text:
\section{Introduction}

Partitioning integers into summands (parts) is a subject of
intensive research in combinatorics, number theory and statistical
physics. If $n$ is a positive integer, then by a partition,
$\lambda$, of $n$, we mean a representation
\begin{equation}\label{partition}
 \lambda: \quad n=\sum_{j=1}^n jm_j,
\end{equation}
 in which $m_j$, called multiplicities of parts $j,
j=1,2,...,n$, are non-negative integers. We use $\Lambda(n)$ to
denote the set of all partitions of $n$ and let
$p(n)=\mid\Lambda(n)\mid$. The number $p(n)$ is determined
asymptotically by the famous partition formula of Hardy and
Ramanujan [8]:
 \begin{equation}\label{hardy}
 p(n)\sim\frac{1}{4n\sqrt{3}}
\exp{\left(\pi\sqrt{\frac{2n}{3}}\right)}, \quad n\to\infty.
\end{equation}
A precise asymptotic expansion for $p(n)$ was found later by
Rademacher [13] (more details may be also found in [2; Chapter
5]). Further on, we assume that, for fixed integer $n\ge 1$, a
partition $\lambda\in\Lambda(n)$ is selected uniformly at random
(uar). In other words, we assign the probability $1/p(n)$ to each
$\lambda\in\Lambda(n)$. In this way, each numerical characteristic
of $\lambda$ can be regarded as a random variable defined on
$\Lambda(n)$.

Corteel et al. [3] proposed and studied three procedures of
sampling parts of a random partition $\lambda\in\Lambda(n)$. They
focused on the multiplicities $\mu_{n,j}=\mu_{n,j}(\lambda),
j=1,2,3,$ of the randomly-selected parts and found their limiting
distributions, as $n\to\infty$, in these three cases of sampling.

Our aim in this paper is to determine asymptotically, as
$n\to\infty$, the distributions of the sizes
$\sigma_{n,j}=\sigma_{n,j}(\lambda)$ of the randomly-chosen parts
for the same sampling procedures. One of these three limiting
distributions is obtained earlier by Fristedt [5]. We present the
proofs of the other two limit theorems in Sections 4 and 6. They
combine probabilistic and analytical techniques, which are briefly
described in Section 3. The main results are also stated there.
Sections 2 and 4 contain some definitions, notations and auxiliary
facts.

\section{Preliminaries}
\setcounter{equation}{0}

 We start with the notation $g(x)$ for the generating function of
 the sequence $\{p(n)\}_{n\ge 1}$. For $\mid x\mid<1$, $g(x)$
 admits the well known representation
 \begin{equation}\label{euler}
 g(x)=1+\sum_{n=1}^\infty p(n)x^n =\prod_{k=1}^\infty (1-x^k)^{-1}
 \end{equation}
 (see e.g. [2; Theorem 1.1]).

 For any $\lambda\in\Lambda(n)$ selected uar, we define the random
 variables
 \begin{equation}\label{alph}
 \alpha_j^{(n)} =\alpha_j^{(n)}(\lambda) =\mbox{the number of
 parts of size } j \mbox { in } \lambda
 \end{equation}
and
 \begin{equation}\label{bet}
\beta_j^{(n)}= \beta_j^{(n)}(\lambda) =\left\{\begin{array}{ll} 1 & \qquad  \mbox {if}\qquad \alpha_j^{(n)}>0, \\
 0 & \qquad \mbox {if}\qquad \alpha_j^{(n)}=0.
 \end{array}\right.
 \end{equation}
 Then, obviously, $Z_n=\sum_{j=1}^n \alpha_j^{(n)}$ equals the
 total number of parts and $Y_n=\sum_{j=1}^n \beta_j^{(n)}$ - the
 number of distinct parts in $\lambda\in\Lambda(n)$. Furthermore,
 for $s$ real and $\ge 1$, we also let
 \begin{equation}\label{ex}
 X_{s,n} =\left\{\begin{array}{ll} \sum_{1\le j\le s} j\alpha_j^{(n)} & \qquad  \mbox {if}\qquad 1\le s\le n, \\
 n & \qquad \mbox {if}\qquad s>n;
 \end{array}\right.
 \end{equation}
 \begin{equation}\label{zi}
 Z_{s,n} =\left\{\begin{array}{ll} \sum_{1\le j\le s} \alpha_j^{(n)} & \qquad  \mbox {if}\qquad 1\le s\le n, \\
 Z_n & \qquad \mbox {if}\qquad s>n;
 \end{array}\right.
 \end{equation}
 \begin{equation}\label{uai}
 Y_{s,n} =\left\{\begin{array}{ll} \sum_{1\le j\le s} \beta_j^{(n)} & \qquad  \mbox {if}\qquad 1\le s\le n, \\
 Y_n & \qquad \mbox {if}\qquad s>n.
 \end{array}\right.
 \end{equation}

 Next, by $\mathbb{E}(C)$ we denote the expected value of a random
 variable $C$ with respect to the uniform probability measure $Pr(.)$ defined on the integer partition space $\Lambda(n)$.
 The following two asymptotic equivalences are well known:
 \begin{equation}\label{euai}
 \mathbb{E}(Y_n)\sim\frac{\sqrt{6n}}{\pi}, \quad n\to\infty,
 \end{equation}
 \begin{equation}\label{ezi}
 \mathbb{E}(Z_n)\sim\frac{\sqrt{6n}}{2\pi}\log{n}, \quad n\to\infty
 \end{equation}
 ((\ref{euai}) was proved by Wilf in [15]; a proof of (\ref{ezi}) can
 be found in [3]).

 We introduce now the joint probability generating functions
 \begin{eqnarray}\label{pgf}
 & & \mathbb{E}(y_1^{\alpha_1^{(n)}}y_2^{\alpha_2^{(n)}}
 ...y_n^{\alpha_n^{(n)}})= \nonumber \\
 & & \sum_{\sum jm_j=n, m_j\in\mathbb{N}_0}
 Pr(\alpha_1^{(n)}=m_1,\alpha_2^{(n)}=m_2,...,\alpha_n^{(n)}=m_n)
 \nonumber \\
 & & \times y_1^{m_1}y_2^{m_2}...y_n^{m_n},
 \end{eqnarray}
 where $\mathbb{N}_0=\{0,1,...\}$.  General enumeration methods imply that
 \begin{equation}\label{gf}
 1+\sum_{n=1}^\infty x^n p(n)
 \mathbb{E}(y_1^{\alpha_1^{(n)}}y_2^{\alpha_2^{(n)}}
 ...y_n^{\alpha_n^{(n)}})
 =\prod_{k=1}^\infty \sum_{m_k\in\mathbb{N}_0} (y_k x^k)^{m_k}
 \end{equation}
 (for more details and proof of (\ref{gf}), see [14; Chapter
 V.5]). Here $x$ and $y_k$ are formal
 variables. Setting in (\ref{pgf}) and (\ref{gf}) $y_k=y^k$ for $1\le k\le [s]$
 and $y_k=1$ for $k>[s]$ ($[s]$ denotes
 the integer part of $s$), from
 (\ref{euler}) and (\ref{ex}) we obtain
 $$
 1+\sum_{n=1}^\infty p(n)x^n \mathbb{E}(y^{X_{s,n}})
 =g(x)\prod_{1\le j\le s} \frac{1-x^j}{1-(xy)^j}.
 $$
 A differentiation with respect to $y$ in this identity leads to
 the formula for the expectations we want:
 \begin{equation}\label{expectation}
 \sum_{n=1}^\infty p(n)x^n\mathbb{E}(X_{s,n}) =g(x)h_s(x),
 \end{equation}
 where
 \begin{equation}\label{hs}
 h_s(x)=\sum_{1\le j\le s}\frac{jx^j}{1-x^j}.
 \end{equation}
 Similarly, using (\ref{bet}), (\ref{uai}) and an argument
 developed by Wilf [15], one can deduce that
 $$
 1+\sum_{n=1}^\infty p(n)x^n \mathbb{E}(y^{Y_{s,n}})
 =g(x)\prod_{1\le j\le s}(1+(y-1)x^j)
 $$
 and, in the same way, that
 $$
\sum_{n=1}^\infty p(n)x^n\mathbb{E}(Y_{s,n})
=g(x)\frac{1-x^{[s]+1}}{1-x}.
 $$
It is shown in [11] that, for $s=t\sqrt{6n}/\pi, 0<t<\infty,$
\begin{equation}\label{euaies}
\mathbb{E}(Y_{s,n})\sim (1-e^{-t})\frac{\sqrt{6n}}{\pi}, \quad
n\to\infty.
\end{equation}
Finally, we notice that the asymptotic behavior of $Z_{s,n}$,
defined by (\ref{zi}), is studied in detail by Fristedt [5]. He
assumed that $s/\sqrt{n}\to 0$ as $n\to\infty$ and showed that
$Z_{s,n}$, appropriately normalized, converges in distribution to
a doubly exponential (extreme value) distributed random variable.
A law of large numbers for $Z_{s,n}$ is also proved.

We now proceed with the description of the sampling procedures
introduced by Corteel et al. [3]. We remind that they are three
two-step procedures that combine the outcomes of two experiments.
Therefore, they lead to three different product  probability
spaces (see e.g. [7; Chapter 1.6]). Since in each procedure we
first sample uar a partition $\lambda\in\Lambda(n)$, the
probability space on $\Lambda(n)$, defined in the Introduction, is
included in each product space. The second steps of sampling are,
however, different and therefore, for each different procedure we
obtain a different product space and different product probability
measure. In what follows next we adopt the common notation
$\mathbb{P}(.)$ for the product probability measure of each
sampling procedure, use notation $\sigma_{n,j}$ for the size of
the randomly-chosen part in procedure $j$ ($j=1,2,3$) and follow
the concept of a product space developed in [7; Chapter 1.6]. We
describe the procedures in terms of events
$\{\lambda\in\Lambda(n)\}$, $\{\sigma_{n,j}\le s\}$ ($s\ge 1$) and
their set product $\{\lambda\in\Lambda(n)\}\times\{\sigma_{n,j}\le
s\}$.

{\it Procedure 1.} Given a partition $\lambda\in\Lambda(n)$ chosen
uar (step 1), we select a part uar among all $Y_n$ different parts
of $\lambda$ (step 2). Hence by the product measure formula [7;
Chapter 1.6] and (\ref{uai}),
$$
\mathbb{P}(\{\lambda\in\Lambda(n)\}\times\{\sigma_{n,1}\le s\})
=Pr(\lambda\in\Lambda(n))\left(\frac{Y_{s,n}}{Y_n}\right)
=\left(\frac{1}{p(n)}\right)\left(\frac{Y_{s,n}}{Y_n}\right).
$$
Summation over all $\lambda\in\Lambda(n)$ yields
\begin{equation}\label{procone}
\mathbb{P}(\sigma_{n,1}\le s)
=\mathbb{E}\left(\frac{Y_{s,n}}{Y_n}\right).
\end{equation}

{\it Procedure 2.} Given a partition $\lambda\in\Lambda(n)$ chosen
uar (step 1), we select a part of it with the probability
proportional to its size and multiplicity (step 2). Recalling
definition (\ref{ex}) of the random variable $X_{s,n}$, we obtain
in a similar way that
$$
\mathbb{P}(\{\lambda\in\Lambda(n)\}\times\{\sigma_{n,2} \le s\})
=\left(\frac{1}{p(n)}\right)\left(\frac{X_{s,n}}{n}\right)
$$
and
\begin{equation}\label{proctwo}
\mathbb{P}(\sigma_{n,2}\le s) =\frac{1}{n}\mathbb{E}(X_{s,n}).
\end{equation}

{\it Procedure 3.} Given a partition $\lambda\in\Lambda(n)$ chosen
uar (step 1), we select a part of it uar among all $Z_n$ parts in
$\lambda$ (without any bias, step 2), i.e.,
\begin{equation} \label{procthree}
\mathbb{P}(\{\lambda\in\Lambda(n)\}\times\{\sigma_{n,3}\le s\})
=\left(\frac{1}{p(n)}\right)\left(\frac{Z_{s,n}}{Z_n}\right),
\end{equation}
where $Z_{s,n}$ is defined by (\ref{zi}), and thus
$$
\mathbb{P}(\sigma_{n,3}\le s)
=\mathbb{E}\left(\frac{Z_{s,n}}{Z_n}\right).
$$

{\it Remark.} Each partition $\lambda\in\Lambda(n)$ has a unique
graphical representation called Ferrers diagram [2; Chapter 1.3].
It is obtained as follows. We use the notation $\lambda_k$ to
denote the $k$th largest part of $\lambda$ for $k$ a positive
integer; if the number of parts $Z_n$ of $\lambda$ is $<k$, then
$\lambda_k=0$. The Ferrers diagram illustrates (\ref{partition})
by a two-dimensional array of dots, composed by $\lambda_1$ dots
in the first (most left) row, $\lambda_2$ dots in the second row,
..., $\lambda_{Z_n}$ dots in the last $Z_n$th row. Therefore, a
Ferrers diagram may be considered as a union of disjoint blocks
(rectangles) of dots with base $j$ and height $\alpha_j^{(n)}$
(the multiplicity of part $j$). In this way, the sampling
probability in Procedure 2 is proportional to the area of the
block to which the chosen part belongs.

 In order to make a comparison between the asymptotic behavior of the typical part size $\sigma_{n,j}$ (see next section) and
 its typical multiplicity $\mu_{n,j}$ for procedure $j$ ($j=1,2,3$),
  we also provide the reader with some results, obtained by Corteel et al. [3]:
 \begin{equation}\label{zipf}
 \lim_{n\to\infty}\mathbb{P}(\mu_{n,1}=m) =\frac{1}{m(m+1)}, \quad
 m\ge 1,
 \end{equation}
 \begin{equation}\label{deb}
 \lim_{n\to\infty}\mathbb{P}(\mu_{n,2}=m) =\frac{6(2m+1)}{\pi^2
 m(m+1)^2},\quad m\ge 1
 \end{equation}
 and
 \begin{equation}\label{limit}
\lim_{n\to\infty}\mathbb{P}\left(\frac{2\log{\mu_{n,3}}}{\log{n}}\le
t\right)=t, \quad 0<t<1.
 \end{equation}

\section{Statement of the Main Results and Brief Description of the Method of Proof}
 \setcounter{equation}{0}

Among other important results on random integer partitions
Fristedt [5; p. 712] has proved the following limit theorem.

\begin{theorem}
Let $0<t<1$. Then, we have
$$
\lim_{n\to\infty}\mathbb{P}\left(\frac{2\log{\sigma_{n,3}}}{\log{n}}\le
t\right)=t.
$$
\end{theorem}

{\it Remark.} A comparison between this result and (\ref{limit})
shows that both $\mu_{n,3}$ and $\sigma_{n,3}$, normalized in the
same manner, have one and the same limiting distribution as
$n\to\infty$. Moreover, these results imply that the proportion of
part sizes not greater than $n^{t/2}$ and the proportion of parts
whose multiplicity is $\le n^{t/2}$ are both approximately equal
to $t\in (0,1)$ as $n\to\infty$.

The main results of this paper are devoted to sampling procedures
1 and 2. In the next sections we prove the following limit
theorems for the randomly-chosen part size $\sigma_{n,j}, j=1,2,$
of a random integer partition.

\begin{theorem}
Let $0<t<\infty$. Then, we have
$$
\lim_{n\to\infty}\mathbb{P}\left(\frac{\pi\sigma_{n,1}}{\sqrt{6n}}
\le t\right) =1-e^{-t}.
$$
\end{theorem}

\begin{theorem}
Let $0<t<\infty$. Then, we have
$$
\lim_{n\to\infty}\mathbb{P}\left(\frac{\pi\sigma_{n,2}}{\sqrt{6n}}
\le t\right) =\frac{6}{\pi^2}\int_0^t\frac{u}{e^u-1}du.
$$
\end{theorem}

{\it Remark.} The limiting distribution in Theorem 3 is expressed
in terms of a Debye function (see e.g. [1; Section 27.1]). We
recall that
$$
\int_0^\infty\frac{u}{e^u-1}du =\frac{\pi^2}{6}.
$$
We also notice that Theorems 2 and 3 show that, for sampling
procedures 1 and 2, the typical size of the randomly-chosen part
is of order $const.\sqrt{n}$, while (\ref{zipf}) and (\ref{deb})
imply that its multiplicity is much smaller and approaches a
discrete random variable as $n\to\infty$.

The rest of the paper contains the proofs of Theorems 2 and 3. In
the proof of Theorem 2 we use Fristedt's conditioning device [5],
which allows us to transfer probability distributions of linear
combinations of the multiplicities $\alpha_j^{(n)}$ into
conditional distributions of the corresponding linear combinations
of independent geometrically distributed random variables. This
allows us to approximate the expectations in (\ref{procone}) and
(\ref{procthree}) by the ratios of the expected values of the
corresponding random variables. The proof of Theorem 3 is based on
a Cauchy integral stemming from (\ref{expectation}),
Hardy-Ramanujan's formula (\ref{hardy}) and Hayman's theorem for
estimating coefficients of admissible power series [9] (see also
[4; Chapter VIII.5]).

\section{Proof of Theorem 2}
\setcounter{equation}{0}

We base our proof on (\ref{procone}) and asymptotic equivalences
(\ref{euai}) and (\ref{euaies}). We follow the argument given in
the proof of Theorem 3 of [3]. To replace the expectation in the
right-hand side of (\ref{procone}) by the ratio
$\mathbb{E}(Y_{s,n})/\mathbb{E}(Y_n)$, we need to study how
unlikely is the event
$$
A_n =\left\{\lambda\in\Lambda(n):\mid\frac{\pi Y_n(\lambda)}
{\sqrt{6n}} -1\mid>\epsilon\right\}, \quad \epsilon>0.
$$
Using Fristedt's method [5], Corteel et al. [3] showed that
\begin{equation}\label{probabeien}
Pr(A_n) \le e^{-c\sqrt{n}}, \quad c=c(\epsilon)>0.
\end{equation}

{\it Remark.} Fristedt's approach [5] is based on the identity
\begin{equation}\label{probabgeom}
Pr(\alpha_j^{(n)}=m_j, j=1,...,n) =Pr\left(\gamma_j=m_j,
j=1,...,n\mid\sum_{j\ge 1} j\gamma_j=n\right),
\end{equation}
where $\{\gamma_j\}_{j\ge 1}$ is a sequence of independent
geometrically distributed random variables, whose distribution is
given by
$$
Pr(\gamma_j=k) =(1-q^j)q^{jk}, \quad k=0,1,...
$$
and $\{m_j\}_{j\ge 1}$ are non-negative integers. Eq.
(\ref{probabgeom}) holds for every fixed $q\in (0,1)$. It is
natural to take $q$ so that $Pr(\sum_{j\ge 1}j\gamma_j=n)$ is as
large as possible. Fristedt's almost optimal choice for $q$ is
$q=e^{-\pi/\sqrt{6n}}$. Then, the bound in (\ref{probabeien}) is
easily obtained using this value of $q$.

Next, we represent the probability in (\ref{procone}) in the
following way
\begin{equation}\label{ind}
\mathbb{P}(\sigma_{n,1}\le s)
=\mathbb{E}\left(\frac{Y_{s,n}}{Y_n}I_{A_n^c}\right)
+\mathbb{E}\left(\frac{Y_{s,n}}{Y_n}I_{A_n}\right),
\end{equation}
where $I_{A_n}$ and $I_{A_n^c}$ denote the indicators of events
$A_n$ and $A_n^c$, respectively. Since, for any $\lambda\in
A_n^c$,
$$
\frac{\pi}{\sqrt{6n}(1+\epsilon)}<\frac{1}{Y_n}<\frac{\pi}{\sqrt{6n}(1-\epsilon)}
$$
if $0<\epsilon<1$, the first summand in (\ref{ind}) is estimated
by
\begin{eqnarray}\label{eienc}
& & \mathbb{E}\left(\frac{Y_{s,n}}{Y_n}I_{A_n^c}\right)
=\frac{\pi}{\sqrt{6n}}(1+O(\epsilon))\mathbb{E}(Y_{s,n}I_{A_n^c})
\nonumber \\
& & =\frac{\pi}{\sqrt{6n}}(1+O(\epsilon)) (\mathbb{E}(Y_{s,n})
-\mathbb{E}(Y_{s,n}I_{A_n})).
\end{eqnarray}
Clearly, with probability $1$, we have $Y_{s,n}\le n$. Hence,
using (\ref{probabeien}), we obtain
 $$
 \mathbb{E}(Y_{s,n}I_{A_n}) =O(nPr(A_n)) =O(ne^{-c\sqrt{n}}).
 $$
 Then, for $s=t\sqrt{6n}/\pi$, by (\ref{euaies}) and
 (\ref{eienc}),
 \begin{eqnarray}
 & & \mathbb{E}\left(\frac{Y_{s,n}}{Y_n}I_{A_n^c}\right)
 =\frac{\pi}{\sqrt{6n}}(1+O(\epsilon))\left(\frac{\sqrt{6n}}{\pi}
 (1-e^{-t})\right)(1+o(1)) \nonumber \\
 & & + O(ne^{-c\sqrt{n}}) =1-e^{-t} +O(\epsilon) +o(1)
 +O(ne^{-c\sqrt{n}}). \nonumber
 \end{eqnarray}
 The second term in the right-hand side of (\ref{ind}) is easily
 estimated using (\ref{probabeien}) since it is not greater than
 $Pr(A_n)$. Consequently, (\ref{ind}) becomes
 $$
 \mathbb{P}(\sigma_{n,1}\le s)
 =\mathbb{P}\left(\frac{\pi\sigma_{n,1}}{\sqrt{6n}}\le t\right)
 =1-e^{-t}+O(\epsilon)+o(1).
 $$
 Letting $n\to\infty$ and then $\epsilon\to 0$, we obtain the
 required result.

 \section{Some Remarks on Meinardus Theorem on Weighted Partitions
 and Hayman Admissibility}
 \setcounter{equation}{0}

 This section presents a brief introduction to the
 analytic combinatorics background needed for the proof of Theorem
 3. The starting point in it is eq. (\ref{proctwo}). It
 requires an asymptotic estimate for $\mathbb{E}(X_{s,n})$. By
 (\ref{expectation}), the coefficient $p(n)\mathbb{E}(X_{s,n})$ of
 $x^n$ can be expressed by a Cauchy integral whose integrand is $g(x)h_s(x)$
 (see also
 (\ref{hs})). Its behavior heavily depends on the analytic properties
 of the partition generating function $g(x)$ whose infinite
 product representation (\ref{euler}) shows that its main
 singularity is at $x=1$ (see [2; Chapter 5]). If the integrand of
 the Cauchy integral was only $g(x)$, then a properly chosen contour of
 integration and a proper asymptotic method (the Hardy-Ramanujan
 circle method or the saddle-point method) would yield Hardy-Ramanujan
 asymptotic formula (\ref{hardy}). It has been
 subsequently generalized in various directions most
notably by Meinardus [10] (see also [2; Chapter 6]) who obtained
the asymptotic of the Taylor coefficients of infinite products of
the form
\begin{equation} \label{product}
\prod_{k=1}^\infty (1-x^k)^{-b_k}
\end{equation}
under certain general assumptions on the sequence of non-negative
numbers $\{b_k\}_{k\ge 1}$. Meinardus approach is based on
considering the Dirichlet generating series
\begin{equation} \label{diofzi}
D(z)=\sum_{k=1}^\infty b_k k^{-z}, \quad z=u+iv.
\end{equation}
Below we briefly describe Meinardus assumptions avoiding their
precise statements as well as some extra notations and concepts.
The first assumption ($M_1$) specifies the domain
$\mathcal{H}=\{z: u\ge -C_0\}, 0<C_0<1,$ in the complex plane, in
which $D(z)$ has an analytic continuation. The second one ($M_2$)
is related to the asymptotic behavior of $D(z)$, whenever $\mid
v\mid\to\infty$. A function of the complex variable $z$ which is
bounded by $O(\mid\Im(z)\mid^{C_1}), 0<C_1<\infty$, in certain
domain of the complex plane is called function of finite order.
Meinardus second condition ($M_2$) requires that $D(z)$ is of
finite order in the whole domain $\mathcal{H}$. Finally, the
Meinardus third condition ($M_3$) implies a bound on the ordinary
generating function of the sequence $\{b_k\}_{k\ge 1}$. It can be
stated in a way simpler than the Meinardus original expression by
the inequality
$$
\sum_{k=1}^\infty b_k e^{-k\omega}\sin^2{(\pi ku)} \ge
C_2\omega^{-\epsilon_1}, \quad 0<\frac{\omega}{2\pi}<\mid
u\mid<\frac{1}{2},
$$
for sufficiently small $\omega$ and some constants $C_2,
\epsilon_1>0$ ($C_2=C_2(\epsilon_1)$) (see [6; p. 310]).

It is
known that Euler partition generating function $g(x)$ (which is
obviously of the form (\ref{product})) satisfies the Meinardus
scheme of conditions ($M_1$)-($M_3$) (see e.g. [2; Theorem 6.3]).

In the asymptotic analysis of the Cauchy integral stemming from
(\ref{expectation}) we apply the saddle-point method using Hayman
admissibility theory [9]; see also [4; Chapter VIII.5]. Hayman
studied a wide class of power series satisfying a set of
relatively mild conditions and established general formulas for
the asymptotic order of their coefficients. To present Hayman's
idea and show how it can be applied in the proof of our Theorem 3,
we need to introduce some auxiliary notations.

We consider here a function $G(x)=\sum_{n=1}^\infty G_n x^n$ that
is analytic for $\mid x\mid<\rho, 0<\rho<\infty$. For $0<r<\rho$,
we let
 \begin{equation}\label{ar}
 a(r)=r\frac{G^\prime(r)}{G(r)},
\end{equation}
\begin{equation}\label{br}
b(r)= r\frac{G^\prime(r)}{G(r)}
+r^2\frac{G^{\prime\prime}(r)}{G(r)}
-r^2\left(\frac{G^\prime(r)}{G(r)}\right).
\end{equation}
In the statement of the Hayman's result we use the terminology
given in [4; Chapter VIII.5]. We assume that $G(x)>0$ for $x\in
(R_0,\rho)\subset (0,\rho)$ and satisfies the following three
conditions.

{\it Capture condition.} $\lim_{r\to\rho} a(r)=\infty$ and
$\lim_{r\to\rho} b(r)=\infty$.

{\it Locality condition.} For some function $\delta=\delta(r)$
defined over $(R_0,\rho)$ and satisfying $0<\delta<\pi$, one has
$$
G(re^{i\theta})\sim G(r)e^{i\theta a(r)-\theta^2 b(r)/2}
$$
as $r\to\rho$, uniformly for $\mid\theta\mid\le\delta(r)$.

{\it Decay condition.}
$$
G(re^{i\theta}) =o\left(\frac{G(r)}{\sqrt{b(r)}}\right)
$$
as $r\to\rho$, uniformly for $\delta(r)\le\theta<\pi$.

{\textbf{Hayman Theorem.} Let $G(x)$ be Hayman admissible function
and $r=r_n$ be the unique solution in the interval $(R_0,\rho)$ of
the equation
\begin{equation}\label{eqar}
a(r)=n.
\end{equation}
Then the Taylor coefficients of $G(x)$ satisfy, as $n\to\infty$,
\begin{equation}\label{hayman}
G_n\sim\frac{G(r_n)}{r_n^n\sqrt{2\pi b(r_n)}}
\end{equation}
 with $b(r_n)$ given
by (\ref{br}).

The proof of Theorem 3 (see next section) is divided into two
parts.

A) Proof of Hayman admissibility for $g(x)$.

B) Obtaining an asymptotic estimate for the Cauchy integral
stemming from (\ref{expectation}).

\section{Proof of Theorem 3}
\setcounter{equation}{0}

{\it Part A.}

First we need to show how Hayman's theorem can be applied to find
the asymptotic behavior of the Taylor coefficients of the
partition generating function $g(x)$. Since in (\ref{euler}) we
have $b_k=1, k\ge 1$, the Dirichlet generating series
(\ref{diofzi}) is $D(z)=\zeta(z)$, where $\zeta$ denotes the
Riemann zeta function. We set in (\ref{ar}) and (\ref{br})
$r=r_n=e^{-d_n}, d_n>0$, where $d_n$ is the unique solution of the
equation
\begin{equation}\label{maineq}
a(e^{-d_n})=n.
\end{equation}
((\ref{maineq}) is an obvious modification of (\ref{eqar}).)
Granovsky et al. [6] showed that the first two Meinardus
conditions imply that the unique solution of (\ref{maineq}) has
the following asymptotic expansion:
\begin{equation}\label{dn}
d_n =\sqrt{\zeta(2)/n} +\frac{\zeta(0)}{2n} +O(n^{-1-\beta})
=\frac{\pi}{\sqrt{6n}}-\frac{1}{4n} +O(n^{-1-\beta}),
\end{equation}
where $\beta>0$ is fixed constant (here we have also used that
$\zeta(0)=-1/2$; see [1; Chapter 23.2]). We also notice that
(\ref{br}) and (\ref{dn}) impliy that
\begin{equation}\label{bdn}
b(e^{-d_n}) =2\zeta(2)d_n^{-3} +O(d_n^{-2}) \sim\frac{\pi^2}{3}
d_n^{-3} \sim\frac{2\sqrt{6}}{\pi} n^{3/2}
\end{equation}
(see [12; Lemma 2.2] with $D(z)=\zeta(z)$). Hence, by
(\ref{maineq}) and (\ref{bdn}), $a(e^{-d_n})\to\infty$ and
$b(e^{-d_n})\to\infty$ as $n\to\infty$, that is, Hayman's
``capture'' condition is satisfied with $r=r_n=e^{-d_n}$. To show
next that Hayman's ``decay'' condition is satisfied by $g(x)$ we
set
\begin{equation}\label{delta}
\delta_n =\frac{d_n^{4/3}}{\Omega(n)}
=\frac{\pi^{4/3}}{(6n)^{2/3}\Omega(n)}
\left(1+O\left(\frac{1}{\sqrt{n}}\right)\right)
\end{equation}
with $d_n$ given by (\ref{dn}), where $\Omega(n)\to\infty$ as
$n\to\infty$ arbitrarily slowly. We can apply now an estimate for
$\mid g(e^{-d_n+i\theta})\mid$ established in a general form in
[12; Lemma 2.4] using all three Meinardus conditions. It states
that there are two positive constants $c_0$ and $\epsilon_0$, such
that, for sufficiently large $n$,
\begin{equation}\label{decay}
\mid g(e^{-d_n+i\theta})\mid \le
g(e^{-d_n})e^{-c_0d_n^{-\epsilon_0}}
\end{equation}
uniformly for $\delta_n\le\mid\theta\mid<\pi$. This, in
combination with (\ref{bdn}), implies that $\mid
g(e^{-d_n+i\theta})\mid=o(g(e^{-d_n})/\sqrt{b(e^{-d_n})})$
uniformly in the same range for $\theta$, which is just Hayman's
``decay'' condition. Finally, by Lemma 2.3 of [12], established
using Meinardus conditions ($M_1$) and ($M_2$), Hayman's
``locality'' condition is also satisfied by $g(x)$. In fact, this
lemma implies in the particular case $D(z)=\zeta(z)$ that
\begin{equation}\label{locality}
e^{-i\theta n}\frac{g(e^{-d_n+i\theta})}{g(e^{-d_n})}
=e^{-\theta^2 b(e^{-d_n})/2}(1+O(1/\Omega^3(n))
\end{equation}
uniformly for $\mid\theta\mid\le\delta_n$, where $b(e^{-d_n})$ and
$\delta_n$ are determined by (\ref{bdn}) and (\ref{delta}),
respectively.
 Hence all conditions of
Hayman's theorem hold and we can apply it with $G_n=p(n),
G(x)=g(x), r_n=e^{-d_n}$ and $\rho=1$ to find that
\begin{equation}\label{asypn}
p(n)\sim\frac{e^{nd_n}g(e^{-d_n})}{\sqrt{2\pi b(e^{-d_n})}}, \quad
n\to\infty.
\end{equation}

{\it Remark.} To show that formula (\ref{asypn}) yields
(\ref{hardy}), one has to replace (\ref{dn}) and (\ref{bdn}) in
the right hand side of (\ref{asypn}).
 The asymptotic of $g(e^{-d_n})$ is
determined by a general lemma due to Meinardus [10] (see also [2;
Lemma 6.1]). Since $\zeta(0)=-1/2$ and
$\zeta^\prime(0)=-\frac{1}{2}\log{(2\pi)}$ (see [1; Chapter
23.2]), in the particular case of $g(e^{-d_n})$ this lemma implies
that
\begin{eqnarray}
& & g(e^{-d_n}) =\exp{(\zeta(2)d_n^{-1} -\zeta(0)\log{d_n}
+\zeta^\prime(0) +O(d_n^{c_1}))} \nonumber \\
& & =\exp{\left(\frac{\pi^2}{6d_n} +\frac{1}{2}\log{d_n}
-\frac{1}{2}\log{(2\pi)} +O(d_n^{c_1})\right)}, \quad n\to\infty,
\nonumber
\end{eqnarray}
 where $0<c_1<1$. The rest of the computation leading to (\ref{hardy}) is based on simple
 algebraic manipulations and cancellations.

 {\it Part B.}

 We are now ready to apply
 Cauchy coefficient formula to (\ref{expectation}). We use the
 circle $x=e^{-d_n+i\theta}, -\pi<\theta\le\pi$, as a contour of
 integration and obtain
  $$
 p(n)\mathbb{E}(X_{s,n}) =\frac{e^{nd_n}}{2\pi}\int_{-\pi}^\pi
 g(e^{-d_n+i\theta}) h_s(e^{-d_n+i\theta}) e^{-i\theta n}d\theta.
 $$
 Then, we break up the range of integration as follows:
 \begin{equation} \label{sumint}
p(n)\mathbb{E}(X_{s,n}) =J_1(s,n) +J_2(s,n),
 \end{equation}
 where
 \begin{equation}\label{jone}
 J_1(s,n) =\frac{e^{nd_n}}{2\pi}\int_{-\delta_n}^{\delta_n}
g(e^{-d_n+i\theta}) h_s(e^{-d_n+i\theta}) d\theta,
\end{equation}
\begin{equation} \label{jtwo}
J_2(s,n)
=\frac{e^{nd_n}}{2\pi}\int_{\delta_n<\mid\theta\mid\le\pi}
g(e^{-d_n+i\theta}) h_s(e^{-d_n+i\theta}) d\theta
\end{equation}
and $\delta_n$ is defined by (\ref{delta}).

To estimate $J_2(s,n)$, for $s=t\sqrt{6n}/\pi, 0<t<\infty$, we
notice that by the definition of Riemann integrals and (\ref{dn}),
\begin{eqnarray}\label{hsest}
& & \mid h_s(e^{-d_n+i\theta})\mid \le \sum_{1\le j\le s}
\frac{je^{-jd_n}}{\mid 1-e^{-jd_n+ij\theta}\mid} \nonumber \\
 & & \le d_n^{-2}\sum_{d_n\le jd_n\le
 sd_n}\frac{jd_ne^{-jd_n}}{1-e^{-jd_n}}d_n \nonumber \\
 & & \sim d_n^{-2}\int_0^t\frac{ue^{-u}}{1-e^{-u}}du=O(d_n^{-2}) =O(n).
\end{eqnarray}
Combining (\ref{dn}), (\ref{bdn}), (\ref{decay}), (\ref{asypn}),
(\ref{jtwo}) and (\ref{hsest}), we obtain
 \begin{eqnarray}\label{jtwoest}
 & & \mid J_2(s,n)\mid \le\frac{e^{nd_n}}{2\pi}\int_{\delta_n<\mid\theta\mid\le\pi}
\mid g(e^{-d_n+i\theta}) h_s(e^{-d_n+i\theta})\mid d\theta
\nonumber \\
 & & =O(e^{nd_n} g(e^{-d_n})ne^{-c_0d_n^{-\epsilon_0}})
 =O\left(\frac{e^{nd_n}g(e^{-d_n})}{\sqrt{b(e^{-d_n})}} n^{1+3/4}
e^{-c_0d_n^{-\epsilon_0}}\right) \nonumber \\
& & =O(p(n)n^{7/4}e^{-c_0d_n^{-\epsilon_0}})
=O(p(n)n^{7/4}e^{-c_2n^{\epsilon_0/2}}) =o(np(n)),
 \end{eqnarray}
where $c_2>0$.

The estimate of $J_1(s,n)$ follows from Hayman's ``locality''
condition (\ref{locality}). We also need to expand $h_s$ by Taylor
formula in the following way:
\begin{eqnarray}\label{hstaylor}
& & h_s(e^{-d_n+i\theta}) =h_s(e^{-d_n})
+O\left(\mid\theta\mid\frac{d}{dx} h_s(x)\mid_{x=e^{-d_n}}\right)
\nonumber \\
& & =h_s(e^{-d_n}) +O\left(\delta_n \frac{d}{dx}
h_s(x)\mid_{x=e^{-d_n}}\right).
\end{eqnarray}
For $s=t\sqrt{6n}/\pi$, we can consider, as previously, the sum
representing $h_s(e^{-d_n})$ as a Riemann sum. So, we can replace
it by the corresponding integral. Thus, by (\ref{hs}) and
(\ref{dn}), we have
\begin{eqnarray}
& & h_s(e^{-d_n}) =d_n^{-2}\sum_{d_n\le jd_n\le t\sqrt{6n}d_n/\pi}
\frac{jd_n e^{-jd_n}}{1-e^{-jd_n}}d_n \nonumber \\
& & \sim d_n^{-2}\int_{d_n}^{\frac{\sqrt{6n}}{\pi}d_n t}
\frac{ue^{-u}}{1-e^{-u}}du =\frac{6n}{\pi^2} \int_0^t
\frac{u}{e^u-1}du +O(\sqrt{n}) \nonumber
 \end{eqnarray}
since $d_n^{-2}=6n/\pi^2+O(\sqrt{n})$. In the same way we can
estimate the first derivative of $h_s$:
\begin{eqnarray}
& & \frac{d}{dx} h_s(x)\mid_{x=e^{-d_n}} =\sum_{1\le j\le s}
\frac{j^2 e^{(j-1)d_n}}{(1-e^{-jd_n})^2}  \sim
d_n^{-3}\sum_{d_n\le jd_n\le t\sqrt{6n}d_n/\pi}
\frac{(jd_n)^2 e^{-jd_n}}{(1-e^{-jd_n})^2}d_n \nonumber \\
& & \sim d_n^{-3}\int_0^t \frac{u^2 e^{-u}}{(1-e^{-u})^2}du
=O(n^{3/2}). \nonumber
\end{eqnarray}
Hence, by (\ref{delta}), the error term in (\ref{hstaylor})
becomes
$$
O\left(\delta_n\frac{d}{dx} h_s(x)\mid_{x=e^{-d_n}}\right)
=O(n^{5/6}/\Omega(n))
$$
and therefore, uniformly for $\mid\theta\mid\le\delta_n$,
$$
h_s(e^{-d_n+i\theta}) =\frac{6n}{\pi^2}\int_0^t\frac{u}{e^u-1}du
+o(n).
$$
Inserting this estimate and (\ref{locality}) into (\ref{jone}) and
applying the asymptotic of the partition function $p(n)$ from
(\ref{asypn}), we obtain
\begin{eqnarray}\label{joneest}
& & J_1(s,n) =\frac{e^{nd_n}g(e^{-d_n})}{2\pi}
\left(\int_{-\delta_n}^{\delta_n} e^{-\theta^2 b(e^{-d_n})/2}
(1+O(1/\Omega^3(n))d\theta\right) \nonumber \\
& & \times\left(\frac{6n}{\pi^2}\int_0^t\frac{u}{e^u-1}du
+o(n)\right) \nonumber \\
& & \sim\frac{e^{nd_n}g(e^{-d_n})}{\sqrt{b(e^{-d_n})}2\pi}
\left(\int_{-\delta_n\sqrt{b(e^{-d_n})}}^{\delta_n\sqrt{b(e^{-d_n})}}
e^{-y^2/2}dy\right)
\left(\frac{6n}{\pi^2}\int_0^t\frac{u}{e^u-1}du\right) \nonumber
\\
& & \sim\frac{e^{nd_n}g(e^{-d_n})}{\sqrt{b(e^{-d_n})}2\pi}
\left(\int_{-\infty}^\infty e^{-y^2/2}dy\right)
\left(\frac{6n}{\pi^2}\int_0^t\frac{u}{e^u-1}du\right) \nonumber
\\
& & =\frac{e^{nd_n}g(e^{-d_n})}{\sqrt{2\pi
b(e^{-d_n})}}\left(\frac{6n}{\pi^2}\int_0^t\frac{u}{e^u-1}du\right)
\nonumber \\
& & \sim
p(n)\left(\frac{6n}{\pi^2}\int_0^t\frac{u}{e^u-1}du\right),
\end{eqnarray}
where for the second asymptotic equivalence we have used
(\ref{bdn}) and (\ref{delta}) in order to get
$$
\delta_n\sqrt{b(e^{-d_n})}\sim
\frac{\pi^{5/6}\sqrt{2}}{6^{1/6}\Omega(n)} n^{1/12}\to\infty
$$
if $\Omega(n)\to\infty$ as $n\to\infty$ not too fast, so that
$\frac{n^{1/12}}{\Omega(n)}\to\infty$. It is now clear that, for
$s=t\sqrt{6n}/\pi$, (\ref{sumint})-(\ref{jtwo}), (\ref{jtwoest})
and (\ref{joneest}) yield
$$
p(n)\mathbb{E}(X_{s,n}) =p(n)\frac{6n}{\pi^2}
\int_0^t\frac{u}{e^u-1}du +o(np(n))
$$
and therefore
$$
\mathbb{E}(X_{s,n})\sim \frac{6n}{\pi^2}
\int_0^t\frac{u}{e^u-1}du.
$$
The result of Theorem 3 follows immediately from (\ref{proctwo}).

\section*{Acknowledgements}

I am grateful the referee for carefully reading the paper and for
his helpful comments.

redpajama_set_name: RedPajamaArXiv


Apple Second Quarter 2013 Results Incoming, Is the Sky Really Falling?
Apple will announce smartphone and tablet sales today that almost any other company (bar Samsung) would be proud of. But, since Apple has built itself and its image up to such heights, they are almost inevitably going to be viewed as a failure by the market.
With only Samsung breathing the same rarefied atmosphere as Apple sales, the likes of Nokia, BlackBerry and Microsoft can only look on with their middling-to-poor smartphone and tablet numbers. For example, Nokia sold 5.6 million Lumias, analysts expect Apple to sell 36 million iPhones. However, just being the best seller isn't enough for the stock market, which needs to see the next big thing and a clear growth line.
Apple has failed to show any such thing, keeping its mouth and its press office firmly closed for months on end. That and reports from Asian suppliers, that can be perceived as signs of sales weakness, have brought the stock price down from its $700 high to a mere $400. That's still up double from 2010's value, so long-term investors aren't exactly crying in their bowl at the soup line.
Perhaps Apple is having troubles, maybe the move to new chips, design processes, and battery and screen technologies are seeing delays (it wouldn't be the first time). But, getting out from under its major technology and hardware partner Samsung's wing seems to be a priority and one that Apple seems happy to pay with dents in the stock price. We should get some answers today (5PM EST) when Apple announces its latest numbers.
Also a factor, now Apple is massively rich, and the partners that it could sign-up for cents for media and content deals in those early years will now be playing very hardball to squeeze Apple back for more than just loose change. All these behind the scenes dealings could be delaying new products and services like iRadio, the Apple TV project and others.
While the numbers are likely to still pretty massive, the market reaction will suggest whether the bubble Apple has ridden with the iPhone and then the iPad since 2010 is over. Lack of device variety (something you could never accuse Samsung of), lack of user interface evolution (iOS 6.0 is six months old, and looked out of date at launch) could all be seeing users turn to other platforms (and if that's what the market believes, then the stock will tumble further).
The critical numbers to look out for are the average revenue per device, with a lower figure showing people are buying lots more iPad mini and older iPhone devices over the latest (and most profitable) models. While Apple, again, won't mind cannibalizing its own sales, and getting on-board millions of emerging market buyers, the stock market likes it when Apple sells millions of big, shiny, new and expensive devices.
The price to earnings ratio will also be of keen interest to market watchers, with Apple's earnings likely to be going down, those magic numbers that propelled it such stock dominance will only come tumbling down further.
On the plus side, the major market movers have likely built in a lot of the potential pad news into today's stock price ahead of the news. While good results from Apple could see the stock rebound, anything less than stellar numbers (which seems most likely) will leave analysts wondering just how much lower they can go before Apple does announce the next-generation products that really could send the stock back up.
While unlikely, and going against tradition, Apple might just drop some future product news into today's earnings call to generate some stories the focus on something other than the numbers, but with most reporters and analysts lined up with canes and clubs, expect today's news to look really ugly.


Turn the sound on. Trust us.
Pittman blasted Slayer's "Raining Blood" as Matthew rained lots of ordinary rain on northern Florida, and headbanged as only he can. His love of metal has gotten him in trouble before; The Orlando Weekly reports he once played "The Star-Spangled Banner" on guitar and was charged with disturbing the peace.
Here's the video in the original context.
Here's video proof of the guitar solo incident.
And here's the original cut of "Raining Blood."
The National Hurricane Center should try that for the next storm.
Blasting Slayer while standing in a hurricane: Stupid or awesome?


text:
Mindset and Mastery
Zettelkasten Book
Full SaaS Developer
The Great Online Game
Fed up with not having the financial freedom to do what you want? Maybe it's time to play The Great Online Game.
Martin Adams
Photo by Tobias Tullius / Unsplash
Do you ever dream of being financially free? I don't mean with lots of money, but to have the freedom that come along with never needing to worry about money.
Freedom comes from no longer needing to trade your time for money. It's where you have enough money that you simply don't need to work again. Where you're not bound but the rules of work, and wearing business attire is a thing of the past. Where you can pick what you want to do and when you want to do it.
Sometimes we refer to this as passive income; the money you make while you sleep. This is where you earn money from value, not time. Freedom allows this.
But alas, this is just a dream. Back to our 9-5 jobs, with responsibilities, bills and debt. Where we can clearly see the ceiling on what we can earn and someone else calls the shots and tells you what you should be doing. Only the lucky few manage to break free, while we're stuck in the status quo.
But what if there was a way to gain this freedom? Something that we can all do, with nearly no risk. Let me introduce you to The Great Online Game.
Welcome to the great online game
The Great Online Game is a game we all can play that lets us achieve financial freedom. It was created by Packy McCormick on his Not Boring newsletter about business strategy and pop culture.
He outlined the game which can be played by billions of connected individuals. The goal is to achieve financial freedom and how well you play determines the rewards you get—whether physical or digital.
The downside to playing is very limited, but the upside is limitless. There is next to zero cost of failure. And best of all, you can play on the side without having to risk your current employment, family security or luxuries.
Alight, I hear you. This is too good to be true. I won't lie, it is too good to be true for a lot of people. But this is not because the game is rigged, but because they don't believe in their potential and put in the work needed to play the game properly.
I personally am playing the game, and finally in 2021, I started to see the momentum of it going in the right direction. I aim to teach you how to do the same.
I appreciate that you still don't know what The Great Online Game is. Simply put, it is your ability to use the Internet to generate a wealth engine via many different means. These may be selling products, building software, educating or entertaining. In some way, all entrepreneurs who have made money online have started seeing the effects of playing The Great Online Game.
What I cover next is the rules of the game, the elements that it consists of and what to expect in return.
To play you must first recognise when you are actively playing The Great Online Game. By actively playing the game, this helps you focus on paying attention, questioning strategies, learning new techniques and experimenting with ideas.
The game rules are simple.
You can play as yourself or under a pseudonym
You play whenever you are
on Twitter, Instagram, YouTube, Discord, etc
working on your project
collaborating with other players
making investments
You choose when to play and for how long
The 4 elements of The Great Online Game
The game can be broken down into 4 key elements:
These are the likes, shares, comments, connections and discussions that you obtain while you play. They are immediate feedback signals to tell you whether you're onto something, or veering off track.
Variable Outcomes
You don't know what will work and what doesn't. Some pieces of content will pop and go viral, some don't. Some news outlets talk about one project, but not others.
Experimentation and consistency is critical to discover what has an impact. So turn up frequently to improve your chances for you to use luck in your favour.
The game also rewards community and cooperation instead of individualism and competition. This means the more you give, the more you deposit into the bank of goodwill. So be curious, give and do it all without expectation.
Sense of Control
You can jump in at any point and don't need any previous reputation or capital to leverage.
The measure of success will be directly related to your skill and effort. YouTube has shown this with the removal of studios sitting between creators and their audience. Everyone can have a voice and be heard. Your skill and your effort will be the deciding factor of whether you get heard or not. This is where the majority fail to realise that it takes dedication, strategy and patience and give up before they see any results.
Fortunately, skill is learnable and effort is a choice.
Connection to the Meta Game
You grow as an individual and your career advances as you evolve playing the game. You level up. You look back at your former self and see how far you've come. You get to meet new people along the way and test out ideas.
However, it's easy to lose sight of the challenges new players have as your start to take on new, larger and more interesting challenges as the game evolves. Ask a YouTuber with 10 million subscribers what it's like to be just starting out, and their world perspective will be very far removed.
By playing the game and using the right strategies, you'll start to see some noticeable changes. These will be:
gaining new skills and knowledge
gaining accountability and trust
Both these are perfect to leverage and to build wealth.
Start by sharing your knowledge, your curiosity and your discoveries on different social platforms.
Do this consistently around 2 years. That sounds a long time, but it should be fun doing what you love.
Pay attention to the problems you can solve and the skills you can learn along the way. See them as gains and celebrate the wins.
But most of all, be authentic, humble and focused.
I will be publishing in-depth strategies to help you play the game. These will include how to find what you love doing, how to create valuable content that builds an audience, how to monetise it and the mindset changes that will allows take it to the next level.
If you are interested in learning more about these strategies, make sure you subscribe to this blog and they'll be delivered to your inbox as they go live.
Welcome to Great Online Game.
The Gravity of Influence: Recognizing Our Responsibilities in the Social Ecosystem
Are you aware of the responsibilities that come with having a fanbase? Learn how to navigate the psychological weight of influence in the digital age and the power of persuasion in social media.
Where you put your roots
Home. It's where you feel safe, comfortable and with people you care deeply about. During a recent trip to South Africa, it allowed me to reflect on what makes a home a home. My wife grew up in South Africa and the moment the plane touched down she felt
A better question to unlock a more ambitious strategy
How you frame you goal can mean the difference between it being failure and achievable
Martin Adams © 2023

redpajama_set_name: RedPajamaCommonCrawl


Breezy today with highs near 50.
Temperatures are not as cold as yesterday morning but it's still cold this morning. We will have a fast warm-up this morning from near 30 to 50 today. The weather story of the day will be the wind with gusts increasing early this morning to 25 mph. It will be breezy all day and night.
We will start to see some spotty showers begin this evening with more rain later on tonight. It will be rainy at times Wendesday with the rain pushing out near 12 a.m. Thursday. As the rain is ending we may see a very brief change to some snowflakes.


Thank you for choosing Perry Township Schools for 2018-19! It is our privilege to educate every student who walks through our doors, and we do not take your trust for granted. In addition to offering every child a 21st century learning experience, we also make school safety a priority, especially in these challenging times. We have several policies and programs in place to help us provide a safe environment to all students, staff, parents, and visitors.
Under the leadership of Chief David Stalcup, our Perry Township School Police Department is a constant presence within our school community. The Department consists of 10 full-time and nine part-time officers who patrol each of our 22 schools daily. They also provide security at school and district events, as deemed necessary. The Department's K-9 unit detects the presence of drugs, alcohol and explosives on school property, as needed.
Our police officers lead "Run. Hide. Fight." training in each of our schools. The "Run. Hide. Fight." response plan teaches educators and students what to do in the event of an active shooter situation. We are required, by law, to perform one of these drills each semester. Teachers and administrators also lead one tornado drill per semester and one fire drill per month.
At each school, entryways are designed so visitors must be admitted to the school office before gaining access to secure areas of the building.
Starting in 2018-19, changes will be made to the way both district employees and visitors access the schools.
Guests will be required to present a driver license or other acceptable form of government-issued identification before signing in.
All district employees will receive a personalized ID badge that will allow front office staff at each school to match the employee's face to the photo on the badge.
Video surveillance camera systems also help administrators monitor activity at each building.
Our safety committee consists of building administrators, central office staff, and School Police who meet monthly to review and revise our safety procedures.
Each building and central office staff member on this committee is a School Safety Specialist through the Indiana Department of Education's School Safety Specialist Academy.
Information on the academy can be found at http://www.doe.in.gov/student-services/safety.
School Safety Specialists must attend two professional development days per year at the Indiana School Safety Academy to remain certified.
School counselors are available to help students navigate academics, mental health, and social challenges. Guidance departments and teachers receive ongoing training on how to recognize students who need support, so they can appropriately intervene.
Social workers are also available to assist students, families, and teachers in addressing problems that could negatively affect the student's emotional growth or academic outcomes.
Our Safety Hotline was created for students to report situations such as drugs, bullying, and threats of violence. The number is 317.789.3905.
This hotline is intended for students who may not feel comfortable reporting an incident to a school official in-person.
We take seriously the safety of all our students. We pride ourselves in being prepared for these rare, but very possible events.


The Terminator is chosen for preservation by the US National Film Registry.
Arnold Schwarzenegger's The Terminator will be added to the US National Film Registry's archive.
The science fiction blockbuster, released in 1984, is one of 25 films that will be preserved based on their cultural and historical significance. The Invisible Man and John Boorman's Deliverance have also been selected.
The Library of Congress's James H. Billington said: "The registry helps this nation understand the diversity of America's film heritage and, just as importantly, the need for its preservation."
The deterioration of film negative has reportedly resulted in the loss of half of the movies produced before 1950 and 90% of those made before 1920.
'Terminator 5' to be a reboot?


text:
export default class Battery{
  constructor(isactive = false){
    this.displayName = "Battery";
    this.dataDescription = this.getDescription();
    this.active = isactive;
  }

  generateData(){
    return new Promise((resolve, reject) => {
      navigator.getBattery().then((batterydata)=>{
        resolve(`${batterydata.level * 100} %`);
      });
    });
  }

  getDescription(){
    return "The navigator.getBattery method returns a battery promise that is resolved in a BatteryManager interface which you can use to interact with the Battery Status API."
  }
}

redpajama_set_name: RedPajamaGithub


We here at the Bowdon Animal Hospital are excited to introduce our newest endeavor....a blog! We will be posting interesting stories, cases, and ideas shared by doctors, technicians, and clients. Be on the look out for resources for you to learn more about preventative medicine for your pet and what our doctors recommend. We hope you enjoy and we look forward to hearing from you and interacting on a cyber level!
Thanks for all u do for us, the clients!


text:
Vali Mayers est une artiste australienne du  (née le  et morte le ), à la personnalité exubérante.

Biographie
Le photographe hollandais Ed van der Elsken réalisa des portraits d'elle.

Elle incarne Ann, l'héroïne du récit photographique Love on the left Bank de ce dernier.

Liens externes

 Site dédié
 Photos
 Site LiveJournal dédié à Vali Myers

Artiste contemporain australien
Plasticienne australienne
Naissance en août 1930
Décès en février 2003
Décès à 72 ans

redpajama_set_name: RedPajamaWikipedia


text:
Illegal fishing is a major threat to Africa's blue economy
in: Africa, Agriculture, Economy, feature, Food
Adam Abdou Hassan, Université de Rouen
The blue economy lies at the heart of globalisation. Ninety percent of international trade takes place via the sea and 95% of global communication relies on underwater networks. The blue economy encompasses all economic activity in and around rivers, lakes, streams, riverbanks, shorelines, groundwater, freshwater, seas and oceans.
The blue economy is mostly unknown, overlooked and underdeveloped in Africa. It could represent a major growth driver.
Its potential is not lost on the African Union (AU), which has made the blue economy one of the priority areas for the next 10 years: the blue economy holds immense potential as a key to a prosperous Africa. The United Nations Economic Commission for Africa (UNECA) and Seychelles Vice President Danny Faure share this outlook, seeing the blue economy as a potential source of blue gold.
But where does it currently stand in Africa? Does it benefit African economies?
Fishing boats returning to shore in Nouakchott, Mauritania.
Michał Huniewicz/Flickr, CC BY
Strategic significance
The strategic significance of the blue economy for Africa is clear. UNECA has gone so far as to publish a policy handbook on the topic, echoing the African Union's position.
According to the handbook, 38 of Africa's 54 countries are coastal states and more than 90% of import-export occurs via the sea. The territorial waters controlled by African nations stretch out over 13 million km², with a continental shelf of about 6.5 million km², including exclusive economic zones. For the AU, the blue economy represents a new frontier of African renaissance.
There are tensions between African nations over the demarcation of maritime borders. But the fishing industry employs nearly 12.3 million Africans. And the blue economy could potentially solve nutritional and food security problems for nearly 200 million Africans. It has the potential to provide vital nutrition through underused resources in fresh and salt water fish.
A grassroots, holistic and collaborative approach would make it possible to establish a blue development strategy, taking into account climate change and sustainable development. But several outside forces jeopardise its success.
The danger of illegal fishing
Blue economic activities are hampered by natural phenomena like storms and rising sea levels. But they are also impeded by human activities such as piracy, and the arms and slave trades. Illegal fishing is one of the most significant threats.
Based on Africa's financial inflows and outflows, UNECA reckons that the continent loses US$42 billion per year through illegal fishing and logging activities.
The plundering of Africa's blue economy, perpetrated by European, Asian and Russian ships, is a reality. In West Africa, the economic loss is estimated at about $700 million per year.
Illegal fishing – untaxed and unregulated – impoverishes nations and triggers cross-border population displacement. According to Professor Jean-François Akandji-Kombé at the University of Paris, the current situation can be explained by the fact that:
The sea as economic entity is a new concept in Africa. For a long time, the continent didn't have the means to exploit marine resources, or the means to assert political power over the seas. There were no seafaring people or nations to speak of in Africa. These people, these nations focused on the land, not the sea.
The European Union has woken up to the magnitude of the systematic plunder of African fishing resources, and its potential political consequences. In recent years, it has been working to establish fishery partnership agreements with African countries like Cape Verde, the Comoros, Côte d'Ivoire, Gabon, Guinea-Bissau, Liberia, Madagascar, Mauritania, Morocco, Mozambique, São Tomé and Príncipe, Senegal and the Seychelles.
China's policy has been remarkably shameless regarding illegal fishing. A number of African states hold Chinese vessels responsible for plundering African maritime resources and have called on the Chinese government to stop illegal fishing in West Africa.
African governments must therefore come up with strategic frameworks if their people are to reap the benefits of this potential bounty.
Translated from the French by Alice Heathwood for Fast for Word.
Adam Abdou Hassan, Enseignant chercheur, Université de Rouen
Previous : How microcredit has hurt the poor and destroyed informal business
Next : Africa needs to develop new ways of stemming illicit financial flows


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Sofia - The wonderful food
If you have read my previous post on Sofia, you will know that Bulgaria has been ruled by Ottomans for in excess of 500 years. That is a long time. Their influence on the cuisine is huge, I would say much larger than the influence that the Mughals had on our food in India. Not only is there the Ottoman influence but there is the Mediterranean influence, not from the Italy/France part of the Mediterranean but more from the Greece Turkey part.
We had 3 dinners in Sofia. Our brief was to eat local cuisine, not Pizza, Pasta, Sushi or Burgers nor exotic French food. And, as is our wont we researched a lot. While in Sofia we went into Supermarkets which are an excellent source of information on what is local. The Concierge at out London Apartment was Bulgarian and was absolutely enthused that we were visiting "my Country". He gave us a list. We had a lot of inputs.
Bulgaria has good wines. These were well priced and we drank some good stuff. With water in Sofia being so good, the Beer was good too. I drank only the local stuff, don't ask me the name, it was written in Cyrillic. Like in Turkey which has Raki and Greece which has Ouzo, herein Bulgaria you have Rakiya. Similar Anise flavoured clear liquor drunk by everyone. No, it is not powerful. Like most spirits it is 40% proof. It is simply a myth that it is more potent. Of course, like in Spirits you have exceptions, but generally it is 40% proof.
The Bulgarians eat all kinds of meat, and all are on offer in restaurants – Beef, Veal, Pork, Chicken, Turkey. Also, on offer is Tripe and Tongue. You get meatballs as well as kebabs all grilled. You have chops also grilled on coal. Roast Chicken is common. Stew's are common.
Yogurt and Cheese plays a large part. Yougurt is served thickened like Labneh with Cucumbers and Dill as a starter. Yogurt soups are popular as is Ayran which is basically "Chaas" or "Lassi" very similar to Turkey. Cucumbers, Cherry Tomato, Olive and Aubergine are most common. Cheeses come in all types using Cow milk as well as Sheep milk. From Cheddar varieties to Blues to a bland Feta everything is available and eaten. Breakfasts are much like Turkish Greek breakfasts with Honey, Olives, Cheeses, Tomato and Cucumber being eaten.
Salads are very good in Sofia. You have a huge selection of salads on every menu. I seriously recommend you start your meal with a salad. Shopska Salad is all pervasive, and is good.
All restaurants have a decent selection of meatless dishes, advertised as such. All restaurants have allergen advise printed on the menus as well as the weight of the dish being served. Useful information when ordering so that you don't over order. Most restaurants have English menus which are happily produced and which are easily comprehensible with virtually no howlers.
So, all in all, food and drink in Sofia is good, with easily navigable menus.
Our first dinner was at Pod Lipite [which translates as Under the Linden Tree] a 5-minute taxi ride from our Hotel. This restaurant is classic Bulgarian and rustic. Every guide book and every internet recommendation will name this place. Pod Lipite has been around from 1926 and they say that the produce for the restaurant comes from the family farm. The décor was plain, as I said rustic, with wooden benches minimally upholstered. Large with nooks and crannies and passages leading to rooms inside. At some point you have a band coming along and singing incomprehensible Bulgarian songs. Large tables were booked by tour groups. You get the picture. The terms Tourist Trap could easily be applied here, with justification. I would say that a vast majority of the punters were non-Bulgarian.
Pod Lipite, written in Cyrillic
The food was disappointing. The Yogurt Salad insipid and under seasoned. My plate of mixed grill had been cooked way in advance, slightly reheated and thoroughly overcooked. HRH the Queen of Kutch ordered a dish of Aubergine with Veal. This came in an earthenware dish which means it should have been baked and come out absolutely nuclear. Alas this was not the case. The dish was good but missed the mark. It was sort of Imam Biyaldi crossed with Eggplant Parmigiana. Careless cooking. Peasant food and we ate like peasants. Poor. Not recommended at all despite what the guide books say. To be honest, this is a place for wild bacchanalian drinking of copious amounts of Rakiya in a large group to be bedded down with heavy stodgy food. That would have probably been fun. We paid about 75 Lev for our meal with wine. That translates to INR 3500/- Cheap very cheap.
The condiments on the table: From top left clockwise: Salt, Dried Thyme, Ground Cumin and Black Pepper
Yogurt salad with Cucumber.
Aubergine with Veal in a Tomato Sauce
Mixed Grill
The Bulgarian Calzone
Disappointed by Pod Lipite the next day we did a little more diligent research. I had seen a restaurant named Grozd during our drive from the airport. This was walking distance from our Hotel so we set about to have a look. It looked great with some swish people having lunch. When we asked to make a dinner reservation, we were told that it was full for a party, and was full the next night too. Both of us immediately thought RAAAACIST! But all was not lost. A few doors down was our second choice Shastlivetsa. This restaurant has some 3 or 4 branches. In we walked to make a reservation for the evening, which we got.
Shastlivetsa was superb. An upmarket place with well-dressed locals forming almost all the diners. The waiter was charming. He said the restaurant was not typically Bulgarian, and on our request steered us to what could be more Bulgarian dishes. The food was a quantum leap over what we had the previous evening. This was not peasant food; this was decent sophisticated food. The Dessert was superb. A Chocolate Fondant or Lava Cake except that instead of Chocolate oozing out you had an Orange Cream. An excellent meal and Shastlivetsa is fully recommended. Our meal cost LEV 113 double of the previous night but still an entirely reasonable INR 5500/- We were pleased.
Restaurant Shastlivetsa in Cyrillic
A brilliant Salad Shopska
Red Pepper stuffed with Bulgarian White Cheese and Truffle. Then Deep fried. Very good.
So called Meatballs. Once again very good and moist.
Iberico Pork Chop.
The excellent Orange Fondant
The last night was at Tavern Izbata located very close to our hotel. We had walked across to see it and being happy with what we saw asked the Concierge to make a booking. We did not want to get into a RAAACIST situation again. Tavern Izbata is in the basement and once again rustic. But the feel seemed far better than at Pod Lipite. The food served was exceptional. The dish of the day being the Veal Tongue, really good. While this food was rustic and Bulgarian, it is clear that it was a different kind of rustic. The presence of breads ["Tandoori Roti"] was new as was the bread with cheese. This was a most pleasant and enjoyable meal. We paid something like Lev 110 which again translates to INR 5500/- Well worth it.
A different looking Shopska Salad.
Local bread with Cheese. Similar to our Cheese Naan.
The fabulous Veal Tongue
Sojuk
Kapama. Pickled cabbage with rice, sausage, chicken, pork and veal meat in earthenware, capped with dough, and baked with fragrant spices
Honestly, I thought that food in Sofia was cheap, in fact much cheaper than in Mumbai. For the volume of protein served coupled with the wine it was certainly cheaper than Mumbai. The food is good and as I have written earlier quite unchallenging for Indian palates, very similar to Turkish food. We ate well.
Sofia Bulgaria - Stunning. All you need to know.
This is a long read. Lots of pictures. All you want to know about Sofia.
Sofia, Bulgaria. We are here for 3 nights.
How many famous Bulgarians do you know? I could not think of any to be honest. I do know that Bulgarians are great Olympic Gold Medal winning weightlifters, wrestlers, shot putters and boxers, but unfortunately all these are anonymous to us Indians. Grigor Dimitrov is a 2017 ATP World Number 3 tennis player. I have heard of him in passing. We can conclude that Bulgaria is an unknown quantity for most of us.
We had read that Sofia the capital of Bulgaria is a great place to visit. After some research we decided to take the plunge. Geographically Bulgaria is pretty far from London. In fact, much to my surprise, it is East of Croatia, and East of Greece. Bulgaria shares a border with Greece and Turkey. The flight is 3 hours, which is relatively long for a European destination. There is a 2-hour time difference between London & Bulgaria. It is that far East of London.
We were booked at the Intercontinental which had recently opened after a change of owners, rebranding from the Radisson Blu and a massive renovation. This was a good property, thoroughly modern and absolutely brilliantly located. A walk from almost all the sights that Sofia has to offer. If you do come to Sofia, stay here or at the Sofia Hotel Balkan which is a Marriot Luxury property, also well located.
Bulgaria is part of the EU, so a Schengen Visa is good to enter. Unfortunately, they don't use the Euro as yet but the Bulgarian Lev. This is a bit of a challenge as getting the currency and getting rid of it is difficult. You waste a lot on either exchange rates or you waste it on buying rubbish at the airport when departing simply to get rid of it. Thank God for those airline promotions of gathering loose change from flyers and giving it to charity.
The city has a decent airport, excellent roads and a decent taxi system. Compared to London, taxis are very cheap. You do need taxis to go to a couple of distant sights and to the airport. There are trams as well as an underground. Language is a bit of a challenge, but the bigger challenge is signage. Almost all signs are in Cyrillic which is as useful to me as Bengali. You have no clue what is written.
We were cheated once by a taxi driver. He spoke enough English to hold his own, did the old Delhi trick of driving around in circles, and then said the meter was 13 Lev which is a lot. On questioning him to show us the meter reading all comprehension of English stopped, immediate fluent Bulgarian commenced. We must have paid 8 Lev extra. Not damaging but upsetting. I point this out as Bulgaria is rated as one of the most corrupt in the EU.
Bulgaria has had a tumultuous history. Somewhat like us in India. Bulgaria has been ruled by a succession of invaders. Originally the locals called Bulgur's held sway. Then the Eastern Romans or Byzantines came in. The Byzantium Empire was then overrun by the Ottomans [the chaps from Turkey] who ruled for over 500 years. The Ottoman Empire itself started to weaken which led to the rise of Bulgarian National movement. This soon became Communist with Russia having a dominant influence in all things. Finally, with the fall in the Berlin Wall in 1989 Bulgaria became a Republic with its own constitution in July 1991. The bottom line is that Bulgaria has stepped out of Communism only in the last 30 years.
The result of these various rulers Bulgaria has a diverse architecture clearly visible.
Bulgaria is a relatively poor country. On the streets of Sofia, you don't see many expensive cars, by expensive I mean upwards of Rs 50 lakhs. Hardly any BMW or Mercedes and certainly no Rolls Royce. I did see a Porsche but that was about it. Most cars are humble. Things are cheap by European standards. Not may fancy upmarket shops, no big brands that I could see, just a few Italian clothes shops. The fact that Bulgaria is a poor country is clearly indicated by (a) most maids and housekeeping staff in London Hotels are Bulgarian and (b) while I hate to write this, newspaper reports show that many of the beggars/tramps in London are Bulgarian.
The sights in Sofia are simply superb. Almost all, as I have written earlier, are located right in the centre of town walking distance from each other and from the Intercontinental Hotel. From the outside these were remarkably well maintained. On the one hand you had the really old Churches, to the more modern palaces and the still more modern Russian/Communist structures. All of them were very impressive. As far as the Churches were concerned, this is similar to what happened in Spain. Churches were converted to Mosques and back to Churches with the change of the "dispensation" as we love to say in India. Other Churches were turned into Hospitals and libraries. Churches were built upon and expanded and the newer construction is clearly visible, especially in the Boyana Church.
St Alexander Nevski Cathedral.
St Nikolas Russian Church
Catedral de Sveta-Nedelya
The Churches are all small compared to many of the magnificent Cathedrals and Churches you have in Europe like in Cologne, Aachen, Paris and so on. Not only are these smaller but they are simple in architecture and construction inside. No stately organs, no statues and elaborate altars. I am unsure whether this is due to the fact that the religion is Christian Orthodoxy or due to the fact that Bulgaria is poor and has been so historically.
There was a Roman walled city in existence in the center of Sofia. This was uncovered when the underground was being constructed. The route of the underground was altered and the excavated are has been preserved and integrated into the stations. Well done. In this area you also have 4 religious monuments standing cheek by jowl. The Bulgarians take great pride in telling you that this shows tolerance and secularism. These 4 are a Synagogue, a Mosque, an Orthodox Church and a Roman Christian Church.
Also, in this area stood a statue of Lenin. In 2000 after the fall of Communism, the Lenin statue was removed and replaced by a huge statue of a lady symbolizing Sofia, as a sort of dedication to the city. Of course, much like in India, much scandal was associated with this statue. The allegation was that the face of the statue was very similar to the face of the Mayors wife. The initial praise of the statue by the Orthodox Church soon turned to criticism as the Owl and the Laurel Wreath held by Sofia were Pagan symbols. Further "Sofia" had her cleavage showing with very prominent nipples. Horrors! The lady could not be a Saint!
Sofia also has hot springs. Around these springs a "Hammam" or public bath was constructed. Very beautiful building. Today the building houses a Museum, not very big nor impressive, unfortunately. The springs still exist and the water is channeled into several outlets. The water is drinkable and comes out warm at 38C. It has a slight metallic taste. People come by and fill large bottles, like our Bisleri 25-liter bottles and take the water home as it has medicinal values. The liver and kidneys are benefited drinking this. I had a few glugs, not feeling better!
Above: 4 pictures of the former Sofia Mineral Baths. Now the Sofia Regional Historical Museum
Buildings in Sofia were schizophrenic. At times you saw beautiful well preserved well maintained old buildings. At times there was simply a sprawl of poorly maintained concrete building much like what we have back home in Mumbai. In fact, these buildings were a lot like Mumbai buildings. Lots of abandoned under construction buildings too, like in Mumbai. I could not discern any pattern of upmarket streets, just a hotch potch of good buildings and poor buildings along the street. Some streets had no good buildings at all, just what seemed to be run down housing. Of course, in the outskirts, i.e. near the airport as well as on the way to the Boyana Church we saw large modern office buildings.
Above: The beautiful
Above: The ugly.
Above: Some Soviet era buildings. All government offices
Much is made of the Boyana Church. This is a UNESCO World Heritage Site located a few kilometers out of town up in the hills. You need to take a taxi to and fro. Ask the cabbie to wait. Now I have a very dim view of something that is branded a UNESCO World Heritage Site. I have always found them subpar, with wheelchair access being the most notable feature. These also happen to be "Note Chaaapne Ka Machines" if you know what I mean. The Boyana Church is small, I mean small. To enter costs 7 Euro or 14 Lev. Only 8 people are allowed to enter at a time for 10 minutes. Crowds are huge. We got there by 10.30am and were informed that we could enter at 1.45pm. We simply abandoned the idea and had a walk around the Church. That took all of 3 minutes. What a huge let down. Let me clarify, on looking on the Internet, the insides of the church are really nothing to get excited about. In retrospect no regrets at all in not going in.
Sofia is famous for its yellow or golden cobblestoned streets. These cobblestones line many of Sofia's roads. They are treacherous when wet or icy, cars skid and pedestrians slip. If you wish to practice orthopedics Sofia is the place to be. The story behind these cobblestones shows once again, the historic corruption in Bulgaria, much like our own Bharat. The King Tsar Ferdinand believed the muddy streets would be beautified with the installation of these gold cobblestones. He claimed that the cobblestones were a gift from the Austro Hungarian King who was his cousin. It turns out that this was bollocks, the State was paying for these cobblestones. Sound familiar? They are now being steadily replaced with the less slippery grey variety.
Sofia, with its years of Communism as well as under the Ottoman Empire, has several rather grim statues. The city has several open spaces and parks. It is a green city and people have lots of benches to sit and relax. It is quite pleasant. These statues, a few of which I have photographed are quite a contrast to the idyllic parks they are located in.
No one plucks flowers for "Pooja".
To conclude, we thoroughly enjoyed Sofia. If you are efficient and methodical in your sightseeing, 2 nights are enough, 3 if you want to be relaxed. Do visit as a second-tier city. In some ways Sofia is like a mini Budapest. These old former Communist cities have a lot to offer. Recommended as a short break.
Celeste - London
Why I like London.


I am subing out to a sweeping company to do my local walmart. I am paying him 1500 a month or 125 per visit to clean the parking lot. I notice he never picks up any of the sand and I personally am not happy. I dont want to loose the contract for Lawn and Snow, so I am thinking about getting into street sweeping and picking up more walmarts. I am wondering what kind of used unit would be Ideal for a walmart lot. I have looked at swharze and Tymco so far at there smaller units. Any advice on getting into it? Units for 10k worth buying?
First of all, I would love to earn $125 per cleaning at Walmart. We are at about 1/2 price.
As for sweeper advice - We have two single engine models and at this time they are our last single engine models. While they vacuum pretty well, we have constant issues between truck repair shop, Izusu and sweeper vendor on where is the problem. When the hydraulic system won't work, we are told it is a truck issue. Take the truck to the shop and they tell you the pump assembly violates the warrently. Finally get them over that and they state the problem is with the vacuum unit. It has been both of their problems at times.
With a 2nd engine, like our other 2 trucks, we know where the problem is. If the back motor is not working the back hydraulics are the problem. I know this won't solve everything, but it will help.
Plus the added benefit of moving slow to pick up gravel and sand in the spring time with a dual engine truck versus going over the debris more times with the single engine.
cost of truck? Compare the cost of the back engine with replacing that front pump. The pump is very expensive and so far the only place to get it has been the sweeper vendor.
Thanks for the worthy advice!! I am sure it would help me on somewhere of my career life.
How much do you all think the average cost for a Wal Mart should be monthly for cleaning. I was thinking of putting some numbers in on some sites around here.
You need to set your prices high enough to stay in business and make a profit.
as a general rule I set my pricing by adding all my overhead plus my wages (including mine) and what I want for profit. Profit is what you can take out of the business without affecting the way it is run. personally I set mine at 1/3 each. 1/3rd overhead, 1/3rd for wages, and 1/3rd for profit.
There are many out there that use a different scale that works for them.
Remember if there is no profit built in, then you would be better off just working for someone else and not have any head aches.


Pixoto Members voted Jim Talbert's image as #1 in the Buildings & Architecture category for the Day of 9/07/18.
Decaying and abandoned home in Kansas with the night sky.
8,011,990 images submitted 472,356,559 votes polled Thanks for contributing and participating.


NewTek Discussions > LightWave 3D Support > General Support > normal map problem?
View Full Version : normal map problem?
I'm using Zbrush to generate some normal maps that sould look like old wood (but I've made only a few attempts); then I exported the maps to Lightwave, and with the node editor I've reached this result.
It's very hard to know anything from just the images.
Is that section smaller in the UV map? That would mean it uses fewer pixels, and therefore is lower resolution.
It's possible that it's just the lighting. With a normal map there are not any real shadows being cast, it's just affecting how much of any light it catches (so how bright or dark). When the surface is facing the light, then it will all be the same brightness, when the surface is at an angle to the light, it will have more dark areas and look like it has more detail.
This is just how normal maps work.
Since you're beginner (at least to forum): did you set color-space correctly?
Normal map cannot be modified by color-space conversion routine.
If you are a beginner, here's some more info: if you're in sRGB color space, you must set your normal maps to Linear in the Image editor. (Regular color textures should be in sRGB of course.) Remember to save the object after making changes.
Also, try flipping the y axis. Normal maps from certain programs need this flipped, although I'm not sure if ZBrush is one of them. (I think GoZ is supposed to apply the correct setting for you.) This doesn't affect the blurring you mentioned but it does affect how the surface catches light.
As for the blurring, check that Mip Map is set to Off in the node. You usually want that off for normal maps.
If the texture res is too low, that might cause blurring too...although with Mip Map disabled, it will probably start aliasing if it's too low rather than blur. In this case, it's better to just use a higher res texture.
Change the colorspace here. My settings were correct in the colorspace section in display properties, but not for the normal map which should be adjusted here. Solved my problems.


CharlYe is a young Swiss singer-songwriter. She has been playing the guitar since she was 8 and started composing and writing songs around the same time. As part of a personal school project, she has released her first EP, "Hide", on December 7th 2017, the date on which she played live for the first time. She then played at the Geneva music festival on June 23rd 2018, and a few months later she made it to the finals of the Swiss festival Vernier sur Rock.
Welcome to her intimist folk universe.


May the Face be with You. Star Wars Funko MyMoji! Coming 2017!
Real life emoticons with the Star Wars Funko MyMoji line!
Welcome to the part 37 and the last of our London Toy Fair series. Stay tuned, because the New York Toy Fair series coming up!
Yes, feel free to ignore my terrible pun in the title. What's that, it's the best pun ever? Oh why thank you! Kidding. The last in the London Toy Fair line are these Star Wars themed MyMojis, should we call that ending on a high note? I'm not the biggest fan of their MyMoji line, i purchased a few from the DC series not long ago and the quality was pretty bad. Are there any MyMoji collectors out there? What are your opinions on them. Seeing as they're still making them they must have a few fans out there, if i had to pick my favorite MyMoji line would be the Disney ones. Glam shot and info coming up!
Marvel Funko MyMojis and Lanyards! Coming January!
Funko ever increasing their reach with these new Marvel Funko MyMojis and Lanyards!
Come this January Funko is adding Marvel to their MyMojis line, they're cute i suppose but i'm not exactly a MyMoji collector! If you are then feel free to let us know, and tell us what draws you to them. I was curious about them at first .. but after Pint Size Heroes came out i switched over to them, the Pint Size Heroes are only a dollar more and much better i feel. Secondly there is two more Lanyards being added! These aren't the first, Funko have quite a range of these official Lanyards now .. i wouldn't collect them but they seem cool to use if you're going to Comic Con or some similar event. Glam shots and info coming up!


text:
Medical Ethics Advisor
<p> The placebo effect eases pain even if research participants know the treatment they are receiving has no medical value whatsoever, according to a recent study.</p>
Placebo effect eases pain — even if participants are aware
Effects are "independent of reported expectations for pain relief"
The placebo effect eases pain even if research participants know the treatment they are receiving has no medical value whatsoever, according to a recent study. Some ethical considerations of treatment include the following:
Reinforcing treatment cues with positive outcomes can create placebo effects that are independent of reported expectations for pain relief.
The placebo effect could not be achieved without the initial conditioning effect, which required nondisclosure of placebo use.
The use of deception or non-disclosure may be necessary in clinical trials.
In some cases, the placebo effect still works even if research participants know the treatment they are receiving to ease pain has no medical value whatsoever, according to a recent study.1
Researchers applied a heating element to research subjects' forearms to induce pain sensations, though not enough to burn the skin. They then applied what the subject thought was an analgesic gel and turned down the temperature.
The participants were divided into two groups — those who received four "conditioning" sessions or only one session. After the researchers revealed the treatment was fake, only the group who received four sessions continued to get pain relief from the placebo.
"These findings suggest that reinforcing treatment cues with positive outcomes can create placebo effects that are independent of reported expectations for pain relief," says the study's lead author, University of Colorado Boulder graduate student Scott Schafer.
The effectiveness of a placebo in providing pain relief is often related to a subject's expectations for pain relief. "While conditioning procedures that pair a placebo with pain relief enhance the subsequent placebo response, they also strengthen expectations for pain relief," notes Schafer.
Conditioning effects on pain are largely thought to be linked to changes in expectations. "By that theory, removing expectations should have eliminated the placebo response," says Schafer. "So, in that sense, these results were quite surprising."
Ben A. Rich, JD, PhD, emeritus professor of medicine (Bioethics) at University of California, Davis School of Medicine, participated in the development of ethical guidelines addressing the use of placebos in pain medicine.
The deceptive use of placebos and the misinterpretation of the placebo response to discredit the patient's pain report are unethical and should be avoided, according to the American Pain Society's 2005 position statement.2 In its 2007 Ethics Charter, the American Academy of Pain Medicine states that placebo use for the treatment of pain in other than research settings is usually considered unjustifiable, both for ethical and clinical reasons.3
In light of these two guidelines, says Rich, "the burden of persuasion that the benefits of the deceptive use of placebos outweigh the risks falls on those who would keep patients or research subjects uninformed — either for their own good, or in the cause of the advancement of medical science."
Rich notes that in the 2015 study, the placebo effect could not be achieved without the initial conditioning effect. This required nondisclosure of placebo use. "There is a potential adverse effect on the trust relationship between clinician and patient or clinical investigator and research subject," says Rich.
The research subjects were surprised with the knowledge that their previous treatments were a placebo. "A fully ethical implementation of the same procedure in clinical practice, such as pain management post-surgery, would necessitate a disclosure of future placebo administration," says Schafer.
However, says Schafer, the exact timing of the shift from a verum drug treatment to a placebo could be hidden in order to preserve the conditioning effect. "This could replicate our effect and maintain the effectiveness of a placebo, even after subjects are sure that the treatment has shifted to a placebo," he says.
Use of placebos in human subjects research is generally ethically acceptable when subjects are informed in advance that they may receive a placebo and they nevertheless consent to participate, says Rich. "The deception or deliberate nondisclosure of otherwise pertinent information, whether or not it involves the use of placebos, is likely to undermine trust in either the patient care or clinical research enterprise," he says.
Studying the placebo effect is an important part of medical science, and involves many complex, ambiguous factors, says Joyce Plaza, MS, MBe, manager of New York City-based Columbia University's IRB. "The use of deception or non-disclosure may be necessary," she adds. Plaza says the use of placebos in research is ethical if one of the following two conditions exists:
For studies involving minimal risk, the use of deception is permissible if the study could not be otherwise conducted, the study presents minimal risk, the rights and welfare of the subjects are not adversely affected, and subjects are fully debriefed after their participation, or at the conclusion of the study, if appropriate.
Placebos may be used in double-blinded clinical studies in which subjects are informed that they may receive the placebo instead of the experimental active treatment, but subjects are also provided standard care in addition to the placebo or active treatment. "The review of the studies by the institutional review board would determine that the ethical responsibilities of the research were met," says Plaza.
In an earlier study, openly administered placebos were found to be as effective as lidocaine at treating pain from irritable bowel syndrome.4 "Critically, in this case subjects were informed about the effectiveness of placebos in treating pain, and were implicitly encouraged to expect the placebos to work," notes Schafer.
In the 2015 study, participants were permitted to stop their participation if they experienced discomfort from the heat, and gave informed consent to participate in a study of treatment effects on pain relief. "All participants were fully debriefed at the conclusion of the study," notes Plaza.
Schafer SM Colloca L, Wager TD. Conditioned placebo analgesia persists when subjects know they are receiving a placebo. J Pain 2015; 16(5):412-420.
AAPM Council on Ethics. Ethics charter from American Academy of Pain Medicine. Pain Med 2005; 6(3):203-212.
American Pain Society Ethics Committee. APS position statement on the use of placebos in pain management. J Pain 2005; 6(4):215-217.
Kaptchuk TJ, Friedlander E, Kelley JM, et al. Placebos without deception: A randomized controlled trial in irritable bowel syndrome. PloS One 2010; 5(12):e15591.
Joyce Plaza, MS, MBe, Institutional Review Board, Columbia University, New York City. Phone: (212) 851-7041. Email: [email protected]
Ben A. Rich, JD, PhD, Emeritus Professor of Medicine (Bioethics), UC Davis School of Medicine. Phone: (916) 734-6010. Fax: (916) 734-1531. Email: [email protected]
Scott Schafer, University of Colorado Boulder. Email: [email protected]
The placebo effect eases pain even if research participants know the treatment they are receiving has no medical value whatsoever, according to a recent study.
Medical Ethics Advisor: Online - $449
Issue Date: October 1, 2015
Aggressive end-of-life care persists in cancer patients
Is patient's POLST form inaccessible to provider?
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As the youngest sprout of the familie, Peter Leppert took over the family business in 1993. Before doing so, he became a trained cook and hotel business economist. He worked as a Chef de partie in Lausanne (Switzerland) and as a Chef poissonnier at the hotel Waldhorn in Ravensburg.
The practice-oriented training of young people is very important to the father of six children. He is a member of the examination board of the IHK Karslruhe for cooks and kitchen chefs. He is a lecturer at the European hotel management school as well. The Hotel Jägersteig trains young people in the professions of Cooks and Restaurant Specialists and provides training places for them.