text:
\section{The invariant torus theorem}

Let $\mathcal{H}$ be the space of germs along $\ensuremath{\mathrm{T}}_0^n := \ensuremath{\mathbb{T}}^n \times
\{0\}$ of real analytic Hamiltonians in $\ensuremath{\mathbb{T}}^n \times \ensuremath{\mathbb{R}}^n =
\{(\theta,r)\}$ ($\ensuremath{\mathbb{T}}^n=\ensuremath{\mathbb{R}}^n/2\pi\ensuremath{\mathbb{Z}}^n$). The vector field associated
with $H \in \mathcal{H}$ is
$$\vec H : \quad \dot\theta = \partial_r H, \quad \dot
r=-\partial_\theta H.$$

For $\alpha\in\ensuremath{\mathbb{R}}^n$, let $\mathcal{K}$ be the affine subspace of
Hamiltonians $K\in \mathcal{H}$ such that $K|_{\ensuremath{\mathrm{T}}_0^n}$ is constant
(i.e.  $\ensuremath{\mathrm{T}}_0^n$ is invariant) and $\vec K|_{\ensuremath{\mathrm{T}}_0^n}=\alpha$. Those
Hamiltonians are characterized by their first order expansion along
$\ensuremath{\mathrm{T}}_0^n$, of the form $c + \alpha \cdot r$ for some $c\in\ensuremath{\mathbb{R}}$, that
is, their expansion is constant with respect to $\theta$ and the
coefficient of $r$ is $\alpha$.

Let
$$\mathrm{D}_{\gamma,\tau} = \{\alpha \in \ensuremath{\mathbb{R}}^n, \; \forall k \in
\ensuremath{\mathbb{Z}}^n \setminus \{0\} \; |k \cdot \alpha| \geq \gamma |k|^{-\tau}\},
\quad |k|:= |k|_1 = |k_1| + \cdots + |k_n|.$$ If $\tau>n-1$, the set
$\cup_{\gamma >0} \mathrm{D}_{\gamma,\tau}$ has full measure
(\cite[p.~83]{Arn63}). See appendix~\ref{app:arithmetics}.

\begin{theorem}[\cite{Kol54,Chi08}]\label{thm:kolmogorov}
  Let $\alpha\in \mathrm{D}_{\gamma,\tau}$ and
  $K^o\in\mathcal{K}$ such that the averaged Hessian
  $$\int_{\ensuremath{\mathbb{T}}^n} \frac{\partial^2 K^o}{\partial r^2}(\theta,0)\, d\theta$$
  is non degenerate. Every $H\in\mathcal{H}$ close to $K^o$ possesses
  an $\alpha$-quasiperiodic invariant torus.
\end{theorem}

This theorem has far reaching consequences. In particular it has led
to a partial answer to the long standing question of the stability of
the Solar system (\cite{Arn64, Fej04, CC07}). See
\cite{Bos86,Sev03,Lla99} for references and background.

Kolmogorov's theorem is a consequence of the following normal form.
Let $\mathcal{G}$ be the space of germs along $\ensuremath{\mathrm{T}}_0^n$ of real
analytic exact symplectomorphisms $G$ in $\ensuremath{\mathbb{T}}^n \times \ensuremath{\mathbb{R}}^n$ of the
following form:
$$G(\theta,r) = (\varphi(\theta), {}^t\varphi'(\theta)^{-1}
(r+\rho(\theta))),$$ where $\varphi$ is a real analytic isomorphism of
$\ensuremath{\mathbb{T}}^n$ fixing the origin, and $\rho$ is an exact $1$-form on $\ensuremath{\mathbb{T}}^n$.

\begin{theorem}[Herman]\label{thm:herman0}
  Let $\alpha\in \mathrm{D}_{\gamma,\tau}$ and $K^o\in\mathcal{K}$.
  For every $H\in\mathcal{H}$ close enough to $K^o$, there exists a
  unique $(K,G,\beta) \in \mathcal{K} \times \mathcal{G} \times \ensuremath{\mathbb{R}}^n$
  close to $(K^o,\id,0)$ such that
  $$H=K \circ G + \beta \cdot r$$
  in some neighborhood of $G^{-1}(\ensuremath{\mathbb{T}}_0^n)$. Moreover, $\beta$ depends
  $C^1$-smoothly on $H$.
\end{theorem}

In other words, the orbits of Hamiltonians $K \in \mathcal{K}$ under
the action of symplectomorphisms of $\mathcal{G}$ locally form a
subspace of finite codimension $n$. The offset $\beta \cdot r$ usually
breaks the dynamical conjugacy between $K$ and $H$; hence Herman's
normal form is of geometrical nature and can be called a \emph{twisted
  conjugacy}. The strategy for deducing the existence of an
$H$-invariant torus (namely, $G^{-1}(\ensuremath{\mathrm{T}}^n_0)$) from that of a
$K$-invariant torus (namely, $\ensuremath{\mathrm{T}}^n_0$) is to show that $\beta$
vanishes on some subset of large measure in some parameter space (in
some cases, the frequency $\alpha$ cannot be fixed and needs to be
varied).

In the paper, $O(r^n)$ will denote the ideal of functions of
$(\theta,r)$ of the $n$-th order with respect to $r$.

\begin{proof}[Proof of theorem~\ref{thm:kolmogorov} assuming
  theorem~\ref{thm:herman0}]
  Let $K^o_2(\theta) := \frac{1}{2} \frac{\partial^2 K^o}{\partial
    r^2}(\theta,0)$. Let $F$ be the analytic function taking values
  among sym\-metric bilinear forms, which solves the cohomological
  equation $\mathcal{L}_\alpha F = K^o_2 - \int_{\ensuremath{\mathbb{T}}^n_0} K^o_2 \,
  d\theta$ (see lemma~\ref{lm:cohomological}), and $\varphi$ be the
  germ along $\ensuremath{\mathrm{T}}^n_0$ of the (well defined) time-one map of the flow
  of the Hamiltonian $F(\theta) \cdot r^2$. The map $\varphi$ is
  symplectic and restricts to the identity on $\ensuremath{\mathrm{T}}_0^n$. At the
  expense of substituting $K^o \circ \varphi$ and $H \circ \varphi$
  for $K^o$ and $H$ respectively, one can thus assume that
  $$K^o = c + \alpha \cdot r + Q \cdot r^2 +
  O(r^3), \quad Q:= \int_{\ensuremath{\mathrm{T}}_0^n} K^o_2(\theta) \, d\theta.$$ The
  germs so obtained from the initial $K^o$ and $H$ are close to one
  another.

  Consider the family of trivial perturbations obtained by translating
  $K^o$ in the direction of actions: $$K^o_R(\theta,r) :=
  K^o(\theta,R+r), \quad R\in\ensuremath{\mathbb{R}}^n,\; \mbox{$R$ small},$$ and its
  approximation obtained by truncating the first order jet of $K^o_R$
  along $\ensuremath{\mathrm{T}}^n_0$ from its terms $O(R^2)$ which possibly depend on
  $\theta$:
  $$\hat K^o_R (\theta,r) := \left(c + \alpha \cdot R \right) +
  \left(\alpha + 2Q \cdot R \right) \cdot r + O (r^2) = K^o_R +
  O(R^2).$$ For the Hamiltonian $\hat K^o_R$, $T^n_0$ is invariant and
  quasiperiodic of frequency $\alpha + 2Q \cdot R$. Hence the Herman
  normal form of $\hat K^o_R$ with respect to the frequency $\alpha$
  is
  $$\hat K^o_R = \left( \hat K^o_R-\hat\beta^o_R \cdot r \right) \circ
  \id + \hat \beta^o_R \cdot r, \quad \hat \beta^o_R:= 2 Q \cdot R.$$
  By assumption the matrix $\left.\frac{\partial \hat\beta^o}{\partial
      R}\right|_{R=0} = 2Q$ is invertible and the map $R \mapsto
  \hat\beta^o(R)$ is a local diffeomorphism.

  Now, theorem~\ref{thm:herman0} asserts the existence of an analogous
  map $R \rightarrow \beta(R)$ for $H_R$, which is a small
  $C^1$-perturbation of $R \mapsto \hat\beta^o(R)$, and thus a local
  diffeomorphism, with a domain having a lower bound locally uniform
  with respect to $H$. Hence if $H$ is close enough to $K^o$ there is
  a unique small $R$ such that $\beta=0$. For this $R$ the equality
  $H_R=K \circ G$ holds, hence the torus obtained by translating
  $G^{-1}(\ensuremath{\mathrm{T}}_0^n)$ by $R$ in the direction of actions is invariant
  and $\alpha$-quasiperiodic for $H$.
\end{proof}

\begin{exercise}
  Simplify this proof when $K^o = K^o(r)$ is integrable.
\end{exercise}

It is the aim of the rest of the paper to prove
theorem~\ref{thm:herman0}, by locally inverting some operator
$$\phi : (K,G,\beta) \mapsto H = K \circ G + \beta \cdot r$$
when $\alpha$ is diophantine.

\section{Complexification and the functional setting}
\label{sec:setting}

\emph{For various sets $U$ and $V$, $\mathcal{A}(U,V)$ will denote the
  set of continuous maps $U \rightarrow V$ which are real analytic on
  the interior $\mathring{U}$, and
  $\mathcal{A}(U):=\mathcal{A}(U,\ensuremath{\mathbb{C}})$.}

Recall notations for the abstract torus and its embedding in the phase
space:
$$\ensuremath{\mathbb{T}}^n = \ensuremath{\mathbb{R}}^n / 2\pi \ensuremath{\mathbb{Z}}^n \quad \mbox{and} \quad \ensuremath{\mathrm{T}}^n_0 = \ensuremath{\mathbb{T}}^n \times
\{0\} \subset \ensuremath{\mathbb{T}}^n \times \ensuremath{\mathbb{R}}^n.$$

Define complex extensions
$$\ensuremath{\mathbb{T}}^n_\ensuremath{\mathbb{C}} = \ensuremath{\mathbb{C}}^n / 2 \pi\ensuremath{\mathbb{Z}}^n \quad \mbox{and} \quad \ensuremath{\mathrm{T}}^n_\ensuremath{\mathbb{C}} = \ensuremath{\mathbb{T}}^n_\ensuremath{\mathbb{C}}
\times \ensuremath{\mathbb{C}}^n$$
as well as bases of neighborhoods
$$\ensuremath{\mathbb{T}}^n_s = \{\theta \in \ensuremath{\mathbb{T}}^n_\ensuremath{\mathbb{C}}, \; \max_{1\leq j\leq n} |\Im
\theta_j| \leq s\} \quad \mbox{and} \quad \ensuremath{\mathrm{T}}^n_s = \{ (\theta,r)\in
\ensuremath{\mathrm{T}}^n_\ensuremath{\mathbb{C}}, \, |(\theta,r)| \leq s\}, $$
with $|(\theta,r)| := \max_{1\leq j \leq n} \max \left( |\Im
  \theta_j|, |r_j| \right)$.

\subsection{Spaces of Hamiltonians}

-- Let $\mathcal{H}_s = \mathcal{A}(\ensuremath{\mathrm{T}}^n_s)$, endowed with the Banach
norm
$$|H|_s := \sup_{(\theta,r)\in \ensuremath{\mathrm{T}}^n_s} |H(\theta,r)|,$$
so that $\mathcal{H}$ be the inductive limit of the spaces
$\mathcal{H}_s$.

-- For $\alpha\in\ensuremath{\mathbb{R}}^n$, let $\mathcal{K}_s$ be the affine
subspace consisting of those $K \in \mathcal{H}_s$ such that
$K(\theta,r) = c + \alpha \cdot r + O(r^2)$ for some $c\in\ensuremath{\mathbb{R}}$.

-- If $G$ is a real analytic isomorphism on some open set of
$\ensuremath{\mathrm{T}}^n_\ensuremath{\mathbb{C}}$ and if $G$ is transverse to $\ensuremath{\mathrm{T}}_s^n$, let
$G^*\mathcal{A}(\ensuremath{\mathrm{T}}_s^n) := \mathcal{A}(G^{-1}(\ensuremath{\mathrm{T}}_s^n))$ be endowed
with the Banach norm
$$|H|_{G,s} := |H \circ G^{-1}|_s.$$

\subsection{Spaces of conjugacies}

\subsubsection{Diffeomorphisms of the torus}

Let $\mathcal{D}_s$ be the space of maps $\varphi\in
\mathcal{A}(\ensuremath{\mathbb{T}}^n_s,\ensuremath{\mathbb{T}}^n_\ensuremath{\mathbb{C}})$ which are analytic isomorphisms from
$\mathring\ensuremath{\mathbb{T}}^n_s$ to their image and which fix the origin.

Let also
$$\chi_s :=\{v \in \mathcal{A}(\ensuremath{\mathbb{T}}^n_s)^n, \;
v(0)=0\}$$ be the space of vector fields on $\ensuremath{\mathbb{T}}^n_s$ which vanish at
$0$, endowed with the Banach norm
$$|v|_s := \max_{\theta\in\ensuremath{\mathbb{T}}^n_s} \max_{1\leq j \leq n}
|v_j(\theta)|.$$
According to corollary~\ref{prop:isom}, the map
$$\sigma \mathring B^\chi_{s+\sigma} := \{v \in \chi_{s+\sigma}, \;
|v|_s < \sigma\} \rightarrow \mathcal{D}_s, \quad v \mapsto \id + v$$
is defined and locally bijective. It endows $\mathcal{D}_s$ with a
local structure of Banach manifold in the neighborhood of the
identity.

We will consider the contragredient action of $\mathcal{D}_s$ on
$\ensuremath{\mathrm{T}}^n_s$ (with values in $\ensuremath{\mathrm{T}}^n_\ensuremath{\mathbb{C}}$)~:
$$\varphi (\theta,r) := (\varphi(\theta),{}^t \varphi'(\theta)^{-1}
\cdot r),$$ in order to linearize the dynamics on the alleged
invariant tori.

\subsubsection{Straightening tori}

Let $\mathcal{B}_s$ be the space of exact one-forms over $\ensuremath{\mathbb{T}}^n_s$,
with
$$|\rho|_s = \max_{\theta\in\ensuremath{\mathbb{T}}^n_s} \max_{1 \leq j \leq n}
|\rho_j(\theta)|, \quad \rho = (\rho_1,...,\rho_n).$$ We will consider
its action on $\ensuremath{\mathrm{T}}^n_s$ by translation of the actions:
$$\rho (\theta,r) := (\theta, r+\rho(\theta)),$$
in order to straighten the perturbed invariant tori.

\subsubsection{Our space of conjugacies}

Let $\mathcal{G}_s = \mathcal{D}_s \times \mathcal{B}_s$,
identified with a space of Hamiltonian symplectomorphisms by
$$(\varphi,\rho)(\theta,r) := \varphi \circ \rho \, (\theta,r) =
(\varphi(\theta), {}^t\varphi'(\theta)^{-1}(r+\rho(\theta))).$$ Endow
its tangent space at the identity $T_{\id}\mathcal{G}_s = \ensuremath{\mathfrak{g}}_s :=
\chi_s \times \mathcal{B}_s$ with the norm
$$|\dot G|_s = |(v,\rho)|_s := \max( |v|_s, |\rho|_s),$$
and its tangent space at $G=(\varphi,\rho)$ with the norm
$$|\delta G|_s := |\delta G \circ G^{-1}|_s, \quad \delta G \in T_G
\mathcal{G}.$$ \emph{Here and elsewhere, the notation $\delta G$, as
  well as similar ones, should be taken as a whole; there is no
  separate $\delta \in \ensuremath{\mathbb{R}}$ in the present paper}.

Also consider the following neighborhoods of the identity:
$$\mathcal{G}_s^\sigma = \left\{G \in \mathcal{G}_s, \; \max_{(\theta,r)
    \in T^n_s} |(\Theta-\theta,R-r)| \leq \sigma, \;
  (\Theta,R)=G(\theta,r)\right\}, \quad \sigma >0.$$

The operators (commuting with inclusions of source and target spaces)
$$\phi_s : E_s:= \mathcal{K}_{s+\sigma} \times
\mathcal{G}_s^\sigma \times \ensuremath{\mathbb{R}}^n \rightarrow \mathcal{H}_s, \quad
(K,G,\beta) \mapsto K \circ G + \beta \cdot r$$ are now defined.

\section{Local twisted conjugacy of Hamiltonians}

\begin{theorem}\label{thm:herman}
  Let $\alpha\in \ensuremath{\mathrm{D}}_{\gamma,\tau}$. For all $0 < s < s+\sigma < 1$,
  $\phi_{s+\sigma}$ has a local inverse: if $|H-K^o|_{s+\sigma}$ is
  small, there is a unique $(K,G,\beta)\in E_{s}$, $|\cdot|_{s}$-close
  to $(K^o,\id,0)$ such that $H=K \circ G + \beta \cdot r$.  Moreover
  $\beta \circ \phi^{-1}$ is a $C^1$-function locally in the
  neighborhood of $K^o$ in $\mathcal{H}_{s+\sigma}$.
\end{theorem}

This entails theorem~\ref{thm:herman0} and itself follows from the
inverse function theorem of appendix~\ref{app:submersion}, from
lemma~\ref{lm:lip} (for the uniqueness) and from
corollary~\ref{cor:smoothness} (for the smoothness of $\beta \circ
\phi^{-1}$). We will now check the two main hypotheses of
appendix~\ref{app:submersion} (one on $\phi'^{-1}$ and one on
$\phi''$).

Let $\mathcal{L}_\alpha$ be the Lie derivative operator in the
direction of the constant vector field $\alpha$~:
$$\mathcal{L}_\alpha : \mathcal{A}(\ensuremath{\mathbb{T}}^n_s) \rightarrow
\mathcal{A}(\ensuremath{\mathbb{T}}^n_s), \quad f \mapsto f' \cdot \alpha = \sum_{1\leq j
  \leq n} \alpha_j \frac{\partial f}{\partial \theta_j}.$$ We will
need the following classical lemma in two instances in the proof of
lemma~\ref{lm:inverse}.

\begin{lemma}[Cohomological equation]\label{lm:cohomological}
  If $g\in \mathcal{A}(\ensuremath{\mathbb{T}}^n_{s+\sigma})$ has $0$-average ($\int_\ensuremath{\mathbb{T}} g
  \, d\theta=0$), there exists a unique function $f\in
  \mathcal{A}(\ensuremath{\mathbb{T}}^n_s)$ of $0$-average such that $\mathcal{L}_\alpha f
  = g$, and there exists a $C_0=C_0(n,\tau)$ such that, for any
  $\sigma$:
  $$|f|_s \leq C_0\gamma^{-1}\sigma^{-\tau-n} |g|_{s+\sigma}.$$
\end{lemma}

\begin{proof}
  Let $g(\theta) = \sum_{k\in \ensuremath{\mathbb{Z}}^n \setminus \{0\}} g_k \, e^{i k
    \cdot \theta}$ be the Fourier expansion of $g$.  The unique formal
  solution to the equation $\mathcal{L}_\alpha f = g$ is given by
  $f(\theta) = \sum_{k\in \ensuremath{\mathbb{Z}}^n \setminus \{0\}} \frac{g_k}{i \, k
    \cdot \alpha} \, e^{i\, k \cdot \theta}$.

  Since $g$ is analytic, its Fourier coefficients decay exponentially:
  we find
  $$|g_k| = \left| \int_{\ensuremath{\mathbb{T}}^n} g(\theta) \, e^{-ik \cdot \theta} \,
    \frac{d\theta}{2\pi} \right| \leq |g|_{s+\sigma} e^{-|k|
    (s+\sigma)}$$ by shifting the torus of integration to a torus $\Im
  \theta_j =\pm (s+\sigma)$.

  Using this estimate and replacing the small denominators $k \cdot
  \alpha$ by the estimate defining the diophantine property of
  $\alpha$, we get
  \begin{eqnarray*}
    |f|_s
    &\leq &\frac{|g|_{s+\sigma}}{\gamma} \sum_k  |k|^\tau \,
    e^{-|k|\, \sigma}\\
    &\leq &\frac{2^n |g|_{s+\sigma}}{\gamma} \sum_{\ell \geq
      1}
    \begin{pmatrix}
      \ell+n-1\\
      \ell
    \end{pmatrix}
    \ell^\tau \, e^{-\ell\, \sigma} \leq \frac{4^n
      |g|_{s+\sigma}}{\gamma \, (n-1)!} \sum_{\ell}
    (\ell+n-1)^{\tau+n-1}\, e^{-\ell \, \sigma},
  \end{eqnarray*}
  where the latter sum is bounded by
  \begin{eqnarray*}
   \int_1^\infty (\ell+n-1)^{\tau+n-1}e^{-(\ell-1)\sigma} \, d\ell
   &=&\sigma^{-\tau-n} e^{n\sigma} \int_{n\sigma}^\infty
   \ell^{\tau+n-1} e^{-\ell} \, d\ell\\
   &<&\sigma^{-\tau-n} e^{n\sigma} \int_0^\infty \ell^{\tau+n-1}
   e^{-\ell} \, d\ell = \sigma^{-\tau-n} e^{n\sigma} \Gamma(\tau+n).
 \end{eqnarray*}
   Hence $f$ belongs to $\mathcal{A}(\ensuremath{\mathbb{T}}^n_s)$ and satisfies the wanted
  estimate.
\end{proof}

We will write $x=(K,G,\beta,c)$, $\delta x = (\delta K, \delta G,
\delta\beta, \delta c)$ and $\delta \hat x = (\delta \hat K, \delta \hat
G, \delta\hat\beta, \delta\hat c)$.

Fix $0 < s < s+\sigma < 1$.

\begin{lemma} \label{lm:inverse} There exists $C'>0$ which is locally
  uniform with respect to $x \in E_s$ in the neighborhood of $G=\id$
  such that the linear map $\phi'(x)$ has an inverse $\phi'(x)^{-1}$
  satisfying
  $$\nm{\phi'(x)^{-1} \cdot \delta H}_{s} \leq
  \sigma^{-\tau-n-1} C' \nm{\delta H}_{G,s+\sigma}.$$
\end{lemma}

\begin{proof}
  A function $\delta H \in G^*\mathcal{A}(T_{s+\sigma})$ being given,
  we want to solve the equation
  $$\delta\phi(x) \cdot \delta x = \delta K \circ G + K' \circ G \cdot
  \delta G + \delta\beta \cdot r + \delta c = \delta H,$$ for the
  unknowns $\delta K\in T_K \mathcal{K}_s \subset
  \mathcal{A}(\ensuremath{\mathrm{T}}^n_s)$, $\delta G \in T_G \mathcal{G}_s$,
  $\delta\beta \in \ensuremath{\mathbb{R}}^n$ and $\delta c \in \ensuremath{\mathbb{R}}$, or, equivalently,
  after composing with $G^{-1}$ to the right,
  $$\delta K + K' \cdot \dot G + \delta\beta \cdot r \circ G^{-1} +
  \delta c = \dot H,$$ where we have set $\dot G:=\delta G \circ
  G^{-1} \in \ensuremath{\mathfrak{g}}_s$ and $\dot H:= \delta H \circ G^{-1} \in
  \mathcal{A}(\ensuremath{\mathrm{T}}^n_s)$.

  More specifically, $G^{-1}$ and $\dot G$ are of the form
  $$G^{-1}(\theta,r) = (\varphi^{-1}(\theta), {}^t\varphi' \circ
  \varphi^{-1}(\theta) \cdot r - \rho \circ \varphi^{-1}(\theta)),
  \quad \dot G = (\dot\varphi, \dot\rho - r \cdot \dot\varphi'),$$
  where $\dot\varphi\in \chi_{s+\sigma}$ and
  $\dot\rho\in \mathcal{B}_{s+\sigma}$,
  and we can expand
  $$K = \alpha \cdot r + K_2(\theta) \cdot r^2 + O(r^3) \quad
  \mbox{and} \quad \dot H = \dot H_0(\theta) + \dot H_1(\theta) \cdot
  r + O(r^2).$$
  The equation becomes
  \begin{multline}
    \label{eq:cohomological}
    \left[\dot \rho \cdot \alpha + \delta c - \rho \circ \varphi^{-1}
      \cdot \delta \beta \right] + r \cdot \left[ -\dot\varphi' \cdot
      \alpha + \varphi' \circ \varphi^{-1} \cdot
      \delta \beta + 2 K_2 \cdot \dot \rho \right] + \\
    \dot K = \dot H + O(r^2),
  \end{multline}
  where the term $O(r^2)$ in the right hand side depends only on $K$
  and $\dot G$, and not on $\dot K$. The equation turns out to be
  triangular in the five unknowns. The existence and uniqueness of a
  solution with the wanted estimate follows from repeated applications
  of lemma~\ref{lm:cohomological} and Cauchy's inequality:

  -- The average over $\ensuremath{\mathrm{T}}_0^n$ of the first order terms with respect
  to $r$ in equation~(\ref{eq:cohomological}) yields
  $$\delta\beta = \left(\int_{\ensuremath{\mathbb{T}}^n} \varphi' \circ \varphi^{-1} \,
    d\theta \right)^{-1} \cdot \int_{\ensuremath{\mathrm{T}}_0^n} \dot H_1 \, d\theta,$$
  which does exist if $\varphi$ is close to the identity
  (proposition~\ref{prop:isom}).

  -- Similarly, the average of the restriction to $\ensuremath{\mathrm{T}}_0^n$
  of~(\ref{eq:cohomological}) yields:
  $$\delta c = \int_{\ensuremath{\mathrm{T}}^n_0} \dot H_0   \, d\theta + \int_{\ensuremath{\mathbb{T}}_0^n}
  \rho \circ \varphi^{-1} \, d\theta \cdot \delta\beta.$$

  -- Next, the restriction to $\ensuremath{\mathrm{T}}_0^n$ of~(\ref{eq:cohomological})
  can be solved uniquely with respect to $\delta\rho$ according to
  lemma~\ref{lm:cohomological} (applied with $\rho=f'$).

  -- The part of degree one can then be solved for $\dot\varphi$
  similarly.

  -- Terms of order $\geq 2$ in $r$ determine $\dot K$.
\end{proof}

\begin{lemma}\label{lm:seconde}
  There exists a constant $C''>0$ which is locally uniform with
  respect to $x \in E_{s+\sigma}$ in the neighborhood of $G=\id$ such
  that the bilinear map $\phi''(x)$ satisfies
  $$\nm{\phi''(x) \cdot \delta x \otimes \delta \hat x}_{G,s}\leq
  \sigma^{-1} C'' \, |\delta x|_{s+\sigma} |\delta\hat
  x|_{s+\sigma}.$$
\end{lemma}

\begin{proof}
  Differentiating $\phi$ twice yields
  $$\phi''(x) \cdot \delta x \otimes \delta \hat x = \delta K' \circ
  G \cdot \delta G + \delta\hat K' \circ G \cdot \delta G + K'' \circ
  G \cdot \delta G \otimes \delta \hat G,$$ whence the estimate.
\end{proof}

\setcounter{section}{0}
\renewcommand{\thesection}{\Alph{section}}
\addcontentsline{toc}{section}{Appendices}
\section{An inverse function theorem}
\label{app:submersion}

Let $E=(E_s)_{0<s<1}$ be a decreasing family of Banach spaces with
increasing norms $| \cdot |_s$, and $\epsilon B^E_s = \{x\in E_s,
\nm{x}_s< \epsilon\}$, $\epsilon>0$, be its balls centered at $0$.

Let $(F_s)$ be an analogous family. Endow $F$ with additional norms
$\nm{\cdot}_{x,s}$, $x\in E_s$, $0<s<1$, satisfying
$$|y|_{0,s} = |y|_s \quad \mbox{and} \quad  \nm{y}_{x',s} \leq
\nm{y}_{x,s+\nm{x'-x}_s}.$$
These norms allow for dealing with composition operators without
artificially loosing some fixed ``width of analyticity'' $\sigma$ at
each step of the Newton algorithm.

Let $\phi : \sigma B^E_{s+\sigma} \rightarrow F_s$, $s<s+\sigma$,
$\phi(0)=0$, be maps commuting with inclusions, twice differentiable,
such that the differential $\phi'(x) : E_{s+\sigma} \rightarrow F_s$
has a right inverse $\phi'(x)^{-1} : F_{s+\sigma} \rightarrow E_s$,
and
$$\left\{
  \begin{array}[c]{ll}
    |\phi'(x)^{-1}\eta|_{s} \leq C' \sigma^{-\tau'}
    |\eta|_{x,s+\sigma}\\
    |\phi''(x) \xi^{\otimes 2}|_{x,s} \leq
    C''\sigma^{-\tau''} |\xi|_{s+\sigma}^2 \quad (\forall s, \sigma,
    x, \xi, \eta)
  \end{array}\right. $$
with $C',C'',\tau',\tau'' \geq 1$. Let $C:=C'C''$ and
$\tau:=\tau'+\tau''$.

\begin{theorem} \label{thm:inversion} $\phi$ is locally surjective
  and, more precisely, for any $s$, $\eta$ and $\sigma$ with $\eta <
  s$,
  $$\epsilon B^F_{s+\sigma} \subset \phi \left( \eta B^E_s \right),
  \quad \epsilon := 2^{-8\tau} C^{-2}\sigma^{2\tau} \eta.$$
\end{theorem}

In other words, $\phi$ has a right-inverse $\psi : \epsilon
B^F_{s+\sigma} \rightarrow \eta B^E_{s}$.

\begin{proof}
  Some numbers $s$, $\eta$ and $\sigma$ and $y \in B^F_{s+\eta}$ being
  given, let
  $$f : \sigma B^E_{s+\eta+\sigma} \rightarrow E_s, \quad x
  \mapsto x + \phi'(x)^{-1} (y-\phi(x))$$
  and
  $$Q : \sigma B^E_{s+\sigma} \times \sigma B^E_{s+\sigma}
  \rightarrow F_s, \quad (x,\hat x) \mapsto  \phi(\hat
  x)-\phi(x)-\phi'(x)(\hat x-x).$$

  \begin{lemma} \label{lm:Q} The function $Q$ satisfies: $\nm{Q(x,\hat
      x)}_{x,s} \leq 2^{-1}C''\sigma^{-\tau''} \nm{\hat
      x-x}_{s+\sigma+|\hat x-x|_s}^2$.
  \end{lemma}

  \begin{proof}[Proof of the lemma]
    Let  $\hat x_t:=(1-t)x + t\hat x$. Taylor's formula yields
    $$Q(x,\hat x) = \int_0^1 (1-t) \, \phi''(\hat x_t) \, (\hat
    x-x)^2\, dt,$$
    hence
    $$\nm{Q(x,\hat x)}_{x,s} \leq \int_0^1 (1-t) \left|\phi''(\hat x_t)
      (\hat x-x)^2\right|_{x,s}\, dt \leq \int_0^1 (1-t)
    \left|\phi''(\hat x_t) (\hat x-x)^2\right|_{\hat x_t,s+|\hat
      x_t-x|_s}\, dt,$$ whence the estimate.
  \end{proof}

  Now, let $s$, $\eta$ and $\sigma$ be fixed, with $\eta < s$ and
  $y\in \epsilon B^F_{s+\sigma}$ for some $\epsilon$. We will see that
  if $\epsilon$ is small enough, the sequence $x_0=0$, $x_n:=f^n(0)$
  is defined for all $n\geq 0$ and converges towards some preimage $x
  \in \eta B^E_s$ of $y$ by $\phi$.

  Let $(\sigma_n)_{n\geq 0}$ be a sequence of positive real numbers
  such that $3 \sum \sigma_n = \sigma$, and $(s_n)_{n\geq 0}$ be the
  sequence decreasing from $s_0:= s+\sigma$ to $s$ defined by
  induction by the formula $s_{n+1} = s_n - 3 \sigma_n$.

  Assuming the existence of $x_0, ..., x_{n+1}$, we see that
  $\phi(x_k) = y + Q(x_{k-1},x_k)$, hence
  $$x_{k+1}-x_k = \phi'(x_k)^{-1}(y-\phi(x_k)) =
  -\phi'(x_k)^{-1}Q(x_{k-1},x_k) \qquad (1\leq k \leq n).$$ Further
  assuming that $|x_{k+1}-x_k|_{s_k} \leq \sigma_k$, the estimate of the
  right inverse and lemma~\ref{lm:Q} entail that
  $$|x_{n+1}-x_n|_{s_{n+1}} \leq c_n |x_n-x_{n-1}|^2_{s_n} \leq
  \cdots \leq c_n c_{n-1}^2 \cdots c_1^{2^{n-1}} \,
  |x_1|_{s_1}^{2^{n-1}}, \quad c_k:=2^{-1}C\sigma_k^{-\tau}.$$ The
  estimate
  $$|x_1|_{s_1} \leq C' (3\sigma_0)^{-\tau'} |y|_{s_0} \leq
  2^{-1}C\sigma_0^{-\tau} \epsilon = c_0\epsilon$$
  and the fact, to be checked later, that $c_k\geq 1$ for all $k\geq
  0$, show~:
  $$|x_{n+1}-x_n|_{s_{n+1}} \leq \left( \epsilon \prod_{k\geq 0}
    c_k^{2^{-k}} \right)^{2^n}.$$ Since $\sum_{n\geq 0} \rho^{2^n}
  \leq 2\rho$ if $2\rho\leq 1$, and using the definition of constants
  $c_k$'s, we get a sufficient condition to have all $x_n$'s defined and
  to have $\sum |x_{n+1}-x_n|_s \leq \eta$:
  \begin{equation}
    \label{eq:epsilon}
    \epsilon = \frac{\eta}{2} \prod_{k\geq 0} c_k^{-2^{-k}} =
    \frac{2\eta}{C^2} \prod_{k\geq 0} \sigma_k^{\tau 2^{-k}}.
  \end{equation}

  Maximizing the upper bound of $\epsilon$ under the constraint
  $3\sum_{n\geq 0} \sigma_n = \sigma$ yields $\sigma_k :=
  \frac{\sigma}{6}2^{-k}$. A posteriori it is straightforward that
  $|x_{n+1}-x_n|_{s_{n}} \leq \sigma_n$ (as earlier assumed to apply
  lemma~\ref{lm:Q}) and $c_n\geq 1$ for all $n\geq 0$. Besides, using
  that $\sum k 2^{-k} = \sum 2^{-k} = 2$ we get
  $$\frac{\eta}{2} \prod_{k\geq 0} c_k^{-2^{-k}} = \frac{\eta}{2}
  \prod_{k\geq 0} \frac{1}{2^{\tau k2^{-k}}} \left( \frac{2}{C} \left(
      \frac{\sigma}{6} \right)^\tau \right)^{2^{-k}} =
  \frac{2\eta}{C^2} \left( \frac{\sigma}{12} \right)^{2\tau} >
  \frac{\sigma^{2\tau} \eta}{2^{8\tau}C^2},$$
  whence the theorem.
\end{proof}

\begin{exercise}
  The domain of $\psi$ contains $\epsilon B^F_S$, $\epsilon =
  2^{-12\tau}\tau^{-1}C^{-2}S^{3\tau}$, for any $S$.
\end{exercise}

\begin{proof}
  The above function $\epsilon(\eta,\sigma) = 2^{-8\tau} C^{-2}
  \sigma^{2\tau} \eta$ attains is maximum with respect
  to $\eta<s$ for $\eta=s$. Besides, under the constraint $s+\sigma =
  S$ the function $\epsilon(s,\sigma)$ attains its maximum when
  $\sigma = 2\tau s$ and $s=\frac{S}{1+2\tau}$. Hence, $S$ being
  fixed, the domain of $\psi$ contains $\epsilon B^F_S$ if
  $$\epsilon < 2^{-8\tau} C^{-2} \frac{S}{1+2\tau} \left( \frac{2\tau
      S}{12(1+2\tau)} \right)^{2\tau}.$$ Given that $S < 1 < \tau$ by
  hypothesis, it suffices that $\epsilon$ be equal to the stated
  value.
\end{proof}

\subsection{Regularity of the right-inverse}
\label{subsec:regularity}

In the proof of theorem~\ref{thm:inversion} we have built right
inverses $\psi : \epsilon B^F_{s+\eta+\sigma} \rightarrow \eta
B^E_{s+\eta}$, of $\phi$, commuting with inclusions. The estimate
given in the statement shows that $\psi$ is continuous at $0$; due to
the invariance of the hypotheses of the theorem by small translations,
$\psi$ is locally continuous.

We further make the following two asumptions:

-- The maps $\phi'(x)^{-1} : F_{s+\sigma} \rightarrow E_s$ are left
(as well as right) inverses (in theorem~\ref{thm:herman} we have
restricted to an adequate class of symplectomorphisms);

-- The scale $(| \cdot |_s)$ of norms of $(E_s)$ satisfies some
interpolation inequality:
$$|x|_{s+\sigma}^2 \leq |x|_s \, |x|_{s+\tilde\sigma} \quad
\mbox{for all $s$, $\sigma$, }
\tilde\sigma=\sigma\left(1+\frac{1}{s}\right)$$ (according to the
remark after corollary~\ref{cor:interpolation}, this estimate is
satisfied in the case of interest to us, since $\sigma + \log
(1+\sigma/s) \leq \tilde\sigma$).

\begin{lemma}[Lipschitz regularity] \label{lm:lip}
  If $\sigma < s$ and $y, \hat y \in \epsilon B^F_{s+\sigma}$ with
  $\epsilon = 2^{-14\tau} C^{-3} \sigma^{3\tau}$,
  $$|\psi(\hat y) - \psi(y)|_s \leq C_L |\hat  y - y|_{s+\sigma},
  \quad C_L = 2C'\sigma^{-\tau'}.$$ In particular, $\psi$ is the
  unique local right inverse of $\phi$, and hence is also its local
  left inverse.
\end{lemma}

\begin{proof}
  Fix $\eta < \zeta < \sigma < s$; the impatient reader can readily
  look at the end of the proof how to choose the auxiliary parameters
  $\eta$ and $\zeta$ more precisely.

  Let $\epsilon = 2^{-8\tau} C^{-2} \zeta^{2\tau}\eta$, and $y,
  \hat y \in \epsilon B^F_{s+\sigma}$. According to
  theorem~\ref{thm:inversion}, $x:=\psi(y)$ and $\hat x:=\psi(\hat y)$
  are in $\eta B^E_{s+\sigma-\zeta}$, provided the condition, to be
  checked later, that $\eta < s+\sigma-\zeta$. In particuliar, we
  will use a priori that
  $$|\hat x - x|_{s+\sigma-\zeta} \leq |\hat x|_{s+\sigma-\zeta} +
  |x|_{s+\sigma-\zeta} \leq 2 \eta.$$

  We have
  \begin{eqnarray*}
    \hat x -x
    &=&\phi'(x)^{-1}\phi'(x)(\hat x -x)\\
    &=&\phi'(x)^{-1}\left( \hat y - y - Q(x,\hat x) \right)
  \end{eqnarray*}
  and, according to the assumed estimate on $\phi'(x)^{-1}$ and to
  lemma~\ref{lm:Q},
  \begin{eqnarray*}
    |\hat x -x|_s
    &\leq&C'\sigma^{-\tau'}|\hat y-y|_{s+\sigma} +
    2^{-1}C\zeta^{-\tau} |\hat x - x|_{s+2\eta+|\hat x-x|_s}^2.
  \end{eqnarray*}
  In the norm index of the last term, we will coarsely bound $|\hat
  x-x|_s$ by $2\eta$. Additionally using the interpolation inequality:
  $$|\hat x - x|_{s+4\eta}^2\leq |\hat x - x|_s |\hat x -
  x|_{s+\tilde\sigma}, \quad \tilde\sigma = 4\eta \left( 1+\frac{1}{s}
  \right),$$
  yields
  $$\left( 1 - 2^{-1}C \zeta^{-\tau} |\hat x - x|_{s+\tilde\sigma}
  \right) |\hat x - x|_s \leq C' \sigma^{-\tau'} |\hat y -
  y|_{s+\sigma}.$$
  Now, we want to choose $\eta$ small enough so that

  -- first, $\tilde\sigma \leq \sigma - \zeta$, which implies $|\hat x
  - x|_{s+\tilde\sigma} \leq 2\eta$. By definition of $\tilde\sigma$,
  it suffices to have $\eta \leq \frac{\sigma-\zeta}{4(1+1/s)}$.

  -- second, $2^{-1}C\zeta^{-\tau} \, 2\eta \leq 1/2$, or $\eta \leq
  \frac{\zeta^\tau}{2C}$, which implies that $2^{-1}C\zeta^{-\tau}
  |\hat x - x|_{s+\tilde\sigma} \leq 1/2$, and hence $|\hat x-x|_{s}
  \leq 2 C'\sigma^{-\tau'}|\hat y-y|_{s+\sigma}$.

  A choice is $\zeta = \frac{\sigma}{2}$ and $\eta =
  \frac{\sigma^\tau}{16C} < s$, whence the value of $\epsilon$ in the
  statement.
\end{proof}

\begin{proposition}[Smoothness] \label{prop:regularity} For every
  $\sigma<s$, there exists $\epsilon,C_1$ such that for every $y, \hat
  y \in \epsilon B^F_{s+\sigma}$,
  $$|\psi(\hat y) - \psi(y) - \phi'(\psi(y))^{-1}(\hat y - y)|_s \leq C_1
  |\hat y - y|_{s+\sigma}^2.$$ Moreover, the map $\psi' : \epsilon
  B^F_{s+\sigma} \rightarrow L(F_{s+\sigma}, E_{s})$ defined locally by
  $\psi'(y) = \phi'(\psi(y))^{-1}$ is continuous.
\end{proposition}

\begin{proof}
  Fix $\epsilon$ as in the previous proof and $y, \hat y\in
  \varepsilon B^F_{s+\sigma}$. Let $x = \psi(y)$, $\eta=\hat y-y$,
  $\xi = \psi(y+\eta) - \psi(y)$ (thus $\eta = \phi(x+\xi)-\phi(x)$),
  and $ \Delta :=\psi(y+\eta) - \psi(y) - \phi'(x)^{-1}\eta$.
  Definitions yield
  \begin{eqnarray*}
    \Delta&=&\phi'(x)^{-1} \left( \phi'(x)\xi - \eta \right)
    =- \phi'(x)^{-1} Q(x,x+\xi).
  \end{eqnarray*}
  Using the estimates on $\phi'(x)^{-1}$ and $Q$ and the latter lemma,
  $$|\Delta|_s \leq C_1 |\eta|_{s+\sigma'}^2$$
  for some $\sigma'$ tending to $0$ when $\sigma$ itself tends to $0$,
  and for some $C_1>0$ depending on $\sigma$. Up the substitution of
  $\sigma$ by $\sigma'$, the estimate is proved.

  The inversion of linear operators between Banach spaces being
  analytic, $y \mapsto \phi(\psi(y))^{-1}$ is continuous in the stated
  sense.
\end{proof}

\begin{corollary}~\label{cor:smoothness}
  If $\pi \in L(E_s,V)$ is a family of linear maps, commuting with
  inclusions, into a fixed Banach space $V$, then $\pi \circ \psi$ is
  $C^1$ and $(\pi \circ \psi)' = \pi \cdot \phi' \circ \psi$.
\end{corollary}

This corollary is used with $\pi : (K,G,\beta) \mapsto \beta$ in the
proof of theorem~\ref{thm:herman}.

\section{Some estimates on analytic isomorphisms}
\label{app:inversion}

In this appendix, we give a quantitative inverse function theorem for
real analytic isomorphisms on $\ensuremath{\mathbb{T}}^n_s$. This is used in
section~\ref{sec:setting}, to parametrize locally $\mathcal{D}_s$ by
vector fields, and, in lemma~\ref{lm:cohomological}, to solve the
cohomological equation for the frequency offset $\delta\beta$.

Recall that we have set $\ensuremath{\mathbb{T}}^n_s := \{ \theta \in \ensuremath{\mathbb{C}}^n/2\pi\ensuremath{\mathbb{Z}}^n, \quad
\max_{1\leq j \leq n} |\Im\theta_j|\leq s\}$. We will denote by $p
: \ensuremath{\mathbb{R}}^n_s := \ensuremath{\mathbb{R}}^n \times i[-s,s]^n \rightarrow \ensuremath{\mathbb{T}}^n_s$ its universal
covering.

\begin{proposition}\label{prop:isom}
  Let $v\in \mathcal{A}(\ensuremath{\mathbb{T}}^n_{s+2\sigma},\ensuremath{\mathbb{C}}^n)$, $|v|_{s+2\sigma} <
  \sigma$. The map $\id + v : \ensuremath{\mathbb{T}}^n_{s+2\sigma} \rightarrow
  \ensuremath{\mathbb{R}}^n_{s+3\sigma}$ induces a map $\varphi : \ensuremath{\mathbb{T}}^n_{s+2\sigma}
  \rightarrow \ensuremath{\mathbb{T}}^n_{s+3\sigma}$ whose restriction $\varphi :
  \ensuremath{\mathbb{T}}^n_{s+\sigma} \rightarrow \ensuremath{\mathbb{T}}^n_{s+2\sigma}$ has a unique right
  inverse $\psi : \ensuremath{\mathbb{T}}^n_{s} \rightarrow \ensuremath{\mathbb{T}}^n_{s+\sigma}$:
  $$\xymatrix{
    \ensuremath{\mathbb{T}}^n_{s+\sigma} \ar^\varphi @{^{(}->} [r] &\ensuremath{\mathbb{T}}^n_{s+2\sigma}\\
    &\ar^\psi @{^{(}->} [ul] \ensuremath{\mathbb{T}}_s^n \ar @{^{(}->} [u] }.$$
  Furthermore,
  $$|\psi-\id|_s \leq |v|_{s+\sigma}$$
  and, provided $2 \sigma^{-1} |v|_{s+2\sigma} \leq 1$,
  $$|\psi'-\id| \leq 2 \sigma^{-1} |v|_{s+2\sigma}.$$
\end{proposition}

\begin{proof}
  Let $\Phi : \ensuremath{\mathbb{R}}^n_{s+2\sigma} \rightarrow \ensuremath{\mathbb{R}}^n_{s+3\sigma}$ be a
  continuous lift of $\id+v$ and $k \in M_n(\ensuremath{\mathbb{Z}})$, $k(l) := \Phi(x+l) -
  \Phi(x)$.
  \begin{enumerate}
  \item \textit{Injectivity of $\Phi : \ensuremath{\mathbb{R}}^n_{s+\sigma} \rightarrow
      \ensuremath{\mathbb{R}}^n_{s+2\sigma}$. }  Suppose that $x, \hat x\in
    \ensuremath{\mathbb{R}}^n_{s+\sigma}$ and $\Phi(x) = \Phi(\hat x)$.  By the mean value
    theorem,
    $$|x-\hat x| = |v(p\hat x) -v(p x)| \leq |v'|_{s+\sigma}
    |x-\hat x|,$$ and, by Cauchy's inequality,
    $$|x-\hat x| \leq \frac{|v|_{s+2\sigma}}{\sigma} |x-\hat x| <
    |\hat x - x|,$$ hence $x=\hat x$.

  \item \textit{Surjectivity of $\Phi$: $\ensuremath{\mathbb{R}}^n_s \subset
      \Phi(\ensuremath{\mathbb{R}}^n_{s+\sigma})$. }  For any given $y\in \ensuremath{\mathbb{R}}^n_s$, the
    contraction
    $$f : \ensuremath{\mathbb{R}}^n_{s+\sigma} \rightarrow \ensuremath{\mathbb{R}}^n_{s+\sigma}, \quad x
    \mapsto y-v(x)$$ has a unique fixed point, which is a pre-image of
    $y$ by $\Phi$.

  \item \textit{Injectivity of $\varphi : \ensuremath{\mathbb{T}}^n_{s+\sigma} \rightarrow
      \ensuremath{\mathbb{T}}^n_{s+2\sigma}$. }  Suppose that $p x$, $p\hat x \in
    \ensuremath{\mathbb{R}}^n_{s+\sigma}$ and $\varphi(p x) = \varphi(p \hat x)$, i.e.
    $\Phi(x) = \Phi(\hat x) + \kappa$ for some $\kappa \in \ensuremath{\mathbb{Z}}^n$. That
    $k$ be in $GL(n,\ensuremath{\mathbb{Z}})$, follows from the invertibility of $\Phi$.
    Hence, $\Phi \left(x-k^{-1}(\kappa) \right) = \Phi(\hat x)$, and,
    due to the injectivity of $\Phi$, $p x = p \hat x$.

  \item \textit{Surjectivity of $\varphi : \ensuremath{\mathbb{T}}^n_s \subset \varphi
      (\ensuremath{\mathbb{T}}^n_{s+\sigma})$. } This is a trivial consequence of that of
    $\Phi$.

  \item \textit{Estimate on $\psi:= \varphi^{-1} : \ensuremath{\mathbb{T}}_s^n \rightarrow
      \ensuremath{\mathbb{T}}_{s+\sigma}^n$. } Note that the wanted estimate on $\psi$ is
    in the sense of $\Psi := \Phi^{-1} : \ensuremath{\mathbb{R}}_s^n \rightarrow
    \ensuremath{\mathbb{R}}_{s+\sigma}^n$. If $y\in \ensuremath{\mathbb{R}}^n_s$,
    $$\Psi(y) - y = - v(p\Psi(y)),$$
    hence $|\Psi- \id|_{s} \leq |v|_{s+\sigma}$.

  \item \textit{Estimate on $\psi'$. } We have $\psi' = \varphi'^{-1}
    \circ \varphi$, where $\varphi'^{-1}(x)$ stands for the inverse of
    the map $\xi \mapsto \varphi'(x) \cdot \xi$. Hence
    $$\psi' - \id = \varphi'^{-1} \circ \varphi - \id,$$
    and, under the assumption that $2\sigma^{-1} |v|_{s+2\sigma} \leq
    1$,
    $$|\psi' - \id|_{s} \leq |\varphi'^{-1}-\id|_{s+\sigma} \leq
    \frac{|v'|_{s+\sigma}}{1-|v'|_{s+\sigma}} \leq
    \frac{\sigma^{-1}|v|_{s+2\sigma}}{1-\sigma^{-1}|v|_{s+2\sigma}}
    \leq 2 \sigma^{-1} |v|_{s+2\sigma}.$$
  \end{enumerate}
\end{proof}

\section{Interpolation of spaces of analytic functions}
\label{app:convexity}

In this section we prove some Hadamard interpolation inequalities,
which are used in~\ref{subsec:regularity}.

Recall that we denote by $\ensuremath{\mathbb{T}}_\ensuremath{\mathbb{C}}^n$ the infinite annulus $\ensuremath{\mathbb{C}}^n
/2\pi\ensuremath{\mathbb{Z}}^n$, by $\ensuremath{\mathbb{T}}_s^n$, $s>0$, the bounded sub-annulus $\{ \theta \in
\ensuremath{\mathbb{T}}_\ensuremath{\mathbb{C}}^n, \; |\Im \theta_j| \leq s, \; j=1...n\}$ and by $\ensuremath{\mathbb{D}}_t^n$,
$t>0$, the polydisc $\{ r\in \ensuremath{\mathbb{C}}^n, \; |r_j| \leq t, \; j=1...n\}$. The
supremum norm of a function $f \in \mathcal{A}(\ensuremath{\mathbb{T}}^n_s \times \ensuremath{\mathbb{D}}^n_t)$
will be denoted by $|f|_{s,t}$.

Let $0 < s_0 \leq s_1$ and $0 < t_0 \leq t_1$ be such that
$$\log \frac{t_1}{t_0} =  s_1-s_0.$$
Let also $0 \leq \rho \leq 1$ and
$$s = (1-\rho) s_0 + \rho s_1 \quad \mbox{and} \quad t =
t_0^{1-\rho} t_1^\rho.$$

\begin{proposition} \label{prop:domains} If $f \in
  \mathcal{A}(\ensuremath{\mathbb{T}}_{s_1}^n \times \ensuremath{\mathbb{D}}_{t_1}^n)$,
  $$|f|_{s,t} \leq |f|_{s_0,t_0}^{1-\rho} \, |f|_{s_1,t_1}^{\rho}.$$
\end{proposition}

\begin{proof}
  Let $\tilde f$ be the function on $\ensuremath{\mathbb{T}}^n_{s_1} \times \ensuremath{\mathbb{D}}^n_{t_1}$,
  constant on $2n$-tori of equations $(\Im \theta,r) =cst$, defined by
  $$\tilde f(\theta,r) = \max_{\mu,\nu\in \ensuremath{\mathbb{T}}^n} \left| f\left((\pm
      \theta_1 + \mu_1,..., \pm \theta_n + \mu_n), \left(r_1 \,
      e^{i\nu_1}, ..., r_n \, e^{i\nu_n}\right)\right)\right|$$ (with
  all possible combinations of signs). Since $\log |f|$ is subharmonic
  and $\ensuremath{\mathbb{T}}^{2n}$ is compact, $\log \tilde f$ too is upper
  semi-continuous. Besides, $\log \tilde f$ satisfies the mean
  inequality, hence is plurisubharmonic.

  By the maximum principle, the restriction of $|f|$ to $\ensuremath{\mathbb{T}}^n_s \times
  \ensuremath{\mathbb{D}}^n_t$ attains its maximum on the distinguished boundary of $\ensuremath{\mathbb{T}}^n_s
  \times \ensuremath{\mathbb{D}}^n_t$. Due to the symmetry of $\tilde f$:
  $$|f|_{s,t} = \tilde f(is\epsilon,t\epsilon), \quad
  \epsilon=(1,...,1).$$

  Now, the function
  $$\varphi(z) := \tilde f(z\epsilon,e^{- (iz+s)}t\epsilon)$$
  is well defined on $\ensuremath{\mathbb{T}}_{s_1}$, for it is constant with respect to
  $\Re z$ and, due to the relations imposed on the norm indices, if
  $|\Im z|\leq s_1$ then $|e^{-(iz+s)}t| \leq e^{s_1-s} t = t_1$.

  The estimate
  $$\log \varphi(z) \leq \frac{s_1-\Im z}{s_1-s_0} \varphi(s_0 i)
  + \frac{\Im z-s_0}{s_1-s_0}\varphi (s_1 i)$$ trivially holds if $\Im
  z = s_0$ or $s_1$, for, as noted above for $j=1$, $e^{s_j-s} t
  = t_j$, $j=0,1$. But note that the left and right hand sides
  respectively are suharmonic and harmonic. Hence the estimate holds
  whenever $s_0 \leq \Im z \leq s_1$, whence the claim for $z=is$.
\end{proof}

Recall that we have let $\ensuremath{\mathrm{T}}_s^n := \ensuremath{\mathbb{T}}_s^n \times \ensuremath{\mathbb{D}}_s^n$, $s>0$, and,
for a function $f \in \mathcal{A}(\ensuremath{\mathrm{T}}_s^n)$, let $|f|_s = |f|_{s,s}$
denote its supremum norm on $\ensuremath{\mathrm{T}}_s^n$. As in the rest of the paper, we
now restrict the discussion to widths of analyticity $\leq 1$.

\begin{corollary}\label{cor:interpolation}
  If $\sigma_1 = -\log \left( 1 - \frac{\sigma_0}{s} \right)$ and $f
  \in \mathcal{A}(\ensuremath{\mathrm{T}}_{s+\sigma_1}^n)$,
  $$|f|_{s}^2 \leq |f|_{s-\sigma_0} |f|_{s+\sigma_1}.$$
\end{corollary}

In~\ref{subsec:regularity}, we will use the equivalent fact that, if
$\tilde\sigma = s + \log \left( 1 + \frac{\sigma}{s} \right)$ and $f
\in \mathcal{A}(\ensuremath{\mathrm{T}}^n_{s+\tilde\sigma})$,
$$|f|_{s+\sigma}^2 \leq |f|_{s} |f|_{s+\tilde\sigma}.$$

\begin{proof}
  In proposition~\ref{prop:domains}, consider the following particular
  case~:
  \begin{itemize}
  \item $\rho = 1/2$. Hence
    $$s = \frac{s_0+s_1}{2} \quad \mbox{and} \quad t =
    \sqrt{t_0t_1}.$$
  \item $s=t$. Hence in particular $t_0 = s\, e^{s_0-s}$ and
    $t_1 = s \, e^{s_1-s}$.
  \end{itemize}
  Then
  $$|f|_s^2 = |f|_{s,s}^2 \leq |f|_{s_0,t_0} |f|_{s_1,t_1}.$$

  We want to determine $\max(s_0,t_0)$ and $\max(s_1,t_1)$. Let
  $\sigma_1 := s-s_0 = s_1-s$. Then $t_0 = s \, e^{-\sigma_1}$ and
  $t_1 = s \, e^{\sigma_1}$. The expression $s + \sigma - se^{\sigma}$
  has the sign of $\sigma$ (in the relevant region $0\leq s+\sigma
  \leq 1$, $0\leq s \leq 1$); by evaluating it at $\sigma=\pm
  \sigma_1$, we see that $s_0 \leq t_0$ and $s_1 \geq t_1$.

  Therefore, since the norm $|\cdot|_{s,t}$ is non-decreasing with
  respect to both $s$ and $t$,
  $$|f|_s^2 \leq |f|_{t_0,t_0} |f|_{s_1,s_1} = |f|_{t_0} |f|_{s_1}$$
  (thus giving up estimates uniform with respect to small values of
  $s$).  By further setting $\sigma_0 = s-t_0 = s\left(1 -
    e^{-\sigma_1} \right)$, we get the wanted estimate, and the
  asserted relation between $\sigma_0$ and $\sigma_1$ is readily
  verified.
\end{proof}

\section{Weaker arithmetic conditions of convergence}

In this section, we look more carefully to the arithmetic conditions
needed for the induction to converge, in the proof of the inverse
function theorem~\ref{thm:inversion}.

A function $\Delta : \ensuremath{\mathbb{N}}_* \rightarrow [1,+\infty[$ being given, define
the set $\mathrm{D}_\Delta$ as the subset of vectors $\alpha \in \ensuremath{\mathbb{R}}^n$
such that
$$|k \cdot \alpha| \geq \frac{(|k|+n-1)^{n-1}}{\Delta(|k|)} \quad (\forall k
\in \ensuremath{\mathbb{Z}}^n \setminus \{0\}).$$ (The function $\Delta$ is just some other
normalization of what is an approximation function in~\cite{Rus75} or
a zone function in \cite{Dumas:2004}.)  For $\mathrm{D}_\Delta$ to be
non empty, trivially we need $\lim_{+\infty} \Delta = +\infty$.

\begin{proposition}\label{prop:arithmetics}
  The conclusions of theorems~\ref{thm:herman}
  and~\ref{thm:kolmogorov} hold of there exist $c>0$ and $\delta \in
  ]0,1[$ such that
  $$\sum_{\ell \geq 1} \Delta(\ell) e^{-\ell / j^2} \leq \exp \left( c\,
    2^{\delta j} \right) \quad \mbox{as } j \rightarrow +\infty.$$
\end{proposition}

\begin{example}
  The Diophantine set $\mathrm{D}_{\gamma,\tau}$ corresponds to a
  polynomially growing function $\Delta$, and to a polynomially
  growing function $\sum_{\ell \geq 1} \Delta(\ell) e^{-\ell
    \,2^{-j}}$. A foriori, $\sum_{\ell \geq 1} \Delta(\ell)
  e^{-\ell/j^2}$ is at most polynomially growing.
\end{example}

\begin{proof}
  Call $L$ the discrete Laplace
  transform of $\Delta$:
  $$L(\sigma) = \sum_{\ell \geq 1} \Delta(\ell) e^{-\ell\sigma},$$
  and assume it is finite for all $\sigma >0$. Patterning the proof of
  lemma~\ref{lm:cohomological}, we get the following generalization.

  \begin{lemma}\label{lm:Cohomology}
    Let $g\in \mathcal{A}(\ensuremath{\mathbb{T}}^n_{s+\sigma})$ having $0$-average. There is
    a unique function $f\in \mathcal{A}(\ensuremath{\mathbb{T}}^n_s)$ of zero average such
    that $\mathcal{L}_\alpha f = g$. This function satisfies
    $$|f|_s \leq C \, L(\sigma) \, |g|_{s+\sigma}, \quad C = \frac{2^n
      e}{(n-1)!}.$$
  \end{lemma}

  (Again, see~\cite{Rus75} for improved estimates. But such an
  improvement is not the crux of our purpose here.)

  Taking up the proof of the inverse fuction theorem of appendix~A
  with our new estimates (see in particular
  equation~(\ref{eq:epsilon})), we see that the Newton algorithm
  converges provided
  $$\sum_{j\geq 0} 2^{-j} \log L(\sigma_j) < \infty,$$
  for some choice of the converging series $\sum \sigma_j$. Choosing
  $\sum\sigma_j = \sum j^{-2}$, we see that it is enough that $\log
  L(\sigma_j) \leq c \, 2^{\delta j}$ for some $c>0$ and $\delta
  \in ]0,1[$, whence the given criterion.
\end{proof}

\section{Comments}

\subsubsection*{Section~1.}

The proof of Kolmogorov's theorem presented here differs from others
chiefly for the following reasons:

-- The seeming detour through Herman's normal form reduces
Kolmogorov's theorem to a functionally well posed inversion problem
(compare with~\cite{Zeh76a,Zeh76b}). This powerful trick consists in
switching the frequency obstruction (obstruction to the conjugacy to
the initial dynamics) from one side of the conjugacy to the other. It
was extensively used in \cite{Mos67}. The remaining, finite
dimensional problem is then to show that the frequency offset $\beta
\in \ensuremath{\mathbb{R}}^n$ may vanish; in general, it is met using a non-degeneracy
hypothesis of one kind or another. Looking backward, this last step is
not the most difficult, but was probably not well understood before
M. Herman in the 80s (see \cite{Russmann:1990} and
\cite{Sevryuk:1999}). The functionnal setting chosen here adapts to
more degenerate cases, including lower dimensional tori, in a
straightforward manner (see~\cite{Fej04}; compare to Herman's prefered
proof for Lagrangian tori, as exposed in~\cite{Bos86}).

-- Classical perturbation series (or some modification of these) have
been shown to converge in some cases (\cite{Siegel:1942} for the
convergence of Schr{\"o}der series in the Siegel problem, see
\cite{Eliasson:1996} for Lindstedt series of Hamiltonians). Direct
methods for proving their convergence are involved because, as
J. Moser noticed in~\cite[p.~149]{Mos67}, these series do not converge
absolutely, and thus the proof of semi-convergence must take into
account compensations or the precise accumulation of small
denominators through a subtle combinatorial analysis. On the other
hand, the perturbation series yielded by the Newton algorithm are
absolutely convergent, provided that one adequatly chooses the width
of analytic spaces at each step of the induction. This was a major
discovery of Kolmogorov. In the first approximation, the series so
obtained can be thought of as obtained by grouping terms of the
classical perturbation series (from step $j$ to step $j+1$, the non
resonant terms of size $\epsilon^{2^j}, \cdots, \epsilon^{2^{j+1}-1}$
are eliminated). The magics is that compensations are taken into
account without noticing, and it would be interesting to understand
how classical and Newton series relate precisely, maybe with mould
calculus.

-- We encapsulate the Newton algorithm in an abstract inverse function
theorem {\`a} la Nash-Moser. The algorithm indeed converges without any
specific hypothesis on the internal structure of the variables. At the
expense of some optimality, ignoring this structure allows for simple
estimates (and control of the bounds) and for solving a whole class of
analogous problems with the same toolbox (lower dimensional tori,
codimension-one tori, Siegel problem, as well as some problems in
singularity theory).

-- The analytic (or Gevrey) category is simpler, in Nash-Moser theory,
than H{\"o}lder or Sobolev categories because the Newton algorithm can be
carried out without intercalating smoothing operators
(cf.~\cite{Ser72,Bos86}).

-- Incidentally, Hadamard interpolation inequalities are simple to
infer for analytic norms because, again, they do not depend on
regularizing operators, as it is shown in appendix~\ref{app:convexity}
(cf.~\cite[Theorem~A.5]{Hor76}).

-- The use of auxiliary norms ($|\cdot|_{G,s}$ in
lemmas~\ref{lm:cohomological} and~\ref{lm:seconde}, $|\cdot|_{x,s}$ in
appendix~\ref{app:submersion}) prevents from artificially loosing, due
to compositions, a fixed width of analyticity at each step of the
Newton algorithm --the domains of analyticity being deformed rather
than shrunk. As a pitfall, the argument of~\cite[Sections~5
and~6]{Jac72} to deduce an analytic function theorem in the smooth
category abstractly from the theorem in the analytic category, does
not apply directly here (see comment below).

\subsubsection*{Section~1. Theorem~\ref{thm:herman}} Herman's normal
form is the Hamiltonian analogue of the normal form of vector fields
on the torus in the neighborhood of Diophantine constant vector fields
(\cite{Arn61,Mos66}). The normal form for Hamiltonians implies the
normal form for vector fields on the torus~\cite[Th{\'e}or{\`e}me~40]{Fej04}
and is actually simpler to prove from the algebraic point of view.

\subsubsection*{Section~3. Lemma~\ref{lm:cohomological}} The estimate
is obtained by bounding the terms of Fourier series one by one. In a
more careful estimate, one should take into account the fact that if
$|k \cdot \alpha|$ is small, then $k' \cdot \alpha$ is not so small
for neighboring $k'$'s. This allows to find the optimal exponent of
$\sigma$, making it independant of the dimension; see
\cite{Mos66a}
and \cite{Rus75}.

\subsubsection*{Appendix~\ref{app:submersion}.
  Theorem~\ref{thm:inversion}}

-- The two competing small parameters $\eta$ and $\sigma$ being fixed,
our choice of the sequence $(\sigma_n)$ maximizes $\epsilon$ for the
Newton algorithm. It does not modify the sequence $(x_k)$ but only the
information we retain from $(x_k)$.

-- In the expression of $\epsilon$, the square exponent of $C$ is
inherent in the quadratic convergence of Newton's algorithm. From this
follows the dependance, in KAM theory, of the size $\epsilon$ of the
allowed perturbation with respect to the small diophantine constant
$\gamma$: $\epsilon = O(\gamma^2)$.

-- The method of \cite{Jac72} (see~\cite{Mos66a} also in order to
deduce an inverse function theorem in the smooth category from its
analogue in the analytic category does not work directly, here. The
idea would be to use Jackson's theorem in approximation theory to
caracterize the H{\"o}lder spaces by their approximation properties in
terms of analytic functions and, then, to find a smooth preimage $x$
by $\phi$ of a smooth function $y$ as the limit of analytic preimages
$x_j$ of analytic approximations $y_j$ of $y$. However, in our
inversion function theorem we require the operator $\phi$ to be
defined only on balls $\sigma B_{s+\sigma}$ with shrinking radii when
$s+\sigma$ tends to $0$. This domain is too small in general to
include all the analytic approximations $y_j$ of a smooth $y$.  Such a
restriction is inherent in the presence of composition operators.
\cite{Jac72} did not have to deal with such operators for the problem
of isometric embeddings. Yet we \emph{could} generalize Jacobowitz's
proof at the expense of making additionnal hypotheses on the form of
our operator $\phi$, which would take into account the specificity of
directions $K$ and $G$, as well as of the real phase space and of its
complex extension.

\subsubsection*{Appendix~\ref{subsec:regularity}}

It is possible to prove that $\psi$ is $C^1$ without additional
asumptions, just by patterning~\cite[p.~626]{Ser72}). Yet the proof
simplifies and the estimates improve under the combined two additional
asumptions. In particular, the existence of a right inverse of
$\phi'(x)$ makes the inverse $\psi$ unique and thus allows to ignore
the way it was built.

\subsubsection*{Appendix~\ref{app:inversion}}

We include this elementary section for the sake of completeness,
although the quantitative estimates are needed only if one wants a
quantitative version of Kolmogorov's theorem, with an explicit value of
$\epsilon$.  A similar proposition (for germs at a point of maps in
$\ensuremath{\mathbb{C}}^n$) is proved in~\cite{Poe01} using a more sophisticated argument
from degree theory.

\subsubsection*{Appendix~\ref{app:convexity}}

In this paragraph, the obtained inequalities generalize the standard
Hada\-mard convexity inequalities. They are optimal and show that
analytic norms are not quite convex with respect to the width of the
complex extensions, due to the geometry of the phase space.
See~\cite[Chap.~8]{Nar71} for more general but less precise
inequalities.

\subsubsection*{Appendix~\ref{app:arithmetics}. Proposition~\ref{prop:arithmetics}}
\label{app:arithmetics}

There are reasons to believe that the so obtained arithmetic condition
is not optimal. Indeed, solving the exact cohomological equation at
each step is inefficient because the small denominators appearing with
intermediate-order harmonics deteriorate the estimates, whereas some
of these harmonics could have a smaller amplitude than the error terms
and thus would better not be taken care of. Even stronger, R{\"u}ssmann
and P{\"o}schel remarkably and recently noticed that at each step it is
worth neglecting part of the low-order harmonics themselves (to some
carefully chosen extent). Then the expense, a worse error term, turns
out to be cheaper than that the gain --namely, the right hand side of
the cohomological equation now has a smaller size over a larger
complex extension. This allows, with a slowly converging sequence of
approximations, to show the persistence of invariant tori under some
arithmetic condition which, in one dimension, is equivalent to the
Brjuno condition; see \cite{Poschel:2009}.

\bigskip \textit{Thank you to P. Bernard, A.  Chenciner, R.
  Krikorian, I. Kupka, D. Sauzin and J.-C. Yoccoz, for illuminating
  discussions, and to A. Albouy and A. Knauf for careful reading and
  correcting.}

\bibliographystyle{abbrvnat}

redpajama_set_name: RedPajamaArXiv


text:
Q: Xpath fetching node by attributes This should be rather a simple task, but for some reasons a part of my xml file hasn't been read properly...
This is the XML Section in question:
<MTX>
...
<MoClass code=".0" modi="XYZ">
.....
</MoClass>
......
</MTX>

Now, i want to fetch the Node by following code:
 $xml = simplexml_load_file(filename)

 //this fails due to invalid predicate...
 $xml->xPath('//MoClass[@code=".0" AND @modi="XYZ"]');

 //but when i reverse the sequence of attributes
 $xml->xPath('//MoClass[@modi="XYZ" AND @code=".0"]');
 //then it works as expected and the node may be processed further...

What is the difference?
 Thanks in advance

A: Xpath is case-sensitive. Use and instead of 'AND'
demo

redpajama_set_name: RedPajamaStackExchange


At the moment, the debt-to-equity of SVB Financial Group (NASDAQ:SIVB) is low, standing at 25.95, a figure that is less than the 61.93 average recorded by the industry. This means that the company is currently holding a debt level at 28.06 B.
Even though there was a rise of 13.64 M in revenue, the company failed to succeed in outperforming the industry average of 10.91%. For the most recent quarter, the net income has jumped by 542.90 M. This strength in their income has affected them and thus increased their earnings to $172.20 M. The 34.04% yoy growth of SIVB's revenue has gone up that of the industry average by -0.41%. For the past 12 months, SVB Financial Group revenue has gone up by 34.45%. The sustained growth in their revenue has helped boost their earnings per share.
The 12-month return on equity has significantly fallen to 20.95 in comparison to the same data for other companies in the same industry. This shows that there is a major weakness within the organization over the past one year. Comparing them to other companies in the industry and the overall Financial sector, the industry average is 16.76 while 16.20 is of the sector.
Also, looking at the price to cash flow of the company and the industry average, the 16.10 ratio of the stock is lower than the industry's 16.92.
SVB Financial Group (NASDAQ:SIVB) has a price-to-earnings ratio of 13.18 which is lower than the 19.55 industry average at the moment. This shows whether the company has what it takes to effectively turn the revenue into profit.
The company's ROA is 1.87 when compared to 2.62 for the stocks operating in the same industry. This can be attributed to the strength recorded in the net income produced by total assets. Comparing it to other companies in the sector, SVB Financial Group ROE is above 16.20 that of both the sector average.
The operating profit margin for SVB Financial Group (SIVB) is 51.65%, a figure which is considered to be strong. It has gone 41.71 from the 45.38 over the past 5 years. In addition to this, their operating margin is 9.94 higher than the industry average.
The net profit margin which stood at 30.60 on average in the past 5 years has jumped to 38.32 in the last 12 months. Added to that, this ratio has surpassed the industry net margin that stands at 32.82.
Analysts meanwhile rate SVB Financial Group (NASDAQ:SIVB) as a strong buy. Still some above discussed indicators of the $12.89B company show strength while others show weakness. There is little evidence at the moment to justify the expectation of the SIVB shares to either perform positively or negatively when compared to other stocks. The primary strengths of SVB Financial Group can be witnessed in its increased revenue, growing earnings per share, higher return on equity, increased operating cash and high net margin. Subsequently, financial analysis have also identified some weak areas that includes high debt, relatively high P/E ratio, lower return on assets and low net margin.


ECO 561 WEEK 6 Challenges of Expansion to a Foreign Location Revise your Week 3 economic analysis using the feedback provided by your facilitator to evaluate the challenges of expanding your company's production to a foreign market. This will require you to revise previously recommended pricing and non-pricing strategies for your company's product based on the selected country's economy's stage in the business cycle and the projected economic conditions. Discuss the evidence that supports your recommendations.
Select a foreign market in which to expand your chosen product.
Write a 1,750-word paper (including any material you include from your Week 3 assignment) detailing your findings. The use of tables and/or charts to display economic data over the time period discussed is highly encouraged.
Describe current global economic conditions and their effect on local macroeconomic indicators in your selected country. Consider forecasts for population growth, GDP growth, GDP per capita growth, export growth, and sales growth.
Evaluate the competitors' existing production in the chosen country.
Evaluate forecast sales in the selected country.
Determine the type of economy that exists in your selected country – closed, mixed, or market. What is the difference between these types of economies and how might this affect your expansion?
Describe how your chosen country's current credit market conditions affect demand for your product or service and your planning or operating decision for your production in that country.
Analyze the role of the selected country's central bank on that country's economy.
Evaluate the availability, education, and job skills of the work force in the selected country. Discuss the additional challenges of international production, such as political stability, availability of government financing or other incentives, threat of capital controls, and exchange rate risks.
Discuss any additional supply chain challenges you anticipate if attempting to sell your product made in your chosen country to countries outside of that market.
Discuss any comparative advantages your company will have over competitors in that country.
Recommend either for or against expanding your company's production into your chosen country based on your research.


text:
Here's What Hedge Funds Think About ScanSource, Inc. (SCSC)
Published on July 5, 2019 at 3:59 pm by Reymerlyn Martin in Hedge Funds,News
Hedge funds are known to underperform the bull markets but that's not because they are bad at investing. Truth be told, most hedge fund managers and other smaller players within this industry are very smart and skilled investors. Of course, they may also make wrong bets in some instances, but no one knows what the future holds and how market participants will react to the bountiful news that floods in each day. Hedge funds underperform because they are hedged. The Standard and Poor's 500 Index returned approximately 12.1% in the first 5 months of this year through May 30th (including dividend payments). Conversely, hedge funds' top 20 large-cap stock picks generated a return of 18.7% during the same 5-month period. An average long/short hedge fund returned only a fraction of this due to the hedges they implement and the large fees they charge. Our research covering the last 18 years indicates that investors can outperform the market by imitating hedge funds' stock picks rather than directly investing in hedge funds. That's why we believe it isn't a waste of time to check out hedge fund sentiment before you invest in a stock like ScanSource, Inc. (NASDAQ:SCSC).
Is ScanSource, Inc. (NASDAQ:SCSC) a healthy stock for your portfolio? Hedge funds are becoming less confident. The number of long hedge fund positions shrunk by 2 lately. Our calculations also showed that SCSC isn't among the 30 most popular stocks among hedge funds. SCSC was in 10 hedge funds' portfolios at the end of March. There were 12 hedge funds in our database with SCSC positions at the end of the previous quarter.
If you'd ask most traders, hedge funds are viewed as worthless, old financial vehicles of years past. While there are over 8000 funds in operation at the moment, We look at the bigwigs of this club, around 750 funds. It is estimated that this group of investors manage most of the smart money's total asset base, and by keeping an eye on their top investments, Insider Monkey has deciphered several investment strategies that have historically defeated the S&P 500 index. Insider Monkey's flagship hedge fund strategy outstripped the S&P 500 index by around 5 percentage points per year since its inception in May 2014 through June 18th. We were able to generate large returns even by identifying short candidates. Our portfolio of short stocks lost 28.2% since February 2017 (through June 18th) even though the market was up nearly 30% during the same period. We just shared a list of 5 short targets in our latest quarterly update and they are already down an average of 8.2% in a month whereas our long picks outperformed the market by 2.5 percentage points in this volatile 5 week period (our long picks also beat the market by 15 percentage points so far this year).
Let's review the fresh hedge fund action surrounding ScanSource, Inc. (NASDAQ:SCSC).
How have hedgies been trading ScanSource, Inc. (NASDAQ:SCSC)?
Heading into the second quarter of 2019, a total of 10 of the hedge funds tracked by Insider Monkey held long positions in this stock, a change of -17% from the fourth quarter of 2018. On the other hand, there were a total of 9 hedge funds with a bullish position in SCSC a year ago. So, let's see which hedge funds were among the top holders of the stock and which hedge funds were making big moves.
More specifically, Pzena Investment Management was the largest shareholder of ScanSource, Inc. (NASDAQ:SCSC), with a stake worth $38.9 million reported as of the end of March. Trailing Pzena Investment Management was AQR Capital Management, which amassed a stake valued at $6.1 million. Two Sigma Advisors, Renaissance Technologies, and D E Shaw were also very fond of the stock, giving the stock large weights in their portfolios.
Seeing as ScanSource, Inc. (NASDAQ:SCSC) has experienced falling interest from the entirety of the hedge funds we track, it's safe to say that there lies a certain "tier" of fund managers who sold off their positions entirely heading into Q3. It's worth mentioning that Israel Englander's Millennium Management dropped the biggest stake of all the hedgies followed by Insider Monkey, totaling an estimated $0.6 million in stock. David Costen Haley's fund, HBK Investments, also dropped its stock, about $0.4 million worth. These bearish behaviors are intriguing to say the least, as total hedge fund interest fell by 2 funds heading into Q3.
Let's go over hedge fund activity in other stocks similar to ScanSource, Inc. (NASDAQ:SCSC). These stocks are SunPower Corporation (NASDAQ:SPWR), Middlesex Water Company (NASDAQ:MSEX), Cheetah Mobile Inc (NYSE:CMCM), and Genesco Inc. (NYSE:GCO). All of these stocks' market caps are closest to SCSC's market cap.
SPWR 9 36580 1
MSEX 9 48181 -2
CMCM 4 5867 0
GCO 19 48420 -3
Average 10.25 34762 -1
As you can see these stocks had an average of 10.25 hedge funds with bullish positions and the average amount invested in these stocks was $35 million. That figure was $50 million in SCSC's case. Genesco Inc. (NYSE:GCO) is the most popular stock in this table. On the other hand Cheetah Mobile Inc (NYSE:CMCM) is the least popular one with only 4 bullish hedge fund positions. ScanSource, Inc. (NASDAQ:SCSC) is not the least popular stock in this group but hedge fund interest is still below average. This is a slightly negative signal and we'd rather spend our time researching stocks that hedge funds are piling on. Our calculations showed that top 20 most popular stocks among hedge funds returned 6.2% in Q2 through June 19th and outperformed the S&P 500 ETF (SPY) by nearly 3 percentage points. Unfortunately SCSC wasn't nearly as popular as these 20 stocks (hedge fund sentiment was quite bearish); SCSC investors were disappointed as the stock returned -10.6% during the same time period and underperformed the market. If you are interested in investing in large cap stocks with huge upside potential, you should check out the top 20 most popular stocks among hedge funds as 13 of these stocks already outperformed the market so far in Q2.
Is Ardmore Shipping Corp (ASC) A Good Stock To Buy? Hedge Funds Have Never Been More Bullish On Atlantic Tele-Network, Inc. (ATNI... Hedge Funds Have Never Been This Bullish On Millendo Therapeutics, Inc. (MLND... Hedge Funds Have Never Been More Bullish On Mesa Laboratories, Inc. (MLAB) Here's What Hedge Funds Think About Tucows Inc. (TCX) Here's What Hedge Funds Think About Controladora Vuela Co Avcn SA CV (...
Is SCSC A Good Investment Right Now? Is SCSC A Good Stock To Buy? NASDAQ:SCSC Yahoo Finance
Here's What Hedge Funds Think About ScanSource, Inc. (SCSC) 13G Filing: Pzena Investment Management and Scansource Inc. (SCSC) Do Hedge Funds Love ScanSource, Inc. (SCSC)? ScanSource, Inc. (SCSC): Are Hedge Funds Right About This Stock? Should You Sell PDF Solutions, Inc. (PDFS)? What Hedge Funds Think About ScanSource, Inc. (SCSC) Avoid ScanSource, Inc. (SCSC)? How to Choose the Best Canadian Credit Card in 4 Simple Steps 25 Gambling Stocks Hedge Funds Are Crazy About in 2019 11 Easiest Orthodontic Residency Programs to Get Into

redpajama_set_name: RedPajamaCommonCrawl


Although this conference was noted on the BAGH calendar a few weeks ago, it's worth reminding the group it's still on. I just received this from San Francisco Ghost Society and it still has openings.
Thanks For Taking The Haunted Haight Walking Tour.
Here is a special invitation to our latest Ghost Conference in Sutter Creek California!
Come investigate and learn about ghosts in the gold rush town! Just $89 per person! Feb 26th and Feb 27th VERY SMALL TEAMS AND WE PROVIDE ALL EQUIPMENT!
If you EVER wanted to go REAL GHOST HUNTING, NOW IS YOUR CHANCE!
You are receiving this email because you purchased tickets for this event.


text:
Take Five ticket worth $22,434 sold in Riverhead
Riverhead Central School District
Two more Riverhead school employees test positive for coronavirus, district will no longer report individual cases to parents
Katie Morosky
Pulaski Street Elementary School in May2019. Photo: Denise Civiletti
Two more staff members at Riverhead Central School District have tested positive for COVID-19, the school district announced on its Facebook page this weekend.
One employee works at Riverhead Middle School. The other works at Pulaski Street School.
No additional information could be provided about the employees or their positions due to to privacy laws, the district said in its announcement.
LIVE UPDATES: Coronavirus coverage, closings and cancellations.
All Riverhead schools have been closed since March 13. Long Island schools have been closed since March 16, and all schools across New York State were closed from March 18 onward.
Schools across the state will remain closed until at least April 15, Gov. Andrew Cuomo announced this weekend.
Riverhead Central School District will no longer be sending out notifications for each individual positive case, the district said last night, since schools have now been closed for two weeks.
"With the increase in testing, the district anticipates additional positive cases within the Riverhead Central School District community," the district said in its announcement.
The first staff member to test positive, an employee at Riverhead Middle School, was announced March 22.
Now more than ever, the survival of quality local journalism depends on your support. Our community faces unprecedented economic disruption, and the future of many small businesses are under threat, including our own. It takes time and resources to provide this service. We are a small family-owned operation, and we will do everything in our power to keep it going. But today more than ever before, we will depend on your support to continue. Support RiverheadLOCAL today. You rely on us to stay informed and we depend on you to make our work possible.
Pulaski Street School
Riverhead Middle School
Katie, winner of the 2016 James Murphy Cub Reporter of the Year award from the L.I. Press Club, is a co-publisher of RiverheadLOCAL. A Riverhead native, she is a 2014 graduate of Stony Brook University. Email Katie


text:
We have over 30,000 local, licensed providers right here.
How Christy Doramus Transitioned From Beauty Publicist to Floral Entrepreneur
By Hilary Sheinbaum | January 7, 2020
Carter Fish
It isn't always roses for Christy Doramus, but most days the florist extraordinaire takes time to smell a bouquet or two. A typical day for the self-proclaimed "Jane of All Trades" might entail a trip to Manhattan's flower district, creating floral arrangements, or chatting with her client Drew Barrymore about Instagram®.
Doramus, 31, is the founder of CMD Creative℠, a Manhattan-based company focused on floral and event design, as well as social media consulting. The Baton Rouge, Louisiana, native graduated from The Art Institute of California®, San Francisco, where she studied Fashion Marketing and Management. After she landed in New York City 10 years ago, she began working as a beauty publicist for Alison Brod's eponymous firm. For almost six years, Doramus worked with big name beauty clients like L'Oréal® Paris and FLOWER BeautyTM by Drew Barrymore.
During her PR experience, Doramus promoted the benefits of skincare products and makeup — so it's no surprise she continues to practice what she preached (including removing her makeup before bed). She's also in tune with her fair skin, meaning she's a big fan of sunscreen and gets her moles checked often by a board-certified dermatologist.
When she eventually decided to branch out on her own four years ago, Doramus started with a floral business. At the time, she primarily created flower crowns, as they were quite popular then — especially to wear at music festivals. Celebrities, influencers, and brands all came calling, excited to sport her designs. While flower crowns aren't as trendy these days, Doramus proved that she and her company have staying power: On the event design front, she's cozy with brands like Kendra Scott®, Sisley® Paris, and L'Oreal Paris. And, her talents as a social media consultant have made her a go-to for Barrymore's new site, FLOWER by Drew®.
"I now work more hours than I ever have in my entire life," said Doramus. "It just doesn't feel like it, because I'm doing so many things that I love."
Doramus is the kind of person you can go to for any type of advice, whether it's floral, beauty, hosting, design, or entrepreneurship-related. But she's more than an encyclopedia of information. Doramus is easy to talk to, well spoken and incredibly approachable (despite having the potential to be mistaken for a model). Read on for our one-on-one chat with Doramus, who discusses everything including dermatology and flower design, where she vacations with her husband, juggling it all, the contents of her makeup bag, and Drew Barrymore.
Spotlyte: How did you first become interested in flowers?
Christy Doramus: Growing up, my mom always taught me different types of flowers. We had vegetable gardens and planted our own flowers, so I've always had an interest in florals.
Spotlyte: Did you have any formal flower education?
CD: As far as the creation of floral crowns or floral design, in general, I'm self-taught. I learned a lot about flower types by asking tons of questions to people in the flower district who worked at the stores versus getting formal education. Most of these individuals have been in the business for their entire lives, because a lot of those businesses are passed down through generations.
Spotlyte: What kinds of things did you learn from being curious?
CD: They really taught me so much about caring for different types of flowers, what's available at different times of the year, water temperatures, price points, and availability.
Spotlyte: In terms of flower design: Does art imitate life? What inspires your designs?
CD: When it comes to creating — utilizing those flowers and creating my own designs — it's all inspired by things that I see in magazines and books, and on social media. At the time when I started making flower crowns, it was something that was gaining popularity with music festivals. It's something that I tried creating on my own, because at the time there were not really many tutorials out there on how to do it.
Spotlyte: What was the first flower crown you created like?
CD: I made it with dental floss. I was in our backyard at this house that we rented, and I pulled a bunch of roses from the rose bush. I was looking around for like a wire or something that I could use. I didn't know if I was going to tie the flowers together, like how you make daisy chains. I found some dental floss, and that's what I used to braid the flowers together. That was my first flower crown. It evolved after that. Over time, I learned better tools.
Spotlyte: What's your best advice for first-time entrepreneurs?
CD: If you're not loving your 9 to 5, then take a step back and figure out what things make you the most happy, and then build a career around that. I think any job can be created out of passion and dedication. It's no walk in the park — I'm not saying it's easy.
Spotlyte: You've maintained a lot of your connections from the past decade. How has this helped you in your current role?
CD: Having been in the [beauty] industry for 10 years, it's so important to remember that you never know when you might work for your old intern. Or, someone that used to work for you might be your boss some day. Fifty percent of the people who hire me now are my old interns or assistants that used to work with me in PR. Over the years, Alison Brod has been one of my biggest supporters.
Spotlyte: Who else has been your biggest cheerleader?
CD: Someone that I've loved working with over the years, and who I find so much inspiration from — and love brainstorming and talking about design and creativity and everything with — is Drew [Barrymore], of course. I love having creative conversations with her. Also, a lot of people that were my colleagues or my interns that I worked with over the years have been really, really super supportive.
Margaret Sullivan Photo
Spotlyte: One of your PR responsibilities was learning about skincare. What is your morning skincare routine?
CD: I have super sensitive skin, so I love June Jacobs® Cooling Cucumber Cleanser. I've also been turned on to Snow Fox SkincareTM Cooling Foam Cleanser, which is really great for sensitive skin. Sometimes I have to do makeup removal at night and in the morning — when I wake up, and there's still mascara.
Spotlyte: What's your evening skincare routine?
CD: I love Bioderma® for makeup removal. It also comes in travel-size, which is nice to throw in your travel bag. A makeup wipe just never gets it all off, so I love Bioderma. I cleanse in the morning and at night. I'm always switching my moisturizer and trying something different. Sunday Riley® TidalTM and SkinCeuticals® Daily Moisture are favorites of mine. And then, I always use Revitabrow®.
Spotlyte: What kind of treatments do you get?
CD: I don't do a lot of spa time, really, but there is this place my husband will take me once a year. It's called the Mayflower Inn [and Spa]℠ in Connecticut, and there's this woman who is amazing at facials. I have been able to do that two or three times.
Also, my friend owns a store called Knockout Beauty® [in New York City], and they also do amazing facials. In the summertime, in the Hamptons, she has Georgia Louise in residence, which is amazing. I haven't been able to go to her yet, but it's on my list.
Yasmin at Yaz Beauty Studio℠ does my eyebrows. She turned me on to Revitabrow serum. It makes your eyebrows grow, and some days I just put that on and add my eyebrow mascara.
Spotlyte: What's in your makeup bag?
CD: For foundation, I either use Jouer® tinted moisturizer or Sisley Paris liquid foundation. I swish on Chanel® or FLOWER Beauty blush. I always, always, always wear a FLOWER Beauty lipstick. My favorite one is called FLOWER Beauty Mix N' MatteTM Lip Duo, and the color is called Tickled PinkTM. It's like the perfect mauve pink color — it works great on every skin tone, but it's also really great for fair people like myself. Really fair skin.
For mascara, I love to wear L'Oréal Paris Voluminous® Carbon Black Mascara or FLOWER Beauty Lash WarriorTM Mascara.
Spotlyte: Since you mentioned fair skin, what sunscreen do you use?
CD: I'm really into Supergoop!® foaming sunscreen. I stay out of the sun as much as possible, if you can't tell, because I look like a ghost.
Spotlyte: Where do you see your dermatologist?
CD: I go to Spring Street Dermatology℠ to get my moles tracked more regularly than most people, because I'm terrified that it's skin cancer.
Spotlyte: What is your stance on medical aesthetics?
CD: Since I've turned 31, I'm starting to notice that my body doesn't bounce back [as well] and neither does my skin, but I still haven't tried anything. I keep thinking that I'll cross that bridge when I get there and decide if it's something for me. I have so many friends who do injectables, and [their results] look great.
[Editor's note: Injectable wrinkle reducers temporarily smooth the look of moderate to severe wrinkles in certain areas of the face, including the forehead, frown lines, and crow's feet; they should not be used more frequently than every three months. Injectable filler is a temporary treatment that adds volume to areas of the face such as the lips, cheeks, and laugh lines. Like any medical treatment, both injectable wrinkle reducers and injectable fillers have potential risks and side effects. Talk to a licensed provider to see if they're right for you. And learn more now by chatting with a trained aesthetic specialist.]
Spotlyte: What do you like to do in your spare time when you're not working or running your business?
CD: My ultimate relaxation is going to my house in Bridgehampton. I love, love, love working in my garden, and just unplugging, and getting off my phone for a few hours on the weekends.
I'm learning a lot from my garden. I have lots of herbs and spices, and I'm trying to grow tomatoes, desperately, but deer keep eating them. It's a big learning curve for me.
I enjoy going to estate sales on the weekends and finding treasures. And reading as many magazines as I possibly can, because I'm such a fan of old-school print magazines. I have so many. Sometimes, at estate sales — if I see a stack of 1990s Architectural Digests® — I'll buy them. I'm like, "Oh my God, I have to have these." Now, I'm just trying to find a place to store them all.
Allergan® may receive commissions for purchases made through links in this article.
Rosacea is Often Misunderstood, According to Dr. Brenda LaTowsky
by Sophie Wirt
|January 9, 2020
Dr. Hadley King on How Selfies Have Caused a Shift In Dermatology
by Hilary Sheinbaum
Meet the Derm of Two Cities, Dr. Jordan Carqueville
How This Dancer and Fitness Influencer Created a Cardio Empire In Under 3 Years
The Powerful Homeopathic Remedy That Transformed Dr. Jo Polich's Career
by Emily Orofino
What One Derm Has Used to Treat and Conceal Her Own Vitiligo
I have read and agree to the Spotlyte Terms and Conditions and Privacy Statement. I confirm that I am 18 years of age. By providing my email address, I agree to receive information from Allergan relating to Spotlyte and other Allergan products and services.
© 2020 ALLERGAN. All rights reserved. All trademarks are the property of their respective owners. Spotlyte is owned and operated by Allergan.
You are now leaving Spotlyte. Links to third-party sites are provided as a courtesy. Allergan does not control or endorse, and is not responsible for, non-Allergan sites, including services or products offered on such sites. Allergan's Privacy Policy and Terms of Use do not apply to third party sites.


I've eyed Stack Overflow Jobs many times as a freelancer, and I just always wondered if your account reputation score/badges were at all linked to the application process. I did search for an answer to this question via Google Search and didn't find a result.
Additionally, if you voluntarily share your Stack Overflow account, I was wondering if employers have in the past taken that into account in the hiring process (though it may be hard to say if they have, either way). This would include looking at the nature of questions you asked and answered and how they relate to the job you applied for.
Employers can see your Stack Overflow reputation, but they're not necessarily looking at it.
We do show your rep score on the employer view of your Developer Story, if you choose to display it. That is, if you add your rep to your Story in the Story editor, and then apply for a job using your Story or declare yourself as "Actively Looking" or "Passively Looking" (so you can be found by recruiters searching our database for candidates), then employers can see your rep.
No one ever asks me about it. If they're in the top percentage for [a given tag], that's another story. If they see in the results that someone is in the top % for a technology they're searching, that's really interesting/exciting to them.
In other words recruiters tend to be less interested in reputation and more interested in your experience with a particular technology, along with your work history, open source contributions, etc. (Of course, that varies from person to person.) That said, in training conversations with recruiters, we do try to discourage them from taking top percentiles too seriously: there are plenty of skilled developers who aren't highly ranked, simply because they don't have the spare time to put in the hundreds of hours it takes to achieve a high rep.
I'd love to give you hard stats, but we haven't run a formal employer survey and sadly we don't have data on which part of the page a recruiter's eyes are looking at. We are working on that technology though, and I'm excited to announce that all new customers will be required to undergo compulsory brain-implant surgery starting in 2018.
Yes they can, and in at least some cases they do.
I can state with certainty that when I interviewed for my current job (which had been advertised here on SO) at least a couple of the people who interviewed me made it clear that they'd looked through my profile and at least some of the answers I'd posted.
In fact, one of the interviews was devoted primarily to (what was at the time) my top-voted answer in a fair amount of detail, and discussing exactly why it was as useful/effective as it was.
Plans for compulsory brain-implant surgery?
Does job application require reputation?


Value A Business What is my Business Worth?
FCBB Will Help You Buy or Sell a Business in Nashville, TN.
Whether your goal is to buy or sell a business in Nashville, it isn't a task for the inexperienced; the process is particularly difficult without the benefit of training and a proper background. With so much at stake, it pays to find a local business broker you can trust – one with up-to-date skills and sharp insights into the industry. So it's no wonder that so many business buyers and sellers in Nashville trust First Choice Business Brokers. Please click here to see our Business for sale Nashville TN list.
Established in 1994, FCBB has grown to become one of the largest business sales organizations in the nation, and here in Nashville, our diverse team of brokers brings insightful experience from areas such as accounting, banking, real estate, and executive search.
We're conveniently located in Nashville, TN, but we're able to serve a much larger area that includes the entire state of Tennessee, as well as Kentucky, Alabama, North Carolina, South Carolina, and Mississippi.
We're large enough to offer our clients the best selection of businesses for sale, yet small enough to give our sellers individual attention. You'll also find that our Nashville team is very discreet. Confidentiality is critical in a business sales transaction, and we aim to complete the sale under the radar of customers, employees, competitors, and suppliers.
If you're a business owner, entrepreneur, strategic buyer, or investor looking for a broker that exceeds your expectations, call First Choice Business Brokers of Nashville at (615) 850-4410 or contact us here . Whether you're buying or selling a business, we're committed to getting you the best options, at the best price, on the best terms.
I can't thank First Choice Business Brokers enough. FCBB had continued faith in the sale of our business and helped us make it happen with the right buyer. Patrick Marsch Jr. is a great broker, and really made us comfortable with our decisions. I wouldn't change our experience for anything. Really, Patrick, thank you so much for your patience and resilience is getting this done. We are very happy with the end product and the perfect buyer!
When we originally bought our business "Four Seasons Candy Company", we didn't use a business broker. Needless to say, that was a costly mistake. So when we decided to sell our business, for sure, we were going to use a business broker even with the commissions. Finding a good business broker is as important as finding a buyer. This time we were lucky because we found Patrick Marsch Jr. with "First Choice Business Brokers". Thanks to Patrick, we found the right buyer for our business. As a team player, Patrick assisted us with a precise, step-by-step analysis that provided us with the confidence that we were offering our business at a full and fair value. Equally important, with patience and resilience, Patrick found us numerous potential buyers and helped us identify the most qualified buyer. With this precise, step-by-step analysis we felt comfortable that the buyer received the best overall deal. If we should ever buy another business or if any of our family or friends should decide to buy or sell a business, we would, without hesitation recommend Mr. Patrick Marsch Jr. with First Choice Business Brokers.
Having a team of business brokers represent me in the sale of my business was the RIGHT choice. Adam and Patrick were able to quickly qualify multiple buyers, and bring a great offer from a buyer who really was the perfect fit for my business. Young, energetic and proactive, the team from FCBB Nashville got the job done and exceeded my expectations. I would recomend their services to anyone.


Cleaning choked drains or repairing leakages are not always easy but costly and time taking too, if it requires pipe replacement.
Drain water reaching the rooms and basement and flowing over the driveway and walkway creating a widespread mess and severe health hazards is the last thing that anyone wants.
The germs and bacteria, as well as the foul smell of drain water, can make the inhabitants sick. Therefore, you should try to avoid drain choking that can cause leakages over time. Especially with indoor drain lines so that repairing is quick, and you can prevent bigger problems.
How Do I Fix Damp or Wet Corners in My Basement?
Drains and drain lines from toilets, kitchens, and bathrooms are vulnerable to clogging depending on the extent and manner of use. Kitchen drains are likely to clog more because the chances of food and vegetable waste along with oil and grease can find its way into the drains despite efforts to prevent it.
Using some filter in the kitchen sink might reduce the chances of solid waste and debris going into the drain, but it does not help to prevent oil and grease passing into the drain lines. Preventing all such items from going into the drain can reduce the chances of pipelines clogging frequently.
Hair accumulation in the bathroom drains is the most common cause of choked drain lines, and there is nothing much you can do about it. Moreover, sanitary pads, toilet paper, hair clips and bobby pins going into the bathroom drain are not unusual, but with some alertness, you can avoid such items find its way into the drain.
While drains clog over a long period due to deposition of debris and sludge in the pipelines, the bigger reason is the negligence of homeowners who abuse drains by allowing unwanted materials are passing into it.
To prevent drain blockages, plumbers advice homeowners to be alert and watchful about what goes into the drain and prevent its entry instead of trying to correct the wrongdoing by undertaking elaborate drain cleaning exercise that causes a lot of inconveniences.
Arrange to segregate the waste generated in kitchens to ensure that insoluble solid materials must never go into the drain. Drains are meant to carry water and liquids only, but you must avoid oils, grease, and chemicals going into the drain because it mixes with the sludge present in water and creates solid chunks that block drains.
Avoid food particles from mixing with water by scrapping the solid waste and collecting it separately. To avoid oils and greases mixing with water clean cooking utensils with detergent thoroughly. Wherever possible, pour out used cooking oil in containers for disposing of like other waste materials.
Similarly, avoid carrying solid materials like hair grips into the bathroom that eliminates its chances of accidentally going into the drain. Make a habit of throwing paper towels, cotton wool and other solid debris in bins.
These efforts should reduce the incidents of drain blockage considerably.


text:
export class WindowMock {
  public navigator: any = {
    language: 'en-US',
    userAgent: 'testing',
  };
  public location: any = {};

  public alert(msg: string): void {
    return;
  }

  public confirm(msg: string): void {
    return;
  }
}

export class WindowMockFrench extends WindowMock {
  constructor() {
    super();
    this.navigator.language = 'fr-US';
  }
}

export class WindowMockNoLanguage extends WindowMock {
  constructor() {
    super();
    this.navigator.language = undefined;
  }
}

redpajama_set_name: RedPajamaGithub


text:
Грасьела Кальдерон Диас Баррига де Жедовский (; 14 июля 1931 — 2 января 2022) — мексиканский ботаник. Известна своим вкладом в систематику растений неотропики.

Биография
Грасьела Кальдерон родилась 14 июля 1931 года в муниципалитете Сальватьерра штата Гуанахуато. Училась биомедицине в Национальном политехническом институте, затем перешла в Национальную школу биологических наук. В 1952 году Кальдерон получила степень бакалавра. В 1953 году работала в Мексиканском институте возобновляемых природных ресурсов, затем перешла в Институт изучения пустынь при Автономном университете Сана-Луиса-Потоси, где преподавала биологию с 1955 по 1957. В 1963 году Кальдерон вернулась в Национальный политехнический институт, где в 1974 году стала профессором. Вместе с мужем, Ежи Жедовским, Кальдерон принимала участие в создании трёхтомного издания Flora fanerogámica del Valle de México. В последние годы жизни Грасьела Кальдерон являлась редактором нескольких научных журналов, в том числе Acta Botanica Mexicana, Chapingo, Boletín del Instituto de Botánica de la Universidad de Guadalajara и Polibotánica.

Род и виды растений, названные в честь Г. Кальдерон

 , «calderonii»

Примечания

Ссылки

Ботаники Мексики
Ботаники по алфавиту
Ботаники XX века
Ботаники XXI века

redpajama_set_name: RedPajamaWikipedia


Our Classic Red Tail Hawk flute in the key of G minor is handcrafted from Aromatic Cedar, which gives the it a soothing, warm, melodic tonal voice. It is inlaid with turquoise cabochons which add to the natural beauty of the flute and enhance the functional art quality of the instrument.
The key of G minor is a very versatile musical key that plays well with many other musical styles such as jazz, blues, contemporary, as well as traditional native style flute playing.
The flute is tuned in the key of "G" minor and can also be played in the relative major diatonic scale in the key of "A#"/"Bb".
This flute is also an excellent accompaniment to on of our bass-tone Contra Bass "G" flute and our high-tone Pocket Flute in the key of "G" flute.
This is a 6-hole flute; we cover the 3rd hole from the top of the flute with a leather cover, which allows it to be a 5-hole flute that can be easily learned to play. This flute is approximately 23 inches (60 cm) long.
Note - flute body designs may vary slightly.


text:
Q: Laravel's query builder JSON selector `field->key` causes syntax error So, I want to query the notifications table in Laravel by comparing a certain ID with the data column. This is how data column looks like:
{
    "Message": "some message",
    "id": 3
}

Now, I need to select all the notifications that have an ID that is equal to 3. Here is how I was trying to do it:
DB::table('notifications')->where('data->id', '3')->get();

But this throws the following error:

SQLSTATE[42000]: Syntax error or access violation: 1064 You have an
  error in your SQL syntax; check the manual that corresponds to your
  MariaDB server version for the right syntax to use near '>'$."id"' =
  ?' at line 1 (SQL: select * from notifications where
  data->'$."id"' = 3)

I am losing my mind here, can anyone help me out?

A: There's nothing wrong with your query. It's your environment.
Problem
Laravel's MySqlGrammar translates the field->key notation in field names (on Laravel side) into field->'$.key'-style extractions (on MySQL side):
/**
 * Wrap the given JSON selector.
 *
 * @param  string  $value
 * @return string
 */
protected function wrapJsonSelector($value)
{
    $path = explode('->', $value);

    $field = $this->wrapValue(array_shift($path));

    $path = collect($path)->map(function ($part) {
        return '"'.$part.'"';
    })->implode('.');

    // Here:
    return sprintf('%s->\'$.%s\'', $field, $path);
}

I just confirmed that MariaDB does not support the -> extraction operator as an alias to the JSON_EXTRACT() function. However, the same query works against a vanilla MySQL 5.7 server.
Assuming this test table:
╔════╤══════════════════╗
║ id │ payload          ║
╟────┼──────────────────╢
║  1 │ {"a": 1, "b": 2} ║
╚════╧══════════════════╝

A query that uses the -> extraction operator:
SELECT payload->"$.b" FROM test;

fails against MariaDB 10.2.8 while it yields a correct 2 against a MySQL 5.7.19 server.
Solutions
The right solution depends on what you're using on production.
Replace MariaDB
If you're using MySQL, replace MariaDB with MySQL in your development env. On a macOS machine managed by homebrew, it'd be as easy as:
brew services stop mysql
brew uninstall mariadb
brew install mysql
brew services start mysql

your data will remain intact.
Rewrite your queries
However, if you're using MariaDB in production, you need to rewrite your queries to use JSON_EXTRACT() function as Elias already mentioned. As you can see you need to be much more verbose with the Laravel API.
The above query would be:
SELECT JSON_EXTRACT(payload, "$.b") FROM test;

A: I made Laravel MariaDB driver for json support. Get it here: ybr-nx/laravel-mariadb

A: The problem is that only MySQL supports the -> operator, then the query will fail on MariaDB, but:
Both MariaDB and MySQL supports the JSON_EXTRACT function, that parses the JSON and get a value from it.
You can solve it by doing a query like this:
SELECT * FROM notifications WHERE JSON_EXTRACT(`notifications.data`, "$.id") = 5

To do it with Laravel's Query Builder, you'll need to use DB::raw method:
DB::table('notifications')->where(DB::raw('JSON_EXTRACT(`notifications.data`, "$.id")'), '=', 3);

Please try it and tell me if go wrong.
Note: The JSON_EXTRACT() is only available on MySQL >= 5.7 or MariaDB >= 10.2.3
Thanks to @Sepehr, I didn't know that the -> was an MySQL Operator, neither that JSON type existed. :-)

A: The issue is with the ->, i think you have a var called $data with some collection in it.
If that so then the correct way is:
DB::table('notifications')->where($data->id, '3')->get();

Or if you have a Model related to notifications table then:
Notification::where($data->id, '3')->get();

If the Model is called Notification (following Eloquent convention)
But if you tried to find all ID's equals 3 then just:
DB::table('notifications')->where('id', '3')->get();


text:
Q: Setting variable as message with discord.py not working My current code:
import random

from discord.ext import commands

class logs(commands.Cog):

def __init__(self, bot):
    self.bot = bot

@commands.command()
async def logs(self, ctx):
    def check(msg):
        return msg.author == ctx.author and msg.channel == ctx.channel
    await ctx.send('Please enter a title.')
    titlemsg = await self.wait_for("message",check=check, timeout=30)
    await ctx.send('Please enter the text.')
    txtmsg = await self.wait_for("message",check=check, timeout=30)
    embed=discord.Embed(title=''+ str(titlemsg) +'', description=''+ str(txtmsg) +'')
    await ctx.send(embed=embed)

What I'm trying to get it to do is this:
User does .logs, bot outputs string "Please enter a title", user inputs a title via messages such as "Title123" for example. It then waits until there's an input, then asks for the text input. Then sends an embed with Title123 as title and Text123 as the text.
However, this is my first time using python or making a bot in discord, and I'm stuck on this error: discord.ext.commands.errors.CommandInvokeError: Command raised an exception: AttributeError: 'logs' object has no attribute 'wait_for'
Edit: This was fixed by changing self.wait_for, to self.bot.wait_for
Now I'm getting this discord.ext.commands.errors.CommandInvokeError: Command raised an exception: NameError: name 'discord' is not defined for the embed = discord.Embed(title=''+ str(titlemsg) +'', description=''+ str(txtmsg) +'') line.


text:
Nedging – wieś w Anglii, w hrabstwie Suffolk, w dystrykcie Babergh, w civil parish Nedging-with-Naughton. W wieś znajduje się kościół. Nedging jest wspomniana w Domesday Book (1086) jako Niedinga. Nedging była parafia w jego własnym imieniu, ale został włączony do Nedging with Naughton w 1935 roku. W 1931 parish liczyła 155 mieszkańców. Znajduje się 17.1 km od Ipswich.

Przypisy

Linki zewnętrzne
 http://www.geograph.org.uk/gridref/TL9948

Wsie w hrabstwie Suffolk
Miejscowości w dystrykcie Babergh
Dawne civil parishes w hrabstwie Suffolk