text:
\begin{document}
\title{\ijde\center\Large\bf Noether's Theorem for Control Problems\\ on Time Scales}
\author{{\bf Agnieszka B. Malinowska} \\
        Faculty of Computer Science\\
Bialystok University of Technology\\
15-351 Bia\l ystok, Poland\\
        {\tt a.malinowska@pb.edu.pl}
\and
        {\bf Moulay Rchid Sidi Ammi}\\
        Department of Mathematics,\\
        AMNEA GROUP\\
Faculty of Sciences and Techniques,\\
Moulay Ismail University, \\
B.P 509, Errachidia, Morocco\\
        {\tt sidiammi@ua.pt}
}
\maketitle
\thispagestyle{empty}

\begin{abstract}
We prove a generalization of Noether's theorem for optimal control problems defined on time scales.
Particularly, our results can be used for discrete-time, quantum, and continuous-time optimal control problems.
The generalization involves a one-parameter family of maps which depend also on the control
and a Lagrangian which is invariant up to an addition of an exact delta differential.
We apply our results to some concrete optimal control problems on an arbitrary time scale.
\end{abstract}

\subj{49K05, 39A12}

\keyw{Time scales, optimal control, conservation laws}

\section{Introduction}

There are close relationships between symmetries and conserved quantities. But how to seek conserved quantities in mechanical systems? Emmy Noether first proposed the famous theorem \cite{noether,ntrans} which is usually formulated in the context of the calculus of variations: it guarantees that the invariance of a variational
integral $\int_a^bL(t,x(t),\dot{x}(t))dt$ with respect to a continuous symmetry transformations that depend on $k$ parameters implies the existence of $k$ conserved quantities along the Euler--Lagrange extremals. Noether's theorem explains conservation laws of mechanical systems: conservation of energy comes from invariance of the system under time translations;
conservation of linear momentum comes from invariance of the system under spacial translations; conservation of angular
momentum reflects invariance with respect to spatial rotations. The result is, however, much more than a theorem.
It is an universal principle, which can be formulated as a precise statement in very different contexts,
and for each such context, under different assumptions \cite{torres2006}. Typically, Noether transformations
are considered to be point-transformations (they are considered to be functions
of coordinates and time), but one can consider more general transformations
depending also on velocities \cite{gelfand,torres}.
In most formulations of Noether's principle, the Noether transformations
keep the integral functional invariant. It is possible,
however, to consider transformations of the problem that keep the invariance of the Lagrangian up to an exact differential,
called a gauge-term \cite{rund,torres2004}. Formulations of Noether's principle are also possible
for optimal control problems \cite{torres,torres1,torres2,torres2006,schaft}.

Time scale calculus is a recent mathematical theory
that unifies two existing approaches to dynamic modelling --- difference and differential equations ---
into a general framework called dynamic models on time scales.
The origins of the idea of time scales calculus date back to the late 1980's when S. Hilger introduced
this notion in his Ph.D. thesis and showed how to
unify continuous time and discrete time dynamical systems \cite{Hilger,Hilger97}. With time this unification
aspect has been supplemented by the extension and generalization features, see e.g., \cite{Agarwal,ahl,Atici06,Atici09,CD:Bohner:2004,livro,loic,MM,zeidan,zeidan2,SA}.
Noether's first theorem has been already extend to the variational calculus on time scales \cite{B:T:08,B:M:T:11}. Therefore, it is natural to ask about a generalization of Noether's theorem for the optimal control problems defined on time scales.
Here we give the answer to this question.

\section{Preliminaries}
\label{sec:prm}
For the convenience of the reader we recall some basic results and notation needed in the sequel.
For the theory of time scales we refer the reader to
\cite{Agarwal,livro,Hilger97}.

A time scale $\T$ is an arbitrary nonempty closed subset of the real
numbers $\R$. The functions $\sigma:\T \rar \T$ and $\rho:\T \rar
\T$ are, respectively, the forward and backward jump operators. The
graininess function on $\T$ is defined by $\mu(t):=\sigma(t)-t$. A point $t\in\mathbb{T}$ is called
right-dense, right-scattered, left-dense, or left-scattered, if
$\sigma(t)=t$, $\sigma(t)>t$, $\rho(t)=t$, or $\rho(t)<t$,
respectively.

Let $\mathbb{T}=[a,b]\cap\mathbb{T}_{0}$ with $a<b$ and
$\mathbb{T}_0$ is a time scale. We define
$\mathbb{T}^\kappa:=\mathbb{T}\backslash(\rho(b),b]$, and
$\mathbb{T}^{\kappa^0} := \mathbb{T}$,
$\mathbb{T}^{\kappa^n}:=\left(\mathbb{T}^{\kappa^{n-1}}\right)^\kappa$
for $n\in\mathbb{N}$.

For a function $f:\T^\kappa\rightarrow \R$ the time scale delta
derivative is denoted by $f^\Delta$ \cite{livro}.
Whenever $f^\Delta$ exists, the following formula holds:
$f^\sigma(t)=f(t)+\mu(t)f^\Delta(t)$,
where we abbreviate $f\circ\sigma$ by $f^\sigma$.
Let $f^{\Delta^{0}} = f$. We define the
$r$th-delta derivative of $f:\mathbb{T}^{\kappa^r}\rightarrow
\R$, $r\in\mathbb{N}$, to be the function $\left(f^{\Delta^{r-1}}\right)^\Delta$,
provided $f^{\Delta^{r-1}}$ is delta differentiable on $\mathbb{T}^{\kappa^r}$.

A function $f:\mathbb{T} \to \mathbb{R}$ is called rd-continuous if
it is continuous at the right-dense points in $\mathbb{T}$ and its
left-sided limits exist at all left-dense points in $\mathbb{T}$. The set of all rd-continuous functions
is denoted by $C_{\mathrm{rd}}$. Similarly, $C^r_{\mathrm{rd}}$ will denote the set of
functions with delta derivatives up to order $r$ belonging to
$C_{\mathrm{rd}}$. A function $f$ is of class $f\in C_{\mathrm{prd}}^r$ if
$f^{\Delta^i}$ is continuous for $i = 0,\ldots,r-1$, and
$f^{\Delta^r}$ exists and is rd-continuous for all, except possibly
at finitely many, $t \in \mathbb{T}^{\kappa^r}$.

A piecewise rd-continuous function $f:\mathbb{T} \to \mathbb{R}$
possess an antiderivative $F$, that is $F^{\Delta}=f$, and in this case the delta
integral is defined by $\int_{c}^{d}f(t)\Delta t=F(d)-F(c)$ for all
$c,d\in\T$. It satisfies
$$\int_t^{\sigma(t)}f(\tau)\Delta\tau=\mu(t)f(t).$$
\begin{example}
\begin{itemize}
\item[a)] If $\T=\R$,
then $\sigma(t)=t=\rho(t)$ and $\mu(t) \equiv 0$ for any $t \in
\R$. The delta derivative, $f^\Delta$, reduces to the standard derivative $f'(t)$ and $\int\limits_{a}^{b}  f(t) \Delta
t=\int\limits_{a}^{b}f(t)dt$, where the integral on the right hand
side is the usual Riemann integral.

\item[b)] If $\T=h\Z$, then $\sigma(t)=t+h$, $\rho(t)=t-h$, and $\mu(t)
\equiv h$ for every $t \in \Z$. The delta derivative, $f^\Delta$, reduces to the $h$-forward difference $\Delta_{h} f(t) = \frac{f(t+h)-f(t)}{h}$ and $\int_{a}^{b}f(t)\Delta t =
\sum_{k=\frac{a}{h}}^{\frac{b}{h}-1} hf(kh)$.

\item[c)] If $\mathbb{T} = q^{\mathbb{N}_0}$, where $q>1$
is a fixed real number, then $\sigma(t) = q t$, $\rho(t) =
q^{-1} t$, and $\mu(t)= (q-1) t$. For the delta derivative and delta integral we get $f^{\Delta}
(t)=\frac{f(q t)-f(t)}{(q-1) t}$ and $\int_{a}^{b}f(t)\Delta t = (q-1) \sum_{t \in [a,b)} t f(t)$, respectively.
\end{itemize}
\end{example}

Let $a,b\in\T$, with $a<b$, and $[a,b]_{\T}:=[a,b]\cap \T$. Consider the nonlinear time scale optimal control problem
\begin{equation}\label{problem:1}
J(x,u) = \int_{a}^{b}
 L(t,x(t),u(t))\Delta t\longrightarrow\min,
\end{equation}
subject to $x\in C^{1}_{\mathrm{prd}}([a,b]_{\T})$, $u\in C_{\mathrm{prd}}([a,\rho (b)]_{\T})$ satisfying
\begin{equation}\label{problem:2}
x^{\Delta}(t)=\varphi(t,x(t),u(t)),\quad t\in [a,\rho (b)]_{\T}.
\end{equation}

\begin{definition}
A pair $(x,u)$ is said to be admissible if it satisfies the dynamic equation \eqref{problem:2}.
\end{definition}

In the next theorem it is assumed that the following regularity conditions hold:
\begin{description}
\item[$(H)$]  The functions $L(t,\cdot,\cdot)$, $\varphi(t,\cdot,\cdot)$ are differentiable in $(x,u)$; and the functions $\frac{\partial L}{\partial x}(t,\cdot,\cdot)$, $\frac{\partial L}{\partial u}(t,\cdot,\cdot)$, $\frac{\partial \varphi}{\partial x}(t,\cdot,\cdot)$, $\frac{\partial \varphi}{\partial u}(t,\cdot,\cdot)$ are continuous at $(x,u)$ uniformly in $t$ and rd-continuous in $t$ for any admissible $(x,u)$; $I+\mu(t)\frac{\partial\varphi}{\partial x}\neq 0 $ for all $t \in[a,\rho (b)]_{\T}$.
\end{description}

\begin{theorem}[See \protect{\cite[Theorem~9.4]{zeidan2}}]
\label{weak:MP}
Assume that $(\bar{x},\bar{u})$ is a weak local minimum for problem \eqref{problem:1}--\eqref{problem:2} such that the assumption $(H)$ holds. Then there exist a constant $\lambda\geq 0$, and a function $\bar{p}\in C^{1}_{\mathrm{prd}}[a,b]_{\T}$, such that $\lambda+\|\bar{p}\|_{C}\neq 0$ (where $\|p\|_{C}:= \max_{t\in [a,\rho (b)]_{\T}}|p(t)|$) and satisfying the following conditions:
\begin{itemize}
\item[(i)] the adjoint equation: for all $t \in [a,\rho (b)]_{\T}$
$$-\bar{p}^{\Delta}(t)=\frac{\partial \bar{\varphi}}{\partial x}\bar{p}^{\sigma}(t)+\lambda \frac{\partial \bar{L}}{\partial x},$$
\item[(ii)] the stationarity condition: for all $t \in [a,\rho (b)]_{\T}$
$$\frac{\partial \bar{\varphi}}{\partial u}\bar{p}^{\sigma}(t)+\lambda \frac{\partial \bar{L}}{\partial u}=0,$$
\end{itemize}
where $\frac{\partial \bar{\varphi}}{\partial x}$, $ \frac{\partial \bar{L}}{\partial x}$, $\frac{\partial \bar{\varphi}}{\partial u}$, $\frac{\partial \bar{L}}{\partial u}$ are evaluated at $(t,\bar{x}(t),\bar{u}(t))$.
\end{theorem}

The Hamiltonian corresponding to problem \eqref{problem:1}--\eqref{problem:2} is defined as follows:
$$H(t,x,u,\lambda,p)=\lambda L(t,x,u)+(p^\sigma)^T\varphi(t,x,u).$$

\begin{remark}
There are two distinct possibilities for the constant $\lambda$ in Theorem~\ref{weak:MP}:
\begin{itemize}
\item[a)] if $\lambda\neq 0$, we say that control $\bar{u}$ is normal, in this situation we may assume that $\lambda=1$;
\item[b)] if $\lambda= 0$, we say that control $\bar{u}$ is abnormal, and in this case the Hamiltonian does not depend on $L$.
\end{itemize}
\end{remark}

For more about normal and abnormal controls, in the context of optimal control problems on time scales, we refer the reader to \cite{loic,zeidan2}

\begin{definition}
We say that $(x,u,\lambda,p)$ with admissible $(x,u)$,
$p\in C_{\mathrm{prd}}^1[a,\rho (b)]_{\T}$, and $\lambda \geq 0$ is an extremal of problem
\eqref{problem:1}--\eqref{problem:2} if the two conditions (i) and (ii) of Theorem~\ref{weak:MP} are satisfied.
\end{definition}

\section{Noether's Theorem without Transformation of Time Variable}

Consider the family of transformations
\begin{equation}\label{tra:fam}
h_s:[a,b]_{\T}\times \R^2\rightarrow \R,
\end{equation}
depending on a parameter $s$, $s\in (-\varepsilon,\varepsilon)$, where  $h_s$ have continuous partial delta derivatives of the first and second order with respect to $s$ and $t$ for all admissible $(x,u)$ (hence, we have equality of mixed partial delta derivatives, see \cite{Bohner+Guseinov2004}), and the value $s=0$ corresponds to the identity transformation: $h_0(t,x,u)=x$ for all $(t,x,u)\in [a,b]_{\T}\times \R^2$. By Taylor's formula we have
\begin{equation}\label{tra:fam:T}
h_s(t,x,u)=h_0(t,x,u)+s\xi(t,x,u)+o(s)=x+s\xi(t,x,u)+o(s),
\end{equation}
where $\xi(t,x,u)=\frac{\partial}{\partial s}h_s(t,x,u)|_{s=0}$.

We define the invariance of \eqref{problem:1}--\eqref{problem:2}  under transformations $h_s(t,x,u)$ up to an exact delta differential as follows.

\begin{definition}\label{def:invariance:1}
 If there exists a function $\Phi_s(t,x,u)$ which has continuous partial delta derivatives of the first and second order with respect to $s$ and $t$ for all admissible $(x,u)$; and for all $s\in (-\varepsilon,\varepsilon)$ and admissible $(x,u)$ there exists a control $u_s(\cdot)\in C_{\mathrm{prd}}[a,\rho (b)]_{\T}$ such that:
\begin{equation}\label{def:invariance:1:1}
L(t,x(t),u(t))+\Phi_s^\Delta(t,x(t),u(t))= L(t,h_s(t,x(t),u(t)),u_s(t)),
\end{equation}
\begin{equation}\label{def:invariance:1:2}
\frac{\Delta}{\Delta t}h_s(t,x(t),u(t))=\varphi(t,h_s(t,x(t),u(t)),u_s(t));
\end{equation}
then problem \eqref{problem:1}--\eqref{problem:2} is invariant under transformation $h_s(t,x,u)$ up to $\Phi_s(t,x,u)$.
\end{definition}

\begin{remark}
We assume that $u_0(\cdot)=u(\cdot)$ and $\frac{\partial}{\partial s}u_s(\cdot)\in C_{\mathrm{prd}}[a,\rho (b)]_{\T}$.
\end{remark}

\begin{remark}
Note that condition \eqref{def:invariance:1:1} is satisfied if and only if
\begin{equation}\label{def:invariance:1:1:1}
\int_{a}^{\beta}\left(L(t,x(t),u(t))+\Phi_s^\Delta(t,x(t),u(t))\right)\Delta t= \int_{a}^{\beta}L(t,h_s(t,x(t),u(t)),u_s(t))\Delta t
\end{equation}
for all $\beta \in [a,b]_{\T}$.
\end{remark}

\begin{theorem}\label{Noether:without:time}
If problem \eqref{problem:1}--\eqref{problem:2} is invariant under transformation $h_s(t,x,u)$ up to $\Phi_s(t,x,u)$, then
\begin{equation}\label{eq:Noether:without:time}
p(t)\frac{\partial}{\partial s}h_s(t,x(t),u(t))|_{s=0}+\lambda\frac{\partial}{\partial s} \Phi_s(t,x(t),u(t))|_{s=0}
\end{equation}
is constant along any extremal of problem \eqref{problem:1}--\eqref{problem:2}.
\end{theorem}

\begin{proof}
Differentiating both sides of equation \eqref{def:invariance:1:1} with respect to $s$, and then setting $s=0$ we obtain
\begin{multline*}
\frac{\partial}{\partial s} \Phi_s^\Delta(t,x(t),u(t))|_{s=0}=\frac{\partial L}{\partial x}(t,x(t),u(t))\frac{\partial }{\partial s}h_s(t,x(t),u(t))|_{s=0}\\+\frac{\partial L}{\partial u}(t,x(t),u(t))\frac{\partial }{\partial s}u_s(t)|_{s=0}.
\end{multline*}
If $(x(\cdot),u(\cdot),\lambda,p(\cdot))$ is an extremal of problem \eqref{problem:1}--\eqref{problem:2}, then it satisfies
$$-p^{\Delta}(t)=\frac{\partial \varphi}{\partial x}p^{\sigma}(t)+\lambda \frac{\partial L}{\partial x}.$$
Therefore,
\begin{equation*}
\lambda\frac{\partial}{\partial s} \Phi_s^\Delta|_{s=0}=-p^\Delta(t)\frac{\partial }{\partial s}h_s|_{s=0}-p^\sigma(t)\frac{\partial \varphi}{\partial x}\frac{\partial }{\partial s}h_s|_{s=0}+\lambda\frac{\partial L}{\partial u}\frac{\partial }{\partial s}u_s(t)|_{s=0}.
\end{equation*}
By the second condition of Theorem~\ref{weak:MP}, we have
\begin{equation*}
\frac{\partial \varphi}{\partial u}p^\sigma(t)\frac{\partial }{\partial s}u_s(t)|_{s=0}+\lambda\frac{\partial L}{\partial u}\frac{\partial }{\partial s}u_s(t)|_{s=0}=0.
\end{equation*}
Combining the last two equations, we arrive to
\begin{equation}\label{prf:1}
\lambda\frac{\partial}{\partial s} \Phi_s^\Delta|_{s=0}+p^\Delta(t)\frac{\partial }{\partial s}h_s|_{s=0}+p^\sigma(t)\frac{\partial \varphi}{\partial x}\frac{\partial }{\partial s}h_s|_{s=0}+\frac{\partial \varphi}{\partial u}p^\sigma(t)\frac{\partial }{\partial s}u_s(t)|_{s=0}=0.
\end{equation}
Observe that, by the assumptions, we have $\frac{\Delta}{\Delta t}\frac{\partial}{\partial s}h_s|_{s=0}=\frac{\partial}{\partial s}\frac{\Delta}{\Delta t}h_s|_{s=0}$. Therefore, from \eqref{def:invariance:1:2} we get
\begin{equation}\label{prf:2}
\frac{\Delta}{\Delta t}\frac{\partial}{\partial s}h_s|_{s=0}=\frac{\partial \varphi}{\partial x}\frac{\partial }{\partial s}h_s|_{s=0}+\frac{\partial \varphi}{\partial u}\frac{\partial }{\partial s}u_s(t)|_{s=0}.
\end{equation}
Inserting \eqref{prf:2} into \eqref{prf:1} we get
\begin{equation}\label{prf:3}
\lambda\frac{\partial}{\partial s} \Phi_s^\Delta|_{s=0}+p^\Delta(t)\frac{\partial }{\partial s}h_s|_{s=0}+ p^\sigma(t)\frac{\Delta}{\Delta t}\frac{\partial}{\partial s}h_s|_{s=0}=0.
\end{equation}
Applying formula $(fg)^\Delta=f^{\sigma}g^{\Delta}+f^{\Delta}g$, we can rewrite \eqref{prf:3} as
\begin{equation*}
\lambda\frac{\partial}{\partial s} \Phi_s^\Delta|_{s=0}+\frac{\Delta}{\Delta t}\left[p(t)\frac{\partial }{\partial s}h_s|_{s=0}\right]=0.
\end{equation*}
Thus,
\begin{equation*}
\frac{\Delta}{\Delta t}\left[\lambda\frac{\partial}{\partial s} \Phi_s|_{s=0}+p(t)\frac{\partial }{\partial s}h_s|_{s=0}\right]=0.
\end{equation*}
Therefore,
\begin{equation*}
\lambda\frac{\partial}{\partial s} \Phi_s|_{s=0}+p(t)\frac{\partial }{\partial s}h_s|_{s=0}
\end{equation*}
is constant along any extremal of problem \eqref{problem:1}--\eqref{problem:2}.
\end{proof}

\begin{example}(Cf. \cite{torres})
Consider the following problem:
\begin{equation}\label{Ex:1:1}
\int_{a}^{b} u^2(t)\Delta t\longrightarrow\min,
\end{equation}
subject to $x\in C^{1}_{\mathrm{prd}}([a,b]_{\T})$, $u\in C_{\mathrm{prd}}([a,\rho (b)]_{\T})$ satisfying
\begin{equation}\label{Ex:1:2}
x^{\Delta}(t)=u(t),\quad t\in [a,\rho (b)]_{\T}.
\end{equation}
This problem is invariant under the transformation $h_s(t,x,u)=x+st$ up to $\Phi_s(t,x,u)=s^2t+2xs$. Indeed, we have
\begin{equation*}
L(u^s)=(u+s)^2=u^2+2us+s^2=u^2+(s^2t+2sx)^\Delta=L(u)+\frac{\Delta}{\Delta t}\Phi_s(t,x,u)
\end{equation*}
and
\begin{equation*}
\frac{\Delta}{\Delta t}h_s(t,x,u)=x^\Delta(t)+s=\varphi(u^s).
\end{equation*}
From Theorem~\ref{Noether:without:time} it follows that $\frac{\Delta}{\Delta t}\left[p(t)t+\lambda 2x(t)\right]=0$ along any extremal of \eqref{Ex:1:1}--\eqref{Ex:1:2}. On the other hand an extremal of \eqref{Ex:1:1}--\eqref{Ex:1:2} should satisfy the following condition:
$p^\Delta (t)=0$ and $p^\sigma(t)+2\lambda u(t)=0$. Therefore, by Theorem~\ref{Noether:without:time}, $\frac{\Delta}{\Delta t}\left[x^\Delta (t)t- x(t)\right]=0$ along any extremal of \eqref{Ex:1:1}--\eqref{Ex:1:2}.
\end{example}

\section{Noether's Theorem with Transformation of Time Variable}

In this subsection we change the time. Thus we consider the optimal control problem on many different time scales. Therefore, we shall assume that problem \eqref{problem:1}--\eqref{problem:2} is defined for all $t\in \R$.
Consider the family of transformations
\begin{equation}\label{tra:fam:2}
h_s(t,x,u)=(h_s^t(t,x,u),h_s^x(t,x,u)):[a,b]_{\T}\times \R^2\rightarrow \R^2,
\end{equation}
depending on a parameter $s$, $s\in (-\varepsilon,\varepsilon)$, where  $h_s^t$, $h_s^x$ have continuous partial delta derivatives of the first and second order with respect to $s$ and $t$ for all admissible $(x,u)$, and the value $s=0$ corresponds to the identity transformation: $(h_0^t(t,x,u),h_0^x(t,x,u))=(t,x)$ for all $(t,x,u)\in [a,b]_{\T}\times \R^2$. By Taylor's formula we have
\begin{eqnarray*}
h_s^t(t,x,u)=h_0^t(t,x,u)+s\zeta(t,x,u)+o(s)=t+s\zeta(t,x,u)+o(s),\\
h_s^x(t,x,u)=h_0^x(t,x,u)+s\xi(t,x,u)+o(s)=x+s\xi(t,x,u)+o(s),
\end{eqnarray*}
where $\zeta(t,x,u)=\frac{\partial}{\partial s}h_s^t(t,x,u)|_{s=0}$ and $\xi(t,x,u)=\frac{\partial}{\partial s}h_s^x(t,x,u)|_{s=0}$.

We assume that for every admissible $(x,u)$ and all $s\in (-\varepsilon,\varepsilon)$ the map $[a,b]\ni t\mapsto \alpha (t):=h_s^t(t,x,u)=t_s\in \R$ is a strictly increasing $C_{\mathrm{rd}}^1$ function and its imagine is again a time scale with the forward shift operator $\bar{\sigma}$, the graininess function $\bar{\mu}$ and the delta derivative $\bar{\Delta}$. Observe that in this case the following holds: $\bar{\sigma}\circ \alpha = \alpha \circ \sigma$ \cite{ahl}.

We define the invariance of \eqref{problem:1}--\eqref{problem:2}  under transformation \eqref{tra:fam:2} up to an exact delta differential as follows.

\begin{definition}\label{def:invariance:2}
Let $I+\bar{\mu}(t_s)\frac{\partial \varphi}{\partial x}\neq 0$ for all
$t_s\in [h_s^t(a,x(a),u(a)), h_s^t(b,x(b),u(b))]^\kappa$. If there exists a function $\Phi_s(t,x,u)$ which has continuous partial delta derivatives of the first and second order with respect to $s$ and $t$ for all admissible $(x,u)$; and for all $s\in (-\varepsilon,\varepsilon)$ and admissible $(x,u)$ there exists a control $u_s(\cdot)\in C_{\mathrm{prd}}[a,\rho (b)]_{\T}$ such that:
\begin{multline}\label{def:invariance:2:1:1}
\int_{h_s^t(a,x(a),u(a))}^{h_s^t(\beta,x(\beta),u(\beta))}L(t_s,h_s^x(t_s,x(t_s),u(t_s)),u_s(t_s))\bar{\Delta}t_s\\
=\int_{a}^{\beta}\left(L(t,x(t),u(t))\Delta t+\Phi_s^\Delta(t,x(t),u(t))\right)\Delta t
\end{multline}
for all $\beta \in [a,b]$;
\begin{equation}\label{def:invariance:2:1:2}
\frac{\bar{\Delta}}{\bar{\Delta} t_s}h_s^x(t_s,x(t_s),u(t_s))=\varphi(t_s,h_s^x(t_s,x(t_s),u(t_s)),u_s(t_s));
\end{equation}
then problem \eqref{problem:1}--\eqref{problem:2} is invariant under transformation \eqref{tra:fam:2} up to $\Phi_s(t,x,u)$.
\end{definition}

\begin{theorem}\label{Noether:with:time}
If problem \eqref{problem:1}--\eqref{problem:2} is invariant under transformation $h_s(t,x,u)$ up to $\Phi_s(t,x,u)$, then
\begin{multline}
\frac{\Delta}{\Delta t}\left[p_k(t)\frac{\partial}{\partial s}h_s^t(t,x(t),u(t))|_{s=0}+p(t)\frac{\partial}{\partial s}h_s^x(t,x(t),u(t))|_{s=0}\right.\\
\left.+\lambda\frac{\partial}{\partial s} \Phi_s(t,x(t),u(t))|_{s=0}\right]=0
\end{multline}
is constant along any extremal of problem \eqref{problem:1}--\eqref{problem:2}.
Moreover, if $\rho \circ \sigma=id_{[a,\rho(b)]_{\T}}$, then
\begin{multline}\label{eq:Noether:with:time}
-H^\rho(t,x(t),u(t),\lambda,p(t))\frac{\partial}{\partial s}h_s^t(t,x(t),u(t))|_{s=0}\\
+p(t)\frac{\partial}{\partial s}h_s^x(t,x(t),u(t))|_{s=0}+\lambda\frac{\partial}{\partial s} \Phi_s(t,x(t),u(t))|_{s=0}
\end{multline}
is constant along any extremal of problem \eqref{problem:1}--\eqref{problem:2}.
\end{theorem}

\begin{proof}
Firstly we show that invariance of \eqref{problem:1}--\eqref{problem:2} in the sense of Definition~\ref{def:invariance:2} is equivalent to invariance of another problem in the sense of Definition~\ref{def:invariance:1}. Let us define
\begin{equation}
\label{problem:3}
\begin{split}
\tilde{J}(k,x,r,u)&=\int_{a}^{b}
\tilde{L}(t;k(t),x(t);r(t),u(t))\Delta t\\
&:=\int_{a}^{b} L(k(t),x(t),u(t))r(t)\Delta t\longrightarrow\min
\end{split}
\end{equation}
subject to
\begin{equation}
\label{problem:4}
\left\{
  \begin{array}{l}
    k^\Delta (t)=r(t)\\
    x^{\Delta}(t)=\tilde{\varphi}(t;k(t),x(t);r(t),u(t)):=\varphi(k(t),x(t),u(t))r(t),\\
  \end{array}
\right.
\end{equation}
$t\in [a,\rho (b)]_{\T}$. The Hamiltonian corresponding to problem \eqref{problem:3}--\eqref{problem:4} has the form
$$\tilde{H}(t,k,x,r,u,\lambda,p_k,p)=\lambda\tilde{L}(t;k,x;r,u)+p_k^{\sigma}r+p^{\sigma}\tilde{\varphi}(t;k,x;r,u).$$
Observe that for $k(t)=t$ we have
$$L(t,x(t),u(t))=\tilde{L}(t;k(t),x(t);r(t),u(t)),$$ so we get $\tilde{J}(k,x,r,u)=J(x,u)$ whenever $k(t)=t$. Moreover, for $k(t)=t$ we have
$$x^{\Delta}(t)=\varphi(t,x(t),u(t))=\tilde{\varphi}(t;k(t),x(t);r(t),u(t)).$$
Now we consider the family of transformations $h_s(t,x,u)=(h_s^t(t,x,u),h_s^x(t,x,u))$. From the invariance of \eqref{problem:1}--\eqref{problem:2}, for $k(t)=t$, we get
\begin{equation}
\begin{split}
\int_{a}^{b}
&\left(\tilde{L}(t;k(t),x(t);r(t),u(t))+\Phi_s^\Delta(t,x(t),u(t))\right)\Delta t\\
&=\int_{a}^{b}
 \left(L(t,x(t),u(t))+\Phi_s^\Delta(t,x(t),u(t))\right)\Delta t\\
 &=\int_{\alpha (a)}^{\alpha (b)}
 L(t_s,h_s^x(t_s,x(t_s),u(t_s)),u_s(t_s))\bar{\Delta}t_s\\
 &=\int_{a}^{b}
L(\alpha (t),h_s^x(\alpha (t),x(\alpha (t)),u(\alpha (t))),u_s(\alpha (t)))\alpha^\Delta (t)\Delta t\\
&=\int_{a}^{b}
\tilde{L}(t;\alpha (t),(h_s^x\circ\alpha) (t,x(t),u(t)),\alpha ^\Delta (t),(u_s\circ \alpha)(t))\Delta t,
\end{split}
\end{equation}
$$\frac{\Delta}{\Delta t}\alpha (k(t))=\alpha ^\Delta (t),$$ and
\begin{equation}
\begin{split}
\frac{\Delta}{\Delta t}(h_s^x\circ \alpha) (t,x(t),u(t))
&=\frac{\bar{\Delta}}{\bar{\Delta} t_s}h_s^x(t_s,x(t_s),u(t_s))\alpha ^\Delta (t)\\
&=\varphi(t_s,h_s^x(t_s,x(t_s),u(t_s)),u_s(t_s))\alpha ^\Delta (t)\\
&=\varphi(\alpha (t),h_s^x(\alpha (t),x(\alpha (t)),u(\alpha (t))),u_s(\alpha (t)))\alpha ^\Delta (t)\\
&=\tilde{\varphi}(t;\alpha (t),(h_s^x\circ \alpha)(t,x(t),u(t));\alpha ^\Delta (t),(u_s\circ\alpha )(t)).
\end{split}
\end{equation}
This means that for $k(t)=t$, if problem \eqref{problem:1}--\eqref{problem:2} is invariant under transformation \eqref{tra:fam:2} up to $\Phi_s(t,x,u)$ in the sense of Definition~\ref{def:invariance:2}, then problem \eqref{problem:3}--\eqref{problem:4} is invariant under transformation \eqref{tra:fam:2} up to $\Phi_s(t,x,u)$ in the sense of Definition~\ref{def:invariance:1}.
Let $(k,x,r,u,\lambda,p_k,p)$ be an extremal of problem \eqref{problem:3}--\eqref{problem:4}. Applying Theorem~\ref{weak:MP} we obtain
$$p^\Delta=-\lambda \frac{\partial \tilde{L} }{\partial x}-p^\sigma\frac{\partial \tilde{\varphi} }{\partial x},$$
$$p_k^\Delta=-\lambda \frac{\partial \tilde{L} }{\partial k} -p^\sigma \frac{\partial \tilde{\varphi} }{\partial k},$$
$$p^\sigma\frac{\partial \tilde{\varphi} }{\partial u}+\lambda\frac{\partial \tilde{L} }{\partial u}=0,$$
$$p_k^\sigma+\lambda\frac{\partial \tilde{L} }{\partial r}+p^\sigma \frac{\partial \tilde{\varphi} }{\partial r}=0.$$
By Theorem~\ref{Noether:without:time}, we get for $k(t)=t$
\begin{multline}
\frac{\Delta}{\Delta t}\left[p_k(t)\frac{\partial}{\partial s}h_s^t(t,x(t),u(t))|_{s=0}+p(t)\frac{\partial}{\partial s}h_s^x(t,x(t),u(t))|_{s=0}\right.\\
\left.+\lambda\frac{\partial}{\partial s} \Phi_s(t,x(t),u(t))|_{s=0}\right]=0.
\end{multline}
Now assume that $\rho \circ \sigma=id_{[a,\rho(b)]_{\T}}$. As $p_k^\sigma=-\lambda L-p^\sigma \varphi$ we have $p_k=-H^\rho$ and condition \eqref{eq:Noether:with:time} holds along any extremal of problem \eqref{problem:1}--\eqref{problem:2} as desired.
\end{proof}

From now on, it is assumed that $\rho \circ \sigma=id_{[a,\rho(b)]_{\T}}$.

\begin{example}[Cf. \cite{gou}]
Consider the following problem:
\begin{equation}
\label{Ex:2:1}
\int_{a}^{b} \left(u_1^2(t)+u_2^2(t)\right)\Delta t\longrightarrow\min,
\end{equation}
subject to $x\in C^{1}_{\mathrm{prd}}([a,b]_{\T})$, $u_1,u_2\in C_{\mathrm{prd}}([a,\rho (b)]_{\T})$ satisfying
\begin{equation}\label{EX:2:2}\left\{
  \begin{array}{l}
    x_1^\Delta (t)=u_1(t)\cos x_3(t),\\
    x_2^{\Delta}(t)=u_1(t)\sin x_3(t),\\
    x_3^\Delta (t)=u_2(t).
  \end{array}
\right.
\end{equation}
The control system \eqref{Ex:2:1}--\eqref{EX:2:2} serves as model for the kinematics of a car \cite{gou,Martin}. This is an autonomous system, so it is invariant under the transformation $h_s^t(t,x,u)=t+s$ up to $\Phi_s\equiv 0$.
From Theorem~\ref{Noether:with:time} it follows that $H\circ\rho$ is constant along any extremal of \eqref{Ex:2:1}--\eqref{EX:2:2}, where
$$H(t,x_1,x_2,x_3,u_1,u_2,\lambda, p_1,p_2,p_3)=\lambda \left(u_1^2+u_2^2\right)+p_1^{\sigma}u_1\cos x_3+p_2^{\sigma}u_1\sin x_3+p_3^{\sigma}u_2.$$
\end{example}

In the next example we apply Theorem~\ref{Noether:with:time} for a problem with abnormal controls.

\begin{example}
Consider the following problem:
\begin{equation}
\label{Ex:3:1}
\int_{0}^{1} u(t)\Delta t\longrightarrow\max,
\end{equation}
subject to $x\in C^{1}_{\mathrm{prd}}([0,1]_{\T})$, $u\in C_{\mathrm{prd}}([0,\rho (1)]_{\T})$ satisfying
\begin{equation}\label{EX:3:2}\left\{
  \begin{array}{l}
    x^\Delta (t)=(u(t)-u^2(t))^2,\\
    x(0)=0,\\
    x(1)=0.
  \end{array}
\right.
\end{equation}
As this system is autonomous, it is invariant under the time transformation $h_s^t(t,x,u)=t+s$ up to $\Phi_s\equiv 0$.
Therefore, by Theorem~\ref{Noether:with:time}, we get that $H\circ\rho$ is constant along any extremal of \eqref{Ex:3:1}--\eqref{EX:3:2}, where
$$H(t,x,u,\lambda, p)=\lambda u+p^{\sigma}(u-u^2)^2.$$ Observe that an extremal of problem \eqref{Ex:3:1}--\eqref{EX:3:2} should satisfy the following conditions:
\begin{equation}\label{Ex:3:3}
p^{\Delta}(t)=0, \quad -4p^{\sigma}(t)u(t)(u(t)-u^2(t))+\lambda=0.
\end{equation}
If we chose $\lambda=0$ and $p=k$, where $k$ is a negative constant, then $(x,u,\lambda,p)=(0,1,0,k)$ or $(x,u,\lambda,p)=(0,0,0,k)$ satisfy conditions \eqref{Ex:3:3}. Note that $x^{\Delta}=(u-u^2)^2\geq0$ and $ x(0)=x(1)=0$ imply that $x^{\Delta}=0$. Hence, if a control is admissible, then we have $u\in\{0,1\}$. In addition, for every admissible $u$, we have
$$\int_{0}^{1} 1\Delta t\geq \int_{0}^{1} u(t)\Delta t.$$
This means that $u=1$ is the optimal control for problem \eqref{Ex:3:1}--\eqref{EX:3:2}.
\end{example}

\begin{remark}
Observe that, for autonomous systems, Theorem~\ref{Noether:with:time} induces that:
\begin{enumerate}
\item $H\circ\rho$ is constant along the optimal path;
\item for $\T=\R$ the well-known result, that the Hamiltonian
$H$ is constant along the optimal path.
\end{enumerate}
\end{remark}

\section*{Acknowledgements}

This work was done during the stay of M. R. Sidi Ammi at the Bialystok University of Technology as a visiting professor.
A. B. Malinowska was supported by the Bialystok
University of Technology grant S/WI/02/2011. The authors are grateful to two anonymous referees
for valuable remarks and comments.
\bibliographystyle{plain}

config: arxiv


TITLE: why $\infty$ +$\infty$, $\infty$ - $\infty$ and 0⋅$\infty$ are left undefined.
QUESTION [1 upvotes]: I'm reading http://en.wikipedia.org/wiki/Riemann_sphere, and having the following question.
1.
What's the mean of symbol $\infty$?Is it a surreal number?
2.
they write

note that ∞ + ∞, ∞ - ∞ and 0 ⋅ ∞ are left undefined.

Why we can't define ∞ + ∞?
why not define  ∞ + ∞=∞

REPLY [4 votes]: The "formal" answer is that $\infty$ is simply an "object" in this new "number system," and that the rules of the system do not permit those "quantities" to be well-defined.
If you are referring to $\infty$ as we usually think of it, the best way to understand these facts is to think in terms of limits. The expressions $\infty - \infty$ and $0\times \infty$ (and several others) are called indeterminate forms because as limits, they can equal anything at all.
Specifically, say we have $f(x)\to \infty$ and $g(x)\to\infty$ as $x\to a$. What is $$\lim_{x\to a}(f(x)-g(x))$$ The answer is that it can be anything at all, depending on what the functions are. And similarly for other indeterminate cases. Having a "rule" that, for example, $\infty-\infty=0$ would lead to erroneous limit calculuations.
You are correct that the "limit law" $\infty+\infty=\infty$ is perfectly valid. But keep in mind that "$\infty$" being discussed here is not the number so much as it is an abstract object, defined to be a part of a number system with certain properties. You could just as well write that "thing" as $x$ or $\text{Turkey}$, it is denoted "$\infty$" because it has certain properties reminiscent of that idea, but it is not the same as infinities seen in limits.


TITLE: Definition of "algebra of sets"
QUESTION [2 upvotes]: Can I use the following definition?
Def.. let be $A$ a set and $B \in \mathscr{P}(\mathscr{P}(A))$, $(A,B)$ is algebra of sets if

$\emptyset \in B$
$\forall X \in B( (A-X) \in B)$
$\forall X,Y \in B((X \cap Y ) \in B)$

It is correct? Thanks in advance!

REPLY [1 votes]: If by "is it correct" you mean "is it equivalent to the normal definition" (closed under pairwise intersections, pairwise unions, contains $\emptyset$ and $A$) then the answer is "yes."
The second bullet asserts it is closed under complements and the third under intersections.
Suppose $S,T\in B$. Since $S\cup T=A\setminus((A\setminus S)\cap (A\setminus T))$, and the right hand side consists of complements and intersections of things in $B$, $S\cup T$ is in $B$ as well.
Finally, $A\setminus\emptyset=A$ is in $B$ also, so you have both identities.


TITLE: form of the group of all the root of unity in $\mathbb{Q}(\zeta^{p^m})$
QUESTION [0 upvotes]: Here is my question (H.W question):

For natural number $m$ and prime number $p\geq3$ , Let $\zeta_{p^m}$ be primitive $p^m$-th root of unity in $\mathbb{C}$.
  Show that the group of roots of unity in $\mathbb{Q}(\zeta^{p^m})$  have the following form :
  { $\pm\zeta^j_{p^m}|0\leq j \lt p^m$ }

I know this is a chyclotomic extentoin and how to compute the minimal polynomial so I tried to prove it by contradiction with Galois theory tools but I got stuck.
I know that { $\pm\zeta^j_{p^m}|0\leq j \lt p^m$ } = $\mathbb{Z}_{2p^m}$ but way this is the only ones ? what about sum of two or three elements from this group ?
thanks

REPLY [2 votes]: Each of the $\pm\zeta_{p^m}^j$ are roots of unity in the field $K=\Bbb Q(\zeta_{p^m})$. If there
was some other roots of unity, say $\zeta_r$, then $r\nmid 2p^m$.
Then $K$ contains the $s$-th roots of unity where $s=\text{lcm}(r,2p^m)>2p^m$.
But the degree of $\Bbb Q(\zeta_s)$ is $\phi(s)>\phi(2p^m)$ as $s=2kp^m$
where $k>1$ is an integer. So $|\Bbb Q(\zeta_s):\Bbb Q|>|K:\Bbb Q|$
so that it's impossible for $\Bbb Q(\zeta_s)\subseteq K$.


TITLE: Is there a quantity measured in kilogram seconds?
QUESTION [7 upvotes]: I'm trying to get a full grasp on the relationship between many of the units that are used in kinetic physics. I've found that it is possible to make a venn-diagram that shows the factors of many of the units: Impulse in the centre, Force and Velocity between two circles each and Mass, Acceleration and Time on the Outside. The only space in the diagram that is missing is mass multiplied by time.
Is there a such thing as a kilogram-second? Is it used for anything in physics?

REPLY [6 votes]: Two examples of the product of mass and duration being quantified over vastly different ranges:

1) It is not entirely uncommon to cost a truck on a tonne.hr basis. Here the payload is measured in tonnes and the trip duration in hours. The example given is:
"if a truck hauls 35 Tonnes and the trip time is 4 hours, the number of tonne$\cdot$hours is 140 (35 x 4).  If the standard cost is USD 3.00 per tonne$\cdot$hour, the truck cost would be USD 420 (= 140 x USD 3.00)."
in SI units a tonne$\cdot$hour would equate to 3.6 million kg$\cdot$seconds.

2) In physics, the product of mass and time appear in the phenomenon "Zitterbewegung". Zitterbewegung provides a physical interpretation for the complex phase factor in the Dirac wave function. It is manifest as a rapid motion of the position of elementary particles that obey the Dirac equation. This motion repeats with a frequency $2mc^2/h$. Therefore, over a timespan $\delta t$, a Dirac particle with mass $m$ will undergo $m \delta t/(h/2c^2)$ 'Zitter cycles'. (It's no coincidence that in essence the same constant $h/2c^2$ measured in SI units kg$\cdot$s appears here as well as in Void's answer.) In other words, when measured in natural units, the product of mass and time equates to the number of 'double Zitter cycles' a Dirac particle will undergo.

REPLY [0 votes]: An example that comes from Compton scattering would be something like
$m_e /\lambda$
would have $kg \cdot m$, $1/\lambda$ is related to the momentum of the photon, but not $kg\cdot s$.
By itself this quantity is not usefull, but it does appear during the calcuations
Edit:
I gave a bad example, edits above


TITLE: What is the approximation obtained with the midpoint method after one step, answer is a function of h?
QUESTION [0 upvotes]: Write down the approximation obtained with the midpoint Runge Kutta method after one step
(the answer will be a function of h).
I am stuck on this part for a first-order system. I'm not sure how to implement my system into the midpoint RK method
The RK midpoint method is
$$ K_1 = f(t_n,y_n) \\ K_2 =f(t_n + \frac{h}{2},y_n +\frac{h}{2}K_1) \\y_{n+1} = y_n + hK_2
$$
The original IVP is
$$ y'' + 2y' + y = te^{-t} - 1    \\y(0) = y'(0) = 0 \\0\leq t \leq1
$$
My system is
$$ y_1' = y_2
$$
$$ y_2' = te^{-t} -1 -2y_2 -y_1
$$
where $y_1(0) = y_2(0) = 0$

REPLY [1 votes]: When you have two equations, $f$ and the $K$s are two position vectors.  Now $y=\begin{pmatrix} y_1\\y_2 \end {pmatrix}$.  You should write your system as a single equation working on two position vectors.  The second argument of $f$ is also a two position vector.  You are given $y_0$ and $t_0$.  Plug them in to get $K_1$.  Plug them and $K_1$ in to get $K_2$
Added:  You may have confusion about the subscripts.  The subscripts $1,2$ are the variables, the subscripts $n$ are the values of the variables at time step $n$.  So at $t=0, y_{10}=y_{20}=0.$  You plug those into $K_{10}=\begin{pmatrix} 0\\-1 \end {pmatrix}$  Then $K_{20}=f\left(\frac h2,\frac h2\begin{pmatrix} 0\\-1 \end {pmatrix}\right)$ and you plug those last into the right side of your equations to compute $K_{20}$


text:
\section{Score tests}
\label{sect:score}
Let $y=(y_{1},\dots,y_{n})$ be an i.i.d. sample from a p.d.f. or p.m.f. $p(x)$. Let also $\mathcal{P}\define\{p(x;\theta): \theta \in \Theta \subseteq \R^{d}\}$ denote a known parametric family of  probability distributions. The unknown function $p(x)$ does not necessarily belong to the parametric family, in other words the parametric model can be misspecified. Following the renown aphorism of \cite{Box1976} \enquote{All models are wrong, but some are useful}, it is widely recognized that in general a (parametric) statistical model cannot be considered exactly correct. See, for example, \cite{White82,Gustafson2001,Wit2012all}, \textsection 1.1.4 by \cite{GLMbook1983}, and p. 2 by \cite{Bickel2015mathematical_II}. Hence it is of particular importance to design methods of statistical inference that are \emph{robust to model misspecification}. In this section we propose \emph{a bootstrap score test procedure} which is valid even in case when the parametric model $\mathcal{P}$ is misspecified.

Let $s(\theta)=s(\theta,y)$ and $I(\theta)$ denote the score function and the Fisher information matrix corresponding to the introduced parametric model
\begin{align*}
s(\theta)\define \sum\nolimits_{i=1}^{n}{\partial} \log p(y_{i};\theta)/{\partial \theta} ,\quad
I(\theta)\define \Var\{ s(\theta)\}.
\end{align*}
We suppose that the standard regularity conditions on the parametric family $\mathcal{P}$ are fulfilled. Let $\theta_{0}\define\mathrm{argmin}_{\theta\in \Theta}\E\log (p(y_{i})/p(y_{i};\theta))$ denote the parameter which corresponds to the  projection of $p(x)$ on the parametric family $\mathcal{P}$ w.r.t. the Kullback-Leibler divergence (also known as the relative entropy).

Consider a simple hypothesis $H_{0}: \theta_{0}=\theta^{\prime}$.  Rao's score test (by \cite{Rao1948large}) for testing $H_{0}$ is based on the following test statistic and its convergence in distribution to $\chi^{2}_{d}$
\begin{equation}
\label{eq:Rtest}
R(\theta^{\prime})\define s(\theta^{\prime})^{T}\{I(\theta^{\prime})\}^{-1} s(\theta^{\prime}) \overset {d \vert H_{0}}{\to} \chi^{2}_{d},\quad n\to \infty,
\end{equation}
provided that matrix $I(\theta^{\prime})$ is p.d. The sign $\overset {d \vert H_{0}}{\to}$ denotes convergence in distribution under $H_{0}$.
  Matrix $I(\theta^{\prime})$ can be calculated explicitly for a known $\theta^{\prime}$ if one assumes that $p(x)\in \mathcal{P}$, i.e. if the parametric model is correct. However, if $p(x)$ does not necessarily belong to the considered parametric class $\mathcal{P}$, then neither $I(\theta^{\prime})$ nor the probability distribution of $s(\theta^{\prime})$ can be calculated in an explicit way under the general assumptions considered here. In this case, the Fisher information matrix $I(\theta)$ is typically estimated using the Hessian of the log-likelihood function $\sum_{i=1}^{n}\log p(y_{i};\theta)$. However the standardization with an empirical Fisher information may considerably reduce the power of the score test for a small sample size $n$ (see \cite{Rao2005} and \cite{Freedman2007can}).

Below we consider {a bootstrap score test} for testing simple hypothesis $H_{0}$, under possible misspecification of the parametric model. Denote
$$\tilde{R}(\theta^{\prime})\define \|s(\theta^{\prime})/\sqrt{n}\|^{2}.$$
 One can consider $s(\theta^{\prime})=\sum\nolimits_{i=1}^{n}X_{i},$ where random vectors $X_{i}\define \partial \log p(y_{i};\theta^{\prime})/{\partial \theta^{\prime}}$ are i.i.d. with $\E X_{i} =0$ under $H_{0}$. Introduce the bootstrap approximations of $s(\theta^{\prime})$ and $\tilde{R}(\theta^{\prime})$:
$$s^{\ast}(\theta^{\prime})\define \sum\nolimits_{i=1}^{n}\Xs_{i},\quad   R^{\ast}(\theta^{\prime})\define \|s^{\ast}(\theta^{\prime})/\sqrt{n}\|^{2},$$
where $\{\Xs_{i}\}_{i=1}^{n}$ are sampled according to Efron's bootstrap scheme \eqref{eq:EB}. Let also
$$t^{\ast}_{\alpha}\define \inf\{t>0: (1-\alpha) \leq \Pb( R^{\ast}(\theta^{\prime}) \leq t)\}$$ denote $(1-\alpha)$-quantile of the bootstrap score statistic for arbitrary $\alpha \in (0,1)$. Suppose that coordinates of vectors $X_{i}= \partial \log p(y_{i};\theta^{\prime})/{\partial \theta^{\prime}}$ satisfy condition \eqref{condit:expm1} with variance factor $\sigma^{2}_{s}>0$. Let also $d$ and $n$ be such that $
\sigma_{s}^{2}(d/\sqrt{n})C_{1\ast} <  \lambda_{\min}(I(\theta^{\prime}))/n$.
 Then Theorem \ref{theorem:bootst} implies the following statement which characterizes accuracy of the bootstrap score test under $H_{0}$.

\begin{theorem}[Bootstrap score test]
\label{theorem:score1}
If the conditions above are fulfilled, it holds
\begin{align*}
\bigl|\P_{H_{0}}\bigl(\tilde{R}(\theta^{\prime}) > t^{\ast}_{\alpha} \bigr) -\alpha\bigr| &\leq \delta_{R},
\end{align*}
where $\delta_{R}$ is analogous to $\delta_{\BCl}$ up to the terms $\sigma^{2}$ and $\Sigma$. A detailed definition of $\delta_{R}$ is given in \eqref{def:deltaR}, see also Remark \ref{remark:delta} for the discussion about its dependence on $d$ and n.
\end{theorem}

The following statement provides a finite sample version of Rao's score test based on \eqref{eq:Rtest} for testing simple hypothesis $H_{0}: \theta_{0}=\theta^{\prime}$. Here we require also the finite 4-th moment of the score in order to apply the higher-order approximation from Theorem \ref{theorem:first}.
\begin{theorem}[Nonasymptotic version of Rao's score test]
\label{theorem:score2}
Suppose that $p(x)\equiv p(x,\theta_{0})$ for some $\theta_{0}\in \Theta$, i.e. there is no misspecification in the considered parametric model.
Let also $\tilde{X}_{i}\define \sqrt{n} \{I(\theta^{\prime})\}^{-1/2}\partial \log p(y_{i};\theta^{\prime})/\partial \theta^{\prime}$ denote the marginal standardized score for the $i$-th observation and $\tilde{\Sigma}\define n^{-1} I(\theta^{\prime})$.  Suppose that $\E |\tilde{X}_{i}^{\otimes 4}|<\infty$,
then the asymptotic poperty \eqref{eq:Rtest}
 for testing $H_{0}: \theta_{0}=\theta^{\prime}$ can be represented in the finite sample form as follows
\begin{align*}
&\sup\nolimits_{\alpha\in(0,1)}\bigl|\P_{H_{0}}\bigl(R(\theta^{\prime}) > q(\alpha;\chi^{2}_{d}) \bigr) -\alpha\bigr| \leq
 (\sqrt{6}\beta^{3})^{-1} \|\E (\tilde{X}_{1}^{\otimes 3})\|_{\mathrm{F}}n^{-1/2}
\\
\nonumber
&
\quad
+  2C_{B,4} \|\tilde{\Sigma}^{-1}\|\|\tilde{\Sigma}\|\bigl\{
(h_{1}(\beta)+(4\beta^{4})^{-1})
\E\|\tilde{X}_{1}\|^{4}+d^{2}+2d
\bigr\}^{1/2}n^{-1/2}
   \\
   \nonumber
   &\quad+
   (2\sqrt{6})^{-1} \bigl\{
h_{1}(\beta) \E \|\tilde{X}_{1}\|^4
   +h_2(\beta)(d^{2}+2d)\bigr\}n^{-1},
\end{align*}
where $q(\alpha;\chi^{2}_{d})$ denotes the $(1-\alpha)$-quantile of $\chi^{2}_{d}$ distribution. The inequality holds for any $\beta\in  (0,1)$, functions $h_{1},h_{2}$ are defined in \eqref{def:h1h2}, constant $C_{B,4}\geq 9.5$ is described in the statement of Theorem \ref{theorem:first}.
\end{theorem}


TITLE: Question about Doob's decomposition and a theorem in Durrett
QUESTION [1 upvotes]: Let $X_n$ be a martingale with $X_0 = 0$ and $EX_n^2<\infty$ for all n. $X_n^2$ is a submartingale and thus $X_n^2 = M_n + A_n$.
I don't understand a step in this theorem: $E(\text{sup}_n |X_n|)\leq 3EA_{\infty}^{1/2}$, where $A_{\infty}=\text{lim}_n A_n$.
Proof: Let $a>0$ and let $N = \text{inf}\{n:A_{n+1}>a^2\}$, then
$$P(\text{sup}_n |X_n|>a)\leq P(N<\infty)+ P(\text{sup}_n |X_{n\wedge N}|>a)$$
I don't understand why {$\text{sup}_n |X_n|>a\} \implies \{N<\infty\}$? Thanks.

REPLY [1 votes]: The inequality follows from the inclusion $(\sup_n |X_n|>a) \subset (N<\infty) \cup (\text{sup}_n |X_{n\wedge N}|>a)$.
That's because $\begin{aligned}[t](\sup_n |X_n|>a)
&= [(\sup_n |X_n|>a)\cap (N<\infty)]\cup [(\sup_n |X_n|>a)\cap (N=\infty)]\\
&=[(\sup_n |X_n|>a)\cap (N<\infty)]\cup [(\sup_n |X_{n\wedge N}|>a)\cap (N=\infty)]\\
&\subset (N<\infty) \cup (\sup_n |X_{n\wedge N}|>a)
\end{aligned}$
The union bound then yields $$P(\sup_n |X_n|>a) \leq P(N<\infty) + (\sup_n |X_{n\wedge N}|>a)$$


TITLE: Adjunction via universal morphisms implies adjunction via unit-counit
QUESTION [1 upvotes]: I know there is a fair amount of similar questions to this one, but none of them deal directly with what I'm struggling with right now. For instance, in this one the right adjoint functor is constructed, while in my situation I already have the pair of functors. In this one the objective is to derive the unit-counit equations, (which is the same I'm trying to prove, essentially), but this question does not have any answers. This question deals with the converse, but that direction I was able to prove.
Goal

Given functors $L \colon C \to D, R \colon D \to C$, a family of morphisms $\eta_X \colon X \to RL(X), \epsilon_Y \colon LR(Y) \to Y$ such that $\forall f \colon L(X) \to Y. \exists ! g \colon X \to R(Y). f= \epsilon_Y \circ L(g) \land g=R(f) \circ \eta_X$ and vice-versa (for every $g$ there is an unique $f$), then $\eta: id_C \to RL, \epsilon: LR \to id_D$ are natural transformations and the natural-transformation-equalities hold.

First I need to prove that $\eta$ is a natural transformation. That is, given a morphism $f \colon X_1 \to X_2$ in $C$, we have $\eta_{X_2} \circ f = RL(f) \circ \eta_{X_1}$.
What I did
Define $g = \eta_{X_2} \circ f$. By hypothesis, there exists an unique $f' \colon L(X_1) \to L(X_2)$ such that $\epsilon_{L(X_2)} \circ L(g) = f'$ and $g= R(f') \circ \eta_{X_1}$. Hence $\eta_{X_2} \circ f = R(f') \circ \eta_{X_1}$. It is sufficient to prove $R(f')=RL(f)$. After some calculations, we have $R(f') = R(\epsilon_{L(X_2)} \circ L(\eta_{X_2})) \circ RL(f)$. How to prove that $R(\epsilon_{L(X_2)} \circ L(\eta_{X_2})) = id_{RL(X_2)}$?

REPLY [1 votes]: Consider the morphism $id_{L(X_2)} \colon L(X_2) \to L(X_2)$. By hypothesis, there exists an unique $g \colon X_2 \to RL(X_2)$ such that $\epsilon_{L(X_2)} \circ L(g) = id_{L(X_2)}$ and $g=id_{RL(X_2)} \circ \eta_{X_2}$. But the only $g$ such that the second equality holds is $g=\eta_{X_2}$, hence $\epsilon_{L(X_2)} \circ L(\eta_{X_2}) = id_{L(X_2)}$.


TITLE: Is any foliation on a 2-torus induced by a suitable flow?
QUESTION [4 upvotes]: Consider the 2-dimensional torus $T^2=\mathbb{R}^2/\mathbb{Z}^2$, and a foliation on it (for example a foliation in circles, maybe the partition of the torus obtained form a Hopf-related map).
I'm wondering if there are some condition on the foliation to be (the union of) the integral curves of a suitable vector field $X_\tau$ defined on the torus...
Note: One can clearly ask something more general (generic group action on a smooth manifold whose orbits are leaves of a given foliation), but I'm really dumb on making good (=well defined) questions so for the moment let's talk about a particular case.
Note 2: I'm not requiring much smoothness for $X_\tau$ just because I suspect that the answer will be "No if you suppose $X_\tau$ is not $C^k$-smooth with $k\ge k_0$".
Thanks a lot!

REPLY [5 votes]: A foliation of a torus need not arise from a vector field, because its tangent line field may be non-orientable.
E.g., representing the torus by a square whose opposite sides are identified, think of a stack of arches (convex downward) filling up the square, except for the vertical sides -- which will constitute a single closed leaf.
In terms of $\mathbb T^2 = \mathbb R^2 / \mathbb Z^2$, this is represented by the real analytic tangent line field defined by
$$L(x,y) := \exp((x + 1/2) \pi i)$$
where antipodal points on the circle of angles are identified.


TITLE: Non-free modules, with a free direct sum
QUESTION [8 upvotes]: I have to find the $M$ and $N$ to be non-free modules over the ring $\mathbb{Z}/6\mathbb{Z}$, such that $M\oplus N$ is free.
My idea was to take $M=2\mathbb{Z}/6\mathbb{Z}$ (That is, the even numbers in the numbers modulo 6), and then take $N=3\mathbb{Z}/6\mathbb{Z}$ (the numbers multiple of three modulo 6).
Then $N$ is not free because the only elements in it are $\{0,3\}$, and none of them is free (if we take $3$, then we could have $r\cdot 3=0$ by letting $r=2$).
Similarly I argue that $M$ is not free. Lastly, it is easy to see that $\mathbb{Z}/6\mathbb{Z}=M\oplus N$ since the only element in common for $N$ and $M$ is zero, and their sum gives the desired result.
Questions; Is the above correct? Which other(easy) examples are there

REPLY [6 votes]: (I'll only consider finitely generated modules here, just for simplicity)
Your ring $R=\mathbb Z/6\mathbb Z$ is isomorphic to the direct product of $R_2=\mathbb Z/2\mathbb Z$ and $R_3=\mathbb Z/3\mathbb Z$, so that $R\cong R_2\times R_3$. A little work will show that every $R$-module is the direct sum of an $R_2$-module and an $R_3$-module, in the obvious sense.
Since $R_2$ and $R_3$ are fields, we know all of their modules: every $R_2$-module is isomorphic to $R_2^n$ for some $n\geq0$, and every $R_3$-module is isomorphic to $R_3^m$ for some $m\geq0$. It follows that every $R$-module is isomorphic to $R_2^n\oplus R_3^m$ for some $n,m\geq0$.
One can see that a module $R_2^n\oplus R_3^m$ is free exactly when $n=m$.
Using this information, you can find all (finitely generated) examples.


TITLE: More variables in constraints than in quadratic objective function
QUESTION [2 upvotes]: Assume we have the following quadratic program
$$\begin{array}{ll} \text{minimize} & \frac{1}{2} x^T Q x \\ \text{subject to} & Ax <b\end{array}$$
This works when the number of constraint variables is equal to the number of objective variables. But let us now assume that
$$x = (x_0,x_1,x_2)$$
but that our objective does not involve $x_0$, in other words we could say that
$$Q=\begin{pmatrix}
0 &0&0\\
0 & Q_{11} & Q_{12}\\
0 & Q_{21} & Q_{22}
\end{pmatrix}$$
The matrix is now not positive definite, and most QP solvers don't work with this. So how should the problem be approached if the minimization function does not involve some variable $x_0$, but the constraint equations do involve this variable?

REPLY [1 votes]: Your current implementation has infinitely many solutions that minimize the cost function (ie any value for $x_0$ that satisfies the constraint). To ensure an unique solution can be computed, either express $x_0$ as a function of $x_1$ and $x_2$ and replace $x_0$ as a variable, or make $Q$ positive definite by setting a very small value for $Q_{0,0}$, like
$$\begin{bmatrix}10^{-8} & 0 & 0 \\ 0 & Q_{1,1} & Q_{1,2} \\  0 & Q_{2,1} & Q_{2,2}\end{bmatrix}$$
That should ensure that there exist an unique solution that minimizes the cost function, while simultaneously weighting $x_0$ as minimal as possible.


TITLE: What is the condition for two extreme points?
QUESTION [0 upvotes]: I am new to the topic of extreme point and trying to get the whole picture. Now I know the extreme theorm that mentions two condition to obtain a critical points what are $f'(x)= 0$ or $f'(x)=$ undefined.
I solved some examples of those on Khan Academy but coming across this example $\mathrm{Let} \, a,b,c,d, \in \, \mathbb{R}, a\neq0, f: \mathbb{R} \longrightarrow \mathbb{R}, f(x)=a\,x^3 + b \,x^2 + c\, x + d.$

(a) What is the condition for two extreme points?
the answer supposed to be (a) $\,b^2>3ac$.

Can I know how and which topic this fall into?

REPLY [0 votes]: if you calculate de $f'(x) = 0$, you will see the condition for the "square root" to be positive (you fall into a quadratic equation), is $$\sqrt{(b^2 - 3ac)},$$
so, to be positive, you'll need $$b^2 > 3ac.$$
After this you can exploit the signals of $a$ and $c$, if you want it.


TITLE: Is this proof that $\kappa^{<\kappa}=\kappa$, when $2^{<\kappa}=\kappa$, correct?
QUESTION [2 upvotes]: Let $\kappa$ be a cardinal number. I want to show that if $\kappa$ is regular and if $2^{<\kappa} = \kappa$ then $\kappa^{<\kappa}= \kappa$. Here is what I got so far:
$$
\begin{align}\kappa^{<\kappa} ~&=~ \left| \bigcup\limits_{\alpha < \kappa}\bigcup\limits_{\beta < \kappa}\beta^\alpha \right| \\&\leq~ \sum\sum\limits_{\alpha, \beta < \kappa} |\beta|^{|\alpha|} \\&\leq~ \sum\sum\limits_{\alpha, \beta < \kappa}2^{|\beta|^{|\alpha|}} \\&=~ \sum\sum\limits_{\alpha, \beta < \kappa}2^{|\beta| \cdot |\alpha|} \\&=~ \sum\limits_{\gamma < \kappa}2^\gamma \\&=~ \kappa \cdot 2^{< \kappa} \\&\hspace{-3pt}\overset{\text{Hyp}}{=} \kappa \cdot \kappa \\&=~ \kappa.
\end{align}$$
But I feel that the first line is wrong...
The following seems right
$$\kappa^{< \kappa} ~=~ \left| \bigcup\limits_{\alpha < \kappa}\kappa^\alpha \right|$$
but I got doubts about
$$\kappa^\alpha ~=~ \bigcup\limits_{\beta < \kappa}\beta^\alpha$$ since $\kappa$ is a regular cardinal.
Any clue?

REPLY [2 votes]: If $\kappa$ is regular, and $\alpha<\kappa$ then every function $f\colon\alpha\to\kappa$ is bounded, so it is in fact a function into some $\beta<\kappa$. So the equality holds.
Assuming that $\kappa$ is regular does in fact finishes the proof. If $\kappa$ is singular the implication is false. For example assuming $\sf GCH$ we have that: $$2^{<\aleph_\omega}=\sup\{2^{\aleph_n}\mid n<\omega\}=\sup\{\aleph_{n+1}\mid n<\omega\}=\aleph_\omega<\aleph_{\omega+1}=\aleph_\omega^{\aleph_0}\leq\aleph_\omega^{<\aleph_\omega}.$$


TITLE: Proving sets are closed under addition
QUESTION [0 upvotes]: Suppose for an example like S={-1,0,1}. I noticed that if you can only add two different elements it would remain closed however if you add the upper or lower bound to itself it would be out of the range of the set. Wouldn't closure under addition just fail hold for all finite sets of real numbers? (other than empty set or 0 set) I don't get what I'm trying to prove here.

REPLY [0 votes]: For any finite set of real numbers, we cannot be sure that always the closure under will not hold true. What you said holds true as long as the result(of the addition of two members of the set) is not part of the set.
However, if we take the case of just the upper, lower bounds with a finite set then we can be sure that the closure will not hold true.


TITLE: Daniell integral vs. Borel measure
QUESTION [2 upvotes]: Let $X$ be a locally compact, Hausdorff, topological space and denote by $\mathcal B_+(X)$ the collection of all Borel-measurable functions from $X$ to $[0,+\infty]$ (extended positive reals).
Suppose that we are given a map
  $$
  I:\mathcal B_+(X) \to [0,+\infty]
  $$
  which is positively homogeneous and satisfies
  $$
  I\big(\sum_{n=1}^\infty f_n\big) =   \sum_{n=1}^\infty I(f_n),
  $$
  for every sequence $\{f_n\}_n$ in $\mathcal B_+(X)$.
Observe that this is essentially the Daniell integral.
Suppose further that $\mu$ is a given regular Borel measure on $X$ such that
  $$
  I(f) = \int_X f\,d\mu,
  $$
  for every non-negative $f$ in $C_c(X)$, the space of all compactly supported continuous functions on $X$.  Does it follow that this identity also holds for every $f$ in $\mathcal B_+(X)$?
Ok to assume that $X$ is metrizable and $\sigma$-compact, if necessary.

REPLY [1 votes]: Here is an answer for $X$ metrizable and $\sigma$-compact. I assume $I$ is finite on compactly supported continuous functions.
Recall that a locally compact metrizable $\sigma$-compact space is Polish (separable and completely metrizable). Moreover, there is an increasing sequence of compact sets $\langle K_n\rangle$ such that $K_n$ is a subset of the interior of $K_{n+1}$ for all $n$ and such that $X=\bigcup_n K_n$. Fix some metric $d$ that completely metrizes $X$.
Clearly, it suffices to show $I(1_A)=\mu(A)$ for every measurable set $A$. By the regularity of $\mu$, one only has to prove this for $A$ compact and, even nicer, for sets of the form $A\cap K_n$ with $A$ compact. Now there is going to be some $\epsilon$ such that $(A\cap K_n)_\epsilon=\{y\in X\mid d(y, A\cap K_n)<\epsilon\}$ is a subset of the interior of $K_{n+1}$. Define $f_n:X\to\mathbb{R}$ by $f_m(x)=\max\{1-md(x,A\cap K_n),0\}$. This is a continuous function and for $m$ larger than $1/\epsilon$, it vanishes outside the compact set $K_{n+1}$. Moreover $\langle f_m\rangle$ decreases pointwise to $1_{A\cap K_n}$. Since $I$ satisfies the dominated convergence theorem, the conclusion follows.


TITLE: Equation of a Tangent to a curve
QUESTION [0 upvotes]: $ f' (x) dx = dy $
Would I be correct in assuming that this is the equation of any tangent line to the curve f (x) ?

REPLY [3 votes]: I disagree with the "no"'s.  You can think of $dx$ and $dy$ as new variables related by $dy = f'(x) dx.$ If you place the origin of the $dx-dy$ plane on the point $(x_0,f(x_0))$, then $dy=f'(x_0) dx$ is indeed the equation of that line (in $dx-dy$ coordinates.)  It's a good way for beginners to think of differentials, because it leads naturally to linear approximations.


text:
\DOC REAL_ARITH

\TYPE {REAL_ARITH : term -> thm}

\SYNOPSIS
Attempt to prove term using basic algebra and linear arithmetic over the reals.

\DESCRIBE
{REAL_ARITH} is the basic tool for proving elementary lemmas about real
equations and inequalities. Given a term, it first applies various
normalizations, eliminating constructs such as {max}, {min} and {abs} by
introducing case splits, splitting over the arms of conditionals and putting
any equations and inequalities into a form {p(x) <><> 0} where {<><>} is an
equality or inequality function and {p(x)} is in a normal form for polynomials
as produced by {REAL_POLY_CONV}. The problem is split into the refutation of
various conjunctions of such subformulas. A refutation of each is attempted
using simple linear inequality reasoning (essentially Fourier-Motzkin
elimination). Note that no non-trivial nonlinear inequality reasoning is
performed (see below).

\FAILURE
Fails if the term is not provable using the algorithm sketched above.

\EXAMPLE
Here is some simple inequality reasoning, showing how constructs like {abs},
{max} and {min} can be handled:
{
  # REAL_ARITH
      `abs(x) < min e d / &2 /\ abs(y) < min e d / &2 ==> abs(x + y) < d + e`;;
  val it : thm =
    |- abs x < min e d / &2 /\ abs y < min e d / &2 ==> abs (x + y) < d + e
}
The following example also involves inequality reasoning, but the initial
algebraic normalization is critical to make the pieces match up:
{
  # REAL_ARITH `(&1 + x) * (&1 - x) * (&1 + x pow 2) < &1 ==> &0 < x pow 4`;;
  val it : thm = |- (&1 + x) * (&1 - x) * (&1 + x pow 2) < &1 ==> &0 < x pow 4
}

\USES
Very convenient for providing elementary lemmas that would otherwise be painful
to prove manually.

\COMMENTS
For nonlinear equational reasoning, use {REAL_RING} or {REAL_FIELD}. For
nonlinear inequality reasoning, there are no powerful rules built into HOL
Light, but the additional derived rules defined in {Examples/sos.ml} and
{Rqe/make.ml} may be useful.

\SEEALSO
ARITH_TAC, INT_ARITH_TAC, REAL_ARITH_TAC, REAL_FIELD, REAL_RING.

\ENDDOC

subset_name: curated