measure of $\Bar{(E \cup E^{-1})^M}$ is $0$ for all $M \geq 1$. The proof is very similar to the proof of Theorem 3.1 in \cite{CCHare2} and therefore is omitted.
\end{rem}

\section{Structure of $N$-PR sets}

In this section, we investigate the structure of $N$-PR sets. We rely heavily on the following structure theorem for general abelian groups.

Notation: Let $p$ be a prime number. The group $\CC(p^\infty)$ is the group of all $p^n$-th roots of unity for $n \geq 1$.

\begin{thm}\label{thm3.1}\cite{structure}
Every abelian group $\Ga$ is isomorphic to a subgroup of
\ali{\bigoplus_{\alpha} \Q_\alpha \oplus \bigoplus_{\beta} \CC(p_\beta^\infty),}
where  $\Q_\alpha$ are copies of $\Q$ and $p_\beta$ are prime numbers.
\end{thm}

Notation: (1) We let $\Ga_0$ be the torsion subgroup of $\Ga$ and $\pi_0: \Ga \ra \Ga / \Ga_0$ be the quotient map. In the notation of \cref{thm3.1}, $\pi_0: \bigoplus_{\alpha} \Q_\alpha \oplus \bigoplus_{\beta} \CC(p_\beta^\infty) \ra \bigoplus_{\alpha} \Q_\alpha$ is the quotient map.

(2) Let $p \in \N$ be a prime number and $n \in \N$. We let $\Ga_{p^n}$ be the subgroup of $\Ga_0$ containing elements whose orders are not a multiple of $p^n$, or equivalently, whose orders are not divisible by $p^n$. Notice $\Ga_{p^n}$ is indeed a subgroup because if $\ga_1$ and $\ga_2$ have orders $m_1$ and $m_2$, neither a multiple of $p^n$, then $\lcm(m_1, m_2)$, the order of $\ga_1\ga_2$, is not a multiple of $p^n$. Let $\pi_{p^n}: \Ga \ra \Ga / \Ga_{p^n}$ be the quotient map.

(3) For $\ga \in \bigoplus_{\alpha \in A} \Q_\alpha \oplus \bigoplus_{\beta \in B} \CC(p_\beta^\infty)$ and $i \in A \cup B$, we let $\Proj_i(\ga)$ be the $i$-th coordinate of $\ga$. In particular, if $\Ga = \bigoplus_{i} \CC(p_i^\infty)$ with $p_i$ distinct, the map $\pi_{p_i}$ can be viewed as $\pi_{p_i} = \Proj_i$.

\begin{prop}\label{prop3.2}
 Let $E \subset \Ga$, $n \in \N$ and $p$ be a prime number. The following are equivalent:

(1) $E$ is $p^n$-PR.

(2) $\pi_{p^{k}}(E)$ is $p^{n+1-k}$-PR and $\pi_{p^k}$ is one-to-one on $E$ for all $1 \leq k \leq n$.

(3) $\pi_{p^{k}}(E)$ is $p^{n+1-k}$-PR and $\pi_{p^k}$ is one-to-one on $E$ for some $1 \leq k \leq n$.
\end{prop}

\begin{proof}
We first show (1) implies (2). Fix $1 \leq k \leq n$ and we suppose $E \subset \Ga$ is $p^n$-PR.  We first claim that $\pi_{p^k}$ is one-to-one on $E$. This is because if $\ga_1, \ga_2 \in E$ have $\ga_1\ga_2^{-1} \in \Ga_{p^k}$, then $\ga_1\ga_2^{-1}$ has finite order that is not divisible by $p^k$. But from \cref{thm2.3}, this implies $E$ is not even $p^k$-PR and therefore contradicts that $E$ is $p^n$-PR. Next, we show that $\pi_{p^k}(E)$ is $p^{n+1-k}$-PR. We let $\ga_1,...,\ga_s \in E$ and $\beta_i := \pi_{p^k}(\ga_i)$ for $1 \leq i \leq s$. Suppose for some $m_1,...,m_s \in \Z$ we have $\prod_{i=1}^s \beta_i^{m_i} = 1$. This means $\prod_{i=1}^s \ga_i^{m_i} \in \Ga_{p^k}$ and therefore there exists $l \in \N$, which is not divisible by $p^k$, such that $\prod_{i=1}^s \ga_i^{lm_i} = 1$. By \cref{thm2.3}, $p^n | lm_i$ and hence $p^{n+1-k} | m_i$. \cref{thm2.3} implies $\pi_{p^k}(E)$ is $p^{n+1-k}$-PR.

 Since (2) implies (3) is obvious, it remains to show (3)  implies (1). We assume $(3)$ holds for some $1 \leq k \leq n$. Let $\ga_1,...,\ga_s \in E$ be distinct and $m_1,...,m_s \in \Z$ such that $\prod_{i=1}^s \ga_i^{m_i} = 1$. Then $\prod_{i=1}^s \pi_{p^k}(\ga_i)^{m_i} = 1$ and the injectivity of $\pi_{p^k}$ implies $\pi_{p^k}(\ga_i)$ are distinct. Since $\pi_{p^k}(E)$ is $p^{n+1-k}$-PR, we have $p^{n+1-k} | m_i$ for all $1 \leq i \leq s$.

If $n+1-k \geq k$, we note that order of the product $\prod_{i=1}^s \ga_i^{m_i/p^{k-1}}$ divides $p^{k-1}$ and therefore $\prod_{i=1}^k \ga_i^{m_i/p^{k-1}} \in \Ga_{p^k}$. This means $\prod_{i=1}^k \pi_{p^k}(\ga_i)^{m_i/p^{k-1}} = 1 \in \Ga/\Ga_{p^k}$. Hence, by \cref{thm2.3}, we have $p^{n+1-k}$ divides $m_i/p^{k-1}$ and this gives $p^n$ divides $m_i$ for all $1 \leq i \leq s$. \cref{thm2.3} gives (1).

Otherwise, we have $n+1-k < k$. Notice that the order of the product $\prod_{i=1}^s \ga_i^{m_i/p^{n+1-k}}$ divides $p^{n+1-k}$. Since $n+1-k < k$, $\prod_{i=1}^k \ga_i^{m_i/p^{n+1-k}} \in \Ga_{p^k}$. As above, we have $p^{n+1-k}$ divides $m_i/p^{n+1-k}$, which means $p^{2(n+1-k)} | m_i$. We continue doing this until we reach $r(n+1-k) \geq k$ for some $r \in \N$. The previous case gives (1).
\end{proof}

\begin{cor}\label{cor3.3}
Suppose $\Ga = \bigoplus \CC(p^\infty)$. Then $p^n$-PR subsets in $\bigoplus \CC(p^\infty)$ are in correspondence with $p$-PR subsets in $ \bigoplus \CC(p^\infty)$ in the following manner: Given a set $E \subset \bigoplus \CC(p^\infty)$, we let $E_{p^n} := \set{\ga^{p^{n-1}} : \ga \in E}$ and $E_{1/p^n} := \set{\xi_\ga : \ga \in E}$, where $\xi_\ga^{p^{n-1}} = \ga$. If $E$ is $p^n$-PR then $E_{p^n}$ is $p$-PR and $|E_{p^n}| = |E|$. If $E$ is $p$-PR then $E_{1/p^n}$ is $p^n$-PR and $|E| = |E_{1/p^n}|$.
\end{cor}
\begin{proof}
Since on $\bigoplus \CC(p^\infty)$, the map $\pi_{p^n}$ can be identified as $\pi_{p^n}(\ga) = \ga^{p^{n-1}}$, this follows from \cref{prop3.2}.
\end{proof}

\begin{cor}\label{cor3.4}
Let $N = p_1^{n_1}...p_k^{n_k}$ for distinct primes $p_i$ and $n_i \in \N$. Then $E$ is $N$-PR if and only if $\pi_{p_i^{n_i}}(E)$ is $p_i$-PR and $\pi_{p_i^{n_i}}$ is one-to-one on $E$ for all $1 \leq i \leq k$.
\end{cor}
\begin{proof}
This follows from \cref{cor2.4} and \cref{prop3.2}.
\end{proof}

We thus have the following result about the structure of $N$-PR sets in the torsion subgroup.

\begin{prop}\label{prop3.5}
Let $E \subset \Ga_0$ be an $N$-PR set and $N = p_1^{n_1}...p_k^{n_k}$ where the $p_i$ are distinct prime numbers. There exist $p_i^{n_i}$-PR sets $E_i \subset \bigoplus \CC(p_i^\infty)$ for $1 \leq i \leq k$, and bijections $f_2: E_1 \ra E_2$, ..., $f_{k}: E_1 \ra E_k$ such that we can represent $E$ as
\ali{E = \set{\ga \oplus f_2(\ga) \oplus ... \oplus f_{k}(\ga) \oplus \beta_\ga : \ga \in E_1}}
for some $\beta_\ga \in \bigoplus_{p \neq p_i \ \forall 1 \leq i \leq k} \CC(p^\infty)$.
\end{prop}
\begin{proof}
For $1 \leq i \leq k$, we let $E_i := \pi_{p_i}(E)$. Since $E$ is $N$-PR, from \cref{prop3.2}, each $E_i$ is $p_i^{n_i}$-PR and the maps $\pi_{p_i}: E \ra E_i$ are injective. For $2 \leq i \leq k$, we define $f_i$ on $E_1$ as $f_i(\pi_{p_1}(\ga)) := \pi_{p_i}(\ga)$ for $\ga \in E$. The injectivity of $\pi_{p_1}$ on $E$ ensures the maps are well-defined. Moreover, the injectivity of $\pi_{p_i}$ implies $f_i$ is injective. Each $f_i$ is clearly surjective by its construction and therefore a bijection. Since each $\ga \in E$ can be represented as
\ali{\ga = \oplus_{i=1}^k \pi_{p_i}(\ga) \oplus \beta_\ga}
for some $\beta_\ga \in  \bigoplus_{p \neq p_i \ \forall 1 \leq i \leq k} \CC(p^\infty)$,  \cref{prop3.5} follows.
\end{proof}

Next, we will discuss the relation between $p$-PR sets and independent sets.

We first note that not every $p$-PR set is an independent set. The following example shows a $p$-PR set whose only independent subsets are singletons.

\begin{exmp}
Consider $E = \set{\ga_n : n \geq 2} \subseteq \bigoplus_{i \geq 1} \CC(p^\infty)$, where $\Proj_1(\ga_n) = 1/p^2$, $\Proj_n(\ga_n) = 1/p$ and $\Proj_k(\ga_n) = 0$ for all $k \neq 1, n$. Here we use additive group operation. $E$ is $p$-PR by \cref{thm2.3}, but $E$ does not contain any independent subsets other than singletons because for all $i \neq j$, $p\ga_i = p\ga_j \neq 1$.
\end{exmp}

Similar to independent sets, we have the following result about the maximum size of a $p$-PR set inside a product group.

\begin{prop}\label{prop4.2}
Suppose $E \subseteq \bigoplus_{i \in B_1} \Q \oplus \bigoplus_{i \in B_2} \CC(p^\infty)$ is $p$-PR. Then $|E| \leq |B_1|+|B_2|$.
\end{prop}
\begin{proof}
We identify each element in $\CC(p^\infty)$ in the form of $a/p^n$ for some $a, n \in \N$ with additive group operation. Hence, we can identify $ \bigoplus_{i \in B_1} \Q \oplus \bigoplus_{i \in B_2} \CC(p^\infty)$ as a subset of the real vector space $\R^{|B_1|+|B_2|}$.

We claim that if $E \subseteq  \bigoplus_{i \in B_1} \Q \oplus \bigoplus_{i \in B_2} \CC(p^\infty)$ is $p$-PR, $E$ is linearly independent in $\R^{|B_1|+|B_2|}$. Indeed, suppose that $E$ is not linearly independent. Then there exist $\set{\ga_i \ | \ 1 \leq i \leq k} \subseteq E$ and $a_1,...,a_k \in \R$ such that $a_1\ga_1 + ... +a_k \ga_k = 0$, while not all $a_i$'s are zero and each $\ga_i \in\R^{|B_1|+|B_2|}$ is identified as above. Since the entries of each $\ga_i$ are in $\Q$, we may assume $a_i \in \Q$ for all $1 \leq i \leq k$. Furthermore, we may assume $a_i \in \Z$. Notice that if $p | a_i$ for all $i$, we may replace $a_i$ by $a_i/p$. Hence, we may find a choice of such $a_i$'s, not all  divisible by $p$. This contradicts that $E$ is $p$-PR by \cref{thm2.3}.

Thus, $|E| \leq  |B_1|+|B_2|$.
\end{proof}

\section{Existence of $N$-PR Sets}

In this section, we show some existence results about $N$-PR sets and that large $N$-PR sets are plentiful. We first prove a lemma.

\begin{lem}\label{lem4.3}
Assume $E \subseteq \bigoplus_{\beta \in B} \Ga_\beta$ is uncountable.

(1) Let $p$ be a prime number. If $\Ga_\beta = \CC(p^\infty)$ for all $\beta \in B$, then $E$ contains a $p$-PR subset of the same cardinality.

(2) If $\Ga_\beta = \Q$ for all $\beta \in B$, then $E$ contains an independent subset of the same cardinality.
\end{lem}
\begin{proof}
Without loss of generality, we further assume for each $\beta \in B$ there exists $\ga \in E$ such that $\Proj_\beta(\ga)$ is non-trivial.
(1) We first prove (1) in the special case that every $\ga \in E$ has order $p$. We consider the collection $\CC$ of subsets of $E$ defined as $A \in \CC$ if for all finite subsets $F \subseteq A$ there exists an arrangement $F = \set{\ga_1,...,\ga_n}$ such that for each $1 \leq k \leq n$ there is some $\beta \in B$ with $\Proj_\beta(\ga_k)$ non-trivial, but $\Proj_\beta(\ga_j)$ trivial for all $1 \leq j \leq k$. We partially order $\CC$ by inclusion and Zorn's Lemma gives a maximal $S \in \CC$.

We claim $|S| = |E|$. Indeed, if $|S| < |E|$, we let $B_1 \subset B$ be given by $\beta \in B_1$ if there exists $\ga \in S$ such that $\Proj_\beta(\ga)$ is non-trivial. We thus have $|B_1| < |B|$. Let $\beta_0 \in B \bs B_1$ and $\ga_0 \in E$ be such that $\Proj_{\beta_0}(\ga_0)$ is non-trivial. Since $\beta_0 \in B \bs B_1$, $\ga_0 \notin S$ and we form the set $S_1 := S \cup \set{\ga_0}$. It is easy to see $S_1 \in \CC$ and this contradicts the maximality of $S$.

Moreover, the construction of $S$, the assumption that every element has order $p$ and \cref{thm2.3} imply $S$ is $p$-PR, which finishes the proof for the special case.

For the general case, we let $E_k$ be defined as the subset in $E$ containing elements of order $p^k$. Then $E = \cup_{k \geq 0} E_k$ and hence there exists a positive integer $K \in \N^+$ such that $|E_K| = |E|$. If $K = 1$, the special case finishes the proof, and therefore we suppose $K > 1$.

We have two cases. The first case is that there exists $1 \leq n_0 < K$ such that $|\pi_{p^{n_0}}(E_K)| = |E_K|$, while $ |\pi_{p^{n_0+1}}(E_K)| < |E_K|$. We let $B_1 \subset B$ be given by $\beta \in B_1$ if there exists $\ga \in E_K$ such that $\Proj_\beta(\ga)$ has order greater or equal to $p^{n_0+1}$. Since $|\pi_{p^{n_0+1}}(E_K)| < |E_K|$, we have $|B_1| < |B| = |E_K|$. For a subset $C \subset B$, we define the projection $\Proj_C: \bigoplus_{\beta \in B} \CC(p^\infty) \ra \bigoplus_{\beta \in C} \CC(p^\infty)$. We have $|\Proj_{B \bs  B_1}(\pi_{p^{n_0}}(E_K))| = |E_K|$. Furthermore, each $\alpha \in \Proj_{B \bs  B_1}(\pi_{p^{n_0}}(E_K))$ has the property that for all $\beta \in B \bs  B_1$, $\Proj_\beta(\alpha)$  has order $p$ or $1$. Thus, the special case gives a $p$-PR set $F \subset  \Proj_{B \bs  B_1}(\pi_{p^{n_0}}(E_K))$ such that $|F| = |\Proj_{B \bs  B_1}(\pi_{p^{n_0}}(E_K))| = |E_K| = |E|$ and the quotient lemma finishes the proof for this case.

The other case is that $|\pi_{p^K}(E_K)| = |E_K|$. Since $\pi_{p^K}(E_K)$ satisfies the special case, the quotient lemma again finishes the proof.

(2) The proof of (2) is similar to the first part of the argument of (1), but much simpler.
\end{proof}

\begin{thm}\label{thm2.8}
Let $E \subset \Ga$ be uncountable. Then there exists a prime number $p$ such that $E$ contains a $p$-PR set of the same cardinality.
\end{thm}
\begin{proof}
Embed $E \subset \bigoplus_{i \in B_0} \Q \oplus \oplus_{j=1}^\infty \bigoplus_{i \in B_j} \CC(p_j^\infty)$, where $(B_j)_{j=0}^\infty$ are index sets and $p_j$ are distinct primes. We assume that for each index $i \in \bigcup_{j=0}^\infty B_j$, there exists $\ga \in E$ such that $\Proj_i(\ga)$ is non-trivial. Since $E$ is uncountable and the groups $\Q$ and $\CC(p_j^\infty)$ are countable, there exists $K \in \N$ such that $|B_K| = |E|$.

 If $K = 0$, from \cref{lem4.3} (2) we may extract an independent set $F \subset \pi_0(E)$ with $|F| = |E|$. If we choose $E' \subset E$ such that $\pi_0$ is one-to-one on $E'$ and $\pi_0(E') = F$, then $E'$ is $N$-PR for all $N \geq 2$ by \cref{lem2.1} and \cref{prop2.2}.

Similarly, if $K \geq 1$, by \cref{lem4.3} (1) we extract a $p_K$-PR subset $F \subset \pi_{p_K}(E)$ with $|F| = |E|$ and therefore obtain a $p_K$-PR subset in $E$ of the same cardinality.
\end{proof}

\begin{rem}
Since any $p$-PR set is a Sidon set, \cref{thm2.8} implies any uncountable subset in $\Ga$ contains a large Sidon set. $I_0$ sets are  more special interpolation sets. A set $E \subset \Ga$ is $I_0$ if every bounded function defined on $E$ can be interpolated as a Fourier transform of a discrete measure. If $p \geq 3$, any $p$-PR set is an $I_0$ set, and therefore in that case any uncountable subset in $\Ga$ contains a large $I_0$ set.
\end{rem}

\begin{thm}\label{cor2.7}
Let $E \subset \Ga$ be uncountable, $p$ be a prime number and $n \in \N$. Then $E$ contains a $p^n$-PR subset of the same cardinality if and only if $|\pi_{p^n}(E)| = |E|$.
\end{thm}
\begin{proof}
If $E$ contains a $p^n$-PR subset $E_1$ of the same cardinality, by \cref{prop3.2} $|\pi_{p^n}(E)| \geq  |\pi_{p^n}(E_1)| = |E_1| = |E|$.

To see the converse, we define the projection
\ali{\pi_1: \bigoplus_{i \in B_1} \Q \oplus \bigoplus_{i \in B_2} \CC(p^\infty) \ra \bigoplus_{i \in B_2} \CC(p^\infty).}
Assume $|\pi_{p^n}(E)| = |E|$. We claim that either $|\pi_0(\pi_{p^n}(E))| = |E|$ or $|\pi_1(\pi_{p^n}(E))| = |E|$.

If $|\pi_0(\pi_{p^n}(E))| < |E|$ and $|\pi_1(\pi_{p^n}(E))| < |E|$, then because $E$ is uncountable,
\ali{|\pi_{p^n}(E)| \leq |\pi_0(\pi_{p^n}(E))||\pi_1(\pi_{p^n}(E))| < |E|,}
which is a contradiction. If $|\pi_0(\pi_{p^n}(E))| = |E|$, we appeal to (2) in \cref{lem4.3} to get a $p^n$-PR set in $\pi_0(\pi_{p^n}(E))$ and \cref{lem2.1} finishes the proof. If $|\pi_1(\pi_{p^n}(E))| = |E|$, then (1) in \cref{lem4.3} similarly finishes the proof.
\end{proof}

\begin{exmp}
If $E$ is countable, \cref{thm2.8} may fail. Let $E = \CC(p^\infty)$. $E$ is countably infinite, but by \cref{prop4.2}, $E$ only contains $p$-PR sets that are singletons.
\end{exmp}

\begin{prop}\label{prop3.6}
Suppose $E = \Ga \subset \bigoplus_{i \in B_1} \Q \oplus \bigoplus_{i \in B_2} \CC(p^\infty)$ is a countably infinite group. Then $E$ contains an infinite $p$-PR set if and only if $E$ contains an infinite independent set.
\end{prop}
\begin{proof}
Assume that all independent sets in $E$ are finite. Zorn's Lemma gives a maximal independent set $S \subset E$, where the partial order is inclusion. Our assumption gives that $S$ is finite and the maximality implies $\ag{S} = E$. Since $S$ is independent, we embed $S$ into $\bigoplus_{\ga \in S} \Ga_\ga$, where $\Ga_\ga = \Q$ if $\ga$ has infinite order and $\Ga_\ga = \CC(p^\infty)$ if $\ga$ has order some power of $p$. Since $S$ is maximal, we may extend  the embedding to $\ag{S} = E$. From \cref{prop4.2}, this implies all $p$-PR sets in $E$ have cardinality at most $|S|$.
The converse is trivial.
\end{proof}

\begin{prop}
Suppose $\Ga$ is an uncountable infinite group and $N=p_1^{m_1}...p_k^{m_k}$ is an integer with prime numbers $p_i$, $1 \leq i \leq k$. Then $\Ga$ contains an $N$-PR set $E$ with $|E| = |\Ga|$ if and only if $|\pi_{p_i^{m_i}}(\Ga)| = |\Ga|$ for all $1 \leq i \leq k$.
\end{prop}
\begin{proof}
If $\Ga$ contains an $N$-PR set $E$ with $|E| = |\Ga|$, then by \cref{cor3.4}  $|\pi_{p_i^{m_i}}(\Ga)| = |\Ga|$ for all $1 \leq i \leq k$.

To see the converse, first of all, if $|\pi_0(\Ga)| = |\Ga|$, since $\Ga$ is uncountable, the conclusion follows by \cref{lem4.3} and \cref{lem2.1}. Thus we may assume $|\pi_0(\Ga)| < |\Ga|$. Hence $|\Ga_0| = |\Ga|$ and we may further assume $\Ga$ is a torsion group. If $|\pi_{p_i^{m_i}}(\Ga)| = |\Ga|$ for all $1 \leq i \leq k$, then by \cref{lem4.3} (1) we let $S_i$ be a subset in $\pi_{p_i^{m_i}}(\Ga)$ such that $S_i$ is $p_i$-PR with $|S_i| = |\Ga|$, $1 \leq i \leq k$. For $1 \leq i \leq k$ we let $J_i \subset \Ga$ be such that $\pi_{p_i^{m_i}}$ is one-to-one on $J_i$ and $\pi_{p_i^{m_i}}(J_i) = S_i$ and moreover, we may assume the order of each $\ga \in J_i$ is a power of $p_i$. Since $|J_i| = |\Ga|$ for all $1 \leq i \leq k$, we let, for $2 \leq i \leq k$, $f_i: J_1 \ra J_i$ be bijections and we form the set
\ali{E := \set{\ga f_2(\ga)...f_k(\ga) : \ga \in J_1}.}
Then $|E| = |\Ga|$.  Since for all $\ga \in J_i$, $1 \leq i \leq k$, the order of $\ga$ is a power of $p_i$, if $\ga \in J_1$, $f_i(\ga) \in \Ga_{p_1^{m_1}}$ for all $2 \leq i \leq k$ and hence
\ali{\pi_{p_1^{m_1}}(\ga f_2(\ga)...f_k(\ga)) = \pi_{p_1^{m_1}}(\ga).}
Similarly,
\ali{ \pi_{p_i^{m_i}}(\ga f_2(\ga)...f_k(\ga)) = \pi_{p_i^{m_i}}(f_i(\ga)),}
for all $2 \leq i \leq k$. Hence, $\pi_{p_i^{m_i}}(E) = \pi_{p_i^{m_i}}(J_i) = S_i$ for $1 \leq i \leq k$. By \cref{cor3.4}, $E$ is $N$-PR.
\end{proof}

\begin{rem}
In \cite{2012paper} the terminology ``$N$-large'' sets is introduced. A set $E \subset \Ga$ is $N$-large if $|Q_{N}(E)| < |E|$ where $Q_{N}: \Ga \ra \Ga / H_{N}$ is the quotient map and $H_{N} \subset \Ga$ is the subgroup of elements of orders divisible by $N$.

Theorem 2.2 in \cite{2012paper} states that if $E \subset \Ga$ and $N$ is the smallest integer for which $E$ is $N$-large, then for all primes powers $p^n$ dividing $N$ there exists a weak $|1-e^{\pi i/p^n}|$-Kronecker subset $F \subset E$ with $|F| = |E|$.

First, we note that the assumption in \cref{cor2.7} is weaker than the assumption in Theorem 2.2 in \cite{2012paper}; specifically, if $E$ is infinite and $N$-large for minimal $N = p^mp_1^{m_1}...p_k^{m_k}$ and $n \leq m$ for some integer $n$, then $|\pi_{p^n}(E)| = |E|$. To see this, we argue by contradiction and assume $|\pi_{p^n}(E)| < |E|$. Let $M := p^{n-1}p_1^{m_1}...p_k^{m_k} < N$. If $\ga \in \Ga_{p^n} \cap H_N$ and $k$ is the order of $\ga$, then $k$ divides $N$ while $p^n$ does not divide $k$. This implies $k$ divides $M$ and hence $\ga \in H_M$. Thus $\Ga_{p^n} \cap H_N = H_M$. As a result, the map
\ali{ T: Q_M(E) &\ra Q_N(E) \times \pi_{p^n}(E) \\
       T(Q_M(\ga)) &= (Q_N(\ga), \pi_{p^n}(\ga)).}
is well-defined and injective. Thus, if $|\pi_{p^n}(E)| < |E|$, then
\ali{ |Q_M(E)| \leq |Q_N(E)| |\pi_{p^n}(E)| < |E| |E| = |E|}
for infinite $E$. Thus $E$ is $M$-large and this contradicts the assumption that $N$ is minimal.

Moreover, recall that $p^n$-PR sets are special weak $|1-e^{\pi i/p^n}|$-Kronecker sets. This shows \cref{cor2.7} improves Theorem 2.2 in \cite{2012paper} when $E$ is uncountable.

\end{rem}

config: arxiv


TITLE: Does it take more energy to open a door when applying force close to the hinge?
QUESTION [20 upvotes]: Assuming an ordinary hinged door (without any springs), would it take more energy to open it when applying force in the middle of the door (point b), rather than at the end of the door (point a), where the door knob is?
"Opening the door" should be interpreted as accelerating the door to a certain rotational speed.

My own answer is no, since the change in force would be proportional to the distance required to open the door and therefore the total energy would remain the same.

REPLY [2 votes]: To get to a final rotational velocity $\Omega$ is takes the same amount of energy regadless of where you push. Why? Well the final kinetic energy is $K=\frac{1}{2} I \Omega^2$ (where $I$ is the mass moment of inertia about the hinge) and this value does not depend on where you push.
This is kind of a boring result.
What does differ is how much you need to push, and how much reaction the hinges provide. If you only push for a small period of time providing with an impulse $J=\int F(t)\,{\rm d}t$ then the hinges would develop a reaction impulse of $R$.
$$\begin{align} J & = \frac{I \Omega}{a} & R = \frac{I \Omega}{a} - m \Omega \frac{\ell}{2} \end{align} $$
where $a$ is the distance of where I push from the hinges and $\ell$ is the width of the door. Given that $I=m \frac{\ell^2}{3}$ it can be seen that when $a=\frac{2}{3} \ell$ there is no reaction on the hinges. That is called the center of percussion (sweet spot).
Now the impulse $J=F \Delta t$ can be viewed as an average force $F$ applied for a small time $\Delta t$. If the time of pushing is fixed, then $$F = m \frac{\Omega \ell^2}{3 a \Delta t} $$ which means the further away from the hinge the less the force (duh!).
Now if the distance $\delta$ by which the force is applied is fixed ($\Delta t = \frac{\delta}{v} = \frac{\delta}{a \Omega}$) the force is $$F = m \frac{\Omega^2 \ell^2}{3 \delta} $$ which does not depend on the distance $a$. This can be explained since force over distance is work, which goes into kinetic energy, since the goal is the same kinetic energy, it takes the same amount of work. If the distance is fixed then the force must be fixed also to get to the same work.
Appendix
The two equations of motion I used are

$J-R = m v_{cm} = m (\Omega \frac{\ell}{2} )$
$(a-\frac{\ell}{2}) J + \frac{\ell}{2} R = m \frac{\ell^2}{12} \Omega $

Figure 1. Door sketch from the top


TITLE: Isn't all light polarised?
QUESTION [4 upvotes]: I apologize if my question does not make sense.(I'm teaching myself microscopy.)
So reading Fundamentals of Light Microscopy and electronic imaging by Douglas&Murphy, at one point the author mentions that light can be described as electromagnetic waves and then gives a picture similar to this one
.
Later on, when describing the "quality of light"(I quote him) he writes:
The kinds of light most frequently referred to in this text include:
•    Monochromatic.  Waves having the same wavelength or vibrational frequency
(the same color).
•    Polarized.  Waves whose E vectors vibrate in planes that are parallel to one
another. The E vectors of rays of sunlight reflected off a sheet of glass are plane
parallel and are said to be linearly polarized.
•    Coherent.  Waves of a given wavelength that maintain the same phase relation
ship while traveling through space (laser light is coherent, monochromatic, and
polarized).
•    Collimated.  Waves having coaxial paths of propagation through space—that is,
without convergence or divergence, but not necessarily having the same wave-
length, phase, or state of polarization. The surface wavefront at any point along
a cross-section of a beam of collimated light is planar and perpendicular to the
axis of propagation.
My confusion stems from the paragraph when the author describes polarized light. In the general form of the waves, given in the picture, the electric field vectors are parallel. So, all light should be polarized, right?
Also, regarding the image describing the waves, does the pink arrow(which is a straight line), which is parallel to the x-axis(Am I right?) describe the path of a photon?

REPLY [4 votes]: Under normal conditions, each photon can be thought of as a purely coherent entity with a definite polarization at all points in momentum (wavenumber) space. I discuss this notion in  more depth in several other answers, notably this one here but the essential idea is this: lone photons propagate following Maxwell's equations (which are pretty much the Dirac equation for a massless, spin 1 particle), so a pure quantum one-photon state is specified by the complex quantum superposition weights of the two cicularly polarization eigenstates at each value of the wavevector $\vec{k}$ in momentum space.
So, at this one-photon level, yes - complicated entanglements aside - each photon interferes only with itself (Dirac's almost correct famous statement) and it has definite polarization, albeit a very complicated one (different superposition weights across a broad spectrum of frequencies).
Most natural sources however output these fundamentally coherent entities in a wide spread of different pure quantum states. So most light is in a mixed state (see also the Wikipedia articles on the Density Matrix and the Wigner's Friend thought experiment) : a classically probabilistic mixture of pure one photon states. This is the easiest way to think of depolarized light, whose classical description gets very involved indeed (Born and Wolf give a whole chapter to it - or at least to the classical description of the highly related notion of the partially coherent light). The full quantum description involving entanglement is also very thorny, but the basic intuitive notion of the mixed state is a much easier way than the classical description to understand depolarization.

Some answers to user Phonon's comment:

"very interesting take! [This post (Physics SE Question "Difference Between Spin and Polarization of a Photon")] may also interest you. (it still lacks a straight answer I'd say.) –"

I think that question shows that "polarization" - like all natural English (or German, Mandarin - any natural language) words - can become a bit vague and hence the need to encode one's ideas in mathematical language. The best definition of polarization I can think of off the top of my head is:

Polarization is a word that, in its strictest sense, applies to pure quantum states; it applies most readily to one-photon Fock states or quantum pure coherent states;
One specifies a one photon Fock state's polarization by defining the complex superposition weights of the left and right angular momentum eigenstates at each and every point $\vec{k}$ in momentum space. So polarization, in general, is a function of the form $\mathbb{R}^3\to\mathbb{C}\times \mathbb{C}$, i.e. two complex valued functions of the momentum space $\mathbb{R}^3$. Note that in general we are not talking about a monochromatic field; monochromatic fields look like spherical shells (defined by $c|\vec{k}| = \omega$) in momentum space. Now;
We can broaden our polarization definition to partially polarized light by thinking of classical mixtures of pure quantum states. So now our partially polarized state becomes a density matrix function $\mathbb{R}^3\to \mathcal{M}(2,\,\mathbb{C})$ assigning a $2\times2$ Hermitian density matrix holding the Stokes parameters for each $\vec{k}\in\mathbb{R}^3$ momentum space.

Now then, the word "spin" has two, related meanings. The first is the spin state of a photon is defined in terms of the superposition weights of the eigenfunctions of the spin observables.  If you consider the spin observable for the component of the angular momentum in the direction of $\vec{k}$ at all points in momentum space and write the photon's state as superpositions of the eigenstates of this observable, then you are indeed computing precisely the polarization as I have defined it above. So in this sense, spin and polarization are pretty much synonymous - or at least that's my understanding of the English (natural language) usage.
The second meaning of the word "spin" is as Lionel Brit's answer to that question discusses: when we say the photon has "spin 1" we are simply saying its state (a solution of Maxwell's equations) transforms in a certain way under the action of the Lorentz group. "Spin 1" names a certain representation of the Lorentz group.


TITLE: a counting problem...
QUESTION [0 upvotes]: Six people sat down along one side of a banquet table completely ignoring their name cards. In how many ways could this have been done so that no person was seated where his/her name card was placed?
I think the solution is 5^6 because each person can go to other 5 chairs that is not seated... However the answer is 265. Can someone explain this to me?

REPLY [1 votes]: To remove this from the queue:
The answer is $!6=265$ where $!n$ denotes the subfactorial function which counts the number of derangements on a set of $n$ elements.  A derangement is a permutation with no fixed points, as is the case with your example of people sitting around the table in any place but their assigned seat.
Various methods exist for counting derangements, perhaps the most accessible in your case is:
$$!6=6!\sum\limits_{i=0}^6\frac{(-1)^i}{i!}$$
which can be derived via inclusion-exclusion on the events that seat $i$ has person $i$ sitting in it.


TITLE: Axioms for the real numbers
QUESTION [0 upvotes]: Suppose we have a field with a total order  which verifies
If $x,y\geq 0$, then $x+y\geq 0$      (here we relax the compatibility property for the addition).
If $x,y\geq 0$, then $xy\geq 0$.
Suppose also that every upper bounded  nonempty set has a supremum.
We don't assume that  $x\leq 0, -x\leq 0$ implies $x=0$.
Is this field the ordered field  of real numbers?

REPLY [4 votes]: The answer is no. E.g., let $R$ be the field of all real numbers with the strict total order $\prec$ defined as follows: for any $x$ and $y$ in $R$ such that $x<y$ (where $<$ is the usual order on $R$), let $x\prec y$ if $y\ge0$, and $y\prec x$ if $y<0$. Then all your conditions on the total order will hold, but $(R,\prec)$ will not be an ordered field.
Details:
We see that the order $\prec$ preserves the usual order among the nonnegative real numbers, but reverses the usual order among the negative real numbers, also keeping the negative numbers less than the nonnegative ones.
Since the $\prec$-nonnegative numbers are the same as the $<$-nonnegative ones, your first two conditions, concerning the addition and multiplication operations, hold.
Now take any nonempty $\prec$-upper-bounded subset $A$ of $R$. If there is a nonnegative number in $A$, then $A$ is $<$-upper-bounded, and $\sup_{\prec} A=\sup_{<} A\in R$; the subscripts $_\prec$ and $_<$ indicate here the order with respect to which the supremum or infimum is taken.
Otherwise, that is, if there is no nonnegative number in $A$, then $A$ is always $\prec$-upper-bounded by $0$. If now $c:=\inf_< A\ne-\infty$, then $c=\sup_\prec A\in R$. Finally, if $c=-\infty$, then $\sup_\prec A=0\in R$.
So, any nonempty $\prec$-upper-bounded subset $A$ of $R$ has $\sup_\prec A\in R$. Thus, your third, and final, condition on the total order holds as well.
However, $(R,\prec)$ is not an ordered field, because $-1\prec-2$ but $-1+1\not\prec-2+1$.


TITLE: Show that this continuous local martingale is a martingale
QUESTION [1 upvotes]: We are given the following SDE:
$$dX_t=X_tdt+\sqrt{2}X_tdB_t, \quad X_0=1,$$
and
$$F(x,t)=e^{-t}x,\quad t\geq0,\; x\in\mathbb{R}.$$
We are asked to apply Ito's formula to $F(t,X_t)$ for $t\geq0$ and determine a continuous local martingale $(M_t)_{t\geq0}$ (starting at $0$) and a continuous bounded variation process $(A_t)_{t\geq0}$ such that $F(t,X_t)=M_t+A_t$ for $t\geq0$.
If I am correct, $M_t=\int_0^tF_x(s,X_s)dX_s=\int_0^te^{-s}ds+\sqrt{2}\int_0^te^{-s}dB_s$, $t\geq0$
Now, we need to show that $M_t$ is a martingale and compute $\langle M,M\rangle_t$ and $\mathbb{E}[e^{-\tau}X_\tau]$ when $\tau=\inf\{t\geq0:X_t=2-t\}$ but I don't know how! Any help would be appreciated!

REPLY [0 votes]: It's not difficult to see that
$$X_t := \exp \left(\sqrt{2} B_t \right)$$
solves the given SDE. (You can either use Itô's formula to check it or use some standard methods for linear SDE's to obtain this solution.) Moreover, by Ito's formula:
$$f(t,X_t)-\underbrace{f(0,x_0)}_{x_0} = \sqrt{2} \int_0^t e^{-s} \cdot X_s \, dB_s + \int_0^t e^{-s} \cdot X_s + (-e^{-s} \cdot X_s) \, ds \\ = \sqrt{2} \int_0^t e^{-s} \cdot e^{\sqrt{2} B_s} \,dB_s \\ \Rightarrow f(t,X_t) = \underbrace{\sqrt{2} \int_0^t e^{\sqrt{2} B_s-s} \,dB_s}_{M_t} + \underbrace{x_0}_{A_t}$$
where $x_0=1$. Let
$$g(s,w) := \sqrt{2} \cdot e^{\sqrt{2} B(s,w)-s}$$
Then $g \in L^2(\lambda_T \otimes \mathbb{P})$, i.e.
$$\int_0^T \int_\Omega g(s,w)^2 \, d\mathbb{P} \, ds <\infty$$
There is a general result which says that this condition implies that $M_t$ is a martingale (and not only a local one). Moreover,
$$\langle M,M \rangle_t = \int_0^t |g(s,w)|^2 \, ds$$
(see René L. Schilling/Lothar Partzsch: "Brownian Motion - An Introduction to stochastic processes", Theorem 14.13).
Concerning the integral $\mathbb{E}(e^{-\tau} \cdot X_\tau)$: Remark that
$$\tau = \inf\{t \geq 0; X_t=2-t\} = \inf\{t \geq 0; \sqrt{2} B_t = \ln(2-t)\}$$
Now let
$$\sigma := 2\tau = \inf\{t \geq 0; \underbrace{\sqrt{2} B_{\frac{t}{2}}}_{=:W_t} = \ln (2-t/2)\}$$
where $(W_t)_{t \geq 0}$ is again a Brownian Motion (scaling property). Thus
$$\mathbb{E}(e^{-\tau} \cdot X_\tau) = \mathbb{E}(e^{-\tau+\sqrt{2} B_\tau}) = \mathbb{E}(e^{-\frac{\sigma}{2}+W_\sigma}) \stackrel{\ast}{=} 1$$
In $(\ast)$ we applied the exponential Wald identity (see remark).
Remark Exponential Wald identity: Let $(W_t)_{t \geq 0}$ a Brownian motion and $\sigma$ a $\mathcal{F}_t^W$-stopping time such that $\mathbb{E}e^{\sigma/2}<\infty$, then $\mathbb{E}(e^{W_\sigma-\frac{\sigma}{2}})=1$. (see René L. Schilling/Lothar Partzsch: "Brownian Motion - An Introduction to stochastic processes", Theorem 5.14)


TITLE: Find $\displaystyle{\lim_{n \to \infty}} \sqrt[n]{n!}$ without Stirling's approximation.
QUESTION [5 upvotes]: Okay I've seen many posts as this one, but in every post it uses Stirling(I can't use it because I've never learnt it), or weird equalities with logs (I also never learnt this) or other methods I'm not allowed to use in my exam.
I have to prove that $\lim n!^{(1/n)}$ = $\infty$, only by using D'Alembert, $n^{th}$ root test, Stolz, squeeze theorem or by playing with inequalities etc
What I thought is that I could prove that $\lim \frac{1}{n!^{(1/n)}} = 0$ (again only using those basic theorems) and that would mean that $\lim n!^{(1/n)} = \infty$ but I'm not sure if that's correct.
Please don't mark as duplicate! I need an answer and I haven't found any post with an answer I can use!!

REPLY [4 votes]: Note that since $n!\ge(n/2)^{n/2}$, we have
$$\sqrt[n]{n!}\ge \sqrt{n/2}$$

REPLY [0 votes]: $n! \ge (\frac{n}{2})^{\frac{n}{2}} \implies (n!)^{1/n} \ge \sqrt{\frac{n}{2}}$


TITLE: Is there any stricy way of eliminating Lagrangian Multipliers or it depends on the specific equations after derivative equations?
QUESTION [0 upvotes]: I am pretty new for the concept of langragian and might be a naive question. After computing the first derivative equation of lagrange version of the function, it is required to get away multipliers but it is a little hasting for me at the moment to see how to do that. Do I need to come up with new method for this elimination for each different problem or is there any intuitive general way to get rid of this multipliers to find the optimal value of objective variable?

REPLY [0 votes]: If there would be a simple "general way to get rid of these multipliers" this way would have been condensed into a theorem, and you would have heard about it.
Whether one should get rid of the multipliers right away depends on the problem at hand.
In many situations where there is just one multiplier $\lambda$ it is possible to express the actually interesting unknowns $x_i$ (resp., $x$, $y$, $z$, etc.) in terms of $\lambda$. The given condition $F(x_1,x_2,\ldots, x_n)=0$ then turns into an equation for $\lambda$ that hopefully can be solved, and finally one obtains the values of the corresponding $x_i$.


text:
\section{3-nets and 3-tangles}

In this section, we will define 3-nets and 3-tangles. They are
generalizations of links and tangles. We will also construct the category
\topf \/ of trivalent  framed tangles, which will allow us to work in
a category-theoretical setting.\\

A 3-net will be something like a ``link with branchings''. To
describe the situation near a trivalent vertex (i.e.\ near a
branching point), we will need the following notion:\\ Let $B$ be
the unit ball in $\R^3$, i.e.\ $\{x\in
\R^3; |x|\leq 1\}$, together with the distinguished subset $T:= \{(t,0,0)|0\leq t\leq1\}\cup
\{(-\frac{1}{2},\frac{\sqrt{3}}{2},0)t|0\leq t\leq 1\}\cup
\{(-\frac{1}{2},-\frac{\sqrt{3}}{2},0)t|0\leq t\leq 1\}$.\\[1em]
\centerline{$B\quad =\quad$\psdiag{7}{21}{1kap1}}

\begin{df}
A {\em framed 3-net} is a subset $N$ of \/ $\R^3$ with a finite
subset $\{t_1,\ldots ,t_n\}\subset N$ such that:
\begin{itemize}
\item[(i)] for every point $t_i$, there exists a
neighbourhood $U_i$ in $\R^3$ and a diffeomorphism
$f_i:U_i\rightarrow B$ with $f_i(N\cap U_i)= T$,
\item[(ii)] $\tilde{N}:= N \backslash \left(\bigcup_{i=1}^n f_i^{-1}(\{x\in B;
|x|<\frac{1}{2}\})\right)$ is an embedded  smooth  closed compact 1-dimensional
manifold,
\end{itemize}
together with:
\begin{itemize}
\item[(iii)] a  smooth vector field on $N$ that is nowhere tangent to $N$ (and in particullary nowhere zero)\footnote{Note that our framing is not a
framing in the classical sense.}.\\
\end{itemize}

  The points
$t_1,\ldots, t_n$ are called {\em trivalent vertices} of $N$;
boundary points $x$ of $\tilde{N}$ with $x\not\in U_i
\:(\forall i)$ are called {\em univalent vertices} of $N$.
\end{df}

Observe that 3-nets without trivalent and univalent vertices are
simply framed links. When we represent a framed 3-net $N$ by a diagram,
the framing of $N$ is given by the blackboard framing\footnote{i.e.\
the vector field assigned to $N$ consists of vectors pointing
upward }.
\newline The  3-net in the following figure is a 3-net
with 7 trivalent
and 3 univalent vertices.\\[1em]
\centerline{\psbild{25}{1kap2a}\quad\quad\quad\quad}

As all 3-nets in this paper will be framed 3-nets, we will usually
omit the word ``framed''.

\begin{df}\label{defeq}
Two 3-nets $N_1$ and $N_2$ are {\em equivalent} if $N_1$ can be
deformed into $N_2$ within the class of framed 3-nets by a smooth
1-parameter family of diffeomorphisms of $\R^3$.\newline A {\em
closed} 3-net is a 3-net without univalent vertices.\newline A
3-net is {\em planar} if it is equivalent to a 3-net $M$ with
$M\subset
\R^2\times \{0\}$ and the vector field assigned to $M$
consists of vectors of the form $(0,0,1)$.
\end{df}

\begin{rmk}
{\em The overview in the introduction intimates that we will obtain
an invariant of closed 3-nets that is composed of Vassiliev
invariants (see also section 6). Vassiliev invariants are usually
defined for oriented links, but we remind the reader that there is
a definition for unoriented links (see e.g.\ [St]):\newline
A link invariant\footnote{i.e.\ a function assigning to
each link an element of an abelian group (usually $\C$) that is
constant on the equivalence classes of framed links} $f$ is a {\it
Vassiliev invariant of type $m$} if for any link $L$, any diagram
$D(L)$ of $L$ and any subset $C$ of the set of crossings of $D(L)$
with cardinality greater than $m$ the following equation holds:
$$\sum_{X\subset C}(-1)^{|X|}f([D(L)_X])=0,$$
where $|X|$ is the cardinality of $X$, $D(L)_X$ is the link diagram
obtained form $D(L)$ by changing all the crossings in $X$, and
$[D(L)_X]$ is a link with diagram $D(L)_X$ (such that the framing
on the link is given by the blackboard framing of the diagram).
\newline
Of course, this definition can be extended to 3-nets. }\end{rmk}
\begin{df}
A {\em (framed) 3-tangle} is a framed 3-net $N$ with $N\subset
[0,1]\times
\R^2$ such that the points of $N$ lying in the planes $\{0\}\times
\R^2$ and $\{1\}\times \R^2$ are exactly the univalent vertices of
$N$ and these lie on one of the lines $\{0\}\times \R\times \{0\}$
and $\{1\}\times\R\times\{0\}$. Additionally,  we require that the
normal plane of $N$ in a univalent vertex $v$ is
parallel to the plane $\{0\}\times\R^2$, and the
vector field assigned to $N$ is $(0,0,1)$ in $v$.\end{df}

\begin{df}\label{deftangeq}
Two 3-tangles $T_1$ and $T_2$ are {\em equivalent} if one can be
deformed into the other within the class of 3-tangles by a smooth
1-parameter family of diffeomorphisms of $\R^3$.\newline A 3-tangle
is {\em planar} if it is equivalent to a 3-tangle $M$ with
$M\subset [0,1]\times\R\times \{0\}$ and the vector field assigned
to $M$ consists of vectors of the form $(0,0,1)$.
\end{df}

Now we will define the category of {\bf t}rivalent, {\bf
p}arenthesized,
{\bf f}ramed tangles \topf. It is an (unoriented) generalization of the category of
non-associative tangles in [BN 2]. \\

First, we will define non-associative words (which will be the objects of
\topf).
\begin{df}
A {\em non-assocative \aw} is a word $w$ in the alphabet
$\{\diamond, ), (\}$ such that $w$ is equal to the empty word,
$\diamond, (w_1)$ or $(w_1w_2)$ where $w_1, w_2$ are
non-associative
\aws.\newline
The {\em length} $l(w)$ of a non-associative word $w$ is the number of symbols
$\diamond$ in $w$.
\end{df}

\begin{ex}\quad\quad\quad\quad\quad $l\left(\diamond((\diamond(\diamond(\diamond\diamond)))\diamond)\right)
= 6$
\end{ex}

\begin{df}
Let \htopf \/ be the monoidal $\C$-category whose objects are
non-as\-so\-cia\-tive words (where the tensor product $w_1\otimes
w_2$ is $(w_1w_2)$ and the unit object is the empty word) and whose
morphisms are freely generated by the following morphisms:
\begin{itemize}
\item[(G1)] A morphism ${\mbox{\pstext{phi}}}_{v,w,x}$ and a morphism
 ${\mbox{\pstext{phi-}}}_{v,w,x}$ for each triple $(v,w,x)$ of
non-associative \aws: The sources of these morphisms are $((vw)x)$
and $(v(wx))$ and their targets are $(v(wx))$ and $((vw)x)$
respectively.\newline Graphically, we represent these morphisms as
the following examples indicate (the parenthesation of source and
target is encoded in the distances between the strands):
\end{itemize}
\begin{tabbing}
xxxxxxxxxxxxxxxxxxxxxxx \= xxxxxxxxxxxxxx \=  \kill
\>${\mbox{\pstext{phi}}}_{ (\diamond\diamond),\diamond,(\diamond\diamond)}\quad\quad$
\>$\leftrightarrow
\quad\quad\mbox{\psdiag{4}{12}{phirel+}}$\\[2ex]
\>${\mbox{\pstext{phi-}}}_{\diamond,(\diamond\diamond),(\diamond\diamond)}\quad\quad$
\>$\leftrightarrow \quad\quad \mbox{\psdiag{4}{12}{phirel-}}$
\end{tabbing}
\begin{itemize}
\item[(G2)]
A morphism ${\mbox{\pstext{Zarg1}}}_{v,w}$ and a morphism
 ${\mbox{\pstext{Zarg5}}_{v,w}}$ for each pair $(v,w)$ of
non-associative \aws: The source of these morphisms is
$(v,w)$  and their target is
$(w,v)$. We will depict them as follows:
\end{itemize}
\begin{tabbing}
xxxxxxxxxxxxxxxxxxxxxxx \= xxxxxxxxxxxxxx \=  \kill
\>${\mbox{\pstext{Zarg1}}}_{\diamond,((\diamond\diamond)\diamond)}\quad\quad$
\>$\leftrightarrow\quad\quad\mbox{\psdiag{3.5}{10.5}{pkrel}}$\\[2ex]
\>${\mbox{\pstext{Zarg5}}}_{(\diamond(\diamond\diamond)),(\diamond\diamond)}\quad\quad$
\>$\leftrightarrow\quad\quad\mbox{\psdiag{4}{12}{nkrel}}$
\end{tabbing}
\begin{itemize}
\item[(G3)]A morphism ${\mbox{\pstext{lbog}}}$ and a morphism \pstext{rbog}
with sources  $(\:)$ and $(\diamond\diamond)$ and targets $(\diamond\diamond)$ and
$(\:)$ respectively; graphically: \end{itemize}
\begin{tabbing}
xxxxxxxxxxxxxxxxxxxxxxx \= xxxxxxxxxxxxxx \=  \kill
\>${\mbox{\pstext{lbog}}}\quad\quad$\>$\leftrightarrow\quad\quad
\mbox{\psdiag{3}{9}{lbog}}$\\[2ex]
\>${\mbox{\pstext{rbog}}}\quad\quad$\>$\leftrightarrow\quad\quad
\mbox{\psdiag{3}{9}{rbog}}$
\end{tabbing}
\begin{itemize}
\item[(G4)]
A morphism ${\mbox{\pstext{trir}}}$ and a morphism ${\mbox{\pstext{tril}}}$ with
sources
$\diamond$ and $(\diamond\diamond)$ and targets $(\diamond\diamond)$ and $\diamond$
respectively; graphically:
\end{itemize}
\begin{tabbing}
xxxxxxxxxxxxxxxxxxxxxxx \= xxxxxxxxxxxxxx \=  \kill
\>${\mbox{\pstext{trir}}}\quad\quad$\>$\leftrightarrow\quad\quad
\mbox{\psdiag{3}{9}{trir}}$\\[2ex]
\>${\mbox{\pstext{tril}}}\quad\quad$\>$\leftrightarrow\quad\quad
\mbox{\psdiag{3}{9}{tril}}$
\end{tabbing}
The graphical representation of the tensor product of two morphisms
is obtained by putting the first above the second, the graphical
representation of the composition $S_2\circ S_1$ is obtained by
glueing the graphical representation of $S_2$ to the one of $S_1$
from the right:
\begin{tabbing}
xxxxxxxxxxxxxxxxxxxxxxx \= xxxxxxxxxxxxxx \=  \kill
\>$S_1 \otimes S_2\quad\quad $\>$\leftrightarrow \quad\quad
\mbox{\psdiag{5}{12}{s1tes2}}$\\[3ex]
\>$S_2 \circ S_1\quad\quad $\>$\leftrightarrow \quad\quad
\mbox{\psdiag{2}{6}{s1coms2}}$
\end{tabbing}
\end{df}

The graphical representations of the morphisms allow us to assign a
3-tangle to each morphism $M$ (namely a 3-tangle $T$ with the
graphical representation of $M$ as diagram, supplied with the
blackboard framing).

\begin{df}
The monoidal $\C$-category \topf \/ is the category whose
objects are the objects of \htopf \/ and whose morphisms are
the equivalence
classes of the morphisms of \htopf \/ under the following
equivalence relation: Two morphisms from $u$ to $w$ are equivalent
if they get assigned equivalent 3-tangles.
\end{df}

\begin{rmk}\label{lokrel} {\em The equivalence relation in the above definition can be
described locally. For morphisms generated by (G1)-(G3) this is done in [BN
2]. If we take the generators in  (G4) as well, we have to add the following
relations for any word $w$:

\begin{tabbing}
\underline{Relation}\hspace{8cm}\= \underline{Graphical
representation}\\[1em]
$(id_\diamond\otimes\mbox{\pstext{rbog}})\,\pstext{phi}_{\diamond,\diamond,\diamond}\,({\mbox{\pstext{trir}}}\otimes
id_\diamond)
= \mbox{\pstext{tril}}$
\>\psdiag{2}{8}{rel3}\\[1em]
$(id_\diamond\otimes
\mbox{\pstext{tril}})\,\pstext{phi}_{\diamond,\diamond,\diamond}\,({\mbox{\pstext{lbog}}}\otimes
id_\diamond)
={\mbox{\pstext{trir}}}$
\>\psdiag{2}{8}{rel4}\\[1em]
$(id_w\otimes{\mbox{\pstext{tril}}}){\mbox{\pstext{Zarg1}}}_{(\diamond\diamond),w}
={\mbox{\pstext{Zarg1}}}_{\diamond,w}({\mbox{\pstext{tril}}}\otimes id_w)$
\>\psdiag{2}{6}{rel1}\\[1em]
$({\mbox{\pstext{tril}}}\otimes id_w){\mbox{\pstext{Zarg1}}}_{w,(\diamond\diamond)}
={\mbox{\pstext{Zarg1}}}_{w,\diamond}(id_w \otimes {\mbox{\pstext{tril}}})$
\>\psdiag{2}{6}{rel2}\\

\end{tabbing}}
\end{rmk}
\begin{rmk}
{\em Observe that any 3-tangle, and in particular any closed 3-net,
is equivalent to a 3-tangle assigned to a morphism of
\topf \/, and so the equivalence classes of 3-tangles correspond
exactly to a basis of the morphisms of \topf. One might
achieve this by taking much simpler generators (e.g.\
only generators without multiple strands),
 but then the local description of the equivalence relation given in remark
\ref{lokrel} would be more
complicated.}
\end{rmk}


TITLE: What is the maximum expected eigenvalue of an $n \times n$ symmetric matrix?
QUESTION [1 upvotes]: Given an $n \times n$ symmetric matrix filled with realizations of a uniform random variable taking values in $[-1, 1]$, determine the maximum expected eigenvalue using analytic methods.

Running multiple MATLAB scripts of a large $n$ matrix gave me a the following plot:

Observe, the elliptic nature of the $100 \times 100$ plot. Intuitively, I suspect the maximum value of the eigenvalues must be $\sqrt{n}$, but I don't know how to determine this value analytically (or even if $\sqrt{n} $ is indeed true).
I've attempted reading Pastur et Al.'s "Eigenvalue distribution for Large Random matrices" But can't make sense of their work.
In short What is the relationship between $n$, dimension of the matrix and the maximum expected eigenvalue?

REPLY [0 votes]: The Correct solution is $n$ opposed to my original intuition of $\sqrt{n}$
For the positive case only
Proof
Observe that
$$ tr(A) = \sum_{i = 1}^{n} a_{ii} = \sum_{i = 1}^{n} \lambda_i $$
Meaning that the sum of the diagonals is simultaneously equal to the sum of the possible eigenvalues in any square matrix.
Now, let $m$ be the largest entry into A (ie. For our case $m = 1$ as all $a_{ij}$ normally randomly chosen $ \in [-1, 1]$
Let $$\lambda _{max} \leq \sum_{i = 1}^{k} \lambda _k = tr(A)$$
$$\implies \lambda _{max} \leq tr(A)$$
$$a_{ii} \leq m$$
Every $a_{ij}$ entry is less than the maximum
$$\implies \sum_{i = 1}^{n} a_{ii} \leq \sum_{i = 1}^{n} m$$
$$\implies tr(A) \leq mn$$
$$\lambda _{max} \leq tr(A) \leq mn$$
$$\lambda _{max} \leq mn$$
$$\tag*{$\blacksquare$}$$
Special thanks to Rahul (in comments) & Raghav Srinivasan of the University of Toronto.
Alternately, observe that intuitively I thought that $\sqrt{n}$ was the correct solution because in a Wigner distribution, the scaling as $n \to \infty$ is $\sqrt{n}$ This can be proven here


TITLE: Prove $\text{rad}(I)/I$ is isomorphic to $\mathfrak{N}(R/I)$
QUESTION [0 upvotes]: I want to know if this is the correct way to do it.
Define $\varphi:\text{rad}(I) \longrightarrow \mathfrak{N}(R/I)$ by $\varphi(r)= r^n+I$,then ker$\varphi = I$, so therefore by the 1st isomorphism theorem, we have:
$$\text{rad}(I)/I \cong \mathfrak{N}(R/I) $$
it satisfies the condition for homomorphism, since
$$\varphi(r_1r_2) = (r_1r_2)^n+I = \varphi(r_1)\varphi(r_2),$$
$$\varphi(r_1+r_2)= (r_1^n+r_2^n) + I = \varphi(r_1)+\varphi(r_2). $$
Do I need to prove something else, or do I miss something? Any help would be appreciate it , thanks

REPLY [2 votes]: $x \in rad(I) \Leftrightarrow x^n \in I,$ for some $n \geq 1.$ So,
$x + I \in rad(I)/I \Leftrightarrow x^n + I = 0$ in $R/I$ for some $n \geq 1 \Leftrightarrow (x + I )^n = 0$ in $R/I$ for some $n \geq 1 \Leftrightarrow x + I \in \mathfrak{N}(R/I).$


TITLE: Find the generating function for this set of strings
QUESTION [1 upvotes]: Let $a(n)$ be the number of $\{0,1\}$-strings of length $n$ which contain no $4$ consecutive $1$'s and no $4$ consecutive $0$'s (don't contain "$0000$" or "$1111$"). Find the generating function for this.
Thank you.

REPLY [1 votes]: Such a string consists of a (possibly empty) sequence of elements from {01, 011, 0111, 001, 0011, 00111, 0001, 00011, 000111} — i.e. {0|00|000}{1|11|111} — optionally preceded by 1, 11 or 111 and optionally followed by 0, 00 or 000. So the generating function is
$$
\frac{(1+z+z^2+z^3)^2}{1-(z+z^2+z^3)^2},
$$
the numerator reflecting the optional prefix and suffix, and the denominator reflecting the sequence.
This is a related question.


TITLE: Prove $\sqrt{n}+\sqrt{n+k}$ is irrational
QUESTION [2 upvotes]: For what $k\in\mathbb N$, $\sqrt{n}+\sqrt{n+k}$ is irrational? ($\forall n\in\mathbb N$)

REPLY [0 votes]: Solution is $k = 2^{2v+1}*j$ where j is odd.
Obviously $k$ cannot equal $0$.
$k$ must/can be an odd number multiplied by an odd power of 2. ($k$ can not be odd.  If $k$ is even, $k$ is not an odd number times an even power of 2.
If $0$ is not considered a natural number than 1 is a possible value for $k$ is the one exception.  (As $\sqrt{n} + \sqrt{n+1}$ is rational $\iff n = 0$).
$\sqrt{n} + \sqrt{n + k} = r = a/b; \gcd(a,b) = 1 \implies n + k = r^2 + n - 2r\sqrt{n} \implies k = r(r - 2\sqrt{n}) \implies k = (a/b)(a/b - 2m); n = m^2$ for some integer $m$.  Which implies $bk = a^2/b - 2am \in \mathbb Z \implies b = 1$.
So for any $k = a(a - 2m)$ we have a possibility of the sum being rational.  Otherwise we don't.
So we can't have $k = 2j + 1$ odd or we'd have $k = k(k - 2j)$ (which would yield $\sqrt{j^2} + \sqrt{j^2 + k = j^2 + 2j + 1}$ being rational.)
We can't have $k = 2^{2v}j$ where $j$ is odd or we'd have $k = 2^vj(2^vj - 2^v(j - 1))$
On the other hand if $k = 2^{2v + 1}j$ where $j$ is odd we can't have $k = a(a - 2m)$.  The powers of 2 just don't add up.


TITLE: Constructing an irreducible polynomial of degree $2$ over $\mathbb{F}_p$
QUESTION [1 upvotes]: I want to construct an irreducible polynomial of degree $2$ over $\mathbb{F}_p$ where $p$ is a prime that can be written as $4k+1$. My attempt is as follow: we can assume that this polynomial is of the form  ${x^2} + ax + b$ for some $a,b \in {\mathbb{F}_p}$. So for all $\lambda  \in {F_p}$, $p$ doesn't divide ${\lambda ^2} + a\lambda  + b$. It follows that ${\lambda ^2}$ is not equal to $a\lambda  + b
\bmod p$. If we can find some $a,b \in {\mathbb{F}_p}$ such that $a\lambda  + b$ is a nonresidue for all $\lambda  \in {F_p}$, it is ok. But I cannot. I wait your response.

REPLY [2 votes]: $x^2-b$ is irreducible mod $p$ iff $b$ is a quadratic non-residue mod $p$.
Since half the classes mod $p$ are quadratic non-residues, there are plenty of choices for $b$.
Explicitly finding one $b$ for a given $p$ is another matter, although it is usually quite small:
see A053760. In particular, the smallest quadratic non-residue mod $p$ is at most $\sqrt{p}+1$ (proof).


TITLE: Proof of $(a+b)^x=a^x+b^x$ implies $x=1$
QUESTION [3 upvotes]: I'd like to prove that for $a,b>0$,
  $$(a+b)^x=a^x+b^x \implies x=1.$$

I tried to study the variation of the map $f:x\mapsto (a+b)^x-a^x-b^x$ but unfortunatly it's not monotone since
$$f'(x)= (\ln (a+b)-\ln (a))a^x + (\ln (a+b)-\ln (b))b^x .$$
Taking $\ln$ in the equation doesn't seems to help either.
I feel like I'm missing something obvious. Any help will be greatly appreciate.

REPLY [0 votes]: You can do this even more simply: let $a=b=1$. Then on the left, $2^x$, and on the right, $1+1$.
EDIT: Though I may have misunderstood your question. I answered it as if you asked "If this holds for all $a,b$, then show that $x=1$". You may be asking instead "If this holds for some $a,b$, then show that $x=1$".


TITLE: (Polar jets) Is it possible that black holes radiate substantial amount of the mass/energy they accrete?
QUESTION [0 upvotes]: Is the amount of energy a black hole emit through polar jets comparable to the mass it accretes or is it negligible? Here I am mostly concerned about the observational findings as opposed to the theoretical ones.
In other words, any mass in the accretion disk has two possible fate: ending up inside the event horizon or escaping through jets. What is the possible probability range of the second happening to each unit of mass?

REPLY [2 votes]: Real-world black holes accrete matter and this matter gets heated before it is devoured by the black hole so it radiates. We may observe this accretion. The radiation coming from this heated matter is qualitatively similar to the radiation in any accretion disk, even in the absence of a black hole.
But the very reason why the observations lead us to be near-certain that there are black holes is that we observe less radiation than what any star-like object would emit. That's because the accreted matter is hiding behind the event horizon. See e.g. this paper from 1997-1998

http://arxiv.org/abs/astro-ph/9706112

about the evidence supporting the existence of the black hole in our galactic center. The models discussed over there are observationally driven.
Concerning your "percentage" question, it is estimated that only 1% of the accreting matter ever crosses the event horizon, see e.g.

http://arstechnica.com/science/2013/08/the-milky-ways-black-hole-like-cookie-monster-loses-more-than-it-eats/

But this number is so tiny only because of an inclusive enough definition of the accretion disk. Within the matter that is really close to the event horizon, a much larger percentage gets devoured.
There is no experimental observation of the Hawking radiation as of today; that is unrealistic for astronomical sizes of black holes and for smaller black holes where it is imaginable, such a discovery would be a huge breakthrough.


TITLE: Dimension of this quotient space
QUESTION [0 upvotes]: Let $c_0=\{ (x_n) : x_n \in \mathbb R, x_n \to 0 \}$, let $M=\{(x_n) \in c_0 : x_0+x_2+ \cdots +x_{10}=0 \}$
Then dimension of $c_0/M$ is ?
Answer is given to be 1.
My attempt: 9 "coordinates" are needed to specify $M$, so $\dim M=9$.
$\dim(c_0/M)=\dim c_0-\dim M$.
How do I find dim $c_0$?

REPLY [4 votes]: Alas, as $c_0$ and $M$ are both infinite-dimensional, your formula
$\dim(c_0/M)=\dim c_0-\dim M$ doesn't work.
The quotient space is $1$-dimensional. This follows from the First Isomorphism Theorem, applied to the map $T:c_0\to\Bbb R$ given by
$T:(x_n)\mapsto x_0+x_2+\cdots+x_{10}$. This is a linear map,
and has image $\Bbb R$. By the First Isomorphism Theorem, $\Bbb R$
is isomorphic to $c_0/\ker T$. Of course the kernel of $T$ is $M$.


TITLE: Confused with equation of plane
QUESTION [0 upvotes]: The equation of plane:
$ Ax + By + Cz = d$
Where $ A \hat i + B \hat j + C \hat k$ is the normal vector $ \vec n$ perpendicular to the plane.
The position vector $ \vec r = x \hat i + y \hat j + z \hat k$
Then,
$ \vec r \cdot \vec n = d$
This last equation says that a position vector to any point on the plane has the same projection along $ \vec n$ because $d$ is a constant.
But I can't see that graphically. Different position vectors to the plane have different slopes, while the normal vector of the plane is always perpendicular and doesn't change, so how does every position vector to any point on the plane have the same projection along $ \vec n$? Thank you.

REPLY [2 votes]: I like to think of it in terms of orthogonality.  Let $(x_0,y_0,z_0)$ be a point on the plane, and consider the position vector $\vec r_0 = x_0 \hat\imath + y_0 \hat\jmath + z_0 \hat k$.  Since the point is on the plane $\vec r_0 \cdot \vec n = d$ too.  So the plane equation becomes
$$
    \vec r \cdot \vec n = \vec r_0 \cdot \vec n \implies (\vec r-\vec r_0)\cdot \vec n = 0
$$
This says that the vector drawn from $(x_0,y_0,z_0)$ to $(x,y,z)$ is perpendicular to $\vec n$.


TITLE: Question about Conditional Expectation
QUESTION [3 upvotes]: I have a seemingly trivial question regarding conditional expectation. Consider $x$ and $y$ be two integrable random variables on Probability space (X, $\Sigma$, $P$) such that $$E(X|Y) =_{a.s} Y$$ and $$E(Y|X) =_{a.s} X$$
Show that $$X=_{a.s} Y$$
The thing looks like a simple proof result, but I don't know how to move on after showing that$$\int_{B} Y = \int_{B} X$$ for all $B \in \sigma(x)$ and $B \in \sigma(y)$
Now any hint on how to proceed? Thank you so much in advance

REPLY [3 votes]: Hints:

For every real number $c$, consider $u(c)=E[X-Y;Y\leqslant c\lt X]$. Use the hypothesis that $E[Y\mid X]=X$ almost surely to show that $u(c)=E[Y-X;c\lt X,c\lt Y]$.
Deduce that, if furthermore $E[X\mid Y]=Y$ almost surely, then $u(c)=-E[Y-X;X\leqslant c\lt Y]$.
Studying the sign of the LHS and the RHS, deduce from the above that $u(c)=0$.
Let $v(c)=P[Y\leqslant c\lt X]$. Deduce from the above that $v(c)=0$ for every real number $c$.
Show that $\int\limits_\mathbb Rv(c)\mathrm dc=E[(X-Y)^+]$.
Deduce that $X\leqslant Y$ almost surely.
Conclude.


TITLE: matlab program help
QUESTION [2 upvotes]: Wanting to write a matlab program to solve the following iteration:
$x^{(k+1)}=b+\alpha\begin{bmatrix}2&1\\1&2\end{bmatrix}x^k,k=0,1,2,\cdots$
where alpha is a real constant. Find the values of alpha for which the method converges. Show the first two iterations for any of this choice of alpha.
I am completely lost.
edit:
no need for a program...
Here is what I have so far. Eigenvalues are 1,3 for the matrix. I took $|\alpha*eigenvalues|>1$  Thus giving me that $(-1/3)<\alpha<1/3$ and $-1<\alpha<1$
I am assuming that I can leave b in the first two iterations, but how would I represent $\alpha$?
thanks for any help.

REPLY [0 votes]: This will get you started:

A = [2, 1; 1, 2];
x = [1; 0]; % put x(0) here
b = [1; 0]; % put b here
alpha = 0.01; % put alpha here
num_iter = 100 % the total number of iterations
for k = 1:num_iter
    x = b + alpha*A*x; % remove the ";" to see result at each step
end
x % print final result
More hints:

$x^{(k+1)} = \alpha^k A^k x^0 + \sum_{n = 0}^{k-1} \alpha^n A^n b $, so you're going to have a problem if $\alpha^kA^k$ has really large entries.
What are the eigenvalues of $A$?
Consider $A = P^{-1}DP$. Then $A^k = P^{-1}D^kP$, so $\alpha^kA^k = P^{-1}(\alpha D)^kP$


TITLE: How to directly show that $\mathbb{Z}_{(p)}$ is a local ring with the unique maximal ideal $p \mathbb{Z}_{(p)}$?
QUESTION [0 upvotes]: I know that $\mathbb{Z}_{(p)}$ is a local ring because it's the localization of $\mathbb{Z}$ over $p$, but is there a direct way to prove that and find its unique maximal ideal?

I've been thinking over this for a while but nothing in the right direction comes to my mind.

REPLY [1 votes]: Every element $a/b$ of that ring (in which $b$ is not divisible by $p$, of course) such that $a$ is not divisible by $p$ is clearly invertible. It follows that the complement of the ideal generated by $p/1$ is entirely composed of invertible elements. This is enough to conclude that the ring is local and that that ideal is the maximal ideal.


TITLE: How to I use Ito's formula to compute quadratic variation?
QUESTION [2 upvotes]: Let B be Brownian motion. Use Itos formula to compute the quadratic
variation of $\left[X_t^i\right]$ for $\left[X_t^1\right]=e^{B_t}$, $\left[X_t^2\right]=\ln(B_t^2+1)$ and $\left[X_t^1\right]=\sin^2B_t+\cos^2B_t.$
I wanted to use this formula: quadratic variation=$\int_{0}^{t} (F'(X_s))^2d\left[X\right]s$, but I dont know how to get the derivative $F'(X_s)$, as there is no $X$ in $e^{B_t}, \ln(B_t^2+1)$ nor $\sin^2B_t+\cos^2B_t$

REPLY [1 votes]: I think the wording in the problem is wrong. We are looking for the "quadratic variation"(co-variation) of the following processes : $\forall t\in \mathbb{R^+}$, $X_t^1=e^{B_t}, X_t^2 = \ln(1+B_t^2), X_t^3=\cos^2(B_t)+\sin^2(B_t)$.
Let's do it for $X_t^1$ :
First, we express $X_t^1$ as an Itô process by applying the Itô's lemma to the function $F(x) = e^x$. In the differential form, we will have :
\begin{equation}
dX_t^1 = d(F(B_t)) = e^{B_t}dB_t + \frac12e^{B_t}dt
\end{equation}
Then we compute the quadratic variation using the informal way:
\begin{equation}
d[X_.^1,X_.^1]_t "=" dX_t^1.dX_t^1 = e^{2B_t}dt
\end{equation}
Therfore, we have $[X^1]_t = \int_0^t e^{2B_s}ds$.
Note that using your definition of the quadratic variation, we have: $\int_0^tF'(B_s)^2d[B]_s = \int_0^t e^{2B_s}ds$


TITLE: Finding Current and Voltage through resistors given overall voltage and resistor values
QUESTION [0 upvotes]: Given a setup like:

Is it possible to find current and voltage through each resistors given resistor values and overall voltage (battery voltage I suppose).
I think $$I_0=I_A=\frac{\Delta V}{R_{total}}$$
I know $I_B+I_C=I_A$ but how do I find $I_B$ if I don't have $\Delta V_B$?

REPLY [1 votes]: Two formulas to understand and internalize:
1) Voltage divider:  For two resistors R1 and R2 in series, the voltage $V_2$ across $R_2$, relative to the total voltage $V_T$ across both resistors, is:
$$\frac{V_2}{V_T}=\frac{1}{1+R_1/R_2} $$
This formula is a direct consequence of the fact that the current through the two series resistors is the same.
2) Current divider:  For two resistors R1 and R2 in parallel, the current $I_2$ through $R_2$, relative to the total current $I_T$ through both resistors, is:
$$\frac{I_2}{I_T}=\frac{1}{1+R_2/R_1} $$
This formula is a direct consequence of the fact that the voltage across the two parallel resistors is the same.
You can always check that you have the correct ratio by letting one resistor get very large relative to the other; for example, in the current divider, as $R_2/R_1$ goes to $\infty$, you expect the current through $R_2$ to go to 0.


TITLE: $\int^{\infty}_{-\infty}e^{-\alpha x^{2}}\frac{P(x)}{Q(x)}dx$
QUESTION [1 upvotes]: What would be the proper way to integrate the expression:
$\int^{\infty}_{-\infty}e^{-\alpha x^{2}}\frac{P(x)}{Q(x)}dx$
If everything is fine, i.e. $\alpha$ is positive, polynomial $Q(x)$ has no real zeroes and the order of $P(x)$ is less than the order of $Q(x)$?
In particular:
$P(x)=1+x$
$Q(x)=(1+(x-a)^2)^s$
$s>0$ is integer, $a$ is real.

REPLY [2 votes]: Let's try using substitution to get the integral into some recognizable form.
$$x=y+a$$
$$I_s(\alpha,a)= e^{- \alpha a^2} \int_{-\infty}^\infty e^{- \alpha y^2} e^{- 2\alpha a y} \frac{1+y+a}{(1+y^2)^s} dy$$
Clearly, we can consider two separate integrals:
$$J_s(\alpha,a)=\int_{-\infty}^\infty e^{- \alpha y^2} \frac{e^{- 2\alpha a y}}{(1+y^2)^s} dy$$
$$Y_s(\alpha,a)=\int_{-\infty}^\infty e^{- \alpha y^2} \frac{e^{- 2\alpha a y} y}{(1+y^2)^s} dy$$
1) Let's start with the first one. Expand the exponential as a series:
$$J_s(\alpha,a)=\sum_{n=0}^\infty \frac{(-1)^n (2 \alpha a)^n}{n!} \int_{-\infty}^\infty e^{- \alpha y^2} \frac{y^n}{(1+y^2)^s} dy$$
The integrals inside the series will be nonzero only for $n=2m$. Then we can write:
$$2\int_0^\infty e^{- \alpha y^2} \frac{y^{2m}}{(1+y^2)^s} dy=\int_0^\infty e^{- \alpha z} \frac{z^{m-1/2}}{(1+z)^s} dz=\Gamma\left(m+\frac12 \right) \mathrm U \left(m+\frac12,m-s+\frac32,\alpha \right)$$
Where $\mathrm U$ is the Tricomi hypergeometric function.
Let's not forget:
$$\Gamma\left(m+\frac12 \right)= 2 \sqrt{\pi} \frac{\Gamma(2m)}{\Gamma(m) 4^m}$$
Simplifying, we obtain:
$$J_s(\alpha,a)=\sqrt{\pi} \sum_{m=0}^\infty \frac{(\alpha a)^{2m}}{m!} \mathrm U \left(m+\frac12,m-s+\frac32,\alpha \right)$$
2) Doing the same series expansion for the second integral:
$$Y_s(\alpha,a)=\sum_{n=0}^\infty \frac{(-1)^n (2 \alpha a)^n}{n!} \int_{-\infty}^\infty e^{- \alpha y^2} \frac{y^{n+1}}{(1+y^2)^s} dy$$
Now we have to take $n=2m+1$, which gives us:
$$2 \int_0^\infty e^{- \alpha y^2} \frac{y^{2m+2}}{(1+y^2)^s} dy=\int_0^\infty e^{- \alpha z} \frac{z^{m+1/2}}{(1+z)^s} dz=\Gamma\left(m+\frac32 \right) \mathrm U \left(m+\frac32,m-s+\frac52,\alpha \right)$$
After simplifications:
$$Y_s(\alpha,a)=-\sqrt{\pi} \sum_{m=0}^\infty \frac{(\alpha a)^{2m+1}}{m!} \mathrm U \left(m+\frac32,m-s+\frac52,\alpha \right)$$
Finally we obtain:

$$I_s(\alpha,a)= \\ =  \sqrt{\pi} e^{- \alpha a^2} \sum_{m=0}^\infty \frac{(\alpha a)^{2m}}{m!} \left((1+a) \mathrm U \left(m+\frac12,m-s+\frac32,\alpha \right)- \\ - \alpha a \mathrm U \left(m+\frac32,m-s+\frac52,\alpha \right)  \right)$$

Note that the series should converge for all reasonable parameters, including $\alpha a>1$. Numerical experiments confirm that the series and the integral are equal.
The Tricomi function in Wolfram language is HypergeometricU[a,b,z].
As an example, let's take $\alpha=3$, $a=2$, $s=1/2$:
NIntegrate[Exp[12-3 x^2] (1+x)/(1+(x-2)^2)^(1/2)/Sqrt[Pi],{x,-Infinity,Infinity},WorkingPrecision->20]
gives 45951.232606275117969.
Sum[36^m/m! (3 HypergeometricU[m+1/2,m+1,3]-6 HypergeometricU[m+3/2,m+2,3]),{m,0,50}]
gives 45951.232606275096333.
The accuracy is good.
Obviously, this is not very useful for actual integral evaluation, because we use special functions, but this is all I managed to do.