TITLE: what happen to the speed of gas particles when the pressure increased?
QUESTION [2 upvotes]: There is gas particles inside a piston, if the piston was pushed inward, the volume will decrease and the pressure will increase, so my question is
What will happen to the speed of the gas particles?

REPLY [1 votes]: If the system is thermally isolated then the pressure and the temperature of the gas will increase.
The reason for this is that as the piston moves inward the rebound speed of the molecules will be larger than the speed before rebound.
After the molecules have collided with others this produces an increase in the overall kinetic energy due to the random motion of the molecules i.e. the gas temperature has increased.
The pressure has increased because the rate of collision between the molecules and the walls has increased and the change in momentum of a molecule when it rebounds has also increased.
If the system has walls which are at a constant temperature then on average the kinetic energy of the molecules does not change but the rate at which the molecules hit the walls increases which produces an increase in pressure.


TITLE: Induction: $\frac1{1 \cdot 5} + \frac1{5 \cdot 9} + \cdots + \frac1{(4n-3)(4n+1)}$
QUESTION [1 upvotes]: For $n \ge 1$, let
  $$a_n = \dfrac1{1 \cdot 5} + \dfrac1{5 \cdot 9} + \cdots + \dfrac1{(4n-3)(4n+1)}.$$
  Guess a simple explicit formula for $a_n$ and prove it by induction.

Hi, I'm trying to answer this question. I was not provided with a solution.

REPLY [0 votes]: $$a_n=\frac{1}{4}\sum_{k=1}^n\left(\frac{1}{4k-3}-\frac{1}{4k+1}\right)=$$
$$=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{4n-3}-\frac{1}{4n+1}\right)=$$
$$=\frac{1}{4}\left(1-\frac{1}{4n+1}\right)=\frac{n}{4n+1}$$


TITLE: Show series of functions is uniformly convergent on $[a,\infty)$ for $a>0$.
QUESTION [1 upvotes]: I'd like to show that
$$ f(x) = \sum_{n=0}^\infty f_n(x) $$
is uniformly convergent on $[a,\infty)$ for $a>0$ where
$$ f_n(x) = \frac{nx}{1+n^4x^2} $$
I have found that $f_n$ has a maximum of $\frac{1}{2n}$ on the interval $[0,\infty)$, specifically at $x=\frac{1}{n^2}$ (except for $f_0(x)=0$ for all $x$). Now this means
$$ f\bigg(\frac{1}{n^2}\bigg) = \sum_{n=0}^\infty \frac{1}{2n} \geq \sum_{n=0}^\infty \frac{nx}{1+n^4x^2} \text{ for all }x\in[0,\infty) $$
I am unsure as to where to go from here.

REPLY [1 votes]: No need to use Calculus: use the inequality $2nx \leq n^{2}+x^{2}$. $$f_n(x) \leq \frac {n^{2}+x^{2}} {2(1+n^{4}x^{2})} \leq \frac {n^{2}+x^{2}} {2n^{4}x^{2}}=\frac 1 {2n^{2}a^{2}} +\frac 1 {2n^{4}}$$ and the series $\sum \frac 1 {2n^{2}a^{2}}$ and $\frac 1 {2n^{4}}$ both converge.


TITLE: The actual period of a pendulum at 90°. Looking for the correct formula
QUESTION [0 upvotes]: Do you have access to any scientific experiment which gives the period of a pendulum when the angle is $90^\circ$:

this article says $T$ varies to about $18\%$ up to $90^\circ,$ so for a seconds pendulum the complete period would be roughly $2.36.$

I d'like to know the ratio at $90^\circ$ with a good approximation, and the span it can vary in relation to air friction, if the bob  is not massive.
Also, I'd like to know the max. speed of the bob at $P_0$ and how it compares with the speed of same ball sliding without friction on a circular surface, would it be the same?

What I am trying to understand here is rather complex, I'll try to explain:
If the block is placed on an air-track or other frictionless surface, the block would acquire a certain speed, momentum and KE in the opposite direction, right? same amount of momentum is lost by the bob and we can find speed loss.

Isn't the same amount of speed lost by the bob when it is attached to a pendulum?
Is this an important factor or may be the most important factor (more relevant than air friction or friction at the pivot) for the increase of the period when the angle is greater than $30^\circ$?
I figured out that this factor can account for about one third of the $18\%$ increase, can you give a more accurate value?

What I am suggesting is that this (loss of horizontal momentum, which is made clear when the bob is not attached to a rod but slides on a block) is the most important factor that makes the actual period increase when the starting point deviates on the vertical direction.

REPLY [1 votes]: REVISED ANSWER
I think you are trying to solve the problem of a small sphere of mass $m$ sliding down the circular face of a wedge of mass $M$. There is no friction between the sphere and the wedge and the wedge is free to move on a horizontal air track.

This problem is similar to block slides on smooth triangular wedge kept on smooth floor.Find velocity of wedge when block reaches bottom except that the flat slant face of the wedge is now the quadrant of a circle. When the sphere reaches the bottom of the slope it is travelling horizontally to the left with velocity $v$ while the wedge is moving to the right with velocity $V$. Energy and momentum are conserved so :
$$mv = MV$$
$$mv^2+MV^2=2mgR$$
therefore
$$v^2=\frac{2gMR}{m+M}$$
$$V^2=\frac{2gmR}{m+M}$$
where $R$ is the radius of the circle traced out by the centre of the small sphere - which is not the same as the radius $R'$ of the circular face of the block : $R=R'-r$ where $r$ is the radius of the sphere.
Note that if $M \gg m$ then $v^2 \approx 2gR$, which is the same as for a pendulum of length $R$. In this case the time of descent will also be the same as the quarter-period of a pendulum of length $R$ and amplitude $\theta_0=\frac{\pi}{2}$. Using the value for $T$ given in this webpage the time of descent is $1.8541\sqrt{\frac{R}{g}}$.
If $m \gg M$ then the bob drops approx. vertically down and the time of descent is $1.4142\sqrt{\frac{R}{g}}$.
If $m$ is comparable with $M$ then the time of descent will lie between the extremes of 1.4142 and 1.8541 in units of $\sqrt{\frac{R}{g}}$. The exact value will be very difficult to calculate - much harder than for the simple pendulum, the exact solution of which involves an elliptic integral for 'large' amplitudes.
ORIGINAL ANSWER
The wikipedia article for the Simple Pendulum gives formulas for the period of the pendulum, in terms of elliptic integrals or as power series, as a function of amplitude. It also gives a very fast-converging method using the arithmetic-geometric mean solution for the elliptic integral.
The speed of the bob at the lowest point is easily calculated using conservation of energy : $v^2=2gh$. Yes, this is the same as for a ball sliding (not rolling) down the inside of a spherical shell.
Your other questions (below your 2nd image) about a block on an air track is not at all clear. How does that relate to the pendulum?


TITLE: Writing $\beta$ and X for a qualitative model
QUESTION [0 upvotes]: Express the following model in matrix form, ie: specify $\beta$ and $X$ so that the model can be written as $Y = X \beta + \epsilon$.
The model $Y_{ij} = \mu_i + \epsilon_i$ where $Y_{ij}$ represents the $j$-th observation observed at level $i$ of a qualitative explanatory variable $T$, for $i=0,...,4$ and $j=1,2$.
I know how to write the vector $\beta$ (just a column of $\mu_0$ to $\mu_4$), but how do you write the matrix X?

REPLY [0 votes]: Write $\tilde{Y}$ as a vector of length $10,$ by stacking $Y_{01},Y_{02},Y_{11},Y_{12},...,Y_{42}.$ Similarly define $\tilde{\varepsilon}.$ You are now  looking for a $10\times 5$ matrix $X$.
$\beta^T=(\mu_0,\mu_1,\mu_2,\mu_3,\mu_4)$
$$X=\begin{bmatrix}1&0&0&0&0\\
1&0&0&0&0\\
0&1&0&0&0\\
0&1&0&0&0\\
0&0&1&0&0\\
0&0&1&0&0\\
0&0&0&1&0\\
0&0&0&1&0\\
0&0&0&0&1\\
0&0&0&0&1\end{bmatrix}$$
Then $\tilde{Y}=X\beta+\tilde{\varepsilon}$ is an expression of $Y_{ij}=\mu_i+\varepsilon_{ij}.$
If instead you had $Y_{ij}=\mu_0+\mu_i+\varepsilon_{ij},$ $\tilde{Y}$ and $\tilde{\varepsilon}$ will have length $8.$ Then
$$X=\begin{bmatrix}1&1&0&0\\
1&1&0&0\\
1&1&0&0\\
1&1&0&0\\
1&0&1&0\\
1&0&1&0\\
1&0&0&1\\
1&0&0&1\end{bmatrix}$$
will be your design matrix so that $\tilde{Y}=X\beta+\tilde{\varepsilon}.$


TITLE: Where am I going wrong integrating $\int x^2(x^3-6)^{34} dx$ by parts?
QUESTION [4 upvotes]: I was helping a friend with homework and was given the fairly straightforward integral:
$$\int x^2(x^3-6)^{34} dx$$
The answer the homework checker accepts as correct baffles me since it has one term with one independent degree:
$$\frac1{105}(x^3-6)^{35}+ C$$
That in no way whatsoever resembles my answer:
$$(x^3-6)^{34}(1/3x^3) - \frac {17}{21} x^{42} + \frac {102}{19} x^{38} + C$$
I feel like there is some weird trick that allows us to avoid integrating by parts, but they wanted us to use integration by parts with $u=x^3-6$. These are the steps I used:
$$
\begin{align}
&u = x^3 - 6\\
&w = u^{34}\\
&\frac {d}{dx} w = \frac {dw}{du} \frac {du}{dx} dx\\
&\cdots = (34u^{33})(3x^2)dx = 102x^{38} - 612x^{35}\\
&dv = x^2 dx\\
&v = 1/3 x^3 + C\\
&I = wv - \int v dw\\
&\cdots= (x^3-6)^{34}(1/3x^3) - I_2\\
&I_2 = \int (1/3 x^3)(102x^{38} - 612x^{35}) dx
\end{align}$$
Regardless of anything else, I can't figure out how you are supposed to have $1$ term with one degree, $35$, in our answer.

EDIT: Just adding what made the accepted answer make sense to me, $$x^2 u^{34} dx = C(u^{34} du)$$ where $C$ is a constant because I wanted to emphasize the general form.

REPLY [3 votes]: $$\int x^2(x^3-6)^{34} dx$$

but they wanted us to use integration by parts with u=x^3-6.

I certainly agree with amwhy's answer.  Integration by parts is an unnecessary complication, since the $u = \left(x^3 - 6\right)$ substitution implies (informally) that $du = 3x^2 dx.$
However, this begs the question: what was the problem composer's intended solution, under the assumption that you (somehow) combine $u = (x^3 - 6)$ with integration by parts?
So, I will take a crack at what the problem composer might have intended.
For convenience, I will omit displaying any constants of integration.
I will use the idea that
$$\left(\int a ~db\right) = ab - \left(\int b ~da\right). \tag1 $$
Set $\displaystyle I = \int x^2(x^3-6)^{34} dx.$
Set $u = (x^3 - 6).$
Set $a = u^{34}.$
Set $db = x^2~dx.$
Then, $\displaystyle I = \int a ~db.$
Following the formula in (1) above,
$\displaystyle I = ab - \int b ~da.$
Since $a = u^{34},$ you have that $da = 34u^{33} du.$
Since $db = x^2 ~dx$, you have that
$$b = \frac{x^3}{3} = \frac{1}{3} (u + 6). \tag2 $$

Edit
I think that arguments could be made, either way that:

This whole approach is daffy, and a substitution attempt should be your first try.  Then, integration by parts would not be used for this problem.

In rebuttal, in addition to reinforcing the correct use of integration by parts, you have that the new Math student is accidentally discovering the convenient result in (2) above, by the pedestrian method of integration by parts.  So, there is the suggestion that if an integration problem does not seem to immediately bend to a substitution, that integration by parts might reveal an insight.

So, $\displaystyle I = u^{34}\left[\frac{1}{3} (u + 6)\right] - J
 ~=~ \left[\frac{u^{35}}{3} + 2u^{34}\right] - J$,
where $\displaystyle J = \int 34u^{33}\left[\frac{1}{3} (u + 6)\right] ~du = \int \left[\frac{34u^{34}}{3} + 68u^{33}\right] ~du.$
You have that $~\displaystyle
J = \frac{34u^{35}}{105} + 2u^{34}.$
Therefore,
$~\displaystyle I = \left[\frac{u^{35}}{3} + 2u^{34}\right] -
\left[\frac{34u^{35}}{105} + 2u^{34}\right] = \frac{u^{35}}{105}
 ~=~ \frac{(x - 6)^{(35)}}{105}.$


text:
\begin{document}

\title{Deadbeat observer: construction via sets}
\author{S. Emre Tuna\footnote{Author is with Department of
Electrical and Electronics Engineering, Middle East Technical
University, 06800 Ankara, Turkey. Email: {\tt
tuna@eee.metu.edu.tr}}} \maketitle

\begin{abstract}
A geometric generalization of discrete-time linear deadbeat
observer is presented. The proposed method to generate a deadbeat
observer for a given nonlinear system is constructive and makes
use of sets that can be computed iteratively. For demonstration,
derivations of observer dynamics are provided for various example
systems. Based on the method, a simple algorithm that computes the
deadbeat gain for a linear system with scalar output is given.
\end{abstract}

\section{Introduction}

Observer design for linear systems is generally acknowledged to be
understood well enough. For discrete-time linear system $x^{+}=Ax$
with output $y=Cx$, Luenberger observer \cite{luenberger64}
dynamics read
\begin{eqnarray}\label{eqn:luenberger}
\xhat^{+}=A\xhat+L(y-C\xhat)
\end{eqnarray}
and designing the observer is nothing but choosing an observer
gain $L$ that places the eigenvalues of matrix $A-LC$ within the
unit circle. Simple and elegant, anyone would hardly doubt that
this construction is {\em the} construction for linear systems.
However, perhaps due arguably to over-elegance of the notation, it
is nontrivial to unearth the true mechanism (if it exists) running
behind Luenberger observer in order to generalize it in some {\em
natural} way for nonlinear systems. In this paper we aim to
provide a geometric interpretation of the righthand side of
\eqref{eqn:luenberger} for the particular case where matrix $A-LC$
is nilpotent, i.e., when the observer is deadbeat. Our
interpretation allows one to construct deadbeat observers for
nonlinear systems provided that certain conditions
(Assumption~\ref{assume:singleton} and
Assumption~\ref{assume:invariance}) hold. We now note and later
demonstrate that when the system is linear those assumptions are
minimal for a deadbeat observer to exist. The literature on
observers accommodates significant results. See, for instance,
\cite{karafyllis07,glad83,moraal95,valcher99,shamma99,fuhrmann06,besancon00,wong04}.

The toy example that we keep in the back of our mind while we
attempt to reach a generalization is the simple case where $A$ is
a rotation matrix in $\Real^{2}$
\begin{eqnarray*}
A =\left[\!\!\begin{array}{rr} \cos\theta &-\sin\theta\\
\sin\theta &\cos\theta
\end{array}\!\!\right]
\end{eqnarray*}
with angle of rotation $\theta$ different from $0$ and $\pi$.
Letting $y=x_{2}$, i.e., $C=[0\ \ 1]$, the deadbeat observer turns
out to be
\begin{eqnarray*}
\xhat^{+}=A\xhat+\left[\!\!\begin{array}{c} \cos2\theta/\sin\theta\\
\sin2\theta/\sin\theta
\end{array}\!\!\right](y-C\xhat)
\end{eqnarray*}
which can be rewritten as
\begin{eqnarray*}
\xhat^{+}=A\left(\xhat+\left[\!\!\begin{array}{c} \cot\theta\\
1
\end{array}\!\!\right](y-C\xhat)\right)
\end{eqnarray*}
Now we state the key observation in this paper: The term in
brackets is the intersection of two equivalence classes (sometimes
called congruence classes \cite{lax96}). Namely,
\begin{eqnarray*}
\xhat+\left[\!\!\begin{array}{c} \cot\theta\\
1
\end{array}\!\!\right](y-C\xhat)=(\xhat+A\,
\nal(C))\cap(x+\nal(C))
\end{eqnarray*}
as shown in Fig.~\ref{fig:intersect}.
\begin{figure}[h]
\begin{center}
\includegraphics[scale=0.6]{intro.eps}
\caption{Intersection of two equivalence
classes.}\label{fig:intersect}
\end{center}
\end{figure}
Based on this observation, one contribution of this paper is
intended to be in showing that such equivalence classes can be
defined even for nonlinear systems of arbitrary order, which in
turn allows one to construct deadbeat observers. There is another
possible contribution that is of more practical nature: We present
a simple algorithm that computes, for linear systems with scalar
output, deadbeat gain $L$ by iteratively intersecting linear
subspaces. (Devising reliable numerical techniques to compute
deadbeat gain for discrete-time linear systems had once been an
active field of research; see, for instance,
\cite{franklin82,lewis82,sugimoto93}.)

The remainder of the paper is organized as follows. Next section
contains some preliminary material. In Section~\ref{sec:def} we
give the formal problem definition. Section~\ref{sec:sets} is
where we describe the sets that we use in construction of the
deadbeat observer. We state and prove the main result in
Section~\ref{sec:main}. An extension of the main result where we
consider the case with input ($x^{+}=f(x,\,u)$) is in
Section~\ref{sec:input}. We provide examples in
Section~\ref{sec:ex}, where we construct deadbeat observers for
two different third order systems. In Section~\ref{sec:alg} we
present an algorithm to compute the deadbeat observer gain for a
linear system with scalar output.

\section{Preliminaries}

Identity matrix is denoted by $I$. Null space and range space of a
matrix $M\in\Real^{m\times n}$ are denoted by $\N(M)$ and $\R(M)$,
respectively. Given map $\mu:\X\to\Y$, $\mu^{-1}(\cdot)$ denotes
the {\em inverse} map in the general sense that for $y\in\Y$,
$\mu^{-1}(y)$ is the set of all $x\in\X$ satisfying $\mu(x)=y$.
That is, we will not need $\mu$ be bijective when talking about
its inverse. Note that $y\notin\mu(\X)$ will imply
$\mu^{-1}(y)=\emptyset$. Linear maps $x\mapsto Mx$ will not be
exempt from this notation. The reader should not think that $M$ is
a nonsingular matrix when we write $M^{-1}y$. (In our case $M$
need even not be square.) For instance, for $M=[0\ \ 0]$ we have
$M^{-1}y=\emptyset$ for $y\neq 0$ and $M^{-1}0=\Real^{2}$. The set
of nonnegative integers is denoted by $\Natural$ and $\Real_{>0}$
denotes the set of strictly positive real numbers.

\section{Problem definition}\label{sec:def}

Consider the following discrete-time system
\begin{subeqnarray}\label{eqn:system}
x^{+}&=&f(x)\\
y&=&h(x)
\end{subeqnarray}
where $x\in\X\subset\Real^{n}$ is the {\em state}, $x^{+}$ is the
state at the next time instant, and $y\in\Y\subset\Real^{m}$ is
the {\em output} or the {\em measurement}. The {\em solution} of
system~\eqref{eqn:system} at time $k\in\Natural$, starting at
initial condition $x\in\X$ is denoted by $\phi(k,\,x)$. Note that
$\phi(0,\,x)=x$ and $\phi(k+1,\,x)=f(\phi(k,\,x))$ for all $x$ and
$k$.

Now consider the following cascade system
\begin{subeqnarray}
x^{+}&=&f(x)\\
\xhat^{+}&\in&g(\xhat,\,h(x))\label{eqn:cascade}
\end{subeqnarray}
We denote {\em a} solution of subsystem~(\ref{eqn:cascade}b) by
$\psi(k,\,\xhat,\,x)$. We then have $\psi(0,\,\xhat,\,x)=\xhat$
and $\psi(k+1,\,\xhat,\,x)\in
g(\psi(k,\,\xhat,\,x),\,h(\phi(k,\,x)))$ for all $x$, $\xhat$, and
$k$. We now use \eqref{eqn:cascade} to define deadbeat observer.

\begin{definition}
Given $g:\X\times\Y\rightrightarrows\X$, system
\begin{eqnarray*}
\xhat^{+}\in g(\xhat,\,y)
\end{eqnarray*}
is said to be a {\em deadbeat observer for
system~\eqref{eqn:system}} if there exists $p\geq 1$ such that
{\em all} solutions of system~\eqref{eqn:cascade} satisfy
\begin{eqnarray*}
\psi(k,\,\xhat,\,x)=\phi(k,\,x)
\end{eqnarray*}
for all $x,\,\xhat\in\X$ and $k\geq p$.
\end{definition}

\begin{definition}
System~\eqref{eqn:system} is said to be {\em deadbeat observable}
if there exists a deadbeat observer for it.
\end{definition}

In this paper we present a procedure to construct a deadbeat
observer for system~\eqref{eqn:system} provided that certain
conditions (Assumption~\ref{assume:singleton} and
Assumption~\ref{assume:invariance}) hold. Our construction will
make use of some sets, which we define in the next section. Before
moving on into the next section, however, we choose to remind the
reader of a standard fact regarding the observability of linear
systems. Then we provide a Lemma~\ref{lem:subspace} as a geometric
equivalent of that well-known result. Lemma~\ref{lem:subspace}
will find use later when we attempt to interpret and display the
generality of the assumptions we will have made.

The following criterion, known as Popov-Belevitch-Hautus (PBH)
test, is an elegant tool for checking (deadbeat) observability.

\begin{proposition}[PBH test]
The linear system
\begin{subeqnarray}\label{eqn:linsystem}
x^{+}&=&Ax\\
y&=&Cx
\end{subeqnarray}
with $A\in\Real^{n\times n}$ and $C\in\Real^{m\times n}$ is
deadbeat observable if and only if
\begin{eqnarray}\label{eqn:PBH}
{\rm rank}\left[\!\!\begin{array}{cc}A-\lambda I\\
C\end{array}\!\!\right]=n\quad \mbox{for all}\quad \lambda\neq 0
\end{eqnarray}
where $\lambda$ is a complex scalar.
\end{proposition}

The below result is a geometric equivalent of PBH test.

\begin{lemma}\label{lem:subspace}
Given $A\in\Real^{n\times n}$ and $C\in\Real^{m\times n}$, let
subspace $\setS_{k}$ of $\Real^{n}$ be defined as
$\setS_{k}:=A\setS_{k-1}\cap\setS_{0}$ for $k=1,\,2,\,\ldots$ with
$\setS_{0}:=\N(C)$. Then system~\eqref{eqn:linsystem} is deadbeat
observable if and only if
\begin{eqnarray}\label{eqn:SET}
\setS_{n}=\{0\}\,.
\end{eqnarray}
\end{lemma}

\begin{proof}
For simplicity we provide the demonstration for the case where
each $\setS_{k}$ is a subspace of $\Complex^{n}$ (over field
$\Complex$). The case $\setS_{k}\subset\Real^{n}$ is a little
longer to prove yet it is true.

We first show \eqref{eqn:SET}$\implies$\eqref{eqn:PBH}. Suppose
\eqref{eqn:PBH} fails. That is, there exists an eigenvector
$v\in\Complex^{n}$ and a nonzero eigenvalue $\lambda\in\Complex$
such that $Av=\lambda v$ and $Cv=0$. Now suppose for some $k$ we
have $v\in\setS_{k}$. Then, since $v$ is an eigenvector with a
nonzero eigenvalue, we can write $v\in A\setS_{k}$. Observe that
$v\in\setS_{0}$ for $Cv=0$. As a result $v\in
A\setS_{k}\cap\setS_{0}=\setS_{k+1}$. By induction therefore we
have $v\in\setS_{k}$ for all $k$, which means that \eqref{eqn:SET}
fails.

Now we demonstrate the other direction
\eqref{eqn:PBH}$\implies$\eqref{eqn:SET}. We first claim that
$\setS_{k+1}\subset\setS_{k}$ for all $k$. We use induction to
justify our claim. Suppose $\setS_{k+1}\subset\setS_{k}$ for some
$k$. Then we can write
\begin{eqnarray*}
\setS_{k+2}
&=&A\setS_{k+1}\cap\setS_{0}\\
&\subset&A\setS_{k}\cap\setS_{0}\\
&=&\setS_{k+1}\,.
\end{eqnarray*}
Since $\setS_{1}\subset\setS_{0}$ our claim is valid. A trivial
implication of our claim then follows:
$\dim\setS_{k+1}\leq\dim\setS_{k}$ for all $k$. Let us now suppose
\eqref{eqn:SET} fails. That is, $\dim\setS_{n}\geq 1$. Note that
$\dim\setS_{0}\leq n$. Therefore $\dim\setS_{n}\geq 1$ and
$\dim\setS_{k+1}\leq\dim\setS_{k}$ imply the existence of some
$\ell\in\{0,\,1,\,\ldots,\,n-1\}$ such that
$\dim\setS_{\ell+1}=\dim\setS_{\ell}\geq 1$. Since
$\setS_{\ell+1}\subset\setS_{\ell}$, both $\setS_{\ell+1}$ and
$\setS_{\ell}$ having the same dimension implies
$\setS_{\ell+1}=\setS_{\ell}$. Hence we obtained
$\setS_{\ell}=A\setS_{\ell}\cap\setS_{0}$ which allows us to write
$\setS_{\ell}\subset A\setS_{\ell}$. Since
$\dim\setS_{\ell}\geq\dim A\setS_{\ell}$ we deduce that
$\setS_{\ell}=A\setS_{\ell}$. Since $\dim\setS_{\ell}\geq 1$,
equality $A\setS_{\ell}=\setS_{\ell}$ implies that there exists an
eigenvector $v\in\setS_{\ell}$ and a nonzero eigenvalue
$\lambda\in\Complex$ such that $Av=\lambda v$. Note also that
$Cv=0$ because $\setS_{\ell}\subset\setS_{0}$. Hence
\eqref{eqn:PBH} fails.
\end{proof}

\begin{remark}\label{rem:dimension}
It is clear from the proof that if \eqref{eqn:SET} fails then
$\dim\setS_{k}\geq 1$ for all $k$.
\end{remark}

\section{Sets}\label{sec:sets}

In this section we define certain sets (more formally, {\em
equivalence classes}) associated with system~\eqref{eqn:system}.
For $x\in\X$ we define
\begin{eqnarray*}
[x]_{0}:=h^{-1}(h(x))\,.
\end{eqnarray*}
Note that when $h(x)=Cx$, where $C\in\Real^{m\times n}$, we have
$[x]_{0}=x+\N(C)$. We then let for $k=0,\,1,\,\ldots$
\begin{eqnarray*}
[x]_{k+1}:=[x]_{k}^{+}\cap[x]_{0}
\end{eqnarray*}
where
\begin{eqnarray*}
[x]_{k}^{+}:=f([f^{-1}(x)]_{k})\,.
\end{eqnarray*}
Note that $[x]_{k}^{+}=\emptyset$ when $x\notin f(\X)$ since then
$f^{-1}(x)=\emptyset$.

\begin{remark}\label{rem:subset}
Note that $[x]_{k+1}\subset[x]_{k}$ and
$[x]^{+}_{k+1}\subset[x]^{+}_{k}$ for all $x$ and $k$.
\end{remark}

The following two assumptions will be invoked in our main theorem.
In hope of making them appear somewhat meaningful and revealing
their generality we provide the conditions that they would boil
down to for linear systems.

\begin{assumption}\label{assume:singleton}
There exists $p\geq 1$ such that, for each $x\in\X$, set
$[x]_{p-1}$ is either singleton or empty set.
\end{assumption}

Assumption~\ref{assume:singleton} is equivalent to deadbeat
observability for linear systems. Below result formalizes this.

\begin{theorem}
Linear system~\eqref{eqn:linsystem} is deadbeat observable if and
only if Assumption~\ref{assume:singleton} holds.
\end{theorem}

\begin{proof}
Let $\setS_{k}$ for $k=0,\,1,\,\ldots$ be defined as in
Lemma~\ref{lem:subspace}. Note then that $[x]_{0}=x+\setS_{0}$. We
claim that the following holds
\begin{eqnarray}\label{eqn:equivalence}
[x]_{k}=\left\{\begin{array}{cl} x+\setS_{k}&\quad\mbox{for}\quad
x\in\R(A^{k})\\
\emptyset&\quad\mbox{for}\quad x\notin\R(A^{k})
\end{array}\right.
\end{eqnarray}
for all $k$. We employ induction to establish our claim. Suppose
\eqref{eqn:equivalence} holds for some $k$. Then we can write
\begin{eqnarray*}
[x]_{k}^{+}
&=&A[A^{-1}x]_{k}\\
&=&A[A^{-1}x\cap\R(A^{k})]_{k}\,.
\end{eqnarray*}
Note that $A^{-1}x\cap\R(A^{k})\neq\emptyset$ if and only if
$x\in\R(A^{k+1})$. Since $[x]_{k+1}=[x]_{k}^{+}\cap[x]_{0}$, we
deduce that $[x]_{k+1}=\emptyset$ for $x\notin\R(A^{k+1})$.
Otherwise if $x\in\R(A^{k+1})$ then there exists some
$\eta\in\R(A^{k})$ such that $A\eta=x$. Using this $\eta$ we can
construct the equality $A^{-1}x=\eta+\N(A)$ and we can write
\begin{eqnarray*}
[x]_{k+1}
&=&[x]_{k}^{+}\cap[x]_{0}\\
&=&A[A^{-1}x]_{k}\cap[x]_{0}\\
&=&A[\eta+\N(A)]_{k}\cap[x]_{0}\\
&=&A(\eta+(\N(A)\cap\R(A^{k}))+\setS_{k})\cap[x]_{0}\\
&=&(A\eta+A\setS_{k})\cap(x+S_{0})\\
&=&(x+A\setS_{k})\cap(x+\setS_{0})\\
&=&x+(A\setS_{k}\cap\setS_{0})\\
&=&x+\setS_{k+1}\,.
\end{eqnarray*}
Since \eqref{eqn:equivalence} holds for $k=0$, our claim is valid.

Now suppose that the system is deadbeat observable. Then by
\eqref{eqn:equivalence} we see that
Assumption~\ref{assume:singleton} holds with $p=n+1$ thanks to
Lemma~\ref{lem:subspace}. If however the system is not deadbeat
observable, then by Remark~\ref{rem:dimension} $\dim \setS_{k}\geq
1$ for all $k$. We deduce by \eqref{eqn:equivalence} therefore
that $[0]_{k}$ can never be singleton nor is it empty. Hence
Assumption~\ref{assume:singleton} must fail.
\end{proof}

\begin{assumption}\label{assume:invariance}
Given $x,\,\xhat\in\X$ and $k$; $\xhat\in[x]_{k}^{+}$ implies
$[\xhat]_{k}^{+}=[x]_{k}^{+}$.
\end{assumption}

\begin{theorem}
Assumption~\ref{assume:invariance} comes for free for linear
system~\eqref{eqn:linsystem}.
\end{theorem}

\begin{proof}
Evident.
\end{proof}
\\

Last we let $[x]_{-1}^{+}:=\X$ and define map
$\pi:\X\times\Y\to\{-1,\,0,\,1,\,\ldots,\,p-2\}$ as
\begin{eqnarray*}
\pi(\xhat,\,y) := \max\,\{-1,\,0,\,1,\,\ldots,\,p-2\}\quad
\mbox{subject to}\quad [\xhat]^{+}_{\pi(\xhat,\,y)}\cap h^{-1}(y)
\neq\emptyset
\end{eqnarray*}
where $p$ is as in Assumption~\ref{assume:singleton}.

\section{The result}\label{sec:main}

Below is our main theorem.

\begin{theorem}\label{thm:main}
Suppose Assumptions~\ref{assume:singleton}-\ref{assume:invariance}
hold. Then system
\begin{eqnarray}\label{eqn:deadbeat}
\hat{x}^{+}\in f([\xhat]^{+}_{\pi(\xhat,\,y)}\cap h^{-1}(y))
\end{eqnarray}
is a deadbeat observer for system~\eqref{eqn:system}.
\end{theorem}

\begin{proof}
We claim the following
\begin{eqnarray}\label{eqn:claim}
\xhat\in [x]_{\ell-1}^{+}\implies \xhat^{+}\in[f(x)]_{\ell}^{+}
\end{eqnarray}
for all $\ell\in\{0,\,1,\,\ldots,\,p-1\}$. Let us prove our claim.
Note that $\xhat\in[x]_{\ell-1}^{+}$ yields
$[\xhat]_{\ell-1}^{+}=[x]_{\ell-1}^{+}$ by
Assumption~\ref{assume:invariance}. Since
$[x]_{\ell-1}^{+}\neq\emptyset$ we have
$[x]_{\ell-1}^{+}\cap[x]_{0}\neq\emptyset$ and, consequently,
$[\xhat]_{\ell-1}^{+}\cap[x]_{0}\neq\emptyset$.
Remark~\ref{rem:subset} then yields
$[\xhat]_{\pi(\xhat,\,h(x))}^{+}\subset[\xhat]_{\ell-1}^{+}$.
Starting from \eqref{eqn:deadbeat} we can proceed as
\begin{eqnarray}\label{eqn:yesnumber}
\hat{x}^{+}
&\in& f([\xhat]^{+}_{\pi(\xhat,\,y)}\cap h^{-1}(y))\nonumber\\
&=& f([\xhat]^{+}_{\pi(\xhat,\,h(x))}\cap h^{-1}(h(x)))\nonumber\\
&\subset& f([\xhat]^{+}_{\ell-1}\cap [x]_{0})\nonumber\\
&=& f([x]^{+}_{\ell-1}\cap [x]_{0})\nonumber\\
&=& f([x]_{\ell})\\
&\subset& f([f^{-1}(f(x))]_{\ell})\nonumber\\
&=& [f(x)]^{+}_{\ell}\,.\nonumber
\end{eqnarray}
Hence \eqref{eqn:claim} holds. In particular,
\eqref{eqn:yesnumber} gives us
\begin{eqnarray}\label{eqn:genco}
\xhat\in [x]_{\ell-1}^{+}\implies \xhat^{+}\in f([x]_{\ell})
\end{eqnarray}
for all $\ell\in\{0,\,1,\,\ldots,\,p-1\}$. Note that $\xhat\in
[x]^{+}_{-1}$ holds for all $x,\,\xhat$. Therefore
\eqref{eqn:claim} and Remark~\ref{rem:subset} imply the existence
of $\ell^{\ast}\in\{0,\,1,\,\ldots,\,p-1\}$ such that
\begin{eqnarray}\label{eqn:dede}
\psi(k,\,\xhat,\,x)\in[\phi(k,\,x)]^{+}_{p-2}
\end{eqnarray}
for all $k\geq\ell^{\ast}$. Also,
Assumption~\ref{assume:singleton} yields us
\begin{eqnarray}\label{eqn:ibo}
[\phi(k,\,x)]_{p-1}=\phi(k,\,x)
\end{eqnarray}
for all $k\geq p-1$. Combining \eqref{eqn:genco},
\eqref{eqn:dede}, and \eqref{eqn:ibo} we can write
\begin{eqnarray*}
\psi(k,\,\xhat,\,x)=\phi(k,\,x)
\end{eqnarray*}
for all $k\geq p$. Hence the result.
\end{proof}

\begin{corollary}\label{cor:foralg}
Consider linear system~\eqref{eqn:linsystem} with
$A\in\Real^{n\times n}$ and $C\in\Real^{1\times n}$. Suppose pair
$(C,\,A)$ is observable\footnote{That is, $\mbox{rank}\ [C^{T}\
A^{T}C^{T}\ \ldots\ A^{(n-1)T}C^{T}]=n$.}. Let $\setS_{k}$ for
$k=0,\,1,\,\ldots$ be defined as in Lemma~\ref{lem:subspace}. Then
system
\begin{eqnarray*}
\xhat^{+}=A((\xhat+A\setS_{n-2})\cap(x+\setS_{0}))
\end{eqnarray*}
is a deadbeat observer for system~\eqref{eqn:linsystem}.
\end{corollary}

\section{System with input}\label{sec:input}

In this section we look at the case where the evolution of system
to be observed is dependent not only on the initial condition but
also on some exogenous signal, which we call the input. To
construct a deadbeat observer for such system we again make use of
sets.

Consider the system
\begin{subeqnarray}\label{eqn:systemwu}
x^{+}&=&f(x,\,u)\\
y&=&h(x)
\end{subeqnarray}
where $u\in\U\subset\Real^{q}$ is the {\em input} or some known
{\em disturbance} (e.g. time). Let ${\bf
u}=(u_{0},\,u_{1},\,\ldots)$, $u_{k}\in\U$, denote an input
sequence. The {\em solution} of system~\eqref{eqn:systemwu} at
time $k$, starting at initial condition $x\in\X$, and having
evolved under the influence of input sequence $\yu$ is denoted by
$\phi(k,\,x,\,\yu)$. Note that $\phi(0,\,x,\,\yu)=x$ and
$\phi(k+1,\,x,\,\yu)=f(\phi(k,\,x,\,\yu),\,u_{k})$ for all $x$,
$\yu$, and $k$.

Now consider the following cascade system
\begin{subeqnarray}
x^{+}&=&f(x,\,u)\\
\xhat^{+}&\in&g(\xhat,\,h(x),\,u)\label{eqn:cascadewu}
\end{subeqnarray}
We denote a solution of subsystem~(\ref{eqn:cascadewu}b) by
$\psi(k,\,\xhat,\,x,\,\yu)$. We then have
$\psi(0,\,\xhat,\,x,\,\yu)=\xhat$ and
$\psi(k+1,\,\xhat,\,x,\,\yu)\in
g(\psi(k,\,\xhat,\,x,\,\yu),\,h(\phi(k,\,x,\,\yu)),\,u_{k})$ for
all $x$, $\xhat$, $\yu$, and $k$.

\begin{definition}
Given $g:\X\times\Y\times\U\rightrightarrows\X$, system
\begin{eqnarray*}
\xhat^{+}\in g(\xhat,\,y,\,u)
\end{eqnarray*}
is said to be a {\em deadbeat observer for
system~\eqref{eqn:systemwu}} if there exists $p\geq 1$ such that
solutions of system~\eqref{eqn:cascadewu} satisfy
\begin{eqnarray*}
\psi(k,\,\xhat,\,x,\,\yu)=\phi(k,\,x,\,\yu)
\end{eqnarray*}
for all $x$, $\xhat$, $\yu$, and $k\geq p$.
\end{definition}

How to define sets $[x]_{k}$ and $[x]_{k}^{+}$ for
system~\eqref{eqn:systemwu} is obvious. We again let
\begin{eqnarray*}
[x]_{0}:=h^{-1}(h(x))\,.
\end{eqnarray*}
and (for $k=0,\,1,\,\ldots$)
\begin{eqnarray*}
[x]_{k+1}:=[x]_{k}^{+}\cap[x]_{0}
\end{eqnarray*}
this time with
\begin{eqnarray*}
[x]_{k}^{+}:=\bigcup_{f(\eta,\,u)=x} f([\eta]_{k},\,u)\,.
\end{eqnarray*}
The following result is a generalization of
Theorem~\ref{thm:main}. (The demonstration is parallel to that of
Theorem~\ref{thm:main} and hence omitted.)
\begin{theorem}\label{thm:mainu}
Suppose Assumptions~\ref{assume:singleton}-\ref{assume:invariance}
hold. Then system
\begin{eqnarray*}
\hat{x}^{+}\in f([\xhat]^{+}_{\pi(\xhat,\,y)}\cap h^{-1}(y),\,u)
\end{eqnarray*}
is a deadbeat observer for system~\eqref{eqn:systemwu}.
\end{theorem}

\section{Examples}\label{sec:ex}

Here, for two third order nonlinear systems, we construct deadbeat
observers. In the first example we study a simple autonomous
homogeneous system and show that the construction yields a
homogeneous observer. Hence our method may be thought to be
somewhat {\em natural} in the vague sense that the observer it
generates inherits certain intrinsic properties of the system. In
the second example we aim to provide a demonstration on observer
construction for a system with input.

\subsection{Homogeneous system}

Consider system~\eqref{eqn:system} with
\begin{eqnarray*}
f(x):=\left[\!\!
\begin{array}{c}
x_{2}\\
x_{3}^{1/3}\\
x_{1}^{3}+x_{2}^{3}
\end{array}\!\!\right]\quad\mbox{and}\quad h(x):=x_{1}
\end{eqnarray*}
where $x=[x_{1}\ x_{2}\ x_{3}]^{T}$. Let $\X=\Real^{3}$ and
$\Y=\Real$. If we let dilation $\Delta_{\lambda}$ be
\begin{eqnarray*}
\Delta_{\lambda}:=\left[\!\!
\begin{array}{ccc}
\lambda&0&0\\
0&\lambda&0\\
0&0&\lambda^{3}
\end{array}\!\!\right]
\end{eqnarray*}
with $\lambda\in\Real$, then we realize that
\begin{eqnarray*}
f(\Delta_{\lambda}x)=\Delta_{\lambda}f(x)\quad\mbox{and}\quad
h(\Delta_{\lambda}x)=\lambda h(x)\,.
\end{eqnarray*}
That is, the system is homogeneous \cite{rinehart09} with respect
to dilation $\Delta$. Before describing the relevant sets
$[x]_{k}$ and $[x]_{k}^{+}$ we want to mention that $f$ is
bijective and its inverse is
\begin{eqnarray}\label{eqn:finv}
f^{-1}(x)=\left[\!\!
\begin{array}{c}
(x_{3}-x_{1}^{3})^{1/3}\\
x_{1}\\
x_{2}^{3}
\end{array}\!\!\right]
\end{eqnarray}
Since $h(x)=x_{1}$ we can write
\begin{eqnarray}\label{eqn:x0}
[x]_{0}=\left\{ \left[\!\!\begin{array}{c} x_{1}\\
\alpha\\
\beta
\end{array}\!\!\right]:\alpha,\,\beta\in\Real
\right\}
\end{eqnarray}
By \eqref{eqn:finv} we can then proceed as
\begin{eqnarray}\label{eqn:x0plus}
[x]_{0}^{+}&=&f([f^{-1}(x)]_{0})\nonumber\\
&=&f\left(\left\{ \left[\!\!\begin{array}{c} (x_{3}-x_{1}^{3})^{1/3}\\
\gamma\\
\delta
\end{array}\!\!\right]:\gamma,\,\delta\in\Real
\right\}\right)\nonumber\\
&=&\left\{ f\left(\left[\!\!\begin{array}{c} (x_{3}-x_{1}^{3})^{1/3}\\
\gamma\\
\delta
\end{array}\!\!\right]\right):\gamma,\,\delta\in\Real
\right\}\nonumber\\
&=&\left\{ \left[\!\!\begin{array}{c} \gamma\\
\delta^{1/3}\\
x_{3}-x_{1}^{3}+\gamma^{3}
\end{array}\!\!\right]:\gamma,\,\delta\in\Real
\right\}
\end{eqnarray}
Recall that $[x]_{1}=[x]_{0}^{+}\cap[x]_{0}$. Therefore
intersecting sets \eqref{eqn:x0} and \eqref{eqn:x0plus} we obtain
\begin{eqnarray*}
[x]_{1}=\left\{ \left[\!\!\begin{array}{c} x_{1}\\
\alpha\\
x_{3}
\end{array}\!\!\right]:\alpha\in\Real
\right\}
\end{eqnarray*}
We can now construct $[x]_{1}^{+}$ as
\begin{eqnarray}\label{eqn:x1plus}
[x]_{1}^{+}&=&f([f^{-1}(x)]_{1})\nonumber\\
&=&f\left(\left\{ \left[\!\!\begin{array}{c} (x_{3}-x_{1}^{3})^{1/3}\\
\gamma\\
x_{2}^{3}
\end{array}\!\!\right]:\gamma\in\Real
\right\}\right)\nonumber\\
&=&\left\{ f\left(\left[\!\!\begin{array}{c} (x_{3}-x_{1}^{3})^{1/3}\\
\gamma\\
x_{2}^{3}
\end{array}\!\!\right]\right):\gamma\in\Real
\right\}\nonumber\\
&=&\left\{ \left[\!\!\begin{array}{c} \gamma\\
x_{2}\\
x_{3}-x_{1}^{3}+\gamma^{3}
\end{array}\!\!\right]:\gamma\in\Real
\right\}
\end{eqnarray}
Now note that sets \eqref{eqn:x0} and \eqref{eqn:x1plus} intersect
at a single point. In particular,
$[x]_{2}=[x]_{1}^{+}\cap[x]_{0}=x$. Therefore
Assumption~\ref{assume:singleton} is satisfied with $p=3$. Observe
also that
\begin{eqnarray*}
[\xhat]_{1}^{+}\cap h^{-1}(y) &=&
\left\{ \left[\!\!\begin{array}{c} \gamma\\
\xhat_{2}\\
\xhat_{3}-\xhat_{1}^{3}+\gamma^{3}
\end{array}\!\!\right]:\gamma\in\Real
\right\}\cap\left\{ \left[\!\!\begin{array}{c} y\\
\alpha\\
\beta
\end{array}\!\!\right]:\alpha,\,\beta\in\Real
\right\}\\
&=&\left[\!\!\begin{array}{c} y\\
\xhat_{2}\\
\xhat_{3}-\xhat_{1}^{3}+y^{3}
\end{array}\!\!\right]
\end{eqnarray*}
which means that $\pi(\xhat,\,y)=p-2=1$ for all $\xhat$ and $y$.
The dynamics of the deadbeat observer then read
\begin{eqnarray*}
\xhat^{+}&=&f([\xhat]_{1}^{+}\cap h^{-1}(y))\\
&=&\left[\!\!\begin{array}{c}
\xhat_{2}\\
(\xhat_{3}-\xhat_{1}^{3}+y^{3})^{1/3}\\
\xhat_{2}^{3}+y^{3}
\end{array}\!\!\right]
\end{eqnarray*}
We finally notice that
\begin{eqnarray*}
f([\Delta_{\lambda}\xhat]_{1}^{+}\cap h^{-1}(\lambda
y))=\Delta_{\lambda}f([\xhat]_{1}^{+}\cap h^{-1}(y))\,.
\end{eqnarray*}
That is, the deadbeat observer also is homogeneous with respect to
dilation $\Delta$.

\subsection{System with input}
Our second example is again a third order system, this time
however with an input. Consider system~\eqref{eqn:systemwu} with
\begin{eqnarray*}
f(x,\,u):=\left[\!\!
\begin{array}{c}
x_{1}x_{2}x_{3}\\
x_{3}/x_{1}\\
\sqrt{x_{1}x_{2}u}
\end{array}\!\!\right]
\quad\mbox{and}\quad h(x):=x_{1}\,.
\end{eqnarray*}
Let $\X=\Real_{>0}^{3}$, $\Y=\Real_{>0}$, and $\U=\Real_{>0}$. Let
us construct the relevant sets $[x]_{k}$ and $[x]_{k}^{+}$. We
begin with $[x]_{0}$.
\begin{eqnarray}\label{eqn:x0nd}
[x]_{0}=\left\{ \left[\!\!\begin{array}{c} x_{1}\\
\alpha\\
\beta
\end{array}\!\!\right]:\alpha,\,\beta>0
\right\}
\end{eqnarray}
Note that $f$ satisfies the following
\begin{eqnarray*}
f\left(\left[\!\!
\begin{array}{c}
x_{1}u/(x_{2}x_{3}^{2})\\
x_{2}x_{3}^{4}/(x_{1}u^{2})\\
x_{1}u/x_{3}^{2}
\end{array}\!\!\right],\,u\right)=x
\end{eqnarray*}
for all $x$ and $u$. Hence we can write
\begin{eqnarray}\label{eqn:x0plusnd}
[x]_{0}^{+}&=&\bigcup_{u\in\U}f\left(\left[\!\!
\begin{array}{c}
x_{1}u/(x_{2}x_{3}^{2})\\
x_{2}x_{3}^{4}/(x_{1}u^{2})\\
x_{1}u/x_{3}^{2}
\end{array}\!\!\right]_{0},\,u\right)\nonumber\\
&=&\bigcup_{u\in\U}f\left(\left\{ \left[\!\!\begin{array}{c} x_{1}u/(x_{2}x_{3}^{2})\\
\gamma\\
\delta
\end{array}\!\!\right]:\gamma,\,\delta>0
\right\},\,u\right)\nonumber\\
&=&\bigcup_{u\in\U}f\left(\left\{ \left[\!\!\begin{array}{c} x_{1}u/(x_{2}x_{3}^{2})\\
x_{2}x_{3}^{4}\gamma/(x_{1}u^{2})\\
x_{1}u\delta/x_{3}^{2}
\end{array}\!\!\right]:\gamma,\,\delta>0
\right\},\,u\right)\nonumber\\
&=&\bigcup_{u\in\U}\left\{ f\left(\left[\!\!\begin{array}{c} x_{1}u/(x_{2}x_{3}^{2})\\
x_{2}x_{3}^{4}\gamma/(x_{1}u^{2})\\
x_{1}u\delta/x_{3}^{2}
\end{array}\!\!\right],\,u\right):\gamma,\,\delta>0
\right\}\nonumber\\
&=&\left\{\left[\!\!\begin{array}{c} x_{1}\gamma\delta\\
x_{2}\delta\\
x_{3}\sqrt{\gamma}
\end{array}\!\!\right] :\gamma,\,\delta>0
\right\}
\end{eqnarray}
Since $[x]_{1}=[x]_{0}^{+}\cap[x]_{0}$, intersecting sets
\eqref{eqn:x0nd} and \eqref{eqn:x0plusnd} we obtain
\begin{eqnarray*}
[x]_{1}=\left\{ \left[\!\!\begin{array}{c} x_{1}\\
x_{2}/\alpha^{2}\\
x_{3}\alpha
\end{array}\!\!\right]:\alpha>0
\right\}
\end{eqnarray*}
We can now construct $[x]_{1}^{+}$ as
\begin{eqnarray}\label{eqn:x1plusnd}
[x]_{1}^{+}&=&\bigcup_{u\in\U}f\left(\left[\!\!
\begin{array}{c}
x_{1}u/(x_{2}x_{3}^{2})\\
x_{2}x_{3}^{4}/(x_{1}u^{2})\\
x_{1}u/x_{3}^{2}
\end{array}\!\!\right]_{1},\,u\right)\nonumber\\
&=&\bigcup_{u\in\U}f\left(\left\{ \left[\!\!\begin{array}{c} x_{1}u/(x_{2}x_{3}^{2})\\
x_{2}x_{3}^{4}/(x_{1}u^{2}\gamma^{2})\\
x_{1}u\gamma/x_{3}^{2}
\end{array}\!\!\right]:\gamma>0
\right\},\,u\right)\nonumber\\
&=&\left\{\left[\!\!\begin{array}{c} x_{1}/\gamma\\
x_{2}\gamma\\
x_{3}/\gamma
\end{array}\!\!\right] :\gamma>0
\right\}
\end{eqnarray}
Now note that sets \eqref{eqn:x0nd} and \eqref{eqn:x1plusnd}
intersect at a single point. In particular,
$[x]_{2}=[x]_{1}^{+}\cap[x]_{0}=x$. Therefore
Assumption~\ref{assume:singleton} is satisfied with $p=3$. Observe
also that
\begin{eqnarray*}
[\xhat]_{1}^{+}\cap h^{-1}(y) &=& \left\{\left[\!\!\begin{array}{c} \xhat_{1}/\gamma\\
\xhat_{2}\gamma\\
\xhat_{3}/\gamma
\end{array}\!\!\right] :\gamma>0
\right\}\cap \left\{ \left[\!\!\begin{array}{c} y\\
\alpha\\
\beta
\end{array}\!\!\right]:\alpha,\,\beta>0
\right\}\\ &=&
\left[\!\!\begin{array}{c} y\\
\xhat_{1}\xhat_{2}/y\\
\xhat_{3}y/\xhat_{1}
\end{array}\!\!\right]
\end{eqnarray*}
which means that $\pi(\xhat,\,y)=p-2=1$ for all $\xhat$ and $y$.
The dynamics of the deadbeat observer then read
\begin{eqnarray*}
\xhat^{+}&=&f([\xhat]_{1}^{+}\cap h^{-1}(y),\,u)\\
&=&\left[\!\!\begin{array}{c}
\xhat_{2}\xhat_{3}y\\
\xhat_{3}/\xhat_{1}\\
\sqrt{\xhat_{1}\xhat_{2}u}
\end{array}\!\!\right]
\end{eqnarray*}

\section{An algorithm for deadbeat gain}\label{sec:alg}

In this section we provide an algorithm to compute the deadbeat
observer gain for a linear system with scalar output. (The
algorithm directly follows from Corollary~\ref{cor:foralg}.)
Namely, given an observable pair $(C,\,A)$ with
$C\in\Real^{1\times n}$ and $A\in\Real^{n\times n}$, we provide a
procedure to compute the gain $L\in\Real^{n\times 1}$ that renders
matrix $A-LC$ nilpotent. Below we let $\nal(\cdot)$ be some
function such that, given matrix $M\in\Real^{m\times n}$ whose
dimension of null space is $k$, $\nal(M)$ is some $n\times k$
matrix whose columns span the null space of $M$.

\begin{algorithm}\label{alg:db}
Given $C\in\Real^{1\times n}$ and $A\in\Real^{n\times n}$, the
following algorithm generates deadbeat gain $L\in\Real^{n\times
1}$.
\begin{eqnarray*}
&& X = \nal(C)\\
&& \mbox{{\bf for}}\quad i = 1:n-2\\
&& \qquad X = \nal\left(\left[
\begin{array}{c}
C\\ \nal((AX)^{T})^{T}
\end{array}
\right]\right)\\
&&\mbox{{\bf end}}\\
&&L_{\rm pre}=AX\\
&&L = \frac{AL_{\rm pre}}{CL_{\rm pre}}
\end{eqnarray*}
\end{algorithm}

For the interested reader we below give a M{\small ATLAB} code.
Exploiting Algorithm~\ref{alg:db}, this code generates a function
(which we named {\tt dbLfun}) whose inputs are matrices $C$ and
$A$. The output of the function, as its name indicates, is the
deadbeat gain $L$.
\begin{verbatim}
function L = dbLfun(C,A)
X = null(C);
for i = 1:length(A)-2
    X = null([C;null((A*X)')']);
end
Lpre = A*X;
L = A*Lpre/(C*Lpre);
\end{verbatim}
One can also use the built-in M{\small ATLAB} function {\tt acker}
to compute the deadbeat gain. We can therefore compare {\tt
dbLfun} with {\tt acker} via a numerical experiment.
Table~\ref{tab:one} gives the experimental results. Number $n$ is
the dimension of the system (that is, the number of columns of $A$
matrix) and the numbers at the bottom row are the percentages of
the cases (among $10^4$ random trials for each $n$) in which {\tt
dbLfun} performed better than {\tt acker}. How we determine which
one is better in a given case is as follows. Given pair $(C,\,A)$,
we let $L_{1}$ be the gain resulting from {\tt dbLfun(C,A)} and
$L_{2}$ be the gain given by {\tt acker(A',C',zeros(n,1))'}. Then
we compare norms $|(A-L_{1}C)^n|$ and $|(A-L_{2}C)^n|$, neither of
which is zero due to round-off errors. The function yielding the
smaller norm is considered to be better.

\begin{table}\caption{Percentages of cases where {\tt dbLfun} performed better than {\tt acker}.}
\begin{center}\label{tab:one}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
n=3 & n=4 & n=5 & n=6 & n=7 & n=8 & n=9 & n=10\\
\hline \%51 & \%60 & \%67 & \%74 & \%80 & \%85 & \%87 & \%91\\
\hline
\end{tabular}
\end{center}
\end{table}

\section{Conclusion}

For nonlinear systems a method to construct a deadbeat observer is
proposed. The resultant observer can be considered as a
generalization of the linear deadbeat observer. The construction
makes use of sets that are generated iteratively. Through such
iterations, observers are derived for two academic examples. Also,
for computing the deadbeat gain for a linear system with scalar
output, an algorithm that works no worse than an already existing
one is given.

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

config: arxiv


TITLE: Do I have to transform the solution into $u(x, y)$?
QUESTION [0 upvotes]: Find the solution of the problem $$u_{xx}(x,y)+u_{yy}(x,y)=0, x^2+y^2>1 \\u=1+3\sin^3 \theta , 0 \leq \theta <2\pi$$ $u$ is bounded.
I have done the following:
$$u(x,y)=v(\rho, \theta) \\ x=\rho \cos \theta \\ y=\rho \sin \theta$$
$$\Rightarrow v_{\rho \rho}(\rho , \theta)+\frac{1}{\rho}v_{\rho}(\rho, \theta)+\frac{1}{\rho^2}v_{\theta \theta}(\rho, \theta)=0, \rho>1 , 0\leq \theta <2\pi \\ v(1, \theta )=1+3 \sin^3 \theta$$
We are looking for the solution of $$v_{\rho \rho}+\frac{1}{\rho}v_{\rho}+\frac{1}{\rho^2}v_{\theta \theta}=0$$ of th form $$v(\rho, \theta )=R(\rho)\Theta(\theta)$$
We have the following two problems: $$\left.\begin{matrix}
\Theta ''(\theta) +\lambda\Theta(\theta)=0, 0 \leq \theta <2 \pi\\
\Theta(0)=\Theta(2 \pi)\\
\Theta '(0)=\Theta '(2 \pi)
\end{matrix}\right\}(1)$$ $$\left.\begin{matrix}
\rho^2 R''(\rho)+\rho R'(\rho)-\lambda R(\rho)=0
\end{matrix}\right\}(2)$$
Solving the problem $(1)$ we get that the eigenvalues are $\lambda_0=0, \lambda_k=k^2, k \in \mathbb{N}$ and the corresponding eigenfunctions are $\Theta_0 (\theta)=1, \Theta_k (\theta)=\cos (k
\theta), \Theta_k (\theta) =\sin (k \theta)$.
For the problem $(2)$ we have the following:
For $\lambda_0=0$ we have $R(\rho)=c_1 \ln \rho +c_2 , \rho>1$. So that the solution is bounded we choose $c_1=0$. So, we have $R_0(\rho)=1$.
For $\lambda_k=k^2$ we have $\rho^2 R''(\rho )+\rho R'(\rho )-k^2 R(\rho )=0$ (Euler).
$R(\rho )=\rho^m$
$\Rightarrow \rho^2 m (m-1) \rho^{m-2}+\rho m \rho^m-k^2 \rho^m=0 \Rightarrow \dots \Rightarrow m^2-k^2=0 \Rightarrow m=\pm k$
Genereal solution: $R_k(\rho )=c_1 \rho^k+c_2 \rho^{-k}, k \in \mathbb{N}, \rho >1$.
So that the soution is bounded we choose $c_1=0$. S, we have $R_k(\rho )=\rho^{-k}$.
So, we have the following $$v(\rho , \theta )=\frac{a_0}{2}+\sum_{k=1}^{\infty}\rho^{-k}(a_k \cos (k \theta)+b_n \sin (k \theta) \\ v(1, \theta )=1+3\sin^3 \theta$$
$$1+3\sin^3 \theta=\frac{a_0}{2}+\sum_{k=1}^{\infty}(a_k \cos (k \theta)+b_n \sin (k \theta))$$
$$a_k=\frac{1}{\pi} \int_0^{2 \pi}(1+3 \sin^3 \theta)\cos (k \theta)d \theta , k=0,1,2 \dots \\ b_k=\frac{1}{\pi}\int_0^{2\pi}(1+3 \sin^3 \theta)\sin (k \theta)d \theta , k=1,2, \dots $$
After calculations I found that $$v(\rho, \theta)=1+\frac{9}{4 \rho}\sin \theta-\frac{3}{4 \rho^3}\sin ( 3\theta)$$
Is this correct??
Do I have to transform it into $u(x, y)$ ?? How could I do that??

REPLY [0 votes]: Yes, your solution is correct.
You do not have to (unless someone asks for it) transform it back to $u(x,y)$, but here is how it is done:
Since $x=\rho\cos\theta$ and $y=\rho\sin \theta$ (and $\rho=\sqrt{x^2+y^2}$), we find that
$$
\sin \theta=y/\sqrt{x^2+y^2}.
$$
Moreover, $\sin 3\theta=3\cos^2\theta\sin\theta-\sin^3\theta$, so
$$
\sin 3\theta=\cdots=\frac{3x^2y-y^3}{(x^2+y^2)^{3/2}}.
$$
I'm sure you can now put the pieces together, to write your solution $u(x,y)$.


TITLE: Are diffeomorphisms a proper subgroup of conformal transformations?
QUESTION [12 upvotes]: The title sums it pretty much. Are all diffeomorphism transformations also conformal transformations?
If the answer is that they are not, what are called the set of diffeomorphisms that are not conformal?
General Relativity is invariant under diffeomorphisms, but it certainly is not invariant under conformal transformations, if conformal transformations where a subgroup of diff, you would have a contradiction. Or I am overlooking something important?

REPLY [4 votes]: I know topic this topic is 6 years old, but having just came across it, I feel the need to clarify some confusion around it, starting from the question, passing through the accepted answer and the ensuing comments.
Most of this confusion sparks from two different but similar notions, that of a conformal transformatrion and that of a conformal isometry.
Let me clarify (I'm following Wald's General Relativity book, appendix D on conformal transformations):
In the context of General Relativity (to which the OP is referring) a conformal transformation of parameter $\Omega:M\to \mathbb{R}$  ( $\Omega(p)>0\forall p$ ) (by other authors called a Weyl transformation) of a spacetime $(M,g)$ gives another spacetime $(M,g^\prime)$ where the metric $g^\prime$ is related to $g$ by $$g^\prime=\Omega^2 g$$ and theories whose predictions are invariant under this operation are said to be conformally invariant, or to posses Weyl symmetry.
According to this definition, a conformal transformation has nothing to do with the group of diffeomorphisms $\phi:M\to M$, in fact whatever atlas (read set of coordinates systems) we use to cover $M$, this is not changed by the conformal transformation (please recall that a manifold $M$ is essentially defined to be topological space with an atlas covering it, while the metric of a spacetime is an additional structure imposed a manifold: changing the metric through a conformal transformation don't do anything to the manifold).
So if the OP is referring to a conformal/Weyl transformation, the answare to its question is: the group of conformal transformations of the metric and the group of diffeomorphisms are two different groups, and neither of the two is a subgroup of the other.
In particular notice that a given tensor $T$ transform under a conformal transformation of the metric only through its functional dependence on the metric: if $T=t(g)$ in the spacetime $(M,g)$, then in the spacetime $(M,g^\prime)$ we have $$T^\prime = t(g^\prime) = t(\Omega^2 g)$$ and the actual relation between $T$ and $T^\prime$ depends on the actual functional form of $t(g)$, rather than on the tensor structure of $T$ (as it would be the case if a conformal transformation was a particular diffeomorphism).
Different is the story of conformal isometries: a conformal isometry is indeed a special kind of diffeomorphism, one for which the the pullback of the metric at a point $p$ is given by $$\phi_\star g_{\phi(p)} = \Omega^2(p) g_p $$Being a conformal isometry a diffeomorphisms, it transforms all tensors according to their tensor structure.
Relating to the OP question, the group of conformal isometries is a subgroup of the group of diffeomorphisms of M.
Regarding the last point in the OP question: General Relativity is indeed not "invariant under conformal transformations", but actually it is not even "invariant under diffeomorphisms" and neither it is "invariant under any isometries". Rather, its equations are tensorial in nature (they are in the form $T=0$ for some tensor $T$), therefor their validity is preserved under the action of all diffeomorphisms, including all isometries and all conformal isometries: $$\phi_\star T=0 \iff T=0$$
Now, according to the principle of general covariance, and if one disregard fermions, the metric $g$ is the only geometrical quantity that can enter the equations of general relativity, but this does not means that each of these equations is invariant under isometries: this is not true even of the Einstein field equation $$G=8\pi T$$ because the Einstein tensor $G$ and the stress energy tensor $T$ are rank-(0,2) tensors.
The last point I want to cover, is the confusion that emerged in the comments regarding the relation between diffeomorphisms and changes of coordinates.
First, a diffeomorphisms $\phi:M\to M$ maps points of $M$ to other points of $M$ (active view point), and is defined to be an homeomorphism preserving the smooth structure of $M$. As such it sends charts and atlases of $M$ to other charts and atlases of $M$ in a smooth and invertible way (if $\varphi : U\subset M \to \mathbb{R}^k$ is a chart, $\phi|_U \circ \varphi$ in another one, etc), thus inducing a change of coordinates (passive view). On the other direction, a smooth change of coordinates is precisely one that sends an atlas covering $M$ to another one in a smooth and invertible manner, thus inducing a smooth and invertible map between points of $M$. So actually diffeomorphisms and coordinate changes are in 1-to-1 correspondence, provided that coordinate changes "do not forget to cover some points of $M$", that is they map atlases to atlases, not merely charts to charts. If a given coordinate change instead "forgets some points", this change of coordinate will only induce a local diffeomorphism.


TITLE: Converting map coordinates of a rotated grid.
QUESTION [1 upvotes]: I'm trying to makes a stellar map, but I'm using a coordinate system different from the standard. I have coordinates in the normal system, but I need them in mine. This equates to rotating the carteesian grid 45 degrees.
Worst case scenario I can just brute force this, but I'd rather just be able to convert with an equation so I'm looking for how to convert a position's x,y coordinate from one grid to another grid when the difference between the two grids are one is spun clockwise so that centerlines are at a 45 degree angle to each other.
side note: Because I don't know if they original coordinates are south or north oriented, the rotation may actually be rotated 225 degrees rather than just 45.

REPLY [1 votes]: Rotation matrices is basicaly what you want.
For rotating a vector
$\begin{pmatrix} x \\ y \end{pmatrix} $
by the angle $\alpha$, just multiply it with the matrix
$\begin{pmatrix} \cos(\alpha) &-\sin(\alpha) \\ \sin(\alpha) &\cos(\alpha) \end{pmatrix}$.
\begin{align}
\begin{pmatrix} x_{new} \\ y_{new} \end{pmatrix} &= \begin{pmatrix} \cos(\alpha) &-\sin(\alpha) \\ \sin(\alpha) &\cos(\alpha) \end{pmatrix} \cdot \begin{pmatrix} x_{old} \\ y_{old} \end{pmatrix} \\
&= \begin{pmatrix} x_{old} \cdot \cos(\alpha) - y_{old} \cdot \sin(\alpha) \\ x_{old} \cdot \sin(\alpha) + y_{old} \cdot \cos(\alpha) \end{pmatrix}
\end{align}
In case you are not familiar with vectors, this means:
\begin{align}
x_{new} &= x_{old} \cdot \cos(\alpha) - y_{old} \cdot \sin(\alpha) \\
y_{new} &= x_{old} \cdot \sin(\alpha) + y_{old} \cdot \cos(\alpha)
\end{align}


TITLE: Supposing $x,y \in \mathbb{Q}^+$, does there exists $n \in \mathbb{Z}^+$ such that $x^n>y$ whenever $x>1$, or $x^n<y$ whenever $0<x<1$?
QUESTION [0 upvotes]: If $x,y$ are rationals and both positive numbers, how can I show:

if $x>1$ there is a positive interger $n$ such that $x^n>y$,
if $0 < x < 1$ there is a positive interger $n$ such that $x^n<y$.

I'd particularly like a proof that uses the binomial theorem.

REPLY [2 votes]: Write $x=(1+t)$, where $t>0$. Then
$$x^n=(1+t)^n=\sum_k \binom nkt^k = 1+nt+\binom n2t^2+\dots \ge 1+nt$$
(so far this is called the Bernoulli inequation). In order that this be $>y$, try to find a corresponding $n\in\Bbb N$.
The easiest if we reduce it to 1. by considering $1/x$ which is $>1$ in this case, and $1/y$ in place of ($y$ in 1).


TITLE: Linear map with Distinct eigenvalues
QUESTION [0 upvotes]: Let $A: \Bbb{R}^2 \rightarrow \Bbb{R}^2$ be a linear transformation with eigenvalues $\frac{2}{3}$ and $\frac{9}{5}$.

Then how to prove there exist a non- zero vector $v \in \Bbb{R}^2$ such that $\vert \vert Av \vert \vert=\vert \vert v \vert \vert$ ?
I know A has distinct eigenvalues, so diagonalizable. How to proceed further?

REPLY [0 votes]: Let $v:=\lambda e_{2/3}+\mu e_{9/5}$ where $e_a$ denotes the Eigenvectors, which are orthogonal.
Then enforce
$$\|v\|^2=\lambda^2+\mu^2=1$$ and $$\|Av\|^2=\frac49\lambda^2+\frac{81}{25}\mu^2=1.$$
This system has positive solutions in $\lambda^2,\mu^2$.


TITLE: Can you simplify dimensional analysis questions before plugging in values?
QUESTION [1 upvotes]: If I were to solve the equation:
$$d=\frac{a^3}{cb^2}$$
and values were given as so: $a=9.7\ \mathrm m$, $b=4.2\ \mathrm s$, and $c=69\ \mathrm{m/s}$.
Why would it be wrong for me to use dimensional analysis first?
Using dimensional analysis you would simplify: $\mathrm{m^3/((m/s)(s^2))}$ to $\mathrm{m^2/s}$.
That is right as far as I know, but when I plug in the values I get $d=.07\ \mathrm{m^2/s}$, that is I plug it in as $(9.7\ \mathrm m^2)/(69\ 4.2\ \mathrm s)$ since I already did dimensional analysis.
However, when I plug in the values as $((9.7^3)(\mathrm m^3))/((69\mathrm m/\mathrm s)((4.2^2)(\mathrm s^2)$ I get the right answer of $d=.75\ \mathrm{m^2/s}$.
I know my dimensional analysis is correct but why is it that I cannot simplify the original equation before plugging in the values (as in I have to plug it in as $9.7^3\ \mathrm m^3$ instead of $9.7\ \mathrm m^2$)?
So essentially in this question I should think of it as for example if I take the equation $5x^2$, then $5x^2$ is equal to $25x^2$? Is that how I should think of it?

REPLY [0 votes]: If you treat units like algebraic, multiplicative factors in every sense, you cannot go wrong. All rules of algebra apply. Realizing this makes matters more simple in every sense.
Just treat units as you would numbers.


text:
\appendix
\section{Butcher tableau for ARS($2,2,2$) and ARS($4,4,3$)}
\label{sec:appendix}
The double Butcher tableau of ARS($2,2,2$) and ARS($4,4,3$) schemes \cite{ascher1997implicit} are respectively
\begin{displaymath}
\begin{array}{c|c c c}
0 & 0&0&0\\
\gamma & \gamma&0&0\\
1 & \delta &1-\delta&0\\
\hline
& \delta &1-\delta&0\\
 \end{array} \ \ \ \ \
\begin{array}{c|c c c}
0 & 0&0&0\\
\gamma & 0&\gamma&0\\
1 &0 &1-\gamma&\gamma\\
\hline
&0 &1-\gamma&\gamma\\
 \end{array}
\quad
\textrm{with}\; \gamma = 1-\frac{1}{\sqrt{2}}, \; \delta = 1-\frac{1}{2\gamma},
\end{displaymath}
and
\begin{displaymath}
\begin{array}{c|c c c c c}
0 & 0&0&0& 0 & 0\\
1/2 &1/2&0&0& 0 &0\\
2/3 &11/18&1/18&0& 0 &0\\
1/2 &5/6&-5/6&1/2& 0 &0\\
1 &1/4&7/4&3/4& -7/4 &0\\
\hline
&1/4&7/4&3/4& -7/4 &0\\
 \end{array} \ \ \ \ \
\begin{array}{c|c c c c c}
0 & 0&0&0& 0 & 0\\
1/2 &0&1/2&0&0 &0\\
2/3 &0&1/6&1/2& 0 &0\\
1/2 &0&-1/2&1/2& 1/2 &0\\
1 &0&3/2&-3/2& 1/2 &1/2\\
\hline
 &0&3/2&-3/2& 1/2 &1/2
 \end{array}~~.
\end{displaymath}