TITLE: proving a hard inequality
QUESTION [1 upvotes]: can someone help me to prove this inequality :
$\left| \sum _{ k=0 }^{ 2n }{ \frac { k }{ k+{ n }^{ 2 } }  } -\sum _{ k=0 }^{ 2n }{ \frac { k }{ { n }^{ 2 } }  }  \right| \le \frac { 4 }{ { n }^{ 2 } } (2n+1)$

REPLY [6 votes]: First note that $1^2 + 2^2 + 3^2 + ... + n^2 =\frac{n(n + 1)(2n + 1)}{6}$
$\left| \sum _{ k=0 }^{ 2n }{ \frac { k }{ k+{ n }^{ 2 } }  } -\sum _{ k=0 }^{ 2n }{ \frac { k }{ { n }^{ 2 } }  }  \right| \leq   \sum _{ k=0 }^{ 2n }\left|\frac { k }{ k+{ n }^{ 2 }} -{ \frac { k }{ { n }^{ 2 } }}\right| = \sum _{ k=0 }^{ 2n }\left|       \frac{k^2}{(k+n^2)n^2}      \right|  \leq  \sum _{ k=0 }^{ 2n } \left|\frac{k^2}{n^4}\right|  = \frac{1}{n^4}\sum _{ k=0 }^{ 2n } k^2 = \frac{1}{n^4}\frac{2n(2n + 1)(4n + 1)}{6} \leq \frac{1}{n^4}\frac{2n(2n + 1)(12n)}{6} = \frac{4(2n + 1)}{n^2} $
(through this estimation the bound can actually be smaller)


TITLE: Prove that the matrix A is diagonalizable if $-4bc < (a-d)^2$
QUESTION [2 upvotes]: I'm trying to prove that matrix A is diagonalizable
if $-4bc < (a - d)^2$ and is not diagonalizable if $-4bc > (a - d)^2$
$A = \begin{bmatrix}
  a & b\\
  c & d
 \end{bmatrix}$
I'm not sure how to start. I was thinking of taking the inverse of A but I'm sure if that's the right approach.

REPLY [2 votes]: I am assuming that you want to work in the field of real numbers. In this case you have the following
1) If the matrix is diagonalizable, then it must have both the roots real
2) If the matrix has eigenvalues that are not repeated,  then it can be diagonalized with the following additional comment: If the eigenvalues are not real, the diagonalization would require complex numbers.
3) The only real matrices with repeated eigenvalues that is also diagonalizable are multiples of the identity matrix, i.e. diagonal matrices with equal values on the diagonal.
To OP:
In one of your comments you have
$$
\lambda^2-(a+b)\lambda+ad+bc$$
It should be
$$
\lambda^2-(a+d)\lambda+ad-bc$$
So for real roots that are distinct, you need
$$
(a+d)^2 -4(ad - bc) = a^2 + 2 a d + d^2 - 4 a d + 4bc = (a-d)^2 + 4 bc \gt 0$$
or
$$ (a-d)^2 \gt -4bc$$
Note $a=d, b=1, c=0$ is not diagonalizable. So you cannot change the $\gt$ to $\ge$.
Also, $a=d, b=c=0$ is diagonalizable. So
$$
\text{If }(a-d)^2 = -4bc, \text{ the matrix is diagonalizable if and only if $b=0$ and $c=0$}$$
** Added in reply to OP **
Let $\Delta$ be the discriminant
There are 3 cases

$\Delta < 0$. Both the eigenvalues are complex so matrix can not be diagonalized.
$\Delta >0$. Both eigenvalues are real and distinct (different). Matrix can be diagonalized
$\Delta=0$. Both eigenvalues are equal. Then there are two cases: Matrix is diagonal, i.e. $b=c=0$. Matrix is not diagonal $b \ne 0$ or $c \ne 0$. In the first case the matrix is diagonalizable (it is already diagonal!). In the second case the matrix cannot be diagonalized.


TITLE: Second derivative exist but first derivative zero
QUESTION [0 upvotes]: Can there be any function $f(x)$ (periodic or nonperiodic) with $f'(x)=0$ but $f''(x) \neq 0$. Is it possible!!

REPLY [5 votes]: If you mean with $f'(x)$ identically equal to $0$ (that is, $0$ everywhere), then no, since $f''(x)$ will then readily be identically $0$, too.
Otherwise, yes. For example, let $f(x)=\cos x.$ Then $f'(x)=0$ exactly when $x$ is an integer multiple of $\pi,$ but $f''(x)=\pm 1$ for all such $x$.

REPLY [0 votes]: No, a simple observation is think of the geometric interpretation of the second derivative. It represents the concavity or curvature. If $f^{'}(x) = 0$ then $f(x) = c$ and clearly there is zero curvature or concavity for a constant function.


TITLE: What is the value of $\sum_{n=0}^{\infty}\frac{(n+1)x^n}{n!}$?
QUESTION [2 upvotes]: I need to find the value of the series $\sum_{n=0}^{\infty}\frac{(n+1)x^n}{n!}$.I've computed its radius of convergence which comes out to be zero.
I'm not getting how to make adjustments in the general terms of the series to get the desired result...

REPLY [1 votes]: $$y=\sum_{n=0}^{\infty}\frac{nx^n}{n!}+\sum_{n=0}^{\infty}\frac{x^n}{n!}$$
$$y=e^x+\sum_{n=0}^{\infty}\frac{nx^n}{n!}$$
$$\frac{y}{x}=\frac{e^x}{x}+\sum_{n=0}^{\infty}\frac{nx^{n-1}}{n!}$$
$$\frac{y}{x}=\frac{e^x}{x}+e^x$$
so
$$y=e^x+xe^x$$


text:
\begin{document}

\maketitle

\begin{abstract}
The exit problem for small perturbations of a dynamical system in a domain is considered. It is assumed that
the unperturbed dynamical system and the domain satisfy the Levinson conditions. We assume that the
random perturbation affects the driving vector field and the initial condition, and each of the components
of the perturbation follows a scaling limit. We derive the joint scaling limit for the random exit time and exit point.
We use this result to study the asymptotics of the exit time for 1-d diffusions conditioned on rare events.
\end{abstract}

\section{Introduction}
\label{Sec: Notation}
Small stochastic perturbations of deterministic dynamical systems
have been studied for several decades, see, e.g., a set of lectures~\cite{Freidlin-lectures:MR1399081} and references therein.
Properties of exit distributions for the resulting diffusions are particularly important. One reason for that is that
one can express the solutions of parabolic and elliptic PDE's containing the generator of the diffusion via the exit
distributions. Another reason is the possibility to use exit distributions for the analysis of the global behavior of the system.
One can cover the state space by several domains and study the process within each of them separately. Using the strong Markov property one
can then treat the exit distribution for one of the domains as the starting distribution for the next one. This approach was vital for the study
of noisy heteroclinic networks in the vanishing noise limit, see~\cite{nhn},\cite{nhn-ds}.

In this note we study a relatively simple situation called the Levinson case (see~\cite[Chapter 2]{Freidlin-lectures:MR1399081}),
where the typical exit happens along a trajectory of the deterministic flow.
 We derive a scaling law for the exit
distribution in the limit of vanishing perturbation assuming that the initial random data as well as both
deterministic and white noise components of the perturbation follow a scaling limit.

We also show that our main result can be used to study rare events for diffusions. We present a 1-dimensional
situation where to reach a certain threshold, the diffusion has to evolve against the deterministic flow. By conditioning
on this unlikely event, we reduce the analysis to the Levinson case.

The paper is organized as follows. In Section~\ref{sec:main-results} we state the main theorem for the Levinson case,
postponing its proof to Section~\ref{sec:proof-levinson}.
In Section~\ref{sec:1-d-diffusion} we state the result on the diffusion conditioned on a rare event and derive it
from the main theorem and some auxiliary statements proven in Section~\ref{sec:aux-1-d}.

\medskip

{\bf Acknowledgment.} Yuri Bakhtin is grateful to NSF for partial support via a CAREER grant DMS-0742424.

\section{Main result}\label{sec:main-results}

We consider a $C^2$-smooth bounded vector field~$b$ in $\R^d$. The unperturbed dynamics is given by
 the deterministic flow $S=(S^t)_{t\in\R}$ generated by $b$:
\[
\frac{d}{dt}S^{t}x_0=b(S^{t}x_0),\quad S^{0}x_0=x_0.
\]

We will introduce three components of perturbations of this deterministic flow. They all depend on
a small parameter $\eps>0$.

The first component is white noise perturbation generated by $\eps\sigma$, where
$\sigma:\R^d \to \R^{d \times d}$ is a $C^2$-smooth bounded matrix valued function.

The second one is $\eps^{\alpha_1}\Psi_\eps$, where $\Psi_\eps$ is
a deterministic Lipschitz vector field on $\R^d$ for each $\eps$,  converging uniformly to
a limiting Lipschitz vector field $\Psi_0$, and $\alpha_1$ is a positive scaling exponent.
These conditions ensure that the stochastic It\^o
equation
\begin{equation}
dX_\eps(t) = \left( b( X_\eps(t) )+\eps^{\alpha_1} \Psi_\eps(X_\eps(t)) \right)dt + \eps \sigma(X_\eps(t))dW \label{eq:PrincipalEquation}
\end{equation}
w.r.t.\ a standard $d$-dimensional Wiener process $W$
has a unique strong solution for any $\eps>0$ and all initial conditions (for a general background on stochastic differential equations see, e.g.,~\cite{Karatzas--Shreve}).

The last component of the perturbation is the initial condition satisfying
\begin{equation}
X_{\epsilon }(0)=x_{0}+\epsilon ^{\alpha_2 }\xi _{\epsilon },\quad\eps>0.
\label{eq:initial_condition}
\end{equation}
Here  $\alpha_2>0$, and $(\xi _{\epsilon })_{\epsilon >0}$ is
a family of random variables independent of $W$, such that for some random
variable $\xi _{0}$, $\xi _{\epsilon }\rightarrow \xi _{0}$ as $\epsilon
\rightarrow 0$ in distribution.

\medskip

Let $M$ be a smooth $C^2$-hypersurface in $\R^d$.  If
\begin{equation*}
\tau _{\epsilon }=\inf \left \{t\ge 0:X_{\epsilon
}(t)\in M \right \},
\end{equation*}
then on $\{\tau_\eps<\infty\}$ we have $X_\eps(\tau_\eps)\in M$. We are going to study
the limiting behavior of $\tau_\eps$ and $X_\eps(\tau_\eps)$ as $\eps\to 0$
under the assumptions above.

Let us describe our assumptions on the joint geometry of the vector field $b$ and the surface $M$.
First we define
\[
T=\inf \left \{t>0: S^tx_0 \in M \right \},
\]
and assume that $0<T<\infty$. Secondly, we denote $z=S^Tx_0\in M$ and assume that
$b(z)$ does not belong to the tangent
hyperplane $T_zM$. In other words, we assume that the positive orbit of $x_0$ intersects $M$ and
the crossing is transversal.

In the case of $\xi_\eps\equiv 0$ and $\Psi \equiv 0$, Levinson's theorem states (see \cite{Levinson:MR0033433},
\cite[Chapter 2]{Freidlin--Wentzell-book}, and \cite[Chapter 2]{Freidlin-lectures:MR1399081}) that
$X_\eps(\tau_\eps)\to z$ in probability as $\eps \to 0$. Levinson worked in the PDE context and
showed how to obtain an expansion for the solution of the corresponding elliptic PDE depending
on the small parameter~$\eps$.
The main result of this note describes the limiting behavior of the correction $(\tau_\eps-T,X_\eps(\tau_\eps)-z)$ and extends  \cite[Theorem 2.3]{Freidlin--Wentzell-book} to the situation with generic
perturbation parameters $\xi_0,\Psi,\alpha_1$, and $\alpha_2$. This extension is essential since, as the analysis in~\cite{nhn}
shows, in the sequential study of entrance-exit distributions for multiple domains one has to consider nontrivial scaling laws for the initial conditions; also, considering nontrivial deterministic perturbations will allow us to study rare events, see
Section~\ref{sec:1-d-diffusion}.

We need more notation. Due to the smoothness of $b$,
\begin{equation}
b(x)=b(y)+Db(y)(x-y)+Q_1(y,x-y),\quad x,y\in\R^d, \label{eqn: b_Taylor}
\end{equation}
where
\begin{equation}
|Q_1(u,v)|\leq K  |v|^2 \label{eqn: bound_Q1},
\end{equation}
 for some constant $K>0$ and any $u,v \in \R^d$.
We denote by $\Phi_x(t)$ the linearization of $S$ along the
orbit of $x$:
\begin{equation}
\frac{d}{dt}\Phi_x(t)=A(t)\Phi_x(t)\text{, \ }\Phi_x(0)=I, \label{eqn: Phi_def}
\end{equation}
where $A(t)=Db(S^tx)$ and $I$ is the identity matrix.

Finally, for any vector $v\in \R^d$, we define $\pi_b v\in \R$ and $\pi_M v\in T_zM$ by
\[
 v=\pi_b v \cdot b(z)+ \pi_M v,
\]
i.e., $\pi_b$ is the (algebraic) projection onto $\Span(b(z))$ along $T_zM$ and
$\pi_M$ is the (geometric) projection onto $T_zM$ along $\Span(b(z))$.

 \begin{theorem}
 \label{thm: Main}  Let $\alpha=\alpha_1 \wedge \alpha_2\wedge 1$, and
\begin{align} \notag
\phi_0 (t) &= \mathbf{1}_{\{\alpha_2= \alpha\}}  \Phi_{x_0}(t) \xi_0 + \mathbf{1}_{\{\alpha_1= \alpha\}}\Phi_{x_0}(t) \int_0^t \Phi_{x_0} (s)^{-1} \Psi_0( S^s x) ds \\
& \quad +\mathbf{1}_{\{1=\alpha\}} \Phi_{x_0}(t)\int_{0}^{t}\Phi_{x_0}^{-1}(s)\sigma(S^s x_0)dW(s), \quad t>0.
\label{eqn: phi_0_def}
\end{align}
Then, in the setting introduced above,
\begin{equation}
 \eps^{-\alpha}(\tau_\eps-T, X_\eps(\tau_\eps)-z){\to} (-\pi_b \phi_0(T), \pi_M \phi_0(T)). \label{eq:main_convergence_statement}
\end{equation}
in distribution.
If additionally we require that $\xi_\eps\to\xi_0$ in probability or that $\alpha_2>\alpha$, then the convergence in \eqref{eq:main_convergence_statement}
is also in probability.
\end{theorem}

\begin{remark}\label{rem:weakening-conditions-by-localization}
\rm The conditions of Theorem~\ref{thm: Main} can be relaxed using the standard localization procedure. In fact, one needs to
require uniform convergence of $\Psi_\eps\to\Psi_0$ and regularity properties of $b$ and $\sigma$
only in some neighborhood of the set $\{S^tx_0:\ 0\le t\le T(x_0)\}$.
\end{remark}

\begin{remark}\rm In applications (see~\cite{nhn},\cite{nhn-ds}), the parameters $\alpha_1$ and $\alpha_2$ can be chosen so that
the r.h.s.\ of~\eqref{eq:main_convergence_statement} is nondegenerate.
\end{remark}

\begin{remark}\rm
In the case where $d=1$, the hypersurface $M$ is just a point. Therefore, $\pi_M$ is identical zero and the only
contentful information Theorem~\ref{thm: Main} provides is the asymptotics of the exit time.
\end{remark}

\section{Conditioned diffusions in 1 dimension}\label{sec:1-d-diffusion}

In this section we apply Theorem~\ref{thm: Main} to the analysis of the exit time of conditioned diffusions in 1-dimensional situation for the large deviation case.

Suppose, for each $\eps>0$, $X_\eps$ is a weak solution of the following SDE:
\begin{align*}
 dX_\eps(t)&=b(X_\eps(t))dt+\eps\sigma(X_\eps(t)) dW(t),\\
 X_\eps(0)&=x_0,
\end{align*}
where $b$ and $\sigma$ are $C^1$ functions on $\R$, such that $b(x)<0$ and $\sigma(x)\ne 0$ for all $x$ in an interval $[a_1,a_2]$
containing $x_0$. We introduce
\[
 \tau_\eps=\inf\{t\ge0:\ X_\eps(t)=a_1\ \text{\rm or}\ a_2\}
\]
and  $B_\eps=\{X_\eps(\tau_\eps)=a_2\}$. Since $b<0$, $B_\eps$ is a rare event since $\lim_{\eps\to0}\Pp(B_\eps)=0$. More precise estimates
on the asymptotic behavior of $\Pp(B_\eps)$ can be obtained in terms of large deviations. However, here we study the diffusion
$X_\eps$ conditioned on the rare event $B_\eps$.

Let $T(x_0)$ denote the time it takes for the solution of $\dot x=-b(x)$ starting at $x_0$ to reach $a_2$:
\[
T(x_0)=-\int_{x_0}^{a_2}\frac{1}{b(x)}dx.
\]

\begin{theorem} \label{thm: conditioned} Conditioned on $B_\eps$, the distribution of $\eps^{-1}(\tau_\eps-T(x_0))$ converges weakly to a centered
Gaussian
distribution with variance
\[
- \int_{x_0}^{a_2} \frac{ \sigma^2 (y) }{ b^3 (y) } dy.
\]
\end{theorem}

To prove this theorem, we will need two auxiliary statements. Their proofs are given in Section~\ref{sec:aux-1-d}.

\begin{lemma} \label{lemma: conditioned_drift}Conditioned on $B_\eps$, the process $X_\eps$ is a diffusion with the same
diffusion coefficient as the unconditioned process, and with the
drift coefficient given by
\[
b_\eps(x)= b(x)+\eps^2\sigma^2(x)\frac{h_\eps(x)}{\int_{a_1}^x h_\eps(y)dy},
\]
where
\begin{equation}
 \label{eq:h_eps}
 h_\eps(x)=\exp\left\{-\frac{2}{\eps^2}\int_{a_1}^x \frac{b(y)}{\sigma^2(y)}dy\right\}.
\end{equation}
\end{lemma}

Further analysis requires understanding the limiting behavior of $b_\eps$. This is the purpose of the next lemma:
\begin{lemma} \label{lemma: b_approx} There is $\delta>0$ such that
\[
\limsup_{\eps \to 0}  \eps^{-2}\left(\sup_{x \in [x_0-\delta, a_2+\delta]} | b_\eps(x) + b (x) | \right) < \infty.
\]
\end{lemma}

\begin{proof}[Proof of Theorem~\ref{thm: conditioned}]
Let us fix $\beta \in (1,2)$. Lemmas~\ref{lemma: conditioned_drift} and~\ref{lemma: b_approx} imply that
$X_\eps$ conditioned on $B_\eps$, up to $\tau_\eps$ satisfies an SDE of the form
\[
d X_\eps (t)= \left( -b(X_\eps(t)) + \eps^{\beta} \Psi_{\eps,\beta} ( X_\eps (t) ) \right) dt + \eps \sigma ( X_\eps (t) ) d\tilde W(t),
\]
for some Brownian Motion $\tilde W$ and with $\Psi_{\eps, \beta} \to 0$ uniformly as $\eps \to 0$. We can assume that after time $\tau_\eps$,
this process still follows the same equation at least up to the time it hits $x_0 -\delta$ or $a_1+\delta$.

So, we can apply Theorem~\ref{thm: Main}  (taking into account Remark~\ref{rem:weakening-conditions-by-localization}) to see that
\begin{equation} \label{eqn: conv_example}
\eps^{-1} ( \tau_\eps - T(x_0) ) \stackrel{\Pp}{\longrightarrow}
 - \frac{1}{b(a_2)}
\Phi_{x_0}(T( x_0) )\int_{0}^{T (x_0)}\Phi_{x_0}^{-1}(s)\sigma(S^s x_0)d\tilde W(s), \quad \eps\to0,
\end{equation}
where $S^t x_0$ is the flow generated by the vector field $-b$, the time $T(x_0)$ solves $S^{T(x_0)}x_0=a_2$, and
$\Phi_{x_0}$ is the linearization of $S$ near the orbit of $x_0$.
The limit is clearly a centered Gaussian random variable. To compute its variance we must first solve
\[
\frac{d}{dt}\Phi_{x_0} (t) = - b^\prime (S^t x_0 ) \Phi_{x_0} (t), \quad \Phi_{x_0} (0)=1.
\]
 The solution to this linear ODE is
\[
\Phi_{x_0} (t)= \exp \left \{- \int_0^t b^\prime (S^s x_0 )ds \right \},
\]
so that after the change of variables $u=S^s x_0$ in the integral, we get
\[
\Phi_{x_0} (t)= \frac{b(S^t x_0)}{b(x_0)}.
\]
Using this expression and It\^o isometry for the limiting random variable in~\eqref{eqn: conv_example}, we get that the variance of such random variable is
\[
\int_0^{T(x_0)} \frac{ \sigma^2 ( S^tx_0 ) }{ b^2 ( S^t x_0 ) } dt.
\]
We can now use the change of variable $u=S^s x_0$ to get the expression in Theorem~\ref{thm: conditioned}.
\end{proof}

\section{Proof of the main result}\label{sec:proof-levinson}

With high probability, at time $T$ the process $X_\eps$ is close to $z$ and the hitting
time~$\tau_\eps$ is close to $T$. The idea of the proof is that while the diffusion is close to~$z$, the process
may be approximated very well by motion with constant velocity~$b(z)$.

We start with a lemma and postpone its proof to the end of this section to keep continuity of the text.

\begin{lemma}
\label{lemma: Bakhtin}
Let $X_\eps$ be the solution of the SDE~\eqref {eq:PrincipalEquation} with initial condition~\eqref{eq:initial_condition}.
Let
\begin{align}
\Theta_\eps (t) &=\eps^{\alpha_2-\alpha}\Phi_{x_0}(t) \xi_\eps+  \eps^{\alpha_1-\alpha}\Phi_{x_0}(t) \int_0^t \Phi_{x_0} (s)^{-1} \Psi_0( S^s x_0) ds \notag \\
&\quad + \eps^{1-\alpha} \Phi_{x_0}(t) \int_0^t \Phi_{x_0} (s)^{-1}\sigma( S^s x_0) dW(s).  \label{eqn: SDE_Theta}
\end{align} Then,
\[
X_\eps(t)=S^tx_0 + \eps^\alpha \phi_\eps (t)
\]
holds almost surely for every $t>0$,
where $\phi_\eps(t)=\Theta_\eps(t)+ r_\eps(t)$, and $r_\eps$ converges to 0 uniformly over compact time intervals in probability.

If $\xi_\eps\to\xi_0$ in distribution, then for any $T>0$,
$\phi_\eps \to \phi_0$  in distribution in $C[0,T]$ equipped with uniform norm,
 where $\phi_0$ is the stochastic process defined in~\eqref{eqn: phi_0_def}.

If $\xi_\eps\to\xi_0$ in probability or $\alpha_2>\alpha$, then  the uniform convergence for $\phi_\eps$
also holds in probability.
\end{lemma}

\begin{remark}\rm
This lemma gives the first-order approximation for $X_\eps(t)$. Higher-order approximations in the spirit of~\cite{Blagoveschenskii:MR0139204}
are also possible. They can be used to refine Theorem~\ref{thm: Main}.
\end{remark}

Our task now is to analyze the process $X_\eps(t)-z$ for $t$ close to $T$. Let us first estimate the deviation of the flow $S$
from the motion with costant velocity~$b(z)$. Let
\begin{equation}
r_\pm (t,x)= S^{\pm t}x-\left( x\pm tb(z) \right),\quad t>0,\ x\in\R^d. \label{eqn: def_r_pm}
\end{equation}
\begin{lemma}
\label{lemma: rectification}
There are constants $C_1$ and $C_2$ so that for any $t>0$ and $x\in\R^d$
\[
\sup_{s\leq t}\left|  r_\pm(s,x) \right| \leq C_1 e^{C_2 t }( t |x-z | + t^2).
\]
\end{lemma}

\begin{proof}
We prove the result for $r_+$. The analysis of $r_-$ is similar since $S^{-t}x$ is the solution to the ODE
\[
\frac{d}{dt} S^{-t}x = -b (S^{-t}x ).
\]
Let $L>0$ be the Lipschitz constant of $b$. The proof follows from the inequalities:
\begin{align*}
\left| r_+(t,x) \right| &\leq \int_0^t \left| b(S^sx)-b(z)\right|ds \\
& \leq L \int_0^t \left| S^sx-z\right|ds \\
& \leq L\int_0^t \left|r_+ (s,x) \right|ds + L\int_0^t \left | x + sb(z)-z\right| ds \\
& \leq L\int_0^t \left|r_+ (s,x) \right|ds + L\int_0^t |x-z|ds + L\int_0^t s|b(z)| ds \\
&\leq  L\int_0^t \left| r_+(s,x) \right|ds + L t |x-z| + t^2 L|b(z)|/2.
\end{align*}
The result follows as an application of Gronwall's lemma.
\end{proof}

\begin{lemma} \label{lemma: X_before_after}
Let $\gamma \in (\alpha/2, \alpha)$. Then, there are two a.s.-continuous stochastic processes $\Gamma_{\eps,\pm}$ such that
\[
\sup_{t\in[0,\eps^\gamma]} |\Gamma_{\eps,\pm} (t)| \stackrel{\Pp}{\longrightarrow} 0, \quad \eps \to 0,
\]
and almost surely for any $t\in[0,\eps^\gamma]$
\begin{equation}
X_\eps (T-t)=z-tb(z)+\eps^\alpha \left( \phi_\eps (T-t) +\Gamma_{\eps,-}(t) \right) \label{eqn: X_before}
\end{equation} and
\begin{equation}
X_\eps (T+t)=z+tb(z)+\eps^\alpha \left( \Phi_z (t) \phi_\eps (T) +\Gamma_{\eps,+}(t) \right). \label{eqn: X_after}
\end{equation}
\end{lemma}
\begin{proof}
Due to Lemma~\ref{lemma: Bakhtin},  the flow property, and~\eqref{eqn: def_r_pm} we have
\begin{align*}
X_\eps (T-t) &= S^{T-t}x_0 + \eps^\alpha \phi_\eps (T-t) \\
&= S^{-t}z + \eps^\alpha \phi_\eps (T-t) \\
&=z-tb(z) + r_-(t,z)  + \eps^\alpha \phi_\eps (T-t).
\end{align*}
The first estimate with $\Gamma_{\eps,-} (t) = \eps^{-\alpha} r_-(t,z)$ follows from Lemma~\ref{lemma: rectification} for $x=z$.

Due to Strong Markov property and Lemma~\ref{lemma: Bakhtin} the process $\tilde X_\eps (t) = X_\eps (t + T)$ is a solution of the initial value problem
\begin{align*}
d\tilde X_\eps(t) &= (b( \tilde X_\eps(t))+\eps^{\alpha_1}\Psi_\eps(\tilde X_\eps(t)))dt + \eps \sigma (\tilde X_\eps(t)) d\tilde W, \\
\tilde X_\eps (0) &= X_\eps (T)=z+\eps^\alpha \phi_\eps (T),
\end{align*}
with respect to the Brownian Motion $\tilde W(t)=W(t+T)-W(T)$. So, again, applying Lemma~\ref{lemma: Bakhtin} to this shifted equation, we obtain $\tilde X_\eps (t)=S^tz+\eps^\alpha \hat \phi_\eps (t)$, where, for $t>0$
\begin{align*}
\hat \phi_\eps (t) = \Phi_z (t) \phi_\eps (T) + \theta_\eps (t),
\end{align*}
and
\[
\theta_\eps (t)= \eps^{1-\alpha}   \Phi_z (t) \int_0^t \Phi_z (s)^{-1}\sigma (S^s z)d\tilde W(s)
+ \eps^{\alpha_1-\alpha}\Phi_z(t) \int_0^t \Phi_z (s)^{-1} \Psi_0( S^s z) ds + \tilde r_\eps (t),
\] where $\tilde r_\eps$ converges to 0 uniformly over compact time intervals in probability.
Then due to~\eqref{eqn: def_r_pm},
\begin{align*}
\tilde X_\eps (t) &=S^tz+\eps^\alpha (\Phi_z (t) \phi_\eps (t) + \theta_\eps (t) ) \\
&= z+tb(z) + r_+(t,z)+\eps^\alpha (\Phi_z (t) \phi_\eps (t) + \theta_\eps (t) ).
\end{align*}
Hence, with $\Gamma_{\eps,+} (t)=\theta_\eps (t) + \eps^{-\alpha} r_+(t,z)$ the result is a consequence of Lemma~\ref{lemma: rectification}.
\end{proof}

\bigskip

Let us now parametrize, locally around $z$, the hypersurface $M$ as a graph of a $C^2$-function $F$ over $T_zM$, i.e.,
$y\mapsto z+y+F(y)\cdot b(z)$ gives a $C^2$-parametrization of a neighborhood of $z$ in $M$ by a neighborhood of $0$ in $T_zM$. Moreover, $DF(0)=0$ so that
$|F(y)|=O(|y|^2)$, $y\to 0$.
With this definition, it is clear that, for $w\in \R^d$ with $w-z$ small enough, $w\in M$ if and only if $\pi_b(w-z)=F(\pi_M(w-z))$.

Let us define
\begin{align*}
\Omega_{1,\eps}&=\bigl\{\tau_\eps = \inf \{  t\geq 0: \pi_b \left(X_\eps(t)-z \right) = F\left( \pi_M \left(X_\eps(t)-z \right)  \right)\}\bigr\},\\
\Omega_{2,\eps}&= \left \{  |\tau_\eps - T| \leq \eps^\gamma \right \},\\
\Omega_\eps&=\Omega_{1,\eps}\cap \Omega_{2,\eps}.
\end{align*}

\begin{lemma}\label{lm:omega_eps_to_1} $\Pp(\Omega_{\eps})\to 1$ as $\eps\to 0$.
\end{lemma}

\begin{proof}
The definition of $F$ and Lemma~\ref{lemma: Bakhtin} imply that  as $\eps\to 0$, $\Pp(\Omega_{1,\eps})\to 1$.

We use~\eqref{eqn: X_after} to conclude that
\[
\pi_b \left( X_\eps (T+\eps^\gamma) -z \right)=\eps^\gamma  \left( 1+ \eps^{\alpha-\gamma} \pi_b \left( \Phi_z(\eps^\gamma) \phi_\eps(T) + \Gamma_{\eps,+} (\eps^\gamma) \right) \right),
\]
and
\[
F \left( \pi_M \left( X_\eps (T+\eps^\gamma) -z \right) \right)=F \left( \eps^\alpha \pi_M \left(  \Phi_z(\eps^\gamma) \phi_\eps(T) + \Gamma_{\eps,+} (\eps^\gamma) \right) \right).
\]
Since  $|F(x)|= O(|x|^2)$, these estimates imply that
\begin{multline*}
\limsup_{\eps \to 0} \Pp\left( \left\{  \tau_\eps > T+\eps^\gamma \right \}\cap \Omega_{1,\eps} \right)\\ \le \limsup_{\eps \to 0} \Pp \left\{  \pi_b \left( X_\eps (T+\eps^\gamma) -z \right)
\le F \left( \pi_M \left( X_\eps (T+\eps^\gamma) -z \right) \right) \right \} =0.
\end{multline*}

It remains to prove
\begin{equation}
\lim_{\eps \to 0} \Pp \left\{  \tau_\eps <T-\eps^\gamma \right \}=0.
\label{eq:rough_lower_estimate_exit_time}
\end{equation}
Let us denote the Hausdorff distance between sets by $d(\cdot,\cdot)$. Then
an obvious estimate
\[
d(\{S^tx_0: 0\le t\le T-\delta\},M)\ge c\delta
\]
 holds true for some $c>0$ and all sufficiently small $\delta>0$. Now~\eqref{eq:rough_lower_estimate_exit_time} follows from Lemma~\ref{lemma: Bakhtin},
and the proof is complete
\end{proof}

\begin{lemma} \label{prop: tau_conv}
Define $\tau_\eps^\prime = \tau_\eps - T $. Then,
\[
\eps^{-\alpha} \tau_\eps^\prime +  \pi_b \phi_\eps (T) \stackrel{\Pp}{\longrightarrow} 0, \quad \eps \to 0.
\]
\end{lemma}
\begin{proof}
 Let us define $A_\eps = \left \{  0 \leq \tau_\eps^\prime  \leq \eps^\gamma \right \}\cap\Omega_{1,\eps}$ and
$B_\eps = \left \{  -\eps^\gamma \leq \tau_\eps^\prime  < 0 \right \}\cap\Omega_{1,\eps}$, so that
$\Omega_\eps = A_\eps \cup B_\eps$.
We can use~\eqref{eqn: X_after} and the definition of $\Omega_{1,\eps}$ to get
\begin{align*}
\one_{A_\eps} \tau_\eps^\prime + \one_{A_\eps} \eps^\alpha  \pi_b \left( \Phi_z (\tau_\eps^\prime) \phi_\eps (T) +\Gamma_{\eps,+}(\tau_\eps^\prime) \right) =
\one_{A_\eps}F\left( \eps^\alpha \pi_M \left( \Phi_z (\tau_\eps^\prime) \phi_\eps (T) +\Gamma_{\eps,+}(\tau_\eps^\prime) \right) \right).
\end{align*}
This implies
\begin{align}
\notag
\one_{A_\eps} \eps^{-\alpha} \tau_\eps^\prime &=\eps^{-\alpha} \one_{A_\eps} F\left( \eps^\alpha \pi_M \left( \Phi_z (\tau_\eps^\prime) \phi_\eps (T) +\Gamma_{\eps,+}(\tau_\eps^\prime) \right) \right)
\\ \notag &\quad-\one_{A_\eps} \pi_b \left( \Phi_z (\tau_\eps^\prime) \phi_\eps (T) +\Gamma_{\eps,+}(\tau_\eps^\prime) \right) \\
\notag &= -\one_{A_\eps} \pi_b \left( \Phi_z (\tau_\eps^\prime) \phi_\eps (T)  \right) + r_{\eps,1} \\
&= -\one_{A_\eps} \pi_b  \phi_\eps (T)  +\one_{A_\eps} \pi_b \left( (I-\Phi_z (\tau_\eps^\prime) )\phi_\eps (T)  \right)+r_{\eps,1}, \label{eqn: tau_on_A}
\end{align}
where $r_{\eps,1}$ is a random variable that converges to $0$ in probability as $\eps \to 0$.

Likewise, since $\tau_\eps = T - (-\tau_\eps^\prime)$ and $\one_{B_\eps} \tau_\eps^\prime \le 0$, we can use ~\eqref{eqn: X_before} and the definition of $\Omega_{1,\eps}$ to see that
\[
\one_{B_\eps}\tau_\eps^\prime + \one_{B_\eps}\eps^\alpha \pi_b \left ( \phi_\eps (T+\tau_\eps^\prime) + \Gamma_{\eps,-} (-\tau_\eps^\prime) \right ) = \one_{B_\eps} F \left (\eps^\alpha\left ( \phi_\eps (T+\tau_\eps^\prime) + \Gamma_{\eps,-} (-\tau_\eps^\prime) \right ) \right ).
\]
Hence, proceeding as before, we see that
\begin{align*}
\one_{B_\eps} \eps^{-\alpha} \tau_\eps^\prime &= -\one_{B_\eps} \pi_b \phi_\eps (T+\tau_\eps^\prime) + r_{\eps,2} \\
&= -\one_{B_\eps} \pi_b \phi_\eps (T)+\one_{B_\eps} \pi_b \left( \phi_\eps (T)-\phi_\eps(T+\tau_\eps^\prime) \right)+r_{\eps,2}
\end{align*}
for some random variable $r_{\eps,2}$ such that $r_{\eps,2} \to 0$ in probability as $\eps \to 0$. Adding this identity and~\eqref{eqn: tau_on_A}, we see that on $\Omega_{\eps}$
\[
 \eps^{-\alpha} \tau_\eps^\prime=-\pi_b \phi_\eps (T)+\one_{A_\eps} \pi_b \left( ( I-\Phi_z (\tau_\eps^\prime) ) \phi_\eps (T)  \right)+\one_{B_\eps} \pi_b \left( \phi_\eps (T)-\phi_\eps(T+\tau_\eps^\prime) \right)+r_{\eps,1}+r_{\eps,2}.
\]
Due to Lemma~\ref{lm:omega_eps_to_1}, to finish the proof it is sufficient to notice that as $\eps \to 0$
\begin{equation}
\label{eqn: aux_1}
\sup_{0\leq t \leq \eps^\gamma } \left| (I-\Phi_z (t) ) \phi_\eps (T)  \right| \stackrel{\Pp}{\longrightarrow} 0,
\end{equation} and
\begin{equation} \label{eqn: aux_2}
\sup_{0\leq t \leq \eps^\gamma } \left| \phi_\eps (T)-\phi_\eps(T+t) \right|\stackrel{\Pp}{\longrightarrow} 0.
\end{equation}

\end{proof}

Lemma~\ref{prop: tau_conv} takes care of the time component in Theorem~\ref{thm: Main}. We shall consider the spatial component now.

 Let $A_\eps$ and $B_\eps$ be as in the proof of Lemma~\ref{prop: tau_conv}. Then, \eqref{eqn: X_after} implies
\begin{multline}
\one_{A_\eps} \left (  X_\eps (\tau_\eps) - z \right)\eps^{-\alpha} = \one_{A_\eps}\eps^{-\alpha} \tau_\eps^\prime b(z) + \one_{A_\eps} \left(  \Phi_z(\tau_\eps^\prime)\phi_\eps(T) + \Gamma_{\eps,+}(\tau_\eps^\prime) \right) \\
= \one_{A_\eps} \left( \eps^{-\alpha}\tau_\eps^\prime b(z) + \phi_\eps (T) \right) + \one_{A_\eps} \left[  (\Phi_z(\tau_\eps^\prime)- I )\phi_\eps(T) + \Gamma_{\eps,+}(\tau_\eps^\prime) \right] \label{eqn: spacial_after}
\end{multline}
Likewise, from~\eqref{eqn: X_before} we get that
\begin{multline}
\one_{B_\eps} \left (  X_\eps (\tau_\eps) - z \right)\eps^{-\alpha} = \one_{B_\eps}\eps^{-\alpha} \tau_\eps^\prime b(z) + \one_{B_\eps} \left(  \phi_\eps(T+\tau_\eps^\prime) + \Gamma_{\eps,-}(-\tau_\eps^\prime) \right) \\
= \one_{B_\eps} \left(\eps^{-\alpha} \tau_\eps^\prime b(z) + \phi_\eps (T) \right) + \one_{B_\eps} \left[  (\phi_\eps(T+\tau_\eps^\prime)-\phi_\eps(T) ) + \Gamma_{\eps,-}(-\tau_\eps^\prime) \right] \label{eqn: spacial_before}.
\end{multline}
Adding~\eqref{eqn: spacial_after} and~\eqref{eqn: spacial_before} and proceding as in the proof of Lemma~\ref{prop: tau_conv} we see that
\[
\left (  X_\eps (\tau_\eps) - z \right)\eps^{-\alpha} - \pi_M \phi_\eps (T)= \left(\eps^{-\alpha} \tau_\eps^\prime +\pi_b \phi_\eps (T) \right)b(z) + \rho_\eps,
\] where, due to~\eqref{eqn: aux_1},~\eqref{eqn: aux_2} and Lemma~\ref{lemma: X_before_after}, $\rho_\eps \to 0$ in probability as $\eps \to 0$.  From this expression and Lemma~\ref{prop: tau_conv} we get that
\[
\left (  X_\eps (\tau_\eps) - z \right)\eps^{-\alpha} - \pi_M \phi_\eps (T) \overset{\Pp}{\longrightarrow} 0, \quad \eps \to 0.
\]
Then, using this and the convergence in Lemma~\ref{prop: tau_conv}
\[
\eps^{-\alpha}(\tau_\eps-T, X_\eps(\tau_\eps)-z)=R_\eps + G(\phi_\eps (T) ),
\]
where $R_\eps$ is a random variable such that $R_\eps \to 0$ in probability as $\eps \to 0$. $G$ is the continuous function $x \mapsto (-\pi_b x, \pi_M x )$. Hence, Theorem~\ref{thm: Main} follows from the convergence in~Lemma~\ref{lemma: Bakhtin}.

It remains to prove Lemma~\ref{lemma: Bakhtin}, the core of the proof of the main result.

\begin{proof}[Proof of Lemma~\ref{lemma: Bakhtin}]
Let $\Delta_\eps^t= X_\eps (t) -S^t x_0$ and note that it satisfies the equation
\[
d\Delta_\eps^t= \left(  \left( b(X_\eps (t))-b(S^t x_0) \right) + \eps^{\alpha_1} \Psi_\eps(X_\eps (t)) \right)dt + \eps \sigma (X_\eps (t) ) dW(t),
\]
with initial condition $\Delta_\eps^0 = \eps^{\alpha_2} \xi_\eps$. We want to study the properties of this equation. We start with the difference in $b$.
Since $b$ is a $C^2$ vector field, we may write
\begin{align}
b(X_\eps (t))-b(S^t x_0) &=Db(S^t x_0) \Delta_\eps^t + Q_1(S^t x_0, \Delta_\eps^t).
 \label{eqn: expansionb}
\end{align}
Also, we can write
\begin{equation} \label{eqn: expansion_Psi}
\Psi_\eps ( X_\eps (t) )=\Psi_0 ( S^t x_0)+Q_2 (S^t x_0,\Delta_\eps^t)+R_\eps(S^t x_0),
\end{equation} and
\begin{equation} \label{eqn: expansionSigma}
\sigma ( X_\eps (t) )=\sigma ( S^t x_0)+Q_3(S^t x_0,\Delta_\eps^t),
\end{equation}
where
\[
 R_\eps(x)=\Psi_\eps(x)-\Psi_0(x)=o(1),\quad \eps\to 0,
\]
uniformly in $x$;
$Q_i:\R^d \times \R^d \to \R^d$, $i=2,3$ satisfies
\begin{equation}
|Q_i (u,v)|\leq K   |v|,\quad u,v\in\R^d  \label{eqn: bound_Q2}.
\end{equation}
We can assume that the constant $K>0$ in~\eqref{eqn: bound_Q1} and~\eqref{eqn: bound_Q2}
is the same for simplicity of notation.

Let $Q=Q_1+ \eps^{\alpha_1} Q_2+\eps^{\alpha_1}R_\eps$. Combine~\eqref{eqn: expansionb}, \eqref{eqn: expansion_Psi}, and \eqref{eqn: expansionSigma} to get
\begin{align} \notag
d\Delta_\eps^t = &\left(  A(t)\Delta_\eps^t + \eps^{\alpha_1} \Psi_0 ( S^t x_0) + Q(S^t x_0,  \Delta_\eps^t) \right)dt \\
&\quad + \eps \left(  \sigma ( S^t x_0 ) +Q_3(S^t x_0,\Delta_\eps^t) \right) dW(t), \label{eqn: SDE_Delta} \\
\Delta_\eps^0 = &\eps^{\alpha_2} \xi_\eps.
\end{align}
Hence, applying Duhamel's principle to ~\eqref{eqn: SDE_Delta} and using~\eqref{eqn: SDE_Theta},  we get
\begin{align} \notag
\Delta_\eps^t & = \eps^\alpha \Theta_\eps (t)+ \Phi_{x_0} (t) \int_0^t \Phi_{x_0} (s)^{-1}Q(S^s x_0,  \Delta_\eps^s )ds\\ \notag
&\quad + \eps \Phi_{x_0} (t) \int_0^t \Phi_{x_0} (s)^{-1}Q_3(S^s x_0, \Delta_\eps^s )dW(s) \\ \label{eqn: expansion_X}
&= \eps^\alpha \Theta_\eps (t) + \Theta'_\eps (t),
\end{align} where $\Theta'_\eps$ is defined by~\eqref{eqn: expansion_X}.
A simple inspection of~\eqref{eqn: SDE_Theta} shows that $\left( \Theta_\eps \right)_{\eps>0}$
 converges in  distribution in $C(0,T)$ to the process $\phi_0 (t)$. This convergence is in probability if $\alpha_2 > {\alpha}$
or $\xi_\eps\to\xi_0$ in probability.
 Therefore, the lemma will follow with $\phi_\eps = \Theta_\eps + \eps^{-\alpha} \Theta'_\eps$ if we show that
\begin{equation} \label{eqn: estimate_Theta'}
\eps^{-\alpha} \sup_{t \leq T} | \Theta'_\eps (t) | \overset{\Pp}{\longrightarrow} 0, \quad \eps \to 0.
\end{equation}
 For any $\delta \in (1/2,1)$, we introduce the stopping time
\[
l_\eps ( \delta )=\inf \left\{ t>0: |\Delta_\eps^t| \geq \eps^{\alpha \delta } \right \}.
\]
Now, $\Theta'_\eps=\Theta'_{\eps,1}+\eps \Theta'_{\eps,2}$, where
\[
\Theta'_{\eps,1} (t) = \Phi_{x_0} (t) \int_0^t \Phi_{x_0} (s)^{-1}Q(S^s x_0, \Delta_\eps^s)ds,
\] and
\[
\Theta'_{\eps,2} (t)=\eps \Phi_{x_0} (t) \int_0^t \Phi_{x_0} (s)^{-1}Q_3(S^s x_0,  \Delta_\eps^s )dW(s).
\]
Bounds~\eqref{eqn: bound_Q1}, and \eqref{eqn: bound_Q2} imply
\begin{equation}
\sup_{t\leq T\wedge l_\eps(\delta) }|\Theta'_{\eps,1} (t) | = O( \eps^{2 \alpha \delta}  + \eps^ {{\alpha_1}+\alpha \delta})+o(\eps^{\alpha_1})=o(\eps^{\alpha}). \label{eqn: Theta_prime_1}
\end{equation}
Likewise,~\eqref{eqn: bound_Q2} for $Q_3$ and BDG inequality imply that for any $\kappa>0$ there is a constant $K_\kappa$ such that
\begin{equation}
\Pp \left \{ \sup_{t\leq T\wedge l_\eps(\delta) }|\Theta'_{\eps,2} (t) | > K_\kappa \eps^ {1+\alpha\delta} \right \} < \kappa \label{eqn: Theta_prime_2}
\end{equation} for all $\eps>0$ small enough. Then, this together with~\eqref{eqn: Theta_prime_1} imply that
\begin{equation}
\eps^{-\alpha\delta} \sup_{t \leq T \wedge l_\eps (\delta)} | \Theta'_\eps (t) | \overset{\Pp}{\longrightarrow} 0, \quad \eps \to 0.
\label{eq:theta-prime-small}
\end{equation}
Then, if $l_\eps (\delta) < T$ we use~\eqref{eqn: expansion_X} to get
\begin{align*}
1 &=\eps^{-\alpha\delta} \sup_{t\leq T \wedge l_\eps (\delta)} | \Delta_\eps^t | \\
& \leq \eps^{\alpha(1-\delta)} \sup_{t\leq  T \wedge l_\eps (\delta)} | \Theta_\eps (t) | +\eps^{-\alpha\delta}
\sup_{t\leq  T \wedge l_\eps (\delta)}  | \Theta'_\eps (t) |.
\end{align*} The r.h.s. converges to 0 in probability due to~\eqref{eq:theta-prime-small} and the tightness of distributions of $\Theta_\eps$. Hence, $\Pp \{ l_\eps (\delta) < T \} \to~0$ as $\eps~\to~0$. Using $T$ instead of $T \wedge l_\eps ( \delta) $ in~\eqref{eqn: Theta_prime_1} and~\eqref{eqn: Theta_prime_2}, we see that with the choice of $\delta > 1/2$, \eqref{eqn: estimate_Theta'} follows and the proof is finished.
\end{proof}

\section{Proof of Lemmas~\ref{lemma: conditioned_drift} and~\ref{lemma: b_approx}}\label{sec:aux-1-d}

\begin{proof}[Proof of Lemma~\ref{lemma: conditioned_drift}] Let us find the generator of the conditioned diffusion. To that end we denote the generator of the original diffusion
by $L_\eps$:
\begin{equation}
\label{eq:generator1}
 L_\eps f(x)=b(x)f'(x)+\frac{\eps^2}{2}\sigma^2(x)f''(x)=\lim_{t\to 0}\frac{\E_x f(X_\eps)-f(x)}{t},
\end{equation}
where $f$ is any bounded $C^2$-function with bounded first two derivatives and $\E_x$ denotes expectation with respect to the
measure $\Pp_x$,
the element of the Markov family describing the Markov process emitted from point $x$.

Let us denote $u_\eps(x)=\Pp_x(B_\eps)$. This function solves the following boundary-value problem for the
backward Kolmogorov equation:
\begin{align*}
L_\eps u_\eps(x)=0,\quad u_\eps(a_1)=0,\quad
u_\eps(a_2)=1.
\end{align*}
Using~\eqref{eq:generator1}, it is easy to check that a unique solution is given by
\[
 u_\eps(x)=\frac{\int_{a_1}^xh_\eps(y)dy}{\int_{a_1}^{a_2}h_\eps(y)dy},
\]
where $h_\eps$ is defined in~\eqref{eq:h_eps}.

Now we can compute the generator $\bar L$ of the conditioned flow. For any smooth and bounded function $f\in C^2$ with
bounded first two derivatives, we can write
\begin{align*}
\E_x[f(X_\eps)|B_\eps]&=u^{-1}(x)\E_x f(X_\eps(t))\one_{B_\eps}
\\&=u_\eps^{-1}(x)\E_x f(X_\eps(t))\one_{B_\eps}\one_{\{\tau_\eps\ge t\}}+R_\eps
\\&=u_\eps^{-1}(x)\E_x\E_x [f(X_\eps(t))\one_{B_\eps}\one_{\{\tau_\eps\ge t\}}|\Fc_t]+R_\eps
\\&=u_\eps^{-1}(x)\E_x f(X_\eps(t)) \Pp_{X_\eps(t)}(B_\eps)+R_\eps
\\&=u_\eps^{-1}(x)\E_x f(X_\eps(t)) u_\eps(X_\eps(t))+R_\eps,
\end{align*}
where
\[
|R_\eps|=u_\eps^{-1}(x)|\E_x f(X_\eps)\one_{B_\eps}\one_{\{\tau_\eps<t\}}|\le C(x)\Pp\{\tau_\eps<t\}=o(t)
\]
for some $C(x)>0$. Therefore, we
obtain
\begin{align*}
\bar L f(x)&=\lim_{t\to 0} \frac{\E_x[f(X_\eps(t))|B_\eps]-f(x)}{t}
\\ &=\lim_{t\to 0}\frac{u_\eps^{-1}(x)\E_x f(X_\eps(t)) u_\eps(X_\eps(t))-f(x)}{t}
\\ &=\frac{1}{u_\eps(x)}\lim_{t\to 0}\frac{\E_x f(X_\eps(t)) u_\eps(X_\eps(t))-f(x)u_\eps(x)}{t}
\\ &= \frac{1}{u_\eps(x)} L(fu_\eps) (x)
\\ &= \left(b(x)+\eps^2\sigma^2(x)\frac{u'_\eps(x)}{u_\eps(x)}\right)f'(x)+\eps^2\frac{\sigma^2(x)}{2}f''(x).
\\ &= \left(b(x)+\eps^2\sigma^2(x)\frac{h_\eps(x)}{\int_{a_1}^x h_\eps(y)dy}\right)f'(x)+\eps^2\frac{\sigma^2(x)}{2}f''(x),
\end{align*}
completing the proof.
\end{proof}

\begin{proof}[Proof of Lemma~\ref{lemma: b_approx}] The proof is a variation of Laplace's method.
Let
\begin{equation}
\Phi (x)=2 \int_{a_1}^x \frac{ b(y)}{ \sigma^2 (y) } dy,\quad x\ge a_1, \label{eqn: def_Phi}
\end{equation}
so that $h_\eps (x) = e^{ -\Phi (x) / \eps^2 }$.  We take any $\beta \in (1,2)$ and break the integral of $h_\eps$ in two parts:
\[
\int_{a_1}^x e^{ -\Phi (y) /\eps^2 } dy = I_{\eps,1} (x) + I_{\eps,2} (x),
\]
where
\begin{equation} \label{eqn: I_eps,1}
I_{\eps,1} (x) =\int_{a_1}^{ x-\eps^\beta} e^{ -\Phi (y) /\eps^2 } dy,
\end{equation}
and
\begin{equation} \label{eqn: I_eps,2}
I_{\eps,2} (x) =  \int_{ x-\eps^\beta} ^ x e^{ -\Phi (y) /\eps^2 } dy.
\end{equation}
The idea is to prove that $I_{\eps,1}$ is exponentially smaller than $I_{\eps,2}$ and then estimate~$I_{\eps,2}$.

We start with some preliminaries for the function $\Phi$. Since both $b$ and $\sigma$ are $C^1$ and $\sigma \neq 0 $ in $[a_1, a_2 ]$ we conclude that $\Phi$ is a $C^2$ function so that  we can find a function $R:\R \times \R \to \R$ and a number $\delta_0>0$ such that for every $x,y \in [a_1,a_2+\delta_0]$, we have the expansion
\begin{equation} \label{eqn: Taylor_phi}
\Phi (y)= \Phi (x) + \Phi ^ \prime (x) (y -x) + R(x, y-x ),
\end{equation}
and
\begin{equation}
|R(x,v)| \leq K_1 |v|^2,\quad x \in [a_1, a_2+\delta_0], v \in \R, \label{eqn: bound_R}
\end{equation}
for some $K_1>0$.

To estimate $I_{\eps,1}$, we introduce
\[J_{\eps,1} (x) = \frac{e^{\Phi(x) / \eps^2}}{\eps^2\sigma^2(x)} I_{\eps, 1 } (x),\quad x \in [a_1, a_2+\delta_0].\]
Since $\Phi$ is decreasing, we have that for some constant $K_2>0$ independent of $x \in [a_1, a_2+\delta_0]$,
\begin{align}
J_{\eps,1} (x) \leq \frac{K_2}{\eps^2}e^ { ( \Phi(x)-\Phi(x-\eps^\beta) )  / \eps^2}. \label{eqn: estimate_2}
\end{align}
Since $\beta<2$ and $\Phi^\prime$ is negative and bounded away from zero, we conclude that there is
$\alpha(\eps)$ such that $\alpha(\eps)=o (\eps ^ 2 )$ as $\eps\to 0$ and
\begin{equation}
\label{eq:sup_estimate-on-J_1}
 \sup_{x \in [a_1, a_2+\delta_0]} J_{\eps,1} (x) \leq \alpha(\eps).
\end{equation}

We now estimate $I_{\eps,2}$. Using expansion~\eqref{eqn: Taylor_phi} and the change of variables $u=- \Phi(x) (y-x)/\eps^2$, we get
\begin{align} \notag
I_{\eps,2}(x)&= e^{ -\Phi (x) /\eps^2 } \int_{ x- \eps^\beta}^x e^{ -\Phi ^ \prime (x) (y -x)/\eps^2 - R(x, y-x ) /\eps^2 } dy \\ \notag
&=-\frac{  \eps^2  } {\Phi^\prime (x) } e^{ -\Phi (x) /\eps^2 }  \int_{\Phi^\prime(x)/\eps^{2-\beta}}^0 e^{ u - R(x,-\eps^2 u/ \Phi^\prime(x) ) /\eps^2 } du \\
&=-\frac{  \eps^2 \sigma^2 (x) } {2 b(x) } e^{ -\Phi (x) /\eps^2 } J_{\eps,2} (x), \label{eqn: Laplace_int}
\end{align}
where we use~\eqref{eqn: def_Phi} to compute the derivative of $\Phi$, and we define $J_{\eps,2}$ by~\eqref{eqn: Laplace_int}.
Hence, combining ~\eqref{eqn: estimate_2} with the definition of $b_\eps$ and~\eqref{eqn: Laplace_int}, we get
\begin{align*}
b_\eps (x) = b(x) + \frac{1}{J_{\eps,1}(x) - \frac{1}{2 b(x)} J_{\eps,2} (x) } .
\end{align*}
Due to \eqref{eq:sup_estimate-on-J_1}, the proof will be complete once we prove that for sufficiently small $\delta>0$,
\[
\limsup_{\eps \to 0}  \eps^{-2}\left(\sup_{x \in [x_0-\delta, a_2+\delta]} | J_{\eps,2} (x) -1 | \right) < \infty.
\]

Note that for any $\delta\in(0,x_0-a_1)$, some constant $K_3=K_3(\delta)>0$ and all $x \in [x_0-\delta,a_2+\delta]$,
\begin{align} \notag
|J_{\eps,2} (x)-1| &=\Bigl| \int_{\Phi^\prime (x)/\eps^{2-\beta} }^0 e^u (1 - e^{ - R(x,-\eps^2 u/ \Phi^\prime(x) ) /\eps^2 } )du  \\ \notag
& \quad +\int_{-\infty}^{\Phi^\prime (x)/\eps^{2-\beta} } e^u du\Bigr| \\
&\leq  \int_{\Phi^\prime (x)/\eps^{2-\beta}}^0 e^u | 1 - e^{ - R(x,-\eps^2 u/ \Phi^\prime(x) ) /\eps^2 } |du  + e^{- K_3/\eps^{2-\beta} }. \label{eqn: approx_int}
\end{align}
Using~\eqref{eqn: bound_R} we see that for some constant $K_4>0$ independent of $x \in [x_0-\delta,a_2+\delta]$ and $u \in \R$,
\[
| R(x,-\eps^2 u/ \Phi^\prime(x) ) |/\eps^2 \leq K_4 \eps^2 u^2.
\]
In particular,
\[
\sup_{x \in [x_0-\delta,a_2+\delta]}\sup_{ u \in [\Phi^\prime (x)/\eps^{2-\beta} , 0] } | R(x,-\eps^2 u/ \Phi^\prime(x) ) |/\eps^2 \leq K_4 \eps^{2 ( \beta -1) }.
\]
Since $\beta>1$, the r.h.s.\ converges to $0$ and we can apply a basic Taylor estimate which implies that for all $\eps>0$ small enough,
\[
 \sup_{x \in [x_0-\delta,a_2+\delta]} \sup_{ u \in [\Phi^\prime (x)/\eps^{2-\beta} , 0] } | 1 - e^{ - R(x,-\eps^2 u/ \Phi^\prime(x) ) /\eps^2 } | \leq K_5 \eps^2 u^2,
\]
for some $K_5>0$. Using this fact in the integral of~\eqref{eqn: approx_int}, we can find a constant $K_6=K_6(\delta)>0$ such that
\[
\sup_{x \in [x_0-\delta, a_2+\delta]}|J_{\eps,2}(x)-1| \leq K_6 \eps^2 + e^{-K_3/\eps^{2-\beta}},
\]
which finishes the proof.
\end{proof}

\bibliographystyle{plain}
\bibliography{happydle}
\end{document}

config: arxiv


TITLE: Examples of (Phi,Gamma)-modules
QUESTION [5 upvotes]: What is the (Phi,Gamma)-module of an elliptic curve over Z_p, expressed by a direct construction ?

REPLY [12 votes]: (By an elliptic curve over $\mathbb Z_p$, I assume you mean an elliptic scheme over spec $\mathbb Z_p$, or which amount to the same, an elliptic curve over $\mathbb Q_p$ which has good reduction.) The Galois representation on $V_p(E)$ is then crystalline, so its $(\phi,\Gamma)$-module $D$ will be triangulate -- in other words, an extension of
two $(\phi,\Gamma)$-modules of dimension $1$, a sub-object $D_1$ and a quotient $D_2=D/D_1$.
As you probably know (since otherwise you would have begun by this case), a
$(\phi,\Gamma)$-module of dimension $1$ is described by a continuous character $\delta: \mathbb Q_p^\ast \rightarrow \mathbb Q_p^\ast$. More precisely, it is always isomorphic
to the Robba ring as a module over the Robba ring, with actions that are twisted from the original one, which  by $\delta_{|\mathbb Z_p^\ast}$ for the $\Gamma$-action and by $\delta(p)$ for the $\phi$-action.
Now let us describe the characters $\delta_1$ and $\delta_2$ corresponding to $D_1$ and $D_2$.
Let $\alpha$ and $\beta$ be the two roots of the polynomial $X^2-a_pX + p =0$, ordered so that $v_p(\alpha) < v_p (\beta)$. Here $a_p$ is as usual $E(\mathbb F_p) -1 - p$, and $v_p$ is the $p$-valuation. So let $\delta_1$ be the character which sends $z \in \mathbb Z^\ast_p$ to $1$ and $p$ to $\alpha$ (so the character $z \mapsto \alpha^{v_p(z)}$ in short),
and let $\delta_2$ be the character which sends $z \in \mathbb Z^\ast_p$ to $1$ and $p$ to $\beta$.
This describes $D_1$ and $D_2$. Am I done? not quite. I need to say which extension
it is. But I have to run, so I will finish later... (later...) So if you go look to the paper
of Colmez on triangulline representations, you will see that there is a result computing the Ext group $Ext^1(D_1,D_2)$ in the category of $(\phi,Gamma)$-modules, and that in our case
the dimension is $1$. So the extension $D$ has two possibility: either it is trivial, or it is non-trivial (and then the proof of that Colmez's theorem give you and explicit description of what it is).
When your elliptic curve $E$ is super singular, $D$ is always the non-trivial extension.
When it is ordinary, it is more complicated. Let $\rho_{E,p}$ be the Galois representation
of $G_{\mathbb Q_p}$ on the Tate module of $E$. It is always reducible, but it may be decomposable or not (with a conjecture saying that it is decomposable if and only if E has CM). Well the extension $D$ is split if and only if $\rho_{E,p}$ is decomposable.
Okay, this describes explicitly $D$. I practice, it is not fundamental to know when $D$ is a split extension or not, the fact that it is an extension of the very concrete
$D_1$ and $D_2$ is enough. If you want to understand all this better, I think that COlmez' paper on triangulline representation is the best place to start.


TITLE: Binomial theorem for prime exponent
QUESTION [4 upvotes]: Could you explain to me why for prime $p$ we have the following?
$$(x+y)^p - (x^p + y^p)= x^p +  \binom{p}{1}x^{p-1}y +  \binom{p}{2}x^{p-2}y^2 +  \binom{p}{p-1}xy^{p-1} + y^p.$$
I found it here: Abstract Algebra - Solutions to Homework (pdf),
solution 5d, page 3.
I would really appreciate your help.
Thank you.

REPLY [3 votes]: As written, what's in the notes is incorrect. We should have
\begin{align*}
(x + y)^p - (x^p + y^p) &= x^p + \begin{pmatrix} p\\ 1\end{pmatrix}x^{p - 1}y + \begin{pmatrix} p\\ 2\end{pmatrix} x^{p - 2}y^2 + \ldots + \begin{pmatrix} p\\ p - 1\end{pmatrix}x y^{p - 1} + y^p - (x^p + y^p)\\
&= \begin{pmatrix} p\\ 1\end{pmatrix}x^{p - 1}y + \begin{pmatrix} p\\ 2\end{pmatrix} x^{p - 2}y^2 + \ldots + \begin{pmatrix} p\\ p - 1\end{pmatrix}x y^{p - 1}\\
&= \sum_{k = 1}^{p-1}\begin{pmatrix}p\\k\end{pmatrix}x^{p - k} y^k.
\end{align*}
I can only assume the writer of the notes got hungry and left in the middle of writing the solutions, and forgot to write a few terms ($\ldots$) upon returning.


TITLE: Evaluate $ \lim_{x \to +\infty}\frac{2x}{(x-3)}$
QUESTION [1 upvotes]: Evaluate the limit as $ \lim_{x \to +\infty}$ and prove the result: $f(x) = \frac{2x}{(x-3)}$
I know that the limit is $2$ but having a hard time to find the right $\epsilon$. Please help

REPLY [1 votes]: To show a limit, you should not find the $\epsilon$. Rather you will be provide with any $\epsilon$ and your goal, in this case, is to find a value $M(\epsilon)$ such that for $x>M(\epsilon)$, we have
$$\left\vert \dfrac{2x}{x-3} - 2 \right\vert < \epsilon$$
To find a candidate $M(\epsilon)$, let us consider $\dfrac{2x}{x-3} - 2$. We have
$$\dfrac{2x}{x-3} - 2 = \dfrac6{x-3}$$ To make this less than $\epsilon$, one candidate $M(\epsilon)$ is $3+\dfrac6{\epsilon}$. Now check if this candidate does indeed satisfy our expectation, i.e., our definition of the limit.


TITLE: Completion of quotient of polynomial ring
QUESTION [5 upvotes]: Hartshorne's Algebraic Geometry uses the following facts on page 35 without proof:

The completion of $(k[x,y]/(y^2-x^2-x^3))_{(x,y)}$ is $k[[x,y]]/(y^2-x^2-x^3)$ and that of $(k[x,y]/(xy))_{(x,y)}$ is $k[[x,y]]/(xy)$.

I studied Atiyah-Macdonald's chapter on completions, but I don't know why the above is true. It might be related to the isomorphism $\hat{M}\cong M\otimes_A \hat{A}$ where $A$ is Noetherian and $M$ is a finitely generated $A$-module, but I'm not sure. I don't have much experience with power series.
Would you please tell me why this is true? Thank you.

REPLY [5 votes]: If $(R,m)$ is Noetherian local, then $m$-adic completion is exact. Moreover $(f)\widehat{R} \cong \widehat{(f)}$ for any $f \in m$. Thus the exact sequence
$$0 \to (f) \to R \to R/(f) \to 0$$ induces an exact sequence
$$0 \to f\widehat{R} \to \widehat{R} \to \widehat{R/(f)} \to 0$$


TITLE: Unique subfield $L$ of $K=\mathbb{Q}(\sqrt[5]{2},\zeta_5)$ such that $[K:L]=5$
QUESTION [0 upvotes]: $K$ is the splitting field of the polynomial $f=x^5-2$, and we need to prove the existence of a unique subfield $L$ such that $[K:L]=5$. It would be nice to use the Galois group here, but it is a semi-direct product which I am unfamilar with. Is there a way to do this without using the Fundamental Theorem of Galois Theory? I know that $[K:\mathbb{Q}]=20$, so the Galois group is a group of order $20$, but I'd like to find a way to do this without invoking the fundamental theorem. Any suggestions?

REPLY [1 votes]: Following the advice of Lubin: one can show using Sylow theory that there is a unique subgroup $P$ of $G=\text{Gal}(K/\mathbb{Q})$ with $|P|=5$. Let $L=\text{Fix}(P)$. Then by Galois Theory, we have $[K:L]=|P|=5$. The uniqueness of $L$ follows from the uniqueness of $P$. Finally, $\mathbb{Q}(\zeta_5)$ is one such extension, so $L=\mathbb{Q}(\zeta_5)$.


TITLE: Isomorphism and partially ordered sets
QUESTION [0 upvotes]: Set $A$ and $B$ partially ordered sets, let $f: A \to B$ an isomorphism. Prove that:
If $C$ is a connected subset of $A$, then  $ f[C] $ is a connected subset of $B$.
Definition connected subset: a subset $C$ of partially ordered set is connected if satisfies the next condition
If $a,b \in C $ and $a \leq x\leq b$, then $x \in C$
Now, my problem is that I don't know where I start, I understand that is a direct proof $p \Rightarrow q$. I can identify my hypothesis. I was thinking this form but I'm not sure let $a,b \in f[C]$ and suppose that $a \leq x \leq b$
$$\Rightarrow$$
$$\Rightarrow$$
$$\Rightarrow$$
I really appreciate that someone can help me with the detailed explanation to be able to identify where I should start the proof

REPLY [0 votes]: Hints:  You start with elements $a,b\in f(C)$ and assume there is some $x$ such that $a\leq x\leq b$.   Now use that $f$ is an isomorphism,  which means it is a bijection, so you can find preimages of each element.  Then recall that the isomorphism preserves ordering which should let you use the connectivity of C.  Finally push back forward into f(C)


TITLE: An algorithm to obtain square of any number.
QUESTION [0 upvotes]: I have discovered a new way to obtain square of any number. I think its a new algorithm to find square of a number than just multiplying it with itself; if its not new then let me know.
Algorithm is as follows:

Divide the number in two parts with one part containing only the number at unit's place say part 'A', and other part say 'B', containing the remaining number.
Now square the number at unit's place. The square will be one of these; {0,1,4,9,16,25,36,49,64,81}. The unit's place digit in this square is the unit's place digit in actual final answer.Write it in the answer. If the square of digit at unit's place is a two digit no like from 16 to 81 in above set; write only the digit at unit's place from this square in the final answer and carry the remaining digit.
Multiply the actual number to be squared by part 'B'(the remaining part than the number at unit's place as described in step 1 ).
Multiply the parts 'A' and 'B'.
Add results of step 3 with results of step 4.
Add the carried digit from step 2 to the sum in prior step, that is step 5.
Now write this sum before the number we wrote at unit's place of final answer in step 2.
This number we now obtain from step 7, is the square of our number.

Example:
Lets find square of 127 by above algorithm;
1. A = 7 and B = 12 here...
2. A^2 = 49 thus final answer will have 9 at units place and 4 is carried to add later.
3. Given number multiplied by 'B' => 127*12 = 1524
4. A*B => 12*7 = 84
5.  1524+84 = 1608  ..... ( step 3 + step 4 )
6. 1608+4 =1612  .... (4 is carried as stated in step 2)
7. Now, 16129 is the answer... (From step 2  and 6)

REPLY [2 votes]: Ok so let's try to square $10b+a$ and we let $a^2=10c+d$ where it is possible that $c=0$.
Step $1$ tells us to identify $a$ and $b$
For step $2$ we have $a^2=10c+d$ and we write $d$ in the final place and remember $c$
Step $3$ gives us $b\cdot (10b+a) =10b^2+ab$
Step $4$ gives us $ab$
Step $5$ gives us $10b^2+2ab$
Step $6$ gives us $10b^2+2ab+c$
In step $7$ "putting the number before" is the same as multiplying by $10$ and adding so we get $10\cdot(10b^2+2ab+c)+d$
To check at step 8 we have $100b^2+20ab+10c+d=100b^2+20ab+b^2=(10b+a)^2$
So your method works. I couldn't say whether it is new, but it requires two multiplications at steps $3$ and $4$, plus the squaring step.

There is an interestingly similar method for speeding up the multiplication of two large numbers. Let $M$ be a large power of $10$, say, so that the numbers $a, b, c, d$ are of comparable size.
Multiplying  $(aM+b)(cM+d)=acM^2+(ad+bc)M+bd$ looks as though it requires the computation of four products $ac, ad, bc, bd$, but if we compute $ac, bd$ and $(a+b)(c+d)$ (three products) we find that $ad+bc=(a+b)(c+d)-ac-bd$. This turns out to be computationally more efficient.
The key thing about a method like yours is not so much whether it works (all sorts of methods work) or whether it is new, but whether it is a practical improvement on existing methods.


TITLE: The Sum of the Odd Divisors of n
QUESTION [5 upvotes]: The sum of the odd divisors of n is $-\sum_{d|n}(-1)^{n/d}d$, and if n is even, then
  $$\sum_{d|n}(-1)^{n/d}d=2\sigma(n/2)-\sigma(n)$$

Could you give me some hints on that?

REPLY [2 votes]: For the first part, you could try writing $n=2^km$, where $m$ is odd. Then the divisors of $n$ are the numbers $d,2d,4d,\dots,2^kd$ where $d$ runs through the odd divisors of $n$, which is to say, through the divisors of $m$. Now you can rewrite your sum to involve a sum on powers of 2.


TITLE: Is this statement equivalent to Goldbach's conjecture
QUESTION [1 upvotes]: Given a number $n\ge 3$, then one of these is true: \begin{equation} \begin{cases}2n = (6m-1)+P, \ \ \ P \in \mathbb P, \  6m-1 \in \mathbb P, \ 6m+1 \in \mathbb P \ \ \ \ (1) \\ 2n-1 \in \mathbb P, \ 2n-1 = P + 6k,\ 6k+1 \in \mathbb P \ \ \ \ (2) \end{cases} \end{equation}
In other words, any even number $2n$ greater than $4$ can be written either as a sum of a prime and another prime which is lower of a pair of twin primes or, if this cannot be done, then $2n-1$ is a prime and it can be written as a sum of a prime and a number divisible by six which is one smaller than a prime.
My justification for this is the Goldbach's conjecture.
Assume every even number greater than $2$ can be written as a sum of two primes. Then list every possible way to write $2n+2$ as sum of two primes: $h = \{P_1+P_2,P_3+P4... \}$
Now $2n = 2n-2$. Then $2n$ can be written as a sum of two primes as well, so there must be either some $P_n$ which is a component of some sum in $h$, for which $P_n-2 \in \mathbb P$ or, if no such prime exists, but we have to get the primes for $2n$ from somewhere, so some $P_n$ must be 3 and if $2n+2=3+P_s$, $2n = 1+P_s$ and $P_s$ can be expressed as a sum of a prime and a number $N$, such that $N+1 \in \mathbb P.$
Assume then that what I stated at the beginning is true. Then $2n$ can be written in one of the two forms and thus $2n+2$ is automatically a sum of two primes.
The other form is necessary, take as an example the smallest number for which the $(1)$ fails: $$38 = 19+19 = 7+31 = 37+1. $$ None of these is a lower of a twin prime pair, but $37$ is a prime and $37 = 31 + 6$ and $6 +1 = 7$ is a prime.

REPLY [6 votes]: Your condition fails for the number 122.
$122 = 61 + 61 = 43 + 79 = 19 + 103 = 13 + 109$, and none of these are the lower of a twin prime pair.  But, $121$ is not a prime, so your condition fails.


TITLE: How to find a matrix $C$ such that $C^{-1}AC$ is in Jordan block form.
QUESTION [2 upvotes]: $A:=\begin{bmatrix}
6 & -1\\
4 & 2
\end{bmatrix}$
Now, just to show I've done some working, at least to find $A$'s eigenvalues and deduced that it's not diagonalisable:

Any hints/advice?
Thanks

REPLY [0 votes]: For a $2 \times 2$ (say complex) matrix $A$ which is not diagonalizable, there is a method to determine the conjugating matrix $C$ such that $C^{-1}AC$ is in Jordan form. The only way that $A$ can fail to be diagonalizable is if the characteristic polynomial of $A$ has a repeated root $s,$ but $A \neq sI.$ Then the only possibility for the Jordan form is $\left( \begin{array}{clcr} s & 1\\0&s \end{array} \right)$. In order to have trace $2s$ and determinant $s^{2},$ the matrix $A$ must either  have the form $\left( \begin{array}{clcr} s+a &  ta\\\frac{-a}{t} & s-a \end{array} \right)$ for some $a,t \neq 0,$ or have one of the forms $\left( \begin{array}{clcr} s & t\\0&s \end{array} \right)$ or $\left( \begin{array}{clcr} s & 0\\t&s \end{array} \right)$ for some non-zero $t.$ We have to solve the equation $AC = C \left( \begin{array}{clcr} s & 1\\0&s \end{array} \right)$ with ${\rm det} C \neq 0.$ This reduces to solving one of the equations $\left( \begin{array}{clcr} a &  ta\\\frac{-a}{t} & -a \end{array} \right)C = C\left( \begin{array}{clcr} 0 & 1\\0&0 \end{array} \right)$ or $\left( \begin{array}{clcr} 0 &  t\\0 & 0 \end{array} \right)C = C\left( \begin{array}{clcr} 0 & 1\\0&0 \end{array} \right)$ or $\left( \begin{array}{clcr} 0 &  0\\t & 0 \end{array} \right)C = C\left( \begin{array}{clcr} 0 & 1\\0&0 \end{array} \right).$ This can always be done.
In your case, you have $s = 4, a = 2$ and $t = \frac{-1}{2}.$


TITLE: show than this PDE can be reduced to heat equation
QUESTION [1 upvotes]: How to reduce this PDE to heat equation
$$x^2G_{xx}=G_t$$
($G_{xx}$ is the 2nd order derivative on $x$, $G_t$ is the 1st derivative on $t$)
We wish to obtain a form such that $G(x,t)=F(U(x,t))$,
when substituted into the original equation we have
$$U_{xx}=U_t$$

REPLY [0 votes]: Let $\begin{cases}x_1=\ln x\\t_1=t\end{cases}$ ,
Then $G_x=G_{x_1}(x_1)_x+G_{t_1}(t_1)_x=\dfrac{G_{x_1}}{x}=e^{-x_1}G_{x_1}$
$G_{xx}=(e^{-x_1}G_{x_1})_x=(e^{-x_1}G_{x_1})_{x_1}(x_1)_x+(e^{-x_1}G_{x_1})_{t_1}(t_1)_x=(e^{-x_1}G_{x_1x_1}-e^{-x_1}G_{x_1})e^{-x_1}=e^{-2x_1}G_{x_1x_1}-e^{-2x_1}G_{x_1}$
$G_t=G_{x_1}(x_1)_t+G_{t_1}(t_1)_t=G_{t_1}$
$\therefore e^{2x_1}(e^{-2x_1}G_{x_1x_1}-e^{-2x_1}G_{x_1})=G_{t_1}$
$G_{x_1x_1}-G_{x_1}=G_{t_1}$
Let $\begin{cases}x_2=x_1-t_1\\t_2=t_1\end{cases}$ ,
Then $G_{x_1}=G_{x_2}(x_2)_{x_1}+G_{t_2}(t_2)_{x_1}=G_{x_2}$
$G_{x_1x_1}=(G_{x_2})_{x_1}=(G_{x_2})_{x_2}(x_2)_{x_1}+(G_{x_2})_{t_2}(t_2)_{x_1}=G_{x_2x_2}$
$G_{t_1}=G_{x_2}(x_2)_{t_1}+G_{t_2}(t_2)_{t_1}=G_{t_2}-G_{x_2}$
$\therefore G_{x_2x_2}-G_{x_2}=G_{t_2}-G_{x_2}$
$G_{x_2x_2}=G_{t_2}$


TITLE: Linear transformation, show that points on a plane mapped back on the same plane.
QUESTION [0 upvotes]: I need help with the second part of this question
Question
The eigenvalues, with their corresponding eigenvectors are as follows (answers to the first part):
λ = 1 , $e_1$ = –2j + k
λ = 2 , $e_2$ = i + j
λ = 3 , $e_3$ = 2i + 2j + k
The answer scheme gives these steps (for the second part):
r = se + tf
A(se + tf) = sAe + tAf = (sλ)e + (tμ)f
I do not understand how λ and μ come about, a step by step guide would be appreciated.

REPLY [0 votes]: By parametric equation of the plane $\Pi$, your question means the set of all points in the plane $r \in \Pi$ written as linear combinations of the two eigenvectors $e$ and $f$. The question asks you to show that for any $r \in \Pi$, $Ar \in \Pi$. The scalars $\lambda, \mu$ are the eigenvalues of $e$ and $f$, respectively.
Let $r = se + tf$. Then, clearly, $r \in \Pi$. Consider $Ar$. Your job is to show that $Ar \in \Pi$ by illustrating that $Ar$ can be written as a linear combination of $e$ and $f$.
$Ar = A(se + tf) = A(se) + A(tf) = sAe + tAf = s(\lambda e) + t(\mu f) = s\lambda e + t\mu f = \beta_1 e + \beta_2 f$.
The $\lambda$ and $\mu$ come from applying the fact that $e$ and $f$ are eigenvectors of $A$ (i.e., $Ae = \lambda e$, etc.).
Therefore, you have shown that $Ar$ can be written as a linear combination of $e$ and $f$, as desired to show that $T$ maps points in $\Pi$ to points in $\Pi$.