text:
\section{Introduction and statement of the main results}\label{s1}

We consider a system of two point masses $m_1$ and $m_2$ moving in a
fixed plane, in which the distance between a point (called pivot)
and $m_1$ and the distance between $m_1$ and $m_2$ are fixed, and
equal to $l_1$ and $l_2$ respectively. We assume the masses do not
interact. We allow gravity to act on the masses $m_1$ and $m_2$.
This system is called the planar {\it double pendulum}.

\smallskip

The position of the double pendulum is determined by the two angles
$\T_1$ and $\T_2$ shown in Figure 1. We consider only the motion in
the vicinity of the equilibrium $\T_1= \T_2= 0$, i.e. we are only
interested in small oscillations around this equilibrium. Expanding
the Lagrangian of this system to second order in $\T_1$ and $\T_2$
and their time derivatives, the corresponding Lagrange equations of
motion are
\begin{equation}\label{pendulum}
\begin{array}{r}
(m_{1}+m_{2})l_{1}\ddot{\theta}_{1}+m_{2}l_{2}\ddot{\theta}_{2}+
(m_{1}+m_{2})g\theta_{1}=0, \vspace{0.2cm}\\
m_{2}l_{2}\ddot{\theta}_{2}+m_{2}l_{1}\ddot{\theta}_{1}+m_{2}g\theta_{2}=0.
\end{array}
\end{equation}
For more details on these equations of motion see \cite{Ir}. Here
the dot denotes derivative with respect to the time $t$.

\begin{figure}
\psfrag{A}{$l_{1}$} \psfrag{B}{$\theta_{1}$}
\psfrag{C}{$\theta_{2}$} \psfrag{D}{$l_{2}$} \psfrag{E}{$m_{1}$}
\psfrag{F}{$m_{2}$} \psfrag{G}{$g$}
\includegraphics[width=3cm]{pendulo.eps}
\vskip 0cm \centerline{} \caption{\small \label{fig} The planar
double pendulum.}
\end{figure}

\smallskip

If we take $m_{1}=m_{2}=m>0$, $l_{1}=l_{2}=l>0$ and divide the above
equations by $ml$, system (\ref{pendulum}) becomes
\begin{equation}\label{pendulum2}
\begin{array}{r}
2\, \ddot{\theta}_{1}+\ddot{\theta}_{2}+2\dfrac{g}{l}\theta_{1}=0,
\vspace{0.2cm}\\
\ddot{\theta}_{1}+\ddot{\theta}_{2}+\dfrac{g}{l}\theta_{2}=0.
\end{array}
\end{equation}

Isolating $\ddot{\theta}_{1}$ and $\ddot{\theta}_{2}$ of equations
(\ref{pendulum2}) and denoting $a=g/l$, we obtain that the equations
of motion of the double pendulum with small oscillations, two equal
masses $m$ and having the two stems the same length $l$ are
\begin{equation}\label{e1}
\begin{array}{l}
\ddot{\T}_{1}=-2a\T_{1}+a\T_{2}, \vspace{0.2cm}\\
\ddot{\T}_{2}=2a\T_{1}-2a\T_{2},
\end{array}
\end{equation}
where $g$ is the acceleration of the gravity.

\smallskip

Taking a new time $\tau$ given by the rescaling $\tau= \sqrt{a}\, t$
the equations of motion \eqref{e1} become
\begin{equation}\label{e1a}
\begin{array}{l}
\T''_{1}=-2\T_{1}+\T_{2}, \vspace{0.2cm}\\
\T''_{2}=2\T_{1}-2\T_{2},
\end{array}
\end{equation}
where now the prime denotes derivative with respect to the new time
$\tau$.

\smallskip

The objective of this paper is to provide a system of nonlinear
equations whose simple zeros provide periodic solutions of the
perturbed planar double pendulum with equations of motion
\begin{equation}\label{e2}
\begin{array}{l}
\T^{''}_{1}=-2\T_{1}+\T_{2}+\e F_1(\tau,\T_1,\T'_1,\T_2,\T'_2), \vspace{0.2cm}\\
\T^{''}_{2}=2\T_{1}-2\T_{2}+\e F_2(\tau,\T_1,\T'_1,\T_2,\T'_2),
\end{array}
\end{equation}
where $\e$ is a small parameter. Here the smooth functions $F_1$ and
$F_2$ define the perturbation. These functions are periodic in
$\tau$ and in resonance $p$:$q$ with some of the periodic solutions
of the unperturbed double pendulum, being $p$ and $q$ positive
integers relatively primes. In order to present our results we need
some preliminary definitions and notations.

\smallskip

The unperturbed system \eqref{e1a} has a unique singular point, the
origin with eigenvalues
\[
\pm \sqrt{2-\sqrt{2}} \, i, \quad \pm \sqrt{2+\sqrt{2}}\, i.
\]
Consequently this system in the phase space $(\T_1,\T'_1,
\T_2,\T'_2)$ has two planes filled of periodic solutions with the
exception of the origin. These periodic solutions have periods
\[
T_1= \dfrac{2\pi}{\sqrt{2-\sqrt{2}}} \quad \mbox{or} \quad T_2=
\dfrac{2\pi}{\sqrt{2+\sqrt{2}}},
\]
according they belong to the plane associated to the eigenvectors
with eigenvalues $\pm \sqrt{2-\sqrt{2}} \, i$ or $\pm
\sqrt{2+\sqrt{2}} \, i$, respectively. We shall study which of these
periodic solutions persist for the perturbed system \eqref{e2} when
the parameter $\e$ is sufficiently small and the perturbed functions
$F_i$ for $i=1,2$ have period either $p T_1/q$, or $p T_2/q$, where
$p$ and $q$ are positive integers relatively prime.

\smallskip

We define the functions:
\begin{equation}\label{e3}
\begin{array}{l}
\CG_1(X_0,Y_0)= \displaystyle \int_0^{pT_1}
\sin\left(\sqrt{2-\sqrt{2}} \, \tau\right) \left(\sqrt{2}\, \bar
F_1+ \bar F_2\right) d\tau, \vspace{0.2cm}\\
\CG_2(X_0,Y_0)= \displaystyle  \displaystyle \int_0^{pT_1} \cos
\left(\sqrt{2-\sqrt{2}} \, \tau\right) \left(\sqrt{2}\, \bar F_1+
\bar F_2\right) d\tau,
\end{array}
\end{equation}
where
\begin{equation}\label{jj}
\bar F_k= F_k(\tau,A_1,B_1,C_1,D_1)
\end{equation}
for $k=1,2$ with
\[
\begin{array}{l}
A_1= \dfrac{1}{\sqrt{4-2 \sqrt{2}}} \left(X_0 \cos
\left(\sqrt{2-\sqrt{2}}\, \tau\right)+Y_0 \sin
\left(\sqrt{2-\sqrt{2}}\, \tau\right)\right), \vspace{0.2cm}\\
B_1= \dfrac{1}{\sqrt{2}} \left(Y_0 \cos \left(\sqrt{2-\sqrt{2}}\,
\tau\right)-X_0\sin \left(\sqrt{2-\sqrt{2}}\, \tau\right) \right), \vspace{0.2cm}\\
C_1= \dfrac{1}{\sqrt{2-\sqrt{2}}} \left(X_0 \cos
\left(\sqrt{2-\sqrt{2}}\, \tau\right)+Y_0
\sin\left(\sqrt{2-\sqrt{2}}\, \tau\right) \right), \vspace{0.2cm}\\
D_1= Y_0 \cos \left(\sqrt{2-\sqrt{2}}\, \tau\right)-X_0 \sin
   \left(\sqrt{2-\sqrt{2}}\, \tau\right).
\end{array}
\]

A zero $(X_0^*,Y_0^*)$ of the nonlinear system
\begin{equation}\label{e4}
\CG_1(X_0,Y_0)=0,\quad \CG_2(X_0,Y_0)=0,
\end{equation}
such that
\[
\det \left(\left.
\dfrac{\p(\CG_1,\CG_2)}{\p(X_0,Y_0)}\right|_{(X_0,Y_0)=
(X_0^*,Y_0^*)}\right) \neq 0,
\]
is called a {\it simple zero} of system \eqref{e4}.

\smallskip

Our main result on the periodic solutions of the perturbed double
pendulum \eqref{e2} which bifurcate from the periodic solutions of
the unperturbed double pendulum \eqref{e1a} with period $T_1$
traveled $p$ times is the following.

\begin{theorem}\label{t1}
Assume that the functions $F_k$ of the perturbed double pendulum
with equations of motion \eqref{e2} are periodic in $\tau$ of period
$pT_1/q$ with $p$ and $q$ positive integers relatively prime. Then
for $\e\neq 0$ sufficiently small and for every simple zero
$(X_0^*,Y_0^*)\neq (0,0)$ of the nonlinear system \eqref{e4}, the
perturbed double pendulum \eqref{e2} has a periodic solution
$(\T_1(\tau,\e),\T_2(\tau,\e))$ tending when $\e\to 0$ to the
periodic solution $(\T_1(\tau),\T_2(\tau))$ run $p$ times of the
unperturbed double pendulum \eqref{e1a} given by
\begin{equation}\label{e5}
\begin{array}{l}
\left( \dfrac{1}{\sqrt{4-2 \sqrt{2}}} \left(X_0^* \cos
\left(\sqrt{2-\sqrt{2}}\, \tau\right)+Y_0^* \sin
\left(\sqrt{2-\sqrt{2}}\, \tau\right)\right), \right. \vspace{0.2cm}\\
\left. \dfrac{1}{\sqrt{2-\sqrt{2}}} \left(X_0^* \cos
\left(\sqrt{2-\sqrt{2}}\, \tau\right)+Y_0^*
\sin\left(\sqrt{2-\sqrt{2}}\, \tau\right) \right) \right).
\end{array}
\end{equation}
\end{theorem}

Theorem \ref{t1} is proved in section \ref{s3}. Its proof is based
in the averaging theory for computing periodic solutions, see the
appendix.

\smallskip

We provide an application of Theorem \ref{t1} in the following
corollary, which will be proved in section \ref{s4}.

\begin{corollary}\label{c1}
If $F_2(\T_1,\T'_1,\T_2,\T'_2)= \left(1-\T_1^2\right) \sin
\left(\sqrt{2-\sqrt{2}}\, t\right)$ and $F_1(\T_1,\T'_1,\T_2,\T'_2)=
0$, then the differential equation \eqref{e2} for $\e\neq 0$
sufficiently small has two periodic solutions
$(\T_1(\tau,\e),\T_2(\tau,\e))$ tending when $\e\to 0$ to the two
periodic solutions $(\T_1(\tau),\T_2(\tau))$ of the unperturbed
double pendulum \eqref{e1a} given by \eqref{e5} with
\[
(X_0^*,Y_0^*)= \left(2 \sqrt{2 \left(2-\sqrt{2}\right)},0   \right)
\,\,\, \mbox{and} \,\,\, (X_0^*,Y_0^*)= \left(0, 2 \sqrt{\frac{2}{3}
\left(2-\sqrt{2}\right)} \right),
\]
respectively.
\end{corollary}

Now we define the functions:
\begin{equation}\label{e3a}
\begin{array}{l}
\CG^1(Z_0,W_0)= \displaystyle \int_0^{pT_2}
\sin\left(\sqrt{2+\sqrt{2}} \, \tau\right) \left(-\sqrt{2}\, \bar
F_1+ \bar F_2\right) d\tau, \vspace{0.2cm}\\
\CG^2(Z_0,W_0)= \displaystyle  \displaystyle \int_0^{pT_2}\cos
\left(\sqrt{2+\sqrt{2}} \, \tau\right) \left(-\sqrt{2}\, \bar F_1+
\bar F_2\right) d\tau,
\end{array}
\end{equation}
where $\bar F_k= F_k(\tau,A_2,B_2,C_2,D_2)$ for $k=1,2$ with
\[
\begin{array}{l}
A_2= -\dfrac{1}{\sqrt{4+2 \sqrt{2}}} \left(Z_0 \cos
\left(\sqrt{2+\sqrt{2}}\, \tau\right)+W_0 \sin
\left(\sqrt{2+\sqrt{2}}\, \tau\right)\right), \vspace{0.2cm}\\
B_2= \dfrac{1}{\sqrt{2}} \left(-W_0 \cos \left(\sqrt{2+\sqrt{2}}\,
\tau\right)+Z_0\sin \left(\sqrt{2+\sqrt{2}}\, \tau\right) \right),
\vspace{0.2cm}\\
C_2= \dfrac{1}{\sqrt{2+\sqrt{2}}} \left(Z_0 \cos
\left(\sqrt{2+\sqrt{2}}\, \tau\right)+W_0
\sin\left(\sqrt{2+\sqrt{2}}\, \tau\right) \right), \vspace{0.2cm}\\
D_2= W_0 \cos \left(\sqrt{2+\sqrt{2}}\, \tau\right)-Z_0 \sin
\left(\sqrt{2+\sqrt{2}}\, \tau\right).
\end{array}
\]
Consider the nonlinear system
\begin{equation}\label{e4a}
\CG^1(Z_0,W_0)=0,\quad \CG^2(Z_0,W_0)=0.
\end{equation}

\smallskip

Our main result on the periodic solutions of the perturbed double
pendulum \eqref{e2} which bifurcate from the periodic solutions of
the unperturbed double pendulum \eqref{e1a} with period $T_2$
traveled $p$ times is the following.

\begin{theorem}\label{t1a}
Assume that the functions $F_k$ of the perturbed double pendulum
with equations of motion \eqref{e2} are periodic in $\tau$ of period
$pT_2/q$ with $p$ and $q$ positive integers relatively prime. Then
for $\e\neq 0$ sufficiently small and for every simple zero
$(Z_0^*,W_0^*)\neq (0,0)$ of the nonlinear system \eqref{e4a}, the
perturbed double pendulum \eqref{e2} has a periodic solution
$(\T_1(\tau,\e),\T_2(\tau,\e))$ tending when $\e\to 0$ to the
periodic solution $(\T_1(\tau),\T_2(\tau))$ run $p$ times of the
unperturbed double pendulum \eqref{e1a} given by
\begin{equation}\label{e5a}
\begin{array}{l}
\left(- \dfrac{1}{\sqrt{4+2 \sqrt{2}}} \left(Z_0^* \cos
\left(\sqrt{2+\sqrt{2}}\, \tau\right)+W_0^* \sin
\left(\sqrt{2+\sqrt{2}}\, \tau\right)\right), \right. \vspace{0.2cm}\\
\left. \dfrac{1}{\sqrt{2+\sqrt{2}}} \left(Z_0^* \cos
\left(\sqrt{2+\sqrt{2}}\, \tau\right)+W_0^*
\sin\left(\sqrt{2+\sqrt{2}}\, \tau\right) \right) \right).
\end{array}
\end{equation}
\end{theorem}

Theorem \ref{t1a} is also proved in section \ref{s3}.

\smallskip

Now we provide an application of Theorem \ref{t1a} in the next
corollary, which will be proved in section \ref{s4}.

\begin{corollary}\label{c2}
If $F_1(\T_1,\T'_1,\T_2,\T'_2)= \T'_2+ \T_1^2 \cos
\left(\sqrt{2+\sqrt{2}}\, t\right)$ and $F_2(\T_1,\T'_1,\T_2,\T'_2)=
0$, then the differential equation \eqref{e2} for $\e\neq 0$
sufficiently small has one periodic solution $(\T_1(\tau,\e),
\T_2(\tau,\e))$ tending when $\e\to 0$ to the periodic solution
$(\T_1(\tau), \T_2(\tau))$ of the unperturbed double pendulum
\eqref{e1a} given by \eqref{e5a} with $(Z_0^*,W_0^*)= \left(0, -8
\left(2+\sqrt{2}\right)\right)$.
\end{corollary}

\section{Proof of Theorems \ref{t1} and \ref{t1a}}\label{s3}

Introducing the variables $(x,y,z,w)= (\T_1,\T'_1,\T_2,\T'_2)$ we
write the differential system of the perturbed double pendulum
\eqref{e2} as a first--order differential system defined in $\R^4$.
Thus we have the differential system
\begin{equation}\label{d1}
\begin{array}{l}
x' = y,\\
y' = -2x+z+\e F_1(\tau,x,y,z,w),\\
z' = w,\\
w' = 2x-2z+\e F_2(\tau,x,y,z,w).
\end{array}
\end{equation}
System \eqref{d1} with $\e=0$ is equivalent to the unperturbed
double pendulum system \eqref{e1a}, called in what follows simply by
the {\it unperturbed system}, otherwise we have the {\it perturbed
system}.

\smallskip

We shall write system \eqref{d1} in such a way that the linear part
at the origin will be in its real normal Jordan form. Then, doing
the change of variables $(\tau,x,y,z,w)\to (\tau,X,Y,Z,W)$ given by
\begin{equation}\label{nn}
\left(
\begin{array}{c}
X\\
Y\\
Z\\
W
\end{array}
\right)= \left(
\begin{array}{cccc}
 \sqrt{1-\dfrac{1}{\sqrt{2}}} & 0 & \dfrac{\sqrt{2-\sqrt{2}}}{2} & 0 \\
 0 & \dfrac{1}{\sqrt{2}} & 0 & \dfrac{1}{2} \\
 -\sqrt{1+\dfrac{1}{\sqrt{2}}} & 0 & \dfrac{\sqrt{2+\sqrt{2}}}{2} & 0 \\
 0 & -\dfrac{1}{\sqrt{2}} & 0 & \dfrac{1}{2}
\end{array}
\right)\left(
\begin{array}{c}
x\\
y\\
z\\
w
\end{array}
\right),
\end{equation}
the differential system \eqref{d1} becomes
\begin{equation}\label{d2}
\begin{array}{l}
X' = \sqrt{2-\sqrt{2}}\, Y, \vspace{0.2cm}\\
Y' = -\sqrt{2-\sqrt{2}}\, X+ \e \dfrac{1}{2} \left(\sqrt{2}\, \tilde
F_1+\tilde F_2\right),\vspace{0.2cm}\\
Z' = \sqrt{2+\sqrt{2}}\, W,\vspace{0.2cm}\\
W' = -\sqrt{2+\sqrt{2}}\, Z+\e \dfrac{1}{2} \left(\tilde
F_2-\sqrt{2}\, \tilde F_1\right),
\end{array}
\end{equation}
where $\tilde F_i(\tau,X,Y,Z,W)=F_i(\tau,\A,\B,\C,\D)$ for $i=1,2$
with
\[
\begin{array}{l}
\A= \dfrac{X}{\sqrt{4-2 \sqrt{2}}}-\dfrac{Z}{\sqrt{2
\left(2+\sqrt{2}\right)}},\vspace{0.2cm}\\
\B= \dfrac{Y-W}{\sqrt{2}},\vspace{0.2cm}\\
\C= \dfrac{X}{\sqrt{2-\sqrt{2}}}+\dfrac{Z}{\sqrt{2+\sqrt{2}}},\vspace{0.2cm}\\
\D= Y+W.
\end{array}
\]
Note that the linear part of the differential system \eqref{d2} at
the origin is in its real normal Jordan form.

\begin{lemma}\label{L1}
The periodic solutions of the differential system \eqref{d2} with
$\e=0$ are
\begin{equation}\label{d3}
\begin{array}{l}
X(\tau)= X_0 \cos \left(\sqrt{2-\sqrt{2}}\, \tau \right) + Y_0 \sin
\left(\sqrt{2-\sqrt{2}}\, \tau \right),\\
Y(\tau)= Y_0 \cos \left(\sqrt{2-\sqrt{2}}\, \tau \right)-X_0 \sin
\left(\sqrt{2-\sqrt{2}}\, \tau \right),\\
Z(\tau)= 0,\\
W(\tau)= 0,
\end{array}
\end{equation}
of period $T_1$, and
\begin{equation}\label{d4}
\begin{array}{l}
X(\tau)= 0,\\
Y(\tau)= 0,\\
Z(\tau)= Z_0 \cos \left(\sqrt{2+\sqrt{2}}\, \tau \right)+W_0 \sin
\left(\sqrt{2+\sqrt{2}}\, \tau \right),\\
W(\tau)= W_0 \cos \left(\sqrt{2+\sqrt{2}}\, \tau\right)-Z_0 \sin
\left(\sqrt{2+\sqrt{2}}\, \tau \right),
\end{array}
\end{equation}
of period $T_2$.
\end{lemma}

\begin{proof}
Since system \eqref{d2} with $\e=0$ is a linear differential system,
the proof follows easily.
\end{proof}

\begin{proof}[Proof of Theorem \ref{t1}]
Assume that the functions $F_k$ of the perturbed double pendulum
with equations of motion \eqref{e2} are periodic in $\tau$ of period
$pT_1/q$ with $p$ and $q$ positive integers relatively prime. Then
we can think that system \eqref{e2} is periodic in $\tau$ of period
$pT_1$. Thinking in this way the differential system and the
periodic solutions \eqref{d3} have the same period $pT_1$.

\smallskip

We shall apply Theorem \ref{tt} of the appendix to the differential
system \eqref{d2}. We note that system \eqref{d2} can be written as
system \eqref{eq:4} taking
\[
{\bf x}=\left(
\begin{array}{c}
X\vspace{0.2cm}\\
Y\vspace{0.2cm}\\
Z\vspace{0.2cm}\\
W
\end{array}
\right), \quad t=\tau, \quad G_0(t,{\bf x})=\left(
\begin{array}{c}
\sqrt{2-\sqrt{2}}\, Y,\vspace{0.2cm}\\
-\sqrt{2-\sqrt{2}}\, X,\vspace{0.2cm}\\
\sqrt{2+\sqrt{2}}\, W,\vspace{0.2cm}\\
-\sqrt{2+\sqrt{2}}\, Z
\end{array}
\right),
\]
\[
G_1(t,{\bf x})=\left(
\begin{array}{c}
0 \vspace{0.2cm}\\
\dfrac{1}{2} \left(\sqrt{2}\, \tilde F_1+\tilde F_2\right) \vspace{0.2cm}\\
0 \vspace{0.2cm}\\
\dfrac{1}{2} \left(\tilde F_2-\sqrt{2}\, \tilde F_1\right)
\end{array}
\right) \quad \mbox{and} \quad G_2(t,{\bf x,\e})=\left(
\begin{array}{c}
0 \vspace{0.2cm}\\
0 \vspace{0.2cm}\\
0 \vspace{0.2cm}\\
0
\end{array}
\right).
\]

We shall study which periodic solutions \eqref{d3} of the
unperturbed system \eqref{d2} with $\e=0$ can be continued to
periodic solutions of the unperturbed system \eqref{d2} for $\e\neq
0$ sufficiently small.

\smallskip

We shall describe the different elements which appear in the
statement of Theorem \ref{tt} in the particular case of the
differential system \eqref{d2}. Thus we have that $\Omega= \R^4$,
$k=2$ and $n=4$. Let $r_1>0$ be arbitrarily small and let $r_2>0$ be
arbitrarily large. We take the open and bounded subset $V$ of the
plane $Z=W=0$ as
\[
V= \{ (X_0,Y_0,0,0)\in \R^4 : r_1< \sqrt{X_0^2+Y_0^2} < r_2 \}.
\]
As usual $\cl(V)$ denotes the closure of $V$. If $\al=(X_0,Y_0)$,
then we can identify $V$ with the set
\[
\{ \al\in \R^2 : r_1< ||\al|| < r_2 \},
\]
here $|| \cdot ||$ denotes the Euclidean norm of $\R^2$. The
function $\be: \cl(V)\to \R^2$ is $\be(\al)= (0,0)$. Therefore, in
our case the set
\[
\mathcal{Z}=\left\{ {\bf z}_{\alpha}=\left( \alpha,
\beta(\alpha)\right),~~\alpha\in \cl(V) \right\}= \{
(X_0,Y_0,0,0)\in \R^4 : r_1\leq \sqrt{X_0^2+Y_0^2} \leq r_2 \}.
\]
Clearly for each ${\bf z}_{\alpha}\in \mathcal{Z}$ we can consider
the periodic solution ${\bf x}(\tau, {\bf z}_{\alpha})= (
X(\tau),Y(\tau)$, $0,0)$ given by \eqref{d3} of period $pT_1$.

\smallskip

Computing the fundamental matrix $M_{{\bf z}_{\alpha}}(\tau)$ of the
linear differential system \eqref{d2} with $\e=0$ associated to the
$T$--periodic solution ${\bf z}_{\alpha}= (X_0,Y_0,0,0)$ such that
$M_{{\bf z}_{\alpha}}(0)$ be the identity of $\R^4$, we get that
$M(\tau)= M_{{\bf z}_{\alpha}}(\tau)$ is equal to
\[
\left(
\begin{array}{cccc}
\cos \left(\sqrt{2-\sqrt{2}}\, \tau\right) & \sin
\left(\sqrt{2-\sqrt{2}}\, \tau\right) & 0 & 0 \\
-\sin \left(\sqrt{2-\sqrt{2}}\, \tau\right) & \cos
\left(\sqrt{2-\sqrt{2}}\, \tau\right) & 0 & 0 \\
0 & 0 & \cos \left(\sqrt{2+\sqrt{2}}\, \tau\right) & \sin
\left(\sqrt{2+\sqrt{2}}\, \tau\right) \\
0 & 0 & -\sin \left(\sqrt{2+\sqrt{2}}\, \tau\right) & \cos
\left(\sqrt{2+\sqrt{2}}\, \tau\right)
\end{array}
\right).
\]
Note that the matrix $M_{{\bf z}_{\alpha}}(\tau)$ does not depend of
the particular periodic solution ${\bf x}(\tau, {\bf z}_{\alpha})$.
Since the matrix
\[
M^{-1}(0)-M^{-1}(pT_1)=\left(
\begin{array}{cccc}
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 2 \sin ^2\left(\sqrt{2}\, \pi \right) & \sin \left(2
 \sqrt{2}\, \pi \right) \\
 0 & 0 & -\sin \left(2 \sqrt{2}\, \pi \right) & 2 \sin
 ^2\left(\sqrt{2}\, \pi \right)
\end{array}
\right),
\]
satisfies the assumptions of statement (ii) of Theorem \ref{tt}
because the determinant
\[
\left|
\begin{array}{ll}
2 \sin ^2\left(\sqrt{2}\, \pi \right) & \sin \left(2 \sqrt{2}\, \pi
\right) \\
-\sin \left(2 \sqrt{2}\, \pi \right) & 2 \sin ^2\left(\sqrt{2}\, \pi
\right)
\end{array}
\right|= 4 \sin ^2\left(\sqrt{2} \pi \right)\neq 0,
\]
we can apply this theorem to system \eqref{d2}.

\smallskip

Now $\xi: \R^4\to \R^2$ is $\xi(X,Y,Z,W)= (X,Y)$. We calculate the
function
\[
\CG(X_0,Y_0)=\CG(\alpha)=\xi\left( \dfrac{1}{pT_1} \displaystyle
\int _0^{pT_1} M_{{\bf z}_{\alpha}}^{-1}(\tau)G_1(\tau,{\bf
x}(\tau,{\bf z}_{\alpha})) d\tau\right),
\]
and we obtain
\[
\left( \begin{array}{c} \CG_1(X_0,Y_0)\vspace{0.2cm}\\
\CG_2(X_0,Y_0)
\end{array}
\right)= \left(
\begin{array}{c}
\displaystyle -\dfrac{\sqrt{2-\sqrt{2}}}{4 \pi } \int_0^{pT_1}
\sin\left(\sqrt{2-\sqrt{2}} \, \tau\right) \left(\sqrt{2}\, \bar
F_1+ \bar F_2\right) d\tau \vspace{0.2cm}\\
\displaystyle \frac{\sqrt{2-\sqrt{2}}}{4 \pi } \int_0^{pT_1}\cos
\left(\sqrt{2-\sqrt{2}} \, \tau\right) \left(\sqrt{2}\, \bar F_1+
\bar F_2\right) d\tau
\end{array}
\right)
\]
where the functions of $\bar F_k$ for $k=1,2$  are the ones given in
\eqref{jj}. Then, by Theorem \ref{tt} we have that for every simple
zero $(X_0^*,Y_0^*)\in V$ of the system of nonlinear functions
\begin{equation}\label{hh}
\CG_1(X_0,Y_0)=0, \quad \CG_2(X_0,Y_0)=0,
\end{equation}
we have a periodic solution $(X,Y,Z,W)(\tau,\e)$ of system
\eqref{d2} such that
\[
(X,Y,Z,W)(\tau,\e)\to (X_0^*,Y_0^*,0,0)\quad \mbox{as $\e\to 0$.}
\]
Note that system \eqref{hh} is equivalent to system \eqref{e4},
because both equations only differs in a non--zero multiplicative
constant.

\smallskip

Going back through the change of coordinates \eqref{nn} we get a
periodic solution $(x,y,z,w)(\tau,\e)$ of system \eqref{d3} such
that
\[
\left(
\begin{array}{c}
x(\tau,\e) \vspace{0.2cm}\\
y(\tau,\e) \vspace{0.2cm}\\
z(\tau,\e) \vspace{0.2cm}\\
w(\tau,\e)
\end{array}
\right)\to \left(
\begin{array}{c}
\dfrac{1}{\sqrt{4-2 \sqrt{2}}} \left(X_0^* \cos
\left(\sqrt{2-\sqrt{2}}\, \tau\right)+Y_0^* \sin
\left(\sqrt{2-\sqrt{2}}\, \tau\right)\right) \vspace{0.2cm}\\
\dfrac{1}{\sqrt{2}} \left(Y_0^* \cos \left(\sqrt{2-\sqrt{2}}\,
\tau\right)-X_0^*\sin \left(\sqrt{2-\sqrt{2}}\, \tau\right) \right) \vspace{0.2cm}\\
\dfrac{1}{\sqrt{2-\sqrt{2}}} \left(X_0^* \cos
\left(\sqrt{2-\sqrt{2}}\, \tau\right)+Y_0^*
\sin\left(\sqrt{2-\sqrt{2}}\, \tau\right) \right) \vspace{0.2cm}\\
Y_0^* \cos \left(\sqrt{2-\sqrt{2}}\, \tau\right)-X_0^* \sin
\left(\sqrt{2-\sqrt{2}}\, \tau\right)
\end{array}
\right)
\]
as $\e\to 0$.

\smallskip

Consequently we obtain a periodic solution $(\T_1,\T_2)(\tau,\e)$ of
system \eqref{e2} such that
\[
(\T_1,\T_2)(\tau,\e)\to\left(
\begin{array}{c}
\dfrac{1}{\sqrt{4-2 \sqrt{2}}} \left(X_0^* \cos
\left(\sqrt{2-\sqrt{2}}\, \tau\right)+Y_0^* \sin
\left(\sqrt{2-\sqrt{2}}\, \tau\right)\right) \vspace{0.2cm}\\
\dfrac{1}{\sqrt{2-\sqrt{2}}} \left(X_0^* \cos
\left(\sqrt{2-\sqrt{2}}\, \tau\right)+Y_0^*
\sin\left(\sqrt{2-\sqrt{2}}\, \tau\right) \right) \vspace{0.2cm}
\end{array}
\right)
\]
as $\e\to 0$. Hence Theorem \ref{t1} is proved.
\end{proof}

\smallskip

\begin{proof}[Proof of Theorem \ref{t1a}]
This proof is completely analogous to the proof of Theorem \ref{t1}.
\end{proof}

\section{Proof of the two corollaries}\label{s4}

\begin{proof}[Proof of Corollary \ref{c1}]
Under the assumptions of Corollary \ref{c1} the nonlinear system
\eqref{e3} becomes
\[
\begin{array}{l}
\CG_1(X_0,Y_0)= -\dfrac{\pi  \left(X_0^2+3 Y_0^2+8
\left(-2+\sqrt{2}\right)\right)}{8 \left(2-\sqrt{2}
\right)^{3/2}}, \vspace{0.2cm}\\
\CG_2(X_0,Y_0)= -\dfrac{\pi  X_0 Y_0}{4
\left(2-\sqrt{2}\right)^{3/2}}.
\end{array}
\]
This system has the following four solutions
\[
(X_0^*,Y_0^*)= \left(\pm 2 \sqrt{2 \left(2-\sqrt{2}\right)},0
\right) \,\,\, \mbox{and} \,\,\, (X_0^*,Y_0^*)= \left(0, \pm 2
\sqrt{\frac{2}{3} \left(2-\sqrt{2}\right)} \right).
\]
But the solutions which differs in a sign are different initial
conditions of the same periodic solution of the unperturbed double
pendulum. Moreover, it easy to check that these solutions are
simple. So, by Theorem \ref{t1} we only have two periodic solutions
of the perturbed double pendulum. This completes the proof of the
corollary.
\end{proof}

\begin{proof}[Proof of Corollary \ref{c2}]
Under the assumptions of Corollary \ref{c2} the nonlinear system
\eqref{e3a} becomes
\[
\begin{array}{l}
\CG^1(Z_0,W_0)= -\dfrac{\pi  \left(\sqrt{\left(10-7 \sqrt{2}\right)
\left(2+\sqrt{2}\right)} W_0-8\right) Z_0}{4 \sqrt{2
\left(2+\sqrt{2}\right)}}, \vspace{0.2cm}\\
\CG^2(Z_0,W_0)= \dfrac{\pi  \left(\sqrt{2} W_0^2-2 W_0^2-16 W_0+3
\sqrt{2} Z_0^2-6 Z_0^2\right)}{8 \sqrt{2 \left(2+\sqrt{2}\right)}}.
\end{array}
\]
This system only has the following two real solutions
\[
((Z_0^*,W_0^*)= \left(0, -8 \left(2+\sqrt{2}\right)\right) \,\,\,
\mbox{and} \,\,\, (Z_0^*,W_0^*)= (0,0).
\]
the other two solutions are not real. But the solution $(0,0)$ is
not valid. Therefore, by Theorem \ref{t1a} we only have one periodic
solution of the perturbed double pendulum. This completes the proof
of the corollary.
\end{proof}

\section*{Appendix: Basic results on averaging theory}\label{ap}

In this appendix we present the basic result from the averaging
theory that we shall need for proving the main results of this
paper.

\smallskip

We consider the problem of the bifurcation of $T$--periodic
solutions from differential systems of the form
\begin{equation}\label{eq:4}
\dot {\bf x}(t)= G_0(t,{\bf x})+\e G_1(t,{\bf x})+\e^2 G_{2}(t,{\bf
x}, \e),
\end{equation}
with $\e=0$ to $\e\not= 0$ sufficiently small. Here the functions
$G_0,G_1: \R \times \Omega \to \R^n$ and $G_{2}:\R\times \Omega
\times (-\e_0,\e_0)\to \R^n$ are $\CC^2$ functions, $T$--periodic in
the first variable, and $\Omega $ is an open subset of $\R^n$. The
main assumption is that the unperturbed system
\begin{equation}\label{eq:5}
\dot {\bf x}(t)= G_0(t,{\bf x}),
\end{equation}
has a submanifold of periodic solutions. A solution of this problem
is given using the averaging theory.

\smallskip

Let ${\bf x}(t,{\bf z},\e)$ be the solution of the system
\eqref{eq:5} such that ${\bf x}(0,{\bf z},\e)= {\bf z}$. We write
the linearization of the unperturbed system along a periodic
solution ${\bf x}(t,{\bf z},0)$ as
\begin{equation}\label{eq:6}
\dot {\bf y}=D_{\bf x}{G_0}(t,{\bf x}(t,{\bf z},0)){\bf y}.
\end{equation}
In what follows we denote by $M_{\bf z}(t)$ some fundamental matrix
of the linear differential system \eqref{eq:6}, and by
$\xi:\R^k\times \R^{n-k}\to \R^k$ the projection of $\R^n$ onto its
first $k$ coordinates; i.e. $\xi(x_1,\ldots,x_n)= (x_1,\ldots,x_k)$.

\smallskip

We assume that there exists a $k$--dimensional submanifold
$\mathcal{Z}$ of $\Omega$ filled with $T$--periodic solutions of
\eqref{eq:5}. Then an answer to the problem of bifurcation of
$T$--periodic solutions from the periodic solutions contained in
$\mathcal{Z}$ for system \eqref{eq:4} is given in the following
result.

\begin{theorem}\label{tt}
Let $V$ be an open and bounded subset of $\R^k$, and let $\beta:
\cl(V)\to \R^{n-k}$ be a $\CC^2$ function. We assume that
\begin{itemize}
\item[(i)] $\mathcal{Z}=\left\{ {\bf z}_{\alpha}=\left( \alpha,
\beta(\alpha)\right),~~\alpha\in \cl(V) \right\}\subset \Omega$ and
that for each ${\bf z}_{\alpha}\in \mathcal{Z}$ the solution ${\bf
x}(t,{\bf z}_{\alpha})$ of \eqref{eq:5} is $T$--periodic;

\item[(ii)] for each ${\bf z}_{\alpha}\in \mathcal{Z}$ there
is a fundamental matrix $M_{{\bf z}_{\alpha}}(t)$ of \eqref{eq:6}
such that the matrix $M_{{\bf z}_{\alpha}}^{-1}(0)- M_{{\bf
z}_{\alpha}}^{-1}(T)$ has in the upper right corner the $k\times
(n-k)$ zero matrix, and in the lower right corner a $(n-k)\times
(n-k)$ matrix $\Delta_{\alpha}$ with $\det(\Delta_{\alpha})\neq 0$.
\end{itemize}
We consider the function $\CG:\cl(V) \to \R^k$
\begin{equation}\label{eq:7}
\CG(\alpha)=\xi\left( \dfrac{1}{T} \int _0^T M_{{\bf
z}_{\alpha}}^{-1}(t)G_1(t,{\bf x}(t,{\bf z}_{\alpha})) dt\right).
\end{equation}
If there exists $a\in V$ with $\CG(a)=0$ and $\displaystyle{\det
\left( \left({d\CG}/{d\alpha}\right)(a)\right)\neq 0}$, then there
is a $T$--periodic solution $\varphi (t,\e)$ of system \eqref{eq:4}
such that $\varphi(0,\e)\to {\bf z}_a$ as $\e\to 0$.
\end{theorem}

Theorem~\ref{tt} goes back to Malkin \cite{Ma} and Roseau \cite{Ro},
for a shorter proof see \cite{BFL}.

\section*{Acknowledgements}

The first author is partially supported by a MICIIN/FEDER grant
MTM2008--03437, by a AGAUR grant number 2009SGR-0410 and by ICREA
Academia. The second author is partially suported by the grant
FAPESP 2011/03896-0 The third author is partially supported by a
FAPESP--BRAZIL grant 2007/06896--5. The first and third authors are
also supported by the joint project CAPES-MECD grant
PHB-2009-0025-PC

redpajama_set_name: RedPajamaArXiv


The Saginaw County Fairground Campground includes 70 clean beautiful campsites. Available not only during Fair week, but spring to fall as well these campsites offer great convenience for exhibitors at the Fair, participants in the many horse shows held on the grounds and those attending the many other events on the grounds. Of course if you're just looking for a place to camp while in the area, you're welcome too.
The campground has hosted camping clubs, jamborees and even those attending wedding receptions in our exhibit hall.
If you would like to spend a night or several with us, please feel free to call the office (989.845.2143) for more information and/or fill out and send in the reservation forms below. Note that during fair week preference is given to those exhibiting livestock and vendors and the spots fill up fast.


text:
import subprocess

cmd = "node hi"
# no block, it start a sub process.
p = subprocess.Popen(cmd , shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)

# and you can block util the cmd execute finish
p.wait()
# or stdout, stderr = p.communicate()

with open("ex.txt", "r") as f:
    stockPrice = f.read();

print stockPrice

redpajama_set_name: RedPajamaGithub


text:
Ayia Napa to install CCTV in public spaces by summer 2020
By Peter Michael September 11, 2019 September 11, 2019 0750
Ayia Napa will be the first municipality in Cyprus to have CCTV cameras installed in public spaces, the mayor's office announced on Wednesday evening, with the cameras expected in the town by summer 2020.
The municipality said that it was decided at Wednesday's cabinet meeting following a proposal made by the town, and that CCTV would help protect public property.
"The municipality's proposal is in line with the recommendations of the Personal Data Commissioner, as it was reviewed by the police as well. Ayia Napa will be the first municipality in Cyprus to install CCTV within the boundaries of the municipality for the protection of public property," the mayor's office said.
The municipality will process the footage of the cameras solely, they said, so as to protect the personal information of the public.
"This is a very important and essential measure, which will significantly improve security and enhance the sense of security in the municipality, as it will act to reduce crime and help with investigations in actions against public property," said the mayor of Ayia Napa, Yiannis Karousos. "Tourists and residents of Ayia Napa municipality will be safe and feel safe."
Ayia Napacctvpicks4public spaces
Bahamian officials say 2,500 people registered as missing in Dorian's wake
Mugabe's body brought home but Zimbabweans unsure of his burial place
Ombudswoman snubs auditor once more
Evie Andreou January 20, 2020

redpajama_set_name: RedPajamaCommonCrawl


PKI only produces repeaters upon receipt of an order as the individual requirements are very different. It is not possible to meet all specifications with one device. When does it make sense to use a repeater? Quite often the transmission power of A/V monitoring units is too low to guarantee large transmission range. This increases the risk of discovery of receivers nearby. The repeater (portfolio size and well camouflaged) receives the A/V signal and transmits it over a distance of several kilometres. It not only enlarges the distance considerably but also helps to overcome constructional and geographic barriers. Suitable for mobile as well as for stationary use.


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• Focuses on the key Super League In - Depth Analysis - BNP Paribas Wealth Management manufacturers, to study the capacity, production, value, market share and development plans in future.


New result was generated from system server. Free download on the result of Separuhku Nano as a video MP3 or audio/song.
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text:
Start-up to help people sell unwanted MP3s
Bopaboo has an online marketplace that enables consumers to sell their unwanted music files. There is no limit to the number of songs offered, but no DRM-wrapped music is allowed.
Greg Sandoval
December 15, 2008 12:47 PM PST
This blog was updated first at 12:30 p.m. PST with some reader feedback and then at 1:40 p.m. PST to include an interview with Bopaboo's CEO.
"Stop illegally sharing, and start legally selling" is the tagline for a start-up that wants to enable music owners to sell their unwanted MP3s.
Ernesto at the blog TorrentFreak has a story about Bopaboo, which has created a digital marketplace where users operate mini download stores.
Bopaboo buyers can search for music in all the usual ways, and the site offers a seller rating to help shoppers learn a merchant's reputation. Sellers register and then are given an MP3 store, where they can upload the music they want to sell. No DRM-wrapped music is allowed, so iTunes owners like me are blocked from selling.
According to TorrentFreak, there is no limit on the number of songs that can be offered. Sellers select their own prices but, of course, must cut Bopaboo a percentage of sales.
This was an idea bound to emerge out of the craze over songs stripped of digital rights management software. People have traded DVDs, mixed tapes, and albums for decades.
Bopaboo says the site is legal, and it sounds like it should be. Don't people own their MP3s? We'll see.
UPDATE: 12:30 p.m. PT: Several of you readers have pointed out that it's going to be tough, if not impossible, to prevent people from reselling songs over and over. This sounds like a fight in the making with the music industry. I've got a few calls in, so I'll further update this post with any objections from the Recording Industry Association of America.
UPDATE: 1:40 p.m. PT: Okay, here's the juice from Bopaboo CEO Alex Meshkin, a 28-year-old who didn't go to college, doesn't know programming but once ran Toyota's Nascar team. Yes, Nascar team. He added that Bopaboo is a Washington D.C.-based company founded a year ago.
I put the most important question to him first.
Q: Why don't you think the recording industry is going to sue you into oblivion?
Meshkin: Obviously, MP3s are very easy to duplicate...It's very difficult to tell the difference between a so-called new copy and a so-called old copy...I can buy a CD and I can rip it and that behavior has basically been endorsed by the music industry. I can resell that CD on Amazon. The industry doesn't have a way to monetize physical goods being traded on the secondary market. The first-sale doctrine protects that right. In the physical world consumers have the right to resell their property and copyright owners can't do anything about that."
Okay, but there's no way that the four largest music labels are going to sit back and let you enable people to sell multiple copies of the same song. Isn't that what people will be able to do using your digital marketplace?
Meshkin: We have the technology in place to prevent you from selling a song more than one time...We take a digital fingerprint through every upload that prevents a user from uploading to our service a track more than one time. Actually we've come up with an algorithm, which is beyond what even (digital filtering company) Gracenote does with song identification.
Have you spoken to the big recording companies or the RIAA?
Talks have sped up...I will be in New York later this week for meetings. The talks so far have been positive and the labels are inclined to look at alternative business models to monetize.
Discuss: Start-up to help people sell unwanted MP3s


text:
Health Officials Optimistic Utah Can Prevent Another Holiday Surge
Dec 23, 2020, 6:50 PM | Updated: 9:13 pm
BY JED BOAL, KSL TV
MURRAY, Utah – Doctors said the people of Utah avoided a serious surge in COVID-19 cases after Thanksgiving, and they hope Utahns will take the same precautions in the days surrounding Christmas and New Year's.
As the long Christmas weekend approached, Dr. Eddie Stenehjem, an infectious disease physician at Intermountain Healthcare, gave Utahns reason to be optimistic just before Christmas but only if people continue to be vigilant about avoiding the virus.
"We're in a nice kind of downward trend of our case counts," said Stenehjem, who cited the data.
Testing, however, has not returned to the levels before Thanksgiving, and the COVID-19 positivity rate Wednesday was 27%.
"That's still really, really high," Stenehjem said. "We really want that to be less than 5%."
That positivity rate indicated to him that testing is missing a lot of active cases. The high number of people in the ICU and dying people also concerned him, but those numbers will drop if case counts continue to decline.
"Fingers crossed that Utah comes together as a community like we did for Thanksgiving," he said. "We didn't see that big surge we were expecting because Utahns took this seriously. I'm hoping we'll see the same thing out of Christmas."
Also encouraging was the ongoing distribution of vaccines. So far, 6,000 Intermountain caregivers have been vaccinated — with nearly twice that number statewide — lightening the mood in the hospital.
"It's a feeling of safety for our caregivers. So, that has helped us tremendously," Stenehjem said.
Pfizer To Supply US With Additional 100M Doses Of Vaccine
24 More Have Died From COVID-19 With Christmas Just Days Away
'Mom's Worth It': US Holiday Travel Surges Despite Outbreak
4-Year-Old Getting Therapy Dog Thanks To Amazing Goldens In Utah County
At the same time, he said, "It is a tough time to be in the hospital. It's a really hard time to both work in the hospital and be a patient in the hospital over the holidays." Especially when they are not allowed to have visitors.
He's encouraging those families to connect with their loved ones in the hospital through FaceTime and to engage as much as possible.
"We know this is a really hard time on people, not just the patients in the hospital, but also their family members," Stenehjem said.
It's also hard for caregivers to be away from their families, caring for high-risk patients.
He said the greatest gift we can give the caregivers is to set a good example by masking up and taking precautions to protect the people around us.


Beyoncé's partnership with Nintendo goes beyond these downloads. She will also be featured in Style Savvy commercials, which will begin airing on Nickelodeon the week prior to this year's Nickelodeon Kids' Choice Awards and during the event on March 27. Beyoncé previously starred in commercials for Nintendo's Rhythm Heaven game.
The new Deréon downloads mark the second time real-life fashions have made their way into the game. Players have also had access to Charlotte Ronson Designer Downloads, which lets players download items from Ronson's fall 2009 line directly into the game.
Style Savvy invites players of any age to express their individual sense of style by managing a virtual clothing boutique. Choosing from more than 10,000 items in a wide variety of styles, designer labels and price points, players can create distinctive looks to satisfy customers with various tastes and budget constraints. Using the simple controls of a Nintendo DS system, players can enjoy Style Savvy anytime, anywhere.


Procurement focuses on security of supply, financial contributions and qualitative requirements while acting in accordance with ethical, environmental, social, and economic principles. These include Bayer's LIFE values and Bayer's commitment to compliance.
Compliance enables us to translate our mission into good business results and is therefore a crucial success factor within Procurement on which we cannot compromise. We expect all our suppliers to conduct themselves in compliance with applicable laws and regulations. Procurement abides by the Corporate Compliance Policy in addition to local legal requirements, contractual obligations and corporate regulations. Focusing on compliant behavior at an early stage of the business, enables us to operate successfully and sustainably.
Through our procurement volume we exert considerable influence on society and the environment in many regions. In 2018, goods and services were procured from 101,188 suppliers in 153 countries for approximately €17.1 billion and recorded in the Group-wide reporting system.


The fiber optic distribution box is widely used as a termination point for the feeder cable to connect with drop cable.The fiber optic cables, splicing, splitting, distribution can be implemented in this box, and meanwhile it provides solid protection and management for the FTTx network building.
2. The material of this box is PC + ABS which can prove more than 10 years working life and wet-proof, water-proof, dust-proof, anti-aging and the protection level is up to IP65.
3. Clamping for feeder cable and drop cable, fiber splicing, fixation, storage, distribution in one.
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6. Cabinet can be installed by the way of wall-mounted suitable for both indoor and outdoor uses.
The grounding device is isolated with the cabinet, isolation resistance is less than 2x104MΩ/500V(DC); Insulation resistance≥2x104MΩ/500V.
The withstand voltage between grounding device and cabinet is no less than 3000V (DC)/min, no puncture, no flash-over, U≥3000V.


text:
BEF Website
Benicia Reads
Benicia High School
Benicia Middle School
Joe Henderson Elementary
Mary Farmar Elementary
Matthew Turner Elementary
Robert Semple Elementary
Maintenance, Operations, & Facilities
Preschool & Child Development
Free & reduced Meal
Open Enrollment & Overflow
Inter District Transfers
Online Registration for Students
What School Will My Child Attend
Attendance / Truancy
Uniform Complaints
Williams Uniform Complaint
Parents Rights & Responsibilities
Wellness Committee website
Site Single Plans (SPSA)
Governing Board Information
Parent Complaints / Forms
Power School Parent
Title IX / Educational Equality
BUSD WebMail
BUSD Van Reservation
Negotiation News
Power School Teacher
Power School Admin
New Technology Director
Jan 9, 2017 | Latest News
I am excited to announce that we have a new Director of Technology and Innovation. Mr. Sebastiano Padoan comes to us with high recommendations from San Ramon Valley Unified School District.
Mr. Padoan earned his undergraduate degree from Cal State University of the East Bay in Applied Mathematics and Computer Science. Mr. Padoan studied and earned a degree in Applied Technology at the Institute of Technology, Venice, Italy. Mr. Padoan began his twelve-year career in education as a Systems Assistant at San Ramon Unified where he rose up the ranks from Systems Assistant to Network Systems Specialist to Supervisor of Technical Support.
Mr. Padoan has a broad skill set in network and device configuration and support, network security, communications systems, education and operating systems and data center operations. He is skilled in the configuration, installation, maintenance, troubleshooting and repair of technology hardware and software. He is equally adept at assisting teachers and students with applications and software.
He comes to Benicia Unified with high recommendations from his previous Superintendent and Supervisor.
"After 17 years at San Ramon Valley Unified, I am thrilled to become part of the team here at Benicia. I look forward to the many exciting innovations, challenges, and opportunities that await me as director of technology and innovation," said Sebastiano.
Update from Superintendent Dr. Young 1/15/2021
Board Meeting 1/14/2021
Update from Superintendent Dr. Young 1/8/2021
Summary for the Upcoming Regular Board Meeting of January 14, 2021
Free meals for all children thru end of school year 2021
ADA Accessible Website Initiative
The Benicia Unified School District is committed to providing accessibility for all. If you experience difficulty with the accessibility of any web pages or documents on our site, please contact us to request this information in an alternative format.
Title IX Page
District programs, activities, and practices shall be free from discrimination, harassment, intimidation, bullying against an individual or group based on race, color, ancestry, nationality, national origin, immigration status, ethnic group identification, ethnicity, age, religion, marital status, pregnancy, false pregnancy, termination of pregnancy, or related recovery, parental status, physical or mental disability, sex, sexual orientation, gender, gender identity, gender expression, or genetic information; a perception of one or more of such characteristics; or association with a person or group with one or more of these actual or perceived characteristics.
The Title IX coordinator for BUSD is the Assistant Superintendent, Dr. Khushwinder Gill. She can be reached at 707-747-8300 or by emailing TitleIX@beniciaunified.org.
Copyright © 2015 - 2021 Benicia Unified School District | 350 East K Street, Benicia, CA 94510 | (707) 747-8300 Phone | (707) 748-0146 Fax | | Title IX Page
BUSD will begin the 2020-21 School Year (Aug. 17) with Full Virtual Learning. In an effort to adhere to the current Solano County Public Health orders of social distancing we will continue to conduct business electronically as much as possible.Click Here for The Latest Info


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The spot will light up once selected and, to teleport, you can pinch your thumb and index finger together at the same time on each hand. The first things you should seek out with Quest are animated short stories sprouting from the early days of Oculus. This might be a given, but for some users that have spent time in VR already they might be inclined to skip these experiences that likely come preloaded on your device. Not only are they great introductions to things like the Guardian system and roomscale wireless VR, but they're just damn good apps in and of themselves. So we've rounded up a list of the 11 best free games and experiences to play and try first on your Oculus Quest or Oculus Quest 2.
As you're probably already aware, Pavlov VR is a realistic first-person multiplayer shooter and it's unlike anything we currently have to play on the quest store. But it's not just people you will be playing as in our quest games! Some of our quests involve our four-legged or finned friends as well. Help the dolphin find his friends and get across the dangerous lagoon in Lagoon Quest, or help Harry the Hamster collect all the pieces of his golden wheel in The Quest for the Golden Wheel.
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The studio is dedicated to creating cinematic virtual reality experiences. Project Terminus is more of an alpha sample experience than a full-on game. But if you enjoy horror survival games, then project terminus is a tremendous short experience to sideload on your oculus quest. Instead of using a joystick or other similar button to move around, you need to use your arms to move through the environment. This mechanic is difficult at first, but once you get used to it, it becomes second nature.
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package org.springframework.security.ldap.authentication;

import org.apache.commons.logging.Log;
import org.apache.commons.logging.LogFactory;
import org.springframework.ldap.NameNotFoundException;
import org.springframework.ldap.NamingException;
import org.springframework.ldap.core.DirContextOperations;
import org.springframework.ldap.core.support.BaseLdapPathContextSource;
import org.springframework.security.authentication.BadCredentialsException;
import org.springframework.security.authentication.UsernamePasswordAuthenticationToken;
import org.springframework.security.authentication.encoding.LdapShaPasswordEncoder;
import org.springframework.security.authentication.encoding.PasswordEncoder;
import org.springframework.security.core.Authentication;
import org.springframework.security.core.userdetails.UsernameNotFoundException;
import org.springframework.security.crypto.codec.Utf8;
import org.springframework.security.ldap.SpringSecurityLdapTemplate;
import org.springframework.util.Assert;

/**
 * An {@link org.springframework.security.ldap.authentication.LdapAuthenticator
 * LdapAuthenticator} which compares the login password with the value stored in the
 * directory using a remote LDAP "compare" operation.
 *
 * <p>
 * If passwords are stored in digest form in the repository, then a suitable
 * {@link PasswordEncoder} implementation must be supplied. By default, passwords are
 * encoded using the {@link LdapShaPasswordEncoder}. Note that compare operations will not
 * work if salted-SHA (SSHA) passwords are used, as it is not possible to know the salt
 * value which is a random byte sequence generated by the directory.
 *
 * @author Luke Taylor
 */
public final class PasswordComparisonAuthenticator extends AbstractLdapAuthenticator {
	// ~ Static fields/initializers
	// =====================================================================================

	private static final Log logger = LogFactory
			.getLog(PasswordComparisonAuthenticator.class);

	// ~ Instance fields
	// ================================================================================================

	private PasswordEncoder passwordEncoder = new LdapShaPasswordEncoder();
	private String passwordAttributeName = "userPassword";
	private boolean usePasswordAttrCompare = false;

	// ~ Constructors
	// ===================================================================================================

	public PasswordComparisonAuthenticator(BaseLdapPathContextSource contextSource) {
		super(contextSource);
	}

	// ~ Methods
	// ========================================================================================================

	public DirContextOperations authenticate(final Authentication authentication) {
		Assert.isInstanceOf(UsernamePasswordAuthenticationToken.class, authentication,
				"Can only process UsernamePasswordAuthenticationToken objects");
		// locate the user and check the password

		DirContextOperations user = null;
		String username = authentication.getName();
		String password = (String) authentication.getCredentials();

		SpringSecurityLdapTemplate ldapTemplate = new SpringSecurityLdapTemplate(
				getContextSource());

		for (String userDn : getUserDns(username)) {
			try {
				user = ldapTemplate.retrieveEntry(userDn, getUserAttributes());
			}
			catch (NameNotFoundException ignore) {
			}
			if (user != null) {
				break;
			}
		}

		if (user == null && getUserSearch() != null) {
			user = getUserSearch().searchForUser(username);
		}

		if (user == null) {
			throw new UsernameNotFoundException("User not found: " + username);
		}

		if (logger.isDebugEnabled()) {
			logger.debug("Performing LDAP compare of password attribute '"
					+ passwordAttributeName + "' for user '" + user.getDn() + "'");
		}

		if (usePasswordAttrCompare && isPasswordAttrCompare(user, password)) {
			return user;
		}
		else if (isLdapPasswordCompare(user, ldapTemplate, password)) {
			return user;
		}
		throw new BadCredentialsException(messages.getMessage(
				"PasswordComparisonAuthenticator.badCredentials", "Bad credentials"));
	}

	private boolean isPasswordAttrCompare(DirContextOperations user, String password) {
		Object passwordAttrValue = user.getObjectAttribute(passwordAttributeName);
		return passwordEncoder.isPasswordValid(new String((byte[]) passwordAttrValue),
				password, null);
	}

	private boolean isLdapPasswordCompare(DirContextOperations user,
			SpringSecurityLdapTemplate ldapTemplate, String password) {
		String encodedPassword = passwordEncoder.encodePassword(password, null);
		byte[] passwordBytes = Utf8.encode(encodedPassword);
		return ldapTemplate.compare(user.getDn().toString(), passwordAttributeName,
				passwordBytes);
	}

	public void setPasswordAttributeName(String passwordAttribute) {
		Assert.hasLength(passwordAttribute,
				"passwordAttributeName must not be empty or null");
		this.passwordAttributeName = passwordAttribute;
	}

	private void setPasswordEncoder(PasswordEncoder passwordEncoder) {
		Assert.notNull(passwordEncoder, "passwordEncoder must not be null.");
		this.passwordEncoder = passwordEncoder;
	}

	public void setPasswordEncoder(Object passwordEncoder) {
		if (passwordEncoder instanceof PasswordEncoder) {
			this.usePasswordAttrCompare = false;
			setPasswordEncoder((PasswordEncoder) passwordEncoder);
			return;
		}

		if (passwordEncoder instanceof org.springframework.security.crypto.password.PasswordEncoder) {
			final org.springframework.security.crypto.password.PasswordEncoder delegate = (org.springframework.security.crypto.password.PasswordEncoder) passwordEncoder;
			setPasswordEncoder(new PasswordEncoder() {
				public String encodePassword(String rawPass, Object salt) {
					checkSalt(salt);
					return delegate.encode(rawPass);
				}

				public boolean isPasswordValid(String encPass, String rawPass, Object salt) {
					checkSalt(salt);
					return delegate.matches(rawPass, encPass);
				}

				private void checkSalt(Object salt) {
					Assert.isNull(salt,
							"Salt value must be null when used with crypto module PasswordEncoder");
				}
			});
			this.usePasswordAttrCompare = true;
			return;
		}

		throw new IllegalArgumentException(
				"passwordEncoder must be a PasswordEncoder instance");
	}
}


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ULA machinists go on strike
by Jeff Foust — May 7, 2018
Machinists demonstrate outside Cape Canaveral Air Force Station, Florida. ULA says the strike won't affect its operations, including preparations for the July 31 launch of NASA's Parker Solar Probe on a Delta 4 Heavy. Credit: Craig Vander Galien
Updated 2:15 p.m. Eastern.
WASHINGTON — Nearly 600 United Launch Alliance workers at three sites voted May 6 to reject the company's latest contract and go on strike.
Machinists who are members of the International Association of Machinists and Aerospace Workers, or IAM, turned down the proposed three-year contract, replacing one that expired May 6, and immediately went on strike. The union announced the strike vote but did not immediately offer additional details, such as the margin of the vote.
The strike affects ULA's main manufacturing facility in Decatur, Alabama, as well as launch sites in Cape Canaveral, Florida, and Vandenberg Air Force Base, California.
Picketing reportedly started immediately after the strike vote. "Be cautious when entering & exiting Cape Canaveral Air Force Station due to union strike activities," the U.S. Air Force's 45th Space Wing, which operates Cape Canaveral, tweeted late May 6. "Anticipate possible delays."
The union's negotiating team had recommended members reject the contract, citing various provisions it deemed unacceptable. Those issues ranged from increased health care costs and contributions to retirement plans to travel pay and procedures. "By voting for this contract, you are letting ULA take away many of your benefits and future money, as well as that of future generations of Machinists," the union advised its members.
"This strike is centered around many issues, but above all has amplified the message that our members want to be valued and respected in the workplace," said Machinists Union International President Robert Martinez Jr. om a May 7 statement. "The full force of 600,000 Machinists Union members across North America stand with our members at ULA."
ULA argued that its proposed contract, which included a ratification bonus of $6,000 and wage increases of 1.5 to 2 percent per year, was "well above market" compared to other aerospace companies.
"We're disappointed that the IAM members rejected ULA's last, best and final offer and voted to strike," Tory Bruno, president and chief executive of ULA, said in a May 6 statement after the strike vote. "We believe our proposed contract is very competitive with other companies. Importantly, ULA's final offer contributes to ULA's long term viability in an increasingly competitive launch business environment."
Prior to the vote, the company said its operations would not be affected by a strike. ULA confirmed that in its May 6 statement, saying it would "implement its strike contingency plans while focusing on meeting its commitments to our customers."
The vote took place a day after ULA's most recent launch, of an Atlas 5 carrying NASA's InSight Mars lander spacecraft. The next ULA mission isn't until the end of July, when a Delta 4 Heavy is scheduled to launch the Parker Solar Probe mission for NASA. The complex trajectory of that mission, which involves a series of Venus flybys to help the spacecraft get closer to the sun, requires that it launch during a window that is open for about three weeks.
Commercial Launch United Launch Alliance


What's this? I think they're called… Blog Posts?
February 21, 2017 February 21, 2017 Joanna V.
I am also a heavy parenthetical user. I offer no mea culpa. Parentheticals are useful.
So where are we? Well, we are 10 months from our next transfer with the US Navy and there have been enough folks expressing an interest in the process that I have agreed (agreed? decided? I'm not sure. I'll go with ENCOURAGED) I have been encouraged to document the process as we go through it. I may (will most likely) have other thoughts about things as we progress and will try to offer as much explanation as I can, with a few caveats.
Caveat 1: I will not provide exact dates for things. If this continues and we get into any kind of military-related timetables, they will not be posted here. These's this thing called Operational Security. It matters.
Caveat 2: I will probably repeat this one a lot, but for the purposes of clarity you should know that my spouse is in the Navy (other military branches have different processes) and Enlisted (Officers in the Navy also have different processes) and active duty (again, reservists have… you get the idea). That being said, this is only my perspective. Yours is probably different. I will regularly remind you that your mileage may vary.
So there we are. I have no additional things to say except that this photo is the most recent one I have of us, taken on the way to brunch on Saturday, and though I didn't notice it until later I'm very entertained by the street signs above our heads. Oh good, now you've seen them, too.
More soon. Probably. Assuming I can put jumbled thoughts into less jumbled paragraphs.
Previous postMy hard drive crashed.


Each year, Florida's pleasing climate and seductive sunshine draws countless Canadian and European travelers looking to escape the long, cold winter. While many rental companies offer special rates for long stays, some also provide planned activities specifically aimed at snowbirds. A diverse array of accommodations, restaurants and attractions for every budget make Florida the ideal place to winter over.
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Siesta Key: Mild winters, exquisite sand, dolphins and shell collecting rank this beach consistently in various critics' top ten list. The quaint village is more affordable than nearby Sarasota, but is close enough to the action should you crave culture.
New Smyrna Beach: Despite its rich history, New Smyrna is relatively free of crowds and the accompanying pollution. Near science museums, art galleries and one of the safest swimming beaches in Florida, this destination definitely has it all.


text:
#5 Stonehill Earns Fifth NCAA Tournament Bid
Head Coach Shelly Morris has led the Skyhawks to their fifth NCAA Tournament bid, the third in her six seasons. (PHOTO BY Emily Reynolds)
Second-seeded Skyhawks will host third-seeded and 7th-ranked Merrimack in NCAA first round on Saturday at W.B. Mason Stadium at noon
INTERACTIVE BRACKET
EASTON, Mass. (November 11, 2013) – Stonehill College, ranked fifth in the most recent Penn Monto/National Field Hockey Coaches Association (NFHCA) Division II poll, has earned its fifth trip to the NCAA Division II Field Hockey Tournament after receiving the No. 2 seed in the North Region and will host a first round matchup at W.B. Mason Stadium on Saturday at noon.
Stonehill (18-3, 8-2 NE-10) will host third-seeded and No. 7-ranked Merrimack College (13-5, 9-1 NE-10) in Saturday's opening round after the Warriors earned their third-straight bid. The Skyhawks are making their third NCAA Tournament appearance under head coach Shelly Morris, and their first since 2010 when they made their second of back-to-back trips to the National Semifinals. It marks the third-straight NCAA bid in which Stonehill will host an NCAA Tournament game at W.B. Mason as it hosted a first round matchup with American International College in 2010 and the 2009 National Championships.
Admission to Saturday's game will be $8.00 for adults and $5.00 for senior citizens and children.
No. 3-ranked LIU Post (20-0, 10-0 NE-10), the Northeast-10 Conference regular season and tournament champions, earned the top seed from the North Region and a bye to the national semifinals which will be played at the USA Field Hockey Training Center at the Virginia Beach SportsPlex in Virginia Beach, Va., on Friday, November 22, at 12:30 p.m.
Top-ranked Shippensburg University (18-1, 10-0 PSAC) earned the top-seed in the South Region and awaits the winner of Saturday's first round matchup between third-seeded and No. 4-ranked West Chester University (14-5, 8-2 PSAC), the two-time defending National Champions, and second-seeded and No. 2-ranked Millersville University (18-2, 8-2 PSAC), the Pennsylvania State Athletic Conference tournament championships, which will be played in Millersville, Pa., at noon. The National Championship game will be played at L.R. Hill Sports Complex on the campus of Old Dominion University in Norfolk, Va., on Sunday, November 24, at 1:15 p.m.
Stonehill has also earned NCAA Tournament bids in 2005, 2006, 2009 and 2010 reaching the national semifinals in 2005, 2009 and 2010. The Skyhawks are a win shy of the single-season program record set with a 19-0 campaign in 2005. Stonehill reached the NE-10 Championship for the fifth time in program history on Sunday, falling to LIU Post, 2-1, in Brookville, N.Y.
Saturday will mark the fourth meeting between Stonehill and Merrimack this season, the third in the last three weeks. The Skyhawks have won two of the three previous meetings this season, including a 3-2 decision in the NE-10 semifinals on Friday, but both wins have come on the road, while the Warriors claimed a 1-0 triumph in their visit to W.B. Mason Stadium on October 29. The Skyhawks are led by sophomore Erika Kelly (Watertown, Mass./Watertown) who has scored a Division II best and NE-10 record 37 goals this season to go with a pair of assists for a program-record 76 points. Her goal total is three shy of the NCAA Division II record of 40 set by Jamie Vanartsdalen from Bloomsburg (Pa.) University in 2008. Senior back Gina Sheehan (Shrewsbury, Mass./Shrewsbury) earned NE-10 Defensive Player of the Year honors to lead the Skyhawks backline.
For the latest on Stonehill Athletics, follow the Skyhawks via social media on Twitter and Facebook. Fans can also download the Stonehill Skyhawks mobile app on iTunes and the Android Market.
2013 NCAA Division II
Field Hockey Championship
#7 Merrimack (13-5, 9-1 NE-10) at #5 Stonehill (18-3, 8-2 NE-10), Noon
#4 West Chester (14-5, 8-2 PSAC) at #2 Millersville (18-2, 8-2 PSAC), Noon
USA Field Hockey Training Center, Virginia Beach, Va.
Merrimack/Stonehill winner vs. #3 LIU Post (20-0, 10-0 NE-10), 12:30 p.m.
West Chester/Millersville winner vs. #1 Shippensburg (18-1, 10-0 PSAC), 3:30 p.m.
L.R. Hill Sports Complex/Old Dominion University, Norfolk, Va.
Semifinals winners, 1:15 p.m.
June 4, 2014 Ingersoll Named Capital One Academic All-American
May 21, 2014 Ingersoll Named NE-10 Woman of the Year Finalist
May 15, 2014 Ingersoll Earns Capital One Academic All-District Honors for Second Straight Year
April 4, 2014 Stonehill Holds Annual Academic Excellence Recognition Breakfast
March 21, 2014 Gouveia Headlines Academic Honors for Stonehill Field Hockey
February 20, 2014 Ingersoll Named NE-10 Scholar-Athlete Sport Excellence Award Winner for Field Hockey
January 31, 2014 Stonehill Places Three on NE-10 Academic All-Conference Field Hockey Team
December 17, 2013 Giordano and Healy Earn Division II All-Rookie Team Honors from SynapseSports.com
December 12, 2013 Stonehill Named Host for 2016 NCAA Division II Field Hockey Championship
December 3, 2013 Four Skyhawks Earn Field Hockey All-America Honors
November 19, 2013 Three Stonehill Seniors Selected to Front Rush/NFHCA Division II North/South Game
November 16, 2013 NE-10 FIRST ROUND: #7 Merrimack Posts Back and Forth Win Over #5 Stonehill, 4-3 (OT)
November 15, 2013 #5 Stonehill Hosts #7 Merrimack with Final Four Bid on the Line
November 13, 2013 BROCKTON ENTERPRISE: "Stonehill opens NCAA field hockey tourney at home"
November 11, 2013 #5 Stonehill Earns Fifth NCAA Tournament Bid
November 10, 2013 NE-10 CHAMPIONSHIP: #3 LIU Post Tops #5 Stonehill for Title, 2-1
November 9, 2013 #5 Stonehill Seeks Second NE-10 Title at #3 LIU Post
November 8, 2013 #5 Stonehill and #7 Merrimack Set to Clash in NE-10 Semifinals
November 8, 2013 NE-10 SEMIFINALS: #5 Stonehill Marches Past #7 Merrimack, 3-2
November 7, 2013 Sheehan Headlines Five All-NE-10 Picks for Stonehill
November 5, 2013 NE-10 FIRST ROUND: #5 Stonehill Strikes Early to Top Southern Connecticut State, 5-0
November 4, 2013 #3 Stonehill Hosts Southern Connecticut State for NE-10 Tournament First Round Matchup
November 2, 2013 #4 LIU Post Tops #3 Stonehill, 4-3 (OT)
October 29, 2013 Genovese Collects NE-10 Goalkeeper of the Week Honors
October 29, 2013 #7 Merrimack Edges #3 Stonehill, 1-0
October 27, 2013 #3 Stonehill Rolls Past Dowling, 7-0
October 24, 2013 Stonehill Tops First NCAA East Regional Ranking of the Season
October 23, 2013 #3 Stonehill Cruises at Franklin Pierce, 6-1
October 21, 2013 BOSTON GLOBE: "Erika Kelly spearheads Stonehill's surge to No. 3 ranking"
October 19, 2013 #3 Stonehill Tops Southern Connecticut in Shootout, 8-4
October 15, 2013 Kelly Earns Third NE-10 Player of the Week Award
October 15, 2013 #3 Stonehill Edges American International, 2-1 (2OT)
October 12, 2013 #4 Stonehill Shuts Out Adelphi, 3-0
October 9, 2013 #4 Stonehill Rallies Again to Top Saint Anselm, 4-3
October 9, 2013 BROCKTON ENTERPRISE: "Stonehill field hockey team off to perfect start"
October 7, 2013 Giordano Named NE-10 Rookie of the Week
October 4, 2013 #4 Stonehill Keeps Rolling Against Assumption, 3-1
October 1, 2013 Kelly Hat Trick Lifts #4 Stonehill Past #9 Bentley, 4-3 (OT)
September 28, 2013 Second Half Surge Sends #4 Stonehill Past Saint Michael's, 5-1
September 25, 2013 #4 Stonehill Defeats Franklin Pierce, 3-1
September 24, 2013 Kelly & Genovese Earn NE-10 Weekly Honors
September 21, 2013 #4 Stonehill Remains Unbeaten at Southern Connecticut, 3-1
September 19, 2013 #4 Stonehill Turns Back #10 American International, 3-1
September 17, 2013 Stonehill Flies Up to #4 in NFHCA Poll
September 17, 2013 #4 Stonehill Shoots Past Merrimack, 3-1
September 14, 2013 #10 Stonehill Fires Past Saint Michael's, 4-2
September 10, 2013 Kelly Named NE-10 Player of the Week
September 8, 2013 #10 Stonehill Rallies Past #9 East Stroudsburg, 6-5
September 7, 2013 #10 Stonehill Downs #3 Bloomsburg, 3-2 (OT)
August 12, 2013 2013 Football Training Camp Schedule


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text:
Q: Integral $\int_0^{\pi/2} \ln(1+\alpha\sin^2 x)\, dx=\pi \ln \frac{1+\sqrt{1+\alpha}}{2}$ $$
I_1:=\int_0^{\pi/2} \ln(1+\alpha\sin^2 x)\, dx=\pi \ln \frac{1+\sqrt{1+\alpha}}{2}, \qquad \alpha \geq -1.
$$
I am trying to prove this integral $I_1$.  We can write
$$
\int_0^{\pi/2} \ln(\alpha(1/\alpha+\sin^2 x))dx=\int_0^{\pi/2} \left(\ln \alpha+\ln (\frac{1}{\alpha}+\sin^2 x)\right)dx=\frac{\pi}{2} \ln \alpha+I_2
$$
where
$$
I_2=\int_0^{\pi/2}\ln (\frac{1}{\alpha}+\sin^2 x) \,dx
$$
however I am not sure what that will do for us....  I also tried differentiating wrt $\alpha$ but didn't get placed.  How can we prove $I_1$ result?  Thanks

A: This is quite similar to Random Variable's solution, just the starting integral is different to make the calculations a bit simpler.
Consider
$$I(b)=\int_0^{\pi/2} \ln(b^2+\sin^2x)\,dx$$
$$\Rightarrow I'(b)=\int_0^{\pi/2} \frac{2b}{b^2+\sin^2x}\,dx=2b\int_0^{\pi/2} \frac{dx}{b^2+\cos^2x}$$
Factor out $\cos^2x$ from the denominator and rewrite $\sec^2x=1+\tan^2x$ to obtain:
$$I'(b)=2b\int_0^{\pi/2} \frac{\sec^2x\,dx}{b^2+1+b^2\tan^2x}\,dx$$
Use the substitution $\tan x=t$ and evaluating the resulting integral is easy so
$$I'(b)=\frac{\pi}{\sqrt{1+b^2}} \Rightarrow I(b)=\pi\ln\left(b+\sqrt{1+b^2}\right)+C$$
For $b=0$, $I(0)=-\pi\ln 2$, hence $C=-\pi\ln2$
$$\Rightarrow \int_0^{\pi/2} \ln(b^2+\sin^2x)\,dx=\pi\ln\left(\frac{b+\sqrt{1+b^2}}{2}\right)$$
Replace $b$ with $1/\sqrt{\alpha}$ and you get:
$$\int_0^{\pi/2} \ln(1+\alpha \sin^2x)\,dx-\frac{\pi}{2}\ln \alpha=\pi\ln\left(\frac{1+\sqrt{1+\alpha}}{2\sqrt{\alpha}}\right)$$
$$\Rightarrow \int_0^{\pi/2} \ln(1+\alpha \sin^2x)\,dx=\pi\ln\left(\frac{1+\sqrt{1+\alpha}}{2}\right)$$
$\blacksquare$

A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin^{2}\pars{x}}\,\dd x
     =\pi\,\ln\pars{1 + \root{1 + \alpha} \over 2}:\ {\large ?}}$

\begin{align}&
\partiald{}{\alpha}\color{#c00000}{\int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin^{2}\pars{x}}\,\dd x}
=\partiald{}{\alpha}\int_{0}^{\pi/2}\ln\pars{1 + \alpha\cos^{2}\pars{x}}\,\dd x
\\[3mm]&=\int_{0}^{\pi/2}{\cos^{2}\pars{x} \over 1 + \alpha\cos^{2}\pars{x}}\,\dd x
={1 \over \alpha}\int_{0}^{\pi/2}
{\bracks{1 + \alpha\cos^{2}\pars{x}} - 1 \over 1 + \alpha\cos^{2}\pars{x}}\,\dd x
\\[3mm]&={\pi \over 2\alpha}-
{1 \over \alpha}\int_{0}^{\pi/2}{\dd x \over 1 + \alpha\cos^{2}\pars{x}}
={\pi \over 2\alpha}-
{1 \over \alpha}\int_{0}^{\pi/2}{\sec^{2}\pars{x}\,\dd x\over
\tan^{2}\pars{x} + 1 + \alpha}
\\[3mm]&={\pi \over 2\alpha} - {1 \over \alpha\root{1 + \alpha}}
\int_{0}^{\infty}{\dd t \over t^{2} + 1}
={\pi \over 2}\pars{{1 \over \alpha} - {1 \over \alpha\root{1 + \alpha}}}
\end{align}

\begin{align}&
\color{#c00000}{\int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin^{2}\pars{x}}\,\dd x}
={\pi \over 2}\
\overbrace{\int_{0}^{\alpha}\pars{{1 \over t} - {1 \over t\root{1 + t}}}\,\dd t}
^{\ds{\mbox{Set}\ x \equiv 1 + \root{1 + t}\ \imp\ t = x^{2} - 2x}}
\\[3mm]&={\pi \over 2}\int_{2}^{1 + \root{1 + a}}{2\,\dd x \over x}
\end{align}

$$\color{#66f}{\large%
\int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin^{2}\pars{x}}\,\dd x
=\pi\,\ln\pars{1 + \root{1 + \alpha} \over 2}}
$$

A: Differentiate $I_2$ (probably easier than $I_1$) with respect to $a$, then use the Weierstrass substitution to transform it into an integral that you can calculate with residues. I will look into it as well.
Edit: you can also integrate by parts to get rid of the log.

A: Let $ \displaystyle I(a) =  \int_{0}^{\pi /2} \ln(1+ a \sin^{2}x) \, dx$.
Then differentiating under the integral sign, $$I'(a) = \int_{0}^{\pi /2} \frac{\sin^{2} x}{1+a \sin^{2} x} \, dx = \int_{0}^{\pi /2} \frac{1}{a+ \csc^{2} x} \, dx .$$
Now let $u = \cot x$.
Then
$$ \begin{align} I'(a) &= \int_{0}^{\infty} \frac{1}{a+1+u^{2}} \frac{1}{1+u^{2}} \, du \\ &= \frac{1}{a} \int_{0}^{\infty} \left(\frac{1}{1+u^{2}} - \frac{1}{1+a+u^{2}} \right) \, du \\ &= \frac{1}{a} \left(\frac{\pi}{2} - \frac{1}{1+a} \int_{0}^{\infty} \frac{1}{1+\frac{u^{2}}{1+a}} \, du \right)  \\ &=\frac{1}{a} \left(\frac{\pi}{2} - \frac{1}{\sqrt{1+a}} \int_{0}^{\infty} \frac{1}{1+v^{2}} \, dv \right)  \\  &= \frac{\pi}{2a} \left(1 - \frac{1}{\sqrt{1+a}} \right). \end{align}$$
Then integrating back,
$$ \begin{align} I(a) &= \frac{\pi}{2} \int \frac{1}{a} \left(1 - \frac{1}{\sqrt{1+a}} \right) \, da \\ &= \frac{\pi}{2} \int \frac{1}{u^{2}-1} \left(1 - \frac{1}{u} \right) 2u \, du \\ &= \pi \int \frac{1}{1+u} \, du \\ &= \pi \ln \left(1+ \sqrt{1+a} \right) + C. \end{align}$$
And since $I(0) = 0$, $C = -\pi \ln 2$.
Therefore,
$$I(a) = \pi \ln \left(\frac{1 +\sqrt{1+a}}{2} \right) .$$

A: Using Feynman's Technique Integration, we first differentiate the integral w.r.t. $\alpha$ and obtain
$$
\begin{aligned}
I^{\prime}(\alpha) &=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{1+\alpha \sin ^{2} x} d x \\
&=\frac{1}{\alpha} \int_{0}^{\frac{\pi}{2}} \frac{1+\alpha \sin ^{2} x-1}{1+\alpha \sin ^{2} x} d x \\
&=\frac{\pi}{2 \alpha}-\frac{1}{\alpha} \int_{0}^{\frac{\pi}{2}} \frac{d x}{1+\alpha \sin ^{2} x} \\
&=\frac{\pi}{2 \alpha}-\frac{1}{\alpha} \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\sec ^{2} x+\alpha \tan ^{2} x} d x \\
&=\frac{\pi}{2 \alpha}-\frac{1}{\alpha} \int_{0}^{\infty} \frac{d t}{1+(1+\alpha) t^{2}}, \text { where } t=\tan x \\
&=\frac{\pi}{2 \alpha}-\frac{1}{\alpha \sqrt{1+\alpha}}\left[\tan ^{-1}(\sqrt{1+\alpha }t)\right]_{0}^{\infty} \\
&=\frac{\pi}{2 \alpha}-\frac{\pi}{2 \alpha \sqrt{1+\alpha}}
\end{aligned}
$$
Integrating give back the integal
$$
\begin{aligned}
I(\alpha) &=\frac{\pi}{2} \ln |\alpha|-\frac{\pi}{2} \int \frac{d \alpha}{\alpha \sqrt{1+\alpha}} \\
&=\frac{\pi}{2} \ln |\alpha|-\pi\int \frac{d \sqrt{1+\alpha}}{(\sqrt{1+\alpha})^{2}-1}\\
&=\frac{\pi}{2}\left[\ln |\alpha|+\ln \left(\frac{\sqrt{1+\alpha}+1}{\sqrt{1+\alpha}-1}\right) \right ]+C\\
&=\frac{\pi}{2}\left[\ln |\alpha|+\ln \left(\frac{\sqrt{1+\alpha}+1)^{2}}{\mid \alpha \mid}\right) \right]+C\\
&=\pi \ln (\sqrt{1+\alpha}+1)+C
\end{aligned}
$$As $I(0)=0$ gives the value $C=-\pi\ln2 $ and hence we can conclude that $$ I(\alpha) =\pi \ln (1+\sqrt{1+\alpha})-\pi\ln 2 =\pi \ln \left(\frac{1+\sqrt{1+\alpha}}{2}\right) .$$

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The Canada Council for the Arts, the UNB Department of English, The Fiddlehead, and the UNB Bookstore are pleased to invite you to a reading by Giller Prize-winning writer Sean Michaels. The event will take place at 8 p.m. on Wednesday, March 23, at the Alumni Lounge, in the Alumni Memorial Building, on the UNB Fredericton campus.
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Sean Michaels was born in Scotland, raised in Ottawa, and eventually settled in Montreal, where he founded Said the Gramophone, one of the earliest music blogs. He has since spent time in Edinburgh and Kraków, written for The Guardian and McSweeney's, toured with rock bands, and received two National Magazine Awards.


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