text:
\begin{document}
\maketitle

\begin{abstract}
Let $G$ be a permutation group acting on a set $V$.  A partition $\pi$ of $V$
is \textsl{distinguishing} if the only element of $G$ that fixes each cell of $\pi$
is the identity.  The \textsl{distinguishing number} of $G$ is the minimum number
of cells in a distinguishing partition.  We prove that if $G$ is a primitive permutation
group and $|V|\ge336$, its distinguishing number is two.
\end{abstract}

\section{Introduction}

Let $G$ be a permutation group acting on a set $V$.  A partition $\pi$ of $V$
is \textsl{distinguishing} if the only element of $G$ that fixes each cell of $\pi$
is the identity.  The \textsl{distinguishing number} of $G$ is the minimum number
of cells in a distinguishing partition.  The study of this parameter was introduced by
Albertson and Collins, see \cite{alco}.
For other recent work on this problem see, for example, the work of M. Chan \cite{chan-max}.

The result of this paper is that if $G$ is primitive and on $V$ and $|V|\ge336$,
then its distinguishing number is two.

We note that Theorem 1 from Seress \cite{seress} yields the same conclusion with
an upper bound of $32$.  The best we can say for this paper is that our methods are
more elementary, and may have some independent interest.

\section{A Lower Bound}

Let $G$ be a permutation group.  The \textsl{degree} of $G$ is just the size of $V$.
The \textsl{minimum degree} of $G$ is the minimum number of points moved by
a non-identity element of $G$.

We need the following lower bound on the order of a permutation group in terms of its
minimum degree.  The argument is based on Cameron \cite{pjc-power}.

\begin{lemma}
\label{lem:pjc}
If $G$ is a permutation group with distinguishing number at least three and the
minimum degree of $G$ is $\de$, then $|G| \ge1+2^{\de/2}$.
\end{lemma}

\proof
Assume $v=|V|$.
We consider the action of $G$ as a group of permutations of the power set of $V$.
We recall that the number of orbits of $G$ on subsets of $V$ is equal to the average number
of subsets fixed by an element of $G$.  If $g\in G$ and $g$ has exactly $k$ orbits,
then it fixes exactly $2^{k}$ subsets.  Since the minimum degree of $G$ is $\de$, we see
that $g$ has at most $v-\de$ orbits of length one, and at most $\de/2$ orbits of length
greater than one.  Therefore $g$ fixes at most $2^{v-\frac{\de}2}$ subsets of $V$,
and the number of orbits of $G$ on subsets is at most
\[
\frac1{|G|}(2^{v}+(|G|-1)2^{v-\frac{\de}2}).
\]

If the distinguishing number of $G$ is greater than two, then any non-empty proper
subset of $V$ must be fixed by at least one non-identity element of $G$, and so the
orbit of any subset has size at most $|G|/2$.  Consequently the number of orbits of
$G$ is at least
\[
\frac{2^{v}}{|G|/2} =\frac{2^{v+1}}{|G|}.
\]
Combining these two bounds and rearranging yields the lemma.\qed

\section{Primitive Groups}

The \textsl{Johnson graph} $J(m,\ell)$ has the $\ell$-subsets of $\{1,\ldots,m\}$
as its vertices, and two $\ell$-subsets are adjacent if their intersection has size $\ell-1$.
The graphs $J(m,\ell)$ and $J(m,m-\ell)$ are isomorphic, and so we will
assume that $2\ell\le m$.  The complete graphs are Johnson graphs (with $\ell=1$).

A subset $S$ of $G$ is a \textsl{block of imprimitivity} for $G$ if for each element
$g$ of $G$,
either $S^{g}=S$ or $S\cap S^{g}=\emptyset$.  Trivial examples of blocks are $\emptyset$,
any singleton from $V$ or $V$ itself.  If $G$ is transitive and there are no other blocks
we say
that $G$ is \textsl{primitive}, otherwise it is \textsl{imprimitive}.  The size of a
non-empty block must divide $|V|$, whence we see that if $G$ is transitive and its degree
is a prime number, then $G$ is primitive.

The \textsl{Cartesian product} $X\cprod Y$ of graphs $X$ and $Y$ has vertex
set $V(X)\times V(Y)$, and vertices $(x_1,y_1)$, $(x_2,y_2)$ in the product
are adjacent if $x_1=y_1$ and $x_2$ is adjacent to $y_2$, or if
$x_2=y_2$ and $x_1$ is adjacent to $y_1$.  We use $X^{\cprod\,n}$
to denote the $n$-th Cartesian power of $X$, the Cartesian product of $n$ copies
of $X$.

For our purposes the main result (Theorem~1) of Guralnick and Magaard \cite{guma} can be summarized as follows.

\begin{theorem}
\label{thm:gurmag}
Let $G$ be a primitive permutation group of degree $v$.  If the minimal degree of $G$
is at most $v/2$, then one the following holds:
\begin{enumerate}[(a)]
\item
$G$ is affine over $GF(2)$ and its minimal degree is $v/2$.
\item
$G$ is a transitive subgroup of the automorphism group of the Cartesian power
of a Johnson graph.
\end{enumerate}
\end{theorem}

The following result comes from Mar\'oti \cite[Theorem~1.1]{maroti}.

\begin{theorem}
\label{thm:maroti}
Let $G$ be a primitive permutation group of degree $v$.  Then one of the following holds:
\begin{enumerate}[(a)]
\item
$G$ is a transitive subgroup of the automorphism group of the Cartesian power of a
Johnson graph.
\item
$v\in\{11,12,23,24\}$ and $G$ is one of the Mathieu groups.
\item
$|G| < v^{1+\lceil\log_{2}v\rceil}$.\qed
\end{enumerate}
\end{theorem}

\section{Distinguishing Primitive Groups}
\label{sec:DPGs}

Imrich and Klavzar prove the following result in \cite[Theorem~1.1]{imkl}.

\begin{theorem}
	\label{thm:imkl}
The distinguishing  number of a Cartesian power of connected graph is two,
except in the cases $K_{2}^{\cprod\,2}$, $K_{3}^{\cprod\,2}$ and $K_{2}^{\cprod\,3}$
(when it is three).
\end{theorem}

If the graph $X$ has $v$ vertices then
\[
\aut{X^{\cprod\,n}} \le\aut{K_{v}^{\cprod\,n}}
\]
and therefore the distinguishing number of $X^{\cprod\,n}$ is bounded by that of
$K_{v}^{\cprod\,n}$.

\begin{lemma}
	\label{lem:jvk}
If $m\ge2\ell$ then the distinguishing number of $J(m,\ell)$ is two unless $\ell=1$
or $\ell=2$ and $m=5$.
\end{lemma}

\proof
Suppose $\ell=2$.  Let $T_{m}$ be the graph constructed from the path on $n$ vertices
by adding an edge joining the second and fourth vertices.  If $m>5$, then $T_{m}$
is asymmetric and its line graph is asymmetric.  It follows that the partition of
the vertices of $L(K_{m})$ into the vertices which represent edges of $T_{m}$
and the vertices which do not is a distinguishing partition for $\aut{J(m,2)}$ when $m\ge6$.

Now suppose $\ell\ge3$.  Let $HP_{m}$ denote the hypergraph with vertex set
$\{1,\ldots,2m\}$ and edges consisting of the $m+1$ triples of consecutive integers
from the vertex set.  Add the edge consisting of the integers $\{2,\ldots,m,m+2\}$.
This modified hyperpath is asymmetric, and it follows that $\aut{J(m,\ell)}$ has
distinguishing number two.\qed

\begin{theorem}
Suppose $G$ is a primitive permutation group of degree $v$ that does not contain $\alt{v}$.
If $v\ge 336$, then the distinguishing number of $G$ is two.
\end{theorem}

\proof
Suppose that the minimum degree of $G$ is at least $v/2$.  Then combining \lref{pjc}
and \tref{maroti} we have
\[
2^{v/4} <v^{1+\lceil\log_{2}v\rceil}.
\]
Taking logarithms of each side of this yields
\[
v/4 <(1+\lceil\log_{2}v\rceil) \log_{2}v.
\]
This inequality fails if $v\ge 336$, and therefore the theorem holds when
the minimum degree of $X$ is at least $v/2$.

If the minimum degree of $G$ is less than $v/2$, then by \tref{gurmag}
$G$ must be a subgroup of the automorphism group of the Cartesian power of a
Johnson graph.  By \tref{imkl}, the distinguishing number of the $n$-th Cartesian power
of $J(m,\ell)$ is two unless $J(m,\ell)$ is $K_{2}$ or $K_{3}$ and $n\le3$,
or unless $n=1$.  Using \lref{jvk} we see that in all these cases $v<336$.
Since the Mathieu groups have degree at most 24, these can also be ignored.\qed

\section{Comments}

Both the theorem of Guralnick and Magaard and the theorem of Ma\'roti depend on the
classification of finite simple groups.  We do have the following important bound due to
Babai which can be used in place of the theorem of Guralnick and Magaard, and whose
proof is entirely combinatorial. (For details, see \cite[Theorem 5.3B]{dimo}.)

\begin{theorem}
\label{thm:babai}
If $G$ is a primitive permutation group of degree $v$ that does not contain $\alt{v}$,
then its minimum degree is greater than $\sqrt{v}/2$.
\end{theorem}

However using this yields a lower bound of order $2^{20}$, much larger than the bound of
336 that we computed.


TITLE: When there exists function $f$ such that for given $g$ we have $f'=g$?
QUESTION [2 upvotes]: I am looking for a theorem that states when function $g: \mathbb R \mapsto \mathbb R$ is a derrivative, i.e. there exists $f$ such that $f'=g$.
What about if we just need this condition almost everywhere?
I have looked at Lebesgue differentiation theorem but I either it is not the way to go or i can't extract it from there. I think I have seen such a theorem few years ago, but I can't recall it.
Maybe I'll provide some background. I am considering the following integral:
$$\int_{\mathbb R} \frac{ - g ( \bar x - y)}{|y|^{\alpha +1}} \phi (y/R) d y$$
Where $g$ is bounded function that is twice-differentiable everywhere except zero. At zero there may be some strange behaviour, not sure how strange. I assume  $\alpha \in (0,2)$ (it has to do with fractional laplacian) and that $\varphi$ is smooth cut-off function thtat is zero at a ball around orgin. I want to write the following:
$$\int_{\mathbb R} \frac{ - g ( \bar x - y)}{|y|^{\alpha +1}} \phi (y/R) d y =  \int_{\mathbb R} \partial_k f(\bar x -y) \frac{\phi (y/R)}{|y|^{\alpha+1}} d y = \int_{\mathbb R}  f(\bar x -y) \partial_k \left[ \frac{\phi (y/R) }{|y|^{\alpha+1}} \right] d y$$
If I explicitly assume that $g$ is a derrivative of $f$ it works. I think I can also works when $g$ is Schwartz function. Am I right? But I want to obtain the most general result, hence my question.

REPLY [2 votes]: Most likely the problem that you have posed will not help at all in the situation you describe, but it is an interesting problem with interesting answers and so, if you approach this as a side topic that offers you a holiday from your real work you might be entertained.
First the problem of when a function is a derivative (pointwise everywhere) is an old problem with no completely satisfying answers.  There are sufficient conditions (continuity of course, bounded and approximately continuous) and there are necessary conditions (Baire one with the Darboux  property, the Denjoy-Clarkson property, Zahorski's properties).  What characterizations exist won't thrill you (even my own characterization, which I will spare you).
The second problem of when a function is an a.e. derivative has quite an interesting answer:

Theorem (Lusin) A necessary and sufficient condition that a function $f$ should be the
  a.e. derivative of a continuous function is that it be measurable and
  a.e. finite.

For all of this and much more you should consult the definitive reference

Andrew M. Bruckner, Differentiation of Real Functions,  CRM Monograph
  Series,  American Mathematical Society, Centre de Recherches
  Mathematiques (1994) ISBN-10: 0821869906, ISBN-13: 978-0821869901

A supplementary article in the Real Analysis Exchange by Andy will also help:

Bruckner, Andrew M.  The problem of characterizing derivatives
  revisited.  Real Anal. Exchange  21  (1995/96),  no. 1, 112--133.


TITLE: (Baby maths) Sum of three positive integers = odd or even? + more info
QUESTION [2 upvotes]: I'm struggling to understand this basic question:
$x$, $y$, and $z$ are positive integers. Is $x+y+z$ even?
Supposedly you can derive the answer, from these two pieces of information separately:
$x-y-z$ is odd,
and;
$xyz$ is odd.
How do you approach this problem? / What do I need to learn?

REPLY [1 votes]: Thanks for all the answers!
I learnt that I should have tried representing the variables differently, as in $2a + 1$ for odd numbers and $2a$ for even numbers.
I guess I have to become more familiar with all those "even + odd = odd" etc.
Also, I think learner's idea was quite clever because of how quick it was.
Some context if you're interested: This question is from an old test and you have roughly two minutes per question, but you don't have to give an exact answer, you just have to deduce whether the information given is insufficient or the least information you can solve it with.


TITLE: Extension of Dirichlet's Arithmetic Progression Theorem
QUESTION [2 upvotes]: Dirichlet's Arithmetic Progression Theorem states that:

Given $a, b\in\mathbb{Z^+}$ with $(a,b)=1$, then $a+kb$ is prime for an infinite number of $k\in\mathbb{Z^+}.$

For any given $a$ and $b$ let $K_{a,b}=\{k\mid a+kb \text{ is prime}\}$.
Also consider another Dirichlet-Valid AP $c+jd$. Restrict $j$ to $j_k\in K_{a,b}$.

Is $c+j_k d$ prime an infinite number of times?

REPLY [7 votes]: Consider the arithmetic progressions $2+3\mathbb N$ and $1+5\mathbb N$ and observe that if $2+3k$ is prime, then $k$ is odd. On the other hand, if $1+5k$ is prime, then $k$ should be even. So, for any $k\in\mathbb N$ the numbers $2+3k$ and $1+5k$ cannot be simultaneously prime.
The same contradiction could be attained on the arithmetic progresions $2+1\cdot \mathbb N$ and $3+1\cdot\mathbb N$.


TITLE: Difference between "necessary" and "necessary but not sufficient"?
QUESTION [4 upvotes]: This is from Discrete Mathematics and Its Applications:

Let $p, q,$ and $r$ be the propositions:
$\quad p:$ Grizzly bears have been seen in the area.
  $\quad q:$ Hiking is safe on the trail.
  $\quad r:$ Berries are ripe along the trail.
Write these propositions using $p,q,$ and $r$ and logical connectives (including negation):

For hiking on the trail to be safe, it is necessary but not sufficient that berries not be ripe along the trail and for grizzly bears not to have been seen in the area.

I read up on necessary and sufficient from here What is the difference between necessary and sufficient conditions?

If $p \to q$ ($p$ implies $q$), then $p$ is a sufficient condition for $q$.
If $\neg p \to \neg q$ (not $p$ implies not $q$), then $p$ is a necessary condition for $q$.

From these two conditions how would you apply necessary but not sufficient?
The way I expressed this is: $$ (\neg r \land \neg p) \to q$$
I mainly got to this because "if $p$ then $q$" is the same as $q$ is necessary for $p$. $q$ in this case would be two conditions — berries not be ripe along the trails and grizzly bears not to have been seen in the area".
How would the necessary but not sufficient clause affect the answer though? Would it make a difference?

REPLY [3 votes]: If we analyse what you wrote, namely $(\neg r \land \neg p) \to q$, then it says "$q$ is necessary for $r$ and $p$ to be false". Equivalently, "$r$ and $p$ being false is sufficient for $q$". That is:

It is necessary that hiking be safe if the berries are not ripe and there are no grizzlies.
It is sufficient that the berries are not ripe and there are no grizzlies for hiking to be safe.

This is not one of the clauses of the given statement. In fact, the second reading is precisely the negation of the second half of the given statement, viz:

For hiking to be safe, it is not sufficient that the berries are not ripe and there are no grizzlies.

Hence, half of our solution is:
$$\neg((\neg r \land \neg p)\to q)$$

For the other part, i.e.:

For hiking to be safe, it is necessary that the berries are not ripe and there are no grizzlies.

let us take a more structured approach. First, replace the relevant parts of the sentence by $p,q,r$:

For $q$, it is necessary that $\neg r$ and $\neg p$.

Using your definition of a necessary statement, this can be symbolised as:
$$\neg(\neg r \land \neg p) \to \neg q$$
Hence in the end, the (or rather, a) symbolic representation of the full sentence becomes:
$$\neg(\neg r \land \neg p) \to \neg q \land \neg((\neg r \land \neg p)\to q)$$
As you rightly remark, $p \to q$ is also a correct interpretation for "$q$ is necessary for $p$", and we could just as well have used that instead of $\neg q \to \neg p$.

The bottom line is that once we can parse a given linguistic construction (case in point, the phrases necessary and sufficient) in symbolism, then we can also combine these with other known logical phrases, such as "not" and "and". Therefore, we can derive a logical interpretation of the phrase "necessary but not sufficient" (since in this context, "but" is to be read as "yet also").


TITLE: What does a flat disk in $\mathbb H^3$ look like?
QUESTION [0 upvotes]: I am trying to understand hyperbolic space better, in particular $\mathbb H^3$ (assume curvature equal to $-1$). If we use the Beltrami–Klein model, the space can be represented by the open unit ball in $\mathbb R^3$ and geodesics appear as straight lines. An open disk like
$$ \{(x,y,0)\in\mathbb R^3: x^2+y^2 < \frac 12\}$$
appears flat, but actually has (I assume) curvature equal to $-1$.
What does a surface look like that is flat in $\mathbb H^3$ ? (e.g. a disk that is flat with respect to the hyperbolic metric).

REPLY [2 votes]: Even in $\mathbb R^3$, surfaces of curvature $0$ can be a little hard to describe in complete generality (I don't like the word "flat" for subsurfaces, because it clashes with other uses of "flatness" in Riemannian geometry). For example developable ruled surfaces in $\mathbb R^3$ have curvature $0$.
So I'll just give some important examples of curvature $0$ surfaces in $\mathbb H^3$, namely the horospheres. I don't know how to describe them in the Beltrami-Klein model, but they are easy to describe in two other models:

Horospheres in the Poincare ball model $\mathbb B^3$ are Euclidean spheres inside the Poincare ball and tangent, at a single point, to the boundary sphere at infinity $\mathbb S^2_\infty$.
Horospheres in the upper half space model $\{(x,y,z) \mid z>0\}$ are of two types: Euclidean spheres in upper half space tangent to the plane $z=0$; and horizontal Euclidean planes of the form $\{(x,y,z_0)\}$ for some constant $z_0>0$.

It's not hard to check that the upper half plane metric restricted to a horizontal Euclidean plane is actually isometric to the ordinary Euclidean plane, and so has Gaussian curvature $0$. For the other types of horospheres, one simple way to verify that their Gaussian curvature equals zero is to produce an isometry of the whole model which takes the given plane to a horizontal Euclidean plane in the upper half space model.


TITLE: Detailed balance and Semi detailed balance in a Markov chain
QUESTION [2 upvotes]: Consider the following transition matrix
$$
   T=
  \left[ {\begin{array}{cccc}
   \frac{1}{3}  & \frac{1}{4} & \frac{1}{5} & \frac{1}{6}\\
   \frac{1}{3} & \frac{1}{4} & \frac{1}{5} &  \frac{2}{6}\\
   \frac{1}{3} & \frac{1}{4} & \frac{3}{5}&  0\\
   0 & \frac{1}{4} & 0& \frac{3}{6}\\
  \end{array} } \right]
$$
of a Markov chain process
$$P_{t+1}=TP_t$$
The matrix $T$ is not symmetric, therefore I conclude that there is no detailed balance, and there is no equilibrium (the process is not reversible).
However, $T_{34}=T_{43}$.
Does this mean that there is semi-detailed balance?
what does it say about the equilibrium of the system?

REPLY [2 votes]: Symmetry of the transition matrix is not a necessary condition for reversibility; consider $$P = \pmatrix{0&1/2&1/2\\1/4&1/2&1/4\\1/4&1/4&1/2}. $$
Since $P$ is irreducible and aperiodic, it has a unique stationary distribution $\pi$ satisfying $\pi=\pi P$. Since $\sum_i \pi_i = 1$ we have the system of equations
\begin{align}
-\pi_1 + 1/4\pi_2 + 1/4\pi_3 &= 0\\
1/2\pi_1 -1/2\pi_2 +1/4\pi_3 &= 0\\
\pi_1 + \pi_2 + \pi_3 &= 1,
\end{align}
which yields $\pi_1=1/5$, $\pi_2=2/5$, $\pi_3=2/5$. Let $P^\star_{ij}=(\pi_i/\pi_j)P_{ij},$ then
$$
P^\star = \pmatrix{0&1/4&1/4\\1/2&1/2&1/4\\1/2&1/4&1/2} = P^T,
$$
so the detailed balance equations are satisfied and $P$ is reversible.
In general, $P$ is reversible if and only if $\Pi * P$ is symmetric, where $\Pi$ is the square matrix whose rows are identically $\pi$ and $*$ denotes component-wise multiplication (see this thesis for more details).
In your example, the stationary distribution for $S:=T^T$ is $$\pi = \pmatrix{33/137&36/137&50/137&18/137}. $$
(The transpose of $T$ is a row-stochastic matrix.)
We compute
$$
\Pi^T * S = \left(
\begin{array}{cccc}
 \frac{33}{137} & \frac{33}{137} & \frac{33}{137} & \frac{33}{137} \\
 \frac{36}{137} & \frac{36}{137} & \frac{36}{137} & \frac{36}{137} \\
 \frac{50}{137} & \frac{50}{137} & \frac{50}{137} & \frac{50}{137} \\
 \frac{18}{137} & \frac{18}{137} & \frac{18}{137} & \frac{18}{137} \\
\end{array}
\right) * \left(
\begin{array}{cccc}
 \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 \\
 \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \\
 \frac{1}{5} & \frac{1}{5} & \frac{3}{5} & 0 \\
 \frac{1}{6} & \frac{1}{3} & 0 & \frac{1}{2} \\
\end{array}
\right) = \left(
\begin{array}{cccc}
 \frac{11}{137} & \frac{11}{137} & \frac{11}{137} & 0 \\
 \frac{9}{137} & \frac{9}{137} & \frac{9}{137} & \frac{9}{137} \\
 \frac{10}{137} & \frac{10}{137} & \frac{30}{137} & 0 \\
 \frac{3}{137} & \frac{6}{137} & 0 & \frac{9}{137} \\
\end{array}
\right),
$$
which is not symmetric, and hence the Markov chain is not reversible.
For Markov chains, the semi-detailed balance condition is precisely the global balance condition $\pi = \pi P$.


TITLE: Prove a finite element space is not conforming
QUESTION [1 upvotes]: Let $\tau=[-1,1]^2$, consider the finite element $\left(\tau, Q_{1}, \Sigma\right)$, $Q=span\{1,x,y,x^2−y^2 \},$ $\Sigma=\{w(−1,0),w(1,0),w(0,−1),w(0,1)\}$.
Show that the unisolvent element leads to a finite element space,
which is not $H^1-$conforming.
I have proved that this is a unisolvent finite element. What should I need to do to prove it leads to a finite element space which is not $H^1-$ conforming.
I know $H^1$-conforming means something like $V_h\subset H^1$.

REPLY [1 votes]: Denote $V_h$ the finite element space. Recall that if $V_h\subset H^1(\Omega)$ and $Q$ consists of continuous functions, then $V_h\subset C(\Omega)$.
Thus, now we only need to show that $V_h$ is not contained in $C(\Omega)$. It's not difficult to find a function in $V_h$ but not continuous.
For example, let us consider $\Omega=\Omega_1\cup \Omega_2$ where $\Omega_1=[-1,0]\times [-1,1]$ and $\Omega_2=[0,1]\times [-1,1]$. Let $v_h\in V_h$ be defined as $v_h|_{\Omega_1}=x$ and $v_h|_{\Omega_2}=x^2-y^2$. Then $v_h|_{\Omega_1}(0,0)=0=v_h|_{\Omega_2}(0,0)$ whcih imples that $v_h\in V_h$. Note that $v_h|_{\Omega_1}(0,-1)=0\neq -1=v_h|_{\Omega_2}(0,-1)$. Thus $v_h\notin C(\Omega)$.


TITLE: Determining the projective closure of a variety.
QUESTION [0 upvotes]: Consider the twisted curve in $\Bbb{C}^3$ denoted by the ideal $\langle x^2-y, z-xy\rangle$
I did it by homogenizing the generators: I got $\langle x^2-yw, zw-xy\rangle$.
The projective closure seems to contain $x=w=0$. Which means the line parametrized by $z,w$ is here in the projective closure. However, my book says that this line does not form part of the closure. Why is that?

REPLY [1 votes]: To homogenize an ideal, you must homogenize a Gröbner basis for it. In M2:
R=QQ[x,y,z]
I=ideal(y-x^2,z-x*y)
gens gb I -- | y2-xz xy-z x2-y |

We see that this adds the generator $y^2-xz$. See this answer.


TITLE: Explanation for an unexpected rainbow
QUESTION [4 upvotes]: Yesterday, I observed an unexpected rainbow in the sky. There was no forecast for rain, neither was it raining anywhere nearby. I have been trying to find an explanation but don't seem to find any. Can someone please explain what this rainbow is?
Note:the colours were way more vivid as compare to the picture I have taken

REPLY [6 votes]: These are tropospheric Iridescent Clouds
According to AtmosphericOptics:

When parts of clouds are thin and have similar size droplets, diffraction can make them shine with colours like a corona. In fact, the colours are essentially corona fragments. The effect is called cloud iridescence or irisation...
The usually delicate colours can be in almost random patches or bands at cloud edges. They are only organised into coronal rings when the droplet size is uniform right across the cloud. The bands and colours change or come and go as the cloud evolves...Iridescence is seen mostly when part of a cloud is forming because then all the droplets have a similar history and consequently have a similar size.

I've saturated the image so the interesting part can be appreciated

And here you have a very similar observation I quickly found by google image search:


TITLE: How does calculating this method of calculating Log on simple Calculator Work?
QUESTION [3 upvotes]: In our finance course , we are allowed to use Simple Calculators that do not have a log function in them. We are provided with Log Tables but we are taught this trick to solve log without using log tables. I am not sure how this works.
Say for example I want to compute the Future value of Rs 100 invested at the rate of 12.31% p.a. compounded continuously.
So the formula for continuous compounding is
    $$A= P.e^{rn} $$
So as a work arround we use the following method:
We solve for $e^{rn}$
Let $x = e^{rn}$
Taking Log on both sides
$ Log x = r.n. Log . e $
r=0.1231 and n= 1 year
$ Log x = 0.1231* 1 * Log 2.71828$
$Log x = 0.1231 * Log 2.71828$
Now on Calculator we type 2.71828 and then we press Square Root symbol 15 times, we get 1.00003051801 {upto 12 digits}
We subtract 1 from it we get  0.00003051801
We divide 0.00003051801 by 0.000070274(This is the constant that we multiply every time to find log)
$ {(1.00003051801-1)\over 0.000070274}  =  0.43427170788 $
Now we substitute this figure in above Equation
$ Log x = 0.1231 * Log 2.71828$
$ Log x = 0.1231 * 0.43427170788$
$ Log x = 0.05345884724 $
Now we take antilog on both sides
$ Antilog (log x)= Antilog (0.05345884724)$
Similary to how we found $Log 2.71828$ we perform the whole series of steps backwards

We first muliply the number with 0.000070274
Then we add back one
Then we sqaure it 15 times

${{{{{{{{{{{{{{{x^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}$
and voila, we get the answer
$x = 1.13099878536$
We substitue x in original Equation
$ A = P.e^{rn} $
$ A = P.x$
$ A = 100 * 1.13099878536$
$ A = 113.099878536$
Even if we ignore that step of multiplying or dividing by "0.000070274" and do sqaure root or sqaure for 12 times rather than 15 times, we still get the same answer.
Can you please explain me how this is working?

REPLY [2 votes]: You are using the fact that $\lim_{n \to \infty}\left(1+\frac an\right)^n=e^a$ and the fact that $\ln(1+y) \approx y$ for $y \ll 1$.
When you type in your approximation for $e$ and take the square root $15$ times, you are taking the $2^{15th}$ root of $e$, which is about $1.000030518$ as you say.  Subtracting $1$ gives you $\ln (e^{2^{-15}})$, which is $2^{-{15}}=\frac 1{32768}\approx 0.000305176.$
The division by $0.00070274$ is immediately reversed by the multiplication, so you may skip both of these steps.
 Multiplying by $r$ and adding $1$ gives you $1+\frac r{32768}$ and squaring $15$ times gives $\left(1+\frac r{32768}\right)^{32768}\approx e^r$
It would be far easier to just divide $r$ by $32768$, add $1$, and square $15$ times.  You could also divide by $2^{12}=4096$, add $1$ and square $12$ times. The first is more accurate because you are using a larger $n$.  It is less accurate because you lose one more digit of $1+\frac rn$ to the fixed precision of your calculator.  Particularly if $r$ is small, the second will dominate and smaller $n$ will be better.


TITLE: Combinatorics in ML: Counting Points with co-ordinates from among a set.
QUESTION [2 upvotes]: I'm trying to re-prove a theorem in the book Understanding Machine Learning: From Theory to Algorithms  by Shalev-Schwartz et. al to aid my understanding of the material. The proof in the book derives a crude bound, and I'd like to derive a tighter one by counting the relevant quantities.
General Version of Problem:
Given a finite domain $X$ of size $M$, I'd like to consider points in the $n$-fold product of X; these points have n-coordinates, each an element of X. There are a total of $M^{n}$ such points.
Now, I'd like to partition these points into $n$ sets. The first set, $S_{1}$, will contain all points that contain exactly 1 unique element of $X$ among its co-ordinates. The second set, $S_{2}$, will contain all points that contain exactly two unique elements, and so on.
I would like to determine $|S_{i}|$ for $1\leq i \leq n$.
$|S_{1}| = M$, since, there are exactly $M$ points with one unique member of $X$ among their co-ordinates. Similarly, $|S_{n}| = \frac{M!}{(M-n)!}$.
Using the inclusion-exclusion formula, I've been able to derive a long and messy formula for $|S_{i}|$. I'm wondering if anyone knows a simple way to do this, or a standard reference where this problem is solved.
Here's an example:
Let $M = 6$, and the members of $X$ are just labelled with integers. Also, suppose $n=3$.
The point (1,1,1)) would belong to $S_{1}$. The point (6,2,2) would belong to $S_{2}$. The point (1,5,3) would belong to $S_{3}$.
This case can be solved explicitly, since $|S_{1}|$ = 6, $|S_{3}|$ = 120, and
$|S_{1}|+|S_{2}|+|S_{3}|$ = 216
I'm not very good at counting, and hope that there is a simple way to do this. Thank you!

REPLY [1 votes]: An $n$-tuple can be interpreted as a function $f : \{1,\dots,n\} \to X$. Namely, $(f(1),f(2),\dots,f(n))$. What you are asking is to count all such functions whose image has $i$ elements. The image is a subset of $X$ and there are $\binom{M}{i}$ subsets with $i$ elements. Then you want a surjective map onto this set.
Surjective functions $g : A \to B$ where $|A| = n$ and $|B| = k$ are counted by Stirling numbers. (See Twelvefold way for details.) Namely there are
$$ k!\left\{ \begin{array}{c} n \\ k \end{array} \right\} = \sum_{j=0}^k (-1)^{k - j} \binom{k}{j}j^n $$
surjective functions $A \to B$. You can also obtain this from inclusion-exclusion.
Therefore
$$ |S_i| = i!\binom{M}{i} \left\{ \begin{array}{c} n \\ i \end{array} \right\} = \binom{M}{i} \sum_{j = 0}^i (-1)^{i - j} \binom{i}{j}j^n. $$


TITLE: Example of a ring such that $R^2\simeq R^3$, but $R\not\simeq R^2$ (as $R$-modules)
QUESTION [13 upvotes]: The usual example of unitary ring without the IBN property is the ring of column finite matrices, and in this case we have $R\simeq R^2$ as (left) $R$-modules. (See also here.) In particular, we have $R^2\simeq R^3$.

I wonder if there is an example of unitary ring (without the IBN property) such that $R^2\simeq R^3$, but $R\not\simeq R^2$ (as $R$-modules).

REPLY [15 votes]: The only other examples of non IBN rings that I am aware of are built using Leavitt path algebras, and they do exactly what you want. It is possible to specify positive integers $n<m$ such that $R^i\ncong R^j$ for $(i,j)$ less than $(n,m)$ in the lexicographic order, but $R^n\cong R^m$ as modules. (The $(m,n)$ pair is called the 'module type' in the Leavitt article.)
Unfortunately, it is a bit unwieldy to reproduce the entire construction here. Here is one paper discussing the construction.
I think the other Abrams reference in the wiki article on IBN must be useful as well (I was the one who put the citation up, but I was a long time ago.)
Here is another article that addresses it much more directly. It cites Leavitt's original result which is what I'm describing (theorem 1.3.3)
For everyone's enjoyment, here's the citation for Leavitt's article:

W. G. Leavitt. The module type of a ring. Trans. Amer. Math. Soc. 103: 113–130, 1962.


TITLE: Does $P(H|E)>P(H)$ imply $P(H|\neg E)<P(H)$?
QUESTION [2 upvotes]: In the book Rationality: From AI to Zombies, in the chapter Absence of Evidence Is Evidence of Absence, Eliezer Yudkowsky (the author) wrote the following:

If $E$ is a binary event and $P(H|E) > P(H)$, i.e., seeing $E$ increases the probability of $H$, then $P(H|¬E) < P(H)$, i.e., failure to observe $E$ decreases the probability of $H$. The probability $P(H)$ is a weighted mix of $P(H|E)$ and $P(H|¬E)$, and necessarily lies between the two.

(By "$\neg E$" Eliezer means "$E^{\complement}$", and by "failure to observe $E$" he means "observing that $E$ is false". Similar terminology was discussed in another question (whose subject is very similar to the subject of this chapter))

According to the law of total probability, $P(H)=P(H|E)P(E)+P(H|\neg E)P(\neg E)$, so indeed as Eliezer (and wikipedia) noted, $P(H)$ is a weighted average of $P(H|E)$ and $P(H|¬E)$. Along with the assumption that $P(H|E) > P(H)$, we can deduce that $P(H|¬E)\le P(H)$.
So I see why $P(H|E)>P(H)$ implies $P(H|\neg E)\le P(H)$.
But does $P(H|E)>P(H)$ imply $P(H|\neg E)<P(H)$ ?
Is the following a valid counterexample, or am I missing something?
Let's say a real number in $[0,1]$ is chosen uniformly at random, let $H$ be the event that $1$ was chosen, and let $E$ be the event that an integer was chosen.
If I understand correctly:

$E$ is a binary event.
$P(H|E)=0.5$(here is a question in which a similar calculation of a conditional probability is discussed)
$P(H)=0=P(H|\neg E)$
($P(H)$ is indeed a weighted average of $P(H|E)$ and $P(H|¬E)$, as $P(E)=0$)
Thus, $P(H|E)>P(H)$ is true, but $P(H|\neg E)<P(H)$ is false.

REPLY [1 votes]: Characteristic for conditional probability $P(A\mid B)$ is the equality:$$P(A\mid B)P(B)=P(A\cap B)$$

Applying that here (i.e. multiplying both sides with $P(E)$), statement $P(H\mid E)>P(H)$ can be converted into:$$P(H\cap E)>P(H)P(E)\tag1$$Similarly statement $P(H\mid E^{\complement})<P(H)$ can be converted into:$$P(H\cap E^{\complement})<P(H)P(E^{\complement})$$or equivalently: $$P(H)-P(H\cap E)<P(H)-P(H)P(E)\tag2$$
It is obvious now that $(1)$ and $(2)$ are equivalent statements.

Your counterexample makes use of an event $E$ that satisfies $P(E)=0$  making it tricky to speak of conditionals like $P(H\mid E)$.
Characteristic for $P(H\mid E)$ is actually the equation $$P(E)P(H\mid E)=P(H\cap E)$$
and nothing more than that.
But in your case $P(E)=P(H\cap E)=0$ so that for any value for $P(H\mid E)$ this will be satisfied.
So there is no ground for setting $P(H\mid E)=0.5$ as you did.


TITLE: Question about a new pseudo-random number generator
QUESTION [3 upvotes]: While investigating non-periodic RNG's (random number generators) for irrational numbers, I came up with a version that actually produces pseudo-random words consisting of $N$ bits, where $N$ is typically a large prime number. Here I explain my RNG. My question is whether it suffers from the same problems as Xorshift RNG's or some other problems. As a starter, the version corresponding to $N=32$ is terrible: its period is $24$. But $N=31$ yields a good generator with a long period and nice statistical properties. In its basic version, it is defined as follows.
Start with a seed $S$. The first random word $B_0$ is $S$. In my case, I picked up the first $N$ binary digits of $\sqrt{2}/2$ for the seed.The $k$-th bit of $B_n$ is denoted as $B_n(k)$. Then $B_{n+1}$ is obtained recursively as follows.

Shifting step: Create the word $C_{n}$ by shifting the bits of $B_{n}$
by $L$ positions as follows: the $k$-th bit of $C_n$ is equal to  $C_n(k)=B_n(\bmod(k+L,N))$ for $k=0,\cdots, N-1$.
Scrambling step: $B_{n+1}(k)=\bmod(B_{n}(N-k-1)+C_{n}(k),2)$ for $k=0,\cdots, N-1$. In other words, $B_{n+1}(k)=\mbox{ XOR}(B_{n}(N-k-1),C_{n}(k))$. Thus the analogy with Xorshift generators.

$L=2$ seems to work best in most cases. For $L=2$ and $N=7, 11$ or $17$, the period is $2^{N-3}-1$. More generally, if $N$ is prime, the period is of the order $2^N$. Of course, there is no way the period could be higher than $2^N$. So prime values of $N$ produce the best generators, though this might not be true for all primes.
Also, the real number $X_n\in [0,1]$ is defined as follows:
$$X_n=\sum_{k=1}^{N} \frac{B_{n}(k-1)}{2^k}.$$
There is a one-to-one mapping between $B_n$ and $X_n$. I studied the patterns in the distribution of successive values of $X_n$ and haven't found any. For instance, unlike other RNG's (see here and follow-up discussion here), the triplets $(X_n,X_{n+1},X_{n+2})$ do not appear to lie in a small number of parallel planes. Successive values of $X_n$ are asymptotically un-correlated. For modern tests (George Marsaglia, 2020) to assess the quality of a RNG, see here and here.
The underlying idea in the design of my generator is this: take a seed consisting of a large number of random bits, such as a the first $N$ binary digits of a normal number in base $2$. Then if you reverse these bits (the binary digits), the new number is a sequence of bits just as random as the previous one, and uncorrelated to the previous number.
Possible improvements
Consider a $q$-order recursion $B_{n}=f(B_{n-1},\cdots,B_{n-q})$ instead of a first-order one as here. Then the period can be of the order $2^{Nq}$. Such an example for a Xorshift generator is provided here by G. Marsaglia, with $q=4$. It uses four seeds. In our case, if we were to use $q$ seeds, you can pick up $q$ irrational numbers that are linearly independent over the set of rational numbers. Their digits sequences are independent from each other (see section 1.3 in this article for a proof). An example (with $q=4$) is the first $N$ binary digits of the following numbers: $\log 2, \frac{\pi}{4}, \frac{\sqrt{2}}{2}$ and $\exp(-\frac{3}{5})$.
Of course, instead of choosing $\sqrt{2}/2$, one might choose  an irrational number impossible to guess, for instance
$$\alpha=\zeta(\sqrt{31}\log 5)\cdot\Gamma(e^{73 \sin 7})+\psi_2\Big(5e^{-11\cos 19}\log(53\pi+\sin 101)\Big)$$
Further improvement is obtained by using $N$ digits of $\alpha$ or $\sqrt{2}/2$ starting at position $M$ in their binary expansion, with $M$ very large and kept secret, rather than $M=0$ as in the code below. If you work with $q$ seeds, choose a different $M$ for each seed.
Source code
It also computes the period. If the period is larger than Niter (in the code) it will return $-1$ for the period: you need to increase Niter accordingly. Use for values of $N$ smaller than 45; to eliminate this problem, get the digits of the seed from a table or use a tool such as this one to get millions of digits for the seed.
#!/usr/bin/perl
$N=31;

$L=2;
$period=-1;
$Niter=50000;

%hash=();

$seed=sqrt(2)/2;

open(OUT,">randx.txt");
print OUT "0\tB";
$x=0;
$word="B";
$s=$seed;
for ($k=0; $k<$N; $k++) {
  $a[$k]=int(2*$s);  # k-th digit of seed
  $s=2*$s-int(2*$s);
  $b[$k]=$a[$k];
  $x+=$b[$k]/(2**($k+1));
  $word=$word."$b[$k]";
  $hash{$word}=0;
  print OUT "$b[$k]";
}
print OUT "\t$x\n";

for ($iter=1; $iter<$Niter; $iter++) {
  print OUT "$iter\tB";
  $x=0;
  for ($k=0; $k<$N; $k++) {
    $c[$k]=$b[($k+$L)%$N];
  }
  $word2="B";
  $nzero=0;
  for ($k=0; $k<$N; $k++) {
    $b[$k]=($c[$k]+$b[$N-$k-1])%2;
    $word2=$word2."$b[$k]";
    $x+=$b[$k]/(2**($k+1));
    print OUT "$b[$k]";
  }
  print OUT "\t$x\n";
  if ($period==-1) {
    if ($hash{$word2} eq "") {
      $hash{$word2}=$iter;
    } else {
      $period=$iter-$hash{$word2};
    }
  }
}
close(OUT);
print "$N $L $period\n";

Note
Obviously, one flaw of all RNG's with $q=1$ (first-order recurrence) is that you never see twice the same word within any period cycle. In true randomness, repetition occurs without causing the cycle to repeat itself entirely. As an example, if you pick up 10 integers randomly between $0$ and $3$, some number MUST appear at least twice.

REPLY [1 votes]: Floating point division varies across platforms especially if the language and the system supports hardware accelerated floating point arithmetic. It is risky to use it in an encryption algorithm standard. You can find a better more secure method if you use a fixed size seed starting at digit m of square root of 2 over 2. The division by 2 is redundant. This is equivalent to 1/(square root of 2). I am pretty sure that this is still irrational but I also think perl will implement these irrational constants differently than some other language. You are giving away too much by using the first N digits of a known irrational constant as part of the standard. It would be harder to attack if you offset true random m digits before sampling the decimal expansion of the irrational. It would be even better to use a hardware TRNG. After that you need to cast the sequence as raw binary data or signed int to eliminate unknown errors resulting from cross platform implementation. I have python libraries that allow very big integers without truncation. If you prefer performance then you are probably working with binary in C.


TITLE: Find the minimum value of $S$
QUESTION [2 upvotes]: Let $a,b,c$ be real numbers greater than 1.
Let $S=\log_a{bc}+\log_b{ac}+\log_c{ba}$
Then find the minimum value of $S$

REPLY [3 votes]: First, we divide the logs of the multiplications into sums:
$\log_a{bc}+\log_b{ac}+\log_c{ba}=\log_a{b}+\log_a{c}+\log_b{a}+log_b{c}+\log_c{a}+\log_c{b}$
After that, we use the $log_a{b}=\frac{\ln{a}}{\ln{b}}$ to further deconstruction:
$\log_a{b}+\log_a{c}+\log_b{a}+log_b{c}+\log_c{a}+\log_c{b}=\frac{\ln{b}}{\ln{a}}+\frac{\ln{c}}{\ln{a}}+\frac{\ln{a}}{\ln{b}}+\frac{\ln{c}}{\ln{b}}+\frac{\ln{a}}{\ln{c}}+\frac{\ln{b}}{\ln{c}}$
Because a, b and c are reals >1, their natural logarithm is a real > 0.
Substituting $x:=\frac{\ln{a}}{\ln{b}}$, $y:=\frac{\ln{a}}{\ln{c}}$, $z:=\frac{\ln{b}}{\ln{c}}$, we are looking for the minimum of $x+\frac{1}{x}+y+\frac{1}{y}+z+\frac{1}{z}$.
The minimum of a number plus its recipe is 2 for positive real numbers [prove is trivial, I give you on need], thus the minimum is $\underline{\underline{6}}$.
P.s. also trivial that we get this minimum if $a=b=c$.


TITLE: Why is this inequality of truncated random variables correct?
QUESTION [0 upvotes]: Given a sequence $X_k$ of $k$ independent random variables, the book am reading says I can form 'truncated' random variables, i.e.
$$X_k(n) = X_k \cdot \textbf{1}_{\{ \omega : |X_k(\omega)| \leq n\}}$$
where $\omega$ means an outcome from the sample space and $\textbf{1}$ is an indicator function.
Define $S_n$ and $\hat{S}_n$ as:
$$
S_n = X_1 + X_2 + ...+X_n
$$
$$
\hat{S}_n = X_1(n) + X_2(n) + ... X_n(n)
$$
so this means that $\hat{S}_n$ is the sum of $n$ truncated random variables.
We also define $m_n$ as:
$$
m_n = \mathbb{E}(X_1(n))
$$
Since the variable distributions (of $X_k$) are the same, we have $m_n=\mathbb{E}(X_k(n))$ for all $k \geq 1$.
The book am reading says that the following inequality is 'obvious', where given $\epsilon > 0$, we have:
$$
P\bigg(\big | \frac{S_n}{n} - m_n \big | \geq \epsilon\bigg) \leq P\bigg(\big | \frac{\hat{S}_n}{n} - m_n \big | \geq \epsilon\bigg) + P\bigg(\hat{S}_n \neq S_n\bigg)
$$
But it's not at all obvious to me and I got lost. How can we have the above inequality? (i.e. we might get some large $X_k$ inside the $S_n$, such that the left term is much larger than the right term which is truncated).

REPLY [1 votes]: The event $[|S_n/n-m_n|\ge\epsilon]$ is contained in the union of the two events $[|\hat S_n/n-m_n|\ge\epsilon]$ and $[\hat S_n\ne S_n]$.  Think: if some outcome is not in $[\hat S_n\ne S_n]$, then $\hat S_n - S_n$, and in that case for $|S_n/n-m_n|\ge\epsilon$ to occur it must also happen that $|\hat S_n/n-m_n|\ge\epsilon$.


text:
\begin{document}

\title[Absence of resonances near the critical line]
{Absence of resonance near the critical line
on asymptotically hyperbolic spaces}
\author[Colin Guillarmou]{Colin Guillarmou}
\address{Laboratoire de Math\'ematiques Jean Leray\\
         UMR 6629 CNRS/Universit\'e de Nantes \\
         2, rue de la Houssini\`ere \\
         BP 92208 \\
         44322 Nantes Cedex 03\\
         France}
     \email{cguillar@math.univ-nantes.fr}

\subjclass[2000]{Primary 58J50, Secondary 35P25}
\maketitle
\begin{abstract}
\noindent As a consequence of a result of Cardoso-Vodev, we show
that the resolvent of the Laplacian on asymptotically hyperbolic
manifolds is analytic in an exponential neighbourhood of the
critical line $\{\Re(\la)=\ndemi\}$. The case of non-trapping
metrics with constant curvature near infinity is also considered:
there is a strip $\{\Re(\la)>\ndemi-\eps\}$ with a finite number
of resonances.

\end{abstract}
\vspace{0.5cm}

\section{Introduction}

The purpose of this note is to give some `free of resonance' regions near the
critical line for the Laplacian on asymptotically hyperbolic manifolds.

An asymptotically hyperbolic manifold is a smooth non-compact
Riemannian manifold $(X,g)$ of dimension $n+1$ which is the
interior of a smooth compact manifold with boundary
$\bar{X}=X\cup\pl\bar{X}$ and such that for all boundary defining
function $x$ of $\bar{X}$ (i.e. $\pl\bar{X}=\{x=0\}$ and
$dx|_{\pl\bar{X}}\not=0$), $x^2g$ extends to a smooth metric on
$\bar{X}$ and $|dx|_{x^2g}=1$ on $\pl\bar{X}$. The metric can then
be expressed in a collar neighbourhood of the boundary
$(0,\eps)_x\x\pl\bar{X}_y$ by
\begin{equation}\label{modelform}
g=\frac{dx^2+h(x,y,dy)}{x^2}
\end{equation}
with $h(x,y,dy)$ a smooth tensor up to the boundary $\{x=0\}$.
It can be seen that $(X,g)$ is a complete manifold with
curvatures approaching $-1$ near the boundary $\pl\bar{X}$
(the boundary is the infinity of $\bar{X}$ with respect to the metric
$g$) and that the hyperbolic convex co-compact quotients
are contained in this class of manifolds.

It is well known that the spectrum of the Laplacian $\Delta_g$ acting on functions splits into
absolutely continuous spectrum $[\frac{n^2}{4},\infty)$ and a
finite set of eigenvalues $\sigma_{pp}(\Delta_g)\subset
(0,\frac{n^2}{4})$. The modified resolvent
\[R(\la):=(\Delta_g-\la(n-\la))^{-1}\]
is then meromorphic on $\{\Re(\la)>\ndemi\}$ with finite rank poles at
each $\la_e$ satisfying $\la_e(n-\la_e)\in\sigma_{pp}(\Delta_g)$.
Mazzeo and Melrose \cite{MM} have constructed the meromorphic extension of
$R(\la)$ to $\cc\setminus \demi(n-\nn)$ with poles of finite multiplicity,
which are called \textsl{resonances}. Physically, the most
interresting resonances are those in a neighbourhood of
the critical line $\{\Re(\la)=\ndemi\}$
(which corresponds to the essential spectrum).
It turns out that their localization is closely related
to the number of geodesics trapped in compact sets of $(X,g)$.
However, a general principle which seems to hold for geometric
scattering on a large class of infinite volume manifolds
is that there exists a free of resonance
region of the form
\[\{\la\in\cc; \Im(\la)\leq e^{-C_1|\la|}, |\Re(\la)|\geq C_2\},\quad C_1,C_2>0\]
when the critical line is the axis $\{\Im(\la)=0\}$.
This was first proved by Burq \cite{BU} in Euclidean scattering
and Vodev \cite{V1} on some surfaces with negative constant curvature near
infinity.
It is worth noting that these results are optimal in the sense that,
in general, the existence of elliptic closed geodesics implies the existence
of resonances which are exponentially close to the critical
line (see \cite{S,P1,P2}).

Here, we deal with the case of asymptotically hyperbolic manifolds:

\begin{theo}\label{main}
Let $(X,g)$ be an asymptotically hyperbolic manifold and $x$ a
boundary defining function,
then there exist $C_1,C_2>0$ such that
the weighted resolvent $x^\demi(\Delta_g-\la(n-\la))^{-1}x^\demi$
extends analytically from $\{\la\in\cc;|\Im(\la)|>C_2, \Re(\la)>\ndemi\}$ to
\begin{equation}\label{exponvois}
\{\la\in\cc; |\Im(\la)|>C_2,
\Re(\la)> \ndemi-e^{-C_1|\la|}\}.
\end{equation}
as bounded operators on $L^2(X):=L^2(X,dvol_g)$.
\end{theo}

The essential ingredients are a uniform bound of the weighted resolvent
norm on the critical line and a sharp parametrix of the
meromorphically continued resolvent.
In our case, the resolvent bound $||\rho R(\la)\rho||\leq Ce^{C|\la|}$
on the critical line has been proved by Cardoso and Vodev
\cite{CV}, $\rho$ being a
weight function decreasing to $0$ near infinity.
To extend analytically $\rho R(\la)\rho$ to the region (\ref{exponvois}),
the main point is to see it as a perturbation
of the resolvent of the Laplacian on a model space $X_0$ which
is sufficiently close to our manifold. A good candidate for $X_0$
is the warped product $(0,\eps)_x\x\pl\bar{X}_y$ equipped with the metric
\[g_0:=\frac{dx^2+h(0,y,dy)}{x^2}\]
and take Dirichlet condition at $x=\eps$, but for technical
reasons we will better use $(0,\infty)\x \pl\bar{X}$ with the
same metric and localize the resolvent near $x=0$ with
cut-off functions (a similar approach is used by Vodev \cite{V1}).
This model resolvent $R_0(\la)$ needs to have an analytic
extension on weighted spaces
in a neighbourhood of the form (\ref{exponvois}), with a norm
bounded by $Ce^{C|\la|}$. The classical resolvent equation
\[R(\la)-R(z)=(\la(n-\la)-z(n-z))R(\la)R(z)\]
in the physical sheet $\{\Re(\la)>\ndemi\}$ and the
approximation of $R(\la)$ by $R_0(\la)$ allow to write
\[\rho R(\la)\rho (1+(\la-z)K(\la,z))=K_1(\la,z)\]
where $K(\la,z), K_1(\la,z)$ are some operators which are expressed
in terms of the
model resolvents, $\rho R(z)\rho$ and some error terms. At last,
the extension properties of $\rho R_0(\la)\rho$ through the
critical line and those of $\rho R(z)\rho$ up to this critical line can
be used to extend $K(\la,z)$ and $K_1(\la,z)$ to $z\in
\{\Re(\la)=\ndemi\}$ and $\la$ in the neighbourhood
(\ref{exponvois}). The bound on the norm of $\rho R(z)\rho$ and
$\rho R_0(\la)\rho$ can then be used to show that
$|\la-z|.||K(\la,z)||\leq \demi$ for $\Re(z)=\ndemi$ and
$|\la-z|\leq C^{-1}e^{-C|z|}$ with $C>0$ large and
independent of $(\la,z)$; this allows to invert holomorphically
$1+(\la-z)K(\la,z)$ and to define $\rho R(\la)\rho$ in
(\ref{exponvois}).

In the case of non-trapping metrics, Vodev \cite{V2}
proved that the norm of the resolvent on the critical
line grows not faster than $C|\la|^{-1}$ when $|\Im(\la)|\to \infty$,
which implies a larger extension of the resolvent through the essential
spectrum. We especially consider the case of non-trapping manifolds
with constant curvature near infinity and obtain:

\begin{theo}\label{hyper}
Let $(X,g)$ be a conformally compact manifold with constant curvature
outside a compact subset and let $x$ be a boundary defining function. If $g$ is
non trapping, there exist $C_1,C_2>0$ such that the weighted
resolvent $x^\demi(\Delta_g-\la(n-\la))^{-1}x^\demi$ extends
analytically from $\{\la\in\cc;|\Im(\la)|>C_2, \Re(\la)>\ndemi\}$ to
\[\{\la\in\cc; |\Im(\la)|>C_2, \Re(\la)> \ndemi-C_1\}\]
as bounded operators on $L^2(X)$.
\end{theo}

Note that this non-trapping condition is not
satisfied for non-elementary convex co-compact
quotients of $\hh^{n+1}$. When $X=\Gamma\backslash\hh^{n+1}$
with $\Gamma$ a non-elementary convex co-compact group of isometries,
better results are available with the help of Selberg's zeta function:
it is well known that there exists a half plane $\Re(\la)>\delta$
with no resonance, where $\delta$ is the dimension of the limit set
(see also a result of Naud \cite{N} in dimension $2$).
This shows that Theorem \ref{hyper} is weak in the sense
that we need the non-trapping assumption but the compensation is that
we do not have to confine ourselves to the rigid class of constant
curvature manifolds.\\

The paper is organized as follows: in Section 1, we recall
Cardoso-Vodev Theorem, then we study our models in Section 2
and finally we give the proof of the results in Section 3.\\

\textbf{Acknowledgements.} I would like to thank Georgi Vodev who
pointed out to me this problem.

\section{Cardoso-Vodev result}

In \cite{CV}, Cardoso and Vodev consider some Riemannian manifolds $X$
with controlled structure near infinity
and they obtain exponential bounds for the weighted resolvent norm on the critical
line. These manifolds have the following properties outside a compact set $Z$
\begin{equation}\label{decomposition}
X\setminus Z \cong \left( [R,\infty)\x
S, g:=dr^2+\sigma(r)\right), \quad R\gg 1,
\end{equation}
where $S$ is a n-dimensional smooth compact manifold, $\cong$
means `isometric' and
$\sigma(r)=\sigma(r,y,dy)$ is a family of metrics on
$S_r:=\{r\}\x S$ which satisfy
\begin{equation}\label{propsigma}
|q(r,y)|\leq C ,\quad
\pl_r q(r,y)\leq Cr^{-1-\delta}, \quad \delta>0,\quad r>R,
\end{equation}
\begin{equation}\label{prop2sigma}
-\pl_r(\sigma^{-1})(r,y,\xi)\geq\frac{C}{r}\sigma^{-1}(r,y,\xi)
,\quad \forall (y,\xi)\in T^*S_r,
\end{equation}
$\sigma^{-1}(r)$ being the
principal symbol of the Laplacian on $(S_r,\sigma(r))$
and $q(r,y)$ is an effective potential defined by
\[q(r,y):=(2^{-1}\pl_r\log\nu)^2+(2\nu)^{-2}\sum_{i,j}\sigma^{ij}\pl_{y_i}\nu
\pl_{y_j}\nu+2^{-1}\nu\Delta_g(\nu^{-1})\]
with $\nu:=(\det(\sigma_{ij}))^{\demi}$. An asymptotically
hyperbolic metric $g=x^{-2}(dx^2+h(x,y,dy))$
can be decomposed as in (\ref{decomposition}) by
putting $x=e^{-r}$ and we get
\[\sigma(r,y,dy)=e^{2r}h(e^{-r},y,dy)\]
with $h(x,y,dy)$ smooth up to $x=0$. In $x$ coordinate, we have
$\nu\in x^{-n}C^\infty(\bar{X})$, $\pl_r=-x\pl_x$ and
$h^{ij}\in C^\infty(\bar{X})$ thus
\[(x\pl_x\log\nu)^2+\nu^{-2}\sum_{i,j}x^2h^{ij}\pl_{y_i}\nu
\pl_{y_j}\nu \in C^\infty(\bar{X})\]
\[\nu\Delta_g\nu^{-1}=\nu\Big(-(x\pl_x)^2-(x\pl_x\log\nu)x\pl_x-
\nu^{-1}\sum_{i,j}\pl_{y_i}(\nu x^2h^{ij}\pl_{y_j})\Big)\nu^{-1}\in
C^\infty(\bar{X})\]
and (\ref{propsigma}) is satisfied for all $\delta>0$. Moreover
we have for all $(y,\xi)\in T^*\pl\bar{X}$
\[\frac{x\pl_x (x^2h^{-1})(x,y,\xi)}{x^2h^{-1}(x,y,\xi)}=
2+x\frac{\pl_x(h^{-1})(x,y,\xi)}{h^{-1}(x,y,\xi)}\geq 1\]
if $x\leq \eps$ with $\eps$ small, and we obtain that
(\ref{prop2sigma}) is satisfied. As a conclusion, asymptotically
hyperbolic manifolds are in the class of manifolds studied by
Cardoso and Vodev \cite{CV} and Vodev \cite{V2}, so their results can
be summarized in that case in the
\begin{theo}\label{cardvod}
Let $(X,g)$ be an asymptotically hyperbolic manifold and $x$
a boundary defining function. There exists
$C>0$ such that the weighted resolvent
$x^\demi(\Delta_g-\la(n-\la))^{-1}x^\demi$
extends continuously from $\{\Re(\la)>\ndemi, |\Im(\la)|\geq 1\}$
to $\{\Re(\la)\geq \ndemi, |\Im(\la)|\geq 1\}$ on $L^2(X)$
and the extension satisfies
\[||x^{\demi}R(\la)x^{\demi}||_{\mc{L}(L^2,H^p)}\leq Ce^{C|\la|},\quad C>0\]
for $p=0,1$, $|\Im(\la)|\geq 1$ and $0\leq\Re(\la)-\ndemi\leq 1$,
where $H^p$ means the p-Sobolev space on $X$ with respect to the metric $g$.
If in addition $g$ is non-trapping we have for $p=0,1$,
$|\Im(\la)|\geq 1$ and $0\leq\Re(\la)-\ndemi\leq 1$
\[||x^{\demi}R(\la)x^{\demi}||_{\mc{L}(L^2,H^p)}\leq C|\Im(\la)|^{-1+p},\quad
C>0.\]
\end{theo}

\section{Two models}

Before giving the models, we recall a few properties of some
differential operators on $X$. Let $(X,g)$ be a conformally compact
manifold and $\Delta_g$ the Riemannian Laplacian. If $x$ is a boundary
defining function and $(y_i)_{i=1,\dots,n}$ some coordinates on
$\pl\bar{X}$, the space $\mc{V}_0(\bar{X})$ of smooth vector
fields on $\bar{X}$ which vanish on $\pl\bar{X}$ is locally
generated by $x\pl_x,x\pl_{y_i}$ for $i=1,\dots,n$ near the
boundary. We denote by $\textrm{Diff}_0^k(\bar{X})$ the space of differential
operators of order $k$ generated by $k$ products of elements of
$\mc{V}_0(\bar{X})$
\[\textrm{Diff}_0^k(\bar{X}):=\underset{0\leq i\leq k}{\textrm{Vect}}
\mc{V}_0(\bar{X})^k,\quad  \mc{V}_0(\bar{X})^0=C^\infty(\bar{X}).\]
For example it is straightforward to check that
\[\Delta_g\in \textrm{Diff}_0^2(\bar{X}).\]
$\Delta_g$ is now considered as the self-adjoint operator obtained
by Friedrichs extension from the Laplacian on $C_0^\infty(X)\subset
L^2(X)=L^2(X,dvol_g)$. For $k\in\rr$, we define the k-Sobolev space by
\[H^k(X):=\mc{D}om((1+\Delta_g)^{\frac{k}{2}}),\]
where $\mc{D}om$ means the domain.
The Sobolev spaces associated to two different
conformally compact metrics are the same
(for instance, it is done for $k=1,2$ by Froese-Hislop
\cite[appendix]{FH} in a more general framework) and
\begin{equation}\label{propdiff0}
\forall D^k\in \textrm{Diff}^k_0(\bar{X}), \quad D^k\in
\mc{L}(H^s(X),H^{s-k}(X)),\quad s\in [0,2].
\end{equation}
Moreover a useful
property of these differential operators is the following
\begin{equation}\label{commutdiff0}
x^{-\alpha}D^kx^\alpha\in \textrm{Diff}_0^k(\bar{X}), \quad
\alpha\in\rr, \quad D^k\in\textrm{Diff}^k_0(\bar{X})
\end{equation}
which is easily seen from the commutator $[x\pl_x,x^\alpha]$
in local charts near $\pl\bar{X}$.\\

Let us now study two models which will be respectively
used for the parametrix construction of the general
case and for the case of constant curvature near infinity.\\

Let $(M,h_0)$ be a Riemannian compact manifold of dimension $n$ and
\begin{equation}\label{model}
X_0:=(0,+\infty)_x\x M, \quad g_0:=x^{-2}(dx^2+h_0).
\end{equation}
Though $(X_0,g_0)$ is not conformally
compact (there is a cusp end when $x\to\infty$),
it has a conformally compact structure near $x=0$. We
could take as model operator the Laplacian on
$((0,1]\x\pl\bar{X},g_0)$ with Dirichlet condition at $x=1$ to
have a conformally compact structure (with boundary), but
we prefer to use $X_0$ since it carries more symmetry and is
therefore easier to study.

As for conformally compact manifolds,
let us denote $\bar{X}_0:=[0,+\infty) \x M$ and
$\textrm{Diff}^k_0(\bar{X}_0)$
the space of smooth differential operators of order $k$
on $\bar{X}_0$ with support in $[0,1]\x M$ and which can be locally
written
\[\sum_{i+|\alpha|\leq k}a_{i,\alpha}(x,y)(x\pl_x)^ix^{|\alpha|}
\pl^\alpha_{y_i}, \quad a_{i,\alpha}\in C^\infty(\bar{X}_0),\]
where $(y_i)_{i=1,\dots,n}$ are some local coordinates on $M$.

By taking the new variable $r=\log x$, it is easy to see that the
Laplacian $\Delta_{g_0}$ is unitarily equivalent to
\[P_0=-\pl_r^2+e^{2r}\Delta_{h_0}+\frac{n^2}{4}\]
on $L^2(\rr\x M, drdvol_{h_0})$.
As before, we define the Sobolev spaces by
\[H^k(X_0):=\mc{D}om((1+\Delta_{g_0})^{\frac{k}{2}})\cong
\mc{D}om((1+P_0)^{\frac{k}{2}}).\]
We first remark that
the arguments given by Froese and Hislop
in \cite[appendix]{FH} prove that for $k=0,1,2$ and $s\in[0,2]$
\begin{equation}\label{propdiff0x}
\forall D^k\in \textrm{Diff}^k_0(\bar{X}_0),\quad
D^k\in \mc{L}(H^{s}(X_0),H^{s-k}(X_0))
\end{equation}
though they do not consider the `cuspidal' part $\{r\in \rr^+\}$.
Of course the case $k=1$ can be directly
obtained from the identity
\begin{equation}\label{sobolevh1}
||(1+P_0)^\demi u||^2_{L^2}=
||\pl_ru||_{L^2}^2+||e^r\Delta^\demi_{h_0}u||^2_{L^2}
+\left(\frac{n^2}{4}+1\right)||u||^2_{L^2}.
\end{equation}

We shall first see how the resolvent of $\Delta_{g_0}$ can be
extended to the non-physical sheet and we will give an upper bound of
its weighted norm.

\begin{lem}\label{casmodel}
Let $x_0<1$, $(X_0,g_0)$ defined in (\ref{model})
and $\rho=\rho(x)$ a smooth function on $X_0$ with support in $\{x<1\}$
such that $\rho(x)=x^\demi$ for $x\leq x_0$. Then the weighted
resolvent
\[\rho R_0(\la)\rho:=\rho(\Delta_{g_0}-\la(n-\la))^{-1}\rho\]
extends analytically from $\{\Re(\la)>\ndemi,|\Im(\la)|>1\}$
to $\{\Re(\la)>\ndemi-\frac{1}{4}, |\Im(\la)|> 1\}$ and it satisfies
\begin{equation}\label{h1bound}
||\pl_\la^q \rho R_0(\la)\rho||_{\mc{L}(L^2,H^p)}\leq
e^{C|\la|}, \quad C>0
\end{equation}
in that region for $q=0,1$, $p=0,1,2$.
\end{lem}
\textsl{Proof}: let us begin by $p=2$. Let $\chi\in C_0^\infty([0,1))$
which is equal to $1$ on $\supp(\rho)$ and $\chi_1\in C_0^\infty([0,1))$
such that $\chi_1=1$ on $\supp(\chi)$. We then have for
$\la_0\in \{\Re(\la)>\ndemi\}$
\[
(\Delta_{g_0}+1)\rho R_0(\la)\rho=
[\Delta_{g_0},\rho]x^{-\demi}\chi(x)x^\demi R_0(\la)\rho+\rho^2+(\Lambda+1)\rho
R_0(\la)\rho,
\]
\[
\chi(x)x^\demi R_0(\la)\rho=R_0(\la_0)\Big([\Delta_{g_0},\chi(x) x^\demi]x^{-\demi}
\chi_1(x)x^\demi R_0(\la)+x^\demi+(\Lambda-\Lambda_0)\chi(x) x^\demi
R_0(\la)\Big)\rho
\]
where $\Lambda:=\la(n-\la),\Lambda_0:=\la_0(n-\la_0)$.
Since $[\Delta_{g_0},\rho]x^{-\demi}\in\textrm{Diff}^1_0(\bar{X}_0)$ and
$[\Delta_{g_0},\chi(x)x^\demi]x^{-\demi}\in\textrm{Diff}^1_0(\bar{X}_0)$
in view of (\ref{commutdiff0}), it suffices to prove (\ref{h1bound})
for $p=0$ and use (\ref{propdiff0x}) to obtain the other cases.

Let us now use $P_0$ instead of $\Delta_{g_0}$.
We have a decomposition induced by the spectral resolution of $\Delta_{h_0}$
\[P_0=\bigoplus_{j\in\nn_0} P^{(j)}_0, \quad
P^{(j)}_0:=-\pl^2_r+e^{2r}\mu^2_j+\frac{n^2}{4}\]
where $(\mu_j^2)_{j\in\nn_0}$ are the eigenvalues of $\Delta_{h_0}$
(counted with multiplicities) associated to
an orthonormal basis of $L^2(M)$ of eigenvectors $(\psi_j)_{j\in\nn_0}$.
If we denote the resolvent of $P^{(j)}_{0}$ on $L^2(\rr,dr)$ by
\[R^{(j)}_{0}(\la):=(P^{(j)}_{0}-\la(n-\la))^{-1}\]
we clearly have for $f\in L^2(X)$ and $f_j:=\cjg
f,\psi_j\cjd_{L^2(M)}$
\begin{equation}\label{formuler}
\rho(\Delta_{g_0}-\la(n-\la))^{-1}\rho f= \sum_{j\in\nn_0}(\rho
R^{(j)}_{0}(\la)\rho f_j)\psi_j.
\end{equation}
Note that for $\mu_j\not=0$ the translation
\[U_j:\left\{
\begin{array}{rcl}
L^2(\rr,dr)&\to& L^2(\rr,dr)\\
f&\to& f(\log(\mu_j)+\bullet)
\end{array}\right.\]
is an isometry and that $U_j^{-1}P^{(j)}_0U_j=Q$ with
$Q:=-\pl^2_r+e^{2r}+\frac{n^2}{4}$. Let us set $k=\la-\ndemi$ for
simplicity. The Green kernel for $Q$ is then easy to find for
$\Re(k)>0$ (see \cite[Ex. 4.15]{T})
\[R_Q(\la;r,t):=(Q-\la(n-\la))^{-1}(r,t)=
-K_{-k}(e^r)I_k(e^t)H(r-t)-I_k(e^r)K_{-k}(e^t)H(t-r)\]
with $H$ the Heaviside function and $I_k,K_k$ the modified Bessel functions
whose integral representations (when they converge) are
\begin{equation}\label{besseli1}
I_k(z)=\frac{2^{1-k}z^k}{\Gamma(k+\demi)\Gamma(\demi)}
\int_0^1(1-u^2)^{k-\demi}\cosh(zu)du
\end{equation}
\begin{equation}\label{besseli2}
I_k(z)=\frac{1}{\pi}
\int_0^\pi e^{z\cos(u)}\cos(ku)du-\frac{\sin(k\pi)}{\pi}\int_0^\infty
e^{-z\cosh(u)-ku}du
\end{equation}
\begin{equation}\label{besselk1}
K_{k}(z)=K_{-k}(z)=\frac{\Gamma(-k+\frac{3}{2})2^{-k+1}}{\Gamma(\demi)}z^{k}
\int_0^\infty \frac{\sin(tz)}{tz}t^2(1+t^2)^{k-\frac{3}{2}}dt\end{equation}
\begin{equation}\label{besselk2}
K_{-k}(z)=\int_0^\infty\cosh(ku)e^{-z\cosh(u)}du.
\end{equation}
Moreover for $\mu_j=0$, the expression of the meromorphically extended
euclidian resolvent kernel is well-known in $\cc\setminus\{0\}$
\[R^{(j)}_0(\la;r,t)=(2k)^{-1}e^{-k|r-t|}\]
and we have for $\Re(\la)>\ndemi-\frac{1}{4}$, $|\Im(\la)|\geq 1$ and
$p=0,1,2$
\begin{equation}\label{mujegal0}
||\pl_r^p \rho(e^\bullet)
R^{(j)}_0(\la)\rho(e^\bullet)||_{\mc{L}(L^2(\rr))}\leq
C\left|\la-\ndemi\right|^{-1+p} \textrm{ if } \mu_j=0. \end{equation}
From
these expressions, one can remark that there is no resonance
except $\ndemi$ for this problem. Without loss of generality, we can
suppose that $\rho(e^r)=e^{\frac{r}{2}}\chi(r)$ where $\chi$ is a smooth
function on $\rr$ such that $\chi(r)=1$ when $r\leq -1$ and
$\chi(r)=0$ when $r\geq 0$.
Since from (\ref{besseli1})-(\ref{besselk2}) we have
\[|I_k(e^t)\rho(e^{t-\log(\mu_j)})|\leq
\left\{\begin{array}{l}
e^{C|k|}e^{e^t}\textrm{ for }t>0\\
 e^{C|k|}e^{\frac{t}{4}}\textrm{ for }t\leq 0
\end{array}\right.\]
and
\[|K_{-k}(e^r)\rho(e^{r-\log(\mu_j)})|\leq
\left\{\begin{array}{l}
e^{C|k|}e^{-e^r} \textrm{ for }r>0\\
e^{C|k|}e^{\frac{r}{4}} \textrm{ for }r\leq 0
\end{array}\right.\]
for $|\Re(k)|\leq \frac{1}{4}$ and $|\Im(k)|\geq 1$, we can easily deduce that
\[|K_{-k}(e^r)\rho(e^{r-\log(\mu_j)})|
\int_{-\infty}^r|I_k(e^t)\rho(e^{t-\log(\mu_j)})|dt\leq e^{C|k|}\]
\[|I_{k}(e^r)\rho(e^{r-\log(\mu_j)})|
\int_{r}^\infty|K_{-k}(e^t)\rho(e^{t-\log(\mu_j)})|dt\leq e^{C|k|}\]
and Schur's lemma implies that
\[||U_j^{-1}(\rho)R_Q(\la)U_j^{-1}(\rho)||_{\mc{L}(L^2)}\leq e^{C|k|}\]
for $|\Re(k)|\leq \frac{1}{4}$, $|\Im(k)|\geq 1$. But since
$U_j^{-1}(\rho(e^{\bullet}))=U_j^{-1}\rho(e^\bullet) U_j$ as
operators and $U_j$ is an isometry, we use
\[R_0^{(j)}(\la)=U_jR_Q(\la)U_j^{-1}\]
to conclude that for $|\Im(\la)|\geq 1$ and
$|\Re(\la)-\ndemi|\leq \frac{1}{4}$
\begin{equation}\label{borneq}
||\rho(e^\bullet) R^{(j)}_0(\la)\rho(e^{\bullet})||_{\mc{L}(L^2)}\leq e^{C|\la|}.
\end{equation}
Finally we combine (\ref{borneq}) with (\ref{mujegal0}) and (\ref{formuler}).
The bound for $q=1$ (one derivative with respect to $\la$) is directly
obtained from the case $q=0$ and Cauchy's formula.
\qed\\

\textsl{Remark}: a better estimate can be obtained
but we do not need it for our purpose.\\

The second model is the hyperbolic space $(\hh^{n+1},g_h)$ with
its usual metric. To see $\hh^{n+1}$ as a conformally compact
manifold we take the Beltrami model
\[\hh^{n+1}=\{m\in\rr^{n+1}; |m|<1\}, \quad g_h=\frac{4dm^2}{(1-|m|^2)^2}\]
and we set for example $x=2(1-|m|)(1+|m|)^{-1}$
as boundary defining function.

\begin{prop}\label{bound:uniform}
The weighted hyperbolic resolvent
\[x^\demi R_h(\la)x^\demi:=x^\demi(\Delta_{g_h}-\la(n-\la))^{-1}x^\demi\]
extends holomophically from $\{\Re(\la)>\ndemi\}$ to
$\{\Re(\la)>\ndemi-\frac{1}{8}\}$ with values
in $\mc{L}(L^2(\hh^{n+1}),H^1(\hh^{n+1}))$ and it satisfies
\begin{equation}\label{normresolvlib}
\left|\left|\pl_\la^q x^\demi R_h(\la)x^\demi\right|\right|_
{\mc{L}(L^2,H^p)}\leq C\left|\la-\ndemi\right|^{-1+p},\quad C>0
\end{equation}
in the same region (with $\la\not=\ndemi$) for $p=0,1$, $q=0,1$.
\end{prop}
\textsl{Proof}: to show these bounds, we consider the wave
operators
\[U_0(t):=\cos\left (t\sqrt{\Delta_{g_h}-\frac{n^2}{4}}\right ),
\quad U_1(t):=\left (\Delta_{g_h}-\frac{n^2}{4}\right )^{-\demi}
\sin\left (t\sqrt{\Delta_{g_h}-\frac{n^2}{4}}\right ).\]
Let $\delta>0$ be small, $\chi\in C_0^{\infty}(\rr^+)$ with support in
$[0,1-\frac{\delta}{2}[$ and which is equal to $1$ on
$[0,1-\delta]$. Then set
\[\chi_t(m):=\chi\left (\frac{4\argth |m|}{t}\right )\]
whose support is included in the hyperbolic ball of radius
$\frac{t}{2}$ and which is equal to $1$ in the hyperbolic ball of radius
$\frac{t(1-\delta)}{2}$.

When $n+1$ is odd, we have by Huygens principle (see \cite{H})
\begin{equation}\label{huygens}
\chi_tU_0(t)\chi_t=0, \quad t>0.
\end{equation}
Now we split $x^\demi$ in the following way
\begin{equation}\label{decomp1}
x^\demi U_0(t)x^\demi=(1-\chi_t)x^\demi U_0(t)(1-\chi_t)x^\demi+
(1-\chi_t)x^\demi U_0(t)\chi_tx^\demi+\chi_t x^\demi U_0(t)(1-\chi_t)x^\demi
\end{equation}
for $t>0$. It is clear that
\begin{equation}\label{norm1}
||\chi_tx^\demi||\leq C, \quad
||U_0(t)||\leq 1
\end{equation}
and remark that
$m\in\supp(1-\chi_t)$ only if $|m|\geq
\tanh\left(\frac{(1-\delta)t}{4}\right )$, that is
\[x=2\frac{1-|m|}{1+|m|}\leq
2\frac{1-\tanh(\frac{t(1-\delta)}{4})}{1+\tanh(\frac{t(1-\delta)}{4})}
=2e^{-\frac{t(1-\delta)}{2}}.\]
So we find
\begin{equation}\label{norm2}
\left|\left|(1-\chi_t)x^\demi\right|\right|\leq Ce^{-\frac{t(1-\delta)}{4}}.
\end{equation}
From (\ref{decomp1}), (\ref{norm1}) and (\ref{norm2}) we deduce that
\[\left|\left|x^\demi U_0(t)x^\demi\right|\right|\leq Ce^{-\frac{t(1-\delta)}{4}}.\]
It remains to use the Laplace transform of $U_0$
\[ \left
(\la-\ndemi\right )x^\demi R_h(\la)x^\demi= \int_0^{\infty}
e^{t(\ndemi-\la)}x^\demi U_0(t)x^\demi dt\]
and (\ref{normresolvlib}) is
proved when $p=0$ and $n+1$ odd by taking $\delta<\demi$.
To deal with the case $n+1$ even, we study
$x^\demi\pl_t U_1(t)x^\demi$ and use $U_0(t)=\pl_t U_1(t)$.
We have
\begin{equation}\label{decomp2}
\chi_t \pl_tU_1(t)\chi_t=\pl_t\left (\chi_t U_1(t)\chi_t\right )
-(\pl_t\chi_t)U_1(t)\chi_t-\chi_t U_1(t)(\pl_t\chi_t)
\end{equation}
and the Schwartz kernel of $U_1(t)$ is (see \cite{H})
\begin{equation}\label{u1}
U_1(t;x,y)=
C_n\left (\sinh^2\Big (\frac{t}{2}\Big )- \sinh^2\Big
(\frac{d_{\hh^{n+1}}(x,y)}{2}\Big )\right )_+^{-\ndemi}
\end{equation}
with $C_n\in\rr$. Hence, by construction
of $\chi_t$, the operators in (\ref{decomp2}) have a smooth kernel
with support in $\{(x,y)\in\hh^{n+1}\x\hh^{n+1},d_{\hh^{n+1}}(x,y)
<t(1-\frac{\delta}{2})\}$ and (\ref{u1}) implies that
there exists $T>0$ such that
\begin{equation}\label{normonde2}
\left|\left|\chi_t \pl_t U_1(t)\chi_t\right|\right|\leq
Ce^{-\frac{nt}{2}}
\end{equation}
for $t\geq T$. At last, we proceed as in
the odd case: we split $x^\demi$ on the support of $\chi_t$ and
outside, which shows (\ref{normresolvlib}) for $p=0$.

Now for $p=1$, it suffices to show that
$Dx^\demi R_h(\la)x^\demi$ extends to
$\{\Re(\la)>\ndemi-\frac{1}{8}\}$ in $\mc{L}(L^2)$
for a finite number of $D\in\textrm{Diff}^1_0(\bbar{\hh}^{n+1})$
which span $\textrm{Diff}^1_0(\bbar{\hh}^{n+1})$ over
$C^\infty(\bbar{\hh}^{n+1})$.
In fact since $Dx^\demi\in x^\demi\textrm{Diff}_0^1(\bbar{\hh}^{n+1})$,
we will study $x^\demi DR_0(\la)x^\demi$.
With $D$ as before, we have for $\Re(\la)>\ndemi$
\begin{equation}\label{transflap2}
x^\demi DR_0(\la)x^\demi=\int_0^\infty
e^{t(\ndemi-\la)}x^\demi DU_1(t)x^\demi dt
\end{equation}
but $||DU_1(t)||\leq C$ in view of the continuity of $D$
from $H^1(\hh^{n+1})$ to $L^2(\hh^{n+1})$. Moreover
\[\chi_tDU_1(t)\chi_t=D\chi_tU_1(t)\chi_t-[D,\chi_t]U_1(t)\chi_t\]
has a smooth kernel with compact support and it is
straightforward to check (from (\ref{u1})) that
there exists $T>0$ such that for all $t\geq T$
\begin{equation}\label{normonde3}
||\chi_tDU_1(t)\chi_t||\leq C e^{-\frac{nt}{2}} \end{equation}
Splitting $x^\demi DU_1(t)x^\demi$ in the same way as
(\ref{decomp1}) and using (\ref{normonde3}), (\ref{transflap2}), one deduces
the bound (\ref{normresolvlib}) for $p=1$.
The case $q=1$ is obtained by the Cauchy formula from $q=0$.
\qed\\

\section{Parametrix construction and proof of the main result}

In this section, we will give the construction of a
parametrix for the resolvent $R(\la)$ of $\Delta_g$ on  an arbitrary
asymptotically hyperbolic manifold $(X,g)$ whose metric is
\[g=x^{-2}(dx^2+h(x))\]
in a collar $(0,\delta)_x\x \pl\bar{X}$ near the boundary,
$h(x)$ being a smooth family of metrics on $\pl\bar{X}$.
Of course this parametrix will only be sufficient
to approach the resolvent in
$\{\Re(\la)>\ndemi-\delta\}$ with $\delta>0$ small.
Roughly, the singularities at $x=0$ and $x'=0$ of the kernel of $R(\la)$
are controlled by those of the resolvent kernel of the
Laplacian induced by the model metric $x^{-2}(dx^2+h(0))$
near the boundary, already studied in the previous section.\\

\textsl{Proof of Theorem \ref{main}}: consider $(X,g)$ an
asymptotically hyperbolic manifold and $R(\la):=(\Delta_g-\la(n-\la))^{-1}$
the resolvent for $\Delta_g$ in the physical sheet
$\{\Re(\la)>\ndemi\}$. Let $V$ be an open subset
in $(X,g)$ isometric to $U:=(0,\delta)_x\x\pl\bar{X}$
(with $\delta>0$) equipped
with the metric $x^{-2}(dx^2+h(x))$ and $i:V\to U$ this
isometry. Note that it is always possible to renormalize $x$ (take
$x'=x\delta^{-1}$) to have $\delta=1$ and it does not change the
structure of the metric, so we suppose $\delta=1$.
We now consider $(X_0,g_0)$ the Riemannian manifold defined in
(\ref{model}) with
\begin{equation}\label{defx0g0}
(M,h_0):=(\pl\bar{X},h(0)).
\end{equation}
Let $\mc{I}_U$ and $\mc{R}_U$ be the following bounded operators
\[\mc{R}_U :\left\{\begin{array}{rcl}
L^2(X_0,dvol_{g_0})&\to& L^2(U,dvol_{g_0})\\
f&\to& f(\iota_U(\bullet))
\end{array}\right.\]
\[\mc{I}_U :\left\{\begin{array}{rcl}
L^2(U,dvol_{g_0})&\to& L^2(X_0,dvol_{g_0})\\
f&\to& \indic_Uf
\end{array}\right.\]
with $\iota_U$ the inclusion $U\subset X_0$ and $\indic_U$ the characteristic
function of $U$. Similarly one can define the operators
$\mc{I}_V$ and $\mc{R}_V$ induced by the inclusion $V\subset X$.
Since $i^*g_0$ and $g$ are quasi-isometric on $V$, we obtain that
$i^*:L^2(U,dvol_{g_0})\to L^2(V,dvol_{g})$ and
$i_*: L^2(V,dvol_{g})\to L^2(U,dvol_{g_0})$ are bounded.
We then set
\[I^*:=\mc{I}_Vi^*\mc{R}_U \in \mc{L}(L^2(X_0,dvol_{g_0}),
L^2(X,dvol_g)),\]
\[I_*:=\mc{I}_Ui_*\mc{R}_V \in\mc{L}(L^2(X,dvol_g),
L^2(X_0,dvol_{g_0})).\]

For $j=1,2,3,4$, let $\psi_j$ a smooth function on $\rr^+$
which is equal to $1$ in $[0,\frac{j}{5}]$ and
to $0$ in $[\frac{j+1}{5},+\infty)$;
$\psi_j$ will also be considered as
a function on $U$ depending only on $x$. With $\til{\psi}_j:=i^*\psi_j$,
we have as operators
\[I_*I^*\psi_j=\psi_j,\quad I^*I_*\til{\psi}_j=\til{\psi}_j.\]
It is easy to check that there exist
$D_R,D_L\in\textrm{Diff}^2_0(\bar{X})$ such that
\[\Delta_g-I^*\Delta_{g_0}\psi_3I_*= xD_R,
\quad \Delta_g-I^*\psi_3\Delta_{g_0}I_*=xD_L.\]
Firstly, we take $\Re(\la)>\ndemi$ and
$R_0(\la)$ is the resolvent considered in Lemma \ref{casmodel}.
Observe that
\begin{equation}\label{restechi1}
(\Delta_{g_0}\psi_3-\la(n-\la))\psi_2R_0(\la)\psi_1=\psi_1+
[\Delta_{g_0},\psi_2]R_0(\la)\psi_1
\end{equation}
since $\psi_3\psi_2=\psi_2$ and $\psi_2\psi_1=\psi_1$.
Let $\chi_1:=1-\til{\psi}_1$
and let $\chi_0$ be a smooth function with compact support on $X$ which
is equal to $1$ on the support of $\chi_1$.
For $\la_0\in \{\Re(\la)>\ndemi\}$ fixed,
$\Lambda:=\la(n-\la)$ and $\Lambda_0:=\la_0(n-\la_0)$ we set
\[R_{0R}(\la):=I^*\psi_2R_0(\la)\psi_1 I_*,\quad
E_R(\la):=R_{0R}(\la)+\chi_0R(\la_0)\chi_1,\]
\[L_R(\la):=[\Delta_g,\chi_0]R(\la_0)\chi_1+
(\Lambda_0-\Lambda)\chi_0R(\la_0)\chi_1+I^*[\Delta_{g_0},\psi_2]R_0(\la)\psi_1I_*
+xD_RR_{0R}(\la),\]
and we deduce from (\ref{restechi1})
\begin{equation}\label{inverseright}
(\Delta_g-\Lambda)E_R(\la)=1+L_R(\la).
\end{equation}
Similarly, we give a left parametrix for $\Delta_g-\Lambda$
\[R_{0L}(\la):=I^*\psi_1R_0(\la)\psi_2I_*,\quad
E_L(\la)=R_{0L}(\la)+\chi_1R(\la_0)\chi_0,\]
\[L_L(\la)=R_{0L}(\la)xD_L+I^*\psi_1R_0(\la)[\psi_2,\Delta_{g_0}]I_*+
\chi_1R(\la_0)[\Delta_g,\chi_0]+
(\Lambda_0-\Lambda)\chi_1R(\la_0)\chi_0,\]
and we have
\begin{equation}\label{inverseleft}
E_L(\la)(\Delta_g-\Lambda)=1+L_L(\la).
\end{equation}
Let $z,\la\in\{s\in\cc;\Re(s)>\ndemi\}$ and $Z:=z(n-z)$, from
(\ref{inverseright}) and (\ref{inverseleft}) we then obtain
\begin{equation}\label{egalpoids}
R(\la)=E_R(\la)-R(\la)L_R(\la),\quad
R(z)=E_L(z)-L_L(z)R(z).
\end{equation}
On the other hand, the resolvent equation
\[R(\la)-R(z)=(\Lambda-Z)R(\la)R(z)\]
combined with the first identity of (\ref{egalpoids}) yield
\begin{equation}\label{paramfinal}
x^\demi R(\la)x^\demi(1+K(\la,z))=K_1(\la,z),
\end{equation}
\[K(\la,z):=(\Lambda-Z)x^{-\demi}L_R(\la)R(z)x^\demi,\quad
K_1(\la,z):=x^\demi R(z)x^\demi+(\Lambda-Z)x^\demi
E_R(\la)R(z)x^\demi.\]
For simplicity, we will now denote by $D^p$ (resp. $D_0^p$)
all differential operator in $\textrm{Diff}^p_0(\bar{X})$
with support in $\supp(\til{\psi}_3)$
(resp. $\textrm{Diff}_0^p(\bar{X}_0)$ with support in
$\supp(\psi_3)$). With these notations,
(\ref{commutdiff0}) is summarized by
\[D^px^\alpha=x^\alpha D^p\]
and we also get
\[D_0^pI_*=I_*D^p, \quad D^pI^*=I^*D_0^p.\]
Using that $\chi_0,\chi_1$ and $[\Delta_{g_0},\psi_2]$
have compact support, we obtain
\begin{equation}\label{lrla}
x^{-\demi}L_R(\la)=\left(D^1+(\Lambda_0-\Lambda)
x^{-1}\chi_0\right)x^\demi
R(\la_0)\chi_1+x^\demi I^*D_0^2R_0(\la)\psi_1I_*,
\end{equation}
\begin{equation}\label{erla}
x^\demi E_R(\la)=x^\demi R_{0R}(\la)+x^\demi\chi_0R(\la_0)\chi_1.
\end{equation}
Similarly, the second indentity of
(\ref{egalpoids}) and the definition of $E_L(\la),L_L(\la)$
imply that
\begin{equation}\label{rxdemi}
\begin{array}{rcl}
R(z)x^\demi&=&I^*\psi_1R_0(z)(I_*x^\demi D^2x^\demi R(z)x^\demi+
\psi_2I_*x^\demi)\\
& &+\chi_1R(\la_0)x^\demi \left(\chi_0+D^1
x^\demi R(z)x^\demi+(Z-\Lambda_0)x^{-\demi}\chi_0R(z)x^\demi\right).
\end{array}
\end{equation}
Combining this last expression with (\ref{lrla}) gives
\[\frac{K(\la,z)}{\Lambda-Z}=\left(D^1+(\Lambda_0-\Lambda)
x^{-1}\chi_0\right)x^\demi R(\la_0)\chi_1R(z)x^\demi\]
\[+x^\demi I^*D_0^2\psi_4R_0(\la)\psi_1I_*x^\demi
x^{-\demi}\chi_1R(\la_0)x^\demi \left(\chi_0+D^1
x^\demi R(z)x^\demi+(Z-\Lambda_0)x^{-\demi}\chi_0R(z)x^\demi\right)\]
\[+ x^\demi I^*D_0^2\psi_4R_0(\la)\psi_1^2R_0(z)\psi_4I_*x^\demi
(D^2x^\demi R(z)x^\demi+\til{\psi}_2).\]
The first two lines of this expression can be extended to
$\{\Re(\la)>\ndemi-\frac{1}{4}\}\cap\{|\Im(\la)|\geq 1\}$
and $\{\Re(z)\geq \ndemi\}\cap\{|\Im(z)|\geq C\}$ as bounded
operators on $L^2(X)$ by using Lemma \ref{casmodel} and
Theorem \ref{cardvod} (to be in the settings of Lemma \ref{casmodel},
we take $\rho:=i_*(x^\demi)\psi_4$ and we remark that
$x^\demi I^*D_0^2\psi_4=I^*D_0^2\rho$ and $\psi_1I_*x^\demi=\rho\psi_1I_*$),
and their $\mc{L}(L^2)$ norms is bounded by
$Ce^{C(|\la|+|z|)}$.

It remains to consider the last line in the expression of
$(\Lambda-Z)^{-1}K(\la,z)$.
Using (\ref{propdiff0}) and Theorem \ref{cardvod}, observe that
\[D^2x^\demi R(z)x^\demi=D^2(\Delta_g+i)^{-1}(D^1x^\demi R(z)x^\demi+x+(i+Z)
x^\demi R(z)x^\demi)\]
can be extended to $\Re(z)=\ndemi$ and $|\Im(z)|\gg 0$ with
$\mc{L}(L^2)$ norm bounded by $Ce^{C|z|}$.
Note that
\begin{eqnarray*}
\rho R_0(\la)\psi_1^2R_0(z)\rho&=&
\rho R_0(\la)\psi_1(\Delta_{g_0}-\Lambda)R_0(\la)R_0(z)
(\Delta_{g_0}-Z)\psi_1R_0(z)\rho\\
&=&\rho(\psi_1+R_0(\la)[\psi_1,\Delta_{g_0}])R_0(\la)R_0(z)
(\psi_1-[\psi_1,\Delta_{g_0}]R_0(z))\rho.
\end{eqnarray*}
Now it is easy to see that it can be rewritten by
\[(\psi_1+\rho R_0(\la)[\psi_1,\Delta_{g_0}]x^{-\demi})
\frac{\rho R_0(\la)\rho -
\rho R_0(z)\rho}{\Lambda-Z}
(\psi_1-x^{-\demi}[\psi_1,\Delta_{g_0}]R_0(z)\rho).\]
Recall that $x^{-\demi}[\psi_1,\Delta_{g_0}]x^{-\demi}
\in\textrm{Diff}^1_0(\bar{X}_0)$ since $[\psi_1,\Delta_{g_0}]$
has compact support in $X_0$.
Using this expression with Lemma \ref{casmodel}, we obtain that
\[x^\demi I^*D_0^2\psi_4R_0(\la)\psi_1^2R_0(z)\psi_4I_*x^\demi
(D^2x^\demi R(z)x^\demi+\til{\psi}_2)\]
extends to $\{\Re(\la)>\ndemi-\frac{1}{4},|\Im(\la)|\geq 1\}$
and $\{\Re(z)\geq \ndemi,|\Im(z)|\geq C\}$ as continuous operators
on $L^2(X)$ with norm  bounded by $Ce^{C(|\la|+|z|)}$.
Fix $z=\ndemi+is$ with $|s|$ large, then all
these estimates prove that
\[||K(\la,z)||_{\mc{L}(L^2)}\leq\left|\Re(\la)-\ndemi\right|Ce^{C|s|}\]
when $\Im(\la)=s$ and $\Re(\la)>\ndemi-\frac{1}{4}$.
Moreover this term is bounded by $\demi$ if
\begin{equation}\label{condition}
\left|\Re(\la)-\ndemi\right|\leq
\demi C^{-1}e^{-C|\Im(\la)|}, \quad \Im(\la)=s
\end{equation}
Since $K(\la,z)$ is holomorphic in $\la$ in $\{\Re(\la)>\ndemi-\frac{1}{4}
,|\Im(\la)|>1\}$ and $C$ does not depend on $s$, one
can invert $(1+K(\la,z))$ holomorphically if $\la$ satisfies
(\ref{condition}).

Finally the term $K_1(\la,z)$ can be treated in a similar way, using
(\ref{erla}) and (\ref{rxdemi}). So the proof is complete in the general case.
\qed\\

When $g$ is non-trapping we could apply the same method
and prove that there exists a free of resonance region
polynomially close to the critical line. We prefer to only write down
the  case of constant curvature near infinity since we obtain a
slightly better result.\\

\textsl{Proof of Theorem \ref{hyper}}: the proof is essentially similar.
When the metric $g$ has constant curvature near
infinity there exist (see \cite{GZ2}) some charts
$(V_j)_{j=1,\dots,M}$ covering a neighbourhood of the boundary
$\pl\bar{X}$ such that each $V_j$ is isometric (note $i_j$ this isometry)
to the open set
\begin{equation}\label{boulehyp}
B:=\{m=(y_1,\dots,y_{n+1})\in\rr^{n+1};y_{n+1}>0, \sum_{i}y_i^2< 1\}
\end{equation}
equipped with the hyperbolic metric
$y^{-2}_{n+1}(\sum_{i}dy_i^2)$. Actually, a parametrix sufficient
for our problem is given in \cite[Prop. 3.1]{GZ2}. Fix
$\la_0\in\{\Re(\la)>\ndemi\}$ and recall that $\Delta_{g_h}$ is the
Laplacian on $\hh^{n+1}$ and $R_h(\la)$ its resolvent studied in
Lemma \ref{bound:uniform}.
Let $\chi_i^j$ be some functions with
support in (\ref{boulehyp}), such that $\chi_1^j=1$ on the support
of $\chi_2^j$ and
\[\chi_i^j(y_1,\dots,y_{n+1})=\phi_i^j(y_1,\dots,y_n)\psi_i^j(y_{n+1})\]
with $\phi_i^j\in C_0^\infty(\rr^n)$ satisfying
$\sum_ji_j^*\phi^j_2=1$ on $\pl\bar{X}$ and $\psi_i^j\in
C_0^\infty([0,1))$ which is equal to $1$ for $y_{n+1}\leq \delta$
with $\delta<1$ small (see \cite{GZ2} or \cite[Lem. 3.2]{PP} for details).
This implies that
$\chi:=1-\sum_{j=1}^Mi_j^*\chi^j_2$ has compact support in $X$.
Let $\chi_0\in C_0^\infty(X)$ such that $\chi_0=1$ on the support
of $\chi$. Finally $I_j^*$ and ${I_j}_*$ are defined as in
the proof of previous theorem with our new isometries $V_j\to B$ and the
inclusions $V_j\subset X$, $B\subset\hh^{n+1}$. We then
apply the proof of previous theorem but we replace $E_R,E_L$ by our
new parametrix
\[E_R(\la):=\chi_0R(\la_0)\chi+\sum_{j=1}^MI_j^*\chi_1^jR_h(\la)\chi_2^j{I_j}_*,
\quad E_L(\la):=\chi
R(\la_0)\chi_0+\sum_{j=1}^MI_j^*\chi_2^jR_h(\la)\chi_1^j{I_j}_*\]
and the error terms are
\[L_R(\la):=[\Delta_{g},\chi_0]R(\la_0)\chi+(\Lambda_0-\Lambda)
\chi_0R(\la_0)\chi+\sum_{j=1}^MI_j^*[\Delta_{g_h},\chi_1^j]R_h(\la)\chi_2^j{I_j}_*,\]
\[L_L(\la):=\chi R(\la_0)[\chi_0,\Delta_g]+(\Lambda_0-\Lambda)
\chi R(\la_0)\chi_0+\sum_{j=1}^MI_j^*\chi_2^j
R_h(\la)[\chi_1^j,\Delta_{g_h}]{I_j}_*\]
where $\Lambda:=\la(n-\la)$, $\Lambda_0:=\la_0(n-\la_0)$. As before,
$x_j:={i_j}_*x$ is a boundary defining function of $\{y_{n+1}=0\}$ in $\bar{B}$.
Moreover it is easy to check (see \cite[Prop. 3.1]{GZ2}) that
\begin{equation}\label{propcom}
x_j^{-\demi}[\Delta_{g_h},\chi_i^j]x_j^{-\demi}\in
\textrm{Diff}_0^1(\bbar{\hh}^{n+1}), \quad i=1,2
\end{equation}
where we consider $B\subset \hh^{n+1}$.
We then have the same as (\ref{paramfinal}) with
\begin{eqnarray*}
\frac{K(\la,z)}{\Lambda-Z}&=&x^{-\demi}\left([\Delta_{g},\chi_0]
R(\la_0)\chi+(\Lambda_0-\Lambda)\chi_0 R(\la_0)\chi+
\sum_{j=1}^MI_j^*[\Delta_{g_h},\chi_1^j]R_h(\la)\chi_2^j{I_j}_*\right)\\
& &\x \Big(\chi R(\la_0)\chi_0+\sum_{j=1}^MI_j^*\chi_2^jR_h(z)
(\chi_1^j{I_j}_*+[\Delta_{g_h},\chi_1^j]{I_j}_*R(z))\\
& &\quad\quad\quad\quad\quad\quad+\chi R(\la_0)[\Delta_{g},\chi_0]R(z)-(\Lambda_0-Z)\chi
R(\la_0)\chi_0R(z)\Big)x^{\demi}
\end{eqnarray*}
for $\la,z$ in the physical sheet. Note that Theorem
\ref{cardvod}, Lemma \ref{bound:uniform} and (\ref{propcom}) prove
that all these products extend to $\Re(z)=\ndemi$ and
$\Re(\la)>\ndemi-\frac{1}{8}$  except maybe
\begin{equation}\label{badterm}
x^{-\demi}\left(\sum_{j=1}^MI_j^*[\Delta_{g_h},\chi_1^j]R_h(\la)\chi_2^j{I_j}_*\right)
\left(\sum_{j=1}^MI_j^*\chi_2^jR_h(\la)(\chi_1^j{I_j}_*+
[\Delta_{g_h},\chi_1^j]{I_j}_*R(z))\right)x^\demi.
\end{equation}
Fix $\la_0=\frac{n}{2}+\frac{1}{8}+is_0$ with $s_0>0$ large and
consider some $(\la,z)$ in
\[\mc{O}:=\left\{(\la,z)\in\cc^2; |\la-\la_0|<\frac{1}{4},  |z-\la_0|<
\frac{1}{4}, \Re(z)\geq\ndemi\right\}.\]
Theorem \ref{cardvod} implies that for $\alpha=-\demi,0$
\begin{equation}\label{boundla0z}
\left|\left|x^{-\demi}[\Delta_{g},\chi_0]R(\la_0)\chi
x^{\alpha}+(\Lambda-\Lambda_0)x^{-\demi}\chi_0 R(\la_0)\chi
x^{\alpha}\right|\right|_{\mc{L}(L^2)}\leq C,
\end{equation}
\begin{equation}\label{boundla0z2}
\left|\left|x^{\alpha}\chi R(\la_0)\chi_0x^\demi
+(Z-\Lambda_0)x^{\alpha}\chi R(\la_0)\chi_0R(z)x^\demi +x^{\alpha}\chi
R(\la_0)[\Delta_{g},\chi_0]R(z)x^\demi\right|\right|_{\mc{L}(L^2)}\leq
Cs_0^{-1}
\end{equation}
and $C$ does not depend on $s_0$. Recall that
$I_j^*, {I_j}_*$ are some isometries
$L^2(V_j)\leftrightarrow L^2(B)$ and $x_j={i_j}_*x$, then
Lemma \ref{bound:uniform} and (\ref{propcom}) lead to
\begin{equation}\label{boundrh}
\left|\left|\sum_{j=1}^MI_j^*x_j^{-\demi}[\Delta_{g_h},\chi_1^j]
R_h(\la)\chi_2^jx_j^\demi {I_j}_*\right|\right|
_{\mc{L}(L^2)}\leq C
\end{equation}
\begin{equation}\label{boundrh2}
\left|\left|\sum_{j=1}^MI_j^*x_j^\demi\chi_2^jR_h(z)x_j^\demi
(\chi_1^j{I_j}_*+x_j^{-\demi}[\Delta_{g_h},\chi_1^j]x_j^{-\demi}{I_j}_*
x^\demi R(z)x^\demi)\right|\right|_{\mc{L}(L^2)}
\leq Cs_0^{-1}
\end{equation}
and it remains to consider (\ref{badterm}), which a priori does not exist in
$\mc{O}$. When $V_j\cap V_i=\emptyset$
it is clear that $\chi_2^j{I_j}_*I_i^*\chi_2^i=0$,
so suppose that $V_j\cap V_i\not=\emptyset$. Using that the isometry
$i_i\circ i_j^{-1}:i_j(V_i\cap V_j)\to i_i(V_i\cap V_j)$ extends to an isometry $I_{ij}$
on $\hh^{n+1}$ and following the proof of
previous theorem, we can see that
\[x_j^\demi R_h(\la)\chi_2^j{I_j}_*I_i^*\chi_2^iR_h(z)x_i^\demi\]
can be expressed for $\la,z$ in the physical sheet $\{\Re(\la)>\ndemi\}$ by
\[(\chi_2^j+x_j^\demi R_h(\la)[\chi_2^j,\Delta_{g_h}]x_j^{-\demi})
\frac{x_j^\demi R_h(\la)\til{x}_j^\demi -x_j^\demi R_h(z)\til{x}_j^\demi}
{\Lambda-Z}(\til{x}_j^{-\demi}[\til{\chi}_2^i,\Delta_{g_h}]R_h(z)\til{x}_j^\demi+\til{\chi}_2^i)I_{ij}^*\]
where $\til{\chi}_2^i:=I_{ij}^*\chi_2^i$, $\til{x}_j:=I_{ij}^*x_i$.
Observe that this operator on $\hh^{n+1}$ extends to $(\la,z)\in\mc{O}$
and satisfies in $\mc{O}$
\begin{equation}\label{boundrlarz}
||x_j^\demi R_h(\la)\chi_2^j{I_j}_*I_i^*\chi_2^iR_h(z)x_i^\demi||_{\mc{L}(L^2,H^1)}
\leq Cs_0^{-1}
\end{equation}
in view of Lemma \ref{bound:uniform} and the following bound (implicitly used
in the proof of previous theorem)
\[\left|\left|\frac{x_j^\demi R_h(\la)\til{x}_j^\demi-x_j^\demi R_h(z)\til{x}_j^\demi}
{\Lambda-Z}\right|\right|_{\mc{L}(L^2,H^1)}\leq
Cs_0^{-1}\sup_{|\la-\la_0|<\frac{1}{4}}||\pl_\la x_j^\demi
R_h(\la)\til{x}_j^\demi||_{\mc{L}(L^2,H^1)}.\]
Combining (\ref{boundla0z})-(\ref{boundrlarz}) with Theorem \ref{cardvod},
one concludes that
\[||K(\la,z)||\leq C|\la-z|\]
for $(\la,z)\in \mc{O}$, hence by taking $z=\ndemi+is_0$ we see
that $1+K(\la,z)$ is holomorphically invertible for
\[\la\in\{\la\in\cc;|\la-z|<\max(C^{-1},\frac{1}{8})\}.\]
Since $C$ does not depend on $s_0$ (thus on $z$), there exists a strip $\{|\Re(\la)-\ndemi|<\eps\}$
where $K(\la,z)$ is invertible except maybe at a finite number of points.
Finally the term $K_1(\la,z)$ defined as in (\ref{paramfinal})
with our new parametrix can be studied similarly.
\qed

config: arxiv


TITLE: Why is the middle third Cantor set written as this?
QUESTION [3 upvotes]: My first question is, is the middle third Cantor set the same as the Cantor set? I've never heard it called the middle third Cantor set.
Secondly, why is this true:
"I’m going to assume that Cantor set here refers to the standard middle-thirds Cantor set $C$ described here. It can be described as the set of real numbers in $[0,1]$ having ternary expansions using only the digits $0$ and $2$, i.e., real numbers of the form $$\sum_{n=1}^\infty \frac{a_n}{3^n},$$ where each $a_n$ is either $0$ or $2$. "
I'm not truly understanding why it would be 0 or 2 for $a_n$. I understand that the numbers from [0,1] can be written in ternary as 0,1, or 2. I don't understand why the "1" is taken out. Is it because we take the middle third out every time and we lose anything that can be written in ternary as "1"?

REPLY [2 votes]: You can build several "essentially" different cantor sets, some have Lebesgue measure 0, some have positive measure. For instance, take the unit interval, remove the "middle fifth" (take a linear function to the interval $[0, 5]$ and remove the preimage of the open $(2,3)$). On step 2, remove the "middle twenty-fifth" of each of the two remaining intervals, etc. At the end you will get a nowhere dense closed set with no isolated points (so, a Cantor set), but its measure will be strictly positive.
The standard Cantor is cbuild by removing one third of each interval at each step, and yes, it relates to the '1' digit in the ternary expansion of the number.


TITLE: Can I work out the variance in batches?
QUESTION [0 upvotes]: So I have a data divided into chunks, and I can only calculate the variance in each of the chunks because of software limitations. But I want to get the variance of the whole data together, not the chunks.
I know the variance is not a linear operator.
I would like the get kind of the average of the variance but this will have to be the same number as If I calculated the variance of the whole data together. Example:
Rolling a dice in 3 groups of 2 rolls I can calculate the variance on each of the groups, so I with this data, I want to calculate the variance of the whole set: rolling a dice 6 times.
Thank you for your help.

REPLY [3 votes]: Refer to the following answer to this question:  How do I combine standard deviations of two groups?
In particular, the final formula
$$s_z^2 = \frac{(n-1) s_x^2 + (m-1) s_y^2}{n+m-1} + \frac{nm(\bar x - \bar y)^2}{(n+m)(n+m-1)}$$
illustrates how to compute the total variance of two samples, one of size $n$, sample mean $\bar x$, and sample variance $s_x^2$, and one of size $m$, sample mean $\bar y$, and sample variance $s_y^2$.  Those are the quantities you need to track.  Also note that the total sample mean is given by the formula $$\bar z = \frac{n \bar x + m \bar y}{n + m}.$$
These formulas readily lend themselves to an extended calculation for any number of groups:

Set $i = 1$.
Compute $n_i$, $\bar x_i$, and $s_{x_i}^2$, the sample size, sample mean, and sample variance of dataset $i$.
Increment $i$.
Repeat Step 2.
Use the above two formulas to compute a new $n_T$, $\bar x_T$, and $s_T^2$ representing the sample size, sample mean, and sample variance of all datasets up to set $i$.
If the last dataset was used to compute the result in step 5, stop.  Otherwise, go to step 3.

Since the original poster has claimed that the formula does not work, I will furnish a numerical example to illustrate.  This example will employ discrete data to match the scenario described in the question, but realizations from a continuous distribution can just as easily be provided.
Let $D_i$ represent dataset $i$.  Then
$$\begin{align*}
D_1 &= \{1, 1, 3, 4, 1, 5, 6, 3, 5, 5\} \\
D_2 &= \{5, 6, 2, 4, 2, 1, 1, 4, 2, 4, 4, 1, 3, 5, 6\} \\
D_3 &= \{3, 2, 6, 4, 1, 5, 2, 1, 3, 1, 5, 2, 2\} \\
D_4 &= \{5, 3, 1, 5, 1\}
\end{align*}$$
Consequently,
$$\begin{array}{|c|c|c|c|}
\hline
i & n_i & \bar x_i & s_{x_i}^2 \\
\hline
1 & 10 & \frac{17}{5} & \frac{18}{5} \\
\hline
2 & 15 & \frac{10}{3} & \frac{65}{21} \\
\hline
3 & 13 & \frac{37}{13} & \frac{73}{26} \\
\hline
4 & 5 & 3 & 4 \\
\hline
\end{array}$$
We now calculate the combined sample sizes, means, and variances of datasets $1$ through $i$:
$$\begin{array}{|c|c|c|c|}
\hline
T & n_T & \bar x_T & \bar s_T^2 \\
\hline
1 & 10 & \frac{17}{5} & \frac{18}{5} \\
\hline
2 & 25 & \frac{84}{25} & \frac{947}{300} \\
\hline
3 & 38 & \frac{121}{38} & \frac{4245}{1406} \\
\hline
4 & 43 & \frac{136}{43} & \frac{2749}{903} \\
\hline
\end{array}$$
The last row represents the total sample size, sample mean, and sample variance for the $4$ combined datasets.
Here is a sample calculation of the aggregate variance of datasets $1$ through $3$:
$$s_T^2 (T = 3) = \frac{(25 - 1)(\frac{947}{300}) + (13 - 1)(\frac{73}{26})}{25 + 13 - 1} + \frac{(25)(13)(\frac{84}{25} - \frac{37}{13})^2}{(25 + 13)(25 + 13 - 1)} = \frac{4245}{1406},$$
which matches the direct calculation based on datasets $D_1, D_2, D_3$.
Finally, Mathematica code to replicate the above computations:
d1 = {1, 1, 3, 4, 1, 5, 6, 3, 5, 5};
d2 = {5, 6, 2, 4, 2, 1, 1, 4, 2, 4, 4, 1, 3, 5, 6};
d3 = {3, 2, 6, 4, 1, 5, 2, 1, 3, 1, 5, 2, 2};
d4 = {5, 3, 1, 5, 1};

stat[x_] := {Length[x], Mean[x], Variance[x]}
data = stat /@ {d1, d2, d3, d4}
var[{n_, x_, sx_}, {m_, y_, sy_}] := {n + m, (n x + m y)/(n + m),
     ((n - 1) sx + (m - 1) sy)/(n + m - 1) + n m (x - y)^2/((n + m) (n + m - 1))}

Rest@FoldList[var[#1, #2] &, {0, 0, 0}, data]

stat[Join[d1, d2, d3, d4]]

In the future, rather than simply asserting that the formula doesn't work, it would be more polite and instructive to provide your own computations showing where you are encountering problems, so that your error can be found.


TITLE: Material Derivate of a vector field
QUESTION [0 upvotes]: Given the vector field $\vec{v}=wr\vec{i}_{\theta}-v_0\vec{i}_r$ (cylindrical-coordinates)
I'm supposed to find the local and convectiv acceleration/derivation
$$\frac{Df}{Dt}=\frac{\partial f}{\partial t}+v\cdot\nabla f$$
For the local derivation I should get $\frac{\partial \vec{v}}{\partial t}=0$?
I'm not sure on how I do the convectiv derivation, It gets messy really fast and I get confused. Do I first take $v\cdot\nabla$ or $\nabla f$? If you could show me with an example, it would be really appriciated!

REPLY [0 votes]: $$D\vec{v}=\partial_{t}\vec{v}+(\vec{v}\cdot\nabla)\vec{v}=$$
$$=\frac{\partial\vec{v}}{\partial{t}}+v_{r}\frac{\partial\vec{v}}{\partial{r}}+v_{\varphi}\frac{1}{r}\frac{\partial\vec{v}}{\partial\varphi}+v_{z}\frac{\partial\vec{v}}{\partial{z}}$$


TITLE: The nth derivative of x[A +Bf(x)]^n
QUESTION [1 upvotes]: In the course of using the Cauchy Residue Theorem for a particular problem, I have a pole of order $n$ that yields the residue
$$
R(z)= \lim_{s\rightarrow z}
\frac{(s-z)^n}{s[A+Bf(s)]^n}
$$
where $A$, $B$, and $z$ which is a zero of $[A+Bf(s)]$ are real constants, $s$ is the Laplace variable, and $f(s)$ is a known infinitely differentiable function of $s$. The indeterminate limit has to be resolved with $n$ applications of L'Hôpital's Rule which requires the $n$th derivative of the term in the denominator of the expression above.
In short therefore, I would like a compact (or at least fairly compact) expression for
$$
\frac{d^n}{dx^n}\Bigl[x\Bigl(A+Bf(x)\Bigr)^n\Bigr]
=
A^n\frac{d^n}{dx^n}\Bigl[x\Bigl(1+rf(x)\Bigr)^n\Bigr]
$$
where $r=B/A$. All of my attempts have resulted in one horrendous book-keeping mess after another. Any help will as always be deeply appreciated.
With a binomial expansion, I've managed to get it down to
$$
\frac{d^n}{dx^n}\Bigl[x\Bigl(A+Bf(x)\Bigr)^n\Bigr]
=
A^n\sum_{k=0}^{n} {n\choose k} r^{n-k}
\frac{d^n}{dx^n}\Bigl[xf(x)\Bigr]^{n-k}
$$
but then each term by term derivative inside appears to open up another monstrosity!
UPDATE: I was WAY overthinking this and thanks to user Raoul pointing it out, the Limit of a power is the power of the limit rule CONSIDERABLY simplifies my life!
https://www.mathdoubts.com/power-rule-of-limits/

REPLY [0 votes]: If $z \neq 0$, your limit is
$$
\lim_{s \to z} \frac{(s-z)^n}{(A+B f(s))^n} = \frac{1}{z} \left ( \lim_{s \to z} \frac{s-z}{A+B f(s)} \right )^n = \frac{1}{zc^n}
$$
where $c$ is the residue of $\frac{1}{A+B f(s)}$ at $z$. If $z = 0$, the limit does not exist.


TITLE: Find the coordinates of point $A'$
QUESTION [0 upvotes]: Find the coordinates of the point $A'$ which is the symmetric point of $A(3,2)$ with respect to the line $(D):2x+y-12=0$

For my trying
I find the length from the point $A(3,2)$ to the line $2x+y-12=0$. $d=\frac{|2\times 3+2-12|}{\sqrt{2^2+1^2}}=\frac{4}{\sqrt{5}}$.
Then I chose a line that through $A(3,2)$ and perpendicular to a line $(D)$.
At this point, I don't know how to do more.

Please kindly help to give me a hint or some ideas about this problem.

REPLY [1 votes]: I chose a line that through $A(3,2)$ and perpendicular to a line $(D)$.

So that's enough. Express the line in parametric equation $
\begin{cases}
x=3+2t\\
y=2+t
\end{cases}$ (we choose the directing vector $(2,1)$ to be parallel th the normal vector of $(D):\, (2,1)\cdot(x,y)-12=0$).
Then just plug $x,y$ from the parametric equation into your distance formula $\frac{|2x+y-12|}{\sqrt{5}}$ and equate to the desired distance:
$$\frac{|2x+y-12|}{\sqrt{5}}=
\frac{|2(3+2t)+(2+t)-12|}{\sqrt{5}}=\frac{4}{\sqrt{5}}$$
$$|5t-4|=4$$
$$5t-4=\pm4$$
$$t=\frac{4\pm4}{5}$$
$$\left[\begin{array}{ll}t=0&\hbox{ -- this is the point A}\\
t=\frac{8}{5}&\hbox{ -- this is not the point A, thus A'}\\\end{array}\right.$$
So substitute the $t=\frac{8}{5}$ back to get $A'=(x,y)=
\left(\frac{31}{5},\frac{18}{5}\right)$.


TITLE: Let $G$ be a group, and suppose that $M$ and $N$ are normal subgroups of $G$. Prove that the intersection of $M$ and $N$ is a subgroup of $G$.
QUESTION [0 upvotes]: How do I find out the intersection of $M$ and $N$? I know $M$ and $N$ both contain the identity of $G$.
Besides, if I prove the intersection between $M$ and $N$ is a subgroup, how could I prove it is a normal subgroup as well?

REPLY [1 votes]: First of all note that $M,N$ subgroups.So $e\in M\cap N$
$\forall x,y \in M\cap N$ we have that $x,y\in M$ and $x,y \in N$ (since $x,y\in M \cap N$)
Then we see that $xy^{-1}\in M$ and $xy^{-1} \in N$ (since $ M $ and $ N$ subgroups)
So $xy^{-1} \in M\cap N$
Thus $M \cap N$ is subgroup
And if $x\in M\cap N$ then $\forall g\in G ,gxg^{-1}\in M$ and $gxg^{-1} \in N$ (since $ M $ and $ N$ are normal subgroups)
So $gxg^{-1}\in M\cap N$
Thus $M\cap N $ is normal subgroup of $G$


TITLE: Big O comparison
QUESTION [0 upvotes]: If $T(n) = \mathcal{O}(n^3)$ Then $T(n) = \mathcal{O}(n^2)$, is this statement right?
Same question for omega, if $T(n) = \Omega(n)$ does that mean $T(n)= \Omega(n^2)$?

REPLY [0 votes]: No for both. Let $T(n) = n^3$ for the first. Then $T(n) = \mathcal{O}(n^3)$ but $T(n) \neq \mathcal{O}(n^2)$. Let $T(n) = n$ for the second. Then $T(n) = \Omega(n)$ but $T(n) \neq \Omega(n^2)$.

Let $T(n) = n^3$. Claim that $T(n) \neq \mathcal{O}(n^2)$. Since $\displaystyle \lim_{n\rightarrow\infty} \dfrac{n^3}{n^2} = \lim_{n\rightarrow\infty} n = \infty \not< \infty$, $n^3 \neq \mathcal{O}(n^2)$. Similarly, $\displaystyle \lim_{n\rightarrow\infty} \dfrac{n}{n^2} = \lim_{n\rightarrow\infty} \dfrac{1}{n} = 0 \not> 0$. So $n \neq \Omega(n^2)$.


TITLE: Find all elements $x \in \mathbb{Z}_{96}$ such that $\left<x\right>=\left<12\right>$
QUESTION [0 upvotes]: Find all elements $x \in \mathbb{Z}_{96}$ such that $\left<x\right>=\left<12\right>$.

I know the answer of this question, the answer is $\left<12,36,60,84\right>$.
However, I am confused that why the answer is not $\left<12,24,36,48,60,72,84\right>$. Can anyone help me?

REPLY [0 votes]: Hint $\: (x)\!=\!(12)$ in $\Bbb Z_{96}\!\!\iff\! (x,96)\!=\!(12)\!\! $ $ \iff\!\! ({\large \frac{x}{12}},8)\! =\! 1\!\! \iff\!\! {\large \frac{x}{12}} \equiv 1,3,5,7 \pmod{\!8}$


TITLE: Spectrum of $S$
QUESTION [1 upvotes]: Let $H$ be a Hilbert space over $\mathbb{C}$, and let $S\in B(H,H)$. Suppose
  that $a$ and $b$ are integers such that $1\leq a<b$, and that we have $S^a=S^b$.
  Show that if $\lambda\in\sigma(S)$, then $\lambda = 0$ or $\lambda^{b-a}=1$.

What I've been able to prove so far:

If $S^k=0$ for some smallest $1\leq k \leq a$, then $Sy=0$ for some
  $y=S^{k-1}x\neq 0$, so $0\in\sigma(S)$.
If $S^a\neq 0$, then $S^{b-a}(S^ax)=S^ax$ for some $y=S^ax\neq 0$, so
  $1\in\sigma(S^{b-a})=\sigma(S)^{b-a}$ by the spectral mapping theorem.
  I also found that $\|S^{b-a}\|=1$.

However, I am not able to conclude that $\sigma(S)$ doesn't contain anything else, or that $S$ and $S^{b-a}$ don't have any other eigenvalues.
Could anyone help me out?

REPLY [3 votes]: By the spectral mapping theorem,
$$
   \{0\} =\sigma(S^b-S^a)=\sigma((S^{b-a}-I)S^{a})\\
        = \{ (\lambda^{b-a}-1)\lambda^{a} : \lambda\in\sigma(A) \}.
$$
If follows that, if $\lambda\in\sigma(A)$, then either $\lambda=0$ or $\lambda^{b-a}=1$.


TITLE: Magnetic field, alternated current and EMF
QUESTION [0 upvotes]: If I apply an alternated current to a solenoid and insert into it a smaller solenoid, I could measure the induced EMF (electromotive force) and study how it changes in relation to the frequency of signal generator.
If I increase the frequency, I think that the magnetic fiel increases and so also the EMF increases, isn't it?
If I insert a metallic cylinder between the two solenoids, the cyclinder "shields" the magnetic field.. but what about EMF? And what about EMF if the cylinder is made by iron?

REPLY [1 votes]: Your question is so confused that it's difficult to answer, so I'll just point out a few things.
What you seem to be describing with a solenoid within a solenoid is a transformer.  That would be a good keyword for more research.
The magnetic field strength is linearly proportional to the current thru the driven solenoid.   Changing the frequency doesn't change this.  The instantaneous magnetic field strength depends on the instantaneous current, whether it had a different value recently or will have a different value in the near future or not.