text:
\section{Introduction}

A considerable efforts was made recently\cite{k_t,N_shk,NN_shk} to extend
so called Grimus-Stockinger theorem,\cite{Grimus} which is the main tool of
the modern theory of neutrino oscillations\cite{Beuthe,NN,ahm} and gives the
leading asymptotic behavior with $R=|\vec{\rm R}| \rightarrow \infty$ for the integral:
\begin{equation}
{\cal J}(\vec{\rm R}) = \int \frac{d^3 q}{(2\pi)^3}
\frac{e^{-i(\q\cdot\vec{\rm R})}\Phi(\q)}{(\q^2 - k^2 - i0)}\sim
\frac{e^{i{k}R}}{4\pi R}\Phi\left(-k\n\right)\left[1+ O(R^{-1/2})\right],
\label{1}
\end{equation}
where: $\vec{\rm R} = R\n$, $\n^2=1$, and the function $\Phi(\q) \in C^3$
decreases at least like $1/\q^2$ together with its first and second derivatives.
In the next section of the present letter it is reminded that the possibility of further
asymptotic expansion and the order of leading correction depend\cite{olv} on
the chosen properties\cite{k_t,N_shk} of this function.
In section 3 for appropriate space of $\Phi(\q)$ a closed formula and simple recurrent
relation for coefficients of asymptotic expansion of ${\cal J}(\vec{\rm R})$ in all
orders of $R^{-s}$ are obtained.
Such extension has particular importance for explanation \cite{N_shk,NN_shk} of
observed \cite{anom} (anti-) neutrino deficit at short distances from the sources
discussed in section 4.
But it may find much more wide implementation in quantum physics and optics, when the
Green function of Helmholtz equation is used.
In sections 4, 5 the space chosen for the functions $\Phi(\q)$ is advocated
on the ground of quantum field theory of wave packets.\cite{NN,k_t2}
Some useful relations and generalization onto $N$- dimension case are placed in
Appendix.

\section{Preliminaries}

In order to understand the physical nature of asymptotic expansion, we notice that for
infinitely differentiable $\Phi(\q)$ uniquely representable by its Taylor expansion
for any finite $|\q|<\infty$: $e^{-i(\q\cdot\vec{\rm R})}\Phi(\q)=
\Phi\left(i\vec{\nabla}_{\vec{\rm R}}\right)e^{-i(\q\cdot\vec{\rm R})}$, and then,
formally:
\begin{equation}
{\cal J}(\vec{\rm R})=
\Phi\left(i\vec{\nabla}_{\vec{\rm R}}\right)\frac{e^{ik R}}{4\pi R}=
\Phi\left(-i\vec{\nabla}_{\xv}\right)
\frac{e^{ik|\vec{\rm R}-\xv|}}{4\pi|\vec{\rm R}-\xv|}\biggr|_{\xv=0},
\label{2}
\end{equation}
where the differential vector operator in spherical basis
$\n,\vec{\eta}_\vartheta,\vec{\eta}_\varphi$ has the following properties:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\n=(\sin\vartheta\cos\varphi,\sin\vartheta\sin\varphi,\cos\vartheta),\quad
\vec{\eta}_\vartheta=\partial_\vartheta\n,\;\;\;
\sin\vartheta \vec{\eta}_\varphi=\partial_\varphi\n,
\label{4} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\vec{\nabla}_{\vec{\rm R}}=\n\partial_R+\frac 1R \vec{\partial}_{\n}, \quad
(\n\cdot\vec{\nabla}_{\vec{\rm R}})=\partial_R, \quad
\vec{\partial}_{\n}=\vec{\eta}_\vartheta\partial_\vartheta+
\frac{\vec{\eta}_\varphi}{\sin\vartheta}\partial_\varphi,
\label{3} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
(\n\cdot\vec{\partial}_{\n})=0,\quad
(\vec{\partial}_{\n}\cdot\n)=2,\quad
(\n\times\vec{\partial}_{\n})^2=\vec{\partial}^2_{\n},\quad
(\n\times\vec{\partial}_{\n})=i\vec{L}_{\n},
\label{5} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
-\vec{\partial}^2_{\n}=\vec{L}^2_{\n}=
2R(\n\cdot\vec{\nabla}_{\vec{\rm R}})+R^2\left(
(\n\cdot\vec{\nabla}_{\vec{\rm R}})^2-\vec{\nabla}^2_{\vec{\rm R}}\right),\;
\mbox{ whence,}
\label{6} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\mbox{for }\;\cos\vartheta=c:\;\;
\vec{L}^2_{\n}=
-\left[\partial_c(1-c^2)\partial_c+(1-c^2)^{-1}\partial^2_\varphi\right]
\equiv {\cal L}_{\n},
\label{7}
\end{eqnarray}
and the well known representation at point $\vec{\rm R}$ for the spherical wave coming
from point $\xv$, as a free Schr\"odinger's 3-dimensional Green function,\cite{T} is
used (see (\ref{A_4})):
\begin{equation}
\frac{e^{ik|\vec{\rm R}-\xv|} }{4\pi|\vec{\rm R}-\xv|}=\!
\int\!\frac{d^3{\rm q}}{(2\pi)^3}\frac{e^{\pm i(\q\cdot(\vec{\rm R}-\xv))}}
{(\q^2-k^2-i0)},
\label{8}
\end{equation}
which for $\xv=0$ satisfys to the equation:
\begin{equation}
\left(-\vec{\nabla}^2_{\vec{\rm R}}-k^2\right)
\frac{e^{\pm ik R}}{4\pi R}=\delta_3(\vec{\rm R}),
\label{9}
\end{equation}
where for $R>0$ $\vec{\nabla}^2_{\vec{\rm R}}$ is given also by (\ref{A_2}) with $N=3$.
Since for $R>0$ the right hand side of this equation is zero, $\delta_3(\vec{\rm R})=0$,
for spherically symmetric case one immediately obtains from (\ref{2}) formally an exact
answer, which really takes place at least for the function $\Psi(k^2)$ regular and
bounded in upper half plane ${\rm Im}\,k\geq 0$ of complex variable $k$:
\begin{equation}
\mbox{when }\;\Phi(\q) = \Psi(q^2),\;\mbox{ for }\;\q^2=q^2, \;\mbox{ then: }\;
{\cal J}(\vec{\rm R}) = \Psi(k^2) e^{ikR}/(4\pi R).
\label{10}
\end{equation}
Therefore the higher order corrections originate only by asymmetry of function
$\Phi(\q)$ relative to the directions of $\q$ in accordance with our previous
result.\cite{k_t}
Indeed, in order to obtain them, we supposed\cite{k_t} that $\Phi(\q)$ and
its first and second derivatives are represented by Fourier-transforms as:
\begin{equation}
\Phi(\q)=\!\int\! d^3{\rm x}\,e^{i(\q\cdot\xv)}\,\phi(\xv), \quad
\vec{\nabla}_q\Phi(\q)=i\!\int\! d^3{\rm x}\,e^{i(\q\cdot\xv)}\,\xv\,\phi(\xv),
\mbox{ and so on.}
\label{11}
\end{equation}
Then by interchanging the order of integration and using the Eq. (\ref{8}) we found
the representation:
\begin{equation}
{\cal J}(\vec{\rm R})=\!
\int\! d^3{\rm x}\,\frac{e^{ik|\vec{\rm R}-\xv|}}{4\pi|\vec{\rm R}-\xv|}\,\phi(\xv).
\label{12}
\end{equation}
Substituting here the expansion, which in the exponential should always contain one
additional order in compare with the number of orders in denominator, for:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
|\vec{\rm R}-\xv|=R\left[1-2\frac{(\n\cdot\xv)}{R}+\frac{\xv^2}{R^2}\right]^{1/2},\;
\;\mbox{ whih: }\; \xv^2-(\n\cdot\xv)^2 \leftrightharpoons (\n\times\xv)^2,
\label{13_0} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
|\vec{\rm R}-\xv|=R-(\n\cdot\xv)+\frac{\xv^2-(\n\cdot\xv)^2}{2R}+\ldots,
\label{13}
\end{eqnarray}
we had the corresponding expansion of integral (\ref{12}) up to $O(R^{-3})$:
\[
{\cal J}(\vec{\rm R})=\frac{e^{ik R}}{4\pi R}
\int d^3{\rm x}\,e^{-ik(\n\cdot\xv)}\phi(\xv)
\left[1+\frac{(\n\cdot\xv)}{R}+\frac{ik}{2R}\left(\xv^2-(\n\cdot\xv)^2\!\right)+\ldots
\right],
\]
that by making use of (\ref{11}) was immediately transcribed as:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
{\cal J}(\vec{\rm R})=\frac{e^{ik R}}{4\pi R}
\left[1-\frac i{R} (\n\cdot\vec{\nabla}_q)+
\frac{ik}{2R}\left((\n\cdot\vec{\nabla}_q)^2-\vec{\nabla}^2_q\right)+\ldots\right]
\Phi(\q)\biggr|_{\q=-k\n},
\label{14}  \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\mbox{with: }\;(\n\cdot\vec{\nabla}_q)\Phi(\q)\bigr|_{\q=-k\n}=
-(\n\cdot\vec{\nabla}_\ka)\Phi(-\ka)=-\partial_k\Phi(-k\n), \;\mbox{ and so on.}
\label{15}
\end{eqnarray}
The next observation is that due to (\ref{15}), with $\ka=k\n$, for $\q = -\ka$,
$\vec{\nabla}_{\q}= -\vec{\nabla}_{\ka}$, from (\ref{6}), (\ref{7}) with
$\vec{\rm R}\mapsto\ka$, one has:
$2k(\n\cdot\vec{\nabla}_{\ka})+k^2\left((\n\cdot\vec{\nabla}_{\ka})^2-
\vec{\nabla}^2_{\ka}\right)={\cal L}_{\n}$, and expression (\ref{14}) takes the following
simple form:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
{\cal J}(\vec{\rm R})=\frac{e^{ik R}}{4\pi R}
\left[1+\frac i{R} (\n\cdot\vec{\nabla}_{\ka})+\frac{i k}{2R}
\left((\n\cdot\vec{\nabla}_{\ka})^2-\vec{\nabla}^2_{\ka}\right)+\ldots\right]\Phi(-\ka)=
\label{16}\\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
=\frac{e^{ik R}}{4\pi R}\left[1+\frac i{2kR}{\cal L}_{\n}+O(R^{-2})\right]\Phi(-k\n).
\label{17}
\end{eqnarray}
In the next section by using the another surprisingly simple way, this result is
generalized onto all orders of $R^{-s}$.
It should be noted, that in spite of the one and the same final results of the both
Eqs. (\ref{14}) and (\ref{16}), (\ref{17}), the substitution like (\ref{13_0}):
$\vec{\nabla}^2_q-(\n\cdot\vec{\nabla}_q)^2\leftrightharpoons (\n\times\vec{\nabla}_q)^2$
is correct till $\q\neq -k\n$ for expressions\cite{k_t,N_shk,NN_shk} like (\ref{14}), but it
becomes incorrect for non-commutative operators $\n$ and $\vec{\nabla}_{\ka}$ in
expressions like (\ref{16}). This difference from Ref. 1-3
 below eventually gives rise an explicit operator expression for higher coefficients of
this asymptotic expansion with arbitrary $s$.
\section{Asymptotic and multipole expansions}

\begin {lemma}
For $\vec{\rm R}=R\n$, $\xv=r\vs$,
$\vs=(\sin\beta\cos\alpha, \sin\beta \sin\alpha,\cos\beta)$,
$|\xv|=r<R$, with operator ${\cal L}_{\n}=\vec{L}^2_{\n}$
(or $\n\mapsto\vs$) defined by Eqs. (\ref{5})--(\ref{7}) and positively defined operator
${\cal L}_{\n}+\frac 14=(\Lambda_{\n}+\frac 12)^2$, so that
$\Lambda_{\n}+\frac 12=\sqrt{{\cal L}_{\n}+\frac 14}$ is positively defined:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\frac{e^{ik|\vec{\rm R}-\xv|}}{4\pi|\vec{\rm R}-\xv|}=
\frac{\chi_{\Lambda_{\n}}(-ikR)}{4\pi R}e^{-ik(\n\cdot\xv)}\sim
\label{18_0} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\sim\frac{e^{ikR}}{4\pi R} \left\{1+\sum^\infty_{s=1}\frac{ \displaystyle
\prod\limits^s_{\mu=1}\left[{\cal L}_{\n}-\mu(\mu-1)\right]}
{s!(-2ikR)^s}\right\} e^{-ik(\n\cdot\xv)}.
\label{18}
\end{eqnarray}
\end {lemma}
\begin {proof}
The expression (\ref{18_0}) is a formal operator rewrite for $R>r$ of
the usual multipole expansion of the Green function\cite{T} (\ref{8}) via the
corresponding expansion of the plane wave,\cite{T} given also by formulas
(8.533), (8.534) of Ref. 12,
 with the formally introduced instead $l$, but not
really appeared operator $l\mapsto\Lambda_{\n}$:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\frac{e^{\pm ik|\vec{\rm R}-\xv|}}{4\pi|\vec{\rm R}-\xv|}=
\frac{1}{kRr}\sum^\infty_{l=0}i^{\mp l}\chi_l(\mp ikR)\,\psi_{l\,0}(kr)
\sum^l_{m=-l}Y^m_l(\n)\overset{*}{Y}\!{}^m_l(\vs),
\label{19} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
e^{\mp i(\ka\cdot\xv)}=
\frac{4\pi}{kr}\sum^\infty_{l=0}i^{\mp l}\,\psi_{l\,0}(kr)
\sum^l_{m=-l}Y^m_l(\n)\overset{*}{Y}\!{}^m_l(\vs).
\label{20}
\end{eqnarray}
Here the spherical functions $Y^m_l(\n)=\langle\n|l\,m\rangle$ and Legendre
polynomials $P_l(\xi)$ for $\xi=c=\cos\vartheta$ or $\xi=(\n\cdot\vs)$,
as eigenfunctions of self-adjoint operator (\ref{6}), (\ref{7}) on the unit sphere,
satisfy to well known orthogonality, parity and completeness
conditions\cite{T,b_e2,vil,b_l} with
delta-function $\delta_\Omega(\n\,,\vs)$ on the unit sphere:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\vec{L}^2_{\n}Y^m_l(\n)=l(l+1)Y^m_l(\n), \quad \quad
\vec{L}^2_{\n}P_l(\xi)=l(l+1)P_l(\xi),
\label{21} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\int d\Omega(\n) \overset{*}{Y}\!{}^m_l(\n)Y^{m^\prime}_j(\n)=
\delta_{lj}\delta_{mm^\prime}, \quad
\sum^\infty_{l=0}\frac{(2l+1)}{4\pi}P_l\left((\n\cdot\vs)\right)=
\delta_\Omega(\n\,,\vs),
\label{22_0} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\sum^l_{m=-l}Y^m_l(\n)\overset{*}{Y}\!{}^m_l(\vs)\equiv
\frac{(2l+1)}{4\pi} P_l\left((\n\cdot\vs)\right), \quad \; (-1)^lY^m_l(\n)=Y^m_l(-\n).
\label{22}
\end{eqnarray}
The solutions $\chi_l(\mp ikr)$, $\psi_{l\,0}(kr)$ of free radial Schr\"oedinger
equation:
\begin{equation}
\left[r^2\left(\frac 1r\,\partial^2_r\,r+k^2\right)\right]\frac{\psi_{l\,0}(kr)}{r}=
l(l+1)\frac{\psi_{l\,0}(kr)}{r},
\label{26}
\end {equation}
are defined by Macdonald $K_\lambda(z)$ and Bessel $J_\lambda(y)$
functions,\cite{olv,T,b_e2,vil,b_l} that for integer $l$ i.e. half integer
$\lambda=l+\frac 12$ are reduced to elementary functions:\cite{olv,b_e2}
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!
\chi_l(b R)\equiv \left(\frac{2bR}{\pi}\right)^{1/2}\!\!
K_{l+{\frac 12}}(bR), \quad
\chi_l(b R) \biggr|_{l=\mathrm{int}}\!\!\! =
e^{-b R}\sum^l_{s=0}\frac{(l+s)!}{s!(l-s)!(2b R)^s},
\label{24} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!
\psi_{l\,0}(kr) \equiv \left(\frac{\pi kr}{2}\right)^{1/2}\!\!\!J_{l+{\frac 12}}(kr)
\equiv \frac{1}{2i}\left[i^{-l}\chi_l(-ikr)-i^l\chi_l(ikr)\right].
\label{24_1}
\end{eqnarray}
The function $K_\lambda(z)$ (\ref{A_1}) is a whole function\cite{olv,b_e2} of
$\lambda^2$, that is the reason, why the above introduced in (\ref{18_0}) well
defined operator $\Lambda_{\n}$ does not appear explicitly.

The expression (\ref{18}) is the known asymptotic series of expression (\ref{18_0}) at
$R\to\infty$, as an infinite asymptotic version\cite{olv,b_e2} of the sum (\ref{24}) for
arbitrary non-integer $l$, $|{\rm arg}(bR)|<3\pi/2$.
It directly results  also from the substitution of finite sums (\ref{22}), (\ref{24}) by
interchanging the order of summations, converted the Eq. (\ref{19}) into the sum:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\frac{e^{ikR}}{4\pi R}\sum^\infty_{s=0}\frac {1}{s!(-2ikR)^s}\,\frac{1}{kr}
\sum^\infty_{l=s}i^{-l} \frac{(l+s)!}{(l-s)!}\,\psi_{l\,0}(kr)
(2l+1) P_l\left((\n\cdot\vs)\right),
\label{27} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\mbox{where : }\;
\frac{(l+s)!}{(l-s)!}=\prod\limits^s_{\mu=1}(l-\mu+1)(l+\mu)=
\prod\limits^s_{\mu=1}\left[l(l+1)-\mu(\mu-1)\right],
\label{28}
\end{eqnarray}
equals to zero for all missing summands with $0\leq l\leq s-1$ automatically and due to
Eqs. (\ref{21}) may be factored out from the sum over $l$ as operator product in the
right hand side of Eq. (\ref{18}). The formal addition of all such missing in fact zero
summands with $0\leq l\leq s-1$ completes then sum over $l$ into the plane wave (\ref{20})
and converts the expression (\ref{27}) into the expansion (\ref{18}).  \\
\noindent
{\sl Remark.} The operator ${\cal L}_{\n}$ in Eq. (\ref{18}) with the
same success may be replaced by operator in square brackets of the left hand side of
Eq. (\ref{26}) or by the same with interchanging $r\rightleftharpoons k$.
\end {proof}
\begin {theorem}
Let $\Phi(\q)\in S(\vec{\rm R}^3_\q)$, the space of functions infinitely
differentiable $\forall\, \q\in\vec{\rm R}^3_\q$, decrease faster than any power of
$1/|\q|$ together with all its derivatives. Then integral ${\cal J}(\vec{\rm R})$
(\ref{1}), (\ref{12}) at $R\to\infty$ admits asymptotic expansion, which has asymptotic
sense\cite{olv} even though $\phi(\xv)$ in Eq. (\ref{11}) has a finite support:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
{\cal J}(\vec{\rm R})\sim \frac{e^{ikR}}{4\pi R} \Phi(-k\n) \left\{
1+\sum^\infty_{s=1}\frac{C_s(k,\n)}{(-2ikR)^s}\right\},\; \mbox{ with:}
\label{29} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\Phi(-k\n) C_s(k,\n)=
\frac{1}{s!}\prod\limits^s_{\mu=1}\left[{\cal L}_{\n}-\mu(\mu-1)\right]\Phi(-k\n),
\;\mbox{ or:}
\label{30} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\Phi(-k\n) C_s(k,\n)=\frac{{\cal L}_{\n}-s(s-1)}{s}\Phi(-k\n)C_{s-1}(k,\n),
\quad   C_0(k,\n)=1,
\label{30_0}
\end{eqnarray}
and which is equivalent to infinite reordering of its multipole expansion:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
{\cal J}(\vec{\rm R})\sim \frac{1}{4\pi R}
\sum^\infty_{j=0}\chi_j(-ikR)\sum^j_{m=-j}B^m_j(k) Y^m_j(\n),
\label{31} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\mbox{with: }\;
\Phi(-k\n) C_s(k,\n)=
\frac{1}{s!}\sum^\infty_{j=s}\frac{(j+s)!}{(j-s)!}\sum^j_{m=-j}\!\!B^m_j(k) Y^m_j(\n),
\label{32} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\mbox{for: }\;
\Phi(-k\n)=\sum^\infty_{j=0}\sum^j_{m=-j}B^m_j(k) Y^m_j(\n), \quad
\phi(\xv)=\!\int\!\frac{d^3{\rm q}}{(2\pi)^3}\Phi(\q)\,e^{-i(\q\cdot\xv)}.
\label{32_1}
\end{eqnarray}
\end {theorem}
\begin {proof}
Since Fourier transformation (\ref{11}) maps the space $S(\vec{\rm R}^3)$
into itself\cite{blt} the function $\phi(\xv)\in S(\vec{\rm R}^3_\xv)$ also and is
represented by the inverse Fourier transform (\ref{32_1}). Let suppose at first that
$\phi(\xv)$ has a finite support at $|\xv|\leq r_0$. Then for $R>r_0$ we can directly
substitute the expressions (\ref{18}) into representation (\ref{12}) of
${\cal J}(\vec{\rm R})$ with the following result after interchange of the order of
summation, differentiation and integration of Fourier transform (\ref{11}),
justified\cite{olv} also for the asymptotic series:
\begin{equation}
{\cal J}(\vec{\rm R})=
\frac{{\chi}_{\Lambda_{\n}}(-ikR)}{4\pi R}\,\Phi(-k\n)\sim
\frac{e^{ikR}}{4\pi R}\left\{1+\sum^\infty_{s=1}\frac{ \displaystyle
\prod\limits^s_{\mu=1}\left[{\cal L}_{\n}-\mu(\mu-1)\right]}
{s!(-2ikR)^s}\right\} \Phi(-k\n).
\label{33}
\end{equation}
This is exactly asymptotic expansion (\ref{29}) with dimensionless coefficients
$C_s(k,\n)$, defined by Eq. (\ref{30}). However that is not the case for the function
$\phi(\xv)$ with infinite support.
Estimating it for $r>R$ as $|\phi(\xv)|<{\rm C}_{\rm M}/r^{\rm M}$ with arbitrary finite
${\rm M}\gg 1$, the two pieces of correction that should be added, are easy estimating
as:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\Delta_R{\cal J}=\!\!\!
\int\limits_{r>R}\!\!\! d^3{\rm x}\frac{e^{ik|\vec{\rm R}-\xv|}}{4\pi|\vec{\rm R}-\xv|}\,
\phi(\xv),\quad
\Delta_R\Phi=-\frac{{\chi}_{\Lambda_{\n}}(-ikR)}{4\pi R}\!\!\!
\int\limits_{r>R}\!\!\!d^3{\rm x}\,e^{-ik(\n\cdot\xv)}\phi(\xv),
\label{34} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
|\Delta_R{\cal J}|<\frac{{\rm C}_{\rm M}}{({\rm M}-2)R^{{\rm M}-2}}, \quad \;\;
|\Delta_R\Phi|<\frac{{\rm C}_{\rm M}}{({\rm M}-3)R^{{\rm M}-2}}\left[1+ O(R^{-1})\right].
\label{34_1}
\end{eqnarray}
Due to this corrections and arbitrariness of ${\rm M}\gg 1$, the expansion (\ref{29}),
(\ref{33}) acquires an additional asymptotic meaning\cite{olv} in comparison with
expansion (\ref{18}). The terms with $s=0,1$ reproduce the results\cite{k_t}
(\ref{10}), (\ref{17}) of previous section.

Coefficients of multipole expansion in (\ref{32_1}) may be defined by the following two
different ways, whose equivalence, for $\xv=r\vs$:
\begin{equation}
B^m_l(k)\equiv\! \int\! d\Omega(\n)\, \Phi(-k\n)\,\overset{*}{Y}\!{}^m_l(\n)=
4\pi \!\int \! d^3{\rm x}\,\frac{\psi_{l\,0}(kr)}{kr}\, i^{-l} \,
\overset{*}{Y}\!{}^m_l(\vs)\,\phi(\xv),
\label{35}
\end{equation}
follows from Eqs. (\ref{11}), (\ref{20})--(\ref{22}) and (\ref{24_0}), with inverse
Fourier transform (\ref{32_1}).

Substitution of multipole expansion of Green function (\ref{19}) into the integral
(\ref{12}) by making use of definition (\ref{35}), after the same steps and under the
same conditions for $\phi(\xv)$ as above leads to the {\bf multipole} expansion
(\ref{31}), which with the help of expansion (\ref{24}) transcribes again as asymptotic
expansion (\ref{29}) with the coefficients given by Eq. (\ref{32}). But the same
expression (\ref{32}) is obtained by direct substitution of (\ref{32_1}) into the
definition (\ref{30}) by means of (\ref{21}), (\ref{28}). This confirms the equivalence
of expansions (\ref{29}) and (\ref{31}). \\
\noindent
{\sl Corollary 1.}
From (\ref{29}) with $C_s=C_s(k,\n)$, for the squared absolute value follows:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
(4\pi R)^2\left|{\cal J}(\vec{\rm R})\right|^2\sim
\nonumber \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\sim|\Phi(-k\n)|^2\left\{1+\sum^\infty_{s=1}
\frac{i^s[C_s+(-1)^s\overset{*}{C}_s]}{(2kR)^s}+
\sum^\infty_{\zeta=2}\frac{i^\zeta}{(2kR)^\zeta}
\sum^{\zeta-1}_{s=1}(-1)^{\zeta-s}C_s\overset{*}{C}_{\zeta-s}\right\}.
\label{37}
\end{eqnarray}
For $\Phi(-\ka)=\overset{*}{\Phi}(-\ka)$: $C_s= \overset{*}{C}_s$, so $s\mapsto 2n$
at the first sum over $s$. But the second internal sum over $1\leq s\leq\zeta-1$ in
(\ref{37}) is equal to itself with the multiplier $(-1)^\zeta$ and thus
$\zeta\mapsto 2n$ also. Dividing further this sum over $1\leq s\leq 2n-1$ into two
parts: with $1\leq s\leq n$ and $n+1\leq s\leq 2n-1$, and puting for the second sum
$s=2n-s^\prime$, one finds:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
(4\pi R)^2\left|{\cal J}(\vec{\rm R})\right|^2 \sim
\Phi^2(-k\n)\left\{1+\sum^\infty_{n=1}\frac{\Upsilon_n}{(2kR)^{2n}}\right\},
\;\mbox{ with: }\; \Upsilon_n=\Upsilon_n(k,\n),
\label{38} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\Upsilon_n=(-1)^n\left[2C_{2n}+\sum^{2n-1}_{s=1}(-1)^{s}C_sC_{2n-s}\right]
=(-1)^n \sum\limits^{2n}_{s=0}(-1)^{s}C_sC_{2n-s},
\label{39}  \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\mbox{or: }\;
\Upsilon_n=
2(-1)^n\sum\limits^{n}_{s=0}(-1)^{s}C_s C_{2n-s}-(C_n)^2, \quad
\Upsilon_1=(C_1)^2-2C_2,
\label{39_0}
\end{eqnarray}
what, due to ``homogeneity'' over $C_s$ relative to index $s$ ``like a power'',
coincides with the relations\cite{N_shk} for dimensional analogs of coefficients
$C_s$ and $\Upsilon_s$. \\
\noindent
{\sl Corollary 2.} Asymtotic expansion of the integral
$\widetilde{\cal J}(\vec{\rm R})={\cal J}(-\vec{\rm R})$ (\ref{1}) is obtained by
substitutions $\n\mapsto -\n$, ${\cal L}_{-\n}={\cal L}_{\n}$ (\ref{7}) in Eqs.
(\ref{29})--(\ref{30_0}), (\ref{33}), (\ref{37})--(\ref{39_0}).\end {proof}

The explicit Eqs. (\ref{29}), (\ref{30}) and recurrent relation (\ref{30_0}) may be
compared with seminumerical calculations\cite{N_shk,NN_shk} of the same expansion,
that for arbitrary $\Phi(\q)\in S(\vec{\rm R}^3_\q)$ have to give the same coefficients
for the same orders $R^{-s}$. For the first correction this is easily seen from
Eqs. (\ref{14})--(\ref{17}) and the Eq. (8) of Ref. 2
respectively with substitution
$\vec{\nabla}^2_q-(\n\cdot\vec{\nabla}_q)^2\leftrightharpoons (\n\times\vec{\nabla}_q)^2$.
The same takes place for the next few corrections computed there.
It is clear that any common multiplicative dependence of function $\Phi(\q)$ on
$|\q|=k$ does not affect the coefficients $C_s(k,\n)$ (\ref{30}) and
$\Upsilon_n(k,\n)$ (\ref{39}).
But there are no ways to trace this property for arbitrary order $s$ with the only
numerically calculable coefficients in Eqs. (41)--(46) of Ref. 2
An important advantage of the results (\ref{29})--(\ref{30_0}) for coefficients of
asymptotic expansion is not only the explicit arbitrariness of their order $s$,
but the elucidating, that in fact they are defined by dependence of $\Phi(\mp k\n)$ on
the {\sl unit vector $\n$ only}. This essentially reduces the number of possible
degrees of freedom and simplifies the further analytical consideration. Indeed, when
$B^m_l(k)=0$ for $l>j$, the dependence (\ref{30_0}), (\ref{32}), (\ref{39_0}) of
$\Upsilon_s$ on function $\Phi(\q)$ (\ref{32_1}) leads to termination of the asymptotic
expansion: $C_s=0=\Upsilon_s$, for $s>j$, which is also clear from Eq. (\ref{31}).
If one has:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\Phi(-k\n) =\!\! \sum^j_{m=-j}\!\! B^m_j(k) Y^m_j(\n),\;\mbox{ then: }\;
C_n=\frac{(j+n)!}{n!(j-n)!},\;\;\Upsilon_1 = 2j(j+1)>0,
\label{40} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\Upsilon_j=(C_j)^2,\quad
\Upsilon_n=\frac{2(-1)^n}{(2n)!}\sum\limits^{n}_{s=0}(-1)^s{\rm C}^s_{2n}
\frac{(j+s)!(j+2n-s)!}{(j-s)!(j-2n+s)!}-(C_n)^2,
\label{40_0}
\end{eqnarray}
where $0\leq n\leq j$, ${\rm max}(0,2n-j)\leq s\leq n$ and ${\rm C}^s_{2n}$ are binomial
coefficients.
\section{Neutrino deficit and wave packets}

At diagrammatic treatment of neutrino
oscillation\cite{N_shk,NN_shk,Grimus,Beuthe,NN,ahm} the function $\Phi(\q)$ in Eq.
(\ref{1}) represents an overlap function ${\cal F}(q)={\cal F}_C(q){\cal F}_D(q)$
defined for $q^\mu=(q^0,\q)$ as a real-valued product of convolutions of the wave packets
of external particles participating in the processes of (anti-) neutrino creation $\{C\}$
and detection $\{D\}$ at respective space-time points.
According to the Feynman rules for respective tree amplitude\cite{Grimus,Beuthe,NN,ahm}
one has to deal with asymptotic expansion of ${\cal J}(\vec{\rm R})$ (\ref{1}) for the
case of antineutrino\cite{Grimus} or with expansion of $\widetilde{\cal J}(\vec{\rm R})$
for the case of neutrino,\cite{Beuthe} where the vector
$\vec{\rm R}=R\n=\Xv_{D}-\Xv_{C}$ for both cases defines the macroscopic geometrical
parameters of observation:\cite{Beuthe} distance $R$ and direction $\n$.
The event rate is obtained to be proportional to $|{\cal J}(\vec{\rm R})|^2$ or
$|\widetilde{\cal J}(\vec{\rm R})|^2$ respectively (see for detail Refs. 2,5,6,7)
that due to (\ref{38}) in the leading and next
to leading orders reads:
\begin{equation}
\propto
\frac{\Phi^2(\mp k\n)}{R^2}\left\{1+\frac{\Upsilon_1(k,\pm\n)}{(2kR)^2}+\ldots\right\}.
\label{41_00}
\end{equation}
Here the absolute value of momentum $k=\sqrt{q^2_0-m^2_j}$ is defined by neutrino mass
$m_j$ and its mean energy $q^0=\mp Q^0$ near its most probable energies
$Q^0_{C}\approx Q^0_{D}$, defined in turn\cite{Grimus,Beuthe,NN,ahm}
by fixed parameters of respective external wave packets, such as their masses $m_a$,
4-vectors of most probable energy-momentum $p_a$ and corresponding widths
$\sigma_a\ll m_a$, because in the plane-wave limit, $\sigma_a\to 0$, $\forall\,a$
the overlap function keeps exact energy-momentum conservation for both vertices, fully
destroying the neutrino-oscillations pattern,\cite{Beuthe}
${\cal F}(q)|_{\sigma_a\to 0}\propto\delta_4(q\pm Q_{C})\delta_4(q\pm Q_{D})$,
where $Q_{C}$ becomes then a full 4-momentum incoming into (anti-) neutrino
propagator, whereas $Q_{D}$ is full its outgoing 4-momentum.
For $\sigma_a>0$ the function ${\cal F}(q)$ can provide thus only approximate  or
``smeared'' conservation\cite{NN} of some mean (anti-) neutrino energy
$Q^0\approx Q^0_{C}\approx Q^0_{D}$ and some mean (anti-) neutrino 3-momentum
$\vec{\rm Q}\approx\vec{\rm Q}_{C}\approx\vec{\rm Q}_{D}$, whose
explicit form depends on the chosen models of wave packets and the used
approximations.\cite{Beuthe,NN,ahm}
So, the function $\Phi(\mp k\n)={\cal F}(q^0,\q)|_{\q=\mp k\n}$ may be more or less
sharply peaked\cite{Beuthe,NN} and varies rapidly: near the point
$k\n\simeq\vec{\rm Q}=|\vec{\rm Q}|\rr$, for the more rough case\cite{Beuthe} (a);
or near the two adjacent points $k\n\simeq\vec{\rm Q}_{C,D}=|\vec{\rm Q}_{C,D}|\rr_{C,D}$,
where $\rr^2_{C,D}=1$, for the more precisely case\cite{NN} (b). For the case (a), up
to fully unessential now multiplicative dependence on $k$ we remain with a
function $\Phi(\mp k\n)\mapsto f(\xi)$ of the one dimensionless variable
$\xi=(\rr\cdot\n)$ only.
For the case (b) without full symmetry relative to vectors $\rr_{C,D}$ we remain
with a function $\Phi(\mp k\n)\mapsto{\cal W}(\xi_{C},\xi_{D})$ of two independent
variables $\xi_{C,D}=(\rr_{C,D}\cdot\n)$, which in Gassian case\cite{Beuthe,ahm} or in
sharply peak approximation\cite{NN} always contains a function
$\Phi(\mp k\n)\propto{\cal H}(\zeta)$ of the one dimensionless variable
$\zeta=(\n \vec{\rm B}\n)>0$ with the dimensionless symmetrical positively defined
tensor $\vec{\rm B}$ built on the components of these vectors, whose traceless part
$\vec{\rm B}_0$ only is in fact necessary.
Some technique with these functions is collected in Appendix B.

When $\Upsilon_1(k,\pm\n)<0$, the last multiplier in expression (\ref{41_00}) leads to
suppression factor to the usual inverse square law for the event rate that reads as:
\begin{equation}
\propto \frac{1}{R^2}\left\{1-\frac{\varrho^2_0}{R^2}\right\},\;\,\mbox{ where: }\;\,
\varrho^2_0=-\, \frac{\overline{\Upsilon}_1(k,\pm\n)}{(2k)^2},
\label{41_01}
\end{equation}
and the averaging $\overline{\Upsilon}_1$ relative to directions of $\rr$ is implied for
the case (a), that in fact approximately puts $\rr=\n$, $\xi=1$. For the case (b) one
has to average relative to the vectors $\rr_{C,D}$ independently, what makes the final
result more model-dependent. According to Refs. 2,3
this suppression factor provided by the higher order corrections to Grimus-Stockinger
formula (\ref{1}) naturally explains the
observed\cite{anom} (anti-) neutrino deficit in reactor and others short-baseline
experiments (see Refs. 2,3,9
and the references therein).
Such explanation does not require any ``new physics'' and perhaps\cite{N_shk} can be
tested experimentally.

Eqs. (\ref{30}), (\ref{39_0}) show that negative sign of coefficient $\Upsilon_1$
originates by the most rapid variation $\Phi(\mp k\n)$ with vector $\n$. Exponential
variations are directly generated by the above limiting properties of the overlap
function and are naturally arising in popular models of wave packets.\cite{Beuthe,NN,ahm}
Since beyond oscillation problem it is not meaningless to leave overlap function be
defined by one wave packet only, any model of wave packet also has to reproduce the
above limiting properties of ${\cal F}_{C,D}(q)$.

In the spirit of Ref. 6
the following form of wave packet in momentum
representation was suggested\cite{k_t2} for $q_\mu \zeta^\mu_a\equiv(q\zeta_a)>0$, with
time-like 4-vector $\zeta^\mu_a$ of mean quantum numbers of wave packet, such that
$\zeta^0_a>0$, $\zeta_{a\mu} \zeta^\mu_a\equiv\zeta^2_a>0$:
\begin{equation}
\zeta_a(p_a,\sigma_a)=p_a g_1(m_a,\sigma_a)+s_a g_2(m_a,\sigma_a),
\quad
\phi^\sigma(\q,\p_a)={N}_\sigma\!\left(m_a,\zeta^2_a\right) e^{-(q\zeta_a)},
\label{41}
\end{equation}
where both the momentum 4-vectors $q^\mu=(E_{\rm q},\q)$, $p^\mu_a=(E_{{\rm p}_a},\p_a)$
are on mass shell: $E_{\rm q}=\sqrt{\q^2+m^2_a}>0$, and so on, $s_a$- is a spin 4-vector,
$g_{1}\gg |g_{2}|$- are some real functions\cite{k_t2} of mass $m_a$ and width
$\sigma_a$, and $\aleph(\tau) m^{-2}_a={N}_\sigma>0$, is normalization
constant with fixed asymptotic behavior\cite{k_t2} as a function of dimensionless
invariant variable $\tau=m_a\sqrt{\zeta^2_a(p_a,\sigma_a)}$, at
$\tau\to\infty$ $(\sigma_a\to 0)$, and at $\tau\to 0$ $(\sigma_a\to\infty)$:
\begin{equation}
\aleph(\tau)\underset{\tau\to\infty}{\longmapsto} 2(2\pi)^{3/2}\tau^{3/2}e^\tau, \;
\mbox{ and }\;
\aleph(\tau)\underset{\tau\to 0}{\longmapsto} \aleph(0)>0.
\label{41_0}
\end{equation}
Coordinate representation of this wave packet with fixed center $x_a$ for scalar case
satisfies Klein-Gordon equation and for $x^\mu=(t,\xv)$ is defined\cite{k_t2} by
Wightman function analytically continued\cite{blt} into the same future tube
($V^+$: $\zeta^0_a>0$, $\zeta^2_a>0$) as:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
F_{p_ax_a}(x)= e^{-i(p_ax_a)}\!\int\!\frac{d^3{\rm q}}{(2\pi)^3 2E_{\rm q}}\,
\phi^\sigma(\q,\p_a)e^{-i(q(x-x_a))}=
\nonumber \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
= (-i)e^{-i(p_ax_a)}{N}_\sigma D^-_{m_a}\!\left(x-x_a-i\zeta_a(p_a,\sigma_a)\right),
\quad (g_2\equiv 0).
\label{45}
\end{eqnarray}
This wave packet conforms with general requirements of quantum field theory\cite{blt}
and due to (\ref{41_0}) admits adequate description\cite{k_t2} of both the limits to the
states localized in momentum and coordinate spaces at $\sigma_a\to 0$ and
$\sigma_a\to\infty$ respectively:
\begin{equation}
\phi^\sigma(\q,\p_a)\underset{\sigma\to 0}{\longmapsto}
(2\pi)^3\,2E_{\rm q}\delta_3(\q-\p_a),\;\quad
\phi^\sigma(\q,\p_a)\underset{\sigma\to\infty}{\longmapsto} { N}_\infty.
\label{45_1}
\end{equation}
Furthermore, because $2(qp_a)=2m^2_a-(q-p_a)^2$, the nonrelativistic limit of
(\ref{41}), (\ref{45}) for $g_1=\sigma^{-2}_a$ exactly reproduces\cite{k_t2}
the usual Gaussian profiles in coordinate and momentum representations
independently of normalization arbitrariness (\ref{41_0}).

To get most simple example preserving oscillations we take an overlap function with
one wave packet (\ref{41}), (\ref{45}) in the one vertex representing the another vertex
by infinitely heavy nucleus in the spirit of Kobzarev model\cite{kbz}. For antineutrino
this may be achieved for example in the Grimus-Stockinger model\cite{Grimus} with
nucleons in creation vertex $\{C\}$ fixed in infinitely heavy nucleus and the bound state
of initial electron in detection vertex $\{D\}$ replaced by above wave packet state for
free electron.
For the case of neutrino one can take the decay $\pi^+\to\mu^++\nu_\mu$ with wave packet
(\ref{41}) for pion state and plane wave for muon in $\{C\}$- vertex taking $\{D\}$-
vertex as in the Kobzarev model.\cite{kbz,Beuthe} The overlap functions\cite{Beuthe}
${\cal F}_{C,D}(-p)$ for the first case (compare with Eq. (11) from
Ref. 4
for $\widetilde{J}_\lambda=$ const), with
$E_{\rm w}=\sqrt{\vec{\rm w}^2+m^2_e}$, are reduced to:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
{\cal F}_{C}(-p)={\rm const}\;\delta(p^0-Q^0_C), \quad Q^0_C=\Delta M-U^0,
\quad Q_D=K-W,
\label{46}  \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
{\cal F}_{D}(-p)=(2\pi)^4\delta\left(p^0+E_{\rm w}-K^0\right)
\frac{\phi^\sigma_e(\vec{\rm w},\vec{\rm W})}{2E_{\rm w}}, \quad
\vec{\rm w}=\vec{\rm K}-\p, \quad W^0=E_{\rm W},
\label{47}
\end{eqnarray}
where: $\Delta M$ is neutron-proton mass difference, $U^\mu=(U^0,\vec{\rm U})$ is
4-momentum of created electron in vertex $\{C\}$; $W^\mu=(W^0,\vec{\rm W})$ is the
most probable 4-mo\-men\-tum of incoming initial electron with
$\zeta_e(m_e,\sigma_e)=g_1W$ in Eq. (\ref{41}), $w^\mu=(E_{\rm w},\vec{\rm w})$, and
$K^\mu=(K^0,\vec{\rm K})$ is the total outgoing 4-momentum carried away by final electron
with final antineutrino in vertex $\{D\}$. Note that 3-vectors $\vec{\rm U}$ and
$\vec{\rm Q}_C$ become indefinite for this case.
The product of these functions belongs to $S(\vec{\rm R}^3_\p)$ only for replaced
argument of second delta-function with $E_{\rm w}\mapsto E_{\rm W}$, which appeares
naturally in sharply peak approximation for
$|\p-\vec{\rm Q}_D|=|\vec{\rm W}-\vec{\rm w}|\ll m_e$, leading to:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
{\cal F}(-p)\approx {\rm const}\;\delta(p^0-Q^0_C)\,\delta(p^0-Q^0_D)
\frac{N_{\sigma_e}}{2W^0}e^{-g_1(wW)},\;\mbox{ where: }\; g_1=\frac{1}{\sigma^2_e},
\label{48} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
(wW)=E_{\vec{\rm w}}W^0-(\vec{\rm w}\!\cdot\!\vec{\rm W})\approx
m^2_e+\frac 12 \left((\p-\vec{\rm Q}_D)^j(\delta^{jl}-{\rm V}^j{\rm V}^l)
(\p-\vec{\rm Q}_D)^l\right)=
\label{49_0} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
=m^2_e+\frac 12 \left\{\vec{\rm Q}^2_D-(\vec{\rm Q}_D\!\cdot\!\vec{\rm V})^2+\p^2-
2\left(\p\!\cdot\!\left[\vec{\rm Q}_D-\vec{\rm V}(\vec{\rm Q}_D\!\cdot\!\vec{\rm V})
\right]\right)-\vec{\rm V}^2(\p\vec{\rm B}\p) \right\},
\label{49} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\mbox{for: }\;
\vec{\rm V}=\frac{\vec{\rm W}}{W^0}=|\vec{\rm V}|\vec{\omega},\quad \vec{\omega}^2=1,
\quad {\rm B}^{jl}=\omega^j\omega^l, \quad
{\rm B}^{jl}_0=\omega^j\omega^l-\frac{\delta^{il}}3,
\label{49_1}
\end{eqnarray}
with vector $\vec{\rm V}$ as a most probable velocity of initial electron in detector.
To define the coefficients $\Upsilon_s(k,\n)$ (\ref{39_0}) of asymptotic expansion
(\ref{38}) by making use of Eqs. (\ref{29})--(\ref{30_0}),  only the last two summands in
the line (\ref{49}) are now necessary, giving exactly the discussed here structure of the
function $\Phi(\q)$ (where due to the absence of vector $\vec{\rm Q}_C$ its role
partially plays vector $\vec{\rm V}$).
For non-relativistic initial electron $|\vec{\rm V}|\ll 1$, and neglecting this velocity,
for $\vec{\rm Q}_D=|\vec{\rm Q}_D|\rr$ and real $\lambda=g_1 k|\vec{\rm Q}_D|>0$,
now it is enough to take (see Appendix B for details and another cases):
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\Phi(\q)=\Phi(-\p)=\Phi(-k\n)\underset{|\vec{\rm V}|\ll 1}{\longmapsto}
e^{\lambda\xi},\quad \xi=(\rr\cdot\n)=\cos\Theta,
\;\mbox{ whence:}
\label{42} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
C_1(k,\n) = 2\lambda \xi-\lambda^2(1-\xi^2), \quad
\Upsilon_1(k,\n)= -4\lambda^2\left[1-(2+\lambda\xi)(1-\xi^2)\right].
\label{43}
\end{eqnarray}
So $\Upsilon_1<0$, if $\lambda\xi<\left[(1-\xi^2)^{-1}-2\right]$, that
$\forall\,\lambda>0$ includes interval $3\pi/4\leq\Theta\leq \pi$,
\begin{equation}
\mbox{which for }\;\lambda\to\infty,\;\mbox{ dilates to: }\;
0\leq \Theta \leq 1/\sqrt{\lambda},\quad
\pi/2+1/\lambda  \leq \Theta \leq \pi.
\label{44_1}
\end{equation}
With the same function (\ref{42}) $\forall\,\lambda<0$ this condition implies interval
$0\leq\Theta\leq \pi/4$,
\begin{equation}
\mbox{dilating to: }\;
0\leq\Theta\leq \pi/2-1/|\lambda|,\quad \pi-1/\sqrt{|\lambda|}\leq\Theta\leq \pi, \;
\mbox{ for }\;|\lambda|\to\infty,
\label{44_0}
\end{equation}
that means the formal interchange in (\ref{44_1}) of forward and backward hemispheres
$\Theta\mapsto\pi-\Theta$. Similarly for the above mentioned case with neutrino, one can
obtain for $\Phi(\q)=\Phi(k\n)$ the same result (\ref{42}) with $\lambda>0$.
Therefore for the both cases $\Upsilon_1(k,\n)<0$ near the forward direction (\ref{44_1})
with the narrow wave packet $\sigma_{e,\pi}\to 0$ and/or with the high energy $p^0=Q^0_D$
of (anti-) neutrino. Since due to Eq. (\ref{49_0}), only this region in fact contributes
to the averaging relative to $\rr$, from (\ref{43}) for the parameter $\varrho_0$
(\ref{41_01}), with $k=\sqrt{(Q^0_D)^2-m^2_j}\approx|\vec{\rm Q}_D|$, and $Q^0_D=Q^0_C$
(\ref{46}), one finds:
\begin{eqnarray}
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\overline{\Upsilon}_1 \simeq -4\lambda^2, \quad
\varrho^2_0\simeq\frac{4\lambda^2}{(2k)^2}=\frac{\vec{\rm Q}^2_D}{\sigma^4_e}, \quad
\varrho_0\approx \frac{k}{\sigma^2_e},
\label{50} \\
&&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\mbox{where: }\;
m_j\leq \sigma_e \ll m_e,\quad k\leq Q^0_C\leq \Delta M-m_e.
\label{51}
\end{eqnarray}
For $k\simeq\Delta M-m_e\approx 0.783\,\mbox{MeV}$ anf
$\sigma_e\sim m_j\simeq 0,2\,\mbox{eV}$ it follows $\varrho_0\simeq 3,86\,\mbox{m}$.
Note that for real $\beta$- decay of heavy nucleus the value of antineutrino energy
$Q^0_C\simeq k$ may be in many times greater.
This qualitative analytical estimation illustrates and elucidates the numerical
calculations\cite{N_shk,NN_shk} with more detailed models, that give the best
value for this parameter $\varrho_0\simeq 3,5\,\mbox{m}$.
The obtained analytical results (\ref{29})--(\ref{30_0}), (\ref{38}), (\ref{39_0}),
together with relations from Appendix B can essentially simplify such calculations
making them probably more transparent and model-independent.

\section{Conclusions}
As shown in Appendix A, following the same way as in Lemma 1 the operator
expansion (\ref{18}) for the Green function of Helmholtz equation may be generalized for
arbitrary $N$- dimensional Euclidean space.
The same concerns the assertion of Theorem 1 about asymptotic expansion for
$N$- dimensional analog of integral (\ref{1}).

The following comments are in order.
Since the main principles of quantum field theory\cite{blt} require the wave packets
belong to $S(\vec{\rm R}^3_{\q})$, the same holds true for their convolution and thus
for $\Phi(\q)$, as was supposed in Theorem 1. This property ensures the possibility
of further asymptotic expansion (\ref{29}), (\ref{33}) of integral (\ref{1}) and defines
the order of leading corrections to Grimus-Stockinger asymptotic,\cite{Grimus} which is
crucial for explanation\cite{N_shk,NN_shk} (\ref{41_01}) of (anti-) neutrino deficit.

For electron in stationary bound state\cite{Grimus} with energy $W^0=E_b$ instead of the
wave packet state the overlap function (\ref{47}), with the also indefinite now vectors
$\vec{\rm W}$, $\vec{\rm K}$ and $\vec{\rm Q}_D$, becomes:
${\cal F}_{D}(-p)=\delta\left(p^0+E_{b}-K^0\right)\widetilde{\psi}_b(\vec{\rm w})
\in C^\infty(\vec{\rm R}^3_{\p})$, where
$|\widetilde{\psi}_b(\vec{\rm K}-\p)|\leq O(|\p|^{-4})$ for $|\p|\to\infty$.
It may be argued following to the Lemma 1 of Ref. 2 and the Theorem of Ref. 4, that this
fourth power fall-off should be enough for the well definiteness of the asymptotic
expansion (\ref{29}) up to the term with $s=2$, and therefore the expansion (\ref{38}),
(\ref{39_0}) up to the term with $n=1$, what preserves Eqs. (\ref{41_00}), (\ref{41_01}).
On the other hand the infinite coherence length of neutrino oscillations for the
stationary case\cite {Grimus} is also the price for the absence of external wave
packets like (\ref{41}), (\ref{45}); only their presence in both the creation $\{C\}$
and detection $\{D\}$ vertices provides a finite value of coherence
length.\cite{Beuthe,NN} So we conclude, that again as in the Ref. 4:
``Therefore the physics under consideration seems to comply naturally with the
mathematical requirements''.

Since the wave packet (\ref{41}) in momentum representation
$\phi^\sigma(\q,\p_a)\in S(\vec{\rm R}^3_{\q})$, the same takes place\cite{blt} for its
coordinate representation (\ref{45}).
So, for fixed $t$, $F_{p_ax_a}(t,\xv)\in S(\vec{\rm R}^3_{\xv})$ obviously with infinite
support and the same holds true for their convolution $\phi(\xv)$ in Eqs. (\ref{32_1})
and (\ref{12}). Therefore the expansion (\ref{33}) acquires also the additional
asymptotic nature, with nonzero corrections (\ref{34}), governed by detailed form of
asymptotic behavior of the function $\phi(\xv)$ itself. These corrections inevitable
introduce an additional dimensional parameter and may also affect the behavior of
${\cal J}(\vec{\rm R})$ at microscopically big but macroscopically small
distances $R$, discussed here and in Ref. 2,3,9.

redpajama_set_name: RedPajamaArXiv


This is a blog about how freedom came for the slaves during the American Civil War, but now and then I indulge in other unrelated subject matter. Such as my post last October poking fun at nervousness in certain quarters of the North in Fall 1861 about British troops in Canada. Today, I intend to indulge in another off-topic post.
February 14, 2012 is the 100th anniversary of Arizona statehood. Arizona became a state on February 14, 1912. I only have been a resident of Arizona for about a year-and-a-half, but my ties to the state go back a long ways. My great-aunt Rose (my father's aunt) came to Arizona before World War II, when as a nurse she contracted tuberculosis. With the help of the state's dry climate she lived long enough to see a cure for the disease and to outlive the rest of her ten brothers and sisters. She was joined eventually by her sister, Mary, and when I was a child, our family would sometimes cross the desert from California to visit them. My maternal grandparents were Canadian snowbirds that found a roost during the winter in a trailer park in Mesa, not far from my great aunts' cluttered Phoenix apartment. So a visit to see my great aunts was usually timed so we could see my mother's parents as well. After a couple of false starts trying to retire in California, my parents finally settled in Mesa, where my mother still lives (my father died a few years ago). Consequently, I have been in and out of the state for years, enough that it was a long a second home to me before finally becoming my first.
But enough of my family's Arizona history. What a lot of people, even state residents are unaware of, is Arizona's Confederate roots. At the start of the Civil War, what is today the states of Arizona and New Mexico, were part of New Mexico Territory. Although still quite isolated and rugged, by 1860, the territory was splitting apart on sectional lines not unlike the rest of the United States. The northern part of New Mexico Territory tended to be loyal to the Union and looked for leadership to the official territorial capital at Santa Fe. However, residents of the southern part of the territory abutting Mexico, some of them from the American South and alienated by federal neglect, hoped to join the Confederacy. On March 16, 1861, they met at Mesilla and voted to secede from the Union, establishing themselves as Arizona Territory. While the nascent Confederate government greeted the news favorably, as it had ambitions to achieve a continental nation from the Atlantic to Pacific, Jefferson Davis and the Confederate Congress had more pressing matters, but finally on February 14, 1862, admitted Arizona as a Confederate territory. Confederate Arizona's existence was short-lived. Union forces defeated an attempt by Confederate troops from Texas to seize New Mexico Territory in Spring 1862 and by that summer the government of Confederate Arizona was forced to flee to Texas where it effectively ceased to exist.
When the U.S. Congress established its own version of Arizona Territory in 1866, it deliberately drew the boundary between it and the rump New Mexico Territory running north to south, so to split the old New Mexico Territory east and west instead north and south such as has been the case during the Civil War (see the map above). However, the Confederates got the last laugh, so to speak. When it selected the date for Arizona statehood, Congress, now with plenty of Southerners in it (which hadn't been the case in 1866), chose the date of February 14, 1912 for Arizona to become the 48th state, fifty years to the date that Arizona had been accepted into the Confederacy. Another example of the South losing the war, but winning the peace.
This puts me in mind of a gentleman from Arizona who once engaged me in some heated discussions in the Civil War Forum. He was one of those angry, League of the South types who claimed both Confederate and Apache heritage. Somehow he had it in his head that the Apache were brothers in arms with Confederates in the Southwest, trying to throw off the yolk of Yankee tyranny. It was a fantasy — legions of Apache warriors joining the ranks of Johnny Rebs — of which he could not be disabused.
He has hard pressed to explain passages I was quoting him from the O.R. — missives from John Baylor, Confederate Governor of Arizona, to Richmond, recommending the best way to deal with the Apache menace. Baylor figured the adult Apache could not be rehabilitated. His idea was to kill all adult men, women and older children among the Apache, and move the small children into American schools where, in a generation, they could be assimilated.
That's another example of an enduring mythology about the Confederacy. Even though the highest Confederate authority in the region was promoting the near-extermination of Apaches, this modern day neo-Confederate is posting treatises on the internet about the mutual respect shared by all the Yankee-hating heroes in the Territories.
Hi David. Thanks for the story. It brings to mind the fact a big reason the small Anglo population in southern New Mexico Territory became alienated from the Union by 1861 was they did not think the U.S. government was doing enough to eliminate the Apache threat. They hoped the Confederacy would do more.
The Arizona story is not unrelated to the larger purposes of your blog. The "secession" of Arizona was not declared by the residents of the southern part of Az, most of whom were Latino, but by a few hundred transplanted Southerners. The declaration, along with the Confederate invasion of the territory, were part of a longer tradition of white Southerners using military force to extend slavery into Latino lands from Nicaragua to New Mexico.
What Professor Shaffer and David talked about is treated in today's NYT Disunion post. Fascinating reading. David shows what is one of the highlights of the Disunion article: racial extermination. Of course, this wasn't something that one section of white America had a monopoly on.
Pretty good history lesson on the southwest. But it doesn't sound like the confederates got the last laugh? Your explanation indicates that the territroy was divided north to south to make states that would dilute the southern influence. If true, then the rest of the country had the last laugh on the southerners.
In return, southerners joined the union on the day they offically decided to fight for slavery. That's not funny – that's sad. It says a lot more about the southern culture of the time of Arizona's statehood.
I don't read the date of Arizona's admission to the Union as the Confederates getting the last laugh. In fact, just the opposite. Congress was figuratively rubbing Confederate sympathizers' faces in the fact that the Confederacy's attempted theft of that territory had been utterly overthrown by the Union. After all, February 14, 1912 is associated totally with Arizona becoming a state of the U.S. Nobody now thinks of the Confederacy in connection with that date.


text:
Q: Revert back to Winter 16 Force.com IDE in eclipse? My target orgs are all still on Winter 16, and I just learned, unfortunately, that I cannot fetch from them after running the updates for Force.Com IDE which will now only access Spring 16 instances.  Does anyone know if I can uninstall and alter the url to grab an earlier version of Force.com IDE?
Thank you for any assistance!

A: You probably want to take a look at .settings/com.salesforce.ide.core.prefs
Here's mine:
eclipse.preferences.version=1
endpointApiVersion=35.0
endpointEnvironment=Sandbox
endpointServer=test.salesforce.com
httpsProtocol=true
ideVersion=35.0
keependpoint=false
metadataFormatVersion=35.0
namespacePrefix=
packageName=unpackaged
preferToolingDeployment=true
projectIdentifier=IDE-4569797042840404286
readTimeout=400
username=me@my.org

I'd guess that you have endpointApiVersion, ideVersion, metadataFormatVersion all set to 36 (Assuming Force.com IDE is in lock-step with api versions)
Try bumping them back to 35, and see what happens.

redpajama_set_name: RedPajamaStackExchange


Sorry to keep you waiting, I'm still trying to get my thoughts together on this subject, so I can write them down. Also, I'm in the process of changing e-mail addresses; the old one isn't working very well since my computer had a major system crash on April 20.
No, you did not miss anything, because I haven't written about Reid's statements yet, but I probably will soon, if he isn't forced to resign because of them. It sounds like you read the my essay called "The Real Enemy". There I declared that our own liberals are a greater enemy than the terrorists. Because I haven't updated that essay in more than a year, Reid isn't in it; nor is our new self-proclaimed "secretary of state," Speaker of the House Nancy Pelosi. If I rewrite it in the near future, I'll probably add those two jokers.
As far as I'm concerned, Senator Reid crossed the line when he declared that "the war is lost." There he stopped being part of the "loyal opposition" and became a traitor, undermining the war effort by denying or delaying funding for the troops. I heard treason defined as "aiding and giving comfort to the enemy," and Reid's words are certainly doing that; look how quickly they appeared in Al Jazeera, and other agents of the enemy's press.
Unquote: Yesterday I was informed that it was the 63rd anniversary of the battle of Slapton Sands, in which the German navy attacked Americans on a beach of England, while they were training for the D-Day invasion. 749 Americans were killed, and the ambush was so embarrassing that nobody talked much about it for more than a generation afterwards (I didn't hear about it until the mid-1990s!). Imagine if we had today's leaders in charge in 1944; after Slapton Sands, would they have allowed D-Day to take place?
Thank you for reading, and come back to visit my website again. Next I plan to upload a timeline, to go with Chapters 1-4 of my Middle Eastern history series.
By the way, I didn't send this picture in my letter, but it's appropriate here. It shows how our current commander in chief isn't the first president with the initials "G.W.", who had somebody of extremely questionable character on his side; George Washington had Benedict Arnold. Click on the picture to see the whole thing.
Not much has happened in the past couple of days, hence the title. On my way home from work yesterday I discovered a road called Lindy Lane, that runs parallel to Liberty Rd. What I found amusing was that near the west end, where I got on, was a sign that said, "Lindy Lane does not stop." Lindy would like that, but unfortunately it's not true; a mile east of there I encountered a stop sign.
Speaking of Lindy, we got two phone calls from her and Adam this week. It looks like Lindy will be able to go back to school as early as June, and they wanted to know if we'll help with the tuition. We probably will, but I told them to check on available scholarships first.
It rained fairly often from Monday to Friday. It also got down into the 40s last night, so it sure doesn't feel like April to Leive & I. I hear they're going through a drought in Florida, and wildfires around Waycross, GA have burned at least 95 square miles. I don't think Lindy and Adam are in danger from that, but it's too bad we can't send our southern neighbors some of the rain we're getting.
Mowed the lawn for the third time today. This time I had to empty the bag three times; the grass seems to be growing faster than ever. No wonder central Kentucky is such good horse (and cattle) country!
Today marks one year since the death of Leive's second brother, Elmo Bendoy. I remember him well; when I visited the Philippines and reached Mindanao, after I claimed my bags I sat down on the floor in the corner of the airport terminal, wondering what to do next, and then Elmo found me. The amazing part is that nobody told him what day I would be in the airport!
Believe it or not, I didn't hear about AIG's "Creation Museum" until last week, though creationism is one of my two hot-button issues (the Middle East is the other one). I guess I've been too busy getting established in my new home. At first, I thought it was something like the traveling exhibit and seminar about creationism put on by a pastor from Christmas, FL. Now I'm hearing it is a $30 million, 60,000-square-foot structure, and it will be opening on May 28. What's more, it is next to Cincinnati's airport, so it will be just 70 miles north of here, on our side of the Kentucky-Ohio border. You can read more about it here, on the Answers In Genesis website.
Apparently Ken Ham picked Cincinnati for the museum site because that puts two-thirds of the United States population within 650 miles of it. Let's see, that will include most (if not all) of the Midwest, the east coast from New York to Georgia, Tennessee and parts of Alabama, Mississippi and Arkansas. They're also offering memberships, which should be a good deal after we get the kids from the Philippines.
Of course I'm tickled pink, because some of the material I used when I wrote "The Genesis Chronicles," ten years ago, came from videotapes Ken Ham made. Also, there isn't much in the way of places around Lexington to take the kids to; the main entertainment we have is horses and college basketball! If we're going to spend the day at anything resembling the theme parks of Orlando, we'll probably have to do it at a Cincinnati attraction. Leive's not thrilled, though; she just told me that she thinks dinosaurs are myths, no more real than the monsters in a science fiction or fantasy movie. I see I've got my work cut out for me in this matter. Perhaps when I visit the museum I'll have to drop her off at the Newport Aquarium, like the time in 1996 when I left her to spend the day at the aquarium in Tampa, while my brother and I went on to an ancient Egyptian exhibition in St. Petersburg.
Today I also heard that while the museum is not open yet, it already had a famous visitor: comedian Bill Maher. I don't have a good feeling about this, because from what I've seen, Maher is not a nice fellow. His former program "Politically Incorrect" was anything but that, he has said that he thinks Christianity is a mental disorder, and while he may not be as obnoxious as Rosie O'Donnell, he seems to be in competition with her for the title of TV's most annoying liberal. No word yet on the real reason why he came, or if the exhibits changed his mind on anything. Wouldn't it be wonderful if they did, turning him into a modern-day version of Sir William Ramsay?
Unquote: I was of the impression that both tomatoes and cucumbers were female, because both have seeds in them. You heard it here first — if our enemies win in the War on Terror, you won't be able to put summer veggies together, and I don't even want to think about what might happen to the melons! It also looks like the delightful cartoon "Veggie Tales" has a problem. The two most prominent Veggie Tales characters are a cucumber named Larry and a tomato named Bob; which one of them is really a "she"?
I don't think I slept at all last night, and I'm not sure why. It could be the warm evening indoors; as I write this, temperatures are 80 upstairs, 78 on the main floor, and 62 in the basement. Or maybe it's the tea I had around 8 PM. I'll have to ask Leive what she put in it.
Last Thursday, I wrote that I didn't like Microsoft as much as I used to, after a Microsoft download forced me to reinstall Windows XP. I forgot to mention it wasn't the first time this happened to me. In 2002 the updates Microsoft insisted I needed caused Windows ME to become unstable on my computer, to the point that I ended up buying a new computer; eventually I also reformatted the hard drive of the old one.
Anyway, since I got my current PC back in order, three more Microsoft utilities have failed. Sync Toy, which keeps files organized in two sets of matching folders, has been giving me error messages from the start. The launch bar which allowed me to run my favorite programs from a line of icons (on the right-hand side of the screen) stopped working right, when I removed one of the icons. Finally, on Saturday I noticed an extreme reluctance to shut down; when I clicked on the button to turn the computer off, it was more likely to lock up.
I ended up making three visits to NoNags.com, my favorite download site, and installed three non-Microsoft programs to do the jobs of the utilities that failed. TreeComp takes care of synchronizing folders, though I have a feeling I won't be doing that as often now. ProgramLaunch replaces the Microsoft launch bar I used to have, and Super Fast Shutdown fixes the problem I had turning the PC off.
Oh well, it's not like I have a vested interest in promoting Bill Gates and his company anymore. I guess I did when I was a teacher, because I taught my students how to use PowerPoint and FrontPage (I still like those programs, by the way). But I can tell they've fallen on hard times since the high-tech industry busted in 2000; whereas in the past they introduced major software upgrades every other year, when they launched Windows XP and Internet Explorer 6 (both in 2001), they were left "as is" until this year. Also, they haven't invited me to any more seminars since October 2001.
Happy Anniversary, Tina & Paul!
I didn't have any big news to report for the last two days. It's warm enough now to go outside without a jacket, and this time it looks like it will stay that way.
I told you about the bird feeder in previous entries; how I put it out for cardinals and ended up getting greedy blackbirds most of the time. From Wal-Mart I bought a sock full of thistle seeds, which is supposed to attract goldfinches. I hung it next to the feeder, and it attracted them all right. Leive thinks the goldfinches are a pretty new attraction in the backyard.
The computer is nearly back to normal, after Thursday's crash. I guess you could say it's working even better, now that I have the older versions of Internet Explorer & Windows Media Player back. The main issue right now is e-mail, so bear with me if I take some time in responding to letters. Since arriving in Kentucky, I have surfed the Web with Insight Broadband, a cable company based in Ohio, but kept Earthlink, the dial-up service I had in Florida. E-mail has been the easiest way to remain in touch with family and friends, and this arrangement meant I didn't have to change my e-mail address. It made sense while I was moving from one place to another, but now the Florida place is finally sold, and I have a feeling I'll be in this house for the rest of my life, unless Lindy has her way about us retiring near her home. Therefore I figured now was the time to see if I can handle e-mail through Insight as easily as I did with Earthlink before Thursday; if so, I'll just use Insight from now on. At this point, I can receive e-mails through my Insight address, but I don't know if I can send letters; no test messages have come back yet.
Last night after church I started putting the third bookcase together. So far, so good, but I hope the missing parts will arrive in the mail soon.


text:
'The Tree of Life' review: Terrence Malick's latest is often profound, sometimes puzzling
Published: May. 27, 2011, 12:01 p.m.
Stephen Whitty | For Inside Jersey
MERIE WALLACEBrad Pitt and Laramie Eppler in "The Tree of Life."
At its literally transcendent best, "The Tree of Life" is a moving illustration of everything the movies can do — turn thoughts into pictures, make dreams into realities, transmute life into art.
At its occasionally perplexing worst, "Tree of Life" is a clear declaration of everything director Terrence Malick chooses not to do — tell clearly linear stories, write conventional scenes, explain.
The two, taken together, do not make for an easy movie. But they do ensure a complex and often beautiful experience.
At its simplest — and there is very little that is simple here — "The Tree of Life" is about a wealthy architect looking back at his childhood in 1950s Texas, where he grew up with a bitter, abusive father and an unconditionally loving mother.
The emotional whiplash of being caught between the two extremes has never left him — and the great tragedy of "The Tree of Life" is that we're there, peeking powerlessly into his past, as we watch the child's joy and innocence slowly slip away.
That is the part of "The Tree of Life" that can be easily explained, and it's superbly acted by Sean Penn, in a small part as the grown Jack, and Brad Pitt in an unusual one as the stern parent. (How stern? He reprimands his sons for calling him "Dad" instead of the more proper "Father.")
It also has two marvelous performances by newer talents — as the mother, Jessica Chastain (who recalls the vulnerable, translucent Sissy Spacek of Malick's "Badlands") and Hunter McCracken as young Jack, his face a summer storm cloud of emotion.
The Tree of Life (PG-13) Fox Searchlight (138 min.) Directed by Terrence Malick. With Brad Pitt, Jessica Chastain, Sean Penn. Now playing in New York; opens in Montclair on June 3. Rating note: The film contains violence.
Stephen Whitty's Review: THREE AND A HALF STARS
But then there's the other, deeper part of "The Tree of Life."
Malick always has had a mystical bent. He once taught philosophy at MIT, and his last two films, "The New World" and "The Thin Red Line," mused at length on man's relationship to the natural world.
"The Tree of Life" goes even deeper, starting with a quotation from the Book of Job and an observation — or maybe just an overheard prayer — from Jack's mother that "The nuns taught us there were two ways through life — the way of nature and the way of grace. You have to choose which one you'll follow."
Trying to choose grace is a preoccupation for this Catholic family. All of them have constant conversations about it with God, or with their better selves, or with the past itself.
Malick, though, makes their yearnings real. "What are we to You," the mother muses, and the film goes into the biggest flashback in cinema history, right to the beginning of the universe — then watches as Earth is formed, life begins, dinosaurs roam.
"How did I lose you?" the grown Jack thinks of his mother, or perhaps his younger self. "Wandered. Forgot you." But then Malick shows Jack literally lost in some biblical desert — and, at the end, meeting his own mother, and youth, on some heavenly beach.
Malick cultists will embrace all of this. Doubters are more likely to feel that — as in his last two films — the director is more interested in abstract images than dialogue, or even drama. It feels, at times, like a movie for people who prefer humanity to actual humans.
Of course some of those charges were lobbed against Stanley Kubrick, too. And in some ways "The Tree of Life" is Malick's "2001" — and not just because he called on that movie's Douglas Trumbull to create another trippy light show or scored it all to classical music.
Both sharply smart and maddeningly elusive, it's yet another film by a great director working at the height of his powers to do something genuinely new. And creating something guaranteed to leave you talking, and arguing, long after the last frame has faded from the screen.

redpajama_set_name: RedPajamaCommonCrawl


text:
Standing Up, Falling Down (2019)
December 5, 2020 December 4, 2020 Often Off Topic
Director: Matt Ratner
Main Cast: Billy Crystal, Ben Schwartz, Grace Gummer, Eloise Mumford, Nate Corddry, Jill Hennessy, Caitlin McGee, Kevin Dunn, Debra Monk
Plot: Mike Wazowski and Sonic the Hedgehog become best friends. Scott (Ben Schwartz) is forced to move back in with his parents as his career as a comedian in LA is struggling. After meeting with Dermatologist Marty (Billy Crystal), the two form an unlikely friendship and help each other with their issues.
My Thoughts: One of my favourite genre fusions has to be the dramedy, a mix of comedy and drama. It was Little Miss Sunshine (2006) that first caught my attention and I haven't turned one down since. Sure, there have been bad ones but the good ones are often excellent rather than average. I'd never heard of Standing Up, Falling Down until I got an email about it but as a huge Parks and Rec fan, I was never going to shy away from a movie with Ben Schwartz as the lead. This movie might not bring anything new to the genre but I adored almost every moment of it.
I don't think I've seen Ben Schwartz in a more serious role, and I hope he lands some more on the back of this because he does it well, and of course he shines best in his funnier scenes. It was Billy Crystal who surprised me the most, though. His character has gone through so much pain but he tries to take it in his stride, improve himself and find the good in every moment. He has one scene in particular that I won't mention for spoilers that really tugged on my heart strings. I haven't watched many of his movies, I have to admit, but I'll make an effort to.
What I loved the most was that there was a pivotal moment in the movie that I should have seen coming, in fact I'm sure you all will have done, but I got caught up in the daily lives of these two characters and just didn't want it to end. I don't know, I think maybe I'm hyping it up a bit too much, but you know when you find a movie with characters you just feel a real connection with? That's me with Standing Up, Falling Down. If you give it a go I hope you enjoy it as much as I did! Oh and one more thing – that Twitter account, @dermatol_OG, is real! Don't look it up until you've finished the movie though.
Best Bit: Skipping past the emotional, spoiler-filled moment, Scott's relationship with his sister is so funny. The banter between them is so real and one of my favourite moments is when Scott turns up at the pretzel stand she works at and ends up terrifying her co-worker.
Worst Bit: I know that Marty did some wrongs in the past but man, he didn't deserve to be treated the way he did by his own son. That scene on the front lawn was so upsetting to watch.
Fun Trivia: I can only find two bits of trivia for this. One is a huge spoiler and the other is a vulgar reference to The Revenant (2015) that I'll spare you the trouble of reading.
My Rating: 4 and a half Jägerbombs which, UNLIKE the characters in this movie, I know how to make. WAY too much energy drink was used in those ones!
2019 Ben Schwartz Billy Crystal Caitlin McGee Debra Monk Eloise Mumford Grace Gummer Jill Hennessy Kevin Dunn Matt Ratner Movie Review Nate Corddry Standing Up Falling Down
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6 thoughts on "Standing Up, Falling Down (2019)"
Pingback: Standing Up, Falling Down (2019) | MovieRob
movierob says:
nice review Allie! and that's for the recommendation even though I didn't like it nearly as much as u did.
*tnx
m. brown says:
I have never even come close to hearing about this one, and you're review definitely has me interested. I mean, Sonic and Mike Wazokski? You really can't go wrong, can you?
Oh, and this moment? What a tease that is! It better deliver!!!! (kidding!)
I had not heard of this but it sounds like something I'd enjoy. Haven't seen Billy Crystal in anything so that's something to look forward to!
I really hope you enjoy it if and when you get round to it. I do think I've probably hyped it up to be more than it is but the characters really resonated with me.


Okay, so did you read Training Ground yet? If not, stop here, go read it already. You do not want to read this one without the back story.
Game Time doesn't pick off where we left Jamie and Emma. No, we fast forward ten years to 2013 and find them on opposites sides of Earth. Jamie is playing for the women's division or Arsenal Ladies Football Club in London. Like many Americans, Jamie has chosen to cross the pond for a chance to keep playing the game she lives for.
Emma has achieved the pinnacle of success, she is a starter for the national team, multiple Olympic medals, and endorsement deals with major sponsors. Emma lives her life in the public eye, easily recognizable. Jamie, on the other hand, is still striving for her chance to be called up to play for her country. While Emma has achieved most of her professional life's goals, she has started to see, that her life also has a major void. She wants what so many of her peers has found, that special person. She knows what true love feels like, she just hasn't found it with anyone else.
Fate and bad luck have literally kept them apart all this time. Just when Jamie has had her chance to play on the national level disaster strikes. First a torn ACL, the second a broken ankle, and with every injury, she must go back, recoup and begin her quest again. With dogged determination, she never gives up, and she keeps striving for her lifetime goal. Jamie has the heart of a lion, and you just can't help but root for her success. Finally, she gets the call. Jamie has been picked for a two-week residency camp in an off-year as the national team begins to find who the 21 players will be that represent the USWNT in 2015 in the pursuit to win the World Cup. Guess who else will be at camp, yup, Emma Blakely. These former best friends will spend 14 days in close proximity, and it's time to move forward.
While Training Ground was painful to some extent, Game Time starts the rebuilding. Accepting that life, missteps, and personal mistakes all took a toll on their relationship, Emma and Jamie can find a way to be what they were first, friends. The foundation is there and with work and communication, these two find their way back to each other. Oh, guess what's still there, yup chemistry, sizzle!
What I loved about this book is there isn't any contrived drama. These two ladies have a lot on their plate as it is. The stressors of being a professional athlete, staying on top of your game, making one of the most select teams in the world are enough right. You will be 100% invested in the journey of these two leading ladies. My heart just about burst when they find their way back to each other, and have their chance at true love. Now I just need the third book ASAP.
Soccer is a passion of mine. I have played most of my life, starting when I was just in the 1st grade. Combine that love with a Kate Christie book. Okay yeah, you have me interested. Even if soccer isn't your passion ,you should give this book a chance for one reason, the author.
Jamie Maxwell is a young high school aged soccer player. She has that single-minded focus of an elite athlete. Soccer, she lives for it, she will sacrifice everything else for the game. How many teenagers are that focused, not many? Combine that with a natural athletic prowess and you have that special combo that just might take you to the top of the game.
Emma Blakeley is exactly like Jamie in the sense that the game defines her. Emma is finishing up her senior year and has already committed to the storied women's soccer program UNC. Emma sees a kinship in Jamie, where the sport they love supersedes everything else. What she also sees, when they randomly bump into each post-tournament is that Jamie radiates sadness. They bond almost instantly seeing a kinship of a fellow athlete but also drawing strength from one another. This instant connection will define the rest their year.
In the beginning, they are strictly friends, sharing their love of the beautiful game, waking up before the sun is up to catch a Premier League match, sharing stories of their day on the pitch. Soccer is the catalyst for more. They become to depend on each other. They become each other's anchors as they deal with some pretty heavy stuff.
This story is jam-packed with drama. You have multiple life defining moments and the reactions and emotions of two young women. My heart ached for both Jamie and Emma, at times individually and simultaneously. To say these two are put through the gamut is an understatement. 2003-2004 will be a time that will forever define the course of these two lives.
Pretty heavy right! If this book was a stand alone, I would be wary to recommend this one. Not because of any flaw in the writing, because it is really good, but it also kind of hurts. While I read romance novels for so many reasons, let's all face the cold hard truth, we want that happy ending. Good news is this is not a stand alone, but one of three books following the lives of soccer's toughest duo. Training Ground is the backstory, vital to understanding how it all began. So you have to read it, and you'll want to, but buy the second part with the first, and go right into it.
In the Company of Women by Kate Christie is both a work of historical fiction as well as a romance. The romance is well developed and had great chemistry, what made the book so very good though is the historical accuracy that went along with it. How many times can you learn about such an important part of history all the while getting a great story too?
The book first introduces us to Caroline "CJ" Jamieson a private in the Women's Army Corps in 1943. CJ a Michigan native was scheduled to be stationed in California, but right at the last minute she gets rerouted to Fort Bliss in El Paso, Texas. CJ is an airplane engine mechanic that recently graduated from University of Michigan, but delayed graduate school to do her part for the war effort.
Upon arriving at Fort Bliss, CJ literally runs into Brady Buchanan. Brady is the public relations section of WAC and hosts the radio program as well as writing for the base newspaper. Their chance meeting sparks an instant friendship and these two become almost inseparable.
Brady and CJ have a wonderful chemistry and it builds nicely. I wanted them together as much as possible, but this book is also full of historical facts of the time, so you have a give and take between the romance and history lesson. I found myself really enjoying this. I thought the information was written in a way that was interesting and engaging. I felt as if I was right there at Fort Bliss. Christie obviously put a lot of research into this novel and to me it paid off in spades. If you are a history buff this is your book, you will not be disappointed. If you have never read Kate Christie's work I will tell you she is one hell of a writer, her books are always great!


Wordpress Memory Limits & Upload Limits - A Solution | CNCPTS - Design, Development & Marketing for the web. We Create Concepts.
Over the past two years I've built a lot of sites in WordPress and time and time again I run into memory limit problems. Despite the active community around WordPress I am never able to get a clear and straight answer on how resolve this problem. You may run into other blog posts explaining how to add a few lines to your .htaccess file or changing a single line in your wp-settings.php file but none of these solutions will work.
So after hours of troubleshooting i'm happy to share a simple, clear and concise solution to WordPress Memory Limits, using a php.ini file.
NOTE: Before starting I suggest checking the current upload limits and memory limits on your site, you can do this by installing TCP! Memory Usage plugin, then go to the PHP tab.
Create a new file in any text editor, Notepad, Textmate or even Text Edit.
Lastly, this file MUST BE UPLOADED TO THE WP-ADMIN directory on your WordPress site.
That's it, I hope this will be helpful to anyone struggling with WordPress Memory Limits or Upload Limits.


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text:
Q: Counting down and then up I'm new to python and trying to run a function that will, given one variable, count down to zero and then up to the original variable. the output should look something like this:
>>> functionname(5)
5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5

so far I've written the code below, but this doesn't count up all the way to the original variable. I guess I need to somehow save the variable in order to refer to it later, but I have no idea how to do that, since python automatically changes n as the function goes on.
def functionname(n):
    n = orginal
    while n > 0:
        print n
        n=n-1
    print n
    if n==0:
        print n
        n=n+1

I would be very grateful for some pointers, as I seem to be completely stuck at the moment.

A: Just count from negative to positive and use math:
def fn(n):
    print ', '.join(str(abs(i)) for i in range(-n, n+1))
fn(5)
# 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5

Pointers:

*

*If you already know the range you want to iterate over, it's much cleaner and more direct to use a for loop instead of a while.

*You can use the range function to generate number sequences.

*You can convert simple for loops into list-comprehensions as I did above.

The "simple" clean implementation of your requirements would look something like this:
def fn(n):
    for i in range(n, 0, -1):
        print i,
    for i in range(n + 1):
        print i,

Other notes:

*

*range can count backwards too

*The end argument to range isn't included in the range itself, that's why the second loop specifies n + 1 as the limit (instead of just n)

*Adding a trailing comma on your print makes it add a space at the end instead of a line-break.

A: Your second block is an if n == 0: (which you know it is since the while loop terminated when n hit 0); presumably you want while n <= 5.
Note that there are nicer ways to accomplish the same thing in Python. For example, using a pair of ranges with itertools.chain to iterate each range one after another allows you to simplify to:
import itertools

def functionname(n):
    for i in itertools.chain(range(n, 0, -1), range(n+1)):
        print i

A: Personally, I'd do something like...
def count(n):
    for x in range(n, -n, -1):
        print(str(abs(x)) + ",")

At the suggestion of dlewin, here's a list comprehension of the same...
def count(n):
    print(','.join(str(abs(x)) for x in range(n, -n, -1)))

A: You need a second while loop that starts at 0 and goes back up to "original".
Do you know about "for" loops yet?  Those are better for counting.

A: Your idea about having original is correct however you are using the assignment operator the wrong way 'round. Also the if n==0 line should be another loop (while or for as suggested by other answers), counting back up to original.
So I'd start with copying the value from n to original like this:
original = n

Hope that helps!

A: You got some bad formatting there. Remember to indent properly for functions and while and if statements.
So first, set n to 5. Then count down from there until you reach 0 with a while loop:
while n != -1:
    print n
    n -= 1

Then after the loop breaks, count back up again and reset n to 0:
n = 0
while n < 6:
    print n
    n += 1


NEW SEASON. The Helena Dress in dusk blue. This off the shoulder dress has a romantic, feminine silhouette in dotted tulle tiered ruffles. Featured with a beautiful statement grosgrain waistband; intricately adorned with shimmering clasp stones and cut beads in fine floral lace motifs - adding a sparkling finish. The perfect dress to make an ultimate style statement.


text:
FOUNDATION_EXPORT double BlipChatVersionNumber;
FOUNDATION_EXPORT const unsigned char BlipChatVersionString[];

redpajama_set_name: RedPajamaGithub


Welcome to Dayton Animal Clinic!
Please help spread the word: We are hiring! If you are a Licensed Veterinary Technician, submit a resume either in person, via mail to 1738 Silver Lake Road, Dayton, VA 22821, or to smanimalclshenvalley@gmail.com. We are looking for someone to join our team!
Dayton Animal Clinic, LLC in Dayton, VA is seeking a full-time Licensed Veterinary Technician to join our established team. Our focus is on small animal medicine and surgery. We are dedicated to our patients and clients and are seeking a technician who is compassionate, organized, detail oriented, friendly, and outgoing that will be a team player.
We are located in the beautiful Shenandoah Valley which is bordered by mountains and rural by nature but on the edge of the smaller city of Harrisonburg which provides cultural and educational opportunities.
The successful candidate must have excellent customer service, technical, communication, and observation skills as well as a positive attitude. We offer above average pay as well as competitive benefits. Candidates must have a valid license to practice as a Licensed Veterinary Technician in the state of Virginia or be able to obtain one before employment begins.
Our clinic hours are Monday - Friday 7:30 am to 5:00 pm and Saturday from 7:30 am to noon. We will consider all applicants regardless of experience level.
Candidates may drop off a resume at Dayton Animal Clinic located at 1738 Silver Lake Road, Dayton, Virginia or send a resume via email to smanimalclshenvalley@gmail.com.
We are a full service animal clinic providing comprehensive healthcare services to pets in Dayton, Bridgewater, Harrisonburg, Rockingham County, Augusta County, and the surrounding areas. Our veterinarians offer a wide variety of medical, surgical, and dental services in our veterinary clinic. We provide complete care for our small animal patients.
Our veterinary hospital is well equipped with advanced equipment and technologies to provide the highest standard of care for your pet. Our facility has the equipment to provide comprehensive in house testing for accurate diagnosis, a surgical suite, a dental suite, a pharmacy, and more. At Dayton Animal Clinic we are committed to providing quality care at every stage of your pet's life. From routine preventive care for your pets to early detection and treatment of a wide range of conditions and diseases, and surgical care. We have the expertise to provide the care your pet needs at every stage of life.
Please browse our website to learn more about our animal clinic and the services we provide for companion animals in Dayton, Bridgewater, Harrisonburg, Rockingham County, Augusta County, and the surrounding areas. Read information in our Pet Library, view videos, and read testimonials. Please call our office today at (540) 879-3833 for all your pet health care needs.


What time ALLEPPEY EXP depart from KANNUR Railway Station?
ALLEPPEY EXP (16308) departs from KANNUR Railway Station at 05:00.
ALLEPPEY EXP reach on day 1 to ALAPPUZHA Railway Station. The arrival time of ALLEPPEY EXP at ALAPPUZHA Railway Station is 13:20.
ALLEPPEY EXP covers 339 km to reach ALAPPUZHA Railway Station at average speed of 41 km/hr. ALLEPPEY EXP passes through 28 stations.


text:
#include <aws/pinpoint/model/SendUsersMessageRequest.h>
#include <aws/core/utils/json/JsonSerializer.h>

#include <utility>

using namespace Aws::Utils::Json;
using namespace Aws::Utils;

namespace Aws
{
namespace Pinpoint
{
namespace Model
{

SendUsersMessageRequest::SendUsersMessageRequest() :
    m_contextHasBeenSet(false),
    m_messageConfigurationHasBeenSet(false),
    m_templateConfigurationHasBeenSet(false),
    m_traceIdHasBeenSet(false),
    m_usersHasBeenSet(false)
{
}

SendUsersMessageRequest::SendUsersMessageRequest(JsonView jsonValue) :
    m_contextHasBeenSet(false),
    m_messageConfigurationHasBeenSet(false),
    m_templateConfigurationHasBeenSet(false),
    m_traceIdHasBeenSet(false),
    m_usersHasBeenSet(false)
{
  *this = jsonValue;
}

SendUsersMessageRequest& SendUsersMessageRequest::operator =(JsonView jsonValue)
{
  if(jsonValue.ValueExists("Context"))
  {
    Aws::Map<Aws::String, JsonView> contextJsonMap = jsonValue.GetObject("Context").GetAllObjects();
    for(auto& contextItem : contextJsonMap)
    {
      m_context[contextItem.first] = contextItem.second.AsString();
    }
    m_contextHasBeenSet = true;
  }

  if(jsonValue.ValueExists("MessageConfiguration"))
  {
    m_messageConfiguration = jsonValue.GetObject("MessageConfiguration");

    m_messageConfigurationHasBeenSet = true;
  }

  if(jsonValue.ValueExists("TemplateConfiguration"))
  {
    m_templateConfiguration = jsonValue.GetObject("TemplateConfiguration");

    m_templateConfigurationHasBeenSet = true;
  }

  if(jsonValue.ValueExists("TraceId"))
  {
    m_traceId = jsonValue.GetString("TraceId");

    m_traceIdHasBeenSet = true;
  }

  if(jsonValue.ValueExists("Users"))
  {
    Aws::Map<Aws::String, JsonView> usersJsonMap = jsonValue.GetObject("Users").GetAllObjects();
    for(auto& usersItem : usersJsonMap)
    {
      m_users[usersItem.first] = usersItem.second.AsObject();
    }
    m_usersHasBeenSet = true;
  }

  return *this;
}

JsonValue SendUsersMessageRequest::Jsonize() const
{
  JsonValue payload;

  if(m_contextHasBeenSet)
  {
   JsonValue contextJsonMap;
   for(auto& contextItem : m_context)
   {
     contextJsonMap.WithString(contextItem.first, contextItem.second);
   }
   payload.WithObject("Context", std::move(contextJsonMap));

  }

  if(m_messageConfigurationHasBeenSet)
  {
   payload.WithObject("MessageConfiguration", m_messageConfiguration.Jsonize());

  }

  if(m_templateConfigurationHasBeenSet)
  {
   payload.WithObject("TemplateConfiguration", m_templateConfiguration.Jsonize());

  }

  if(m_traceIdHasBeenSet)
  {
   payload.WithString("TraceId", m_traceId);

  }

  if(m_usersHasBeenSet)
  {
   JsonValue usersJsonMap;
   for(auto& usersItem : m_users)
   {
     usersJsonMap.WithObject(usersItem.first, usersItem.second.Jsonize());
   }
   payload.WithObject("Users", std::move(usersJsonMap));

  }

  return payload;
}

} // namespace Model
} // namespace Pinpoint
} // namespace Aws


Page Associate Principal and healthcare expert Dan Killebrew has been appointed to the Board of Directors of a new organization, Health Care Institute (HCI) North Texas.
Page Associate Principal and healthcare expert Dan Killebrew has just been honored with an appointment to the Board of Directors of a new organization, Health Care Institute (HCI) North Texas. This new entity will join HCI Chicago and HCI Midwest as a regional focus for healthcare facility professionals and Dan has accepted the Board position of Program Chair.
HCI, an IFMA Alliance Partner, strives to enhance the skills of facility management professionals and further the overall profession through networking events and educational seminars that allow facilities managers to problem solve with healthcare architects, designers, engineers, contractors and administrators. Thus, Dan will have a key role in ensuring that its programming meets members' needs.
Dan has extensive experience in serving on industry boards, most recently the Health Industry Council in North Texas from 2006 – 2014. Similar to HCI, the council's primary objective was to create a forum for education, networking, and growth of any business related to the healthcare industry of Health. Prior to that he served first as Commissioner of Practice and then Treasurer for the board of AIA Dallas Chapter.
To learn more about HCI North Texas, click here.
To see the announcement of the launch of HCI North Texas and its new board members, click here.


Healthy boundaries facilitate empowerment rather than rescuing. One supports authentic power, the other perpetuates victim consciousness. It is perfectly appropriate and beautiful to help one another, in fact, we love you for your tender hearts and caring nature! We just urge you to make sure it is for the greatest good of all involved.
A good rule of thumb is to make sure that you never become more invested in another's wellness than they are in their own, and to never become more invested in another's wellness than you are in your own. Support and encouragement uplift, while rescuing or doing for others, what they can very well do for themselves, reinforces the illusion of their powerlessness.
We cannot stress the importance of creating a safe and nurturing environment for yourself that supports your alignment, growth, self expression, and purpose. You are on the planet to be uniquely you and to make the contributions only you can make. Surround yourself with people who remind you of your truth, who encourage you in your beauty and beingness, who celebrate you in all your you-ness!


How Magazine App can help you grow?
Every magazine publisher wants to grow. This is particularly hard in todays economy and with a limited resources. Fortunately there is an easy way to significantly enlarge markets that your company address – and this is mobile. While going mobile may sound tough it really isn't with a help of Magazine App. Publishers can test mobile markets spending very little money comparing to building custom solution. Therefore reach new people and grow!
Why Publishers should use Magazine App?
Within couple of months of Apple releasing the original iPad publishers realize that tablets are not a threat to them. Quite the opposite – they saw this new category of devices as a way to reach more markets, more target groups and therefore sell more. One thing that was missing in 2010 is now also solved: software. Whole new category of Magazine App were developed to help publishers focus on their job better.
When Apple introduced the iPad in 2010 very few small publishers thought they can compete with big titles. What has happened since then surprised experts and media all around the World: rise of Magazine App helped mostly small publishers, while big media corporations fall behind or got stuck with custom build solutions that aged fairly quickly.


I wax nostalgic a bit today about my cradle Catholic years with Lent and Easter, but having never experienced the Triduum liturgies until just a few years ago. We linger over Holy Thursday, Good Friday, and the Easter Vigil, and the meaning behind those liturgies. To be continued last week since I got so into this!
What are your experiences with the Holy Week liturgies? Did you participate in them as a child, or only as an adult like me? I would love to hear from you!


text:
Deleted 'Jeepers Creepers 2' Dead Soldiers Scene Revealed in Full
While doing my daily horror hunt I came across this deleted scene from Victor Salva's Jeepers Creepers 2, which was cut and not included on any DVD or Blu-Ray. I haven't checked out the Scream Factory edition so I don't know for sure if the scene has been added on that version or not, but for those who have never seen it, here it is in full.
I really wish this scene was left in the film, it's so eerie and gives a real sense of fear. The soundtrack used in this scene really accompanies the death and terror of what lies ahead, and what hell those soldiers went through.
You can check out the full deleted scene below!
Did you like Jeepers Creepers 2? Let us know your thoughts on the film.
Related Topics:Blu-RayDVDJeepers Creepers 2Jonathan BreckLuke EdwardsRay WiseScream FactoryVictor Salva
Trailer For Darren Lynn Bousman's New Horror Film 'Abattoir'
Doom Head Opens Up Murder School in New Clip From Rob Zombie's '31'
'Krampus: The Naughty Cut' 4K Ultra HD Blu-Ray Gets December Release in the US from Scream Factory
Watch the Alternate "Dr. Death" Opening to 'Halloween 5: The Revenge of Michael Myers [Video]
'Halloween 5': Original "Doctor Death" Opening Will Be Included on the Scream Factory Blu-Ray Release
Scream Factory: Wes Craven's 'Deadly Friend' Brings the Chaos to Blu-Ray for the First Time This October!
Scream Factory Releasing 'Elvira's Haunted Hills' on Collector's Edition Blu-Ray in the US This September