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\begin{document}
\large
\title{On
a form of degree $d$ in $2d+1$ variables ($d\geq 4$)}
\author{Manoj Verma}
\address{MANOJ VERMA: Institute of Mathematical Sciences,
CIT Campus, Taramani, Chennai 600113, India}
\email{mverma@imsc.res.in}

\begin{abstract} For $k\geq 2$, we derive an asymptotic formula for the number of zeros of the forms
\begin{center}
$\prod_{i=1}^{k}(x_{2i-1}^2+x_{2i}^2)+\prod_{i=1}^{k}(x_{2k+2i-1}^2+x_{2k+2i}^2)-x_{4k+1}^{2k}$
\end{center}
and
\begin{center}
$x_1\prod_{i=1}^{k}(x_{2i}^2+x_{2i+1}^2)+x_{2k+2}\prod_{i=1}^{k}(x_{2k+2i+1}^2+x_{2k+2i+2}^2)-x_{4k+3}^{2k+1}$
\end{center} in the box $1\leq x_i\leq P$ using the circle method.\\\\
Mathematics Subject Classification 2010: Primary 11D45; Secondary 11D85, 11P55.
\end{abstract}
\maketitle
\Section {Introduction}
\label{introduction}
\noindent The problems of representation of positive integers and of non-trivial
representation of zero by forms with integral coefficients have attracted attention of many mathematicians since the beginning of the 20th century. Most of the work in this area has been carried out using the circle method. The square-root barrier implies that the circle method is unlikely to yield an asymptotic formula in case of a form  of degree $d$ with less than $2d+1$ variables. In 1962 Birch, Davenport and Lewis \cite{BDL} found an asymptotic formula for the number representations of zero in a box $|x_i|\leq P$ by the cubic forms in seven variables
of the type
$$N_1(x_1,x_2,x_3)+N_2(x_4,x_5,x_6)+a_7x_7^3$$ where $N_1$ and $N_2$ are norm
forms of some cubic field extensions of $\mathbb Q$ and $a_7$ is a nonzero integer. Incidentally, whereas an asymptotic formula for the number of representations of an integer as a sum of eight cubes has been established by Vaughan \cite{Vaughan1}, an asymptotic formula in the case of a sum of seven cubes seems to be out of reach as of now. However, a lower bound of the expected order of magnitude has been established by Vaughan \cite{Vaughan3}.
A few more families of cubic forms in seven variables have been dealt with.
However, for $d \geq 4$ no example of an asymptotic formula for the number of representations of either zero or of large positive integers by a form of degree $d$ in $2d+1$ variables seems to exist in the literature. In this paper, for each $d\geq 4$, we establish an asymptotic formula for the number of zeros of a particular form of degree $d$ in $2d+1$ variables in a suitable expanding box using the circle method. It turns out that our proof goes through in cases $d=1, 2, 3$ as well and retrieves previously known results in those cases. For $d$ even, say $d=2k$, $k\geq 1$, we look at the number zeros of the form
\begin{equation*} \label{e:one1e}
f({\bf x})=f(x_1, \ldots, x_{4k+1})=\prod_{i=1}^{k}(x_{2i-1}^2+x_{2i}^2)+\prod_{i=1}^{k}(x_{2k+2i-1}^2+x_{2k+2i}^2)-x_{4k+1}^{2k}
\end{equation*}
while for $d$ odd, say $d=2k+1$, $k\geq 0$, we look at the number of zeros of the form
\begin{equation*} \label{e:one1o}
f({\bf x})=
f(x_1, \ldots, x_{4k+3})=
x_1\prod_{i=1}^{k}(x_{2i}^2+x_{2i+1}^2)+x_{2k+2}\prod_{i=1}^{k}(x_{2k+2i+1}^2+x_{2k+2i+2}^2)-x_{4k+3}^{2k+1}
\end{equation*}
in a box $1\leq x_i\leq P$.
When convenient we shall abbreviate and write
\begin{equation} \label{e:one1}
f({\bf x})=
f(x_1, \ldots, x_{2d+1})=
f_d(x_1, \ldots, x_d)+f_d(x_{d+1}, \ldots, x_{2d})-x_{2d+1}^d
\end{equation}
where
\begin{equation*}
f_d(x_1, \ldots, x_d)=\left\{ \begin{array}{ll}
\prod_{i=1}^{k}(x_{2i-1}^2+x_{2i}^2) & \mbox{ if $d$ is even, $d=2k$, $k\geq 1$}\\
x_1\prod_{i=1}^{k}(x_{2i}^2+x_{2i+1}^2) & \mbox{ if $d$ is odd, $d=2k+1$, $k\geq 0$}
\end{array}
\right..
\end{equation*}
We could define $f_d=f_d(x_1, \ldots, x_d)$ recursively: $f_1(x_1)=x_1$, $f_2(x_1, x_2)=x_1^2+x_2^2$, and $f_d(x_1, \ldots, x_d)=(x_{d-1}^2+x_{d}^2)\cdot f_{d-2}(x_1, \ldots, x_{d-2})$ for $d\geq 2.$\\
\indent Taking $x_{d-2}=x_{2d-2}=x_{d-4}=x_{2d-4}=\ldots=1$,\, $x_{d-3}=x_{2d-3}=x_{d-5}=x_{2d-5}=\ldots =0$ and applying the four square theorem we see that this form represents every integer infinitely many times; our interest lies in establishing the asymptotic formula for the number of representations of zero with $1\leq x_i\leq P$. Only minor changes are required to prove the expected asymptotic formula for the number of representations of a large positive integer $N$ by this form (or by the form obtained by changing the $-$ sign in the front of $x_{2d+1}^{d}$ to $+$) with the variables restricted so that $1\leq x_i \leq \lfloor N^{\frac{1}{d}} \rfloor $.\\
\indent Note that if $d\geq 3$ then the form $f$ is singular (e.g., consider the point (1, 0, 0, \ldots, 0)) (so Deligne's estimate for complete exponential sums is not applicable) and its $h$-invariant is 3. (The $h$-invariant of a form $F$, denoted $h(F)$, is defined as the smallest integer $h$ such that $F$ can be written in the form
$$F({\bf x})=A_1({\bf x})B_1({\bf x})+\ldots+A_h({\bf x})B_h({\bf x})$$
where $A_i({\bf x})$ and $B_i({\bf x})$ are forms of positive degrees having rational coefficients.)\\
\indent In what follows, $\varepsilon$ can take any positive real value in any statement in which it appears, $e(\alpha)=e^{2\pi i \alpha}$, the symbols $\ll$ and $O$ have their usual meanings with the implicit constants depending at most on $\varepsilon$ and $d$. If $A$ is a set, $\#A$ denotes the cardinality of $A.$ If $n, m$ are positive integers, $d(n)$ denotes the number of positive divisors of $n$ and $d_m(n)$ denotes the number of (ordered) factorizations of $n$ with $m$ positive integral factors. For a real number $x$, $\lfloor x \rfloor$ and $\lceil x \rceil$ denote the greatest integer less than or equal to $x$ and the smallest integer greater than or equal to $x$ respectively.
\begin{thm} \label{t:one1}
The number of zeros  $R(0;P)$ of the form $f$ in (\ref{e:one1}) with $1\leq x_i \leq P$ satisfies
\begin{equation*}
R(0;P)=P^{d+1}\mathfrak{S}J+O(P^{d+1-\frac{1}{1+5\cdot 2^{d-1}}+\varepsilon})
\end{equation*}
where $${\mathfrak{S}}=\sum_{q=1}^{\infty}\sum_{\substack{a=1\\(a,q)=1}}
^{q}
q^{-2d-1}S(q,a) \mbox{ with }S(q,a)=\sum_{{\bf z} \bmod q} e\left( \frac{a}{q}f({\bf z})\right),$$
and
$$J=\int_{-\infty}^{\infty}\left( \int_{[0,1]^{2d+1}}e(\gamma f({\bf \xi}))\,d{\bf
\xi} \right) d\gamma;$$
$\mathfrak{S}$ and $J$ are positive.
\end{thm}
\begin{rem}
One can also consider the problem of finding an asymptotic formula for the number of zeros of the form $f$ inside the box $|x_i|\leq P$ but then the contribution from those ``degenerate" zeros of $f$ that lie inside the box $|x_i|\leq P$ and also in the set
\begin{equation*}
U=\{{\bf x}\in \mathbb{Z}^{2d+1}: f_d(x_1, \ldots, x_d)=f_d(x_{d+1}, \ldots, x_{2d})=x_{2d+1}= 0\}
\end{equation*}
would be $2^{2d-2}P^{2d-2}+O_{d}(P^{2d-3})$ in case $d$ is odd and $2^{2d-6}d^2P^{2d-4}+O_{d}(P^{2d-5})$ in case $d$ is even and thus the contribution from the zeros of $f$ with at least one of the $x_i$ equal to zero would be $\gg P^{d+2}$ if $d\geq 5$ and cannot be determined precisely via the circle method. On the other hand, the number of zeros of $f$ with $|x_i|\leq P$ but none of the $x_i$ equal to zero is $2^{2d+1}$ times the number of zeros with $1\leq x_i\leq P$ in case $d$ is even, and is $2^{2d-1}$ times the sum of the number of zeros with $1\leq x_i\leq P$ of the three forms
$$f_d(x_1, \ldots, x_d)+f_d(x_{d+1}, \ldots, x_{2d})-x_{2d+1}^d,$$
$$-f_d(x_1, \ldots, x_d)+f_d(x_{d+1}, \ldots, x_{2d})-x_{2d+1}^d$$
and
$$f_d(x_1, \ldots, x_d)-f_d(x_{d+1}, \ldots, x_{2d})-x_{2d+1}^d;$$
theorem 1 applies to the latter two forms as well.
\end{rem}
\Section{The Minor Arcs}
\label{minorarcs}
\noindent Let $P$ be a large positive integer, $I=\{1, 2, \ldots, P\}$, $B=I^{2d+1}$ and
\begin{equation} \label{e:two1}
F(\alpha)= \sum_{{\bf x}\in B} e(\alpha f({\bf x})).
\end{equation}
Then the number of zeros of the form $f$ inside the box $1\leq x_i\leq P$ is
\begin{equation} \label{e:two2}
R(0;P)=\int_{0}^{1} F(\alpha)\,d\alpha=\int_{0}^{1} F_1(\alpha)^2 F_2(\alpha)\,d\alpha
\end{equation}
where
$$F_1(\alpha)=\sum_{x_1, \ldots, x_{d} \in I} e(\alpha f_d(x_1, \ldots, x_{d}))$$
and
$$F_2(\alpha)=\sum_{x \in I} e(-\alpha x^{d}).$$
\indent Let $\delta$ be a sufficiently small positive real number to be specified
later. For $1 \leq q \leq P^{\delta},\; 1 \leq a \leq q,\; (a,q)=1,$ define the major arcs $\mathfrak{M}(q,a)$ by
\begin{equation*}
\mathfrak{M}(q,a)=\left\{\alpha:\left| \alpha
- \frac{a}{q}\right| \leq P^{\delta-d}\right\}
\end{equation*}
and let $\mathfrak{M}$ be the union of the $\mathfrak{M}(q,a)$. The set $\mathfrak{m}=(P^{\delta - d}, 1+P^{\delta - d}]\backslash \mathfrak{M}$ forms the minor arcs.
\begin{lem} \label{l:two1} We have
\begin{equation*}
\int_{0}^{1} \left| F_1(\alpha)\right|^2 d\alpha \ll P^{d+\varepsilon}.
\end{equation*}
\end{lem}
\begin{proof} We assume that $d$ is even, $d=2k$, $k\geq 1$; the proof is similar in case $d$ is odd. We have
\begin{equation*}
\int_{0}^{1}|F_1(\alpha)|^2 d\alpha
=\sum_{0\leq n \leq 2^kP^{d}}a(n)^2
\end{equation*}
where \begin{equation*}
a(n) = \#\{(x_1, \ldots, x_{d})\in I^{d}: f_d(x_1, \ldots, x_{d})=n\} \mbox{ for }
n\geq 0.
\end{equation*}
Clearly $a(0)=1$ and $\sum_{0\leq n \leq 2^kP^{d}}a(n)=P^{d}$. For $n\geq 1$, in any solution of $\prod_{i=1}^{k}(x_{2i-1}^2+x_{2i}^2)=n$ we must have $x_{2i-1}^2+x_{2i}^2=n_i$ ($1\leq i\leq k$) for some factorization $n=n_1\ldots n_k$ of $n$ with $k$ (positive) factors. For a positive integer $r$, the number of solutions of the equation $s^2+t^2=r$ in non-negative integers  $s, t$ is $\leq 8d(r) \ll r^{\varepsilon}$. Thus, for $1 \leq n\leq 2^kP^{d}$ we have
\begin{equation*} a(n)\ll \sum_{n_1\ldots n_k=n}n_1^{\varepsilon}\ldots n_k^{\varepsilon}=d_{k}(n)n^{\varepsilon}\ll n^{\varepsilon}n^{\varepsilon}\ll P^{2d\varepsilon}.
\end{equation*}
Hence
\begin{equation*}
\int_{0}^{1} |F_1(\alpha)|^2 d\alpha = 1+\sum_{1\leq n \leq 2^kP^{d}}a(n)^2
\ll 1+P^{2d\varepsilon}\sum_{1\leq n \leq 2^kP^{d}}a(n)\ll P^{d+2d\varepsilon}.
\qedhere
\end{equation*}
\end{proof}
\begin{lem} \label{l:two2}
We have
\begin{equation*}
\int_{\mathfrak{m}}F(\alpha)\,d\alpha \ll P^{d+1+\varepsilon-\frac{\delta}{2^{d-1}}}.
\end{equation*}
\end{lem}
\begin{proof}
For $\alpha \in \mathfrak{m}$,
by Dirichlet's theorem on Diophantine approximation, there exist $a, q$ with
$(a,q) =1, q\leq P^{d-\delta}$ and $\left| \alpha
- \frac{a}{q}\right| \leq q^{-1}P^{\delta-d} < q^{-2}.$ Since
$\alpha \in \mathfrak{m}\, \subseteq (P^{\delta - d},
1-P^{\delta - d})$, we must have $1\leq a < q$ and $q
>P^{\delta}$ (otherwise $\alpha$ would be in $\mathfrak{M}(q,a)$). By Weyl's
inequality (See \cite{DBook} or \cite{VaughanBook}),
$$F_2(\alpha)\ll P^{1+\varepsilon-\frac{\delta}{2^{d-1}}}.$$
This, with (\ref{e:two2}) and lemma \ref{l:two1} gives
\begin{eqnarray*}
\left| \int_{\mathfrak{m}}F(\alpha)\,d\alpha \right|
&= & \left| \int_{\mathfrak{m}}F_1^2(\alpha) F_2(\alpha)\,d\alpha \right|
\leq \left( \sup_{\alpha\in\mathfrak{m}}|F_2(\alpha)|
\right) \left( \int_{\mathfrak{m}} |F_1(\alpha)|^2\,d\alpha \right)\\
&\leq & \left( \sup_{\alpha\in\mathfrak{m}}|F_2(\alpha)| \right)
\left( \int_{0}^{1} |F_1(\alpha)|^2 d\alpha \right)
\ll P^{d+1+\varepsilon-\frac{\delta}{2^{d-1}}}. \qedhere
\end{eqnarray*}
\end{proof}
\Section{The Major Arcs}
\label{majorarcs}
\noindent To deal with the major arcs we need to obtain an approximation to
the generating function $F(\alpha)$, (\ref{e:two1}), in terms of the auxiliary functions
\begin{equation} \label{e:three1}
S(q,a)=\sum_{{\bf z} \bmod q} e\left( \frac{a}{q}f({\bf z})\right)
\end{equation}
and
\begin{equation} \label{e:three2}
I(\beta)=\int_{[0,P]^{2d+1}}e(\beta f({\bf \xi}))\,d{\bf \xi}.
\end{equation}
\begin{lem} \label{l:three1}
For $\alpha \in \mathfrak{M}(q,a),$ writing $\alpha = (a/q)+\beta,$ we have
\begin{equation*}
F(\alpha)=q^{-2d-1} S(q,a)I(\beta)+O(P^{2d+2\delta}).
\end{equation*}
\end{lem}
\begin{proof} We have
\begin{eqnarray*}
F\left( \frac{a}{q}+\beta \right) &=& \sum_{{\bf x}\in B}e\left( \left( \frac{a}{q}+\beta \right) f({\bf x})\right)\\
&=& \sum_{{\bf z} \bmod q} \ \sum_{{\bf y}:q{\bf y}+{\bf z}\in B}e\left( \left( \frac{a}{q}+\beta \right) f(q{\bf y}+{\bf z})\right)\\
&=& \sum_{{\bf z} \bmod q} \ \sum_{{\bf y}:q{\bf y}+{\bf z}\in B}e\left( \frac{a}{q}f(q{\bf y}+{\bf z})\right) e(\beta f(q{\bf y}+{\bf z}))\\
&=& q^{-4k-1}\sum_{{\bf z} \bmod q} e\left( \frac{a}{q}f({\bf z})\right) \sum_{{\bf
y}:q{\bf y}+{\bf z}\in B}q^{2d+1}e(\beta f(q{\bf y}+{\bf z})).
\end{eqnarray*}
The inner sum is a Riemann sum of the function $e(\beta f({\bf \xi}))$ over a $2d+1$-dimensional cuboid which depends on ${\bf z}$.
We want to replace it by the integral over the $2d+1$-dimensional cube
$[0,P]^{2d+1}$
and estimate the error in doing so.
Firstly, any first-order partial derivative of $e(\beta f({\bf \xi}))$ is of the form $2\pi i\beta g({\bf \xi})e(\beta f({\bf \xi}))$
where $g({\bf \xi})$ is a degree $d-1$ form (in $2d+1$ real variables)
whose coefficients are bounded and which itself is, therefore, $O(P^{d-1})$ on the
cube $[0,P]^{2d+1}$.
Therefore, $e(\beta f({\bf \xi}))$ does not vary by more that $O(|\beta| q P^{d-1})$ on a cube of side $q$.
There are $q^{2d+1}$ terms in the outer sum and $\ll (P/q)^{2d+1}$ terms in each inner sum so the error in replacing the sum by the integral over the corresponding
cuboid is $O(|\beta| q P^{d-1} q^{2d+1}(P/q)^{2d+1})=O(|\beta|q P^{3d})$.
Secondly, each side of this cuboid depends on ${\bf z}$ but is within length $q$
of the sides of $[0,P]^{2d+1}$.
Since the integrand is bounded by 1 in absolute value, the error in replacing the region of integration by $[0,P]^{2d+1}$ is $O(q P^{2d})$.
Since we have $q\leq P^{\delta}$ and $|\beta|\leq P^{\delta-d}$, the total error is $O(P^{2d+2\delta})+O(P^{2d+\delta})=O(P^{2d+2\delta})$.
Hence
$$F\left( \frac{a}{q}+\beta \right)= q^{-2d-1}\sum_{{\bf z} \bmod q} e\left( \frac{a}{q}f({\bf z})\right) \left(I(\beta)+O(P^{2d+2\delta})\right)$$
which gives the lemma. \qedhere
\end{proof}
\begin{lem} \label{l:three2}
We have
$$\int_{\mathfrak{M}} F(\alpha)\,d\alpha=P^{d+1}\mathfrak{S}(P^{\delta})J(P^{\delta})+O(P^{d+5\delta})$$
where
\begin{equation} \label{e:three3}
{\mathfrak{S}}(Q)=\sum_{q\leq Q}\sum_{
\substack{
a=1\\
(a,q)=1}
}
^{q}
q^{-2d-1}S(q,a)
\end{equation}
and
\begin{equation} \label{e:three4}
J(\mu)=\int_{|\gamma|<\mu}\left( \int_{[0,1]^{2d+1}}e(\gamma f({\bf \xi}))\,d{\bf
\xi} \right)d\gamma.
\end{equation}
\end{lem}
\begin{proof} Note that
${\mathfrak{M}}(q,a)=[(a/q)-P^{\delta-d},(a/q)+P^{\delta-d}]$ has length $2P^{\delta-d}$.
The change of variable \mbox{$\alpha =(a/q)+\beta$} and lemma \ref{l:three1} give
$$\int_{{\mathfrak{M}}(q,a)} F(\alpha)\,d\alpha=
 q^{-2d-1}S(q,a)\int_{-P^{\delta-d}}^{P^{\delta-d}}I(\beta)\,d\beta+O(P^{d+3\delta}).$$
Summing over $1 \leq q\leq P^{\delta},\; 1 \leq a \leq q,\; (a,q)=1,$
we get
$$\int_{\mathfrak{M}} F(\alpha) \,d\alpha=\mathfrak{S}(P^{\delta})\int_{-P^{\delta-d}}^{P^{\delta-d}}I(\beta)
\,d\beta+O(P^{d+5\delta}).$$
Applying the change of variable $\gamma=P^{d}\beta$, the integral in the above
equation becomes
$$P^{-d}\int_{-P^{\delta}}^{P^{\delta}}I(P^{-d}\gamma)\,d\gamma.$$
Since $f$ is a homogeneous function of degree $d$, the change of variable ${\bf \xi} '=P^{-1}{\bf \xi}$ gives
\begin{eqnarray*}
(P^{-d}\gamma)&=&\int_{[0,P]^{2d+1}}e(\gamma P^{-d} f({\bf \xi}))\,d{\bf \xi}
=\int_{[0,P]^{2d+1}}e(\gamma f({P^{-1}\bf \xi}))\,d{\bf \xi}\\
&=& P^{2d+1}\int_{{[0,1]^{2d+1}}}e(\gamma
 f({\bf \xi '}))\,d{\bf \xi '}
\end{eqnarray*}
which completes the proof of the lemma.
\end{proof}
\Section{The Singular Integral}
\label{singularintegral}
\noindent Note that the inner integral appearing in the definition of $J(\mu)$,
(\ref{e:three4}), is bounded trivially by 1.
Applying the change of variable ${\bf \xi'} = \gamma^{1/2k}{\bf \xi}$ we see that
$$\int_{[0,1]^{2d+1}}e(\gamma
f({\bf \xi}))\,d{\bf \xi}= \gamma^{-\frac{2d+1}{d}}\int_{\gamma^{\frac{1}{d}}[0,1]^{2d+1}}e(f({\bf
\xi'}))\,d{\bf \xi'}.$$
We think of the last integral as an iterated integral and apply Lemma 7.3 of
Vaughan \cite{VaughanBook} to each iteration to get the bound
\begin{equation} \label{e:four1}
\int_{[0,1]^{2d+1}}e(\gamma f({\bf \xi}))d{\bf \xi} \ll \mbox{min}
(1,\gamma^{-\frac{2d+1}{d}}).
\end{equation}
\begin{lem} \label{l:four1}
For $\mu\geq 1$,
\begin{equation*}
J(\mu) = J+O(\mu^{-\frac{d+1}{d}})
\end{equation*}
where
\begin{equation*}
J=\int_{-\infty}^{\infty}\left( \int_{[0,1]^{2d+1}}e(\gamma f({\bf \xi}))\,d{\bf
\xi} \right)d\gamma >0.
\end{equation*}
\end{lem}
\begin{proof}
The convergence of $J(\mu)$ to $J$ as $\mu \rightarrow \infty$ at the rate asserted follows from (\ref{e:four1}). To show that $J>0$, we imitate \cite[Chapter 4]{DBook}. Writing $\underline{\alpha}_1$ for $(\alpha_1, \ldots, \alpha_{d})$ and $\underline{\alpha}_2$ for $(\alpha_{d+1}, \ldots, \alpha_{2d})$,
\begin{equation*}
J=\lim_{\lambda\rightarrow\infty}\int_{-\lambda}^{\lambda}\left(\int_{[0,1]^{2d+1}}e(\gamma (f_d(\underline{\alpha}_1)+f_d(\underline{\alpha}_2)-\alpha_{2d+1}^{d})\,d\underline{\alpha}_1\,d\underline{\alpha}_2\,d\alpha_{2d+1}\right)d\gamma.
\end{equation*}
Since $\int_{0}^{1}e(-\gamma \alpha_{2d+1}^{d})\,d\alpha_{2d+1}=\int_{0}^{1}\frac{1}{d}\alpha^{\frac{1}{d}-1}e(-\gamma \alpha)\,d\alpha$ and $\int_{-\lambda}^{\lambda}e(\gamma \mu)\,d\gamma=\sin(2\pi\lambda\mu)/\pi\mu$ for any real numbers $\lambda$ and $\mu$,
\begin{equation*}
J=\frac{1}{d}\lim_{\lambda\rightarrow\infty}\int_{[0,1]^{2d+1}}\alpha^{\frac{1}{d}-1}\left(\frac{\sin (2\pi \lambda (f_d(\underline{\alpha}_1)+f_d(\underline{\alpha}_2)-\alpha))}{\pi (f_d(\underline{\alpha}_1)+f_d(\underline{\alpha}_2)-\alpha)}\right)d\underline{\alpha}_1\,d\underline{\alpha}_2\,d\alpha.
\end{equation*}
Putting $\beta=f_d(\underline{\alpha}_1)+f_d(\underline{\alpha}_2)-\alpha$ so that $\alpha=f_d(\underline{\alpha}_1)+f_d(\underline{\alpha}_2)-\beta$,
\begin{equation*}
J=\frac{1}{d}\lim_{\lambda\rightarrow\infty}\int_{-1}^{2^{\lfloor d/2 \rfloor +1}}\phi(\beta)\frac{\sin (2\pi \lambda \beta)}{\pi \beta}\,d\beta
\end{equation*}
where
\begin{equation*}
\phi(\beta)=\frac{1}{d}\int_{\{\beta <f_d(\underline{\alpha}_1)+f_d(\underline{\alpha}_2)<\beta+1\}\cap [0,1]^{2d}}
(f_d(\underline{\alpha}_1)+f_d(\underline{\alpha}_2)-\beta)^{\frac{1}{d}-1}\,d\underline{\alpha}_1\,d\underline{\alpha}_2.
\end{equation*}
Fourier's integral theorem for a finite interval is applicable and gives
\begin{equation*}
J=\phi(0)=\int_{\{0<f_d(\underline{\alpha}_1)+f_d(\underline{\alpha}_2)<1\}\cap [0,1]^{2d}}
(f_d(\underline{\alpha}_1)+f_d(\underline{\alpha}_2))^{\frac{1}{d}-1}\,d\underline{\alpha}_1\,d\underline{\alpha}_2.
\end{equation*}
Since the region of integration is a non-empty open set and the integrand is positive throughout the region of integration, $J>0$.
\end{proof}
\Section{The Singular Series}
\label{singularseries}
\noindent From the definition of ${\mathfrak{S}}(Q)$, (\ref{e:three3}),
\begin{equation} \label{e:five1}
{\mathfrak{S}}(Q)=\sum_{q\leq Q}S(q)
=\sum_{q\leq Q} q^{-2d-1}\sum_{\substack{a=1\\(a,q)=1}}^{q}S_1(q,a)^2S_2(q,a)
\end{equation}
where$$S_1(q,a)=\sum_{x_1, \ldots, x_{d}=1}^{q} e \left(\frac{a}{q}f_d(x_{1}, \ldots, x_{d})\right),\mbox{ }
S_2(q,a)=\sum_{x = 1}^{q} e\left( -\frac{a}{q}x^{d}\right)$$
and $$S(q)=q^{-2d-1}\sum_{\substack{a=1\\(a,q)=1}}^{q}S_1(q,a)^2S_2(q,a).$$
\noindent $S(q)$ is a multiplicative function of $q$.
In fact, if $(q_1,q_2)=(a,q_1q_2)=1$ and we choose $a_1,a_2$ so that $a_2q_1+a_1q_2=1$
then for $i=1,2
$\begin{equation} \label{e:five2}
S_i(q_1q_2,a)=S_i(q_1q_2,a_1q_2+a_2q_1)=S_i(q_1,a_1)S_i(q_2,a_2).
\end{equation}
This follows from the fact that as $r_1$ and $r_2$ run through
a complete (or reduced) set of residues modulo $q_1$ and $q_2$ respectively, $r_2q_1+r_1q_2$ runs through
a complete (or reduced) set of residues modulo $q_1q_2$.
Thus it suffices to study $S_i(q,a)$ when $q$ is a prime power and $(a,q)=1$. For a prime power $p^l$ and $p\nmid a$ we have
\begin{equation} \label{e:five3}
S_1(p^l,a)=\sum_{u=1}^{p^l}N_{d-2}(p,l,u)\left(\sum_{y=1}^{p^l}e\left(\frac{auy^2}{p^l}\right)\sum_{z=1}^{p^l}e\left(\frac{auz^2}{p^l}\right)\right).
\end{equation}
where
\begin{equation*}
N_{d-2}(p,l,u)=\#\{(x_1, \ldots, x_{d-2}): 1\leq x_1, \ldots, x_{d-2} \leq p^{l}, f_{d-2}(x_{1}, \ldots, x_{d-2})\equiv u \,(\mbox{mod }p^l)\}.
\end{equation*}
For a fixed $u$ with $p^j\|u$, writing $u= p^ju_1$ and dividing the sum over $y$ (or $z$) into $p^j$ subsums by grouping $p^{l-j}$ consecutive terms
together we get
\begin{equation*}
\sum_{y=1}^{p^l}e\left(\frac{auy^2}{p^l}\right)
=\sum_{z=1}^{p^l}e\left(\frac{auz^2}{p^l}\right)
=\sum_{v_1=1}^{p^j}\sum_{v_2=1}^{p^{l-j}}
e\left(\frac{au_1(p^{l-j}v_1+v_2)^2}{p^{l-j}}\right)=p^j\sum_{v_2=1}^{p^{l-j}}
e\left(\frac{au_1v_2^2}{p^{l-j}}\right).
\end{equation*}
The sum over $v_2$ is a quadratic Gauss sum modulo $p^{l-j}$ and its value is well known (e.g., see \cite[Chapter 9]{MontgomeryVaughan1}). We find that
\begin{equation*}
\sum_{y=1}^{p^l}e\left(\frac{auy^2}{p^l}\right)\sum_{z=1}^{p^l}e\left(\frac{auz^2}{p^l}\right)=\left\{\begin{array}{ll}
p^{2l} & \mbox{if } j=l\\
p^{l+j} & \mbox{if } p \equiv 1 \, (\mbox{mod }4)\\
(-1)^{l+j}p^{l+j} & \mbox{if } p \equiv 3 \, (\mbox{mod }4)\\
0 & \mbox{if } p=2,\,  j=l-1\\
2\cdot \sqrt{-1}\cdot 2^{l+j} & \mbox{if } p=2,\, j\leq l-2
\end{array}
=G(p,l,j)
\right.,
\end{equation*}
say, depends on $p, l$ and $j$ but not on $a$ or $u_1$. Thus, from (\ref{e:five3}),
\begin{equation} \label{e:five4}
S_1(p^l,a)=p^{2l}M_{d-2}(p,l,l)+\sum_{j=0}^{l-1}M^{*}_{d-2}(p,l,j)G(p,l,j)
\end{equation}
where, for $d\geq 3$, $l\geq 1$, $0\leq j \leq l,$
\begin{equation*}
M_{d-2}(p,l,j):=\#\{(x_1, \ldots, x_{d-2}): 1\leq x_1, \ldots, x_{d-2} \leq p^{l},\, p^j|f_{d-2}(x_{1}, \ldots, x_{d-2})\}.
\end{equation*}
and, for $d\geq 3$, $l\geq 1$, $0\leq j \leq l-1$,
\begin{eqnarray*}
M^{*}_{d-2}(p,l,j)&:=&\#\{(x_1, \ldots, x_{d-2}): 1\leq x_1, \ldots, x_{d-2} \leq p^{l},\, p^j\|f_{d-2}(x_{1}, \ldots, x_{d-2})\}\\
&=&M_{d-2}(p,l,j)-M_{d-2}(p,l,j+1).
\end{eqnarray*}
The equation (\ref{e:five4}) also holds for $d=1$ and $d=2$ if we define
\begin{equation*}
M_{-1}(p,l,j)=0 \mbox{ for } 0\leq j\leq l, \, M_{-1}^{*}(p,l,j)=0 \mbox{ for } 0\leq j\leq l-1,
\end{equation*}
\begin{equation*}
f_0=1,\, M_0(p,l,j)=\left\{
\begin{array}{ll}
1 \mbox{ if } j=0\\
0 \mbox{ if } 1\leq j\leq l
\end{array} \right.
\mbox{ and }
M_0^{*}(p,l,j)=\left\{
\begin{array}{ll}
1 \mbox{ if } j=0\\
0 \mbox{ if } 1\leq j\leq l-1
\end{array} \right..
\end{equation*}
It is clear that
\begin{equation*}
M_1(p,l,j)= p^{l-j} \mbox{ for } 0\leq j\leq l
\mbox{ and }
M_1^{*}(p,l,j)=p^{l-j}(1-p^{-1}) \mbox{ for } 0\leq j\leq l-1.
\end{equation*}
The treatment of the singular series is trivial in cases $d=1, 2, 3$ now; we dispose them off.
When $d=1$ we have
\begin{equation} \label{e:fived1}
S_1(p^l,a)=0;\mbox{ } S_1(q,a)=S_2(q,a)=S(q)= 1 \mbox{ if } q=1, \, 0 \mbox{ if } q> 1;\mbox{ } {\mathfrak S}=1.
\end{equation}
When $d=2$ we have
\begin{equation} \label{e:fived2}
S_1(p^l,a)=G(p,l,0);\mbox{ } |S_1(q,a)|\leq 2q;\mbox{ }|S_2(q,a)|\leq \sqrt{2q};\mbox{ } |S(q)|\leq 4\sqrt{2}q^{-\frac{3}{2}}.
\end{equation}
When $d=3$ we have
\begin{equation} \label{e:fived3}
S_1(p^l,a)|< (l+1)p^{2l};\mbox{ } |S_1(q,a)|< q^2d(q);\mbox{ }|S_2(q,a)|\ll q^{2/3};\mbox{ } |S(q)|\ll q^{-\frac{4}{3}+\varepsilon}.
\end{equation}
\indent To deal with the cases $d\geq 4$, we first calculate $M_2(p,l,j)=\#\{(w,x): 1\leq w, x \leq p^{l},\, p^j|w^2+x^2\}$ for $1\leq j \leq l$. We do so by dividing the set $\{(w,x): 1\leq w, x \leq p^{l},\, p^j|w^2+x^2\}$ according to the highest power of $p$ dividing $w$. Thus, for $1\leq j \leq l$,
\begin{eqnarray*}
M_2(p,l,j)&=&\#\{(w,x): 1\leq w, x \leq p^{l},\,p^{\lfloor \frac{j+1}{2} \rfloor}|w,\, p^j|w^2+x^2\}\\
&&+\sum_{m=0}^{\lfloor \frac{j-1}{2} \rfloor}
\#\{(w,x): 1\leq w, x \leq p^{l},\,p^m\|w,\, p^j|w^2+x^2\}\\
&=&p^{2l-2\lfloor \frac{j+1}{2} \rfloor}+\sum_{m=0}^{\lfloor \frac{j-1}{2} \rfloor}
\#\{(w',x'): 1\leq w', x' \leq p^{l-m},\,p\nmid w',\, p^{j-2m}|w'^2+x'^2\}.
\end{eqnarray*}
Since $-1$ has two square roots modulo $p^{j-2m}$ if $p\equiv 1 \pmod{4}$, one square root modulo $p^{j-2m}$ if $p^{j-2m}=2$, and no square root in other cases, the $m^{th}$ term in the sum equals $p^{l-m}(1-\frac{1}{p})\cdot 2\cdot p^{(l-m)-(j-2m)}=2p^{2l-j}(1-\frac{1}{p})$  if $p\equiv 1 \pmod{4}$, $2^{2l-j-1}$ if $p^{j-2m}=2$ and
zero otherwise. Thus, for $1\leq j \leq l$,
$$M_2(p,l,j)=
\left\{\begin{array}{ll}
p^{2l-2\lfloor \frac{j+1}{2} \rfloor}+2 \lfloor \frac{j+1}{2} \rfloor p^{2l-j}(1-\frac{1}{p}) & \mbox{if } p \equiv 1 \, (\mbox{mod }4)\\
p^{2l-2\lfloor \frac{j+1}{2} \rfloor} & \mbox{if } p \equiv 3 \, (\mbox{mod }4)\\
2^{2l-j} & \mbox{if } p=2
\end{array}
\right..$$
Hence, for $0\leq j\leq l-1$
$$M^{*}_2(p,l,j)=
\left\{\begin{array}{ll}
(j+1)p^{2l-j}(1-\frac{1}{p})^2 & \mbox{if } p \equiv 1 \, (\mbox{mod }4)\\
p^{2l-j}(1-\frac{1}{p^2}) & \mbox{if } p \equiv 3 \, (\mbox{mod }4) \mbox{ and $j$ is even} \\
0 & \mbox{if } p \equiv 3 \, (\mbox{mod }4) \mbox{ and $j$ is odd}\\
2^{2l-j-1} & \mbox{if } p=2
\end{array}
\right..$$
Thus for $0\leq j \leq l-1$ we have $M^{*}_2(2,l,j)=2^{2l-1-j}$, $M^{*}_2(p,l,j)<p^{2l-j}$ if $p \equiv 3(\mbox{mod }4)$, and $M^{*}_2(p,l,j)<(j+1)p^{2l-j}$ for all $p$. Also, $M_2(p,l,l)\leq p^{l}$ if $p=2$ or $p \equiv 3(\mbox{mod }4)$, and $M_2(p,l,l)<(l+1)p^{l}$ for all $p$. To estimate $S_1(q, a)$ we consider the cases of even $d$ and odd $d$ separately.\\
\indent Now suppose that $d$ is even and $d\geq 4$, say $d=2k$ and $k\geq 2$. Then, for $0\leq j \leq l-1$,
\begin{equation*}
M^{*}_{2k-2}(p,l,j)=\sum_{\substack{0\leq j_i\leq l-1\\ j_1+\ldots +j_{k-1}=j}} \prod_{i=1}^{k-1}M^{*}_2(p,l,j_i)
\end{equation*}
Thus
\begin{eqnarray*}
M^{*}_{2k-2}(2,l,j) &=& \sum_{\substack{0\leq j_i\leq l-1\\ j_1+\ldots +j_{k-1}=j}} \prod_{i=1}^{k-1}2^{2l-1-j_i}\\
&=& \sum_{\substack{0\leq j_i\leq l-1\\ j_1+\ldots +j_{k-1}=j}} 2^{(k-1)(2l-1)-j}\\
& \leq & l^{k-1}2^{(k-1)(2l-1)-j} \leq l^{k-1}2^{2kl-2l-1-j}
.\end{eqnarray*}
If $p \equiv 3(\mbox{mod }4)$, then
\begin{eqnarray*}
M^{*}_{2k-2}(p,l,j)&< &\sum_{\substack{0\leq j_i\leq l-1\\ j_1+\ldots +j_{k-1}=j}} \prod_{i=1}^{k-1}p^{2l-j_i}\\
&=& \sum_{\substack{0\leq j_i\leq l-1\\ j_1+\ldots +j_{k-1}=j}} p^{2(k-1)l-j}\\
& \leq & l^{k-1}p^{2(k-1)l-j}=l^{k-1}p^{2kl-2l-j
.\end{eqnarray*}
If $p \equiv 1(\mbox{mod }4)$, then
\begin{eqnarray*}
M^{*}_{2k-2}(p,l,j)&<&\sum_{\substack{0\leq j_i\leq l-1\\ j_1+\ldots +j_{k-1}=j}} \prod_{i=1}^{k-1}(j_i+1)p^{2l-j_i}\\
&=& p^{2(k-1)l-j}\sum_{\substack{0\leq j_i\leq l-1\\ j_1+\ldots +j_{k-1}=j}} \prod_{i=1}^{k-1}(j_i+1)\\
& < & l^{k-1}p^{2(k-1)l-j}\sum_{\substack{0\leq j_i\leq l-1\\ j_1+\ldots +j_{k-1}=j}}1\\
& < & l^{k-1}l^{k-1}p^{2(k-1)l-j}=l^{2k-2}p^{2kl-2l-j
.\end{eqnarray*}
Dividing the $(2k-2)$-tuples $(x_1, \ldots, x_{2k-2})$ with $p^l|f_{2k-2}(x_1, \ldots, x_{2k-2})$ into the ones with one of the factors $x_{2i-1}^2+x_{2i}^2$ divisible by $p^l$ and the ones with none of the factors $x_{2i-1}^2+x_{2i}^2$ divisible by $p^l$ we see that for any prime $p$
\begin{eqnarray*}
M_{2k-2}(p,l,l)&\leq &\binom{k-1}{1} M_2(p,l,l)\cdot (p^{2l})^{k-2}+\sum_{\substack{0\leq j_i\leq l-1\\ j_1+\ldots +j_{k-1}\geq l}} \prod_{i=1}^{k-1}M^{*}_2(p,l,j_i)\\
& < &(k-1)(l+1)p^l\cdot p^{2kl-4l}+l^{2k-2}p^{2kl-2l-l}\\
& < &(l+1)^{2k-2}p^{2kl-3l}
.\end{eqnarray*}
(The last inequality follows since $(k-1)(l+1)+l^{2k-2}\leq (k-1)(2l)+l^{2k-2}=\binom{2k-2}{1}\cdot l+l^{2k-2}<(l+1)^{2k-2}.$)
Thus from (\ref{e:five4}),
\begin{equation*}
|S_1(p^l,a)|<(l+1)(l+1)^{2k-2}p^{2kl-l}=(p^l)^{2k-1}d(p^l)^{2k-1}.
\end{equation*}
Thus from (\ref{e:five2}) for even $d\geq 4$ and $(a,q)=1,$
\begin{equation} \label{e:five6}
|S_1(q,a)|\leq q^{2k-1}d(q)^{2k-1} \ll q^{2k-1+\varepsilon}=q^{d-1+\varepsilon}.
\end{equation}
\indent Now suppose that $d$ is odd and $d\geq 5$, say $d=2k+1$ and $k\geq 2$. Note that there is one value of $x_1$ modulo $p^l$ that is divisible by $p^l$ while for $0\leq j \leq l-1$ there are $p^{l-j}(1-p^{-1})$ values of $x_1$ modulo $p^l$ that are exactly divisible by $p^j$. Hence, for $0\leq j \leq l-1$,
\begin{equation*}
M^{*}_{2k-1}(p,l,j)=\sum_{\substack{0\leq j_i\leq l-1\\ j_0+\ldots +j_{k-1}=j}} p^{l-j_{0}}(1-p^{-1})\prod_{i=1}^{k-1}M^{*}_2(p,l,j_i).
\end{equation*}
Thus
\begin{eqnarray*}
M^{*}_{2k-1}(2,l,j) &=& \sum_{\substack{0\leq j_i\leq l-1\\ j_0+\ldots +j_{k-1}=j}} 2^{l-j_{0}}(1-2^{-1})\prod_{i=1}^{k-1}2^{2l-1-j_i}\\
&=& \sum_{\substack{0\leq j_i\leq l-1\\ j_0+\ldots +j_{k-1}=j}} \frac{1}{2}\cdot 2^{l+(k-1)(2l-1)-j}\\
& \leq & l^{k}2^{l+2kl-2l-(k-1)-j-1} \leq l^{k}2^{2kl-l-j-2}
.\end{eqnarray*}
If $p \equiv 3(\mbox{mod }4)$, then
\begin{eqnarray*}
M^{*}_{2k-1}(p,l,j)&<&\sum_{\substack{0\leq j_i\leq l-1\\ j_0+\ldots +j_{k-1}=j}} p^{l-j_{0}}(1-p^{-1})\prod_{i=1}^{k-1}p^{2l-j_i}\\
&=&\sum_{\substack{0\leq j_i\leq l-1\\ j_0+\ldots +j_{k-1}=j}} (1-p^{-1})p^{(2k-1)l-j}\\
&<&l^{k}p^{(2k-1)l-j}=l^{k}p^{2kl-l-j
.\end{eqnarray*}
If $p \equiv 1(\mbox{mod }4)$, then
\begin{eqnarray*}
M^{*}_{2k-1}(p,l,j)& < &\sum_{\substack{0\leq j_i\leq l-1\\ j_0+\ldots +j_{k-1}=j}} p^{l-j_{0}}(1-p^{-1})\prod_{i=1}^{k-1}(j_i+1)p^{2l-j_i}\\
&<&(1-p^{-1})p^{(2k-1)l-j}\sum_{\substack{0\leq j_i\leq l-1\\ j_0+\ldots +j_{k-1}=j}} \prod_{i=1}^{k-1}(j_i+1)\\
&<&l^{k-1}p^{(2k-1)l-j}\sum_{\substack{0\leq j_i\leq l-1\\ j_0+\ldots +j_{k-1}=j}}1\\ &<&l^{k}l^{k-1}p^{(2k-1)l-j}=l^{2k-1}p^{2kl-l-j
.\end{eqnarray*}
Dividing the $(2k-1)$-tuples $(x_1, \ldots, x_{2k-1})$ with $p^l|f_{2k-2}(x_1, \ldots, x_{2k-1})$ into the ones with $x_1$ divisible by $p^l$, the ones with one of the factors $x_{2i}^2+x_{2i+1}^2$ divisible by $p^l$ and the ones with none of the factors of $f_{2k-1}$ divisible by $p^l$, we see that for any prime $p$
\begin{eqnarray*}
M_{2k-1}(p,l,l)&\leq & 1\cdot p^{(2k-2)l}+\binom{k-1}{1} M_2(p,l,l)\cdot p^{(2k-3)l}\\
& &+\sum_{\substack{0\leq j_i\leq l-1\\ j_0+\ldots +j_{k-1}\geq l}} p^{l-j_{0}}(1-p^{-1})\prod_{i=1}^{k-1}M^{*}_2(p,l,j_i)\\
&<&p^{2kl-2l}+(k-1)(l+1)\cdot p^{2kl-2l}+l^{2k-1}p^{2kl-l-l}\\
&<& (l+1)^{2k-1}p^{2kl-2l}
.\end{eqnarray*}
(The last inequality follows since $1+(k-1)(l+1)+l^{2k-2}\leq 1+(k-1)(2l)+l^{2k-2}<1+\binom{2k-1}{1}\cdot l+l^{2k-1}<(l+1)^{2k-1}.$)
Thus from (\ref{e:five4}),
\begin{equation*}
|S_1(p^l,a)|<(l+1)(l+1)^{2k-1}p^{2kl}=(p^l)^{2k}d(p^l)^{2k}.
\end{equation*}
Thus from (\ref{e:five2}) for odd $d\geq 5$ and $(a,q)=1,$
\begin{equation} \label{e:five7}
|S_1(q,a)|\leq q^{2k}d(q)^{2k} \ll q^{2k+\varepsilon}=q^{d-1+\varepsilon}.
\end{equation}\\
\indent It is well known from the study of Waring's problem that for $(a,q)=1$ (See \cite{DBook} or \cite{VaughanBook}),
\begin{equation*}
|S_2(q,a)|\ll q^{1-\frac{1}{d}}.
\end{equation*}
Thus, from (\ref{e:fived1})-(\ref{e:five7}) and (\ref{e:five1}),
\begin{equation}  \label{e:five5}
|S(q)|\ll q^{-1-\frac{1}{d}+\varepsilon}.
\end{equation}
\begin{lem} \label{l:five1}
The series \begin{equation*}
{\mathfrak{S}}=\sum_{q=1}^{\infty}S(q)
\end{equation*} converges
absolutely, ${\mathfrak{S}}>0$, and for any $Q\geq 1$ we have
\begin{equation*}
{\mathfrak{S}}(Q)= {\mathfrak{S}}+O(Q^{-\frac{1}{d}+\varepsilon}).
\end{equation*}
\end{lem}
\begin{proof} The convergence at the asserted rate follows from (\ref{e:five5}). The argument in \cite[Chapter 17]{DBook} applies to forms of any degree and shows that the positivity of ${\mathfrak{S}}$ would follow from the existence of a non-singular zero of the form $f.$ The point $(x_1, \ldots, x_{2d+1})$ with $x_1=x_3=\ldots =x_{2 \lfloor (d-1)/2 \rfloor+1}=x_{2d+1}=1,$ and the rest of the $x_i$ equal to zero is a non-singular zero of $f$.
\end{proof}
\Section{Completion of the Proof of Theorem 1}
\label{completion}
\noindent From (\ref{e:two2}) and lemmas \ref{l:two2}, \ref{l:three2}, \ref{l:four1}, \ref{l:five1} we get
\begin{eqnarray*}
R(0;P)&=&P^{d+1}(\mathfrak{S}+O(P^{-\frac{\delta}{d}+\varepsilon}))(J+O(P^{-\frac{d+1}{d}\delta}))+O(P^{d+5\delta})+O(P^{d+1+\varepsilon-\frac{\delta}{2^{d-1}}})\\
&=&P^{d+1}\mathfrak{S}J+O(P^{d+1+\varepsilon-\frac{\delta}{2^{d-1}}}+P^{d+5\delta}).
\end{eqnarray*}
Taking $\delta=\frac{2^{d-1}}{1+5\cdot 2^{d-1}}$ completes the proof of Theorem 1.\\\\
\begin{small} {\bf Acknowledgments.} I would like to thank Prof. Robert C. Vaughan for suggesting me the case $d=4$ of this problem and Sean Prendiville for asking me if I could do something similar for all $d$.\end{small}
\\\\

redpajama_set_name: RedPajamaArXiv


text:
Turkey Region Maps
Dalaman Map
" Dalaman amazes everyone with breathtaking views ! "
Immaculate coast , attracts tourists with historic beauty …
Dalaman from October to April is the best time to wearing … But thanks to the spa facilities in the region a preferred avenue Dalaman in winter …
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There are many historical ruins around Dalaman . Kapıdağı Lidai , Lissay , Cry and Cleopatra Baths , cheers at Kalyn and tombs, at Gökdağ Lycian rock caves , Kayadibi the Lycian rock tombs must-see among the ruins … Kavacık and Bozbel Hippokome the ancient city 's .
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Near about 50 DM per person prices …
Gocek, Fethiye, Marmaris , Koycegiz .
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Safranbolu Map
General Information "World heritage town sırtlayan " Name in this region, the saffron grown from plants Safranbolu , a UNESCO " World Heritage " list located...
Pamukkale Map
General Information " The unique nature , full of health with the warm waters of health tourism center since Roman times ... " Naturally occurring white...
Olympos Map
General Information " Bellerophon , Pegasos, Zeus and remote Lycian : Olympos " Finike road and the road extending from the arch at the bottom of...
Christopher February 4, 2021 At 5:29 pm
Die Orte, die wir auf unserer vorherigen Reise gesehen haben, sollten unbedingt besucht werden, Sie haben unsere Erinnerungen aufgefrischt.
Dawson February 5, 2021 At 5:51 am
Hello, Dalaman Map nice sharing.
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Je me demande pourquoi je suis venu en Turquie jusqu'à présent. Je viendrai à la première occasion.
Deborah February 9, 2021 At 10:00 am
Foster February 11, 2021 At 12:37 pm
Michelle February 13, 2021 At 1:15 pm
There are dream-like landscapes in Turkey.
Clark February 13, 2021 At 8:38 pm
Margaret February 16, 2021 At 6:34 am
Baker February 18, 2021 At 3:04 am
I will come at the first opportunity.
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Es gibt einige wirklich coole Orte.
Jennifer February 18, 2021 At 8:04 am
What are the places to visit in Karabük?
Margaret February 18, 2021 At 11:54 am
Grazie per aver fornito così tante informazioni sulla Turchia, condivido.
Laura February 18, 2021 At 12:01 pm
On the first occasion we planned a trip to Turkey.
Baker February 21, 2021 At 7:42 pm
David February 26, 2021 At 4:32 pm
Baker March 6, 2021 At 3:15 pm
Edward March 18, 2021 At 10:32 pm
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Je pense qu'il y a beaucoup de beaux endroits en Turquie.
Baker May 20, 2021 At 6:14 pm
Edward May 24, 2021 At 10:39 pm
Es gibt viele schöne Resorts in der Türkei.
William May 27, 2021 At 5:24 am
In der Türkei gibt es traumhafte Landschaften.
Jennifer July 27, 2021 At 7:21 pm
Ich bin mit meiner Familie gekommen, es hat uns sehr gut gefallen, wir werden wiederkommen.
Elizabeth July 27, 2021 At 7:24 pm
Sì sono d'accordo, ci sono tanti posti bellissimi.
Anderson July 27, 2021 At 7:26 pm
Deborah July 27, 2021 At 8:13 pm
Carter July 30, 2021 At 7:46 am
Bishop September 11, 2021 At 8:10 pm
Laura September 14, 2021 At 9:06 pm
Michael September 28, 2021 At 7:13 pm
Мы обязательно приедем в этом году с друзьями.
Carol October 2, 2021 At 10:50 am
Clark October 26, 2021 At 11:14 pm
Mary October 26, 2021 At 11:14 pm
Christopher October 28, 2021 At 9:09 am
Margaret November 6, 2021 At 6:58 am
Bishop November 6, 2021 At 7:38 am
Mary January 22, 2022 At 8:21 am
Helen January 23, 2022 At 7:31 pm
Что нужно сделать, чтобы провести отпуск в Турции? Я зашел на ваш сайт и получаю много информации, когда спрашиваю.
Campbell January 31, 2022 At 8:52 pm
Edward February 1, 2022 At 4:11 pm
Mark February 2, 2022 At 1:45 pm
I think I should go to more places.
Sarah February 9, 2022 At 6:44 am
Ich frage mich, warum ich bis jetzt in die Türkei gekommen bin. Ich werde bei der ersten Gelegenheit kommen.
Helen February 9, 2022 At 12:49 pm
Foster February 16, 2022 At 3:43 am
Michelle February 24, 2022 At 12:25 pm
Hi, thank you Dalaman Map.
Elizabeth February 27, 2022 At 1:37 pm
J'ai visité de nombreux endroits et j'ai beaucoup aimé.
Edward February 28, 2022 At 8:03 pm
Alla Turchia piace molto il sogno di aprire una porta.
Sarah February 28, 2022 At 10:18 pm
Brown March 1, 2022 At 1:55 am
Christopher March 1, 2022 At 2:26 pm
Я спрашиваю, почему я приехал в Турцию до сих пор. Приду при первой же возможности.
Carol March 7, 2022 At 8:50 pm
There are many beautiful landscapes in Turkey.
Anthony March 16, 2022 At 2:13 am
Christopher March 16, 2022 At 4:47 pm
Black April 1, 2022 At 1:16 pm
Carol April 2, 2022 At 12:58 pm
Sarah April 3, 2022 At 8:51 pm
Kimberly April 4, 2022 At 12:35 am
Kenneth April 5, 2022 At 3:41 pm
Michelle April 18, 2022 At 1:34 am
Maria May 14, 2022 At 11:00 am
Laura May 14, 2022 At 8:23 pm
Nous sommes allés à Antalya en Turquie et nous avons beaucoup aimé.
Mark May 24, 2022 At 2:33 pm
Linda May 26, 2022 At 1:13 pm
I just want to Turkey's holiday resort.
Joseph June 1, 2022 At 12:25 am
Brian June 4, 2022 At 6:46 am
Вы можете экономно отдохнуть в Турции.
Patricia June 5, 2022 At 10:02 pm
What should be done to make your holiday in Turkey I came to your site and I get a lot of information when asking.
John June 13, 2022 At 12:28 am
Linda June 16, 2022 At 4:21 am
Was sollte getan werden, um Ihren Urlaub in der Türkei zu gestalten? Ich bin auf Ihre Website gekommen und erhalte viele Informationen, wenn ich Sie frage.
John July 2, 2022 At 6:35 pm
Joseph July 3, 2022 At 5:25 am
Мы ездили в Измир в Турции, и нам очень понравилось.
Laura July 4, 2022 At 7:07 pm
Betty July 6, 2022 At 8:26 am
Clark July 9, 2022 At 1:42 am
Margaret July 9, 2022 At 1:43 am
Robert July 11, 2022 At 5:00 pm
Mary July 15, 2022 At 5:22 am
Maria July 23, 2022 At 7:01 pm
Bishop July 29, 2022 At 12:56 pm
Nancy July 30, 2022 At 8:46 am
What are the places to visit in Artvin?
Burns August 9, 2022 At 7:50 am
What are the places to visit in Zonguldak?
Baker August 14, 2022 At 11:19 pm
Ruth August 20, 2022 At 11:17 pm
В Турции много красивых курортов.
Joseph August 29, 2022 At 9:20 pm
There are some really cool places Dalaman Map.
David September 1, 2022 At 6:20 pm
David September 3, 2022 At 10:58 pm
Kimberly September 4, 2022 At 5:57 pm
Bell September 6, 2022 At 1:35 pm
Lina September 13, 2022 At 9:48 am
Daniel September 15, 2022 At 2:26 pm
Mary September 27, 2022 At 7:34 am
Hallo, schönes Teilen.
Carol October 4, 2022 At 9:59 am
Bilecik in Turkey, I like the city very much.
Brown October 4, 2022 At 6:47 pm
Helen October 5, 2022 At 5:08 am
Anthony October 5, 2022 At 12:56 pm
Baker October 5, 2022 At 2:08 pm
Christopher October 6, 2022 At 6:59 pm
Ciao, bella condivisione.
Edward October 7, 2022 At 4:00 pm
Karen October 11, 2022 At 2:41 pm
Non sapevo che la Turchia fosse così bella.
Black October 17, 2022 At 3:30 am
Michael October 28, 2022 At 5:02 pm
Edward November 5, 2022 At 5:23 am
Christopher November 11, 2022 At 12:55 pm
Barnes November 16, 2022 At 3:22 am
Ruth November 16, 2022 At 5:01 am
Helen November 21, 2022 At 3:34 am
Jennifer November 23, 2022 At 3:52 pm
Ich liebe es, historische Orte zu besuchen.
Edward November 23, 2022 At 3:53 pm
Nancy November 23, 2022 At 3:53 pm
Je ne savais pas que la Turquie était si belle.
Sharon November 24, 2022 At 1:03 am
Я не знал, что Турция такая красивая.
Sharon November 29, 2022 At 4:48 pm
Brian December 7, 2022 At 2:05 am
Lisa December 8, 2022 At 11:21 pm
Maria December 10, 2022 At 9:54 am
Richard December 15, 2022 At 9:55 pm
Kevın December 17, 2022 At 6:39 am
Sharon January 1, 2023 At 10:17 pm
What are the places to visit in Balıkesir?
Sharon January 2, 2023 At 2:42 pm
John January 4, 2023 At 3:54 am
Ich möchte wirklich in die Türkei gehen.
Michelle January 7, 2023 At 7:50 pm
Lina January 13, 2023 At 1:50 pm
Paul January 19, 2023 At 7:30 am
Michelle January 29, 2023 At 5:38 pm
Laura January 30, 2023 At 1:22 am
Nous sommes allés à Izmir en Turquie et nous avons beaucoup aimé.

redpajama_set_name: RedPajamaCommonCrawl


Guangzhou Zoemir Trading Co., Ltd. • Chińscy eksporterzy • BigChina - import z Chin, import Chiny, biznes i Chiny, chinskie produkty.
Strona Główna > Chińscy eksporterzy > Guangzhou Zoemir Trading Co., Ltd.
ZOEMIR specialized manufacturer of cosmetic packaging in China. Our products are sold to more than 30 countries and regions around the world.
WE has integrated all the processes of production from raw material procurement, manufacturing of containers, to the logistic of final product delivery. We use modern equipment and our qualified technicians monitor every stage of production closely to ensure the satisfactory quality for customers.
We have delivered cosmetic packaging items including lipstick cases, compacts for the eye and face, containers for mascara, eyeliner and lip-gloss. We invite you to review our broad range of products. You will be able to find a unique combination, with in our existing range of standard items, which will give your product that 'custom-made look'.
We have a strong team and a strict QC management system.We have very professional design team and business team. They are responsible for the design of the pattern and shape on products, and give some useful suggestions on collocations; According to the different market demand, they will give the most scientific suggestions on products.
We are innovative, providing packaging designs and decorative options in a wide range of colors that help turn a standard item into a unique, elegant and personalized article. This is achieved by the use of metallizing, lacquering, soft touch effects, special coloring and finishing to name a few. We are also always open to implement new ideas and techniques for our customers.
Welcome to contact us, we are expecting your visit and inquiry!
Nazwa firmy: Guangzhou Zoemir Trading Co., Ltd.
Adres: A-812 Citycoin Building No.238 Xingangdong Rd.haizhu Dist.


Warrington Borough Council is a Unitary Authority in England. It contains approximately 92,898 residential properties subject to Council Tax, and approximately 6,514	commercial properties subject to Business Rates.
The most common Council Tax band in Warrington is A, and the median Council Tax band is B.
The average (mean) Council Tax value in Warrington is £1,297.58.
Amounts are the average across Warrington, and do not take acount of parish precepts where applicable.
Although Warrington Borough Council sends out the bills and collects the money, it is not responsible for the entire amount that you owe.


We have a state of art infrastructure facility that is spread across a wide area of 7500 square feet. This unit is equipped with sophisticated machinery and products are made as per the latest technologies. We have a team of experts to maintain this unit. These experts use premium quality raw materials to manufacture these products. They are made as per the specifications of the clients.
Our Area is spreaded across the area of 7500 sq. ft. We have well equipped with hi-tech machines. Our production capacity is 50 ton per month.
We have with us a capacious warehousing unit which effectively store our entire range of LDPE, HM/HDPE Rolls & Bags. & Well equipped with all the requisite facilities, this unit enable our products to be stored in an appropriate, safe and secure manner. In order to ensure easy storage and retrieval of the products, we have divided the warehousing unit into different departments as per the category name of the products. Further, the entire unit is taken care of by our skilled warehousing personnel who make sure that all the products are free from pests and rodents.
In addition to this, we also have an advanced packaging unit which is manned by our packaging experts. These professionals make use of high quality plastic and corrugated boxes to pack our entire range of products, thereby ensuring safe transit.


This past Tuesday, our family marked a small but significant milestone in its life. Mark and Micaela turned in to their teachers their choices for electives for this coming school year. It is the first time they have actually had a choice in which classes they are going to take. This year brings an important transition for them as they move from elementary school to Junior High.
We all experience key moments of transition in our lives: promotions from one part of our education to another, graduation, our first full-time job, marriage, the birth of a child, the day when our last child leaves home, the day that we retire. Each of these major milestones leads us to step back from our usual routine, the things we do day after day usually without thinking much about them, and to take a look at where we are in our lives as a whole. And sometimes, that process can enable us see important realities that we might have forgotten, or maybe missed altogether in our whirl of routine activities. They were there all along; we just did not notice them. Times of transition can help us to see anew.
Places can do that, too. During the eight or so years that Judy and I were "empty-nesters," we took the opportunity to visit and hike in a variety of National Parks, both in the U.S. and in Canada. My personal favorite was Waterton-Glacier International Peace Park, which spans the border between the two countries, part of it lying in Montana and part in Alberta.
One morning, we decided to spend the day hiking a trail that begins at Logan Pass; at more than 6600 feet, it is the highest point along Going-to-the-Sun Road. The trail is called the Highline Trail and runs along the west side of the Continental Divide. It was still fairly early when we started; and, a little way out, we found ourselves hiking through a cloud. We could see the trail clearly, along with our immediate surroundings, so we were not in any real danger; but we could not see very far to either side or up ahead.
Suddenly, we rounded a corner and stepped out of the cloud. The view was breathtaking. We could see literally for miles. There were the tops of the Rockies, maybe 50 yards at most to our right, and the valley far, far below to our left. Up ahead was a gorgeous Alpine meadow, filled with wildflowers. And along the trail, just beyond the point where we stood, two long-horned sheep took a quick glance at us and then scampered down the mountainside. This magnificent scenery had been there the whole time, but we had not been able to see it because of the cloud. Now, for far too brief a time, it was all ours to take in and to be embraced by its awesome beauty. We now could see what had been there all along, but what we had not seen before.
Today's gospel reading is about seeing what was there all along, but what had been hidden from the disciples' sight. Peter, James and John were among the first people who had begun to follow Jesus. They had presumably been with him for quite a while. Yet until this moment, they had not seen a central aspect of who he was, a central part of his identity.
Six days before the event narrated in our gospel reading, Jesus' disciples had been with him at Caesarea Philippi, and Peter had affirmed the first part of that identity, calling Jesus "the Messiah," "the Christ." Now today, here on the mountain, a voice from a cloud completed that description: "This is my Son, the Beloved." The reality of who Jesus was had been with them all along, but they just had not seen it. The way that St. Mark's paints their awakening is ironic. The disciples had been figuratively in a cloud ever since they had become his disciples, and had not been able to see who he really was. Now, enveloped in a real, physical cloud there on the mountain, they could at last see clearly.
What are those times and places in life in which we, at least for a moment, step out of a cloud and see clearly the presence and life-giving love of God in our lives? Sometimes, that experience comes during particular events, such as the transitions in life that I mentioned earlier. Sometimes, it happens because we are in a particular place. Celtic spirituality speaks of those times and places as the "thin places" of life: the places where the separation between God and our everyday lives is thin, where we are so close to God that we can catch at least a glimpse of that greater reality.
There are special times and places in every life that allow us to look through the clouds that surround us and find ourselves, at least briefly, within a far greater world: the world as God sees it. Unfortunately we do not always take time to notice them, to identify them and to acknowledge them. That is the first step.
But the key issue, when we do notice them, is what we choose to do with those experiences. Do we savor them for a moment and then go back to living life the way we did before? Or do we allow those experiences to transform our way of seeing on an ongoing basis? Do we allow them to make a permanent change in us?
Do you suppose that that was the way they thought about it: nobody out of the ordinary, only Jesus? Or could it be that, because they had had such an extraordinary experience there on the mountain, they had now come to a fuller realization of who Jesus was: a deeper understanding of the one who walked among them every day: the one who was both the Messiah and God's Son, the Beloved?
Where are the "thin places" in our lives: those times and places where we are so close to God that we can almost reach out and touch our Creator? Do we take time to notice them? And, if we do, do we allow them to be just a nice experience, relegated to our past? Or do we allow them to change us, to transform us, to make a lasting difference in our lives?
This morning, as we conclude this Epiphany season, we will once again be using, in today's Eucharistic Prayer, the Preface for the Epiphany. It is one that gives thanks to God "because in the mystery of the Word made flesh, you have caused a new light to shine in our hearts, to give the knowledge of your glory in the face of your Son Jesus Christ our Lord." May that light shine in our hearts and in our lives, helping us to see in new ways: helping us to see the God who is always close to us, helping us to catch sight of and work for the fullness of the reign of God which, in Jesus, has come near.


The pre-europeans were held this week in my home country just couple of hours drive from my home. It took a while to collect all the equipment I would need for my XC competition so it is needless to say that I have not missed this chance to take part in this pre-european, Hungarian Fai 2 open.
Many thanks to Flytec for equipping me with the Flytec Element so that I don't have my equipment as an excuse. Also thanks to my paragliding club Vertigo for the huge support financial and non financial. I could see my phone on the cockpit and seeing them chatting and following the live track was one of the best things that could happen to my motivation and determination.
First day was cancelled due to rain. Second day, there was a task but it got cancelled right away due to rain on the route.
Third day there was a 55 something km task, my first task ever.
I had problems with the start gate, apparently I set it to enter cylinder instead of exit. Like that was not enough I got to base with 10 minutes to go and then lost all the high. So when the gate was opened I had to climb and than to start the race.
But other than that everything went smoothly I managed to catch few people and got to see the goal smiley face on the new Flytec Element.
After all that self confidence I was ready to take task 2.
Task 2 (61 km) started a lot better than the previous day. I managed to climb with the first group, Enzos and Boomerangs all around me, It was one of the highlights of the comp for sure. However when the gate opened they still managed to escape me. I bombed out after 2 turn points or 44 km, did everything good I was in a group the guys managed to climb I didn't. I was low, the thermals were bent and inconsistent, more skill was required, still great fun and a big smile on my face.
Task 3 was one of my most serious fights ever. Reaching TP 1 was very difficult for me. I almost bombed out, had a low save, made many good and bad decisions about the line, had to push in strong wind to catch the turn point. After I got to it I had another low save but this time next turn point was downwind and I finally made base. From then on next turn points were easy but I couldn't catch anything and had to land 5 km to Goal. I went through all kinds of things on this flight it lasted 3 hours and 40 minutes, I was exhausted after it, a unforgettable experience indeed.
Huge thanks to Stamat who was always there helping me, It was a pleasure and a privilege to compete with him, Thank you for showing me how not to give up. Can't wait to compete with even more friends around me.
In this three days I was flying for 9h & 20 min and covered 160 km, I can only say that this things can get addictive.


As a business owner, and particularly as a female business owner with lots of insecurities and seemingly endless doubts I often think "what if"….
What if my website was like that one?
What if my business had more exposure?
What if my customers really saw how much I cared?
What if I had more likes?
What if I had more followers?
What if more people commented?
What if I had the skills?
What if I was smarter?
What if I was better at business?
It's exhausting to think of all the "what ifs". What if you used that same energy and invested it into moving forward.
Here's what I'm asking… in that moment where you feel yourself going into the "what if" space, stop and hold on for just 5 seconds more. Mel Robbins shared her insights on Goalcast about her 5-second-rule…. In the clip she sheds light on the difference in using the 5 seconds to change the way you think about life and how to step into your space and own it!
You can move from idea into action! Check the clip out below…. just brilliant!
It's time to stop focussing on the things that we can't control and time to focus on the things that we can and move forward, 5 seconds at a time.
5 – 4 – 3 – 2 – 1 … are you ready?


Minyan Dorshei Derekh in Philadelphia has taken an innovative approach to this Purim tradition by making communal mishlo'ach manot packages for each megillah reading attendee. Community members donate time, money and items to create the gift baskets and most of the money goes to tzedakah.
Rabbis Fred Dobb and Richard Hirsh look at the development of the philosophy and practice of kashrut from Biblical times to the present and what a Reconstructionist approach to kashrut can mean for us today. An episode of Heart, Mind and Spirit.
This is a transcript from edtalk, the listserv of Reconstructionist educators. In 2001, education intern Sarah Rubin asked for ideas about how to teach about the role of obligation in Judaism—particularly in the context of bar/bat mitzvah. What obligations do we have as individuals? What obligations does the Jewish community have to us?
This is a one-page guide that explains the structure, content and choreography of a Shabbat morning service. It is especially useful for a Bar/Bat Mitzvah or other special event when there will be many guests.
This policy describes the process of becoming a Bar/Bat Mitzvah and the requirements that must be fulfilled.
This lesson helps acquaint children and adults with the idea and practice of Shabbat.
Prayer for the State of Israel as it appears in the Kol Haneshama Siddur.


text:
Q: Format of .dex files for Android 2.1 (Eclair), i.e. API level 7 I'm writing an assembler for Android/Dalvik .dex files. The official Android docs seem to only describe the newest version of the .dex format (apparently 038 as of May 2019), with a few short notes describing differences back to version 035. A quick glimpse into Android sources seems to suggest, that version 035 of the format corresponds to Android API level 13, meaning Android 3.2.x (Honeycomb). I own a device with Android 2.1 (Eclair), which apparently means API level 7. I would like to be able to learn how to emit correct .dex and .apk files for that device, to hopefully add support for those to my assembler.
Where can I find information that could help me emit correct .dex files for Android 2.1 (Eclair), i.e. API level 7? I don't even know what's the .dex format version string for that API level!
For "bonus points", I would really love to find some good reference explaining the differences in .dex format between all API levels (at least back to API level 7). I would be also interested if there are any substantial differences in the format of the binary .xml and resource files (as emitted by the aapt tool) between Android API levels. I assume that there's a high chance the .apk format itself, as well as the signing algorithm, hopefully didn't change at all, given that it's the same format used to sign Java .jar files (i.e. a regular .zip with a few simple special files in the META-INF/ subdirectory). But the .dex format spec (including bytecodes) is probably the most important part, or at least a good starting point.

A: https://source.android.com/devices/tech/dalvik/dex-format has some quick blurbs on the differences between the versions.
Note: Support for version 037 of the format was added in the Android 7.0
release. Prior to version 037 most versions of Android have used version 035 of
the format. The only difference between versions 035 and 037 is the addition of
default methods and the adjustment of the invoke.

Note: Support for version 038 of the format was added in the Android 8.0
release. Version 038 added new bytecodes (invoke-polymorphic and invoke-
custom) and data for method handles.

Although, note that the newest version is 039, which doesn't seem to be mentioned in the doc. iirc, 039 added the invoke-custom/range, const-method-handle and const-method-type instructions.
Other sources of information include looking at the changelog of those documents:
https://android.googlesource.com/platform/docs/source.android.com/+log/refs/heads/master/en/devices/tech/dalvik/dex-format.html
https://android.googlesource.com/platform/docs/source.android.com/+log/refs/heads/master/en/devices/tech/dalvik/dalvik-bytecode.html
And prior to that, before the files were moved:
https://android.googlesource.com/platform/docs/source.android.com/+log/a3b748b40bab557fb47fe5a48a5bfb642837fb05/src/devices/tech/dalvik/dex-format.jd
https://android.googlesource.com/platform/docs/source.android.com/+log/a3b748b40bab557fb47fe5a48a5bfb642837fb05/src/devices/tech/dalvik/dalvik-bytecode.jd
Also, you should be able to glean some info from the smali source. It has the min/max api levels, where applicable, for every instruction.
For the instructions with the min/max set as an art version, you can use this mapping to map back to an api level.

redpajama_set_name: RedPajamaStackExchange


Luxembourg-Baku, 4 December 2017 – Silk Way West Airlines is the latest airline to go live with CHAMP's Weight & Balance load planning solution to serve their fleet of Boeing 747 freighters after receiving an Azerbaijan CAA approval. The choice of the Baku-based all-cargo airline to award the contract for Weight & Balance software to CHAMP Cargosystems re-affirms their strong relationship with CHAMP.
Serving more than 50 destinations worldwide, Silk Way West Airlines was searching for a credible IT solution that would allow their staff to perform their load planning faster, with greater accuracy, whilst optimizing the fuel burn of their aircraft.
In addition to Weight & Balance, Silk Way West Airlines are using CHAMP's Traxon Global Customs, Cargospot Airline and Cargospot Revenue applications.
CHAMP's Weight & Balance is currently the most advanced load planning software in the market, with functionality that encompasses an auto-load option with built-in logic to ensure effective off-loading on multiple flight-leg operations. The solution enables airlines to efficiently and safely plan all freighter aircraft loads, however complex, and helps load masters to keep an aircraft within permissible center of gravity limits throughout the flight.
The software not only ensures compliance with all operational and safety requirements, but also enables comprehensive audit trails and reporting, providing further business intelligence into the fleet's overall performance.


Glenn Foot clashes with Sam Omidi in his four-round points win.
Glenn Foot's British and Commonwealth title fight against Robbie Davies Jnr could take place on the undercard of another Lewis Ritson-headlined night in Newcastle.
The British Boxing Board of Control have ordered the duo to fight for the vacant British super-lightweight title before the end of October.
Phil Jeffries with Darren Surtees.
A statement on their website reads: "The stewards decided to put out the above contest to purse bids to be submitted in accordance with the terms and conditions attached to the board's head office by 12:00noon on Wednesday, 11th July 2018, the contest to take place by the end of October 2018."
And the Echo has learned that Foot's clash with Davies Jnr could take place as one of the chief supports on another massive night of boxing at Newcastle's Metro Radio Arena, with Forest Hall's British lightweight title holder Ritson, one of the most exciting talents on the UK boxing scene, topping the bill.
Foot, who won the Commonwealth strap after stopping Jason Easton in Glasgow in March, came through a four-rounder on Saturday night on the Ritson and Josh Kelly undercard.
His fight was meant to be a six-rounder against Manchester's Sam Omidi, but was cut to four, three minute rounds at the demand of TV coverage. The fight was shown live on Sky Sports.
Foot's promoter Phil Jeffries, said: "Glenn is up for the fight.
"He fell out of love with the sport a bit but he has that fire back again and he knows he can win this title.
"Wins can breed confidence and you could see from the way Glenn enjoyed himself up there he has the love back.
"Boxing is booming in the North East with the likes of Lewis Ritson. The sport is on the up in the region.
"Glenn could be on a November bill with Ritson headlining."
Meanwhile, Jeffries' event at Sunderland AFC's Stadium of Light on Sunday, July 1 is taking shape.
Hartlepool super lightweight Luke Cope is out again after a debut win in Darlington last month.
Redcar brothers Tom and Joe Hill, Trimdon's Terry Tyres, Lewis Ritson's training partner Terry Wilkinson, Birtley's Joe Laws, Gateshead's Chris Pilkington, 7-0 Darren Surtees, of County Durham, and middleweight Chris Atkinson have all been named on the card.
County Durham's Tommy Ward could yet be added to that stacked bill on Wearside.
Doors open at 1pm on the day.
Rnd 1: Foot all over Omidi like a rash. We know he loves a scrap and he's in the mood tonight. Not messing about.
Rnd 2: Foot using every trick in the book. A right good scrap. The #Sunderland man living every minute of it as he goes head-to-head with Omidi after the bell!
Rnd 3: Another in the bank for Foot. Good work out this.
Rnd 4: Some heavy right hands from Foot. Came for a fight tonight and to be fair Omidi obliged. Four rounds all to Glenn , you'd think. Onwards to the British title.
Glenn Foot 40 Sam Omidi 36: Comfortable night's work for the Marley Pots fighter. Now for Josh Kelly and his Commonwealth title bout with Kris George. It was all about Newcastle earlier now it's the Sunderland lads' time to shine!


text:
Q: C# sql instancing, is it possible to list all sql connections not disposed of within the context of an object instance? I'm getting some really daft errors relating to a block of code that i know for absolute certain cleans up after itself.
essentially i have an N-tier application framework that makes a call in to a "data provider" that in turn talks to a SQl database.
I want to say to the data provider "show me all connections that have been used and not yet disposed of" so i can if need be "force disposal" of any connections that should not be open.
If i can't do this in C# is there some alternative on the Sql server that lets me get some insight in to whats going on?
EDIT :
Okin this instance the code was being called from any location in an application based on a business object instance that calls a single location in an API layer.
The API layer exposes an interface that it calls methods on.
I was hoping to have either the object that implements the interface or the API layer look at the call stack and decide weather to wait, throw an exception or make the call (open a new connection).
The problem i found was that I had a business object like say "person" that had a property called "vehicle" and that was an object of type vehicle which contained a property called "owner" this of course referred back to the parent object.
Now because of bad coding on my part the system went in to a sort of loop creating nested instance after nested instance of the 2 objects.
So the fix was simple enough ... lazy load at at least 1 of the 2 levels or remove the circular reference altogether.
However I would still like to look at my call stack at runtime and ask the question ... is this going to cause a problem if i try to create an instance of this business object?
This then led me on to thinking about the caching (to be implemented in the API layer later) ... if this was in place that problem of too many connections goes away but is replaced with a much bigger "im gonna munch all your ram" problem.
So I got thinking ... why cant i analyse the call stack?
The answer ...
I shouldn't have to if i write good code ... but i still want to :)

A: Don't know about C# but in Sql Server you can use that queries for monitoring opened connections:
1.Get list of databases and number of connections
SELECT
    DB_NAME(dbid) as DBName,
    COUNT(dbid) as NumberOfConnections,
    loginame as LoginName
FROM
    sys.sysprocesses
WHERE
    dbid > 0
GROUP BY
    dbid, loginame

2.Get all connections for user - sp_who 'userName'
3.Get info for connection DBCC INPUTBUFFER(1)

A: Regardless of this being possible, you are trying to solve the problem in a wrong place - if you do not mind me saying this. In a multi-threaded environment, how would you know if a the work of the connection has finished and it can be safely closed?
Whoever owns a resource is responsible to release it. So I suggest focusing on releasing the resource in the class that owns it.
However, to answer your question, ADO.NET does not provide a central registry so that you could look at the open connections - also perhaps for a good reason since it could drive bad design. There could be some API that can be called at the connection pooling level but it could be database specific.
If you really have to do it, nothing stops you to manually register a connection with a central registry when it gets open and de-register when it is closed. This could be implemented as singleton pattern - but as I said, I advise against it.

A: Check for circular references to prevent crazy errors :)
Don't ask stupid questions that expect microsoft to allow for bad coding to be "checked at runtime".


In spite of many days of heavy smoke from the fires burning around us, the Landmark Dragons are moving forward faster than ever. They did stop for a day and help out with a service project. An ooportunity to donate they tops of pop cans to the Ronald McDonald House, cans were donated and the Dragons made quick work removing the tabs from the cans and smashing them. One of our own facultymembers' family will directly benefit from this as their family members recieve cancer treatment. What a great way to be a dragon?
The Dragons are quickly making cuttings of a variety of plants to supply the greenhouse with all kinds of vines, flowers, and grasses that will be used for next year's production of baskets, planters, etc. We sell to the public in the springtime. This is a great way to save on costs related to our greenhouse production. Also, it is a great opportunity to learn propagating skills.
Another project recently finished is the donation of seeds packet to the greenhouse. Vegetables, perrenials, and flower seeds of all kinds. There were separated by variety and cataloged for use immediately for spring planting. Many varities of perrentials were sown to winter over, making them reading for bloom next year.
October came so quickly at Landmark. Heavy rains did not stop the final harvest of all the planter boxes outside the greenhouse. Tomatoes, peppers, strawberries, all were gathered before the frost comes. Clean up and maintenance of the garden area is vital, preparing it once again for spring planting. What a great group of kids eager to pull up vines, clean-up, and drain water lines.
As a side note: Poinsettias are turning red and succulent dish gardens are selling nicely. Many more already for the upcoming holiday season. Great job Dragons!


text:
Meeting Name: County Council Agenda status: Approved
Meeting date/time: 5/22/2019 5:00 PM Minutes status: Draft
Meeting location: Regular Meeting, Fourth Floor, Gold Room
Meeting Items (25)
11090-19 1 Proclamation Proclamation declaring May 22, 2019 to be "Sean Rovers Day" in Allegheny County. Not available Not available
11091-19 1 Proclamation Proclamation declaring May 2019 to be Brain Tumor Awareness Month in Allegheny County. Not available Not available
11092-19 1 Proclamation Proclamation recognizing Director James P. Reardon upon the momentous occasion of his retirement from Allegheny County. Not available Not available
11093-19 1 Proclamation Proclamation recognizing all the vital services provided by Allegheny County Department of Public Works. Not available Not available
11094-19 1 Proclamation Proclamation declaring the week May 19 - 25, 2019 as National Public Works Week. Not available Not available
11095-19 1 Proclamation Certificates of Recognitions presented to the sign crafters with the Allegheny County Department of Public Works in recognition of being awarded the Exemplary Employee award. Not available Not available
11096-19 1 Proclamation Certificates of Achievement presented to the 2019 Pittsburgh's 50 Finest. Not available Not available
11097-19 1 Proclamation Certificate of Recognition presented to the Borough of Castle Shannon for its Community Day celebration on Sunday, May 26, 2019. Not available Not available
11098-19 1 Proclamation Certificate of Recognition presented to the City of Clairton for hosting its annual Memorial Day Celebration & Parade on Monday, May 27, 2019. Not available Not available
11099-19 1 Proclamation Proclamation congratulating Borough of Castle Shannon as it celebrates 100 years since its incorporation. Not available Not available
11100-19 1 Proclamation Certificate of Recognition presented to Mr. and Mrs. Ronald Jones upon the momentous occasion of their 66th wedding anniversary. Not available Not available
11101-19 1 Proclamation Proclamation congratulating Conco Services Corporation for being named the 2019 Mid-Atlantic Region and Western Pennsylvania Exporter of the Year by the U.S. Small Business Administration. Not available Not available
11102-19 1 Proclamation Certificate of Recognition presented to Mr. and Mrs. Ralph and Mary Jo LeDonne upon the momentous occasion of their 50th wedding anniversary. Not available Not available
11103-19 1 Proclamation Certificate of Recognition presented to Mrs. Stella Chiappino Radley upon the momentous occasion of her 90th birthday. Not available Not available
11104-19 1 Floor Motion Motion to approve the minutes of the April 9, 2019 regular meeting of Council. Not available Not available
11088-19 1 Appointment Approving the appointment of Heather Wechter to serve as a member of the Council of Friends organization (South Park) for a term to expire on May 23, 2021. Not available Not available
11089-19 1 Appointment Approving the reappointment of Julie Derence to serve as a member of the Community College of Allegheny County Board of Trustees for a term effective through December 31, 2024. Not available Not available
11017-19 1 Appointment Approving the appointment of Mayor Richard Lattanzi to serve as a member of the Community College of Allegheny County Board of Trustees for a term effective through December 31, 2021. Not available Not available
11019-19 1 Appointment Approving the appointment of Dannai Wilson, M.S., to serve as a member of the Children, Youth and Families Advisory Board for a term effective from January 1, 2019 through December 31, 2021. Not available Not available
11020-19 1 Appointment Approving the reappointment of Judith Thomas Horgan to serve as a member of the Children, Youth and Families Advisory Board for a term effective from January 1, 2019 through December 31, 2021. Not available Not available
11043-19 1 Appointment Approving the reappointment of Barbara Wolvovitz to serve as a member of the Mental Health/Intellectual Disability Advisory Board for a term effective through December 31, 2021. Not available Not available
11072-19 1 Resolution A Resolution of the County of Allegheny amending the Grants and Special Accounts Budget for 2019 (Submission #02-19). Not available Not available
11071-19 1 Ordinance An Ordinance of the County of Allegheny, Commonwealth of Pennsylvania, authorizing Manhunt Productions Inc. to use certain County-owned real and personal property in connection with filming a television series. Not available Not available
11105-19 1 Ordinance An Ordinance of the County of Allegheny, Commonwealth of Pennsylvania, authorizing Riddle Bomb Pictures LLC to use certain County-owned real and personal property in connection with filming a motion picture. Not available Not available
11106-19 1 Communication Communication summarizing approved Executive actions from April 1 through April 30, 2019. Not available Not available


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text:
Did AARP Sell Out Seniors?
By Barbara Marquand Insure.com
Three Republican congressmen are asking the IRS to investigate whether AARP, the powerful senior lobby and health care reform supporter, should be stripped of its tax-exempt status.
At issue is whether there's a conflict of interest between the group's mission to advocate forseniors and the money it makes from endorsing insurance products.
As a result of provisions in the health care reform law, AARP stands to make an additional $1 billion through royalties on insurance products branded with the organization's name over the next 10 years, according to a 29-page report released March 30 by Republican members of the U.S. House Ways and Means Committee.
"AARP enjoys a privileged tax-exempt status, but in many cases AARP resembles a for-profit entity," said Ways and Means Oversight Subcommittee Chairman Charles Boustany of Lousiana in a statement.Wally Herger of California, chairman of the Ways and Means Health Subcommittee, and Rep. Dave Reichert of Washington prepared the report.
AARP Counters Criticism
In an April 1 joint hearing of the Ways and Means Health and Oversight Subcommittees, AARP CEO Barry Rand said the organization was disappointed in the substance and title of the report, "Behind the Veil: The AARP America Doesn't Know."
"There is no veil," Rand told subcommittee members. "Quite frankly, we disagree with each of the conclusions drawn in this one-sided report."
Democrats said the report was politically motivated, noting that Republicans lauded AARP's endorsement of the Medicare Modernization Act of 2003, which created a private Medicare drug benefit.
"But now, since AARP worked to help enact health reform and will surely oppose Republican plans to convert Medicare to a voucher, privatize Social Security and block grant Medicaid, Republicans want to bring them down," said Health Subcommittee Ranking Member Pete Stark of California in a statement.
AARP, which had revenues of $1.4 billion in 2009, makes money from royalty payments for products it endorses, membership dues, publication advertising and grants, and uses the proceeds to support its lobbying efforts and programs, such as free tax counseling for seniors. The Republican report takes issue with the organization's reliance on royalty income from insurers, which it says comprised almost 46 percent of AARP's revenue in 2009, and its support of health care reform, which could add to profits for the group.
Insurance Sales Resulting from Health Care Reform
Among many provisions, health care reform will cut government subsidies of Medicare Advantage plans, which provide Medicare coverage for hospital and outpatient care and additional services. The Obama administration projects the subsidy cut will prompt seniors to buy Medigap policies, which are supplemental health insurance plans that pay for costs not covered by Medicare Parts A and B.
AARP earns the largest portion of royalty income from Medicare-related insurance plans offered by UnitedHealthcare, including Medicare Advantage and Medigap policies. But financial agreements are structured differently for each, the report says. AARP earns a flat royalty fee for endorsing the insurer's Medicare Advantage plan and a per capita fee for each AARP-branded Medigap policy sold by UnitedHealthcare.
When seniors migrate to Medigap plans, the report estimates AARP will earn $55 million to $166 million in 2014 alone.
But in a conference call with reporters, AARP officials said revenues don't drive the group's policy positions, and that it had not studied how health care reform would impact demand for Medicare Advantage and Medigap plans. AARP supported the end of subsidies for Medicare Advantage plans to protect the long-term solvency of Medicare, said David Certner, the group's legal counsel and legislative policy director.
About three-fourths of Medicare beneficiaries pay higher premiums in order to subsidize the one-quarter of beneficiaries enrolled in Medicare Advantage plans, he said.
IRS Investigation Possible
It's difficult to project how long an investigation would take if the IRS launches an inquiry.
"It's certainly not something that will happen in a week," says New York University School of Law professor Jill Manny, who runs the National Center on Philanthropy and the Law.
Should the IRS take action, it could pursue an examination, which would involve requesting and reviewing information from AARP, or a full-blown audit. Regardless, stringent confidentiality rules will prevent the IRS from revealing if it's conducting an investigation.
"The public will not know anything about the investigation until it is complete, unless AARP reveals information," Manny says.
In testimony before the House Ways and Means Health Subcommittee, tax law expert William "Bill" Josephson said he would need more time to study the complex legal and accounting issues in the report to express an opinion, but he thinks further investigation is warranted by the committee, the IRS and the Government Accountability Office.
The original article can be found at Insure.com:Did AARP sell out seniors?


Registration is limited to approved students in the Medical Front Office program. This class is included in the Medical Front Office Certificate Level I block schedule program (BLCK-2130-21351).
Orientation: August 31 at 4:00 p.m., Rm. A423. For more information email okerr@dcccd.edu or call 214-860-2280.
Email instructor okerr@dcccd.edu by the first class day for orientation information.


Two researchers who recently named the first new species of mushroom from the UC Berkeley campus in more than 30 years are emphasizing the need for continued green and open space on campus, as well as a full-fledged catalog of all North American mushroom species.
Else Vellinga and Nhu Nguyen were ecstatic last year when they found and described Helvella dryophila, a beautiful black "elfin saddle" associated with oaks on Observatory Hill, an open space area next to the C.V. Starr East Asian Library. The mushroom is edible, but doesn't taste very good and may be poisonous if not cooked properly. They published a paper about their finding in Mycologia.
The find illustrates the importance of maintaining and retaining open space and healthy vegetation on Cal's 1,232-acre urban campus, the researchers said. The discovery also demonstrates the need for a definitive catalog of mushrooms in North America, a longtime goal of Professor Tom Bruns of the Department of Plant & Microbial Biology at UC Berkeley.
Vellinga currently works as a Research Associate in the UC herbarium and in the Bruns Lab, and Nguyen was a graduate student in the Bruns Lab, before graduating and going to Professor Peter Kennedy's Lab at the University of Minnesota as a postdoctoral scholar.
Bruns is one of the world's foremost experts on fungal ecology. The research done in his lab regarding fungi that are symbiotic on tree roots is at the cutting-edge. For years, Bruns has been trying to get funding in order to put together a catalog of North America mushroom species; this project is tangential to his main work in fungal ecology, and is a necessary step for the field of fungal biology.
As of 2014, a total of 12 mushroom species have been described from campus. The other 11 species were described as new between 1907 and 1985.
The UC Berkeley campus serves as the "type locality" for these species, meaning that the type collection (representative specimen) for these species was made on campus, and that any other mushrooms found of the same species would be compared to this collection. "Type localities have elevated scientific value because it is where the representative specimen was found," said Nguyen.
"Many mushroom species in California, and across North America, are mistakenly known by European names, but with the advent of DNA sequencing and more precise identification, it was discovered that our native mushrooms are entirely different," Nguyen said.
"Once we figured out that the California elfin saddles were completely different from the European elfin saddles, we focused on the group and described two new species from California," Nguyen said.
The other species described from the same paper, Helvella vespertina, is a larger mushroom. Its type locality is Salt Point State Park, a productive and diverse mushroom habitat in Sonoma County. Both species can be found in Southern California to the Oregon border, and interestingly on our very campus.
"What was happening is that these mushrooms were parading around under another species' name," Vellinga said. "We can still find mushroom species that are new to science, right here on campus." What has enabled the team to identify the mushroom correctly is DNA sequence analysis, the current definitive measure for all species identification.
Bruns and Vellinga have recorded about 150 species of mushroom from the UC Berkeley campus (including the 12 species that have their type locality on campus). On the list are edibles such as chanterelles and porcini, and deadly species such as the death cap (Amanita phalloides). Researchers issue a strong note of caution when searching for mushrooms. In addition to the obvious – make sure they're not poisonous – also watch out for pesticide contamination when hunting for mushrooms in the cities and on lawns.
Researchers agree that along with identifying new species of mushrooms and continuing to hunt for them, a definitive catalog of all mushrooms in North America is crucial.
"We're desperately trying to catalog life on earth, and fungi are a huge part of that. Biological diversity is important to the future of the planet, especially in the face of changes in climate, invasive species and other threats. The richness that we have on the planet is important to identify and maintain, and knowing what exists now and has existed in the past gives scientists information to stabilize ecosystems," Nguyen said.
"Within the entire continent of North America there is not a single regional, state or local catalogue of the macrofungal species that presents a credible account of the species present and their distributions," Bruns mentioned.
It's not just a problem for fungal biologists or for North America, Bruns said.
"The problem is most acute for plant scientists who need to identify the mycorrhizal (symbiotic relationship) or pathogenic (disease-causing) fungal associates of their plants, and for fungal ecologists who need a global view of diversity," Bruns said.
With these three ingredients, researchers are ready to produce the said mycoflora. The first step towards this effort is the recently funded Advanced Digitalization of Biological Collections project that is digitizing the data for 1.3 million collection records of macrofungi housed in 35 institutions in the US including UC Berkeley.
The goal is to bring that "dark data" that resides in separate locations into one centralized database. However, the project does not fund the steps necessary to produce a mycoflora and Bruns has been working with other mycologists in North America to push for funding.
Funding is still needed because many species remain undescribed. Many described species need to have DNA sequences determined, an organized way to have the information accessible, and an efficient method to provide identification tool. Helvella dryophila is a good example to witness the importance of these described problems.
The researchers agree that if we can find a new species right in our own academic backyard there are many more that remain to be discovered. New collections will be needed for this reason and are also needed to provide better descriptions to obtain DNA sequences. The older specimens that are the basis of the digitalization project often lack good descriptions and photographs, and as they age it becomes more difficult to obtain DNA sequences from them.
Citizen scientists are an important part of the mycoflora project.
Debbie Viess, a trained biologist and "citizen scientist" said she "fell in love with fungi" and has 20 years of experience and training hunting for mushrooms.
"Many species are just waiting to be described or documented in some way," Viess said.
Cataloging mushrooms has some similarity to bird counts done by the Audubon Society in North America, moth surveys in the UK, and urban plant surveys in France. All of these efforts utilize large numbers of citizen scientists. But one difference is that in mushroom hunting, actual specimens must be found and catalogued, which are usually stored at public gardens or university centers like the Herbarium at UC Berkeley.
The problem may seem like an issue that would only concern specialists, but it has a much broader impact as it directly affects the way we think about the environment, biogeography, conservation, and invasive species. From wood-rotting fungi that clear the forest floors of dead wood, to the truffles in our food, mushrooms and other fungi play a critical role in the lives of plants and animals, including the health and welfare of humans. Understanding fungal biodiversity will be critical in analyzing the effects of habitat change, nutrient cycling in ecosystems and much more.


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katie Scarfe brought the characters to life by using different intonations and styles of speaking, for each person. i enjoyed this enough to listen three times and I originally only bought it for bedside listening as I find it hard to sleep. So, unfortunately I listened intently and couldn't sleep haha so it became a daytime listen instead. Thanks this was brilliant.
Would you listen to The King's Concubine again? Why?
I would listen again because it's such an intricate tale that there are sure to be things you miss with the first hearing.
The characters are well fleshed out.
What about Sophie Aldred's performance did you like?
Pleasant engaging voice conveyed the feelings of the characters well.
The Kings death bed scene had me crying too..
Also, when they took everything she had accrued legitimately and banished her.
Interesting story about a woman I had hardly had knowledge of prior to this.