TITLE: On the prefactor in the path integral formulation
QUESTION [2 upvotes]: The propagator $K$ from ($x_a,t_a$) to ($x_b,t_b$), as defined by Gottfried, can be written as
$$ K(b,a) = F(t_b-t_a)\exp\left(\frac{i}{\hbar}S_{c}(b,a)\right) $$
where $S_c$ is the classical action and $F(t_b-t_a)$ is the integral over all paths from the origin and back, during the interval $t_b-t_a$, and is known as the prefactor in some literatures.
I've noticed that the prefactor for both the "regular" harmonic oscillator and the driven oscillator (with any arbitrary forcing $f(t)$) is the same,
$$ F(t_b-t_a) = \sqrt{\frac{m\omega}{2\pi i\hbar \sin\omega(t_b-t_a)}} .$$
Is there any physical or mathematical reason for this? How can I justify this must be the case for the driven oscillator, without going through a 10 page calculation using Feynman's trick of exploiting the composition law and the fact that $F$ is the propagator from the origin to the origin?

REPLY [3 votes]: There is certainly a mathematical reason:

The external force $f$ appears in the term linear in the position variable $x$ of the action $S[x]$.
When we split the path integral variable $x~=~x_{\rm cl}+ y$ in a classical path $x_{\rm cl}$ plus quantum fluctuation $y$, we know that the action $S[x]~=~S[x_{\rm cl}]+ S_q[y]$ has no term linear in the fluctuations (because the Taylor coefficient $\frac{\delta S}{\delta x}|_{x=x_{\rm cl}}~=~0$ of the linear term is the classical EOM).
The quantum action $S_q[y]$ does not depend on the classical path $x_{\rm cl}$ since the action $S[x]$ for the harmonic oscillator has at most quadratic terms in $x$. (This is important since the classical path $x_{\rm cl}$ depends implicitly on the external force $f$.)
Hence the quantum action $S_q[y]$ and the corresponding path integral
$$ F(t_b-t_a) = \int_0^0 {\cal D}y~e^{\frac{i}{\hbar}S_q[y]} $$
cannot depend on the external force $f$.

References:

R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals, 1965; Problem 3.11.


TITLE: Problem-solving with ratios and similar triangles.
QUESTION [1 upvotes]: The following question is designed to test problem-solving and reasoning skills with ratios and similar triangles. Without using Pythagoras Theorem or Trigonometry this question is supposed to be solvable. But how?
I'm trying to figure out if there is an error in this question. Can this be answered using ratio and scale factor only?
A string 50m long is pegged to the ground and tied to the top of a flag pole. It just touches the head of Maureen, who is 5 meters away from the point where the string is held to the ground. If Maureen is 1.5 meters tall, show that the height, h, of the flagpole is 14.37m

REPLY [0 votes]: Here is my take on solving the problem without an appeal to Pythagoras.
Referring to the first figure below, it's obvious from the similarity of triangles that ${H}/{h} = (5+x)/{5}, \text{that is, }$
$$
H=h+ \frac{xh}{5} \tag{1}
$$
From the figure of the square formed with four triangles congruent to the triangle in the problem, the following is obvious:
$$
50^2 = 2H(5+x) + (5+x-H)^2
$$
Substituting (1) in the above equation we get a quadratic in $x$:
$$
1.09x^2 + 10.9x - 2471.75 = 0.
$$
The positive solution is $x=42.882$, from which the result follows.
This proof refers to the well known arrangement of right triangles used to prove Pythagoras theorem, however, the theorem is not referenced in any way.


TITLE: Why isn't $H^*(\mathbb{R}P^\infty,\mathbb{F}_2)\cong \mathbb{F}_2[[x]]$?
QUESTION [13 upvotes]: We just computed in class a few days ago that $$H^*(\mathbb{R}P^n,\mathbb{F}_2)\cong\mathbb{F}_2[x]/(x^{n+1}),$$ and it was mentioned that $H^*(\mathbb{R}P^\infty,\mathbb{F}_2)\cong \mathbb{F}_2[x]$, but I  (optimistically?) assumed that the cohomology ring functor would turn limits into colimits, and so $$\mathbb{R}P^\infty=\lim\limits_{\longrightarrow}\;\mathbb{R}P^n$$ would mean that we'd get the formal power series ring $$\lim\limits_{\longleftarrow}\;\mathbb{F}_2[x]/(x^{n+1})=\mathbb{F}_2[[x]].$$
My professor said that the reason we get $\mathbb{F}_2[x]$ is just that the cohomology ring, being the direct sum of the cohomology groups, can't have non-zero elements in every degree, which certainly makes sense.
Even though that should settle the matter, for some reason, I'm still having a bit of trouble making this make click for me. Is the explanation just that the cohomology ring functor doesn't act as nicely as I'd hoped? Is there an intuitive explanation of what's going on?

REPLY [1 votes]: This is probably something that should be a comment, mainly because I'm not sure it is correct. Feel free to tell me it is rubbish!
1) I have definitely seen it written $H^*(\mathbb{R} P^\infty; \mathbb{F}_2) \simeq \mathbb{F}_2[\![ x ]\!]$. For example see Jacob Lurie's notes here.
2) Cohomlogy doesn't play well with inverse limits in general. There is the Milnor exact sequence, which is this case should give:
$$0 \to \text{lim}^1 H^{\ast-1}(\mathbb{R} P^\infty;\mathbb{F}_2) \to H^*(\mathbb{R} P^\infty;\mathbb{F}_2) \to \underset{n}{\text{lim}} H^*(\mathbb{R}P^\infty;\mathbb{F}_2) \to 0$$
where $\text{lim}^1$ is the first dervied functor of the inverse limit.
In this case the $\text{lim}^1$ terms should all be zero by the Mittag-Leffler criteria (which follows since the maps $H^*(\mathbb{R} P^{n+1};\mathbb{F}_2)\to H^*(\mathbb{R} P^{n};\mathbb{F}_2)$ are surjective). Thus we can conclude that $$H^*(\mathbb{R} P^\infty; \mathbb{F}_2) \simeq \mathbb{F}_2[\![ x ]\!]$$
As an aside this is the calculation usually given to show $E^*(\mathbb{C} P^\infty) \simeq E_*[\![ x ]\!]$ for any complex oriented cohomology theory $E$, which basically sets up the whole relationship between complex oriented cohomology theories and formal group laws.
Edit: Please see the comments by Mariano below.
Possibly also this Math Overflow link is relevant


TITLE: If Z has a normal distribution with mean 0 and variance $\sigma^{2}$, and $Y=Z^{2}$, what would the density function of Y be?
QUESTION [0 upvotes]: How would I go about finding this density function?
Thanks

REPLY [2 votes]: The cumulative distribution function for $Y$ is:
$$
F_Y(y)=\operatorname{Prob}(Y<y)=\operatorname{Prob}(-\sqrt{y}<z<\sqrt{y})=F_Z(\sqrt{y})-F_Z(-\sqrt{y})=2F_Z(\sqrt{y})-1
$$
Now to get the density differentiate with respect to $y$, and of course to differentiate $2F_Z(\sqrt{y})$ you use the chain rule and the fact that the derivative of $F_Z(x)$ with respect to $x$ is $f_Z(x)$ the density of $Z$, which you know as $Z\sim N(0,1)$.


text:
\begin{document}

\title{Linear Algebra in the vector space of intervals $\overline{\mathbb{IR}
}$}
\author{Nicolas GOZE }
\date{}
\maketitle

\begin{abstract}
In a previous paper, we have given an algebraic model to the set of intervals. Here, we apply this model in a linear frame.
We define a notion of diagonalization of square matrices whose coefficients are intervals. But in this case, with respect to the real case,
a matrix of order $n$ could have more than $n$ eigenvalues (the set of intervals is not factorial). We consider a notion of
central eigenvalues permits to describe criterium of diagonalization. As application, we define a notion of Exponential mapping.
\end{abstract}
\section{The associative algebra $\overline{\mathbb{IR}}$}

In \cite{GN-R}, we have given a representation of the set of intervals in
terms of associative algebra. More precisely, we define on the set $\mathbb{
IR}$ of intervals of $\mathbb{R}$ a $\mathbb{R}$-vector space structure.
Next we embed $\mathbb{IR}$ in a $4$-dimensional associative algebra. This
embedding permits to describe a unique distributive multiplication which
contains all the possible results of the usual product of intervals and the
monotony property is always conserved. Moreover, this new product is minimal with
respect the distributivity and the monotony properties.

\noindent In this section, we present briefly this construction (for more
details, see \cite{GN-R}).

\subsection{Vector space structure on $\mathbb{IR}$}

Let $\mathbb{IR}$ be the set of intervals of $\mathbb{R}$, that is
\begin{equation*}
\mathbb{IR=\{[}a,b],a,b\in \mathbb{R}\text{, }a\leq b\}.
\end{equation*}
This set is provided with a semi-group structure that we can complete as
follow: we consider the equivalence relation on $\mathbb{IR}\times \mathbb{IR
}$:
\begin{equation*}
(X,Y)\sim (Z,T)\Longleftrightarrow X+T=Y+Z
\end{equation*}
for all $X,Y,Z,T\in \mathbb{IR}$. The quotient set is denoted by $\overline{
\mathbb{IR}}$. The addition of intervals is compatible with this equivalence
relation :
\begin{equation*}
\overline{(X,Y)}+\overline{(z,t)}=\overline{(x+z,y+t)}
\end{equation*}
where $\overline{(x,y)}$ is the equivalence class of $(X,Y).$ The unit is $
\overline{0}=\{(X,X),X\in \mathbb{IR\}}$ and each element has an inverse
\begin{equation*}
\smallsetminus \overline{(X,Y)}=\overline{(Y,X)}.
\end{equation*}
Then $(\overline{\mathbb{IR}},+)$ is a commutative group. We prove also in
\cite{GN-R}, that any equivalence class admits a canonical representant of
type $(K,0)$ or $(0,K)$ or $(a,a)$, with $K\in \mathbb{IR}$ and $a=[a,a],\
a\in \mathbb{R}$. We provides the group $(\overline{\mathbb{IR}},+)$ with a
real vector space structure, the external product being given by
\begin{equation*}
\left\{
\begin{array}{c}
\alpha \cdot \overline{(K,0)}\text{\ }=\overline{(\alpha K,0)}\text{ } \\
\alpha \cdot \overline{(0,K)}\text{\ }=\overline{(0,\alpha K)}\text{ }
\end{array}
\right.
\end{equation*}
if $\alpha >0.$ If $\alpha <0$ we put $\beta =-\alpha $ and
\begin{equation*}
\left\{
\begin{array}{c}
\alpha \cdot \overline{(K,0)}\text{\ }=\overline{(0,\beta K)}\text{ } \\
\alpha \cdot \overline{(0,K)}\text{\ }=\overline{(\beta K,0)}\text{ }
\end{array}
\right. .
\end{equation*}
The triplet $(\overline{\mathbb{IR}},+,\cdot )$ is a real vector space.

\medskip

\noindent\textbf{Remark.} To simplify notations, we write $(K,0)$ or $(0,K)$
in place of $\overline{(K,0)}$ or $\overline{(0,K)}$.

\subsection{The associative algebra $\mathcal{A}_4$}

Recall that by an algebra we mean a real vector space with an associative
ring structure. Consider the $4$-dimensional associative algebra whose
product in a basis $\{e_{1},e_{2},e_{3},e_{4}\}$ is given by
\begin{equation*}
\begin{tabular}{|l|l|l|l|l|}
\hline
& $e_{1}$ & $e_{2}$ & $e_{3}$ & $e_{4}$ \\ \hline
$e_{1}$ & $e_{1}$ & $0$ & $0$ & $e_{4}$ \\ \hline
$e_{2}$ & $0$ & $e_{2}$ & $e_{3}$ & $0$ \\ \hline
$e_{3}$ & $0$ & $e_{3}$ & $e_{2}$ & $0$ \\ \hline
$e_{4}$ & $e_{4}$ & $0$ & $0$ & $e_{1}$ \\ \hline
\end{tabular}
.
\end{equation*}
The unit is the vector $e_{1}+e_{2}.$ This algebra is a direct sum of two
ideals: $\mathcal{A}_{4}=I_{1}+I_{2}$ where $I_{1}$ is generated by $e_{1}$
and $e_{4}$ and $I_{2}$ is generated by $e_{2}$ and $e_{3}.$ It is not an
integral domain, that is, we have divisors of $0.$ For example $e_{1}\cdot
e_{2}=0.$ The ring $\mathcal{A}_{4}$ is principal that is every ideal is
generated by one element. The cartesian expression of this product is, for $
x=(x_{1},x_{2},x_{3},x_{4})$ and $y=(y_{1},y_{2},y_{3},y_{4})$ in $\mathcal{A
}_{4}$:
\begin{equation*}
x\cdot
y=(x_{1}y_{1}+x_{4}y_{4},x_{2}y_{2}+x_{3}y_{3},x_{3}y_{2}+x_{2}y_{3},x_{4}y_{1}+x_{1}y_{4}).
\end{equation*}
The multiplicative group $\mathcal{A}_{4}^{\ast }$ \ of invertible elements
is the set of elements $x=(x_{1},x_{2},x_{3},x_{4})$ such that
\begin{equation*}
\left\{
\begin{array}{c}
x_{4}\neq \pm x_{1}, \\
x_{3}\neq \pm x_{2}.
\end{array}
\right.
\end{equation*}
If $x\in $ $\mathcal{A}_{4}^{\ast }$ we have:
\begin{equation*}
x^{-1}=\left( \frac{x_{1}}{x_{1}^{2}-x_{4}^{2}},\frac{x_{2}}{
x_{2}^{2}-x_{3}^{2}},\frac{x_{3}}{x_{2}^{2}-x_{3}^{2}},\frac{x_{4}}{
x_{1}^{2}-x_{4}^{2}}\right) .
\end{equation*}

\subsection{A product in $\overline{\mathbb{IR}}$}

We define a correspondence between $\mathbb{IR}$ and $\mathcal{A}$. Let $
\varphi$ be the map
\begin{equation*}
\varphi : \overline{\mathbb{IR}} \longrightarrow \mathcal{A}_{4}
\end{equation*}
given by
\begin{equation*}
\varphi (K,0)=\left\{
\begin{array}{l}
(x_{1},x_{2},0,0)\text{ if }x_{1},x_{2}\geq 0, \\
(0,x_{2},-x_{1},0)\text{ if }x_{1}\leq 0\text{ and }x_{2}\geq 0, \\
(0,0,-x_{1},-x_{2})\text{ if }x_{1},x_{2}\leq 0,
\end{array}
\right.
\end{equation*}
and
\begin{equation*}
\overline{\varphi }(0,K)=-\overline{\varphi }(K,0).
\end{equation*}
for any $K=[x_{1},x_{2}]$. Then we have
\begin{equation*}
\begin{tabular}{lll}
$\overline{\varphi }(0,K)$ & $=$ & $\left\{
\begin{array}{l}
-x_{1},-x_{2},0,0)\text{ if }x_{1}\geq 0, \\
(0,-x_{2},-x_{1},0)\text{ if }x_{1}x_{2}\leq 0, \\
(0,0,-x_{1},-x_{2})\text{ if }x_{2}\leq 0.
\end{array}
\right. $
\end{tabular}
\end{equation*}
Thus the image of $\overline{\mathbb{IR}}$ in $\mathcal{A}_{4}$ is
constituted of the elements
\begin{equation*}
\left\{
\begin{array}{l}
(x_{1},x_{2},0,0)\text{ with }0\leq x_{1}\leq x_{2}\text{ which corresponds
to }([x_{1},x_{2}],0), \\
(0,x_{2},-x_{1},0)\text{ with }x_{1}\leq 0\leq x_{2}\text{ which corresponds
to }([x_{1},x_{2}],0), \\
(0,0,-x_{1},-x_{2})\text{ with }x_{1}\leq x_{2}\leq 0\text{ which
corresponds to }([x_{1},x_{2}],0), \\
(-x_{1},-x_{2},0,0)\text{with }0\leq x_{1}\leq x_{2}\text{ which corresponds
to }(0,[x_{1},x_{2}]), \\
(0,-x_{2},x_{1},0)\text{ with }x_{1}\leq 0\leq x_{2}\text{ which corresponds
to }(0,[x_{1},x_{2}]), \\
(0,0,x_{1},x_{2})\text{ with }x_{1}\leq x_{2}\leq 0\text{ which corresponds
to }(0,[x_{1},x_{2}]).
\end{array}
\right.
\end{equation*}
The map $\overline{\varphi }:$ $\overline{\mathbb{IR}}\mathbb{
\longrightarrow }\mathcal{A}_{4}$ \ is not linear. We introduce in $\mathcal{
A}_{4}$ the following equivalence relation $\mathcal{R}$ given by
\begin{equation*}
(x_{1},x_{2},x_{3},x_{4})\sim (y_{1},y_{2},y_{3},y_{4})\Longleftrightarrow
\left\{
\begin{array}{c}
x_{1}-y_{1}=x_{3}-y_{3}, \\
x_{2}-y_{2}=x_{4}-y_{4}
\end{array}
\right.
\end{equation*}
and consider the map
\begin{equation*}
\overline{\overline{\varphi }}:\overline{\mathbb{IR}}\longrightarrow
\overline{\mathcal{A}_{4}}=\frac{\mathcal{A}_{4}}{\mathcal{R}}
\end{equation*}
given by $\overline{\overline{\varphi }}=\Pi \circ \overline{\varphi }$
where $\Pi $ is a canonical projection. This map is surjective. In fact we
have the correspondence

\begin{itemize}
\item $x_{1}-x_{3}\geq 0$, $x_{2}-x_{4}\geq 0$, $x_{1}-x_{3}\leq x_{2}-x_{4}$
\begin{equation*}
(x_{1},x_{2},x_{3},x_{4})\sim (x_{1}-x_{3},x_{2}-x_{4},0,0)=\overline{
\varphi }([x_{1}-x_{3},x_{2}-x_{4}],0).
\end{equation*}

\item $x_{1}-x_{3}\geq 0$, $x_{2}-x_{4}\geq 0$, $x_{1}-x_{3}\geq x_{2}-x_{4}$
\begin{equation*}
(x_{1},x_{2},x_{3},x_{4})\sim (0,0,x_{3}-x_{1},x_{4}-x_{2})=\overline{
\varphi }(0,[x_{3}-x_{1},x_{4}-x_{2}]).
\end{equation*}

\item $x_{1}-x_{3}\geq 0$, $x_{2}-x_{4}\leq 0$
\begin{equation*}
(x_{1},x_{2},x_{3},x_{4})\sim (0,x_{2}-x_{4},x_{3}-x_{1},,0)=\overline{
\varphi }(0,[x_{3}-x_{1},x_{4}-x_{2}]).
\end{equation*}

\item $x_{1}-x_{3}\leq 0$, $x_{2}-x_{4}\geq 0$
\begin{equation*}
(x_{1},x_{2},x_{3},x_{4})\sim (0,x_{2}-x_{4},x_{3}-x_{1},,0)=\overline{
\varphi }([x_{3}-x_{1},x_{2}-x_{4}],0).
\end{equation*}

\item $x_{1}-x_{3}\leq 0$, $x_{2}-x_{4}\leq 0$, $x_{1}-x_{3}\geq x_{2}-x_{4}$
\begin{equation*}
(x_{1},x_{2},x_{3},x_{4})\sim (x_{1}-x_{3},x_{2}-x_{4},0,0)=\overline{
\varphi }(0,[x_{3}-x_{1},x_{4}-x_{2}]).
\end{equation*}

\item $x_{1}-x_{3}\leq 0$, $x_{2}-x_{4}\leq 0$, $x_{1}-x_{3}\leq
x_{2}-x_{4}(x_{1},x_{2},x_{3},x_{4})\sim (0,0,x_{3}-x_{1},x_{4}-x_{2})=
\overline{\varphi }([x_{3}-x_{1},x_{2}-x_{4}],0).$
\end{itemize}

\noindent This correspondence defines a map
\begin{equation*}
\psi :\overline{\mathcal{A}_{4}}\longrightarrow \overline{\mathbb{IR}}.
\end{equation*}
In the following, to simplify notation, we write $\overline{\varphi }$
instead of $\overline{\overline{\varphi }}.$

\begin{definition}
For any $\mathcal{X},\mathcal{X}^{\prime }\in \overline{\mathbb{IR}}$, we
put
\begin{equation*}
\mathcal{X}\bullet \mathcal{X}^{\prime }=\psi (\overline{\overline{\varphi }(
\mathcal{X})}\bullet \overline{\overline{\varphi }(\mathcal{X}^{\prime })}).
\end{equation*}
\end{definition}

This multiplication is distributive with respect the the addition. In fact
\begin{equation*}
(\mathcal{X}_{1}+\mathcal{X}_{2})\bullet \mathcal{X}^{\prime }=\psi (
\overline{\overline{\varphi }(\mathcal{X}_{1}+\mathcal{X}_{2})}\bullet
\overline{\overline{\varphi }(\mathcal{X}^{\prime })}).
\end{equation*}
Suppose that $\overline{\varphi }(\mathcal{X}_{1}+\mathcal{X}_{2})\neq
\overline{\varphi }(\mathcal{X}_{1})+\overline{\varphi }(\mathcal{X}_{2}).$
In this case this means that $\overline{\varphi }(\mathcal{X}_{1})+\overline{
\varphi }(\mathcal{X}_{2})\notin \ Im \overline{\varphi }.$ But by
construction $\overline{\overline{\varphi }(\mathcal{X}_{1}+\mathcal{X}_{2})}
\in \ Im \overline{\varphi }$ and this coincides with $\overline{\varphi }(
\mathcal{X}_{1}+\mathcal{X}_{2}).$ For numerical application of this product, see (\cite[GN-R]).

\section{The module $gl(n,\overline{\mathbb{IR}})$}

Let $gl(n,\overline{\mathbb{IR}})$ be the set of square matrices of order $n$
whose elements are in $\overline{\mathbb{IR}}.$ A matrice of $gl(n,\overline{\mathbb{IR}})$ is  denoted by
\begin{equation*}
A=\left( \mathcal{X}_{ij}\right) _{i,j=1,\cdots ,n}
\end{equation*}
with $\mathcal{X}_{ij}=(K_{ij},0)$ or $(0,K_{ij}).$ It is clear that $gl(n,
\overline{\mathbb{IR}})$ is a real vector space. We define a product on it
puting
\begin{equation*}
A\cdot B=\left( \mathcal{X}_{ij}\right) \cdot \left( Y_{ij}\right) =\left(
Z_{ij}\right)
\end{equation*}
with $Z_{ij}=\sum\limits_{k=1}^{n}\mathcal{X}_{ik}\cdot Y_{ik}.$ This last
product being the associative product on $\overline{\mathbb{IR}}.$ Thus $
gl(n,\overline{\mathbb{IR}})$ is an associative algebra.

\begin{defi}
A matrice $A\in gl(n,\overline{\mathbb{IR}})$ is called inversible if its
determinant, computed by the Cramer rule, is an inversible element in $\overline{\mathbb{IR}}$.
\end{defi}

\noindent Recall that the group $\overline{\mathbb{IR}}$ of
inversible elements contain
\begin{equation*}
\mathcal{X}_{i}=(K_{i},0)\text{ or }(0,K_{i})
\end{equation*}
with $0\notin K_{i}.$ To compute the determinant, we use the classical
formula of Cramer.

\bigskip

\noindent \textbf{Example 1. }Let us consider the matrix
\begin{equation*}
M=\left(
\begin{array}{cc}
\lbrack 1,2] & [-1.3] \\
\lbrack -1,3] & [1,2]
\end{array}
\right)
\end{equation*}
Thus
\begin{eqnarray*}
\det B_1 &=&([1,2],0)([1,2],0)\smallsetminus ([-1,3],0)([-1,3],0) \\
&=&([1,4],0)\smallsetminus ([-3,9],0) \\
&=&([0,[-4,5]) \\
&=&\smallsetminus ([-4,5],0).
\end{eqnarray*}
As $([-4,5],0)$ is not an inversible element of $\overline{\mathbb{IR}},$
the matrix $B_1$ is not inversible.

\bigskip

\noindent \textbf{Example 2. }Now if
\begin{equation*}
B_2=\left(
\begin{array}{cc}
\lbrack 1,2] & [-1.3] \\
\lbrack -1,3] & [1,7]
\end{array}
\right)
\end{equation*}
then, by the similar computation, we obtain
\begin{equation*}
\det B_2=\smallsetminus ([-7,-4],0)
\end{equation*}
and $B_2$ is invertible.

\medskip

\begin{definition}
If $A$ is an invertible matrix on $gl(n,\overline{\mathbb{IR}})$, the
inverse matrix $A^{-1}$ of $A$ is given by
\begin{equation*}
A\cdot A^{-1}=Id
\end{equation*}
where
\begin{equation*}
Id=\left(
\begin{array}{cccc}
1 & 0 & \cdots & 0 \\
0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1
\end{array}
\right)
\end{equation*}
with $1=([1,1],0)$ and $0=([0,0],0).$
\end{definition}

The determination of $A^{-1}$ can be computed using the classical rules.

\medskip

\noindent \textbf{Example. }If we consider the invertible matrix $B_2$, we
obtain
\begin{equation*}
B_2^{-1}=[\frac{1}{7},\frac{1}{4}]\left(
\begin{array}{cc}
\lbrack 1,7] & \smallsetminus \lbrack -1.3] \\
\smallsetminus \lbrack -1,3] & [1,2]
\end{array}
\right) .
\end{equation*}
Let us verify that $B_2B_2^{-1}=Id$. Using the product on $\overline{\mathbb{
IR}} $ we obtain
\begin{equation*}
B_2B_2^{-1}=[\frac{1}{7},\frac{1}{4}]\left(
\begin{array}{cc}
\lbrack 1,2] & [-1.3] \\
\lbrack -1,3] & [1,7]
\end{array}
\right) \cdot \left(
\begin{array}{cc}
\lbrack 1,7] & \smallsetminus \lbrack -1.3] \\
\smallsetminus \lbrack -1,3] & [1,2]
\end{array}
\right) .
\end{equation*}
The coefficient in place $(1,1)$ is
\begin{equation*}
a_{11}=[\frac{1}{7},\frac{1}{4}]([1,2][1,7]+[-1.3](\smallsetminus \lbrack
-1,3])).
\end{equation*}
>From the definition of the product (see section 1), this element is
\begin{eqnarray*}
a_{11} &=&(\frac{1}{7},\frac{1}{4},0,0)((1,2,0,0)(1,7,0,0)-(0,3,1,0)(0,3,1,0)
\\
&=&(\frac{1}{7},\frac{1}{4},0,0)((1,14,0,0)-(0,10,6,0)) \\
&=&(\frac{1}{7},\frac{1}{4},0,0)(1,4,-6,0) \\
&=&(\frac{1}{7},\frac{1}{4},0,0)(7,4,0,0) \\
&=&(1,1,0,0)
\end{eqnarray*}
which corresponds to $[1,1]$. Similarly we have $a_{12}=a_{21}=(0,0,0,0)$
and $a_{22}=(1,1,0,0).$ Thus $B_2B_2^{-1}=Id.$

\section{Diagonalization}

\bigskip

\subsection{\protect\bigskip Eigenvalues and central eigenvalues}

Let $A$ be in $gl(n,\overline{\mathbb{IR}}).$ An eigenvalue of $A$ is an
element $\mathcal{X}\in \overline{\mathbb{IR}}$ such that there exists a
vector $\mathcal{V}\neq 0\in \overline{\mathbb{IR}}^{n}$ with
\begin{equation*}
A\cdot ^{t}\mathcal{V}=\mathcal{X}\cdot ^{t}\mathcal{V}.
\end{equation*}
Thus $\mathcal{X}$ is a root of the characteristical polynomial with
coefficients in the ring $\overline{\mathbb{IR}}$
\begin{equation*}
C_{A}(\mathcal{X})=\det (A-\mathcal{X}I)=0.
\end{equation*}
\textbf{Example.} Let
\begin{equation*}
B_3=\left(
\begin{array}{cc}
\lbrack 1,2] & [1,2] \\
\lbrack 1,3] & [2,5]
\end{array}
\right) .
\end{equation*}
We have
\begin{equation*}
B_3-\mathcal{X}I=\left(
\begin{array}{cc}
\lbrack 1,2]\smallsetminus \mathcal{X} & [1,2] \\
\lbrack 1,3] & [2,5]\smallsetminus \mathcal{X}
\end{array}
\right)
\end{equation*}
and
\begin{eqnarray*}
\det (B_3-\mathcal{X}I) &=&([1,2]\smallsetminus \mathcal{X}
)([2,5]\smallsetminus \mathcal{X})-[1,3][1,2] \\
&=&[2,10]-\mathcal{X}[2,5]-\mathcal{X}[1,2]+(\smallsetminus \mathcal{X}
)(\smallsetminus \mathcal{X})-[1,6] \\
&=&(\smallsetminus \mathcal{X})(\smallsetminus \mathcal{X})-\mathcal{X}
[3,7]+[1,4].
\end{eqnarray*}
Let $\mathcal{X}=([x,y],0).$ It is represented in $\mathcal{A}_{4}$ by $
(x,y,0,0)$ or $(0,y,x,0)$ or $(0,0,x,y)=-(x,y,0,0).$

\bigskip

\textbf{First case}: \ $\det (B_3-\mathcal{X}
I)=(x^{2},y^{2},0,0)-(3x,7y,0,0)+(1,4,0,0)=(x^{2}-3x+1,y^{2}-7y+4,0,0).$
Then $\det (B_3-\mathcal{X}I)=0$ implies
\begin{equation*}
\left\{
\begin{array}{c}
x^{2}-3x+1=0, \\
y^{2}-7y+4=0,
\end{array}
\right.
\end{equation*}
that is
\begin{equation*}
\left\{
\begin{array}{c}
x=\dfrac{3\pm \sqrt{5}}{2}, \\
y=\dfrac{7\pm \sqrt{33}}{2}.
\end{array}
\right.
\end{equation*}
We obtain
\begin{equation*}
\left\{
\begin{array}{c}
\mathcal{X}_{1}=([\dfrac{3+\sqrt{5}}{2},\dfrac{7+\sqrt{33}}{2}],0), \\
\mathcal{X}_{2}=([\dfrac{3-\sqrt{5}}{2},\dfrac{7+\sqrt{33}}{2}],0), \\
\mathcal{X}_{3}=([\dfrac{3-\sqrt{5}}{2},\dfrac{7-\sqrt{33}}{2}],0). \\
\end{array}
\right.
\end{equation*}

\textbf{Second case: } $\det (B_3-\mathcal{X}
I)=(0,y^{2}+x^{2},2xy,0)-(0,7y,7x,0)+(1,4,0,0)=(1,y^{2}+x^{2}-7y+4,2xy-7x,0).
$ Then $\det (B_3-\mathcal{X}I)=0$ implies
\begin{equation*}
\left\{
\begin{array}{c}
1-2xy+7x=0, \\
y^{2}+x^{2}-7y+4=0.
\end{array}
\right.
\end{equation*}
This gives
\begin{equation*}
4y^{4}-56y^{3}+261y^{2}-455y+197=0.
\end{equation*}
We have the following solutions
\begin{equation*}
(x;y)=\{(-2,8;3,32),(2,9;3,67),(-0,17;0,63),(0,17;6,37)\}.
\end{equation*}
We obtain the eigenvalues
\begin{equation*}
\left\{
\begin{array}{c}
\mathcal{X}_{4}=([-2,8,\ 3.32],0), \\
\mathcal{X}_{5}=([-0.17,\ 0.63],0). \\
\end{array}
\right.
\end{equation*}

\qquad

\textbf{Third} \textbf{\ case}: $\det (B_3-\mathcal{X}
I)=(x^{2},y^{2},0,0)+(3x,7y,0,0)+(1,4,0,0)=(x^{2}+3x+1,y^{2}+7y+4,0,0).$
Then $\det (B_3-\mathcal{X}I)=0$ implies
\begin{equation*}
\left\{
\begin{array}{c}
x=\dfrac{-3\pm \sqrt{5}}{2}, \\
y=\dfrac{-7\pm \sqrt{33}}{2},
\end{array}
\right.
\end{equation*}
then
\begin{equation*}
\begin{array}{c}
\mathcal{X}_{6}=([\dfrac{-3-\sqrt{5}}{2},\dfrac{-7+\sqrt{33}}{2}],0). \\
\end{array}
\end{equation*}
We obtain six eigenvalues.

\bigskip

\noindent\textbf{Remark. }To compute the interval-eigenvalues of a matrix $A$
, we have to find the roots of the characteristical polynomial of $A$. But
this polynomial is with coefficients in $\overline{\mathbb{IR}}$ (or $
\mathcal{A}_{4}$) and this set is not a field neither a factorial ring. Then
it is natural to meet some special results (e.g if we consider the second
degree polynomial $X^{2}-1$ with coefficients in $\frac{\mathbb{Z}}{8\mathbb{
Z}}$ which is not factorial, it admits four roots,$1,3,5,7.$) In our example
we finds $6$ roots. Now if we consider the real matrix whose coefficients
are the centers of interval-coefficients of $B_3$, that is
\begin{equation*}
c_{B_3}=\left(
\begin{array}{cc}
1.5 & 1.5 \\
2 & 3.5
\end{array}
\right)
\end{equation*}
then the eigenvalues of $c_{B_3}$ are $4.5$ and $0.5$ which are closed to
the center of $\mathcal{X}_{1}$ and $\mathcal{X}_{3}.$ We call these
eigenvalues, the central eigenvalues.

\begin{definition}
Let $A$ be a matrix in $gl(n,\overline{\mathbb{IR}}).$ Lat $A_{c}$ be the
real matrix whose elements are the center of the intervals of $A$. We say
that an eigenvalue of $A$ is a central eigenvalue if its center is (close
to) an eigenvalue of $A_{c}$.
\end{definition}

\bigskip \noindent\textbf{Remark.} The determination of negative eigenvalues
that is of type $(0,K)$ is similar. Nevertheless we have to consider only
matrices with positive entries thus we studies only the positive
eigenvalues. The negative eigenvalues do not correspond to physical entities.

\subsection{Eigenvectors, eigenspaces}

Now we will look the problem of reduction of an interval matrix. Recall that
the characteristical polynomial is with coefficient in a non factorial ring.
This is the bigest change with respect the classical real linear algebra.

\begin{defi}
Let $A$ a square matrix with coeffcients in $\overline{\mathbb{IR}}$. If $
\mathcal{X}$ is an eigenvalue of $A$, then every vector $\mathcal{V}\in
\overline{\mathbb{IR}}^{n}$ satisfying $A^{t}\mathcal{V=X}^{t}\mathcal{V}$
is an eigenvector associated with $\mathcal{X}.$
\end{defi}

Let $E_{\mathcal{X}}$ be the set \
\begin{equation*}
E_{\mathcal{X}}=\{\mathcal{V}\in \overline{\mathbb{IR}}^{n}\text{ such that }
A^{t}\mathcal{V=X}^{t}\mathcal{V}\}.
\end{equation*}
Then $E_{\mathcal{X}}$ is a $\mathbb{R}$-subspace of $\overline{\mathbb{IR}}
^{n}$ where $n$ is the order of the matrix $A$. It is also a $\overline{
\mathbb{IR}}$ submodule of $\overline{\mathbb{IR}}^{n}$.

\begin{propo}
Let $\mathcal{X}_{1}$ and $\mathcal{X}_{2}$ be two distinguish eigenvalues
of $A $. Then $E_{\mathcal{X}_{1}}\cap E_{\mathcal{X}_{2}}=\{0\}.$
\end{propo}

\textit{Proof}\textbf{. }Let $\mathcal{V}$ be in $E_{\mathcal{X}_{1}}\cap E_{
\mathcal{X}_{2}}$. We have
\begin{eqnarray*}
A^{t}\mathcal{V} &\mathcal{=}&\mathcal{X}_{1}\mathcal{V}, \\
A^{t}\mathcal{V} &\mathcal{=}&\mathcal{X}_{2}\mathcal{V}.
\end{eqnarray*}
This $\mathcal{X}_{1}\mathcal{V}\smallsetminus \mathcal{X}_{2}\mathcal{V}=(
\mathcal{X}_{1}\smallsetminus \mathcal{X}_{2})\mathcal{V}=0$. As $\overline{
\mathbb{IR}}$ is without zero divisor, we have $\mathcal{X}
_{1}\smallsetminus \mathcal{X}_{2}=0$ or $\mathcal{V}=0$. We deduce $E_{
\mathcal{X}_{1}}\cap E_{\mathcal{X}_{2}}=\{0\}.$

\bigskip

\begin{propo}
Let $C_{A}(\mathcal{X})$ be the characteristical polynomial of $A$. If the
real polynomial $C_{C_{A}}(\mathcal{X})$ associated with the central matrix
of $A$ is a product of factor of degree $1$, then $C_{A}(\mathcal{X})$
admits a factorization on $\overline{\mathbb{IR}}$
\end{propo}

We have seen that $C_{A}(\mathcal{X})$ can be have more than degree($C_{A}(
\mathcal{X}$) roots. If $\mathcal{X}_{1},\cdots ,\mathcal{X}_{n}$ are the
central roots, we have the decomposition
\begin{equation*}
C_{A}(\mathcal{X})=a_{n}\prod\limits_{i=1}^{n}(\mathcal{X}\smallsetminus
\mathcal{X}_{i}).
\end{equation*}

\medskip

\noindent\textbf{Example. }If we consider the matrix
\begin{equation*}
B_3=\left(
\begin{array}{cc}
\lbrack 1,2] & [1,2] \\
\lbrack 1,3] & [2,5]
\end{array}
\right) .
\end{equation*}
then $C_{B_3}(\mathcal{X})$ admits $\mathcal{X}_{1},\cdots ,\mathcal{X}_{6}$
as positive roots. The central eigenvalues are $\mathcal{X}_{1}$ and $
\mathcal{X}_{3}$ and we have
\begin{equation*}
\det (B_3\smallsetminus \mathcal{X}I)=(\mathcal{X}\smallsetminus \mathcal{X}
_{1})(\mathcal{X}\smallsetminus \mathcal{X}_{3}).
\end{equation*}
If we consider the roots $\mathcal{X}_{2}=([\dfrac{3-\sqrt{5}}{2},\dfrac{7+
\sqrt{33}}{2}],0)$, and if we assume that $C_{B_3s}(\mathcal{X})=(\mathcal{X}
\smallsetminus \mathcal{X}_{2})(\mathcal{X}\smallsetminus Y)$, we obtain
\begin{equation*}
Y=(3,7,\frac{3-\sqrt{5}}{2},\frac{7+\sqrt{33}}{2})=(\frac{3+\sqrt{5}}{2},
\frac{7-\sqrt{33}}{2},0,0)
\end{equation*}
which does not correspond to a positive eigenvalue.

\bigskip

\begin{theorem}
For any $n$-ule of roots $(\mathcal{X}_{1},\cdots ,\mathcal{X}_{n})$ such
that $C_{A}(\mathcal{X})=a_{n}\prod\limits_{i=1}^{n}(\mathcal{X}
\smallsetminus \mathcal{X}_{i})$, and if for any $i=1,\cdots ,n$ the
dimension of $E_{\mathcal{X}_{i}}$ coincides with the multiplicity of $
\mathcal{X}_{i},$ then we have the vectorial decomposition $\overline{
\mathbb{IR}}^{n}=\oplus _{i\in I}E_{\mathcal{X}_{i}}$ where the roots $
\mathcal{X}_{i},i\in I$ are pairwise distinguish.
\end{theorem}

\bigskip

\noindent\textbf{Example.} Let us compute the eigenspaces of $B_3$
associated to the central eigenvalues.

\begin{itemize}
\item $\mathcal{X}_{1}=([\dfrac{3+\sqrt{5}}{2},\dfrac{7+\sqrt{33}}{2}],0).$

\noindent Let $V=\left(
\begin{array}{c}
V_{1} \\
V_{2}
\end{array}
\right) \in \overline{\mathbb{IR}}.$ Then
\begin{equation*}
(A-\mathcal{X}_{1}I)\left(
\begin{array}{c}
V_{1} \\
V_{2}
\end{array}
\right) =0
\end{equation*}
is equivalent to
\begin{equation*}
\left(
\begin{array}{ll}
(0,[\dfrac{1+\sqrt{5}}{2},\dfrac{3+\sqrt{33}}{2}])V_{1} & ([1,2],0)V_{2} \\
([1.3],0)V_{1} & (0,[\dfrac{-1+\sqrt{5}}{2},\dfrac{-3+\sqrt{33}}{2}])V_{2}
\end{array}
\right) =0
\end{equation*}
that is
\begin{equation*}
\left\{
\begin{array}{c}
\smallsetminus \lbrack \dfrac{1+\sqrt{5}}{2},\dfrac{3+\sqrt{33}}{2}
]V_{1}+[1,2]V_{2}=0, \\
\lbrack 1.3]V_{1}\smallsetminus \lbrack \dfrac{-1+\sqrt{5}}{2},\dfrac{-3+
\sqrt{33}}{2}]V_{2}=0.
\end{array}
\right.
\end{equation*}
This gives
\begin{equation*}
V_{2}=\frac{[\dfrac{1+\sqrt{5}}{2},\dfrac{3+\sqrt{33}}{2}]V_{1}}{[1,2]}.
\end{equation*}
If we choose $V_{1}=([1,1],0)$ we have
\begin{equation*}
\begin{array}{llll}
\medskip V_{2} & = & \frac{[\dfrac{1+\sqrt{5}}{2},\dfrac{3+\sqrt{33}}{2}]}{
[1,2]} &  \\
\medskip  & = & [\dfrac{-1+\sqrt{5}}{2},\dfrac{-3+\sqrt{33}}{2}]\bullet
(\smallsetminus \lbrack -1,\frac{-1}{2}]) &  \\
& = & \smallsetminus ([-\frac{3+\sqrt{33}}{2},-\frac{1+\sqrt{5}}{4}]) &
\end{array}
\end{equation*}
Thus the $\mathcal{X}_{1}$-eigenvectors are of the form
\begin{equation*}
V=\left(
\begin{array}{c}
([1,1],0) \\
\smallsetminus ([-\frac{3+\sqrt{33}}{2},-\frac{1+\sqrt{5}}{4}],0)
\end{array}
\right) .
\end{equation*}
\end{itemize}

\noindent\textbf{Remark. }We can choose $V_{1}$ such that all tje
coordinatez of $V$ are positive. For example if $V_{1}=[1,2]$ then $V=\left(
\begin{array}{c}
([1,2],0) \\
([\dfrac{1+\sqrt{5}}{2},\dfrac{3+\sqrt{33}}{2}],0)
\end{array}
\right) $

\begin{itemize}
\item $\mathcal{X}_{3}=([\dfrac{3-\sqrt{5}}{2},\dfrac{7-\sqrt{33}}{2}],0).$

\noindent Let $V=\left(
\begin{array}{c}
V_{1} \\
V_{2}
\end{array}
\right) \in \overline{\mathbb{IR}}.$ Then
\begin{equation*}
(A-\mathcal{X}_{1}I)\left(
\begin{array}{c}
V_{1} \\
V_{2}
\end{array}
\right) =0
\end{equation*}
is equivalent to
\begin{equation*}
\left(
\begin{array}{ll}
([\dfrac{-1+\sqrt{5}}{2},\dfrac{-3+\sqrt{33}}{2}],0)V_{1} & ([1,2],0)V_{2}
\\
([1.3],0)V_{1} & ([\dfrac{1+\sqrt{5}}{2},\dfrac{3+\sqrt{33}}{2}],0)V_{2}
\end{array}
\right) =0
\end{equation*}
that is
\begin{equation*}
\left\{
\begin{array}{c}
\lbrack \dfrac{-1+\sqrt{5}}{2},\dfrac{-3+\sqrt{33}}{2}]V_{1}+[1,2]V_{2}=0,
\\
\lbrack 1.3]V_{1}+[\dfrac{1+\sqrt{5}}{2},\dfrac{-3+\sqrt{33}}{2}]V_{2}=0.
\end{array}
\right.
\end{equation*}
This gives
\begin{equation*}
V_{2}=\frac{\smallsetminus ([\dfrac{-1+\sqrt{5}}{2},\dfrac{-3+\sqrt{33}}{2}
])V_{1}}{[1,2]}.
\end{equation*}
If we choose $V_{1}=([1,1],0)$ we have
\begin{equation*}
\begin{array}{llll}
\medskip V_{2} & = & \frac{\smallsetminus \lbrack \dfrac{-1+\sqrt{5}}{2},
\dfrac{-3+\sqrt{33}}{2}]}{[1,2]} &  \\
\medskip  & = & \smallsetminus \lbrack \dfrac{-1+\sqrt{5}}{2},\dfrac{-3+
\sqrt{33}}{2}]\bullet (\smallsetminus \lbrack -1,\frac{-1}{2}]) &  \\
& = & ([\frac{3-\sqrt{33}}{2},\frac{1-\sqrt{5}}{4}]) &
\end{array}
\end{equation*}
Thus the $\mathcal{X}_{3}$-eigenvectors are of the form
\begin{equation*}
V=\left(
\begin{array}{c}
([1,1],0) \\
([\frac{3-\sqrt{33}}{2},\frac{1-\sqrt{5}}{4}],0)
\end{array}
\right) .
\end{equation*}
\end{itemize}

\section{The Exponential map}

We define the exponential map
$$Exp :gl(n,\overline{\mathbb{IR}})\longrightarrow gl(n,\overline{\mathbb{IR}}) $$
in a classical way by series expansions. If the matrix $A$ is diagonalizable, then
$$D=P^{-1}AP$$
is diagonal and $Exp(A)$ is a diagonal matrix whose diagonal element are the exponential of the eigenvalues.

\noindent {\bf Example.} Let
\begin{equation*}
B_3=\left(
\begin{array}{cc}
\lbrack 1,2] & [1,2] \\
\lbrack 1,3] & [2,5]
\end{array}
\right) .
\end{equation*}
The central eigenvalues are
\begin{equation*}
\left\{
\begin{array}{c}
\mathcal{X}_{1} =([\dfrac{3+\sqrt{5}}{2},\dfrac{7+\sqrt{33}}{2}],0), \\
\mathcal{X}_{3}=([\dfrac{3-\sqrt{5}}{2},\dfrac{7-\sqrt{33}}{2}],0). \\
\end{array}
\right.
\end{equation*} and we have
\begin{equation*}
D=\left(
\begin{array}{cc}
\mathcal{X}_{1} & 0 \\
0 & \mathcal{X}_{3}
\end{array}
\right)
\end{equation*}
with
\begin{equation*}
P=\left(
\begin{array}{cc}
([1,1],0) & ([1,1],0) \\
\smallsetminus ([-\frac{3+\sqrt{33}}{2},-\frac{1+\sqrt{5}}{4}],0)& ([\frac{3-\sqrt{33}}{2},\frac{1-\sqrt{5}}{4}],0)
\end{array}
\right) .
\end{equation*}
We deduce
$$Exp(B_3)=P.\left(
\begin{array}{cc}
 ([exp(\dfrac{3+\sqrt{5}}{2}),exp(\dfrac{7+\sqrt{33}}{2}],0)) & 0 \\
0 & ([exp(\dfrac{3-\sqrt{5}}{2}),exp(\dfrac{7-\sqrt{33}}{2}],0))
\end{array}
\right).P^{-1}$$
In a forthcomming paper, we apply this calculus to solve linear differential system.

config: arxiv


TITLE: Basic geometry in logic
QUESTION [2 upvotes]: I'm trying to solve the following exercise, which just seems to express a geometric system in terms of symbols. (I edited the exercise slightly)
Let $\mathcal{L}_{\text {inc }}$ be a first order language that consists of only relation symbols $\{P, L, I\}$ where $P$ and $L$ are unary relation symbols and $I$ is a binary relation symbol. An incidence structure $\mathfrak{A}$ is an $\mathcal{L}_{\text {inc }}$ -structure where the elements of $P^{\mathfrak{A}}$ are interpreted as points, elements of $L^{\mathfrak{A}}$ are interpreted as lines and $I^{\mathfrak{A}} \subseteq P^{\mathfrak{A}} \times L^{\mathfrak{A}}$ is the incidents relation. When $(p, l) \in I^{\mathfrak{A}}$ then we say point $p$ is incident with line $l$. Consider the following set of axioms:
(A0) $P^{2}$ and $L^{\mathfrak{A}}$ partitions the domain of $\mathfrak{A}$ i.e. every element of the domain is either a point or a line but not both;
(A1) Any two distinct points are incident with exactly one line;
Express $\mathrm{A} 0, \mathrm{~A} 1$ as $L_{i n c}$ -sentences.
Here is my solution:
$\mathrm{A} 0:\forall v(((P(v) \vee L(v)) \wedge (\lnot(P(v) \land L(v)))) .$
$\mathrm{A} 1:\forall v_{1}, v_{2}\left(\left(P\left(v_{1}\right) \wedge P\left(v_{2}\right)\right) \land \left(\exists v_{3}\left(L\left(v_{3}\right) \wedge I\left(v_{1}, v_{3}\right) \wedge I\left(v_{2}, v_{3}\right) \wedge\left(\left(\forall v_{4}\left(L\left(v_{4}\right) \wedge I\left(v_{1}, v_{4}\right) \wedge I\left(v_{2}, v_{4}\right)\right) \rightarrow\right.\right.\right.\right.\right.$
$\left.\left.\left.v_{3} \doteq v_{4}\right)\right)\right)$
The official solution is:
$\mathrm{A} 0:(\forall v(P(v) \vee L(v))) \wedge \forall v((P(v) \rightarrow \neg L(v)) \wedge(L(v) \rightarrow \neg P(v))) .$
$\mathrm{A} 1:\forall v_{1}, v_{2}\left(\left(P\left(v_{1}\right) \wedge P\left(v_{2}\right)\right) \rightarrow\left(\exists v_{3}\left(L\left(v_{3}\right) \wedge I\left(v_{1}, v_{3}\right) \wedge I\left(v_{2}, v_{3}\right) \wedge\left(\left(\forall v_{4}\left(L\left(v_{4}\right) \wedge I\left(v_{1}, v_{4}\right) \wedge I\left(v_{2}, v_{4}\right)\right) \rightarrow\right.\right.\right.\right.\right.$
$\left.\left.\left.v_{3} \doteq v_{4}\right)\right)\right)$
The question is, is my solution correct? I think A0 is correct, but I'm concerned about A1, especially because I used $\land$ instead of $\rightarrow$. If my solution is wrong, can you help me understand why? Thanks.

REPLY [2 votes]: A0 looks good: it translates to "any v is either a point or a line and isn't both a point and a line"
A1 will not be true if $v_1$ or $v_2$ is a line.


TITLE: Questions regarding Rings.
QUESTION [2 upvotes]: I barely passed abstract algebra when I was in college, and 3 years later I bought a book and studied on my own.  And currently I am having trouble with Rings with certain conditions.

Let $\mathbb R$ be the field of real numbers and $\mathbb R[x]$ the ring of polynomials in $x$ with coefficients in $\mathbb R$.
Condition: The subset $S$ is all polynomials whose degree is an even integer, together with the zero polynomial.
I would like to show that this $S$ is not a subring.
My understanding is that to show that a set with two binary operations $+$ and $\cdot$ is a ring iff:
(a) $\langle S,+\rangle$ is an abelian group;
(b) $\langle S,\cdot\rangle$ is a semigroup. ($\cdot$ is associative);
(c) The distributive law holds.
I could be spending few minutes checking every single condition, but this is a problem from the GRE and I cannot do that.
I think it satisfies (b) because multiplying even degree polynomials results in another even degree polynomial.
I don't think (a) is a problem because the degree of the polynomials shouldn't seem to affect if it is an abelian additive group...
Is it (c) ? or am I missing something?

Another question about a ring $T$ such that $s = s^2, \forall s \in T$.
I want to argue that since $s^2-s = 0 \Rightarrow s = 0 \text{,} 1$, so this is the ring $\mathbb Z_2$.
But my instinct tells me that this is not true.
The problem asks me:

Which of the following are true:
I. $s+s=0,\forall s \in T$
II. $(s+t)^2=s^2+t^2, \text{for each} s,t\in T$
III. $T$ is commutative.

I guessed yes to everything according to my assumption that $T \equiv \mathbb Z_2$, only to be disappointed.

REPLY [1 votes]: We have $x^2 \in S$ and also $x^2 + x \in S$. But...

As for your second problem, you have established that for all $s \in T$:
$$s(s-1) = 0$$
and you have assumed that this implies $s = 0$ or $s-1 =0$. This suggests that you read up the definition of a zero divisor. General rings (especially non-commutative ones) can behave in "strange" ways.

REPLY [0 votes]: Every subset of an algebra which is closed under its operations has the same signature - is a subalgebra of the same type. Because $\Bbb{R}$ is an integral domain, $\Bbb{R}[x]$ is, too. This insures the sum of degrees of two polynomials is the degree of their product, so the algebra is indeed closed under multiplication. For it to be closed under addition, there must be no elements $r$ of the ground ring ($\Bbb{R}$) such that $r+...+r=0,$ as by distributivity this means $x^2r+...+x^2r=0.$ Thus one can add $x^2r$ to $x^2+x$ some number of times to get $x$. $\Bbb{R}$ is of characteristic 0, and $\Bbb{R}[x]$ is freely generated over it, so this is not a problem. But subtraction is also possible, and $x^2$ can be subtracted from $x^2+x$ to get a degree-0 polynomial. However, restricted to positive leading coefficients, it would be closed as a rig - a ring without negative elements, or a monoid that distributes over a commutative monoid.


TITLE: Example showing set of periodic functions from $\mathbb{R}$ to $\mathbb{R}$ is not subspace of $\mathbb{R}^{\mathbb{R}}$.
QUESTION [3 upvotes]: Is set of periodic functions from $\mathbb{R}$ to $\mathbb{R}$ subspace of $\mathbb{R}^{\mathbb{R}}$?

My answer is No. I give the following counterexample:
We consider periodic functions $f(x),g(x)$ so that $f(x)=x$ for $x \in [0,\sqrt{2})$ and $f(x)=f(x+\sqrt{2})$, function $g(x)=x$ for $x \in [0,\sqrt{3})$ and $g(x)=g(x+\sqrt{3})$ then $f+g$ is not periodic.
Indeed, assume that the function $f(x)+g(x)$ is periodic with length $T$. Note that if $T/\sqrt{2}$ is not an integer then $f(x)<f(x+T)$, similar to $T/\sqrt{3}$. Thus, since $T/ \sqrt{2}$ or $T/ \sqrt{3}$ can't be both integers so $f(x)+g(x)<f(x+T)+g(x+T)$, a contradiction. Thus, $f+g$ is non-periodic function. This yields that set of periodic function from $\mathbb{R}$ to $\mathbb{R}$ is not a subspace of $\mathbb{R}^{\mathbb{R}}$.
My questions are

Is this counterexample correct?
Are there any other examples?

REPLY [3 votes]: Your argument that $f+g$ is not periodic is not quite correct.  For instance, if $T=1$ then $T/\sqrt{2}$ is not an integer, but $f(1)=1>f(1+T)=2-\sqrt{2}$.  However, you can fix your argument by specifically looking at $x=0$: then $f(x)+g(x)=0+0=0$, but at least one of $f(x+T)$ and $g(x+T)$ must be positive since $T$ cannot be an integer multiple of both $\sqrt{2}$ and $\sqrt{3}$.  So for $x=0$, you must have $f(x)+g(x)<f(x+T)+g(x+T)$.
More generally, a similar argument shows that if $f$ and $g$ are periodic with periods $a$ and $b$ respectively where $a/b$ is irrational, and $f$ and $g$ both achieve their minimum values only at integer multiples of $a$ and $b$ respectively, then $f+g$ is not periodic.


TITLE: Minimal distance between quadratic function and point
QUESTION [1 upvotes]: I have a function or line $R\rightarrow R^n$
$$ y_i = f(x) = {-b_i \pm \sqrt{-4 \cdot a_i\cdot c_i + b_i^2+4\cdot c_i \cdot x}\over 2 \cdot c_i}$$
where the parameter vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are known. For any value x, I will get a n-dimensional point out. (The equation should look familiar, meaning x(y) are quadratic functions.)
I would like to find for an arbitrary point $\mathbf{p}$ the closest point on this line $f$, and its corresponding $x$.
What I did so far was to differentiate the euclidean norm $\Delta = \sum_i (y_i(x) - p_i)^2 $ and setting it to $0$, yielding, after dividing away non-zero terms:
$$2\cdot \sum_i {p_i + {b_i\over 2 \cdot c_i }\over \sqrt{4 \cdot c_i\cdot  x - 4 \cdot a_i\cdot c_i+ b_i^2}} - {1\over2\cdot c_i} \overset{!}{=} 0 $$
Now I'm stuck. Does an analytic solution for finding $x$ exist?

REPLY [0 votes]: I doubt this has any closed form solution. Here's why. The difficult part of the formula you have is simplifying an expression of the form $\frac1{\sqrt x} + \frac1{\sqrt y} + \frac1{\sqrt z}$, and some additional terms, so that there is a common denominator with no radical in it. Once you get there, you have an expression of the form $\sqrt a + \sqrt b + \sqrt c$, for which there is at least some hope.
However, to get there, you cannot really avoid converting to the expression $\frac{yz\sqrt{x}+xz\sqrt{y}+xy\sqrt{z}}{xyz}$. Similar things work for longer sums than three, but unfortunately, this is where the good luck stops. We multiply both sides by $xyz$ and what we find is that there are degree-three polynomials on each side of the equality sign, in addition to things in square roots. So the smallest polynomial you can hope to have here is degree six.
Furthermore, there is almost no relationship between the coefficients. Yes, it all looks nice and symmetric when we're doing the heuristics, but $x$, $y$, and $z$ are all nontrivially linear in the variable and there are also constant terms floating around out there. Therefore, Abel is going to come in and say that you probably cannot solve this polynomial algebraically.
Sorry about that :(


text:
\section{Proof sketch of structural result}
\label{sec:structural}
The proofs of the lemmas in this section can be found in the Appendix A.
Consider the mutual information proxy when the values of $X_u$, $X_I$, and $X_S$ are fixed:
\begin{align*}
&\nu_{u,I|S}(x_u,x_I|x_S)\\
&\quad = \left| \mathbb{P}(X_u=x_u,X_I=x_I|X_S=x_S) - \mathbb{P}(X_u=x_u|X_S=x_S) \mathbb{P}(X_I=x_I|X_S=x_S) \right|.
\end{align*}

We first establish a version of Theorem \ref{thm:nu_bound_alone} for $\nu_{u,I|S}(x_u,x_I|x_S)$, and then generalize it to $\nu_{u,I|S}$.

In Lemma \ref{lemma:expr_nu}, we express $\nu_{u,I|S}(x_u,x_I|x_S)$ as a sum over configurations of latent variables $U$ connected to observed variables in $I$. Note that $|U| \leq s$, so the summation is over at most $2^s$ terms.

\begin{lemma}
\label{lemma:expr_nu}
Fix observed variable $u$ and subsets of observed variables $I$ and $S$, such that all three are disjoint. Suppose that $I$ is a subset of the MRF neighborhood of $u$. Then
\[\nu_{u,I|S}(x_{u},x_{I}|x_S)= \left| \sum_{q_U \in \{-1,1\}^{|U|}} \left(\sum_{q_{\sim U} \in \{-1,1\}^{m - |U|}} \bar{f}(q, x_u, x_S)\right) \prod_{i \in I} \sigma(2x_i(J^{(i)}\cdot q + h_i)) \right|\]
for some function $\bar{f}$, where $U$ is the set of latent variables connected to observed variables in $I$, $J^{(i)}$ is the $i$-th row of $J$, and the entries of $q_{\sim U}$ in the expression $J^{(i)} \cdot q$ are arbitrary.
\end{lemma}

Lemma \ref{lemma_non_cancel_abstract} gives a generic non-cancellation result for expressions of the form $\left|\sum_{i=1}^m a_i \prod_{j=1}^n x_{i,j}\right|$.
Then, Lemma \ref{lemma_non_cancel_nu_indiv_cond} applies this result to the form of $\nu_{u,I|S}(x_u,x_I|x_S)$ in Lemma \ref{lemma:expr_nu}, and guarantees the existence of a subset $I' \subseteq I$ with $|I'| \leq 2^s$ such that $\nu_{u,I'|S}(x_u,x_{I'}|x_S)$ is within a bounded factor of $\nu_{u,I|S}(x_u,x_I|x_S)$.

\begin{lemma}
\label{lemma_non_cancel_abstract}
Let $x_{1,1}, ..., x_{m,n} \in [-1, 1]$, with $n > m$. Then, for any $a \in \mathbb{R}^m$, there exists a subset $S \subseteq [n]$ with $|S| \leq m$ such that
\[\left| \sum_{i=1}^m a_i \prod_{j \in S} x_{i,j} \right| \geq \frac{1}{4^m} \left(\frac{1}{n}\right)^{m(m+1)} \left| \sum_{i=1}^m a_i \prod_{j=1}^{n} x_{i,j} \right|.\]
\end{lemma}

We remark that, in this general form, Lemma \ref{lemma_non_cancel_abstract} is optimal in the size of the subset that it guarantees not to be cancelled. That is, there are examples with $\sum_{i=1}^m a_i \prod_{j=1}^n x_{i,j} \neq 0$ but $\sum_{i=1}^m a_i \prod_{j \in S} x_{i,j} = 0$ for all subsets $S \subseteq [n]$ with $|S| \leq m - 1$. See the Appendix A for a more detailed discussion.

\begin{lemma}
\label{lemma_non_cancel_nu_indiv_cond}
Fix observed variable $u$ and subsets of observed variables $I$ and $S$, such that all three are disjoint. Suppose that $I$ is a subset of the MRF neighborhood of $u$. Fix any assignments $x_u$, $x_I$, and $x_S$. Then there exists a subset $I' \subseteq I$ with $|I'| \leq 2^s$ such that
\[\nu_{u, I'|S}(x_u, x_{I'}|x_S) \geq \frac{1}{4^{2^s}} \left(\frac{1}{|I|}\right)^{2^s(2^s+1)} \nu_{u, I | S}(x_u, x_{I} | x_S)\]
where $x_{I'}$ agrees with $x_I$.
\end{lemma}

Finally, Lemma \ref{lemma_non_cancel_nu_cond_full} extends the result about $\nu_{u, I | S}(x_u, x_{I} | x_S)$ to a result about $\nu_{u,I|S}$. The difficulty lies in the fact that the subset $I'$ guaranateed to exist in Lemma \ref{lemma_non_cancel_nu_indiv_cond} may be different for different configurations $(x_u, x_I, x_S)$. Nevertheless, the number of subsets $I'$ with $|I'| \leq 2^s$ is smaller than the number of configurations $(x_u, x_I, x_S)$, so we obtain a viable bound via the pigeonhole principle.

\begin{lemma}
\label{lemma_non_cancel_nu_cond_full}
Fix observed variable $u$ and subsets of observed variables $I$ and $S$, such that all three are disjoint. Suppose that $I$ is a subset of the MRF neighborhood of $u$. Then there exists a subset $I' \subseteq I$ with $|I'| \leq 2^s$ such that
\[\nu_{u, I' | S} \geq\frac{1}{(4|I|)^{2^s}} \left( \frac{1}{|I|} \right)^{2^s(2^s+1)} \nu_{u, I | S}.\]
\end{lemma}

This result completes the proof of Theorem \ref{thm:nu_bound_alone}.