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\title{\bf{$I$}-divergence and \bf{$I^*$}-divergence in cone metric spaces}

\author{Amar Kumar Banerjee and Anirban Paul}

\address[A.K.Banerjee]{Department of Mathematics, The University of Burdwan, Golapbag, Burdwan-713104, West Bengal, India.}

\email{ akbanerjee1971@gmail.com, akbanerjee@math.buruniv.ac.in }

\address[A.Paul]{Department of Mathematics, The University of Burdwan, Golapbag, Burdwan-713104, West Bengal, India.}

\email{ paulanirban310@gmail.com, anirbanpaul310math@gmail.com}

\begin{abstract} In this paper we have studied the ideas of $I$-divergence and $I^*$-divergence of sequences in cone metric spaces. We have investigated the relationship between $I$\text-divergence and $I^*$-divergence and their equivalence under certain condition. Further we prove a decomposition theorem for $I$-convergent sequences in a cone normed space.
\end{abstract}
\maketitle
\author{}

\textbf{Key words and phrases:}
$I$-divergence, $I^*$-divergence, Cone metric spaces, Cone normed spaces, Decomposition theorem, $I$-convergence.   \\

\textbf {(2010) AMS subject classification :} 54A20, 40A35  \\

\section{Introduction}
The idea of statistical convergence of sequences of real number were introduced by Steinhaus \cite{q} and Fast \cite{f} as a generalizations of ordinary convergence of sequences of real numbers. Later many more works were done in this direction \cite{B, g, p}. In 2001, P. Kostryko et al \cite{h} introduced the idea of $I$-convergence of real sequences using the ideals of the set of natural numbers as a generalization of statistical convergence. Later in 2005, Lahiri and Das \cite{j} studied the same in a topological space and then many works were carried out in this direction \cite{2, 4, d, i}. In 2017, Banerjee and Mondal \cite{3} studied the same for double sequences in a topological space.\\

The concepts of divergent sequences of real numbers was generalized to statistically divergent sequences of real numbers by Macaj and Salat in \cite{l}. Recently Das and Ghosal in \cite{c} introduced the notion of $I$ and $I^*$-divergence of sequences in a metric space where the condition (AP) has been used as necessary and sufficient condition to prove the equivalence of $I$ and $I^*$-divergence.\\

In 2007, Nabiev et al \cite{m} established a decomposition theorem for $I$-convergent sequences in a linear metric space. The concept of cone metric spaces is a generalization of the notion of usual metric spaces where the distance between two points is given by an element of a Banach space endowed with suitable partial ordering. After the initial work on cone metric spaces space by Guang and Xian \cite{k}, a lot of work have been done on this structure. In this paper we proceed in a different direction and extend the notion of $I$ and $I^*$-divergence in a cone metric space.\\

The idea of cone normed spaces was given by M. Eshaghi Gordrji et al \cite{g*} which is a generalization of normed spaces. In section \ref{sec} we have proved the decomposition theorem of $I$-convergence  sequences in a cone normed space and have studied its consequences, where the methods of proof are not analogous as in the case of a metric space.
\section{Preliminaries}

The following definitions and notion will be needed.\\

\begin{defn} \cite{t} Let $X\neq\phi$. A family $I \subset 2^X$ of subsets of $X$ is said to be an ideal in $X$ provided the following conditions holds:\\
\\
$(i)\: A, B \in I \Rightarrow A\cup B\in I$\\
$(ii) \: \:A\in I, B\subseteq A \Rightarrow B\in I$
\end{defn}
Note that $\phi\in I$ follows from the condition(ii). An ideal $I$ is called nontrivial if $I \neq \left\{\phi\right\}$ and $X\not\in I$. $I$ is said to be admissible if $\left\{x\right\}\in I$ for each $x\in X$ \cite{t**}.
\begin{defn} \cite{t*} Let $X\neq\phi$. A non-empty family $F\subset 2^X$ is said to be a filter on $X$ if the following are satisfied,\\
$(i)\: \phi\not\in F $\\
$(ii)\: A, B \in F \:\Rightarrow \:A\cap B\in F $\\
$(iii)A\in F, A\subset B\: \Rightarrow \: B\in F$
\end{defn}
Clearly if $I$ is a nontrivial ideal then the family of sets $F(I)=\left\{M\subset X : \text{there exists}\: A\in I, M=X\setminus A\right\} $ is a filter in $X$. It is called the filter associated with the ideal $I$.

\begin{defn} \cite{h,t**} Let $I\subset 2^\mathbb{N}$ be a proper ideal in $\mathbb{N}$ and $(X,d)$ be a metric space. The sequence $\left\{x_n\right\}_{n\in \mathbb{N}}$ of elements of $X$ is said to be $I$-convergent to $x\in X$ if for each $\varepsilon>0$ the set $A(\varepsilon)=\left\{n\in \mathbb{N}:d(x_n,x)\geq \varepsilon\right\}$ belongs to $I$.
\end{defn}

\begin{defn} \cite{h}\label{1} An admissible ideal $I \subset 2^\mathbb{N}$ is said to satisfy the condition (AP) if for every countable family of mutual disjoint sets $\left\{A_1,A_2,A_3,\cdots\right\}$ belonging to $I$ there exists a countable family of sets $\left\{B_1,B_2,B_3,\cdots\right\}$ such that $A_j\Delta B_j$ is a finite set for each $j\in\mathbb{N}$ and $B=\displaystyle{\bigcup_{j=1}^{\infty}}B_j\in I$.
\end{defn}
Note that $B_j\in I$ for each $j\in \mathbb{N}$.\\

The concepts of $I^*$-convergence which is closely related to the $I$-convergence has been given in \cite{h} as follows:\\
\begin{defn} \cite{h} The sequence $\left\{x_n\right\}_{n\in\mathbb{N}}$ of elements of X is said to be $\text{I}^*$\text-convergent to $x\in X$ if and only if there exists a set $M\in F(I)$, $ M=\left\{m_1<m_2 \cdots <m_k\cdots \right\}\subseteq \mathbb{N}$ such that $\displaystyle{\lim_{k\to\infty}}d(x,x_{m_k})=0$.
\end{defn}

In \cite{h} it is seen that $I$ and $I^*$-convergence are equivalent for an admissible ideal with the property (AP).\\

Now we recall the following definitions.
\begin{defn} \cite{1} A sequence $\left\{x_n\right\}_{n\in \mathbb{N}}$ in a metric space $(X,d)$ is said to be divergent (or properly divergent) if there exists an element $x \in X$ such that $d(x,x_n)\to \infty$ as $n\to\infty$.\\

\end{defn}
Note that a divergent sequence in a metric spaces can not have any convergent subsequence.
\begin{defn} \cite{c} A sequence $\left\{x_n\right\}_{n\in \mathbb{N}}$ in  a metric space $(X,d)$ is said to be $I$\text-divergent if there exists an element $x\in X$ such that for any positive real number G, $A(x,G)=\left\{n \in \mathbb{N}:d(x,x_n)\leq G\right\}\in I$
\end{defn}
\begin{defn} \cite{c} A sequence in a metric space $(X,d)$ is said to be $I^*$-divergent if there exists $M\in F(I)$ i,e $\mathbb{N}\setminus M\in I$ such that $\left\{x_n\right\}_{n\in M}$ is divergent. That is there exists at least one $x\in X$ such that $\displaystyle{\lim_{n\to\infty}}d(x,x_n)=\infty$
\end{defn}
We now state the decomposition theorem from \cite{m} for $I$-convergent sequences.\\

\begin{thm} \cite{m}
Let $(X,d)$ be a linear metric space, $x=(x_n)\in X$ and $I\subset 2^\mathbb{N}$ be an admissible ideal with the property (AP). Then the following are equivalent:\\
(a)\: $\text{I}\text -lim x_n=\xi$\\
(b)\: There exists $y=(y_n)\in X$ and $z=(z_n)\in X$ such that $x=y + z $, $\displaystyle{\lim_{n\to\infty}}d(y_n,\xi)=0$ and $\text{supp z}\in I$, where $\text{supp z}=\left\{n\in \mathbb{N}:z_n\neq \theta_X\right\}$ and $\theta_X$ is the zero element of $X$.

\end{thm}

We now recall the following basic concepts from \cite{k} which will be needed throughout the paper. $E$ will always denote a real Banach space with null vector $\theta_E$ and P be a subsets of $E$. P is called cone if and only if \\
(i) P is closed and $P\neq \left\{\theta_E\right\}$\\
(ii) $a, b\in \mathbb{R} \text{,} \: a, b\geq 0 \text{,} \:x,y\in P \Rightarrow ax+by \in \text{P}$\\
(iii) $x\in \text{P}$ and $-x\in \text{P}$ $\Rightarrow$ $x=\theta_E$\\

Given  cone $\text{P}\subset E$, we define a partial ordering $\leq$ with respect to P by $x\leq y$ if and only if $y -x \in \text{P}$. We shall write $x < y$ to indicate that $x\leq y$ but $x\neq y$, while $x<< y$ will stand for $y - x \in \text{Int P}$, Int P denotes the interior of P.\\

The cone P is called normal if there is a number $k>0$ such that for all $x, y \in E$\\
$\theta_E\leq x \leq y$ implies $\norm x \leq k\norm y$.\\
The least positive numbers satisfying above is called the normal constant of P.\\
The cone is called regular if every increasing sequence which is bounded from above is convergent. That is if $\left\{x_n\right\}$ is a sequence such that $$x_1\leq x_2\leq \cdots \leq x_n\leq \cdots \leq y$$ for some $y \in E$, there is $x\in E$ such that $\norm {x_n - x }\to 0 \: (n \to \infty)$. Equivalently the cone P is regular if and only if every decreasing sequence which is bounded from below is convergent. It is well known that a regular cone is a normal cone.\\

In the following we always suppose $E$ is a real Banach space, with null element $\theta_E$, P is a cone with $\text{Int P} \neq \phi$ and $\leq$ is a partial ordering with respect to P.\\
\begin{defn} \cite{k} Let $X$ be a non-empty set. Suppose the mapping $d : X\times X \mapsto E $ satisfies\\
(d1) $\theta_E< d(x,y)$ for all $x, y \in X$ and $d(x,y)=\theta_E$ \:if and only if\: $x=y$\\
(d2) $d(x , y)=d(y , x)$ for all $ x, y \in X$\\
(d3) $d(x , y)\leq d(x , z) + d(y , z) $ for all $x, y, z \in X $.\\
Then $d$ is called a cone metric and $(X,d)$ is called a cone metric space. It is obvious that cone metric spaces generalize metric spaces. Several examples of cone metric space are seen in \cite{k}.
\end{defn}

\begin{defn} \cite{k} Let $(X,d)$ be a cone metric space. Let $\left\{x_n\right\}$ be a sequence in $X$ and $x\in X$. If for every $c\in E$ with $\theta_E<<c$ there is $N$ such that for every $n>N$, $d(x,x_n)<< c$, then $\left\{x_n\right\}$ is said to be convergent and $\left\{x_n\right\}$  converges to $x$ and $x$ is the limit of $\left\{x_n\right\}$ . We denote this by $\displaystyle{\lim_{n\to\infty}}x_n=x$ or $x_n\to x \: (n\to\infty)$
\end{defn}
\begin{lem} \cite{k} Let $(X,d)$ be a cone metric space. P be a normal cone with normal constant $k$. Let $\left\{x_n\right\}$ be a sequence in $X$. Then $\left\{x_n\right\}$ converges to $x$ if and only if $d(x,x_n)\to \theta_E \: (n\to\infty)$.
\end{lem}
We now consider the following definitions. \\

\begin{defn} \cite{s} Let $X$ be a real vector space . Suppose that the mapping $\norm . _\text{P} : X\mapsto E $ be such that
\begin{align*}
&(i)\norm {.}_\text{P} > \theta_E\: \text{for all} \: x\in X \text{and} \norm . _\text{P}= \theta_E \:\text{if and only if} \: x=\theta_X\\
&(ii) \norm {\alpha x}_\text{P} = |\alpha|\norm{x}_{\text{P}} \text{for all} \:x\in X\text{ and}\: \alpha \in \mathbb{R}\\
&(iii) \norm {x + y}_{\text{P}} \leq \norm{x}_{\text{P}} + \norm{y}_{\text{P}}, \text{for all }x, y\in X.
\end{align*}

Then $\norm{.}_{\text{P}}$ is called a cone norm on $X$ and $(\norm{.}_{\text{P}}, X)$ is called cone normed space.\\
\end{defn}
It is easy to show that every normed space is cone normed space by putting $E=\mathbb{R}$, $\text{P}=[0,\infty)$.\\
\begin{rem} \cite{s} Let $(X,\norm {.}_{\text{P}})$ be a cone normed space, set $d(x , y) =\norm {x - y}_{\text{P}}$, it is easy to show that $(X,d)$ is a cone metric space, $d$ is called `` the cone metric induced by the cone norm $\norm {.}_{\text{P}}$".
\end{rem}
Now we consider the following theorem as in \cite{s}.
\begin{thm} \cite{s} The cone metric $d$ induced by a cone norm on a cone normed space satisfies \\
(i) $d(x+a,y+a) = d(x,y), \: x, y, a\in X$\\
(ii) $d(\alpha x, \alpha y )=|\alpha | d(x , y), \: x, y\in X,\: \alpha \in \mathbb{R}$

\end{thm}

Now we consider the following definition from \cite{n}.\\
\begin{defn} \cite{n} Let $(X, d)$ be a cone metric space. Let $\left\{x_n\right\}_{n\in \mathbb{N}}$ be a sequence in $X$ and let $x \in X$. If for every $c\in E$ with $\theta_E << c$, the set $\left\{ n\in \mathbb{N} : c - d(x,x_n)\not\in \text{Int P}\right\} \in I$ then $\left\{x_n\right\}_{n\in \mathbb{N}}$ is said to be $I$-convergent to $x$ and we write $I\text-\displaystyle{\lim _n} x_n=x$
\end{defn}
\begin{defn} \cite{n} A sequence  $\left\{x_n\right\}_{n\in \mathbb{N}}$ in $X$ is said to be $I^*$-convergent to $x\in X$ if and only if there exists a set $M \in F(I), M=\left\{m_1<m_2<\cdots <m_k<\cdots\right\}$ such that $\displaystyle{\lim_{k\to \infty}}x_{m_k}= x$ i,e for every $c\in E$ with $\theta_E<<c$, there exists $p\in \mathbb{N}$ such that $ c - d(x,x_{m_k})\in \text{Int P}$ for all $k\geq p$.
\end{defn}

It is known \cite{r} that any cone metric space is first countable Hausdorff topological space with the topology induced by the open balls defined as usual for each element $z$ in $X$ and for each element $c$ in Int P. So as in \cite{j} it can be seen that $I^*$-convergence always implies $I$-convergence but the converse is not true. The two concepts are equivalent if and only if the ideal $I$ has the condition (AP).\\

Now we consider the following theorems as in  \cite{5}.
\begin{thm} \cite{5}
Let $E$ be a real Banach space with cone P. If $x_0\in \text{Int P}$ and $c(>0)\in \mathbb{R}$ then $c x_0\in \text{Int P}$.
\end{thm}
\begin{thm} \cite{5}
Let $E$ be a real Banach space and P be a cone in $E$, if $x_0\in \text{P}$ and $y_0\in \text{Int P}$ then $ x_0 + y_0 \in \text{Int P}$
\end{thm}

\section{$I$-divergence and $I^*$-divergence}

Throughout the paper $X$ stands for a cone metric space with normal cone P and the cone metric $d : X\times X \mapsto E$, $\theta_E$ stands for the null vector of $E$, $I$ for nontrivial admissible ideal of $\mathbb{N}$, the set of natural numbers unless otherwise stated.

\begin{defn} A sequence $\left\{x_n\right\}_{n\in \mathbb{N}}$ in a cone metric space $(X,d)$ is said to be divergent (or properly divergent) if there exists an element $x\in X$ such that for any $c \in E$ with $\theta_E<< c$ there exists $n_0 \in \mathbb{N}$ such that $d(x,x_n) - c\in \:\text{Int P}$ for all $n\geq n_0$.
\end{defn}
\begin{exmp}Let $E=\mathbb{R}^{2}$ with usual norm, $\text{P}=\left\{(x,y)\in E : x,y \geq 0\right\}\subset \mathbb{R}^{2}$. $X=\mathbb{R}^{2}$ and $d: X \times X \mapsto E$ such that $d(x,y )=( |x_1 - y_1 | , |x_2 - y_2| )$, where $x=(x_1 ,x_2)$, $y=(y_1,y_2)$. Let us define a sequence $\left\{x_n\right\}_{n\in \mathbb{N}}$ as follows $x_n=(n,n)$. Then there exists $x=(0,0)$ such that for any $c\in E $ with $\theta_E<<c$ there exists $n_0=\text{max}\left\{c_1, c_2\right\} + 1 \in \mathbb{N}$, where $c=(c_1, c_2)$ such that $d(x , x_n) - c =(n,n) -c \in \:\text{Int P}$ for all $n\geq n_0$.
\end{exmp}
Now we have the following theorem related to divergent sequence in a cone metric space.\\
\begin{thm} A divergent sequence can not have a convergent subsequence in a cone metric space.
\end{thm}
\begin{proof}
Let $\left\{x_n\right\}_{n\in \mathbb{N}}$ be a sequence in a cone metric space which is divergent. Let $c\in E$ with $\theta_E<< c$. Since  $\left\{x_n\right\}_{n\in \mathbb{N}}$ is divergent there exists an element $x\in X$ such that for this $c\in E$ there exists $n_0\in \mathbb{N}$ such that $d(x,x_n) - c \in \:\text{Int P}$ when $n\geq n_0$. Now if possible let $\left\{x_{n_k}\right\}_{k\in \mathbb{N}}$ be a subsequence of  $\left\{x_n\right\}_{n\in \mathbb{N}}$ converging to some $x \in X$. Then for this $c\in E$ we have a $p \in \mathbb{N}$ such that $d(x,x_{n_k})<< c$ for all $n_k \geq n_p$. Let r = max $\left\{n_0, n_p\right\}$. Then $d(x,x_r) - c \in \: \text{Int P}$ and $ c - d(x,x_r)\in \: \text{Int P}$ i,e $-(d(x,x_r) - c)\in \: \text{Int P}$. So $\theta_E \in \: \text{Int P}$, which is a contradiction. Hence the result follows.
\end{proof}
\begin{defn} A sequence  $\left\{x_n\right\}_{n\in \mathbb{N}}$ in a cone metric space $(X,d)$ is said to be $I$-divergent if there exists an element $x\in X$ such that for any $c\in E$ with $\theta_E<< c$, $A(x,c)=\left\{ n\in \mathbb{N} : c -d(x,x_n)\in \:\text{Int P}\right\} \\
\in I$.
\end{defn}
\begin{defn} A sequence  $\left\{x_n\right\}_{n\in \mathbb{N}}$ in a cone metric space $(X,d)$ is said to be $I^*$-divergent if there exists an $M\in F(I)$ (i,e $\mathbb{N}\setminus M \in I$ ) such that  $\left\{x_n\right\}_{n\in \mathbb{N}}$ is divergent i,e there exists an element $x\in X$ such that for any $c\in E$ with $\theta_E<< c$ there exists $n_0\in \mathbb{N}$ such that $d(x,x_n) - c\in \:\text{Int P}$ $\forall n \geq n_0, n\in M$.
\end{defn}
\begin{thm}\label{th2} Let $I$ be an admissible ideal. If $\left\{x_n\right\}_{n\in \mathbb{N}}$ is $I^*$ -divergent then  $\left\{x_n\right\}_{n\in \mathbb{N}}$ is $I$-divergent.
\end{thm}
\begin{proof}
Since  $\left\{x_n\right\}_{n\in \mathbb{N}}$ is $I^*$-divergent, there exists an $M\in F(I) \:  (i,e \mathbb{N}\setminus M\in I)$ such that  $\left\{x_n\right\}_{n\in M}$ is divergent. That is there exists an $x\in X$ such that for any $c\in E$ with $\theta_E<< c$ $\exists$ \: $n_0\in \mathbb{N}$ such that $d(x,x_n) - c\in \:\text{Int P}$ for all $n\geq n_0 $,$n\in M$. Now the set $A(x,c)=\left\{ n\in \mathbb{N}: c - d(x,x_n)\in \: \text{Int P}\right\}\subset (\mathbb{N}\setminus M)  \cup \left\{1,2,\cdots, n_0\right\}$. Since $I$ is an admissible ideal the set $\mathbb{N}\setminus M \cup\left\{1,2,\cdots,n_0\right\}\in I$. So $A(x,c)\in I$. This implies that $\left\{x_n\right\}_{n\in \mathbb{N}}$ is $I$-divergent.
\end{proof}
The converse of the above theorem is not always true. The following is one such example.\\

\begin{exmp}
Let $\mathbb{N}=\displaystyle{\bigcup_{j\in \mathbb{N}}}\Delta_j$ be a decomposition of $\mathbb{N}$ such that $\Delta_j$ is infinite and $\Delta_i \cap\Delta_j=\phi$ if $i\neq j$. Let $I$ be the class of all those subsets of $\mathbb{N}$ that intersects with only finite numbers of $\Delta_i$'s. Then $I$ is a non-trivial admissible ideal of $\mathbb{N}$. Now let us consider $X=\mathbb{R}$, $E=\mathbb{R}^2$, with usual norm and $d(x,y )=( |x - y| , |x - y|)$, $P=\left\{ (x , y) : x,y \geq 0\right\}$. Then $(X,d)$ is a cone metric space. Now let us construct a sequence as follows: $y_i = n$ if $i \in \Delta_n$. Now for any $c\in E$ with $\theta_E<< c$ the set $\left\{ i \in \mathbb{N} : c - d(0,y_i)\in \: \text{Int P}\right\}$ is contained in the union of finite numbers of $\Delta_j$'s and hence belongs to $I$. So $\left\{y_i\right\}_{i\in \mathbb{N}}$ is $I$-divergent. Next we shall show that $\left\{y_i\right\}_{i\in \mathbb{N}}$ is not $I^*$-divergent. If possible assume that $\left\{y_i\right\}_{i\in \mathbb{N}}$ is $I^*$-divergent. Then by definition there is a $M\in F(I)$ such that  $\left\{y_i\right\}_{i\in M}$ is divergent. Since $\mathbb{N}\setminus M\in I$ so there exists a $l \in \mathbb{N}$ such that $\mathbb{N}\setminus M \subset \Delta_1 \cup\Delta_2\cup\cdots\cup\Delta_l$. [For, if no such $l$ is found then $\mathbb{N}\setminus M$ intersects  infinitely many $\Delta_j$ which contradicts the fact that $\mathbb{N}\setminus M \in I $]. But then $\Delta_i \subset M \: \forall i > l$. In particular $\Delta_{l+1}\subset M$. But this implies that  $\left\{y_i\right\}_{i\in \Delta_{l+1}}$ is a constant subsequence of  $\left\{y_i\right\}_{i\in M}$ which converges to $l+1$. This contradicts the fact that $\left\{y_i\right\}_{i\in M}$ is divergent.
\end{exmp}

\begin{thm}\label{th1} If $I$ is an admissible ideal with property (AP) then any $I$-divergent sequence $\left\{x_n\right\}_{n\in \mathbb{N}}$ in $X$ is $I^*$-divergent if the following additional condition (C) holds for the cone P of a real Banach space $E$,\\
 (C) : For any $c\in E$ and $\eta\in \text{Int P}$ there exists a natural number $K$ such that $c<< K\eta$.
\end{thm}
\begin{proof}
First suppose that $I$ satisfies the condition (AP). Since  $\left\{x_n\right\}_{n\in \mathbb{N}}$ is $I$-divergent, so there exists an element $x\in X$ such that for any $c\in E$ with $\theta_E <<c$,  $A(x,c)=\left\{n\in \mathbb{N} : c - d(x,x_n)\in \: \text{Int P}\right\}\in I$ i,e for any $c\in \text{Int P}$, $A(x,c)=\left\{n\in \mathbb{N} : d(x,x_n) << c\right\}\in I$. Now let us choose a $\eta\in E$ with $\theta_E<< \eta$. For all $k\geq 2$, let\begin{align*}&A_1 = \left\{n\in \mathbb{N} : d(x,x_n) << \eta\right\}\\
&A_2 = \left\{n\in \mathbb{N} : \eta \leq d(x,x_n) << 2\eta\right\}\\
&A_3 = \left\{ n\in \mathbb{N} : 2\eta \leq d(x,x_n)<< 3\eta\right\}\\
&\vdots\\
&A_k = \left\{ n\in \mathbb{N} : (k - 1)\eta \leq d(x,x_n)<< k\eta\right\}.
\end{align*} \\
Since for any $k\geq 2$, $k\eta\in \text{Int P}$ as $\eta\in \text{Int P}$, it follows that $\left\{n\in\mathbb{N} : d(x,x_n) << k\eta \right\}
\in I$. Since $A_k \subset \left\{n\in\mathbb{N} : d(x,x_n) << k\eta \right\}$, $A_k\in I$ \: $\forall \: k\geq 2$; also $A_1\in I$. Now suppose $A_i \cap A_j \neq \phi$. Without loss of generality suppose $i< j$, then $i+1\leq j$. If $j= i + 1$ then $A_i=\left\{ n\in \mathbb{N}: (i - 1) \eta \leq d(x,x_n) << i\eta\right\}$ and $A_j = A_{i+1} = \left\{ n \in \mathbb{N}: i\eta \leq d(x,x_n)<< (i+1)\eta\right\}$. Let $m\in A_i\cap A_j$ then $d(x,x_m)<< i\eta$ and $i\eta\leq d(x,x_m)$ $\Rightarrow$ $i\eta - d(x,x_m) \in \text{ Int P}$ and $d(x,x_m) - i\eta \in \text{P}$ $\Rightarrow$ $\left\{( i\eta - d(x,x_m)) + (d(x,x_m) - i\eta)  \right\}\in \text{Int P}$ $\Rightarrow$ $\theta_E \in \text{Int P}$, a contradiction. Again if $i + 1 < j$ then $(j - i- 1)> 0$, so $\eta(j - i -1)\in \text{Int P}$ i,e $(j - 1)\eta - i\eta \in \text{Int P}$ and hence $i\eta << (j-1)\eta$. Let $m\in A_i \cap A_j$. Now $m\in A_i$ $\Rightarrow$ $d(x, x_m) << i\eta$ which implies that $d(x,x_m) << i\eta << (j-1)\eta $ $\Rightarrow$ $ d(x,x_m) << (j -1)\eta$. Again $m\in A_j$ $\Rightarrow$ $(j - 1)\eta \leq d(x,x_m)$  $\Rightarrow$ $d(x,x_m) - (j-1)\eta \in \text{P}$. Hence $\left\{ ((j -1 )\eta - d(x,x_m)) + ( d(x,x_m) - (j -1 )\eta)\right\}=\theta_E\in \text{Int P}$, a contradiction. Thus we get a collection of mutual disjoint sets $\left\{A_i\right\}_{i\in \mathbb{N}}$ with $A_i\in I$ for all $i\in \mathbb{N}$. By the condition (AP) there exists a family of sets $\left\{B_i\right\}_{i\in \mathbb{N}}$ such that $A_i\Delta B_i$ is finite for all $i$'s and $B=\displaystyle{\bigcup_{i\in\mathbb{N}}}B_i\in I$. Let $M=\mathbb{N}\setminus B $ then $M \in F(I)$. Let us take any $c\in E$ with $\theta_E<< c$. Since the cone P satisfied the condition (C), choose $k\in \mathbb{N}$ such that $c<< k\eta$. Then $ \left\{n\in \mathbb{N}: d(x,x_n) << c\right\} \subset A_1\cup A_2\cup \cdots \cup A_k$. Since $A_i \Delta B_i$ is finite, so there exists $n_0\in \mathbb{N}$ such that $ (\displaystyle{\bigcup_{i=1}^{k}} B_i) \cap \left\{ n\in \mathbb{N} : n\geq n_0\right\} =(\displaystyle{\bigcup_{i=1}^{k}}A_i) \cap\left\{n\in\mathbb{N} : n\geq n_0\right\}$. Clearly if $n\geq n_0$ and $n\in M$ then $ n\not\in \displaystyle{\bigcup_{i=1}^{k}}B_i \Rightarrow n\not\in  \displaystyle{\bigcup _{i=1}^{k}}A_i$. Therefore $d(x,x_n)\geq k\eta>>c$. This implies $d(x,x_n)>>c$. Thus  $\left\{x_n\right\}_{n\in M}$ is $I^*$-divergent.
\end{proof}
\begin{note}
Note that  the condition (C) stated in the theorem \ref{th1} is precisely the Archimedean condition for $E$ with cone P, also note that there are Banach space and corresponding cones where this additional condition (C) holds as shown in the following example.
\end{note}
\begin{exmp}
Let $E=\mathbb{R}^2$, equipped with usual norm. Let $\text{P}=\left\{ (x,y) : x,y \geq 0\right\}$. Let $y=(y_1, y_2)$ be any element in $E$ and $x=(x_1, x_2)\in \text{Int P}$. Let $r_1=\max\left\{ |y_1|, |y_2|\right\} $ and $r_2=\min\left\{x_1, x_2\right\}$. Therefore $r_2 > 0$ and $x_1 \geq r_2$ and $x_2\geq r_2$. Also $|y_1| \leq r_1$, $|y_2| \leq r_1$. So $y_1 \leq r_1$ and $y_2 \leq r_1$. Choose an natural number $n$ by the Archimedean property of $\mathbb{R}$, such that $n r_2 > r_1$. So $n x_1\geq n r_2 > r_1 \geq y_1$ and $n x_2 \geq n r_2 >r_1 \geq y_2$. So $n (x_1, x_2) - (y_1, y_2) = (n x_1 - y_1 , n x_2 - y_2) \in \text{Int P}$, since $n x_1 - y_1 > 0$ and $n x_2 - y_2> 0$. Hence $n (x_1, x_2) >> (y_1, y_2)$.
\end{exmp}
\begin{thm}\label{th3} Let $(X,d)$ be a cone metric space containing at least one divergent sequence. If every $I$-divergent sequence  $\left\{y_n\right\}_{n\in \mathbb{N}}$ is $I^*$-divergent then $I$ satisfies the condition (AP).
\end{thm}
\begin{proof}
Let $\left\{x_n\right\}_{n\in \mathbb{N}}$ be a divergent sequence in $X$. Then there exists an element $x\in X$ such that for any $c\in E$ with $\theta_E<<c$ there exists $k\in \mathbb{N}$ such that $d(x,x_n) - c\: \in \text{Int P}$ for all $n\geq k$. Suppose $\left\{A_i : i=1,2,3,\cdots\right\}$ is a sequence of mutually disjoint non-empty sets from $I$. Define a sequence $\left\{y_n\right\}_{n\in \mathbb{N}}$ as follows: $y_n = x_j $ if $n\in A_j$ and $y_n = x_n$ if $n\not\in A_j$ for any $j\in \mathbb{N}$. Let $c\in E$ be any element such that $\theta_E<< c$. Now $A(x,c)=\left\{n\in \mathbb{N} : d(x,y_n)<< c\right\} \subset A_1\cup A_2\cup A_3\cup \cdots \cup A_k \cup\left\{ 1,2,\cdots,k\right\}\in I$. So $\left\{y_n\right\}_{n\in \mathbb{N}}$ is $I$-divergent. By our assumption $\left\{y_n\right\}_{n\in \mathbb{N}}$ is $I^*$-divergent. So there exists $M\subset \mathbb{N}$ such that $M\in F(I)$ and $\left\{y_n\right\}_{n\in M}$ is divergent. Let $B=\mathbb{N}\setminus M$. Then $B\in I$. Put $B_j=A_j \cap B$ for all $j\in \mathbb{N}$. Since $\displaystyle{\bigcup_{j\in \mathbb{N}}}B_j \subset B$; $\displaystyle{\bigcup_{j\in \mathbb{N}}}B_j \in I$. Then $A_j\cap M$ is a finite set. For, if possible let $A_j \cap M$ is not finite set, then $M$ must contain an infinite sequence of elements $\left\{m_k\right\}$. So $y_{m_k}=x_j$ for all $k\in \mathbb{N}$, which forms a convergent sub-sequence of  $\left\{y_n\right\}_{n\in M}$. But this contradicts the fact that  $\left\{y_n\right\}_{n\in M}$ is divergent. Hence $A_i\Delta B_i=A_i\setminus B_i =A_i \cap B_i^{c}=  A_i \cap (M\cup A_i^{c})=A_i \cap M$, which is a finite subset of $\mathbb{N}$. This proves that $I$ satisfies the condition (AP).
\end{proof}
\begin{cor}
If the condition (C) stated in the theorem \ref{th1} holds for a real Banach space and its corresponding cone. Then for an admissible ideal $I$ of $\mathbb{N}$ a $I$-divergent sequence is $I^*$-divergent if and only if $I$ satisfies the condition (AP).
\end{cor}
\begin{proof}
Follows from the theorem \ref{th1} and theorem \ref{th3}.
\end{proof}
\section{The decomposition theorem}\label{sec}
In this section we prove a decomposition theorem for $I$-convergent sequence in cone normed spaces.
\begin{thm}\label{thDe}
Let $(X, \norm{.}_{\text{P}})$ be a cone normed space, P be a normal cone and $(X,d)$ be a cone metric space where $d$ is the cone metric induced by the cone norm $\norm{.}_\text{P}$. Also let $x=(x_n)$ be a sequence in $X$ and $I\subset 2^\mathbb{N}$ be an admissible ideal with the property (AP). Then the following condition are equivalent:\\
(a) $I\text-\lim x_n=\xi $\\
(b) There exists sequences $y=(y_n)\in X$ and $z=(z_n)$ in $X$ such that $x= y + z $, $d(y_n,\xi) \to \theta_E$ as $n\to\infty$ and $\text{supp z}\in I$, where $\text{supp z}=\left\{n\in \mathbb{N} : z_n\neq \theta_X \right\}$, where $\theta_X$ is the zero element of $X$.
\end{thm}
\begin{proof}
Let $I\text-\displaystyle{\lim_{n\to \infty}}x_n=\xi$. Since $I$ has the property (AP), we conclude that there exists a set $M\in F(I)$, $M=\left\{m_1<m_2<\cdots<m_k <\cdots\right\}$ such that $\displaystyle{\lim_{k \to\infty}}x_{m_k}=\xi$. Now we define a sequence $y=(y_n)$ in X as follows:
 \begin{equation}\label{eq3}
y_n=
\begin{cases}x_n       \hspace{2cm}  \: n\in M.\\
\xi                     \hspace{2cm}  \:n\in \mathbb{N}\setminus M.
\end{cases}
\end{equation}
Thus we can see that $\displaystyle{\lim_{n\to \infty}}y_n=\xi$. Since P is normal so $d(y_n,\xi)\to \theta_E$ as $n\to \infty$. Further put $z_n= x_n - y_n$, $n\in\mathbb{N}$. Since $\left\{k\in \mathbb{N} : x_k\neq y_k\right\} \subset \mathbb{N}\setminus M \in I$. We have $\left\{k\in \mathbb{N} : z_k\neq \theta_X \right\} \in I$ . So it follows that $\text{supp z}\in I$. Also $x=y + z$.\\

Conversely, Suppose there exists two sequence $y=(y_n)\in X $ and $z=(z_n)\in X$ such that $x= y + z$, $d(y_n, \xi )\to \theta_E$ as $n\to \infty$ and $\text{supp z}\in I$. We will prove that $I\text-\displaystyle{\lim_{n \to \infty}} x_n=\xi$. Define a subset $M$ of $\mathbb{N}$ by $M=\left\{m\in \mathbb{N}: z_m\neq \theta_X\right\}$. Now since $\text{supp z}=\left\{m\in \mathbb{N}: z_m\neq \theta_X\right\}\in I$, we have $M\in F(I)$; hence $x_n=y_n$ if $n\in M$. Therefore there exists a set $M=\left\{m_1<m_2<\cdots\right\}\text{,}\: M\in F(I)$ such that $\displaystyle{\lim_{k\to\infty}}x_{m_k}=\xi$. Now as $I$ satisfies the condition (AP), therefore we can say that $I\text-\displaystyle{\lim_{n\to\infty}}x_n=\xi$. Hence the proof is complete.
\end{proof}
\begin{lem}\label{lem1}
Let $(X,\norm{.}_{\text{P}})$ be a cone normed space, P be a normal cone and $(X,d)$ be a cone metric space where $d$ is the cone metric induced by the cone norm $\norm{.}_\text{P}$. Then if $I\text-\lim x_n=\xi$, $I\text-\lim y_n=\eta$ then $I\text-\lim (x_n + y_n)=\xi + \eta$
\end{lem}
\begin{proof}
Since $I\text-\lim x_n=\xi$ and $I\text-\lim y_n=\eta$. So for any $c\in E$ with $\theta_E<< c$ \\
$\left\{n\in \mathbb{N}: \frac{c}{2} - d(x_n,\xi) \not\in \:\text{Int P}\right\}\in I\:\text{and}\: \left\{n\in \mathbb{N}: \frac{c}{2} - d(y_n,\eta) \not \in \:\text{Int P} \right\}\in I$.\\
Now we claim that
\begin{equation}\label{eq2}
\begin{split}
\left\{n\in \mathbb{N} : c - d(x_n + y_ n, \xi + \eta) \not \in \:\text{Int P}\right\}\subset \left\{n\in \mathbb{N}: \frac{c}{2} - d(x_n,\xi) \not\in \:\text{Int P}\right\} \\
\cup \left\{n\in \mathbb{N}: \frac{c}{2} - d(y_n,\eta) \not \in \:\text{Int P} \right\}.
\end{split}
\end{equation} For, let $m\not \in \left\{n\in \mathbb{N}: \frac{c}{2} - d(x_n,\xi) \not\in \:\text{Int P}\right\} \cup \left\{n\in \mathbb{N}: \frac{c}{2} - d(y_n,\eta) \not \in \:\text{Int P} \right\}$. Then $ \frac{c}{2} - d(x_m,\xi) \in \:\text{Int P}$ and $\frac{c}{2} - d(y_m , \eta)\in \:\text{Int P}$. So $ c - ( d(x_m, \xi) + d(y_m, \eta)) \in \:\text{Int P}$ that is $d(x_m,\xi) + d(y_m,\eta) << c$. Now, \begin{equation}\label{eq1}
d(x_m + y_m , \xi + \eta ) \leq d(x_m + y_m , x_m + \eta) + d(x_m + \eta , \xi +\eta)
\end{equation}
Now for the cone metric $d$ induced by a cone norm on a cone normed space satisfies $d(x + c , y + c) = d(x,y)$. Thus (\ref{eq1}) implies $d(x_m + y_m , \xi +\eta)\leq d(y_m , \eta) + d(x_m , \xi)$. So we get $d(x_m + y_m , \xi + \eta)\leq d(y_m, \eta)+ d(x_m , \xi) << c$. Hence $c - d(x_m + y_m , \xi + \eta) \in \: \text{Int P}$. Thus $m\not \in \left\{n \in \mathbb{N} : c - d(x_n + y_n , \xi + \eta)\not \in \:\text{int P}\right\}$. Therefore our claim is true. Now the right hand sides of (\ref{eq2}) belongs to $I$. So left hand of (\ref{eq2}) also belongs to $I$. So, $I\text-\lim( x_n + y_n)=\xi + \eta $
\end{proof}
\begin{cor}
$I\text-\displaystyle{\lim_{n\to\infty}}x_n=\xi$ if and only if there exists $(y_n)\in X$ and $(z_n)\in X$ such that $x_n= y_n + z_n$, $d(y_n , \xi)\to \theta_E$ as $n\to\infty$ and $I\text-\displaystyle{\lim_{n\to\infty}}z_n=\theta_X$
\end{cor}
\begin{proof}
Let $z_n= x_n - y_n$, where $(y_n)$ is the sequence defined by (\ref{eq3}). Then $d(y_n , \xi)\to \theta_E$ as $n\to \infty$. Also $I\text-\displaystyle{\lim_{n\to\infty}}z_n = \theta_X$; as for any $c\in E$ with $\theta_E<< c$, $\left\{n\in\mathbb{N} : c - d(z_n , \theta_X) \not \in \:\text{Int P}\right\} \subset \text{supp z} \in I$.\\
Further let $x_n= y_n + z_n $, where $d(y_n , \xi )\to \theta_E$ as $n\to \infty$ and $I\text-\displaystyle{\lim_{n\to\infty}}z_n = \theta_X$. Since $I\text-\displaystyle{\lim_{n\to\infty}}y_n =\xi$, then by lemma (\ref{lem1}) we get $\text{I}\text-\displaystyle{\lim_{n\to\infty}}x_n=\xi$
\end{proof}
\begin{rem}\label{rem2}
From the proof of the theorem (\ref{thDe}), it is clear that (b) implies (a) even if the ideal $I$ have not the property (AP), as $I^*$-convergence implies $I$-convergence. In fact, let $x_n = y_n + z_n $ ,$ d(y_n, \xi) \to \theta_E$ as $n\to\infty$ and $\text{supp z}\in I$, where $I$ is an admissible ideal which has not the property (AP). Since $\left\{n\in\mathbb{N}: c - d(z_n, \theta_X) \not \in \text{Int P}\right\} \subset \left\{n\in \mathbb{N}: z_n\neq \theta_X\right\}\in I$ for any $c\in E$ with $\theta_E<< c$; we have $I\text-\displaystyle{\lim_{n\to\infty}}z_n= \theta_X$. Hence $I\text-\displaystyle{\lim_{n\to\infty}}x_n=\xi$
\end{rem}


TITLE: Why does putting the eigenvectors as columns in a matrix give us the diagonalizing matrix?
QUESTION [5 upvotes]: For $A$ of $n \times n$ if we have $n$ eigenvectors,  we can put them as columns in a matrix and get the diagonalizing matrix - why does it work?

REPLY [0 votes]: Follows, I think, a pretty straightforward and simple way to see this:
First, let $B$ be any $n \times m$ matrix with columns $B_1, B_2, \ldots, B_m$, so that we may write $B$ in columnar form as
$B = [B_1 \; B_2 \; \ldots \; B_m]; \tag 1$
then is it easy to see that
$AB = [AB_1 \; AB_2 \; \ldots \; AB_m]; \tag 2$
now if $E_1, E_2, \ldots, E_n$ are linearly independent eigenvectors of $A$, that is, if we have
$AE_i = \mu_i E_i, \tag 3$
for the scalar eigenvalues $\mu_i$, $1 \le i \le n$, then the linear independence of the $E_i$ implies that the $n \times n$ matrix
$E = [E_1 \; E_2 \; \ldots \; E_n] \tag 4$
is invertible; that is, there exists an $n \times n$ matrix $E^{-1}$ such that
$E^{-1}E = E^{-1}[E_1 \; E_2 \; \ldots E_n] = I; \tag 5$
since
$E^{-1}[E_1 \; E_2 \; \ldots E_n] = [E^{-1}E_1 \; E^{-1}E_2 \; \ldots E^{-1}E_n], \tag 6$
we may infer from (5) and (6) that
$E E_i = \mathbf e_i = \begin{pmatrix} \delta_{i1} \end{pmatrix},  \tag 7$
i.e., $EE_i = \mathbf e_i$ is the column vector whose $i$-th row is $1$ with all other rows (entries) equal to $0$.  Now in accord with (2) and (3) we also have
$AE = A [E_1 \; E_2 \; \ldots \; E_n]$
$= [AE_1 \; AE_2 \; \ldots \; AE_n] = [\mu_1 E_1 \; \mu_2 E_2 \; \ldots \; \mu_n E_n], \tag 8$
whence
$E^{-1}AE = E^{-1}[\mu_1 E_1 \; \mu_2 E_2 \; \ldots \; \mu_n E_n]$
$= [\mu_1 E^{-1}E_1 \; \mu_2 E^{-1}E_2 \; \ldots \; \mu_n E^{-1}E_n]$
$= [\mu_1 \mathbf e_1 \; \mu_2 \mathbf e_2 \ldots, \mu_n \mathbf e_n] = \text{diag}(\mu_1, \mu_2, \ldots, \mu_n), \tag 9$
the $n \times n$ diagonal matrix which has the eigenvalues $\mu_i$ of $A$ along its main diagonal an zeroes elsewhere.  We thus see that $E$ is the diagonalizing matrix for $A$.
Now as our OP Alon asked, "why does it work?"  Scrutiny of the above discussion reveals two essential factors which contribute to the success of this program:  first, left multiplication of the eigenvector matrix $E$ by $A$ multiplies each column separately, thus in accord with the eigen-equation (3) each column $E_i$ is merely multiplied by the corresponding scalar $\mu_i$; and second, multiplication of such columns by $E^{_1}$ returns the corresponding column of the identity matrix $I$ multiplied by the corresponding $\mu_i$; in this way the matrix $\text{diag}(\mu_1, \mu_2, \ldots, \mu_n)$ is the net result of these operations acting in concert; and thus $A$ is diagonalized by the eigenvector matrix $E$.
Note Added in Edit; Sunday 19 January 2020 9:56 PM PST:  The attentive reader may have observed that, while the assumption that $A$ has $n$ eigenvectors is explicit in the text of the problem itself, no mention is there made of their linear independence; but that in fact I have introduced this concept ca. (2)-(4) in my answer; in this sense I have added an assumption to the question as stated by our OP Alon.  In fact, this assumption is essential if $E = [E_1 \; E_2 \; \ldots \; E_n]$ is to be a diagonalizing matrix for $A$, since it is equivalent to the existence of $E^{-1}$.  Finally, we note that without the hypothesis of linear independence, it is possible for $A$ to have and infinite number of eigenvectors; indeed, taking $A = aI$, for any vector $V$ we find $AV = aIV = aV$; thus every $V$ is an eigenvector corresponding to $a$; however, the number of linearly independent eigevectors of $A$ is $n = \text{size}(A)$.  End of Note.


TITLE: Argument principle for Riemann surfaces
QUESTION [0 upvotes]: I am basically looking for a good reference for an extension of Cauchy's argument principle on Riemann surfaces. For meromorphic functions on the complex plane, the theorem is the following (hope I get it right).
Theorem (Cauhcy's argument principle). Let $U$ be a simply connected open set of $\mathbb{C}$, let $f:U\to \mathbb{C}$ be a meromorphic function on $U$, let $S\subset U$ be the set of zeros ans poles of $f$, $S$ finite. Then for any  closed contour $\gamma: [0,1]\to \mathbb{C}$, that does not intersect itself, and such that $\gamma ([0,1])\subset U\setminus S$, we have
$$ \frac{1}{2\mathrm{i}\pi} \int_\gamma \frac{f'(z)}{f(z)}\mathrm{d}z = Z(f,\gamma) - P(f,\gamma), $$
where $Z(f,\gamma)$ (respectively $P(f,\gamma)$) is the number of zeros of $f$, counted with multiplicity, that lie within the bounded component of $\mathbb{C}\setminus \gamma ([0,1])$ (resp. the number of poles).
My question is: How exactly does this result extend to the case where $f$ is a meromorphic function on a compact Riemann surface of any genus? I have read that somehow it does, and that it implies that the number of zeros equals the number of poles. But I cannot manage to find a good reference for that precise statement. Does anybody know some?

REPLY [0 votes]: Any standard introdutory book on Riemann surfaces may discuss this.
See, for example, Rick Miranda's book page 124.


TITLE: Airflow through two different diameter outlet
QUESTION [0 upvotes]: A fixed airflow travels through a duct that would be split into two outlets. One of the outlets has a smaller diameter, which the other has a larger diameter, will the larger diameter outlet has a high airflow rate than the smaller diameter outlet?

REPLY [0 votes]: We can not tell exactly that rate of air flows from the two different nozzles be equal. It depends on the speeds of air  coming from them. If we assume them to be equal, then obviously the larger one has high air flow rate.


TITLE: Order of convergence of Fixed Point Iteration
QUESTION [0 upvotes]: I have to determine the order of convergence of this iteration specification but I don't know how to do this in $\mathbb{R}^n$. (So far, I've only done this in $\mathbb{R}$).
Any help is highly appreciated (and sorry for my bad english).
$x^{(k+1)} = \begin{pmatrix}
0 & -1/2 & -1/4\\
-1/2 & 0 & 0\\
0 & -1/2 & 0
\end {pmatrix}x^{(k)}+\begin{pmatrix} 3/4 \\ -2 \\ -1/2 \end{pmatrix}$
for $k \geq 0$
$x^{(k)} = \begin{pmatrix} x_1^{(k)} \\ x_2^{(k)} \\ x_3^{(k)} \end{pmatrix}$

REPLY [1 votes]: Your iteration is special case of the stationary iteration $$x_{n+1} = Gx_n + f$$ which can occasionally be used to solve the linear system $$x = Gx +f.$$ The initial guess $x_0$ must be selected by the user, but $x_0 = 0$ is a perfectly acceptable choice. If $\|G\|_2<1$, then $I-G$ is nonsingular, and the sequence $\{x_n\}_{n=0}^\infty$ is convergent and $$x_n \rightarrow x = (I-G)^{-1} f, \quad n \rightarrow \infty, \quad n \in \mathbb{N}$$
regardless of the choice of $x_0$. If $x_0 = 0$, then by induction on $n$ you can establish that $$x_n = \sum_{j=0}^{n-1} G^j f.$$ It follows, that $$\|x-x_n\|_2 \leq \left \| \sum_{j=n}^\infty G^j f \right\|_2 \leq \|G\|_2^n\|x\|_2 = \epsilon_n.$$ It follows that $x_n \rightarrow x$ at least linearly as $n \rightarrow \infty$ and $n \in \mathbb{N}$ because
$$ \frac{\epsilon_{n+1}}{\epsilon_n} = \frac{ \|G\|_2^{n+1} }{\|G\|_2^{n}} = \|G\|_2 \rightarrow \|G\|_2 < 1.$$
You are free to replace the $2$-norm with any other norm induced by a vector norm. In particular, you can use the $\infty$-norm for which your particular matrix $G$ satisfies $\|G\|_\infty = \frac{3}{4}$.
In any case, the order of convergence is at least $p=1$.


TITLE: Finding a number
QUESTION [0 upvotes]: Every digit has to contain $0-9$. Some possible combinations are:
$98765432$
$87654321$
I am trying by trial and error but I want to know how to solve this by strategy. Any guides?
Also what would strictly decreasing mean?

REPLY [2 votes]: The searched number is composed of the 8 digits decreasing and is divisible by 10, so his last digit is $0$.
He is also divisible by 20, so his dozen digit is neither 1 nor 3, so it is $2$
We just need to remove another digit, and we know the final number is also divisible by 9,
$0+2+3+4+5+6+7+8+9=44=36+8$, so the searched number is $97654320$

REPLY [2 votes]: It has to end with a 0, because it's divisible by 10.
Its digits have to sum up to a multiple of 9 because it's divisible by 9.
It's divisible by 4 so its last two digits can't be "10" or "30".
Therefore they must be "20". (Can't be "40" because there are only two digits missing).
Therefore, the sum of first 6 digits must be of form $3k+1$.
Sum of numbers from $3$ to $9$ is $42=14\cdot 3$.
So we must be missing a $2, 5$ or $8$.
Now you only have 3 numbers left to check. That's easy enough to consider the problem solved.


TITLE: $X$~Exp$(\lambda)$ with $Y=cX$ - find the cdf for $Y$
QUESTION [0 upvotes]: I have this question that I am not sure on how to deal with the $Y=cX$ part, can anyone help?
$X$ is a continuous random variable with $X$~$Exp(\lambda)$ and let $Y=cX$ where $c > 0$ is a fixed constant. I must find the probability density function (pdf) for $Y$ using two different methods:
1) Find the distribution function (cdf) for $Y$ and then differentiate it to find the pdf for $Y$.
I am fine with differentiation, I just am not sure how to find the cdf for $Y$
2) Noting that the function $g(x) = cx$ is strictly increasing and differentiable, find the pdf for $Y$ directly from the pdf for $X$
All I know is that you can write $X$~$\exp(\lambda)$ as:
$$f_X (x)=\left\{ \begin{array}{ll}
         0 & \mbox{$x \le 0$};\\
        \lambda e^{(-\lambda x)} & \mbox{$x < 0$}.\end{array} \right.$$
But I do not know what I can do with this (if anything).

REPLY [1 votes]: Hint
CDF of random variable $Y$ is $P(Y \leq y)$, so we can write $P(Y \leq y) = P(cX \leq y) = P(X \leq \frac{y}{c})$
For the second method you can use $f_Y(y)=\frac{f_X(g^{-1}(y))}{g'(g^{-1}(y))}$
For more information about the formula this may be helpful: http://www.cs.unm.edu/~williams/cs530/gst2.pdf


TITLE: Equation for Angle Around an Ellipse given semi-major (a), semi-minor (b), and length of radius (r)?
QUESTION [0 upvotes]: What is the equation to find the angle $\theta$ around an ellipse if given only horizontal semi-axis $a$, vertical semi-axis $b$, and just a length for radius $r$? I've also attached an image that graphically shows the problem.

I'm stuck trying to rearrange and isolate $\theta$ for solving using the equation:
$$r = \frac{ab}{\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}}$$
Thank you for any help.

REPLY [0 votes]: For easy calculation... if reciprocals of $(a,b,r)$, are  computed as $(a',b',r')$ then for non-circular ellipses
$$ \frac{\cos2 \theta }{2}= \dfrac { (2 r'^2 - a'^2  - b'^2)}{(a'^2-b'^2)}$$
which solve to four angles.


TITLE: Prove existence of divergent sequence in Newton's method
QUESTION [2 upvotes]: Given $ f(x)=x^3-1$, how to prove that there exists a sequence of initial values  $x_{0,1}>x_{0,2}>x_{0,3}>...$ where $x_{0,1}=0,x_{0,2}=-2^{1/3}$, such that the sequence produced by Newton's method is divergent?

REPLY [1 votes]: If we start at $x_0=0$, then the next member of the sequence is $-\frac{f(x_0)}{f'(x_0)}=-\infty$, so the method fails.
The line $y=kx$ is tangent to the curve at $x$ if $k=3x^2$ and $kx=x^3-1$, so $x=-\left(\frac{1}{2}\right)^{1/3}$. So if we take $x_0=-\left(\frac{1}{2}\right)^{1/3}$ we are also in trouble because the next approximation will be $x=0$ and then $-\infty$.
This process can be continued indefinitely, having found a bad point $x=-h$ we find a point on the curve whose tangent passes through the point $(-h,0)$. So we need a point $(-H,-H^3-1)$ such that the tangent there passes through $(-h,0)$. The gradient at $x=-H$ is $3H^2$, so the tangent is $y+H^3+1=3H^2(x+H)$. For this to pass through $(-h,0)$ we need $H^3+1=3H^2(H-h)$ or $2H^3-3hH^2-1=0$.
Put $g(x)=2x^3-3hx^2-1$. Then $g(h)=-h^3-1<0$ but the leading coefficient is positive, so there must be a root $H>h$. So we can find an infinite sequence of bad starting points.


TITLE: evaluate the following integral
QUESTION [4 upvotes]: $$
\int J_0(x)\sin x~{\rm d}x
$$
Where $J_0$ is Bessel function of first kind of order $0$
This what I tried
$$
\int J_0(x)\sin x~{\rm d}x= -J_0(x) \cos x - \int J_0'(x)\cos x~{\rm d}x
$$
$$
J_0'(x)=-J_1(x)
$$
$$
\int J_0(x)\sin x ~{\rm}x= -J_0(x) \cos x -(J_1(x)\sin x - \int J_1'(x)\sin x~{\rm d}x)
$$
$$
\int J_0(x)\sin x~{\rm d}x=-J_0(x) \cos x - J_1(x) \sin x +\left(\int J_0(x)\sin x~{\rm d}x + \int(\sin x/ x) J_1(x)~{\rm d}x\right)
$$
But this won't help to evaluate it, is there any other method?

REPLY [0 votes]: I know that I answered this question after so long time. But the main fact is I have a solution in another way and that's why I'm  here. And I think this new way will help people a lot to find out the solution.
..................................................................................................................................................................
\begin{equation}I=\int J_0(x) \sin(x) dx\\
=\dfrac 1{2i}\left(\int J_0(x) e^{ix}-\int J_0(x) e^{-ix}\right)dx\\
=\dfrac 1{2i}\left(\int J_0(x) e^{ix}~dx+\int J_0(-y) e^{iy}~dy\right)\\
\left[\text{Second part change the integration variable to $y=-x$}\right]\\
\\
=\dfrac 1{2i}\left(\int J_0(x) e^{ix}~dx+\int J_0(y) e^{iy}~dy\right)\\
\left[\text{By the parity of the Bessel function}\right]\\
=\dfrac 1{2i}\left\{e^{ix}x(J_0(x)−iJ_1(x))+e^{iy}y(J_0(y)−iJ_1(y))\right\}+c\\
\left[\text{where $~c~$ is a constant}\right]\\
=\dfrac 1{2i}\left\{e^{ix}x(J_0(x)−iJ_1(x))-e^{-ix}x(J_0(-x)−iJ_1(-x))\right\}+c\\
=\dfrac 1{2i}\left\{e^{ix}x(J_0(x)−iJ_1(x))-e^{-ix}x(J_0(x)+iJ_1(x))\right\}+c\\
\left[\text{By the parity of the Bessel function}\right]\\
=\dfrac x{2i}\left\{J_0(x)(e^{ix}-e^{-ix})-iJ_1(x)(e^{ix}+e^{-ix})\right\}+c\\
=x[\sin x · J_0(x) − \cos x · J_1(x)]+c\\
\\\end{equation}
Hence $$\boxed{\int J_0(x) \sin(x) dx=x\left\{\sin x · J_0(x) − \cos x · J_1(x)\right\}+c}$$where $~c~$ is a constant.
..................................................................................................................................................................

Parity of the Bessel function : $~J_n(-x)=(-1)^nJ_n(x)~$
$$\int J_0(x) e^{ix}~dx=x~e^{ix}\left\{J_0(x)−iJ_1(x)\right\}$$


TITLE: The 'prime logarithm'
QUESTION [3 upvotes]: Lately I've been thinking about the functional equation $$f(ab) = f(a) + f(b)$$ but not in the usual sense where continuity or differentiability are assumed. It's clear that $f(1) = 0$, by letting $a = b = 1$. It is also clear, that given any natural number $n = p_1^{a_1}p_2^{a_2}\cdots p_m^{a_m}$ that $f(n) = a_1f(p_1) + a_2f(p_2) + ... + a_mf(p_m)$.
Consider the function $\ell:\mathbb N\setminus\{0\}\to\mathbb N$ such that it is a solution to the functional equation, and $\ell(p) = p$ when $p$ is prime.
Some example values:
$$\ell(1) = 0\\\ell(2) = 2\\\ell(15) = \ell(5) + \ell(3) = 8\\\ell(264) = 3\ell(2) + \ell(3) + \ell(11) = 20$$
Here is a plot of the function for $1 \leq n \leq 1000$:

So my question is, does this function have a name? Does it have any useful purpose, or interesting properties? It seems we can extend the domain of this function to the positive rational numbers, making the range $\mathbb Z$ rather than $\mathbb N$.
Some values if we extend to rational numbers:
$$\ell\left(\frac12\right) = -2\\\ell\left(\frac{7}{9}\right) = \ell(7) - 2\ell(3) = 1 \\\ell\left(\frac{42}{110}\right) = \ell(2) + \ell(3) + \ell(7) - \ell(11)-\ell(5) - \ell(2) = -6$$
Notice how the common factors cancelled out in the last example without having to simplify the fraction, showing that $\ell$ is well defined for all $n \in \mathbb Q^+\setminus\{0\}$.
* I made the graph quickly on excel, please feel free to replace it with a better one.

REPLY [1 votes]: Long comment:
The commutative monoid $(\mathbb{N}\setminus\{0\},\times,1)$ is freely generated (as a commutative monoid) by the prime numbers. In other words, for any commutative monoid $X$ and any function $f:\mathbb{P} \rightarrow X$, there's a unique homomorphism $\hat{f}:(\mathbb{N}\setminus\{0\},\times,1) \rightarrow X$ such that the restriction of $\hat{f}$ to $\mathbb{P}$ is $f.$

Your function is $\hat{f}$ for the special case where $X = (\mathbb{N},+,0)$ and $f:\mathbb{P} \rightarrow X$ is defined by $f(p) = p$.
If instead of adding prime factors, you just want $\hat{f}$ to count them, take $f(p) = 1$ instead.
If you want $\hat{f}$ to compute the radical of a number, take $X = (\mathbb{N},\mathrm{lcm},1)$ and $f(p) = p$.
If you want $\hat{f}$ to tell you the largest prime in the factorization, take $X=(\mathbb{P},\mathrm{max},2)$ or $X = (\mathbb{N},\mathrm{max},0)$, either one, and define $f(p) = p$.

I find these kinds of functions super-cool, by the way.
Also, note that $(\mathbb{Q}_{>0},\times,1)$ is the commutative group freely generated by $\mathbb{P}$. So you can do similar stuff in that context.

Addendum.
Also, note that for all $a,b \geq 2$, we have $a+b \leq ab$. Hence if $\hat{f}$ is the function you're interested in, then $\hat{f}(n) \leq n$. So letting $n \geq 2$ denote be fixed, we see that the sequence $$(n,\hat{f}(n),\hat{f}(\hat{f}((n)),\ldots)$$ is eventually constant. Two obvious questions are:

How many steps before its constant?
What is the final value of this sequence? (I think it will always be prime.)

For example, applying $\hat{f}$ repeatedly to $15$, we get:
$$15 \mapsto 8 \mapsto 6 \mapsto 5 \mapsto 5 \mapsto 5 \mapsto \cdots$$
so it takes $3$ steps and we stabilize at $5$.
Kind of cool, huh? This gives us a potentially interesting way of assigning to each element of $\mathbb{N}_{\geq 2}$ a prime number (e.g. to $15$ we assign $5$.) We might write $\hat{f}^{\infty}(15) = 5$ or something like that. Perhaps this number tells us something important about the structure of the number we started with.


text:
\begin{document}

\maketitle

\begin{abstract}
We study analogue source transmission over MIMO block-fading channels with receiver-only channel state information. Unlike previous work which considers the end-to-end expected distortion as a figure of merit, we study the distortion outage probability. We first consider the well known transmitter informed bound, which yields a benchmark lower bound to the distortion outage probability of any coding scheme. We next compare the results with source-channel separation. The key difference from the expected distortion approach is that if the channel code rate is chosen appropriately, source-channel separation can not only achieve the same diversity exponent, but also the same distortion outage probability as the transmitter informed lower bound.
\end{abstract}

\section{Introduction}
\label{sec:Introduction}

The block-fading channel was introduced in \cite{ozarow1994} in order to model delay-limited transmission over slowly varying wireless communications channels. In the channel, each codeword spans only a finite and fixed number $N$ of independent fading blocks. Practical scenarios include OFDM and frequency hopping for low-mobility wireless scenarios. Under this setup, it follows that the Shannon capacity of this channel is zero since there is an irreducible probability that a given transmission rate $\Rc$ is not supported by a particular channel realisation \cite{ozarow1994,biglieri1998}. In particular, a communication outage occurs whenever the instantaneous mutual information is less than the target data rate we wish to communicate at \cite{ozarow1994,biglieri1998}. As shown in \cite{malkamaki1999cdb}, the outage probability is the natural fundamental limit of the channel.

An important reliability metric over block-fading channels is the SNR exponent or outage diversity, defined as the high-SNR slope of the outage probability in a log-log scale.

Inspired by the work by Laneman {\em et al.} \cite{laneman2005}, the end-to-end expected distortion has been studied to characterise the performance of continuous or analogue source transmission over outage-limited multiple-antenna fading channels \cite{caire2007,deniz2007,deniz2008,tim2008}. The above works consider the SNR exponent of the end-to-end expected distortion (where the expectation is also taken over the fading) as a performance metric for a number of joint source-channel coding schemes. In particular, when the expected distortion is considered, these references illustrate the suboptimality of source-channel separation. In order to improve the performance, a number of joint-source channel schemes based on hybrid analogue-digital or multi-layered coding have been proposed \cite{caire2007,deniz2007,deniz2008,tim2008}.

The expected distortion is the natural performance metric for {\em ergodic} fading channels, or channels with no stringent delay constraints. However, for outage-limited channels or channels with stringent delay constraints, the expected distortion fails to characterise the true end-to-end performance of such wireless systems, and thus it might not be the appropriate performance metric. In this paper, we take a different approach to the same problem. In particular, our contribution is the study of the distortion outage probability instead of the expected distortion as a figure of merit for system performance. The distortion outage probability is defined as the probability that the instantaneous distortion (a random variable that depends on the channel realisation and SNR) is larger than a target quality-of-service (QoS) distortion, i.e.,
\begin{equation}
\label{eq:eq44}
P_{\rm out}( \snr,\bar{D})=\Pr\left\{D(\Hm, \snr)>\bar{D}\right\}
\end{equation}
where $D(\Hm,\snr)$ is the instantaneous distortion achieved at SNR $\snr$ for a given $\nt\times \nr$ multiple-input multiple-output (MIMO) channel realisation $\Hm$, and $\bar{D}$ is the target distortion level characterizing the acceptable QoS of the systems. We are particularly interested in the distortion-outage SNR exponents.

In this work, similarly to previous works \cite{laneman2005,caire2007,deniz2007,deniz2008,tim2008} we first study a lower bound on the distortion outage performance, i.e., the transmitter informed bound. This bound assumes perfect channel state information at the transmitter (CSIT), and allows to adapt the joint source-channel code to the instantaneous channel conditions. In particular, we find out the relationships between this lower bound on the distortion outage probability and the information outage probability \cite{ozarow1994,biglieri1998}, as well as the corresponding SNR exponents.

We next consider source-channel separation \cite{cover}, and show that the separation scheme achieves the same SNR exponent as the transmitter informed bound. We also show that, when the channel coding rate is chosen appropriately, then the separation scheme yields the same distortion outage probability (not only the exponent) as the transmitter informed lower bound. This result, rather surprising a priori --separation is known to be largely suboptimal when the expected distortion SNR exponent is considered \cite{laneman2005,caire2007,deniz2007,deniz2008,tim2008}-- shows that separation can be optimal, when the distortion outage probability is used as a figure of merit for system design.

\section{System Model}
\label{sec:SystemModel}

\subsection{Channel Model}
\label{sec:ChannelModel}
We consider a MIMO block-fading channel model with $N$ fading blocks, $\nt$ transmit and $\nr$ receive antennas, and block  length $L$. The channel model is expressed as
\begin{equation}
\Ym_i=\sqrt{\frac{\snr}{\nt}}\Hm_i\Xm_i+\Zm_i,~~ i=1,\dotsc,N
\label{eq:eq3}
\end{equation}
where $\Hm_i\in\mathbb{C}^{\nr\times \nt}$, $\Xm_i\in\mathbb{C}^{\nt\times L}$, $\Ym_i\in\mathbb{C}^{\nr\times L}$, and $\Zm_i\in \CC^{\nr\times L}$ are the channel matrix, transmitted, received and AWGN signals corresponding to block $i$. We assume that the entries of $\Hm_i$ and $\Zm_i$ are independently circularly symmetric complex Gaussian with zero mean and unit variance $\sim\mathcal{N}_{\mathbb{C}}(0,1)$. We define the space-time codewords as $\Xm=\left[\Xm_1,\ldots,\Xm_N\right]$, and we assume they are normalized in energy, i.e., satisfying $\frac{1}{\nt NL}\trace\left(\EE\left[\Xm^H\Xm\right]\right)\leq 1$. The input and output alphabets are denoted by $\Xc^{\nt}$ and $\Yc^{\nr}$, respectively. We consider both random codes constructed using Gaussian and discrete channel inputs (PSK, QAM). For discrete channel inputs we define $m=\log_2|\Xc|$.

We define $\Hm=\text{diag}(\Hm_1,\ldots,\Hm_N)$, assumed to be known perfectly to the receiver. For simplicity we assume that the entries of $\Hm_i$ are i.i.d. $\sim\mathcal{N}_{\mathbb{C}}(0,1)$ (Rayleigh fading), so that $\frac{1}{\nt \nr}\trace\left(\EE\left[\Hm_i^H\Hm_i\right]\right)\leq 1$ and the average SNR per receive antenna is $\snr$. We assume that the transmitter knows the statistics of the channel, but not the channel realisation. Let
\begin{align}
&I_{\Hm}(\snr)
&=\frac{1}{N}\sum_{i=1}^N\mathbb{E}\left[\log_2\frac{P_{\Ym_i|\Xm_i,\Hm_i}(\Ym_i| \Xm_i,\Hm_i)}{P_{\Ym_i|\Hm_i}(\Ym_i|\Hm_i)}\biggl | \Hm_i\right]
\end{align}
denote the instantaneous mutual information of the channel, for a given channel realisation $\Hm$.

\subsection{Joint Source-Channel Coding}
\label{sec:JointSourceChannelCode}
We consider transmission of analogue sources over the MIMO block-fading channel described in Section \ref{sec:ChannelModel}. Consider a real continuous source that outputs $K$-length vector $\sv\in\mathbb{R}^K$. A $K$-to-$(\nt\times NL)$ joint source-channel encoder is a mapping $\phi:\mathbb{R}^K\rightarrow\mathbb{C}^{\nt\times NL}$ that maps blocks of $K$ source symbols $\sv\in\mathbb{R}^K$ onto length $NL$ space-time channel codewords $\Xm=\left[\Xm_1,\ldots,\Xm_N\right]$. At the receiver end, the corresponding source-channel decoder is a mapping $\varphi:\mathbb{C}^{\nr\times NL}\rightarrow\mathbb{R}^K$ that, for every channel realisation $\Hm$, maps the channel output $\Ym=\left[\Ym_1,\ldots,\Ym_N\right]$ into $\hat{\sv}\in \CC^K$, a reconstruction of the block of source symbols. In order to make explicit the dependencies on $\snr$ and $\Hm$, we denote the reconstructed block of symbols $\hat{\sv}(\snr,\Hm)$. The bandwidth ratio of the code is defined as
\begin{equation}
b\triangleq\frac{NL}{K} ~~\text{channel uses per source symbol}
\end{equation}
The bandwidth ratio can also be expressed as $b=W_c/W_s$, where $W_s,W_c$ are the source and channel bandwidths, respectively. The source rate is denoted by $\Rs=1/W_s$ and the channel rate is denoted by $\Rc=1/W_c$. The average quadratic distortion for a fixed $\Hm$ is given by
\begin{equation}
D(\Hm,\snr)=\frac{1}{K}\mathbb{E}\left[\left|\sv-\hat{\sv}(\snr,\Hm)\right|^2 \bigl|\Hm\right]
\end{equation}
where expectation is with respect to $\sv$, $\hat{\sv}$ and the channel noise, but depends on $\snr$ and on the channel realisation $\Hm$.

Mirroring results from channel coding for block-fading channels \cite{ozarow1994,biglieri1998,malkamaki1999cdb}, we define distortion outage probability
\begin{equation}
P_{\rm out}( \snr,\bar{D})\triangleq\Pr\left\{D(\Hm, \snr)>\bar{D}\right\}.
\label{eq:pout_def}
\end{equation}

We consider a family of joint source-channel coding schemes $\left\{\mathcal{C}_b\right\}$ of bandwidth ratio $b$. The distortion outage probability SNR exponent of the family is defined as
\begin{equation}
d_{\rm out}^*(b,\bar{D})=\sup_{\mathcal{C}_b}\lim_{\snr\rightarrow\infty}\frac{-\log P_{\rm out}( \snr,\bar{D})}{\log\snr}.
\end{equation}
in the forthcoming sections we study the distortion outage probability and the corresponding SNR exponents, and we compare them to those obtained using the expected distortion as a figure-of-merit.

\section{Informed Transmitter}
\label{sec:InformedTransmitter}

We now study the distortion outage probability exponent for the transmitter informed bound which assumes availability of channel state information at the transmitter (CSIT). Hence, the transmitter can choose the coding rate $\Rc(\Hm)$ equal to the instantaneous mutual information of the $N$-block MIMO fading channel, and the source rate $\Rs=\Rc(\Hm)b$. As shown in \cite{laneman2005,caire2007,deniz2007,deniz2008,tim2008}, this scheme is pointwise optimal for each $\Hm$, and its distortion outage probability is a lower bound on the minimum achievable distortion outage probability for any system of the original channel. In particular, the transmitter informed bound selects the channel coding rate $\Rc(\Hm)=I_{\Hm}(\snr)$.
Then, the instantaneous end-to-end distortion for a given channel realisation with a Gaussian source of unit variance and an informed transmitter is
\begin{equation}
D(\Hm,\snr) = 2^{-2bI_{\Hm}(\snr)}.
\label{eq:eq8}
\end{equation}
Substituting Equation \eqref{eq:eq8} into Equation \eqref{eq:pout_def}, we can write the transmitter informed bound on the distortion outage probability as
\begin{align}
P_{\rm out}(\snr,\bar{D})

&=\Pr\left\{I_{\Hm}(\snr)<-\frac{\log_2\bar{D}}{2b}\right\}
\label{eq:eq47}
\end{align}
which shows that the transmitter informed bound on the distortion outage probability can be written as the information outage probability $\Pr\{I_\Hm(\snr)<R\}$ \cite{ozarow1994,biglieri1998} evaluated at target rate $R=-\log_2(\bar{D})/2b$.

We next examine the behavior of the SNR exponent. Following closely the arguments in \cite{lizhong2003}, we have the following result.
\begin{theorem}
\label{thm:thm1}
The SNR exponents of the transmitter informed lower bound for any fixed bandwidth ratio $b>0$, any fixed target distortion level $0\leq \bar{D}\leq 1$ are given by
\begin{equation}
d_{\rm out}^{\rm tx}(b,\bar{D}) = \nt\nr N
\label{eq:tx_bound_exp_infor_gi}
\end{equation}
for Gaussian channel inputs, while for discrete channel inputs
\begin{equation}
d_{\rm out}^{\rm tx}(b,\bar{D})=
\nr\left(1+\left\lfloor N\left(\nt-\frac{\Rs(\bar{D})}{b\,m}\right)\right\rfloor\right)
\label{eq:tx_bound_exp_infor_di}
\end{equation}
where $\Rs(\bar{D})\triangleq-\frac{\log_2(\bar{D})}{2}$ is the rate-distortion region of the source evaluated at $\bar D$.
\end{theorem}
\begin{proof}
The transmitter informed lower bound on the distortion outage probability can be written as the information outage probability \cite{ozarow1994,biglieri1998} evaluated at $R=-\frac{\log_2(\bar{D})}{2b}$ (see Eq. \eqref{eq:eq47}). For Gaussian inputs, the SNR exponent of the information outage probability is $\nt\nr N$ for $R>0$. Since $0\leq \bar{D}\leq 1$,
$-\frac{\log_2(\bar{D})}{2b}$ is positive. Then,  the resulting SNR exponent is also $\nt\nr N$. For discrete inputs we have that the SNR exponent of the information outage probability is given by the Singleton bound $\nr\left(1+\left\lfloor N\left( \nt-\frac{R}{m}\right)\right\rfloor\right)$ for $0\leq R\leq m$ \cite{NguyenThesis2009,nguyen2009}. Then, the resulting SNR exponent with discrete inputs is given by $\nr\left(1+\left\lfloor N\left( \nt-\frac{R_{\rm s}(\bar{D})}{b\,m}\right)\right\rfloor\right)$ for $R_{\rm c}(\bar{D})=-\frac{\log_2(\bar{D})}{2b}\in\left[0,m\right]$.
\end{proof}

It is important to note that, since the  transmitter informed lower bound on the distortion outage probability has the exponents given by Theorem \ref{thm:thm1}, the SNR exponents of any coding scheme will be upper bounded by Eqs. \eqref{eq:tx_bound_exp_infor_gi} and \eqref{eq:tx_bound_exp_infor_di}.

\begin{remark}[Diversity-Multiplexing Tradeoff]
\label{rem:dmt}
The results of Theorem \ref{thm:thm1} for Gaussian channel inputs can be generalized to a family of joint source and channel codes whose rate increases with $\snr$. In particular, letting $\Rs = r_{\rm s}\log\snr$ and $\Rc = r_{\rm c}\log\snr$ with $b=\frac{ r_{\rm s}}{ r_{\rm c}}$ results in a diversity-multiplexing tradeoff $d_{\rm out}^{\rm tx}(b,\bar{D},r_{\rm c})$ given by the piecewise linear function joining the points ($r_{\rm c},d(r_{\rm c})$)
\begin{equation}
d(r_{\rm c}) = N (\nt-r_{\rm c})(\nr -r_{\rm c})
\end{equation}
achieving the result of Theorem \ref{thm:thm1} for $r_{\rm c}=0$ \cite{lizhong2003}.
\end{remark}

\begin{remark}[Comparison with Expected Distortion]
The SNR exponent of the informed transmitter lower bound with Gaussian channel inputs when the expected distortion is used as performance metric is given by \cite{deniz2008},
\begin{equation}
d_{\rm exp}^{\rm tx}=N\sum_{i=1}^{\min(\nr,\nt)}\min\left\{\frac{2b}{N},2i-1+\left|\nt-\nr\right|\right\}
\end{equation}
In \figref{fig:comp1}, we illustrate the SNR exponents of the transmitter informed bound for Gaussian channel inputs, from both distortion-outage and expected distortion perspectives. We observe that the distortion-outage exponent is always larger (for small bandwidth ratios) or equal to the expected distortion exponent. In \figref{fig:comp2}, we compare the distortion-outage SNR exponents of Gaussian random codes with that of discrete inputs. We observe that full diversity ($N\nr\nt$) is achieved when the we have a large bandwidth ratio for all inputs, and that a larger constellation size results in a larger support with full diversity. Note that Singleton bound is valid when $R_c\leq m$, hence we can obtain a bound on $b$, which is $b\geq-\frac{\log_2\bar{D}}{2m}$. For $b$ smaller than this threshold, the exponent is zero.

\begin{figure}
	\centering
		\includegraphics[width=1\columnwidth]{exponentGaussian}
	\caption{Distortion-outage vs. expected distortion SNR exponents in a $4\times 4$ MIMO block-fading channel with $N=2$, Gaussian inputs and $\bar{D}=0.05$. Solid lines correspond to the distortion-outage exponents, while dashed and dash dotted lines correspond to expected distortion exponents.
	}
	\label{fig:comp1}
 \vspace{-3mm}
 \end{figure}

 \begin{figure}
	\centering
		\includegraphics[width=1\columnwidth]{exponentGaussianqpsk16qam}
	\caption{Distortion-outage SNR exponents in a $4\times 4$ MIMO block-fading channel with $N=2$, different channel inputs and $\bar{D}=0.05$.

	}
	\label{fig:comp2}
 \vspace{-3mm}
 \end{figure}
\end{remark}

\section{Separation}
\label{sec:Seperation}

A source-channel coding separation scheme consists of the concatenation of a fixed-length block source encoder $\phi_{\rm s}:\RR^K\to \RR^{K}$, of rate $\Rs$ bits per source sample, with a space-time channel encoder $\phi_{\rm c}:\RR^{K}\to \CC^{\nt \times NL}$ of rate $\Rc$ bits per channel use. Source and space-time coding rates are related through the bandwidth ratio as $\Rs=\Rc b$. Let $D_{\rm s}(\Rs)$ denote the distortion-rate function of the source and $P_{\rm e}( \snr,\Hm)$ denote the error probability of the channel code for a particular $\snr$ and channel realisation $\Hm$. Following \cite{hochwald1997}, \cite[Ch. 7]{viterbi1979pdc}, we write the distortion achieved by the separation scheme for a fixed channel realisation $\Hm$ as
\begin{align}
&D_{\rm sep}(\Hm,\snr)\notag\\
 &= D(\snr\,|\,\Hm,\text{no channel error})\Pr\{\text{no channel error}\,|\,\Hm\}\nonumber\\
 &+ D(\snr\,|\,\text{channel error},\Hm)\Pr\{\text{channel error}\,|\,\Hm\}.
 \label{eq:sep_channel_form}

\end{align}
Following \cite{hochwald1997}\cite[Ch. 7]{viterbi1979pdc}, we can upperbound \eqref{eq:sep_channel_form} as
\begin{align}
D_{\rm sep}(\Hm,\snr)

&\leq D_{\rm s}(\Rs)+ d_0P_e( \snr,\Hm)
\label{eq:sep_dist_bound}
\end{align}
where $d_0$ is a bound to the mean MSE distortion and $d_0^2$ upperbounds its variance \cite[Sec. 7.5]{viterbi1979pdc}.

Since the channel realisation is unknown to the transmitter, the average distortion when there is no channel error is the distortion-rate function of the source code, and that it does not depend on $\Hm$ nor $\snr$.

Using Gallager's error exponents for channel coding, we further upperbound \eqref{eq:sep_dist_bound} as \cite{gallager1968ita}
\begin{equation}
D_{\rm sep}(\Hm,\snr)\leq D_{\rm s}(\Rs)+ d_0 2^{-nE_{\rm r}(\Rc,\Hm)}
\label{eq:d_exp}
\end{equation}
where $n=NL$ is the codeword length,
\begin{equation}
E_{\rm r}(\Rc,\Hm)=\sup_{0\leq\rho\leq1} E_0(\rho,\Hm)-\rho \Rc
\end{equation}
is the random coding error exponent, and $E_0(\rho,\Hm)$

is the Gallager function \cite{gallager1968ita}.
According to Gallager's noisy channel coding theorem, the random coding error exponent $E_{\rm r}(\Rc,\Hm)>0$ whenever $\Rc<I_{\Hm}(\snr)$ and $E_{\rm r}(\Rc,\Hm)=0$ when $\Rc\geq I_{\Hm}(\snr)$ \cite{gallager1968ita}.

It is also clear from \eqref{eq:d_exp}, that any separation scheme will achieve a distortion that is upperbounded by
\begin{equation}
D_{\rm sep}(\Hm,\snr)< D_{\rm s}(\Rs)+ d_0.
\label{eq:d_upbnd}
\end{equation}
Furthermore, for large $n$ we have that
\begin{equation}
\lim_{n\rightarrow\infty} P_{\rm e}(\snr,\Hm)=\begin{cases}1 &\Rc\geq I_{\Hm}(\snr) \\
0 &\Rc<I_{\Hm}(\snr)\end{cases}
\end{equation}
Therefore, in the limit for large $n$, we obtain that the distortion obtained with separation can be upper bounded as
\begin{equation}
D_{\rm sep}(\Hm,\snr)\leq D_{\rm s}(\Rs)+ d_0\openone\left\{I_{\Hm}(\snr)\leq \Rc\right\}.
\label{eq:eq78}
\end{equation}
where $\openone\{\Ec\}$ is the indicator function of the event $\Ec$.

The corresponding distortion outage probability is therefore simply bounded as
\begin{align}
P_{\rm out}^{\rm sep}&(\snr,\bar{D})=\Pr\left\{D_{\rm sep}(\Hm,\snr)>\bar{D}\right\}\nonumber\\
&\leq \Pr\Bigl\{D_{\rm s}(b\Rc)+ d_0\openone{\left\{I_{\Hm}(\Xm;\Ym)\leq \Rc\right\}}>\bar{D}\Bigr\}.
\label{eq:eq79}
\end{align}

We have the following result.
\begin{theorem}
\label{thm:thm2}
The distortion outage SNR exponent of a tandem separation scheme is given by
\begin{equation}
d_{\rm out}^{\rm sep}(\Rc,b,\bar{D})=N\nt\nr
\label{eq:eq_sep_exp_gi}
\end{equation}
for Gaussian channel inputs, while for discrete channel inputs
\begin{equation}
d_{\rm out}^{\rm sep}(\Rc,b,\bar{D})=
\nr\left(1+\left\lfloor N\left(\nt-\frac{\Rc}{m}\right)\right\rfloor\right).
\label{eq:eq_sep_exp_di}
\end{equation}
\end{theorem}
\begin{proof}
From \eqref{eq:eq79} we have
\begin{align}
P_{\rm out}^{\rm sep}(\snr,\bar{D})&\leq\Pr\Bigl\{D_{\rm s}(b\Rc)+d_0\openone\left\{I_{\Hm}(\snr)\leq \Rc\right\}>\bar{D}\Bigr\}\nonumber\\
&=\Pr\left\{\openone\left\{I_{\Hm}(\snr)\leq \Rc\right\}>\frac{\bar{D}-D_{\rm s}(b\Rc)}{d_0}\right\}.
\label{eq:eq80}
\end{align}
We note that the quantity $\frac{\bar{D}-D_{\rm s}(b\Rc)}{d_0}\in[0,1)$, since
\begin{align}
D_{\rm s}(b\Rc)\leq \bar{D}< D_{\rm s}(b\Rc)+ d_0.
\label{eq:eq_d_range}
\end{align}
Then, since the  indicator function takes only the values 0 or 1, we rewrite \eqref{eq:eq80} as
\begin{align}
P_{\rm out}^{\rm sep}(\snr,\bar{D})\leq&\Pr\left\{\openone\left\{I_{\Hm}(\snr)\leq \Rc\right\}>\frac{\bar{D}-D_{\rm s}(b\Rc)}{d_0}\right\}\nonumber\\
&=\Pr\left\{I_{\Hm}(\snr)\leq \Rc\right\},
\label{eq:eq_sep_outage}
\end{align}
which is exactly the information outage probability of the MIMO block fading channel when the channel coding rate equals to $\Rc$. Hence, the result follows from \cite{lizhong2003,NguyenThesis2009,nguyen2009}.
\end{proof}

\begin{remark}
From the above result and Remark \ref{rem:dmt} it is clear that the same diversity-multiplexing tradeoff will be achieved in the case of separation as well.
\end{remark}

\indent From \eqref{eq:eq_d_range}, we find that for each $\bar{D}$, there is a range of values for coding rate $\Rc$ that we can use to achieve the exponents in \eqref{eq:eq_sep_exp_gi} and \eqref{eq:eq_sep_exp_di}.
\begin{align}
\bar{D}-d_0<&2^{-2b\Rc}\leq\bar{D}
\end{align}
which readily implies $\frac{1}{b}\Rs(\bar{D})\leq \Rc<\frac{1}{b}\Rs(\bar{D}-d_0)$,
where $\frac{1}{b}\Rs(\bar{D})=-\frac{\log_2\bar{D}}{2b}$ for the real Gaussian source.
Equation \eqref{eq:eq_sep_outage} implies that when $\Rc=\frac{1}{b}\Rs(\bar{D})$, the distortion outage probability for separation can be upper bounded by
\begin{equation}
P_{\rm out}^{\rm sep}(\snr, \bar{D})\leq\Pr\left\{I_{\Hm}(\snr)\leq-\frac{\log_2\bar{D}}{2b}\right\}
\end{equation}
which coincides with the transmitter informed bound, and hence achieves the minimum possible distortion outage probability making separation with this particular choice of the channel coding rate distortion-outage optimal.

\begin{remark}[Comparison with Expected Distortion]
The expected distortion exponent for separation scheme for $N$ block-fading channel is given by \cite{caire2007}, for $\frac{1}{b}\in\left[\frac{2(j-1)}{d^*(j-1)},\frac{2j}{d^*(j)}\right)$
\begin{align}
d_{\rm exp}^{\rm sep}=N\frac{2b(jd^*(j-1)-(j-1)d^*(j))}{2b+d^*(j-1)-d^*(j)}
\end{align}
for $j=1,\ldots,\min(\nr,\nt)$. Where $d^*(k)$ is the optimal tradeoff curve given by $d^*(k)=N(\nr-k)(\nt-k)$.

\end{remark}
We observe from \figref{fig:comp1} that the distortion-outage exponent of separation exhibits a large gain over its expected distortion exponent counterpart for all bandwidth ratios. Remark that Theorem 2 shows that the distortion outage exponents are equal to those of the transmitter informed bound.

\begin{figure}
	\centering
		\includegraphics[width=1\columnwidth]{fig1}
	\caption{Transmitter informed lower bound and separation upper bound on the distortion outage probability with Gaussian source and channel inputs, $b=2$, $d_0=0.5$ in a $2\times2$ MIMO system. In this case, $\Rc^*=\frac{1}{b}\Rs(\bar{D})=1.08$.}
	\label{fig:mimo2by2_5}
 \vspace{-3mm}
 \end{figure}

\begin{figure}[htb]
 \centering
 \includegraphics[width=1\columnwidth]{paperISITdiscrete2}
 \caption{Transmitter informed lower bound and separation upper bound on the distortion outage probability with Gaussian source and BPSK channel inputs in a $2\times2$ system with $N=2$, $b=1.5$ and $d_0=0.5$. In this case, $\Rc^*=\frac{1}{b} \Rs(\bar D) = 1.353$.
 }
 \label{fig:mimo2by2_6}
 \vspace{-3mm}
\end{figure}

As examples, we show in \figref{fig:mimo2by2_5} the distortion outage probability of a $2\times2$ MIMO block-fading channel with i.i.d. Rayleigh fading with $N=1,2$ and $\bar D = 0.05$, for Gaussian source and channel inputs. As predicted by our results, the transmitter informed lower bound for distortion outage probability using informed transmitter always has a slope that equals to $N\nt\nr$ and it is independent of channel coding rate $\Rc$, target distortion $\bar{D}$ and bandwidth ratio $b$. The figure validates our results that the SNR exponent of distortion outage probability of separation scheme is equal to $N\nt\nr$. We also note that when the channel coding rate is chosen to be $\Rc=\frac{1}{b}R_{\mathcal{Q}}(\bar{D})$, the resulting distortion outage probability upperbound matches the transmitter informed bound. We also have shown in \figref{fig:mimo2by2_6} the distortion outage probability of a $2\times2$ MIMO block-fading channel with $N=2$ for $\bar{D}=0.06$ with BPSK. We observe an exponent of 4 when $\Rc^*=1.353$, and 2 when $\Rc=1.7$, as predicted by the  Singleton bound. We remark that for high $\Rc$, there is a significant loss in distortion outage probability (not only in gain, but also in exponent) due to the Singleton bound.

\section{Conclusions}
\label{sec:Conclusions}

We have revisited analogue source transmission over MIMO block-fading channels and proposed the distortion outage probability as a new performance metric for system design. We have argued that the distortion outage probability is the natural performance metric for delay-limited channels. We have derived the SNR exponents for both Gaussian and coded modulation inputs. We have shown that the distortion-outage SNR exponents are always larger than the expected distortion exponents, in both, transmitter informed bound and separation.  We have furthermore shown that source-channel separation can not only achieve the SNR exponent of the transmitter informed bound, but also the actual distortion outage probability, when the channel coding rate is chosen appropriately.

\newpage
\bibliographystyle{IEEE}
\bibliography{outagedistortion}

\end{document}

config: arxiv


TITLE: Proof about floor function: $\lfloor x\rfloor+\lfloor x + \frac{1}{n} \rfloor+\cdots + \lfloor x + \frac{(n-1)}{n} \rfloor = \lfloor nx \rfloor$
QUESTION [0 upvotes]: How can we prove that $$\left\lfloor x\right\rfloor + \left\lfloor x + \frac{1}{n} \right\rfloor + \left\lfloor x + \frac{2}{n} \right\rfloor + \cdots + \left \lfloor x + \frac{(n-1)}{n} \right\rfloor = \left\lfloor nx \right \rfloor$$ Where $\lfloor x\rfloor$ denotes greatest integer less than or equal to $x$. I was able to prove it when $n = 2$ or $3$. Please see this link.
But I can't prove it generally. I tried it using Principle of mathematical Induction but couldn't. Can someone prove it using some other way using properties of greatest integer function?

REPLY [3 votes]: Denoting the left hand side with $f(x)$, the right hand side with $g(x)$, show that $f(x+\frac1n)=f(x)+1$ and $g(x+\frac1n)=g(x)+1$. Also show $f(x)=0=g(x)$ for $0\le x<\frac 1n$. Then show by backward and forward induction on $k\in \Bbb Z$ that $f(x)=g(x)$ for all $x$ with $\frac kn\le x<\frac{k+1}n$.
Alternatively, write $x=k+y$ with $k=[x]\in\Bbb Z$ and $y\in[0,1)$, then write $ny=m+z$ with $m=[ny]\in\{0,1,\ldots,n-1\}$ and $z\in[0,1)$ and see what happens if you plug in $x=k+\frac1n m+\frac1nz$.


TITLE: Conditional probability on hypergeometric distribution
QUESTION [0 upvotes]: I'm working through a practice question in combinatorics and my first attempt at counting through a particular problem has yielded the wrong answer (I have a copy of the actual solutions, for reference) but I'm not quite sure why.
The problem I'm doing is

An urn contains $4$ red balls and $3$ white balls, and we draw three balls from the urn without replacement. What is the conditional probability that all $3$ balls are red, given that the third ball drawn is red?

The worked answers I have do it in the standard way by letting $Y$ be a $\{R,W\}^3$-valued random variable and calculating
$$\frac{\mathbb{P}\left(Y = (R,R,R)\right)}{\mathbb{P}\left(Y = (R,R,R)\right) + \mathbb{P}\left(Y = (R,W,R)\right) + \mathbb{P}\left(Y=(W,R,R)\right) + \mathbb{P}\left(Y=(W,W,R) \right)}=\frac{1}{5}.$$
Now, I suppose my question is: isn't this equivalent to just asking what the probability is of drawing two red balls out of the urn without replacement? The way I intuitively wanted to treat this question is by designating a random variable $X\sim \mathcal{H}_{2,(4,3)}$, where this denotes the hypergeometric distribution modelling drawing $2$ items out of a population with $4$ successes and $3$ failures, without replacement.
Then is this problem not the same as simply calculating $\mathbb{P}\left(X = (2,0)\right) = \frac{2}{7}$ as the probability that the first two balls we draw are red? Clearly, we get different answers, but I do not understand why. What part of my intuitions were off about trying to use the hypergeometric distribution this way? Is there a subtlety to conditional probability that I am not picking up on?

REPLY [1 votes]: A route that comes close to your second approach is the following:
Let there be $7$ spots (one for each ball) and on spot $3$ place a red ball. Then from the remaining balls ($3$ reds and $3$ whites) select $2$ balls that are placed on spot $1$ and $2$.
The probability that the selected balls are both red is:$$\frac{\binom32\binom30}{\binom62}=\frac15$$
Here identify the balls landing on the spots $1,2,3,$ as the $3$ balls that were drawn from the urn.