TITLE: Formal way to evaluate this limit
QUESTION [5 upvotes]: I've some doubt about the formal way to evaluate the following limit:
$$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n^2}}$$
I know that
$$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n}}=e^{-1}$$
my calculus teacher says that we can't evaluate the limit by pieces, so we can't (in general) say that
$$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n^2}}=\lim_{n\to\ +\infty}{\left(e^{-1}\right)^{n}}=0$$ even if in this case it is the right answer.
So I'm asking a formal way to solve this, my approach is:
Say that the limit of a product is the product of the limits, so I can say that
$$\lim_{n\to\ +\infty}{\left(1-\frac{1}{n}\right)^{n^2}}=\lim_{n\to\ +\infty}{e^{n^2\ln\left(1-\frac{1}{n}\right)}}=e^{\lim_{n\to\ +\infty}{n^2\ln\left(1-\frac{1}{n}\right)}}=e^{\lim_{n\to\ +\infty}{n^2}{\lim_{n\to\ +\infty}\ln\left(1-\frac{1}{n}\right)}}=e^{\lim_{n\to\ +\infty}{-n}}=0$$
Can I do this (I've used the continuity of the exponential function) or am I still doing the limit by pieces when I've substituted $ln\left(1-\frac{1}{n}\right)$ with $-\frac{1}{n}$?
Furthermore, can I use the theorem of the product even if there is the indeterminate form $0\cdot(+\infty$)?
Thanks for your time.

REPLY [0 votes]: Late answer since it was tagged as duplicate.
An elementary way without any $e$ or logarithm could be:
\begin{eqnarray*}\left(1-\frac{1}{n}\right)^{n^2}
& = & \frac 1{\left(1+\frac{1}{n-1}\right)^{n^2}}\\
& \stackrel{\text{binomial formula}}{\leq} & \frac 1{\frac{n^2}{n-1}}\\
& < & \frac 1 n \stackrel{n\to \infty}{\longrightarrow} 0
\end{eqnarray*}


TITLE: Remove cycles from digraph without losing information (aka howto 'dagify' a digraph)
QUESTION [3 upvotes]: Given a digraph (directed graph) $G = (V, E)$, i'm looking for simply and reasonable ways of transforming $G$ into a DAG (directed acyclic graph) without "losing information".
One way of doing this is perhaps to partition $G$ into strongly connected components and consider the DAG induced on components (in a natural way...). I don't like this (for example, it would require the heavy-duty computation of the strongly connected components).
I've come up with something which goes a follows. For a positive integer $L$, define the digraph $G^{(L)} = (V, E^{(L)})$ by
$$E^{(L)} = \{(x, y) \in V^2 | \exists \text{ a path }x \rightarrow \ldots \rightarrow y \text{ of length } \ge L \}.
$$
Thus $(x, y) \in E^{(L)}$ iff it is possible to go from $x$ to $y$ in $L$ or more steps. In particular, $G^{(1)} = \bar{G}$, the transitive closure of $G$, defined by saying $(x, y)$ is an edge of $\bar{G}$ iff there is a directed path from $x$ and $y$ in $G$.
Finally, define $l(G) = \inf \{L | G^{(L)}\text{ is a DAG}\}$ and $G^{(*)} := G^{(l(G))}$, the 'dagification' of $G$.
Now, for simplicity, assume that the set of nodes $V$ is finite. The following observations are immediate:

$l(G)$ is always finite (in fact $l(G) \le |V|^2$)
$G \subseteq G^{(1)} \supseteq G^{(2)} \supseteq \ldots \supseteq G^{(l(G))}$
If $G = C_n$ is the directed cycle on $n$ nodes, then $l(G) = n$ and $G^{(*)} = \emptyset$.

Are there other interesting things which can be said ? For example,

when do we have $l(G) = 1$ ? Or equivalently, when is $\bar{G}$ a DAG ?
Can we (tightly) bound $l(G)$ in terms of $|V|$ and $|E|$ for a reasonably large / interesting family of digraphs ?
Is there a simple relationship between $G$ and $G^{(*)}$.
Is $G^{(*)}$ (conceptually) similar to any known objects in domain of markov chains ?

What do i mean by "information loss" ?
Well, I've not yet been able to precisely formulate what I mean by "information loss" in a reasonable way. However, I'll remark that squashing $C_n$ to $\emptyset$ is no loss of information since $C_n$ really contains no information except for the number $n$. Ultimately, for any digraph $G$, I'd want to transform it to a DAG in which cycles are shrunk to $\emptyset$, connected components are shrunk to single points, etc. As you see, it's not really yet clear in my mind... I hope to make this clearer very soon.

REPLY [2 votes]: I think I can answer the first two parts of your question.
$l(G)=1$ if and only if $G$ is acyclic. Indeed, as you mentioned $G^{(1)}$ is the transitive closure of $G$. Strongly connected components in $G$ correspond to complete subgraphs in the transitive closure, so $G$ is acyclic if and only if the transitive closure of $G$ is acyclic.
For a relationship between $G^{L}$ and $G^{L+1}$, the best I’ve got is that $G^{L+1}$ is a subgraph of $G^{L}$. Indeed, if a path of length $\geq L+1$ exists between teo vertices, than surely a path of length $\geq L$ exists. I’m not sure this could be considered “nice” though.
You mention in your question that $l(G)\leq n^2$(where $n$ is the order of your graph), but it seems that we can get $l(G)\leq n$ since a path that uses every vertex is length $n-1$, and thus we do not have longer paths. Thus $G^{(n)}$ is empty for any digraph $G$ on $n$ vertices.


TITLE: Figure out what to multiply a number by to get it to double after x times.
QUESTION [3 upvotes]: I'm trying to figure out if there is a way to calculate what to multiply a number by to get it to double after $x$ times. For example, the number $100$, would be $200$ after multiplying it by $m$ $7$ times. I'm not sure of a way to do this aside from trying to manually find a number, but that is tedious and inaccurate. Any help would be appreciated, thanks.

REPLY [0 votes]: Assume $x$ is the original number, which when multiplied by $m$, $y$ times, gets doubled.
$x×m^y=2x\\
m^y=2\\
y\log(m)=\log(2)\\
\log(m)=\dfrac{\log(2)}{y}\\
m=10^\left({\dfrac{log(2)}{y}}\right)$


TITLE: Find the micro-organism growth after a certain amount of minutes
QUESTION [0 upvotes]: For an experiment,
I put in a cell growing medium, 1 bacterium of type A and 3 bacteria of type B and every minute I add 1 bacterium of type A and 3 bacteria of type B. So, after 1 minute it means that my cell growing medium will contains 6 bacteria of type A and 9 bacteria of type B. Both bacteria grows at different speed.
The first bacterium of type A is growing at $A(t) = 4t+1$
The second bacterium of type B is growing at $B(t) = 2^t$
I am trying to figure the total amount of bacteria I will have after 5 minutes.
Just to sum up what I have so far:

For A:

t = 0, $[A]$
t = 1, $4*1+1=5$ then I add $1$, $6$
t = 2, $4*6+1=25$ then I add $1$, $26$
t = 3, $4*26+1=105$ then I add $1$, $126$

For B:

t = 0, $[BBB]$
t = 1, $2^3=8$ -- not valid
t = 2, can't figure that one.

I'm really curious to see the answer and how to solve this kind of problem.
Thanks

REPLY [1 votes]: Solution with new data values in mind:
You mean $A(t) = 4 A(t-1) + 1$ is how they grow, and then you add $1$, so in fact $$A(t) = 4A(t-1) + 2$$
Similarly, $$B(t) = 2^t B(t-1) + 3$$
This gives $B(1) = 9$ and $B(2) = 39$, not $2051$.
Solving the recurrence relationships gives:
$$ A(t) = \frac{5}{3} 4^t - \frac{2}{3} = \frac{1}{3} \left(5 \cdot 4^t - 2\right)$$
I have seen the explicit solution for $B(t)$ using Mathematica and it isn't pretty.  You'd be better off just looking at the recurrence equation but anyway...
$$ B(t) = 2^{\frac{1}{2} \left(t^2+t-2\right)} \left(\sum _{k=0}^{t-1} 3\ 2^{-\frac{1}{2}k(k+3)}+6\right) $$


TITLE: Derivation of equation of the straight line through two points $x_1$ and $x_2$ in $\mathbb{R}^N$ (in convex function definition).
QUESTION [2 upvotes]: I am getting started on reading convex optimization. One equation that is being used to represent traveling from one point to another in a straight line in a convex set is:
$$y = (1-\theta) x_1 + \theta x_2$$ for two points $ x_1 \neq x_2$ where  $x_1, x_2 \in \mathbb{R}^N$.
I think I have an intuitive understanding of how this is a straight line, but I am trying to derive is from the usual equation of straight line going through two points. $y = y_1 + m(x-x_1)$ where $m = \frac{y_1 - y_2}{x_1-x_2}$. Could anyone convince me how the first equation was derived and what the parameter $\theta$ means or represents?
Thanks a lot.

REPLY [1 votes]: If $x, y\in \mathbb{R}^n$, then
$$
L := \{ x + \theta(y-x) \mid \theta \in \mathbb{R} \}
$$
represents the line passing through $x$ (when $\theta = 0$) and $y$ (when $\theta = 1$) in the direction of the vector $y-x$.
Notice that $x + \theta(y-x) = (1-\theta)x + \theta y$ and
$$
L(x,y) := \{ (1-\theta)x + \theta y \mid \theta \in [0,1] \}
$$
is the line-segment containing $x$ and $y$.
This can be visualized in $\mathbb{R}^2$ using the parallelogram law:


TITLE: Given the adjoint $\mathrm{adj}(A)$, how do you find $\det(A)$ and $A^{-1}$?
QUESTION [1 upvotes]: Given $\mathrm{adj}(A)$ where $A$ is an $n\times n$ matrix, how do you find the value of $\det(A)$ and $A^{-1}$?

REPLY [1 votes]: Hint
$$
\left( \begin{array}{ccc}
\text{det}(A) &  &  0\\
 & \ddots &  \\
0 &  & \text{det}(A)
\end{array} \right)=\text{det}(A) \cdot \text{I}_n = A \cdot \text{adj}(A)
$$
and
$$
A^{-1}= \frac{1}{\text{det}(A)}\text{adj}(A)
$$


TITLE: Characteristic Function of Multivariate Normal
QUESTION [0 upvotes]: I am reading through the lecture notes of a statistics class, and it describes several steps on how to transfer the characteristic function of a standard normal into the characteristic function of a multivariate normal. Most of the steps I have no problem with, but there is one that seems slightly too big a leap. It says:
$$X\in N(0,I) \rightarrow \phi_X(t)=\exp\left(-\frac{1}{2}t^Tt\right)$$ which is fair enough, but then immediately jumps to:
$$X\in N(\mu,\Sigma) \rightarrow \phi_X(t)=\exp\left(it^T\mu-\frac{1}{2}t^T\Sigma t\right)$$
I can see how this makes sense given the characteristic function of the normal distribution, but would this step not have to be substantiated a little bit more? Or is there some property that I am missing, that allows for this step to be easily made?

REPLY [1 votes]: If $X\sim\mathsf{Norm}(\mu,\Sigma)$ then a matrix $A$ exists such that $X=\mu+AU$ where $U\sim\mathsf{Norm}(0,I)$  and $\Sigma=AA^T$.
Then: $$\phi_X(t)=\mathbb Ee^{it^T(\mu+AU)}=e^{it^T\mu}\mathbb Ee^{it^TAU}=e^{it^T\mu}\phi_{U}(A^Tt)=e^{it^T\mu}e^{-\frac12t^TAA^Tt}=e^{it^T\mu-\frac12t^T\Sigma t}$$


TITLE: Given that $(g \circ f)$ is invertible, can we conclude that $f$ and $g$ are invertible?
QUESTION [0 upvotes]: Given that $(g \circ f)$ is invertible, can we conclude that $f$ and $g$ are invertible?
I have previously proved that if $f$ and $g$ are invertible, then $g \circ f$ is invertible, however I am not sure if the converse of this statement is true.

REPLY [1 votes]: The simple example by Nitrogen already tells you the answer is no. But you can say some things: if $g \circ f$ is injective, $f$ is injective: otherwise $x \neq y$ in the domain of $f$ exist with $f(x) = f(y)$ but then also $(g \circ f)(x) = g(f(x)) = g(f(y)) = (g \circ f)(y)$ contradicting the injectivity of $g \circ f$.
Also, if $g \circ f$ is surjective, so is $g$: let $z$ be in the codomain of $g$ (which is also the codomain of $g \circ f$). Some $x$ in the domain of $f$ exists such that $(g \circ f)(x) = z = g(f(x))$. But then $f(x)$ witnesses that $g$ assumes $z$ as a value (and $f(x)$ is in the domain of $g$).
So if $g \circ f$ is bijective, $f$ is injective and $g$ is surjective, which is true in the simple counterexample with $x \in X, |X| \ge 2$: $f: \{x\} \rightarrow X, f(x) = x$ and $g: X \rightarrow \{x\}, \forall p \in X g(p) = x$.


TITLE: what is wrong with my derivative?!!!
QUESTION [0 upvotes]: I have a question which want to derivative $\left(\frac{3x-2}{5x}\right)$
what I get is $\left(\frac{3\left(5x\right)-5\left(3x-2\right)}{\left(5x\right)^2}\right)$
then $\left(\frac{15x-15x+10}{\left(5x\right)^2}\right)$
and finally
$\left(\frac{10}{\left(5x\right)^2}\right)$
BUT the correct answer is $\frac{2}{5x^2}$
how it's come?!!

REPLY [7 votes]: You are not wrong because $$\frac{10}{(5x)^2} = \frac{2}{5x^2} $$


TITLE: Where does entropy-increasing waste heat go during non-reversible processes?
QUESTION [1 upvotes]: The first law of thermodynamics states that energy is conserved. $\Delta U = Q - W$ (internal energy change is the difference between external energy supplied and work done on the environment).
The second law of thermodynamics states that entropy is never decreasing and is only constant in reversible thermodynamic processes. In a non-reversible thermodynamic process, the entropy of the universe is increased.
Suppose we have an abstract 1-dimensional cellular automata universe consisting of 0s (empty space) and 1s (energy/atoms) but obeying the first and second laws of thermodynamics. The 1s are drawn together via "gravity" forces:
t0: 0010000100
t1: 0001001000
t2: 0000110000

Unlike microscopic time-reversible phenomena, let us assume that the collision of the two 1s generates entropy.
Q1: Would the final state of this universe be cycling through all possible $10 \choose 2$ states, as that is the maximum entropy configuration?
Q2: If the answer to Q1 is yes, then that implies a set of dynamics that encourage an equilibrium distribution of two 1s over the 10 position. If we have any less or more than 2 1s, wouldn't that violate conservation of energy? (the total energy of the starting system is 2).
Q3: Where does entropy "go" when it is produced? In this toy universe I have no way to embody any other non-zero energy than "1". If we choose to introduce a new letter "H" as an "entropy" particle, does this violate the first law of thermodynamics? (since the new energy of the system is 1 + 1 + H). Does entropy occupy space?
t1: 0001001000
t2: 000011H000

Or is it a "hidden state" a cell that follows energy around?
The bigger question here is that when I think of living things perform some thermodynamically irreversible work to lower their entropy, I am wondering where that entropy "goes".  If entropy is created but mass and energy are conserved, how do we end up with heat "for free" without changing total energy?
t1: 0001001000
h1: 0000000000

t2: 0000110000
h2: 00000H0000

Q4: Are random dynamics required here in order to eventually reach a state of maximum entropy? One way to model this would be to treat H as "non-useful work" that fills up the universe and potentially creates more H when it interacts with non-H cells.
Q5: Are there any limitations to this 1D universe in my understanding of how the first and second law of thermodynamics works?

REPLY [0 votes]: This is an answer to the title of the question

Where does entropy-increasing waste heat go during non-reversible processes?

The "waste heat " goes to black body radiation. This radiation is only modeled correctly with quantum mechanics.
A lot of the energy will go into the kinetic energy of the particles in the medium studied, raising the temperature.
The simple model you have in the content seems to me to be reversible.


text:
\begin{document}
\def\C{{\mathbb C}}
\def\R{{\mathbb R}}
\def\N{{\mathbb N}}
\def\Z{{\mathbb Z}}
\def\T{{\mathbb T}}
\def\Q{{\mathbb Q}}
\def\SP{{\mathbb S}}
\def\d{{\partial}}
\def\mc{{\mathcal H}}
\def\1b{{\mathbb I}}
\def\tr{{\rm tr}\:}
\def\pv{\partial_x V}

\title[Spectral problems]
{ Spectral problems for operators with crossed magnetic and
electric fields}
\author[M. Dimassi]{Mouez Dimassi}
\author[V. Petkov]{Vesselin Petkov}

\address {D\'epartement de Math\'ematiques,
Universit\'e Paris 13, Avenue J.-B. Clement, 93430 Villetaneuse,
France}
\email{dimassi@math.univ-paris13.fr}
\address {Universit\'e Bordeaux I, Institut de Math\'ematiques de Bordeaux,  351, Cours de la Lib\'eration, 33405  Talence, France}
\email{petkov@math.u-bordeaux1.fr}
\thanks{The second author was partially supported by the ANR project NONAa}
\thanks{2000 {\it Mathematics Subject Classification:} Primary 35P25; Secondary 35Q40}
\maketitle

\hspace{5 cm} {\it In memory of Pierre Duclos}\\

\begin{abstract} We obtain a representation formula for the derivative of the spectral shift function $\xi(\lambda; B, \epsilon)$ related to the operators $H_0(B,\epsilon) = (D_x - By)^2 + D_y^2 + \epsilon  x$ and $H(B, \epsilon) = H_0(B, \epsilon) + V(x,y), \: B > 0, \epsilon > 0$.  We prove that the operator $H(B, \epsilon)$ has at most a finite number of embedded eigenvalues on $\R$ which is a step to the proof of the conjecture of absence of embedded eigenvalues of $H$ in $\R.$ Applying the formula for $\xi'(\lambda, B, \epsilon)$, we obtain a semiclassical asymptotics of the spectral shift function related to the operators $H_0(h) = (hD_x - By)^2 + h^2D_y^2 + \epsilon x$ and $H(h) = H_0(h) + V(x,y).$

\end{abstract}

\section{Introduction}
\renewcommand{\theequation}{\arabic{section}.\arabic{equation}}
\setcounter{equation}{0}

Consider the two-dimensional Schr\"odinger operator with
homogeneous magnetic and electric fields

$$H = H(B, \epsilon) = H_0(B, \epsilon) + V(x, y),\: D_x = -\ii \partial_x,\: D_y = -\ii \partial_y,$$
 where
$$H_0 = H_0(B, \epsilon) = (D_x - By)^2 + D_y^2 + \epsilon x.$$
Here $B > 0$ and $\epsilon > 0$ are proportional to the strength
of the homogeneous magnetic and electric fields and $V(x, y)$ is a
$L^\infty(\R^2)$ real valued function satisfying the estimates
\begin{equation}\label{eq:1.1}
| V(x, y)|  \leq C(1 +
|x|)^{-2 -\delta }(1 + |y|)^{-1- \delta },
\delta > 0.
\end{equation}
For $\epsilon \not= 0$ we have $\sigma_{\rm ess}(H_0(B, \epsilon))
= \sigma_{\rm ess} (H(B, \epsilon)) = \R$. On the other hand, for decreasing potentials $V$ it is possible to have embedded eigenvalues $\lambda \in \R$
and this situation is quite different from that with $\epsilon =
0$ when the spectrum of $H(B, 0)$ is formed by eigenvalues with
finite multiplicities which may accumulate only to Landau levels
$\lambda_n = (2n +1)B,\: n \in \N$ (see \cite{I},
\cite{MR}, \cite{RW} and the references cited there). The analysis of the spectral
properties of $H$ and the existence of resonances have been
studied in \cite{FK1}, \cite{FK2}, \cite{DP2} under the assumption
that $V(x,y)$ admits a holomorphic extension in the $x$- variable
into a domain
$$\Gamma_{\delta_0} = \{z \in \C:\: 0 \leq |\Im z| \leq \delta_0\}.$$
On the other hand, without any assumption on the analyticity of $V(x,y)$,  it was proved in \cite{DP2} that the operator $(H - z)^{-1} - (H_0 - z)^{-1}$ for $z \in \C,\: \Im z \neq 0,$ is trace class. Thus, following the general setup \cite{K}, \cite{Ya},  we may define the spectral shift function $\xi(\lambda)=
\xi(\lambda; B, \epsilon)$ related to $H_0(B, \epsilon)$ and $H(B,
\epsilon)$ by
$$\la \xi', f \ra = {\rm tr} \Bigl(f(H) - f(H_0)\Bigr),\: f \in C_0^{\infty}(\R).$$
By this formula $\xi(\lambda)$ is defined modulo a constant but for the analysis of the derivative $\xi'(\lambda)$ this is not important. For the analysis of the behavior of $\xi(\lambda; B, \epsilon)$ it is important to have
 a representation of the derivative $\xi'(\lambda; B, \epsilon)$. Such representation has been obtained in \cite{DP2} for strong magnetic fields $B \to +\infty$ under the  assumption that $V(x,y)$ admits an analytic continuation in $x$-direction.\\

In this paper we consider the operator $H$ without {\it any
assumption} on the analytic continuation of $V(x,y)$ and without
the {\it restriction} $B \to +\infty$. For such potentials we cannot use the techniques in \cite{FK1}, \cite{FK2} and \cite{DP2} related to the resonances of the perturbed problem.
    Our purpose is to study $\xi'(\lambda; B, \epsilon)$ and the
existence of embedded eigenvalues of $H$.  The key point in this direction is the following

\begin{thm} Let $V, \partial_xV\in L^\infty(\R^2;\R)$ and  assume that $(1.1)$ holds for $V$ and $\partial_xV$.
 Then for every $f \in C_0^{\infty}(\R)$ and $\epsilon \not= 0$ we have
\begin{equation} \label{eq:1.2}
\tr\: \Bigl(f(H) - f(H_0)\Bigr) = - \frac{1}{\epsilon} \tr\:
\Bigl((\d_x V) f(H)\Bigr).
\end{equation}
\end{thm}
Notice that in (\ref{eq:1.2}) by $\partial_x V$ we mean the operator of multiplication by $\partial_x V$.
The formula (\ref{eq:1.2}) has been proved by D. Robert and X. P. Wang
\cite{RW2} for Stark Hamiltonians in absence of magnetic field ($B = 0$). In fact, the
result in \cite{RW2} says that
\begin{equation} \label{eq:1.3}
\xi'(\lambda; 0, \epsilon) = -\frac{1}{\epsilon}\int_{\R^2} \partial_x V(x, y)
\frac{\partial e}{\partial \lambda}(x,y, x, y; \lambda, 0,
\epsilon) dxdy,
\end{equation}
where $e(., .; \lambda, 0, \epsilon)$ is the spectral function of
$H(0, \epsilon).$
On the other hand, the spectral shift function in \cite{RW2} is related to the trace of the {\it time delay} operator $T(\lambda)$ defined via the corresponding scattering matrix $S(\lambda)$ (see \cite{RW1}). The presence of magnetic filed $B \not= 0$ and Stark potential lead to some serious difficulties to follow this way. Recently, Theorem 1 has been established by the authors in \cite{DP3} but the proof in \cite{DP3} is technical, long and based on the trace class properties of the operators
\begin{equation} \label{eq:1.4}
\psi (H \pm \ii)^{-N}, \: \partial_x \circ \psi (H \pm \ii)^{-N},\: (H \pm \ii)\partial_x \circ \psi (H \pm \ii)^{-N -2}
\end{equation}
 with $\psi \in C_0^{\infty}(\R)$ and $N \geq 2.$ The idea is to use the commutators with the operators $\chi_R \partial_x$, where $\chi_R(x, y) = \chi\Bigl(\frac{x}{R}, \frac{y}{R}\Bigr)$ and $\chi \in C_0^{\infty}(\R^2)$ is a cut-off such that $\chi =1$ for $|(x, y)| \leq 1.$ One shows that
\begin{equation} \label{eq:1.5}
 \tr\Bigl([\chi_R \partial_x, H] f(H) - [\chi_R \partial_x, H_0] f(H_0)\Bigr) = 0
\end{equation}
and next we are going to examine the limit $R \to \infty$ of the trace of the operators in (\ref{eq:1.5}). The commutators with $\partial_x$ and the presence of magnetic field lead to operators involving $D_x - By$ and this is one of the main difference with the case $B = 0.$ To overcome this difficulty we used in \cite{DP3} the trace class operators (\ref{eq:1.4}) which led to technical problems. On the other hand, the operator $\partial_x$ is often used for operators with Stark potential $\epsilon x$ and this influenced our approach in \cite{DP3}. One of the goal of this work is to present a new shorter and elegant proof of Theorem 1. The new idea is to apply the shift operator $U_{\tau}: f(x, y) \longrightarrow f(x + \tau, y)$ instead of $\partial_x$. In Proposition 1 we show that
$$\tr \Bigl([U_{\tau}, H]f(H) - [U_{\tau}, H]f(H_0)\Bigr) = 0.$$
The proof of the later equality is much easier than that of (\ref{eq:1.5}) and we don't need the trace class properties of the operators (\ref{eq:1.4}). Moreover, applying the operator $U_{\tau}$, we may generalize the result of Theorem 1 for Schr\"odinger operators $(D_x - C(y))^2 + D_y^2 + \epsilon x + V(x, y)$ with variable magnetic filed as well as for operators with magnetic potentials in $\R^n, n \geq 3.$

The second question examined in this work is the existence of
embedded real eigenvalues of $H$. In the physical literature one conjectures that for
$\epsilon \not= 0$ there are no  embedded eigenvalues. We established in \cite{DP3} a weaker result saying that in every interval $[a, b]$ we may have at most a finite number of embedded eigenvalues
with finite multiplicities. Under the assumption for analytic
continuation of $V$ it was proved in \cite{FK1} that in some finite interval $[\alpha(B, \epsilon), \beta(B, \epsilon)]$ there are no resonances $z$ of $H(B, \epsilon)$ with $\Re z \notin
[\alpha(B, \epsilon), \beta(B, \epsilon)]$. Since the real resonances $z$ coincide with the eigenvalues of $H(B, \epsilon)$, we obtain some information for the embedded eigenvalues. We prove in Section 3  without the condition of analytic continuation of $V(x, y)$ that $H$ has no embedded eigenvalues outside an interval  $[\alpha(B, \epsilon), \beta(B, \epsilon)]$. Combining this with the result in \cite{DP3}, we conclude that $H$ has at most a finite number of embedded eigenvalues. Finally, applying the representation formula for the derivative of the spectral shift function $\xi_h(\lambda) = \xi_h(\lambda, B, \epsilon)$ related to the operators $H_0(h) = (hD_x - By)^2 + h^2D_y^2 + \epsilon x$ and $H(h) = H_0(h) + V(x, y)$, we obtain a semiclassical  asymptotics of $\xi_h(\lambda)$ as $h \searrow 0$ uniformly with respect to $\lambda \in [E_0, E_1]$ under some assumptions on the critical values of the symbol of $H(h).$

\section{Representation of the spectral shift function}
\renewcommand{\theequation}{\arabic{section}.\arabic{equation}}
\setcounter{equation}{0}
\def\uh{U_{\tau}}

We suppose without loss of generality that $B = \epsilon = 1.$ Set $\la z \ra = (1 + |z|^2)^{1/2}.$ For reader convenience we recall the following lemma proved in \cite{DP3}
\begin{lem}
Let $\delta>0$ and let $k_j(x,y)=\la x\ra^{-j(1+\delta)}\la y\ra^{-j({1\over 2}+\delta)}, j=1,2$. The operators
$G_2:=k_2(H_0+\ii)^{-2},\,\,G_2^*$,
$($resp. $G_1:=k_1(H_0+\ii)^{-1},\,\,\, G_1^*)$, are trace class $($resp. Hilbert-Schmidt$)$.
\end{lem}
As an application of Lemma 1 recall that Proposition 1 in \cite{DP3} says that for $g \in C_0^{\infty}(\R)$ the operators $V g(H)$ and $V g(H_0)$ are trace class. Obviously, the same is true for $V(x + \tau, y)g(H)$ and we will use this fact below.
Consider the shift operator
$$\uh: f(x, y) \longrightarrow f(x + \tau, y).$$
 Let $H_0 = (D_x - y)^2 + D_y + x, H = H_0 + V(x, y).$ It is clear that
$$[\uh, H_0] u = \uh H_0 u - H_0 \uh u = \uh (xu) - x\uh u = \tau\uh u,$$
hence $[\uh, H_0] = \tau \uh.$ Next
$$[\uh, V] = \uh(Vu) - V \uh u = V(x + \tau)\uh u - V \uh u = \Bigl(V(x+\tau, y) - V(x, y)\Bigr)\uh u.$$
Thus given a function $f \in C_0^{\infty}(\R)$, we get
$$[\uh, H] f(H) - [\uh, H_0] f(H_0) = \Bigl[ \tau + (V(x +\tau, y) - V(x, y))\Bigr] \uh f(H) - \tau \uh f(H_0)$$
$$= \tau \uh \Bigl(f(H) - f(H_0)\Bigr) + \Bigl(V(x +\tau, y) - V(x, y)\Bigr) \uh f(H).$$
\begin{prop} We have the equality
\begin{equation} \label{eq:2.1}
{\rm tr} \Bigl( [\uh, H] f(H) - [\uh, H_0] f(H_0)\Bigr) = 0.
\end{equation}

\end{prop}
\begin{proof} We write
$${\rm tr}\: \Bigl[ \uh H f(H) - \uh H_0 f(H_0) + H_0 \uh f(H_0) - H \uh f(H)\Bigr]$$
$$ = {\rm tr}\: \uh \Bigl(H f(H) - H_0 f(H_0)\Bigr) + {\rm tr} \Bigl( H_0 \uh f(H_0) - H \uh f(H)\Bigr) = (I) + (II).$$
For the term $(I)$, by using the cyclicity of the trace, we have
\begin{equation} \label{eq:2.2}
(I) = {\rm tr}\: \Bigl((H f(H) - H_0 f(H_0)) \uh\Bigr) = {\rm tr}\: \Bigl(f(H) H - f(H_0) H_0\Bigr) \uh.
\end{equation}
On the other hand,
$$(II) = {\rm tr}\: \Bigl((H_0- H) \uh f(H_0) \Bigr) + {\rm tr}\: \Bigl[ H \uh \Bigl(f(H_0) - f(H)\Bigr) \Bigr] = (II_1) + (II_2).$$
and we justify below the trace class properties of the operators $(II_1)$ and $(II_2)$.
For $(II_1)$ we write
$$- (II_1) = V \uh f(H_0) = \uh [ \uh^{-1} V \uh ] f(H_0) = \uh V(x - \tau, y) f(H_0)$$
and the operator on the right hand side is trace class.

It is easy to see that the operator $\Bigl(f(H_0) - f(H)\Bigr)(H + \ii)$ is trace class since
$$\Bigl(f(H_0) - f(H)\Bigr)(H + \ii)= \Bigl[f(H_0)(H_0 + \ii) - f(H)(H + \ii)\Bigr] + f(H_0) V,$$
where on the right hand side we have a sum of two trace class operators. The same argument shows that the operator $H (f(H_0) - f(H))$ is trace class. Next we show that the operator $H (f(H_0) - f(H)) (H + \ii)$ is trace class.
To do this, we write
$$H (f(H_0) - f(H)) (H + \ii) = \Bigl(H_0 f(H_0) (H_0 + \ii) - H f(H)(H + \ii)\Bigr) + V f(H_0) (H_0 + \ii)$$
$$+ V f(H_0) V + H_0 f(H_0) V$$
and the four operators on the right hand side are trace class. This implies that
$H\uh (f(H_0) - f(H))(H + \ii)$ is trace class since the commutator $[H , \uh]$ is a bounded operator.
After these preparations we write
$$(II_2) = H \uh (f(H_0) - f(H)) = \uh H (f(H_0) - f(H)) + [H, \uh] (f(H_0) - f(H))$$
which obviously is trace class. Exploiting the trace class properties, we can write
$$(II_2) = {\rm tr}\: \Bigl[ H \uh (f(H_0) - f(H)(H + \ii)(H + \ii)^{-1}\Bigr]$$
$$ = {\rm tr}\: \Bigl[ (H + \ii)^{-1}H \uh (f(H_0) - f(H))(H + \ii)\Bigr]$$
$$ = {\rm tr}\: \Bigl((f(H_0) - f(H))(H+ \ii)(H + \ii)^{-1} H \uh\Bigr) = {\rm tr}\: \Bigl((f(H_0) - f(H)) H \uh\Bigr).$$
Combining the above expressions, we get
$$(I) + (II_1) + (II_2) = {\rm tr}\: \Bigl((H_0 - H) \uh f(H_0)\Bigr) + {\rm tr}\: \Bigl(f(H_0)(H - H_0) \uh\Bigr)$$
$$= {\rm tr}\: \Bigl(-V \uh f(H_0)\Bigr) + {\rm tr}\: \Bigl(\uh f(H_0) V\Bigr).$$
It remains to show that ${\rm tr}\: \Bigl(V \uh f(H_0)\Bigr) = {\rm tr}\:\Bigl(\uh f(H_0) V \Bigr).$ To do this, choose a function $\chi \in C_0^{\infty}(\R^2)$ such that $\chi = 1$ for $|(x,y)|\leq 1$. For
$R > 0$ set
$$\chi_R(x, y) = \chi\Bigl(\frac{x}{R}, \frac{y}{R}\Bigr)$$
and consider
$${\rm tr}\:\Bigl(V \uh f(H_0) \chi_R\Bigr) = {\rm tr}\: \Bigl(\uh f(H_0) V \chi_R\Bigr).$$
The operator $\chi_R$ converges strongly to identity as $ R \to \infty$ and applying the well known property of trace class operators (see for instance, Proposition 1 in \cite{DP3}), we conclude that
$${\rm tr}\:\Bigl(V \uh f(H_0)\Bigr) = {\rm tr}\: \Bigl(\uh f(H_0) V \Bigr)$$
and the proof is complete.
\end{proof}
{\bf Proof of Theorem 1.} According to Proposition 1, we have
\begin{equation} \label{eq:2.3}
{\rm tr}\:\Bigl(\uh (f(H) - f(H_0)\Bigr) = - {\rm tr}\: \Bigl(\frac{V(x + \tau, y) - V(x, y)}{\tau} \uh f(H)\Bigr).
\end{equation}
We take the limit $ \tau \to 0$ and observe that
$$\uh \longrightarrow I,\: \frac{V(x + \tau, y) - V(x, y)}{\tau} \uh  \longrightarrow \partial_x V$$
 strongly. Since $(f(H) - f(H_0))$ is a trace class operator, applying once more the property of trace class operators, we get
$$\lim_{\tau \to 0} {\rm tr}\: \Bigl(\uh (f(H) - f(H_0)\Bigr) = {\rm tr}\: (f(H) - f(H_0)).$$
To treat the limit $\tau \to 0$ in the right hand term of (\ref{eq:2.3}), consider the function,
$$g_{\delta}(x,y) = \la x \ra ^{-2-\delta} \la y \ra ^{-1-\delta}$$
 $\delta > 0$ being the constant of (1.1). Following Lemma 1, the operator $g_{\delta}(H_0 + \ii)^{-2}$ is trace class. Hence
$$g_{\delta} f(H) = g_{\delta} (f(H) - f(H_0)) + g_{\delta}(H_0 + \ii)^{-2} (H_0 + \ii)^2 f(H_0)$$
is also a trace class operator.

 To treat the limit $\tau \to 0,$ we use the representation
$$\Bigl(\frac{V(x + \tau, y) - V(x, y)}{\tau} g_{\delta}^{-1}\Bigr) \Bigl[ g_{\delta}\uh g_{\delta}^{-1}] g_{\delta} f(H).$$
The operators in the brackets $\Bigl(...\Bigr)$, $[...]$ converge strongly as $\tau \to 0$ to $(\partial_x V) g_{\delta}^{-1}$ and $I$, respectively. Letting $\tau \to 0$, we obtain
$$\lim_{\tau \to 0} {\rm tr}\: \Bigl(\frac{V(x + \tau, y) - V(x, y)}{\tau}\Bigr) \uh f(H) = {\rm tr}\: \Bigl((\partial_x V) f(H)\Bigr)$$
 and the proof is complete.
\begin{rem} The proof of Theorem $1$ works for operators $M =(D_x - C(y))^2 + D_y^2 + \epsilon x + V(x, y)$ with non-linear $C(y)$ assuming that we have an analog of Lemma $1$ for $H$ and $H_0$ replaced by $M$ and $M_0 = (D_x - C(y))^2 + D_y^2 + \epsilon x,$ respectively. Also we may examine the operators in $\R^3$ having the form
$$\Bigl(D_x + \frac{B}{2}y\Bigr)^2 + \Bigl(D_y - \frac{B}{2}x\Bigr)^2 + D_z^2 + \epsilon z + V(x, y, z)$$
applying the shift operator $U_{\tau}: \:f(x, y, z) \longrightarrow f(x, y , z + \tau).$ Some operators with magnetic potentials and Stark potential in $\R^n,\: n \geq 3,$ can be investigated by the same approach.

\end{rem}

Now consider the operators $H_0(h) = (hD_x - By)^2 + h^2D_y^2 + \epsilon x,\: H(h) = H_0(h) + V(x, y),\: h > 0.$
Under the assumption (1.1) for $V(x, y)$ we have the statement of Lemma 1 for $H_0$ replaced by $H_0(h)$. Moreover, the operators $Vg(H(h))$ and $Vg(H_0(h))$ are trace class for every $g \in C_0^{\infty}(\R).$ Thus for every $f \in C_0^{\infty}(\R)$ the operator $f(H(h)) - f(H_0(h))$ is trace class and we can define the spectral shift function $\xi_h = \xi_h(\lambda, B , \epsilon)$ modulo constant by the formula
$$\langle \xi_h', f \rangle = {\rm tr}\: \Bigl(f(H(h) - f(H_0)\Bigr),\: f \in C_0^{\infty}(\R).$$
Under the assumption of Theorem 1, we obtain repeating the proof of (1.2) the representation
\begin{equation} \label{eq:2.4}
{\rm tr}\Bigl(f(H(h)) - f(H_0(h))\Bigr) = -\frac{1}{\epsilon} {\rm tr} \Bigr((\partial_x V) f(H(h))\Bigr).
\end{equation}

\section{ Embedded eigenvalues of $H$}

In this section we use the notation
$$L = H(0) = (D_x - By)^2 + D_y^2 + \epsilon x.$$

 Our purpose is to prove the following

\begin{thm}
There exists $C > 0$ such that $H$ has no eigenvalues $\lambda,\: |\lambda| \geq C.$
\end{thm}

\begin{proof}
First notice that for every function $f \in C_0^{\infty}(\R)$ we have

\begin{equation}\label{eq:3.1}
f(H)[\partial_x,H]f(H)= \epsilon f^2(H) + f(H)\partial_{x}V f(H).
\end{equation}
We will show the absence of embedded eigenvalues $\lambda > C > 0$. The case $\lambda < - C$ can be treated by the same argument. Assume that there exists a sequence of eigenvalues $\lambda_n \longrightarrow + \infty,\: \lambda_{n+1} > \lambda_n + 1,\: \forall n$ and let $H\varphi_n = \lambda_n \varphi_n,\: n \in \N$ with $(\varphi_i, \varphi_j) = \delta_{i,j}.$ Choose cut-off functions $f_n(t) \in C_0^{\infty}(\R)$ so that $f_n(\lambda_n) = 1,\: 0 \leq f_n(t) \leq 1$ and $f_n(t) = 0$ for $|t - \lambda_n| \geq 1/2$. It is clear that $f_n(H) \varphi_n = \varphi_n$ and
$$(\varphi_n, f_n(H)[\partial_x, H]f_n(H)\varphi_n) = 0,\: \forall n \in\N.$$
We wish to prove that for $n$ large enough we have
\begin{equation} \label{eq:3.2}
\Bigl|(\varphi_n, f_n(H) \pv f_n(H) \varphi_n)\Bigr| = \Bigl|(\varphi_n, \pv f_n(H) \varphi_n)\Bigr| \leq  \epsilon/2
\end{equation}
which leads to a contradiction with (\ref{eq:3.1}) since $(\varphi_n, f^2_n(H) \varphi_n) = 1.$ Consider the operator
$$ f_n(H) = -\frac{1}{\pi}\int_{W_n}\bar{\partial} \tilde{f}_n(z)(z - H)^{-1} L(dz),$$
where $\tilde{f}_n(z)$ is an almost analytic continuation of $f_n$ with supp $\tilde{f}_n(z) \subset W_n$, $W_n = \{z \in \C:\: |z - \lambda_n| \leq 2/3\}$ is a complex neighborhood of $\lambda_n$ and
$$\bar{\partial}\tilde{f}_n(z) = {\mathcal O}(|\Im z|^{\infty})$$
 uniformly with respect to $n$. Here $L(dz)$ is the Lebesgue measure in $\C.$ We write
$$(\varphi_n, \pv f_n(H)\varphi_n) = -\frac{1}{\pi}  \int_{W_n\cap \{|\Im z| \leq \eta\}}  \bar{\partial}\tilde{f}_n(z)(\varphi_n, (\pv) (z - H)^{-1}\varphi_n) L(dz)$$
$$  -\frac{1}{\pi}  \int_{W_n\cap \{|\Im z| > \eta\}}  \bar{\partial}\tilde{f}_n(z)(\varphi_n, (\pv - V_0)(z - H)^{-1}\varphi_n) L(dz) $$
$$ -\frac{1}{\pi}  \int_{W_n\cap \{|\Im z| > \eta\}}  \bar{\partial}\tilde{f}_n(z)(\varphi_n,  V_0(z - H)^{-1}\varphi_n) L(dz) = R_n + Q_n + S_n,$$
where $V_0(x, y) \in C_0^{\infty}(\R^2).$
We choose $\eta > 0$ small enough to arrange $|R_n| \leq \epsilon/6$ for all $n \in \N.$ Next we fix $0 < \eta < 1$ and we will estimate $Q_n$  and $S_n.$
For the resolvent $(z - L)^{-1}$ we will exploit the following
\begin{prop} $($\cite{FK2}$)$  Let $f,\: g$ be bounded functions with compact support in $\R^2$. Then for every compact ${\mathcal K} \subset \R \setminus \{0\}$ we have
$$\lim_{\lambda \to \pm \infty}  \|f (\lambda + \ii \gamma - L)^{-1} g\| = 0$$
uniformly for $\gamma \in {\mathcal K}.$
\end{prop}
We choose $V_0$ so that $\|\pv - V_0\|$ is sufficiently small in order to arrange $|Q_n| \leq \epsilon/6,\: \forall n \in \N.$
Now we pass to the estimation of $S_n.$ We have
\begin{equation} \label{eq:3.3}
V_0(z - H)^{-1} = V_0(z - L)^{-1} + V_0(z - L)^{-1}(V - V_1) (z - H)^{-1} + V_0(z - L)^{-1} V_1 (z - H)^{-1}.
\end{equation}
 We replace  $V_0(z - H)^{-1}$ in $S_n$ by the right hand side (\ref{eq:3.3}) choosing $V_1 \in C_0^{\infty}(\R^2).$
For the term involving $(V - V_1)$ in (\ref{eq:3.3}) we take $V_1$ so that $\|V - V_1\|$ is small enough, to obtain a term bounded by $\epsilon/18.$ Next we fix the potentials $V_0,\: V_1$ with compact support. By Proposition 2 setting $z = \lambda + \ii\gamma,\: \eta \leq |\gamma| \leq 1,$ we get
$$\|\bar{\partial}\tilde{f}_n(z)V_0( \lambda + \ii \gamma - L)^{-1} V_1 (H - z)^{-1}\|\leq C_2 \eta^{-1}\|V_0(\lambda + \ii \gamma - L)^{-1} V_1\| \leq \frac{9}{4 \pi^2}\frac{\epsilon}{18}$$
for $\Re z = \lambda \geq C_{\epsilon, \eta}.$ We choose $n \geq n_0 = n_0(\epsilon, \eta)$, so that $\Re z \geq C_{\epsilon, \eta}$ for $z \in W_n$ and $n \geq n_0$. Thus we can estimate the term involving $V_0(z - L)^{-1} V_1$ in (\ref{eq:3.3}) by $\epsilon/18.$ It remains to deal with the term containing $V_0 (z - L)^{-1}.$ Let $\psi(x, y) \in C_0^{\infty}(\R^2)$ be a cut-off function such that $\psi = 1$ on the support of $V_0$. We write
$$\psi V_0 (z - L)^{-1} = V_0 (z - L)^{-1} \psi - V_0 (z - L)^{-1} [(D_x - By)^2 + D_y^2, \psi] (z -L)^{-1}$$
$$ = V_0 (z - L)^{-1} \psi - V_0 (z - L)^{-1} \psi_1 [(D_x - By)^2 + D_y^2, \psi] (z -L)^{-1},$$
where $\psi_1$ is a cut-off function equal to 1 on the support of $\psi.$ For $n$ large enough we will have $\Re z = \lambda \geq C'_{\epsilon, \eta}$ for $z \in {\rm supp}\: W_n$ and can treat $V_0 (z - L)^{-1} \psi$ and $V_0 (z - L)^{-1}\psi_1$ as above. On the other hand,
\begin{equation} \label{eq:3.4}
[(D_x - By)^2 + D_y^2, \psi] = -2 \ii \partial_x \psi (D_x - By) - 2 \ii\partial_y \psi D_y - \Delta_{x, y} \psi
\end{equation}
and the operators $\partial_x \psi(D_x - By) (z - L)^{-1}$ and $\partial_y \psi D_y (z - L)^{-1}$ are bounded by $C\eta^{-1}$ for $|\Im z| \geq \eta.$  Indeed, we have
$$(z - L) = (\ii - L)^{-1}[ I +  (\ii - z) (z - L)^{-1}]$$
and it suffices to show that $\partial_x \psi(D_x - By) (\ii - L)^{-1}$ and $\partial_y \psi D_y (\ii - L)^{-1}$ are bounded.
Next, $(\ii - L)^{-1}$ is a pseudodifferential operator
and the symbol of the pseudodifferential operator $(D_x - By)(\ii - L)^{-1}$ becomes
$$\frac{\xi - By}{\ii - (\xi - By)^2 - \eta^2 - \epsilon x} -\frac{\ii B\eta}{(\ii - (\xi - By)^2 - \eta^2 - \epsilon x)^2}.$$
From the well known results for the $L^2$ boundedness of pseudodifferential operators (see \cite{DS}) we deduce that (\ref{eq:3.4}) is bounded by $C|\Im z|^{-1}.$ Consequently, applying Proposition 2 once more, we can arrange the norm of the operator
$$V_0 (z - L)^{-1} \psi_1 [(D_x - By)^2 + D_y^2, \psi] (z -L)^{-1}$$
to be sufficiently small for $z \in W_n,\: |\Im z | \geq \eta$ and $n \geq n_1 > n_0$.
Combining this with the previous estimates, we get
$|S_n| \leq \epsilon/6$, hence $|R_n + Q_n + S_n| \leq \epsilon/2$ for $n$ large enough.
This implies (\ref{eq:3.2}) and the proof is complete.
\end{proof}
\begin{cor} Assume in addition to $(1.1)$ that $\partial_x^2 V \in C_0(\R^2) \cap L^{\infty}(\R^2)$. Then $H$ has at most finite number of embedded eigenvalues in $\R.$
\end{cor}
This result follows from Theorem 2 and Corollary $1$ in \cite{DP3} which guarantees  that $H$ has at most finite number of embedded eigenvalues in every interval $[a, b] \subset \R.$  The conjecture is that $H$ has no embedded eigenvalues on $\R$.

\section{Asymptotics of the spectral shift function}

 Our purpose in this section is to apply Theorem 1 and (2.4) to give a Weyl type  asymptotics with optimal remainder estimates for the spectral shift function $\xi_h(\lambda):=\xi(\lambda;H(h), H_0(h))$  corresponding to the operators
$$H(h)=(h D_x- y)^2+h^2D_y^2+ x,\:H_0(h) = H(h) + V(x, y),\: h > 0.$$
For simplicity of the exposition in this section we assume that $B = \epsilon = 1.$
Let $p_2(x, y, \zeta, \eta) = (\zeta - y)^2 + \eta^2 + x + V(x, y).$
For the analysis of $\xi_h(\lambda)$ we need the following theorems.

\begin{thm}
Let $\psi\in C^ \infty_0(\R^2)$ and let $f\in C^\infty_0(]0,+\infty[;\R)$. Then we have
\begin{equation}\label{eq:4.1}
{\rm tr}\Bigl[ \psi f(H(h))\Bigr]\sim\sum_{j=0}^ \infty a_j h^{j-2},\,\, h\searrow 0,
\end{equation}
with
\begin{equation}\label{eq:4.2}
a_0={1\over (2\pi)^ 2} \iint\psi(x,y) f(p_2(x,y,\zeta, \eta)) dx dyd\zeta d\eta.
\end{equation}
\end{thm}

\medskip

\begin {thm}  Assume that $\psi\in C^\infty_0(\R^2)$.  Let   $f \in C^\infty_0(\lbrack E_0, E_1 \lbrack)$ and $\theta\in C^\infty_0(\rbrack-{1\over C_0}, {1\over C_0}\lbrack;\:{\R}),\:\: \theta = 1$ in a neighborhood of $0$. Assume that if $p_2(x,y, \zeta, \eta) = \tau,\: \tau \in [E_0, E_1]$, then $dp_2 \neq 0.$ Then  there exists $C_0>0$ such that for all $N, m\in \N$ there exists $h_0>0$ such that
\begin{equation} \label{eq:4.3}
{\tr}\Bigl(\psi\breve\theta_h(\tau-H(h))f(H(h))\Bigr)
=(2 \pi h)^{-2} \Bigl( f(\tau)\sum_{j=0}^{N -1}\gamma_j(\tau) h^j +{\mathcal O}(h^N\langle \tau\rangle^{-m})\Bigr),
\end{equation}
uniformly with respect to $\tau \in \R$ and  $h\in ]0,h_0]$,
where
$$\gamma_0(\tau) = -(2\pi \ii)^{-1}\int\int_{\R^{4}}
\psi(x,y)\Bigl((\tau + \ii 0 - p_2(x,y,\zeta, \eta))^{-1}- (\tau - \ii 0 - p_2(x,y, \zeta, \eta))^{-1}\Bigr) dxdy d\zeta d\eta.$$
Here $$\breve\theta_h(\tau)= (2\pi h)^{-1} \int e^{i\tau t/h} \theta(t)dt.$$

\end{thm}
\medskip

{\it Proof of Theorem $3$ and Theorem $4$.} Here and below $\psi \prec \varphi$ means that $\varphi(x)=1$ on the support of $\psi$.
Let $G$  $\in C^{\infty}_0(\R^2)$ with  $\psi  \prec  G$.
 Introduce the operator
$$ \tilde H(h) =(hD_x-G(x,y)y)^2+h^2D_y^2+G(x,y)x + V(x,y),$$
and set
$$I ={\tr}\Bigl[ \psi \Bigl(f(H(h))-f(\tilde H(h)\Bigr)\Bigr].$$
 Let $\tilde{f}(z) \in C_0^{\infty}(\C)$ be an almost analytic continuation of $f$ with $\bar{\partial}_z \tilde{f}(z) ={\mathcal O}(\vert \Im z\vert^\infty)$ .
 From Helffer-Sj\"ostrand formula it follows    that
$$
I =\frac{1}{\pi} \int  \bar{\partial}_z \tilde f(z)
\tr\Bigl[\psi \Bigl((z-\tilde H(h))^{-1}- (z- H(h))^{-1}\Bigr) \Bigr] L(dz),$$
where $L(dz)$ denotes the Lebesgue measure on $\C.$

 Let $\psi_1\in C^\infty(\R^2)$ be a function with $ \psi_1=1$ near
${\rm supp}\:\:(1-G)$ and $\psi_1=0$ near ${\rm supp}\:\psi$, and let $\tilde \psi\in C^\infty_0(\R^2)$ be equal to  one near ${\rm supp}(\nabla\psi_1)$ and $\tilde\psi=0$
near ${\rm supp}\:\psi$. A simple calculus shows that $\tilde H(h)-H(h)= \psi_1 (\tilde H(h)-H(h))$ and $[\tilde H(h), \psi_1]=\tilde\psi[\tilde H(h), \psi_1]\tilde H$.
Then
\begin{equation}\label{eq:4.4}
\psi\Bigl((z-\tilde H(h))^{-1}-(z-H(h))^{-1}\Bigr)= \psi
(z- \tilde H(h))^{-1} \psi_1(\tilde H(h)-H(h))(z- H(h))^{-1}
\end{equation}
$$=\psi(z-\tilde H(h))^{-1}\tilde\psi[\tilde H(h),\psi_1](z-\tilde H(h))^{-1}  (\tilde H(h)-H(h)) (z-H(h))^{-1}.$$
Let $\chi_1,...,\chi_N \in C_0^\infty(\R^2;\lbrack 0,1\rbrack)$
with $\psi_1\prec \chi_1\prec...\prec \chi_N$ and $\chi_i\tilde\psi=0,\: i = 1,...,N$.
By using  the equalities $\chi_1 \psi_1 =...=\chi_N\psi_1=\psi_1,\: \chi_k\: \tilde \psi=0$,  $\chi_{k-1}[\chi_k,\tilde H(h)]=0$ and the fact that
 $$[\chi_k,(z-\tilde H(h))^{-1}]=
(z-\tilde H(h))^{-1} [\chi_k,\tilde H(h)](z-\tilde H(h))^{-1},$$
 we get
$$       \psi(z-\tilde H(h))^{-1}\tilde\psi[\tilde H(h),\psi_1]  $$
$$ =\psi(z-\tilde H(h))^{-1}
[\chi_1,\tilde H(h)](z-\tilde H(h))^{-1}...[\chi_N,\tilde H(h)]
(z-\tilde H(h))^{-1} \tilde\psi[\tilde H(h),\psi_1]     =:L_N(h).$$

Here
$$L_N(h) ={\mathcal O}_N(1)\Bigl( {h^N\over \vert\Im z\vert^N}\Bigr): H^{s}(\R^2)\rightarrow H^{s+N}(\R^2),$$
where we equip $H^N(\R^2)$ with the $h$-dependent norm $\Vert \langle hD\rangle^N u\Vert_{L^2}$.  Choose  $N > 2$ and let $s=-N$.
According to  Theorem 9.4 of \cite{DS}, we have
$$\Bigl\|\Bigr(-h^2\Delta+ 1\Bigr)^{-N/2}\tilde\psi\Bigr\|_{\tr}={\mathcal O}(h^{-2}).$$
Then
\begin{equation} \label{eq:4.5}
\|    \psi(z-\tilde H(h))^{-1}\tilde\psi[\tilde H(h),\psi_1]  \tilde{\psi}\|_{\tr} =
 \Bigl\| L_{N}(h)     \Bigr(-h^2\Delta + 1\Bigr)^{N/2}  \Bigr(-h^2\Delta + 1\Bigr)^{-N/2}
  \tilde\psi \Bigr\|_{\tr}
\end{equation}
$$\leq C \Bigl\|\Bigr(-h^2\Delta+ 1\Bigr)^{-N/2}  \tilde\psi\Bigr\|_{\tr} \Bigr({h^N\over \vert\Im z\vert^N}\Bigr) \leq C_1 \Bigl({h^{N-2}\over \vert\Im z\vert^N}\Bigr).$$
Combining this with (4.4) and using the fact that
$$\Vert (z-\tilde H(h))^{-1}  (\tilde H(h)-H(h)) (z-H(h))^{-1}\Vert=\Vert (z-\tilde H(h))^{-1}-(z-H(h))^{-1}\Vert=
{\mathcal O} \Bigl(\vert \Im z\vert^{-1}\Bigr),$$
we obtain
$$\Bigl\|\psi\Bigl((z-\tilde H(h))^{-1}-(z-H(h))^{-1}\Bigr)\Bigr\|_{{\rm tr}}={\mathcal O} \Bigl({h^{N-2}\over \vert\Im z\vert^{N+1}}\Bigr).$$
Since  $\bar{\partial}_z \tilde{f}(z) = {\mathcal O}(|\Im z|^\infty)$, we have
$$I= {\mathcal O}(h^\infty).$$

Summing up, we have proved that
\begin{equation} \label{eq:4.6}
{\tr}\Bigl(\psi f(H(h))\Bigr)
={\tr}\Bigl(\psi f(\tilde H(h))\Bigr) +{\mathcal O}(h^\infty).
\end{equation}
In the same way, we obtain
\begin{equation}\label{eq:4.7}
{\tr}\Bigl(\psi\breve\theta_h(\tau-H(h))f(H(h))\Bigr)
={\tr}\Bigl( \psi
 \breve\theta_h(\tau-\tilde H(h))f(\tilde H
(h))\Bigr) + {\mathcal O}(h^\infty).
\end{equation}

The operator $\tilde H(h)$ is elliptic semi-bounded $h$-pseudodifferential operator, so Theorem 3 and Theorem 4  follow from the $h$-pseudodifferential calculus and the analysis of elliptic operators in Chapters 8, 9, 12 in \cite{DS} (see also \cite{R}). The calculus of the leading terms is given by  trivial modification of the argument of Section 7 in  \cite{DP1} and we omit the details.
\hfill{$\Box$}\\
\begin{rem}
Notice that $dp_2\not=0$ on $p_2=\tau$ is equivalent to
\begin{equation} \label{eq:4.8}
 \nabla_{x,y}(x+V(x,y))\not=0,\,\,\,{\rm on}\,\,\,\, \{(x,y);\,\,\, x+V(x,y)=\tau\}.
\end{equation}
\end{rem}

Now we will apply  Theorem 3 and Theorem 4  to obtain a
Weyl-type asymptotics for $\xi_h(\lambda)$ when $h \searrow 0.$

\begin{thm}
Assume that $V\in C^\infty_0(\R^2)$ and suppose that $(\ref{eq:4.8})$  holds for  $\tau =\lambda_1,\lambda_2$.
Then there exists $h_0 > 0$ such that for $h \in ]0,h_0]$ we have
\begin{equation}\label{eq:4.9}
\xi_h(\lambda_2)-\xi_h(\lambda_1)=(2\pi h)^{-2} (c_0(\lambda_2)-c_0(\lambda_1))+{\mathcal O}(h^{-1}),
\end{equation}

where
\begin{equation}\label{eq:4.10}
c_0(\lambda)= -\pi \:\int_{\R^2} \partial_x V(x,y)(\lambda - x - V(x,y))_{+} dxdy.
\end{equation}
\end{thm}

\begin{proof}

Choose a large constant $M$ such that
$$M\geq \Vert \partial_x V\Vert_\infty+1.$$
Let $\psi\in C^\infty_0(\R^2;[0,1])$ with $\partial_x V \prec  \psi^2$.
According to (\ref{eq:2.4}), by using the cyclicity of the trace, we get
$$\langle \xi_h', f \rangle = {\rm tr}\: \Bigl(f(H(h)) - f(H_0(h)\Bigr) = -{\rm tr}\Bigl((\partial_x V) f(H(h))\Bigr)$$
$$={\rm tr}\Bigl((M-\partial_x V )^{1/2} \psi f(H(h) \psi (M-\partial_x V )^{1/2} \Bigr)-M
{\rm tr}\Bigl( \psi f(H(h))\psi\Bigr)$$
$$=:\langle \xi'_1,f\rangle-\langle \xi'_2,f\rangle.$$
Since
$$f \rightarrow {\rm tr}\Bigl((M-\partial_{x_1}V )^{1/2} \psi f(H(h) \psi (M-\partial_{x_1}V )^{1/2} \Bigr)$$
 and
$$f \rightarrow M{\rm tr}\Bigl( \psi f(H(h))\psi\Bigr)$$
are positive functions for $f\geq 0$, we deduce that the functions $\lambda \rightarrow \xi_i(\lambda), \: i = 1,2$ are monotonic.

Consequently, we may apply Tauberian arguments for the analysis of the asymptotics of $\xi_i(\lambda), \: i = 1,2.$
We treat below $\xi_2(\lambda).$ Let $\varphi \in C_0^{\infty}(\R),\: \varphi \geq 0,$  and suppose that  (\ref{eq:4.8}) holds for all
$\tau\in {\rm supp}\: \varphi.$ Consider the function
$$F_{\varphi}(\lambda) = \int_{-\infty}^\lambda \xi_2'(\mu) \varphi(\mu) d\mu.$$
Applying (4.3) with $N=1$ and $m=2$, we obtain
\begin{equation}\label{eq:4.11}
\frac{d}{d\lambda} (\breve{\theta}_h \ast F_{\varphi})(\lambda)= \int \breve{\theta}_h(\lambda - \mu) \xi_2'(\mu) \varphi(\mu) d\mu =(2\pi h)^{-2} \Bigl(\varphi(\lambda)\gamma_0(\lambda)+{\mathcal O}\Bigl(\frac{h}{ \langle \lambda\rangle^2}\Bigr)\Bigr).
\end{equation}

We integrate from $-\infty $ to $\lambda$  and we get
\begin{equation}\label{eq:4.12}
\int\Bigl(\int_{-\infty}^\lambda \breve{\theta}_h(\lambda'-\mu) d\lambda' \Bigr)\xi'_2(\mu) \varphi(\mu) d\mu
\end{equation}
$$= {1\over (2\pi h)^2} \Bigl( \int\int_{ p_2\leq \lambda} M \psi^2(x,y) \varphi(p_2) dxdyd\eta d\zeta+{\mathcal O}(h)\Bigr).$$
In the following we choose $\theta\in C^\infty_0(\R)$ with $\breve{\theta}_h\geq 0$. Let $h\breve{\theta}_h(0) = \frac{1}{2\pi}\int_\R \theta(u) du \geq 2C_1 > 0.$
Therefore, it follows that there exist $C_2 > 0$ such that
 $$\vert t\vert <{h\over C_2}\Longrightarrow  h\breve\theta_h(t)\geq C_1.$$
Combining this with the fact that $\breve\theta_h\geq 0$, and using $\langle \xi'_2, f\rangle\geq 0$ for $f\geq 0$,  we obtain
\begin{equation}\label{eq:4.13}
C_1 \int_{\{\vert \lambda -\mu\vert<{h\over C_0}\}} \xi'_2(\mu)\varphi(\mu)d\mu \leq
h\int_{\{\vert \lambda -\mu\vert<{h\over C_0}\}}\breve{\theta}_h(\lambda - \mu) \xi_2'(\mu) \varphi(\mu) d\mu$$
$$\leq h \int_{\R}\breve{\theta}_h(\lambda - \mu) \xi_2'(\mu) \varphi(\mu) d\mu=
h\frac{d}{d\lambda} (\breve{\theta}_h \ast F_{\varphi})(\lambda) ={\mathcal O}(h^{-1}),
\end{equation}
uniformly with respect to  $\lambda\in\R$. On the other hand, a simple calculus shows that
\begin{equation}\label{eq:4.14}
\int_{-\infty}^\lambda \breve{\theta}_h(\lambda'-\mu) d\lambda' =
\int_{-\infty}^{{\lambda-\mu\over h}}  \breve{\theta}_1(t) dt=
{\bf 1}_{]-\infty, \lambda[}(\mu) +
{\mathcal  O}\Bigl(\big\langle {\frac{\lambda-\mu}{h}}\big\rangle^{-\infty}\Bigr).
\end{equation}
Indeed, for $\mu < \lambda$ and all $k \in \N$ we have
$$\int_{-\infty}^{\frac{\lambda - \mu}{h}} \breve{\theta}_1(t) dt -1 = -\int_{\frac{\lambda - \mu}{h}}^{\infty} t^k\breve{\theta}_1(t) \frac{1}{t^k}dt$$
and
$$\int_{\frac{\lambda - \mu}{h}}^{\infty} t^k\breve{\theta}_1(t) \frac{1}{t^k}dt
 \leq \Bigl(\frac{\lambda - \mu}{h}\Bigr)^{-k} \int_{\R} t^k \breve{\theta}_1(t) dt.$$
A similar argument works for $\mu > \lambda.$
From (\ref{eq:4.13}) we have for $k \geq 2$ the estimate
\begin{eqnarray}\label{eq:4.15}
\int_{\R}\big \langle \frac{\lambda-\mu}{h}\big\rangle^{-k}\xi_2'(\mu) \varphi(\mu)d\mu = \sum_{m = -\infty}^{\infty}\int_{\frac{m}{C_0} \leq \frac{\mu - \lambda}{h} < \frac{m+1}{C_0}} \big\langle \frac{\lambda-\mu}{h}\big\rangle^{-k}\xi_2'(\mu) \varphi(\mu)d\mu\\ \nonumber
\leq \sum_{m = 0}^{\infty} \Bigl(1 + \Bigl(\frac{m}{C_0}\Bigr)^2\Bigr)^{-k/2} \int_{\lambda + \frac{m h}{C_0}}^{\lambda + \frac{(m+1)h}{C_0}}\xi_2'(\mu) \varphi(\mu) d\mu\\ \nonumber
+ \sum_{m =-\infty}^{-1} \Bigl(1 + \Bigl(\frac{|m +1|}{C_0}\Bigr)^2\Bigr)^{-k/2} \int_{\lambda + \frac{m h}{C_0}}^{\lambda + \frac{(m+1)h}{C_0}}\xi_2'(\mu) \varphi(\mu) d\mu
\leq  \sum_{m = -\infty}^{\infty}  \frac{1}{(C_0 + |m|)^k} {\mathcal O}(h^{-1}),
\end{eqnarray}
where in the last inequality at the right hand side we used the fact that (\ref{eq:4.13}) holds uniformly with respect to $\lambda \in \R$ and we can estimate the integrals involving $\xi_2'(\mu) \varphi(\mu)$ by ${\mathcal O}(h^{-1})$ uniformly with respect to $m$.

Inserting the right hand side of (\ref{eq:4.14})   in the left hand side of (\ref{eq:4.12})  and using (\ref{eq:4.15}), we get
$$ F_{\varphi}(\lambda) =  (2\pi h)^{-2} \Bigl( \int\int_{ p_2\leq \lambda}  M\psi^2(x,y) \varphi(p_2) dxdyd\eta d\zeta+{\mathcal O}(h)\Bigr).$$

We apply the same argument for $\xi_1(h)$ and introduce the function
$$G_{\varphi}(\lambda) = \int_{-\infty}^\lambda \xi_1'(\mu)\varphi(\mu) d\mu.$$
 Replacing the function $\psi$ by $(M -\partial_x V)^{1/2} \psi$, we
get
$$G_{\varphi}(\lambda) =
{1\over (2\pi h)^2} \Bigl( \int\int_{ p_2\leq \lambda} ( M- \partial_x V)\psi^2(x,y) \varphi(p_2) dxdyd\eta d\zeta+{\mathcal O}(h)\Bigr).$$
Since $\xi_h = \xi_1 - \xi_2$, the above results yield
\begin{equation}\label{eq:4.16}
M_\varphi(\lambda)=\int_{-\infty}^\lambda \xi_h'(\mu) \varphi(\mu) d\mu=
{1\over (2\pi h)^2} \Bigl( \int\int_{ p_2\leq \lambda}  -\partial_xV(x,y) \varphi(p_2) dxdyd\eta d\zeta+{\mathcal O}(h)\Bigr).
\end{equation}
Now, we are ready to prove Theorem 5. Assume that $\lambda_1<\lambda_2$, and let $\epsilon>0$ be small enough. Let
$\varphi_1, \varphi_2, \varphi_3\in C^\infty_0(]\lambda_1-\epsilon,\lambda_2+\epsilon[)$ with $\varphi_1+\varphi_2+\varphi_3=1$ on $[\lambda_1,\lambda_2]$,
 ${\rm supp}\:\varphi_1\subset
]\lambda_1-\epsilon,\lambda_1+\epsilon[$, ${\rm supp}\:\varphi_2\subset
]\lambda_2-\epsilon,\lambda_2+\epsilon[$ and  ${\rm supp}\:\varphi_3\subset
]\lambda_1,\lambda_2[$. We choose $\epsilon$ small enough so that (\ref{eq:4.8})
  holds for all $\tau\in ]\lambda_1-\epsilon,\lambda_1+\epsilon[\:\cup \:]\lambda_2-\epsilon,\lambda_2+\epsilon[$.
We write
$$\xi_h(\lambda_2)-\xi_h(\lambda_1)=\int_{\lambda_1}^{\lambda_2} (\varphi_1+\varphi_2+\varphi_3)(\lambda) \xi_h'(\lambda)d\lambda$$
$$= M_{\varphi_2}(\lambda_2) + M_{\varphi_1}(\lambda_2) -M_{\varphi_2}(\lambda_1) -M_{\varphi_1}(\lambda_1) -{\rm tr}(\partial_x V\varphi_3(H)),$$
where for the function $\varphi_3$ we have exploited (\ref{eq:2.4}). Next for the term involving $\varphi_3$ we
apply Theorem 3 and obtain
$${\rm tr}\: (\pv \varphi_3(H)) = \frac{1}{(2 \pi h)^{2}} \int\int \partial_x V \varphi_3(p_2) dx dy d\zeta d\eta + {\mathcal O}(h^{-1}).$$
For $M_{\varphi_{1}}(\lambda_i)$ and $M_{\varphi_{2}}(\lambda_i),\: i = 1,2,$ we exploit the above argument and we deduce the asymptotics taking into account (\ref{eq:4.16}). Summing the terms involving $\varphi_j,\:j = 1,2,3$, we conclude that
$$\xi_h(\lambda_2) - \xi_h(\lambda_1) = (2 \pi h)^{-2} d(\lambda_2, \lambda_1) + {\mathcal O}(h^{-1}).$$
For the leading term we have
$$d(\lambda_2, \lambda_1) = \int\int_{\lambda_1 \leq p_2 \leq \lambda_2} -\pv (x,y) \Bigl(\varphi_1(p_2) + \varphi_2(p_2)+ \varphi_3(p_2)\Bigr) dx dy d\zeta d\eta $$
 $$= -\int\int_{ p_2 \leq \lambda_2} \pv (x, y)  dx dy d\zeta d\eta + \int\int_{p_2 \leq \lambda_1} \pv (x, y) dx dy d\zeta d\eta.$$
Finally, notice that
$$c_0(\lambda) = -\int\int_{p_2 \leq \lambda} \pv(x, y) dx dy d\zeta d\eta = - \int_{\R^2} \pv(x, y)\Bigl(\int_{(\zeta - y)^2 + \eta^2 \leq (\lambda - x - V(x,y))_+} d\zeta d\eta\Bigr)dxdy$$
$$ =-\pi \int_{\R^2} \pv(x, y) (\lambda - x - V(x, y))_+ dx dy$$
and the proof of Theorem 5 is complete.
\end{proof}
\begin{rem}
If $\lambda \gg 1$ is large enough (resp. $\lambda \ll -1$) then on ${\rm supp}\:(\partial_x V)$, we have
$$(\lambda-x-V)_+=\lambda-x-V,\,\,\, ({\rm resp.}\,\, (\lambda-x-V)_+=0).$$
Consequently,
$$c_0(\lambda)=- \pi\int_{\R^2} V(x,y)dxdy,\,\,\,\,{\rm for}\,\,\,  \lambda \gg 1,$$
and
$$c_0(\lambda)=0,\,\,\, {\rm for}\,\,\, \lambda \ll -1.$$
In particular, if we normalize $\xi_h(\lambda)$ by $\lim_{\lambda \rightarrow -\infty}\xi_h(\lambda)=0$, we get
$$\xi_h(\lambda)=(2\pi h)^{-2} c_0(\lambda)+{\mathcal O}(h^{-1}).$$
\end{rem}

\begin{rem} The results of this section can be generalized for potentials $V(x, y)$ for which there exists $\delta_1 \in \R$ such that supp $V \subset \{(x, y) \in \R^2:\: x \geq \delta_1\}$ by using the techniques in \cite{DP1}. For simplicity we treated the case of $V \in C_0^{\infty}(\R^2)$ to avoid the complications caused by the calculus of pseudodifferential operators.

\end{rem}

{\footnotesize

config: arxiv


TITLE: Finding a coprime of a general magnitude.
QUESTION [4 upvotes]: I have an arbitrary number $x$. I would like to compute a number that is coprime to $x$ that's close(ish) to the square root of $x$. I don't need to find them all, and factoring $x$ is expensive. I just need one number.  I could also check a few primes near the square root of $x$, but computing primes and storing primes is expensive.
Constant time and space, preferably.

REPLY [3 votes]: Since calculating the gcd of two arbitrary number isn't expensive, you can practically brute force it.
Algorithm:

Generate a random number n.
Find its integer-square-root r.
Check if gcd(n,r)=1. If yes, return r and exit.
If no, add 1 to r, and repeat step 3.

Python:
import random
import time
start = time.time()
bit = 512 # even
num = 1
for _ in range(bit): num = num * 2 + random.randrange(2)
end = time.time()
print("random number:")
print(num)
print("time for generating random number:       %f"%(end-start))

# integer square root (wiki)
start = time.time()
res = 0
bit = 2**bit
while bit:
    if num >= res + bit:
        num -= res + bit
        res = (res >> 1) + bit
    else:
        res >>= 1
    bit >>= 2
end = time.time()
print("time for generating itneger square root: %f"%(end-start))

def gcd(a,b):
    while a: a,b = b%a,a
    return b

start = time.time()
while gcd(res,num) != 1: res += 1
end = time.time()
print("time for generating co-prime number:     %f"%(end-start))

print(res)

Does it in fraction of a second.
Try it online!
Sample output:
random number:
16956998685025525617332092680088906859010597516989049974644188276809460728386128015966080402491132114558031760245789600047216269236212122151725198496639367
time for generating random number:       0.000913
time for generating itneger square root: 0.000193
time for generating co-prime number:     0.000031
130219041176878297644835828972023265387463111246248427493495319607240172982284


TITLE: Linearization of Autonomous Equations
QUESTION [0 upvotes]: Consider $y'=y(y-50)(y-100)$
(a) Let $y_e$ be the stable equilibrium solution and let $u=y-y_e$, rewrite the equation as a differential equation for $u$.

$y_e=50, u=y-50, u'=(u-50)(u)(u+50)$

(b) What is the linearized equation near $u=0$, and what is its general solution?

Near $u=0$: $u'=-2500u$. General solution: $u=Ce^{-2500t}$

(c) If $y(0)=40$, estimate $y(1)$ and $y(2)$ be linearizing near the equilibrium solution.
I am stuck on part (c) as I don't understand how we can linearize near $0$ if the equilibrium solution is at $50$. Did I make a mistake in part (a) or (b)?

REPLY [1 votes]: Yes $u(t) = u_0 e^{-2500 t}$ is a good approximation for when $u_0$ small.
So treat $u_0 = y_0 - 50 = -10$. This isn't small so the linearized solution might be rather bad. We can plug it anyway and get
$$
u(1)=-10 e^{-2500}\\
y(1)=50-10 e^{-2500}\\
u(2)=-10 e^{-5000}\\
y(2)=50-10 e^{-5000}\\
$$
40 is closer to 50 than 0 so there is hope for this approximate solution, but you should check.


TITLE: Determining if ellipse is fully inside another ellipse
QUESTION [4 upvotes]: I have two ellipses which are both not rotated, but have different sizes. I want to know if the smaller ellipse 1 is fully inside the larger ellipse 0. In my special case both ellipses have the same axis ratio, so $a_0/b_0=a_1/b_1$, but I doubt this makes a difference for the calculation.

I have started to work on a solution, but I am not so sure if this is the right approach and if it will ultimately lead to success. Here is what I have come up with so far:
The radius of ellipse 0 can be calculated for any point on the ellipse as
$$
r_0=\sqrt{(x-x_0)^2+(y-y_0)^2}
$$
The general equation for an ellipse gives a relation between $x$ and $y$ for points on the ellipse.
$$
\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1
$$
$$
(y-y_0)^2=b*\sqrt{1-\frac{(x-x_0)^2}{a^2}}
$$
This can be used to calculate radius of both ellipses depending on the x coordinate.
$$
r_0 = \sqrt{(x-x_0) + b_0^2(1-\frac{(x-x_0)^2}{a_0^2})}
$$
$$
r_1 = \sqrt{(x-x_1) + b_1^2(1-\frac{(x-x_1)^2}{a_1^2})}
$$
To calculate the distance $s$ between both ellipses at a certain $x$ coordinate, we simply have to add the distance $d_{01}$ between the center points to radius 1 and subtract this from radius 0.
$$
s=r0 - (d_{01} + r1)
$$
where
$$
d_{01} = \sqrt{(x_1-x_0)^2 + (y1-x_0)^2}
$$
Now I have an equation where $s$ only depends on $x$. My idea is to get the minimum of $s$ through the first derivative. If the minimum is non-negative, ellipse 1 is inside or on ellipse 0. If it is negative, part of ellipse 1 is outside ellipse 0.
Here are my questions:

Am I on the right way so far?
Is calculating the minimum for s the right approach?
Do I have a chance to actually calculate the minimum of $s$?

Note: Ther is a similar question here, but I think my case is much simpler because the ellipses are not rotated similar.

REPLY [4 votes]: Because they have the same axis ratio and are orientated the same, the easy solution is to multiply all y-coordinates by $\frac{a}{b}$. Then you have two circles.
All points on the large circle are within $a_0$ of its centre whereas all points on the small circle are within $a_l+\sqrt{(x_l-x_0)^2+\frac{a^2}{b^2}(y_l-y_0)^2}$.


TITLE: Finding the norm of a dual space (and then applying it to an element function)
QUESTION [0 upvotes]: Let $V=C([-1,1],\mathbb R)$ be the space of continuous functions from $[-1, 1]$ to $\mathbb R$ equipped with the norm $\|f\|_{\infty}=\sup_{x\in\mathbb R}|f(x)|$. Consider the mapping $L:V\to \mathbb R$ defined by
  $$L(f):=\int_{-1}^1\sin(\pi t)f(t)\mathrm d t\qquad\forall f\in V.$$
  Show that $L\in V'$ and find its norm $\|L\|_{V'}$.

In the above, $V'$ refers to the dual space of $V$. I struggled with this question. Firstly, I wasn't sure how $\|\cdot\|_{V'}$ is defined. Is it the supremum norm of $V'$? Secondly, when I tried to visualize $\sup_{f\in V}|L(f)|$, I thought of $f$ as being a function with a steep gradient, with $f(0)=0$, $f(t)>0$ when $t>0$ and $f(t)<0$ when $t<0$ so that $\sin(\pi t)f(t)$ is positive. The steeper the gradient of $f$, the greater the area underneath $\sin(\pi t)f(t)$, without limit. So I could not find $\sup_{f\in V}|L(f)|$.
In a nutshell, how am I supposed to know how the norm $\|\cdot\|_{V'}$ is defined?
Edit: Why was I required to show that $L\in V'$? I ignored that because I thought it was obvious, but another question in the tutorial asked the same thing. How does one "show" that?

REPLY [0 votes]: As $V$ is a normed vector space, we associate to each $L \in V'$, the number $\left\| L \right\| = \sup\limits_{x\in V, \left\|x\right\| \le 1} L(x)$.
This number is called the norm of $L$ in $V'$, as it makes $V'$ a normed space.
P.S., one can check that $L\in V'$ by verifying the linearity of the addition and scalar multiplication.