text:
\section{Introduction}

Let $a > 1$ be a positive integer and let $\mu$ be a finitely supported measure on $\mathbf{Z}$. Assume that $\gcd(\opr{supp} \mu - \opr{supp} \mu) = 1$, i.e., that $\mu$ is not supported on a coset of a proper subgroup. Define a random process $(X_n)_{n\geq 0}$ on $\mathbf{Z}$ by $X_0 = 0$ and
\[
  X_{n+1} = a X_n + b_n,
\]
where $b_1, b_2, \dots$ are independent and distributed according to $\mu$. Let $\mu_n$ be the law of $X_n$. In other words, $\mu_n$ is defined by the recursion
\begin{align*}
  \mu_0 &= \delta_0, \\
  \mu_n &= \mu * (m_a)_* \mu_{n-1} \qquad (n\geq 1),
\end{align*}
where $m_a(x) = ax$ is the multiplication-by-$a$ map.

Now for any positive integer $q$ consider the reduction $X_n \bmod q$. The mixing time of this random process, loosely speaking the smallest $n$ such that
\[
  \|\mu_n \bmod q - u_q\| = o(1),
\]
where $u_q$ is the uniform measure on $\mathbf{Z} / q\mathbf{Z}$ and $\|\cdot\|$ denotes $l^1$ norm, was first studied in detail by Chung, Diaconis, and Graham~\cite{CDG} (motivated by pseudorandom number generation). In the prototypical case $a = 2$, $\mu = u_{\{-1, 0, 1\}}$ (uniform on $\{-1, 0, 1\}$), the following bounds have been proved.
\begin{itemize}
  \item If $n \geq (c_1 + \varepsilon) \log_2 q$, where
  \[
    c_1 = \left( 1 - \log_2 \left( \frac{5 + \sqrt{17}}{9} \right)\right)^{-1} \approx 1.02,
  \]
  then
  \[
    \frac12 \|\mu_n \bmod q - u_q\| \leq \varepsilon
  \]
  for almost all odd $q$ (Chung--Diaconis--Graham~\cite{CDG}).
  \item If $n \leq c_2 \log_2 q$, where
  \[
    c_2 = \frac1{-\frac12 - \frac4{18} \log_2(\frac{4}{36}) - \frac{5}{18} \log_2(\frac{5}{36})} \approx 1.004,
  \]
  then
  \[
    \frac12 \|\mu_n \bmod q - u_q\| \to 1
  \]
  as $q \to\infty$ (Hildebrand~\cite{hildebrand}).
  For further work in this direction, see \cite{Hil2}, \cite{Nev} and the references therein.
\end{itemize}
Thus the mixing time of $X_n \bmod q$ is a slight constant multiple larger than $\log_2 q$, for typical $q$. (Chung, Diaconis, and Graham also show that there are infinitely many odd $q$ for which the mixing time exceeds $c \log q \log \log q$.)

We advance the following explanation of this phenomenon. Defining the entropy of any measure $\mu$ by
\[
  H(\mu) = \sum_x \mu(\{x\}) \log \mu(\{x\})^{-1},
\]
the \emph{asymptotic entropy} (or \emph{entropy rate}) of $(X_n)_{n\geq0}$ is defined by
\[
  H(a, \mu) = \lim_{n\to\infty} \frac1n H(\mu_n).
\]
It is well known that the limit exists, and that
\[
  0 < H(a, \mu) \leq \min(\log a, H(\mu)).
\]
In the case of $\mu = u_{\{-1, 0, 1\}}$, a computation shows that the entropy $H(2, \mu)$ is slightly less than $\log 2$. On the other hand, $X_n \bmod q$ cannot equidistribute before the entropy of $\mu_n$ reaches $(1-o(1))\log q$. Hence it takes slightly more than $(\log q) / (\log 2)$ steps for $X_n \bmod q$ to mix.

In general, the entropy $H(a, \mu)$ and the mixing time of $X_n \bmod q$ are related by our main theorem, provided that $H > \frac12 \log a$.

\begin{theorem} \label{mainthm}
Let $a$ and $\mu$ be given. Let $H = H(a, \mu)$. The following hold for all $\d>0$.
\begin{enumerate}[label=(\roman*)]
  \item If $Hn \leq (1-\delta) \log q$ then
  \[
    \frac{1}{2}\|\mu_n \bmod q - u_q\| = 1 - o(1).
  \]
  \item Assume $H > \frac12 \log a$. Then for a proportion $1-o(1)$ of primes $q$ such that $\log q \leq (1-\delta) Hn$, we have
  \[
    \|\mu_n \bmod q - u_q \| = o(1).
  \]
  The same holds for a proportion $1-o(1)$ of the composite $q$ such that $(q, a) = 1$.
\end{enumerate}
\end{theorem}

We have already sketched the proof of the first part. The proof of the second part consists of the following ideas:
\begin{enumerate}
  \item By a version of the Shannon--McMillan--Breiman theorem (a.k.a.\ asymptotic equipartition), there is a set $S$ of $\mu_n$-measure $1-o(1)$ such that
  \[
    \mu_n(\{x\}) = \exp(-(H + o(1)) n)
  \]
  for all $x \in S$. In particular, there is a measure $\nu_n$ (the conditional measure on $S$) such that $\|\mu_n - \nu_n\| = o(1)$ and \[\log \|\nu_n\|_2^{-2} = (H + o(1))n.\]
  \item By the large sieve, for most primes $q \asymp |S|$, the $l^2$ norm of $\nu_n \bmod q$ is not much larger than the $l^2$ norm of $\nu_n$. Here we use the hypothesis that $H > \frac{1}{2} \log a$.
  \item Therefore for most primes $q \asymp |S| = \exp((H + o(1))n)$, $\mu_n \bmod q$ is close to uniform, so the mixing time modulo $q$ is at most $n$.
\end{enumerate}
The proof in the composite case also uses the large sieve, but in a more involved way.

We believe that the hypothesis $H > \frac12 \log a$ is unnecessary.  It might be possible to relax this to $H \ge \frac12 \log a$ by a suitable modification of our proof. We briefly sketch some ideas towards this in Section \ref{sc:OP}, but we do not pursue this in detail. The problem of removing the hypothesis entirely is related to some open questions about the large sieve: it is believed that the only examples for which the large sieve is sharp are somehow quadratic (Green--Harper~\cite{green--harper}), but the support of $\mu_n$ cannot be anything like quadratic. Although we have not succeeded in removing this hypothesis, we can adapt Gallagher's larger sieve to prove the following bound which is weaker by a factor of $2$, and holds only for a looser sense of ``almost all''.

\begin{theorem} \label{larger-sieve-thm}
Let $\varepsilon > 0$, and let $\primes_n$ be the set of primes $p$ such that
\[
  \log p \leq (1/2-\varepsilon) Hn
\]
and
\[
  \|\mu_n \bmod p - u_p \| \geq \varepsilon.
\]
Then
\[
  \sum_{{p \in \primes_n}} \frac{\log p}{p} \ll_\varepsilon n^{1/2}.
\]
\end{theorem}

By contrast, Mertens' first theorem states
\[
  \sum_{\log p \leq X} \frac{\log p}{p} = X + O(1).
\]
Thus the theorem asserts $\primes_n$ has asymptotically zero density among primes $p$ such that $\log p \leq (1/2-\varepsilon)Hn$ when the primes are weighted by $(\log p)/p$. Equivalently, the mixing time of $X_n \bmod p$ is at most $(2+o(1)) H^{-1} \log p$ for almost all primes in almost all dyadic intervals.

The entropy $H(a, \mu)$ is equivalent, by reversal and rescaling, to the entropy of the self-similar Cantor-like measure $\lambda$ on $\mathbf{R}$ defined by
\[
  \lambda = \mu * (m_{1/a})_* \lambda,
\]
where $m_{1/a}(x) = x/a$. This general class of measures includes the case of Bernoulli convolutions, which were studied extensively by Erd\H{o}s, Garsia, and others. In the particular case $a=2, \mu = u_{\{-1, 0, 1\}}$, Hare, Hare, Morris, and Shen~\cite{hhms}*{Theorem~3.1} prove that
\[
  \frac{H}{\log 2} = \log 3 - \frac{3}{2} L(1/3),
\]
where $L$ is the analytic function in the unit disk defined by
\[
  L(z) = (1 - 3z)^2 \sum_{n=1}^\infty z^n \sum_{\substack{1 \leq i < j \\ \gcd(i, j) = 1 \\ e(i, j) = n}} j \log j,
\]
where $e(i, j)$ is the number of subtractions made by the Euclidean algorithm applied to $(i, j)$ (the singularity $z = 1/3$ is removable). They deduce
\[
  \frac{H}{\log 2} = 0.9887658714\dots.
\]
Since in particular this is greater than $0.5$, Theorem~\ref{mainthm} applies, so the mixing time of $X_n \bmod q$ is almost always (and never less than)
\[
  (1.01136176816\dots) \log_2 q.
\]

\subsection{Notation}

We write $\|\mu\|$ for the standard norm for a complex measure $\mu$, so that the maximum distance between two probability measures $\mu$ and $\nu$ is $2$ (many authors divide this norm by $2$).
When we write $\|\mu\|_2$ or $H(\mu)$ we are implicitly identifying $\mu$ as the function $\mu(\{x\})$, its density with respect to the discrete measure, so that
\begin{align*}
  \|\mu\|_2^2 &= \sum_x \mu(\{x\})^2, \\
  H(\mu) &= \sum_x \mu(\{x\}) \log \mu(\{x\})^{-1}.
\end{align*}

Likewise we normalize the Fourier transform with the discrete measure on the physical side $\mathbf{Z}$ and the uniform measure on the frequency side $\mathbf{R}/\mathbf{Z}$; thus
\begin{align*}
  \muhat(\xi) &= \sum_{x \in \mathbf{Z}} \mu(\{x\}) e(-\xi x) = \int e(-\xi x) \, d\mu(x), \\
  \mu(\{x\}) &= \int_\xi \muhat{}(\xi) e(\xi x) \, d\xi,
\end{align*}
where as usual $e(x) = e^{i 2 \pi x}$.
We keep this normalization when we reduce modulo $q$, so the Fourier transform of $\mu \bmod q$ is just $\muhat(r/q)$ $(0\leq r < q)$. The inversion formula modulo $q$ is
\[
  (\mu \bmod q)(\{x\}) = \mu(x + q\mathbf{Z}) = \frac1q \sum_{r=0}^{q-1} \muhat(r/q) e(rx/q).
\]
We note that the above choice of definition of the Fourier transform differs from that in \cite{CDG} in a sign.

We use standard asymptotic notation from analytic number theory where expedient: $X \ll Y$ or equivalently $X = O(Y)$ means $X \leq CY$ for some constant $C>0$; $X \asymp Y$ means $X \ll Y$ and $Y \ll X$; $X\sim Y$ means $X/Y\to 1$ as $n\to \infty$. Constants $C, c$, etc, are usually allowed to depend on $\mu$ and $a$.

\subsection{Organization of the paper}

We discuss the Shannon--McMillan--Breiman theorem for random walks (and the rate of convergence) in Section~\ref{sc:SMB}.
The lower bound in Theorem~\ref{mainthm} is proved in Section~\ref{sc:lower}.
In Section~\ref{sc:small-moduli}, we revisit an argument in \cite{CDG} to prove a crude upper bound on the mixing time which is valid for all moduli.
This will be used in later sections in the proof of the upper bound on the mixing time in Theorem~\ref{mainthm} to treat small divisors of composite moduli.
In Sections~\ref{sc:large-sieve} and \ref{sc:alternative}, we give two different proofs of the upper bound in Theorem~\ref{mainthm}.
We give the proof of Theorem~\ref{larger-sieve-thm} in Section~\ref{sc:larger-sieve}.
We conclude the paper with some open problems in Section~\ref{sc:OP}.

\section{SMB via Efron--Stein}\label{sc:SMB}

We need a version of the Shannon--McMillan--Breiman theorem which states that
\[
  \mu_n(\{x\}) = \exp(-H(\mu_n) + o(n))
\]
for asymptotically $\mu_n$-almost-all $x$.

\begin{theorem} \label{SMB}
There is a constant $C$ depending only on $\mu$ such that
\[
  \mu_n\left( \{ x: |-\log \mu_n(\{x\}) - H(\mu_n)| \geq \a n \}\right) \leq \frac{C}{\a^2 n}
\]
for all $n\geq 1$ and $\a > 0$.
\end{theorem}

The Shannon--McMillan--Breiman theorem for a random walk on a group is due to
Derriennic~\cite{Der-sub-additive}*{Section IV}, and Kaimanovich and Vershik~\cite{KV-RW}*{Theorem 2.1}.
For our application, we need an estimate for the rate of convergence, and for this reason we give a proof.
Our proof exploits special features of the walk and is based on the Efron--Stein inequality.

Fix some $n\geq 1$, and let $b_1, \dots, b_n$ be independent $\mu$-distributed random variables. For $m \in \mathbf{Z}$ and $j \in \{1, \dots, n\}$, let
\begin{align*}
  g(m) &= \mathbf{P}\left(\sum_{i=1}^n b_i a^i = m\right),\\
  g_j(m) &= \mathbf{P}\left(\sum_{i\neq j} b_ia^i = m\right).
\end{align*}
(Note that $g(x) = \mu_n(\{x\})$.) For $x_1, \dots, x_n \in \mathbf{Z}$ define
\begin{align*}
  f(x_1,\ldots,x_n) &=\log g\left(\sum_{i=1}^n x_ia^i\right)\\
  f_j(x_1,\ldots,x_n) &=\log g_j\left(\sum_{i\neq j} x_ia^i\right).
\end{align*}
Our aim is show that the random variable
\[
  Z = f(b_1, \dots, b_n).
\]
is concentrated around its mean
\[
  \mathbf{E}[Z] = -H(\mu_n).
\]
We will apply the Efron--Stein inequality (in the asymmetric case due to Steele~\cite{steele--efron--stein}). For $j\in\{1,\ldots,n\}$, let
\[
  Z_j = f(b_1, \dots, b_{j-1}, b_j', b_{j+1}, \dots, b_n),
\]
where $b_j'$ is an independent copy of $b_j$. The Efron--Stein inequality states
\begin{equation} \label{efron--stein}
  \opr{Var}[Z]\le\frac12\sum_{j=1}^{n}\mathbf{E}[(Z-Z_j)^2].
\end{equation}

\begin{lemma}\label{lem:variance}
For $j \in \{1,\dots,n\}$ we have
\[
  \mathbf{E}[(Z-f_j(b_1,\dots,b_n))^2]
  = \mathbf{E}[(Z_j-f_j(b_1,\dots,b_n))^2]
  \le C,
\]
where $C$ is a constant depending only on $\mu$. In particular,
\[
  \mathbf{E}[(Z - Z_j)^2] \leq 4C.
\]
\end{lemma}
\begin{proof}
The equality in the claim holds because $f_j$ does not depend on the $j$th variable and $b_j'$ is an independent copy of $b_j$ (so in fact $Z - f_j(b_1, \dots, b_n)$ and $Z_j - f_j(b_1, \dots, b_n)$ are identically distributed). The last statement will follow by
\begin{align*}
  \mathbf{E}[(Z - Z_j)^2]
  &\leq 2 \left( \mathbf{E}[(Z - f_j(b_1, \dots, b_n))^2] + \mathbf{E}[(Z_j - f_j(b_1, \dots, b_n))^2] \right) \\
  &\leq 4C.
\end{align*}
Hence it suffices to prove the main claim that
\[
  \mathbf{E}[(Z - f_j(b_1, \dots, b_n))^2] \leq C.
\]

Let $m\in\mathbf{Z}$ and $x\in\opr{supp} b_j$. First,
\[
  g(m+xa^j)\ge\mathbf{P}(b_j=x)\mathbf{P}\left(\sum_{i\neq j}b_ja^j=m\right)\ge c g_j(m),
\]
where $c>0$ is a constant depending only on $\mu$, namely,
\[
  c = \min_{x \in \opr{supp} \mu} \mu(\{x\}).
\]
Second,
\begin{align*}
  g(m+xa^j)
  &= \sum_{y \in \opr{supp} b_j} \mathbf{P}(b_j=y)\mathbf{P}\left(\sum_{i\neq j}b_ia^i=m+(x-y)a^j\right)\\
  &\le \max_{y\in\opr{supp} b_j} g_j(m+(x-y)a^j).
\end{align*}
Thus, for $(x_1, \dots, x_n) \in (\opr{supp} \mu)^n$,
\[
  f(x_1,\ldots,x_n) \ge f_j(x_1,\ldots,x_n)+\log c,
\]
and if
\[
  f(x_1,\ldots,x_n) \ge f_j(x_1,\ldots,x_n)+t
\]
for some $t\ge 0$ then
\[
  \sum_{i\neq j} x_ia^i \in M_t,
\]
where
\[
  M_t = \{m\in\mathbf{Z}:\exists x\in\opr{supp} (b_1-b_2): g_j(m)\le e^{-t} g_j(m+xa^j)\}.
\]

We show that
\begin{equation}\label{eq:ESclaim}
  \mathbf{P}\left(\sum_{i\neq j} b_ia^i \in M_t\right)
  = \sum_{m\in M_t} g_j(m)\le|\opr{supp} (b_1-b_2)| e^{-t}.
\end{equation}
We consider the set
\[
  M_{t,x} = \{m\in\mathbf{Z}: g_j(m)\le e^{-t} g_j(m+xa^j)\}
\]
separately for each $x\in\opr{supp} (b_1-b_2)$. Since $\sum g_j = 1$ we have
\[
  \sum_{m \in M_{t, x}} g_j(m) \leq e^{-t} \sum_{m \in M_{t,x}} g_j(m + xa^j) \leq e^{-t}.
\]
Summing over $x \in \opr{supp}(b_1 - b_2)$, we get the claim \eqref{eq:ESclaim}.

Combining our estimates we have
\begin{equation} \label{Z-tail-bound}
  \mathbf{P}(|Z-f_j(b_1,\dots,b_n)|>t)< C e^{-t}
\end{equation}
for each $t>C$, where $C$ is a constant depending only on $\mu$. Thus indeed
\begin{align*}
  \mathbf{E}[(Z - f_j(b_1, \dots, b_n))^2]
  &= 2 \int_0^\infty t\mathbf{P}(|Z - f_j(b_1, \dots, b_n)| > t) \, dt \\
  &\leq C'.
\end{align*}
This completes the proof of the lemma.
\end{proof}

We can now prove Theorem~\ref{SMB}. By Lemma~\ref{lem:variance} and the Efron--Stein inequality \eqref{efron--stein}, we have $\opr{Var}[Z] \leq 2Cn$. Hence by Chebyshev's inequality
\[
  \mathbf{P}(|Z - \mathbf{E}[Z]| \geq \a n) \leq \frac{2C}{\a^2 n}.
\]
By definition, $Z$ has the same distribution as $\log \mu_n(\{X_n\})$ and $\mathbf{E}[Z] = -H(\mu_n)$. This proves the theorem.

\begin{corollary} \label{cor:nu_n-exists}
Let $\a = \a(n) > 0$, with $\a^2 n \to \infty$. There is a probability measure $\nu_n$ such that
\begin{enumerate}
  \item $\|\mu_n - \nu_n\| \ll 1 / (\a^2 n)$,
  \item $\opr{supp} \nu_n \subset \opr{supp} \mu_n$,
  \item $\nu_n(\{x\}) = \exp(-H(\mu_n) + O(\alpha n))$ for every $x \in \opr{supp} \nu_n$.
\end{enumerate}
\end{corollary}

The corollary follows by defining
\begin{align*}
  &S = \{x : |-\log\mu_n(\{x\}) - H(\mu_n)| \leq \a n\}, \\
  &\nu_n(A) = \mu_n(A \cap S) / \mu_n(S).
\end{align*}
By Theorem~\ref{SMB},
\[
  \|\mu_n - \nu_n\| = 2\mu_n(S^c) \ll \frac1{\alpha^2n}.
\]

\begin{remark}
The key estimate \eqref{Z-tail-bound} is a decent tail bound, so it may be possible to get something much stronger with more sophisticated concentration inequalities. See for example the book of Boucheron, Lugosi, and Massart~\cite{blmbook}*{Section~6.9}. Persuing this could lead to an improved estimate for the size of the cut-off window in the mixing of $X_n \bmod q$.
\end{remark}

\section{The lower bound}\label{sc:lower}

In this section we prove Theorem~\ref{mainthm}(i), which states that the mixing time of $X_n \bmod q$ is at least $H^{-1} \log q$ for all sufficiently large $q$. More precisely, if $H n \leq (1-\delta) \log q$ for some constant $\delta > 0$ then
\[
  \frac12 \|\mu_n \bmod q - u_q\| \to 1 \qquad (q \to \infty).
\]
By Corollary~\ref{cor:nu_n-exists}, there is a measure $\nu_n$ such that
\begin{align*}
  &\|\mu_n - \nu_n\| \ll 1 / (\a^2 n), \\
  &|\opr{supp} \nu_n| \leq \exp(H(\mu_n) + O(\a n)).
\end{align*}
From the bound on $|\opr{supp} \nu_n|$,
\begin{align*}
  \frac12 \|\nu_n \bmod q - u_q\|
  &\geq u_q((\opr{supp} \nu_n \bmod q)^c) \\
  &\geq 1 - \exp(H(\mu_n) + O(\alpha n) - \log q).
\end{align*}
Now $H(\mu_n) = Hn + o(n)$, and $\log q - H n \geq \delta \log q \geq \delta H n$. Hence, if $\alpha$ is a constant sufficiently smaller than $\delta H$,
\[
  \frac12 \|\nu_n \bmod q - u_q\| = 1 - e^{-cn},
\]
so
\[
  \frac12 \|\mu_n \bmod q - u_q\| = 1 - O(1/n).
\]

\begin{remark}
To prove just that $\|\mu_n \bmod q - u_q\|$ does not tend to zero, we do not need the SMB theorem. Suppose $H(\mu_n) \leq (1-\delta) \log q$. By Markov's inequality,
\[
  \mu_n(\{x : \log \mu_n(\{x\})^{-1} > \log q - 10 \}) \leq \frac{H(\mu_n)}{\log q - 10} \leq 1-\delta + o(1).
\]
Therefore $\mu_n(\{x\}) \geq e^{10} / q$ on a set $S$ of $\mu_n$-measure at least $\delta + o(1)$, so
\[
  \frac12 \| \mu_n \bmod q - u_q \| \geq (1 - e^{-10}) (\delta + o(1)).
\]
\end{remark}

\section{Small moduli}\label{sc:small-moduli}

In the next section we will need to know that $\mu_n \bmod q \approx u_q$ for any modulus $q$ such that $(q, a) = 1$ and $q = o(\log n / \log \log n)$. In other words, the mixing time of $X_n \bmod q$ is at most $O(\log q \log \log q)$ for all $q$. This was proved in \cite{CDG}*{Section~4} in the model case $a = 2$, $\mu = u_{\{-1, 0, 1\}}$, which was later generalized by Hildebrand in \cite{Hil-thesis}*{Chapter 3}
and \cite{Hil0}*{Theorem 2}. We state the result in the form we will use it.

\begin{lemma} \label{lem:small-moduli}
Assume $(q, a) = 1$, and that $n > C \log q \log \log q$ for a sufficiently large constant $C$ (depending on $\mu$ and $a$). Then
\[
  q\| \mu_n \bmod q - u_q \|_2^2 \leq e^{-c n / \log q}
\]
for some constant $c>0$.
\end{lemma}

While all the ideas required for the proof are already in \cites{CDG, Hil-thesis, Hil0}, we include the proof for the reader's convenience.

\begin{proof}
Since by assumption $\mu$ is not supported on a coset of a proper subgroup, $\muhat(\xi)=1$ only if $\xi = 0$ (for $\xi \in \mathbf{R}/\mathbf{Z}$). Moreover, for a $\mu$-distributed random variable $X$,
\begin{align*}
  \muhat'(0) &= \partial_{\xi=0} \mathbf{E} e(\xi X) = i 2 \pi \mathbf{E} X, \\
  \muhat''(0) &= \partial^2_{\xi=0} \mathbf{E} e(\xi X) = -(2 \pi)^2 \mathbf{E} X^2 < 0,
\end{align*}
so there is some constant $c>0$ such that
\[
  |\muhat(\xi)| \leq e^{-c |\xi|^2},
\]
where $|\xi|$ denote the distance from $\xi$ to $0$ in $\mathbf{R}/\mathbf{Z}$.

In general, for $n \geq 1$ we have
\begin{equation} \label{muhatn-formula}
  \muhat[n](\xi) = \prod_{i=0}^{n-1} \muhat(a^i \xi).
\end{equation}
Let
\[
  \xi = 0.\xi_1\xi_2\xi_3\dots \qquad (0 \leq \xi_i \leq a-1)
\]
be the base-$a$ expansion of $\xi$. Then
\[
  |\muhat[n](\xi)| \leq e^{-c' A_n(\xi)},
\]
where $A_n(\xi)$ is the number of indices $i \in \{1, \dots, n-1\}$ such that we do \emph{not} have
\[
  \xi_i = \xi_{i+1} \in \{0, a-1\},
\]
as in all other cases we have $|a^{i-1} \xi| \geq 1/a^2$.

Let $n_0 = \ceil{\log_a q}$ and $n = kn_0$ for some $k\in\mathbf{Z}_{>0}$. By H\"older's inequality,
\begin{align*}
  q\|\mu_n \bmod q - u_q \|_2^2
  &= \sum_{r\neq 0} |\muhat[n](r/q)|^2 \\
  &= \sum_{r\neq 0} |\muhat[n_0](r/q)|^2 \cdots |\muhat[n_0](a^{(k-1)n_0} r/q)|^2 \\
  &\leq \sum_{r\neq 0} |\muhat[n_0](r/q)|^{2k}.
\end{align*}
Note $|r/q - r'/q| \geq 1/q \geq a^{-n_0}$ for $r \neq r'$, and in particular $|r/q| \geq a^{-n_0}$ for $r \neq 0$. Thus every $r/q$ is identified by its first $n_0$ digits $\xi_1, \dots, \xi_{n_0}$, and unless $r=0$ we cannot have $\xi_1 = \cdots = \xi_{n_0} \in \{0, a-1\}$. The number of $\xi = 0.\xi_1\cdots\xi_{n_0}$ with $A_{n_0}(\xi) = A$ is bounded by
\[
  \binom{n_0}{A} a^{A+1},
\]
as unless $i$ counted by $A_{n_0}(\xi)$ then $\xi_{i+1} = \xi_i$. Hence
\begin{align*}
  q\|\mu_n \bmod q - u_q \|_2^2
  &\leq \sum_{\xi = 0.\xi_1 \cdots \xi_{n_0} \neq 0} e^{-c' A_{n_0}(\xi) k} \\
  &\leq \sum_{A=1}^{n_0} \binom{n_0}{A} a^{A+1} e^{-c'A k} \\
  &= a((1 + a e^{-c' k})^{n_0} - 1) \\
  &\leq a(\exp(a e^{-c'k} n_0) - 1).
\end{align*}
Provided that $k > C \log \log q$ for a sufficiently large constant $C$, this is bounded by
\[
  O(a^2 e^{-c' k} n_0) \le e^{-c'' n / \log q},
\]
as claimed.
\end{proof}

\section{The large sieve argument}\label{sc:large-sieve}
The purpose of this section is to prove Theorem~\ref{mainthm}(ii).

\def\mathcal{M}{\mathcal{M}}
\newcommand{\piop}[1]{\pi_{#1}}
\newcommand{\Pop}[2][q_0]{P_{#2}^{(#1)}}

For any group $G$, let $\mathcal{M}(G)$ be the space of complex measures on $G$. For $q_1 \mid q$, we consider $\mathcal{M}(\mathbf{Z}/q_1\mathbf{Z})$ to be the subspace of $\mathcal{M}(\mathbf{Z}/q\mathbf{Z})$ or $\mathcal{M}(\mathbf{Z})$ consisting of measures which are uniform on fibres mod $q_1$. For integers $q_0 \mid q$, we define two projection operators
\[
  \piop{q}, \Pop{q} : \mathcal{M}(\mathbf{Z}) \to \mathcal{M}(\mathbf{Z}/q\mathbf{Z}).
\]
The easiest way to give their definition is in terms of Fourier multipliers: for $(r,s) = 1$, $s \mid q$,
\begin{align*}
  \widehat{\piop{q} \nu}(r/s)
  &= \widehat{\nu}(r/s) 1_{s \mid q} \\
  \widehat{\Pop{q} \nu}(r/s)
  &= \widehat{\nu}(r/s) 1_{[s, q_0] = q}.
\end{align*}
Notice that
\begin{align*}
  \piop{q} \nu &= \nu \bmod q \\
  &= \sum_{s: q_0 \mid s \mid q} \Pop{s} \nu,\\
  \Pop{q} \nu &= \sum_{s : q_0 \mid s \mid q} \mu(s / q_0) \piop{s} \nu,
\end{align*}
where $\mu$ is the M\"obius function.
From these identities, we deduce that $\|\piop{q}\| = 1$ and $\|\Pop{q}\| \leq d(q/q_0)$,
where $\|\cdot\|$ stands for the operator norm with respect to the $l^1$ norm, and $d(\cdot)$
stands for the divisor function.

\begin{lemma} \label{lem:large-sieve}
Assume $2H > (1+\varepsilon) \log a$. Let $Q$ be large, and let $n$ be large enough that
\[
  H(\mu_n) > (1 + \varepsilon) \log Q.
\]
Then for all but at most $Q^{1-\varepsilon/3}$ many $q \in [Q/2, Q]$, and for all $q_0 < q^{\varepsilon / 3}, q_0 \mid q$, we have
\[
  \|\Pop{q} \mu_n\| \ll \frac{\|\Pop{q}\|}{\log q}.
\]
\end{lemma}

Note that if $q$ is prime then $\Pop[1]{q} = \pi_q - \pi_1$, and
\[
  \Pop[1]{q} \mu_n = \mu_n \bmod q - u_q.
\]
Thus the prime case of Theorem~\ref{mainthm}(ii) follows immediately. The composite case will require more work, which we explain at the end of this section.

The proof of the lemma relies on the large sieve inequality, which we recall now.
Letting $q$ be an integer, we write $R(q)$ for the residues modulo $q$ that are coprime to $q$.

\begin{theorem}[\cite{friedlander--iwaniec--opera}*{Theorem 9.3}]\label{th:large-sieve}
Let $f:\mathbf{Z}\to\mathbf{C}$ be a function supported on $[-N,N]$ for some $N\in\mathbf{Z}_{>0}$.
Then
\[
\sum_{q\le Q}\sum_{r\in R(q)} |\widehat f(r/q)|^2\le (Q^2+2N)\sum_n |f(n)|^2.
\]
\end{theorem}

We record the following corollary.
\begin{corollary}\label{cr:large-sieve}
Let $q_0,Q,N\in\mathbf{Z}_{>0}$, and let $\nu$ be a probability measure supported on $\mathbf{Z}\cap[-N,N]$.
Then
\[
\sum_{\substack{q \in [Q/2, Q]\\ q_0|q}} q\|P_q^{(q_0)}\nu\|_2^2\ll(Q^2+N)\|\nu\|_2^2.
\]
\end{corollary}

\begin{proof}
By definition, we have
\[
q\|P_q^{(q_0)}\nu\|_2^2=\sum |\widehat\nu(r/s)|^2,
\]
where the summation on the right extends over all fractions $r/s$
such that $s \mid q$, $[s,q_0]=q$ and $r\in R(s)$.
We sum this up for $q \in [Q/2, Q]$ and $q_0 \mid q$, and note that each fraction $r/s$ will appear
at most once on the right hand side.
Therefore, we can invoke Theorem~\ref{th:large-sieve}, which yields the desired inequality.
\end{proof}

\begin{proof}[Proof of Lemma \ref{lem:large-sieve}]
Note that $\|\Pop{q}\mu_n\|$ is monotonic in $n$, as $\Pop{q}$ commutes with convolution:
\begin{align*}
  \|\Pop{q}\mu_{n+1}\|
  &= \|\Pop{q} (\mu_n * (m_{a^n})_* \mu)\|\\
  &= \|(\Pop{q}\mu_n) * (m_{a^n})_* \mu\|\\
  &\leq \|\Pop{q}\mu_n\|.
\end{align*}
Hence, by reducing $n$ if necessary, and by assuming $Q$ is large enough, we may assume that
\[
  H(\mu_n) \sim H n \sim (1 + \varepsilon) \log Q.
\]
In particular, since $2H > (1+\varepsilon) \log a$, we have $Q^2 > a^n$.

By Corollary~\ref{cor:nu_n-exists}, there is a measure $\nu_n$ such that $\|\mu_n - \nu_n\| \ll \a^{-2}/n$ and
\[
  \log \|\nu_n\|_2^{-2} = H(\mu_n) + O(\a n) \sim (1 + \varepsilon) \log Q.
\]
Fix any $q_0 < Q^{\varepsilon / 3}$. The measures $\Pop{q} \nu_n$ for various $q < Q, q_0 \mid q$, have disjoint Fourier supports. Therefore, by the large sieve, Corollary \ref{cr:large-sieve},
\[
  \sum_{\substack{q \in [Q/2, Q] \\ q_0 \mid q}} q \|\Pop{q} \nu_n\|_2^2 \ll (a^n + Q^2) \|\nu_n\|_2^2 \ll Q^{1-\varepsilon}.
\]
Hence with at most $Q^{1-2\varepsilon/3}$ exceptions we have
\[
  q \|\Pop{q} \nu_n\|_2^2 \ll Q^{-\varepsilon/3}.
\]
Since there are at most $Q^{1-2\varepsilon/3}$ exceptions for each $q_0 < Q^{\varepsilon / 3}$, there are at most $Q^{1-\varepsilon/3}$ exceptions in all.

For unexceptional $q$ we have, for any $q_0 < Q^{\varepsilon / 3}, q_0 \mid q$,
\begin{align*}
  \|\Pop{q} \mu_n\|
  &\leq \|\Pop{q}(\mu_n - \nu_n)\| + \|\Pop{q} \nu_n\|\\
  &\leq \|\Pop{q}(\mu_n - \nu_n)\| + q^{1/2} \|\Pop{q} \nu_n\|_2\\
  &\ll \|\Pop{q}\| \alpha^{-2} n^{-1} + Q^{-\varepsilon/6}.
\end{align*}
If $\alpha$ is a sufficiently small constant we deduce
\[
  \|\Pop{q} \mu_n\| \ll \frac{\|\Pop{q}\|}{n} \asymp \frac{\|\Pop{q}\|}{\log Q},
\]
as claimed.
\end{proof}

For the proof of the composite case of Theorem~\ref{mainthm}(ii), we need one more ingredient, a result about the prime factorization of a typical integer.

\begin{lemma}\label{lm:divisors}
All but $o(Q)$ integers $q\in[1,Q]$ can be written as
\[
q=p_1p_2\cdots p_k q_0,
\]
where $p_1>\cdots>p_k$ are primes and
\begin{align*}
  k &\asymp \log \log \log Q,\\
  \log q_0 &\in \left[\frac{\log Q}{(\log \log Q)^3}, \frac{\log Q}{(\log \log Q)^2}\right].
\end{align*}
\end{lemma}

(The exponents $2$ and $3$ are not special: we could replace the interval for $\log q_0$ with any logarithmically long interval, adjusting $k$ as necessary.)

This follows easily from well-known results on the ``anatomy of integers''.
For the sake of completeness we give the proof.

\begin{proof}
We first apply \cite{HT-divisors}*{Theorem~07}, which states the following.
Let $x\ge z\ge y\ge 2$ be integers. Writing $\Theta(x,y,z)$ for the number of positive integers $n\le x$
such that the product of those prime factors of $n$ (allowing for multiplicities) that are less than $y$
exceeds $z$, we have
\[
\Theta(x,y,z)\ll x\exp\Big( -c_0\frac{\log z}{\log y}\Big),
\]
where $c_0$ and the implied constant are absolute.
We take
\begin{align*}
  x &= Q,\\
  \log y &= \log Q / (\log\log Q)^{2.5}, \\
  \log z &= \log Q / (\log\log Q)^2.
\end{align*}
The conclusion is that for all but $o(Q)$ integers $q\in[1,Q]$ we have
\[
  q=p_1p_2\cdots p_k q_0,
\]
where $p_1\ge\cdots\ge p_k$ are all the prime factors $p$ of $q$ such that
\[
  \log p > \log Q/(\log\log Q)^{2.5},
\]
and
\[
  \log q_0 \le \frac{\log Q}{(\log \log Q)^2}.
\]

The number of integers $q\in[1,Q]$ that are divisible by $p^2$ for some prime $p>\exp(\log Q/(\log\log Q)^3)$ can be estimated by
\[
\sum_{p=\exp(\log Q/(\log\log Q)^3)}^Q Q/p^2\le o(Q).
\]
Therefore we have $p_1 > \cdots > p_k$ almost surely.

Next, for any set of primes $T$, with harmonic sum
\[
  H(T) = \sum_{p \in T} \frac1p,
\]
if $\Omega(q, T)$ is the number of prime factors of $q$ in $T$, counted with multiplicity, then for all but $o(Q)$ integers $q \leq Q$ we have
\begin{equation} \label{OmegaT-estimate}
  \Omega(q, T) = H(T) + O(H(T)^{1/2} \log H(T))
\end{equation}
(and $\log$ can be replaced by any function tending to infinity). Indeed for any $k \geq 0$ we have a Poisson-type bound of the form
\[
  |\{q \leq Q : \Omega(q, T) = k\}| \ll Q e^{-H(T)} \frac{H(T)^k}{k!} \left(1 + \frac{k}{H(T)}\right)
\]
(see for example \cite{HT-divisors}*{Section~0.5}, \cite{ford-notes}*{Theorem 2.13}, or \cite{tudesq}*{Th\'eor\`eme 1}), and \eqref{OmegaT-estimate} follows by comparison with the Poisson distribution.

By \eqref{OmegaT-estimate} with
\[
  T_1 = \{p~\text{prime} : \log Q / (\log \log Q)^{2.5} < \log p \leq \log Q\},
\]
we find that
\[
  k \sim H(T_1) = 2.5 \log \log \log Q + O(1).
\]
Similarly, by \eqref{OmegaT-estimate} with
\[
  T_2 = \{p~\text{prime} : \log Q / (\log \log Q)^{3} < \log p \leq \log Q / (\log \log Q)^{2.5}\},
\]
we find that $q_0$ is almost surely divisible by some $p \in T_2$ (in fact some $c \log \log \log Q$ many $p$), so in particular
\[
  \log q_0 \geq \log Q / (\log \log Q)^3.
\]
This completes the proof.
\end{proof}

Let $E_1$ be the set of all $q$ which are exceptional in the sense of Lemma~\ref{lem:large-sieve}. Let $E_2$ be the set of all $q \leq Q$ which have some divisor $d \in E_1$ with $d > \log Q$. Then, by Lemma~\ref{lem:large-sieve} we have
\[
  |E_1 \cap [Q_1/2, Q_1]| \leq Q_1^{1-\varepsilon/3} \qquad (\log Q \leq Q_1 \leq 2Q),
\]
so
\begin{align*}
  |E_2|
  &\leq \sum_{d \in E_1 \cap [\log Q, Q]} \frac{Q}{d} \\
  &\ll \sum_{\substack{Q_1 \in [\log Q, 2Q] \\ \text{dyadic} }} \frac{Q}{Q_1^{\varepsilon/3}} \\
  &\ll_\varepsilon \frac{Q}{(\log Q)^{\varepsilon / 3}}.
\end{align*}
Hence $E_2$ is almost empty.

A typical integer $q$ can be written
\[
  q = p_1 p_2 \cdots p_k q_0
\]
as in Lemma \ref{lm:divisors}. Note that
\[
  d(q/q_0) = 2^k \ll \log \log q.
\]
We have a decomposition
\begin{align*}
  \mu_n \bmod q
  &= \sum_{s : q_0 \mid s \mid q} \Pop{s} \mu_n \\
  &= \piop{q_0} \mu_n + \sum_{\substack{s : q_0 \mid s \mid q \\ s \neq q_0}} \Pop{s} \mu_n.
\end{align*}
Hence
\[
  \|\mu_n \bmod q - u_q\|
  \leq \|\mu_n \bmod q_0 - u_{q_0}\| + \sum_{\substack{s : q_0 \mid s \mid q \\ s \neq q_0}} \|\Pop{s} \mu_n\|.
\]

If $q \notin E_2$ then each of the divisors $s$ with $q_0 \mid s \mid q$ is unexceptional in the sense of Lemma~\ref{lem:large-sieve}. There are two cases. If $q_0 < s^{\varepsilon / 3}$ then Lemma~\ref{lem:large-sieve} applies, so
\[
  \|\Pop{s}\mu_n\| \ll \frac{\|\Pop{s}\|}{\log s} \leq \frac{d(s/q_0)}{\log q_0}.
\]
If on the other hand $q_0 \geq s^{\varepsilon / 3}$, $q_0 \neq s$, then
\begin{align*}
  \Pop{s}\mu_n
  &= \sum_{q_1 : q_0 \mid q_1 \mid s} \mu(q_1 / q_0) \piop{q_1} \mu_n \\
  &= \sum_{q_1 : q_0 \mid q_1 \mid s} \mu(q_1 / q_0) (\piop{q_1} \mu_n - u_{q_1}),
\end{align*}
so, by Lemma~\ref{lem:small-moduli},
\[
  \|\Pop{s}\mu_n\| \leq d(s/q_0) \max_{q_1 \leq s} \|\mu_n \bmod q_1 - u_{q_1}\| \leq d(s/q_0) e^{-cn / \log s}.
\]
Note $\log s \ll \varepsilon^{-1} \log q_0$. Similarly,
\[
  \|\mu_n \bmod q_0 - u_{q_0}\| \leq e^{-c n / \log q_0}.
\]

Putting the above estimates together, we have
\begin{align*}
  \|\mu_n\bmod q - u_q \|
  &\leq d(q / q_0) e^{-c\varepsilon n / \log q_0} + \frac{d(q/q_0)^2}{ \log q_0 } \\
  &\leq e^{-c \varepsilon (\log \log q)^2} + \frac{(\log \log q)^5}{\log q}.
\end{align*}
This completes the proof of Theorem~\ref{mainthm}(ii).

\section{Alternative argument based on sophisticated pruning}\label{sc:alternative}

In this section, we present an alternative argument to prove Theorem~\ref{mainthm}(ii) that is closer
to that in \cite{CDG}*{Section~6}.
This also relies on the large sieve inequality, but it avoids the use of the projection operators
introduced in the previous section.

For each $n$, let $\nu_n$ be a measure as in Corollary~\ref{cor:nu_n-exists} with $\alpha = 1/\log n$. Thus
\begin{align*}
  & \| \mu_n - \nu_n\| = n^{-1 + o(1)}, \\
  & \log \|\nu_n\|_2^{-2} = H n + o(n).
\end{align*}
Let $\pi_n$ be the following measure:
\begin{align*}
  \pi_n =& \mu_{w_0}
  *[(m_{a^{w_0}})_* \nu_{w_1-C} *(m_{a^{w_0 + w_1-C}})_*\mu_C]\\
  &*[(m_{a^{w_0 + w_1}})_* \nu_{w_2-C} * (m_{a^{w_0 + w_1+w_2-C}})_*\mu_C]*\cdots\\
  &*[(m_{a^{w_0 + \cdots + w_{k-1}}})_* \nu_{w_k-C}*(m_{a^{w_0 +\cdots+ w_k-C}})_*\mu_C],
\end{align*}
where $w_0 = \floor{n / \log n}$, $C$ is a suitably large integer depending on $a$ and $\mu$ such that
\[
\opr{supp}\nu_{w-C}\subset[-a^w/2,a^w/2],
\]
and $w_1, \dots, w_k$ are chosen as in the following lemma.

\begin{lemma}
We can choose $w_1, \dots, w_k \geq n^{0.1}$ such that
\begin{enumerate}
  \item $k \sim (\log n)^2$,
  \item $w_1 + \cdots + w_k = n - w_0$, and
  \item for every $w \in [n^{0.2}, n]$ there is some $i$ such that \[w_1 + \cdots + w_i \in [(1-o(1))w, w].\]
\end{enumerate}
\end{lemma}
\begin{proof}
Let $k \sim (\log n)^2$, and let $a$ be such that
\[
  \exp((1+a) \log (n-w_0)/(k+a))\sim n^{0.15}.
\]
Choose $w_1, \dots, w_k$ so that
\[
  w_1 + \cdots + w_i = \floor{\exp((i+a) \log(n-w_0)/(k+a))}.
\]
Then
\[
  \frac{w_1 + \cdots + w_{i+1}}{w_1 + \cdots + w_i} \sim e^{1 / \log n} = 1 + O(1 / \log n),
\]
so every $w \in \{1, \dots, n\}$ can be approximated as claimed.
\end{proof}

It follows that
\[
  \|\mu_n - \pi_n\| \leq \sum_{i=1}^k \| \mu_{w_i-C} - \nu_{w_i-C}\| \ll n^{-0.1 + o(1)},
\]
and we also have the following property.
\begin{lemma}\label{lm:pi}
For every $w \in [n^{0.2}, n]$, we have a factorization
\[
  \pi_n = \mu_{w_0} * (m_{a^{w_0}})_* \pi * \pi'
\]
for some probability measures $\pi, \pi'$ such that $\opr{supp} \pi \subset \opr{supp} \mu_w$ and
\[
  \log \|\pi\|_2^{-2} = Hw + o(w).
\]
\end{lemma}
\begin{proof}
We take
\begin{align*}
\pi=& \nu_{w_1-C}*(m_{a^{w_1}})_* \nu_{w_2-C} * \cdots * (m_{a^{w_1 + \cdots + w_{i-1}}})_* \nu_{w_i-C}\\
\pi'=&(m_{a^{w_0+w_1 + \cdots + w_{i}}})_*\nu_{w_{i+1}-C} * \cdots * (m_{a^{w_0+w_1 + \cdots + w_{k-1}}})_* \nu_{w_k-C}\\
&*(m_{a^{w_0 + w_1-C}})_*\mu_C*\cdots*(m_{a^{w_0 +\cdots+ w_k-C}})_*\mu_C
\end{align*}
for a suitable choice of $i$ so that
\[
w_1+\cdots+w_i\in[(1-o(1))w,w].
\]
It follows by construction that $\opr{supp} \pi\subset\opr{supp} \mu_w$.

By the choice of $C$ in the definition of $\pi$,
it follows that the map
\begin{align*}
&\opr{supp} \nu_{w_1-C}\times\cdots
\times\opr{supp}  (m_{a^{w_1 + \cdots + w_{i-1}}})_* \nu_{w_i-C}&\to&\mathbf{Z}\\
&(x_1,\ldots,x_i) &\mapsto& x_1+\ldots+x_i
\end{align*}
is injective, so
\begin{align*}
  \|\pi\|_2^2
  &=\|\nu_{w_1-C}\|_2^2\cdots\|\nu_{w_i-C}\|_2^2 \\
  &=e^{-H(w_1-C) + o(w_1)} \cdots e^{-H(w_i-C) + o(w_i)} \\
  &=e^{-Hw + o(w) + iCH}
\end{align*}
and the claim follows since $w\geq n^{0.2}$ and $i \leq n^{0.1}$.
\end{proof}

\begin{proposition} \label{CDG-like-prop}
Assume $2H \geq (1+\varepsilon)\log a$ and $Hn \geq (1+\varepsilon) \log Q$. Then
\[
  \sum_{q \leq Q} q \|\pi_n \bmod q - u_q\|_2^2 \leq Q \exp( - c (\log Q)^{0.49}).
\]
\end{proposition}

The exponent $0.49$ can actually be improved to $0.5$. To do this, instead of $w_0 = \floor{n / \log n}$, take $w_0$ to be a sufficiently small (depending on $\varepsilon$) constant multiple of $n$. We omit the details.

\begin{proof}
Write $Q = a^t$. We want to bound
\begin{align*}
  \sum_{q \in (a^{t-1}, a^t)} q \|\pi_n \bmod q - u_q\|_2^2
  &= \sum_{q \in (a^{t-1}, a^t)} \sum_{0 < k < q} |\pihat[n](k/q)|^2 \\
  &= \sum_{w=1}^t \sum_{r, s} M(r/s) |\pihat[n](r/s)|^2,
\end{align*}
where the last sum runs over all $r, s$ with
\begin{align*}
  &s \in (a^{w-1}, a^w), \\
  &(a, s) = 1, \\
  &r \in R(s) = \{r \bmod s : (r, s) = 1\},
\end{align*}
and $M(r/s)$ is the number of $k/q = r/s$ with $q \in (a^{t-1}, a^t)$ and $0 < k < q$. In other words, $M(r/s)$ is the number of multiples of $s$, coprime with $a$, in the range $(a^{t-1}, a^t)$. We have
\[
  M(r/s) \leq a^t / s,
\]
so our sum is bounded by
\[
  a^t \sum_{w=1}^t \sum_{r, s} \frac1s |\pihat[n](r/s)|^2.
\]
We split this sum into two parts depending on the size of $w$: either $w \leq n^{1/2} / \log n$ or $w > n^{1/2} / \log n$.

Consider first the lower range $w \leq n^{1/2} / \log n$. Since $\mu$ is not supported on a coset of a subgroup, $\muhat{}(\xi) = 1$ only if $\xi = 0$. Find $c > 0$ such that
\[
  |\muhat{}(\xi)|^2 \leq e^{-c} \qquad (\xi \notin (-1/a, 1/a)).
\]
For any $r \in R(s)$, $s \in (a^{w-1}, a^w)$, $r/s \notin (-a^{-w}, a^{-w})$, so for some $j \leq w-1$ we have $a^j r/s \notin (-1/a, 1/a)$, so
\[
  |\muhat[w](r/s)| = \prod_{j=0}^{w-1} |\muhat(a^j r/s)| \leq e^{-c}.
\]
Hence
\[
  |\muhat[n'w](r/s)|^2 \leq \left(\max_{r' \in R(s)} |\muhat[w](r'/s)|^2\right)^{n'} \leq e^{-cn'}
\]
for any $n'\in\mathbf{Z}_{>0}$.
Therefore
\[
  \sum_{s \in (a^{w-1}, a^w)} \frac1s \sum_{r \in R(s)} |\muhat[n'w](r/s)|^2 \leq a^{w} e^{-c n'}.
\]
Now because $\mu_{w_0}$ is a factor of $\pi_n$ we also have
\[
  \sum_{s \in (a^{w-1}, a^w)} \frac1s \sum_{r \in R(s)} |\pihat[n](r/s)|^2 \leq a^{w} e^{-c n'},
\]
provided that $n'w \leq w_0 \sim n / \log n$. Take $n' \sim w^{-1} n / \log n$. Then $a^w e^{-cn'}$ is negligible for $w < c'(n / \log n)^{1/2}$ for a sufficiently small constant $c'$.

Now consider the upper range $w > n^{1/2} / \log n$. Let $w' = \min(2w, n)$. By construction, $\pi_n$ has a factor $(m_{a^{w_0}})_* \pi$ such that
\[
  \opr{supp} \pi \subset \opr{supp} \mu_{w'} \subset [-O(a^{2w}), O(a^{2w})]
\]
and
\[
  \log\|\pi\|_2^{-2} = Hw' + o(w)
\]
by Lemma \ref{lm:pi}.
Therefore
\[
  |\pihat[n](r/s)|^2 \leq |\pihat(a^{w_0} r/s)|^2,
\]
and by the large sieve (Theorem \ref{th:large-sieve}),
\[
  \sum_{s \in (a^{w-1}, a^w)} \frac1s \sum_{r \in R(s)} |\pihat(r/s)|^2
  \ll a^w \|\pi\|_2^2 = a^w e^{-Hw' + o(w)}.
\]
By hypothesis
\begin{align*}
  2H &\geq (1+\varepsilon) \log a, \\
  Hn &\geq (1+\varepsilon) t \log a,
\end{align*}
so
\[
  H w' \geq (1+\varepsilon) w \log a.
\]
Hence $a^w e^{-Hw' + o(w)}$ is bounded by $e^{-c w}$ for some $c>0$. Thus the contribution from this case is negligible too.
\end{proof}

We have proved that
\[
  \sum_{q \leq Q} q \|\pi_n \bmod q - u_q\|_2^2 \leq Q \exp(-c (\log Q)^{0.49}),
\]
provided that $Hn > (1+\varepsilon) \log Q$. Hence apart from some
\[
  Q \exp(-c (\log Q)^{0.49})
\]
exceptions $q \leq Q$ we have
\[
  q \|\pi_n \bmod q - u_q\|_2^2 = \exp(-c (\log Q)^{0.49}),
\]
and hence
\begin{align*}
  \|\mu_n \bmod q - u_q\|
  &\leq \|\mu_n - \pi_n\| + q^{1/2} \|\pi_n \bmod q - u_q\|_2 \\
  &= (\log q)^{-c}.
\end{align*}
This proves Theorem~\ref{mainthm}(ii) (both prime and composite cases).

\section{The larger sieve argument}\label{sc:larger-sieve}

The purpose of this section is to prove Theorem \ref{larger-sieve-thm}.
The following argument is based on Gallagher's larger sieve as described in \cite{friedlander--iwaniec--opera}*{Chapter~9.7}. By Corollary~\ref{cor:nu_n-exists}, there is a measure $\nu_n$ such that
\begin{align*}
  & \|\mu_n - \nu_n\| = o(1), \\
  & \|\nu_n\|_2^{-2} = \exp((H+o(1))n).
\end{align*}
Let $Q = \|\nu_n\|_2^{-2}$. Note that
\[
  \|\nu_n \bmod p\|_2^2 = \frac1{Q} + \sum_{x \neq y} \nu_n(x) \nu_n(y) 1_{p \mid (x - y)}.
\]
Summing over primes $p \leq X$ with weight $\log p$, we have
\[
  \sum_{p \leq X} \|\nu_n \bmod p\|_2^2 \log p
  \leq \frac1{Q} \sum_{p \leq X} \log p + \sum_{x \neq y} \nu_n(x) \nu_n(y) \sum_{p \mid (x-y)} \log p.
\]
The first sum is $O(X)$. Since $\nu_n$ is supported on $[-O(a^n), O(a^n)]$, for $x, y \in \opr{supp} \nu_n$ we have
\[
  \sum_{p \mid (x - y)} \log p \leq \log |x-y| \ll n.
\]
Thus
\[
  \sum_{p \leq X} \|\nu_n \bmod p\|_2^2 \log p \ll \frac{X}{Q} + n.
\]
Let $B_1$ be the set of primes $p \leq Q$ such that
\[
  p \|\nu_n \bmod p\|_2^2 > \delta n^{1/2}.
\]
(Later we will take $\delta$ to be a sufficiently small constant.) Then it follows that
\[
  \sum_{p \in B_1} \frac{\log p}{p} \ll \delta^{-1} n^{1/2}.
\]
In the rest of the argument we assume $p \leq Q$ and $p \notin B_1$, i.e.,
\[
  p \|\nu_n \bmod p\|_2^2 \leq \delta n^{1/2}.
\]

Now we use a Fourier multiplicity argument, as in \cite{BV--ax+b}*{Section~3}. Define
\[
  \nu_n^{(m)} = \frac1{m+1} \sum_{i=0}^{m} \mu_i * (m_{a^i})_* \nu_n * (m_{a^{i+n}})_* \mu_{m-i}.
\]
Then
\[
  \|\mu_{n+m} - \nu_n^{(m)}\| \leq \|\mu_n - \nu_n\| = o(1),
\]
and
\[
  \hat{\nu_n^{(m)}}(\xi)
  = \frac1{m+1} \sum_{i=0}^m \muhat[i](\xi) \hat{\nu_n}(a^i \xi) \muhat[m-i](a^{i+n} \xi),
\]
so
\begin{align*}
  |\hat{\nu_n^{(m)}}(\xi)|
  &= \frac1{m+1} \sum_{i=0}^m |\muhat[i](\xi)| |\hat{\nu_n}(a^i \xi)| |\muhat[m-i](a^{i+n} \xi)| \\
  &\leq \frac1{m+1} \sum_{i=0}^m |\hat{\nu_n}(a^i \xi)| .
\end{align*}
By Cauchy--Schwarz,
\[
  |\hat{\nu_n^{(m)}}(\xi)|^2 \leq \frac1{m+1} \sum_{i=0}^m |\hat{\nu_n}(a^i \xi)|^2.
\]
Hence, provided $a$ has order at least $m+1$ in $\mathbf{F}_p^\times$,
\[
  \max_{r \neq 0} |\hat{\nu_n^{(m)}}(r/p)|^2 \leq \frac1{m+1} \sum_{r\neq 0} |\hat{\nu_n}(r/p)|^2 \leq \frac{\delta n^{1/2}}{m+1}.
\]
Applying this with $m = \delta n$, we get
\begin{equation} \label{notB2}
  \max_{r \neq 0} |\hat{\nu_n^{(m)}}(r/p)|^2 \leq n^{-1/2}.
\end{equation}

Let $B_2$ be the set of primes $p \leq Q$ such that $a$ does not have order at least $m+1$ in $\mathbf{F}_p^\times$. For $p \in B_2$, $p$ divides $a^i - 1$ for some $i \leq m$, so $|B_2| \leq m^2$. Thus
\[
  \sum_{p \in B_2} \frac{\log p}{p}
  \ll \sum_{p \leq m^2} \frac{\log p}{p}
  + |B_2| \frac{\log m}{m^2}
  \ll \log m.
\]
Thus we may assume $p \notin B_2$, so \eqref{notB2} holds.

Finally, observe that
\[
  \|\mu_{2n+m} - \nu_n * (m_{a^n})_*\nu_n^{(m)}\| \leq 2 \|\mu_n - \nu_n\| = o(1),
\]
and
\begin{align*}
  q\|\nu_n * (m_{a^n})_*\nu_n^{(m)} \bmod q - u_q\|_2^2
  &= \sum_{r\neq 0} |\hat{\nu_n}(r/p)|^2 |\hat{\nu_n^{(m)}}(a^nr/p)|^2 \\
  &\leq q\|\nu_n \bmod p\|_2^2 \max_{r\neq 0} |\hat{\nu_n^{(m)}}(r/p)|^2 \\
  &\leq \delta.
\end{align*}
Thus
\begin{align*}
  \|\mu_{2n+m} \bmod p - u_p\|
  &\leq \|\mu_{2n+m} - \nu_n * (m_{a^n})_*\nu_n^{(m)}\|\\
  &\qquad + p^{1/2}\|\nu_n * (m_{a^n})_*\nu_n^{(m)} \bmod p - u_p\|_2 \\
  &\leq \delta + o(1).
\end{align*}
In other words, for every $p \notin B_1 \cup B_2$ there is some
\[
  n' = 2n + m = (2 + \delta + o(1)) H^{-1} \log Q
\]
such that $\|\mu_{n'} \bmod q - u_q\| \leq \delta + o(1)$. Taking $\delta$ smaller than $\varepsilon$, we deduce Theorem~\ref{larger-sieve-thm}.

\section{Open problems}\label{sc:OP}

It would be very interesting to understand the mixing time of the random walk if we drop the condition $H>\frac12\log a$ in
Theorem~\ref{mainthm}(ii).
We do not think this condition is necessary, but it is required in our proof so that we can apply the large sieve.
More specifically, we formulate the following open problem.

\begin{OP}
Let $\mu=\frac{1}{2}(\d_0+\d_1)$ and let $a\ge 2$ be an integer.
Is the mixing time of $X_n \bmod q$ equal to $(1+o(1))\log_2 q$ for almost all $q$ such that $(q, a)=1$?
\end{OP}

For $a=2$, the answer is yes trivially, because $X_n$ is uniformly distributed in the interval $[0,2^n-1]\cap \mathbf{Z}$.
The $a=3$ case is covered by our Theorem~\ref{mainthm}, and the answer is yes.
The $a=4$ case is not covered by our Theorem~\ref{mainthm}, but the answer is still yes, as can be seen by a modification
of our argument, which we sketch bellow.
The $a\ge 5$ case appears to be beyond the reach of the large sieve and we only have the bounds from Theorem~\ref{larger-sieve-thm}.

For the $a=4$ case of the above problem we modify the proof of Theorem~\ref{mainthm}(ii) as follows.
For simplicity, we only discuss the case of prime moduli.
Fix some $n\in\mathbf{Z}_{>0}$ and also some $\varepsilon=\varepsilon(n)>0$ to be specified below.
Write
\begin{align*}
\eta_1=&\mu_{\varepsilon n},\\
\eta_2=&(m_{4^{\varepsilon n}})_*\mu_{(1-\varepsilon)n}\\
\eta_3=&(m_{4^{n}})_*\mu_{\varepsilon n}.
\end{align*}
We choose $\varepsilon$ in such a way that $\varepsilon n$ is an integer,
$\varepsilon$ slowly goes to $0$ as $n$ grows.
Observe that
\begin{align*}
\|\eta_1*\eta_1*\eta_2\|_2^2<& 2^{-(1+c\varepsilon_1)n},\\
\opr{supp}(\eta_1*\eta_1*\eta_2)<&[-C 4^n,C4^n]
\end{align*}
with some absolute constants $c,C$.
Now we can apply the large sieve for the measure $\eta_1*\eta_1*\eta_2$
and conclude that
\[
\sum_{r\in R(p)}|\widehat\eta_1(r/p)|^2\cdot|\widehat\eta_1(r/p)|^2\cdot|\widehat\eta_2(r/p)|^2\ll 2^n\cdot2^{-(1+c\varepsilon_1)n}=2^{-c\varepsilon_1 n}
\]
for most primes $p$ in the range $[2^{n-1},2^n]$.
Now the claimed mixing time bound follows from
\begin{align*}
\|\mu_{(1+\varepsilon)n} \bmod p -u_p\|_2^2
=&\sum_{r\in R(p)}|\widehat\eta_1(r/p)|^2\cdot|\widehat\eta_2(r/p)|^2\cdot|\widehat\eta_3(r/p)|^2\\
\le&\Big(\sum_{r\in R(p)}|\widehat\eta_1(r/p)|^4\cdot|\widehat\eta_2(r/p)|^2\Big)^{1/2}\\
&\times\Big(\sum_{r\in R(p)}|\widehat\eta_3(r/p)|^4\cdot|\widehat\eta_2(r/p)|^2\Big)^{1/2}\\
\ll&2^{-c\varepsilon_1 n}.
\end{align*}
The second factor in the second to last line can be estimated similarly to the first one as we did above.

It would be also interesting  to obtain a better understanding of the exceptional moduli in Theorem~\ref{mainthm}.
We suggest the following two problems in this direction.

\begin{OP}
Fix some $\mu$ and $a\in\mathbf{Z}_{\ge 2}$.
Suppose that the base $a$ expansion of some $q$ is generic in a suitable sense.
Is the mixing time of $X_n \bmod q$ equal to $(1+o(1))H^{-1}\log q$?
\end{OP}

This question is somewhat vague.
It is motivated by the counter-example of $q=a^n-1$, for which the mixing time
is $O(\log q\log\log q)$ as was demonstrated in \cite{CDG}.

\begin{OP}
Fix some $\mu$, and write
$\{X_n^{(a)}\}_{n\in\mathbf{Z}_{>0}}$ for the random walk defined in the introduction with the parameter $a$.
Is it true that for each $q$ with $2,3\nmid q$ at least one of the following holds:
\begin{itemize}
\item the mixing time of $X_n^{(2)} \bmod q$ equals $(1+o(1))H(\mu,2)^{-1}\log q$,
\item the mixing time of $X_n^{(3)} \bmod q$ equals $(1+o(1))H(\mu,3)^{-1}\log q$?
\end{itemize}
\end{OP}

This question is motivated by a series of problems posed by Furstenberg on the joint action of
$x\mapsto 2 x$ and $x\mapsto 3 x$ on $\mathbf{R}/\mathbf{Z}$.

\section*{Acknowledgements}

We thank Kevin Ford for discussions related to Lemma~\ref{lm:divisors}.
We thank Martin Hildebrand and Max Wenqiang Xu for comments and suggestions on an earlier version of this paper.
We are grateful to the anonymous referee for carefully reading the paper and for helpful comments.

redpajama_set_name: RedPajamaArXiv


The Law Alumni Association (LAA) is an integral part of the educational, intellectual and social life at New York University School of Law. Upon graduation, all alumni automatically become members of the LAA and will receive invitations to all events sponsored by the LAA, including the Fall Conference, held annually in November; the Alumni Luncheon, held in January; and various recent graduate events.
The Law Alumni of Color Association (LACA), formerly known as the Black, Latino, Asian Pacific American Law Alumni Association, serves as the umbrella organization for all of NYU School of Law's alumni of color. In 2016, the association changed its name to more wholly represent all of its members. Founded over 40 years ago, LACA promotes and supports the professional development and general advancement of the Law School's alumni and students of color, and seeks to more fully involve alumni, students and faculty of color in the educational, social, and intellectual life of the Law School. This organization further serves as a powerful vehicle through which professional and personal ties are cultivated.


text:
The trial of Bill Cosby
Sentencing in Bill Cosby's sex assault case set for Sept. 24
Michael Sisak, Associated Press
Actor and comedian Bill Cosby reacts while being notified a verdict was in in his sexual assault retrial, Thursday, April, 26, 2018, at the Montgomery County Courthouse in Norristown, Pa. A jury convicted the "Cosby Show" star of three counts of aggravated indecent assault on Thursday. The guilty verdict came less than a year after another jury deadlocked on the charges. (Mark Makela/Pool Photo via AP)
Bill Cosby will be sentenced Sept. 24, almost five months to the day after he was convicted of sexual assault, a judge said Tuesday.
Lawyers for the comedian, who turns 81 in July and faces the prospect of the rest of his life in prison, had asked Judge Steven O'Neill to delay sentencing until the end of the year.
Cosby's convictions on three counts of aggravated indecent assault will likely be combined into one charge that carries a standard sentence of five to 10 years in prison.
Cosby's spokesman did not immediately respond to a message left seeking comment.
The man once revered as America's Dad has been a prisoner in his suburban Philadelphia mansion since his April 26 conviction and he hasn't been heard from since lashing out at a prosecutor in court that day.
O'Neill ordered Cosby be outfitted with a GPS monitoring bracelet and said he needed permission to leave the home, where jurors concluded he drugged and molested Andrea Constand in January 2004.
Even then, Cosby can go out only to visit his lawyers or the doctor.
Before sentencing, Cosby must undergo an assessment to determine if he is a sexually violent predator. Because of his conviction, he will also be required to register as a sex offender under Megan's Law.
Constand, a former Temple University women's basketball administrator, is expected to speak at sentencing. O'Neill said in a court order the hearing could take two days.
Constand testified at Cosby's trial that the comedian knocked her out with three blue pills he called "your friends" and then penetrated her with his fingers as she lay immobilized, unable to resist or say no.
Cosby claimed the encounter was consensual, saying he gave her the cold and allergy medicine Benadryl to relax. His lawyers have vowed to appeal, but that can't happen until Cosby is sentenced.
In Pennsylvania, defendants are typically sentenced within 90 days of a conviction. O'Neill initially said he would schedule Cosby's sentencing for July. His court order did not explain why he moved it to September.
Constand's testimony and that of five other accusers allowed prosecutors to uncloak Cosby — who solidified his reputation as a family man by playing Dr. Cliff Huxtable on "The Cosby Show" — as a manipulative predator who used his built-in trust to trick women into taking powerful intoxicants so he could violate them.
One woman pointedly called Cosby a "serial rapist," and another asked him through her tears, "You remember, don't you, Mr. Cosby?"
O'Neill has yet to rule on whether he'll identify the jurors who convicted Cosby, even though the state Supreme Court has long held that juror names should be made public under the First Amendment.
Prosecutors urged O'Neill to consider a "cooling off" before they're disclosed, citing a 90-day wait in the Casey Anthony case.
A lawyer for The Associated Press and other news media outlets said that was different because jurors were threatened after acquitting the Florida mother in 2011 of charges she killed her young daughter.
The Associated Press does not typically identify people who say they are victims of sexual assault unless they grant permission, as Constand has done.
Follow Sisak at twitter.com/mikesisak
Cosby Unraveled
Cosby Unraveled explores Cosby's beginnings in the projects of North Philadelphia, the heights of his influence, and the causes of his downfall.
Part of the series
View the series
Bill Cosby loses appeal of sex assault conviction
A Pennsylvania appeals court rejected Bill Cosby's bid to overturn his sexual assault conviction Tuesday over the trial judge's decision to let five other accusers testify.
Bill Cosby agrees to settle 7 defamation lawsuits
Filings in federal court in Springfield, Mass. show the two sides have negotiated a settlement since Cosby went to prison last fall in a separate Pa. sexual assault case.
Key prosecutor says allegation of racism in Cosby case 'completely offensive'
Entertainer Bill Cosby has been sentenced to three to 10 years in state prison.
Listen 7:59

redpajama_set_name: RedPajamaCommonCrawl


text:
Android Fans Want Better Battery Life for Google Pixel 3
5 Oct 2018, 2 p.m.
In a recent PCMag survey of 2,500 participants, we found that 64 percent of people want Google to unveil a Pixel phone with a longer battery life than previous Google phones.
Consumers eyeing Google's new Pixel lineup are hoping for a device that will last all day and night.
Google is set to unveil its next-gen Pixel devices on Oct. 9. In a recent PCMag survey of 2,500 participants, we found that 64 percent of people want a Pixel 3 with a longer battery life than previous Google phones.
In our tests of the Pixel 2 and 2 XL, the Pixel 2's 2,700mAh battery provided 8 hours and 48 minutes of LTE streaming, which is pretty good for a phone its size. The Pixel 2 XL came in at a more impressive 9 hours and 25 minutes, but it has more room inside for a 3,520mAh battery.
SEE ALSO: The Best UK Mobile Phone Deals for January 2020
Next on consumers' Pixel 3 wish list is wireless charging; 38 percent want it on Google's new phones. The Pixel 2 and 2 XL supported fast charging but did not have wireless charging. Now that even Apple has jumped on the wireless charging bandwagon with its latest iPhones, it stands to reason that Google will do the same. Rumors suggest it's on tap for the Pixel 3, but we'll have to wait and see.
At number three, the desire for a better camera (28 percent) barely surpasses the desire for a larger display (27 percent). These days, specced out phone cameras are all the rage; LG's new V40 ThinQ has five. But so are large phones. Apple's iPhone XS Max is just that at 6.5 inches, so the new Pixels have some room to grow. The Pixel 2 is just 5 inches, while the XL 2 comes in at 6 inches.
The Pixel lineup, meanwhile, is usually thought of as a phone for the Android faithful. The stripped-down, stock Android experience lets developers and enthusiasts play around with Google's newest OS, in this case Android 9.0 Pie, without having to deal with bloatware added by Android phone makers like Samsung or LG.
But average consumers like Pixels, too. Of the 59 percent of Android users who prefer Google smartphones to other Android smartphones, 36 percent said price was the main reason. The Pixel 2 and XL started at $650 and $850, respectively, which definitely places it below the Note 9 and Apple's iPhone X lineup, both of which top $1,000 depending on configuration.
Another 25 percent prefer Google phones because of camera features, while 19 percent like the company's clean design, 11 percent like the phones' speed, and 9 percent appreciate their capacity.
Will everyone get what they want? PCMag will be at Google's event on Oct. 9 so stay tuned for all the details.
Google Pixel 2 ...
Google Android 9.0 ...
Apple iPhone XS ...
Apple Launches Free iPhone Battery Case Replacement Program
Apple Touts Camera Improvements on iPhone 11 Lineup
iPhone 11 Case Suggests Apple Pencil Support
OnePlus, Google See Big Smartphone Sales in the US in Q2
Chloe Albanesius Executive Editor, News & Features
Chloe Albanesius has been with PCMag.com since April 2007, most recently as Executive Editor for News and Features. Prior to that, she worked for a year covering financial IT on Wall Street for Incisive Media. From 2002 to 2005, Chloe covered technology policy for The National Journal's Technology Daily in Washington, DC. She has held internships at NBC's Meet the Press, washingtonpost.com, the Tate Gallery press office in London, Roll Call, and Congressional Quarterly. She graduated with a bachelor's degree in journalism from American University in Washington, D.C. See Full Bio
The Worst 'Best of CES' Picks From the 2010s
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Everything Coming to Netflix in January


text:
About The Audiobook
HIV/AIDS in Africa
Sophie Okonedo
Sophie Okonedo is an acclaimed film, television and stage actress born in London. A graduate of Cambridge University, Okonedo trained at the Royal Academy of Dramatic Art.
Okonedo was nominated for Best Supporting Actress at the 2005 Academy Awards for her role as Tatiana Rusesabagina in the Rwandan genocide film Hotel Rwanda. She received an NAACP Image Award nomination for the same role, and won the award in 2007 for her role in the HBO miniseries Tsunami: The Aftermath. Her other film credits include The Secret Life of Bees, Dirty Pretty Things, BBC's adaptation of Oliver Twist, and the upcoming Skin.
Okonedo's stage credits include The Vagina Monologues, The Arabian Nights, and Serious Money. She has worked with several theatre companies, including the Royal Shakespeare Company, London's Royal Court Theatre, and London's National Theatre.
The Hare and the Tree Spirit
Nelson Mandela's Favorite African Folktales is an extraordinary audiobook featuring Gillian Anderson, Benjamin Bratt, LeVar Burton, Ricardo Chavira, Don Cheadle, Matt Damon, Whoopi Goldberg, Sean…
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Nelson Mandela's Favorite African Folktales
An Audiobook benefitting children in South Africa orphaned and impacted by HIV/AIDS
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No matter how tremendously positive you may be, it's gonna be very hard to maintain your positive stance if you need to work together with a negative person. This is just like multiplication. Minus multiplied by Plus cannot be plus, but always be minus. No matter how big number you may have, it naturally turns out to be minus if multiplied by minus.
Then how can you be positive ? I think this is the question that we all need to think.....because this answer is nothing but the way to peace.
To keep away from negativity may be good, but it's just a temporarily solution. Without facing each other, true peace never come.
I think the only way for peace is to deny your belief what you think you are right. You need to realize you may not be right. And another person also need to realize she or he may not be right. If we can realize we may not be right equally, then it cannot be conflict or war. This is just like Minus multiplied by Minus can turn out to be Plus.
I know this is not as easy as Math, but just like thought on "roundabout" I want to work for peace.


Video of a suspected terror attack at an office building complex in Nairobi, Kenya, may have captured a US Navy SEAL on a secretive mission to combat Islamic militants in Africa. Watch for the alleged US Navy SEAL rescuing a woman at the 30 second mark. Read more.
In March, 36-year-old SEAL Jonathan Grant was in a car accident. His wife, 32-year-old Laura Browning Grant, left their.


text:
package com.venky.swf.routing;

import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.net.URL;
import java.util.HashSet;
import java.util.Iterator;
import java.util.List;
import java.util.Set;
import java.util.jar.JarEntry;
import java.util.jar.JarFile;
import java.util.regex.Pattern;

import com.venky.core.io.ByteArrayInputStream;
import com.venky.core.string.StringUtil;
import com.venky.swf.routing.Config;

public class SWFClassLoader extends ClassLoader {
    private Set<File> pathsHandled = new HashSet<>();

    public SWFClassLoader(ClassLoader parent) {
        super(parent);
        List<URL> watchedUrls = Config.instance().getResourceBaseUrls();
        for (URL watchedUrl : watchedUrls) {
            pathsHandled.add(getFile(watchedUrl, ""));
        }
    }

    public File getFile(URL watchedUrl, String loc) {
        String path = watchedUrl.getPath();

        if (watchedUrl.getProtocol().equals("jar")) {
            return new File(path.substring("file:".length(),
                    path.lastIndexOf("!")));
        } else if (watchedUrl.getProtocol().equals("file")) {
            return new File(path.substring(0, path.length() - loc.length()));
        } else {
            throw new RuntimeException("Don't know how to load Class from url:"
                    + watchedUrl.toString());
        }
    }
    Set<Pattern> exceptionClasses = new HashSet<Pattern>(){{
        add(Pattern.compile("com\\.venky\\..*\\._[^\\.]*"));
        add(Pattern.compile("com\\.venky\\.swf\\.routing\\..*"));
        add(Pattern.compile("com\\.venky\\.core\\.log\\..*"));
        add(Pattern.compile("javax\\.servlet\\..*"));
    }};

    public Class<?> loadClass(String name,boolean resolve) throws ClassNotFoundException {
        /*
        import com.venky.swf.db._IDatabase;
        import com.venky.swf.path._IPath;
        import com.venky.swf.views._IView;
        import javax.servlet.ServletException;
        import javax.servlet.http.HttpServletRequest;
        import javax.servlet.http.HttpServletResponse;
        import jakarta.servlet.http.HttpSession;

         */
        boolean matches = false;
        for (Iterator<Pattern> pi = exceptionClasses.iterator() ; pi.hasNext()  && !matches ; ){
            Pattern p = pi.next();
            matches = matches || p.matcher(name).matches();
        }
        if (matches){
            return super.loadClass(name,resolve);
        }
        synchronized (getClassLoadingLock(name)) {
            // First, check if the class has already been loaded
            Class<?> c = findLoadedClass(name);
            if (c != null){
                return c;
            }
            String loc = name.replace('.', '/') + ".class";
            for (File file : pathsHandled) {
                try {
                    byte[] clazzByte = null;
                    if (file.isDirectory()) {
                        File classFile = new File(file, loc);
                        if (!classFile.exists()) {
                            continue;
                        }
                        clazzByte = StringUtil.readBytes(new FileInputStream(classFile));
                    } else {
                        JarFile jf = new JarFile(file);
                        JarEntry entry = jf.getJarEntry(loc);
                        if (entry == null) {
                            continue;
                        }
                        clazzByte = StringUtil.readBytes(jf.getInputStream(entry));
                        jf.close();
                    }
                    return defineClass(name, clazzByte, 0, clazzByte.length);
                } catch (IOException e) {
                    continue;
                }
            }
            return super.loadClass(name,resolve);
        }
    }

    public InputStream getResourceAsStream(String name) {
        for (File file : pathsHandled) {
            try {
                byte[] clazzByte = null;
                if (file.isDirectory()) {
                    File classFile = new File(file, name);
                    if (!classFile.exists()) {
                        continue;
                    }
                    clazzByte = StringUtil.readBytes(new FileInputStream(classFile));
                } else {
                    JarFile jf = new JarFile(file);
                    JarEntry entry = jf.getJarEntry(name);
                    if (entry == null) {
                        continue;
                    }
                    clazzByte = StringUtil.readBytes(jf.getInputStream(entry));
                    jf.close();
                }
                return new ByteArrayInputStream(clazzByte);
            } catch (IOException e) {
                continue;
            }
        }
        return super.getResourceAsStream(name);
    }
}

redpajama_set_name: RedPajamaGithub


Bund integrity testing is fundamental because the purpose of a bund is to provide containment for tanks or drums in the case of a leak. It is therefore important that bund integrity testing be carried out before a bund enters service and then assessed for integrity on a regular basis thereafter.
Bund integrity testing by means of a hydrostatic test should always be carried out on new bunds before they enter service.
For in-service bunds, the method of bund integrity testing depends on the regulator. In Ireland, the EPA's preference is for hydrostatic testing unless there are valid safety or practicality issues in which case a visual inspection may be carried out by a Chartered Engineer.
In the UK, the COMAH-Containment-Policy states that hydrostatic testing of in-service bunds is not best practice due to dangers in tank stability and also because water can be forced under the base of the tank which will lead to corrosion problems. Visual inspections are therefore preferred in the UK for bund integrity testing.
The procedure for hydrostatic bund integrity testing requires the bund to be filled with water and the drop in water level to be measured over a defined period. There are two standards for hydrostatic testing, BS8007 and CIRIA 163 with the latter standard being a much shorter test.
Bund integrity testing by visual inspection entails the examination of the bund for cracks, holes, and other defects which may cause a loss of integrity. Examples of defects include exposed rebar, unsealed pipes passing through bund walls and unplugged shuttering bolt holes. In both Ireland and the UK the inspection must be by a competent person such as a Chartered Engineer.
Environmental Efficiency carries hundreds of bund visual assessments every year across the UK and Ireland. Where necessary Environmental Efficiency can fill bunds with water and dispose of afterwards.


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Health & Safety Signs
Keep Safe Distance Sign Blue/White 300mm x 450mm - Polypropylene
Maintain Social Distancing At All Times Sign Red/White 300mm x 450mm - Polypropylene
This Premise Has Space Sign 300mm x 450mm - Polypropylene
Keep Safe Distance Sign White/Blue 300mm x 450mm - Polypropylene
Stop The Spread Sign 300mm x 450mm - Polypropylene
Social Distancing Sign 300mm x 450mm - Polypropylene
Avoid Injury Storage Height 450mm x 600mm - Polypropylene
Avoid Injury Don't Twist 450mm x 600mm - Polypropylene
Avoid Injury Use Mechanical Means 450mm x 600mm - Polypropylene
Avoid Injury Carry Loads With a Straight Back 450mm x 600mm - Polypropylene
Avoid Injury Straight Back and Knees Bent 450mm x 600mm - Polypropylene
Avoid Injury Do Not Jump 450mm x 600mm - Polypropylene
/Safety & PPE/Signs/Information Signs/Health & Safety Signs 20355 19620 Health & Safety Signs https://www.signet.net.au/safety-ppe/signs/information-signs/health-safety-signs.html 1 5
Health & Safety Signs provide reminders to staff to help prevent injuries and accidents from occurring in the workplace. Signet's range of Health & Safety Signs aim to identify poor technique and work practices, that may occur when performing tasks.
What size and material are your Health & Safety Signs?
All of our Health & Safety Signs are made from high quality polypropylene which is strong, resilient and durable. Signet's Health & Safety Signs come in various sizes including 600mm x 450mm and 300mm x 450mm.
Why is it important to have work place Health & Safety signs?
Injuries and accidents can come at a great cost for a business, both in terms of time and money. Providing reminders to staff on the correct practices in the workplace can ensure the right procedures are being followed, lowering the risk of an accident or injury occurring.
Signet's Own
Signet's Own Polypropylene Safety Sign
Carton length
400mm and above
Carton width


text:
Home|Using Nintex|Case Studies|New Belgium Brewing Co. crafts solution to overcome unexpected challenges
New Belgium Brewing Co. crafts solution to overcome unexpected challenges
Who'd have thought that a brewer with operations in Colorado and North Carolina would have to comply with privacy regulations promulgated in California? New Belgium Brewing learned that would be the case at the end of 2018, when the California Consumer Privacy Act (CCPA) was passed with provisions covering California consumers doing business with out-of-state companies—like New Belgium Brewing. The compliance details that New Belgium Brewing needed weren't available until the following July, and the company had only five months to develop a compliance system by the law's deadline of December 2019. Failure to do so could have meant big fines.
With help from Nintex, New Belgium Brewing developed a solution in just one month. Its solution standardizes and automates the way the company handles privacy requests from its California customers, enabling the company to meet those requests twice as fast as the law requires and leaving employees with plenty of time for their top priority: crafting great beer.
Fast, easy solution created in just one month
Solution enables brewer to respond to consumers twice as fast as the law requires
Flexible technology is always on hand to address unexpected challenges
New Belgium Brewing Company
newbelgium.com
Capabilities Used
Nintex Workflow Cloud®
New Belgium strives to be a role model for other businesses in its impact on the communities it serves and the planet itself. The craft beer maker has breweries in Ft. Collins, Colorado and Ashville, North Carolina.
What they needed
New Belgium Brewing Co. had five months to come up with an accurate, standardized way to respond to privacy requests from California residents—or face big fines.
How they did it
The company turned to the same technology it has used for 12 years: Nintex. It created a streamlined, automated workflow solution that facilitates every step of the process, in and out of the cloud, within the company and between it and the consumer.
The Swiss army knife of technologies
Using Nintex, New Belgium Brewing had previously streamlined, automated workflow solutions for everything from human resources and sales to volunteer management, planning product launches, and tracking beer shelf life. "We began using Nintex 12 years ago," says Tye Eyden, Collaboration Business Systems Analyst at New Belgium Brewing. "We use it on-prem and in the cloud, for solutions within a department and throughout the company. The more our people experience Nintex, the more they come to us to propose new ways to use it."
The CCPA-compliance solution is among the broadest Nintex-based solutions at New Belgium Brewing, engaging the company's marketing, sales, environment health and risk, HR, and even corporate sustainability departments.
The process kicks off when California-based consumers ask the company what personally identifiable information it holds on them through a form built on Nintex Workflow Cloud. Once submitted, the data is stored in SharePoint Online and a workflow routes the request to each department. Each department identifies the consumer's information in their database and forwards their responses. The consumer is then sent an email that lists the stored information and gives them the option to request its deletion. Once they reply, the workflow routes their request back for each department to take the appropriate action and once completed, a confirmation email is sent. All these steps are captured by Nintex, stored in SharePoint, and are immediately available in case of audit.
We have other tools; we're a big Microsoft shop. But I know I can do it simpler and easier, and get it done faster, with Nintex.
Tye Eyden, Collaboration Business Systems Analyst, New Belgium Brewing
Related success stories
New Belgium Brewing Co. gets beer to market faster with Nintex Workflow for SharePoint
Saves weeks or months per new brew launch Launching a new beer involves multiple departments and teams at New Belgium Brewing Co–each with its own set of tasks and processes to follow. About 20%, or 100 of 570, of New Belgium's employees are involved each launch. New Belgium traditionally relied on Outlook, meetings, and SharePoint […]
Avara Foods improves health and safety reporting
Avara Foods supplies poultry products to some of Britain's largest supermarkets and restaurant chains. Avara Foods has been using K2 Software for over 20 years to streamline processes and deliver ongoing efficiencies for the business. Avara Foods uses the Nintex K2 Five as a catalyst to deliver a suite of health and safety apps for […]


In separate incidents, a man and a woman in South Jersey are facing charges for injuring cops with their motor vehicles. 2 men were charged after being involved in a physical fight with police or sheriff's officers.
A 36-year-old Galloway man drove into the cops during a traffic stop and escaped from the accident spot. The motor vehicle was found at a short distance from where it hit the cops and after a search of the vehicle, officers found heroin in it. The suspect was taken into custody by the police from his apartment complex.
Another 36-year-old woman failed to stop at a stop sign and collided with the patrol car which left a Gloucester County police officer with multiple injuries. There were 3 children riding in the motor vehicle with the woman for which she was charged with drunken driving and child endangerment.
A 41-year-old Newark man was sentenced to 7 years in state prison as he resisted arrest and didn't obey the instructions given by Orange police to stay away from his motor vehicle as it was being towed after the crash. He has 5 felony convictions, including one for cocaine and heroin possession related to that incident.
A Jersey City resident pleaded guilty to a disorderly conduct charge for his involvement in a physical fight with 2 Essex County sheriff's officers. The incident happened during the sentencing of a man who killed his brother.
A Tom River used car dealer is sentenced to 3 years in jail for not paying the amount of $220,000 in sales tax; the amount had to be paid in 4 years.
Another North Jersey man faces possible time in jail for allegedly not cooperating with law enforcement officers. A 27-year-old was shot 2 times while stealing 2 motor vehicles and he sent a state trooper out of a moving motor vehicle. A state grand jury indicted the Newark resident on multiple charges.
A 38-year-old Elmwood Park man is accused of the death of 2 men as he provided them heroin. The Passaic County Prosecutor's Office charged him with 2 counts of strict liability for drug induced-death because he was the reason of both deaths.
According to authorities, an on-off duty Irvington police officer assisted in taking an Edison bank robber and his accomplice into custody as the sergeant saw a man running out of the bank and into a waiting getaway motor vehicle.


Let Us Setup Your Site!
Before getting started, if you don't want the hassle of installing and setting up everything yourself, or just don't have the time, then we can do it for you. We have services available to quickly install, setup and configure both WordPress and your theme without you having to lift a finger!
So, if you want us to do all the hard work for you then click here to find out more. If you've already purchased an installation and setup service then you can submit an installation request here.
To successfully setup your site you'll need to source web hosting, as well as complete installation of WordPress and the Flexr theme. These tasks are covered in the following sections.
Before you can install WordPress and Flexr you'll need to source your own web hosting. You can choose from any web host that supports WordPress (which is nearly all web hosts these days). If you don't already have hosting then we recommend TerraNetwork web hosting. Click the banner image to the right to get started.
If you don't go with TerraNetwork then please make sure your web host meets the minimum requirements for running WordPress.
If you don't already have WordPress installed on your site then you'll need to do this before you can install the Flexr theme. Most web hosts allow you to install WordPress directly from your website control panel – which is the easiest method by far. Contact your web host to see if they support this.
If your host does not allow you to install WordPress from the control panel then you'll have to install it yourself manually. Fortunately it is a simple process and there are a few steps involved. Click here for instructions on installing WordPress manually.
If you have existing website content (i.e. images and text) prior to installing WordPress then you'll need to fully back-up this data, otherwise it could get overwritten during the installation process. Once this is done it's a good idea to delete all site files before installing WordPress so you don't have any conflicts between your old and new site.
Note: Please install the WPGO Metro plugin (download from your WPGO Themes account), as described in your purchase email, BEFORE working through this document as it includess all the custom post types, widgets, and shortcodes described below.
Once WordPress has been installed it's time to get Flexr installed and activated. You'll be glad to know that this only needs to be done manually once, as future theme updates occur via automatic notifications in your WordPress admin (see the section Flexr Automatic Updates below for more details).
Download the theme zip file from your wpgothemes.com account page to your computer.
Log into your WordPress site and go to Appearance > Themes.
Click on the Add New button to go to the Add Themes page, then click the Upload Theme button.
Follow the prompts to upload the theme zip file you downloaded to your computer in step 1.
Once successfully uploaded, activate your theme.
Working through the rest of this document will help you to get to grips with customizing Flexr. If you have ANY questions, suggestions, or general comments we would love to hear from you. Your feedback is very important to us!
Include the new logo placement customizer settings.
One of the first things you'll want to do is to add a logo to your site. To do this go to Appearance > Customize. Expand the Site Identity section and click on Add Logo to select your logo image. You can add a logo from the WordPress media library or upload an image from your computer. Once selected, your logo will automatically appear in the preview window.
You can also customize how the logo is rendered in relation to the site title, and tagline via the Logo Placement drop down box. The default is to display the logo above the site title, and tagline (if they are defined). You can change this so that the logo is placed next to the site title (but still above the tagline), or is placed next to site title and tagline.
Click on Save & Publish if you want to make the change publicly viewable.
Flexr fully supports the use of Retina images, including using a Retina image for your logo. Simply make sure that you upload two images for the Retina image you want to display. The only two caveats are the images MUST be in the same folder on your server, and the Retina image has to have an identifier in the filename. e.g. logo.png, and logo@2x.png.
If you make sure to always upload the normal and Retina images at the same time then you won't run into any problems.
The Retina.js script included with Flexr will auto detect the device Flexr is being viewed on and display the Retina image version where appropriate. If no Retina image exists then the normal image is displayed.
Go to Appearance > Menus to create your custom menus. You can create as many menus as you wish, but they will only appear on the front end of your site if they are assigned to a theme location.
Look for the check boxes underneath your custom menu to assign menus to specific locations, or click the Manage Locations tab at the top of the page.
Each item added to a menu can also be prefixed with an icon. Go to Appearance > Menus and expand an individual menu item. In the Specify Menu Icon Name text box enter a valid font icon name.
Make sure the icon name is the same as it appears on the Font Awesome site. Hit the Enter key or click the Save Menu button to save your changes.
You can even hide menu labels if you want to show icons only. To do this simply prefix a menu label with a double underscore. e.g. If you had a menu item with the label Home then replace this with __Home to hide it on the front end.
When hiding menu labels make sure you enter a valid icon name in the Specify Menu Icon Name field otherwise your menu item won't be visible. Also, menu labels must never be blank or the whole menu item will be deleted when you try to save the menu!
You can add widgets to sidebars, the header, the footer, or to pages that use the "Widgetized Page" page template.
Go to Appearance > Widgets and drag the widgets (on the left) to widget areas (on the right). You will need to expand a widget area before you can drag a widget into it.
Once added to a widget area, you can edit the widget settings to your requirements. View the relevant area on the front end of your site to see the widget in action!
A brand new feature available in all our themes apart from Minn is Header Templates. This allows you to specify exactly how the header is rendered on your site for every page. Please click here for full details.
Go to Customizer > Colors and use the color pickers to alter color settings. Click on the default button for each color picker to revert to the default value.
Go to Appearance > Flexr Options to view the theme options page. Here you can see links to theme support, this setup page, our Twitter account, and your WPGO Themes account area.
You can also add styles to the "Custom CSS" text box to customize your site. This is a very convenient way to quickly add CSS customizations without having to use a WordPress child theme.
Add your theme license key to actively receive automatic theme update notifications.
The following section covers some misc features.
Favicon support. Go to Customizer > Site Identity > Site Icon to select a favicon image.
Header/footer code insertion (e.g. to insert Google Analytics code). Go to Settings > WPGO Metro to add code that will be added to your site header/footer.
Since version 3.4, WordPress has included a very cool feature called the WordPress customizer. For themes that support it the customizer allows you to quickly make design changes in real-time and preview them prior to making them publicly viewable.
On the front end of your site (when logged in as admin) click the 'Customize' drop down menu link from the WordPress toolbar.
By clicking on Appearance > Customize in the WordPress admin.
Upload a logo to the site header.
Edit the site title and tagline.
Toggle the visibility of the site title and tagline.
Change sites colors via multiple color picker controls.
Set global column layout for all pages (can be overridden on individual pages).
Set global sidebar width for all pages.
Choose any Google font and apply it to CSS selector(s) of your choice.
Set the site background image.
Set the custom menus for the Main, and Top menu locations.
Set the static pages used for the homepage and the blogs page.
Customize the links displayed in the footer.
If you plan to make significant changes to theme styles or need to implement PHP/JS code modifications then using a child theme is the recommended way to customize your site. The main advantage of using a child theme is to protect your customizations during theme updates. Otherwise, if you only altered the main theme files then all changes would be lost during theme updates.
Note: Changes to your site made via the theme customizer or theme options won't be affected during a theme update.
We provide a sample Flexr child theme to get you started, which you can download via your wpgothemes.com account page. The child theme is installed and activated in exactly the same way as the Flexr parent theme.
Once your child theme is installed and activated, go to Appearance > Editor to make CSS customizations in style.css, and add custom PHP code in functions.php. It is also possible to accomplish more advanced tasks by copying theme template files from the parent theme to the child theme. This forces WordPress to use the template file in your child theme instead of the parent theme, so you are safe to make modifications to it that won't be lost during theme updates. To copy a theme template file you will need to do it via your web host control panel or FTP software. As mentioned earlier, any changes to your site you make via the child theme will be preserved during updates to the parent Flexr theme.
Even with a child theme active you can still use the Custom CSS text box in Flexr theme options to make tweaks to your site styles. These Custom CSS style rules are rendered AFTER the parent/child theme styles, and so will ALWAYS have priority. This is done on purpose to allow you to use the Custom CSS theme option to tweak styles on your site whether you are using a child theme or not.
Once Flexr has been installed and activated you'll see an update notice appear every so often in WordPress admin prompting you to install the latest version. Before you proceed with the update you can optionally view details about the new version first to decide whether you wish to install it.
In general, it is recommended that you ALWAYS update to the latest theme version when available, for security reasons.
By the same token, you should always update other plugins you have installed to the latest versions, as well as WordPress itself whenever updates become available.
Note: You'll only be prompted to update Flexr if you have first entered an active theme license key in theme options. You can find your license key via your wpgothemes.com account page.
Custom Post Types – Enable a range of powerful WordPress custom post types such as testimonials, portfolios, and slides.
Several widgets and shortcodes to enhance functionality.
Sidebar Manager – Create custom sidebars to display on specific pages.
Note: WPGO Metro works with ANY WPGO theme so if you have installed it before for another WPGO theme then you do not need to install it again. Once installed, and the license key entered, WPGO Metro will receive update notifications automatically.
The sections below give more details about WPGO Metro features.
There are four powerful WordPress custom post types including Slides, Portfolio, Documents, and Testimonials. The slides CPT, for example, allows you easily build flexible custom slides for use in the content slider widget. There are also various associated shortcodes to accompany these Custom Post Types.
Create unlimited custom sidebars for your site, and display them in any order on posts/pages. This plugin makes it easy to have unique sidebars on every page on your site!
You can use shortcodes in any post or page. Take a look at this Flexr demo page to see examples of how to use the available shortcodes.
Displays the main content and sidebar according to your chosen layout.
Displays a contact form. The form will be sent to the e-mail address in Settings > General.
Adds styles and functionality for a FAQ page.
Create any landing page you like with this page template as it collapses all margins and padding to give you full control over the page layout.
Displays your portfolio items. Includes options for small, medium, or large images, plus a choice of portfolio group display. You will first need to specify at least one portfolio item and one portfolio group.
Displays a sitemap of your pages in a single list.
Allows widgets to be displayed inside a WordPress page. To use this effectively you will need to use sidebar manager to select which sidebars (and in what order) are displayed.
With Flexr you can easily manage the number of columns displayed on any page. In the theme customizer you can specify the 'global' column layout that will be used for EVERY page by default (the default setting is 2-columns, with sidebar on the right of the page). You can also select a different column layout for each individual page if you need to. On the page editor select the 'Column Layout' dropdown in the 'Page Display Options' meta box and choose a new column layout.
Sometimes it is handy to be able to hide page titles (the title displayed on the page, not the browser page title). This is easily done in Flexr. In the 'Page Display Options' meta box select the 'Hide single Page title' check box and click the update button to save changes. When you view the page the title will now be hidden.
Don't forget if you start to install Flexr but then decide you'd prefer us to do all the work, then we can do that for you.


I always thought I just hate it the color white and did not realize I had Leukophobia- fear of the color white.
One day when I was reading the Internet a discovered this phobia and decided I needed to do something to overcome it.
Your book, Energy Therapy was a big help in overcoming my phobia. I have since been able to accept the color white, and not be as judgmental about it as I once was.
I do appreciate you having such a good book to read, understand, and utilize in overcoming my fear. Thank you very much.


text:
FactoryGirl.define do
  factory :user do
    first_name { Faker::Name.first_name }
    last_name  { Faker::Name.last_name }

    after(:create) do |user|
      create_list(:email, 3, user: user)
      create_list(:phone, 2, user: user)
    end
  end
end


Since 1975, the Black Wharton Undergraduate Association has existed as a student led organization serving the African-American community at the University of Pennsylvania. As both an academic support and professional development organization, BWUA has had a profound impact on its members. In addition to supporting our members in gaining industry knowledge and networking skills, BWUA can be credited with directly helping its members gain access to internships and full-time jobs in some of America's most competitive and elite Fortune 500 companies.
BWUA has a substantial alumni base spanning across the country in varied array of industries. Both the talent and motivation of this group of individuals highlight the success of BWUA's mission. The current BWUA board realizes that having a good relationship with alumni is advantageous for ensuring both the mentorship of its members and the sustainability of the organization. As a result, we are making increasing efforts to maintain our alumni network and increase alumni interactions with the organization. Through semi-annual newsletters, we intend to keep alumni aware of the detailed progress of BWUA events and students' individual successes. Furthermore, our multiple alumni receptions every year give alumni the opportunity to return to Penn and meet BWUA's current members.
If you are interested in getting more involved with BWUA's alumni initiatives or have any questions, comments, or recommendations, please contact our Director of Alumni Relations Dawit Wondimagegn at dawitw@wharton.upenn.edu.
Thank you! Please encourage your classmates to join the database as well.
Congratulations on successfully completing your journey here at Penn. You have achieved and excelled during your short time here and stand as examples of excellence for many of us coming behind to look up to. With that being said, I ask that you please fill out our alumni survey!
Black Wharton understands the tremendous value of its alumni network. We want to keep our members up to date on your current endeavors and latest accomplishments as we work to improve the existing relationship between you and the organization. We would also like to serve you in continually connecting you with your classmates as well as older BWUA alumni as you embark on your professional journey.
3. Allow us to better keep track of our alumni to send invitations for events such as HEMM Conference, Alumni Mixers (Penn Club and beyond), and our annual G.A.L.A.
4. Feature you in our Alumni Spotlights, making our members aware of the different activities our alumni are involved in.
Additionally, I ask that you take a quick second to join our LinkedIn group. In this new digital age of recruitment and networking, we strive to make the BWUA LinkedIn page a valuable resource for all of you as you head into the workforce or graduate school. Simply type "Black Wharton Undergraduate Association Alumni" in the LinkedIn search bar and you will be able to find and join the group (direct link: https://www.linkedin.com/groups/1615007 ).
We see you all as an essential core to BWUA and view this as a great opportunity to strengthen our relationship with you. Each and every one of us has benefited from those who came before, and this is a great opportunity to help the younger students who will come after you. I wish you all the best for your future and have no doubt your accomplishments will soon tally up! As always, please feel free to reach out, BWUA will always be here to help in any way we can!


While Android is today's top platform in terms of smartphone numbers, tablets are an altogether different proposition. The Apple iPad still leads the game in terms of tablet use and mobile access for tablets. But Google's Nexus 7 may already be making a dent on Apple's sales, although in a small way.
Chitika reports that Google's tablet offering still has a long way to go to the top. Apple's iPad (various models) still enjoys 91% of all web traffic from tablets. The rest of the pie is shared among the likes of Samsung's Galaxy tab, along with other Android tablets and the BlackBerry Playbook. The Nexus 7 itself has a 0.33% share of tablet traffic. But don't be fooled by the paltry figures, as the Nexus 7 is inching its way up, with a 135% growth since July.
Still, iPad figures are difficult to beat. Cnet notes that Apple's offering still leads the game, and Android has not yet succeeded in grabbing a significant market share from iOS in the tablet market. "In order to make the graph readable, we had to measure other tablets on a 'per 100 iPad impressions' scale," said a Chitika spokesperson in an interview.
The Chitika figures might indicate that Apple may be on the decline, albeit quite slow, with a 0.34% decrease in web traffic in the same period. But with the launch of the latest Apple iPad Mini, should Android tablet makers be concerned that Apple will continue to reign supreme in this market?


text:
Q: What is VBS UCASE function doing to Japanese? In order to avoid case conflicts comparing strings on an ASP classic site, some inherited code converts all strings with UCASE() first.  This seems to work well across languages ... except Japanese.  Here's a simple example on a Japanese string.  I've provided the UrlEncoded values to make it clear how little is changing behind the scenes:
Server.UrlEncode("戦艦帝国") = %E6%88%A6%E8%89%A6%E5%B8%9D%E5%9B%BD
UCASE("戦艦帝国") = ƈ�ȉ�Ÿ�ś�
Server.UrlEncode(UCASE("戦艦帝国")) = %C6%88%A6%C8%89%A6%C5%B8%9D%C5%9B%BD

So is UCASE doing anything sensible with this Japanese string?  Or is its behavior buggy, undefined, or known to be incompatible with Japanese?
(LCASE leaves the sample string alone.  But now I'm wary of switching all comparisons to LCASE because I don't know if it bungles other non-western languages that do work with UCASE....)

A: https://msdn.microsoft.com/en-us/library/1systdcy(v=vs.84).aspx

Only lowercase letters are converted to uppercase; all uppercase letters and non-letter characters remain unchanged.

https://en.wikipedia.org/wiki/Letter_case

Most Western languages (particularly those with writing systems based on the Latin, Cyrillic, Greek, Coptic, and Armenian alphabets) use letter cases in their written form as an aid to clarity. Scripts using two separate cases are also called bicameral scripts. Many other writing systems make no distinction between majuscules and minuscules – a system called unicameral script or unicase.

"lowercase or uppercase letters" does not apply in Chinese-Japanese-Korean languages, hence, the output of UCase() should remain unchanged.

redpajama_set_name: RedPajamaStackExchange


cats – Page 10 – Hello Adorable!
The perfect gift for your all laser-loving cat-owning friends. Designed/created by TinyMeat and available to buy at BuyOlympia.
Don't mistake this for a cardboard cat house. This is a "Cat Chalet!" Love the deer antlers at the entry way. A lovely touch. You can buy this online at Plasticashop.com or go to their awesome brick and mortar store in L.A.
Hats off to the ad agency who came up with this cheeky (and no doubt viral) ad campaign for the New York State Lottery's "Sweet Million" game.
What Are They Putting in the Pet Food?
Unlikely inter-species friendships are documented all over the internet but I think this is the largest number of different animals I've seen all together. Honestly, um, how the heck did this happen?
Not to add to overpopulation of cute cat photos swirling around the internet… but, c'mon. How could I pass this up? Via Digg.
Rarely captured on film, the elusive ritual of a dog and cat sharing a milkshake has finally been immortalized in this mousepad in never-before-seen clarity.
Available to buy at myfavoritemirror.com. Designed/illustrated by the super-talented Kathleen Habbley.
This hysterical 1-minute cat re-enactment of LOST almost makes up for my disappointment with the LOST series finale a couple of nights ago. Almost.
At any rate, it's talking cats; it's LOST-related. How can you go wrong?