text:
\section{Introduction}
The radial Dunkl process associated with a finite reflection group is a diffusion valued in a Weyl chamber and depends on a set of non negative parameters called multiplicity values (\cite{CDGRVY}, Chapters II and III). Its infinitesimal generator is the differential part of the so-called Dunkl-Laplace operator subject to normal reflections at the boundary of the chamber. In the rank-one case, the radial Dunkl process is nothing else but a Bessel process and if all the multiplicities values vanish, then it reduces to the reflected Brownian motion in the Weyl chamber. Another quite interesting instance of the radial Dunkl process is the Brownian motion conditioned to stay in the interior of the Weyl chamber, due to its close connections to the representation theory of semi-simple finite-dimensional Lie algebras (\cite{BBO1}, \cite{BBO2}).

For small multiplicity values, the radial Dunkl process hits almost surely the boundary of the Weyl chamber. In this case, the tail distribution of the first hitting time of the boundary can be computed from the absolute-continuity relations existing between the process distributions corresponding to different sets of multiplicity values (see \cite{CDGRVY}, p.208). In particular, this probability is expressed, for the infinite families of irreducible root systems $A,B,C,D$, through multivariate confluent hypergeometric function (\cite{Demni}). For dihedral root systems, the tail distribution is given by a series of Jacobi polynomials and one-variable confluent hypergeometric functions, which generalize the expansions derived in \cite{Deb} for the planar Brownian motion in dihedral wedges (in \cite{Demni0}, the formulas were only given for even dihedral groups and similar ones may be derived for odd dihedral groups). Note in passing that combinatorial techniques were used in \cite{Dou-Oco} where the tail distribution of the first exit time from Weyl chambers by a Brownian motion was represented through pfaffians.

Back to the radial Dunkl process in dihedral wedges, an integral representation of the density of the reciprocal of the first hitting time of the boundary was obtained in \cite{Demni1} for the special angle $\pi/4$ and equal multiplicity values. The obtained integral has the merit to be obviously positive in contrary to its original series representation, and allows for proving an analogue of Dufresne's result on the first hitting time of zero by a Bessel process (see e.g. \cite{MY}) and others due to Vakeroudis and Yor on the exit time from a cone by a planar Brownian motion (\cite{VR}, \cite{Vak-Yor}). It is also more suited to derive asymptotics of this density yet we are intended to investigate them here. In this paper, we shall rather extend the aforementioned integral representation to all even dihedral groups with equal multiplicity values. Our proof of this extension is different from the one written in \cite{Demni1} and appeal to an identity proved in \cite{Del-Dem} and satisfied by ultraspherical polynomials (see eq. \eqref{IdGeg} below). Compared to the integral obtained for the wedge of angle $\pi/4$, the domain of integration is  the multidimensional standard simplex and its integrand still involves a one-variable confluent hypergeometric function yet with a more complicated argument. When the radial Dunkl process starts at a point lying on the bisector of the dihedral wedge, we express the weighted Laplace transform of the density through the fourth Lauricella function, thereby extending to all even dihedral groups Corollary 3.5 proved in \cite{Demni1}. As to the extension Corollary 3.3 in \cite{Demni1}, it is straightforward and we will only allude to it without details in a aside remark. Using an analytic analogy between the infinitesimal generators of radial Dunkl processes corresponding on one side to odd dihedral groups and on the other side to even dihedral groups with equal multiplicity values, we derive an analogous integral representation for odd dihedral wedges where this time, the integrand involves rather a Fox-Wright confluent hypergeometric function. The weighted Laplace transform is then computed but is not expressed through a multivariable hypergeometric series. Let us finally stress that the choice of equal multiplicity values is not a restriction and is rather made for sake of simplicity since the formulas are already cumbersome. Nonetheless, we hope dealing with the case of different multiplicity values in a future paper.

The paper is organized as follows. The next section recalls some facts on reflection groups and systems, as well as on the radial Dunkl process in dihedral wedges. In the third section, we recall the definitions and some properties of various special functions occurring in the paper. The fourth section is devoted to the derivation of the integral representation of the density for even dihedral wedges and equal multiplicity values, and of its weighted Laplace transform. In the last section, we derive analogous results for odd dihedral wedges.

\section{Reminder: Dunkl processes in dihedral wedges}
\subsection{Dihedral systems and groups}
For general facts on root systems and reflection groups, we refer the reader to the monographs \cite{Dun-Xu} and \cite{Hum}. The dihedral group, $\mathcal{D}_2(n), n \geq 3$, consists of orthogonal transformations leaving invariant a regular $n$-sided polygon centered at the origin. Actually, it contains $n$ rotations:
\begin{equation*}
r_j: z \mapsto ze^{2i\pi j/n}, j =1, \quad \ldots, n,
\end{equation*}
and $n$ reflections:
\begin{equation*}
s_j: z \mapsto \overline{z}e^{2i\pi j/n}, \quad j =1, \ldots, n,
\end{equation*}
where $z \in \mathbb{C}$ and $i = \sqrt{-1}$ is the complex imaginary unit. As a finite reflection group, $\mathcal{D}_2(n)$ corresponds to the dihedral root system $I_2(n)$, which may be defined as the set of planar vectors:
\begin{equation*}
I_2(n) := \{\pm i e^{i\pi l/n},\, 1 \leq l \leq n\}.
\end{equation*}
To the following choice of positive root system
\begin{equation*}
\{- i e^{i\pi l/n},\, 1 \leq l \leq n\},
\end{equation*}
corresponds a cone, named the positive Weyl chamber $C$, and is the wedge of dihedral angle $\pi/n$ parametrized in polar coordinates as:
\begin{equation*}
C := \{(r, \theta), \, r > 0, \, 0 < \theta < \pi/n\}.
\end{equation*}
On the other hand, the dihedral group acts on its root systems in a natural way and this action gives rise to two orbits when $n := 2p, p\geq 2,$ (the roots forming the diagonals and those forming the lines joining the midpoints of the polygon) while it is transitive otherwise. Accordingly, we shall assign to the two orbits non negative real parameters $k_0,k_1,$ respectively when $n$ is even and only one non negative real parameter $k$ otherwise (these parameters are known in the realm of Lie groups as multiplicity values). With these data, we are now ready to recall the definition of the radial Dunkl process associated with dihedral groups (see \cite{Demni0} for further details).

\subsection{Radial Dunkl processes in dihedral wedges}
The radial Dunkl process $X$ associated with a given reduced root system is a $\overline{C}$-valued diffusion reflected at the boundary $\partial C$ (see \cite{CDGRVY} for further details). In particular, for odd dihedral groups $\mathcal{D}_2(n), n \geq 3$, its infinitesimal generator acts on smooth functions supported in $\overline{C}$ as (\cite{Demni0}):
\begin{equation}\label{G1}
\frac{1}{2}\left[\partial_r^2  + \frac{2nk +1}{r}\partial_r\right] + \frac{1}{r^2}\left[\frac{\partial_{\theta}^2}{2} +nk\cot(n\theta)\partial_{\theta}\right],
\end{equation}
subject to Neumann boundary conditions. While for even dihedral groups $\mathcal{D}_2(2p)$, the action of the infinitesimal generator of $X$ is given by:
\begin{equation}\label{G2}
\frac{1}{2}\left[\partial_r^2  + \frac{2p(k_0+k_1) +1}{r}\partial_r\right]  + \frac{1}{r^2}\left[\frac{\partial_{\theta}^2}{2} +p(k_0\cot(p\theta) - k_1\tan(p\theta))\partial_{\theta}\right].
\end{equation}
These polar decompositions show that the Euclidean norm of $X$ is a Bessel process and that the cosine of its angular part is, up to a time change, a real Jacobi process on the interval $[0,\pi/n]$ (\cite{Rev-Yor}, \cite{Demni0}). This last observation was the key ingredient in the study  of the first hitting time $T_0$ of $\partial C$ by $X$:
\begin{equation*}
T_0 := \inf\{t \geq 0, \, X_t \in \partial C\}.
\end{equation*}
In this respect, recall from \cite{Chy}, Proposition 6 and from \cite{Dem}, Proposition 1, that $T_0$ is almost surely finite if and only if (at least) one of the multiplicity values is strictly less than $1/2$. In particular, let $n = 2p, p \geq 2, x = \rho e^{i\phi} \in C$ and assume $(k_0, k_1) \in (1/2, 1]$. Then, the radial Dunkl process $X$ starting at $X_0 = x$ and associated with the multiplicity values $(1-k_0, 1-k_1)$ hits almost surely the boundary of the dihedral wedge of angle $\pi/(2p)$. As such, the corresponding reciprocal of $T_0$
\begin{equation*}
V_0 := \frac{\rho^2}{2T_0}
\end{equation*}
is a almost surely positive random variable. A similar statement holds true for odd dihedral wedges.

\section{Special functions}
In this section, we record the definitions of various special functions occurring in the remainder of the paper as well as some of their properties we will need in our subsequent computations. The reader is referred for instance to \cite{AAR}, \cite{Er}, \cite{Erd}, \cite{Man-Sri}.
We start with the Gamma integral:
\begin{equation*}
\Gamma(z) = \int_0^{\infty} e^{-u}u^{z-1} du, \quad \Re(z) > 0,
\end{equation*}
and the Legendre duplication formula:
\begin{equation}\label{Leg}
\sqrt{\pi}\Gamma(2z+1) = 2^{2z-1}\Gamma\left(z+\frac{1}{2}\right)\Gamma(z+1).
\end{equation}
Then, we recall the Pochhammer symbol:
\begin{equation*}
(a)_k = (a+k-1)\dots(a+1)a, \quad a \in \mathbb{C}, \, k \in \mathbb{N},
\end{equation*}
which may be written when $a > 0$ as:
\begin{equation*}
(a)_k = \frac{\Gamma(a+k)}{\Gamma(a)}.
\end{equation*}
Next comes the generalized hypergeometric function defined by the series:
\begin{equation*}
{}_rF_q((a_i, 1 \leq i \leq r), (c_j, 1 \leq j \leq q); z) = \sum_{m \geq 0}\frac{\prod_{i=1}^r(a_i)_m}{\prod_{j=1}^q(c_j)_m}\frac{z^m}{m!},
\end{equation*}
provided it converges absolutely. Here, an empty product equals one and the parameters $(a_i, 1 \leq i \leq r)$ are complex numbers while $(c_j, 1 \leq j \leq q) \in \mathbb{C} \setminus -\mathbb{N}$. If $a_i = -n \in - \mathbb{N}$ for some $1 \leq i \leq r$,
then the hypergeometric series terminates and as such, reduces  to a polynomial of degree $n$. For instance, the $j$-th Gegenbauer polynomial of parameter $\lambda$ is defined through the Gauss hypergeometric function ${}_2F_1$:
\begin{equation}\label{GegJac}
C_j^{(\lambda)}(z) := \frac{(2\lambda)_j}{j!}{}_2F_1\left(-j, j+2\lambda, \lambda+\frac{1}{2}, \frac{1-z}{2}\right), \quad \lambda \neq 0.
\end{equation}
For this polynomial and this hypergeometric function, the following argument transformations hold:
\begin{equation}\label{Sym}
C_j^{(\lambda)}(-z) = (-1)^j C_j^{(\lambda)}(z),
\end{equation}
\begin{equation}\label{ET}
{}_2F_1(a,c_1, c_2; z) = (1-z)^{-a} {}_2F_1\left(a, c_2-c_1, c_2; \frac{z}{z-1}\right), \quad |\arg(1-z)| < \pi.
\end{equation}
Now, the confluent hypergeometric series ${}_1F_1$ converges in the whole complex plane:
\begin{equation}\label{ExpConf}
{}_1F_1(a,c, z) = \sum_{j \geq 0} \frac{(a)_j}{(c)_j} \frac{z^j}{j!},
\end{equation}
and admits the following integral representation valid for $\Re(c) > \Re(a) > 0$:
\begin{align}
{}_1F_1\left(a, c, z\right) = \frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1e^{uz}u^{a-1}(1-u)^{c-a-1}du. \label{Euler}
\end{align}
We shall also need the fourth Lauricella function $F_D^{(p-1)}$ in $p-1, p \geq 2$ variables:
\begin{equation}\label{Laur}
F_D^{(p-1)}(a, d_1, \ldots, d_{p-1}, d_p; z_1, \ldots, z_{p-1}) := \sum_{m_1, \ldots, m_{p-1} \geq 0} \frac{(a)_{m_1+\ldots+m_{p-1}}}{(d_p)_{m_1+\ldots+m_{p-1}}}\prod_{s=1}^{p-1} (d_s)_{m_s} z_s^{m_s},
\end{equation}
which converges for $|z_s| < 1, 1 \leq s \leq p-1$, where $a, d_1, \ldots, d_{p-1} \in \mathbb{C}$ and $d_p \in \mathbb{C} \setminus \mathbb{N}$. When $\Re(d_s) > 0, 1 \leq s \leq p,$ it admits the following Euler-type integral representation
(see e.g. \cite{Lij-Reg}, eq. (2.1)):
\begin{multline}\label{IRD}
F_D^{(p-1)}(a, d_1, \ldots, d_{p-1}, d_1 + \ldots + d_{p-1} + d_p; z_1, \ldots, z_{p-1}) = \frac{\Gamma(a+d_1+\ldots+d_{p-1})}{\Gamma(a)\prod_{s=1}^{p-1}\Gamma(d_s)}\int_{\Sigma_p} du_1 \ldots du_{p-1}
\\ \left(1-u_1z_1-\ldots - u_{p-1}z_{p-1}\right)^{-a} \prod_{s=1}^{p}u_s^{d_s-1},
\end{multline}
where
\begin{equation*}
\Sigma_p := \{(u_1, \ldots, u_p), \, u_1, \ldots, u_p \geq 0, \, u_1 + \ldots + u_p = 1\}
\end{equation*}
is the standard simplex in $\mathbb{R}^p$. Finally, we recall the Dirichlet integral: for any $\beta_1, \ldots, \beta_p > 0$,
\begin{equation}\label{Dir}
\int_{\Sigma_p} du_1 \ldots du_{p-1}\prod_{s=1}^p u_s^{\beta_s-1} = \frac{\Gamma(\beta_1) \ldots \Gamma(\beta_p)}{\Gamma(\beta_1+\ldots+\beta_p)},
\end{equation}
and we denote
\begin{equation*}
\mu^{s}(du) = \frac{\Gamma(s+1/2)}{\sqrt{\pi}\Gamma(s)}(1-u^2)^{s-1} {\bf 1}_{[-1,1]}(u)du, \quad s > 0,
\end{equation*}
the symmetric Beta distribution.

\section{Integral representation of the density of $V_0$: even dihedral wedges and equal multiplicity values}
Let $k_0 = k_1 \equiv k \in (1/2,1]$ and consider the radial Dunkl process in an even dihedral wedge of angle $\pi/(2p), p \geq 2,$ and associated with $(1-k, 1-k)$. In \cite{Demni1}, it was shown that\footnote{The result is also valid for $p=1$} the density of $V_0$ is, up to a normalizing constant, the even part of the following series (see eq. (3.2), p.146):
\begin{equation}\label{Serie1}
\sin^{2\nu}(2p\phi) e^{-v}v^{2p\nu-1}\sum_{j \geq 0}\frac{\Gamma(p(j+1))}{\Gamma(2p(j+k))}v^{pj}{}_1F_1\left(p(j+1), 2p(j+k)+1, v\right)C_{j}^{(k)}(\cos(2p\phi)),
\end{equation}
where
\begin{equation*}
v > 0, \quad \nu := k - \frac{1}{2}.
\end{equation*}
For the particular value $p =2$, the following integral representation of \eqref{Serie1} was obtained in \cite{Demni1}, Lemma 3.2:
\begin{multline}\label{Formula}
\sum_{j \geq 0}\frac{\Gamma(2(j+1))}{\Gamma(4(j+k))}v^{2j}{}_1F_1\left(2(j+1), 4(j+k) + 1, v\right)C_{j}^{(k)}(\cos(4\phi)) = \frac{1}{\Gamma(4k)} \\ \int {}_1F_1\left(2, 2\nu+\frac{3}{2}; \frac{v(1-\cos(2\phi)u)}{2}\right)\mu^{k}(du).
\end{multline}
The extension of \eqref{Formula} to all values $p \geq 2$ is as follows:

\begin{pro}
For any $p \geq 2$,
\begin{multline}\label{Formula1}
\sum_{j \geq 0}\frac{\Gamma(p(j+1))}{\Gamma(2p(j+k))}v^{pj}{}_1F_1\left(p(j+1), 2p(j+k)+1, v\right)C_{j}^{(k)}(\cos(2p\phi))  = \frac{\Gamma(p)\Gamma(pk)}{[\Gamma(k)]^p\Gamma(2kp)} \int_{\Sigma_p}du_1 \ldots du_{p-1} \prod_{s=1}^p u_s^{k-1}
\\ {}_1F_1 \left(p, pk+\frac{1}{2}; v \sum_{s=1}^p u_s \cos^{2}\left(\phi + \frac{s\pi}{p}\right)\right).
\end{multline}
\end{pro}

\begin{proof}
From \eqref{ExpConf}, we readily derive
\begin{align*}
\frac{\Gamma(p(j+1))}{\Gamma(2p(j+k))}{}_1F_1\left(p(j+1), 2p(j+k)+1, v\right) & = 2p(j+k) \sum_{m \geq 0} \frac{\Gamma(a_j+m)}{\Gamma(b_j+m+1)} \frac{v^m}{m!},
\end{align*}
whence
\begin{multline*}
\sum_{j \geq 0}\frac{\Gamma(p(j+1))}{\Gamma(2p(j+k))}v^{pj}{}_1F_1\left(p(j+1), 2p(j+k)+1, v\right)C_{j}^{(k)}(\cos(2p\phi)) =  (2p) \sum_{j, m\geq 0}  (j+k) \\
\frac{\Gamma(pj+m + p)}{\Gamma(2pj+m + 2pk +1)m!} v^{pj+m} C_{j}^{(k)}(\cos(2p\phi))
\\ = \sum_{N \geq 0}v^N \Gamma(N+p) \sum_{\substack{m,j \geq 0 \\ N =pj+m}} \frac{2p(j+k)}{\Gamma(2pj+m + 2pk +1)m!}  C_{j}^{(k)}(\cos(2p\phi)).
\end{multline*}
Now, recall from \cite{Del-Dem}, Proposition 1,  the following identity: for any integers $M \geq 0, q \geq 1$, any real numbers $k > 0, \xi \in [0,\pi]$, we have:
\begin{equation}\label{IdGeg}
\sum_{\substack{m,j \geq 0 \\ M = 2m+qj}}\frac{q(j+k)}{m!\Gamma(q(j+k)+m+1)}C_j^{(k)}(\cos \xi) = \frac{2^M}{\Gamma(M+qk)} \sum_{\substack{j_1, \ldots, j_q \geq 0 \\ j_1+\cdots j_q = M}}
(k)_{j_1}\ldots (k)_{j_{q}} \frac{[b_{1,q}(\xi)]^{j_1}}{j_1!}\cdots  \frac{[b_{q,q}(\xi)]^{j_{q}}}{j_{q}!},
\end{equation}
where
\begin{equation*}
b_{s,q}(\xi) := \cos\left(\frac{\xi + 2s\pi}{q}\right), \quad s = 1, \ldots, q.
\end{equation*}
In particular, if $M = 2N, q = 2p, \xi = 2p\phi$, then \eqref{IdGeg} specializes to:
\begin{align*}
\sum_{\substack{m,j \geq 0 \\ N = m+pj}}\frac{2p(j+k)}{m!\Gamma(2p(j+k)+m+1)}C_j^{(k)}(\cos(2p\phi)) = \frac{2^{2N}}{\Gamma(2N+2pk)} \sum_{\substack{j_1, \ldots, j_{2p} \geq 0 \\ j_1+\cdots j_{2p} = 2N}}
\prod_{s=1}^{2p} \frac{(k)_{j_s}}{j_s!} \cos^{j_s} \left(\phi + \frac{s\pi}{p}\right).
\end{align*}
As a result,
\begin{multline*}
\sum_{j \geq 0}\frac{\Gamma(p(j+1))}{\Gamma(2p(j+k))}v^{pj}{}_1F_1\left(p(j+1), 2p(j+k)+1, v\right) C_{j}^{(k)}(\cos(2p\phi)) = \sum_{N \geq 0}v^N  \frac{2^{2N}\Gamma(N+p)}{\Gamma(2N+2pk)}
\\ \sum_{\substack{j_1, \ldots, j_{2p} \geq 0 \\ j_1+\cdots j_{2p} = 2N}}  \prod_{s=1}^{2p} \frac{(k)_{j_s}}{j_s!} \cos^{j_s}\left(\phi + \frac{s\pi}{p}\right)
 = \frac{\Gamma(p)}{\Gamma(2pk)} \sum_{N \geq 0}v^N\frac{(p)_N}{(pk)_N(pk+(1/2))_N} \\
  \sum_{\substack{j_1, \ldots, j_{2p} \geq 0 \\ j_1+\cdots j_{2p} = 2N}}  \prod_{s=1}^{2p} \frac{(k)_{j_s}}{j_s!} \cos^{j_s}\left(\phi + \frac{s\pi}{p}\right).
\end{multline*}
Now, since $\cos(a+\pi) = -\cos(a)$, then the finite sum in last equality may be written as:
\begin{equation}\label{Sum1}
\sum_{\substack{j_1, \ldots, j_{2p} \geq 0 \\ j_1+\cdots j_{2p} = 2N}} (-1)^{j_1+\dots+j_p} \prod_{s=1}^{2p} \frac{(k)_{j_s}}{j_s!}  \prod_{s=1}^p \cos^{j_s + j_{s+p}}\left(\phi + \frac{s\pi}{p}\right),
\end{equation}
which may be further simplified as follows. Consider a $2p$-tuple of integers in the sum \eqref{Sum1} arranged in $p$ pairs:
\begin{equation*}
(j_1,j_{p+1}), \ldots, (j_p, j_{2p}), \quad \sum_{s=1}^{2p} j_s = 2N,
\end{equation*}
and assume that one pair consists of an even and an odd integers. Then, the $2p$-tuple obtained from the previous one by permuting the even and the odd integers in the given pair (while keeping the other pairs unchanged) cancels the latter since the expression
\begin{equation*}
\prod_{s=1}^{2p} \frac{(k)_{j_s}}{j_s!}  \prod_{s=1}^p \cos^{j_s + j_{s+p}}\left(\phi + \frac{s\pi}{p}\right)
\end{equation*}
is the same for both tuples. As a matter of fact, the only tuples that remains in \eqref{Sum1} after cancellations are those for which $j_s+j_{s+p}$ is even, therefore
\begin{multline}\label{Sum2}
\sum_{\substack{j_1, \ldots, j_{2p} \geq 0 \\ j_1+\cdots j_{2p} = 2N}} (-1)^{j_1+\dots+j_p} \prod_{s=1}^{2p} \frac{(k)_{j_s}}{j_s!}  \prod_{s=1}^p \cos^{j_s + j_{s+p}}\left(\phi + \frac{s\pi}{p}\right) =
\sum_{\substack{j_1, \ldots, j_{2p} \geq 0 \\ j_1+\cdots j_{2p} = 2N \\ j_s+j_{s+p} \textrm{is even}}}  (-1)^{j_1+\dots+j_p} \\ \prod_{s=1}^{2p} \frac{(k)_{j_s}}{j_s!}  \prod_{s=1}^p \cos^{j_s+j_{s+p}}\left(\phi + \frac{s\pi}{p}\right).
\end{multline}
Setting $m_s = j_s+j_{s+p}$ and using the identity
\begin{equation*}
\sum_{j_s = 0}^{2m_s} (-1)^{j_s} \frac{(k)_{j_s}(k)_{2m_s-j_s}}{j_s!(2m_s - j_s)!} = \frac{(k)_{m_s}}{m_s!},
\end{equation*}
which is readily checked by equating generating functions of both sides, then the right-hand side of \eqref{Sum2} may be written as
\begin{align*}
\sum_{\substack{m_1, \ldots, m_{p} \geq 0 \\ m_1+\cdots m_{p} = N}} \prod_{s=1}^p \cos^{2m_s}\left(\phi + \frac{s\pi}{p}\right)  \sum_{j_s = 0}^{2m_s} (-1)^{j_s} \frac{(k)_{j_s}(k)_{2m_s-j_s}}{j_s!(2m_s - j_s)!} =
\sum_{\substack{m_1, \ldots, m_{p} \geq 0 \\ m_1+\cdots m_{p} = N}} \prod_{s=1}^p \cos^{2m_s}\left(\phi + \frac{s\pi}{p}\right) \frac{(k)_{m_s}}{m_s!}.
\end{align*}

Consequently,
\begin{multline*}
\sum_{j \geq 0}\frac{\Gamma(p(j+1))}{\Gamma(2p(j+k))}v^{pj}{}_1F_1\left(p(j+1), 2p(j+k)+1, v\right) C_{j}^{(k)}(\cos(2p\phi))  = \frac{\Gamma(p)}{\Gamma(2pk)}\sum_{N \geq 0}v^N\frac{(p)_N}{(pk)_N(pk+(1/2))_N}  \\
\sum_{\substack{m_1, \ldots, m_{p} \geq 0 \\ m_1+\cdots m_{p} = N}} \prod_{s=1}^p \cos^{2m_s}\left(\phi + \frac{s\pi}{p}\right) \frac{(k)_{m_s}}{m_s!}
= \frac{\Gamma(p)}{\Gamma(2pk)}\sum_{m_1, \ldots, m_{p} \geq 0 }\frac{(p)_{m_1+\dots+m_p}}{(pk)_{m_1+\dots+m_p}(pk+(1/2))_{m_1+\dots+m_p}} \\
\prod_{s=1}^p \frac{(k)_{m_s}}{m_s!} \left\{v\cos \left(\phi + \frac{s\pi}{p}\right)\right\}^{2m_s}.
\end{multline*}
Finally, substituting $\beta_s = k+ m_s$ in \eqref{Dir} and using the multinomial theorem:
\begin{equation*}
(a_1+\ldots +a_p)^N = \sum_{m_1+\ldots + m_p = N} \frac{N!}{m_1!\ldots m_p!} \prod_{s=1}^p a_s^{m_s},
 \end{equation*}
together with \eqref{ExpConf}, we end up with :
\begin{multline*}
\frac{\Gamma(p)}{\Gamma(2pk)}\sum_{m_1, \ldots, m_{p} \geq 0 }\frac{(p)_{m_1+\dots+m_p}}{(pk)_{m_1+\dots+m_p}(pk+(1/2))_{m_1+\dots+m_p}}
\prod_{s=1}^p \frac{(k)_{m_s}}{m_s!} \left\{v\cos^2 \left(\phi + \frac{s\pi}{p}\right)\right\}^{m_s} = \frac{\Gamma(p)\Gamma(pk)}{[\Gamma(k)]^p\Gamma(2pk)}
\\ \int_{\Sigma_p} du_1\ldots du_{p-1} \sum_{m_1, \ldots, m_{p} \geq 0 }\frac{(p)_{m_1+\dots+m_p}}{(pk+(1/2))_{m_1+\dots+m_p}}
\prod_{s=1}^{p} \frac{u_s^{k+m_s-1}}{m_s!} \left\{v\cos^2 \left(\phi + \frac{s\pi}{p}\right)\right\}^{m_s}
\\ = \frac{\Gamma(p)\Gamma(pk)}{[\Gamma(k)]^p\Gamma(2pk)}
 \int_{\Sigma_p} du_1\ldots du_{p-1} \prod_{s=1}^{p} u_s^{k-1}\sum_{N \geq 0}\frac{(p)_N}{(pk+(1/2))_N} \\
 \sum_{m_1+\ldots + m_p= N}  \prod_{s=1}^{p} \frac{u_s^{m_s}}{m_s!} \left\{v\cos^2 \left(\phi + \frac{s\pi}{p}\right)\right\}^{m_s}
 \\ = \frac{\Gamma(p)\Gamma(pk)}{[\Gamma(k)]^p\Gamma(2pk)}
 \int_{\Sigma_p} du_1\ldots du_{p-1} \prod_{s=1}^p u_s^{k-1}\sum_{N \geq 0}\frac{(p)_N}{(pk+(1/2))_N} \frac{1}{N!}\left\{v\sum_{s=1}^pu_s\cos^2 \left(\phi + \frac{s\pi}{p}\right)\right\}^N.
\end{multline*}

Using \eqref{ExpConf}, the proposition is proved.
\end{proof}

\begin{rem}
If $p=2$, then \eqref{Formula1} simplifies to
\begin{multline*}
\frac{\Gamma(2k)}{[\Gamma(k)]^2\Gamma(4k)} \int_0^1u^{k-1}(1-u)^{k-1}  {}_1F_1 \left(2, 2k+\frac{1}{2}; v \left\{u \sin^{2}\left(\phi \right) + (1-u) \cos^2(\phi)\right\}\right) du = \frac{\Gamma(2k)}{[\Gamma(k)]^2\Gamma(4k)}
\\ \int_0^1u^{k-1}(1-u)^{k-1}  {}_1F_1 \left(2, 2k+\frac{1}{2}; v \left\{\frac{1+ (1-2u) \cos(2\phi)}{2}\right\}\right) du.
\end{multline*}
Performing the variable change $u \mapsto (1-u)/2$ in this integral and using Legendre duplication formula \eqref{Leg}, one recovers \eqref{Formula} (recall $k = \nu + (1/2))$.
\end{rem}

Keeping in mind \eqref{Serie1}, we get:
\begin{cor}\label{Cor1}
Let $k \in (1/2,1]$ and $p \geq 2$. Then, the density of $V_0$ admits the following integral representation (up to a normalizing constant depending only on $p,k$):
\begin{multline*}
\sin^{2\nu}(2p\phi) e^{-v}v^{2p\nu-1} \int_{\Sigma_p} du_1 \ldots du_{p-1} \prod_{s=1}^p u_s^{k-1}
\left\{ {}_1F_1 \left(p, p\nu+\frac{(p+1)}{2}; v \sum_{s=1}^p u_s \cos^{2}\left(\phi + \frac{s\pi }{p}\right)\right) \right. + \\ \left.{}_1F_1 \left(p, p\nu+\frac{(p+1)}{2}; v \sum_{s=1}^p u_s \cos^{2}\left(\phi - \frac{(2s-1)\pi }{2p}\right)\right)\right\}.
\end{multline*}
\end{cor}
\begin{proof}
As claimed in the beginning of this section, the density of $V_0$ is half of the sum of both series:
\begin{equation*}
\sum_{j \geq 0}(\pm 1)^j\frac{\Gamma(p(j+1))}{\Gamma(2p(j+k))}v^{pj}{}_1F_1\left(p(j+1), 2p(j+k)+1, v\right)C_{j}^{(k)}(\cos(2p\phi)).
\end{equation*}
Using the symmetry relation \eqref{Sym}, we get:
\begin{equation*}
(-1)^jC_{j}^{(k)}(\cos(2p\phi)) = C_{j}^{(k)}\left(\cos\left(2p\left(\frac{\pi}{2p} - \phi\right)\right)\right),
\end{equation*}
whence
\begin{multline}\label{Formula2}
\sum_{j \geq 0}\frac{\Gamma(p(j+1))}{\Gamma(2p(j+k))}v^{pj}{}_1F_1\left(p(j+1), 2p(j+k), v\right)(-1)^jC_{j}^{(k)}(\cos(2p\phi))  = \frac{\Gamma(p)\Gamma(pk)}{[\Gamma(k)]^p\Gamma(2kp)} \int_{\Sigma_p}du_1\ldots du_{p-1}
\\ \prod_{s=1}^p u_s^{k-1} {}_1F_1 \left(p, pk+\frac{1}{2}; v \sum_{s=1}^p u_s \cos^{2}\left(\phi - \frac{\pi(2s+1)}{2p} \right)\right).
\end{multline}
Since
\begin{equation*}
\cos^{2}\left(\phi - \frac{\pi(2p+1)}{2p} \right) = \cos^{2}\left(\phi  - \frac{\pi}{2p} \right),
\end{equation*}
then equality \eqref{Formula2} is rewritten:
\begin{multline}\label{Formula3}
\sum_{j \geq 0}\frac{\Gamma(p(j+1))}{\Gamma(2p(j+k))}v^{pj}{}_1F_1\left(p(j+1), 2p(j+k), v\right)(-1)^jC_{j}^{(k)}(\cos(2p\phi))  = \frac{\Gamma(p)\Gamma(pk)}{[\Gamma(k)]^p\Gamma(2kp)} \int_{\Sigma_p}du_1\ldots du_{p-1}
\\ \prod_{s=1}^p u_s^{k-1} {}_1F_1 \left(p, pk+\frac{1}{2}; v \sum_{s=1}^p u_s \cos^{2}\left(\phi - \frac{\pi(2s-1)}{2p} \right)\right),
\end{multline}
and the corollary is proved.
\end{proof}

\begin{rem}
With \eqref{Formula} in hands, one readily extends Corollary 3.3 in \cite{Demni1} to all $p \geq 2$ provided
\begin{equation*}
p(k -1) + \frac{1}{2} > 0,
\end{equation*}
ensuring the validity of the Euler-type integral representation \eqref{Euler}. As to the extension of Corollary 3.5 in \cite{Demni1}, it goes as follows:
\end{rem}

\begin{cor}\label{Cor2}
Assume the radial Dunkl process starts at a point lying on the bisector of the Weyl chamber:
\begin{equation*}
x = \rho e^{i\pi/(4p)}, \quad \rho > 0.
\end{equation*}
Denote $\mathbb{E}_{\rho, \pi/(4p)}^{(-\nu)}$ the distribution of this process associated with the common multiplicity value $(1-k)$. Then, for any $y > 0$,
\begin{multline*}
\mathbb{E}_{\rho, \pi/(4p)}^{(-\nu)}\left(V^{(p+1)/2- p\nu} e^{-yV_0}\right) \propto \frac{\sin^{2\nu}(2p\phi)}{(1+y)^{p(\nu-1)+(p+1)/2}} \left[y + \sin^2\left(\frac{\pi}{4p}\right)\right]^{-p}
\\ F_D^{(p-1)}\left(p, \underbrace{k, \ldots, k}_{p-1}, pk; - \frac{\sin[3\pi/(2p)] \sin[(\pi/p)]}{y+\sin^2(\pi/(4p))}, \ldots, -\frac{\sin[(2p+1)\pi/(2p)] \sin[((p-1)\pi/p)]}{y+\sin^2(\pi/(4p))}\right),
\end{multline*}
where the coefficient of proportionality only depends on $k,p$.
\end{cor}
\begin{proof}
The choice $\phi = \pi/(4p)$ ensures the equality:
\begin{equation*}
\sum_{s=1}^p u_s \cos^{2}\left(\frac{\pi}{4p} + \frac{s\pi }{p}\right) = \sum_{s=1}^p u_s \cos^2\left( \frac{(4s+1)\pi}{4p} \right)  = \sum_{s=1}^p u_s \cos^{2}\left(\frac{\pi}{4p} - \frac{\pi(2s-1)}{2p}\right),
\end{equation*}
so that both confluent hypergeometric functions displayed in Corollary \ref{Cor1} coincide. Now, set
\begin{equation*}
M_p(u_1, \ldots, u_p) := \sum_{s=1}^p u_s \cos^{2}\left(\frac{(4s+1)\pi}{4p}\right).
\end{equation*}
Then, the Gamma integral together with the generalized binomial Theorem entail:
\begin{multline*}
\int_0^{\infty} v^{p\nu + (p-1)/2} e^{-v(1+y)} {}_1F_1 \left(p, p\nu+\frac{(p+1)}{2}; vM_p(u_1, \ldots, u_p)\right) = \frac{\Gamma(p\nu+(p+1)/2)}{(1+y)^{p\nu+(p+1)/2}} \sum_{N = 0}^{\infty} \frac{(p)_N}{N!} \\
\left[\frac{M_p(u_1, \ldots, u_p)}{(1+y)}\right]^N  = \frac{1}{(1+y)^{p(\nu-1)+(p-3)/2}}\frac{1}{[1+y - M_p(u_1, \ldots, u_p)]^p}.
\end{multline*}
Next, we compute
\begin{align*}
1+y - M_p(u_1, \ldots, u_p) & = 1+y - \sum_{s=1}^{p-1} u_s\cos^{2}\left(\frac{(4s+1)\pi}{4p}\right) - (1-u_1-\ldots - u_{p-1})\cos^2\left(\frac{\pi}{4p}\right)
\\& = y + \sin^2\left(\frac{\pi}{4p}\right) - \sum_{s=1}^{p-1}u_s\left\{\cos^{2}\left(\frac{(4s+1)\pi}{4p}\right) - \cos^2\left(\frac{\pi}{4p}\right)\right\}
\\& = y + \sin^2\left(\frac{\pi}{4p}\right) - \frac{1}{2}\sum_{s=1}^{p-1}u_s\left\{ \cos\left(\frac{(4s+1)\pi}{2p}\right) - \cos\left(\frac{\pi}{2p}\right)\right\}
\\& = y + \sin^2\left(\frac{\pi}{4p}\right) - \sum_{s=1}^{p-1}u_s \sin\left(\frac{(2s+1)\pi}{2p}\right)\sin\left(-\frac{s\pi}{p}\right)
\\& = \left[y + \sin^2\left(\frac{\pi}{4p}\right)\right]^{-1} \left\{1- \sum_{s=1}^{p-1}u_s \frac{\sin[(2s+1)\pi/(2p)] \sin[-(s\pi/p)]}{y+\sin^2(\pi/(4p))}\right\}.
\end{align*}
Using \eqref{IRD}, we are done.
\end{proof}
\begin{rem}
If $p=2$, then
\begin{equation*}
\sin^2\left(\frac{\pi}{8}\right) = \frac{\sqrt{2}-1}{2\sqrt{2}}
\end{equation*}
and the Lauricella function ${}F_D^{(1)}$ coincides with the Gauss hypergeometric function:
\begin{equation*}
{}_2F_1\left(2, k, 2k; \frac{2}{1- \sqrt{2}(1+2y)}\right).
\end{equation*}
Using the argument transformation \eqref{ET}, this hypergeometric function is transformed into:
\begin{equation*}
{}_2F_1\left(2, k, 2k; \frac{2}{1- \sqrt{2}(1+2y)}\right) = \frac{(1-\sqrt{2}(1+2y))^2}{( 1+\sqrt{2}(1+2y))^2} {}_2F_1\left(2, k, 2k; \frac{2}{1+ \sqrt{2}(1+2y)}\right).
\end{equation*}
As a result,
\begin{equation*}
\mathbb{E}_{\rho, \pi/(8)}^{(-\nu)}\left(v^{(3)/2- 2\nu} e^{-yV_0}\right) \propto \frac{\sin^{2\nu}(4\phi)}{(1+y)^{2\nu-(1/2)}}\frac{1}{( 1+\sqrt{2}(1+2y))^2} {}_2F_1\left(2, k, 2k; \frac{2}{1+ \sqrt{2}(1+2y)}\right)
\end{equation*}
which is the expression derived in the proof of Corollary 3.5 in \cite{Demni1}.
\end{rem}

\section{Integral representation of the density of $V_0$: odd dihedral wedges}
Our previous reasoning applies to odd dihedral groups. Indeed, notice that if $k_0 = k_1$, then the infinitesimal generator \eqref{G2} reduces to
\begin{equation*}
\frac{1}{2}\left[\partial_r^2  + \frac{4pk_0 +1}{r}\partial_r\right]  + \frac{1}{r^2}\left[\frac{\partial_{\theta}^2}{2} +2pk_0\cot(2p\theta) \right], \quad \theta \in [0, \pi/(2p)],
\end{equation*}
which is exactly the infinitesimal generator \eqref{G1} after the identifications $k_0 \leftrightarrow k, n \leftrightarrow 2p$. Since the tail distribution of $T_0$ starting at $x$ is the unique solution of the heat equation with appropriate boundary conditions (see e.g. \cite{Deb}), then this probability has the same expression in both cases (i.e., corresponding to odd dihedral wedges on the one side and to even ones with equal multiplicity values on the other side) under the previous identifications. Of course, one can also mimic the derivation of the tail distribution written in \cite{Demni0}, paragraph 7.1 and retrieve the same expression. Hence, if $n \geq 3$ is odd and $k \in (1/2,1]$, then the radial Dunkl process associated with the dihedral group $\mathcal{D}_2(n)$ and corresponding to the multiplicity value $1-k \in [0,1/2)$ hits $\partial C$ almost surely and the density of $V_0$ is (up to a normalizing constant) the even part of the following series:
\begin{equation}\label{Density0}
\sin^{2\nu}(n\phi) e^{-v}v^{n\nu-1}\sum_{j \geq 0}\frac{\Gamma(n(j+1)/2)}{\Gamma(n(j+k))}(\sqrt{v})^{j}{}_1F_1\left(n(j+1)/2, n(j+k)+1, v\right)C_{j}^{(k)}(\cos(n\phi)), \quad v > 0.
\end{equation}
Using again \eqref{ExpConf} followed by \eqref{IdGeg}, we may rewrite \eqref{Density0} as:
\begin{multline*}
\sum_{j \geq 0}\frac{\Gamma(n(j+1)/2)}{\Gamma(n(j+k))} (\sqrt{v})^{j} {}_1F_1\left(n(j+1)/2, n(j+k)+1, v\right)C_{j}^{(k)}(\cos(n\phi)) = \sum_{j, m \geq 0} n(j+k) \\
\frac{\Gamma(n(j+1)/2+m)}{m!\Gamma(nj+m+nk+1)} (\sqrt{v})^{nj+2m}C_{j}^{(k)}(\cos(n\phi))
= \sum_{N \geq 0} (\sqrt{v)}^N\Gamma((N+n)/2)
\\ \sum_{N = nj+2m} \frac{n(j+k)}{m!\Gamma(nj+m+nk+1)} C_{j}^{(k)}(\cos(n\phi))
= \sum_{N \geq 0} (2\sqrt{v})^N\frac{\Gamma((N+n)/2))}{\Gamma(N+nk)}
\\ \sum_{\substack{j_1, \ldots, j_n \geq 0 \\ j_1+\cdots j_n = N}}  (k)_{j_1}\ldots (k)_{j_{n}} \frac{[b_{1,n}(n\phi)]^{j_1}}{j_1!}\cdots  \frac{[b_{n,n}(n\phi)]^{j_{n}}}{j_{n}!}.
\end{multline*}
Finally, the Dirichlet integral \eqref{Dir} and the multinomial Theorem lead again to:
\begin{pro}
Let $n \geq 3$ be an integer and $k > 0$. Then, for any $v > 0$,
\begin{multline*}
\sum_{j \geq 0}\frac{\Gamma(n(j+1)/2)}{\Gamma(n(j+k))}v^{j/2}{}_1F_1\left(n(j+1)/2, n(j+k)+1, v\right)C_{j}^{(k)}(\cos(n\phi)) = \frac{1}{[\Gamma(k)]^n} \int_{\Sigma_n} du_1\ldots du_{n-1} \prod_{s=1}^n u_s^{k-1}
\\ \sum_{N \geq 0} \frac{\Gamma((N+n)/2))}{(nk)_N} \frac{(2\sqrt{v})^N}{N!} \left[\sum_{s=1}^n u_s\cos\left(\phi + \frac{2s\pi}{n}\right)\right]^N.
\end{multline*}
\end{pro}

The series occurring in the integrand is a Fox-Wright confluent hypergeometric function ${}_1\Psi_1$ (\cite{Man-Sri}):
\begin{equation*}
\sum_{N \geq 0} \frac{\Gamma((N+n)/2) )}{(nk)_N} \frac{z^n}{N!} = \Gamma(nk) {}_1\Psi_1 \left[((n/2), (1/2)), (nk, 1); z\right], \quad z \in \mathbb{C}.
\end{equation*}
Moreover, one obtains a similar integral representation for the density of $V_0$ along the same lines written in the proof of Corollary \ref{Cor1}:
\begin{cor}
Let $n$ be an odd integer and $k \in (1/2,1]$. Then, the density of $V_0$ admits the following integral representation (up to a constant depending only on $n,k$):
\begin{multline*}
\sin^{2\nu}(n\phi) e^{-v}v^{n\nu-1} \int_{\Sigma_n} du_1\ldots du_{n-1} \prod_{s=1}^n u_s^{k-1}\left\{{}_1\Psi_1 \left[((n/2), (1/2)), (nk, 1); 2\sqrt{v} \sum_{s=1}^n u_s\cos\left(\phi + \frac{2s\pi}{n}\right)\right]\right.
 \\ \left. + {}_1\Psi_1 \left[((n/2), (1/2)), (nk, 1); 2\sqrt{v} \sum_{s=1}^n u_s\cos\left(\phi - \frac{(2s-1)\pi}{n}\right)\right]\right\}.
\end{multline*}
\end{cor}
As to the analogue of Corollary \ref{Cor2}, we similarly notice that
\begin{equation*}
\sum_{s=1}^n u_s\cos\left(\frac{\pi}{2n}+ \frac{2s\pi}{n}\right) = \sum_{s=1}^n u_s\cos\left(\frac{\pi}{2n} - \frac{(2s-1)\pi}{n}\right)
\end{equation*}
and integrate the density specialized to $\phi = \pi/(2n)$ with respect to the Gamma weight $v^{(n+1)/2 - n\nu} e^{-yv}$ for a given $y > 0$. Using the Gauss duplication formula, the resulting weighted Laplace transform is then given by:
\begin{cor}
Let $n \geq 3$ be an odd integer and $k > 0$. If the radial Dunkl process associated with the multiplicity value $(1-k)$ starts at $\rho e^{i\pi/(2n)}$, then
\begin{multline*}
\mathbb{E}_{\rho, \pi/(2n)}^{-\nu}\left(V_0^{(n+1)/2 -n\nu} e^{-yV_0} \right) \propto \frac{1}{(1+y)^{(n+1)/2}} \int_{\Sigma_n} du_1\ldots du_{n-1} \prod_{s=1}^n u_s^{k-1} \\ {}_1F_1 \left[n, nk; \frac{1}{\sqrt{1+y}}\sum_{s=1}^n u_s\cos\left(\frac{(4s+1)\pi}{2n}\right)\right]
\end{multline*}
where the coefficient of proportionality depends only on $n,k$.
\end{cor}

redpajama_set_name: RedPajamaArXiv


A consultant with decades of experience in the recycling industry, Aaron joins the Partnership with the coolest job title yet. As Data Architect, Aaron will work with the various groups across The Recycling Partnership to oversee the development and integration of better national recycling data.
Aaron has over 10 years of experience in the recycling and recovery world at RRS as the leader of their analysis team. There, he helped in all aspects of material recovery, assisting communities to improve their programs and corporations set better recovery strategies. He also dabbles in enterprise supply chain analytic software, and in the past installed solar panels and even did hydrology modeling. He currently lives in Ann Arbor with his wife, two boys and their cattle dog. However, his heart is in the mountains of Colorado where he visits frequently and hopes to return to live soon.


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text:
Social and emotional wellbeing: early years
Public health guideline [PH40] Published date: October 2012
Introduction: scope and purpose of this guidance
2 Public health need and practice
3 Considerations
5 Recommendations for research
Appendix A Membership of the Public Health Interventions Advisory Committee (PHIAC), the NICE project team and external contractors
Appendix B Summary of the methods used to develop this guidance
Appendix C The evidence
Appendix D Gaps in the evidence
Appendix E Supporting documents
Benefits of social and emotional wellbeing
Factors that impact on social and emotional wellbeing
Government policy puts a significant emphasis on early intervention services to ensure all children have the best possible start in life. The aim is to address the inequalities in health and life chances that exist between children living in disadvantaged circumstances and those in better-off families.
The importance of social and emotional wellbeing in relation to healthy child development is set out in a joint Department for Education and Department of Health publication, 'Supporting families in the foundation years' (2011). The primary aim of the foundation years (years 0–5) is defined as: 'promoting a child's physical, emotional, cognitive and social development so that all children have a fair chance to succeed at school and in later life'.
In addition the new 'Statutory framework for the early years foundation stage' (Department for Education 2012a) makes personal, social and emotional development a cornerstone of early years learning and education.
Other relevant policy documents and related reviews include:
'Fair society, healthy lives' (Marmot Review Team 2010).
'Healthy child programme: pregnancy and the first five years of life' (DH 2009).
'Healthy lives, healthy people: our strategy for public health in England' (DH 2010a).
'Healthy lives, healthy people: update and way forward' (DH 2011).
'No health without mental health: a cross-government mental health outcomes strategy for people of all ages' (HM Government 2011).
'Support and aspiration: a new approach to special educational needs' (Department for Education 2011a).
'The early years: foundations for life, health and learning' (Tickell 2011).
'The importance of teaching' (Department for Education 2010).
Social and emotional wellbeing is important in its own right, but it also provides the basis for future health and life chances.
Poor social and emotional capabilities increase the likelihood of antisocial behaviour and mental health problems, substance misuse, teenage pregnancy, poor educational attainment and involvement in criminal activity. For example, aggressive behaviour at the age of 8 is a predictor of criminal behaviour, arrests, convictions, traffic offences, spouse abuse and punitive treatment of their own children (Farrington et al. 2006).
The child's relationship with their mother (or main carer) has a major impact on social and emotional development. In turn, the mother's ability to provide a nurturing relationship is dependent on her own emotional and social wellbeing and intellectual development – and on her living circumstances. The latter includes family environment, social networks and employment status (Shonkoff and Phillips 2000).
Most parents living in poor social circumstances provide a loving and nurturing environment, despite many difficulties. However, children living in a disadvantaged family are more likely to be exposed to adverse factors such as parental substance misuse and mental illness, or neglect, abuse and domestic violence. Consequently, they are more likely to experience emotional and behavioural problems that can impact on their development and opportunities in life (Farrington et al. 2006; Shonkoff and Phillips 2000).
For example, measures of 'school readiness' show that the poorest 20% of children are more likely to display conduct problems at age 5, compared to children from more affluent backgrounds (Sabates and Dex 2012; Waldfogel and Washbrook 2008).
There are less opportunities after the preschool period to close the gap in behavioural, social and educational outcomes (Allen 2011; Field 2010).
Services that support families and children during their early years are generally not well coordinated and integrated either at the strategic or local level (Allen 2011a; Field 2010; Munro 2011; Tickell 2010).
The level and quality of early childcare and education services varies, with the most disadvantaged children likely to get the worse provision (Ofsted 2010). In addition, only an estimated 50% of children aged 2 and 2½ years in England are being assessed as part of the Healthy Child Programme – and not all women are being offered antenatal and parenting support services (Care Quality Commission 2010; DH 2010b).
The approaches and interventions used to address specific problems (such as abuse, maternal mental health problems and poor parenting) also vary widely and, while some interventions have been proven to be effective and cost effective, others have not. Where evidence-based interventions are used, they are not always being implemented effectively (Allen 2011a; Field 2010).
There is limited UK data on the indicators that provide an overall measure of the social and emotional wellbeing of children aged under 5 years. Independent reviews recommend that measures should be developed to assess children's cognitive, physical and emotional development at ages 3 and 5 years (Allen 2011b; Field 2010; Tickell 2011).
Early intervention can provide a good return on investment (Knapp et al. 2011). For example, an evaluation of the US-based Nurse-Family Partnership estimated that the programme made savings by the time the children of high-risk families had reached the age of 15. These savings, which were over five times the cost of the programme itself, resulted from reduced expenditure in the welfare and criminal justice systems, higher tax revenues and improved physical and mental health (Karoly et al. 2005). (The cost effectiveness of the UK Family Nurse Partnership (FNP) model is currently being investigated as part of the FNP trial.)
The cost of not intervening to ensure (or improve) the social and emotional wellbeing of children and their families are significant, for both them and wider society (Aked et al. 2009). For example, by the age of 28, the cumulative costs for public services are much higher when supporting someone with a conduct disorder, compared to providing services for someone with no such problems (Scott et al. 2001).

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"Bullock somehow manages to infuse serious beauty and poetry into the very real voice of this remarkably likeable character. Funny and intensely felt, this novel holds a magnifying glass up to some of the ugliest parts of small-town life, and yet reminds us that there is still kindness, beauty, and humor in the world after all." —Ashley Cowger, author of Peter Never Came
"The voice that describes this marvelous seaside place is a fountain of delights." —Robert Pope, author of Jack's Universe
Click to view John's reading from the novel: https://youtu.be/SU9nW-71LG0
Click to view John's interview with Carter Taylor Seaton on Chapters: https://youtu.be/JJffaeQtZX4
NOTE: Sheila-Na-Gig fiction submissions are by invitation only. Please do not send us fiction at this time.
Remembering Sheila-Na-Gig
Sheila-Na-Gig online: News


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Yingting Shih's Review Summary: Here is a lotus park with lots of flowers. You can sit in the pavilion on the lotus pond to enjoy the lotus. Best of all, you can visit without paying, and the park also offers free lotus tea for you to drink. It is a good place for a family trip.


My Beloved is thinking about going gluten free again. To me this means I have been perusing all the new gluten free cookbooks that have come out recently (I love the library). I got this from Cooking For Isiah. I have to admit that I have already tried different spices and baking times. I personally like more crunchy but my Beloved likes less. He thinks these are fabulous snack foods.
Put the oil on a baking sheet. Spread it around and then pour the chickpeas over it. Sprinkle with salt, pepper, and cayenne. Toss or stir about to cover. Bake for 20 minutes.
Roasted yumminess. My Beloved said these got him through the eighteen hour days he was working. They were originally for salads. Just a nice change to croutons if you if do not have to worry about gluten. The ancho/paprika/chipolte option is more savory then spicy which really surprised me!


text:
Pac-12 Players Issue Unity Demands
A Players Tribune piece lists an ambitious set of conditions for moving forward with the 2020 season
By Max Vrooman@UWDP_maxvroom Aug 2, 2020, 8:49am PDT
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Photo by Kevin Abele/Icon Sportswire via Getty Images
In an unprecedented step within college athletics, a substantial group of Pac-12 athletes have come together and put forth a list of demands that must be met in order for them to not boycott the 2020 season. The complete list was published this morning in The Players' Tribune. Several of the players driving the movement also spoke with Sports Illustrated.
The first section is about COVID-19 and includes the ability to opt-out of the season for pandemic-related reasons without scholarship, a clause that essentially all pro leagues have adopted in their restarts. It also lists prohibiting waivers that excuse the school of any COVID-19 liability and for safety standards coming from a 3rd party be instituted and approved of by the players.
Additional demands were made about protecting sports from elimination during the pandemic and resulting financial crisis, taking steps to address racial injustice and systemic racism in college athletics, to provide health insurance for players' injury-related health insurance for 6 years post-college, and to allow both Name, Image, and Likeness rights as well as revenue sharing with the players.
Before even addressing the feasibility of the items listed we should first acknowledge the potential impact that this could have on the college sports landscape. The NCAA has long had no incentive to change a system that allows them to keep all of the overhead by paying players significantly less than the free market would allow in the forms of scholarships and stipends. The recent Name, Image, and Likeness legislation that began in California was one of the first to force the NCAA to change parts of their amateurism model, albeit kicking and screaming.
However, outside of national legislation making their actions outright illegal the NCAA still has no reason for major change. Northwestern attempted to unionize back in 2014 but was just a single team (and frankly not one that drives a lot of the revenue). Now, what appears to be at least a substantial portion of if not an entire conference is threatening to boycott.
In order for the college athletics model to function it absolutely needs to generate TV revenue and needs to get back to generating fan ticket revenue as soon as possible given the state of the pandemic. Now more than ever the system is realizing how devastating losing a year of that revenue can be. And the players have realized it too which is why they feel empowered enough to take this step.
This kind of a boycott only works if it includes a large enough proportion of the total player pool. If this were 5 players on each team then the Pac-12 and the NCAA would have no reason to cooperate and could just let them sit out. While we don't know the exact scope of how many players have signed on just yet it does appear based on the SI article to be in the hundreds which would suggest 15+ players per team. If over half of the starters in the conference threaten to sit out does the Pac-12 let that happen?
The only two players that we currently know of that publicly signed on to the boycott from the University of Washington were Ty Jones and Joe Tryon. Oregon's Jevon Holland who is viewed as a potential All-American and 1st round NFL Draft pick also is one of the 12 players made available for comments from the media.
"We're not your entertainment, we're human beings," Oregon safety Jevon Holland told SI. "Just like you would help your family, we want to help our mother, father, grandmother, everyone."
Comments from Ty Jones and Joe Tryon from the Pac-12 players' press release. It's unclear how many other UW players might be involved. pic.twitter.com/T5Ol0QNJUx
— Christian Caple (@ChristianCaple) August 2, 2020
Given Larry Scott's track record it's hard to imagine him making the best choice especially since one of the listed demands is that he make substantially less than his current $3.5 salary.
The COVID-19 demands seem to be the easiest and most likely to happen. If a player doesn't feel safe then they absolutely should not be asked to put themselves at risk for the school's monetary benefit. Similarly the fact that there doesn't appear to be a well-thought out system for trying to make things as safe as possible is a giant red flag and should be done no matter what.
A bubble is much less feasible for college football than the NBA but even if the NBA hadn't gone that route, you can tell they thought through every detail. The current spread through several MLB clubhouses shows what happens when you just let things ride. It's self-defeating for the conference not to work with health care professionals to increase safety or there's no chance of a season happening anyways. And if you have those precautions in place then liability becomes a much lower concern.
The demands related to racial injustice are also easier to address. Creating a civic-engagement task force to address racial injustice issues and a Black College Athlete Summit are absolutely doable and if the players want them to exist then there's no reason not to provide those platforms.
Where the sticking point will be and how this ultimately plays out depends on the players' reactions when they don't get their listed demands with regards to the money. Some of it is partly because of the impracticality. The Pac-12 conference doesn't directly control all of the revenue streams throughout college football. Even if they became a perfectly willing advocate for players' rights to the NCAA they'd just be laughed out of the room by the other conferences. 50% revenue sharing with the players is not going to happen. Name, Image, and Likeness rights are being discussed in Congress and appear to be on the way regardless of this boycott. The majority of college endowments legally may only be used for specific purposes and are not available to support athletic departments in financial trouble.
But Larry Scott has control over his own salary. Each Pac-12 member institution has control over their coaches' salaries and contract terms. This is ultimately where the fate of the season will be determined. If the players get everything they want related to COVID and permanent racial injustice actions will they call that good enough?
The initial starting point in any negotiation involves aiming higher than you'll ultimately be willing to settle. There is a 0% chance that every one of these things happens over the next 7 weeks and the season starts on time with the players completely happy. The players know that.
We'll find out though over the coming days and weeks whether the Pac-12 has any inclination to cooperate, which (if any) demands the players are willing to back down on, and whether there's no season to boycott anyways because public health officials force its cancellation.
However, if the players truly insist that they won't play unless Larry Scott is making far less than his $3.5 million salary? Expect to either see no Pac-12 player suiting up in the fall or a starting lineup that involves a lot of walk-ons.
EDIT 2:15p
Some well-stated thoughts from Elijah Molden on the subject:
pic.twitter.com/lxmMJ1Pb0G
— Elijah Molden (@e_mold3n) August 2, 2020


Want to visit over the Easter Holidays? Passes ordinarily take up to 5 working days to arrive and with the bank holidays upon us, there are fewer working days so we advise you to head to an attraction to purchase your Merlin Annual Pass at the same great price.
You can also choose to pay on a monthly basis with our new Merlin Monthly Membership available exclusively online. Not included as part of Easter Sale.
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Choose from either a Merlin Pass Monthly Membership, available exclusively online or buy a Merlin Annual Pass. Whether you pay monthly or in one go you'll get the same great benefits and we'll post your Pass directly to you.
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London & Resort Theme Park date exclusions and other restrictions apply to the Standard Merlin Annual Pass. See calendar below for more details.
The V.I.P. Merlin Annual Pass excludes entry into Falls Creek and Mount Hotham plus any new attraction unless deemed otherwise by Merlin.
Lanyard, card holder, a pop badge, 3 x £5 off Fastrack voucher for one use at some of our Theme Parks and 3 Share the Fun vouchers!
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Passholders have up to 8 weeks from the date of their expiry to renew their Merlin Annual Passes at the renewal rate. You can renew online or at your home attraction. Please note print and post is not available for renewing Passholders.
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Anyone signed up to the V.I.P. waiting list will be contacted should a pass become available. Payment can be made straight away or held for up to 2 weeks. Payment must be paid in full and cannot be held past the 2 weeks. Once the payment has been made we will send out your V.I.P. Pack. Please note it can take up to three weeks for your pack to reach you. V.I.P. passes cannot be purchased online or by contacting us directly. Our waiting list is currently unavailable due to high demand.


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The U.S. Department of Labor has released its early jobs report. The findings: Unemployment has fallen to 6.1 percent — its lowest level since September 2008. In addition, about 288,000 new domestic jobs were added in June. These statistics exceed economists' expectations, but they can be misleading for several reasons.
The Department of Labor's unemployment rate only includes people actively seeking employment. The headcount excludes discouraged people — including many teenagers — who've given up their job searches, and underemployed individuals who've accepted lower-level or part-time positions to make ends meet. So, it may be that unemployment figures are going down for the wrong reasons.
The Department of Labor publishes six unemployment rates. The official unemployment rate is also known as the U-3 rate. A more realistic metric, known as the U-6 or "real" unemployment rate, includes all unemployed people, regardless of whether they're underemployed or actively seeking jobs. It currently stands at 12.1 percent. That's much higher than the official U-3 unemployment rate. But the U-6 rate has also fallen to its lowest level since 2008.
In June 2013, the official (U-3) unemployment rate was 7.5 percent and the real (U-6) unemployment rate was 14.6 percent. So, the employment market appears to be slowly improving overall.
Nationwide statistics overlook variances from state to state. Some states are experiencing unemployment rates above (or below) the national average. And in some states, there are more (or fewer) jobs than before the recession (see the right-hand box).
For example, in Texas, there are currently about 1 million more jobs than before the recession, thanks to a boom in oil and natural gas. North Dakota also reports below-average unemployment due to job growth in mining and utilities. Although the District of Columbia added many professional, health service, hospitality and education jobs, its unemployment rate (7.5 percent in May) remains above the national average. In Nevada and Arizona, analysts blame above-average unemployment rates on fewer jobs in the construction sector since the start of the recession.
Evaluate openings. Just because someone leaves, it doesn't necessarily mean you need to replace him or her immediately. Review the individual's responsibilities. It's possible that some tasks can be reassigned to another employee, who's bored with his or her current role or ready to take on more senior-level responsibilities. By reallocating the workload, you may be able to hire a less experienced or part-time replacement for less money.
Also consider whether an independent contractor might fit the position. You'll lose control over the worker — but you'll also spend less on payroll taxes and fringe benefits. If you're unfamiliar with the differences between employees and independent contractors, consult with your tax professional to ensure compliance with IRS rules. Misclassifications can lead to payroll tax deficiencies, penalties and interest. You also may be liable for employee benefits that should have been provided but were not, if you classify a worker incorrectly.
Set your budget. Decide how much you're willing to spend on salaries and benefits, as well as on recruiting new candidates. But don't just accept the status quo. If you're replacing someone who's been with the company for decades, you may be able to find a less expensive replacement with fresh ideas. But if you're an entrepreneur hoping to raise the company to the next level, you may seek a more experienced candidate who knows how to implement formal policies and procedures.
For new positions, research the market to determine how much other companies pay for similar positions and which benefits and perks are most valued by employees. If your compensation package is attractive, the most qualified candidates will show interest in the opening and apply.
Define your requirements. With social media, it's easy to post a job and receive hundreds of applications. Make sure your job description is specific enough to weed out under-qualified or over-qualified candidates. Recruiters can help your human resource department write job descriptions — including education, experience and salary requirements — and narrow your search down to, say, a dozen qualified candidates. In addition, recruiters can conduct initial telephone interviews to screen applicants. This can save significant time and minimize frustration in-house.
When you're finally ready for face-to-face interviews, take a pragmatic, modern approach. Today's applicants search the Internet for sample resumes and cover letters — and "borrow" from them. They can also read up on sample interview questions — and learn which responses most impress potential employers. Therefore, someone who interviews well or looks good on paper isn't necessarily the best fit for your company.
One idea is to start the process by checking references, including any people in your network who know the candidate but aren't listed as references on their job application. The social media site LinkedIn is a good way to see if you know someone who's connected to an applicant. Hearing someone's praise — or reticence — about an applicant is a good indicator of his or her competence. Most employers reserve reference checks until they're about to make an offer, but reversing the process might speed things up and produce better candidates.
Another idea — that works for almost every level of employment — is a hands-on test. Pick a couple of tasks or sample problems to see first-hand how the candidate will perform. For example, if you're hiring a marketing manager, ask the applicant to write a press release. Ask a sales candidate to run through a mock sales call. Or give a hotel manager applicant the opportunity to walk through your lobby and guest rooms to compile a list of recommended improvements. These practical tests can quickly reveal which applicants will be the best employees, not the best job applicants.
It's easy to get excited about your top choice and to tire of the hiring process, which might take months to complete. But it's imperative to follow through with every step, including background and credit checks. If your organization has ever been the victim of employee theft or fraud, you know these are more than formalities, especially for applicants who will have access to cash and financial accounting records.
The formal hiring process can be overwhelming, but rash decisions can be costly. If you need to fire someone later on, it can lead to unemployment claims, lost training time and productivity, additional recruiting costs, higher unemployment insurance rates in the future — and potential legal claims from the employee. So, take your time to pick people who are the right fit for your corporate culture and available positions.


Beat cream cheese and 1/2 cup butter at medium speed with an electric mixer until creamy; gradually add flour, beating well.
Divide dough in half, and shape each portion into a ball; cover and chill 1 hour.
Melt 3 tablespoons butter in a large skillet. Add mushrooms and onion; sauté until tender. Stir in sour cream and next 3 ingredients; set aside.
Roll 1 dough portion to 1/8-inch thickness on a lightly floured surface; cut with a 2 1/2-inch round cutter, and place on greased baking sheets. Repeat procedure with remaining dough.
Spoon 1 teaspoon mushroom mixture onto half of each dough circle. Moisten edges with beaten egg, and fold dough over filling. Press edges with a fork to seal; prick tops. Brush turnovers with beaten egg.
Bake at 450° for 8 to 10 minutes or until golden.


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The purpose of life is to experience your experiences. Think about it. . .
Many of us move mindlessly from experience to experience and fail to be aware of the essential nature of the experience. We rush through tasks, conversations, projects, thinking we need to get on to the next thing. Such rushing causes us to miss the richness, the beauty, and the potential that lie at the heart of every experience.
When we construct a life that is based on rushing and urgency, we can begin to feel empty inside and mistakenly believe that if we simply change what we are experiencing all will be well again. And so we rush after yet another new experience, trying to gobble up life so that we can feel satisfied, purposeful, or happy.
What is the antidote to this grasping after experience and then not really being present to the very experience we wanted so desperately?
This is the topic of the workshop we're offering this coming Saturday called Foundational Practices for Living Happily, Creatively, and Productively. I want to share with you some of the reasons we can work very, very hard at creating a life, and yet still feel dissatisfied. We'll talk about the thought patterns and obstacles that get in our way, and we'll also examine the habits of mind and the daily practices that will create the conditions for your unique flourishing in this world.
We'll meet from 1:00 to 4:30 on Saturday, September 10, to explore these ideas and to construct your own set of foundational practices for healthy living. Join us for an inspiring few hours that may help you to more fully experience your experiences!
Please contact me at lezlielaws.com to register for this event. A $20. reservation fee will hold a place for you.


The Lads in Their Hundreds , a setting by Alfred Edward Housman, was composed by George Butterworth in 1911. It is the fifth song of his song cycle, Six Songs from A Shropshire Lad.
It is quite tragic to think of how much more beautiful music Butterworth could have still written, had he not died during World War I, aged thirty. Very few works for Butterworth remain: he did not have the time to write many compositions and before leaving for war, he destroyed most of them except a few. Among them, his song cycle, Six Songs from A Shropshire Lad, written in 1911, is remarkable for its uniqueness and musical style. It sets poems all part of A Shropshire Lad written by A. E. Housman; Butterworth in 1912 will set another cycle of 5 poems from that same book in his song cycle, Bredon Hill and Other Songs.
The lads in their hundreds is the fifth song of his song cycle, Six songs from A Shropshire Lad. It follows melancholic considerations on mortality (Loveliest of trees, N.1), others on how unwise one can be at 21 year's old (When I Was One-and-Twenty, N.2), allusions to the Greek myth of Narcisse (Look Not in My Eyes, N.3) and a forceful acclaim of carefreeness (Think no more, lad, N.4).
The lads in their hundreds is remarkable for its minimalist conception. The melody intertwined with a lovely piano ritornello repeats at each verse. The spares harmonies set focus on the text. The frequent ternary bar changes from 6/8 to 9/8 fit the prosody to its closest and create a soft elegant musical flow.
Lyribox, the popular online platform for classical sheet music, accompaniment and ipa translations brings the best by G. Butterworth by offering the sheet music, accompaniments and translations to the fans gathered around the site, for the best price that could be found online. The sheet music, accompaniments and translations provided by Lyribox for are of highest quality and accuracy.
The Lads in Their Hundreds originated was composed by George Butterworth . The original lyrics are in English. The melodious note arrangement of The Lads in Their Hundreds ranges from pianissimo (very soft) mellow notes to forte (loud) notes carrying the player and the audience through an array of vivid emotions. The sweet soft beginning of the music piece gradually increases and decreases the loudness in smooth transitions. This music piece is decorated with phrases, slurs and an arpeggio. A perfect composition for piano players to try with an intermediate knowledge in music. Students can make use of this sweet music piece to improve their knowledge in music expressions, ornaments and sight-reading, while all music lovers could enjoy this brilliant extract whole-heartedly. The Lads in Their Hundreds sheet music produced by Lyribox is clear, precise and is transcribed in English by Lyribox. By purchasing The Lads in Their Hundreds sheet music from us, you can simply sing along as you play this masterpiece on your piano. Unlike many sheet music sold online with all black notations, sheet music we offer you is marked using three distinct colors. While piano notation is illustrated in black, the music expressions and the lyrics are given in red and blue respectively. Get The Lads in Their Hundreds by G. Butterworth sheet music from Lyribox today to enjoy and share this wonderful music composition with your friends and family.
The Lads in Their Hundreds piano accompaniment is offered by Lyribox is mellow and smooth. It feeds the poem an underlying pulse, adding a sense of propulsion to it. Meanwhile the vocal line flows in a dreamy leisure manner forcing the audience into a calming and languid mood. The vocal line of The Lads in Their Hundreds by Lyribox has the potential to provide slight pitch support for the voice's sometimes unusual intervals, enhancing the fullness of it. This accompaniment is not at all demanding. Using these rich The Lads in Their Hundreds accompaniments you could create one of the most brilliant performances that is pregnant with layers and layers of expressions. Purchase the The Lads in Their Hundreds piano accompaniment together with the vocal line accompaniment from Lyribox today!


Stony Run Brew House on a Friday evening. Jamie Kinsley Photo.
Two things I love – history and beer – are now available at Stony Run Brew house on East Market Street.
Opening just five months ago, the brew house has a full dinner menu and about a dozen hand-crafted and house-made beers. I recommend the York City Pretzel Bites and Vegetable Spring Rolls paired with Emily's dream Mango Milkshake IPA.
Up until June, the trio had other full-time jobs. March was a logistics manager for a construction company, March II was an operations manager for ES3's automation and robotics department and AAA Southern PA, and Elmiger first got into the brew business when he opened Bailee's Homebrew and Wine Supplies, located right next door to the brewery.
Bob Elmiger has an interesting life story. He worked construction for over 42 years and played international-level rugby until the age of 64. Caring for the adult education, he taught people who have been underprivileged such as prisoners and non-traditional students in Baltimore City trade skills. To add another layer, Elmiger is an ordained minister.
While March operates as the lead brewer, March II performs the business operations while Elmiger functions as the general manager.
Tim March, owner and lead brewer at Stony Run, chats with a patron. Jamie Kinsley Photo.
Malt for sale at Bailee's Homebrew and Wine Supplies. Malt sourced locally. Jamie Kinsley Photo.
March home brewed beer for quite a few years. His eight years of experience encapsulated all his passions: chemistry, flavors, and getting to drink his experiment.
Stony Run tries to locally source its products and its been quite successful. Most items come from the region including grain from Deer Creek Malthouse in PA and hops from Carlisle and Ritz Hop Yard in Hellam.
Stony Run Brew House's old world interior design. Jamie Kinsley Photo.
What attracts patrons to Stony Run is its location. "We're off the beaten path," March told me. "Some people may not want to go downtown. People have been waiting for something like this to open up between York and Lancaster." Due to its immense success since it opened in 2018, the brewery is looking for a satellite location in York's downtown; however, its only in the investigative stages.
Bailee, Bob Elmiger's dog, goes to work everyday at Baliee's Homebrew and Wine Supplies. Jamie Kinsley Photo.
This friendly attitude was laced throughout my entire visit to the brewery. As a researcher/blogger/journalist, sometimes people are hesitant to allow me into their lives, understandably. However, March and Elmiger were extremely warm and welcoming. And, this treatment was not unique to me. I watched as they greeted customers, chatted with people at the bar, and constantly checked on customers.
Now, I could be a little biased because, as my friends know, the animal-lover in me could have fallen in love with Bailee, the affectionate golden lab who accompanies Elmiger to work everyday.
Stony Run Brew House has 14 beers on tap in addition to two guest taps. March matches the style of the beer with a name that reflects York. "The names just come to me," said March. He went to the York County History Center's archives to get as many historical facts as he could from the 1700s, 1800s, and 1900s and applies them to his beers.
Camp Security Redcoat IPA – Named after an American Revolutionary War Prison Camp in York from 1781. March felt his intentionally, British-flavored red IPA reflects the stockade and living quarters built on a local farm where British privates and noncommissioned officers were held. Camp Security was abandoned in 1783 when the war ended.
POW Camp Security Site. YDR Photo.
Three Mile Run Kolsch Style Ale – Named after a waterway crossing in Springettsbury Township under East Market Street, Northern Way, and Eastern Boulevard. The Three Mile Run got its name due to its distance – three miles – from this point to the square in York as well as a neighboring inn called Three Mile House.
Street view of Three Mile Run. Stephen Smith Photo.
Lumber rafts on the lower Susquehanna River. Stephen Smith Photo.
Hidden Treasure Belgian Golden Strong Ale – In 1778, James Rankin, a Yorker, is said to have buried a large amount of gold and silver on his Springettsbury Township property. Remaining a British loyalist during the Revolutionary War, he was arrested and jailed in Yorktown as a "traitor." After escaping from jail, he fled to England and his buried treasure was never found.
Newspaper from 1908 covering alleged buried treasure. Stephen Smith Photo.
Even Stony Run Brew House got its name from the York landscape. Stony Run, a creek, originates up at Rocky Ridge Park. And, the Rocky Ridge Amber Ale happens to be March's favorite beer on tap – an amber ale with a malty caramel flavor.
A list of Stony Run Brew house's beers on tap. Most are named after local events, people, and places. Jamie Kinsley Photo.
To the owners of Stony Run Brew House, supporting the community is important. Elmiger told me that the brewery "sort of takes from the community when you have a business. That's why we believe in giving back." For example, with every beer kit it sells, it makes a donation to the SPCA.
Additionally, two times a year it holds fundraising events for good causes including 4AydenStrong Foundation. It plans to keep its eyes and ears open for children in the area who may have health issues so it can raise money for them, their families, and research.
Elmiger and March explained their dedication to providing quality food, exceptional beer, a fun environment, and good service.
The brewery and restaurant is open Tuesday through Thursday 11am to 9pm, Friday and Saturday 11am to 11pm, and Sunday 12pm to 7pm.
Its plans for the future include a building expansion, beer garden, event room, and the downtown location.
While visiting Stony Run Brew house, I tried the Slow & Easy Belgian Wheat Ale, Rocky Ridge Amber Ale, Talus 40 Volume Blonde Ale, and Emily's Dream Mango Milkshake IPA. Jamie Kinsley Photo.
Nice post … the Stony Run Brew House has a neat "History on Tap" concept. Tim March contacted me in 2017 about their intent to honor local history in naming their beers. He referenced some of my YorksPast posts … thanks for sharing those links.
Your years of research and quality blogging made it easy for me! Thanks, Stephen.


Now here's a statement piece if ever there was one! Made from sterling silver and with a diamond set in 18ct gold, this piece is made to turn heads! The textured and high polished finish creates an added dimension, and the piece hangs beautifully on a multistrand neckwire. Approximately 43mm long.


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Home Entertainment Starbucks And Netflix Have Teamed Up To Create A Book Club-Inspired Entertainment...
Starbucks And Netflix Have Teamed Up To Create A Book Club-Inspired Entertainment Series
October 16, 2021 1:26 am EDT
Netflix is prepared to discuss literature over more than a cup of coffee. Netflix said on Wednesday that it will start the Netflix Book Club, where members will be the first to learn about new novels, films, and series adaptations, as well as get unique access to the adaptation process for each book. Netflix and Starbucks will collaborate to bring the book club to life with a social series called But Have You Read the Book?
Uzo Aduba will become the series' first presenter, revealing regular book choices which will be translated by the streamer. Aduba will meet with the actors, producers, and authors to discuss the book adaptation process, which will be more than simply a cup of coffee at Starbucks.
The first monthly pick for the Netflix Book Club will be Nella Larsen's 1929 novel Passing, which will be accessible on the streaming service on November 10th. Tessa Thompson and Ruth Negga play two childhood friends who meet after a long separation in Rebecca Hall's film. Their reunion eventually throws their carefully manufactured worlds into question, with one "passing" for white.
During this year, Netflix stated that its recent adaptation of literary classics have increased print sales, with half of the top ten titles on the New York Times best-seller list for print and ebook fiction being linked to a Netflix series.
Following the release of Bridgerton on Netflix, author Julia Quinn's The Duke and I, a historical romance that inspired the first season of Netflix's Shonda Rhimes' blockbuster series, soared to the top of the best-seller list 21 years after its original publication. Meanwhile, with the publication of Anya Taylor-The Joy of Queen's Gambit, Walter Tevis' 1983 novel re-entered the New York Times best-seller list for the first time since its original publication.
The trailers for A.J. Finn's The Woman in the Window, starring Amy Adams; Jojo Moyes' The Last Letter From Your Lover, starring Felicity Jones and Shai Woodley; R.L. Stine's Fear Street Trilogy; Caroline Kepnes' You and Robyn Carr's Virgin River, season three; Amy Poehler's adaptation of Jennifer Mathieu's Moxie; and Stephanie Land's Maid.
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Next articleNetflix And UNESCO Have Launched A Competition To Find The Next Generation Of African Filmmakers
Hi there, My self Aniket Jain. A virtual seo expert, content writer, and a digital marketer. Works with lots of passion, hardly dedicated for what's best and what's next ✌. Hope u all like my work ✌


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The US soldier who caught Saddam broke his silence… The 20-year secret has been revealed!
hurriyet.com.tr / FOREIGN NEWS
Creation Date: January 02, 2023 12:11
The soldier who captured Saddam Hussein, who disappeared after the Second Gulf War and was wanted for a while by the US-led coalition forces, broke the silence. With the statement of the US soldier, a 20-year secret was deciphered.
Iraqi leader who was overthrown in 2003 Saddam HusseinNew information has emerged about the operation in which .
After the US-led coalition forces entered the capital, Baghdad, the soldier who caught Saddam, who almost disappeared into the background, made remarkable statements.
With the statements of Kevin Holland, the secret of 20 years has been deciphered.
According to the news in the British Daily Mail newspaper, Holland stated that they went to the region upon a tip and told the unknowns of the operation.
Stating that a Texas soldier pulled Saddam out of his hiding place, Holland said that they had detained the ousted leader, considering the possibility that he had a weapon with him.
THE FIRST WORD SADDAM HEAR: BUSH HAS A GREETING
Stating that they tied Saddam down later, Holland said that they told him "President Bush is sending his greetings".
Stating that Saddam told them that he was open to negotiation in English, Holland said that they answered him, 'It is too late now'.
Stating that the footage they took was published in the US press, Holland replied when asked how the ousted leader survived in a hole for so long; 'He didn't stay there long.'
Holland added that Saddam had two bodyguards in the region.
MEET WITH POLITICS EARLY
Born in Tikrit in 1937, Saddam's acquaintance with politics dates back to his early youth. In those days, Saddam, who lost himself to the nationalist-libertarian and anti-imperialist wind that dominated the Arab world, joined the Baath Party at a young age. He attempted a failed coup in 1956. After the end of the monarchy, he played an important role in an assassination organization created to kill Prime Minister Abdul Karim Hassam. However, this event came to light and Saddam was forced to flee the country.
When the Baath Party came to power in 1963, he returned to his country. Meanwhile, he married his cousin Sacide and had five children, two boys and three girls. However, over the years, the differences between him and the Baath Party began to deepen. When the conflicts got worse, Saddam was imprisoned.
Coup Escaped Jail
The coup in 1968 also saved Saddam from prison. Rapidly rising in the party, Saddam entered the Revolutionary Council, one of the most important structures of the Baath, thanks to his uncompromising determination and firmness. Over time, he consolidated his position and became the main source of power behind the power of President Ahmed Hasan Bekri. In 1979, he 'dropped the curtain' by seizing power in a coup d'etat. First, he launched a ruthless 'extermination' campaign against his opponents.
Since then, Saddam has based his power on repressive methods based on a powerful intelligence network. He never hesitated to kill anyone who raised his voice. Sometimes these extermination campaigns turned into 'genocide' against the whole city, as in the case of Halabja.
THE PEOPLE PAYED THE BILL OF POWER AGGRESSION
In 1980, Saddam thought he saw an opportunity that would carry him to the leadership of the Arab world and make him indispensable in the eyes of the West. The Islamic Revolution in Iran was in full swing. The relations of the Khomeini regime with the West, especially the USA, were deteriorating gradually, and Iran was perceived as a threat to the entire region with its 'revolution export' policy. Based on this determination, Saddam declared war on Iran. His calculations were based on the fact that he would easily get the support of the West in this war and that Iran, which had had turbulent days, would not be able to resist much. In the early days of the war, Iraqi troops captured an important water area, Shatt al-Arab.
But Iran turned out to be more formidable than Saddam had anticipated. And 8 years of war resulted in the death of hundreds of thousands of people. The economies of both countries were also destroyed. When the war was over, both sides were at the starting point.
SADDAM TO THE END
Knowing the power of oil very well because it was the only power he had, Saddam turned his attention to Kuwait when he couldn't get the profit he hoped for from the Iran War. On August 2, 1990, Saddam's troops invaded Kuwait.
The other Gulf countries, alarmed by the invasion of Kuwait, approached the West. Saudi Arabian territory was opened to multinational power. When efforts to persuade Saddam to withdraw were also inconclusive, the war began on January 17. Saddam described the war as 'the mother of all wars' but had to retreat in the face of international forces and a life began in Iraq under the shadow of embargoes.
During the years of embargoes, while Saddam and those around him continued their comfortable lives, the Iraqi people living on one of the world's largest oil reserves remained in hunger and misery.
SECOND GULF WAR
After the Second Gulf War, Saddam, his family and closest men disappeared. While Saddam's staff were captured one by one, Saddam's sons, Uday and Qusay, who were notorious for their cruelty, were killed. Coalition forces intensified their search in and around Tikrit, the birthplace of Saddam. The tapes that appeared from time to time, in which Saddam announced that he was alive and that he was organizing the resistance, USA It was causing demoralization in the forces led by him. However, on Saturday, December 14, 2003, the news that "Saddam was caught", which was first claimed as an allegation, became final with the publication of the images taken while he was undergoing medical examination in the afternoon.
Saddam was executed by hanging in 2008.
Turned to spirituality! The famous actor's love of God made everyone cry


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Q: Kivy add new vertex to rotation figure I have some points(vertex) that rotate in 3d, but how can I add to this rotation vertex more points?
Example took wrom Kivy's monkey rotation example! I want when to click def on_touch_down and change coordinates of vertex , but nothing happends , how to reload init or mae be another way to do this
from kivy.app import App
from kivy.clock import Clock
from kivy.core.window import Window
from kivy.uix.widget import Widget
from kivy.resources import resource_find
from kivy.graphics.transformation import Matrix
from kivy.graphics.opengl import *
from kivy.graphics import *
from objloader import ObjFile

class Renderer(Widget):
    a = [0.35, -0.41, -0.31, ----------->coord of vertex
      0.22, -0.41, 0.18,
     -0.50, -0.41, 0.18,
      0.35, 0.58, 0.68]

    def __init__(self, **kwargs):
        self.canvas = RenderContext(compute_normal_mat=True)
        self.canvas.shader.source = resource_find('simple.glsl')
        super(Renderer, self).__init__(**kwargs)
        with self.canvas:
            self.cb = Callback(self.setup_gl_context)
            PushMatrix()
            self.setup_scene()
            PopMatrix()
            self.cb = Callback(self.reset_gl_context)
        Clock.schedule_interval(self.update_glsl, 1 / 60.)

    def setup_gl_context(self, *args):
        glEnable(GL_DEPTH_TEST)

    def reset_gl_context(self, *args):
        glDisable(GL_DEPTH_TEST)

    def update_glsl(self, delta):
        asp = self.width / float(self.height)
        proj = Matrix().view_clip(-asp, asp, -1, 1, 1, 100, 1)
        self.canvas['projection_mat'] = proj
        self.canvas['diffuse_light'] = (1.0, 1.0, 0.8)
        self.canvas['ambient_light'] = (0.1, 0.1, 0.1)
        self.rot.angle += delta * 20
    def on_touch_down(self, touch):
        Renderer.a=[0.3,0.3,0.3]   ----------------> here I change coord
        print (Renderer.a)
    def setup_scene(self):
        Color(1, 1, 1, 1)
        PushMatrix()
        Translate(0, 0, -3)
        self.rot = Rotate(1, 0, 1, 0)
        self.b =  [0,1,2,3,4,5,6,7,8,9]
        UpdateNormalMatrix()
        self.mesh = Mesh(
             vertices=Renderer.a,    -------------->here use
             indices=self.b,
             fmt=[(b'v_pos', 3, 'float')],
             mode='points',
         )
        PopMatrix()

class RendererApp(App):
    def build(self):
        return Renderer()

if __name__ == "__main__":
    RendererApp().run()

A: Your code sets Renderer.a, but this doesn't do anything special. You need to instead change the vertices attribute of self.mesh, as this is the object representing the actual drawing on the canvas.

redpajama_set_name: RedPajamaStackExchange


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Economists with platform for independent analyses of the Corona crisis
Insight Team
By Vilde Blomhoff Pedersen
Leading Norwegian economists have recently launched a platform for independent analyses of the Corona crisis. Katrine V. Løken and Kjell G. Salvanes from FAIR are two of the initiators.
Katrine Vellesen Løken
Professor, Deputy Head of Department
FAIR - Center for Empirical Labor Economics
Kjell Gunnar Salvanes
Ten leading economists in Norway are starting a platform for coordination, discussion, and dissemination of various independent analyses of the Corona crisis. The initiators believe this platform will be useful for journalists, authorities, ministries (Health, Finance, Central Bank, etc), as well as the newly established expert commission on COVID-19, who all look for relevant analyses.
"We hope that gathering knowledge on the economic policy and the economic consequences of COVID-19 in one place, will make it easier for the policy makers and politicians to find information and advice from the country's leading economists", says two of the initiators Katrine V. Løken and Kjell G. Salvanes, from FAIR.
The webpage will provide op-ed pieces, economic scientific analyses, including short pieces, discussion forums, a list of policy relevant research questions and webinars on key topics.
To maximize public impact in Norway, the web page's main language is Norwegian, but contributions in English are of course equally welcome.
The initiators are:
Katrine V. Løken (NHH), Kjell Gunnar Salvanes (NHH), Torfinn Harding (NHH), Annette Alstadsæter (NMBU), Edwin Leuven (UiO), Kjell Erik Lommerud (UiB), Gisle Natvik (BI), Oddbjørn Raaum (Frischsenteret), Kjetil Storesletten (UiO) & Fredrik Wulfsberg (Oslomet)
The platform "Corona virus and the Norwegian Economy"


ProPath's Dermatopathology Fellowship is a one-year, ACGME-accredited program. The overall goal is to provide an organized, comprehensive educational experience that will expose the fellow to both the theoretical and practical applications of dermatopathology. Averaged over the year, eight months of the fellowship is devoted to dermatopathology training, with the remaining four months focused on clinical dermatology (for pathologists) or anatomic pathology (for dermatologists), to insure a comprehensive knowledge base in both disciplines.
Candidates must be board-eligible or board-certified by either the American Board of Dermatology or the American Board of Pathology by the anticipated start date of the fellowship.
The academic year begins July 1, 2020 and ends June 30, 2021.
Please download the application below, fill it out completely and send it with a copy of your curriculum vitae and a one-page letter of interest to the below address. Three recommendation letters should also be sent directly from the author to the address listed below. One recommendation must be from your current or most recent program director. For specific instructions, please refer to the program application available at the link below.


Twice the Target for the same cost as a standard bag.
4 Loops for hanging or stabe enough for free standing.
Suitable for shoots up to 60 lbs .
PLEASE NOTE: Due to size and weight of these bosses, please call to confirm delivery costs.


This Playtime walk-in shower is the simple yet elegant solution to your showering needs. Simply walk in and enjoy the refreshing cascade of water - no doors or moving parts required.
Give yourself a truly elegant showering experience with this spacious and stunning enclosure. The Playtime Walk-Through Shower 900 and Integrated Head will give your bathroom a modern and cutting-edge feel to it, while the open frameless enclosure will give your bathroom a feeling of space and minimalism. The simple design will take centre stage of any contemporary-feel bathroom.
The enclosure measures at 2000mm tall and 826mm wide, and is made using durable 8mm glass. It comes with an integrated showerhead that holds the sheet of glass in place, giving you a stunningly seamless design. The enclosure can be fitted to one of our shower trays, or simply added to some of our shower tiles to give a truly walk-in feel.


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Q: APEX 3.2: Button to download interactive report data to Excel My requirements are to create two buttons.  One button that will download all of the data in an interactive report (Select * from DB_TABLE).  The other button will download the interactive report data based on filter items (Select * from DB_TABLE where Name = :P1_Name)
I thought about utilizing javascript:gReport.controls.download(); but that didn't seem to work.
Any ideas or sites you can point me to?  Many thanks in advance.
And no, I don't know why they want two separate buttons to download the report instead of using the 'Download' button in the interactive report menu.

A: Set the Request value of the button to CSV, so your link looks like f?p=181:9:3839908094418724:CSV::::
This will only perform the default CSV download though. There is however no built-in way to get the query from an IR. What i mean by that is that an IR which has filters applied.
If you only want to download results which are filtered through page items in the where clause, and don't bother with any IR filters or such, you could set up a report on another page, and have your link for a second button target that page (with CSV in request, but do use a redirect). The browser will prompt you with a save dialog, but you'll not end up on the page.


The power of a Ti or Gio, but in a smaller package, sized just right for your budget, space or tour. Gio @5 includes an integral, articulating multi-touch display and supports up to two external displays, which can also be multitouch. The back-lit keypad's layout is identical to Gio and the Eos Programming Wing. Gio @5 can integrate into an Eos family system with ease, acting as a primary, back-up, or client console in combination with the rest of the family. All of the accessories available for Eos Ti, Gio and Ion service Gio @5 as well.
Ships With: Gio @5 consoles ship with a dust cover, two active Display Port to DVI adapters, mouse, external backlit alpha-numeric keyboard, locking power cord and one LED Littlelite.