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\begin{document}

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\title[$\theta-\phi-$contraction on $\left(\alpha,\eta \right)-$complete rectangular $b-$
 metric spaces]{$\theta-\phi-$contraction on $\left(\alpha,\eta \right)-$complete rectangular $b-$
 metric spaces}

\author[Abdelkarim Kari, Mohamed Rossafi, El Miloudi Marhrani and Mohamed Aamri]{Abdelkarim Kari$^2$, Mohamed Rossafi$^{1*}$,  El Miloudi Marhrani$^2$ {and} Mohamed Aamri$^2$  }

\address{$^{1}$Department of Mathematics, University of Ibn Tofail, B.P. 133, Kenitra, Morocco}
\email{\textcolor[rgb]{0.00,0.00,0.84}{rossafimohamed@gmail.com}}
\address{$^{2}$Laboratory of Algebra, Analysis and Applications Faculty of Sciences Ben M’Sik, HassanII University, Casablanca, Morocco}
\email{\textcolor[rgb]{0.00,0.00,0.84}{abdkrimkariprofes@gmail.com}}
\email{\textcolor[rgb]{0.00,0.00,0.84}{fsb.marhrani@gmail.com}}
\email{\textcolor[rgb]{0.00,0.00,0.84}{aamrimohamed82@gmail.com}}

\keywords{fixed point, $\theta-\phi-$contraction, $\left(\alpha ,\eta \right)-$complete rectangular $b-$
 metric spaces.}

\date{
\newline \indent $^{*}$Corresponding author}

\begin{abstract}
In this paper, we present some fixed point results for generalized $\theta -\phi -$contraction in the framework of $\left( \alpha ,\eta \right)-$compete rectangular $b-$metric spaces. Further, we establish some fixed point theorems for this type of mappings defined on such spaces. The presented extend, generalize and improve existing results in the literature. Moreover, to support our main results we give an illustrative example.
\end{abstract} \maketitle

\section{Introduction}
The well known Banach contraction theory is one of the methods used, which states that if $ (X,d) $ is a complete metric space and  $\ T:X\rightarrow X$ is self-mapping with contraction, then $ T $ has a unique fixed point \cite{BAN}.
 \\In 2000, Branciari \cite{BRA} introduced the notion of generalized metric spaces, such as the triangle inequality is replaced by the inequality $d(x, y) \leq d(x,u) + d(u, v) + d(v, y)$ for all pairwise distinct points $x$, $y$, $u$, $v \in X$. Since then, several results have been proposed by many mathematicians on such spaces( see  \cite{BAR,Li,DAS,DSA,JS,KS}.\\
 The concept of metric space, as an ambient space in fixed point theory, has been generalized in several directions. Such as, $b- $ metric spaces \cite{CZ} and generalized metric spaces.\\
Combining conditions used for definitions of b-metric and generalized metric spaces. Roshan et al \cite{RO} announced the notion of rectangular b-metric space.\\
Hussain and et al \cite{HU} introduced the concept of $ \alpha-\eta- $complete rectangular $ b- $metric space and proved certain results of fixed point theory on such spaces.\\
In this paper, we provide some fixed point results for generalized $\theta -\phi -$contraction in the framework of $\left( \alpha ,\eta \right)-$compete rectangular $b-$metric spaces, also we give two examples to support our results.

\section{Preliminaries}
\begin{definition}
\cite{RO}. Let $X$ be a nonempty set, $s\geq 1$ be a given real number, and let
d: $X\times X\rightarrow \left[ 0,+\infty \right[ $
be a mapping such that for all $x,y$ $\in X$ and all distinct points $u,v\in
X,$ each  distinct from $x$ and $y$:\\
\item[1.] $d\left(x, y\right) =0,$ if only if $x=y;$\\
\item[2.] $d\left(x, y\right) =d\left(y, x\right);$\\
\item[3.] $d\left(x, y\right) \leq $ $s\left[d\left( x,u\right)+d\left(u, v\right) +d\left(v, y\right) \right] $ $\left( b-rectan gular\ inequality \right) .$\\

Then $\left(X, d\right) $ is called a $b-$rectangular metric space.
\end{definition}
\begin{lemma}
\cite{RO}. Let $\left(X, d\right) $ be a rectangular b-metric space.\\
\begin{itemize}
\item[(a)]  Suppose that sequences $\left( x_{n}\right) $ and $\left(y_{n}\right) $ in $X$ are such that $x_{n}\rightarrow x$ and $y_{n}\rightarrow y$ as $n\rightarrow \infty ,$ with $x\neq y,$ $x_{n}\neq x$ and $y_{n}\neq y$ for all $n\in \mathbb{N}.$ Then we have
\begin{equation*}
\frac{1}{s}d\left( x,y\right) \leq \lim_{n\rightarrow \infty }\inf d\left(x_{n},y_{n}\right) \leq \lim_{n\rightarrow \infty }\sup d\left(x_{n},y_{n}\right) \leq sd\left( x,y\right).\\
\end{equation*}
\item[(b)]  if $y\in X$ and $\left( x_{n}\right) $ is a Cauchy sequence in $X$ with $x_{n}\neq x_{m}$ for any $m,n\in \mathbb{N},$ $m\neq n,$converging to $x\neq y,$ then
\begin{equation*}
\frac{1}{s}d\left( x,y\right) \leq \lim_{n\rightarrow \infty }\inf d\left(x_{n},y\right) \leq \lim_{n\rightarrow \infty }\sup d\left( x_{n},y\right)\leq sd\left( x,y\right),
\end{equation*}
for all $x\in X.$\\
\end{itemize}
\end{lemma}
Zheng et al \cite{ZH} introduced a new type of contractions called $\theta -\phi $-contractions in metric spaces and proved a new fixed point theorems for such mapping.\\
\begin{definition}
\cite{JA} We denote by $\Theta _{\text{ }}$ the set of functions $\theta :\left] 0,\infty \right[ \rightarrow \left] 1,\infty \right[,$\\
satisfying the following conditions:\\
\item[$\left( \theta _{1}\right) $] $\theta$ is increasing ;\\
\item[$\left( \theta _{2}\right) $]  for each sequence $\left( x_{n}\right) $ $\in \left] 0,\infty \right[ ,\lim_{n\rightarrow \infty }\theta \left(x_{n}\right) =1\Leftrightarrow $ $\lim_{n\rightarrow \infty }x_{n} =0;$\\
\item[$\left( \theta _{3}\right) $] $\theta$ is continuous on $\left] 0,\infty \right[$.\\
\end{definition}
\begin{definition}
\cite{ZH} We denote by $\Phi $ the set of functions $\phi $ : $
\left[ 1,\infty \right[ $ $\rightarrow $ $\left[ 1,\infty \right[ $\\
satisfying the following conditions:\\
\item[$\left( \Phi _{1}\right) $] $\phi $ : $\left[ 1,\infty \right[ $ $\rightarrow \left[ 1,\infty \right[ $ is non-decreasing;\\
\item[$\left( \Phi _{2}\right) $] for each $\ t>1$, $lim_{n\rightarrow \infty}\phi ^{n}(t)=1$;\\
\item[$\left( \Phi _{3}\right) $] $\phi $ is continuous on $\left[ 1,\infty \right[ $.\\
\end{definition}
\begin{lemma}
\cite{ZH}. If $\phi \in \Phi $, then $\phi (1)=1$ and $
\varphi (t)<t$ for each $t>1$.\\
\end{lemma}
In 2014 Hussain et al. \cite{H} introduced a weaker notion than the concept of completeness and called it $ \alpha $-completeness for a metric spaces.
	\begin{definition}
\cite{H}. Let $T:X\rightarrow X$ and $\alpha,\eta $ :$X\times X\rightarrow \left[ 0,+\infty \right[.$ We say that $T$
is a triangular $\left( \alpha ,\eta \right) -$admissible mapping if
\end{definition}
$\left( T_{1}\right) $ $\alpha \left( x,y\right) \geq 1$ $\Rightarrow $ $\alpha \left( Tx,Ty\right) \geq 1,$ $x,y\in X$;\\
$\left( T_{2}\right) $ $\eta \left( x,y\right) \leq 1$ $\Rightarrow $ $\eta\left( Tx,Ty\right) \leq 1,$ $x,y\in X$;\\
\bigskip $\left( T_{3}\right) \left\{ \begin{array}{c}\alpha \left( x,y\right) \geq 1 \\
\alpha \left( y,z\right) \geq 1\end{array}\right. \Rightarrow \alpha \left( x,z\right) \geq 1$ for all $x,y,z\in X$;\\
$\left( T_{4}\right) \left\{ \begin{array}{c}\eta \left( x,y\right) \leq 1 \\ \eta \left( y,z\right) \leq 1\end{array}\right. \Rightarrow \eta \left( x,z\right) \leq 1$ for all $x,y,z\in X$.\\
\begin{definition}
\cite{H}. Let $\left( X,d\right) $ be a b-rectangular metric space and let $\alpha ,\eta $ :$X\times X\rightarrow \left[ 0,+\infty \right[
$ be two mappings. The space is said to be:\\
\item[(a)] T is $\alpha -$continuous mapping on $\left( X,d\right) ,$ if for given
point $x\in X$ and sequence $\left( x_{n}\right) $ in $X,$
$x_{n}\rightarrow x$ and $\alpha \left( x_{n},x_{n+1}\right) \geq 1$ for all $n\in \mathbb{N},$ imply that $Tx_{n}\rightarrow Tx$ $.$\\
\item[(b)] T is $\eta $ sub$-$continuous mapping on $\left( X,d\right) ,$ if for
given point $x\in X$ and\\ sequence $\left( x_{n}\right) $ in $X,$ $x_{n}\rightarrow x$ and $\eta \left( x_{n},x_{n+1}\right) \leq 1$ for all $ n\in \mathbb{N},$ imply that T$x_{n}\rightarrow Tx$$.$\\
\item[(c)] T is $\left( \alpha ,\eta \right) $ $-$continuous mapping on $\left(X,d\right) ,$ if for given point $x\in X$ and\\ sequence $\left( x_{n}\right) $ in $X,$ $x_{n}\rightarrow x$ and $\alpha \left( x_{n},x_{n+1}\right) \geq 1$ or $ \eta \left( x_{n},x_{n+1}\right) \leq 1$ for all $n\in \mathbb{N},$ imply that $Tx_{n}\rightarrow Tx$$.$
\end{definition}
The following definitions was given by Hussain et al \cite{HU}.
\begin{definition}
\cite{HU}. Let $d\left( X,d\right) $ be a rectangular b-metric space and let
 $\alpha ,\eta $ :$X\times X\rightarrow \left[ 0,+\infty \right[ $ be two mappings. The space $ X $ is said to be:\\
\item[(a)] $\alpha -$complete$,$ if every Cauchy sequence $\left( x_{n}\right) $ in $X$ with $\alpha \left( x_{n},x_{n+1}\right) \geq 1$ for all $n\in \mathbb{N},$ converges in $X.$\\
\item[(b)] $\eta -\sup -$complete$,$ if every Cauchy sequence $\left( x_{n}\right) $ in $X$ with $\eta \left( x_{n},x_{n+1}\right) \leq 1$ for all $n\in  \mathbb{N},$ converges in $X.$\\
\item[(c)] $\left( \alpha ,\eta \right)-$complete$,$ if every Cauchy sequence $ \left( x_{n}\right) $ in $X$ with $\alpha \left( x_{n},x_{n+1}\right) \geq 1 $ or $\eta \left( x_{n},x_{n+1}\right) \leq 1$ for all $n\in \mathbb{N},$ converges in $X.$\\
\end{definition}
\begin{definition}
\cite{HU}. Let $\left( X,d\right) $ be a rectangular b-metric space and let $\alpha ,\eta
$ :$X\times X\rightarrow \left[ 0,+\infty \right[ $ be two mappings. The
space $ X $ is said to be:\\
\item[(a)] $\left( X,d\right) $ is $\alpha $ -regular, if $x_{n}\rightarrow x$,
where $\alpha \left( x_{n},x_{n+1}\right) \geq 1$ for all $n\in \mathbb{N},$ implies $\alpha \left( x_{n},x\right) \geq 1$ for all $n\in \mathbb{N}.$\\
\item[(b)]$\left( X,d\right) $ is $\eta -$sub -regular, if $x_{n}\rightarrow x$,
where $\eta \left( x_{n},x_{n+1}\right) \leq 1$ for all $n\in \mathbb{N},$ implies $\eta \left( x_{n},x\right) \leq 1$ for all $n\in \mathbb{N}.$\\
\item[(b)] $\left( X,d\right) $ is $\left( \alpha ,\eta \right) $-regular, if $x_{n}\rightarrow x$,
where  $\alpha \left( x_{n},x_{n+1}\right) \geq 1$ or $
\eta \left( x_{n},x_{n+1}\right) \leq 1$ for all $n\in \mathbb{N},$ imply that $\alpha \left( x_{n},x\right) \geq 1$ or $\eta \left( x_{n},x\right) \leq 1$ for all $n\in \mathbb{N}.$\\
\end{definition}
\section{main results}
\begin{definition}
 Let $d\left( X,d\right) $ be a $ (\alpha, \eta) $- rectangular b-metric space with parameter $s>1$ and let $T$ be a self mapping on $\ X$. Suppose that $\alpha ,\eta
:X\times X\rightarrow \left[ 0,+\infty \right[ $ are two functions. We say that $T$ is an $\left( \alpha ,\eta \right) -\theta -\phi -$contraction, if for all $x,y\in X$ with $\left( \alpha \left( x,y\right) \geq 1\text{ or }\eta \left( x,y\right) \leq 1\right) $ and $d\left( Tx,Ty\right) >0$ we have
\begin{equation}
\theta \left( s^{2}.d\left( Tx,Ty\right) \right) \leq \phi\left[ \theta \left( \beta_{1}d\left( x,y\right) +\beta _{2}d\left( Tx,x\right) +\beta _{3}d\left(Ty,y\right) +\beta _{4}d\left( y,Tx\right) \right)\right]
\end{equation}
where $\theta \in \Theta ,\ \phi \in \Phi$, $\beta _{i}\geq0$ for $i\in\{1,2,3,4\},$ $\sum\limits_{\substack{i=0 }}^{i=4} {\beta_i}\leq1$  and  $\beta _{3}<\frac{1}{s}.$
\end{definition}
\begin{definition}
Let $(X,d)$ be a $ (\alpha, \eta) $-complete rectangular $b-$metric space and $T:X\rightarrow X$ \ be a mapping.\\
\item[(1)]$T$ is said to be a $\theta -\phi -$
Kannan- type contraction if there exists $\theta \in \Theta $ and $\phi \in \Phi $ with $\alpha \left( x,y\right) \geq 1$ or $\eta \left(x,y\right) \leq 1$ for any $x,y\in X,$ $Tx\neq Ty$, we have\\
\begin{equation*}
\theta \left[ s^{2}d\left( Tx,Ty\right) \right] \leq \phi \left[ \theta\left(\frac{\left( d\left( Tx,x\right) +d\left( Ty,y\right) \right) }{2s}\right) \right].\\
\end{equation*}
\item[(2)]$T$ is said to be a $\theta -\phi -$ Reich-type contraction \ if there exists $\theta \in \Theta $ and $\phi \in \Phi $ with $\alpha \left( x,y\right) \geq 1$ or $\eta \left( x,y\right) \leq 1$ for any $x,y\in X,$ $Tx\neq Ty$, we have\\
\begin{equation*}
\theta \left[s^{2} d\left( Tx,Ty\right) \right]\leq \phi \left[ \theta \left( \frac{d\left( x,y\right) +d\left( Tx,x\right)+d\left( Ty,y\right) }{3s}\right) \right].\\
\end{equation*}
\item[(3)]$T$ is said to be a Kannan type mapping, that is, if there exists $\alpha \in \left] 0,\frac{1}{2s}\right[ $ with $\alpha \left(x,y\right) \geq 1$ or $\eta \left( x,y\right) \leq 1$ for any $x,y\in X,$ $Tx\neq Ty$, we have\\
\begin{equation*}
s^{2}.d\left( Tx,Ty\right) \leq \alpha \left( d\left( Tx,x\right) +d\left(y,Ty\right) \right).\\
\end{equation*}
\item[(4)]$T$ is said to be a Reich type mapping, that is, if there
exists $\lambda \in \left] 0,\frac{1}{3s}\right[ $ with $\alpha \left(x,y\right) \geq 1$ or $\eta \left( x,y\right) \leq 1$ for any $x,y\in X,$ $Tx\neq Ty$, we have
\begin{equation*}
 s^{2}.d\left( Tx,Ty\right)\leq \lambda\left[  d\left( x,y\right) +d\left( Tx,x\right)+d\left( Ty,y\right)\right]  .\\
\end{equation*}
\end{definition}
\begin{theorem}
Let $\left( X,d\right) $ be a $\left( \alpha ,\eta \right) -$complete rectangular b-metric and let\\ $\alpha ,\eta :X\times X\rightarrow \left[0,+\infty \right[ $ be two functions. Let $T:X\times X\rightarrow X$ be a self mapping satisfying following
conditions:\\
\item[(i)] $T$ is a triangular $\left( \alpha ,\eta \right) -$admissible mapping;\\
\item[(ii)] $T$ is an $\left( \alpha ,\eta \right)-\theta -\phi -$contraction;\\
\item[(iii)] there exists $x_{0}\in X$ such that $\alpha \left(x_{0},Tx_{0}\right) \geq 1$ or $\eta \left( x_{0},Tx_{0}\right)\leq 1;$\\
\item[(iv)] $T$ is a $\left( \alpha ,\eta \right) -$continuous.\\

Then T has a fixed point. Moreover, $T$ has a unique fixed point when $\alpha\left( x,y\right) \geq 1$ or $\eta \left( x, y\right) \leq 1$ for all $x,y\in X.$\\
\end{theorem}
\begin{proof}
 Let $x_{0}\in X$ such that  $\alpha \left(x_{0},Tx_{0}\right) \geq 1$ or $\eta \left( x_{0},Tx_{0}\right) \leq 1.$\\
Define a sequence $\left\lbrace  x_{n}\right\rbrace  $ by $x_{n}=T^{n}x_{0}=Tx_{n-1}.$ Since T is a triangular $\left( \alpha,\eta \right) -$admissible mapping, then
 $\alpha \left( x_{0},x_{1}\right) =\alpha \left( x_{0},Tx_{0}\right) \geq 1\Rightarrow \alpha \left( Tx_{0},Tx_{1}\right) \geq 1=\alpha \left( x_{1},x_{2}\right) $ or $\eta \left( x_{0},x_{1}\right) =\eta \left( x_{0},Tx_{0}\right) \leq 1\Rightarrow \alpha \left( Tx_{0},Tx_{1}\right) \leq 1=\alpha \left( x_{1},x_{2}\right)$\\
Continuing this process we have $\alpha \left( x_{n-1},x_{n}\right) \geq 1$ or $\eta \left( x_{n-1},x_{n}\right) \leq 1,$
for all $n\in \mathbb{N}.$ By $\left( T_{3}\right) $ and $\left( T_{4}\right) ,$ one has.
\begin{equation}
\alpha \left( x_{m},x_{n}\right) \geq 1\text{ or }\eta \left(x_{m},x_{n}\right) \leq 1,\text{ \ \ \ \ \ }\forall m,n\in \mathbb{N},\text{ }m\neq n.
\end{equation}
Suppose that there exists $n_{0}$ $\in \mathbb{N}$ such that $x_{n_{0}}=Tx_{n_{0}}.$ Then $x_{n_{0}}$ is a fixed point of $T$
and the prove is finished. Hence, we assume that $x_{n}\neq Tx_{n}$,
i.e. $d\left( x_{n-1},x_{n}\right) >0$ for all $\ n\in \mathbb{N}.$ We have
\begin{equation}
x_{n}\neq x_{m},\ \forall m,n\in \mathbb{N},m\neq n.
\end{equation}
Indeed, suppose that $x_{n}=x_{m}$ for some $m=n+k>n,$ so we have\\
$$x_{n+1}=Tx_{n}=Tx_{m}=x_{m+1}.$$ \\
Denote $ d_{m}=d\left( x_{m},x_{m+1}\right).$
Then, $\left( 3.1\right) $ and lemma $\left( {2.5}\right) $ implies that
\begin{equation*}
\theta \left( d_{n}\right) =\theta \left( d_{m}\right)\leq\theta \left(s^{2}d_{m}\right) =\theta \left( s^{2}d\left( Tx_{m-1},Tx_{m}\right) \right)\leq \phi \left( \theta \left( \beta _{1}d_{m-1}+\beta _{2}d_{m-1}+\beta_{3}d_{m}\right) \right)
\end{equation*}
\begin{equation*}
<\theta \left( \beta _{1}d_{m-1}+\beta _{2}d_{m-1}+\beta _{3}d_{m}\right).\\
\end{equation*}
As $\theta $ is increasing, so
$$d_{n}=d_{m}<\beta _{1}d_{m-1}+\beta_{2}d_{m-1}+\beta _{3}d_{m}.$$
Hence
$$d_{m}<\frac{\beta _{1}+\beta _{2}}{1-\beta _{3}}d_{m-1}.$$\\
Since $$\beta _{1}+\beta _{2}+\beta _{3}\leq 1.$$\\ Thus $$d_{m}<d_{m-1}.$$ Continuing this process, we
can prove that $d_{n}=d_{m}<d_{n},$ which is a Contradiction. Thus, in  follows, we can assume that $\left( 3.2\right) $ and $\left( 3.3\right) $ hold.\\
We shall prove that
\begin{equation}
\lim\limits_{n \rightarrow +\infty}d\left( x_{n},x_{n+1}\right) =0\text{ \ and \ }\lim\limits_{n \rightarrow +\infty}d\left( n_{n},x_{n+2}\right) =0.
\end{equation}
Since $T$ $\ $is  $\left( \alpha ,\eta \right) -\theta -\phi -$ contraction, we get
\begin{align*}
   \theta \left( d_{n}\right)      &  = \theta \left( d\left( Tx_{n-1},Tx_{n}\right)\right) \\
                          & \leq \theta \left( s^{2}d\left( Tx_{n-1},Tx_{n}\right) \right)\\
                          & \leq \phi \left( \theta \left( \beta _{1}d_{n-1}+\beta _{2}d_{n-1}+\beta_{3}d_{n}\right) \right)\\
                          & <\theta \left( \beta _{1}d_{n-1}+\beta_{2}d_{n-1}+\beta _{3}d_{n}\right).\\
\end{align*}
Since $\theta $ is increasing, we deduce that $d_{n}<\beta _{1}d_{n-1}+\beta
_{2}d_{n-1}+\beta _{3}d_{n},$ thus
\begin{equation}
d_{n}<\frac{\beta _{1}+\beta _{2}}{1-\beta _{3}}d_{n-1}.
\end{equation}
Since $\frac{\beta _{1}+\beta _{2}}{1-\beta _{3}}\leq 1.$ Then\\
\begin{equation}
d_{n}<d_{n-1}.
\end{equation}
Therefore $ d(x{_n},x_{n+1}) $ is monotone strictly decreasing sequence of non negative real numbers. Consequently, there exists $ \alpha \geq 0 $, such that such that
$$ \lim\limits_{n \rightarrow +\infty}d\left( x_{n},x_{n+1}\right) =\alpha .$$\\
which again by $\left( 3.1\right), \left( 3.6\right) $ and property of $\left( \theta \right),$ we have
\begin{equation*}
1<\theta \left( d_{n}\right) \leq \phi \left( \theta \left( \beta_{1}d_{n-1}+\beta _{2}d_{n-1}+\beta _{3}d_{n}\right) \right)
\end{equation*}
\begin{equation}
\leq\phi \left( \theta \left( d_{n-1}\right) \right) \leq \phi ^{2}\left( \theta\left( d_{n-2}\right) \right) \leq...\leq\phi ^{n}\left( \theta \left(d_{0}\right) \right) =\phi ^{n}\left( \theta \left( d\left(x_{0},x_{1}\right) \right) \right).
\end{equation}
By taking the limit as $n$ $\rightarrow $ $\infty $ in $\left( 3.7\right) $ and using $\left( \Phi _{2}\right) $ , we have
\begin{equation*}
1\leq \lim\limits_{n \rightarrow +\infty}\theta \left( d\left( x_{n},x_{n+1}\right)\right) \leq \phi ^{n}\left( \theta \left( d\left( x_{0},x_{1}\right)\right) \right).\\
\end{equation*}
Then $\lim\limits_{n \rightarrow +\infty}\theta \left( d\left( x_{n},x_{n+1}\right)
\right) =1$,
by $ \Theta_2,$ we obtain
\begin{equation}
\lim\limits_{n \rightarrow +\infty}d\left( x_{n},x_{n+1}\right) =0.
\end{equation}
On the other hand,\\
$\theta \left( s^{2}d\left( x_{n},x_{n+2}\right) \right) \leq \phi \left[ \theta \left( \beta _{1}d\left( x_{n-1},x_{n+1}\right) +\beta _{2}d\left(x_{n-1},x_{n}\right) +\beta _{3}d\left( x_{n+1},x_{n+2}\right) +\beta_{4}d\left( x_{n+1},x_{n}\right) \right) \right]  $ \\
$\leq \phi \left[  \theta \left( \begin{array}{c}
s\beta _{1}d\left( x_{n-1},x_{n+2}\right) +s\beta _{1}d\left(x_{n+2},x_{n}\right) +s\beta _{1}d\left( x_{n},x_{n+1}\right) +\beta_{2}d\left( x_{n-1},x_{n}\right)  \\
+\beta _{3}d\left( x_{n+1},x_{n+2}\right) +\beta _{4}d\left(x_{n+1},x_{n}\right) \end{array}\right)\right]  $\\
 $\leq \phi \left[  \theta \left( \begin{array}{c}s^{2}\beta _{1}d\left( x_{n-1},x_{n}\right) +s^{2}\beta _{1}d\left(x_{n},x_{n+1}\right) +s^{2}\beta _{1}d\left( x_{n+1},x_{n+2}\right) +s\beta_{1}d\left( x_{n+2},x_{n}\right)  \\
+s\beta _{1}d\left( x_{n},x_{n+1}\right) +\beta _{2}d\left(x_{n-1},x_{n}\right) +\beta _{3}d\left( x_{n+1},x_{n+2}\right) +\beta
_{4}d\left( x_{n+1},x_{n}\right) \end{array}\right) \right].$\\
By $ \theta_1 $ and Lemma $ 2.5$, we obtain
\begin{equation*}
s^{2}d\left( x_{n},x_{n+2}\right) <s^{2}\beta _{1}d\left(x_{n-1},x_{n}\right) +s^{2}\beta _{1}d\left( x_{n},x_{n+1}\right)+s^{2}\beta _{1}d\left( x_{n+1},x_{n+2}\right) +s\beta _{1}d\left(x_{n+2},x_{n}\right)
\end{equation*}
\begin{equation*}
+s\beta _{1}d\left( x_{n},x_{n+1}\right) +\beta _{2}d\left(x_{n-1},x_{n}\right) +\beta _{3}d\left( x_{n+1},x_{n+2}\right) +\beta
_{4}d\left( x_{n+1},x_{n}\right),
\end{equation*}
Therefore,
\begin{equation*}
\left( s^{2}-s\beta _{1}\right) d\left( x_{n},x_{n+2}\right) <s^{2}\beta_{1}d\left( x_{n-1},x_{n}\right) +s^{2}\beta _{1}d\left(
x_{n},x_{n+1}\right) +s\beta _{1}d\left( x_{n+1},x_{n+2}\right)
\end{equation*}
\begin{equation}
+s\beta _{1}d\left( x_{n},x_{n+1}\right) +\beta _{2}d\left(x_{n-1},x_{n}\right) +\beta _{3}d\left( x_{n+1},x_{n+2}\right) +\beta
_{4}d\left( x_{n+1},x_{n}\right).
\end{equation}
Taking the limit as $\ n$ $\rightarrow \infty $ in $\left( 3.9\right) $ and using $\left( 3.8\right) ,$ since $ s^{2}-s\beta _{1}>0$ , we have
\begin{equation*}
\lim\limits_{n \rightarrow +\infty}d\left( x_{n},x_{n+2}\right) =0.\\
\end{equation*}
Hence $\left( 3.4\right) $ is proved.\\
Next, we show that $\left\lbrace  x_{n}\right\rbrace  $ is an Cauchy sequence in\ $X,$
if otherwise there exists an $\varepsilon $ $>0$ for which we can find sequences of positive integers \ $\left\lbrace  x_{n_{\left( k\right) }}\right\rbrace _k $ and $\left\lbrace x_{m_{\left( k\right) }}\right\rbrace _k $ of  $\left( x_{n}\right) $ such that,for all positive integers
$k$, $n_{\left( k\right) }>m_{\left( k\right) }>k,$
\begin{equation}
d\left(x_{m_{\left( k\right) }},x_{n_{\left( k\right) }}\right) \geq \varepsilon
\end{equation}
and
\begin{equation}
\text{ }d\left( x_{m_{\left( k\right) }},x_{n_{\left( k\right) -1}}\right)
<\varepsilon
\end{equation}
From $\left( 3.10\right) $ and using the rectangular inequality, we get
\begin{equation}
\varepsilon \leq d\left( x_{m_{\left( k\right) }},x_{n_{\left( k\right)}}\right) \leq sd\left( x_{m_{\left( k\right) }},x_{m_{\left( k\right)+1}}\right) +sd\left( x_{m_{\left( k\right) +1}},x_{n_{\left( k\right)+1}}\right) +sd\left( x_{n_{\left( k\right) +1}},x_{n_{\left( k\right)}}\right).\\
\end{equation}
Taking the upper limit as $k$ $\rightarrow \infty $ in $ (3.12) $ and using $\left(3.4\right),$ we get
\begin{equation}
\frac{\varepsilon }{s}\lim\limits_{n \rightarrow +\infty}\sup d\left(x_{m_{\left( k\right) +1}},x_{n_{\left( k\right) +1}}\right).
\end{equation}
 Moreover,
\begin{equation*}
d\left( x_{m_{\left( k\right) }},x_{n_{\left( k\right) }}\right) \leq sd\left( x_{m_{\left( k\right) }},x_{n_{\left( k\right) -1}}\right)+sd\left( x_{n_{\left( k\right) -1}},x_{n_{\left( k\right) +1}}\right)+sd\left( x_{n_{\left( k\right)+1}},x_{n_{\left( k\right) }}\right).\\
\end{equation*}
Then, from $\left( 3.11\right) $ and $\left( 3.8\right),$ we get
\begin{equation}
\lim\limits_{n \rightarrow +\infty}\sup d\left( x_{m_{\left( k\right) }},x_{n_{\left(
k\right) }}\right) \leq s\varepsilon.
\end{equation}
On the other hand we have
\begin{equation}
d\left( x_{n_{\left( k\right) }},x_{m_{\left( k\right) +1}}\right) \leq sd\left( x_{n_{\left( k\right) }},x_{n_{\left( k\right) -1}}\right)+sd\left( x_{n_{\left( k\right) -1}},x_{m_{\left( k\right) }}\right)+sd\left( x_{x_{m_{\left( k\right) }}},x_{m_{\left( k\right) +1}}\right).
\end{equation}
Then , from $\left( 3.11\right) $ and $\left( 3.8\right) $, we get
\begin{equation}
\lim\limits_{n \rightarrow +\infty}\sup d\left( x_{n_{\left( k\right) }},x_{m_{\left(
k\right) +1}}\right) \leq s\varepsilon.
\end{equation}
Applying $\left( 3.1\right)$ with $ x=  x_{m_{\left( k\right) }}$ and $ y= x_{n_{\left( k\right) }} $, we have
\begin{equation*}
\theta \left( s^{2}d\left( x_{m_{\left( k\right) +1}},x_{n_{\left( k\right)+1}}\right) \right) =\theta \left( s^{2}d\left( Tx_{m_{\left( k\right)}},Tx_{n_{\left( k\right) }}\right) \right)
\end{equation*}
\begin{equation}
\leq \phi \left( \theta \left( \begin{array}{c}\beta _{1}d\left( x_{m_{\left( k\right) }},x_{n_{\left( k\right) }}\right)
+\beta _{2}d\left( x_{m_{\left( k\right) }},Tx_{m_{\left( k\right) }}\right) \\
+\beta _{3}d\left( x_{n_{\left( k\right) }},Tx_{n_{\left( k\right) }}\right)+\beta _{4}d\left( x_{n_{\left( k\right) }},Tx_{m_{\left( k\right) }}\right) \end{array}\right) \right).
\end{equation}
Now taking the upper limit as $k\rightarrow $ $\infty $ in $\left(3.17\right) $ and using $\left( \theta _{1}\right) ,\left( \theta_{3}\right) ,\left( \phi _{3}\right) $,  $ \left(3.13\right), $ $\left( 3.8\right) ,\left(3.14\right) ,\left( 3.16\right) $ and Lemma $\left( {2.5}\right) $ we have
\begin{equation}
\theta \left( s^{2}.\frac{\varepsilon }{s}\right) =\theta \left( \varepsilon .s\right) \leq \theta \left( s^{2}\lim\limits_{k \rightarrow +\infty}\sup d\left( x_{m_{\left( k\right) +1}},x_{n_{\left( k\right) +1}}\right) \right)
\end{equation}
\begin{equation*}
\leq \phi \left( \theta \left( \begin{array}{c}\beta _{1}\lim\limits_{k \rightarrow +\infty}\sup d\left( x_{m_{\left( k\right)
}},x_{n_{\left( k\right) }}\right) +\beta _{2}\lim\limits_{k \rightarrow +\infty}\sup
d\left( x_{m_{\left( k\right) }},Tx_{m_{\left( k\right) }}\right)  \\
+\beta _{3}\lim\limits_{k \rightarrow +\infty}\sup d\left( x_{n_{\left( k\right)
}},Tx_{n_{\left( k\right) }}\right) +\beta _{4}\lim\limits_{k \rightarrow +\infty}\sup
d\left( x_{n_{\left( k\right) }},Tx_{m_{\left( k\right) }}\right) \end{array}\right) \right)
\end{equation*}
\begin{equation}
\leq\phi \left( \theta \left( \beta _{1}s\varepsilon +\beta_{4}s\varepsilon \right) \right)  <\theta \left(s\varepsilon \left( \beta _{1}+\beta _{4}\right) \right).
\end{equation}
Therefore
$\varepsilon .s<s\varepsilon \left( \beta _{1}+\beta _{4}\right)$ implies $s<\beta _{1}+\beta _{4}.$ Which is a contradiction.\\ Consequently, $\left\lbrace  x_{n}\right\rbrace  $ is a Cauchy sequence in $ \alpha-\eta- $complete rectangular b- metric space $\left( X,d\right).$
Since $\alpha \left( x_{n-1},x_{n}\right) \geq 1$ or $\eta \left( x_{n-1},x_{n}\right)\leq 1,$ for all $n\in \mathbb{N}$.\\
 This implies that the sequence $\left\lbrace  x_{n}\right\rbrace  $ is converges to some $z\in X$.
Suppose that\\ $z\neq Tz$. Then, we have all the assumption of Lemma 2.2 and $T$ is $\left( \alpha ,\eta \right)-$continuous, then  $Tx_{n}\rightarrow Tz  $ as $ n\rightarrow \infty$.
Therefore,
\begin{equation*}
\frac{1}{s}d\left( z,Tz\right) \leq\lim\limits_{n \rightarrow +\infty}\sup d\left(
x_{n},Tx_{n}\right) =0.
\end{equation*}
Hence we have $d\left( z,Tz\right) =0$ and so $Tz=z.$ Thus $z$ is a fixed point of $T.$\\
Uniqueness.\\Let $z,u\in $ Fix $\left( T\right) $ where $z\neq u$ and $\alpha \left(z,u\right) \geq 1$ or $\eta \left( z,u\right) \leq 1.$\\
Applying $\left(3.1\right) $ with $ x=z $ and $ y=u $ we have
\begin{align*}
\theta \left( d\left( z,u\right) \right)  &=\theta \left( d\left(Tz,Tu\right) \right)\\
&\leq \theta \left( s^{2}d\left( Tz,Tu\right) \right)\\
&\leq \phi \left( \theta \left( \beta _{1}d\left( z,u\right) +\beta_{2}d\left( z,Tz\right) +\beta _{3}d\left( u,Tu\right) +\beta _{4}d\left(Tz,u\right) \right) \right)\\
&\leq \phi \left( \theta \left( \beta _{1}d\left( z,u\right)  +\beta _{4}d\left(Tz,u\right) \right) \right)\\
& \leq \phi \left( \theta \left( d\left( z,u\right) \right) \right).
\end{align*}
Since $ \theta $ is increasing, Therefore
\begin{equation*}
d\left( z,u\right) <d\left( z,u\right).
\end{equation*}
Which is a contradiction. Hence, $z=u$ and $T$ has a unique fixed point.
\end{proof}
 Recall that a self-mapping $ T $ is said to have the property $ P $, if $ Fix(T)=Fix( T^n) $ for every $ n\in \mathbb{N} $
\begin{theorem}
 \bigskip Let $\alpha ,\eta $ : $X\times X\rightarrow \mathbb{R}^{+}$ be two function and let $\left( X,d\right) $ be an $\left( \alpha,\eta \right) -$complete rectangular b-metric space. Let $T:X\rightarrow X$ be a mapping satisfying the following conditions:\\
\item[(i)] $T$ is a triangular $\left( \alpha ,\eta \right) -$ admissible mapping;\\
\item[(ii)] $T$ is an $\left( \alpha ,\eta \right) -\theta -\phi -$contraction;\\
\item[(iii)] $\alpha \left( z,Tz\right) \geq 1$ or $\eta \left(z,Tz\right) \leq 1,$ for all $z\in $ Fix $\left( T\right) .$\\
Then $\ T$ has the property $P.$\\
\end{theorem}
\begin{proof}
Let $z$ $\in $ Fix $\left( T^{n}\right) $ for some fixed $n>1$. As $\alpha \left( z,Tz\right) \geq 1$ or $\eta \left( z,Tz\right) \leq 1$ \\and  $T$ is a triangular $\left( \alpha ,\eta \right) $-admissible mapping, then\\
\begin{equation*}
\alpha \left( Tz,T^{2}z\right) \geq1 \text{ or }  \eta \left( T^{2}z,Tz\right)\leq 1.
\end{equation*}
Continuing this process, we have
\begin{equation*}
\alpha \left( T^{n}z,T^{n+1}z\right) \geq 1\text{ or }\eta \left(T^{n}z,T^{n+1}z\right) \leq 1,
\end{equation*}
for all $n\in \mathbb{N}$. By $\left( T_3\right)$ and $\left( T_4\right)$  we get
\begin{equation*}
\alpha \left( T^{m}z,T^{n}z\right) \geq 1\text{ or }\eta \left(
T^{m}z,T^{n}z\right) \leq 1,\text{ \ }\forall \text{ }m,n\in \mathbb{N},\text{ }n\neq m.
\end{equation*}
Assume that $z\notin $ Fix $\left( T\right) ,$i.e. $d\left( z,Tz\right) >0.$\\
Applying $\left(3.1\right) $ with $ x=T^{n-1}z $ and $ y=z $, we get
\begin{equation*}
d\left( z,Tz\right) =d\left( T^{n}z,Tz\right) =d\left( TT^{n-1}z,Tz\right)\leq s^{2}d\left(TT^{n-1}z,Tz\right),
\end{equation*}
which implies that
\begin{align*}
\theta(d\left(TT^{n-1}z,Tz\right))&\leq \phi\left[ \theta\left( \beta _{1}d\left( T^{n-1}z,z\right) +\beta _{2}d\left(T^{n-1}z,T^{n}z\right) +\beta _{3}d\left( z,Tz\right) +\beta _{4}d\left(
z,T^{n}z\right)\right) \right]\\
&< \theta\left[ \beta _{1}d\left( T^{n-1}z,z\right) +\beta _{2}d\left(T^{n-1}z,T^{n}z\right) +\beta _{3}d\left( z,Tz\right) +\beta _{4}d\left(z,T^{n}z\right)\right]\\
&=\theta\left[ \beta _{1}d\left( T^{n-1}z,T^{n}z\right) +\beta _{2}d\left(T^{n-1}z,T^{n}z\right) +\beta _{3}d\left( z,Tz\right)\right].
\end{align*}
Since $ \theta $ is increasing, Therefore,
\begin{equation*}
 d\left( z,Tz\right) <\frac{\beta _{1}+\beta _{2}}{1-\beta _{3}}d\left( T^{n-1}z,T^{n}z\right) \leq d\left(T^{n-1}z,T^{n}z\right),
\end{equation*}
which is a contradiction as $d\left( T^{n-1}z,T^{n}z\right) \rightarrow 0$ and $d\left( z,Tz\right) >0.$
\end{proof}
Assuming the following conditions, we prove that Theorem $ 3.3 $ still hold for $ T $ not necessarily continuous:
\begin{theorem}
Let $\alpha,\eta :$ $X\times X\rightarrow \mathbb{R} ^{+}$ be two functions and let $d\left( X,d\right) $ be an $\left( \alpha,\eta \right) -$complete rectangular b-metric space.\\
Let $T:X\rightarrow X$ be a mapping satisfying the following assertions:\\
\item[(i)]$T$ is triangular $\left( \alpha ,\eta \right) -$admissible;\\
\item[(ii)] $T$ is $\left( \alpha ,\eta \right) -\theta
-\phi -$contraction;\\
\item[(iii)] there exists $x_{0}\in X$ such that $\alpha \left(x_{0},Tx_{0}\right) \geq 1$ or $\eta \left( x_{0},Tx_{0}\right) \leq 1;$\\
\item[(iv)] $\left( X,d\right) $ is an $\left( \alpha ,\eta \right) $-regular rectangular b-metric space.\\

Then $T$ has a fixed point. Moreover, $T$ has a unique fixed point whenever $\\
\alpha \left( z,u\right) \geq 1$ or $\eta \left( z,u\right) \leq 1$ for all $z,u\in Fix\left( T\right).$
\end{theorem}
\begin{proof}
Let x$_{0}\in X$ such that $\alpha \left(x_{0},Tx_{0}\right) \geq 1$ or $\eta \left( x_{0},Tx_{0}\right) \leq 1$.
 Similar to the proof of Theorem 3.3, we can conclude that
\begin{equation*}
\left( \alpha \left( x_{n},x_{n+1}\right) \geq 1\text{ or }\eta \left(x_{n},x_{n+1}\right) \leq 1\right),\text{ and }x_{n}\rightarrow z\text{ as }n\rightarrow \infty,
\end{equation*}
where $x_{n+1}=Tx_{n}.$ \\
From $\left( iv\right)$ $ \alpha \left( x_{n+1},z\right) \geq 1\text{ or }\eta \left( x_{n+1},z\right)\leq 1 $, hold for $n\in \mathbb{N}.$\\
Suppose that $Tz=x_{n_{0+1}}=Tx_{n_{0}}$ for some $n_{0}\in \mathbb{N}.$ From the  Theorem 3.3 we know that the members of the sequence $\left\lbrace x_{n}\right\rbrace  $ are distinct. Hence, we have $Tz\neq Tx_{n},$ i.e. $d\left(Tz,Tx_{n}\right) >0$ for all $n>n_{0}.$ Thus, we can apply $\left(3.1\right) $, to $x_{n}$ and $z$ for all $n>n_{0}$ to get
\begin{align*}
\theta \left( d\left( Tx_{n},Tz\right) \right) &\leq \theta \left(s^{2}d\left( Tx_{n},Tz\right) \right)\\
& \leq \phi \left( \theta \left( \beta_{1}d\left( x_{n},z\right) +\beta _{2}d\left( x_{n},Tx_{n}\right) +\beta_{3}d\left( z,Tz\right) +\beta _{4}d\left( z,Tx_{n}\right) \right) \right).
\end{align*}
By Lemma $\left( {2.5}\right) $ and $\left( \theta _{1}\right) $, we obtain
\begin{equation}
d\left( Tx_{n},Tz\right) <\left( \beta _{1}d\left( x_{n},z\right) +\beta_{2}d\left( x_{n},Tx_{n}\right) +\beta _{3}d\left( z,Tz\right) +\beta_{4}d\left( z,Tx_{n}\right) \right).
\end{equation}
By taking the limit as $n$ $\rightarrow $ $\infty $ in $\left( 3.20\right) $, we have
$$ \lim_{n\rightarrow \infty }\sup d\left(Tx_{n},Tz\right) \leq \beta _{3}d\left( z,Tz\right). $$
Assume that $z\neq Tz.$ Then, from lemma 2.2,
\begin{equation*}
\frac{1}{s}d\left( z,Tz\right) \leq \lim_{n\rightarrow \infty }\sup d\left(
Tx_{n},Tz\right) \leq \beta _{3}d\left( z,Tz\right).
\end{equation*}
by assumption $\beta _{3}<\frac{1}{s},$ we have $d\left( z,Tz\right) =0
$ and so $z=Tz.$ Thus, $z$ is a fixed point of $T.$\\
The proof of the uniqueness is similarly to that of Theorem 3.3.\\
\end{proof}
Above Theorems. If we take $\phi \left( t\right) =t^{k},$ for some fixed $k\in
\left] 0,1\right[ ,$ \\
where $\beta _{1}=1\	and \ \beta _{2}=\beta _{3}=\beta _{4}=0.$
We obtain the following extension of  Jamshaid et al, result $\left(\text{Theorem 2.2}\right)$  \cite{JA}  of $\left( \alpha ,\eta \right) -$ complete rectangular $b$-metric space.
\begin{corollary}
Let $\alpha ,\eta :X\times X\rightarrow \left[ 0,+\infty \right[ $ be two functions, $d\left( X,d\right) $ be an $\left( \alpha ,\eta \right) -$ complete rectangular $b$-metric space and let $T:X\rightarrow X$ be self-mapping. Suppose for all $x,y\in X$ with $\alpha \left( x,y\right) \geq 1$ or $\eta \left( x,y\right) \leq 1$ and $d\left( Tx,Ty\right) >0$\\
 we have \\
\begin{equation*}
\theta\left[ s^{2} d\left( Tx,Ty\right) \leq \left[ \theta \left( d\left( x,y\right)\right]
\right) \right] ^{k},
\end{equation*}
where $\ \theta \in \Theta $ and $k\in \left] 0,1\right[$. If the mapping $ T $ satisfying the following assertions:\\
point, if \\
\item[(i)] $T$ is a triangular $\left( \alpha ,\eta \right)-$admissible mapping;\\
\item[(ii)] there exists $x_{0}\in X$ such that $\alpha \left(x_{0},Tx_{0}\right) \geq 1$ or $\eta \left( x_{0},Tx_{0}\right) \leq 1;$\\
\item[(iii)] $T$ is $\left( \alpha,\eta \right) $ continuous; or \\
\item[(iv)] is an $\left( \alpha ,\eta \right) -$regular rectangular $b-$metric space.\\
Then $T$ has a fixed point. Moreover, $T$ has a unique fixed point whenever $\\
\alpha \left( z,u\right) \geq 1$ or $\eta \left( z,u\right) \leq 1$ for all $z,u\in Fix\left( T\right).$
\end{corollary}
\begin{proof}
Let $\phi \left( t\right)$ $=t^{k }$, we prove that  $T$ is an $\left( \alpha ,\eta \right)
-\theta -\phi -$contraction, Hence T satisfies in assumption of Theorem $ 3.3 $ or Theorem $ 3.5 $ and is the unique fixed point of $
T.$\\
\end{proof}
It follows from Theorem 3.3, we obtain the following fixed point theorems
for $\theta -\phi $- Kannan-type contraction and $\theta -\phi $ Reich-type contraction.\\
\begin{theorem}
Let $\left( X,d\right) $ be a $\left( \alpha ,\eta \right) -$complete rectangular $b$-metric space and let $\alpha ,\eta :X\times X\rightarrow \left[ 0,+\infty \right[ $ be two functions. Let $T:X\times X\rightarrow X$ be a self mapping satisfying following conditions:\\
\item[(i)]  $T$ is a triangular $\left( \alpha ,\eta \right) -$ admissible mapping;\\
\item[(ii)] $T$ is a $\left( \alpha ,\eta \right) -\theta -\phi -$Kannan-type contraction;\\
\item[(iii)] $T$ there exists $x_{0}\in X$ such that $\alpha \left(x_{0},Tx_{0}\right) \geq 1$ or $\eta \left( x_{0},Tx_{0}\right) \leq 1;$\\
\item[(iv)] $T$is a $\left( \alpha ,\eta \right) -$continuous.\\
Then T has a fixed point. Moreover, $T$ has a unique fixed point when $ \alpha \left( x,y\right) \geq 1$ or $\eta \left( x,y\right) \leq 1$ for all $x,y\in X.$\\
\end{theorem}
\begin{proof}
If $T$ is a $\left( \alpha ,\eta \right) -\theta -\phi -$ Kannan-type contraction, thus
there exists $\theta \in \Theta $ and $\phi \in \Phi $ with $\alpha \left( x,y\right) \geq 1$ or $\eta \left(x,y\right) \leq 1$ for any $x,y\in X,$ $Tx\neq Ty$, we have\\
\begin{equation*}
\theta \left[ s^{2}d\left( Tx,Ty\right) \right] \leq \phi \left[ \theta\left(\frac{\left( d\left( Tx,x\right) +d\left( Ty,y\right) \right) }{2s}\right) \right].\\
\end{equation*}
Therefore,
\begin{equation*}
\theta \left[ s^{2}d\left( Tx,Ty\right) \right] \leq \phi \left[ \theta\left( ( \beta _{1}d\left( x,y\right) + \beta _{2}d\left( Tx,x\right)+\beta _{3}d\left( Ty,y\right)+ \beta _{4}d\left( Tx,y\right)\right) \right],\\
\end{equation*}
where $\beta _{1}=\beta _{4}=0,$ $\beta _{2}=\beta _{3}=\frac{1}{2s},$ which implies that $T$ is a $\left( \alpha ,\eta \right) -\theta -\phi $ contraction
Therefore, from the Theorem $\left( {3.3}\right).$ $T$ has a unique fixed point.\\
\end{proof}
\begin{theorem}
Let $\left( X,d\right) $ be a $\left( \alpha ,\eta \right) -$complete rectangular $b$-metric space and let $\alpha ,\eta :X\times X\rightarrow \left[ 0,+\infty \right[ $ be two functions. Let $T:X\times X\rightarrow X$ be a self mapping satisfying following conditions:\\
\item[(i)] $T$ is a triangular $\left( \alpha ,\eta \right)-$ admissible mapping;\\
\item[(ii)] $T$ is a $\left( \alpha ,\eta \right) -\theta -\phi -$ Reich-type contraction;\\
\item[(iii)] $T$ there exists $x_{0}\in X$ such that $\alpha \left(x_{0},Tx_{0}\right) \geq 1$ or $\eta \left( x_{0},Tx_{0}\right) \leq 1;$\\
\item[(iv)] $T$ is a $\left( \alpha ,\eta \right) -$continuous.\\
Then T has a fixed point. Moreover, $T$ has a unique fixed point when\\ $\alpha \left( x,y\right) \geq 1$ or $\eta \left( x,y\right) \leq 1$ for all $x,y\in X.$\\
\end{theorem}
\begin{proof}
If $T$ is a $\left( \alpha ,\eta \right) -\theta -\phi -$ Reich-type contraction, thus
there exists $\theta \in \Theta $ and $\phi \in \Phi $ with $\alpha \left( x,y\right) \geq 1$ or $\eta \left(x,y\right) \leq 1$ for any $x,y\in X,$ $Tx\neq Ty$, we have
$$\theta \left[s^{2} d\left( Tx,Ty\right) \right] \leq \phi \left[ \theta\left(\frac{\left( d\left(x,y\right)+ d\left( Tx,x\right) +d\left( Ty,y\right) \right) }{3s}\right) \right].$$
Therefore,
\begin{equation*}
\theta \left[ s^{2}d\left( Tx,Ty\right) \right] \leq \phi \left[ \theta\left(
( \beta _{1}d\left( x,y\right) + \beta _{2}d\left( Tx,x\right)+\beta _{3}d\left( Ty,y\right)+ \beta _{4}d\left( Tx,y\right)\right) \right],
\end{equation*}
where $\beta _{1}=$ $\beta _{2}=\beta _{3}=\frac{1}{3s}\ and \ \beta _{4}=0$, which implies that $T$ is a $\left( \alpha ,\eta \right) -\theta -\phi $ contraction. Therefore, from the Theorem $\left( {3.3}\right).$ $T$ has a unique fixed
point.\\
\end{proof}
\begin{corollary}
Let $\left( X,d\right) $ be a $\left( \alpha ,\eta \right) -$complete rectangular $b$-metric space and let $\alpha ,\eta :X\times X\rightarrow \left[ 0,+\infty \right[ $ be two functions. Let $T:X\times X\rightarrow X$ be a self mapping satisfying following conditions:\\
\item[(i)] $T$ is a triangular $\left( \alpha ,\eta \right) -$ admissible mapping;\\
\item[(ii)] $T$ is a $\left( \alpha ,\eta \right) -$ Kannan type mapping,\\
\item[(iii)] $T$ there exists $x_{0}\in X$ such that $\alpha \left(x_{0},Tx_{0}\right) \geq 1$ or $\eta \left( x_{0},Tx_{0}\right) \leq 1;$\\
\item[(iv)] $T$is a $\left( \alpha ,\eta \right) -$continuous.\\

Then T has a fixed point. Moreover, $T$ has a unique fixed point when\\ $\alpha \left( x,y\right) \geq 1$ or $\eta \left( x,y\right) \leq 1$ for all $x,y\in X.$ Then $T$ has a unique fixed point $x\in X.$
\end{corollary}
\begin{proof}
Let $\theta (t)=$ $e^{t}$ for all $t$ $\in $ $\left] 0,+\infty \right[ $, and $\phi \left( t\right) =t^{2s\alpha }$ for all $t$ $\in \left[ 1,+\infty
\right[ $.\\ It is obvious that $\theta \in \Theta$ and  $\phi \in \Phi$. We prove that $T$ is a $\theta -\phi $- Kannan-type contraction.\\
\begin{align*}
\theta \left(s^{2} d\left( Tx,Ty\right) \right)&=e^{\displaystyle s^{2}d\left( Tx,Ty\right)}\\
&\leq e^{\displaystyle \alpha \left( d\left( Tx,x\right) +d\left( y,Ty\right) \right)}\\
&=\displaystyle {e}^{\displaystyle 2s\alpha \left( \frac{d\left( Tx,x\right) +d\left( y,Ty\right) }{2s}\right) }\\
&=\left[ e^{\displaystyle \left( \frac{d\left( Tx,x\right) +d\left( y,Ty\right) }{2s}\right) }\right] ^{2s\alpha }\\
&=\phi \left[ \theta  \left( \frac{d\left( Tx,x\right) +d\left(y,Ty\right) }{2s} \right) \right].
\end{align*}
Therefore, from Theorem 3.3, $T$ has a unique fixed point $x\in X$\\
\end{proof}
\begin{corollary}
Let $\left( X,d\right) $ be a $\left( \alpha ,\eta \right) -$complete rectangular $b$-metric space and let $\alpha ,\eta :X\times X\rightarrow \left[ 0,+\infty \right[ $ be two functions. Let $T:X\times X\rightarrow X$ be a self mapping satisfying following conditions:\\
\item[(i)] $T$ is a triangular $\left( \alpha ,\eta \right) -$admissible mapping;\\
\item[(ii)] $T$ is a $\left( \alpha ,\eta \right)-$Reich type mapping,\\
\item[(iii)] $T$ there exists $x_{0}\in X$ such that $\alpha \left(x_{0},Tx_{0}\right) \geq 1$ or $\eta \left( x_{0},Tx_{0}\right) \leq 1;$\\
\item[(iv)] $T$is a $\left( \alpha ,\eta \right) -$continuous.\\

Then T has a fixed point. Moreover, $T$ has a unique fixed point when \\$\alpha \left( x,y\right) \geq 1$ or $\eta \left( x,y\right) \leq 1$ for all $x,y\in X.$\\
\end{corollary}
\begin{proof}
Let $\theta (t)=$ $e^{t}$ for all t $\in $ $\left] 0,+\infty \right[ $, and $\phi \left( t\right) =t^{3s\lambda }$
for all t $\in \left[ 1,+\infty \right[ $.\\ We prove that T is a $\theta -\phi $- Reich-type contraction.
\begin{align*}
\theta \left( s^{2}d\left( Tx,Ty\right) \right)& =e^{\displaystyle s^{2}d\left( Tx,Ty\right) }\\
&\leq \displaystyle e^{\displaystyle3\displaystyle  \lambda s \frac{\left( d\left( x,y\right) +d\left( Tx,x\right) +d\left(y,Ty\right) \right) }{3s}}\\
&=\left[\displaystyle e^{\displaystyle  \frac{\left( d\left( x,y\right) +d\left( Tx,x\right)
+d\left( y,Ty\right) \right) }{3s}}\right] ^{3\lambda s }
\end{align*}
\begin{equation*}
=\phi \left[ \theta \left( \left( \frac{\left( d\left( x,y\right) +d\left(
Tx,x\right) +d\left( y,Ty\right) \right) }{3}\right) \right) \right] .
\end{equation*}
Therefore, from Theorem 3.3, $T$ has a unique fixed point $x\in X.$\\
\end{proof}
\begin{example}
	\bigskip Consider the set $X=\left\{ 1,2,3,4\right\} .$ It is easy to check
	that the mapping $d:X\times X\rightarrow \left[ 0,+\infty \right[ $ given by\\
	\item[(i)] $d\left( x,y\right) =d\left( y,x\right) ,$ $d\left(
	x,x\right) =0$ for all $\ x,y\in X,$\\
	\item[(ii)] $d\left( 1,2\right) =\frac{1}{24},$ $d\left( 1,3\right)
	=3,$ $d\left( 1,4\right) =4,$\\
	\item[(iii)] $d\left( 2,3\right) =5,$ $d\left( 2,4\right) =6,$ $
	d\left( 3,4\right) =18.$\\

	Clearly $\left( X,d\right) $ is a rectangular $b-$metric space with parameter $
	s=2.$\\

	Define mapping $T:X\rightarrow X$ and $\alpha ,\eta :X\times X\rightarrow \left[ 0,+\infty \right[ $ by\\
	\begin{equation*}
	\left\lbrace
	\begin{aligned}
	T(1)	&=1\\
	T(2)	&=1\\
	T(3)	&=1\\
	T(4)	&=2
	\end{aligned}
	\right.
	\end{equation*}
	and
	$$ \alpha \left( x,y\right) =\frac{x+y}{\max \left\{ x,y\right\} },$$ \ $$
	\eta \left( x,y\right) =\frac{\left\vert x-y\right\vert }{\max \left\{x,y\right\} }.$$\\

	Then, $T$ is an $\left( \alpha ,\eta \right) -$continuous triangular $\left(\alpha ,\eta \right) -$admissible mapping.\\

	Let $\theta \left( t\right) =\sqrt{t}+1,$ $\phi \left( t\right) =\frac{2t+1}{3}$ and $\beta _{1}=\frac{4}{10},$ $\beta _{2}=\frac{1}{10},$ $\beta _{3}=\frac{3}{10},$ $\beta _{4}=\frac{2}{10}.$\\
	It obvious that $\theta \in \Theta $ and $\phi \in\Phi .$\\
	Evidently, ($\left( \alpha \left( x,y\right) \geq 1\text{ or }\left( x,y\right) \leq 1\right) $ and $d\left( Tx,Ty\right) >0$ are when\\
	$\left\{ x,y\right\} =\left\{ 1,4\right\} ,\left\{ x,y\right\} =\left\{2,4\right\} ,$ or $\left\{ x,y\right\} =\left\{ 3,4\right\} $. Consider the
	following four possibilities:\\
	For $x=1,$ $y=4.$ then
	\begin{equation*}
	\theta \left( s^{2}.d\left( T1,T4\right) \right) =\sqrt{\frac{1}{6}}+1= 1.4,\\
	\end{equation*}
	and
	\begin{equation*}
	\phi \left( \theta \left( \beta _{1}d\left( 1,4\right) +\beta _{2}d\left(
	1,T1\right) \right) +\beta _{3}d\left( 4,T4\right) +\beta _{4}d\left(
	4,T1\right) \right) \\
	=\phi \left( \theta \left( \frac{21}{5}\right) \right) =2.36\text{,}\\
	\end{equation*}
	then
	\begin{equation*}
	\theta \left( s^{2}.d\left( T1,T4\right) \right) \leq \phi \left( \theta
	\left( \beta _{1}d\left( 1,4\right) +\beta _{2}d\left( 1,T1\right) \right)
	+\beta _{3}d\left( 4,T4\right) +\beta _{4}d\left( 4,T1\right) \right).\\
	\end{equation*}

	For $x=2,$ $y=4.$ then

	\begin{equation*}
	\theta \left( s^{2}.d\left( T2,T4\right) \right) =\sqrt{\frac{1}{6}}+1=1.4,\\
	\end{equation*}

	and
	\begin{eqnarray*}
		&&\phi \left( \theta \left( \beta _{1}d\left( 2,4\right) +\beta _{2}d\left(
		2,T2\right) \right) +\beta _{3}d\left( 4,T4\right) +\beta _{4}d\left(
		4,T2\right) \right) \\
		&=&\phi \left( \theta \left( 3.65\right) \right) =2.27\text{,}\\
	\end{eqnarray*}

	then
	\begin{equation*}
	\theta \left( s^{2}.d\left( T2,T4\right) \right) \leq \phi \left( \theta
	\left( \beta _{1}d\left( 2,4\right) +\beta _{2}d\left( 2,T2\right) \right)
	+\beta _{3}d\left( 4,T4\right) +\beta _{4}d\left( 4,T2\right) \right).\\
	\end{equation*}

	For $x=3,$ $y=4.$ then

	\begin{equation*}
	\theta \left( s^{2}.d\left( T3,T4\right) \right) =\sqrt{\frac{1}{6}}+1=1.4,\\
	\end{equation*}

	and
	\begin{equation*}
	\phi \left( \theta \left( \beta _{1}d\left( 3,4\right) +\beta _{2}d\left(
	3,T3\right) \right) +\beta _{3}d\left( 4,T4\right) +\beta _{4}d\left(
	4,T3\right) \right) \\
	=\phi \left( \theta \left( 10.1\right) \right) =3.13\text{,}\\
	\end{equation*}
	then
	\begin{equation*}
	\theta \left( s^{2}.d\left( T3,T4\right) \right) \leq \phi \left( \theta
	\left( \beta _{1}d\left( 3,4\right) +\beta _{2}d\left( 3,T3\right) \right)
	+\beta _{3}d\left( 4,T4\right) +\beta _{4}d\left( 4,T3\right) \right) .\\
	\end{equation*}
	Hence $  T $ satisfying assumption of Theorem (3.3) and 1 is the unique fixed point of $ T $.\\
\end{example}
\begin{example}
	Let $ X=A\cup B $, where $ A=\lbrace \frac{1}{n}:n\in\lbrace 2,3,4,5,6,7\rbrace \rbrace $ and $ B=\left[1,2 \right]  $. Define $ d:X\times X\rightarrow \left[0,+\infty \right[  $ as follows:
	\begin{equation*}
	\left\lbrace
	\begin{aligned}
	d(x, y) &=d(y, x)\ for \ all \  x,y\in X;\\
	d(x, y) &=0\Leftrightarrow y= x.\\
	\end{aligned}
	\right.
	\end{equation*}
	and
	\begin{equation*}
	\left\lbrace
	\begin{aligned}
	d\left( \frac{1}{2},\frac{1}{3}\right) =d\left( \frac{1}{4},\frac{1}{5}\right) =d\left( \frac{1}{6},\frac{1}{7}\right) 	&=0.05\\
	d\left( \frac{1}{2},\frac{1}{4}\right) =d\left( \frac{1}{3},\frac{1}{7}\right) =d\left( \frac{1}{5},\frac{1}{6}\right) 	&=0.08\\
	d\left( \frac{1}{2},\frac{1}{6}\right) =d\left( \frac{1}{3},\frac{1}{4}\right) =d\left( \frac{1}{5},\frac{1}{7}\right) 	&=0.4\\
	d\left( \frac{1}{2},\frac{1}{5}\right) =d\left( \frac{1}{3},\frac{1}{6}\right) =d\left( \frac{1}{4},\frac{1}{7}\right) 	&=0.24\\
	d\left( \frac{1}{2},\frac{1}{7}\right) =d\left( \frac{1}{3},\frac{1}{5}\right) =d\left( \frac{1}{4},\frac{1}{6}\right) 	&=0.15\\
	d\left( x,y\right) =\left( \vert x-y\vert\right) ^{2} \ otherwise.
	\end{aligned}
	\right.
	\end{equation*}
	Then $ (X,d) $ is a rectangular b-metric space with coefficient s=3. However we have the following:
	\item[1)] $ (X,d) $ is not a  metric space, as $d\left( \frac{1}{5},\frac{1}{7}\right)=0.4>0.29=d\left( \frac{1}{5},\frac{1}{4}\right)+d\left( \frac{1}{4},\frac{1}{7}\right) 	$.
	\item[2)] $ (X,d) $ is not a  b-metric space for s=3, as $d\left( \frac{1}{3},\frac{1}{4}\right)=0.4>0.39=3\left[ d\left( \frac{1}{3},\frac{1}{2}\right)+d\left( \frac{1}{2},\frac{1}{4}\right)\right]  	$.
	\item[3)] $ (X,d) $ is not a rectangular metric space, as $d\left( \frac{1}{5},\frac{1}{7}\right)=0.4>0.28=d\left( \frac{1}{5},\frac{1}{4}\right)+d\left( \frac{1}{4},\frac{1}{2}\right)+d\left( \frac{1}{2},\frac{1}{7}\right)$.\\
	Define mapping $T:X\rightarrow X$ and $\alpha ,\eta :X\times X\rightarrow \left[ 0,+\infty \right[ $ by\\
	\begin{equation*}
	T(x)=\left\lbrace
	\begin{aligned}
	\sqrt[6]{x}	& \ if \ x\in \left[1,2 \right]  \\
	& 1  \ if \ x\in A.\\
	\end{aligned}
	\right.
	\end{equation*}
	and
	\begin{equation*}
	\alpha(x,y)=\left\lbrace
	\begin{aligned}
	& sinh(x+y)	 \ if \ x,y\in \left[1,2 \right]  \\
	& \frac{1}{e^{x+y}}	\ otherwise,\\
	\end{aligned}
	\right.
	\end{equation*}
	\begin{equation*}
	\eta(x,y)=\left\lbrace
	\begin{aligned}
	&\frac{x+y}{4}	 \ if \ x,y\in \left[1,2 \right]  \\
	& 1+e^{-(x+y)}	 \ otherwise. \\
	\end{aligned}
	\right.
	\end{equation*}
	Then, $T$ is an $\left( \alpha ,\eta \right) -$continuous triangular $\left(\alpha ,\eta \right) -$admissible mapping.\\
	Let $\theta \left( t\right) =\sqrt{t}+1,$ $\phi \left( t\right) =\frac{t+1}{2}$ and taking $\beta _{1}=1,$ $\beta _{2}=\beta _{3}=\beta _{3}=0.$ It obvious that $\theta \in \Theta $ and $\phi \in\Phi .$\\
	Evidently, ($\left( \alpha \left( x,y\right) \geq 1\text{ or }\left( x,y\right) \leq 1\right) $ and $d\left( Tx,Ty\right) >0$ are when $\left\{ x,y\right\}\in \left[1,2 \right]  $ with $x\neq y$.\\
	Consider two cases :\\
	case 1: $ x>y $.
	$$ \theta(s^{2} d(Tx,Ty)= 3\left( \sqrt[6]{x}-\sqrt[6]{y}\right) +1$$
	and
	$$ \phi\left[ \theta( d(x,y))\right] =\frac{x-y}{2}+1.$$
	On the other hand
	\begin{align*}
	\theta(s^{2} d(Tx,Ty)- \phi\left[ \theta( d(x,y))\right]& =\frac{6\left( \sqrt[6]{x}-\sqrt[6]{y}\right)-( x-y)}{2}\\
	&=\frac{1}{2}\left(\left( \sqrt[6]{x}-\sqrt[6]{y}\right) \right)\left[  6-\left( \sqrt[6]{x{^5}}+\sqrt[6]{x{^4}y}+\sqrt[6]{x{^3}y{^2}}+\sqrt[6]{x{^2}y{^3}}+\sqrt[6]{xy{^4}}+\sqrt[6]{y{^5}}\right) \right].
	\end{align*}
	Since $ x,y\in \left[1,2 \right]  $, then
	$$\left[  6-\left( \sqrt[6]{x{^5}}+\sqrt[6]{x{^4}y}+\sqrt[6]{x{^3}y{^2}}+\sqrt[6]{x{^2}y{^3}}+\sqrt[6]{xy{^4}}+\sqrt[6]{y{^5}}\right) \right]\leq 0  .$$
	Which implies that
	\begin{align*}
	\theta(s^{2} d(Tx,Ty) &\leq  \phi\left[ \theta( d(x,y))\right]\\
	&= \phi\left[ \theta(\beta _{1} d(x,y))+\beta _{2} d(x,Tx)+\beta _{3} d(y,Ty)+\beta _{4} d(y,Tx)\right]
	\end{align*}
	case 2: $ y>x $.
	$$ \theta(s^{2} d(Tx,Ty)= 3\left( \sqrt[6]{y}-\sqrt[6]{x}\right) +1$$
	and
	$$ \phi\left[ \theta( d(x,y))\right] =\frac{y-x}{2}+1.$$
	Similarly of case 2, we conclude that
	\begin{align*}
	\theta(s^{2} d(Tx,Ty) &\leq \phi\left[ \theta(\beta _{1} d(x,y))+\beta _{2} d(x,Tx)+\beta _{3} d(y,Ty)+\beta _{4} d(y,Tx)\right]
	\end{align*}
	Hence, the condition $ (3.1) $ is satisfied. Therefore, $ T $ has a unique fixed point $ z=1 $.
\end{example}
\bibliographystyle{amsplain}


TITLE: Improving Newton Iteration
QUESTION [0 upvotes]: What are the options for a Newton Iteration that starts jumping between two values and never converges?
I am projecting 3D points on the spherical UV surface and approaching the poles the issue arise. Is the issue releated to the projection problem (U, V tangents get smaller) or to the Newton-Raphson implementation?
Thanks.

REPLY [1 votes]: I would indeed suspect numerically unstable reciprocals of tiny derivatives as you get close to the pole, in case you are iterating in the $UV$ space and solutions are close to poles. You could try iterating in a rotated $U'V'$ space so that derivatives would behave better.
Can you perhaps describe your problem more thoroughly, i.e. what is the function of which root you are finding?


TITLE: How many ways can the school choose a President Vice President?
QUESTION [1 upvotes]: There are n >= 4 students. The school  has a Board of Directors, consisting of one president and three vice-presidents. The entire board consists of four distinct students. How can I prove that $$n\binom{n-1}{3} = (n-3)(\binom{n}{3})$$ by couting in two different ways, the number of ways to choose a board of director?
I know that out of 4 student there is 4 way to choose a president because anyone of the students can be choosen, which leave 3 students and out of those three student there is only 1 way of choosing a vice-president because the board consit of three vice-president. I can see that the equation also equals to 4=4 but how do i prove it by counting in two different ways?

REPLY [1 votes]: First choose the vice presidents. There are $n$ selections for the first choice, $n-1$ for the second, and $n-2$ for the third. Since these are all taken in sequence, we multiply these numbers. Since the order of the group of three that we choose doesn't matter (there isn't a first-place vice president, second-place vice president, third-place vice president, in other words), we then cut out the number of ways we can arrange the group of three of our choice, leaving us $$\frac{n(n-1)(n-2)}{3*2*1}=\frac{n!}{(n-3)!3!}=\binom{n}{3}$$ ways to choose three vice presidents.
There are now $n-3$ people remaining, from which we must now choose a president. Putting this all together, we have $$(n-3)\binom{n}{3}=\frac{n(n-1)(n-2)(n-3)}{3*2*1}$$ ways to choose one president and three vice presidents.
Reverse the order in which you choose, i.e. now choose one president first, meaning $n$ choices, then from the group of $n-1$ remaining, make three choices (again, the order doesn't matter for the choices since the vice presidents are not distinguished from one another). You then have $$(n)\binom{n-1}{3}=n*\frac{(n-1)!}{(n-3-1)!3!}$$ ways to choose a president followed by three vice presidents. I will leave the final details to you to show that the above equation is equal to our first total.


TITLE: Showing the limit of given function does not exist
QUESTION [0 upvotes]: Given $f:\mathbb{R}\to \mathbb{R}$ such that $\mid f(x)\mid\geq M$ for some $M > 0$, $\forall x\in \mathbb{R}$.
I need to show that $\lim_{x\to 0} \frac{f(x)}{\sqrt[3]{x}}$ does not exist.
My try :
Assume limit exist, say L.
$\lim_{x\to 0} \frac{f(x)}{\sqrt[3]{x}}=\frac{\lim_{x\to 0} f(x)}{\lim_{x\to 0} \sqrt[3]{x}}=L$
$\Rightarrow$ $\lim_{x\to 0} f(x)=L.\lim_{x\to 0} \sqrt[3]{x}=0$ , contradiction to the fact that $\mid f(x)\mid \geq M, \forall x$

REPLY [1 votes]: Summarizing a set of comments below the OP, the only mistake in your proof is in the problematic expression $\lim_{x\to0}f(x)\over\lim_{x\to0}\sqrt[3]x$, which is of the utterly meaningless form $\text{possibly undefined}\over 0$. But it's an expression you don't need at all; it's enough to say $\lim_{x\to0}{f(x)\over\sqrt[3]x}=L$. That equation is all it takes to express the assumption that the limit, call it $L$, exists.  As for the denouement, what you've written is basically correct, but it might be clearer to write it as
$$\lim_{x\to0}f(x)=\lim_{x\to0}\left({f(x)\over\sqrt[3]x}\sqrt[3]x\right)=\left(\lim_{x\to0}{f(x)\over\sqrt[3]x}\right)\cdot\left(\lim_{x\to0}\sqrt[3]x\right)=L\cdot0=0$$
The take home message here, in terms of learning to write proofs, is to look at each and every assertion you make and ask yourself if it's both meaningful and true. If an assertion fails on either criterion, you then need to ask yourself what's it doing there? In this case, the answer is Nothing!


TITLE: Prove : $p$(n│even number of ODD parts)=$p$(n│distinct parts ,number of ODD parts is even )
QUESTION [1 upvotes]: I'm trying to prove the following Integer Partition claim :
$p$(n│even number of ODD parts) = $p$(n│distinct parts ,number of ODD parts is even) .
So I tried to prove a stronger claim :
$p$(n│number of odd parts) = $p$(n│number of distinct parts) :
Using generating functions :
DISTINCT PARTS =
$$(1+x)\cdot(1+x^2 )\cdot(1+x^3 )\cdot(1+x^4 )\cdot\cdot\cdot\cdot$$
=
$$\frac {1-x^2} {1-x} \cdot \frac {1-x^4} {1-x^2} \cdot \frac {1-x^6} {1-x^3} \cdot \frac {1-x^8} {1-x^4}\cdot\cdot\cdot\cdot$$
$$ \frac {1} {1-x} \cdot \frac {1} {1-x^3} \cdot \frac {1} {1-x^5} \cdot \frac {1} {1-x^7} \cdot\cdot\cdot\cdot
$$
= ODD PARTS
But I'm not sure regarding the proof , does it really prove that even number of odd parts equals distinct parts ,where number of ODD parts is even ?
Thanks

REPLY [2 votes]: Given the rewritten descriptions and comments, I think this equality is not really so much a case where some new clever manipulation of generating functions or other argument is necessary, but is really an exercise to see if one can unravel the definitions. It really comes down to the already proved statement that the number of partitions into odd summands equals the number of partitions into distinct summands.
A look at either side shows the requirement of an even number of odd parts, so for either side to be nonzero, $n$ must be even. [In a proof one would say "suppose $n$ is odd, then both sides are zero because..."]
Now if $n$ is even, and we look at the right side, which is partitions of $n$ into distinct parts, it follows even without saying so that the number of odd parts will automatically be even. So the right side (in the $n$ even case) boils down simply to the number of partitions of $n$ into distinct parts.
Now look at the left side. We again can see that the number of odd parts must automatically be even, since $n$ itself is even. So here again the extra term "even number of" may be dropped, and the left side is simply the number of partitions of the (even) number $n$ into odd parts. [At the same time this makes it clear the left side should not allow even summands at all, otherwise the count on the left will exceed the right side.]
So the overall statement holds, but is really just a re-work of the already mentioned (and shown in the OP) statement about odd parts and distinct parts.


TITLE: Is $\int_{\mathbb{R}} \int_{\mathbb{R}^n} \alpha w(t) e(\alpha (a_1t_1 + \dotsb + a_n t_n)) dt\,d \alpha = 0$?
QUESTION [1 upvotes]: Let $a_i$ be a nonzero real number for each $1 \leq i \leq n$. $w$ a smooth nonnegative with compact support. I would like to understand the following integral.
$$
I = \int_{\mathbb{R}} \int_{\mathbb{R}^n} \alpha w(t) e(\alpha (a_1t_1 + \dotsb + a_n t_n)) dt\,d \alpha.
$$
If the factor $\alpha$ were not present in the integrand, then the contribution comes from where $\alpha = 0$, but with this additional factor, I was thinking that perhaps $I$ is $0$? or something with very small absolute value. Any comment or suggestions to estimate this integral would be appreciated!
PS. $e(x) = e^{2 \pi i x}$.

REPLY [1 votes]: Example for divergence.  $n=1$, $a_1=1$, $w(t) = \mathbf1_{[0,1]}(t)$.  Note $w$ is not "smooth"; it is not even continuous.
Then
$$
\int_{\mathbb{R}^n} \alpha w(t) e(\alpha (a_1t_1 + \dotsb + a_n t_n)) dt
=\int_0^1(\alpha\exp(2\pi i(\alpha t))\;dt
=\frac{i}{2\pi}\big(1-e^{i 2 \pi \alpha}\big)
\\
\int_{-K}^K \frac{i}{2\pi}\big(1-e^{i 2 \pi \alpha}\big)\;d\alpha
=\frac{i}{2\pi^2}\big(2\pi K -\sin(2\pi K)\big)
\\
\lim_{K\to+\infty} \frac{i}{2\pi^2}\big(2\pi K -\sin(2\pi K)\big) = +i\infty
$$

According to Carlo's argment, we do have convergence when $w$ is $C^1$ with compact support.


TITLE: Amount of ways I can pick cards in a deck.
QUESTION [1 upvotes]: Question: Using all of the face cards in a standard deck of playing cards, in how many ways can I make three piles of five cards.
Thanks :) Been trying for a while now and I can't do it..

REPLY [2 votes]: You have 15 different cards. To make the first pile, you choose 5 of them, and there are ${15 \choose 5} = 3003$ ways to do that.
You now have 10 cards left. To make the second pile, you choose 5 of them, and there are ${10 \choose 5} = 252$ ways to do that.
You now have 5 cards left, and there's only 1 way to choose all of them to make the third pile.
So there are $3003 \times 252 \times 1 = 756756$ ways to pick your three piles.
However, there is a complication - this assumes that it's important which pile is the first pile, which is the second, and which is the third. If you just care about splitting the cards into three piles, then you have to remove the repetition that comes from being able to swap around which pile is which. In fact, each way of dividing the cards has 6 equivalent ways of organising the piles around, so our count of different ways to pick the piles is 6 times too big. Thus, if order of piles doesn't matter, there are $756756 \div 6 = 126126$ ways to do that.


TITLE: How to notate a subtraction of a matrix columns with a vector?
QUESTION [1 upvotes]: According to matrix algebra, I cannot subtract a vector from a matrix, because they don't have the same size.

I want to notate that every column of $R_i^t$ is subtracted by vector $R_f^t$, but I don't want to either merge them into 1 matrix, nor change the vector to matrix in order to emphasise the nature of the variables.

REPLY [2 votes]: Given matrix $\mathbf A \in \mathbb R^{n \times p}$, we can subtract vector $\mathbf b \in \mathbb R^n$ from all $p$ columns of $\bf A$ via $\color{blue}{\mathbf A - \mathbf b {\bf 1}_p^{\top}}$.


TITLE: How can the D'Alembertian of a field be interpreted intuitively?
QUESTION [2 upvotes]: The D'Alembertian operator is defined as
$$
\Box = g^{\nu\mu}\nabla_\nu\nabla_\mu
$$
For the Minkowski metric in Cartesian coordinates that is
$$
\Box=\frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2} - \frac{\partial^2}{\partial z^2}
$$
Can it be intuitively described, just as a gradient or divergence, curl or Laplacian may be?
I'm looking for something similar to the interpretation of a Laplacian given in this question and answer.

REPLY [3 votes]: The operator is just $\partial_t^2-\nabla^2$. So it is the difference between a "temporal laplacian" and a "spatial laplacian". Since laplacian measures curvature, this is basically telling you the difference in curvature between the spatial and temporal variation of the field.
One reason this comes up in physics is in describing elastic sheets under tension. In an elastic sheet, if there is (spatial) curvature at a point, the tension in the sheet will pull the point in order to flatten out the curvature. Thus the point feels a force in the same direction as the curvature. So by newtons second law, the point on the sheet will accelerate, that is, have a second time derivative, in the direction of the curvature. This is why you would expect the difference in the "temporal laplacian" and "spatial laplacian" to be zero.
If this operator is non-zero, then it means the temporal and spatial variations are inconsistent with each other, and it looks like there is an external force acting on the point in your elastic sheet.


TITLE: This integral... Please help (those that look like seven are zs)
QUESTION [2 upvotes]: $$
 \int_0^4 z \left(\sqrt{16 - z^2} - \sqrt{25 - z^2} + 3 \right) dz
$$
I'm studying for measure theory but i got stuck on this integral. I tried solving it by parts but i get stuck over the same pattern on functions in the new integral.

REPLY [4 votes]: $$2\pi \int_0^4 z \left(\sqrt{16 - z^2} - \sqrt{25 - z^2} + 3 \right) \, dz$$
is same as
$$2\pi \left( \int_0^4 z \left(\sqrt{16 - z^2} \right) - \int_0^4 z\left(\sqrt{25 - z^2}\right) + \left(\int_0^43 \right) \right)\, dz$$
Use substitutions $z=4cos\theta$ in the first integral and a similar one in the second integral.


TITLE: On the frequency of Prime Factor 2 with upper bound n
QUESTION [0 upvotes]: I was wondering if this is true:

$ \sum_{l=2}^n \max(2^{x} | l) < n$
  For $n > 4,x \in \mathbb N$

(I hope I got that right), the Sum denotes the frequency of Prime Factor 2 in l < n.
Is it true, that the frequency of Prime Factor 2 in all integers starting from 2 to n, is always less than the value of n?
Thanks

REPLY [2 votes]: Yes, it is true.  As $1$ has no factors of $2$ we can start your sum at $1$ with no change.  Now note that the number of even numbers up to $n$ is $\lfloor \frac n2 \rfloor$, each contributing at least $1$ to your sum.  The numbers that are multiples of $4$, of which there are $\lfloor \frac n4 \rfloor$ contribute another $1$, the $\lfloor \frac n8 \rfloor$ that are multiples of $8$ contribute another and so on.  The sum of all of these is
$$\sum_{k=1}^{\log_2 n}\left  \lfloor \frac n{2^k} \right \rfloor \le \sum_{k=1}^{\log_2 n}\frac n{2^k}=n \sum_{k=1}^{\log_2 n}\frac 1{2^k}\lt n$$
where we know the last sum is equal to $1$ if we continue it to infinity and we have deleted the trailing terms.  This indicates that the closest we can get is when $n$ is a power of $2$, and in fact your sum will be $n-1$ in that case.


TITLE: Lattice projections
QUESTION [3 upvotes]: I imagine the following result is folklore
Theorem. Those $k$-dimensional subspaces $\zeta \subset \mathbb{R}^n$ $(1 \leq k \leq n-1)$ for which the orthogonal projection of the lattice $\mathbb{Z}^n$ onto $\zeta$ is dense in $\zeta$ form a subset of full measure in the Grassmannian $G_k(\mathbb{R}^n)$.
What would be a (good) reference? Is there a "standard" proof?
Below in the answers I'll give a simple proof of my own that just uses a simple result in the geometry of numbers (an immediate corollary of van der Corput's extension of Minkowski's first theorem in the geometry of numbers):
If $C \subset \mathbb{R}^n$ is a $0$-symmetric closed convex set with infinite volume, then it contains an infinite number of integer points.

REPLY [1 votes]: Disclaimer: this is not an answer to the question, but a long comment I wish to keep separate. The proof of the result in the question follows immediately from the two lemmas below.
Definition. Two subspaces $L, K \subset V$ of a finite-dimensional spaces $V$ are said to have minimal intersection if the dimension of their intersection is the least possible compatible with their dimensions and the dimension of $V$.
Lemma 1. If every subspace of $\mathbb{R}^n$ that is spanned by integer vectors has minimal intersection with the orthogonal complement of a $k$-dimensional subspace $\zeta$ $(1 \leq k \leq n-1)$, then the orthogonal projection of $\mathbb{Z}^n$ onto $\zeta$ is dense in $\zeta$.
Proof. Let $D_\epsilon \subset \zeta$ be the closed disc of radius $\epsilon > 0$ centered at the origin, and let $C_\epsilon$ be the cylinder $\pi_\zeta^{-1}(D_\epsilon)$, where $\pi_\zeta : \mathbb{R}^n \longmapsto \zeta$ is the orthogonal projection onto $\zeta$. Note that $C_\epsilon$ is a $0$-symmetric, closed, convex set with infinite volume and hence contains an infinite number of
integer vectors.
Claim. The dimension of the subspace $\eta_\epsilon$ spanned by the (infinite) set of all integer vectors in $C_\epsilon$ is greater than $k$, the dimension of the subspace $\zeta$.
Indeed, $\eta_\epsilon$ is spanned by integer vectors and if its dimension were less than or equal to $k$, then by hypothesis it would intersect $\zeta^\bot$ only at the origin. This would imply that the intersection of $\eta_\epsilon$ and the cylinder $C_\epsilon$ is compact and hence it could not contain an infinite number of integer vectors. This would contradict the definition of $\eta_\epsilon$.
Notice now that if the dimension of $\eta_\epsilon$ is greater than $k$, then the minimal intersection hypothesis implies that its projection onto $\zeta$ is surjective. Hence, we can find integer vectors $\mathbf{m}_1$, $\mathbf{m}_2$, $\dots$, $\mathbf{m}_k$ in $C_\epsilon$ such that their projections form a basis of $\zeta$. The norm of these projections is less than or equal to $\epsilon$ and so the lattice spanned by these vectors is a subset of $\pi_\zeta(\mathbb{Z}^n)$ which is $2 \epsilon$ close to every point in $\zeta$.  Since $\epsilon > 0$ was arbitrary, we conclude that $\pi_\zeta(\mathbb{Z}^n)$ is dense in $\zeta$.
Lemma 2. The set of $k$-dimensional subspaces of $\mathbb{R}^n$ that have minimal intersection with every subspace spanned by integer vectors has full measure in the Grassmannian $G_k(\mathbb{R}^n)$.
Proof. Fix a dimension $p$ between $1$ and $n-1$ and a $p$-dimensional subspace $\eta$. Consider the set of all $(n-k)$-dimensional subspaces $\zeta^\bot$ that do not intersect $\eta$ minimally. This is a finite union of lower-dimensional submaifolds and hence has measure zero in $G_{n-k}(\mathbb{R^n})$. The union of all these sets as $\eta$ ranges over the countable set of all subspaces spanned by integer vectors has still measure zero and so its complement has full measure.


TITLE: Linear algebra - Prove that for any isomorphism there is an "identity basis(?)"
QUESTION [5 upvotes]: I'd appreciate if you could help me with this question.
What I thought about so far is showing that for any basis $B$, we know that $[T]_B$ is invertible so there are $k$ elementary matrices such that $E^k*...*E^1*[T]_B=I$.
I'm not sure if it's a good direction. Here is the question:
Let $V$ be a vector space over $F$. Prove that for any isomorphism $T : V \rightarrow F^n$, there exists a basis $B$ for $V$ such that $T=Is_B$, meaning $T(v)=[v]_B$
Note: if you see a solution I'd be happy if you could write a hint in addition to your solution, so i can try it myself first.
Also, I'm not a native english speaker so if anything is not clear please tell me and I'll fix it.

REPLY [1 votes]: Hint
Let $B=\{ v_1,..., v_n \}$ be the basis you seek. Since $T(v)=[v]_B$ for all $v$ yoou must have for $v=v_i \in B$:
$$T(v_i)=[v_i]_{B}$$
Now use the fact that
$$[v_i]_B=(0,0,..., 0,1 ,0,..0)=e_i$$


TITLE: Characteristic polynomial of linear operator and matrix
QUESTION [0 upvotes]: Does the characteristic polynomial of a linear operator always equal to the characteristic polynomial of its representation under some basis $\mathcal{B}$?
That is $c_{\tau}(x) = c_{[\tau]_{\mathcal{B}}}(x)$?
Another question is why the characteristic polynomial of a symmetric linear operator may have complex root? (we know symmetric matrix always has real eigenvalues.) Is there some example?
The characteristic polynomial is defined by product of elementary divisor for $\tau$ .
The symmetric operator means $(\tau v,w) = (v,\tau w)$ that is adjoint for real bilinear form.
$[\tau]_{\mathcal{B}}$ is the matrix that columns are coordinate of $\tau(b_i)$ under basis $\mathcal{B}$

REPLY [1 votes]: Yes: the characteristic polynomial of a linear operator always equal to the characteristic polynomial of its representation under any choice basis $\mathcal B$.   The fact that this is so is a direct consequence of the structure theorem for PIDs (which presumably you have seen if you are doing anything with "elementary divisors").  In particular, there is a direct correspondence between elementary divisors of an operator and elementary divisors of the associated matrix.
For a more thorough discussion, see this wikipedia page.
Regarding your second question: if our vector space is over the field $\Bbb R$ and the bilinear form is positive definite, then "symmetric operators" will correspond to "symmetric (real) matrices", which means that the spectral theorem applies and the roots of the characteristic polynomial must be real.  For other bilinear forms, we can make no such guarantee.
An example of a "symmetric" operator that fails to have real eigenvalues.  Consider the bilinear form over $\Bbb R^2$ given by
$$
(x,y) = x_1y_2 + x_2 y_1.
$$
Relative to this bilinear form, we find that the operator $x \mapsto Ax$ where
$$
A = \pmatrix{0&-1\\1&0}
$$
is "symmetric", but has no real eigenvalues.


TITLE: The image of orthonormal basis under compact operator
QUESTION [4 upvotes]: I need a help to prove that statement: if $\{e_n\}$ an orthonormal basis in Hilbert space $H$ and $A$ is a compact operator from $H$ to $H$, then $Ae_n\rightarrow 0$. Thx for any help.

REPLY [2 votes]: I think the simplest proof goes as follows. Assume that $(Ae_n)$ does not converge to $0$ strongly. Then, by the compactness of $A$, there is a subsequence of $(Ae_n)$ that converges strongly to some nonzero vector $v\neq 0$. On the other hand, by Bessel's inequality, $(e_n)$ converges to $0$ weakly, hence $(Ae_n)$ also converges to $0$ weakly. As strong convergence implies weak convergence, we conclude that $v=0$, a contradiction.


TITLE: Regularity of convex sets in $\mathbb{R}^n$
QUESTION [11 upvotes]: The following result is Proposition 2.4.3 in [1]:

Theorem. Let $K\subset\mathbb{R}^n$ be a bounded convex set with the non-empty interior. Then $\partial K\in C^{1,1}$ if and only if
there is $r>0$ such that $K$ is the unioin of balls of radius $r$.

Question. Do you know who is the author of this result?
Hörmander does not provide any reference so it could be that his book is the first place where it was proved. However, the statement seems so classical that I tend to believe the result had been known before.
It is very likely that I will refer to this result in my research and I would like to give a proper reference to it.
Edit. It seems that there are earlier references than the one in the answer, see the comments below the answer. That means the questions might still have a better answer.
[1] L. Hörmander,
Notions of convexity.
Progress in Mathematics, 127. Birkhäuser Boston, Inc., Boston, MA, 1994.

REPLY [5 votes]: To my knowledge, this is the first place where I saw the result being claimed:
C. O. Kiselman. Regularity classes for operations in convexity theory. Kodai Math. J. 15. 1992.
In particular on the first page the author states:

To  describe   the  simplest   case   of   our  results,   let  $A$   be  a  compact  set  in  $R^n$. If  the  boundary  of  $A$   is   of   class   $C^{1,1}$,  then  $A$   is  a  union  of   Euclidean  balls   with radii   bounded   from   below;   if   $A$   is  convex,   the  converse   holds.


text:
The following examples consist of nonlinearly coupled systems that are simulated with help of input zero mean unit variance normal uncorrelated innovations $w_i(n)$. All simulations ($10,000$ realizations each) were preceded by an initial a burn-in period of $10,000$ data points to avoid transient phenomena. Estimation results are examined as a function of $n_s = \{ 32, 64, 128, 256, 512, 1024, 2048 \}$ with $\alpha = 1\%$ significance.

For brevity Example 1 is carried out in full detail whereas approach performance for the other ones is gauged mostly through the computation of observed detection rates except for Examples 4 and 5 which also portray model order choice criterion behaviour.

Simulations results are displayed as function of realization length $n_s$ assuming $\alpha = 1\%$ in testing \eqref{eq:null_hypo}.

\subsection{Example 1}
\label{subsec:example1}

\input{example1}

\subsection{Example 2}
\label{subsec:example2}

\input{example2}

\subsection{Example 3}
\label{subsec:example3}

\input{example3}

\subsection{Example 4}
\label{subsec:example4}

\input{example4}

\subsection{Example 5}
\label{subsec:example5}

\input{example5}

config: arxiv


TITLE: Solve for $\alpha$, $z^* ( B + \alpha A )^{-1} A ( B + \alpha A )^{-1} z = c$, where ${\rm rank}(A)=1$,$B$ full rank matrix
QUESTION [1 upvotes]: Problem (I raised a similar problem, but in this problem $A$ and $B$ are additionally Hermitian):
Let $A \in M_{n,n}(\mathbb{C})$ be a rank one matrix and Hermitian, $B \in M_{n,n} (\mathbb{C})$ be a full rank matrix and Hermitian, $z \in M_{n,1} (\mathbb{C})$ be a column vector, and $c \in \mathbb{R}$ are given, then
Solve for $\alpha \in \mathbb{R}$ (which is non-negative, i.e., $\alpha > 0$) such that
  \begin{align}
z^* \left( B^* + \alpha A^* \right)^{-1} A \ \left( B + \alpha A \right)^{-1} z &= c \\
z^* \left( B + \alpha A \right)^{-1} A \ \left( B + \alpha A \right)^{-1} z &= c \ \hspace{5mm} \textrm{since } A \textrm{ and } B \textrm{ are Hermitian}.
\end{align}
Is it possible to obtain a closed-form solution for $\alpha$?

NOTE:

If $B = I$ and $A$ is Hermitian then one can solve the problem in closed form. However, I don't know yet how to solve for any Hermitian matrix $B$.

REPLY [1 votes]: Since $A$ is rank one and Hermitian, then we can write
\begin{equation}
 A = aa^H
\end{equation}
So the equation is now
\begin{align}
z^H \left( B^H + \alpha A^H \right)^{-1} aa^H \ \left( B + \alpha A \right)^{-1} z =c
\end{align}
which is
\begin{equation}
 \vert a^H(B + \alpha A)^{-1}z \vert^2 = c
\end{equation}
Let's solve
\begin{equation}
 a^H(B + \alpha A)^{-1}z = \sqrt{c}e^{i \theta} = d
\end{equation}
for $\theta \in \mathbb{R}$. This means in your solution, you will have a free $\theta$ parameter.
Now, let's apply the Matrix Inversion Lemma and since $B$ is invertible, we can say:
\begin{equation}
 (B + \alpha aa^H)^{-1}
 =
 B^{-1} - \alpha \frac{B^{-1}AB^{-1}}{1 + \alpha a^H B^{-1} a}
\end{equation}
Plugging this we get
\begin{equation}
 a^H(B^{-1} - \alpha \frac{B^{-1}AB^{-1}}{1 + \alpha a^H B^{-1} a})z =  d
\end{equation}
which is
\begin{equation}
 a^H B^{-1} z
 -
 \alpha \frac{a^HB^{-1}AB^{-1}z}{1 + \alpha a^H B^{-1} a}
 =
  d
\end{equation}
which is
\begin{equation}
 [a^H B^{-1} z][1 + \alpha a^H B^{-1} a] - \alpha[a^HB^{-1}AB^{-1}z]
 =
  d[1 + \alpha a^H B^{-1} a]
\end{equation}
Let
\begin{align}
 \beta_1 &= a^H B^{-1} z \\
 \beta_2 &= a^H B^{-1} a\\
 \beta_3 &= a^HB^{-1}AB^{-1}z = \beta_2\beta_1
\end{align}
Rearranging, we get
$$\alpha=\frac{\beta_1 - d}{d\beta_2}$$


TITLE: Find the radius of convergence of $\sum_{n=1}^{\infty}{n!x^{n!}}$
QUESTION [0 upvotes]: Find the radius of convergence of $\sum_{n=1}^{\infty}{n!x^{n!}}$. Should I look at this series as: $\sum_{n=1}^{\infty}({n!x^{(n-1)!})x^{n}}$? I am really confues here. In addition, any attempt to compute any kind of "lim sup" fails. I would appreciate your help in this.

REPLY [1 votes]: For each $ n\in \mathbb {N}$, let  $a_n=n$ if $n=k!$ for some  $k\in \mathbb {N} $ and $a_n=0$ otherwise. Then the radius of convergence of  $\sum\limits_{n=1}^{\infty}n!x^{n!}= \sum\limits_{n=1}^{\infty}a_nx^n $ is $$ \dfrac{1}{\limsup \sqrt[n]{a_n}}=\dfrac{1}{\lim \limits_{k}\sqrt[k!]{k!}}. $$ Since $\langle \sqrt[k!]{k!}\rangle $ is a sub sequence of $\langle \sqrt[n]{n}\rangle $ and $\lim\limits_{n}\sqrt[n]{n}=1$ we have that $\lim\limits_{k}\sqrt[k!]{k!}=1$. Therefore the radius of convergence is $1$.


TITLE: How is it that a single transmission line can be modeled by a two port network?
QUESTION [1 upvotes]: I am studying electric power distribution. I've read in Kirtley's Electric Power Principles (2010) that a transmission line can be modeled by the 'pi' model. Which is a two port network with series conductance and parallel capacitance. But a transmission line (single phase) is just one cable, with current going through it. How can this be modeled by a two port network?
Thanks.

REPLY [0 votes]: A power transmission line is, indeed, one cable, but the current that flows into the cable has a return path via the ground. Thus you have two terminals (first port) at the power station and two terminals (second port) where the load is located. Therefore you can consider the power transmission line as a two-port network.


TITLE: Integrability of Derivative of a Continuous Function
QUESTION [2 upvotes]: Let $f$ be continuous on $[a,b]$ and has finite derivative a.e. on $[a,b]$.
Let $f_n(x)=n[f(x+1/n)-f(x)] $ s.t. $f_n$ be uniformly integrable on $[a,b]$.
I want to show : $f'$ is Lebesgue integrable.
(I noted that $f_n\rightarrow f'$ pointwise.)

REPLY [0 votes]: By the Mean Value Theorem, for each positive integer $n$ and for each $x \in [a,b]$,
there exists $z_{n,x} \in (x, x+1) $ so that
$$
f_n(x)= f'(z_{n,x}).
$$
It follows that $|f_n(x)| \leq M$ for all $n$ and for all $x\in [a,b]$.
Since $\{f_n\}$ converges pointwise to $f'$ on $[a,b]$, the Bounded Convergence Theorem yields
$$
\int_a^b f' = \lim_{n \to \infty } \int_a^b f_n
$$
which shows that $f'$ is Lebesgue integrable.