text:
\section{Proof of Theorem \ref{TeoWaveBeam}}\label{sec:Proofthm1}

The goal of this section is to prove Theorem \ref{TeoWaveBeam}. The key point of the proof is to construct symbolic dynamics (an infinite symbols Smale horseshoe) for the resonant model \eqref{calham0} which has been derived from the equations \eqref{Hartree}, \eqref{Wave}, \eqref{Beam}. In Theorem \ref{thm:SplittingHomo2Rect} we have constructed transverse homoclinic orbits to saddles for \eqref{calham0}. It is well known that the intersection of invariant manifolds of critical points in flows do not always lead to the existence of symbolic dynamics (see, for instance, \cite{Devaney78}).
Therefore, the first step of the proof is to

obtain transverse homoclinic points to certain periodic orbits. This is done in Section \ref{sec:Thm1:MelnikovPO}. Then, following \cite{Moser01}, in Section \ref{sec:Thm1:SymbDyn} we construct an invariant set of (a suitable Poincar\'e map of) the flow associated to the Hamiltonian \eqref{calham0} whose dynamics is conjugated to a shift of infinite symbols (see Section \ref{sec:heuristics}). Finally in Section \ref{sec:ProofThm1:Application} we complete the proofs of Theorem  \ref{TeoWaveBeam} by checking that the non-degeneracy conditions imposed on \eqref{calham0} are satisfied for the resonant models obtained from the PDEs \eqref{Wave}, \eqref{Hartree} and \eqref{Beam}.

\subsection{Transversality of invariant manifolds of periodic orbits}\label{sec:Thm1:MelnikovPO}
The main result in this section is the following.

\begin{prop}\label{prop:SplittingPO2Rect}
Consider the Hamiltonian \eqref{calham0} and assume that \eqref{tildekappas} holds.
Then there exists $\varepsilon_0>0$ such that for all $\varepsilon\in (0, \varepsilon_0)$ there exists $h_0=h_0(\varepsilon)>0$ such that for all $h\in (0, h_0)$,
\begin{enumerate}

\item[$(i)$] The hyperbolic  periodic orbits $\mathtt{P}^{\pm, 0 }_h$ in \eqref{def:PO:ext}
persist and have period $\mathtt{T}_h$. That is, Hamiltonian \eqref{calham0} has
hyperbolic periodic orbits  $\mathtt{P}^{\pm, 0 }_{h,\eps}$ that are
$\mathcal{O}(\eps)$--close to $\mathtt{P}^{\pm, 0}_h$.
\item[$(ii)$] The invariant manifolds   $W^u(\mathtt{P}^{+, 0 }_{h,\eps})$ and
$W^s(\mathtt{P}^{-, 0 }_{h,\eps})$ intersect transversally along orbits  (within the
energy level).
\end{enumerate}
\end{prop}

Note that in the coordinates introduced in \eqref{def:cartesian}, the periodic orbits $\mathtt{P}^{+, 0 }_{h,\eps}$ and $\mathtt{P}^{-, 0 }_{h,\eps}$ blow down to the same periodic orbit, which we denote by $\mathtt{P}^{0 }_{h,\eps}$. In the coordinates  \eqref{def:cartesian}, Proposition \ref{prop:SplittingPO2Rect} can be restated as that the  manifolds $W^u(\mathtt{P}^{0 }_{h,\eps})$ and $W^s(\mathtt{P}^{0 }_{h,\eps})$ intersect transversally
within the energy level.

\begin{proof}[Proof of Proposition \ref{prop:SplittingPO2Rect}]
To prove $(i)$ is more convenient to use the cartesian coordinates $\{x_j,y_j\}$ in \eqref{def:cartesian} and therefore Hamiltonian $\widetilde H_\Res$ in \eqref{redhamNCartesian} (with $N=2$) to avoid the blow up of $K_j=0$. Then, the invariant subspace $\{K_2=0\}$ corresponds to $\{x_2=y_2=0\}$. The Hamiltonian on this invariant subspace is given by
\begin{equation*}

\begin{aligned}
\mathcal{H}( x_1, 0,y_1, 0)=& \frac{1}{2}\left(3x_1^2-y_1^2\right) - \frac{1}{4}
\left(3x_1^2-y_1^2\right)\left(x_1^2+y_1^2\right)\\
&+\varepsilon\Big[ \frac{1}{2} a_{1} \left(x_1^2+y_1^2\right)
+\frac{1}{4} b_{1} \left(x_1^2+y_1^2\right)^2  +\frac{1}{4}c_1
\left(x_1^2+y_1^2\right)\left(2-x_1^2-y_1^2\right)\Big].
\end{aligned}

\end{equation*}

This Hamiltonian is integrable both for $\eps=0$ and  $\varepsilon>0$ and has the saddle $(0,0)$ at the energy level. Integrability and the particular form of $\mathcal{H}$ implies that the energy levels close to zero are given by periodic orbits. These periodic orbits are $\eps$-close to those of the unperturbed problem

(see \eqref{calham0}).

To  prove $(ii)$ we proceed as in Section \ref{sec:resonantmodel} by doing approximations of several Melnikov functions and using an auxiliary parameter $\de$. We follow the notation of Section \ref{sec:HomoMelnikov}, In particular,  we consider the Hamiltonians $\mathbf{H}_0$, $\mathbf{H}_1$ in \eqref{bfham0}, which taking $\de=\eps$ also define the Hamiltonian $\mathcal{H}$.

By the particular form of  Hamiltonian $\mathbf{H}_0^{(j)}$,
$j=1,2$ (see \eqref{bfham0}, it can be easily checked that it  has the saddles

$(\pm \Psi_*,0)$ (they correspond to $x_1=y_1=0$ in the blow down coordinates \eqref{def:cartesian}. These saddles are connected by the homoclinic orbits $\gamma_0^{(j)}$, $j=1, 2$, introduced in \eqref{gammazero}.

Let $h>0$ small, then the Hamiltonian $\mathbf{H}_0$ possesses

the hyperbolic periodic orbits

\[

\mathtt{P}^{\pm,0}_{\delta,h} =\left\{\big(\gamma_{\delta,h}^{(1)}(\tau),\pm \Psi_*,0 \big) : \tau\in\mathbb{R}\right\},

\]
where $\gamma_{\delta,h}^{(k)}$ is the time parametrization of the periodic orbit defined by
$\{\mathbf{H}_0^{(k)}=h\}$ (see Figure \ref{fig:PhaseSpaceHomoclinic}).

When $\varepsilon = 0$, the homoclinic manifold
$W^u_0(\mathtt{P}_{\delta,h}^{+,0})\equiv W^s_0(\mathtt{P}_{\delta,h}^{-,0})$ is
parameterized by
\begin{equation}\label{man}

\Gamma_{\delta,h,0}(\vec{\tau}) := (\gamma_{\delta,h}^{(1)}(\tau_1),\gamma_{0}^{(2)}(\tau_2)).

\end{equation}

\begin{remark}\label{rem:uniformconv}
The periodic orbits $\mathtt{P}^{\pm,0}_{\delta,h}$ converge pointwise for any fixed $\tau$  to $(\gamma_0^{(1)}(\tau),\pm \Psi_*,0)$ as $h\to 0$.
Similarly, for fixed $\tau$, the parametrization $\Gamma_{\delta,h,0}(\vec{\tau})$ converges
to $\gamma_0(\vec{\tau})$ in \eqref{gammazero} as $h\to 0$.

\end{remark}
 When $\varepsilon>0$, the periodic orbits  $\mathtt{P}_{\delta,h}^{\pm,0}$ persist
 .

Direct application of Melnikov Theory, as in Section
\ref{sec:resonantmodel}, ensures the following. There exists $\de_0>0$,
$\eps_0>0$ small enough such that for any $\de\in(0,\de_0)$ and $\eps\in
(0,\eps_0)$ small enough, the distance function between the invariant manifolds
$W^u(\mathtt{P}^{+, 0 }_{h,\eps})$ and $W^s(\mathtt{P}^{-, 0 }_{h,\eps})$ in a well
chosen transversal section is given by
\[
 d(\vec{\tau})=\eps
\frac{\mathcal{M}_h(\vec{\tau})}{\lVert\nabla\mathbf{H}_0^{(2)}(\gamma_{0}^{(2)}
(\tau_2)) \rVert}+\OO\left(\eps^2\right),
\]
where the error $\OO(\eps^2)$ is uniform in $\de$ and $h$ and $\mathcal{M}_h$ is the
Melnikov function given by
\begin{equation}\label{melfunz2}
\mathcal{M}_h(\tau_1,\tau_2)=\int_\mathbb{\R}\left\{\mathbf{H}_0^{(2)},\mathbf{H}
_1\right\} (\Gamma_{\delta,h,0}(\tau_1+t,\tau_2+t))dt.
\end{equation}
We note that as $t\to\pm\infty$ the $K_2$-component of $\Gamma_{\delta,h,0}(\tau_1+t,\tau_2+t)$ goes exponentially fast to zero, then by a direct computation it is easy to see that
\[
\lim_{\tau_2 \to \pm\infty} \{ \mathbf{H}_0^{(2)}, \mathbf{H}_1 \}(\gamma_{\delta,h}^{(1)}(\tau_1),\gamma_{0}^{(2)}(\tau_2))=0
\]
with exponentially fast convergence and \eqref{melfunz2} is well defined.
To obtain the non-degeneracy of the zeros of the Melnikov function, we compare
\eqref{melfunz2} to the Melnikov function \eqref{melfunz} associated to the homoclinic
orbits to the saddles $\mathfrak{e}_{\pm}^{(0)}$.

Let us consider the reduced Melnikov functions $\mathcal{M}_h^{(0)}(\tau_0)=\mathcal{M}_h(\tau_1, \tau_2)$ and $\mathcal{M}_0^{(0)}(\tau_0)=\mathcal{M}_0(\tau_1, \tau_2)$, where $\tau_0:=\tau_1-\tau_2$ (recall \eqref{melfunz}).

\begin{lem}\label{lem:periodicomo}
Let $\mathbf{K}\subset\mathbb{R}$ be a closed interval and
 let us define
\[
\mathfrak{O}_h(\tau_0):=\mathcal{M}_0^{(0)}(\tau_0)-\mathcal{M}_h^{(0)}(\tau_0).
\]
There exists $\de_0>0$ small such that $\forall \de\in (0, \de_0)$ there exist $h_0=h_0(\de)$, positive and small, such that $\forall h\in (0, h_0)$ there exists $\nu_*\in (0, 1)$ such that the following holds
\begin{equation*}

 \lVert \mathfrak{O}_h \rVert_{C^0(\mathbf{K})}+ \lVert \partial_{\tau_0} \mathfrak{O}_h \rVert_{C^0(\mathbf{K})} \lesssim \de^{\nu_*}.
\end{equation*}
\end{lem}

\begin{proof}
We consider the splitting
\begin{align}
\mathfrak{O}_h(\tau_0)=& \int_{\lvert t \rvert>\mathtt{c}|\ln\de|}  \{ \mathbf{H}_0^{(2)}, \mathbf{H}_1 \}(\gamma_0(t,\tau_0+t))- \{ \mathbf{H}_0^{(2)}, \mathbf{H}_1 \}(\Gamma_{\delta,h,0}(t,\tau_0+t))\,dt
\label{pippo1} \\ \label{pippo2}
&+\int_{\lvert t \rvert\le \mathtt{c}|\ln\de|}  \{ \mathbf{H}_0^{(2)}, \mathbf{H}_1 \}(\gamma_0(t,\tau_0+t))- \{ \mathbf{H}_0^{(2)}, \mathbf{H}_1 \}(\Gamma_{\delta,h,0}(t,\tau_0+t))\,dt,
\end{align}
where $\mathtt{c}$ is some positive constant. We observe that
\[
\{ \mathbf{H}_0^{(2)}, \mathbf{H}_1\}_{|_{\{K_1=0\}}}=0 \qquad \mbox{and} \qquad \mathtt{P}_{\delta, h}^{\pm,0}\subset \{ K_2=0\}.
\]
Then, by the exponential convergence of the flow to the hyperbolic saddles $\mathfrak{e}^{\pm}_0$ and the hyperbolic periodic orbits $\mathtt{P}_{\delta, h}^{\pm,0}$,
 the term on the r. h. s. of \eqref{pippo1} is bounded by $C\de^{\mathtt{d}}$ where $C, \mathtt{d}>0$ are two constants independent of $h$. Let us call $\mathcal{I}:=[-\mathtt{c}\lvert \ln\de \rvert, \mathtt{c}\lvert \ln\de\rvert]$. By Remark \ref{rem:uniformconv} we have
\[
\lim_{h\to 0} \sup_{(\tau_0, t)\in \mathbf{K}\times \mathcal{I}} \lvert \gamma_0(t,\tau_0+t)-\Gamma_{\delta,h,0}(t,\tau_0+t) \rvert=\lim_{h\to 0} \sup_{(\tau_0, t)\in \mathbf{K}\times \mathcal{I}} \lvert  (\gamma_0^{(1)}(t)-\gamma_{\delta, h}^{(1)}(t),0) \rvert=0.
\]
Hence there exists $h_0=h_0(\de)>0$ such that if $h\in (0, h_0)$ then \eqref{pippo2} is bounded, up to constant factors, by $\de$. The derivative $\partial_{\tau_0} \mathfrak{O}_h$ has the expression \eqref{pippo1}, \eqref{pippo2} with the double Poisson $\{ \mathbf{H}_0^{(2)}, \{ \mathbf{H}_0^{(2)}, \mathbf{H}_1 \} \}$ instead of $\{ \mathbf{H}_0^{(2)}, \mathbf{H}_1 \}$. Clearly it is still true that this Poisson vanishes at $\{K_2=0\}$. Then one can repeat the same argument to get a bound as for $\mathfrak{O}_h$.
\end{proof}

Lemma \ref{lem:periodicomo} and Proposition \ref{propfond} imply that $\mathcal{M}_h$ has a non-degenerate zero. Then, proceeding as in Section \ref{sec:HomoMelnikov} and taking $\eps=\de$ one obtains transverse heteroclinic orbits between the periodic orbits $\mathtt{P}^{+, 0 }_{h,\eps}$ and $\mathtt{P}^{-, 0 }_{h,\eps}$.
This
completes the proof of Proposition \ref{prop:SplittingPO2Rect}.
\end{proof}

\subsection{Symbolic dynamics of infinite symbols}\label{sec:Thm1:SymbDyn}

To construct symbolic dynamics for Hamiltonian \eqref{calham0} we consider a
section transverse to the flow, within a given energy level, and the associated
Poincar\'e map.

We proceed as
in \cite{Moser01}. Fix $0<\eps\ll1$ and $0<h\ll 1$.  We build an invariant set with points
arbitrarily close
to the transverse homoclinic orbit to the $\mathtt{T}_h$-periodic orbit
$\mathtt{P}^{0 }_{h,\eps}$ obtained in Proposition \ref{prop:SplittingPO2Rect}.

We define the section within the energy level $\{\mathcal{H}=h\}$,
\begin{equation}\label{def:SectionSh}
\mathcal{S}_h=\{K_2=1/2,\,\, \psi_2\in (-2\pi/3-m,-2\pi/3+m),\,\,
\mathcal{H}(\Psi_1,\Psi_2,K_1,K_2)=h \},
\end{equation}
for some small $m>0$ (see Figure~\ref{fig:horseshoe}). This section is transverse to the unperturbed flow
($\eps=0$) and therefore also transverse, for $\eps>0$ small enough, to the perturbed one. In particular, by Proposition \ref{prop:SplittingPO2Rect}, it contains points in $W^u(\mathtt{P}^{0 }_{h,\eps})\cap W^s(\mathtt{P}^{0 }_{h,\eps})$ (classical perturbative arguments ensure that the perturbed invariant manifolds are $\OO(\eps)$ to the unperturbed ones).

Denote by $\Phi^t_\mathcal{H}$ the flow associated to the Hamiltonian \eqref{calham0}. For a point
$z\in\mathcal{S}_h$, we define $T(z)>0$ the first (forward) return time of the trajectory $\Phi^t_\mathcal{H}(z)$ to this
section whenever it is defined. For those points whose forward
trajectory never hits again $\mathcal{S}_h$ we can take $T(z)=+\infty$. Note that
this happens in particular for the points in $W^s(\mathtt{P}^{-, 0 }_{h,\eps})$
(note that by the perturbative results in Section \ref{sec:Thm1:MelnikovPO} this
intersection is not empty).

Then, we define the open set $\mathcal{U}\subset\mathcal{S}_h$ as
$\mathcal{U}=\{z\in\mathcal{S}_h:T(z)<+\infty\}$ and the associated Poincar\'e
map $\mathtt{P}\colon \mathcal{U}\subset\mathcal{S}_h\to\mathcal{S}_h$ defined by
\[
\mathtt{P}(z):=\Phi^{T(z)}_{\mathcal{H}} (z).
\]

\begin{figure}
\begin{center}
\includegraphics[width=1.\textwidth]{section_coord_new.png}
\end{center}
\caption{The periodic orbit in the $(\Psi_1,K_1)$-plane, its invariant manifolds and the section $\mathcal{S}_h$.}
\label{fig:horseshoe}
\end{figure}

\begin{prop}[Existence  of Horseshoe]\label{horseshoeteo}
 Assume \eqref{tildekappas}.

Then there exists $\varepsilon_0>0$ such that $\forall \varepsilon\in (0,
\varepsilon_0)$ the Poincar\'e map $\mathtt{P}$ possesses an invariant set
$Y\subset\mathcal{U}$ whose dynamics is conjugated to the infinite symbols
shift. Namely, there exists a homeomorphism $h:\Sigma\to Y$, where
\[
\Sigma=\N^\Z=\left\{\{\omega_k\}_{k\in\Z}:\omega_k\in\N\right\},
\]
such that $\mathtt{P}|_Y=h\circ\sigma\circ h^{-1}$ where
$\sigma:\Sigma\to\Sigma$ is the shift, that is
\[
 (\sigma\omega)_k=\omega_{k+1},\quad k\in\Z.
\]
Moreover, $h^{-1}$ can be defined as follows. Fix $z^*\in Y$ and define
$\omega^*=h^{-1}(z^*)$. Associated to $z$ one can define the sequence of
hitting times
\[
t_0=0,\qquad t_k=T\left(\mathtt{P}^{k-1}(z)\right)\quad\text{for}\quad k\geq
1,\qquad  t_k=T\left(\mathtt{P}^{k}(z)\right)\quad\text{for}\quad k\leq -1.
\]
Then, there exists $C^*\in\mathbb{N}$ independent of $z^*$ such that
\begin{equation}\label{def:omegastar}
 \omega^*_k=\left\lfloor\frac{t_k-t_{k-1}}{\mathtt{T}_h}\right\rfloor-C^*
\end{equation}
where $\mathtt{T}_h$ is the period of the periodic orbit $\mathtt{P}^{\pm, 0
}_{h,\eps}$.

 \end{prop}

This proposition gives symbolic dynamics for a Poincar\'e map associated to
Hamiltonian \eqref{calham0}. Note that it is constructed in a way that higher
symbols in $\Sigma$ imply longer return times. In particular those can be
unbounded. The proof of this proposition follows the same lines as the
construction of symbolic dynamics done  by Moser in Chapter 3 of
\cite{Moser01}. Note that the natural $C^*$ in \eqref{def:omegastar} is just to normalize and have as symbols $\N$ (since the horseshoe is build close the homoclinic orbit, the hitting times satisfy $|t_k-t_{k-1}|\gg 1$).

We remark that condition \eqref{tildekappas} is necessary, indeed
the term that breaks the integrability in the Hamiltonian \eqref{calham0} has
the form
$d_{12} \,K_1\,K_2$ (see for instance \eqref{bfham0},
\eqref{trueperturbation}). Hence if the condition \eqref{tildekappas} does not
hold then the Hamiltonian \eqref{calham0} is integrable.

\subsection{Application to the Wave, Beam and Hartree
equations}\label{sec:ProofThm1:Application}

To proof  Theorem \ref{TeoWaveBeam} (and also the result for the Hartree equation \eqref{Hartree}) by applying Proposition \ref{horseshoeteo},  one needs to
check that the condition  \eqref{tildekappas} is satisfied by the resonant models
derived from the Hartree, Beam and Wave equations.

To thus end, recall the definitions \eqref{def:commonham}, \eqref{def:Cj1}, \eqref{def:Cj2} and the symplectic reduction performed in Section \ref{sectionsymplecticreduction}.
Next lemmas check condition \eqref{tildekappas}
under the hypotheses considered for these three equations.

\begin{lem}\label{lemma:WaveBeamMelnikovCond}
Let us consider Hamiltonian \eqref{calham0} associated to either the Wave
equation \eqref{Wave} or the Beam equation \eqref{Beam} and to a set
$\Lambda$ satisfying Proposition \ref{prop:LambdaSetWaveHom} . Then, the
condition  \eqref{tildekappas} is satisfied.
\end{lem}

\begin{lem}\label{lemma:HartreeMelnikovCond}
Let us consider Hamiltonian \eqref{calham0} associated to the Hartree equation
\eqref{Hartree} with a potential $V$ as in \eqref{def:potential} and to a set $\Lambda$
satisfying Proposition \ref{prop:LambdaSetHartreeHet}
Then for a generic choice of the $\{\gamma_{n}\}_{n\in\Lambda}$, the condition  \eqref{tildekappas}  is satisfied.
\end{lem}

These lemmas, together with Proposition \ref{horseshoeteo}, complete the proof of Item $(i)$ of Theorem \ref{thm:geometric}.

\begin{proof}[Proof of Lemma \ref{lemma:WaveBeamMelnikovCond}]
Recall \eqref{def:g}, \eqref{coeffN}, \eqref{def:coeffmatrixA}, \eqref{def:Cj2}.
For the Wave and Beam equations \eqref{Wave}, \eqref{Beam},
\begin{equation*}

 d_{12}=\frac{3}{32 \mathtt{g}}\,\sum_{\substack{1 \leq i \leq 4 \\ 5 \leq j \leq 8}}
\frac{(-1)^{i+j}}{|n_i|^{\kappa} |n_j|^\kappa}=\frac{3}{32 \mathtt{g}}\left(\sum_{i=1}^4\frac{(-1)^i}{|n_i|^{\kappa}}\right)\left(\sum_{j=5}^8\frac{(-1)^j}{|n_j|^{\kappa}}\right)

 \end{equation*}
where $\kappa=1$ for the Wave equation and  $\kappa=2$ for the Beam equation.

We write $d_{12}$ in a different form. To this end, we introduce the following notations. For each finite set of indexes
$I=\{i_1,\ldots,i_n\} \subset \{ 1, \dots, 8 \}$ and for any pair of positive
integers $i_1, i_2 \in \{ 1, \dots, 8 \}$, we define
\begin{equation}\label{def:IndexesPiDelta}
\prod^{I}= \frac{1}{\prod_{k=1}^n |n_{i_k}|^\kappa},\qquad
\Delta_{i_1,i_2}= |n_{i_1}|^\kappa - |n_{i_2}|^\kappa.
\end{equation}

Using the  identities

\[\prod^{i,j} - \prod^{i,k} = \prod^{i,j,k} \Delta_{k,j},\qquad

\prod^{i,j,k} - \prod^{l,j,k} = \prod^{i,j,k,l} \Delta_{l,i},\]
and the fact that the resonance relations (see \eqref{def:Abh},\eqref{def:Aw})
imply  $\Delta_{1,2}
= \Delta_{4,3}$, $\Delta_{5,6} = \Delta_{8,7}$,

one can see that

\begin{equation}\label{eq:numbeam3}
d_{12}= \Delta_{2,1} \Delta_{6,5} \left( |n_3|^\kappa |n_4|^\kappa - |n_1|^\kappa |n_2|^\kappa
\right) \left( |n_7|^\kappa |n_8|^\kappa - |n_5|^\kappa |n_6|^\kappa \right)
\prod^{1,2,3,4,5,6,7,8}.
\end{equation}

Therefore $d_{12}$ vanishes if one of the following
conditions holds
\begin{equation}\label{NotAdmissibleBeam}
|n_1| = |n_2|,\qquad|n_5| = |n_6|,\qquad
 |n_1||n_2| = |n_3||n_4|,\qquad
|n_5||n_6| = |n_7||n_8|.
\end{equation}
\begin{remark}
We point out that the conditions \eqref{NotAdmissibleBeam} do not involve at the same time modes belonging to two different $4$-tuple resonances.
\end{remark}
Condition \eqref{eq:Wave} implies that the two first
conditions cannot be satisfied.
We check now that under the hypotheses of Proposition  \ref{prop:LambdaSetWaveHom}, one has
\begin{equation}\label{def:CondExtraForPk}
|n_1||n_2| \neq |n_3||n_4|
\end{equation}
(the condition $|n_5||n_6| \neq |n_7||n_8|$ can be checked analogously).

We start with the Beam equation, that is $\kappa=2$. Arguing by contradiction,
assume that $n_1,n_2,n_3,n_4$ satisfy  $|n_1||n_2| = |n_3||n_4|$, \eqref{eq:Wave} and the resonance condition
\begin{equation}\label{cond:resforPk}
 |n_1|^2-|n_2|^2=-|n_3|^2+|n_4|^2
\end{equation}

The resonance relation can be written as
\begin{align*}
(|n_1|-|n_2|)(|n_1|+|n_2|) &= (|n_4|-|n_3|)(|n_4|+|n_3|)
\end{align*}
Squaring each side, one has
\begin{align*}
(|n_1|^2 + |n_2|^2)^2 - 4 |n_1|^2 |n_2|^2= (|n_3|^2 + |n_4|^2)^2 - 4 |n_3|^2 |n_4|^2.

\end{align*}
Therefore, since we are assuming $|n_1||n_2| = |n_3||n_4|$,  we get $|n_1|^2 +
|n_2|^2= |n_3|^2 +
|n_4|^2$, which combined with the resonance relation \eqref{cond:resforPk} leads to $|n_2|^2=
|n_3|^2$,
which contradicts assumption \eqref{eq:Wave}.

For the Wave equation \eqref{Wave}, that is $\kappa=1$, one can proceed
analogously, arguing by contradiction. Assume that
$n_1,\ldots,n_4$ satisfy \eqref{eq:Wave},
the resonance condition
\[
 |n_1|-|n_2|=-|n_3|+|n_4|
\]
and $|n_1||n_2| = |n_3||n_4|$. Squaring the resonance condition and using this
last assumption, one has

\[|n_1|^2 + |n_2|^2 = |n_3|^2 + |n_4|^2.\]

Multiplying both sides by $|n_4|^2$ and using again $|n_1||n_2| = |n_3||n_4|$
one obtains

$(|n_1|^2 -|n_4|^2)(|n_4|^2 - |n_2|^2) = 0$,

which contradicts \eqref{eq:Wave}.
\end{proof}

\begin{proof}[Proof of Lemma \ref{lemma:HartreeMelnikovCond}]
Recall \eqref{coeffN}, \eqref{def:coeffmatrixA}.
For the Hartree equation \eqref{Hartree}, $d_{12}$ is of the form
\[
 d_{12}=\sum_{k\in I}\al_k V_k
\]
where $\al_k\neq 0$ and
\[
I:=\{ k\in\mathbb{Z}^2 : k=n_i-n_j \,\,\mbox{for some}\,\,n_i\in\mathcal{R}_1, \,\,n_j\in\mathcal{R}_2\}.
\]
We observe that the cardinality of $I$ is bounded by $4N(4N-1)/2$. Therefore, by condition
\eqref{nondegLambda}, $d_{12}$ is a polynomial in the $4N(4N-1)/2$ variables $\gamma_k$, $k\in I$. Such polynomial is not identically zero because if we set one of the $\gamma_k$'s equal to one and all the others at zero then $d_{12}\neq 0$.

\end{proof}

\subsection{End of the proof of Theorem \ref{TeoWaveBeam}}

 Lemmas \ref{lemma:WaveBeamMelnikovCond}, \ref{lemma:HartreeMelnikovCond} imply that condition \eqref{tildekappas} holds and, therefore,  Proposition \ref{horseshoeteo} can be applied to the resonant models associated to the Wave \eqref{Wave}, Beam \eqref{Beam} and Hartree \eqref{Hartree} equations. This proposition gives certain orbits of these resonant models. These orbits will be the first order (up to changes of coordinates) of orbits of equations \eqref{Wave}, \eqref{Beam} and \eqref{Hartree}.

Fix $0<\eps\ll 1$ and $0<h\ll 1$ and consider the periodic orbit $\mathtt{P}^{0 }_{h,\eps}$ given by Proposition \ref{prop:SplittingPO2Rect}, which has period $\mathtt{T}_h$. By Proposition \ref{horseshoeteo} there exist a set  $Y\subset\mathcal{S}_h$ which is an invariant hyperbolic set (a Smale horseshoe) for the Poincar\'e map associated to the Hamiltonian $\mathcal{H}$ in \eqref{calham0}. This set can be built arbitrarily close to homoclinic points of  $\mathtt{P}^{0 }_{h,\eps}$. Fix $\omega\in\Sigma$  such that $|\omega_k|\geq M_0\,\mathtt{T}_{h}$, where $M_0$ satisfies $M_0\gtrsim\log\eps$ and $\mathtt{T}_h$ is the period of the periodic orbit $\mathtt{P}^{0}_{h, \varepsilon}$.
Then, Proposition \ref{horseshoeteo} ensures that there exists an orbit $\gamma(t)$  of $\mathcal{H}$ with initial condition in $Y$,
\[
\gamma(t):=(\Psi_1(t), \Psi_2(t), K_1(t), K_2(t)), \quad t\in [0, T] \quad \mbox{for some}\,\,\,T>0,
\]
 which satisfies the following. There exists a sequence of times $\{t_k\}_{k\in\ZZ}$ satisfying \eqref{def:omegastar} such that $\gamma(t_k)\in \mathcal{S}_h$ where $\mathcal{S}_h$ is the section defined in \eqref{def:SectionSh}. Note that, by \eqref{def:omegastar}, the times $t_k$ satisfy
\[
 t_{k+1}=t_{k}+ \mathtt{T}_h (\omega^*_k+C^*+\theta_k)\qquad \text{ for some }\quad \theta_k\in (0,1) \quad \text{ and }C^*\in\N.
\]
By construction, there exists another sequence of times $\{\bar t_k\}_{k\in\ZZ}$ with $\bar t_k\in (t_k,t_{k+1})$ such that $\gamma(\bar t_k)$ satisfies
\[
 K_2(\bar t_k)=\frac{1}{2},\qquad  \left|\Psi_2(\bar t_k)-\frac{2\pi}{3}\right|\ll 1.
\]

The Smale horseshoe, can be built arbitrarily close to the invariant manifolds of $\mathtt{P}^{0}_{h, \varepsilon}$ and therefore, one can ensure that there exist intervals
\begin{itemize}
\item  $I_k\subset (\overline{t_k},t_{k+1})$  such that, for $t\in I_k$, $\ga(t)$ belongs to a $\eps$-neighborhood of  $\mathtt{P}^{0}_{h, \varepsilon}$;
\item  $J_k\subset (t_k, \overline{t_k})$  such that for $t\in J_k$ the orbit $\gamma(t)$ belongs to a $\OO(\eps)$-neighborhood of $K_2=1$, since the homoclinic orbit obtained in Proposition \ref{prop:SplittingPO2Rect} have points $\OO(\eps)$-close to $K_2=1$.

\end{itemize}
This behavior implies estimates \eqref{eq:TheoRandomTimeOsc1} and \eqref{eq:TheoRandomTimeOsc2} in Theorem \ref{TeoWaveBeam}, once we undo the symplectic reductions, the changes of coordinates and we add the error terms as it is  explained below.

By Proposition \ref{prop:SplittingPO2Rect} the parameterization of the periodic orbit  $\mathtt{P}^{0}_{h, \varepsilon}$ is $\varepsilon$-close to \eqref{def:PO:ext}, hence we have that
\[
 K_1(t)=Q(t)+\tilde{R}_2(t)

\]
where $Q(t)$ is the time parameterization of  $\mathtt{P}^{0 }_{h,\eps}$ and thus is   $\mathtt{T}_h$-periodic,  and $\sup_{t\in [0, T]} |\tilde{R}_2(t)|\le \varepsilon$.

By the symplectic reduction performed in Section \ref{sectionsymplecticreduction} there exists $r(t)$ solution of $H_{\Res}$ in \eqref{Hres} with Fourier support $\Lambda$ such that
\[
|r_{n_1}(t)|^2=| K_{1}(t) |^2, \qquad |r_{n_5}(t)|^2=| K_{2}(t) |^2,\qquad \text{for}\qquad t\in [0,T].\]
This can be seen using Remark \ref{rem:Kcartcoord}, which gives also the behavior of the other actions.

Since the solutions of $H_{\Res}$ are invariant under the scaling \eqref{def:scaling}, we can consider $r^{\delta}(t):=\delta r(\delta^2 t)$. Then, $r^{\delta}(t)$ is also a solution of $H_{\Res}$ for  $t\in [0, \delta^{-2} T]$.

Now it only remains to obtain an orbit for the equations \ref{Wave}, \ref{Beam} and \ref{Hartree} which is close (up to certain changes of coordinates) to $r^{\delta}(t)$. First step is to apply Proposition \ref{approximationargument}. It ensures that there exists $0<\delta_2\ll 1$ such that for all $\delta\in (0, \delta_2)$, there exists a solution $w(t)$ of $H\circ \Gamma\circ \Psi=H_{\Res}+\mathcal{R}'$   such that $w(t)=r^{\delta}(t)+\widetilde{R}(t)$ with $\widetilde{R}(0)=0$, $\| \widetilde{R}(t) \|_{\rho}\lesssim \delta^2$ for $t\in [0, \delta^{-2} T]$. We note that,  by Item $(ii)$ of Proposition \ref{WBNF}, the Birkhoff map $\Gamma$ is $\delta^3$-close to the identity. Finally the transformations \eqref{rotatingcoord1} and \eqref{fase} preserve the modulus of the Fourier coefficients. The last change of coordinates that one has to apply (for the Wave \eqref{Wave} and Beam \eqref{Beam} equations) is  passing from complex coordinates \eqref{def:complexcoordinates} to the original ones. We remark that by \eqref{eq:Wave} if $n_i\in\Lambda$ then $-n_i\notin \Lambda$. Thus
  \[
  u_{n_i}=\frac{1}{\sqrt{2 |j|}} \Psi_{n_i} \qquad n_i\in\Lambda.
  \]


TITLE: Prove $\int_{0}^{\infty} \psi^{(2)} (1+x) \ln (x) \, dx = \zeta'(2) + \zeta(2)$
QUESTION [8 upvotes]: I'm looking for alternate methods of proving the following result:

$$\int_{0}^{\infty} \psi^{(2)} (1+x) \ln (x) \, dx = \frac{\pi^2}{6} \left( \gamma + \ln (2\pi)-12 \ln A +1\right) = \zeta'(2) + \zeta(2)$$
Where $\psi$ is the polygamma function, $A$ is the Glaisher-Kinkelin constant and $\gamma$ is the Euler-Mascheroni constant.

Here is how I originally proved it:
Begin with the known result $$\int_{0}^{\infty} \frac{x \ln x}{e^{2 \pi x}-1} \, dx = \frac{1}{2} \zeta'(-1)=\frac{1}{24}-\frac{1}{2} \ln A$$
then use the following well-known property of the Laplace transform
$$\int_{0}^{\infty} f(x) g(x) \, dx = \int_{0}^{\infty} \left( \mathcal{L} f\right) (y) \left( \mathcal{L}^{-1} g \right) (y) \, dy$$
Let $f(x) = \frac{x^2}{e^{2 \pi x} - 1}$ and $g(x) = \frac{\ln x}{x}$ then:
$$\left(\mathcal{L} f\right) (y) = -\frac{1}{8 \pi^3} \psi^{(2)} \left(1+\frac{y}{2\pi}\right)$$
$$\left(\mathcal{L}^{-1} g\right)(y) = -\gamma - \ln (y)$$
$$\implies \frac{1}{24} - \frac{1}{2} \ln A = \frac{1}{8\pi^3}\int_{0}^{\infty} \gamma \, \psi^{(2)} \left(1+\frac{y}{2\pi}\right) \, dy + \frac{1}{8\pi^3} \int_{0}^{\infty} \psi^{(2)} \left(1+\frac{y}{2\pi}\right) \ln (y) \, dy$$
Proceed with the substitution $y \mapsto 2\pi x \implies dy = 2 \pi\,dx$
$$\implies \frac{1+\gamma}{24} - \frac{1}{2} \ln A = \frac{1}{4\pi^2} \int_{0}^{\infty} \psi^{(2)} \left(1+x\right) \ln (2 \pi x) \, dx$$
Now separate $\ln (2\pi x) = \ln (2 \pi) + \ln (x)$ and the result quickly follows.
I was wondering if there are any alternative methods that don't rely on the initial $\frac{1}{2} \zeta'(-1)$ result.

Motivation:
I've been experimenting with the following identity:
$$\int_{0}^{\infty} \frac{x^{2n-1}}{e^{2\pi x}-1}\, dx = \frac{(-1)^{n-1} B_{2n}}{4n}\,\text{ for } n \in \mathbb{N}$$
I noticed that if one takes the partial derivative with respect to $n$ and then proceeds to take the limit as $n \to 1$, then we arrive at our $-\frac{1}{2} \zeta'(-1)$ result. This means it is possible for one to define the 'derivative' of the Bernoulli numbers at $n=1$.
More precisely:
$$B_{2}^{\prime} = \frac{1}{4} - 2 \ln A$$
is this result already known, and if so, are any other Bernoulli derivatives definable in a similar way using the Glaisher-Kinkelin constant? (I know it is possible to do it with Glaisher-Kinkelin-like constants such as the Bendersky-Adamchik constants due to their similar connection to hyperfactorials, $\zeta'(-2)$ and $\zeta(3)$ for example). I'm not sure what meaning one can give to the 'derivative' of the Bernoulli numbers, but figure 2 of this paper seems to roughly agree with my result.

REPLY [4 votes]: For another solution using the complete beta function:
Using the same expansion
$$\psi^{(2)}(x) = - 2 \sum_{n=0}^{\infty} \frac{1}{(n+x)^{3}},$$
we have
$$ \begin{align} \int_{0}^{\infty} \psi^{(2)}(x+1) \ln (x) \, \mathrm dx &= -2\int_{0}^{\infty}\ln(x) \sum_{n=0}^{\infty} \frac{1}{(n+x+1)^{3}} \, \,  dx \\ &= - 2 \sum_{n=0}^{\infty} \int_{0}^{\infty} \frac{\ln (x)}{(n+x+1)^{3}} \,  dx.\\
=& - 2 \sum_{n=0}^{\infty} \frac{1}{(n+1)^3} \int_{0}^{\infty}  \frac{\ln (x)}{(1+\frac{x}{(n+1)})^{3}} \,  dx\\
=& - 2 \sum_{n=0}^{\infty} \frac{1}{(n+1)^3} \int_{0}^{\infty}  \frac{\left(\frac{d}{dt}\Big|_{t=0+}x^t\right)}{(1+\frac{x}{(n+1)})^{3}} \,  dx\\
=& \frac{d}{dt}\Big|_{t=0+} - 2 \sum_{n=0}^{\infty} \frac{1}{(n+1)^3} \int_{0}^{\infty}  \frac{x^t}{(1+\frac{x}{(n+1)})^{3}}dx \\
=& \frac{d}{dt}\Big|_{t=0+} - 2 \sum_{n=0}^{\infty} \frac{(n+1)^t}{(n+1)^2} \int_{0}^{\infty}  \frac{w^t}{(1+w)^{3}}dw \quad \\
=& \frac{d}{dt}\Big|_{t=0+} - 2 \sum_{n=0}^{\infty} \frac{(n+1)^t}{(n+1)^2}B(t+1,2-t)\\
=& \frac{d}{dt}\Big|_{t=0+}  \sum_{n=0}^{\infty} \frac{(n+1)^t}{(n+1)^2}\pi (t-1)t \csc(\pi t)\\
=& \sum_{n=0}^{\infty} \frac{1}{(n+1)^2} - \sum_{n=0}^{\infty} \frac{\ln(n+1)}{(n+1)^2}\\
=& \zeta(2)+\zeta'(2)
\end{align}\\$$


TITLE: Show that $A$ is not path-connected.
QUESTION [0 upvotes]: I have the following problem:
In $\mathbb{R}^2$ with the usual topology consider the set $A= \{(x,y)\in\mathbb{R}^2:x\in\mathbb{R}\setminus\mathbb{Q},y\geq 0\}\cup\{(x,y)\in\mathbb{R}^2:x\in\mathbb{Q},y<0\}$. Show that the set $A$ is not path connected.
I have an attempt but I believe its wrong.
My Attempt: By contradiction. Suppose that A is path connected, the for $(q_1,-1)$ and $(q_2,-1)$ with $q_1,q_2\in\mathbb{Q}$, exists a path $f$ continuous and satisfies that $f(0)=(q_1,-1)$ and $f(1)=(q_2,-1)$. Since $f$ is continuous, $h:=\Pi_1\circ f$ is continuous where $\Pi_1$ is the proyection on $x$.
We have that $h(0)=(q_1,0)$ and $h(1)=(q_2,0)$ but $h(0),h(1)\notin A$ and then, $h$ is not continuos, and so, $f$ is not continuos. Therefore, A is not path connected.
I hope you can help me to fix it if or give me an idea of how can I prove this, please.
Thanks!!!

REPLY [0 votes]: Let $h:[0,1]\to\mathbb{R}$ be the projection onto the first coordinate of the path $f$ that is connecting $(q_1,-1)$ and $(q_2,-1)$ with $q_1,q_2\in\mathbb{Q}$.
Consider the set $V=h^{-1}(\{q_1\})$. It is closed, since $h$ is continuous.
Let $t_0\in V$. If in every neighborhood $U$ of $t_0$ we had points $t\in U$ where $h$ takes other values different from $q_1$, since $h$ is continuous, $h$ would take on $U$ all the values in the interval from $q_1$ to $h(t)$. Therefore, it would take irrational values. At those points $f$ is in the upper half plane. So, $f(t_0)$ is in the open lower half plane and in every neighborhood of $t_0$ there are points where $f$ is in the upper half plane. That cannot be, since $f$ is continuous. Therefore, $h$ is locally constant at $t_0$. Hence $V$ is open.
Since $V$ is non-empty, closed and open, and $[0,1]$ is connected, then $V=[0,1]$.
Therefore, $h([0,1])=\{q_1\}$. This contradicts that $h(1)=q_1\neq q_1$.


TITLE: Finding set B when A and C is given along with relation .
QUESTION [0 upvotes]: Set $C$ contains all the elements that can be represented using sum of an element of $A$ and an element of $B$.
In such a case if we have the values of set $A$ and set $C$, how can we find out the set $B$.
For eg :
If set $A=\{1,2\}$
and set $C=\{3,4,5\}$,
then set $B$ would be $B=\{2,3\}$.
I can work this around for small sets, but I fail to understand this for bigger sets, is there a global solution for this ?

REPLY [1 votes]: I'm afraid you might have to go with a brute force solution.
You can easily find $\min(B)$ and $\max(B)$ by taking $\min(B)=\min(C)-\min(A)$ and $\max(B)=\max(C)-\max(A)$. After that I see no other way but looking at all $\min(B)<n<\max(B)$ to see whether $A+n\subset C$. If $A+n\subset C$ then $n\in B$, if $A+n\subsetneq C$ then $n\not\in B$. Note that this gives the biggest possible $B$.
There might be a more elegant solution out there, and there might be much better solutions out there for "nice" sets, until then I would just program a quick for loop.


TITLE: What is the meaning of subtracting from the identity matrix?
QUESTION [6 upvotes]: If I subtract the matrix $A$ from the identity matrix $I$, $I - A$, is there a meaning to the resulting matrix perhaps given some conditions like invertibility or symmetry? For example,
$$
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
- \begin{bmatrix} a & b \\ c & d \end{bmatrix}
= \begin{bmatrix} 1-a & -b \\ -c & 1-d \end{bmatrix}
$$
For future reference, is there a good reference to lookup answers to such questions when I can't derive the answer on my own?

REPLY [4 votes]: If there exists $n\in\mathbb{N}$ with $A^{n}=0$, then $I−A$ is invertible (try to prove this yourself before reading why!).
The reason: multiply out the brackets in the expression
$$(I−A)(A^{n−1}+A^{n−2}+\cdots+A+I)$$
and see that you get $I$. It follows that $A^{n−1}+A^{n−2}+\cdots+A+I$ is the inverse of $I−A$ in this case.
By the way, a matrix $A$ with $A^{n}=0$ for some $n\in\mathbb{N}$ is called nilpotent.

By the way, the fact presented above can be used to give a proof of the (basic) fact that all eigenvalues of a nilpotent matrix must equal zero. In particular, this means that the determinant and trace of a nilpotent matrix are always $0$.
If $A$ is nilpotent then so is $\frac{1}{\lambda}A$ for any $\lambda\neq0$. Hence $I-\frac{1}{\lambda}A=\frac{1}{\lambda}(\lambda I-A)$ has an inverse. It follows that $\det{(\lambda I-A)}\neq0$ whenever $\lambda\neq0$ and $A$ is nilpotent, so the characteristic polynomial only has $0$ as a root.
So considering $I-A$ can actually come in handy to prove things.


TITLE: $f(x)$ is irreducible over finite field $F_p$ , does $f(x)$ split over $F_p / \langle f(x) \rangle$?
QUESTION [3 upvotes]: I want to know whether it is possible to generalize the next thing exercise.
$x^4 + x + 1$ is irreducible over $F_2[x]$ , so $F_2[x]/\langle x^4 +x+1\rangle = span_{F_2} \{1,\bar{x} , \bar{x}^2,\bar{x}^3\} \cong F_{2^4}$. So $\bar{x}^4 + \bar{x} +1 = 0$ in $F_2[x]/\langle x^4+x+1\rangle$ and due to the fact that in $F_2$ $(a+b)^2 = a^2 + b^2$ ,we get that $\bar{x}^8 + \bar{x}^2 +1 = 0=(\bar{x}^4 + \bar{x} +1)^2 = 0$ so $\bar{x}^2$ is also a root. And similarly $\bar{x}^4 , \bar{x}^8$ are roots. A simple check gives that those are 4 different roots, thus indeed the polynomial splits.
So my question,
is there a way to use the fact that $F_{p^n}$ is the splliting field of $x^{p^n}-x$ over $F_p$ and the fact that for every $f(x)$ irreducible polynomial over $F_p[x]$ from degree $n$ , $F_p[x]/ \langle p(x) \rangle \cong F_{p^n}$ to show that $f(x)$  splits over $F_p[x]/ \langle f(x) \rangle$ ?

REPLY [3 votes]: Yes, your conjecture is true and can be proven without Galois theory (@commenters):

Theorem 1. Let $K$ be a finite field. Let $f \in K\left[x\right]$ be an irreducible polynomial. Let $L$ be the field $K\left[t\right] / \left(f\left(t\right)\right)$. Then, $f$ factors into linear polynomials over the field $L$.

Note that I have used two different indeterminates $x$ and $t$ here in order to avoid the awkwardness of having a polynomial in $x$ over a ring that already sort-of contains an $x$ in it. This is either necessary or redundant depending on the pedantic minutiae of your definition of polynomial rings; in either case, it does not hurt.
Proof of Theorem 1. Let $q = \left|K\right|$. It is well-known that for any commutative $K$-algebra $A$, the map $A \to A,\ a \mapsto a^q$ is a $K$-algebra endomorphism of $A$ (since any $a, b \in A$ satisfy $\left(a+b\right)^q = a^q + b^q$ and $\left(ab\right)^q = a^q b^q$, and since each $\lambda \in K$ satisfies $\lambda^q = \lambda$). This endomorphism will be denoted by $F_A$, and is called the Frobenius endomorphism of $A$.
Let $n$ be the degree of $f$. For each polynomial $g \in K\left[t\right]$, let $\overline{g}$ be the projection of $g \in K\left[t\right]$ onto the quotient ring $K\left[t\right] / \left(f\left(t\right)\right) = L$. Let $u = \overline{t} \in L$. Let $F$ be the map $F_L$. Thus, $F$ sends each $a \in L$ to $a^q$ (by the definition of $F_L$). Hence, for each nonnegative integer $i$ and each $a \in L$, we have
\begin{align}
F^i\left(a\right) = a^{q^i}
\label{darij1.pf.t1.Fi}
\tag{1}
\end{align}
(indeed, this can be proven by induction on $i$, using the previous sentence in the induction step).
We shall show that $F^0\left(u\right), F^1\left(u\right), \ldots, F^{n-1}\left(u\right)$ are $n$ distinct roots of $f$ in $L$. This will quickly yield that $f$ factors into linear polynomials over $L$. Let us proceed in three steps:

Claim 1: The $n$ elements $F^0\left(u\right), F^1\left(u\right), \ldots, F^{n-1}\left(u\right)$ of $L$ are distinct.

Proof of Claim 1. Assume the contrary. Thus, there exist two elements $i$ and $j$ of $\left\{0,1,\ldots,n-1\right\}$ such that $i<j$ and $F^i\left(u\right) = F^j\left(u\right)$. Consider these $i$ and $j$. We have $i < j$, thus $q^i < q^j$ (since $q = \left|K\right| > 1$).
Let $a \in L$ be arbitrary. We are going to prove that $a^{q^j} - a^{q^i} = 0$.
Indeed, we have $a \in L = K\left[t\right] / \left(f\left(t\right)\right)$; in other words, $a = \overline{s}$ for some polynomial $s \in K\left[t\right]$. Consider this $s$. But $s = s\left(t\right)$, so that $\overline{s} = \overline{s\left(t\right)} = s\left(\overline{t}\right)$ (since $s$ is a polynomial with coefficients in $K$). In view of $u = \overline{t}$, this rewrites as $\overline{s} = s\left(u\right)$. Hence, $a = \overline{s} = s\left(u\right)$. But $F^i$ is a $K$-algebra homomorphism (since $F$ is a $K$-algebra homomorphism), and thus commutes with the application of any polynomial over $K$. Hence, in particular, $F^i$ commutes with $s$. Thus, $F^i\left(s\left(u\right)\right) = s\left(F^i\left(u\right)\right)$. Now, applying $F^i$ to both sides of the equality $a = s\left(u\right)$, we find
\begin{align}
F^i\left(a\right) = F^i\left(s\left(u\right)\right) = s\left(F^i\left(u\right)\right) .
\end{align}
Comparing this with \eqref{darij1.pf.t1.Fi}, we obtain $a^{q^i} = s\left(F^i\left(u\right)\right)$.
The same argument (applied to $j$ instead of $i$) yields $a^{q^j} = s\left(F^j\left(u\right)\right)$. Hence,
\begin{align}
a^{q^i} = s\left(\underbrace{F^i\left(u\right)}_{= F^j\left(u\right)}\right)
= s\left(F^j\left(u\right)\right) = a^{q^j} .
\end{align}
In other words, $a^{q^j} - a^{q^i} = 0$.
Now, forget that we fixed $a$. We thus have shown that $a^{q^j} - a^{q^i} = 0$ for each $a \in L$. In other words, each $a \in L$ is a root of the polynomial $x^{q^j} - x^{q^i} \in K\left[x\right]$. This polynomial is nonzero (since $q^i < q^j$, so that the terms $x^{q^j}$ and $x^{q^i}$ do not cancel) and has degree $q^j$ (since $q^i < q^j$), and thus has at most $q^j$ many roots in $L$ (since any nonzero polynomial over $L$ of degree $d$ has at most $d$ many roots in $L$). Since each $a \in L$ is a root of this polynomial, we thus conclude that there are at most $q^j$ many $a \in L$. In other words, $\left|L\right| \leq q^j$.
But $L = K\left[t\right] / \left(f\left(t\right)\right)$ is a $K$-vector space of dimension $n$ (since the polynomial $f$ has degree $n$), and thus has size $\left|K\right|^n = q^n$ (since $\left|K\right| = q$). Hence, $\left|L\right| = q^n$, so that $q^n = \left|L\right| \leq q^j$ and therefore $n \leq j$. This contradicts $j \in \left\{0,1,\ldots,n-1\right\}$. This contradiction shows that our assumption was wrong. Hence, Claim 1 is proven.

Claim 2: The $n$ elements $F^0\left(u\right), F^1\left(u\right), \ldots, F^{n-1}\left(u\right)$ of $L$ are roots of $f$.

Proof of Claim 2. From $u = \overline{t}$, we obtain $f\left(u\right) = f\left(\overline{t}\right) = \overline{f\left(t\right)}$ (since $f$ is a polynomial with coefficients in $K$). Thus, $f\left(u\right) = \overline{f\left(t\right)} = 0$ (since $f\left(t\right) \equiv 0 \mod f\left(t\right)$).
Now, let $i \in \left\{0,1,\ldots,n-1\right\}$ be arbitrary. Then, $F^i$ is a $K$-algebra homomorphism (since $F$ is a $K$-algebra homomorphism), and thus commutes with the application of any polynomial over $K$. Hence, in particular, $F^i$ commutes with $f$. Thus, $F^i\left(f\left(u\right)\right) = f\left(F^i\left(u\right)\right)$. Hence, $f\left(F^i\left(u\right)\right) = F^i\left(\underbrace{f\left(u\right)}_{=0}\right) = F^i\left(0\right) = 0$ (again since $F^i$ is a $K$-algebra homomorphism). In other words, $F^i\left(u\right)$ is a root of $f$.
Now, forget that we fixed $i$. We thus have shown that $F^i\left(u\right)$ is a root of $f$ for each $i \in \left\{0,1,\ldots,n-1\right\}$. In other words, the $n$ elements $F^0\left(u\right), F^1\left(u\right), \ldots, F^{n-1}\left(u\right)$ of $L$ are roots of $f$. This proves Claim 2.
We are now almost done proving Theorem 1. Claim 2 shows that the $n$ elements $F^0\left(u\right), F^1\left(u\right), \ldots, F^{n-1}\left(u\right)$ of $L$ are roots of $f$. Since these $n$ elements are distinct (by Claim 1), we thus conclude that the polynomial $f$ has (at least) $n$ distinct roots over $L$. But this polynomial $f$ has degree $n$; hence, if it has $n$ distinct roots over $L$, then it factors into linear polynomials over $L$. So we have proven that $f$ factors into linear polynomials over $L$. This proves Theorem 1. $\blacksquare$


TITLE: Square root of normal distribution
QUESTION [11 upvotes]: Let $X$ and $Y$ be independent random variates with the same probability distribution, $P(x)$. Assuming that the product $Z=XY$ is a random variate with normal distribution, say $$f_Z(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} x^2}$$ what is the form of $P(x)$? Does it exist in closed form?
EDIT: I just realized that a product distribution is necessarily divergent at zero, which means that the normal distribution cannot be a product distribution (at least when $P(x)$ is interpreted as a normal function; maybe there is a generalized function that solves the problem?). Do you agree?

REPLY [10 votes]: First:  No helvio, the density of the product of two iid random variables (r.v.'s) is not necessarily divergent at 0. E.g., take the product of two iid r.v.'s each having a Gamma distribution with shape parameter 2.
Second: Let $U:=\ln|Z|$ and $W:=\ln|X|$. Then the characteristic function (c.f.) $f$ of $U$ is given by the formula $f(t)=2^{i t/2} \Gamma((1+i t)/2)/\sqrt{\pi}$ for real $t$. So, the c.f. of $W$ (say $g$) would be a square root of $f$, and then the density (say $q$) of $W$ would be given by the formula $q(x)=\frac1{2\pi}\,\int_{-\infty }^{\infty } g(t) e^{-i t x} \, dt$  for real $x$. I have tried to evaluate the latter integral numerically, and that seems to suggest that $q(3)<0$; however, I have not done a rigorous estimate of the approximation error.
On the other hand, numerical experiments using Bochner's theorem on the positive definite functions suggest that $g$ is a c.f.! If so, then the corresponding r.v. $W$ exists, and then it suffices to let $X$ have the distribution of $\varepsilon e^W$, where $\varepsilon$ is a Rademacher r.v. independent of $W$.
Addendum: The calculation of $q(3)$ mentioned previously was likely mistaken, as I forgot to make sure that $g(t)$ be taken to be the value of the square root of $z:=f(t)$ on the Riemann surface for $\sqrt z$, so as for $g$ to be a continuous function -- as it must be.
Another approach to the problem: Note that $\mu_j:=E|Z|^j=2^{j/2} \Gamma \left(\frac{j+1}{2}\right)/\sqrt{\pi }$ and then one would have $\nu_j:=E|X|^j=\sqrt{\mu_j}$ for $j>0$. To prove that such a desired r.v. $X$ exists, it is enough to show that for all nonnegative integers $n$ the determinants, say $d_n$ and $e_n$, of the matrices $(\nu_{i+j})_{i,j=0}^n$ and $(\nu_{i+j+1})_{i,j=0}^n$ are nonnegative (see e.g. Ch. V in Kreǐn, M. G. and Nudel'man, A. A.,
The Markov moment problem and extremal problems). These determinants are indeed positive for $n\le9$. One can hopefully find a recursion for $d_n$ and $e_n$ to show that they are indeed nonnegative for all $n$. Note here that $\mu_j$ is an integer or an integral multiple of $\sqrt{2/\pi}$.


TITLE: What tools do you use to give presentations?
QUESTION [25 upvotes]: I was wondering what tools that people in this field (theoretical computer science) use to create presentations. Since a great deal of computer science is not just writing papers but also giving presentations I thought that this would be an important soft question. This is inspired by the previous question what tools do you use to write papers.  The most common that I have seen are as follows.

Beamer
Microsoft PowerPoint
LaTeX
GraphViz

I was wondering if there are any other tricks that I am missing?

REPLY [1 votes]: For drawing it is also convenient to use Inkscape instead of tikz: it can embed latex code with the plugin Textext, and you can directly draw what you want instead of starting worrying about coordinates.
I use both depending on the particular task at hand.

REPLY [0 votes]: Another powerful option for drawing graphics is Asymptote.  It is a language, not a GUI, and LaTeX is baked in.  If you are familiar with Knuth's MetaPost then think of it as like that but with the drawing constructs extended to 3D and with a more familiar syntax, in that you usually produce stand-alone graphics that you then import into the document.  It is Freely available, under current development, and comes with TeX Live or MiKTeX.
For folks who know TikZ, one contrast is that there you usually include the drawing generation as part of the document, which adds complexity, and here the 3D stuff is arguably more powerful.
There is a great tutorial, and a large number of packages, including an unofficial graph package that I find useful.


TITLE: I need a function that converts a permutation to its transposition
QUESTION [0 upvotes]: If I have the permutation: $(a, b, c,..., n)$ how do I find the transposition of this? (AKA the number of "swaps" to get from this: $(1, 2, 3,..., N)$ to that: $(a, b, c,..., n)$.
So:
$f(3, 2, 1) = (13)$
$f(5, 3, 4, 2, 1) = (15)(23)(24)$
$f(8, 6, 4, 2, 7, 5, 3, 1) = (18)(26)(25)(27)(23)(24)$
Any help would be appreciated, thanks.
Note - I know that $(15)(23)(24)$ can be written as $(15)(234)$ but I would prefer the former so I can easily tell if the permutation is even or odd.

REPLY [2 votes]: Note that every permutation can be written as the product of distinct cycles & each cycle can be written as the product of transpositions as follows
\begin{eqnarray*}
(1234 \cdots n) =(12)(13)(14) \cdots (1n).
\end{eqnarray*}


TITLE: Finding the maximum order of an element in a product group
QUESTION [0 upvotes]: I'm trying to find the maximum order of an element in $\mathbb{Z}_{4}\times \mathbb{Z}_{8}$. I found a previous similar thread (link). But I can't seem to understand how they calculated the order.
I know that $o((g,h))=lcm(o(g),o(h))$ but how does it help to calculate the maximum order? From the thread I learn to calculate:
$$ (1 + 4 \mathbb Z, 1 + 8 \mathbb Z) \in \mathbb Z_4 \times \mathbb Z_{8} $$
But what does it gives to me?

REPLY [2 votes]: The order of an element in the product is the lcm of its two components. But you know the possible orders of the elements of $\mathbb Z_4$ and $\mathbb Z_8$ because they are cylic. Namely, $1$, $2$ and $4$ are the possible orders in $\mathbb Z_4$, and $1$, $2$, $4$, $8$ are the possible orders in $\mathbb Z_8$.
It remains to see that the maximum lcm we can get from the possible pairs is $8$, which is the order of any pair $(x,1)$ for $x\in \mathbb Z_4$.


TITLE: Finding the solid angle subtended at a viewer's eyes by a movie screen
QUESTION [2 upvotes]: I've got to calculate which is the seating row of a movie theater which has the greatest angle of vision. To calculate this, I would like to consider the movie screen as having two dimensions (length and width).
While I was searching for methods to solve this, I found that Paul Nahin used the concept of solid angles to calculate something that is similar to my problem. However, in his problem, he solved for a person standing directly in front of the center of a painting, as shown here:

I would like to know if you guys have any idea on how to solve for my case, since in a movie theater, there is an inclination in the seats and there is even a small distance between the screen and the first seating row, as shown here:

REPLY [0 votes]: Without making any approximations, this problem became quite involved, but it does look doable, although the length of the process makes me suspicious that I made a mistake somewhere. If so, please point it out. But for what it's worth, here goes:
To begin, I sketched the following pictures showing a cross-sectional view and a bird's eye view of the problem and labeled the relevant lengths and angles (I assumed that the best seat within a row is the middle seat):

So the question can be stated "What is the value of $x$ that maximizes the solid angle of the screen?"
Before we can answer that, we need to figure out how to compute, for a given $x$, what the solid angle measure of the view is. In the diagram I made, $\theta$ refers to the angle of elevation of our gaze (i.e. if you imagine encasing your head within a globe, then $\theta$ would be the latitude you're looking at), and $\phi$ refers to the azimuth (i.e. the "longitude" of our gaze). So the question is, what is the region of all gazes (i.e. look directions) that wind up on a point on the screen?
You'd think at first that such a region would be a rectangle: the screen corresponds, in angle space, to a region $\theta_\textrm{bottom} \leq \theta \leq \theta_\textrm{top}$, and $\phi_\textrm{left} \leq \phi \leq \phi_\textrm{right}$, but this is not true. The trouble is that the elevation angle $\theta$ of, say, the top of the screen varies as we change $\phi$. You can see why this is if you imagine an infinitely wide screen. If you're looking at the middle-top of the screen and you shift your gaze to the right, your gaze will drop in elevation according to the effects perspective, finally disappearing into a vanishing point on the horizon.$\newcommand{\thetatop}{\theta_\mathrm{top}}$
Hence that is why I depicted $\theta_\textrm{top}$ in my picture as a function of the azimuth $\phi$, and can be computed using trigonometry. It's clear that $\tan\thetatop(\phi) = (c+s-y)/d(\phi)$, hence
$$\thetatop(\phi) = \arctan \frac{c+s-y}{d(\phi)} $$
$d(\phi)$ can be computed using the second picture. We see that $\cos \phi = (x+L)/d(\phi)$, so $d(\phi) = (x+L)/\cos \phi$. Hence
$$ \thetatop(\phi) = \arctan\left(\frac{c+s-y}{x+L} \cos \phi \right) $$
Likewise, $\theta_\mathrm{bottom}$ is similar, except we subtract $c$ (the screen height) from the numerator. That is,
$$ \theta_\mathrm{bottom}(\phi) = \arctan\left(\frac{s-y}{x+L} \cos \phi \right) $$
So we know how the $\theta$ boundaries vary against $\phi$. What are the $\phi$ boundaries? They are whatever $\phi$ values correspond to the left and right edges of the screen. Since we're in the middle seat, they will simply be negatives of each other. Let's name them $\pm \phi_w$. This can be readily computed from the second picture since
$$ \tan \phi_w = \frac{w}{x+L} $$
So
$$ \phi_w = \arctan\left(\frac{w}{x+L} \right) $$
With this figured out, we now know that the region of all possible gazes is
\begin{align}
\theta_\mathrm{bottom}(\phi) &\leq \theta \leq \theta_\mathrm{top}(\phi) \\
-\phi_w &\leq \phi \leq \phi_w
\end{align}
Thus we can set up the following double integral to compute the solid angle measure:$\newcommand{\thetabot}{{\theta_\mathrm{bot}}} \newcommand{\sat}{\operatorname{sat}}$
\begin{align}
\Omega = \iint d\Omega &= \int_{-\phi_w}^{\phi_w} \int_{\thetabot(\phi)}^{\thetatop(\phi)} \cos \theta \, d\theta\,d\phi \\
&= \int_{-\phi_w}^{\phi_w} \big[\sin \theta \big]_{\thetabot(\phi)}^{\thetatop(\phi)} \, d\phi \\
&= \int_{-\phi_w}^{\phi_w} \left[\sat\left(\frac{c+s-y}{x+L} \cos\phi \right) - \sat\left(\frac{s-y}{x+L} \cos\phi \right) \right] \, d\phi
\end{align}
where $\sat(t)$ is the sine-of-arctangent function, defined this way:
$$ \sat(t) := \sin(\arctan t) = \frac{t}{\sqrt{t^2+1}} $$
Because the integrand is even with respect to $\phi$, we can rewrite the integral as
$$ \Omega = 2\int_0^{\phi_w} \left[\sat\left(\frac{c+s-y}{x+L} \cos\phi \right) - \sat\left(\frac{s-y}{x+L} \cos\phi \right) \right] \, d\phi $$
which (with the help of a computer) I'm able to evaluate as
$$ 2\arcsin\left[\sat\left(\frac{c+s-y}{x+L} \right) \sin \phi_w \right] - 2\arcsin\left[\sat\left(\frac{s-y}{x+L} \right) \sin \phi_w \right] $$
I did not write it explicitly, but remember that $y$ is actually dependent on $x$. Specifically $\tan \alpha = y/x$. Hence $y = x \tan \alpha$. Since $\tan\alpha$ is a constant, I'll denote it by $m$, so we have $y = mx$. Also note that $\phi_w$ is dependent on $x$ too. If we substitute it in, we see we get another $\sat$ function. So our final expression (explicitly showing all the dependencies on $x$) becomes
$$ \Omega(x) = 2\arcsin\left[\sat\left(\frac{c+s-mx}{x+L} \right) \sat\left(\frac{w}{x+L} \right) \right] - 2\arcsin\left[\sat\left(\frac{s-mx}{x+L} \right) \sat\left(\frac{w}{x+L} \right) \right] $$
All that's left to do now is to maximize this function with respect to $x$. Doing that explicitly by taking the derivative and setting it equal to zero is probably too hard (I tried). So your best bet at this point is to do it numerically.


TITLE: Expected value for $\min\{|x_j|\}$ of normal distribution sample
QUESTION [1 upvotes]: Let $\{x_1, x_2, \ldots, x_n\} $ is sample of the size $n$ of  normally distributed random value:
$$X \sim N(0,1).\tag{1}$$
How to find expected value for $$m_n = \min_{1\le j \le n}\{|x_j|\}.\tag{2}$$
I have just experimental data in the table:
\begin{array}{|l|l|}
\hline
n & E(m_n) \\
\hline
1 & \approx 0.795 \\
2 & \approx 0.467 \\
5 & \approx 0.215 \\
10 & \approx 0.115 \\
100 & \approx 0.0124 \\
1000 & \approx 0.00126 \\
\hline
\end{array}
from which we can derive that
$$E(m_n) = \frac{c(n)}{n},\tag{3}$$
where roughly $0.5<c(n)<1.5$; or
$$E(m_n) = \frac{c(n)}{n+1},\tag{3'}$$
where roughly $1.2<c(n)<1.6$.
So is there any formula (or more accurate estimation) for numerator $c(n)$ in $(3)$ or $(3')$?

REPLY [2 votes]: We can find the expected value of a nonnegative random variable $X$ with $\mathbb{E}(X)=\int_{0}^{\infty}\mathbb{P}(X>x)\mathrm{d}x.$ The event $\{m_{n}>x\}$ is just the intersection of the events $\{|x_{i}|>x\},$ so assuming this is an iid sample, we have $\mathbb{P}(m_{n}>x)=\prod_{i=1}^{n}\mathbb{P}(|x_{i}|>x)=\mathbb{P}(|x_{1}|>x)^{n}=(1-\mathrm{erf}(x/\sqrt{2}))^{n}.$
Then one way to write this expected value is $$E(m_{n})=\int_{0}^{\infty}(1-\mathrm{erf}(x/\sqrt{2}))^{n}\,\mathrm{d}x=\int_{0}^{\infty}\mathrm{erfc}(x/\sqrt{2})^{n}\,\mathrm{d}x.$$
Using the bound $\mathrm{erfc}(x)\leq e^{-x^{2}}$ for all $x>0,$ we see that $E(m_{n})\leq\int_{0}^{\infty}e^{-nx^{2}/2}\mathrm{d}x$, and by symmetry of this function, this equals $\frac{1}{2}\int_{-\infty}^{\infty}e^{-nx^{2}/2}\mathrm{d}x$. Recognizing the integrand as the density of a $\mathcal{N}(0,1/n)$ random variable, we obtain $$E(m_{n})\leq\frac{1}{2}\sqrt{\frac{2\pi}{n}}\int_{-\infty}^{\infty}\sqrt{\frac{n}{2\pi}}e^{-x^{2}/(2/n)}\,\mathrm{d}x=\sqrt{\frac{\pi}{2n}}.$$ This is weaker than the estimate you made from your calculations, but has been rigorously proved. There are many bounds on the complementary error function that could be brought to bear to obtain tighter estimates and lower bounds, but this is a simple one that gives some evidence in favor of what you've already stated.


TITLE: Quadratic estimation error differential entropy
QUESTION [2 upvotes]: I am reading the book "Elements of information theory" by thomas M. Cover and Joy A. Thomas, second edition. In page 255 of the book there is a theorem:
For any random variables $X$ and estimator $\hat{X}$, $E(X - \hat{X})^2 \geq \frac{1}{2\pi e}e^{2h(X)}$.
The proof uses at its core the fact that $var(X) \geq \frac{1}{2\pi e}e^{2h(X)}$.
Edit: I include here the proof:
$E(X-\hat{X})^2 \geq \min_{\hat{X}}E(X-\hat{X})^2 = E(X-E(X))^2 = var(X) \geq \frac{1}{2\pi e}e^{2h(X)}$
Right after the proof there is a corollary that says: Given side information $Y$ and estimator $\hat{X}(Y)$, it follows that: $E(X - \hat{X}(Y))^2 \geq \frac{1}{2\pi e}e^{2h(X|Y)}$, however, no further explanation is supplied for how to derive this corollary from the previous theorem.
I assume that if I fight the equations for a while I might be able to derive the corollary from something, however I can't manage to find a simple way to derive it from the previous theorem. I wonder if I miss something very simple here.
So far my best attempt was to split $(X - \hat{X}(Y))$ into $(X - Y) + (Y - \hat{X}(Y))$.

REPLY [0 votes]: $$h(X|Y) = h(X-E(X|Y)|Y),$$ since shift (as opposed to scaling) does not change differential entropy. Now apply previous result (and Jensen's inequality) in the following fashion:
$$ h(X|Y) = h(X-\hat X|Y) \leq E\left(\frac 12 \log (2\pi e E( (X - \hat{X})^2|Y)) \right) \leq \frac 12 \log (2\pi e E(X-\hat{X})^2)).$$


TITLE: Prove the existence of minimal height of a convex polygon
QUESTION [2 upvotes]: Suppose we have a polygon in $\mathbb{R}^n$. Obviously, we can always trap the polygon into two parallel hyperplanes perpendicular to any given direction. Now the question is, how to prove the existence of the two hyperplanes whose distance is minimum among all parallel hyperplanes sandwiching the polygon?
I think this should be a silly question and basic calculus knowledge should be enough, but I tried a lot and just couldn't make the question mathematically rigorous. There is the same question asked here https://stackoverflow.com/questions/23644372/rotating-a-polygon-to-its-minimal-height, but it is about the algorithm solution, which is not what I want.
Anyone can give me a rigorous proof? Thanks in advance!

REPLY [0 votes]: Let $P$ be the polygon, which is compact.
Let $\phi(h) = \max_{x,y \in P} \langle h , x-y \rangle$. It is straightforward to show that $\phi$ is continuous.
The set $S=\partial B(0,1)$ is compact, hence $\phi$ is minimised on $S$ at some $\hat{h} \in S$.
Now choose $x,y \in P$ such that $\phi(\hat{h}) = \langle \hat{h} , x-y \rangle$. Then the requisite hyperplanes are
$H_1 = \{ z | \langle \hat{h} , z-x \rangle = 0 \}$ and
$H_2 = \{ z | \langle \hat{h} , z-y \rangle = 0 \}$.
Addendum: To see that $\phi$ is continuous, suppose $h_n \to h$, and let $K$ be any subsequence (that is, $K$ is an infinite subset of $\mathbb{N}$). Let
$x_n,y_n \in P$ such that $\phi(h_n) = \langle h_n , x_n-y_n \rangle$. Since
$P$ is compact, there is some subsequence $K' \subset K$ such that $x_n \overset{K'} {\rightarrow} x$,  $y_n \overset{K'}{\rightarrow} y$, and so
$\langle h_n , x_n-y_n \rangle \overset{K'} {\rightarrow} \langle h , x-y \rangle \le \phi(h)$. Since $\langle h_n , x_n-y_n \rangle \ge \langle h_n , w-z \rangle$ for any $w,z \in P$, we see that
$\langle h , x-y \rangle \ge \langle h , w-z \rangle$ for any $w,z \in P$,
and so $\phi(h) = \langle h , x-y \rangle$, that is
$\phi(h_n) \overset{K'}{\rightarrow} \phi(h)$. Since $K$ was an arbitrary subsequence, we have $\phi(h_n) {\rightarrow} \phi(h)$.


TITLE: Showing stability of non-constant matrix (first order) which method?
QUESTION [0 upvotes]: Given a  matrix $A \in \Bbb R^{3 \times 3}$ , which is non-constant of the form $$y'(t)=A(t)y(t)$$
  I want to check the stability of the trivial solution.

My problem is the following:
If $$y(t)=\begin{pmatrix}y_1(t) \\ y_2(t) \\ y_3(t)\end{pmatrix}$$
My matrix consists not  of $t$ but $y_1,y_2,y_3$ and I wanted to know how do I show this ?
I mean I can't calculate the eigenvalues because of the $y_i's$, this method is only for constant matrices as far as I know.
Edit:
My matrix looks like:
$$\begin{pmatrix}2(1-y_2-y_1) & 1 & 0 \\ -y_2 & -1 & -2 \\ 1&0 & -1\end{pmatrix}$$

REPLY [0 votes]: The linearization of the system at the origin gives the equation
$$
\begin{pmatrix}y_1'\\y_2'\\y_3'\end{pmatrix}=
\begin{pmatrix}2 & 1 & 0 \\ 0 & -1 & -2 \\ 1&0 & -1\end{pmatrix}
\begin{pmatrix}y_1\\y_2\\y_3\end{pmatrix}.
$$
The eigenvalues of the matrix are $0$ and $\pm\sqrt3$. Since one of the eigenvalues is positive (or if you prefer, since one of the eigenvalues has positive real part), the origin is unstable. So the original system has an unstable manifold and the origin is again unstable.


TITLE: Integral representation of a limit
QUESTION [4 upvotes]: Let
$$
\tilde{\mathbb C}={\mathbb C}\smallsetminus (-\infty,0]
$$
the complex plane without the negative real axis. Let $V$ denote the set of all holomorphic functions $f:\tilde{\mathbb C}\to{ \mathbb C}$ such that
$$
f^+(0)=\lim_{\lambda\searrow 0}f(\lambda)
$$
exists.
If $K\subset \{\mathrm{Re}(z)>0\}$ is a compact subset which is not finite, then a given $f\in V$ is uniquely determined by its restriction to $K$. Therefore it is reasonable to ask:
Does there exists a compact set $K$ and a complex valued Radon measure $\mu$ on $K$ such that
$$
f^+(0)=\int_K f(\lambda)\,d\mu(\lambda)
$$
holds for every $f\in V$?

REPLY [1 votes]: Here is a simple proof without Hahn-Banach. I assume that the complex Radon measure is finite. Then your representation would imply
$$|\lim_{\lambda\to 0}f(\lambda)|\leq C\max_{K}|f(z)|,$$
which is evidently not true.


TITLE: Help With Difficult Integration
QUESTION [0 upvotes]: I'm working through a homework problem and came to the point where I needed to evaluate this integral.
$$\int_{-(z^2 - 2z)}^{z^2 - 2z}    \dfrac{-1}{\alpha z} \dfrac{1}{\sqrt{ \dfrac{2}{m} \bigg (E - z^2 - 2z - \dfrac{l^2 \alpha^2}{2m(\ln(z))^2} \bigg )} } dz$$
I'll give some background on the problem quickly. I'm trying to find the equation of motion of a particle of mass $m$ in the 1D Morse potential,
$$V(x) = D \big ( e^{-2\alpha x} - 2e^{-\alpha x} \big ).$$
In consulting with my professor, he gave me the hint of setting  $z = e^{-\alpha x}$ and solving for $x$. This gives $x = -\frac{1}{\alpha}\ln(z)$. He then suggested I use an equation from Goldstein's Classical Mechanics, which was
$$ t = \int_{r_0}^r \dfrac{dr}{\sqrt{\dfrac{2}{m}\big ( E - V - \dfrac{l^2}{2mr^2} \big )}}$$
I did as he suggesting using $r=x$. I chose my integration bounds to be the turning points of the potential, $E = \pm V(x) = \pm(z^2 -2z)$. Now I'm stuck with this seemingly horrendous integral that I don't know how to solve. Once I get this, it's a simple matter of inverting that equation from $t(x)$ to $x(t)$ to get my equation of motion.
Any help would be greatly appreciated.
Thank you!

REPLY [1 votes]: There are some issues with your integral, like how $z$ is both present as the dummy variable and in the limit of integration. If I understand correctly, $z$ is real, so how can you integrate something with $\ln z$ over a negative domain? So instead, I'll address the modified integral from the comments. We need
$$ \int_{-(z^2-2z)}^{(z^2-2z)}\frac{1}{\alpha t\sqrt{\frac2m(E-t^2+2t)}}dt. $$
Notice that the argument of the square root in the denominator can be expressed as $(\frac{2E}m+1)-(t-1)^2$, allowing an easy trigonometric substitution to complete the integral. Then, just integrate your resulting integral just like you typically would in an elementary calculus course. (I have a feeling that you got your signs wrong somewhere, since the limits of integration do not work out nicely at all, but since I'm not acquainted with the physics of the problem I can't say for certain. Either way, this is probably the best method.)


TITLE: The range of a continuous even function
QUESTION [0 upvotes]: Is there any rule that says that this range cannot be $\mathbb{R}$? Apparently there is, but then what about functions like $\log |x|$?

REPLY [3 votes]: Consider $f(x) = x\sin x$. It is easy to check that $f$ is even.
Also, $f$ is continuous and obtains arbitrarily large positive and negative values. You can use the intermediate value theorem to show that every real number is in the range of $f$.
Hence, $f$ is a continuous even function whose range is $\mathbb{R}$. Therefore, there is no rule that a continuous even function cannot have all of $\mathbb{R}$ as its range.


TITLE: Resolvent of Dirichlet Laplacian via Fourier Transform
QUESTION [5 upvotes]: For $-\Delta$ considered as a self-adjoint operator on $L^2(\mathbb{R}^d)$, one may write it's resolvent as the Fourier multiplier $g(\xi)=(|\xi|^2-z)^{-1}$. Now let $\Omega\subset \mathbb{R}^d$ be a bounded domain and let $-\Delta^\Omega$ be the Dirichlet Laplacian on $\Omega$, defined, for instance, via the Friedrichs Extension Theorem. Is there an expression for $(-\Delta^\Omega-z)^{-1}$ in terms of the operator $g(\xi)$ and some multiplication operator $f(x)$? If not, can we approximate $(-\Delta^\Omega-z)^{-1}$ by operators of the form $f(x)g(\xi)$? It seems that something like $f(x)g(\xi)f(x)$ might work for $f$ some bump function of $\Omega$, but the fact that $-\Delta^\Omega$ has to be defined abstractly leaves me unsure of the details.
My motivation is that I would like to prove that $-\Delta^\Omega$ has compact resolvent from the fact that an operator of the form $f(x)g(\xi)$ is compact if $f$ and $g$ vanish at $\infty$. Since one usually proves that the resolvent is compact from the Rellich embedding theorem and one can prove this theorem from the above compactness criterion, I have reason to believe this should work.

REPLY [0 votes]: Try writing the Fourier multiplier as an expression involving the heat kernel:
$$
         \int_0^{\infty}e^{-t(|\xi|^2-z)}dt=\frac{1}{|\xi|^2-z}.
$$


TITLE: Probability of exponential distribution less than normal distribution.
QUESTION [2 upvotes]: $X$ is the exponential distribution $e^{-x}$, $Y$ is the normal distribution with mean $0$ and variance $1$. $X$ and $Y$ are independent. How do I find $P(X<Y+1)$?
My attempt:
$$\int_{-1}^{\infty} F_X(y+1) f_Y(y)    dy$$
$$=\int_{-1}^{\infty} (1-e^{-(y+1)}) f_Y(y)    dy$$
$$=\phi(1)-\int_{-1}^{\infty} (e^{-(y+1)}) f_Y(y)    dy$$
Am I on the right track? How do I proceed from here?

REPLY [2 votes]: $$\phi(1)-\int_{-1}^{\infty} (e^{-(y+1)}) f_Y(y)    dy $$
$$ = \phi(1)-\frac{1}{\sqrt{2\pi}}\int_{-1}^{\infty} e^{-\frac{1}{2}(y^2+2y+2)}dy$$
$$ =  \phi(1)-\frac{e^{-\frac 1 2}}{\sqrt{2\pi}}\int_{0}^{\infty} e^{-\frac{1}{2}t^2}  dt $$
where $t=y+1$.
$$ = \phi(1) - \frac{e^{-\frac 1 2}}{2}$$


TITLE: What is the distribution of the number of inversions for a random permutation?
QUESTION [1 upvotes]: I am solving some questions in Introduction to Probability Models by Sheldon Ross as part of my preparation for my upcoming exams. I am stuck on this problem number 62 from Chapter 2.
If $\{a_i\}$ is an ascending ordered set of numbers. Then $N_i$ is the number of terms $a_k \ ;\ k<i$ such that $a_k$ appears after $a_i$ in the permutation.

My attempt
So I could see that $N_i < i$, since we only have that many terms that can cause an inversion for $a_i$.
I tried solving for small values of $i$. We can see $P(N_1 = 0) = 1$ since no $a_k$ exists such that $k<1$, since $a_1$ is the smallest number.
$P(N_2 = 0) = \frac{((n-1)+(n-2)+\cdots+1)(n-2)!}{n!} = \frac{1}{2}$
So we can fix $a_2$'s position in the permutation and notice there are $n-1$ free positions for $a_1$. The more to the right we fix the position the less possible places $a_1$ can take and still only cause $0$ inversions. The rest $n-2$ numbers can take any of the other places so that gives us the number of favorable permutations. The denominator is the total number of possible permuatations.
Now for the case of $N_i=r$, we have $i-1$ numbers that are potential candidates for an inversion.
To get exactly $r$ inversions we need to select which $r$ positions of the available $k$ is taken by these numbers. Then which $r$ of the $i-1$ numbers comes after $a_i$ in the permutation. We can now permute the positions of these $r$ numbers among themselves and the rest $(n-r)$ terms can also take any position they want.
We now sum over all the possible $k$'s which is determined by fixing the position of $a_i$.
$P(N_i = r) = \frac{\sum_{k=r}^{n-i-r} {k \choose r}{i-1 \choose r} (n-r)! r!}{n!}$
But this seems overly complex and wrong. Can you help with finding the distribution of $Ni$? Also I planned to show all $n$ random variables are independent (part a) by first showing that $N_i$ is independent to $N_{i-1}, N_{i-2}, \cdots, N_1$.

REPLY [3 votes]: When considering $N_i$, you only care about the relative positions of $a_1, \ldots, a_i$; you can ignore $a_{i+1}, \ldots, a_n$ and consider the $i!$ permutations of $a_1, \ldots, a_i$. The event that $N_i = r$ is the same as $a_i$ appearing in position $i-r$ among these $i$ objects, which is $\frac{1}{i}$. So $N_i$ is uniform over $\{0, 1, \ldots, i-1\}$.

The above argument is actually rigorous. But if you want to count everything explicitly, here is a more detialed explanation. The event $N_i = r$ means that exactly $r$ of $a_1, \ldots, a_{i-1}$ appears after $a_i$. In particular, this means $i-1-r$ of $a_1, \ldots, a_{i-1}$ appear before $a_i$ and $r$ of them appear after $a_i$. So among these $i$ objects, $a_i$ appears in the $i-r$ position, and there are $(i-1)!$ ways to position the remaining $a_1, \ldots, a_{i-1}$.
We have so far counted the number of ways to order $a_1, \ldots, a_i$ among themselves. Now let's consider all $n$ objects. There are $\binom{n}{i}$ ways to choose $i$ positions for $a_1, \ldots, a_i$; the remaining $n-i$ positions will be filled by $a_{i+1}, \ldots, a_n$, and there are $(n-i)!$ ways to order these objects.
Combining everything and dividing by $n!$ yields
$$\frac{1}{n!}  (i-1)! \binom{n}{i} (n-i)!
= \frac{1}{n!} (i-1)! \frac{n!}{i!}
= \frac{1}{i}.$$


text:
Here we gather smaller statements that are used multiple times in \autoref{sec:semi_positivity}, \autoref{sec:nefness_threshold} and \autoref{sec:positivity}.

\subsection{Normality of total spaces}

\begin{lemma}
\label{lem:reducing_reduced_fibers}
In the situation of \autoref{notation:CM_semi_positivity}, there exists a finite morphism from  a smooth projective curve $\tau :S \to T$ such that if $g : Y \to S$ is the normalized pullback of $f$ (so the normalization of the pullback) and $\pi : Y \to X$ the induced morphism, then $g$ has reduced fibers, and there is an effective $\bQ$-divisor $\Gamma$ on $Y$ such that
\begin{enumerate}
 \item \label{itm:reducing_reduced_fibers:canonical_bundle_formula} $\pi^* (K_{X/T} + \Delta) = K_{Y/S} + \Gamma$,
 \item \label{itm:reducing_reduced_fibers:CM} $\lambda_g = \sigma^* \lambda$, where $\lambda_g$ is the CM line bundle for $g$.
\end{enumerate}
\end{lemma}

\begin{proof}
Let $\tau$ be any finite cover such that at the closed points $t \in T$ over which the fiber $X_t$ is non reduced, the ramification order of $\tau$ is divisible by all the multiplicities of all the components of $\tau$. Then, $g$ will have reduced fibers, and \autoref{sec:base_change_relative_canonical_smooth_base}
 implies the existence of $\Gamma$ (denoted by $\Delta_Z$ there). Finally, \autoref{prop:relative_canonical_base_change_normal}.\autoref{itm:relative_canonical_base_change:base_change} yields point \autoref{itm:reducing_reduced_fibers:canonical_bundle_formula}, and \autoref{prop:CM_base_change}.\autoref{itm:CM_base_change:smooth_base} yields point \autoref{itm:reducing_reduced_fibers:CM}.

\end{proof}

\begin{lemma}
\label{lem:normal_stable_pullback_fiber_product}
If $f : X \to T$ is a surjective morphism from a normal variety to a smooth projective curves with reduced fibers, $m>0$ is an integer and $ \tau : S \to T$ is a finite morphism from another smooth curve, then
\begin{enumerate}
\item \label{itm:normal_stable_pullback_fiber_product:pullback} $X \times_T S$ is normal, and
\item \label{itm:normal_stable_pullback_fiber_product:fiber_product} $X^{(m)}$ is normal (see \autoref{sec:product_notation} for the product notation).
\end{enumerate}

\end{lemma}

\begin{proof}
First we note that $f$ is flat and hence so is $f^{(m)} : X^{(m)} \to T$ by induction on $m$ and the stability of flatness under base-change.

We know that a variety $Z$ is normal if and only if  it is $S_2$ and $R_1$. In the particular case, when $Z$ maps to a smooth curve $U$ via  a flat morphism $g$, then $Z$ is $S_2$ if and only if the general fibers of $g$ are $S_2$ and the special ones are $S_1$ (so without embedded points) \cite[6.3.1]{Grothendieck_Elements_de_geometrie_algebrique_IV_II} \cite[12.2.4.i]{Grothendieck_Elements_de_geometrie_algebrique_IV_III}, and it is $R_1$ if the general fibers are $R_1$ and the special ones are $R_0$ (so reduced) \cite[12.2.4.ii]{Grothendieck_Elements_de_geometrie_algebrique_IV_III}. It is immediate then that this characterization of $S_2$ and $R_1$ propagates both to fiber powers and to base-changes.
\end{proof}

\subsection{Semi-positivity engine}

\begin{proposition}
\label{prop:semi_positivity_engine_downstairs}
Let $f : (X, \Delta) \to T$  be a surjective morphism from a normal, projective pair to a smooth curve such that $(X_t,\Delta_t)$ is klt for general $t \in T$ (as $X_t$ is normal for $t \in T$ general, $\Delta$ is $\bQ$-Cartier at the codimension $1$ points of $X_t$, and hence $\Delta_t$ makes sense), and let $L$ be a Cartier divisor on $X$ such that $L- K_{X/T} - \Delta$ is an $f$-ample and nef $\bQ$-Cartier divisor. Then, $f_* \sO_X(L)$ is a nef vector bundle.
\end{proposition}

\begin{proof}
According to \autoref{lem:reducing_reduced_fibers} we may assume that the fibers of $f$ are reduced. According to \cite[Lem 3.4]{Patakfalvi_Semi_positivity_in_positive_characteristics}, it is enough to prove that for all integers $m>0$, the following vector bundle is generated at a general $t \in T$ by global sections:
\begin{equation*}
\omega_T (2t) \otimes \bigotimes_{i=1}^m f_* \sO_X(L) \cong

\underbrace{f^{(m)}_* \sO_{X^{(m)}} \left( L^{(m)} + \left( f^{(m)} \right)^* K_T + 2 X^{(m)}_t \right)}_{\textrm{\cite[Lem 3.6]{Kovacs_Patakfalvi_Projectivity_of_the_moduli_space_of_stable_log_varieties_and_subadditvity_of_log_Kodaira_dimension}, and see \autoref{sec:product_notation} for the fiber product notation}}
.
\end{equation*}
For that it is enough to prove that the natural restriction homomorphism $
H^0\left( X^{(m)}, N \right) \to H^0\left( X_t^{(m)}, N_t \right)$
is surjective, where
\begin{equation*}
 N:=L^{(m)} + \left( f^{(m)} \right)^* K_T + 2 X^{(m)}_t  = K_{X^{(m)}} +  \Delta^{(m)} +  (L - K_{X/T} - \Delta)^{(m)} + 2 X_t^{(m)}.
\end{equation*}
We note here that  according to \autoref{lem:normal_stable_pullback_fiber_product}, $X^{(m)}$ is normal. Furthermore, $K_{X^{(m)}} +  \Delta^{(m)} = (K_{X/T} + \Delta)^{(m)} +  \left( f^{(m)} \right)^* K_T$ is $\bQ$-Cartier. We also note that the only generality property of $t$ that we use below is that $X_t$ is normal, $X_t \subsetneq \Supp \Delta_t$ and  $(X_t, \Delta_t)$ is klt. Hence, at this point, we fix a $t$ with such properties.

Set $\sI:=\sJ_{\left( X^{(m)}, \Delta^{(m)}\right)}$, where $\sJ$ denotes the multiplier ideal of the corresponding pair. Then for the above surjectivity the next diagram, the top row of which is exact, shows that it is enough to prove the vanishing of $H^1\left( X^{(m)}, \sI \otimes \sO_{X^{(m)}} \left(  N - X_t^{(m)}\right)\right)$.
\begin{equation*}
\xymatrix@C=10pt@R=10pt{
 H^0\left( X^{(m)}, \sI \otimes   \sO_{X^{(m)}}\left( N \right)\right) \ar[r] \ar@{^(->}[d]

&

H^0\left( X^{(m)}_t,  N|_{X^{(m)}_t}\right) \ar[r]

& H^1\left( X^{(m)}, \sI \otimes \sO_{X^{(m)}}\left(  N - X^{(m)}_t\right)\right)  \\

H^0\left( X^{(m)},   N \right) \ar[ur]
}
\end{equation*}
We note that here we used that $\left(X_t^{(m)}, \Delta_t^{(m)} \right)$ is klt by \autoref{lem:product}, and hence by inversion of adjunction \cite[Thm 5.50]{Kollar_Mori_Birational_geometry_of_algebraic_varieties} so does $\left(X^{(m)}, \Delta^{(m)} \right)$ in a neighborhood of $X_t^{(m)}$. This then implies that $\sI$ is trivial in a neighborhood of $X_t^{(m)}$.

We conclude by noting that the above cohomology vanishing is given by Nadel-vanishing as
\begin{equation*}
 N -  X_t^{(m)} = K_{X^{(m)}} +  \Delta^{(m)} +  \underbrace{(L - K_{X/T} - \Delta)^{(m)} +  X_t^{(m)}}_{\textrm{ample}}.
\end{equation*}

\end{proof}

\begin{corollary}
\label{cor:semi_positivity_engine_upstairs}
Let $f : (X, \Delta) \to T$  be a surjective morphism from a normal, projective pair to a smooth curve such that $(X_t,\Delta_t)$ is klt for some (or equivalently general) $t \in T$, and let $L$ be a $\bQ$-Cartier divisor on $X$  such that
\begin{enumerate}
\item \label{itm:semi_positivity_engine_upstairs:sections_non_zero}  $L_t$ is globally generated for $t \in T$ general,
 \item $L  - K_{X/T} - \Delta$ is an $f$-ample and nef $\bQ$-Cartier divisor, and

 \item there is  a $\bQ$-Cartier divisor $N$ on $T$ such that $L + f^* N $ is Cartier.
\end{enumerate}
Then $L $ is nef.
\end{corollary}

\begin{proof}
 According to \autoref{lem:reducing_reduced_fibers} we may assume that the fibers of $f$ are reduced, and by further pullback (using \autoref{lem:normal_stable_pullback_fiber_product}.\autoref{itm:normal_stable_pullback_fiber_product:fiber_product}) we may also assume that $N$ is Cartier, whence $L$ is also Cartier. Then, we may apply \autoref{prop:semi_positivity_engine_downstairs} yielding that $f_* \sO_X(L)$ is nef.
Then, our proof is concluded by the natural homomorphism $f^* f_* \sO_X(L) \to \sO_X(L)$, which is surjective over a non-empty open set of $T$ according to assumption \autoref{itm:semi_positivity_engine_upstairs:sections_non_zero} and the fact that cohomology and base change always holds over an opens set.
 \end{proof}

config: arxiv


TITLE: Given a complex function in $x$ and $y$ change into $z$, where $z=x+iy$
QUESTION [1 upvotes]: I have $$f(x,y)=\sinh{x}\sin{y}+e^y\sin{x}+i(e^y\cos{x}-\cosh{x}\cos{y}+C)$$ for $C$ a constant.
Now if I want to write this in terms of the complex variable $z$ I've heard that I can just set $x=z$ and $y=0$. Is this true? If so, this makes this terribly simple. We are given that the above equation is analytic.

REPLY [2 votes]: $$g(z)=f(\Re(z),\Im(z))$$
If $z$ is real, then
$$g(z)=f(z,0)$$
Since $g$ and $z\rightarrow f(z,0)$ are both analytic and they're equal on the real line. Then they're equal on the complex plane.