text:
\section{Introduction} \label{farey_intro}

In \cite{Weyl1} Hermann Weyl developed a general and far-reaching theory for the equidistribution of sequences modulo 1, which is discussed from a historical point of view in Stammbach's paper \cite{Stammbach}. Especially Weyl's result that for real $t$ the sequence
$
t ,\, 2t ,\, 3t,\,\ldots
$
is equidistributed modulo 1 if and only if $t$ is irrational
can be found in \cite[\S 1]{Weyl1}. This means that
\begin{equation*}
\lim \limits_{N \to \infty} \frac{1}{N}
\#\left\{nt -\lfloor nt \rfloor \in [a,b] : n \leq N \right\}=b-a
\end{equation*}
holds for all subintervals $[a,b] \subseteq [0,1]$
if and only if $t \in \mathbb{R} \setminus \mathbb{Q}$.
Here $\#M$ denotes the number of elements of a finite set $M$.
This generalization of Kronecker's theorem \cite[Chapter XXIII, Theorem 438]{hw} is an important result in number theory. We have only mentioned its one dimensional version, but the higher dimensional case is also treated in Weyl's paper.

Now we put
\begin{equation}\label{Sxdef}
S(x,t)=\sum \limits_{k \leq x}
\left(kt -\lfloor kt \rfloor-\frac12\right)
\end{equation}
for $x \geq 0$ and $t \in \mathbb{R}$.
If the sequence $(n t)_{n \in \mathbb{N}}$ is
"well distributed" modulo 1 for irrational $t$, then $|S(x,t)|/x$ should be "small"
for $x$ large enough.

In \cite[equation (2), p. 80]{Ost1922} Ostrowski used the continued fraction expansion $t=\langle \lambda_0,\lambda_1,\lambda_2,\ldots \rangle$ for irrational $t$ and presented a very efficient calculation of $S(n,t)$ with $n \in \mathbb{N}_0$, he namely obtained a simple iterative procedure using
at most $\mathcal{O}(\log n)$ steps for $n \to \infty$, uniformly in $t \in \mathbb{R}$. We have summarized his result in Theorem \ref{ostrowski} of the paper on hand.
From this theorem he derived an estimate for $S(n,t)$ in the case of irrational
$t \in \mathbb{R}$ which depends on the choice of $t$.
Especially if $(\lambda_k)_{k \in \mathbb{N}_0}$ is a bounded sequence, then we say that $t$ has \textit{bounded partial quotients}, and have in this case from Ostrowski's paper
\begin{equation}\label{ostopt}
|S(n,t)| \leq C(t) \log n \,,\quad n \geq 2\,,
\end{equation}
with a constant $C(t)>0$ depending on $t$\,. Ostrowski also showed that this
gives the best possible result, answering an open question posed by Hardy and Littlewood.\\

In \cite{Lang1966} and \cite[III,\S 1]{Lang1995} Lang obtained
for every fixed $\varepsilon > 0$ that
\begin{equation}\label{lang_fast_ueberall}
|S(n,t)| \leq\left(\log n \right)^{2+\varepsilon} \quad
\mbox{~for ~} n \geq n_0(t,\varepsilon)
\end{equation}
for almost all $t \in \mathbb{R}$ with a constant $n_0(t,\varepsilon) \in \mathbb{N}$.
Let $\alpha$ be an irrational real number and $g \geq 1$ be an increasing function,
defined for sufficiently large positive numbers. Due to Lang
\cite[II,\S 1]{Lang1995} the number $\alpha$ is of type $\leq g$ if for all sufficiently large numbers $B$, there exists a solution in relatively prime integers $q,p$ of the inequalities
$$
|q\alpha-p|<1/q\,, \quad B/g(B)\leq q <B\,.
$$
After Corollary 2 in \cite[II,\S 3]{Lang1995}, where Lang studied the quantitative connection between Weyl's equidistribution modulo 1 for the sequence $t,2t,3t,\ldots$ and the type of the irrational number $t$, he mentioned the work of Ostrowski \cite{Ost1922} and Behnke \cite{Behnke} and wrote:
"Instead of working with the type as we have defined it, however,
these last-mentioned authors worked with a less efficient way of determining the approximation behaviour of $\alpha$ with respect to $p/q$, whence followed weaker results and more complicated proofs." \\

Though Lang's theory gives Ostrowski's estimate \eqref{ostopt}
for all real irrational numbers $t$ with bounded partial quotients,
see \cite[II, \S 2, Theorem 6 and III,\S 1, Theorem 1]{Lang1995}, as well as
estimate \eqref{lang_fast_ueberall} for almost all $t \in \mathbb{R}$, Lang did not use
Ostrowski's efficient formula for the calculation of $S(n,t)$\,.
We will see in Section \ref{dirichlet_section} of the paper on hand that Ostrowski's formula can be used as well in order to derive estimate \eqref{lang_fast_ueberall} for almost all $t \in \mathbb{R}$, without working with the type defined in \cite[II,\S 1]{Lang1995}. For this purpose we will present the general and useful Theorem \ref{kettenabschaetz}, which will be derived in Section \ref{sawtooth} from the elementary theory of continued fractions. Our resulting new Theorems \ref{B0_mass_almost_everywhere},
\ref{B0_mass} now have the advantage to provide an explicit form for those sets of $t$-values
which satisfy crucial estimates of $S(n,t)$.\\
If $\Theta:[1,\infty) \to [1,\infty)$ is any monotonically increasing function
with $\begin{displaystyle}\lim \limits_{n \to \infty} \Theta(n)=\infty\end{displaystyle}$\,,
then Theorem \ref{B0_mass} gives the inequality
$
|S(n,t)| \leq 2 \log^2(n) \Theta(n)
$
uniformly for all $n \geq 3$ and all $t \in \mathcal{M}_n$ for a sequence of sets
$\mathcal{M}_n \subseteq [0,1]$ with $\lim \limits_{n \to \infty}|\mathcal{M}_n|=1$.
Here $|\mathcal{M}_n|$ denotes the Lebesgue-measure of $\mathcal{M}_n$.
On the other hand Theorem \ref{Bx_L2} states that
$$
\left(\int \limits_{0}^{1}S(n,t)^2\,dt\right)^{1/2}=\mathcal{O}(\sqrt{n})
\quad \text{for~}n \to \infty
$$
gives the true order of magnitude for the $L_2(0,1)$-norm of $S(n,\cdot)$.
If $\Theta$ increases slowly then the values of $S(n,t)$ with $t$ in the unit-interval $[0,1]$
which give the major contribution to the $L_2(0,1)$-norm have their pre-images only in the small complements $[0,1] \setminus \mathcal{M}_n$. We see that $n_0(t,\varepsilon)$
in estimate \eqref{lang_fast_ueberall}
depends substantially on the choice of $t$.
Moreover, a new representation formula for $B_n(t)=S(n,t)/n$
given in Section \ref{sawtooth}, Theorem \ref{Bx_thm} will also give an alternative proof of Ostrowski's estimate \eqref{ostopt} if $t$ has bounded partial quotients.
In this way we summarize and refine the corresponding results given by Ostrowski and Lang,
respectively.\\

For $n \in \mathbb{N}$ and $N = \sum\limits_{k=1}^{n}\varphi(k)$
the Farey sequence $\mathcal{F}_n$ of order $n$ consists of all reduced and ordered fractions
\begin{equation*}
\frac{0}{1}=\frac{a_{0,n}}{b_{0,n}} < \frac{a_{1,n}}{b_{1,n}} < \frac{a_{2,n}}{b_{2,n}}
< \ldots < \frac{a_{N,n}}{b_{N,n}}=\frac{1}{1}
\end{equation*}
with $1 \leq b_{\alpha,n} \leq n$ for $\alpha = 0,1, \ldots,N$.
By $\mathcal{F}^{ext}_n$ we denote the extension of $\mathcal{F}_n$
consisting of all reduced and ordered fractions $\frac{a}{b}$ with $a \in \mathbb{Z}$
and $b \in \mathbb{N}$, $b \leq n$.

In the former paper \cite{Ku4} we have studied 1-periodic functions
$\Phi_n : \mathbb{R} \to \mathbb{R}$ which are related to the Farey sequence $\mathcal{F}_n$,
based on the theory developed in \cite{Ku1, Ku2, Ku3} for related functions.
For $k \in \mathbb{N}$ and $x>0$ the 1-periodic functions $q_k, \Phi_x : \mathbb{R} \to \mathbb{R}$ are
\begin{equation}\label{familien}
\begin{split}
q_k(t) &=-\sum \limits_{d | k} \mu(d)\,
\beta \left(\frac{kt}{d}\right) ~\,
\mbox{with}~\, \beta(t)=t-\lfloor t \rfloor - \frac12\,,\\
\Phi_x\left(t \right) &= \frac{1}{x} \sum \limits_{k \leq x}
q_k(t)=- \frac{1}{x} \sum \limits_{j \leq x}
\sum \limits_{k \leq x/j} \mu(k) \beta\left(jt \right)\,.\\
\end{split}
\end{equation}
The functions $\Phi_x$ determine the number of Farey-fractions in prescribed intervals.
More precisely, $t\sum\limits_{k \leq n}\varphi(k)+n\Phi_n(t)+\frac12$ gives the number
of fractions of $\mathcal{F}^{ext}_n$ in the interval $[0,t]$ for $t \geq 0$ and $n \in \mathbb{N}$.
Moreover, there is a connection between the functions $S(x,t)$ and $\Phi_x(t)$
via the Mellin-transform and the Riemann-zeta function, namely the relation
\begin{equation*}
\int \limits_{1}^{\infty} \frac{S(x,t)}{x^{s+1}}dx
=-\zeta(s)\,\int \limits_{1}^{\infty} \frac{\Phi_x(t)}{x^s}dx\,,
\end{equation*}
valid for $\Re(s)>1$ and any fixed $t \in \mathbb{R}$.
We will use it in a modified form in Theorem \ref{F_thm}.

In contrast to Ostrowski's approach using elementary evaluations of $S(n,t)$
for real values of $t$, Hecke \cite{Hecke} considered the case of special quadratic irrational numbers $t$, studied the analytical properties of the corresponding Dirichlet series
\begin{equation}\label{Hecke_series}
\sum \limits_{m=1}^{\infty}\frac{mt-\lfloor mt \rfloor-\frac12}{m^s}=
s\int \limits_{1}^{\infty} \frac{S(x,t)}{x^{s+1}}dx
\end{equation}
and obtained its meromorphic continuation to the whole complex plane,
including the location of poles. Hecke could use his analytical method to derive estimates
for $S(n,t)$, but he did not obtain
Ostrowski's optimal result \eqref{ostopt}
for real irrationalities $t$ with bounded partial quotients.

For positive irrational numbers $t$
Sourmelidis \cite{Sourmelidis} studied analytical relations
between the Dirichlet series in \eqref{Hecke_series} and the so called Beatty zeta-functions and Sturmian Dirichlet series.

For $x>0$ we set
\begin{equation*}
r_x =\frac{1}{x} \sum \limits_{k \leq x} \varphi(k)\,,
\quad
s_x = \sum \limits_{k \leq x} \frac{\varphi(k)}{k} \,,
\end{equation*}
and define the \textit{continuous} and odd function $h : \mathbb{R} \to \mathbb{R}$ by
\begin{equation*}
h(x)=\begin{cases}
0 &\ \text{for}\ x =0 \,,\\
3x/\pi^2+r_x-s_x &\ \text{for}\ x >0 \,,\\
-h(-x) &\ \text{for}\ x < 0\, .\\
\end{cases}
\end{equation*}
Then we obtained in \cite[Theorem 2.2]{Ku4}
for any fixed reduced fraction $a/b$ with $a \in \mathbb{Z}$ and $b \in \mathbb{N}$
and any $x_*>0$ that for $n \to \infty$
$$
\tilde{h}_{a,b}(n,x)=-b \, \Phi_n\left(\frac{a}{b}+\frac{x}{bn}\right)
$$
converges uniformly to $h(x)$ for $-x_* \leq x \leq x_*$.
For this reason we have called $h$ a limit function.
It follows from \cite[Theorem 1]{Mont} with an absolute constant $c>0$
for $x\geq 2$ that
$\begin{displaystyle}
h(x) = \mathcal{O}\left( e^{-c\sqrt{\log x}}\right)\,.
\end{displaystyle}$
Plots of this limit function are presented in
Section \ref{appendix}, Figures \ref{h25eps},\ref{h50eps},\ref{h500eps}.

In Section \ref{sawtooth} we introduce another limit function ${\tilde \eta} : \mathbb{R} \to \mathbb{R}$ by ${\tilde \eta}(0)=-\frac12$ and
\begin{equation*}
{\tilde \eta}(x)=
\frac{(x-\lfloor x \rfloor)(x-\lfloor x \rfloor-1)}{2x}
 \quad \text{for}\ x \in \mathbb{R} \setminus \{0\}\,,
\end{equation*}
and obtain from Theorem \ref{rescaled_limit} for $B_n(t)=S(n,t)/n$
analogous to \cite[Theorem 2.2]{Ku4} the new result that for $n \to \infty$
$$
\tilde{\eta}_{a,b}(n,x)= b \, B_n\left(\frac{a}{b}+\frac{x}{bn}\right)
$$
converges uniformly to $\tilde{\eta}(x)$ for $-x_* \leq x \leq x_*$.
A plot of  ${\tilde \eta}(x)$ for $-8 \leq x \leq 8$ is given in
Section \ref{appendix}, Figure \ref{eta8eps}.
Now Theorem \ref{Bx_thm}(b) follows from part (a) and leads to
the formula \eqref{Bsequence}, which bears a strong resemblance
to that in Ostrowski's theorem \ref{ostrowski} and gives an alternative proof for Ostrowski's estimate \eqref{ostopt} if $t$ has bounded partial quotients.
Hence it would be interesting to know whether there is a deeper reason for this analogy.

\section{Sums with sawtooth functions}\label{sawtooth}

With the sawtooth function $\beta(t)=t-\lfloor t \rfloor -\frac12$ we define
for $x>0$ the 1-periodic
functions $B_{x}: \mathbb{R} \to \mathbb{R}$ by
\begin{equation}\label{Bxdef}
B_{x}\left(t \right)= \frac{1}{x}\sum \limits_{k \leq x} \beta(kt)\,.
\end{equation}
The function $\beta(t)$ has jumps of height $-1$ exactly at integer numbers $t \in \mathbb{Z}$
but is continuous elsewhere.
Let $a/b$ with $a \in \mathbb{Z}$, $b \in \mathbb{N}$ be any reduced fraction with denominator $b \leq x$.

By $\begin{displaystyle} u^{\pm}(t)=\lim \limits_{\varepsilon \downarrow 0}
u(t \pm \varepsilon) \end{displaystyle}$
we denote the one-handed limits of a real- or complex valued function $u$ with respect to the real variable $t$.

Then the height of the jump of $B_x$ at $a/b$ is given by
\begin{equation}\label{Bxjump}
B_x^+(a/b)-B_x^-(a/b)=-\frac{1}{x}\left\lfloor\frac{x}{b}\right\rfloor\,.
\end{equation}
We introduce the function ${\tilde \eta} : \mathbb{R} \to \mathbb{R}$ given by ${\tilde \eta}(0)=-\frac12$ and
\begin{equation}\label{etalimit}
{\tilde \eta}(x)=
\frac{(x-\lfloor x \rfloor)(x-\lfloor x \rfloor-1)}{2x}
 \quad \text{for}\ x \in \mathbb{R} \setminus \{0\}\,.
\end{equation}
The function ${\tilde \eta}$ is continuous apart from the zero-point with derivative
\begin{equation}\label{etader}
{\tilde \eta}'(x)= \frac12-\frac{\lfloor x \rfloor (\lfloor x \rfloor+1)}{2x^2}\quad \text{for} ~
x \in \mathbb{R} \setminus \mathbb{Z}\,.
\end{equation}

In the following theorem we assume that $\frac{a}{b} < \frac{a^*}{b^*}$
are conse\-cu\-tive reduced fractions in the extended Farey sequence $\mathcal{F}^{ext}_b$
of order $b \leq n$ with $b,b^*,n \in \mathbb{N}$.

\begin{thm}\label{Bx_thm}
\begin{itemize}
\item[(a)]
For $0<x\leq n/b^*$ we have
\begin{equation*}
\begin{split}
&B_{n}\left(\frac{a}{b}+\frac{x}{bn}\right)
	=B_{n}\left(\frac{a}{b}\right)+\frac{1}{2b}+\frac{{\tilde \eta}(x)}{b}
	-\frac{x}{n}\, B_{x}^-\left(\frac{n/x-b^*}{b}\right)\\
&+\frac{x}{2bn} +\frac{1}{n}\,\sum \limits_{k \leq x}\beta\left(\frac{n-kb^*}{b}\right)\,.\\
\end{split}
\end{equation*}
\item[(b)] For $0<t \leq 1$ and $n \in \mathbb{N}$ we have
\begin{equation*}
B_{n}\left(t\right)
	={\tilde \eta}(tn)
	-\frac{\lfloor tn \rfloor}{n}\, B_{\lfloor tn \rfloor}^-\left(\frac{1}{t}-
	\left \lfloor \frac{1}{t} \right \rfloor\right)
+\frac{tn-\lfloor tn \rfloor}{2n}\,.
\end{equation*}
\end{itemize}
\end{thm}

\begin{proof} Since (b) follows from (a) in the special case $a=0$, $a^*=b^*=b=1$,
it is sufficient to prove (a). We define for $0 < x \leq n/b^*$:
	\begin{equation}\label{rntilde}
	\begin{split}
	R_n\left(\frac{a}{b},x\right) &=-b\left(
	B_{n}\left(\frac{a}{b}+\frac{x}{bn}\right)-
	B_{n}\left(\frac{a}{b}\right)\right)\\
	&+\frac12+{\tilde \eta}(x)-\frac{bx}{n}\, B_{x}^-\left(\frac{n/x-b^*}{b}\right)+\frac{x}{2n}\,.\\
		\end{split}
	\end{equation}

	We use \eqref{Bxdef}, \eqref{etader} and obtain, except of the discrete set of jump discontinuities
	of $R_n$, its derivative
	\begin{equation*}
	\begin{split}
	 \frac{d}{dx} R_n\left(\frac{a}{b},x\right) &=-b\cdot\frac{n+1}{2}\cdot \frac{1}{bn}
		+\frac{1}{2}-\frac{\lfloor x \rfloor (\lfloor x \rfloor+1)}{2x^2}\\
		&-\frac{b}{n}\,\frac{d}{dx}
		\left(
		 x \, B_{x}^-\left(\frac{n/x-b^*}{b}\right)
		\right)+\frac{1}{2n}\\
		& = -\frac{\lfloor x \rfloor (\lfloor x \rfloor+1)}{2x^2}-\frac{b}{n}\,\frac{d}{dx}
		\sum \limits_{k \leq \lfloor x \rfloor}
		\beta^-\left(k \, \frac{n/x-b^*}{b}\right)\\
		& = -\frac{\lfloor x \rfloor (\lfloor x \rfloor+1)}{2x^2}-\frac{b}{n}
		\sum \limits_{k \leq \lfloor x \rfloor}k
		\cdot \frac{n}{b} \cdot \left(-\frac{1}{x^2}\right)=0\,.\\
		\end{split}
	\end{equation*}
	Note that $B_n=B_n^+$ and $R_n=R^+_{n}$.
  We deduce from \cite[Theorem 2.2]{Ku3} for any $x$ in the interval
	$0 < x \leq n/b^*$  that
	$\begin{displaystyle} a/b+x/(bn)\end{displaystyle}$
	is a jump discontinuity of $B_{n}$ if and only if
	$\begin{displaystyle}(n/x-b^*)/b\end{displaystyle}$
	is a jump discontinuity of $B_{x}$. Let
	$$
	q =\frac{a'}{b'}=\frac{n/x_+(q)-b^*}{b}
	$$
	be any reduced fraction $a'/b' \in  \mathcal{F}^{ext}_{\lfloor x_+(q) \rfloor}$
	from \cite[Theorem 2.2(b)]{Ku3}. We use \eqref{Bxjump} and have
	\begin{equation}\label{phin_sprung}
    -b (B_{n}^{+}-B_{n}^{-})\left(\frac{a}{b}+\frac{x_+(q)}{nb}\right)
		=	\frac{b}{n} \left\lfloor \frac{n}{b^*b'+ba'} \right\rfloor
		=	\frac{b}{n} \left\lfloor \frac{x_+(q)}{b'} \right\rfloor\,.
	\end{equation}

	First we consider the case that $x_+(q)$ is a {\it non-integer} number.
	Using again \eqref{Bxjump}
	we obtain
	\begin{equation*}
	\begin{split}
   & \lim \limits_{x \,\downarrow \,x_+(q)}
     \left(x \,B_{x}^-\left(\frac{n/x-b^*}{b}\right)
		\right)-
		    \lim \limits_{x \,\uparrow \, x_+(q)}
     \left(x \,B_{x}^-\left(\frac{n/x-b^*}{b}\right)
		\right)\\
		&=x_+(q)\cdot\left(B_{x_+(q)}^- -B_{x_+(q)}^+ \right)\left(\frac{a'}{b'}\right)
		=\left\lfloor \frac{x_+(q)}{b'} \right\rfloor\,,
		\end{split}
  \end{equation*}
 taking into account that $(n/x-b^*)/b$ is monotonically decreasing with respect to $x$.
 In this case we obtain from \eqref{phin_sprung}
	\begin{equation*}
     \left(R_n^{+}-R_n^{-}\right)\left(\frac{a}{b},x_+(q)\right)
		=	\frac{b}{n} \left\lfloor \frac{x_+(q)}{b'} \right\rfloor
		-\frac{b}{n} \left\lfloor \frac{x_+(q)}{b'} \right\rfloor=0\,,
	\end{equation*}
 which implies that $R_n$ is free from jumps at non-integer arguments $x$.
 It remains to calculate the jumps of $R_n$ at any {\it integer argument}
$k$ with $0<k \leq n/b^*$.
 Here we also have to take care of the jump in $B_{x}=B_{k}$
 with respect to the index $x=k$, and conclude
\begin{equation}\label{sprung_fall2}
	\begin{split}
   & \lim \limits_{x \,\downarrow \,k}
     \left(x \,B_{x}^-\left(\frac{n/x-b^*}{b}\right)
		\right)-
		    \lim \limits_{x \,\uparrow \, k}
     \left(x \,B_{x}^-\left(\frac{n/x-b^*}{b}\right)
		\right)\\
		&=\lim \limits_{\varepsilon \,\downarrow \,0}
		\left[
		\sum \limits_{j \leq k}\beta^-\left( j \cdot\frac{\frac{n}{k+\varepsilon}-b^*}{b}\right)
		-
		\sum \limits_{j < k}\beta^-\left(j \cdot \frac{\frac{n}{k-\varepsilon}-b^*}{b}\right)
		\right]\\
		&=\sum \limits_{j \leq k}\beta^- \left( j \cdot \frac{n/k-b^*}{b}\right)
		-\sum \limits_{j \leq k}\beta^+ \left(j \cdot \frac{n/k-b^*}{b}\right)+
		\beta \left(\frac{n-kb^*}{b} \right)\\
		&=\beta \left(\frac{n-kb^*}{b} \right)-
		k\left(B_{k}^{+}-B_{k}^{-} \right)\left(\frac{n/k-b^*}{b} \right)\,.
		\end{split}
  \end{equation}

Using \eqref{rntilde}, \eqref{sprung_fall2} we obtain
\begin{equation*}
	\begin{split}
     \left(R_n^{+}-R_n^{-}\right)\left(\frac{a}{b},k \right)
		&=-\frac{b}{n}\,\beta \left(\frac{n-kb^*}{b}\right)\\
     &	-b\, ( B_{n}^{+}-
      B_{n}^{-}) \left( \frac{a}{b}+\frac{k}{bn}\right)\\
      &+\frac{b}{n}k\, ( B_{k}^{+}-
      B_{k}^{-}) \left(\frac{n/k-b^*}{b}\right)\,.\\
  \end{split}
	\end{equation*}
Due to \eqref{phin_sprung} and \cite[Theorem 2.2]{Ku3} the second and third terms
on the right-hand side cancel each other.

We conclude that $R_n$ is a step function with respect to $x$ for a given fraction
$a/b$ which has jumps of heigth
\begin{equation*}
R_n^+\left(\frac{a}{b},k \right) -R_n^-\left(\frac{a}{b},k \right)
=-\frac{b}{n}\,\beta \left(\frac{n-kb^*}{b}\right)
\end{equation*}
only at integer numbers $k$ with $0<k \leq n/b^*$.
To complete the proof of the theorem we only have to note that
$\begin{displaystyle}
\lim \limits_{\varepsilon \, \downarrow \, 0} R_n(a/b,\varepsilon)=0\,.
\end{displaystyle}$
\end{proof}

\begin{thm}\label{Bx_L2}
For $x \to \infty$ we have with the $L_2(0,1)$-Norm $\|\cdot\|_2$
$$
\|B_{x}\|_2=\mathcal{O}\left(\frac{1}{\sqrt{x}}\right)\,.
$$
On the other hand we have a constant $C>0$ with
$$
\|B_{x}\|_2 \geq\frac{C}{\sqrt{x}} \quad \text{for~}x \geq 1\,.
$$
\end{thm}
\begin{proof}
Since
$\begin{displaystyle}
\int \limits_0^1\beta(mx)\beta(nx)dx = \frac{(m,n)^2}{12 mn}
\end{displaystyle}$ for all $m,n \in \mathbb{N}$, we obtain
\begin{equation*}
\begin{split}
\|B_{x}\|_2^2 &=\frac{1}{12x^2}\sum \limits_{m,n \leq x}\frac{(m,n)^2}{mn}
=\frac{1}{12x^2}\sum \limits_{d \leq x}\sum \limits_{\underset{(m,n)=d}{m,n\leq x\,:}}\frac{d^2}{mn}\\
&=\frac{1}{12x^2}\sum \limits_{d \leq x}\sum \limits_{\underset{(j,k)=1}{j,k\leq x/d\,:}}\frac{1}{jk}
\leq \frac{1}{12x^2}\sum \limits_{d \leq x}\sum \limits_{j,k\leq x/d}\frac{1}{jk}\\
&\leq \frac{1}{12x^2}\sum \limits_{d \leq x}\left(\log(x/d)+2\right)^2
=\mathcal{O}\left(\frac{1}{x}\right) \quad\text{for}~x \to \infty\\
\end{split}
\end{equation*}
from Euler's summation formula, regarding that
$$
\int \limits_{1}^{x}\left(\log(x/t)+2 \right)^2\,dt=10(x-1)-6\log(x)-\log(x)^2=\mathcal{O}\left(x \right)\,,
\quad x \geq 1\,.
$$
To complete the proof we note that
\begin{equation*}
\|B_{x}\|_2^2 =\frac{1}{12x^2}\sum \limits_{d \leq x}
\sum \limits_{\underset{(j,k)=1}{j,k\leq x/d\,:}}\frac{1}{jk}\geq
\frac{1}{12x^2}\sum \limits_{d \leq x}1=\frac{\lfloor x \rfloor}{12x^2}\,.
\end{equation*}
\end{proof}

The next two theorems employ the elementary theory of continuous fractions.
We will use them to derive estimates for $B_n(t)$
with $t$ in certain subsets $\mathcal{M}_n , \tilde{\mathcal{M}}_n \subset (0,1)$ and
$\lim \limits_{n \to \infty} |\mathcal{M}_n|=
\lim \limits_{n \to \infty}|\tilde{\mathcal{M}}_n| =1$.

First we recall some basic facts and notations about continued fractions.
For $\lambda_0 \in \mathbb{R}$ and $\lambda_1,\lambda_2,\ldots,\lambda_m>0$
the finite continued fraction $\langle \lambda_0,\lambda_1,\ldots,\lambda_m \rangle$
is defined recursively by $\langle \lambda_0\rangle=\lambda_0$,
$\langle \lambda_0,\lambda_1\rangle=\lambda_0+1/\lambda_1$ and
\begin{equation*}
\langle \lambda_0,\lambda_1,\ldots,\lambda_m \rangle=
\langle \lambda_0,\ldots,\lambda_{m-2},\lambda_{m-1}+1/\lambda_m \rangle\,,
\quad m\geq 2\,.
\end{equation*}
Moreover, if $\lambda_j\geq 1$ is given for all $j \in \mathbb{N}$, then the limit
\begin{equation*}
\lim \limits_{m \to \infty}\langle \lambda_0,\lambda_1,\ldots,\lambda_m \rangle=
\langle \lambda_0,\lambda_1,\lambda_2\ldots\rangle
\end{equation*}
exists and defines an infinite continued fraction. Especially for integer numbers
$\lambda_0 \in \mathbb{Z}$ and  $\lambda_1,\lambda_2,\ldots \in \mathbb{N}$ we obtain a unique
representation
\begin{equation}\label{kette0}
t=\langle \lambda_0,\lambda_1,\lambda_2\ldots\rangle
\end{equation}
for all $t \in \mathbb{R} \setminus \mathbb{Q}$ in terms of an infinite continued fraction.
Here the coefficients $\lambda_j$ are obtained as follows:
For given  $t \in \mathbb{R} \setminus \mathbb{Q}$ we put
\begin{equation}\label{kette1}
\vartheta_0=t\,, \quad \vartheta_j=\frac{1}{\vartheta_{j-1}-\lfloor \vartheta_{j-1}\rfloor}>1\,, \quad j \in \mathbb{N}\,.
\end{equation}
We may also write $\vartheta_j =\vartheta_j(t)$ in order to
indicate that the quantities $\vartheta_j$ depend on the fixed number $t$.
We have
\begin{equation}\label{kette2}
\lambda_0= \lfloor t \rfloor\,,~\lambda_j = \lfloor \vartheta_j \rfloor \quad \text{and~}
t=\langle \lambda_0,\ldots,\lambda_{j-1},\vartheta_j\rangle \quad  \text{for ~ all ~} j \in \mathbb{N}\,.
\end{equation}

The following theorem is due to A. Ostrowski.
It allows a very efficient calculation of the values $B_n(t)$
in terms of the continued fraction expansion of $t$.

\begin{thm}\label{ostrowski}Ostrowski \cite[equation (2), p. 80]{Ost1922}\\
Put $S(n,t)=\sum \limits_{k \leq n}\beta(kt)=nB_n(t)$ for $n \in \mathbb{N}_0$ and $t \in \mathbb{R}$.
Given are the continued fraction expansion $t=\langle\lambda_0,\lambda_1,\lambda_2,\ldots\rangle$
of any fixed $t \in \mathbb{R} \setminus \mathbb{Q}$ and $n \in \mathbb{N}$. Then there is exactly
one index $j_* \in \mathbb{N}$ with $b_{j_*} \leq n < b_{j_*+1}$, where
$\begin{displaystyle}
a_k/b_k = \langle \lambda_0,\ldots,\lambda_{k-1} \rangle
\end{displaystyle}$
are reduced fractions $a_k/b_k$ and $k,b_k \in \mathbb{N}$. Put
$$
n'=n-b_{j_*}\left \lfloor \frac{n}{b_{j_*}}\right \rfloor\,.
$$
Then we have
\begin{equation}\label{Sost1}
 S(n,t)=S(n',t)+\frac{(-1)^{j_*}}{2}\left \lfloor \frac{n}{b_{j_*}}\right \rfloor
\left(1-\rho_{j_*} (n+n'+1) \right)
\end{equation}
with $\rho_{j_*}  = |b_{j_*} t-a_{j_*} |$ and
\begin{equation}\label{Sost2}
\left \lfloor \frac{n}{b_{j_*}}\right \rfloor \leq \lambda_{j_*}\,,\quad
0 < \left|1-\rho_{j_*}(n+n'+1) \right| <1\,.
\end{equation}
\end{thm}

Following Ostrowski's strategy we note two important conclusions.
We fix any number $t = \langle \lambda_0,\lambda_1,\lambda_2,\ldots \rangle \in \mathbb{R} \setminus \mathbb{Q}$
and apply Ostrowski's theorem \ref{ostrowski} successively, starting with the calculation
of $S(n,t)$ and $\begin{displaystyle} |S(n,t)|\leq |S(n',t)| +\lambda_{j_*}/2\end{displaystyle}$\,.
If $n'=0$, then $S(n',t)=0$, and we are done. Otherwise we replace $n$ by the reduced number $n'$
with $0<n'<b_{j_*}$ and apply Ostrowski's theorem again, and so on. For the final calculation
of $S(n,t)$ we need at most ${j_*}$ applications of the recursion formula and conclude from
\eqref{Sost1}, \eqref{Sost2} that
\begin{equation}\label{Snfinal}
 n|B_n(t)| = |S(n,t)| \leq \frac12 \, \sum \limits_{k=1}^{j_*} \lambda_k \,.
\end{equation}
From $b_0=0$, $b_1=1$ and $b_{j+1}=b_{j-1}+\lambda_j b_j$ for $j \in \mathbb{N}$
we obtain $b_{j+1}\geq 2b_{j-1}$, and hence for all $j \geq 3$ that
$\begin{displaystyle}
b_j \geq 2^{\frac{j-1}{2}}\,.
\end{displaystyle}$
Since $n \geq b_{j_*}$, we obtain without restrictions on $j_*$ for $n \geq 3$ that
$\begin{displaystyle}
n \geq 2^{\frac{j_*-1}{2}}
\end{displaystyle}$ and
\begin{equation}\label{jstarabschaetz}
\begin{split}
j_* \leq 1+\frac{2}{\log 2 }\log n  \leq \left(1 + \frac{2}{2/3}\right)\log n = 4 \log n
\,, \quad n \geq 3\,.
\end{split}
\end{equation}

We will see that \eqref{Snfinal} and \eqref{jstarabschaetz} have important conclusions,
an immediate consequence is Ostrowski's estimate \eqref{ostopt}
for irrational numbers $t$ with bounded partial quotients,
but first shed new light on these estimates by using Theorem \ref{Bx_thm}(b)
instead of Theorem \ref{ostrowski}. We put $J_*=(0,1)\setminus \mathbb{Q}$ and fix
any $t \in J_*$ and $n \in \mathbb{N}$. The sequence
\begin{equation}\label{tsequence}
t_0 =t\,, \quad t_{j}=\frac{1}{t_{j-1}} - \left \lfloor\frac{1}{t_{j-1}} \right \rfloor
\end{equation}
with $j \in \mathbb{N}$ is infinite, whereas the corresponding sequence of non-negative integer numbers
\begin{equation*
n_0 =n\,, \quad n_{j} = \lfloor t_{j-1} n_{j-1} \rfloor
\end{equation*}
is strictly decreasing and terminates if $n_{j}=0$.
Therefore $n_{j'}=0$ for some index $j'\in \mathbb{N}$.
We assume $1 \leq j<j'$ and distinguish the two cases $0<t_{j-1}<1/2$ and $1/2<t_{j-1}<1$.
In the first case we have $n_{j+1}<n_{j} < n_{j-1}/2$,
and in the second case again
$$n_{j+1}=
\lfloor t_{j}\lfloor t_{j-1} n_{j-1}\rfloor\rfloor
< t_{j} t_{j-1} n_{j-1} = (1-t_{j-1})n_{j-1} < n_{j-1}/2\,.
$$
If $j'$ is odd, then
$$
n=n_0 \geq 2^{\frac{j'-1}{2}}n_{j'-1} \geq 2^{\frac{j'-1}{2}}\,,
$$
otherwise
$$
n=n_0 \geq n_1 \geq 2^{\frac{j'-2}{2}}n_{j'-1} \geq 2^{\frac{j'-2}{2}}\,,
$$
and $n \geq 2^{\frac{j'-2}{2}}$ in both cases. Therefore
\begin{equation}\label{jstrich}
j' \leq 2+ \frac{2}{\log 2} \log n \leq \left(1+ \frac{2}{\log 2}\right) \log n
\leq 4 \log n\,,\quad n \geq 8\,.
\end{equation}
Estimate \eqref{jstrich} bears a strong resemblance with \eqref{jstarabschaetz}.
Now it follows from Theorem \ref{Bx_thm}(b) that
\begin{equation}\label{Bsequence}
B_n(t) = \sum \limits_{j=0}^{j'-1}(-1)^j
\left(
\frac{n_j}{n} {\tilde \eta}(t_j n_j)
+ \frac{t_j n_j-\lfloor t_j n_j \rfloor}{2n}
\right)\,.
\end{equation}
For the sequence in \eqref{tsequence} we have $\vartheta_{j+1}t_{j}=1$ for all $j \in \mathbb{N}_0$,
and we obtain from the definition \eqref{etalimit} of ${\tilde \eta}$ that
\begin{equation*}
\frac{n_j}{n} {\tilde \eta}(t_j n_j)+ \frac{t_j n_j-\lfloor t_j n_j \rfloor}{2n}
=-\frac{t_j n_j-\lfloor t_j n_j \rfloor}{2n}
\left\{
\vartheta_{j+1}\left(1-\left(t_j n_j-\lfloor t_j n_j \rfloor\right)\right)-1
\right\}\,.
\end{equation*}
We see from \eqref{Bsequence} with \eqref{kette1} and \eqref{kette2} that
\begin{equation}\label{Bestimate_modified}
\left | B_n(t)\right | \leq \sum \limits_{j=0}^{j'-1}
\frac{\max(1,\vartheta_{j+1}-1)}{2n}
\leq \frac{1}{2n}\sum \limits_{k=1}^{j'}\lambda_k\,.
\end{equation}
We finally conclude that estimates \eqref{Bestimate_modified} and
\eqref{Snfinal} are equivalent. Hence Theorem \ref{Bx_thm}(b) may be used as well
for an efficient calculation and estimation of $B_n(t)$ and $S(n,t)$.

\begin{thm}\label{kettenabschaetz}
Given are integer numbers $\alpha_1,\ldots,\alpha_m \in \mathbb{N}$.
We put $J_*=(0,1)\setminus \mathbb{Q}$.
Using the notations from \eqref{kette0}, \eqref{kette1}, \eqref{kette2}
we obtain for the measure $| \mathcal{M}|$ of the set
$\begin{displaystyle}
\mathcal{M}=\left\{t \in J_*\,:\,\vartheta_j < \alpha_j \quad\text{for~all~} j=1,\ldots,m\right\}
\end{displaystyle}$
the estimates
\begin{equation*}
\prod \limits_{j=1}^m \left(1-\frac{1}{\alpha_j} \right)^2
\leq | \mathcal{M}| \leq \prod \limits_{j=1}^m \left(1-\frac{1}{\alpha_j} \right)\,.
\end{equation*}
\end{thm}
\begin{proof}
The desired result is valid for $m=1$ with
$\begin{displaystyle}
\mathcal{M}=\left\{t \in J_*\,:\,1/t < \alpha_1\right\}
\end{displaystyle}$
and
$\begin{displaystyle}
|\mathcal{M}|=1-1/\alpha_1\,.
\end{displaystyle}$
Assume that the statement of the theorem is already true for a given $m \in \mathbb{N}$.
We prescribe $\alpha_{m+1} \in \mathbb{N}$ and will use induction to prove the statement for $m+1$.

For all $j \in \mathbb{N}$ and general given numbers $\lambda_0 \in \mathbb{R}$
and $\lambda_1,\ldots,\lambda_{j-1}>0$
we put for $1\leq k<j$:
\begin{equation}\label{kette3}
\begin{split}
 a_0=1\,,~  a_1=\lambda_0\,,~ & a_{k+1}=a_{k-1}+\lambda_k a_k\,,\\
 b_0=0\,,~  b_1=1\,,~ & b_{k+1}=b_{k-1}+\lambda_k b_k\,.\\
\end{split}
\end{equation}
We have
\begin{equation}\label{kette4}
\begin{split}
& \langle \lambda_0,\lambda_1, \ldots \lambda_{j-1},x \rangle -
\langle \lambda_0,\lambda_1, \ldots \lambda_{j-1},x' \rangle\\
&=
\frac{(-1)^j (x-x')}{(b_jx+b_{j-1})(b_jx'+b_{j-1})}\quad \text{for ~}x,x'>0\,.
\end{split}
\end{equation}
Especially for $\lambda_0=0$ and {\it integer numbers} $\lambda_1,\ldots,\lambda_j \in \mathbb{N}$
we define the set $J(\lambda_1,\ldots,\lambda_j)$
consisting of all $t \in J_*$ between the two rational numbers
$\langle 0,\lambda_1, \ldots ,\lambda_{j-1},\lambda_{j}\rangle$ and
$\langle 0,\lambda_1, \ldots ,\lambda_{j-1},\lambda_{j}+1 \rangle$.

It follows from \eqref{kette3},\eqref{kette4} and all $j \in \mathbb{N}$  that
\begin{equation}\label{kette5}
|J(\lambda_1,\ldots,\lambda_j)|=
\frac{1}{(b_j(\lambda_j+1)+b_{j-1})(b_j \lambda_j +b_{j-1})}\,.
\end{equation}
The sets $J(k)=(1/(k+1),1/k)\setminus \mathbb{Q}$ with $k \in \mathbb{N}$
form a partition of $J_*=(0,1)\setminus \mathbb{Q}$.
More general, it follows from \eqref{kette1}, \eqref{kette2} that
for fixed numbers $\lambda_1,\ldots,\lambda_j \in \mathbb{N}$
the pairwise disjoint sets $J(\lambda_1,\ldots,\lambda_j,k)$ with $k \in \mathbb{N}$
form a partition of the set $J(\lambda_1,\ldots,\lambda_j)$.
We conclude by induction with respect to $j$ that the pairwise disjoint sets
$J(\lambda_1,\ldots,\lambda_j)$ with $(\lambda_1,\ldots,\lambda_j) \in \mathbb{N}^j$
also form a partition of $J_*$.

Now we put $j=m$ and distinguish two cases, $m$ odd and $m$ even, respectively.
In both cases, $m$ odd or $m$ even, the union
\begin{equation*}
\bigcup \limits_{k=1}^{\alpha_{m+1}-1}J(\lambda_1,\ldots,\lambda_m,k)
\end{equation*}
is the set of all numbers $t \in J_*$ with $\lfloor \vartheta_j \rfloor = \lambda_j$
for $j=1,\ldots,m$ such that $\vartheta_{m+1} < \alpha_{m+1}$.
We define the set
\begin{equation*}
\mathcal{M}'=\left\{t \in J_*\,:\,\vartheta_j < \alpha_j \quad\text{for~all~} j=1,\ldots,m+1\right\}
\end{equation*}
and conclude
\begin{equation}\label{mstrich}
|\mathcal{M}'| = \sum \limits_{\lambda_1=1}^{\alpha_{1}-1}
\sum \limits_{\lambda_2=1}^{\alpha_{2}-1}
\cdots
\sum \limits_{\lambda_m=1}^{\alpha_{m}-1}
\sum \limits_{k=1}^{\alpha_{m+1}-1}
|J(\lambda_1,\ldots,\lambda_m,k)|\,.
\end{equation}
It also follows from our induction hypothesis that
\begin{equation}\label{hypo}
\begin{split}
&\prod \limits_{j=1}^m \left(1-\frac{1}{\alpha_j} \right)^2\\
\leq&\sum \limits_{\lambda_1=1}^{\alpha_{1}-1}
\sum \limits_{\lambda_2=1}^{\alpha_{2}-1}
\cdots
\sum \limits_{\lambda_m=1}^{\alpha_{m}-1}
|J(\lambda_1,\ldots,\lambda_m)| \leq \prod \limits_{j=1}^m \left(1-\frac{1}{\alpha_j} \right)\,.\\
\end{split}
\end{equation}
We evaluate the inner sum in \eqref{mstrich}, and obtain for odd values of $m$
the telescopic sum
\begin{equation*}
\begin{split}
&\sum \limits_{k=1}^{\alpha_{m+1}-1}
|J(\lambda_1,\ldots,\lambda_m,k)|\\
&=
\sum \limits_{k=1}^{\alpha_{m+1}-1}
\left(
\langle 0,\lambda_1,\ldots,\lambda_m,k+1\rangle -\langle 0,\lambda_1,\ldots,\lambda_m,k\rangle
\right)\\
&=\langle 0,\lambda_1,\ldots,\lambda_m,\alpha_{m+1}\rangle -\langle 0,\lambda_1,\ldots,\lambda_m,1\rangle\\
&=\langle 0,\lambda_1,\ldots,\lambda_{m-1},\lambda_m+1/\alpha_{m+1}\rangle -
\langle 0,\lambda_1,\ldots,\lambda_{m-1},\lambda_m+1\rangle\,.\\
\end{split}
\end{equation*}
Apart from a minus sign on the right hand side we get the same result for even values of $m$,
and hence from \eqref{kette4} with $j=m$ in both cases
\begin{equation}\label{tele}
\begin{split}
&\sum \limits_{k=1}^{\alpha_{m+1}-1}
|J(\lambda_1,\ldots,\lambda_m,k)|\\
&=
\frac{1-\frac{1}{\alpha_{m+1}}}{(b_m(\lambda_m+1)+b_{m-1})
(b_m(\lambda_m+\frac{1}{\alpha_{m+1}})+b_{m-1})}\,.\\
\end{split}
\end{equation}
Using $\lambda_m\geq 1$ we have
\begin{equation*}
\frac{\left(1-\frac{1}{\alpha_{m+1}}\right)^2}{b_m \lambda_m+b_{m-1}}
\leq
\frac{1-\frac{1}{\alpha_{m+1}}}{
b_m(\lambda_m+\frac{1}{\alpha_{m+1}})+b_{m-1}}
\leq
\frac{1-\frac{1}{\alpha_{m+1}}}{b_m \lambda_m+b_{m-1}}\,,
\end{equation*}
and obtain from \eqref{tele} and \eqref{kette5} with $j=m$ that
\begin{equation}\label{innerschaetz}
\begin{split}
&|J(\lambda_1,\ldots,\lambda_m)| \left(1-\frac{1}{\alpha_{m+1}}\right)^2\\
&\leq \sum \limits_{k=1}^{\alpha_{m+1}-1}|J(\lambda_1,\ldots,\lambda_m,k)|
\leq |J(\lambda_1,\ldots,\lambda_m)| \left(1-\frac{1}{\alpha_{m+1}}\right)\,.\\
\end{split}
\end{equation}
The theorem follows from \eqref{mstrich}, \eqref{hypo} and \eqref{innerschaetz}.
\end{proof}
\begin{rem}
Since $\alpha_j \in \mathbb{N}$ for $j \leq m$, the conditions $\vartheta_j < \alpha_j$
in the definition of the set $\mathcal{M}$ may likewise be replaced
by the equivalent conditions $\lambda_j \leq \alpha_j-1$,
where $\lambda_j$ are the coefficients in the continued fraction expansion of $t$,
see \eqref{kette1}, \eqref{kette2}\,.
\end{rem}

\section{Dirichlet series related to Farey sequences} \label{dirichlet_section}
We define the sawtooth function $\beta_0:\mathbb{R} \to \mathbb{R}$ by
\begin{equation*}
\beta_{0}(x)=\begin{cases}
x-\lfloor x \rfloor-\frac12
 &\ \text{for}\ x \in \mathbb{R} \setminus{\mathbb{Z}}\, ,\\
0 &\ \text{for}\ x \in \mathbb{Z}\,.\\
\end{cases}
\end{equation*}

With $x>0$ the 1-periodic
function $B_{x,0}: \mathbb{R} \to \mathbb{R}$ is
the arithmetic mean of $B_{x}^-$, $B_{x}^+=B_x$, see \eqref{Bxdef}, hence
\begin{equation}\label{Bx0def}
B_{x,0}\left(t \right)= \frac{1}{x}\sum \limits_{k \leq x} \beta_{0}(kt)
=\frac12\left(B_{x}^-(t) + B_{x}^+(t) \right)\,.
\end{equation}

\begin{lem}\label{B0_rational}
For (relatively prime) numbers $a \in \mathbb{Z}$ and $b \in \mathbb{N}$ we have
$$
|B_{x,0}(a/b)| \leq \frac{b}{x}
$$
for all $x>0$.
\end{lem}
\begin{proof}
Without loss of generality we may assume that $a \in \mathbb{Z}$ and $b \in \mathbb{N}$
are relatively prime and that $b\geq2$, since $B_{x,0}(0)=0$. For $m \in \mathbb{N}$ we define
$$ t_{a/b}(m)=1+2b\beta\left(\frac{am}{b}\right)=
2am-2b\left\lfloor\frac{am}{b}\right\rfloor-b+1\,,$$
and obtain from \cite[(7) in the proof of Lemma 2.2]{Ku2} for all $x>0$ that
\begin{equation}\label{tabschaetz}
\left|\sum \limits_{m \leq x} t_{a/b}(m)\right| \leq b(b+1)\,.
\end{equation}
Using \eqref{Bxjump} we obtain
\begin{equation*}
\begin{split}
B_{x,0}(a/b)&=
B_x(a/b)+\frac{1}{2x}\left\lfloor\frac{x}{b}\right\rfloor\\
&=\frac{1}{2bx}\sum \limits_{m \leq x}\left(t_{a/b}(m)-1 \right)+\frac{1}{2x}\left\lfloor\frac{x}{b}\right\rfloor\\
&=\frac{1}{2bx}\sum \limits_{m \leq x} t_{a/b}(m)+
\frac{1}{2x}\left(\left\lfloor\frac{\lfloor x \rfloor}{b}\right\rfloor-\frac{\lfloor x \rfloor}{b}\right)\,,\\
\end{split}
\end{equation*}
hence we see from \eqref{tabschaetz} with $b \geq 2$ that
\begin{equation*}
\left| B_{x,0}(a/b) \right| \leq \frac{b+1}{2x}+\frac{1}{2x} \leq \frac{b}{x}\,.
\end{equation*}
\end{proof}

Using Theorem \ref{Bx_thm}(a), Lemma \ref{B0_rational}, \eqref{Bx0def}, \eqref{Bxjump}
and for $x \in \mathbb{R}$ the symmetry relationship
\begin{equation*}
B_n\left(\frac{a}{b}-\frac{x}{bn}\right)=-B_{n}^-\left(\frac{b-a}{b}+\frac{x}{bn}\right)\,,
\end{equation*}
we obtain the following result, which has the counterparts \cite[Theorem 3.2]{Ku2}
and \cite[Theorem 2.2]{Ku4} in the theory of Farey-fractions:

\begin{thm}\label{rescaled_limit}
Assume that $a/b \in \mathcal{F}^{ext}_n$ and put
$$
\tilde{\eta}_{a,b}(n,x)=b\,B_n\left(\frac{a}{b}+\frac{x}{bn}\right) \,,
\qquad x \in \mathbb{R}\,.
$$
Then for $n \to \infty$ the sequence of functions $\tilde{\eta}_{a,b}(n,\cdot)$ converges uniformly
on each interval $[-x_*,x_*]$, $x_*>0$ fixed, to the limit function $\tilde{\eta}$
in \eqref{etalimit}.
\end{thm}

For the following two results we apply Theorem \ref{ostrowski}
and recall \eqref{Snfinal}, \eqref{jstarabschaetz}.
Due to Theorem \ref{rescaled_limit} the functions $B_n$ cannot converge
uniformly to zero on any given interval. Instead we have the following

\begin{thm}\label{B0_mass}
Let $\Theta : [1,\infty) \to [1,\infty)$ be monotonically increasing with
$\begin{displaystyle} \lim \limits_{n \to \infty} \Theta(n)=\infty\,.\end{displaystyle}$

We fix $n \in \mathbb{N}$, $\varepsilon >0$, put $m=\lfloor 4 \log n\rfloor$,
use \eqref{kette1}, recall $J_*=(0,1) \setminus \mathbb{Q}$ and define
$$\mathcal{M}_n=
\left\{t \in J_*\,:\,\vartheta_j(t) <1+\lfloor\Theta(n)\log n\rfloor \quad
\mbox{for~all ~ } j=1,\ldots,m \right\}\,.
$$
Then $\begin{displaystyle}  \lim \limits_{n \to \infty}|\mathcal{M}_n|=1\end{displaystyle}$
and
\begin{equation}
\label{betteresti}
\left| B_{n,0}(t) \right|=\left| B_{n}(t) \right| \leq 2 \, \frac{\log^2 n}{n}\,\Theta(n)
\end{equation}
for all $n \geq 3$ and all $t \in \mathcal{M}_n$.
\end{thm}
\begin{proof}
We apply Ostrowski's theorem on any number $t \in \mathcal{M}_n$ with continued fraction expansion
$t = \langle 0,\lambda_1,\lambda_2,\ldots \rangle$ and
obtain $j_* \leq m$ from \eqref{jstarabschaetz}, since $j_*$ is an integer number.
From $j_* \leq m$ and $t \in \mathcal{M}_n$
we conclude that $\lambda_k \leq \Theta(n) \log n$ for $k=1,\ldots,j_*$,
and the desired inequality follows with \eqref{Snfinal}\,.
The first statement follows from Theorem \ref{kettenabschaetz} via
\begin{equation*}
\begin{split}
|\mathcal{M}_n| &\geq \left(1-\frac{1}{ 1+\lfloor\Theta(n)\log n\rfloor} \right)^{2m}\\
&\geq \left(1-\frac{1}{\Theta(n)\log n} \right)^{2m}
\geq \left(1-\frac{1}{\Theta(n)\log n} \right)^{8\log n}\,,\\
\end{split}
\end{equation*}
since the right-hand side tends to $1$ for $n \to \infty$.
\end{proof}

\begin{rem} The sets $\mathcal{M}_n$ in the previous theorem are chosen in such a way that
the large values $B_n(t)$ from the peaks of the rescaled limit function
around  the rational numbers with small denominators
predicted by Theorem \ref{rescaled_limit} can only occur in the small complements
$J_* \setminus \mathcal{M}_n$ of these sets.
However, the quality of the estimates of the values $B_n(t)$
on the sets $\mathcal{M}_n$ depends on the
different choices of the growing function $\Theta$. For example,
$\Theta(n)=1+\log\left(1+\log n\right)$ gives a much smaller bound
than $\Theta(n)=16 \sqrt{n}/(4+\log n)^2$, whereas the latter choice leads to
a much smaller value of $|J_* \setminus \mathcal{M}_n|=1-|\mathcal{M}_n|$\,.
\end{rem}

\begin{thm}\label{B0_mass_almost_everywhere}
Let $\Theta : [1,\infty) \to [1,\infty)$ be monotonically increasing with
$\begin{displaystyle} \lim \limits_{n \to \infty} \Theta(n)=\infty\,.\end{displaystyle}$
We fix $n \in \mathbb{N}$, $\varepsilon >0$, use \eqref{kette1}, recall $J_*=(0,1) \setminus \mathbb{Q}$ and put
$$\tilde{\mathcal{M}_n}=
\left\{t \in J_*\,:\,\vartheta_j(t) < 1+\lfloor\Theta(n)j^{1+\varepsilon}\rfloor \quad\text{for~all~} j \in \mathbb{N} \right\}\,.
$$
Then $|\tilde{\mathcal{M}}|=1$ for
$\tilde{\mathcal{M}}= \bigcup \limits_{n=1}^{\infty}\tilde{\mathcal{M}_n}$,
and for all $t \in \tilde{\mathcal{M}}$ there exists an index $n_0=n_0(t,\varepsilon)$ with
$$\left| B_{n,0}(t) \right|=\left| B_{n}(t) \right|
\leq \frac{(4\log n)^{2+\varepsilon}}{2n}\,\Theta(n)
\quad \mbox{for ~ all ~} n \geq n_0\,.
$$
The complement $J_* \setminus \tilde{\mathcal{M}}$ is an uncountable null set
which is dense in the unit interval $(0,1)$.
\end{thm}
\begin{proof}
The function $\Theta$ is monotonically increasing, hence
$
\tilde{\mathcal{M}_1} \subseteq \tilde{\mathcal{M}_2}\subseteq \tilde{\mathcal{M}_3}\ldots\,,
$
and we have
\begin{equation}\label{kettenmass1}
|\tilde{\mathcal{M}}|=
\lim \limits_{n \to \infty} |\tilde{\mathcal{M}_n}|\,.
\end{equation}
For all $n,k \in \mathbb{N}$ we define
$$\tilde{\mathcal{M}_{n,k}}=
\left\{t \in J_*\,:\,\vartheta_j(t) < 1+\lfloor\Theta(n)j^{1+\varepsilon}\rfloor
\quad\text{for~all~} j =1,\ldots,k \right\}\,.
$$
Then
$\tilde{\mathcal{M}_n}= \bigcap \limits_{k=1}^{\infty}\tilde{\mathcal{M}_{n,k}}$
and
\begin{equation}\label{kettenmass2}
|\tilde{\mathcal{M}_n}|=
\lim \limits_{k \to \infty} |\tilde{\mathcal{M}_{n,k}}|
\end{equation}
from
$
\tilde{\mathcal{M}_{n,1}} \supseteq \tilde{\mathcal{M}_{n,2}}\supseteq \tilde{\mathcal{M}_{n,3}}\ldots\,.
$
It follows from Theorem \ref{kettenabschaetz} for all $n,k \in \mathbb{N}$ that
\begin{equation*}
|\tilde{\mathcal{M}_{n,k}}|
\geq \prod \limits_{j=1}^{k}\left(1-\frac{1}{\Theta(n)j^{1+\varepsilon}} \right)^2
\geq \prod \limits_{j=1}^{\infty}\left(1-\frac{1}{\Theta(n)j^{1+\varepsilon}} \right)^2\,.
\end{equation*}
The product on the right-hand side is independent of $k$ and converges to $1$
for $n \to \infty$, hence $|\tilde{\mathcal{M}}|=1$ from
\eqref{kettenmass1}, \eqref{kettenmass2}\,.
Each rational number in the interval $(0,1)$ is arbitrary close to a member of
the complement $J_* \setminus \tilde{\mathcal{M}}$, and the complement contains
all $t=\langle 0,\lambda_1,\lambda_2,\lambda_3,\ldots \rangle$
for which $(\lambda_j)_{j \in \mathbb{N}}$ increases faster then any polynomial.
We conclude that $J_* \setminus \tilde{\mathcal{M}}$
is an uncountable null set which is dense in the unit interval $(0,1)$.
Now we choose $t \in \tilde{\mathcal{M}}$ and obtain $n_0 \in \mathbb{N}$
with $t \in \tilde{\mathcal{M}}_{n_0}$. Then $t \in \tilde{\mathcal{M}}_{n}$
for all $n \geq n_0$, and we may assume that $n_0 \geq 3$.
Note that $n_0$ may depend on $t$ as well as on $\varepsilon$.
We have $t = \langle 0,\lambda_1,\lambda_2,\lambda_3,\ldots \rangle$ and
$$
 \lambda_j \leq \Theta(n)j^{1+\varepsilon}
$$
for all $n \geq n_0$ and all $j \in \mathbb{N}$.
We finally obtain from \eqref{Snfinal}, \eqref{jstarabschaetz} that
$$
n|B_n(t)| = |S(n,t)| \leq \frac12 \, \sum \limits_{k=1}^{j_*} \lambda_k
\leq \frac{j_*}{2}\Theta(n)j_*^{1+\varepsilon}
\leq \frac12 \Theta(n) (4\log n)^{2+\varepsilon}\,, \quad n \geq n_0\,.
$$
\end{proof}

\begin{rem}
We replace $\varepsilon$ by $\varepsilon/2$,
choose $\Theta(n)=1+\log\left(1+\log n\right)$ in the previous theorem
and obtain the following result of Lang, see \cite{Lang1966}
and \cite[III,\S 1]{Lang1995} for more details:
For $\varepsilon > 0$ and almost all $t \in \mathbb{R}$ we have
\begin{equation*}
|S(n,t)| \leq\left(\log n \right)^{2+\varepsilon} \quad
\mbox{~for ~} n \geq n_0(t,\varepsilon)
\end{equation*}
with a constant $n_0(t,\varepsilon) \in \mathbb{N}$.
Here the sum $S(n,t)$ is given by \eqref{Sxdef}\,.
This doesn't contradict Theorem \ref{Bx_L2}, because the pointwise estimates of $S(n,t)$ and
$B_n(t)$ in Theorem \ref{B0_mass_almost_everywhere} are only valid for sufficiently large values of $n \geq n_0(t,\varepsilon)$, depending on the choice of $t$ and $\varepsilon$.

We conclude from Theorem \ref{B0_mass} that the major contribution of $\|B_n\|_2$
comes from the small complement of $\mathcal{M}_n$. Indeed, the crucial point
in Theorem \ref{B0_mass} is that it holds for \textit{all} $n \geq 3$, but
not so much the fact that the upper bound in estimate \eqref{betteresti} is slightly better
than that in Theorem \ref{B0_mass_almost_everywhere}.
\end{rem}

For $k \in \mathbb{N}$ and $x>0$ the 1-periodic
functions $q_{k,0},\Phi_{x,0} : \mathbb{R} \to \mathbb{R}$
corresponding to \eqref{familien} are
defined as follows:
\begin{equation*}
q_{k,0}(t) =-\sum \limits_{d | k} \mu(d)\,
\beta_{0} \left(\frac{kt}{d}\right) ~\,,
\end{equation*}
\begin{equation*}
\Phi_{x,0}\left(t \right)= \frac{1}{x}\,\sum \limits_{k \leq x} q_{k,0}(t)
=- \frac{1}{x}\sum \limits_{j \leq x}
\sum \limits_{k \leq x/j}\mu(k)\,\beta_{0}\left(jt \right)\,.
\end{equation*}

In the half-plane $H = \{ s \in \mathbb{C}\,:\,\Re(s)>1\}$
the parameter-dependent Dirichlet series $F_{\beta}, F_{q} : \mathbb{R} \times H \to \mathbb{C}$ are given by
\begin{equation*}
F_{\beta}(t,s) =\sum \limits_{k=1}^{\infty} \frac{\beta_{0}(kt)}{k^s} \,,\quad
F_{q}(t,s) =\sum \limits_{k=1}^{\infty} \frac{q_{k,0}(t)}{k^s} \,.
\end{equation*}
Now Theorem \ref{B0_mass_almost_everywhere} and \eqref{Hecke_series} immediately gives
\begin{thm}\label{F_thm}
For $t \in \mathbb{R}$ and $\Re(s)>1$ we have with absolutely convergent series and integrals
\begin{itemize}
\item[(a)]
\begin{equation*}
\begin{split}
\frac{1}{s}\,F_{\beta}(t,s)=\int \limits_{1}^{\infty} B_{x,0}(t)\frac{dx}{x^s}\,,\quad
\frac{1}{s}\,F_{q}(t,s)=\int \limits_{1}^{\infty} \Phi_{x,0}(t)\frac{dx}{x^s}\,.\\
\end{split}
\end{equation*}
\item[(b)]
\begin{equation*}
F_{q}(t,s)=-\frac{1}{\zeta(s)}\,F_{\beta}(t,s)\,.
\end{equation*}
\end{itemize}
For almost all $t$ the function $F_{\beta}(t,\cdot)$
has an analytic continuation to the half-plane $\Re(s)>0$\,.
\end{thm}

\section{Appendix: Plots of the limit functions $h$ and $\tilde{\eta}$}\label{appendix}
\begin{center}
\begin{figure}[H]
\includegraphics[width=0.7\textwidth]{h25.eps}
\caption{Plot of $h(x)$ for $-25 \leq x \leq 25$\,. \label{h25eps}}
\end{figure}
\end{center}
\begin{center}
\begin{figure}[H]
\includegraphics[width=0.7\textwidth]{h50.eps}
\caption{Plot of $h(x)$ for $25 \leq x \leq 50$\,.  \label{h50eps}}
\end{figure}
\end{center}
\begin{center}
\begin{figure}[H]
\includegraphics[width=0.7\textwidth]{h500.eps}
\caption{Plot of $h(x)$ for $50 \leq x \leq 500$\,. \label{h500eps}}
\end{figure}
\end{center}
\begin{center}
\begin{figure}[H]
\includegraphics[width=0.7\textwidth]{eta8.eps}
\caption{Plot of $\tilde{\eta}(x)$ for $-8 \leq x \leq 8$\,. \label{eta8eps}}
\end{figure}
\end{center}

redpajama_set_name: RedPajamaArXiv


Hex #43a036 or rgb(67,160,54) in RGB space is the tint of green. It has hue angle of 112.64 degrees, value = 63 and saturation = 66. #43a036 can be obtained by mixing 4 colors: 5 drops of yellow, 4 drops of cyan, 5 drops of green, 5 drops of black. Nearest safe hex - #339933. Below you can see the block with #43a036 color and its structure: in procentage ratio and in drops of pigments. Click "TRY" button to move #43a036 to the mixer and play with it.
The funnel shows the process of mixing multiple colors step by step. Here you can see the mix of 26.3% of yellow, 21.1% of cyan, 26.3% of green, 26.3% of black colors.
We find few colors similar to #43a036. For more information about color - click on the circle.
Below you can see the model of the mix of #43a036 with pure water. The more water added to the mixture, the mixture will be less saturated.


Found that special home and now you're ready for an inspection? I would be honored to be your inspector and provide you with the information you need to make an informed investment. I will perform a thorough inspection, provide a written report and take the time necessary to discuss the report and answer any questions you may have. Chances are, most of your inspection will be related to maintenance recommendations and minor imperfections, which are good to know about.
Anything in these categories should be addressed as soon as possible. Your Real Estate Professional can assist you with addressing these items. It is important to maintain perspective, since some minor defects and imperfections are normal.
Home ownership is both a joyful experience and an important responsibility. The development of an annual maintenance plan will help identify and manage home maintenance tasks before major issues develop. Maintenance inspections are instrumental in keeping your family safe and your home in top condition for years to come. I welcome the opportunity to discuss with you how I can assist with developing this plan.
I offer you the option of a follow-up inspection to ensure the summary of deficiencies you are expecting to be corrected have been completed and are acceptable. This inspection does not replace the final walk-thru prior to closing. I recommend you make every effort to make a final walk-thru before closing, which I'm certain your Real Estate Professional will suggest as well.
Are you selling your home? Let me inspect it before you even list it. A Move-In Certified™ home has been pre-inspected, which means that the seller can confirm that there are no major systems in need of immediate repair or replacement, and no known safety hazards. A Move-In Certified™ Seller Inspection informs you of any defects or problems with your home so that you can address them before prospective buyers discover them. You can then take the time you need to obtain reasonable repair estimates. Show prospective buyers that you are dealing in good faith. Avoid 11th-hour negotiations and delays, and better justify your asking price by having your home pre-inspected now.
First-time home ownership can be exciting, but also very intimidating. The many facets of finding and buying a home can seem overwhelming and can sometimes feel like an emotional rollercoaster. The home inspection process is an important and enlightening step in understanding more about home ownership. I will spend the time necessary reviewing the inspection report with you to ensure you understand the results of the inspection and will be happy to share some general guidelines on maintaining your home.
Following my evaluation, I will compile my findings in an easy to read report, which will include descriptions and digital photos of any issues I've discovered. Your report will be emailed to you within 24 hours. My work for you is not finished until you understand everything in the report.


During my class on Data Structures and Algorithms, one of my assignments was to use a stack to make a function parser. The algorithm was given, and it handled variables, addition, subtraction, multiplication, and division. I was inspired to expand on this parser and use it to implement a Function class, where a math function was represented by a Java object. This allowed for functions to find the derivative and Taylor Series of a function, among other things.
The ideas for the project expanded, and graphing the rectangular functions, as well as polar and parametric functions were added. A side project was also added where you can view and zoom in on the Mandelbrot set. This is in a separate main method.


text:
Micro Machines World Series - PlayStation 4
Model:D1382
Experience an updated classic with Micro Machines World Series for the PlayStation 4. Gorgeous HD visuals and a new battle arena take the familiar locations of the series to a new level. Micro Machines World Series lets you race against friends or join online competitions with up to 12 players.
Battle Mode introduces all-new Battle Arenas which allow you to wreak mass destruction on a miniature scale against your friends or AI
Lets you work together using your vehicles' unique skills in modes such as "Capture the Flag" and "King of the Hill"
Classic Racing mode retains the manic social gameplay of the beloved series
The series are updated with stunning HD visuals showcased in the classic locations, such as garden, kitchen, workshop and many other places to race through with a vast array of miniature vehicles
For 1 up to 4 players on the same screen or multiplayer support for up to 12 over LAN or Internet
Micro Machines World Series
E10+ (Everyone 10+)
ESRB Descriptors
Certifications & Listings
User rating, 3.7 out of 5 stars with 48 reviews.
5 star rating. 20 reviews520
Page 1 Showing 1-8 of 48 reviews
ConsumerMan
Was not expecting much...
For the price I was thinking this game would fail to impress but I was a huge fan of the original Micro Machines for NES back in 1991. I know you old school gamers remember the gold cartridge from Codemasters! That game was a classic! So, I had to check this game out and sure enough, it even has little facts about the original game and some other fun little reminders from the OG sprinkled throughout the game. This game is not only nostalgic, it's cool. They took advantage of modern capabilities within the game without losing the touch of the original game. Sure it is simple, but with added battle abilities and online play, this game delivers a lot of bang for the buck in my opinion. A must own game for any fan of the original and a game the new generation of kids can create nostalgic memories of the future with as well. Up to 4 players on a single console as well as PS4 Pro enhanced fun for under $30 can't be beat!
It's not an easy game at all. There are a number of levels and different things for you to do. The view is kind of like controlling an R/C car. Unfortunately you cannot change the view. There's no way to make it easy for a little one. I would recommend it as a challenge though. Just because it says Micro Machines, doesn't mean it's easy.
Cjp920
Surprisingly fun
Actually not a bad game. The battles and races are fun and actually a little challenging. Mostly an online game which is fine unless your looking for a story mode.
Posted 3 years ago . Owned for 1 week when reviewed.
My kids like this game. It's about what you expect. As long as this is low price it's not a bad game.
Baddad
Good game
Good game if you like 'mini' cars. The update HD visuals look great plus you can race against your friends. I give it a 4 out of 5
Dopey
Good...except...
Posted 3 years ago . Owned for less than 1 week when reviewed.
Too much of the game requires PS4 account. What happened to just playing a game.
Nice multi-player game
I was looking for a multi-player game and there are not so many options. This game is a nice try to play with family.
Fun and entertaining with lot's of action and car racing and shooting
Show More Customer Reviews

redpajama_set_name: RedPajamaCommonCrawl


Peter WEAVER b. 1822~656 at New York656. d. bef 1858657.
Mercy BUTLER b. 183134 at Steuben Co., New York38. d. 190634. bd. Bradford Cem., Bradford, NY34.
Married656 1847~656. 1865 Census NY, Bradford lists Mercy Barkley (wife of Franklin Barkley at that time) as married 2 times. Mary H. Weaver is listed as child in the household. Assume Mercy married 1st WEAVER. Possibility that Franklin Barkley married 1st WEAVER!
Mercy's maiden name and Peter Weaver's first name is taken from Nettie W Weaver's death listing1061. Mercy married first about 1847 Peter Weaver. That date is assumed from the date of their child Emily, age 2, as she appears in the 1850 census for Bradford, NY, listed with Peter and Mercy Weaver. Mercy and Peter had children: Emily, born ~1848; Mary Helen, born April 1851; and Henrietta (Nettie), born ~1854. Peter apparently died before 1858, the assumed marriage date for Mercy Butler Weaver and Franklin Barkley. Mary Helen and Nettie are listed with the surname "Barkley" in 1860 and 1865. But, in the 1870 and 1880 censuses, they're listed as "Weaver." In the 1865 census, Mercy is listed as being married 2 times and the mother of 7 children. Those children were 3 by her first husband Peter Weaver, as mentioned above, and 4 by Franklin Barkley, listed herein.
Child 3. HenriettaAlthough Henrietta ("Nettie") is listed in the 1860 census as "Henrietta Barkley," she is listed in the 1875 census for Bradford in Franklin and Mercy's household as "Nettie Weaver," and, in 1880, is listed as the step-daughter of Franklin Barkley.
1. Emily WEAVER656 b. 1848~656 at New York656.
2. Mary Helen WEAVER38 b. 1851 Apr1083 at Steuben Co., NY38.
3. Henrietta Nettie WEAVER56, 657 b. 1854~657 at Steuben Co., NY38.


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BOYBAND The Wanted are edging closer to cracking the notoriously difficult American music market.
The band, including Tonge Moor's Tom Parker and former Bolton School pupil Max George, played a gig at Central Park as part of Good Morning America's series of summer concerts last week.
Dressed in black and white, the lads performed their hits, including 'Glad You Came', 'All Time Low', 'I Found You' and 'We Own The Night' to a screaming audience in New York's historic landmark.
The band have been trying their luck over the pond since 2012 when they were given a walk-on part on hit TV show Entourage.


1. Jed Ortmeyer netted the game-tying and game-winning goals in the third period to send Houston to a 3-1 win over Binghamton in Game 1 of the Calder Cup Finals.
2. Colton Gillies chipped in a pair of assists in the winning effort for Houston.
3. Defenseman Marco Scandella also posted two helpers as the Aeros took a 1-0 series lead.
Twenty-four hours after coming away with a 3-1 victory in Game 1 of the 2011 Calder Cup Finals, Houston looks to hold serve at home again tonight when the puck drops for Game 2 against Binghamton… The Aeros entered the third period last night down by a goal but scored three times in the third to take the win, improving to 3-5 this postseason when trailing after 40 minutes… Houston has now outscored opponents 22-12 in the third period during these playoffs… The Senators saw their eight-game road winning streak end – one shy of the league record of nine established by the 1998 Philadelphia Phantoms – and Binghamton was limited to just one goal for the first time since its postseason opener, a 2-1 loss at Manchester on Apr. 14… Aeros forward Jed Ortmeyer tied the game up at 1:47 of the third period and then notched the game-winner at 6:09, with former first-round draft pick Colton Gillies (16th overall by Minnesota in 2007) and defenseman Marco Scandella earning assists on both goals… Ortmeyer, who registered his fifth and sixth goals of the postseason, now has four goals in the last four games… Gillies has picked up four helpers over in his last three games and moved up to third place on the Aeros in scoring with 12 points – including a team-high seven goals – in 19 playoff contests… Gillies, however, is without a goal in six straight outings since scoring in Game 2 of the conference finals vs. Hamilton… Scandella, a former second-round draft pick by Minnesota who missed Houston's first four playoff games this spring due to injury, recorded his first multiple-point effort since picking up three assists vs. Abbotsford on Mar. 6… Aeros rookie goaltender Matt Hackett (13-6, 2.48, .901) earned the victory in Game 1 with 23 saves on 24 shots, including a perfect 16-for-16 showing over the final two periods… It marked the 11th time in 19 appearances during this playoff run that Hackett limited an opponent to two or fewer goals, but he had surrendered three or more goals in each of his previous four outings… Binghamton scored first for the 10th time in 18 playoff games when rookie Bobby Butler got on the board at 16:17 of the opening period off a feed from fellow first-year pro Colin Greening… Butler now leads all AHL rookies – and ranks third overall – with 11 goals (11-3-14) this postseason after combining for 32 markers in a regular season which was split between Binghamton and parent-club Ottawa… The 24-year-old Butler has now scored in four of his last five games (4-1-5) and five of his last seven (5-1-6)… He has registered single goals in 11 of the club's 18 playoff games overall, yet to record more than one in the same game… Greening's assist snapped a 10-game points drought which dated back to a two-assist effort in Game 7 of the Senators' division semifinal series at Manchester on Apr. 23… B-Sens rookie goaltender Robin Lehner stopped 38 of 41 shots in net but was still saddled with just his third loss of the playoffs (10-3, 2.38, .933)… Lehner has made at least 30 saves in four straight appearances and 11 of his 14 outings overall this spring; in two of the three exceptions, the 19-year-old started but was pulled early in favor of Barry Brust… AHL leading playoff scorer Ryan Potulny (14-11-25) was held without a point for just the fourth time in 18 playoff games this spring… The only other time during this playoff run that the Senators dropped a Game 1 – in the first round vs. Manchester – they also played Game 2 the very next day and came away with a 4-3 overtime win on Apr. 15 to even the series.
Just as Jed Ortmeyer's two third-period goals delivered a much-needed boost to the Aeros in their 3-1 win Friday night in Game 1 of the Calder Cup finals at Toyota Center, their late surge also delivered a pointed message to the Binghamton Senators. The Senators, winners of their past five playoff games but coming off a nine-day layoff after clinching the Eastern Conference title, led 1-0 entering the final period but were outshot 14-7 down the stretch and consistently outplayed over the final 20 minutes.
The Houston Aeros scored twice early in the third period and added a late empty-netter to snap the Binghamton Senators' five-game win streak and capture a 3-1 victory in Game 1 of the Calder Cup Finals on Friday night at the Toyota Center. "I have no idea who just played out there," B-Sens coach Kurt Kleinendorst said. "That wasn't my team. We didn't win a battle. You don't win games if you don't win battles."
Binghamton Senators head coach Kurt Kleinendorst says his team got what it deserved in Game 1 of the American Hockey League's Calder Cup final. Jed Ortmeyer scored twice early in the third and Matt Hackett made 23 saves Friday as the Houston Aeros beat Binghamton 3-1. Warren Peters added an empty-net goal for Houston, which leads the series 1-0 and hosts Game 2 on Saturday. "If we win that game, then we think playing poorly and winning is a good combination, so it might be the best thing that happened to us," Kleinendorst said.
Aeros coach Mike Yeo was Aeros captain during the 1999 Turner Cup finals. He was injured during the playoffs, but he led the team to its best regular-season finish, and the Aeros capped that season with a seven-game win over Orlando. Yeo won a Stanley Cup as an assistant coach for the Pittsburgh Penguins in 2009, one year after finishing as a runner-up to the Detroit Red Wings. Now, he is just three wins from his third championship. "This is why you do it," Yeo said. "This is why you go through the grind of 80 games during the regular season."
All of the Binghamton Senators are excited to be playing in the Calder Cup Finals. Barry Brust and Corey Locke are a little more excited than most about where they get to play. Brust and Locke each returned to Houston for the first time since playing for the Aeros earlier in their careers. Brust spent three seasons with the Aeros before joining the B-Sens this season. Locke played the 2008-09 season with the current Western Conference champs.
Ryan Kesler refers to teammates Kevin Bieksa and Alex Burrows as his brothers. "Not like Hank and Danny brothers," he says of the Sedin twins. "We're like brothers that fight. We get into our arguments." But they always settle them and have been now for about seven years, ever since that day the three of them, all green as grass pros, became teammates with the American Hockey League's Manitoba Moose. Bieksa remembers it like it was yesterday.
Last night in the Stanley Cup Playoffs, Nathan Horton snapped a scoreless tie with 12:27 gone in the third period as Boston captured a 1-0 win over Tampa Bay in Game 7 of the Eastern Conference Finals… David Krejci and 2001 AHL All-Star Andrew Ference picked up assists on the goal, and goaltender Tim Thomas posted a 24-save shutout.
52 – Total goals allowed by both Binghamton (in 18 games) and Houston (19 games) so far this postseason.
7-4 – Houston's playoff record when surrendering the game's first goal, a mark which was improved on Friday in Game 1 of the Calder Cup Finals.
May 28, 2004 – Jed Ortmeyer scores two goals as Hartford defeats Wilkes-Barre/Scranton, 4-1, in Game 6 of the Eastern Conference Finals, squaring the series at three games apiece.
May 28, 2003 – Houston opens the Calder Cup Finals with a 2-1 win in Hamilton behind goals from Stephane Veilleux and Jean-Guy Trudel.
May 28, 2001 – At a sold-out Harbour Station, the Saint John Flames win the Calder Cup with a 1-0 win over Wilkes-Barre/Scranton.
May 28, 2000 – Brad Smyth scores twice in Hartford's 3-1 win over Rochester in Game 3 of the Calder Cup Finals.


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text:
Nogle History, Family Crest & Coats of Arms
The surname Nogle is derived from the Norman surname de Angulo. The Gaelic form of this surname is de Nógla.
Early Origins of the Nogle family
The surname Nogle was first found in at Milford Haven, Pembrokeshire, SW Wales, where they held a family seat from very early times and were granted lands by Duke William of Normandy, their liege Lord, for their distinguished assistance at the Battle of Hastings in 1066 A.D. The Nogle surname arrived Cork, where Gilbert D'Angulo accompanied Strongbow into Ireland in 1172.
Early History of the Nogle family
This web page shows only a small excerpt of our Nogle research. Another 59 words (4 lines of text) covering the years 1689, 1757, 1830, 1508, 1536, 1541, 1719, 1784, 1636, 1699, 1686, 1691 and 1689 are included under the topic Early Nogle History in all our PDF Extended History products and printed products wherever possible.
Nogle Spelling Variations
Church officials and medieval scribes often spelled early surnames as they sounded. This practice often resulted in many spelling variations of even a single name. Early versions of the name Nogle included: Nagle, Nangle, Neagle, Naigle, Naegle, Naigel, Nagell, Nagele, Naegell, Naigel, Naigell, Nagale, Naigall, Nanegle, Nangel, Nangell, Nangale, Naingale, Naingel, Naingle, O'Nagel, O'Nagle, O'Naigle and many more.
Early Notables of the Nogle family (pre 1700)
Notable amongst the family up to this time was Thomas Nangle, 15th Baron of Navan; and his son, John Nangle, 16th Baron of Navan (died before 1508), an Irish nobleman and courageous soldier who fought with distinction at the Battle of Knockdoe; Richard Nangle D.D., Irish prelate of the Provincial of the Order of Saint Augustine, Bishop of Clonfert (1536-1541); Nano...
Another 61 words (4 lines of text) are included under the topic Early Nogle Notables in all our PDF Extended History products and printed products wherever possible.
Nogle migration to the United States +
Ireland went through one of the most devastating periods in its history with the arrival of the Great Potato Famine of the 1840s. Many also lost their lives from typhus, fever and dysentery. And poverty was the general rule as tenant farmers were often evicted because they could not pay the high rents. Emigration to North America gave hundreds of families a chance at a life where work, freedom, and land ownership were all possible. For those who made the long journey, it meant hope and survival. The Irish emigration to British North America and the United States opened up the gates of industry, commerce, education and the arts. Early immigration and passenger lists have shown many Irish people bearing the name Nogle:
Nogle Settlers in United States in the 19th Century
Joh. Nogle, aged 19, who landed in America, in 1892
Alice Nogle, aged 23, who settled in America, in 1894
Am Nogle, aged 19, who immigrated to the United States, in 1894
Mary Nogle, aged 57, who landed in America, in 1894
James C. Nogle, aged 45, who immigrated to America from London, England, in 1908
Clarra Nogle, aged 51, who landed in America from Melbourne, Australia, in 1924
John T. Nogle, aged 59, who immigrated to the United States from Melbourne, Australia, in 1924
Family seat
Great Potato Famine
The Nogle Motto +
Motto: Non vox sed votum
Motto Translation: Not in voice but a wish.
Nogle (Irish)


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High School Musical 3 El Capitan
July 10, 2019July 10, 2019 admin
Canniable Corpse Lead Singer May 25, 2019 · Swedish band Amon Amarth followed, invited the crowd to a "Viking party" complete with the prow of a Viking ship center stage
The modern-day Bel Air High School. and musical performances with students and alumni that are scheduled through December. "We're able to get our alumni to really participate in [the celebration],".
Olivia Curcuru, who lives in Phoenix, Arizona, sustained a spinal cord injury at the age of 3. Although. ever to climb the.
The third film in the musical fantasy franchise, which also stars Dove Cameron, Sofia Carson and Booboo Stewart as high.
By Trent Mallonee and many of his Los Banos High teammates have been hunting a Sac-Joaquin Section championship since they.
School visits require advance registration and a separate fee of $135 for each route. Departure and return times vary by route, but in general, groups will depart the hotel between 7-8 AM and return.
That heat didn't stop a squad of El Capitan High School football players were gearing up for a summertime. As of around 4 p.m., it was stopped at 3 acres and was 50 percent contained, according to.
He had a stretch of 401/3 innings. event in high school. She was optional all-around champion three times in four years. Tessa Anderson, West Hills, Sr. Gabi Howze, La Costa Canyon, So. Megan Lee,
"I hope he's my doctor one day if that what he wants to do," said El Capitan boys water polo and swimming coach Rodd Parker, who's coached Ma in high school. "We went 3-1 in the tournament. We.
The second phase with a grant of AED 1,836,750 (USD 500,000) is expected to benefit around 300 additional students and.
La Costa Canyon, 3:15 p.m. at Rancho Bernardo High School •BaseballHilltop Tournament, Rancho Bernardo vs. El Capitan, 3 p.m. at El Capitan High School Poway vs. Escondido, 3 p.m. at Poway high School.
A couple of years ago, El Capitan. "Between golf and school, that's about all I do," she said. "And really, I enjoy golf more than school, although I know it should be the other way around." Not to.
The Harris Center is an important community venue for the finest local arts producers and presenters in the capital region,
Variety's Jay Weissberg described it as, "A mainstream rap musical served up with generous helpings of. the first person to ever free solo climb Yosemite's 3,000 foot high El Capitan wall. With no.
Red Orchestra Ii Barrel Burst BRDM-2 armored scout car and the flank armor of a BMP infantry fighting vehicle. The M651E1 tear gas grenade carried a payload of CS gas.
Ramona High School won three. gave the Bulldogs a 3-1 Valley League record while Ramona ended the tournament with an overall 17-10 season mark. Ramona's Oct. 22 tournament game was an 8-5 loss to.
The musical. March 2 and March 3 Tickets are $15 for adults, $13 for seniors and $10 for students and children, available through brownpapertickets.com or at St. Mary's High School's main office,
He turned El Capitan into a mathematical equation. "I like to think the risk, the chance of me falling, is extremely low while the consequence is extremely high," he says. Maybe it's going back to.
The movie is about rock climber Alex Honnold, who performed a free climb of El Capitan in 2017. It usually takes a rock. Pleasant Grove moved to 8-3 in Region 4 play with one game remaining.
Azeri Pop Music
Canniable Corpse Lead Singer


Houston, Texas. The Houston Roofing Cost Report is a concise report on everything you need to know about the cost of roofing in the Houston area.
Includes necessary manhours. Cost also includes asphalt shingles, delivery, underlayment, standard flashing, and ridge vents. Does not include shingle removal, hauling, and disposal.
Manhours for metal roof installation are included in estimate. Cost takes into account metal roofing materials, typical flashing, and underlayment. Additional charge for tearing off existing roof and dump fees.
Labor for wood shake roof installation is included. Cost does include wood shake, delivery, underlayment, flashing, stack covers and roof vents. Typically there is an increased cost for removal of existing roof material and waste disposal.
Estimated quote for roof cleaning is included. Includes supplies and cleanup. Cost does not include zinc strips, anti-mold spray, or high pitch roofs.
Includes the price of roof inspection. Cost accounts for conditions report and photographs. Does not include roof repairs or structural analysis.
Price takes into account the cost of roof heating cables installation. No additional charge for de-icer cables (self-regulating; zig-zag pattern layout) and necessary fasteners. Excludes extended warranties, upgrading the electrical service, steep roofs, and three-story homes.
Includes asphalt roll roof installation labor cost. There is no additional charge for rolled roof material and flashing. Price does not take into account roof tear off and disposal fees.
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Includes pricing for slate roof installation. Cost takes into account slate material, flashing and roof underlayment. Excludes roof tear off, disposal, and structural analysis.
We base our billing the fact that not all jobs are created equal. 2 exact jobs with same square footage could cost 2 different amounts. 1 could be maintained and the other could be neglected. Our philosophy is to wow you with both service and price.
The type of roof, plus the steepness and accessibility are all major factors in the estimates.
Roof inspections are included with a full General Home Inspection at no additional cost.
Price includes manhours for asphalt roofing installation. Also includes shingles, felt underlayment, vents, and cleanup. Cost estimate excludes tearing off existing roof and debris removal.
Includes the cost of asphalt roof installation. Price includes asphalt shingles, delivery, underlayment, standard flashing, and ridge vents. Estimate does not take into account shingle removal, hauling, and disposal.
This asphalt roof installation estimate takes into account labor. Includes asphalt shingles, delivery, underlayment, standard flashing, and ridge vents. Price estimate excludes shingle removal, hauling, and disposal.
Cost estimate includes standing seam roof installation labor. Includes material, vents, underlayment, flashing, and product shipping. There is an additional charge for removal of existing roof and disposing of waste.
Estimate takes into account the cost of metal roof installation. Price does include material, delivery, underlayment, flashing, and vents. Excludes tearing off existing roof and dump fees.
Cost accounts for the price of standing seam roof installation. Does include material, vents, underlayment, flashing, and product shipping. Price does not take into account removal of existing roof and disposing of waste.
This price includes installation of wood shake. No additional charge for wood shake, underlayment, flashing, product shipping, and roof vents. Cost does not account for tear off of existing material and debris removal.
This is an estimated quote for wood shake roof installation. Price also includes wood shake, delivery, underlayment, flashing, stack covers and roof vents. Does not include removal of existing roof material and waste disposal.
Cost of wood shake roof installation is included. Price estimate includes wood shake, delivery, underlayment, flashing, stack covers and roof vents. Does not include removal of existing roof material and waste disposal.
This is a general estimate for roof cleaning. Cost does include cleaning supplies and materials. Cost does not include zinc strips, anti-mold spray, or high pitch roofs.
This roof cleaning cost estimate considers the cost of labor. Price takes into account cleaning supplies, equipment, and cleanup. Excludes excessively steep roofs that require additional safety equipment.
Price range takes into account roof cleaning labor. Price also includes cleaning supplies, equipment, and cleanup. Items that are not included: excessively steep roofs that require additional safety equipment.
Price range takes into account roofing inspection labor. Does include report on findings. Cost does not include repairing roof, structural analysis and load testing.
This price includes roof inspection. No additional charge for roof report and photos. Cost does not account for roof repairs or structural analysis.
Cost estimate includes roof inspection labor. Includes roof report and photos. This estimate does not include roof repairs or structural analysis.
This is a general price for de-icer roof heating cable installation. Cost includes roof heating cables, mounting hardware, and cleanup. Excludes roof repairs, high pitch roofs, lift rental, electrical panel upgrades, and homes with 3 or more stories.
Price of roof heating cables installation is included in the estimate. Estimate also includes self-regulating cables, clips, and mounting brackets. Does not include extended warranties, upgrading the electrical service, steep roofs, and three-story homes.
This roof heating cables installation estimate takes into account labor. Includes self-regulating cables, clips, and mounting brackets. Price estimate excludes extended warranties, upgrading the electrical service, steep roofs, and three-story homes.
Cost includes labor. Includes material, delivery, vent and chimney flashing. Does not account for roof tear off and disposal fees.
This estimated cost includes rolled asphalt roof installation labor. Price estimate includes roof material, flashing, delivery, and cleanup. Items excluded: removing existing roof and hauling.
This quote takes into account the cost of asphalt roll roof installation. It includes material, delivery, vent and chimney flashing. Price excludes roof tear off and disposal fees.
Labor for tile roof installation is included. Cost does include tile, material delivery, underlayment, minor flashing, and vents. Typically there is an increased cost for structural analysis, existing tile removal, hauling, and disposal.
Cost assumes a typical ceramic tile roofing installation. Estimate accounts for tile (ceramic/terra cotta), felt underlayment, roof vents, flashing installation, and cleanup. Price does not account for tearing off existing tile roof, removing debris, and analysis of roof structure.
Estimated quote for tile roof installation is included. Includes tile, material delivery, underlayment, minor flashing, and vents. Cost does not include structural analysis, existing tile removal, hauling, and disposal.
Price estimate includes slate roof installation. It also includes slate shingles, delivery, underlayment, and flashing. Cost does not take into account roof tear off, disposal, and structural analysis.
Labor for natural slate roof installation is included in the price. Accounts for cost of material, flashing, slate delivery, underlayment, and cleanup Cost does not account for removing existing roof and analyzing structure.
This is an average slate roof installation price. Also includes slate shingles, delivery, underlayment, and flashing. Additional cost for roof tear off, disposal, and structural analysis.
Oscar Arguijo, All World Stone & Design LLC.
Roof inspections are included in full inspection at no extra charge. Tile roofs are excluded.
All World Stone & Design LLC.


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Sasha Barrese Net Worth
in Richest Celebrities › Actors
Sasha Barrese Net Worth:
Sasha Barrese net worth: Sasha Barrese is an American actress who has a net worth of $4 million. Sasha Barrese was born in Maui, Hawaii in April 1981. She is best known for starring as Tracy Billings in The Hangover film series. Barrese starred as Sally Franklin in the television series Run of the House from 2003 to 2004.
From 2004 to 2005 she starred as Caitlin Mansfield in the TV series LAX. Barrese had a recurring role in the series Just Shoot Me! from 2002 to 2003 as Tess. She has starred in the movies American Pie, Hellraiser: Inferno, Legally Blonde, The Ring, The Hangover, Let Me In, The Hangover Part II, and The Hangover Part III. She has also appeared in episodes of the TV series Undressed, Boy Meets World, CSI: Miami, Leverage, and more.
Betsy Randle Net Worth
Jessy Schram Net Worth
Lela Loren Net Worth
Sasha Barrese
Date of Birth: Apr 24, 1981 (40 years old)
Profession: Actor, Model
Nationality: United States of America
All net worths are calculated using data drawn from public sources. When provided, we also incorporate private tips and feedback received from the celebrities or their representatives. While we work diligently to ensure that our numbers are as accurate as possible, unless otherwise indicated they are only estimates. We welcome all corrections and feedback using the button below.
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The University of Gloucestershire features in a new research paper by the Environmental Association of Universities and Colleges (EAUC) looking at a whole-institution approach to sustainability.
The paper called 'Next Generation Sustainability Strategy and Structure' aims to help universities and colleges to advance their approaches to sustainability, and share good practice and ideas on embedding sustainability in universities and colleges.
Read the full report including Gloucestershire's case study here.


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Q: Admob interstitial ads on 2.2 I want to implement interstitial (full screen ads) from Admob in my game, but I want to support Android 2.2.
Now admob SDK go with Google play services and require 2.3+ android. There is services for froyo (2.2), but in this library there is now admob support (as I know)
Is there any way to implement interstitial ads from admob in 2.2?
Thanks

A: No and quite honestly you should consider the business value of attempting to support Android 2.2 since it has such a minute (and falling0 market share.

redpajama_set_name: RedPajamaStackExchange


We are proud to announce that this Dec 16th, 2:00 PM to 4:00 PM we will be hosting an Open House at 7157 10th Avenue in the Edmonds neighborhood, Burnaby. This is an opportunity to visit this excellent Half Duplex for sale in beautiful Edmonds.
We are proud to announce that this Dec 9th, 2:00 PM to 4:00 PM we will be hosting an Open House at 1515 East 6th Avenue in Woodland Terrace in the Commercial Drive neighborhood, Vancouver, BC. This is an opportunity to visit this excellent Apartment for sale in beautiful Commercial Drive.


With so many fans eager to get their hands on RiRi's lingerie, the Savage x Fenty website employed an entry waiting list — and that's before anyone could even see the products. And, with many finding their spot on hold for more than an hour, Twitter started to explode.
The results, as you can probably imagine, were pretty savage. Browse the best reactions below.
When you about to checkout with a cart full of @SavageXFenty but the website kicks you out ??
Did you manage to cop any Fenty x Savage?
Now take a peek inside Kanye's Yeezy headquarters in Calabasas.


2018 Team Hair Peace Charities Prizes!
Join the Hair Peace Charities Team and dedicate your participation and fundraising on behalf of those undergoing all types of cancer treatment, and are in need of a wig due to hair loss from chemotherapy. YOUR fundraising will help us further our mission of helping women and girls receive financial assistance towards the purchase of a wig. Hair Peace assists women in the 11 county region of Western Pennsylvania. We also hold support group meetings in the North and South Hills of Pittsburgh open to women with all types of cancer. With your help Hair Peace Charities can reach our goal!
Now it's time to get started in just a few steps!
1. Click the "Join" button to register.
2. Start fundraising and meet your goal! Ask friends and family for donations.
3. Get a free registration! Once you reach the fundraising goal you are all set to enjoy the run.
For more information or fundraising tips, please contact Debbie Lanzarotta by email at d.lanzarotta@aol.com.
In Pennsylvania, most insurance companies do not cover the cost of a wig. This is why radio personality Bonny Diver, after being diagnosed with breast cancer, started Hair Peace Charities. This non-profit provides financial assistance of up to $150 towards the purchase of a wig. Funds are available to those living in the area codes 412 and 724. HPC also offers monthly supports meeting, in the North and South Hills, for women diagnosed with all types of cancer.
If you know someone going through cancer in need of assistance, have them call Bonny at 412-327-5177.


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Kim Fluxgold
I think it's probably safe to say that face masks are here to stay, but for how long, well that's pretty much anybody's guess?
For the past 6 years going into any large public spaces has not been an easy task for me. For the better part of 6 years now I've managed to avoid going to most Grocery Stores, Walmart and Malls as much as humanly possible but when the need arises nowadays I put on a face mask to help protect myself and others.
Since it looks as though face masks are here to stay for the foreseeable future most (but definitely not all) people have begun to embrace our new norm as more than just a necessity to help curb the spread of Covid-19 but from what I can see they have quickly become one of the hottest selling fashion accessories around and for many people they are also becoming the perfect way to show Self-Expression.
With face masks now being a way of life for most of us, many individuals who were once afraid to express their true authentic self are now using this as an opportunity to show the world who they really are even if the irony is not lost as we are having to do so by covering up our faces.
Self-Expression doesn't necessarily have to be something creative it just has to be you. It's a way of transforming our thoughts and feelings into some other form. It allows us to display our individuality, reflect on our beliefs and needs and it's also a great way to help validate oneself.
As I continue on my difficult journey toward finding Mental Wellness and Self-Discovery I can now do so while sporting my brand new fashion accessory which I have just added to my collection. It is a true and authentic expression of who I am, it is a reflection of what I so passionately advocate for and it is a gentle way to remind each and every passerby that no matter what; "You Are Enough".
***this is the logo I had created a few years ago which has not only become my signature and the name of my Blog but it has also become the backdrop for my Facebook page "youareenough" and several other projects***
Please continue to follow my journey at: https://youareenough712.wordpress.com
Kim Fluxgold is a wife and mother to 3 beautiful young adult children. She is a Blogger, a Mental Health Advocate and the author of the children's book Where Did Mommy's Smile Go? which aims to teach parents and caregivers how to help children cope with and understand their feelings when someone they love is suffering with depression.
How Do You Sabotage…Your Self-Sabotage?
Ah, September. The start of the autumn season when we fall back to regular routines, winding down from summer vacation — whether it's back to
EPISODE 002 – If Broken Crayons Still Colour, Why Does No One Choose Em?
Once upon a time………… There lived a girl. A girl who had gone to church all her life-so 'church' was all she knew. She did
EPISODE 003 – Crazy Sh*t People Have Told Us
In this episode, Shauna & Leah talk about: The things people have said to them in their careers & how these things have affected their
How to Live in Flow with Your Cycle
For years, I cursed, hated, dreaded, and cried over my period, every cycle. This phase caused me SO much pain, embarrassment, disgust. It tortured me,


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Corey Myers, 3B - D-Backs
By Robert Sherry
Corey Myers: The fourth pick overall pick in the 1999 draft by the Arizona Diamondbacks, Myers looked like a pretty large reach to go so high and his play in 2000 has done basically nothing to change many people's view. Some of the struggles might be due to him converting from shortstop over to third base this spring.
Once signed, Myers received a $2 million signing bonus and was assigned to Rookle-level Missoula. He helped in calming down the overrated talks by batting .276 with five homers and 44 runs batted in in 272 at-bats.
However, he this season has bee a struggle for this 19-year-old. In the spring he was converted over to third and was assigned to Single-A South Bend in the Midwest League. He struggled at the plate (.125, 4 RBI in 64 AB) and committed 14 errors (in 19 games) in the field. Myers was sent back to the Osprey and to see if he can settle homself down, but the struggles have continued as he has batted .218 with 5 home runs, 38 runs batted in, 14 doubles, 2 triples, 31 runs scored, 17 walks 51 strikeouts, and 4 steals in 216 at-bats.
Myers is still young so he does have a lot of time left, but he needs to bounce back next season and show an improvement at the plate, especially in making more contact. If he can settle in at third and improve at the plate, maybe the D-backs will have their eventual successor to Matt Williams. But that is a long ways away right now.


Sozo Family Fitness – Fitness for your family.
We are a welcoming neighborhood gym with terrific equipment, personal training packages, and a wide array of classes including Cycle, Zumba, Yoga, and more!
You can see pricing here with no obligation to join. It's on the second screen after entering your info.
We are fully equipped with all the strength and cardio equipment you will need.
Enjoy some self-care time as we provide a safe and fun area for your child.
Group classes like Cycle, Zumba, Yoga, and more. Exercise alongside your neighbors and reach your goals together.
Age-specific classes: fun, engaging classes for kids designed to introduce and encourage active lifestyles.
At Sozo we are a welcoming, neighborhood gym where individuals and families will be encouraged, get healthy, and become whole.
Sozo Fitness has trained me from the inside out, meaning, my mindset and thoughts were extremely self destructive and negative. Kara is working with me to have positive self talk and to enjoy the small wins. Sozo has also helped me with better food choices by educating me on how food is our fuel so eating certain food at certain times of my day are important. Kara and her team are a step above the typical gym setting type of people. They are relational and caring making me feel like they are a family and once I started, I am part of the family now too.
Sozo has given me the self assurance and confidence I thought was gone for good! Their encouragement and belief in me and has given me the belief that I CAN and I AM! I have learned to trust my journey and understand it is different from everyone else's! Sozo is more than just another gym it's like going home, being with family, feeling that warm and fuzzy! If you thought you couldn't do it....or I'm too old ...too out of shape ...don't have enough time.. So did I! But I took that leap of faith and gave them a call ...it truly changed my life!
Active families are healthy families.
You're never too old or too young to start being active.
Going to the gym should be something you look forward to!
That if you call us or come by you will see the difference!
Looking for ultra-private, one-on-one training? Check out our other location, Sozo Personal Fitness.