text:
\section{Introduction}

Let $\{\mu_{n}\}$ be a sequence of complex numbers and
$\mathcal{L}\colon \mathbb{C}[x] \rightarrow\mathbb{C}$ be a linear
functional def\/ined by
\begin{gather*}
\mathcal{L}\big( x^{n}\big) =\mu_{n},\qquad n=0,1,\ldots.
\end{gather*}
Then, $\mathcal{L}$ is called the \emph{moment functional} determined by the
formal moment sequence $\{\mu_{n}\}$. The number~$\mu_{n}$ is
called the \emph{moment} of order~$n$. The task of f\/inding an explicit
representation for the functional $\mathcal{L}$ is called a \emph{moment
problem} \cite{MR0184042,MR0458081,MR0008438}.

A sequence $\{\Pi_{n}(x)\} \subset\mathbb{C}[x]$, with $\deg( \Pi_{n}) =n$ is called an
\emph{orthogonal polynomial sequence} with respect to~$\mathcal{L}$ provided
that~\cite{MR0481884}
\begin{gather*}
\mathcal{L} ( \Pi_{n}\Pi_{m} ) =K_{n}\delta_{n,m},\qquad
n,m=0,1,\ldots,
\end{gather*}
where $K_{n}\neq0$ and $\delta_{n,m}$ is Kronecker's delta.

The moments play a fundamental role in the theory of orthogonal polynomials
since, among other results, we have the determinantal representation
\begin{gather*}
\Pi_{n}(x) =C_{n}\left\vert
\begin{matrix}
\mu_{0} & \mu_{1} & \cdots & \mu_{n}\\
\mu_{1} & \mu_{2} & \cdots & \mu_{n+1}\\
\vdots & \vdots & \ddots & \vdots\\
\mu_{n-1} & \mu_{n} & \cdots & \mu_{2n-1}\\
1 & x & \cdots & x^{n}
\end{matrix}
\right\vert ,
\end{gather*}
for some normalization constant $C_{n}\neq0$, with
\begin{gather*}
\left\vert
\begin{matrix}
\mu_{0} & \mu_{1} & \cdots & \mu_{n}\\
\mu_{1} & \mu_{2} & \cdots & \mu_{n+1}\\
\vdots & \vdots & \ddots & \vdots\\
\mu_{n} & \mu_{n+1} & \cdots & \mu_{2n}
\end{matrix}
\right\vert \neq0,\qquad n=0,1,\ldots.
\end{gather*}
Given their importance, it is very striking that they are not explicitly
listed in the standard books on orthogonal polynomials, or even in
encyclopedic texts such as~\cite{MR2656096}. In fact, the only place where we
found a comprehensive enumeration of the moments of classical orthogonal
polynomials was the recent article~\cite{MR3286553}, based on the results
obtained in the Ph.D.\ Thesis of the f\/irst author~\cite{Njionou}\footnote{We
thank one of the referees for stressing this fact.}.

In this paper, we focus our attention on linear functionals def\/ined by
\begin{gather}
\mathcal{L}(f) =
{\displaystyle\sum\limits_{x=0}^{\infty}}
f(x)\rho\big( x;\vec{\alpha},\vec{\beta},c\big) ,
\label{L}
\end{gather}
where the weight function $\rho( x;\vec{\alpha}
,\vec{\beta},c) $ is of the form
\begin{gather}
\rho\big( x;\vec{\alpha},\vec{\beta},c\big)
=\frac{( \vec{\alpha})_{x}}{(
\vec{\beta}+1)_{x}}\frac{c^{x}}{x!}, \label{weight}
\end{gather}
with
\begin{gather*}
 ( \vec{\alpha} )_{x} = ( \alpha_{1})
_{x} ( \alpha_{2} )_{x}\cdots ( \alpha_{p} )_{x},\qquad
 ( \vec{\beta}+1 )_{x} = ( \beta_{1}+1 )
_{x} ( \beta_{2}+1 )_{x}\cdots ( \beta_{q}+1 )_{x},
\end{gather*}
and $(a)_{x}$ denotes the Pochhammer symbol (also called
shifted or rising factorial) def\/ined by~\cite[(5.2.4)]{MR2723248}
\begin{gather*}
(a)_{0} =1, \qquad (a)_{x} =a ( a+1 ) \cdots ( a+x-1 ),\qquad x\in\mathbb{N},
\end{gather*}
or by
\begin{gather*}
(a)_{x}=\frac{\Gamma ( a+x ) }{\Gamma (a ) },\qquad a+x\neq0,-1,\ldots,
\end{gather*}
and $\Gamma(z) $ is the Gamma function. Unless stated otherwise,
we always assume that
\begin{gather*}
\beta_{i}>-1,\qquad 1\leq i\leq q,
\end{gather*}
and we will use the notation
\begin{gather*}
\vec{\alpha} =\alpha_{1},\alpha_{1},\ldots,\alpha_{p},\qquad
\vec{\beta}+1 =\beta_{1}+1,\beta_{2}+1,\ldots,\beta_{q}+1.
\end{gather*}

Note that we have
\begin{gather*}
\frac{\rho ( x;\vec{\alpha},\vec{\beta},c )
}{\rho ( x;\vec{\alpha},\vec{\beta},c )
}=\frac{\eta(x) }{\phi ( x+1 ) },
\end{gather*}
with%
\begin{align}\label{phi-eta}
\phi(x) =x ( x+\beta_{1} ) ( x+\beta
_{2} ) \cdots ( x+\beta_{q} ) ,\qquad
\eta(x) =c ( x+\alpha_{1} ) ( x+\alpha
_{2} ) \cdots ( x+\alpha_{p} ) .
\end{align}
Hence, the weight function $\rho ( x;\vec{\alpha
},\vec{\beta},c ) $ satisf\/ies the \emph{Pearson equation} (see~\cite{Pearson} or \cite[(6.3)]{MR1761401})
\begin{gather*}
\Delta ( \phi\rho ) = ( \eta-\phi ) \rho,
\end{gather*}
where
\begin{gather*}
\Delta f(x)=f(x+1)-f(x)
\end{gather*}
is the forward dif\/ference operator. If we def\/ine $s$ by
\begin{gather*}
s=\max\big\{ \deg(\phi) -2,\deg(\phi-\eta)
-1\big\} ,
\end{gather*}
the sequence of polynomials orthogonal with respect to~$\mathcal{L}$ is
called \emph{semiclassical} of class~$s$~\cite{MR2936305,MR1270222}.

Weight functions of the form~(\ref{weight}) are also related to discrete
probability distributions (Poisson, Pascal, binomial, hypergeometric, etc.)~\cite{MR1224449}\footnote{We thank one of the editors for suggesting this reference.}. The Generalized Hypergeometric probability distributions were
studied by Adrienne W.~Kemp in her Ph.D.\ Thesis~\cite{Kemp}\footnote{Unfortunately, we haven't been able to obtain a copy of it.}. An
excellent re\-ference outlining the connections between the theory of
probability and orthogonal polynomials is~\cite{MR1761401}.

In \cite{MR3055670}, it was pointed out that since
\begin{gather}
\mu_{n}(c) =
\sum\limits_{x=0}^{\infty}
x^{n}\rho\big( x;\vec{\alpha},\vec{\beta},c\big) ,
\label{mun def}
\end{gather}
one has
\begin{gather}
\mu_{n+1}(c) =
\sum\limits_{x=0}^{\infty}
x^{n}x\rho\big( x;\vec{\alpha},\vec{\beta},c\big)
=\vartheta\mu_{n}(c) , \label{reqmu}
\end{gather}
where the dif\/ferential operator $\vartheta$ is def\/ined by \cite[(16.8.2)]{MR2723248}%
\begin{gather}
\vartheta f(c)=c\frac{df}{dc}. \label{theta}
\end{gather}
Successive applications of (\ref{reqmu}) give
\begin{gather*}
\mu_{n}=\vartheta^{n}\mu_{0},
\end{gather*}
and it follows that the f\/irst moment~$\mu_{0}$ determines the whole sequence
$\{ \mu_{n}\}$. If we use the operational formula~\cite{MR2798171}
\begin{gather*}
\vartheta^{k}=\sum\limits_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
c^{k}\frac{d^{k}}{dc^{k}},
\end{gather*}
we have
\begin{gather}
\mu_{n}(c) =
\sum\limits_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
c^{k}\frac{d^{k}\mu_{0}}{dc^{k}}, \label{munD}
\end{gather}
where $\genfrac{\{}{\}}{0pt}{}{n}{k}$ denote the Stirling numbers of the second kind def\/ined by~\cite[(26.8)]{MR2723248}
\begin{gather}
\genfrac{\{}{\}}{0pt}{}{n}{k}
=\frac{1}{k!}
{\displaystyle\sum\limits_{j=0}^{k}}
\binom{k}{j}(-1) ^{k-j}j^{n}. \label{stirling}
\end{gather}

From (\ref{weight}), we have
\begin{gather}
\mu_{0}(c) =
 \sum\limits_{x=0}^{\infty}
\rho\big( x;\vec{\alpha},\vec{\beta},c\big)
= {}_{p}F_{q}\left[
\begin{matrix}
\vec{\alpha}\\
\vec{\beta}
\end{matrix}
;c\right] , \label{mu0}
\end{gather}
where ${}_{p}F_{q}$ is the generalized hypergeometric function~\cite[(16.2.1)]{MR2723248}. Depending on the values of~$p$ and~$q$, we have to consider three
dif\/ferent cases:
\begin{enumerate}\itemsep=0pt
\item If $p<q+1$, $\mu_{0}(c) $ is an entire function of $c$.

\item If $p=q+1$, $\mu_{0}(c) $ is analytic inside the unit disk
$\vert c\vert <1$.

\item If $p>q+1$, the series~(\ref{mu0}) diverges for $c\neq0$, unless one or
more of the top parameters $\alpha_{i}$ is a negative integer. If we take
$\alpha_{1}=-N$, with $N\in\mathbb{N}$, then $\mu_{0}(c)$
becomes a polynomial of degree~$N$.
\end{enumerate}

Using the formula \cite[(16.3.1)]{MR2723248}
\begin{gather*}
\frac{d^{n}}{dc^{n}}\,{}_{p}F_{q}\left[
\begin{matrix}
\vec{a}\\
\vec{b}
\end{matrix}
;c\right] =\frac{( \vec{a})_{n}}{(
\vec{b})_{n}}\,{}_{p}F_{q}\left[
\begin{matrix}
\vec{a}+n\\
\vec{b}+n
\end{matrix}
;c\right] ,
\end{gather*}
in (\ref{munD}), we have
\begin{gather}
\mu_{n}(c) =
\sum\limits_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
c^{k}\frac{( \vec{\alpha})_{k}}{(
\vec{\beta}+1)_{k}}\,{}_{p}F_{q}\left[
\begin{matrix}
\vec{\alpha}+k\\
\vec{\beta}+1+k
\end{matrix}
;c\right] . \label{mun gen}
\end{gather}
Although~(\ref{mun gen}) seems to give an explicit formula for the moments,
this type of sums (to our knowledge) can't be evaluated in closed form.

An alternative is to consider \emph{generalized moments}, def\/ined by~\cite{MR1688958}
\begin{gather*}
\nu_{n}=\mathcal{L} ( \varphi_{n}) ,
\end{gather*}
where $\{ \varphi_{n}\} \subset\mathbb{C}[x] $,
with $\deg(\varphi_{n}) =n$. Choosing\footnote{As suggested by
one of the anonymous referees.}
\begin{gather*}
\varphi_{n}(x) =(x-n+1)_{n},
\end{gather*}
we get%
\begin{gather*}
\nu_{n}(c) =
\sum\limits_{x=0}^{\infty}
(x-n+1)_{n}\frac{( \vec{\alpha})_{x}
}{( \vec{\beta}+1)_{x}}\frac{c^{x}}{x!}=\frac{(
\vec{\alpha})_{k}}{( \vec{\beta}+1)
_{k}}\,{}_{p}F_{q}\left[
\begin{matrix}
\vec{\alpha}+k\\
\vec{\beta}+1+k
\end{matrix}
;c\right] .
\end{gather*}
Since \cite[(26.8.10)]{MR2723248}
\begin{gather}
x^{n}=
\sum\limits_{k=1}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
(x-k+1)_{k}, \label{inversion1}
\end{gather}
we have
\begin{gather*}
\mu_{n}(c) =\mathcal{L}\big( x^{n}\big) =
\sum\limits_{k=1}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
\mathcal{L}( \varphi_{k}) =
\sum\limits_{k=1}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
\nu_{k}(c) ,
\end{gather*}
and we recover~(\ref{mun gen}).

In a series of papers \cite{MR2364955,MR2401156,MR2427672,MR2741218,MR2966475,MR3277948}, we studied polynomial
solutions of dif\/ferential-dif\/ference equations of the form%
\begin{gather}
P_{n+1}(x)=A_{n}(x)P_{n}^{\prime}(x)+B_{n}(x)P_{n}(x),\qquad n\geq0,
\label{eq:diff2}
\end{gather}
where $P_{0}(x)=1$, and $A_{n}(x)$, $B_{n}(x)$ are polynomials of degree at
most $2$ and $1$ respectively. In this article, we consider some extensions of~(\ref{eq:diff2}) to the multidimensional case, with $P_{n}(x)$ replaced by a~vector $\vec{P}_{n}(x)$, and $B_{n}(x)$ replaced by
a matrix~$\mathbf{B}_{n}(x)$.

In \cite{MR1451259} and~\cite{MR1627382}, it was shown that some families of
orthogonal polynomials can be represented as moments of probability measures.
In this work, we do the opposite and express the mo\-ments~$\mu_{n}(c)$ in terms of some polynomials~$\vec{P}_{n}(c)$.

The paper is organized as follows: in Section~\ref{section2}, we derive a
dif\/ferential-dif\/ference equation for the polynomials $\vec{P}_{n}(c) $ associated to the moments~$\mu_{n}(c)$.
We also f\/ind formulas for the exponential generating functions and Stieltjes
transforms of $\mu_{n}(c) $ and $\vec{P}_{n}\left(
c\right) $. In Section~\ref{section3}, we apply our results to most of the families of
discrete semiclassical orthogonal polynomials of class~1 studied in~\cite{MR3227440}, except for limiting and $c=1$ cases. Finally, in Section~\ref{section4}, we outline some conclusions and possible future directions.

\section{Main results}\label{section2}

The function $\mu_{0}(c) $ satisf\/ies the dif\/ferential equation
\cite[(16.8.3)]{MR2723248}
\begin{gather}
\big[ \vartheta ( \vartheta+\beta_{1} ) \cdots (
\vartheta+\beta_{q} ) -c ( \vartheta+\alpha_{1} )
\cdots ( \vartheta+\alpha_{p} ) \big] \mu_{0}=0, \label{ODE}
\end{gather}
where $\vartheta$ was def\/ined in~(\ref{theta}). We can rewrite the ODE~(\ref{ODE}) as
\begin{gather}
\vartheta^{q+1}\mu_{0} =
\sum\limits_{k=0}^{q}
\sigma_{k}(c) \vartheta^{k}\mu_{0},\qquad q>p-1,\nonumber\\
(1-c) \vartheta^{q+1}\mu_{0} =
\sum\limits_{k=0}^{q}
\sigma_{k}(c) \vartheta^{k}\mu_{0},\qquad q=p-1,\label{diffmu0}\\
c\vartheta^{p}\mu_{0} =
\sum\limits_{k=0}^{p-1}
\sigma_{k}(c) \vartheta^{k}\mu_{0},\qquad q<p-1,\nonumber
\end{gather}
where the coef\/f\/icients $\sigma_{k}(c) $ are linear functions of~$c$.

Introducing the quantities
\begin{gather*}
\vec{\mu}(c) =
\begin{bmatrix}
\mu_{0}(c) \\
\mu_{1}(c) \\
\vdots\\
\mu_{\xi}(c)
\end{bmatrix}
,\qquad\xi=\max \{ p-1,q \} ,
\end{gather*}
and
\begin{gather*}
(\lambda,\tau) =
\begin{cases}
(1,0) , & q+1>p, \\
(1,-1) , & q+1=p, \\
(0,1) , & q+1<p,
\end{cases}
\end{gather*}
we can rewrite (\ref{diffmu0}) as%
\begin{gather*}
( \lambda+c\tau) \mu_{\xi+1}=
\sum\limits_{k=0}^{\xi}
\sigma_{k}(c) \mu_{k}.
\end{gather*}
If we def\/ine the $(\xi+1) \times(\xi+1) $ matrix
$\mathbf{M}(c) $ by
\begin{gather*}
\mathbf{M}_{i,j}=
\begin{cases}
\sigma_{j}, & i=\xi+1,\quad 0\leq j\leq\xi,\\
\lambda+c\tau, & j=i+1,\quad 0\leq i\leq\xi-1,\\
0, & \text{otherwise},
\end{cases}

\end{gather*}
we get%
\begin{gather}
\mathbf{M}\,\vec{\mu}= ( \lambda+c\tau ) \vartheta
 \vec{\mu}.\label{Mmu}
\end{gather}
We can now state our main result.

\begin{proposition}
\label{main}Let the $(\xi+1)$-vector polynomials
$\vec{P}_{n}(c) $ be defined by
\begin{gather}
\vec{P}_{0}(c) =
\begin{bmatrix}
1\\
0\\
\vdots\\
0
\end{bmatrix}
\label{P0}
\end{gather}
and
\begin{gather}
\vec{P}_{n+1}=c ( \lambda+c\tau ) \frac
{d \vec{P}_{n}}{dc}+\big( \mathbf{M}^{T}-n\tau c\mathbf{I}
\big) \vec{P}_{n},\qquad n=0,1,\ldots,\label{reqPn}
\end{gather}
where $\mathbf{I}$ is the $(\xi+1) \times(\xi+1) $
identity matrix. Then,
\begin{gather}
\mu_{n}(c) =(\lambda+c\tau) ^{-n}
 \vec{P}_{n}(c) \cdot\vec{\mu} (c) ,\qquad n=0,1,\ldots. \label{Pmu}
\end{gather}
\end{proposition}

\begin{proof}
Using (\ref{Mmu}) and (\ref{reqPn}), we have
\begin{gather*}
\vec{P}_{n+1}\cdot\vec{\mu} = ( \lambda
+c\tau ) \vartheta \vec{P}_{n}\cdot\vec{\mu
}+\big( \mathbf{M}^{T}-n\tau c\mathbf{I}\big) \vec{P}
_{n}\cdot\vec{\mu}\\
\hphantom{\vec{P}_{n+1}\cdot\vec{\mu}}{}
 = ( \lambda+c\tau ) \vartheta \vec{P}_{n}
\cdot\vec{\mu}+\vec{P}_{n}\cdot ( \mathbf{M}-n\tau
c\mathbf{I} ) \vec{\mu}\\
\hphantom{\vec{P}_{n+1}\cdot\vec{\mu}}{}
 = ( \lambda+c\tau ) \vartheta \vec{P}_{n}
\cdot\vec{\mu}+\vec{P}_{n}\cdot ( \lambda
+c\tau ) \vartheta \vec{\mu} -n\tau c \vec{P}_{n}\cdot\vec{\mu}.
\end{gather*}
Multiplying by $( \lambda+c\tau) ^{-n-1}$ we get
\begin{gather*}
(\lambda+c\tau) ^{-n-1} \vec{P}_{n+1}\cdot
\vec{\mu} =( \lambda+c\tau) ^{-n}\vartheta
\vec{P}_{n}\cdot\vec{\mu}+\vec{P}_{n}
\cdot(\lambda+c\tau) ^{-n}\vartheta \vec{\mu} -n\tau c(\lambda+c\tau) ^{-n-1} \vec{P}_{n}
\cdot\vec{\mu}\\
\hphantom{(\lambda+c\tau) ^{-n-1} \vec{P}_{n+1}\cdot \vec{\mu}}{}
 =\vartheta\big[ (\lambda+c\tau) ^{-n} \vec{P}_{n}\cdot\vec{\mu}\big] .
\end{gather*}
Thus, the sequence
\begin{gather*}
r_{n}=(\lambda+c\tau) ^{-n}\vec{P}_{n}\cdot \vec{\mu}
\end{gather*}
satisf\/ies the recurrence $r_{n+1}=\vartheta r_{n}$ with initial condition
\begin{gather*}
r_{0}=\vec{P}_{0}\cdot\vec{\mu}=\mu_{0}.
\end{gather*}
From (\ref{reqmu}), we conclude that $r_{n}=\mu_{n}$.
\end{proof}

\subsection{Generating functions}\label{section2.1}

Let's consider the exponential generating function for the moments, def\/ined by
the formal power series
\begin{gather*}
G_{\mu}(c,w)=
\sum\limits_{n=0}^{\infty}
\mu_{n}(c) \frac{w^{n}}{n!}.
\end{gather*}
Given that
\begin{gather*}
\mu_{n+1}=c\mu_{n}^{\prime},
\end{gather*}
we have \cite{MR2172781}
\begin{gather*}
\frac{\partial G_{\mu}}{\partial w}=c\frac{\partial G_{\mu}}{\partial c},
\end{gather*}
with general solution
\begin{gather*}
G_{\mu}(c,w)=H\big( ce^{w}\big)
\end{gather*}
for some function $H(w)$. But since
\begin{gather*}
G_{\mu}(c,0)=\mu_{0}(c) ,
\end{gather*}
we conclude that
\begin{gather}
G_{\mu}(c,w)=\mu_{0}\big( ce^{w}\big) . \label{Gmu}
\end{gather}

Using (\ref{Pmu}), we see that the exponential generating function for the
polynomials $\vec{P}_{n}(c)$
\begin{gather*}
\vec{G}_{P}(c,w)=
\sum\limits_{n=0}^{\infty}
\vec{P}_{n}(c) \frac{w^{n}}{n!}
\end{gather*}
satisf\/ies
\begin{gather*}
G_{\mu}( c,( \lambda+\tau c) w) =\vec{G}_{P}(c,w)\cdot\vec{\mu}(c) ,
\end{gather*}
or, using (\ref{Gmu}),
\begin{gather}
\vec{G}_{P}(c,w)\cdot\vec{\mu}(c) =\mu
_{0}\big( ce^{( \lambda+\tau c) w}\big) . \label{GPn}
\end{gather}

\subsection{Stieltjes transform}\label{section2.2}

A dif\/ferent type of generating function for the moments that is very important
in the theory of orthogonal polynomials is the Stieltjes transform (or~$Z$
transform), that can be def\/ined by the formal Laurent series
\begin{gather*}
S_{\mu}(c,z)=
\sum\limits_{n=0}^{\infty}
\frac{\mu_{n}(c) }{z^{n+1}}
\end{gather*}
We have
\begin{gather*}
S_{\mu}(c,z) =
\sum\limits_{n=0}^{\infty}
\frac{\mu_{n}(c) }{z^{n+1}}=
\sum\limits_{n=-1}^{\infty}
\frac{\mu_{n+1}(c) }{z^{n+2}}=\frac{1}{z}
\sum\limits_{n=-1}^{\infty}
\frac{\mu_{n+1}(c) }{z^{n+1}}
 =\frac{1}{z}\left[ \mu_{0}(c) +
\sum\limits_{n=0}^{\infty}
\frac{\mu_{n+1}(c) }{z^{n+1}}\right] .
\end{gather*}
Hence,
\begin{gather}
\sum\limits_{n=0}^{\infty}
\frac{\mu_{n+1}(c) }{z^{n+1}}=zS_{\mu}(c,z)-\mu_{0}(c) \label{shift}
\end{gather}
and using (\ref{reqmu}), we get
\begin{gather}
c\frac{\partial S_{\mu}}{\partial c}=zS_{\mu}-\mu_{0}(c).\label{PDES}
\end{gather}
From (\ref{mun def}), we have
\begin{gather*}
S_{\mu}(0,z)=\frac{1}{z}
\end{gather*}
and solving (\ref{PDES}) we obtain
\begin{gather*}
S_{\mu}(c,z)=-c^{z}
\int_{0}^{c}
\frac{\mu_{0}(x) }{x^{z+1}}dx=-
\int_{0}^{1}
\frac{\mu_{0}(ct) }{t^{z+1}}dt.
\end{gather*}
Using the recurrence relation for the Gamma function, we can write%
\begin{gather*}
-
\int_{0}^{1}
\frac{\mu_{0}(ct) }{t^{z+1}}dt=\frac{1}{z}\frac{\Gamma(
1-z) }{\Gamma(1) \Gamma(-z)}
\int_{0}^{1}
\frac{\mu_{0}(ct) }{t^{z+1}}dt,
\end{gather*}
and therefore
\begin{gather}
S_{\mu}(c,z)=\frac{1}{z}\,{}_{p+1}F_{q+1}\left[
\begin{matrix}
-z,\ \vec{\alpha}\\
1-z,\ \vec{\beta}+1
\end{matrix}
;c\right] ,\label{stiltjes}
\end{gather}
where we have used the integral representation \cite[(16.5.2)]{MR2723248}
\begin{gather*}
{}_{p+1}F_{q+1}\left[
\begin{matrix}
\alpha_{0},\ \vec{\alpha}\\
\beta_{0},\ \vec{\beta}
\end{matrix}
;c\right]
 =\frac{\Gamma ( \beta_{0}) }{\Gamma( \alpha_{0})
\Gamma( \beta_{0}-\alpha_{0}) }
\int_{0}^{1}
t^{\alpha_{0}-1}( 1-t) ^{\beta_{0}-\alpha_{0}-1}\,{}_{p}
F_{q}\left[
\begin{matrix}
\vec{\alpha}\\
\vec{\beta}
\end{matrix}
;ct\right] dt.
\end{gather*}
We derived (\ref{stiltjes}) in \cite{MH} using a dif\/ferent approach.

If we def\/ine the Stieltjes transform of $\vec{P}_{n}(c) $ by
\begin{gather*}
\vec{S}_{P}(c,z) =
\sum\limits_{n=0}^{\infty}
\frac{\vec{P}_{n}(c) }{z^{n+1}},
\end{gather*}
then it follows from (\ref{Pmu}) that
\begin{gather*}
\frac{1}{\lambda+\tau c}S_{\mu}\left( c,\frac{z}{\lambda+\tau c}\right)
=\vec{S}_{P}(c,z) \cdot\vec{\mu}(c) ,
\end{gather*}
or
\begin{gather}
\vec{S}_{P}(c,z) \cdot\vec{\mu} (
c ) =\frac{1}{z}\,{}_{p+1}F_{q+1}\left[
\begin{matrix}
-\frac{z}{\lambda+\tau c},\ \vec{\alpha}\\
1-\frac{z}{\lambda+\tau c},\ \vec{\beta}+1
\end{matrix}
;c\right] . \label{laplaceG}
\end{gather}

\section{Examples}\label{section3}

\strut In \cite{MR3227440} we studied all families of semiclassical
polynomials of class $s\leq1$ orthogonal with respect to~(\ref{L}). When
$s=0$, we have three canonical cases (the \emph{discrete classical
polynomials}):
$$
\begin{tabular}
[c]{|c|c|c|}\hline
$\deg(\eta) $ & $\deg(\phi) $ & \\\hline
$0$ & $1$ & Charlier\\\hline
$1$ & $1$ & Meixner\\\hline
$2$ & $2$ & Hahn\\\hline
\end{tabular}
$$
where $\phi(x) $ and $\eta(x) $ were def\/ined in~(\ref{phi-eta}). These polynomials are associated with the Poisson, Pascal,
and hypergeometric probability distributions, and results about their moments
have appeared in many places before (see~\cite{MR1224449}, for instance).

In \cite{MR3286553}, the authors used (\ref{inversion1}) and inversion
formulas of the form
\begin{gather*}
(x-n+1)_{n}=\sum\limits_{k=0}^{n}
c_{n,k}Q_{k}(x) ,
\end{gather*}
to derive expressions for the moments and exponential generating functions of
the discrete classical polynomials. Stieltjes' transforms were not considered.

When $s=1$, we obtained f\/ive cases:%
$$
\begin{tabular}
[c]{|c|c|c|}\hline
$\deg(\eta) $ & $\deg(\phi) $ & \\\hline
$0$ & $2$ & generalized Charlier\\\hline
$1$ & $2$ & generalized Meixner\\\hline
$2$ & $1$ & generalized Krawtchouk\\\hline
$2$ & $2$ & generalized Hahn of type I\\\hline
$3$ & $3$ & generalized Hahn of type II\\\hline
\end{tabular}
$$
The moments of these polynomials have not (to our knowledge) been studied before.

\subsection{Charlier}\label{section3.1}

The Charlier polynomials \cite[(18.20.8)]{MR2723248}
\begin{gather*}
C_{n}(x;c) ={}_{2}F_{0}\left(
\begin{matrix}
-n,-x\\
-
\end{matrix}
;-\frac{1}{c}\right)
\end{gather*}
are orthogonal with respect to the weight function \cite{MR2300274}
\begin{gather*}
\rho_{C}(x) =\frac{c^{x}}{x!},\qquad c>0.
\end{gather*}
In this case, we have
\begin{gather*}
\mu_{0}(c) ={}_{0}F_{0}\left[
\begin{matrix}
-\\
-
\end{matrix}
;c\right] =e^{c},
\end{gather*}
and hence
\begin{gather*}
\xi=0,\qquad(\lambda,\tau) =(1,0) .
\end{gather*}
From (\ref{ODE}) we see that $\mu_{0}(c) $ satisf\/ies the ODE
\begin{gather*}
(\vartheta-c) \mu_{0}=0,
\end{gather*}
which implies $\mu_{1}=c\mu_{0}$. Thus, $\sigma_{0}(c) =c$, and
Proposition~\ref{main} gives
\begin{gather*}
\mu_{n}(c) =P_{n}(c) \mu_{0}(c) ,
\end{gather*}
with $P_{n}(c) $ def\/ined by $P_{0}(c) =1$ and
\begin{gather}
P_{n+1}=c\frac{dP_{n}}{dc}+cP_{n}. \label{Bell}
\end{gather}
The polynomials satisfying~(\ref{Bell}) are known as Bell (or Touchard, or
exponential) polynomials. It is well known that they have the explicit
representation~\cite{MR2583007}
\begin{gather}
P_{n}(c) =
\sum\limits_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
c^{k}, \label{bell}
\end{gather}
and therefore (see also \cite[equation~(44)]{MR3286553})
\begin{gather*}
\mu_{n}(c) =e^{c}
\sum\limits_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
c^{k},
\end{gather*}
in agreement with (\ref{mun gen}).

Using (\ref{Gmu}) and (\ref{GPn}), we see that the generating functions of
$\mu_{n}(c) $ and $P_{n}(c) $ are given by (see
also \cite[equation~(50)]{MR3286553})
\begin{gather*}
G_{\mu}(c,w)=\mu_{0}\big( ce^{w}\big) =e^{ce^{w}},
\end{gather*}
and
\begin{gather*}
G_{P}(c,w)=\frac{e^{ce^{w}}}{e^{c}}=e^{c( e^{w}-1) }.
\end{gather*}

\subsection{Meixner}\label{section3.2}

The Meixner polynomials \cite[(18.20.7)]{MR2723248}
\begin{gather*}
M_{n}(x;\alpha,c) ={}_{2}F_{1}\left(
\begin{matrix}
-n,-x\\
\alpha
\end{matrix}
;1-\frac{1}{c}\right)
\end{gather*}
are orthogonal with respect to the weight function \cite[(18.19)]{MR2723248}
\begin{gather}
\rho_{M}(x;\alpha,c) =(\alpha)_{x}\frac{c^{x}
}{x!},\qquad 0<c<1,\qquad\alpha>0. \label{rhomeixner}
\end{gather}
In this case, we have%
\begin{gather}
\mu_{0}^{M}(c;\alpha) ={}_{1}F_{0}\left[
\begin{matrix}
\alpha\\
-
\end{matrix}
;c\right] =(1-c) ^{-\alpha}, \label{mu0meixner}
\end{gather}
and hence
\begin{gather*}
\xi=0,\qquad(\lambda,\tau) =(1,-1) .
\end{gather*}
From (\ref{ODE}) we see that $\mu_{0}^{M}(c;\alpha) $ satisf\/ies
the ODE%
\begin{gather*}
[ \vartheta-c(\vartheta+\alpha) ] \mu_{0}^{M}=0,
\end{gather*}
which implies%
\begin{gather*}
(1-c) \mu_{1}^{M}=\alpha c\mu_{0}^{M}.
\end{gather*}
Thus, $\sigma_{0}(c) =\alpha c$, and Proposition \ref{main} gives
\begin{gather}
\mu_{n}^{M}(c;\alpha) =(1-c) ^{-n} P_{n}
^{M}(c) \mu_{0}^{M}(c;\alpha) , \label{mu meixner}
\end{gather}
with $P_{n}^{M}(c) $ def\/ined by $P_{0}^{M}(c) =1$
and (see also \cite[equation~(5.68)]{Njionou})
\begin{gather}
P_{n+1}^{M}=c(1-c) \frac{dP_{n}^{M}}{dc}+ ( \alpha +n) cP_{n}^{M}. \label{DDMeixner}
\end{gather}

From (\ref{mun gen}), we have (see also \cite[equation~(43)]{MR3286553})
\begin{gather*}
\mu_{n}^{M}(c;\alpha) =
\sum\limits_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
c^{k}(\alpha)_{k} (1-c) ^{-(\alpha+k) },
\end{gather*}
and therefore
\begin{gather}
P_{n}^{M}(c) =
\sum\limits_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
(\alpha)_{k} c^{k}(1-c) ^{n-k}. \label{meixner}
\end{gather}

Using (\ref{Gmu}) and (\ref{GPn}), we see that the generating functions of
$\mu_{n}^{M}(c;\alpha) $ and $P_{n}^{M}(c) $ are
given by (see also \cite[equation~(49)]{MR3286553})
\begin{gather*}
G_{\mu}(c,w)=\mu_{0}\big( ce^{w}\big) =\big( 1-ce^{w}\big)
^{-\alpha},
\end{gather*}
and
\begin{gather*}
G_{P}(c,w)=\frac{\big[ 1-ce^{(1-c) w}\big] ^{-\alpha}
}{(1-c) ^{-\alpha}}=\left[ \frac{1-ce^{(1-c) w}}{1-c}\right] ^{-\alpha}.
\end{gather*}

\subsubsection{Krawtchouk polynomials}

\strut The Krawtchouk polynomials \cite[(18.20.6)]{MR2723248}
\begin{gather*}
K_{n} ( x;\mathfrak{p,}N ) ={}_{2}F_{1}\left(
\begin{matrix}
-n,-x\\
-N
\end{matrix}
;\frac{1}{\mathfrak{p}}\right) ,
\end{gather*}
are orthogonal with respect to the weight function \cite{MR2390273}
\begin{gather*}
\rho_{K} ( x;\mathfrak{p,}N ) =\binom{N}{x}\mathfrak{p}^{x} (
1-\mathfrak{p} ) ^{N-x},\qquad N\in\mathbb{N},\qquad 0<\mathfrak{p}<1.
\end{gather*}
It is related to the weight function of the Meixner polynomials~(\ref{rhomeixner}) by
\begin{gather}
\rho_{K} ( x;\mathfrak{p,}N ) = ( 1-\mathfrak{p} )
^{N}\frac{( -N)_{x}}{x!}\left( \frac{\mathfrak{p}}
{\mathfrak{p}-1}\right) ^{x}=( 1-\mathfrak{p}) ^{N}\rho
_{M}\left( x;-N,\frac{\mathfrak{p}}{\mathfrak{p}-1}\right) .\label{rhokrawt}
\end{gather}

Using (\ref{rhokrawt}) in (\ref{mu0meixner}), we obtain
\begin{gather*}
\mu_{0}^{K} ( \mathfrak{p} ) = ( 1-\mathfrak{p} )
^{N}\mu_{0}^{M}\left( -N,\frac{\mathfrak{p}}{\mathfrak{p}-1}\right) =1.
\end{gather*}
and using (\ref{mu meixner}) and (\ref{meixner}), we get $\mu_{n}^{K} (
\mathfrak{p} ) = P_{n}^{K} ( \mathfrak{p} ) $, with (see
also \cite[equation~(42)]{MR3286553})
\begin{gather*}
P_{n}^{K} ( \mathfrak{p} ) =
 \sum\limits_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
\binom{N}{k}k! \mathfrak{p}^{k}.
\end{gather*}
From (\ref{DDMeixner}) and (\ref{rhokrawt}), we have
\begin{gather*}
P_{n+1}^{K}=\mathfrak{p} ( 1-\mathfrak{p} ) \frac{dP_{n}^{K}
}{d\mathfrak{p}}+N\mathfrak{p}P_{n}^{K}.
\end{gather*}

To obtain the generating function of $P_{n}^{K}( \mathfrak{p})$,
we can use \cite[(26.8.12)]{MR2723248}
\begin{gather}
\sum\limits_{n=0}^{\infty}
\genfrac{\{}{\}}{0pt}{}{n}{k}
\frac{w^{n}}{n!}=\frac{( e^{w}-1) ^{k}}{k!}, \label{gen Stirling}
\end{gather}
and we get (see also \cite[equation~(48)]{MR3286553})
\begin{gather*}
G_{P}(\mathfrak{p},w)=\big[ 1+\big( e^{w}-1\big) \mathfrak{p}\big]^{N}.
\end{gather*}

\subsection{Generalized Charlier}\label{section3.3}

These polynomials are orthogonal with respect to the weight function~\cite{MR2996960}
\begin{gather*}
\rho(x;\beta,c) =\frac{1}{(\beta+1)_{x}}
\frac{c^{x}}{x!},\qquad c>0,\qquad \beta>-1.
\end{gather*}
In this case, we have
\begin{gather*}
\mu_{0}(c) ={}_{0}F_{1}\left[
\begin{matrix}
-\\
\beta+1
\end{matrix}
;c\right] =c^{-\frac{\beta}{2}}\Gamma(\beta+1) I_{\beta}\big(
2\sqrt{c}\big) ,
\end{gather*}
where $I_{\beta}(z) $ is the modif\/ied Bessel function of the
f\/irst kind \cite[(10.39.9)]{MR2723248}, and hence
\begin{gather*}
\xi=1,\qquad(\lambda,\tau) =(1,0) .
\end{gather*}
From (\ref{ODE}) we see that $\mu_{0}(c) $ satisf\/ies the ODE
\begin{gather*}
\vartheta ( \vartheta+\beta ) \mu_{0}-c\mu_{0}=0,
\end{gather*}
which implies
\begin{gather*}
\mu_{2}=c\mu_{0}-\beta\mu_{1},
\end{gather*}
and therefore
\begin{gather*}
\sigma_{0}(c) =c,\qquad\sigma_{1}(c) =-\beta.
\end{gather*}
Proposition \ref{main} gives
\begin{gather*}
\mu_{n}(c) =\vec{P}_{n}(c) \cdot\vec{\mu}(c) ,
\end{gather*}
with $\vec{P}_{n}(c) $ def\/ined by~(\ref{P0}) and
\begin{gather}
\vec{P}_{n+1}=c\frac{d \vec{P}_{n}}{dc}+
\begin{bmatrix}
0 & c\\
1 & -\beta
\end{bmatrix}
 \vec{P}_{n}.\label{dd gen charlier}
\end{gather}

To obtain the Stieltjes transform of $\vec{P}_{n}(c)$, let's write
\begin{gather*}
\vec{P}_{n}(c) =\left[
\begin{matrix}
Q_{n}(c) \\
R_{n}(c)
\end{matrix}
\right] ,\qquad\vec{S}_{P}(c,z) =\left[
\begin{matrix}
U(c,z)\\
V(c,z)
\end{matrix}
\right] .
\end{gather*}
From (\ref{dd gen charlier}), we have
\begin{gather*}
Q_{n+1} =cQ_{n}^{\prime}+cR_{n},\qquad
R_{n+1} =cR_{n}^{\prime}-\beta R_{n}+Q_{n},
\end{gather*}
with $Q_{0}=1$, $R_{0}=0$. Using~(\ref{shift}), we get
\begin{gather}\label{UV laplace}
zU-1 =c\frac{\partial U}{\partial c}+cV, \qquad
zV =c\frac{\partial V}{\partial c}-\beta V+U.
\end{gather}
The solution of the system (\ref{UV laplace}) is given by
\begin{gather*}
U(c,z) =\frac{1}{z}\,{}_{1}F_{2}\left[
\begin{matrix}
1\\
1-z,\ -\beta-z
\end{matrix}
;c\right] ,\qquad
V(c,z) =\frac{1}{z ( z+\beta ) }\,{}_{1}
F_{2}\left[
\begin{matrix}
1\\
1-z,\ 1-\beta-z
\end{matrix};c\right] .
\end{gather*}

In this case
\begin{gather*}
\mu_{0}(c) ={}_{0}F_{1}\left[
\begin{matrix}
-\\
\beta+1
\end{matrix}
;c\right] ,\qquad
\mu_{1}(c) =\frac{c}{\beta+1}\,{}_{0}F_{1}\left[
\begin{matrix}
-\\
\beta+2
\end{matrix}
;c\right] ,
\end{gather*}
and since
\begin{gather*}
 \frac{1}{z}\,{}_{1}F_{2}\left[
\begin{matrix}
1\\
1-z,\ -\beta-z
\end{matrix}
;c\right] \,{}_{0}F_{1}\left[
\begin{matrix}
-\\
\beta+1
\end{matrix}
;c\right] \\
 \qquad{} +\frac{1}{z(z+\beta) }\,{}_{1}F_{2}\left[
\begin{matrix}
1\\
1-z,\ 1-\beta-z
\end{matrix}
;c\right] \frac{c}{\beta+1}\,{}_{0}F_{1}\left[
\begin{matrix}
-\\
\beta+2
\end{matrix}
;c\right]  =\frac{1}{z}\,{}_{1}F_{2}\left[
\begin{matrix}
-z\\
1-z,\ \beta+1
\end{matrix}
;c\right] ,
\end{gather*}
we see that (\ref{laplaceG}) is satisf\/ied.

\subsection{Generalized Meixner}\label{section3.4}

\strut These polynomials are orthogonal with respect to the weight function~\cite{MR2749070}
\begin{gather*}
\rho( x;\alpha,\beta,c) =\frac{(\alpha)_{x}
}{(\beta+1)_{x}}\frac{c^{x}}{x!},\qquad c,\alpha>0,\qquad
\beta>-1.
\end{gather*}
In this case, we have
\begin{gather*}
\mu_{0}(c) ={}_{1}F_{1}\left[
\begin{matrix}
\alpha\\
\beta+1
\end{matrix}
;c\right] =M ( \alpha,\beta+1,c ) ,
\end{gather*}
where $M(a,b,c) $ is Kummer's (conf\/luent hypergeometric)
function \cite[(13.2.2)]{MR2723248}, and hence
\begin{gather*}
\xi=1,\qquad(\lambda,\tau) =(1,0) .
\end{gather*}
From (\ref{ODE}) we see that $\mu_{0}(c) $ satisf\/ies the ODE
\begin{gather*}
\vartheta ( \vartheta+\beta ) \mu_{0}-c ( \vartheta +\alpha ) \mu_{0}=0
\end{gather*}
which implies
\begin{gather*}
\mu_{2}=c\alpha\mu_{0}+(c-\beta) \mu_{1},
\end{gather*}
and therefore
\begin{gather*}
\sigma_{0}(c) =\alpha c,\qquad\sigma_{1}(c) =c-\beta.
\end{gather*}
Proposition \ref{main} gives
\begin{gather*}
\mu_{n}(c) =\vec{P}_{n}(c) \cdot\vec{\mu}(c) ,
\end{gather*}
with $\vec{P}_{n}(c) $ def\/ined by~(\ref{P0}) and
\begin{gather*}
\vec{P}_{n+1}=c\frac{d \vec{P}_{n}}{dc}+ \begin{bmatrix}
0 & \alpha c\\
1 & c-\beta
\end{bmatrix}
\vec{P}_{n}
\end{gather*}

To obtain the Stieltjes transform of $\vec{P}_{n}(c)$, let's write
\begin{gather*}
\vec{P}_{n}(c) =\left[
\begin{matrix}
Q_{n}(c) \\
R_{n}(c)
\end{matrix}
\right] ,\qquad\vec{S}_{P}(c,z) =\left[
\begin{matrix}
U(c,z)\\
V(c,z)
\end{matrix}
\right] .
\end{gather*}
From (\ref{dd gen charlier}), we have
\begin{gather*}
Q_{n+1} =cQ_{n}^{\prime}+\alpha cR_{n},\qquad
R_{n+1} =cR_{n}^{\prime}+(c-\beta) R_{n}+Q_{n},
\end{gather*}
with $Q_{0}=1$, $R_{0}=0$. Using (\ref{shift}), we get
\begin{gather}\label{UV laplace 1}
zU-1 =c\frac{\partial U}{\partial c}+\alpha cV,\qquad
zV =c\frac{\partial V}{\partial c}+(c-\beta) V+U.
\end{gather}

If we represent the functions $U$, $V$ as
\begin{gather*}
U(c,z)=
\sum\limits_{n=0}^{\infty}
u_{n}(z) c^{n},\qquad V(c,z)=
\sum\limits_{n=0}^{\infty}
v_{n}(z) c^{n},
\end{gather*}
then (\ref{UV laplace 1}) gives
\begin{gather}\label{req un vn}
zu_{n}-\delta_{n,0} =nu_{n}+\alpha v_{n-1},\qquad
zv_{n} =nv_{n}+v_{n-1}-\beta v_{n}+u_{n}.
\end{gather}
We have (assuming that $u_{n}=v_{n}=0$ for $n<0)$
\begin{gather*}
u_{0}=\frac{1}{z},\qquad v_{0}=\frac{1}{z(z+\beta) },
\end{gather*}
and%
\begin{gather}
u_{n}=(z+\beta-n) v_{n}-v_{n-1}. \label{un}
\end{gather}
Using (\ref{un}) in (\ref{req un vn}), we obtain
\begin{gather*}
v_{n}=\frac{z+\alpha-n}{( z-n) (z+\beta-n) }v_{n-1},\qquad n=1,2,\ldots,
\end{gather*}
and therefore
\begin{gather*}
v_{n}=\frac{(-1) ^{n}}{z(z+\beta) }\frac{(
1-\alpha-z)_{n}}{( 1-z)_{n}( 1-\beta-z)
_{n}},\qquad n=0,1,\ldots.
\end{gather*}
From (\ref{un}), it follows that
\begin{gather*}
u_{n}=\alpha\frac{(-1) ^{n}}{z(z+\alpha) }
\frac{(-\alpha-z)_{n}}{(1-z)_{n}(-\beta-z)_{n}},\qquad n=1,2,\ldots.
\end{gather*}
Thus,
\begin{gather*}
U(c,z) =\frac{1}{z}+\frac{\alpha}{z(z+\alpha)
}
\sum\limits_{n=1}^{\infty}
\frac{(-\alpha-z)_{n}}{(1-z)_{n}(
-\beta-z)_{n}}(-c) ^{n}\\
\hphantom{U(c,z)}{} =\frac{1}{z}-\frac{\alpha}{z(z+\alpha) }+\frac{\alpha
}{z(z+\alpha) }
\sum\limits_{n=0}^{\infty}
\frac{(-\alpha-z)_{n}}{(1-z)_{n}(
-\beta-z)_{n}}(-c) ^{n}\\
\hphantom{U(c,z)}{}
 =\frac{1}{z+\alpha}+\frac{\alpha}{z(z+\alpha) }\,{}_{2}F_{2}\left[
\begin{matrix}
1,-\alpha-z\\
1-z,\ -\beta-z
\end{matrix}
;-c\right] ,
\end{gather*}
and
\begin{gather*}
V(c,z) =\frac{1}{z(z+\beta) }\,{}_{2}F_{2}\left[
\begin{matrix}
1,\ 1-\alpha-z\\
1-z,\ 1-\beta-z
\end{matrix}
;-c\right] .
\end{gather*}

In this case
\begin{gather*}
\mu_{0}(c) ={}_{1}F_{1}\left[
\begin{matrix}
\alpha\\
\beta+1
\end{matrix}
;c\right] ,\qquad
\mu_{1}(c) =\frac{\alpha c}{\beta+1}\,{}_{1}F_{1}\left[
\begin{matrix}
\alpha+1\\
\beta+2
\end{matrix}
;c\right] ,
\end{gather*}
and since
\begin{gather*}
 \left( \frac{1}{z+\alpha}+\frac{\alpha}{z(z+\alpha) }
\,{}_{2}F_{2}\left[
\begin{matrix}
1,-\alpha-z\\
1-z,\ -\beta-z
\end{matrix}
;-c\right] \right) \,{}_{1}F_{1}\left[
\begin{matrix}
\alpha\\
\beta+1
\end{matrix}
;c\right] \\
\! \qquad {} +\frac{1}{z(z+\beta) }\,{}_{2}F_{2}\left[
\begin{matrix}
1,\ 1-\alpha-z\\
1-z,\ 1-\beta-z
\end{matrix}
;-c\right] \frac{\alpha c}{\beta+1}\,{}_{1}F_{1}\left[
\begin{matrix}
\alpha+1\\
\beta+2
\end{matrix}
;c\right]   =\frac{1}{z}\,{}_{2}F_{2}\left[
\begin{matrix}
-z,\ \alpha\\
1-z,\ \beta+1
\end{matrix}
;c\right] ,
\end{gather*}
we see that (\ref{laplaceG}) is satisf\/ied.

\subsection{Generalized Krawtchouk}\label{section3.5}

\strut These polynomials are orthogonal with respect to the weight function~\cite{MR3227440}
\begin{gather*}
\rho(x;\alpha,N,c) =(\alpha)_{x} (
-N )_{x}\frac{c^{x}}{x!},\qquad c<0,\qquad\alpha>0,\qquad N\in
\mathbb{N}.
\end{gather*}
In this case, we have
\begin{gather*}
\mu_{0}(c) ={}_{2}F_{0}\left[
\begin{matrix}
\alpha,\ -N\\
-
\end{matrix}
;c\right] =C_{N}\big({-}\alpha;-c^{-1}\big) ,
\end{gather*}
where $C_{N}(x;\mu) $ is the Charlier polynomial
\cite{MR2300274}, and hence%
\begin{gather*}
\xi=1,\qquad(\lambda,\tau) =(0,1) .
\end{gather*}
From (\ref{ODE}) we see that $\mu_{0}(c) $ satisf\/ies the ODE
\begin{gather*}
\vartheta\mu_{0}-c(\vartheta+\alpha) (\vartheta-N)
\mu_{0}=0,
\end{gather*}
which implies
\begin{gather*}
c\mu_{2}=\alpha Nc\mu_{0}+(Nc-\alpha c+1) \mu_{1},
\end{gather*}
and therefore%
\begin{gather*}
\sigma_{0}(c) =\alpha Nc,\qquad\sigma_{1}(c)
=(N-\alpha) c+1.
\end{gather*}
Proposition \ref{main} gives
\begin{gather*}
\mu_{n}(c) =c^{-n}\vec{P}_{n}(c)
\cdot\vec{\mu}(c) ,
\end{gather*}
with $\vec{P}_{n}(c) $ def\/ined by (\ref{P0}) and
\begin{gather*}
\vec{P}_{n+1}=c^{2}\frac{d \vec{P}_{n}}{dc}+
\begin{bmatrix}
-nc & \alpha Nc\\
c & \left( N-\alpha-n\right) c+1
\end{bmatrix}
\vec{P}_{n}(c) .
\end{gather*}

\subsection{Generalized Hahn polynomials of type I}\label{section3.6}

These polynomials are orthogonal with respect to the weight function~\cite{MR3227440}
\begin{gather}
\rho( x;\alpha_{1},\alpha_{2},\beta;c) =\frac{( \alpha
_{1})_{x}( \alpha_{2})_{x}}{(\beta+1)
_{x}}\frac{c^{x}}{x!}, \label{rho gen hahn}
\end{gather}
where $0<c<1$, $\alpha_{1},\alpha_{2}>0$, $\beta>-1$. In this case, we have%
\begin{gather*}
\mu_{0}(c) = {}_{2}F_{1}\left[
\begin{matrix}
\alpha_{1},\ \alpha_{2}\\
\beta+1
\end{matrix}
;c\right] ,
\end{gather*}
and hence%
\begin{gather*}
\xi=1,\qquad(\lambda,\tau) =(1,-1) .
\end{gather*}
From (\ref{ODE}) we see that $\mu_{0}(c) $ satisf\/ies the ODE
\begin{gather*}
\vartheta ( \vartheta+\beta ) \mu_{0}-c ( \vartheta+\alpha
_{1} ) ( \vartheta+\alpha_{2} ) \mu_{0}=0,
\end{gather*}
which implies
\begin{gather*}
(1-c) \mu_{2}=\allowbreak\alpha_{1}\alpha_{2}c\mu_{0}+\big[
 ( \alpha_{1}+\alpha_{2} ) c-\beta\big] \mu_{1},
\end{gather*}
and therefore
\begin{gather*}
\sigma_{0}(c) =\alpha_{1}\alpha_{2}c,\qquad\sigma_{1} (
c ) = ( \alpha_{1}+\alpha_{2} ) c-\beta.
\end{gather*}
Proposition~\ref{main} gives
\begin{gather*}
\mu_{n}(c) =(1-c) ^{-n} \vec{P}_{n}(c) \cdot\vec{\mu}(c) ,
\end{gather*}
with $\vec{P}_{n}(c) $ def\/ined by~(\ref{P0}) and
\begin{gather*}
 \vec{P}_{n+1}=c(1-c) \frac{d \vec{P}
_{n}}{dc}+
\begin{bmatrix}
nc & \alpha_{1}\alpha_{2}(1-c) c\\
 1-c & ( \alpha_{1}+\alpha_{2}+n ) c-\beta
\end{bmatrix}
\vec{P}_{n}.
\end{gather*}

\subsection{Hahn polynomials}\label{section3.7}

The Hahn polynomials \cite[(18.20.5)]{MR2723248}
\begin{gather*}
H_{n}(x;\alpha,\beta,N) ={}_{3}F_{2}\left(
\begin{matrix}
-n,-x,n+\alpha+\beta+1\\
-N,\alpha+1
\end{matrix}
;1\right)
\end{gather*}
are orthogonal with respect to the weight function \cite[(18.19)]{MR2723248}
\begin{gather*}
\rho_{H}(x;\alpha,\beta,N) =\frac{(\alpha+1)_{x}}{x!}\frac{(\beta+1)_{N-x}}{(N-x)!},\qquad
\alpha,\beta\notin[-N,-1] ,\qquad N\in\mathbb{N},
\end{gather*}
or
\begin{gather*}
\rho_{H}(x;\alpha,\beta,N) =\binom{\alpha+x}{x}\binom{\beta
+N-x}{N-x}.
\end{gather*}
The relation between $\rho_{H}(x;\alpha,\beta,N) $ and the
weight function of the generalized Hahn polyno\-mials~(\ref{rho gen hahn}) is
given by
\begin{gather*}
\rho_{H}(x;\alpha,\beta,N) =\frac{(\beta+1)_{N}}{N!}\rho ( x;\alpha+1,-N,-N-\beta-1;1 ) .
\end{gather*}

From (\ref{mun gen}), we get (see also \cite[equation~(41)]{MR3286553})
\begin{gather*}
\mu_{n}^{H}=
\sum\limits_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
(\alpha+1)_{k} \frac{( \alpha+\beta+2+k)_{N-k}}{( N-k) !},
\end{gather*}
where we have used
\begin{gather*}
\mu_{0}^{H}=\frac{(\alpha+\beta+2)_{N}}{N!}.
\end{gather*}

To obtain the generating function of $\mu_{n}^{H}$, we can use~(\ref{gen Stirling}), and we get
\begin{gather*}
G_{\mu^{H}}(z)=\frac{( \alpha+\beta+2)_{N}}{N!}\,{}_{2}
F_{1}\left[
\begin{matrix}
-N,\ \alpha+1\\
\alpha+\beta+2
\end{matrix}
;1-e^{z}\right] .
\end{gather*}
This generating function seems not to have been considered before.

\begin{remark}
Except for the Bell polynomials (\ref{bell}), all the other families
$\vec{P}_{n}(c) $ associated to the moments $\mu
_{n}(c) $ seem to be new.
\end{remark}

\begin{remark}
Stieltjes transforms of the generalized Krawtchouk and generalized Hahn
polynomials of type~I have been omitted because of the complexity of the formulas.
\end{remark}

\section{Conclusion}\label{section4}

We have developed a technique for computing the moments of weight functions of
hypergeometric type. We have shown that the moments are linear combinations of
the f\/irst $\xi+1$ moments with polynomial coef\/f\/icients in the parameter~$c$.
We have also constructed generating functions for both the moments and the
polynomials associated with them.

All the results in Sections~\ref{section3.3}--\ref{section3.5} are new, and give ef\/f\/icient ways of
computing the moments of the discrete semiclassical polynomials of class~$1$.
The same method can be used to f\/ind the moments of polynomials of class~$s>1$.

In a previous work \cite{MR3277948}, we found the asymptotic zero distribution
of polynomial families satisfying f\/irst-order dif\/ferential-recurrence
relations of the form~(\ref{eq:diff2}). It would be interesting to know if our
results could be extended to include the polynomials $\vec{P}_{n}(c)$ studied in this paper.

\subsection*{Acknowledgements}

We want to thank the anonymous referees and the editors of this
special issue for providing us with extremely valuable suggestions that helped
us to greatly improve our f\/irst draft of the paper.

\pdfbookmark[1]{References}{ref}

redpajama_set_name: RedPajamaArXiv


November is the next Disney retreat at Coronado Springs.
Still putting up hay and mowing the pastures.
Amy has taken a position with the new ARC POWER grant funded Local Food Shed program as Senior Coordinator of Sheep, Lamb and Goats.
This entry was posted in Podcasts and tagged farm, foodie, knitting, spinning, travel on September 6, 2017 by Amy Manko.


text:
Q: Probability - Birthday paradox Let us assume every child in the world has a random and uniform favorite number between $1$ and $m$, and also has a different random and uniform unfavorite number between $1$ and $m$.
Denote $E_{k,m}$ the event in which, in a class with $k$ children there's a pair of children
who chose exactly the same numbers (it doesn't matter favorite/unfavorite).
Prove that for any $\alpha>1$ it holds that:
$\lim _{m\rightarrow \infty }\left( \min _{k\in(m^{\alpha},\infty)} \mathbb{P}(E_{k,m})\right) = 1$.
I though about working with $E_{k,m}^{c}$ as done in the birthday paradox but I'm stuck on it and don't have any leads on how to show it

redpajama_set_name: RedPajamaStackExchange


Many people are afraid they do not have the skills necessary to complete their home improvement projects. Home improvement does not need to be difficult when you have the right information before you begin it. Use these tips to get started with your own home improvement projects. Visit https://radonmannen.no/radonmembran for more details about radon prevention.
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To minimize the "peek-through" of walls behind paneling, paint stripes on the walls prior to putting up paneling. Regardless of your skill, it seems inevitable that some of the old wall will show through. To make this less noticeable, measure off the areas in which the panels will join before your do the installation. Use paint that matches the paneling.
Don't buy furnishings with busy patterns. You may love loud florals or crazy geometric shapes, but they can be a challenge to decorate around. Buy solid-colored furniture that can be easily complemented with other shades. Experiment with different pattern combinations with your pillows and throws.
As you know, lots of people would like to take on home improvement projects, but give up when they realize how challenging they are. Getting started in improving your home can be surprisingly simple if you start small and do your homework. The tips in this article can be an excellent springboard to launch your motivation for home improvement.


text:
Kathy Griffin continues her feud with Anderson Cooper by throwing shade about his late mother Gloria Vanderbilt: 'I knew her better'
Controversial newspaper columnist is SLAMMED by rival writer at his farewell party who accuses him of sexism and 'having a taste for child prostitutes' - before he blasts back accusing her of telling 'f**king lies'
'You are sexting her!' Former convict gets into an EXPLOSIVE and violent argument with her husband after finding a text from another woman on his cellphone while on a romantic trip
Cut into pieces for profit: Brutal image shows mutilated elephant with its tusks and trunk cut off by ivory poachers in Botswana where hunting ban was lifted in May
Kim Kardashian ditches Kimono brand name following fierce backlash over shapewear line and pressure from Japanese mayor
Kim faced criticism after announcing her Kimono shapewear line last Tuesday
She initially stood firm amid the backlash, telling the New York Times on Thursday that she would not be changing the name
Kyoto mayor Daisaku Kadokawa released a letter on Friday urging the reality star to 'reconsider' the name
By Monday morning she had caved under the mounting pressure, telling fans she was changing Kimono name
She is yet to reveal the new name of her 'Solutionwear' line
In April she faced backlash for wearing a traditional Indian maang tikka head piece for a church service
In June 2018, she was slammed for wearing tight, cornrow-style braids to the MTV Movie and TV Awards show
By Heidi Parker For Dailymail.com
Kim Kardashian has announced she is dropping the brand name Kimono following fierce backlash and accusations of cultural insensitivity.
The 38-year-old made the revelation on Instagram on Monday morning, telling her fans that after 'careful thought and consideration' she was changing the name of her 'Solutionwear' shapewear line. She is yet to reveal the new name.
She said she had launched the Kimono line with the 'best intentions', but is committed to 'listening, learning and growing.'
While the criticism of her Kimono name came swiftly after she unveiled her new company last Tuesday, Kim initially stood firm and told the New York Times on Thursday that she would not be changing the name.
But the outrage mounted, with Kyoto mayor Daisaku Kadokawa releasing a letter on Friday urging her 'reconsider' the name.
She finally caved into pressure, telling her fans on Instagram on Monday: 'Being an entrepreneur and my own boss has been one of the most rewarding challenges I've been blessed with in my life.
U turn: Kim Kardashian announced on Monday that she is changing the name of her Kimono Solutionwwear following fierce backlash. She is yet to reveal the new name
'What's made it possible for me after all of these years has been the direct line of communication with my fans and the public,' added the mother of North, Saint, Chicago and Psalm.
'I am always listening, learning and growing - I so appreciate the passion and varied perspectives that people bring to me.'
Khloe Kardashian 'didn't really get why' baby daddy Tristan... Love Island SPOILER 'I'm ready for ALL the men':... Serena Williams nails sporty chic in a casual 'mom' hat as... Kourtney Kardashian stuns in sheer black top with jeans as...
She added: 'When I announced the name of my shapewear line, I did so with the best intentions in mind.
'My brands and products are built with inclusivity and diversity at their core and after careful thought and consideration, I will be launching my Solutionwear brand under a new name. I will be in touch soon. Thank you for your understanding and support always.'
Many felt Kim's trademarked Kimono brand disrespected the traditional 15th century kimono clothing.
Some also took issue with the fact that the reality star was focusing on under-garments, with actual kimonos meant to be an outer layer clothing item.
One social media user wrote: 'I feel very sad that the name ''Kimono'' is being used to something completely different from what we Japanese know about it. Kimono is Japanese traditional clothes and we are very proud of its history and culture. I'm sorry but I feel this name choice is simply ignorant. #KimOhNo.'
Pressure: Kyoto mayor Daisaku Kadokawa released a letter on Friday urging Kim to 'reconsider' the Kimono name
Dating back to 15th century Japan, the kimono, which literally means 'thing to wear on the shoulders', is a T-shaped wrapped garment with set loose long sleeves, tied with a sash.
Considered the national dress of Japan, it is now mostly worn during special occasions such as weddings and coming-of-age ceremonies, and is mostly worn by women.
Those who tend to wear a kimono the most on a daily basis are older men and women, geisha and sumo wrestlers, who are required to wear traditional Japanese dress whenever appearing in public.
In his open letter to Kim on Friday, Kyoto mayor Daisaku Kadokawa noted that kimono culture has been 'passed down with care.'
Controversy: Many felt Kim's trademarked Kimono brand disrespected the traditional 15th century kimono clothing (a Japanese kimono pictured right)
'Kimono is a traditional ethnic dress fostered in our rich nature and history with our predecessors' tireless endeavors and studies,' Kadokawa said in the letter, 'and it is a culture that has been cherished and passed down with care in our living.'
The mayor relished the detail and craftsmanship that goes into making kimonos before adding that they symbolize the 'beauty, spirits and values of Japanese.'
Kadokawa also noted that international tourists have flocked to Kyoto and other cities in Japan to sport their kimonos, which he deemed was 'proof' that the clothing was 'loved by people around the world.'
Kardashian trademarked the Kimono brand in the US last year under the names 'Kimono Body', 'Kimono Intimates' and 'Kimono World'.
This is not the first time the reality star has been accused of cultural appropriation.
In April she faced backlash for wearing a traditional Indian maang tikka head piece for a church service.
She shared her outfit on social media with the caption 'Sunday Service Vibe', but some Instagram users were unimpressed with her apparent mishmash of cultural items.
Kim accessorized her flowing white dress with gold Indian-inspired jewelry, including the head piece typically worn at times of celebration.
In June 2018, Kim Kardashian was slammed for wearing tight, cornrow-style braids to the MTV Movie and TV Awards show
One user said that 'culture is not a costume' and that it was wrong for the media mogul to be wearing the ornament.
Someone else explained the cultural significance of the head piece and wrote in depth about how Kim had appropriated her culture.
'For those of you who don't know, the maang tikka is a ceremonial head piece traditionally worn by the bride. Wearing white in India is traditionally reserved for funerals,' another explained.
'On top of that she wore this to a Christian service. She probably thought it was just jewelry, but when that piece has literally been used for centuries wouldn't you kinda think there must be some sort of significant meaning held towards it?'
And in June 2018, she was slammed for wearing tight, cornrow-style braids to the MTV Movie and TV Awards show.
Earlier the same year she sparked outrage when she posted photographs on Instagram of her hair in braids with white beads at the ends.
In a Snapchat that angered the public, Kim called the braids 'Bo Derek braids'. She was referencing the actress' appearance in the 1979 film, 10.
People were outraged by the comment because they thought it was 'disrespectful' to the culture that first modeled the hairstyle.
The star also caused a stir when she straightened her biracial daughter North's naturally curly hair.
Kim Kardashian West's Kimono underwear meets Japanese backlash - BBC News
Kim Kardashian West Not First To Use Kimono Name and Won't Produce Garment
Kim Kardashian changes Kimono name following backlash

redpajama_set_name: RedPajamaCommonCrawl


Firework Festival - Winter is Gone! Let's Party On!
LoD has been blazing the charts since its release four months ago. The acclaimed multiplayer action RPG has been breaking all expectations and that's all thanks to the fans. After just receiving a huge update with tons of new content, we will offer some fantastic events in this March!
Login Bonus: Log in every day during Event period to collect rewards from the Event page.
Special Titles: During the Firework Events, after launching of them, you will earn a special title. Once the server launches enough fireworks, every player will be rewarded.
Mystic Chest Festival: available daily at 12:00~14:00 and 17:00~19:00 at server time during festival. Don't miss out! Blow up chests to get great rewards. During Event period, clear main quests to get Firework Tokens, which can be used to exchange great rewards from the shop!
During the event time, we will offer a great discount on the new fashions Celestial Heart, Royal Assassin, and Heavy Metal for limited days.
Participate in various community events and share your festival memories to gain special rewards! Please stay tuned for more news regarding Legacy of Discord – Furious Wings on the official Facebook page and website.


Despite the tumultuous session, Senate yesterday condemned the alleged militarisation of the electoral process and the inconsistent application of Electoral Laws by the Independent National Electoral Commission (INEC) in the country.
The upper chamber averred that if the trend of using extreme military force witnessed in the 2019 general elections is not halted, it is capable of posing serious threats to democracy or its survival in the country.
The senate however directed its committee on INEC to carry out thorough investigation on all perceived inconsistent application of Electoral Laws by the electoral body in the general elections.
This is just as it urged President Muhammadu Buhari to give assent to the recent Amendment to the Electoral Act to ensure a level playing field and adoption of equal standards in our national elections for a strong and peaceful democracy in Nigeria.
It also urged INEC to in future elections; ensure unrestricted and consistent application of all Electoral Laws without bias to a candidate or a party.
These were fallouts of a motion moved to that effect by Senator Dino Melaye ( PDP Kogi West) and co-sponsored by seven other senators on the platform of the Peoples Democratic Party ( PDP) .
Senator Melaye had in his lead debate on the motion titled: " The militarisation of the Nigerian Electoral Process and the inconsistent application of Electoral Laws by the Independent National Electoral Commission ( INEC)", submitted that the alleged ugly trend witnessed in the 2019 general elections must not be allowed to happen in future elections. He particularly cited alleged cases of such ugly incidences of militarisation of electoral process in Kogi and Rivers states.
According to him, while the Aide de Camp ( ADC) to the Kogi State governor , Alhaji Yahaya Bello, held members of opposition parties to ransom during the elections, the military, particularly men of the Nigerian Army did same to voters and INEC officers in Rivers State.
"This extreme militarisation of a democratic electoral process, and the inconsistent application of Electoral Laws by INEC in matters of national elections pose serious threats to our democracy, and has security implications that must be nipped in the bud. Persuaded that the nation is on the edge of a precipice, and our democracy can be saved for future posterity, if only we can build strong institutions that can operate within established laws, and with our military forces restricted to their traditional roles of defending the nation", he said.
Though the Senate Leader, Ahmed Lawan ( APC Yobe North) , seconded the motion on the grounds of required legislative intervention in strengthening the nation's electoral process , but trouble started when the Senate Minority Leader, Biodun Olujimi ( PDP Ekiti South), made her contributions.
Debate on the motion assumed partisan dimension when Olujimi in her contributions alleged that rigging was legalised in the elections and votes buying was the order of the day.
Angered by her submission , the senate leader , raised order 53(4) of the senate's standing rules for the Senate President, Bukola Saraki to call her to order.
Even efforts made by the Senate President , Bukola Saraki to make senators debate the motion dispassionately and not along partisan lines proved abortive which made him to hurriedly put the four prayers on the motion to voice votes .
Partisan colouration of debates on the motion was further demonstrated by both the PDP and the APC senators during the voice voting as the PDP senators shouted ayes, to all the four prayers while their APC counterparts , shouted nay, with the Senate President ruling in favour of the ayes.
"I feel very sad that I was denied opportunity to make contribution to a motion in which wide allegations on what did not happen were made against the military in Rivers State by a senator from another state," he said.


text:
China's film market becomes a challenging one for foreign filmmakers 28/05/2013 04:00 (By Etienne Jean de la PERLE)
China's film market becomes a challenging one for foreign filmmakers
The film market in China has become a valuable source of income for many foreign filmmakers and production companies in the US and elsewhere. The main issue as to the acceptance of new productions is the country's censorship rules.
It is common knowledge that China's political regime imposes certain restrictions on the country's film market. Foreign filmmakers therefore have to find ways to circumvent these restrictions so as to share their work with the Chinese and potentially hit the jackpot whilst doing it. <?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />
Cultural differences between countries have always been one of the world's biggest issues as far as efficient communication is concerned. What happens when such differences affects the importation of new productions onto the Chinese film market? Django Unchained's pulling from the cinemas in China seemed to suggest that crude violence and opposition against the government was a no-no.
Cannes film market
However, the film's reinstatement a month later with some minor alterations (three minutes worth of scenes of torture and nudity were censored) has shown that the rules of censorship could be quite arbitrary. According to Pen Kang, a researcher at the Hong Kong Baptist University of Film the successful movies on China's film market are mostly Hollywood blockbusters. This was recently proved by Disney/Marvel's Iron Man 3.
There are however certain exceptions. Cloud Atlas is the most recent one. Despite being a very successful blockbuster worldwide, Chinese officials cut 40 minutes of the 172 minutes film rendering its moral quasi-obsolete. Eric Garandeau, director of France's National Cinema Centre (CNC), says that for a film to be awarded a place on China's film market, it must steer clear from the "three Ts'". In his own words, "You mustn't include the 'three Ts' -- Tibet, Tiananmen and Taiwan -- and you can't have any story that involves ghosts and sex."
Typos: chyna, chiana, flim market, film makret
Cloud Atlas official website
Image credit: © AFP PHOTO / VALERY HACHE


The group will reprise their seminal debut album Love Comes Close for two special performances in 2019.
Today, Cold Cave have unveiled a music video for the EP's second single and title track, which is streaming now on YouTube.
Look for more news from Cold Cave to surface soon.
* Top photo by: Travis Shinn.


text:
Four-year-old and his father critically wounded in a South Philadelphia shooting
Police investigate the scene after a 30-year-old man and his four-year-old son were critically wounded in a shooting on the 2700 block of Daly Terrace in South Philadelphia Monday night. Photos for the GunCrisis Reporting Project by Joseph Kaczmarek.
Police said the gunman fired at least nine shots into the victims' minivan, according to the Philadelphia Inquirer, but that a nine-year-old girl who was also inside was not struck.
The boy was reported in critical but stable condition Monday night at the Children's Hospital of Philadelphia and his father was reported critical but stable at the Hospital of the University of Pennsylvania.
Police said they believe that the boy's father was the intended target, according to the Philadelphia Daily News.
According to 6ABC, medics were called to the scene at about 7:20 p.m. and found the father shot in the back and grazed in the head while his son suffered a gunshot to the stomach.
Police believe this shooting may be connected to a murder on December 1, according to a report from NBC10.
Police are searching for the suspect who they say was wearing a black hoodie and was last seen headed west on Daly Terrace, according to CBS3.
Posted in Crime Scenes, Incidents, News, Photos
← Police find second body inside same home, one day after man was found shot to death Saturday in Tioga
Weekend review: Gun violence leaves seven dead and eight wounded across Philadelphia →


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The village of Lesseux is a small village located north east of France. The town of Lesseux is located in the department of Vosges of the french region Lorraine.	The town of Lesseux is located in the township of Saint-Dié-des-Vosges-Est part of the district of Saint-Dié-des-Vosges. The area code for Lesseux is 88268 (also known as code INSEE), and the Lesseux zip code is 88490.
The altitude of the city hall of Lesseux is approximately 410 meters. The Lesseux surface is 2.94 km ². The latitude and longitude of Lesseux are 48.284 degrees North and 7.081 degrees East. Nearby cities and towns of Lesseux are : Combrimont (88490) at 1.54 km, Frapelle (88490) at 1.66 km, Lusse (88490) at 1.95 km, Le Beulay (88490) at 2.12 km, Bertrimoutier (88520) at 2.53 km, Provenchères-sur-Fave (88490) at 2.80 km, Neuvillers-sur-Fave (88100) at 3.19 km, Raves (88520) at 3.40 km.
The population of Lesseux was 110 in 1999, 142 in 2006 and 147 in 2007. The population density of Lesseux is 50.00 inhabitants per km². The number of housing of Lesseux was 73 in 2007. These homes of Lesseux consist of 57 main residences, 12 second or occasional homes and 4 vacant homes.
Here are maps and information of the close municipalities near Lesseux.
Search hotels near Lesseux : Book your hotel now !
At right you can find the localization of Lesseux on the map of France. Below, this is the satellite map of Lesseux. A road map, and maps templates of Lesseux are available here : "road map of Lesseux". This map show Lesseux seen by the satellite of Google Map. To see the streets of Lesseux or move on another zone, use the buttons "zoom" and "map" on top of this dynamic map.
To search hotels, housings, tourist information office, administrations or other services, use the Google search integrated to the map on the following page : "map Lesseux".
This is the last weather forecast for Lesseux collected by the nearest observation station of Saint-Dié-des-Vosges.
The latest weather data for Lesseux were collected Thursday, 18 April 2019 at 19:51 from the nearest observation station of Saint-Dié-des-Vosges.


text:
Fed is world's greatest hedge fund, Buffett says
Billionaire investor Warren Buffett compared the US Federal Reserve to a hedge fund because of the central bank's ability to profit from bond purchases, while accumulating a balance sheet of more than US$3 trillion.
"The Fed is the greatest hedge fund in history," Buffett told students on Thursday at Georgetown University in Washington.
It is generating "US$80 billion or US$90 billion a year probably" in revenue for the US government, he said. "And that wasn't the case a few years back."
The central bank has been buying US$85 billion of bonds a month to help the US recover as it emerges from the deepest economic slump since the Great Depression. Fed Chairman Ben Bernanke and other Fed policymakers unexpectedly opted this week to sustain that pace of asset purchases instead of tapering it, saying they need to see more signs of lasting improvement in the economy.
The Fed "is under no pressure, none whatsoever to have to deleverage," Buffett said. "So it can pick its time and if you have somebody wise there — and I think Bernanke is wise and I certainly expect his successor to be — it can be handled. But it is something that's never quite been done on this scale. It will be interesting to watch."
Bernanke steered the Fed through the 2008 crisis, and his term ends in January next year. Fed Vice Chairman Janet Yellen is a leading candidate to replace Bernanke after former US treasury secretary Lawrence Summers withdrew his name from consideration.


Chelsea star, Eden Hazard, has expressed his belied that they can be genuine title challengers this season.
The Blues have opened their campaign with three straight wins and are joint top.
However, Maurizio Sarri's men are still regarded as outsiders by the bookies in the title race.
But Hazard says they are getting better and more confident with every game.
"We need to keep this momentum. We want to be at the top at the end of the season.
"We know the league is not finished, it's just the beginning but confidence is coming, goals is coming, so we just look forward for the next game.
"We have made a good start and we have also shown our character and now we have to keep going," Hazard told Sky Sports.


text:
Lil Nas X's "Montero (Call Me by Your Name)" Debuts at No. 1
ByBrenton Blanchet
Photo by Mike Coppola/FilmMagic via Getty Images
After one of the most interesting track rollouts and certainly the most impressive pole slide to the depths of hell ever recorded on film, Lil Nas X just scored his second No. 1 single on the Billboard Hot 100 Monday with "Montero (Call Me By Your Name)."
The chart champion, who flaunts the title of most consecutive weeks at No. 1 thanks to "Old Town Road" and its historic 19-week run in 2019, is finally back on top, dethroning Justin Bieber's "Peaches" in the process.
Nas' latest chart victory comes as its music video—featuring the musician throwing some ass on Satan—and his Nike Satan kicks set the internet on fire all week. Nas has surely handled any and all conservative commentary masterfully, from calling certain pundits "shitty pants" to inspiring TikTok trends featuring his now iconic pole slide. But beyond any clapbacks, the song's meaning is an important one to Nas, so surely seeing the track top the charts is a major win for him.
In a note to his 14-year-old self during the song's release, Nas wrote "I know we promised to never be 'that' type of gay person, I know we promised to die with the secret, but this will open doors for many other queer people to simply exist."
"You see this is very scary for me, people will be angry, they will say I'm pushing an agenda," he continued. "But the truth is, I am. The agenda to make people stay the fuck out of other people's lives and stop dictating who they should be."
Lil Nas X's "Montero" Eyes No. 1 Debut on Billboard Hot 100
Lil Nas X Responds to Backlash Over Provocative "Montero" Video: 'I Am An Adult'
The Lil Nas X Outrage Is Exposing Hypocrites
Outside of "Old Town Road" and now "Montero," Nas has had three other entries on the chart: "Panini," which hit No. 5 in 2019, "Rodeo," which peaked at No. 22, and "Holiday," which peaked at No. 37 last November.
Reacting to the news on Twitter, Nas wrote that he was told he would "never have a hit again," as he thanked his fans and team.
"Y'all told a 19 year old who had just escaped the lowest point in his life he would never have a hit again," he wrote. "You told him to stop while he's ahead. He could've gave up. But 4 multi platiunum songs and 2 #1's later, he's still here."
Nas has now spent 20 weeks at No. 1, as "Montero" earned him 46.9 million U.S. streams and sold 21,000 first-week downloads, according to MRC Data.
Last week, Complex's Joe La Puma linked with Nas for the latest episode of Sneaker Stories, where the two talked about Nas' first pair of Jordans and the his latest controversial kick collab with MSCHF.
"People already demonize who I am and put me in a painting of, 'OK, he's evil, he's doing this, he's doing that,'" Nas said. "So it's like, you know what? You know what? I'll take that, I'll be that, and I'm going to make the best of it."
Join the conversation on Complex today!
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text:
The Clare Necessities
My 2021 in Theatre
What a year we have just had. The virus we all thought would be over in a few months is still with us almost two years on. It has affected us all in many ways, from physically to mentally and for Theatre it has been for some devastating.
For me personally, the year started with a variety of streamed shows which kept my theatre heart beating if only lightly but the return of live and in the room theatre brought me such happiness and joy, if not a good deal of nervousness and trepidation.
For me, nothing is better than sitting in a theatre, whether it be alone or with friends and family and the lights go down, the overture starts and the show begins. So to return to the theatre after eighteen months away was absolutely glorious. My return came in the shape of RENT at the Hope Mill Theatre in Manchester. The show itself was absolutely amazing and fantastic, the cast were incredible and everything about it was brilliant but for me what made it one of my most favourite experiences in theatre this year was that I got to see it with some of my favourite people, people that I gained as friends (almost theatre family) through three lockdowns and a global pandemic. They have literally made the last two years liveable and without them I'd hate to think where I would be.
September saw me finally being formally introduced to one of my favourite musicals of the year and maybe the decade, Amelie the Musical. This show was drip fed to me throughout many a zoom quiz and through months of listening to the Original London Cast recording but nothing could have prepared me to sit in a theatre and watch it in person, so much so that in the space of two weeks I watched it twice. I do wish that I'd seen it more because my love for this show is so strong but the memories of the show that I have will live with me for a very long time.
Out of London, I was able to see a few things more locally on UK tours including Everybody's Talking about Jamie, School of Rock and Groan Ups which were lovely to see and all of which I'd one hundred per cent see again and recommend to anyone to watch if they're coming anywhere near you.
Frozen the Musical currently playing at the Theatre Royal, Drury Lane is definitely up there as one of my favourite shows I've seen in 2021, the complete spectacle of the show from the costumes to the sets, the cast and the front of house staff too it was so amazing to finally see the show after the pandemic had pushed back our visit to Arendelle. From the moment you first step in you can feel the air thick with Disney magic and it is glorious, I definitely want to return next year with my friend Elle and her daughter so they can witness the magic too.
There is final show that I wanted to mention (if this post has not been enough theatre for you all, check out my YouTube Channel The Theatre Necessities, in which I will be uploading a complete recap of all the theatre that I saw this year very soon) and that is What's New Pussycat at the Birmingham Rep theatre as it was the show surprised me the most this year. I had no expectations or knowledge really of the show prior to taking my seat but it was absolutely fantastic. What's New Pussycat uses the songs of the Welsh rather famous singer Tom Jones to tell the story of another Tom Jones, based on the novel by Henry Fielding called The History of Tom Jones, a Foundling. It sounds confusing but it's very easy to follow and a great watch that I hope gets as many awards as it is nominated for.
I can't wait to see what 2022 has in store for me and for theatre and I can't wait to see what the year has ahead.
Found in: 2021, Theatre, year in review
Welcome to The Clare Necessities. A happy home for all who visit. Come in, sit down, take a break and look around.
REVIEW|Everybody's Talking About Jamie UK Tour
BOOK REVIEW| The Two Lives of Lydia Bird by Josie Silver
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Damon: "What changed your mind Elena?" Elena: "I didn't wanna see wewe get hurt, okay! I was...I was worried about you."
Favorite Movie: POTC, the notebook, the proposal, ect.
Favorite Musician: lifehouse, nickelback, onerepublic, katy perry, lady gaga, ect.


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text:
With Packers' Davante Adams, sure hands are the asset
Green Bay Packers rookie wide receiver Davante Adams is trying to replace James Jones this season. Credit: Mark Hoffman
By Tyler Dunne of the Journal Sentinel
Green Bay — Through the one-handed acrobatics and Houdini acts in traffic, Davante Adams needed a nickname.
"Megatron" and "Baby Megatron" were tossed around first. Then, "Optimus Prime."
As Fresno State's quarterback, the leader, Derek Carr decided "Optimus Prime" — a robot character from the "Transformers" movie franchise — needed to stick.
"He couldn't be Megatron," Carr said, "because that was taken by a special receiver. I won't say the name. I called him Optimus Prime because he reminds me a lot of that guy.
"He didn't get a tattoo or anything. But I tried to talk him into it."
Adams sidesteps any perceived parallels to Detroit's Calvin Johnson. Adams is the 6-foot-1 second-round pick. The rookie. Johnson is, you know, the 6-foot-5 Megatron. If a cartoon-themed nickname does stick in the pros, it'll be because of Adams' vise-strong yet baby-soft hands. No receiver in major college football caught more passes (131) than Adams last season.
These are 9-inch, basketball-weathered claws of death.
Throughout the NFL draft, the Green Bay Packers repeated that they were after natural pass catchers. The 53rd overall pick. Not scorching fast, not necessarily tall, those hands must remain automatic.
"It becomes natural," Adams said. "It's second nature. When you put a lot of work into it, it becomes easy.
"I put a lot into my craft. You have to make sure you're the best at whatever it is, and I pride myself on having good hands."
When September flipped to October last fall, Adams wasn't on a 131 track. Far from it.
After an uninspired 42-37 win over hapless Hawaii, Fresno State coach Tim DeRuyter called out the receiver. That game, Adams had five receptions, a bad drop and only 22 yards.
So DeRuyter told local reporters that Adams "has got to play harder," that "the issue was effort" and "I think (Adams) is embarrassed by it." He added that Adams played like he was ill even though he didn't think the receiver was.
When, in fact, Adams did have the flu. The wideout was hurting all week. During practice, Carr remembers the receiver looking like he was going to throw up at any moment.
He played. He played poorly. No hard feelings, Adams says.
"They coached me hard," Adams said, "and I wouldn't expect anything other than that."
Over his next two games — wins over Idaho and UNLV — Adams dominated with 24 receptions for 406 yards and seven touchdowns. The turnaround was a microcosm of Adams' approach to the game, to a ball zipping his way at 60 mph. He stays in tunnel vision.
Whether it's the flu or a pestering cornerback, he finds a level of Zen, of calm.
The result is total concentration.
"I just block everything out," Adams said. "It's not even something that I do. When the ball's in the air, everything goes quiet. That's how it is. I don't really think about it much. That's how I play Tunnel vision."
Asked to recall a drop by Adams in high school, Adams' old Palo Alto coach pauses for six, seven seconds. Palo Alto rolled to a 14-0 state championship season in Adams' senior year. Yet in that state title game, there was one questionable call along the sideline. Adams believed he caught it, cleanly, feet in bounds. The official claimed he bobbled it.
Incomplete.
This is the closest to an actual drop that Earl Hansen can remember.
"The ball never hit the ground, and he went out of bounds," Hansen said. "The call probably would have been reversed if we had instant replay.
"Why don't you ask him about that bobble. I guarantee you he'll remember."
He does. Adams still says it should have been a catch.
Of course, Palo Alto wouldn't have gone undefeated that 2010 season if, in a preceding playoff game, Adams didn't somehow snare a touchdown pass in the corner of the end zone with 21 seconds left, too. In a monsoon of a game — "it was pouring, pouring," Hansen said — Adams' strong grip kept a dream alive.
To Hansen, this was the rebounder in Adams. The coach always encouraged receivers to play hoops. Adams averaged 15 points, eight rebounds and seven assists as a junior and then 9.4, 5.8 and 5.4 as a senior. Along the way, he grew to naturally attack the ball at its highest point. To him, the rebounding directly translates.
It's no surprise Adams admired former NBA star Dennis Rodman. Any loose ball was Rodman's loose ball.
"When you go up for the rebound, you can't wait for the ball to come down," Adams said. "You have to go get the ball at the highest point. That's how it is in football. If you want to win those jump balls and those 50-50 balls, you've got to go up and get it."
As a result, Carr said Adams "very rarely" used his body to make a catch. The only times his top receiver did cradle the ball against his chest was when he sneaked a quick throw on him, surprised him. He can't think of a body catch at the top of his head.
Said Carr, "I'm scared to see what he can be in a couple years."
"So effortless"
The Optimus Prime-wide catching radius. The ability to reroute on the fly. The cosmic knowledge of the offense.
Carr describes Adams as a quarterback's best friend.
Green Bay is getting, he said, "one of, if not the best, receiver in the draft." Adams' hands, Rodman tenacity and 39½-inch vertical allowed the quarterback to gamble.
"I knew that if I was throwing it to Davante," Carr said, "I just had to put it somewhere in his vicinity and he was going to come down with it. It could be underthrown. It could be way too wide. Or it could be 13 feet in the air and he'll go up in the air and grab it. It won't be a problem for him."
For example, Carr continued, every fade he lofted Adams' direction. Like Green Bay's offense, a fade can snap into a back-shoulder bullet any moment.
That's why criticism of Fresno State's spread offense — often painted an easy-does-it slot machine — bothers Carr. He insists the Bulldogs had "pro concepts" built in. Routes demanded decision-making in real time. Fresno State simply operated much faster and against inferior competition.
In this case, back-shoulder chemistry with Aaron Rodgers shouldn't take two, three years. With Adams, Carr learned he could throw high and back-shoulder. With Adams, there's "a lot of room to miss."
The week after the Hawaii game, Carr called a "go" route for Adams. Get vertical. Both read that the cornerback was sitting atop the route — no audibles were called, no signals made — and Carr went back shoulder.
Adams dorsiflexed his body nearly 360 degrees like Jordy Nelson and plucked the one-handed touchdown pass above his head. As Carr punctuated, "I don't care who you're playing — there's not a lot of people who can do that."
He pauses.
"It's so effortless," he said, "it almost upset me even though we scored."
Jones out, Adams in
Now Carr is in Oakland with James Jones. And Adams is in Green Bay with Rodgers.
The question: Can Adams replace Jones?
Adams grasps the challenge. The two worked out together at Fresno State. Jones' close friend and former receivers coach Keith Williams was Adams' first position coach at Fresno State. Jones was a common sight. Carr does see similarities.
Like Jones, the veteran who gritted through a bad knee and cracked ribs last season, Adams prefers a physical game. His own training includes a heavy dose of boxing.
"I could still beat him up," Carr said. "But he has that boxing background. I see a lot of that physicality. Kind of the Bay Area in both of them. That's how they are."
In Green Bay, it always starts with the hands.
Through the 2010 Super Bowl run, Jones dropped a string of potential 50-, 60-yard touchdown catches. In 2012, he had three drops on 108 targets. Stickum-sure hands must be the basis of Adams' rise in Green Bay. That's what accelerates trust and timing with Rodgers.
Back at Fresno State, Adams would catch 50-100 balls at the JUGS machine after practice and then refine specific routes with Carr. They even perfected that one-handed, back-shoulder catch.
So, of course, the first ball thrown to Adams at Friday's rookie camp practice was a drop. The process begins. The "Optimus Prime" nickname is on hold for now. And, really, Adams says nobody calls him that anymore.
In Green Bay, he's got to re-prove himself.
"I think I fit right into what they have going on here," Adams said. "I can come in and contribute right away."
About Tyler Dunne
Tyler Dunne covers the Green Bay Packers. He has been on the beat since 2011, winning awards with the Pro Football Writers of America and Milwaukee Press Club.
@TyDunne
tdunne@journalsentinel.com
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Olympic gold medalist James Cracknell is no stranger to a challenge, having rowed the Atlantic, cycled across the Arctic, run an ultramarathon across the Sahara Desert (not to mention winning two Olympic Gold Medals). Cracknell is now spearheading the Duracell Power Me campaign, which aims to motivate people to get more out of their lives.
The following video gives a taste of what he's in for as he takes on a series of challenges in the London Eye, one of the UK's best known landmarks. Watch James cycle, box, lift weights, work out with a balance ball, do circuits and finish with a Duracell Bunny workout, which takes him a total of three hours as the London Eye slowly rotates.
Sponsored post: Mark Cavendish is up for a new kind of challenge, are you?


text:
Q: Incorrect Data Being Saved From AJAX Drop Down Menus I've been trying to sort out an issue I have with how one of my fields of data is being saved to a mySQL database. The issue I have is that instead of the text value being saved, the 'id' for the text value is being saved and I've no idea why.
The field is called 'detectorname' and the text value for this is selected via a drop down menu, which in turn, opens a second drop down menu called searchheadname. Both drop down menus are populated via PHP and AJAX scripts and show the correct text values when being selected by the user.
I appreciate that I haven't included any code, there is an awful lot and I have to admit I wasn't sure which sections would be of use.
I just wondered whether someone could perhaps help me out with this please and let me know which parts of the code they would like me to post.
UPDATE
HTML Form & AJAX Script
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">

        <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
        <title>Location</title>
        <link rel="stylesheet" href="css/style.css" type="text/css" media="all" />
    link rel="stylesheet"   href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.8.6/themes/base/jquery-ui.css"
        type="text/css" media="all" />

                <form name="addfindstolocation" method='POST'>

                        <input name="userid" type="text" id="userid" value="1"/>
                        <input name="locationid" type="text" id="locationid"  value="2"/>

Click on the map to place the marker for the find that has been made and drag until the precise location has been found.

Find OSGB36 Latitude Co-ordinate

                                    <input name="findosgb36lat" type="text" id="findosgb36lat" size="20" />

Find OSGB36 Longitude Co-ordinate

                                    <input name="findosgb36lon" type="text" id="findosgb36lon" size="20" />

Date of Trip

                                    <input name="dateoftrip" type="text" id="dateoftrip" size="10" />

Find Category

                                    <?php
                                    mysql_connect("hostname", "username", "password") or die("Connection Failed");
                                    mysql_select_db("dbname")or die("Connection Failed");
                                    $query = "SELECT * FROM findcategories";
                                    $result = mysql_query($query);
                                    ?>
                                    <select name="findcategory" id="findcategory">
                                        <option value=''>Select One</option>
                                        <?php
                                        while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
                                        ?>
                                        <option value="<?php echo $line['findcategory'];?>"> <?php echo $line['findcategory'];?> </option>

                                        <?php
                                        }
                                        ?>
                                    </select>

                                    Find Name

                                        <input name="findname" type="text" id="findname" size="35" />

                                    Find Description

                                        <input name="finddescription" type="text" id="finddescription" size="150" />

                                    Detector Used

                                        <select name=detectors id="detectorid" onchange="AjaxFunction(this.value);">
                                            <option value=''>Select One</option>
                                            <?
                                            require "phpfile.php";// connection to database
                                            $q=mysql_query("SELECT * from detectors WHERE userid='1'ORDER BY 'detectorname' ASC");
                                            while($n=mysql_fetch_array($q)){
                                            echo "<option value=$n[detectorid]>$n[detectorname]</option>";
                                            }

                                            ?>
                                      </select>

                                    Search Head Used

                                        <select name=searchheads id="searchheadid">
                                        </select>

                                    Detector Settings

                                        <textarea name="detectorsettings" cols="50" rows="12" id="detectorsettings"></textarea>

                                    PAS Ref. (if known)

                                        <input name="pasref" type="text" id="pasref" size="9" />

Additional Comments

                                        <textarea name="additionalcomments" cols="50" rows="12" id="additionalcomments"></textarea>

                                    <input name="submit" type="submit" onclick="MM_callJS('savedata()')" value="Submit" />
                                </form>

**'Searchheadetails.php' File (Called in file above)**

     <?
$detectorid=$_GET['detectorid'];
require "config.php";
$q=mysql_query("SELECT * FROM searchheads WHERE detectorid='$detectorid' ORDER BY 'searchheadname' ASC");
echo mysql_error();
$myarray=array();
$str="";
while($nt=mysql_fetch_array($q)){
$str=$str . "\"$nt[searchheadname]\"".",";
}
$str=substr($str,0,(strLen($str)-1)); // Removing the last char , from the string
echo "new Array($str)";

?>

A: I would try changing the detectorid to detectorname in one line of the code below:

<select name=detectors id="detectorid" onchange="AjaxFunction(this.value);">
<option value=''>Select One</option>
<?
require "phpfile.php";// connection to database
$q=mysql_query("SELECT * from detectors
  WHERE userid='1'ORDER BY 'detectorname' ASC");
while($n=mysql_fetch_array($q)){
echo "<option value=$n[detectorname]>$n[detectorname]</option>";
//***I changed the line above***  TODO:Remove this comment
}
?>
</select>


The number of employees of Polk County, IA was 4,537 for media in 2017.
This dataset contains the quarterly employment and wage data for the counties and statewide by two and three-digit sector in Iowa. This data is a census collected from unemployment insurance-covered employment in the state of Iowa.
The dataset contains information on IPERS-covered wages, employer contributions and employee contributions by membership group, employer and fiscal year. An employer may have more than one membership group. Count of employers is before modifications due to GASB 68.
The Iowa Insurance Division regulates and supervises the business of insurance in Iowa. This dataset provides a list of insurance companies licensed to do business in Iowa.
The Iowa Industrial New Jobs Training (260E) program provides employers expanding Iowa's workforce with new employee training. The 260E program is designed to increase worker productivity and company profitability, and is administered by Iowa's 15 community colleges and financed through bonds sold by the colleges. Depending on wages paid, the participating businesses divert 1.5% or 3% of the Iowa state withholding taxes generated by the new positions to the community college to retire the bonds. Businesses may also be eligible to receive reimbursement for their on-the-job training expenses, and/or corporate tax credits.
 This dataset lists 260E contracts open in or after 2012, Qtr 2, and includes information on: the administering community college, participating employer, location of employment, training expenses, and employment information.
 More on Iowa's 260E program.


I posted a question a few days ago but believe I put up the wrong log file which wouldn't have shown off the main problems I'm having. To keep it short, basically when I go to launch the engine/editor through the launcher, the loading splash screen appears for a few seconds before disappearing...that's it. Nothing appears, no obvious error message, just nothing.
Would really appreciate any help as I've tested on another PC and it had no issues but I really need it on my home PC.
See if there are any solutions listed on the link below that may help. If you still can't fix the issue, could you also post your Verbose Logs? Directions on how to find those is listed towards the bottom of that wiki link.
Our apologies, the above link is for Launcher problems, but your launcher sounds to be working fine and the issue is with the Editor.
Can you ensure that your graphics card is updated to the latest drivers?
Thanks for replying. Yes I believe the problem is editor related but I am going to attach another log to this message which I believe is launcher related. I've double checked and appear to be running the latest drivers for my video card (Geforce 340.52).
The only thing that is different now after my most recent re-install is that it sits on the editor loading splash screen for a lot longer sometimes than it used to. As though it's making some progress but still nothing comes as a result.
LogWindowsTextInputMethodSystem:Error: Terminating failed while unadvising the language notification sink from the TSF source.
But really I have no idea. I included a log from the launcher though. Thanks.
NOSdog, I can't help you with the Launcher problem, but I'm curious about the crash log you posted in the question (the Editor log file).
Maybe a silly question, but does your home PC have a network adapter installed and is it enabled?
Also, can you tell me whether the Editor crashes consistently with the log you posted, or only sometimes? Is the error at the bottom of the log always the same?
Yes my computer does have a network adapter which is enabled and working as far as I can tell. I disabled and re-enabled it as well but it made no difference.
My crashing issues are also very consistent, I've attached a picture which compares the last few lines from 3 logs taken from completely different days and except for one minor and probably insignificant difference on one line they all appear to be exactly the same error.
Hmm, sadness. I have not seen this particular problem before. Would you be able to compile the Editor from source code, so we can debug this in Visual Studio? Unfortunately, the error messages do not contain enough details as to what exactly is going wrong.
I'm compiling as I type this, hopefully I haven't screwed up the process and will post results once it's done.
It will be faster to iterate in chat than here, thanks!
Thanks I'll use that from now on but I just thought I'd post a screenshot of the first immediate problem I ran into when trying to launch/debug the compiled build. This message pops up almost instantly when trying to launch.
Holy crap I got it working!!
After googling the above debugging error, someone had posted an answer for a similar problem that involved running cmd.exe and entering the following two lines.
I was rebuilding the source code version at the time but decided to try launching through the launcher just for kicks and what do you know it worked!
Thanks for all the help/advice, hopefully this isn't a one off and it's sorted my issues for good.


Every Monday and Friday we have a very active Luncheon Club for senior members of our community. We currently have 180 members enlisted on the Club although the actual number attending is around 60-70. We are grateful to the vibrant team of volunteer caterers who prepare an excellent variety of delicious dishes. This activity is subsidised by GHS but senior members contribute a nominal amount towards the lunch.
Community members, on an auspicious day or for a celebration, sponsor the lunch. Senior members are treated to a party during the Diwali festival.


text:
Watch this video and more on The Criterion Channel
The War Room • 1h 36m
Directed by Chris Hegedus and D. A. Pennebaker • 1993 • United States
The 1992 presidential election was a triumph not only for Bill Clinton but also for the new breed of strategists who guided him to the White House—and changed the face of politics in the process. For this thrilling, behind-closed-doors account of that campaign, renowned cinema verité filmmakers Chris Hegedus and D. A. Pennebaker captured the brainstorming and bull sessions of Clinton's crack team of consultants—especially James Carville and George Stephanopoulos, who became media stars in their own right as they injected a savvy, youthful spirit and spontaneity into the process of campaigning. Fleet-footed and entertaining, THE WAR ROOM is a vivid document of a political moment whose truths ("It's the economy, stupid!") still ring in our ears.
Up Next in The War Room
William J. Clinton Foundation Panel, ...
In 2011, the William J. Clinton Foundation hosted a discussion in Arkansas to mark the twentieth anniversary of Clinton's candidacy for president. In the following excerpts from that event, veterans of the campaign discuss the trials of New Hampshire, and an unexpected audience member—President C...
Return of the War Room
In Chris Hegedus and D. A. Pennebaker's 2008 follow-up feature to THE WAR ROOM, advisers James Carville, George Stephanopoulos, and Mary Matalin, among others, reflect on the revolutionary 1992 campaign. In addition, Mark McKinnon and Frank Luntz, members of the new generation of campaign operati...
Making THE WAR ROOM
In this interview, shot in 2011, THE WAR ROOM directors Chris Hegedus and D. A. Pennebaker and producers R. J. Cutler and Wendy Ettinger discuss the process of gaining access to, and then filming, a presidential campaign.


When I think of Midsummer, images of bonfires and solstice celebrations spring to mind, followed by the symbolic Mediæval custom of paying a red rose "in the time of roses" on Quarter Day, but today I am remembering a less uplifting event, following on from cannon-fire and the red of blood – the Surrender of the City of Oxford during the English Civil War, on 24 June in 1646 – and one besieging Parliamentary soldier who did not live to see the day.
Respectively writing on 19 April and 02 May 1648, Surgeon Valentine Nickson and Captain Edward Lewis use identical phrasing – down to the spelling – "to Certifie all whome itt may Concerne that Joseph Cannon being a Souldier in Major Generall Skippons Troope of horse" "euer since the New Moddell was slaine" "before Oxford June 9th 1646".
Although Joseph himself has left no footprint in the indexed national historical record, with the New Model Army having been formed in 1645, we know that as a soldier in Philip Skippon's Troop he would have fought in the Battle of Naseby before his death the following year during the third Siege of Oxford, while the Treaty which eventually issued in the Royalists' surrender of the city was being negotiated.
Two weeks too late for Joseph, but thanks to The National Archives providing readers with a means for 'tagging' catalogue descriptions, his family's descendants can now locate his appearance in these State Papers at the bottom of the page, and his service has left a footprint after all.