TITLE: $\mathfrak{a\subseteq b}$ and $\mathfrak{bc}=\lambda D$ then there is an ideal $\mathfrak{b}'$ such $\mathfrak{a=bb'}$
QUESTION [1 upvotes]: Let be $D$ a commutative domain, $\mathfrak{a,b,c}\subseteq D$ ideals. Show that:

if $\mathfrak{a\subseteq b}$ and $\mathfrak{bc}=\lambda D$ then there is an ideal $\mathfrak{b}'$ such $\mathfrak{a=bb'}$

How to work this?
My work:
Set $\mathfrak{b}'=\{d\in D: bd\in\mathfrak{a}, \forall b\in\mathfrak{b}\}$. is clear to see that $\mathfrak{b}'$ is and ideal and $\mathfrak{b \subseteq a} $ but i thinks that I'm missing something because not using the fact that $\mathfrak{b}$ is invertible ($\mathfrak{bc}=\lambda D$)
What is the idea behind this exercise? I'm in a number theory course and we have make properties that intuition tells me that ideals works as numbers but I can't figure out the intuition behind this problem.

REPLY [1 votes]: okey. I've worked and found the solution:
$$\mathfrak{ac\subseteq bc=}\lambda D$$
* then exists some $\mathfrak{b}'$ ideal such $\mathfrak{ac}=\lambda \mathfrak{b}'$
since $D$ is commutative, $$\mathfrak{abc}=\lambda\mathfrak{ b b}'$$
$$\mathfrak{a}\lambda D=\lambda\mathfrak{bb'}\Rightarrow \mathfrak{a}D=\mathfrak{a=bb'}$$

(*)
Let be $D$ a commutative domain, $\lambda\in D\setminus \{0\}$ and $A\subset D$ an ideal. Then

if $A\subset \lambda D$ then exists $A'\subset D$ such $A=\lambda A'$

Proof
$$A' = \{ x \in D : \lambda x \in A\}.$$
Now it remains to see that $A'$ has the required properties. We need to verify $A = \lambda A'$. By definition, we have $\lambda A' \subset A$. And since by assumption $A \subset \lambda D$, for every $a\in A$ there is a $d\in D$ with $a = \lambda d$. That implies $\lambda d \in A$, and hence $d \in A'$, so we also have $A \subset \lambda A'$.


text:
\begin{document}
\mainmatter

\title{A new way to classify 2D higher order quantum superintegrable systems}

\titlerunning{Quantum superintegrable systems }

\author{Bjorn K. Berntson\inst{1} \and Ian Marquette\inst{2}
 \and Willard Miller, Jr.\inst{3} }

\authorrunning{Bjorn K. Berntson et al.}

\tocauthor{Ivar Ekeland, Roger Temam, Jeffrey Dean, David Grove,
Craig Chambers, Kim B. Bruce, and Elisa Bertino}

\institute{Department of Mathematics, KTH Royal Institute of Technology,
Stockholm, Sweden\\
\email{bbernts@kth.se}
\and
School of Mathematics and Physics, The University of Queensland,\\ Brisbane, Australia\\
\email{i.marquette@uq.edu.au},\\ WWW home page:
\texttt{https://smp.uq.edu.au/profile/211/ian-marquette}
\and
 School of Mathematics, University of Minnesota,\\con
Minneapolis, Minnesota, U.S.A.\\
\email{mille003@math.umn.edu},\\ WWW home page:
\texttt{http://www-users.math.umn.edu/~mille003}}

\maketitle

\begin{abstract}
We  revise a method by Kalnins, Kress and Miller (2010) for constructing a canonical form for symmetry operators of
arbitrary order for the Schr\"odinger eigenvalue equation $H\Psi \equiv (\Delta_2 +V)\Psi=E\Psi$ on any
2D Riemannian manifold, real or complex, that admits a separation of variables in some orthogonal
coordinate system.  We apply the method, as an example,  to  revisit the Tremblay and Winternitz (2010) derivation of
 the Painlev\'e VI potential for a 3rd order superintegrable  flat space system that separates  in polar coordinates and, as new results,  we give a listing of the possible potentials on the 2-sphere that separate in spherical coordinates and all 2-hyperbolic (two-sheet) potentials separating in horocyclic coordinates. In particular, we  show that the Painlev\'e VI potential also appears for a 3rd order superintegrable system on the 2-sphere that separates in spherical coordinates, as well as a 3rd order superintegrable system on the 2-hyperboloid that separates in spherical coordinates and one that separates in horocyclic coordinates. Our aim is  to develop tools for analysis and classification of higher order superintegrable systems on any 2D Riemannian space, not just Euclidean space.
\keywords{quantum superintegrable systems, Painlev\'e VI equation, Weierstrass equation}
\end{abstract}

\section{Introduction}

In the paper \cite{KKM2010} the authors constructed a canonical form for symmetry operators of any order in 2D and used it to give the first  proof of the superintegrability  of the  quantum Tremblay, Turbiner,
 and Winternitz (TTW) system \cite{TTW} in polar coordinates, for all rational values of the parameter $k$. In the original method the various potentials were given and the problem was the construction of higher order symmetry operators that would verify superintegrability. The method was highly algebraic and required the solution of systems of difference equations on a lattice. Here, we consider   an arbitrary space admitting a separation in some orthogonal coordinate system (hence admitting a 2nd order symmetry operator), and search for all potentials $V$ for which the Schr\"odinger  equation admits an additional  independent symmetry operator of order higher than 2. Now the problem reduces to solving a system of partial differential equations.

 In \S\ref{section2} we give a brief introduction to the method and then in \S\ref{section3} we specialize it to 3rd order superintegrable systems. More details can be found in \cite{KKM2010} and \cite{BMM2020}. In \S\ref{section4}   we treat a few important examples. We revisit the Tremblay and Winternitz derivation of
 the Painlev\'e VI potential for a 3rd order superintegrable  flat space system that separates  in polar coordinates, \cite{TW}, and we show among other new  results that the Painlev\'e VI potential also appears for a 3rd order superintegrable system on the 2-sphere that separates in spherical coordinates, as well as a 3rd order superintegrable system on the 2-hyperboloid that separates in horocyclic coordinates.

In \S\ref{section5} we classify all systems on the complex 2-sphere that admit a 3rd order symmetry operator and separate in spherical coordinates. For some of the cases where the system is 3rd order superintegrable we work out the symmetry algebra generated by the Hamiltonian $H$, the 2nd order symmetry operator $A$ associated with separation in spherical coordinates and the 3rd order symmetry operator $B$. In \S\ref{section6} we analyse the cases where the operators $A$ and $B$ are algebraically dependent and relate them to the Weierstrass $\wp$-function and its degenerations. In \S\ref{section7} we study superintegrable systems $\{H,A,B\}$ that are algebraically dependent but with $B$ of order higher than 3.  In \S\ref{section8} we classify 3rd order superintegrable systems on the two-sheet 2-hyperboloid that separate in
horocyclic coordinates. \S\ref{section9} is devoted to discussion.
Some of our principal results were announced in the proceedings paper \cite{BMM2020}, but here we give much more detail.

\section{The method} \label{section2}
We consider a Schr\"odinger equation on a 2D real or complex Riemannian manifold with Laplace-Beltrami operator $\Delta_2$ and potential $V$:
\begin{equation} \label{TIS}H\Psi\equiv (-\frac{\hbar^2}{2}\Delta_2+V)\Psi=E\Psi \end{equation}
that also  admits an orthogonal  separation of variables.
If $\{u_1,u_2\}$ is the  orthogonal separable coordinate system   the corresponding Schr\"odinger
operator can always be put in the  form
\begin{equation} \label{canform}
H=-\frac{\hbar^2}{2}
\Delta_2+V(u_1,u_2)=\end{equation}
\[\frac{1}{f_1(u_1)+f_2(u_2)}\left(-\frac{\hbar^2}{2}\partial^2_{u_1}-\frac{\hbar^2}{2}\partial^2_{u_2}+V_1(u_1)+V_2(u_2)
\right)
\]
and, due to the separability, there is the second-order symmetry operator
\begin{equation} \label{Lequation}
L= \frac{f_2(u_2)}{f_1(u_1)+f_2(u_2)}\left(-\frac{\hbar^2}{2}\partial^2_{u_1}+V_1(u_1)\right)\end{equation}
\[-\frac{f_1(u_1)}{f_1(u_1)+f_2(u_2)}\left(-\frac{\hbar^2}{2}
\partial^2_{u_2}+V_2(u_2)\right),
\]
i.e., $
[H,L]=0$.
We  look for a  partial differential symmetry   operator of arbitrary order  ${\tilde L}(H,L,u_1,u_2)$ that satisfies
\begin{equation}
[H,{\tilde L}]= 0.\label{newconditions1}
\end{equation}
We require that the symmetry operator take the standard form
\begin{align}\label{standardLform}
{\tilde
L}= \sum_{j,k} &\left( A^{j,k}(u_1,u_2)\partial_{u_1u_2}-B^{j,k}(u_1,u_2)\partial_{u_1}
-C^{j,k}(u_1,u_2)\partial_{u_2}\right. \\
&\left. + D^{j,k}(u_1,u_2) \nonumber
\right)H^jL^k.
\end{align}

This can always be done. More details of the derivation can be found in \cite{KKM2010} and \cite{BMM2020}.

In this view we can write
\begin{equation}\label{generalLform}
{\tilde
L}(H,L_2,u_1,u_2)=A(u_1,u_2)\partial_{u_1 u_2}-B(u_1,u_2)\partial_{u_1}
-C(u_1,u_2)\partial_{u_2}
+ D(u_1,u_2),
\end{equation}
and consider $\tilde  L$ as an at most second-order  order differential operator in $u_1,u_2$ that is analytic in the parameters $H,L$.
Then the above system of equations can be written in the more compact form {\small
\begin{equation} \label{partialxyHL}
\partial_{u_1}^2A+\partial_{u_2}^2A-2\partial_{u_2}B-2\partial_{u_1}C
=0,
\end{equation}
\begin{equation} \label{partialxHL}
\frac{\hbar^2}{2}(\partial_{u_1}^2B+\partial_{u_2}^2B)-2\partial_{u_2}A\,V_2-\hbar^2\partial_{u_1}D-AV'_2+(2\partial_{u_2}A\,f_2+Af'_2)H-2\partial_{u_2}A\,L_2=0,
\end{equation}
\begin{equation} \label{partialyHL}
\frac{\hbar^2}{2}(\partial_{u_1}^2C+\partial_{u_2}^2C)-2\partial_{u_1}AV_1-\hbar^2\partial_{u_2}D-AV'_1+(2\partial_{u_1}A\,f_1+Af'_1)H+2\partial_{u_1}A\,L_2=0,
\end{equation}
\begin{equation}\label{constanttermHL}
-\frac{\hbar^2}{2}(\partial_{u_1}^2D+\partial_{u_2}^2D)+2\partial_{u_1}B\,V_1+2\partial_{u_2}C\,V_2+BV'_1+Cv'_2
\end{equation}
\[
-(2\partial_{u_1}B\,f_1+2\partial_{u_2}C\,f_2+Bf'_1
+Cf'_2)H+(-2\partial_{u_1}B+2\partial_{u_2}C)\,L_2=0.
\] }

We can view (\ref{partialxyHL}) as an equation for $A,B,C$ and (\ref{partialxHL}), (\ref{partialyHL}) as the defining equations for $\partial_{u_1}D, \partial_{u_2}D$.
Then $\tilde L$ is $\hat L$ with the terms in $H$ and $L$ interpreted as (\ref{standardLform}) and considered as partial differential operators.

We can simplify this system by noting that there are two functions\hfill\break $F(u_1,u_2,H,L)$, $G(u_1,u_2,H,L)$ such
that (\ref{partialxyHL}) is satisfied by
\begin{equation}\label{ABCtoFG}
A=F,\qquad B=\frac12 \partial_{u_2}F+\partial_{u_1}G,\qquad C=\frac12\partial_{u_1} F-\partial_{u_2}G.
\end{equation}
Then the integrability condition for (\ref{partialxHL}), (\ref{partialyHL}) is (with the shorthand notation $\partial_{u_j}F=F_j$, $\partial_{u_j}\partial_{u_\ell}F=F_{j\ell}$, etc., for $F$ and $G$),

{\small
\begin{align}&\hbar^2 G_{1222}+\frac14 \hbar^2 F_{2222}-2F_{22}(V_2-f_2H+L_2)
-3F_{2}(V'_2-f_2'H)-F(V''_2-f''_2H) \label{eqn1}\\
 &=-\hbar^2 G_{1112} +\frac14 \hbar^2 F_{1111}-2F_{11}(V_1-f_1H-L_2)
-3F_{1}(V'_1-f'_1H)-F(V''_1-f''_1H),\nonumber \end{align} }
and equation (\ref{constanttermHL}) becomes  {\small
\begin{align}
&\frac14 \hbar^2 F_{1112}-2F_{12}(V_1-f_1H)-F_{1}(V'_2-f'_2H)+\frac14 \hbar^2
G_{1111} \label{eqn2}\\
&-2G_{11}(V_1-f_1H-L_2)-G_{1}(V'_1-f'_1H)=-\frac14\hbar^2F_{1222}+2F_{12}(V_2-f_2H) \nonumber \\
&+F_{2}(V'_1-f'_1H)+\frac14 \hbar^2
G_{2222}-
2G_{22}(V_2-f_2H+L_2)-G_{2}(V'_2-f'_2H).\nonumber\end{align}
We remark that any solution of (\ref{eqn1}), (\ref{eqn2}) with $A,B,C$ not identically $0$ corresponds to a symmetry operator that does not commute with $L$, hence is algebraically  independent of the symmetries $H, L$.

\section{3rd order superintegrability}\label{section3}
To show how equations (\ref{eqn1}) and (\ref{eqn2}) can be used to find potentials for superintegrable systems, we  provide detailed derivations of the determining equations for 3rd order superintegrability.  First we note that the most general 3rd order operator must be of the form (\ref{standardLform}) with  {\small
\begin{eqnarray*} &A&=A^0(x,y),\quad B=B^0(x,y)+B^H(x,y) H+B^L(x,y)L,\\
&C&=C^0(x,y)+C^H(x,y) H+C^L(x,y)L,\
 D=D^0(x,y)+D^H(x,y) H+D^L(x,y)L,\end{eqnarray*} }
 or, in view of (\ref{ABCtoFG}),
\begin{equation}\label{eqn3} F(x,y)=F^0(x,y),\quad G(x,y)=G^0(x,y)+G^H(x,y)H +G^L(x,y)L.\end{equation}

Substituting (\ref{eqn3}) into (\ref{eqn1}), (\ref{eqn2}) and noting that the coefficients of independent powers of $H$ and $L$ in these expressions must vanish, we obtain 9 equations, (the first 3 from (\ref{eqn1}) and the next 6 from (\ref{eqn2})):

\begin{eqnarray*}&0=&-6V_1'F^0_1+6V_2'F^0_2-4V_1F^0_{11}+4V_2F^0_{22}-2\hbar^2G^0_{1112}-2\hbar^2G^0_{1222}\\
&&+2F^0V_2''-2F^0V_1'',\\
&0=& F^0_{11}+F^0_{22},\\
&0=&-\hbar^2G^H_{1112}-\hbar^2G^H_{1222}+3f_1'F^0_1-3f_2'F^0_2+2f_1F^0_{11}-2f_2F^0_{22}-F^0f_2''+F^0f_1'',\\
&0=&V_2'F^0_1+V_1'F^0_2+V_1'G^0_1-V_2'G^0_2+2F^0_{12}V_2+2F^0_{12}V_1+2V_1G^0_{11}-2V_2G^0_{22}-\\
&&\frac14\hbar^2G^0_{1111}+\frac14\hbar^2G^0_{2222},\\
&0=&V_1'G^L_1-V_2'G^L_2+2V_1G^L_{11}-2G^0_{11}-2V_2G^L_{22}-2G^0_{22},\\
&0=& G^L_{11}+G^L_{22},\\
&0=& -f_2'F^0_1-f_1'F^0_2+V_1'G^H_1-f_1'G^0_1-V_2'G^H_2+f_2'G^0_2-2F^0_{12}f_2-2F^0_{12}f_1+2V_1G^H_{11}\\
&&-2f_1G^0_{11}-2V_2G^H_{22}+2f_2G^0_{22}-\frac14\hbar^2G^H_{1111}+\frac14\hbar^2G^H_{2222},\\
&0=&-f_1'G^L_{1}+f_2'G^L_{2}+2f_2G^L_{22}-2f_1G^L_{11}-2G^H_{11}-2G^H_{22},\\
&0=&-f_1'G^H_1+f_2'G^H_2+2f_2G^H_{22}-2f_1G^H_{11}.\\
\end{eqnarray*}

\section{Some examples (mostly new)}\label{section4} For our first examples we are particularly interested in potentials with nonlinear defining equations. First, we show that we can obtain the result of Tremblay and Winternitz \cite{TW} that the  quantum system separating in polar coordinates in 2D Euclidean space admits potentials that are expressed in terms of the sixth Painlev\'e transcendent or in terms of the Weierstrass $\wp$-function. To do this we must put the system in the canonical form (\ref{canform}). The separable polar coordinates are $(x,y)=(r\cos(\theta),r\sin(\theta))$. For the canonical form we have $r=\exp(u_1),\ \theta =u_2$. Thus, $f_1(u_1)=\exp(2u_1)$ and $f_2(u_2)=0$. We know that these Painlev\'e VI can appear only if the potential depends  on the angular variable alone, so we set $V_1(u_1)=0$.  Since we want only systems that satisfy nonlinear equations alone, whenever an explicit linear equation for the potential appears, we require that it vanish identically. We have the freedom to replace the angular variable $u_2$ by $u_2+c$ for some real constant $c$ to simplify the expressions. Also we can rescale the answer.

We obtain a solution of the equations in \S\ref{section3} as
\begin{eqnarray*} F^0 &=& 4\hbar^2\exp(-u_1)\sin(u_2),\quad G^L= -8 \exp(-u_1)\cos(u_2)+a_4 u_2+a_3,\\  G^0 &=& -U_1(u_2)\exp(-u_1)+U_2(u_2),\quad
 G^H=a_5,\end{eqnarray*}
subject to the conditions
\begin{eqnarray}\label{cond1}&0=&a_4\frac{dV_2}{du_2}+2\frac{d^2U_2}{du_2^2},\\ \label{cond2}&0=& \hbar^2\frac{d^4U_2}{du_2^4}+4a_4\frac{dV_2}{du_2}V_2-4\frac{dV_2}{du_2}\frac{dU_2}{du_2},\\
\label{cond3}&0=&8V_2\cos(u_2)+4\frac{dV_2}{du_2}\sin(u_2)-\frac{d^2U_1}{du_2^2}- U_1,\\
\label{cond4}&0=&\frac{dV_2}{du_2}\frac{dU_1}{du_2}-\hbar^2\frac{d^3V_2}{du_2^3}\sin(u_2)-4\hbar^2 \frac{d^2V_2}{du_2^2}\cos(u_2)\\
&&+2\sin(u_2)(\hbar^2+4V_2)\frac{dV_2}{du_2}+2V_2\left(6\hbar^2\cos(u_2)+8V_2\cos(u_2)-U_1\right) \nonumber
.\end{eqnarray}

There are basically two cases to consider: \begin{enumerate} \item $a_4=0$. \\ \medskip

Then condition (\ref{cond1}) says that $U_2$ is linear in $u_2$. Thus condition (\ref{cond2}) is a linear equation for $V_2(y)$ which must vanish. Then condition (\ref{cond3}) can be solved for $U_1(y)$ and the result substituted into condition (\ref{cond4}) to obtain an equation for $V_2(u_2)$. After some manipulation (using the fact that $V_2$ is unchanged by transformations $W\to W+c$, where $c$ is a constant), we obtain an equation characterizing Painlev\'e VI, in agreement with \cite{TW}, equation (4.27):
\begin{equation} \label{PVI} \hbar^2\left(\sin(u_2) \frac{d^4W}{du_2^4}+4\cos(u_2)\frac{d^3W}{du_2^3}-6\sin(u_2)\frac{d^2W}{du_2^2}-4\cos(u_2)\frac{dW}{du_2}\right)   \end{equation}
\[ -12\sin(u_2)\frac{dW}{du_2}\frac{d^2W}{du_2^2}-4\cos(u_2)W\frac{d^2W}{du_2^2}-4(\beta_1\sin(u_2)-\beta_2\cos(u_2))\frac{d^2W}{du_2^2}\]
\[-16\cos(u_2)(\frac{dW}{du_2})^2+8\sin(u_2)W\frac{dW}{du_2}-8(\beta_1\cos(u_2)+\beta_2\sin(u_2))\frac{dW}{du_2}=0\] Here $V_2(u_2)=\frac{dW(u_2)}{du_2}$.
\item $a_4 \ne 0$. \medskip\\   Solving condition (\ref{cond1}) for $V_2(u_2)$ and substituting the result and (\ref{cond1}) into (\ref{cond2}) we obtain the equation that characterizes the Weierstrass $\wp$-function (in fact it is a translated and rescaled version):
\begin{equation} \label{Weierstrass} \hbar^2 \frac{d^3V_2}{du_2^3}-12\frac{dV_2}{du_2}V_2+12a_1\frac{dV_2}{du_2}=0,\end{equation}
where $a_1$ is an arbitrary constant. Thus $V_2(u_2)=\hbar^2\wp(u_2-u_{2,0};g_2,g_3)+a_1$, where $u_{2,0}$, $g_2$, and $g_3$ are arbitrary constants. As shown in \cite{TW} this solution is subject to the compatibility condition (\ref{cond3}) and (\ref{cond4}), which leads to a complicated nonlinear differential equation for $V_2(u_2)$.
\end{enumerate}

Now we consider the analogous system on the 2-sphere, separable in spherical coordinates. Here $s_1=\sin(\theta)\cos(\phi),\ s_2=\sin(\theta)\sin(\phi),s_3=\cos(\theta)$ with $s_1^2+s_2^2+s_3^2=1$. This system is in canonical form with coordinates $\{u_1,u_2\}$ where
\begin{equation}\label{cansp} \sin(\theta)=(\cosh(u_1))^{-1},\ \phi=u_2,\ f_1(u_1)=(\cosh(u_1))^{-2},\ f_2(u_2)=0. \end{equation} As before we look for solutions such that $V_1(u_1)=0$ and $V_2(u_2)$ satisfies a nonlinear equation only.

The computation is very similar to that for the Euclidean space example. We obtain the solution
\begin{equation} \label{nonlinearsoln} F^0 = 4\hbar^2\cosh(u_1)\sin(u_2),\ G^L= 8\sinh(u_1)\cos(u_2)+a_{4}y+a_3,\end{equation}
\[G^0 =  \sinh(u_1)\ U_1(u_2)+U_2(u_2),\
 G^H=a_5,\]
 subject to the conditions (\ref{cond1}-\ref{cond4}), exactly the same as for Euclidean space.
  Thus the system on the 2-sphere also admits Painlev\'e VI and special Weierstrass $\wp$-function potentials for 3rd order superintegrability. It is clear from these results that these systems in Euclidean space can be obtained as B\^ocher contractions, \cite[Chapter 15]{Bocher}, of the corresponding systems on the 2-sphere.

\bigskip

Next we consider spherical coordinates  on the hyperboloid $s_1^2-s_2^2-s_3^2=1$,
\[ s_1=\cosh(x),\ s_2=\sinh(x)\cos(\phi),\ s_3=\sinh(x)\sin(\phi).\]
For the canonical form we find
\[ \tanh \bigg(\frac{u_1}{2}\bigg)=\exp(x),\ u_2=\phi,\quad  f_1(u_1)=\frac{1}{\sinh^2(u_1)},\ f_2(u_2)=0,\]
 and we look for solutions such that $V_1(u_1)=0$ and $V_2(u_2)$ satisfies only a nonlinear equation. We obtain the solution
\[ F^0 = 4\hbar^2\sin(u_2)\sinh(u_1),\ G^L= 8\cos(u_2)\cosh(u_1)+a_{4}u_2+a_3,\]
\[G^0 =  \cosh(u_1)\ U_1(u_2)+U_2(u_2),\
 G^H=a_5,\]
 subject to the conditions (\ref{cond1})-(\ref{cond4}), again exactly the same as for flat space. Thus the system on the 2-hyperboloid  admits Painlev\'e VI and special Weierstrass $\wp$-function potentials for 3rd order superintegrability.

For our next example we consider horocyclic coordinates $\{u_1,u_2\}$ on the hyperboloid $s_1^2-s_2^2-s_3^2=1$, e.g. \cite[Section 7.7]{KKM2018}:
 \begin{equation}\label{horcoords} s_1=\frac12\bigg(u_1+\frac{u_2^2+1}{u_1}\bigg), \ s_2=\frac12\bigg(u_1+\frac{u_2^2-1}{u_1}\bigg), \ s_3=\frac{u_2}{u_1}.\end{equation}
These coordinates are separable and the canonical system is defined by $f_1(u_1)=1/u_1^2$, $f_1(u_2)=0$. We look for systems such that $V_1(u_1)=0$, in analogy with our first three examples.

We obtain the solution
\begin{eqnarray*}F^0 &=&- \frac12 a_8 \hbar^2u_1,\quad G^L= \frac{u_1^2(a_8 u_2+a_9)}{2}-\frac{a_8u_2^3}{6}-\frac{a_9u_2^2}{2}+a_{10}u_2,\\
G^0& =& \frac{u_1^2}{2}U_1(u_2)+U_2(u_2),\quad
 G^H=a_7,\end{eqnarray*}
 subject to the conditions
 \begin{eqnarray}&0=& a_8\frac{dV_2}{du_2}+2\frac{d^2U_1}{du_2^2},\label{hcond1}\\
 &0=& \frac12\hbar^2 a_8\frac{d^3V_2}{du_2^3}-4a_8\frac{dV_2}{du_2}V_2+4\frac{dV_2}{du_2}\frac{dU_1}{du_2},\label{hcond2}\\
  &0=&(2a_{10}-2a_9 u_2 -a_8u_2^2)\frac{dV_2}{du_2}-4(a_9+a_8 u_2)V_2+4U_1+4\frac{d^2U_2}{du_2^2},\label{hcond3}\\
 &0=&   -2\hbar^2 a_8 u_2^2\frac{dV_2}{du_2}+16(a_9+a_8 u_2)V_2^2-4(2a_{10}-2a_9u_2+a_8 u_2^2)\frac{dV_2}{du_2}V_2 \label{hcond4} \\
   &&\quad +\frac{\hbar^2}{2}(2a_{10}-2a_9 u_2-a_8u_2^2)\frac{d^3V_2}{du_2^3}-4\hbar^2(a_9+a_8 u_2)   \nonumber \\
   &&\quad -16V_2U_1+8\frac{dV_2}{du_2}\frac{dU_2}{du_2}.   \nonumber \end{eqnarray}
   There are again two basic cases here: \begin{enumerate} \item $a_8=0$. \\

Then conditions (\ref{hcond1}) and (\ref{hcond2}) say that $U_1$ is a constant: $U_1(u_2)=d_1$.  Then condition (\ref{hcond3}) can be solved for $U_2(u_2)$ and the result substituted into condition (\ref{hcond4}) to obtain an equation for $V_2(u_2)$:
\begin{equation}\label{horocyclicnonlinear}   -4a_9(\frac{dW}{du_2})^2+\left((-3a_9u_2+3a_{10})\frac{d^2W}{du_2^2} +4d_1\right)\frac{dW}{du_2}+\end{equation}
\[(-a_9 W+2d_1u_2-2d_3)\frac{d^2W}{du_2^2}
+\hbar^2a_9\frac{d^3W}{du_2^3}-\frac14\hbar^2(a_{10}-a_9u_2)\frac{d^4W}{du_2^4}=0,\]
where $ V_2(u_2)=\frac{dW(u_2)}{du_2}$.

With the integrating factor $\mu(u_2)=u_2 a_{9}-a_{10}$, the fourth order nonlinear differential equation admits the following first integral
\begin{eqnarray} J&=&\frac{1}{4}\hbar^{2}( a_{10}-a_{9} u_2)^{2} \frac{d^{3}W}{du_2^{3}} -\frac{\hbar^2}2(a_{10}a_{9}\hbar^{2}-a_{9}^{2}\hbar^{2})\frac{d^{2}W}{du_2^{2}} \\
&&-\frac32 ( a_{10}^{2} -2 a_{10}a_{9} u_2- a_{9}^{2} u_2^{2}) \bigg(\frac{dW}{du_2}\bigg)^{2}+  ( a_{10}a_{9}-a_{9}^{2} u_2) W \frac{dW}{du_2}\nonumber \\
&&+( 2 a_{10} d_{3} -\frac{a_{9}^{2} \hbar^{2}}{2} - 2 a_{10} d_{1} u_2 - 2 a_{9} d_{3} u_2 + 2 a_{9} d_{1} u_2^{2} ) \frac{dW}{du} \nonumber\\
&&+\frac{1}{2}a_{9}^{2}W^{2}+ 2 a_{9} d_{3} W - 2 a_{10} d_{1}. \nonumber \end{eqnarray}

We now consider two subcases.

\begin{enumerate}

\item $a_9\neq 0$. Using the following transformation of the independent and dependent variables
\begin{equation}
W(u_2) =    u_2  \tilde{W}(u_2) + b + c u_2^2 + d u_2^4, \quad z = -\frac{a_{10}}{a_{9}} + u
\end{equation}
and further using
\begin{equation}y = z^{2} \end{equation}
and using the constraint $d_3 = \frac{a_{10}d_{1}}{a_{9}}$ (for $a_{9} \neq 0$), we obtain a third order differential equation can be related to the Chazy I equation. This equation appears in the classification, \cite[Eq. A.3]{Cosgrove2000b}. It has the form
\[  \tilde{W}''' =-\frac{2}{f^{2}(y)} ( 3 c_{1} y ( y \tilde{W}' - \tilde{W})^{2} + c_{2} ( y \tilde{W}' -\tilde{W}) ( 3 y \tilde{W}'-\tilde{W})\]
\[+ c_{3} \tilde{W}' ( 3 y \tilde{W}' -2 \tilde{W}) + 3 c_{4} (\tilde{W}')^{2} + 2c_{5} y (y \tilde{W}'-\tilde{W}) \]
\[ + c_{6} ( 2 y \tilde{W}'-\tilde{W}) + 2c_{7} \tilde{W}' + c_{8} y +c_{9})- \frac{f'}{f} \tilde{W}'', \]
where
\[ f(y)= c_{1} y^{3} +c_{2} y^{2} +c_{3} y +c_{4}\]
Our case corresponds to the following choice of parameters:
\begin{eqnarray*}
 &&c_{1} = c_{2} =0 , \quad c_{3} = 2 a_{9}, \quad c_{4}= c_{5} = 0, \quad c_{6} = -2 (2 a_{9} c + d_{1}), \\
  &&c_{7} = \frac{1}{4} a_{9} (8 b - a_{9}),\quad  c_{8} = 2 (a_{9} c^{2} + c d_{1}),   \\
 &&c_9=\frac{1}{2} (-8 a_{9} b c - 4 b d_{1} - a_{9}^{2} c),\quad
 d = 0, \quad a_{9}= - \hbar^{2}.
 \end{eqnarray*}

This equation can be further integrated, \cite[Eq. A.21]{Cosgrove2000b} and the resulting equation takes the following form
\begin{eqnarray}
 (\tilde{W}'')^{2}&=&-\frac{4}{f^{2}}\big( c_{1} ( x \tilde{W}'- \tilde{W})^{2}+ c_{2} \tilde{W}' ( y \tilde{W}'-\tilde{W})^{2}+c_{3} (\tilde{W}')^{2} (y \tilde{W}'-\tilde{W}) \nonumber\\
 && + c_{4} (\tilde{W}')^{3} +c_{5}(y \tilde{W}' -\tilde{W})^{2}+ c_{6}\tilde{W}'(y\tilde{W}'-\tilde{W}) +c_{7} (\tilde{W}')^{2} \nonumber \\
 &&+c_{8} (y\tilde{W}'-\tilde{W})+c_{9}\tilde{W}' +c_{10}\big)
 \end{eqnarray}
where $c_{10}$ is an integration constant. This equation is known as SD-I and was discussed in \cite[Eq. 4.9]{Cosgrove1993}. It has 6 subcases: SD-Ia, SD-Ib, SD-Ic, SD-Id, SD-Ie, SD-If.
Here the non zero parameters are $c_{3}$, $c_{6}$, $c_{6}$, $c_{8}$ , $c_{9}$, according to \cite{Cosgrove2000b}. As described in \cite[Eq. 5.5]{Cosgrove1993} the equation is related to Painlev\'{e} III and Painlev\'{e} V equations. Explicit formulas are given in \cite[Eqs. 5.26-5.40]{Cosgrove1993}.

\item $a_{9}=0$. We obtain the third order equation
\[-2 d_{1} W + ( 2  d_{3} - 2  d_{1} u_2) W'
-\frac{3}{2} a_{10} W'^{2} + \frac{1}{4} a_{10} \hbar^{2} W'''=0,\]
which is a special case of \cite[Eq. 83]{MSW}:
\begin{eqnarray}
&& \alpha \hbar^{2} W''' - 6 \alpha W'^{2} - 4 ( c_{1} u^{3} - \alpha c_{2} u^{2}+b_{1} u +b_{0} )W' \nonumber\\
&&- 4 ( 3 c_{1} u^{2}- 2 c_{2} \alpha u +b_{1})W  -\frac{2}{3} c_{2}^{2} \alpha u^{4}
 + 4 ( \frac{1}{3} c_{2} b_{1} -c_{1}a_{2})u^{3} \nonumber \\
 &&-2( 3 c_{1} a_{1} -2 c_{2}b_{0})u^{2} + k_{2}u +k_{4}=0.
\end{eqnarray}
This equation admits the first integral
\begin{eqnarray}
J=&&\alpha \hbar^{2} W' -\alpha W^{2} - ( \alpha a_{1} - 2(b_{0}-\alpha a_{2}) u - 2 b_{1} u^{2} -\frac{2}{3} \alpha c_{2} u^{3}- 2c_{1}u^{4})W \nonumber\\
&& -\frac{1}{9} \alpha c_{2}^{2} u^{6} + \frac{1}{6} ( 3 a_{2} c_{1} -b_{1}c_{2}) u^{5} + (\frac{2}{3}(\alpha a_{2}c_{2}-b_{0}c_{2}) +a_{1}c_{1})u^{4} \nonumber \\
&& + (\frac{4}{3}\alpha a_{1} c_{2}-\frac{k_{2}}{4})u^{3}-\frac{1}{8}(\alpha k_{1} +4 k_{4})u^{2} +k_{5}u -\alpha \frac{k_{3}}{2} =0.
\end{eqnarray}
Then, a Cole-Hopf transformation $W=-\hbar^{2} \frac{U'}{U}$ gives a second order linear equation for $U$.

\end{enumerate}

\item $a_8\ne 0$. \\ \medskip  Here we can solve (\ref{hcond1}) for $V_2(u_2)$ and
substitute the result into (\ref{hcond2}) to obtain the equation
\begin{equation}\label{hcond5}\hbar^2\frac{d^3V_2}{du_2^3}-12V_2\frac{dV_2}{du_2}+12a_1\frac{dV_2}{du_2}=0, \end{equation}
where $a_1$ is an arbitrary constant.  Solutions of (\ref{hcond5}) are further subject to the requirement that a solution $U_2(u_2)$ of equations (\ref{hcond3}) and (\ref{hcond4}) exists. The general solution of
(\ref{hcond5}) is $V_2(u_2)=\hbar^2 \wp(u_2-u_{2,0};g_2,g_3)+a_1$, where $u_{2,0}$, $g_2$ and $g_3$ are arbitrary constants.

\end{enumerate}
\section{Classification of systems on the 2-sphere separating in spherical coordinates}\label{section5}
We use the coordinates (\ref{cansp}) with $x=\sin(\theta),\ y=\phi$ and list the systems that admit a 3rd order symmetry operator.
 They fall into 4 classes:

 \begin{enumerate} \item Systems that are 2nd order superintegrable.

  \bigskip

 These systems are all known and they are classical (all parameters in the potential are arbitrary).
  \item Systems that are neither 2nd or 3rd order superintegrable.

  \bigskip

  These are special cases of 2nd order superintegrable systems, except that they depend on $\hbar$, so they can't be scaled.

  \item Systems with algebraically dependent generators $A$ and $B$.
  \item Systems that are truly 3rd order superintegrable.
  \end{enumerate}

\subsection{Systems that are 2nd order superintegrable}

    \begin{equation}\label{system 27} V_1(x)=\frac{c_1\sqrt{1-x^2}}{x}+c_2,\quad V_2(y)=\frac{c_3}{\cos^2(y)}+\frac{c_4\sin(y)}{\cos^2(y)}.\end{equation}

 This is the 2nd  order superintegrable system $S_7$, \cite{KKMP2001}.
    \begin{equation}\label{system 1}V_1(x)=\frac{c_1}{1-x^2}+c_2,\ V_2(y)=\frac{c_3}{\cos^2(y)}+\frac{c_4}{\sin^2(y)}.\end{equation}
  This is the 2nd order superintegrable system $S_9$, \cite{KKMP2001}. It is characterized by the fact that all of the parameters $\alpha_j$ and $\beta_{jk}$ are zero.
  These are the only 2nd order 4-parameter superintegrable systems.

  \begin{equation}\label{system 30} V_1(x)=0,\quad V_2(y)=c\exp(-2iy).\end{equation}
This complex potential system admits a 3rd order symmetry and also a 1st order symmetry, so it is 2nd order superintegrable, the 2nd order system $S_5$, \cite{KKMP2001}.
 It is PT-symmetric so the energy eigenvalues are real.

  All other 2nd order systems are special cases of these. We will omit all classical special cases of these systems and include only purely quantum special cases.

\begin{enumerate}\item Quantum special cases.
  \begin{itemize}

 \item \begin{equation}\label{system 9} V_1(x)=\frac{\hbar^2}{c(1-x^2)}, \quad V_2(y)=0.\end{equation}
 This is a special case of $S_9$, \cite{KKMP2001},  but only quantum. The system admits a 1st order symmetry, so it is 2nd order superintegrable.
  \item \begin{equation}\label{system 18} V_1(x)=0,\quad V_2(y)=\frac{27 \hbar^2}{\cos^2(y)}.\end{equation}
 This system is  2nd order superintegrable and only quantum, a special case of $S_9$. It admits a 1st order symmetry.
\end{itemize}

\subsection{Systems that are  3rd order superintegrable but do not generate a cubic algebra}

   \begin{itemize}\item  \begin{equation}\label{system13}V_1(x)=0,\quad V_2(y)= \frac{-3\hbar^2(4\cos^4(y)-3)}{\cos^2(y)(2\cos^2(y)-3)^2}.\end{equation}
       This system is truly 3rd order superintegrable but only quantum.  The generators $\{H,A,B,C\}$ where $C=[A,B]$ do not close under commutation to
       form a cubic algebra. This is a special case of an isospectral deformation of the trigonometric Scarf potential, \cite{Quesne}. The bound state spectrum of $A$ is
\[ \lambda = 2 \hbar^2(\frac12 + \nu) ^2 ,\quad \nu=0, 1, 2, \ldots,\] and the $A$ eigenfunctions are proportional to Jacobi exceptional orthogonal  polynomials of degree $\nu+2$ in $\sin(y)$.
\end{itemize}
 \end{enumerate}

\subsection{Systems with  algebraically dependent generators $A$ and $B$}
These systems are not 2nd order superintegrable and $C=[A,B]=0$. Thus the set $\{H,A,B\}$ is commutative so (by Burchnall-Chaundy theory), \cite{BurchnallChaundi}  there must be  a algebraic relation between $A$ and $B$ that we can compute. This permits us to exhibit an explicit first integral for the eigenfunctions of $A$. Thus these systems are special. We give the details for our first example. The other cases are similar.
    \begin{itemize}
     \item \begin{equation}\label{system 25} V_1(x)=0,\quad V_2(y)=\frac{4 \hbar^2}{\cos^2(2y)},\end{equation}
 Here $B$ is the formally self-adjoint operator
 \begin{equation}\label{B25} B=i\hbar^3\partial_y^3+8i\hbar^3\frac{2\cos^4(y)-2\cos^2(y)-1}{(2\cos(y)^2-1)^2}\partial_y-\frac{48i\hbar^3\sin(y)\cos(y)}{(2\cos(y)^2-1)^3}.\end{equation}
  The relationship is
 \[  A^3-
 \frac18 B^2-4 \hbar^2 A^2 +4 \hbar^4 A =0.\]
 \end{itemize}

  However, since $C=0$, the formally self-adjoint operators $B$ and $A$ admit  common eigenfunctions $g(y)$:
 \[ Bg=\mu g,\quad  Ag=\lambda g,\quad \lambda^3-\frac18\mu^2-4\hbar^2\lambda^2+4\hbar^4\lambda=0.\]
 Solving for $\mu$  we find
 \[ \mu=\pm 2\sqrt{2}(\lambda-2\hbar^2)\sqrt{\lambda}.\]
   Now consider the equation
 \[ (Bg-\mu g)+a\frac{d}{dy}(Ag-\lambda g)=0 \]
for  any constant $a$. Choosing $a=2i\hbar$ we eliminate the 3rd derivative term in $y$ and obtain the 1st order differential equation
\begin{align} \label{A25} &-\frac{2i\hbar}{g(y)}\frac{d g(y)}{dy}= \\
&\frac{-8\cos^6(y)\mu+16i\cos(y)\sin(y)\hbar^3+12\cos^4(y)\mu-6\cos^2(y)\mu+\mu}{(2\cos^2(y)-1)(8\cos^4(y)\hbar^2-4\cos^4(y)\lambda-8\cos^2(y)\hbar^2+4\cos^2(y)\lambda-\lambda)}, \nonumber
\end{align}
into which we substitute the two possibilities for $\mu$. These are explicit first integrals for the 2nd order differential equations satisfied by the eigenfunctions of $A$.

\medskip

 {\bf Another example}:
 \begin{itemize}
     \item \begin{equation}\label{system 15} V_1(x)=0,\quad V_2(y)=\frac{N(y)}{D(y)},\end{equation}
       where {\small
       \begin{eqnarray*} N(y)&=&\\
      &-& 4\hbar^2\left(2\sqrt{3}\sin(y)\cos^3(y)+2\cos^4(y)+18\sqrt{3}\cos(y)\sin(y)\right.\\
      && \left. -21\cos^2(y)-81\right),\\
        D(y)&=&\\
        &16&\cos^6(y)+72\sqrt{3}\sin(y)\cos^3(y)+48\cos^4(y)+108\sqrt{3}\cos(y)\sin(y)\\
        &-&207\cos^2(y)+243.
       \end{eqnarray*}}
  This system admits a 3rd order symmetry $B$, but then $C=0$. Thus $A$ and $B$ are algebraically dependent and the system is not 3rd order superintegrable. The algebraic relationship between $A$ and $B$ is
  \[A^3-\frac{1}{8}B^2-\hbar^2 A^2+\frac{i\sqrt{3}\hbar^3}{3} B +\frac{\hbar^4}{4} A+\frac23\hbar^6 =0.\]
  \end{itemize}
  {The general $C=0$ case}
  \begin{itemize}
     \item \begin{equation}\label{system31} V_1(x)\ {\rm arbitrary}, \
\hbar^2\bigg(-\frac{dV_2}{dy} \frac{d^4V_2}{dy^4}+\frac{d^2V_2}{dy^2}\frac{d^3V_2}{dy^3}\bigg)+12\bigg(\frac{dV_2}{dy}\bigg)^3 = 0.\end{equation}
 This differential equation admits the Weierstrass $\wp$-function as a solution. The system admits a 3rd order symmetry provided the additional condition
 \begin{equation} \label{system31a} \hbar^3\frac{d^3V_2}{dy^3}-12\hbar V_2\frac{dV_2}{dy}-4z_1\frac{dV_2}{dy}=0 \end{equation}
 is satisfied for some constant $z_1$. However, this condition is always satisfied since (\ref{system31a}) is a first integral of (\ref{system31}).  Equation (\ref{system31a}) implies $C=0$, so that $A$ and $B$ are algebraically dependent. Indeed, they obey the relation
 \[ -B^2+A^3+\alpha A^2=0,\]
 where
 \[-i B=\ \frac{3\hbar}{2}\frac{dV_2}{dy}-\hbar^3\partial_y^3+\left(3\hbar V_2+\alpha\hbar\right)\partial_y.\]
  Thus this general system is not 3rd order superintegrable.
 \end{itemize}

There are also  elementary function solutions of the nonlinear defining equation, none of which lead to 3rd order superintegrability.
 In particular the following systems are solutions:
\begin{itemize} \item \begin{equation}\label{system 33} V_2(y)=\hbar^2\left(\frac{1}{\sin^2(y)}+\frac{1}{\cos^2(y)}\right),\,\end{equation}
 This can be considered as a 0-parameter potential  within $S_9$. It has a 2-parameter 3rd order symmetry, $\beta_{31},\alpha_3$,  but if $\beta_{31}\ne0$ the   system doesn't close to a cubic algebra. If $\beta_{31}=0,\alpha_3 \ne 0$ we find $C=0$. Then $A$ and $B$ are related by
 \[ A^3-\frac18 B^2-4 \hbar^2 A^2 +4 \hbar^4 A=0.\]
 \end{itemize}
 \begin{itemize}
 \item \begin{equation}\label{system 34} V_2(y)=\frac{4\hbar^2}{(2\sin^2(y)-1)^2}.\end{equation} This system admits a 3rd order symmetry but $C=0$ so $A$ and $B$ are algebraically dependent:
 \[ A^3-\frac18 B^2-4 \hbar^2 A^2 +4 \hbar^4 A =0.\]
 \item
 \begin{equation}\label{system35} V_2(y) = \frac{\hbar^2(4b^2+c^2)}{(2b\cos^2(y)\pm c\sin(y)\cos(y)-b)^2}, \end{equation}
 This can be obtained from the Weierstrass $\wp$-function solution.
 The system admits a 3rd order symmetry but $C=0$ so $A$ and $B$ are algebraically dependent. The relationship between $A$ and $B$ is
 \[ A^3-\frac18 B^2-4 \hbar^2 A^2 +i
\frac{48 b^3 \hbar^3+12 b c^2 \hbar^3}{c^3}
 B +4\hbar^4 A +288\frac{(4b^2+c^2)^2 b^2\hbar^6}{c^6}=0.\]
\end{itemize}

\subsection{Systems that are truly 3rd order superintegrable}

In addition to the nonlinear solution (\ref{nonlinearsoln}) we have:
    \begin{itemize}
     \item   \begin{equation}\label{system 20} V_1(x)=0,\quad V_2(y)=\frac{c}{(4\cos^2(y)-3)^2\cos^2(y)}.\end{equation}

This system is truly 3rd order superintegrable.
 It satisfies a cubic algebra of the form.
\begin{equation}\label{[A,C]} [A,C]=\alpha A^2+\beta\{A,B\} +\gamma A+ \delta B+ \epsilon,\end{equation}
\begin{equation}\label{[B,C]} [B,C]= \mu A^3+\nu A^2-\beta B^2-\alpha \{A,B\}+\xi A-\gamma B+\zeta,\end{equation}
where
\[ \gamma=\gamma_0+\gamma_1 H,\quad \delta=\delta_0+\delta_1 H,\quad \epsilon=\epsilon_0+\epsilon_1 H+\epsilon_2 H^2,\]
\[ \nu=\nu_0+\nu_1 H,\quad \xi=\xi_0+\xi_1 H+\xi_2 H^2,\quad \zeta=\zeta_0+\zeta_1 H+\zeta_2 H^2+\zeta_3 H^3.\]
\end{itemize}

 The final closure relation is
 \begin{eqnarray} C^2=
  &&\frac{9\hbar^2}{2}\{A,B^2\}+576\hbar^2 A^4+9\hbar^2 BAB -576\hbar^2 H^3A+\nonumber\\ &&1728\hbar^2H^2A^2-1728\hbar^2HA^3-\frac{243\hbar^2}{4}B^2+(16704\hbar^4-576c\hbar^2)A^3\nonumber\\
  &&+(10080\hbar^4-1728c\hbar^2)H^2A+(-26784\hbar^4+1728c\hbar^2)HA^2\nonumber\\
  &&+576c\hbar^2 H^3+(27792\hbar^6-8928c\hbar^4)A^2-2304c\hbar^4H^2\nonumber\\
  &&+(-17280\hbar^6+11232c\hbar^4)HA+1728c\hbar^6H+(2592\hbar^8-5760c\hbar^6)A.\nonumber
 \end{eqnarray}
 The possible spectra for $H$ can be computed from these algebraic relations.

\section{Systems with  algebraically dependent generators as degenerations of $\wp$-potentials}
\label{section6}
The systems with algebraically dependent generators $A$ and $B$ can all be obained as degenerations of $\wp$-potentials.
The Weierstrass $\wp$-function with invariants $g_2,g_3$ satisfies the 1st order, nonlinear differential equation
\begin{equation}\label{deq_1}
(\wp')^2=4\wp^3-g_2\wp-g_3.
\end{equation}
We can parameterize the invariants $g_2$ and $g_3$ as
\begin{equation}
g_2=2(e_1^2+e_2^2+e_3^2),\quad g_3=4e_1e_2e_3,
\end{equation}
so that
\begin{equation}\label{deq_2}
(\wp')^2=4(\wp-e_1)(\wp-e_2)(\wp-e_2),\quad e_1+e_2+e_3=0.
\end{equation}
When two of the roots $e_i$ coincide, (\ref{deq_2}) can be integrated in terms of elementary functions. We take $e_1=e_2$ so that $e_3=-2e_1$. The general solution of
\begin{equation}\label{deq_3}
(\wp')^2=4(\wp-e_1)^2(\wp+2e_1)
\end{equation}
is
\begin{equation}\label{hyperbolic_degeneration}
\wp(z-z_0;12e_1^2,-8e_1^3)=3e_1\mathrm{csch}^2\big(\sqrt{3e_1}(z-z_0)\big)+e_1
\end{equation}
or, making the replacement $e_1\to-e_1$,
\begin{equation}\label{trig_degeneration}
\wp(z-z_0;12e_1^2,-8e_1^3)=3e_1\mathrm{csc}^2\big(\sqrt{3e_1}(z-z_0)\big)-e_1.
\end{equation}
Either of these expressions degenerates to a rational solution as $e_1\to 0$:
\begin{equation}\label{rational_degeneration}
\wp(z-z_0;0,0)=\frac{1}{(z-z_0)^2}.
\end{equation}

\subsection{Potentials from third-order superintegrability}

On the $2$-sphere we have the potential
\begin{equation}
V_2(u_2)=\hbar^2\wp( u_2-u_{2,0};g_2,g_3)+a_1
\end{equation}
Only the trigonometric degeneration is relevant here:
\begin{equation}
V_2(u_2)=\hbar^2\kappa^2\mathrm{csc}^2(\kappa( u_2-u_{2,0}))+\frac{\kappa^2}3+a_1,
\end{equation}
for an arbitrary real parameter $\kappa$.

On the $2$-hyperboloid we have the potential
\begin{equation}
V_2(u_2)=\hbar^2\wp( u_2-u_{2,0};g_2,g_3)+a_1.
\end{equation}
There are three degenerate potentials:
\begin{enumerate}
    \item $V_2(u_2)=\hbar^2\kappa^2\mathrm{csch}^2(\kappa( u_2-u_{2,0}))+\frac{\kappa^2}3+a_1$
    \item $V_2(u_2)=\hbar^2\kappa^2\mathrm{csc}^2(\kappa( u_2-u_{2,0}))+\frac{\kappa^2}3+a_1$
    \item $V_2(u_2)=\hbar^2 (u_2-u_{2,0})^{-2}+a_1$,
\end{enumerate}
where $\kappa$ is an arbitrary complex parameter.

\section{Higher order FD-superintegrable systems}\label{section7}
    A standard superintegrable system    on an $n$-dimensional Riemannian manifold (real or complex) is a system that possesses $2n-1$ functionally independent constants of the motion in the classical case and $2n-1$ algebraically independent symmetry operators in the operator case.
    If a system possess $2n-1$ linearly independent symmetries but they are not functionally (or algebraically) independent, we will call it {\it Functionally Dependent - superintegrable}, (or FD-superintegrable for short).   (These are to be distinguished from Functionally Linearly Independent systems, \cite{BKM2020}, of which there is only one example in two dimensions)

 In this paper we have found a number of FD-superintegrable systems of a rather simple type: $C=0$. For these systems the operators $A$ and $B$ commute. Further, they are ordinary differential operators in the variable $y$ alone. The Abelian algebra generated by $\{H,A,B\}$ does have structure because $A$ and $B$ satisfy an algebraic equation. It is clear, moreover that the machinery constructed here will work almost unchanged on any manifold with metric of the form $ds^2= dx^2+F(x) dy^2$, not just those of constant curvature.

Consider the 1D Hamiltonian
\[A= -\frac{\hbar^2}{2} \frac{d^2}{dy^2}f(y)+V_0(y)f(y)\]
and the symmetry operator
\[B=W_0(y)+W_1(y)\frac{d}{dy}+W_2(y)\frac{d^2}{dy^2}+W_3(y)\frac{d^3}{dy^3}.\]
We require that $[A,B]=0$ and that $B$ is linearly independent of $A$.
The solution, unique up to a constant factor, is
\begin{equation} B=\frac{d^3}{dy^3}-\frac{-b_1\hbar^2+3V_0(y)}{\hbar^2} \frac{d}{dy}-\frac{3}{2\hbar^2}\frac{dV_0(y)}{dy}, \end{equation}
subject to the condition
\begin{equation}\label{system31A} -4b_1\frac{dV_0}{dy}\hbar^2-\frac{d^3V_0}{dy^3}\hbar^2+12V_0\frac{dV_0}{dy}=0.\end{equation}
This can be identified with equation (\ref{system31a}), so the treatment of (\ref{system31}) carries over immediately to this 1D system.

To summarize, the differential equation
\begin{equation}\hbar^2\bigg(-\frac{dV_2}{dy} \frac{d^4V_2}{dy^4}+\frac{d^2V_2}{dy^2}\frac{d^3V_2}{dy^3}\bigg)+12\bigg(\frac{dV_2}{dy}\bigg)^3 = 0. \label{4th_order_ode}\end{equation}
can be integrated in terms of the Weierstrass $\wp$-function. This equation passes the Painlev\'{e} test with movable double poles as the only singularities. We can find the general solution.

This construction can be extended to general $n$. We present an example that is pertinent to 5th order FD-superintegrability  for the 2-sphere. The 5th order operator that commutes with $A$ is
\begin{align}
 B=& \frac{d^5}{dy^5}+c_4\frac{d^4}{dy^4}-\frac{5V_0}{\hbar^2}\frac{d^3}{dy^3}
 -\frac{(-2c_2\hbar^2+8c_4V_0+15\frac{dV_0}{dy})}{2\hbar^2}\frac{d^2}{dy^2}-\\
 &\frac{-4c_1\hbar^4+16c_4\frac{dV_0}{dy}\hbar^2+25\frac{d^2V_0}{dy^2}\hbar^2-30V_0^2}{4\hbar^4}\frac{d}{dy}-\nonumber\\
 &\frac{-8c_0\hbar^4+16c_4\frac{d^2V_0}{dy^2}\hbar^2+16c_2\hbar^2V_0+15\frac{d^3V_0}{dy^3}\hbar^2-32V_0^2c_4-60V_0\frac{dV_0}{dy}}{8\hbar^4},\nonumber
\end{align}
  subject to the condition
  \begin{eqnarray} \label{5thordercond}16c_1\frac{dV_0}{dy}\hbar^4+\frac{d^5V_0}{dy^5}\hbar^4-20V_0\frac{d^3V_0}{dy^3}\hbar^2-40\frac{dV_0}{dy}\frac{d^2V_0}{dy^2}\hbar^2+120V_0^2\frac{dV_0}{dy}=0.
  \end{eqnarray}
    Here, $A$ and $B$ are related by the expression
    \begin{eqnarray}\label{algrel} B^2+a_1A^5+a_2BA^2+a_3A^4+a_4BA+a_5A^3+a_6B+a_7A^2+a_8A=0,\end{eqnarray}
   where
  \[ a_1=\frac{32}{\hbar^{10}},\ a_2=-\frac{8c_4}{\hbar^4},\ a_3=\frac{16c_4^2}{\hbar^8},\ a_4=\frac{4c_2}{\hbar^2},\]
  \[ a_5=\frac{c_1-c_2c_4}{\hbar^6},\ a_6=-2c_0,\ a_7=\frac{-k_1}{2\hbar^{10}},\]
and $V_0(y)$ satisfies the equation
\begin{eqnarray}\label{5thordercond2} 48V_0^5&&+(64c_1\hbar^4+80\hbar^2\frac{d^2V_0}{dy^2}V_0^3+(-192c_4c_0\hbar^6-96c_2^2\hbar^6- \\
&&120\hbar^2\frac{dV_0}{dy}^2-12k_1)V_0^2+(-20\frac{d^2V_0}{dy^2})^2\hbar^4+32\hbar^6c_1\frac{d^2V_0}{dy^2}) \nonumber\\
&&-96\hbar^8(a_8\hbar^2+4c_0c_2-2c_1^2))V_0+(-16\hbar^6c_1+4\hbar^4\frac{d^2V_0}{dy^2})\frac{dV_0}{dy}^2\nonumber\\
&&-2\hbar^2(16c_0c_4\hbar^6+8c_2^2\hbar^6+k_1)\frac{d^2V_0}{dy^2}+128c_0c_1c_4\hbar^{10}+64c_1c_2^2\hbar^{10}\nonumber\\
&&-64c_0^2\hbar^{10}+(\frac{d^3V_0}{dy^3})^2\hbar^6+8c_1\hbar^4k_1-a_{10}=0\nonumber.\end{eqnarray}
Here (\ref{5thordercond2}) is an integrated form of (\ref{5thordercond}). See \cite{BurchnallChaundi} for the theory behind these results.

However, since $[A,B]=0$, the  formal operators $B$ and $A$ admit common eigenfunctions $g(y)$:
 \[ Bg=\mu g,\quad  Ag=\lambda g,\]
and
\begin{eqnarray}\label{algrel1}\mu^2+a_1\lambda^5+a_\mu\lambda^2+a_3\lambda^4+a_4\mu\lambda+a_5\lambda^3+a_6\mu+a_7\lambda^2+a_8\lambda=0,\end{eqnarray}
a quadratic equation for $\mu$ as a function of $\lambda$. Differentiating the equation $Ag=\lambda g=0$ repeatedly we can derive linear equations for $\frac{d^5 g}{dy^5}, \frac{d^4 g}{dy^4}, \frac{d^3 g}{dy^3}, \frac{d^2 g}{dy^2}$, as functions of $\frac{dg}{dy},g$. Substituting these results into the equation $Bg-\mu g=0$ we obtain a first order differential equation for $g(y)$. Thus this construction enables one to express the eigenfunctions of $A$ in terms of a single integral.

 The equation (\ref{5thordercond}) passes the Painleve test. It appears in the  Cosgrove list, \cite{Cosgrove2000} and is equivalent to
 \[ u^{(5)}=20 u u^{(3)}+40 u' u'' -120 u^2 u'+\alpha u'\]
 via the change of variables
 \[ V_0(y)=u(x),\quad y=\hbar x,\quad c_1=-\frac{\hbar^5}{16}\alpha.\]
 This is Cosgrove’s equation Fif-III \cite[Eq. 2.71]{Cosgrove2000} with $\lambda=\kappa=0$.
The equation can be integrated once to
\[ u ^{(4)} = 20 u u '' + 10(u ' )^2 - 40u^ 3 + \alpha y + \beta.\]
This is Cosgrove's equation F-V \cite[Eq. 1.7]{Cosgrove2000} with $\kappa  = 0$.
 The general solution is in terms of genus two hyperelliptic functions, but there are particular solutions in terms of elliptic functions and degenerations. This Burchnal-Chaundy construction  occurs in many fields, see for example the connection with the KdV hierarchy, equation \cite[Eq. 2.11]{GRT}. Here we are pointing out its significance in superintegrability theory.

\medskip

For all systems with $C=0$ we can use this method to reduce the solution of the $A$ eigenvalue equation to a single quadrature. The method also extends to systems where $B$ is of arbitrarily high order, $A$ is 2nd order and $C=0$, and always yields stucture equations relating $A$ and $B$ and a solution for the eigenvalues of $A$ up to a single quadrature.

  \section{The 2-hyperboloid in horocyclic coordinates} \label{section8}

We classify 3rd order superintegrable systems on the 2-sheet hyperboloid $s_1^2-s_2^2-s_3^2=1$ with horocyclic coordinates. In this section, we set $\hbar=\sqrt{2}i$ for simplicity. Then, with horocyclic coordinates $u_1,u_2$, (\ref{horcoords}), a separable Hamiltonian takes the form
\begin{equation}
H=u_2^2\left(\partial_{u_1}^2+\partial_{u_2}^2\right)+u_2^2(V_1(u_1)+V_2(u_2)).
\end{equation}
and
\begin{equation}
A=\partial_{u_1}^2+V_1(u_1)
\end{equation}

We obtain the following systems.

 \subsection{Systems that are 2nd order superintegrable}

 \begin{itemize}
 \item
 \[ V_1(u)=c_1+c_2 u_1+4c_3 u_1^2,\quad V_2(v)=c_3 u_2^2 \]
 This is the 2nd order superintegrable system $S1$, \cite{KKMP2001}.
 \item
  \[ V_1(u)=\frac{c_1}{u_1^2}+c_2 +c_3 u_1^2,\quad V_2(v)=c_3 u_2^2 \]
  This is the 2nd order superintegrable system $S2$, \cite{KKMP2001}.
 \end{itemize}

 \subsection{Systems with  algebraically dependent generators $A$ and $B$.}

 {\bf A truly 3rd order system, but no cubic algebra.}

\medskip

  Here, $V_2(v)=0$ and $V_1(u)$ satisfies the nonlinear equation
\begin{eqnarray}
&&8V_1+4u_1V_1'+6cV_1V_1'+cV_1'''=0 \nonumber
\end{eqnarray}
It admits the first integral
\[ V_1''=-\frac{4V_1^3c^2+16V_1^2cu_1-c^2(V_1')^2+16u_1^2V_1-4V_1'c+2c_1c}{2c(cV_1+2u_1)}. \]
This equation does not satisfy the Painlev\'e property.
The structure equation is
\begin{equation}
[A,B]=2A=C
\end{equation}
which gives no information about the eigenvalues of the Hamiltonian, but shows that $A$ is a raising operator for the eigenvalues of $B$. Here,
\[ B = \frac34cV_1 +\left(  u_1 +\frac32 cV_1\right)\partial_{u_1} +  u_2\partial_{u_2}+c\partial_{u_1}^3.\]
 {

{\bf Systems with  $C=0$}

\medskip

  There are formally the same functionally superintegrable solutions for the hyperboloid as for the sphere, except that now $V_2(u_2)$ is arbitrary, and the ranges of the variables $u_1,u_2$ differ from those of $x,y$.

 \subsection{Systems that are truly 3rd order superintegrable.}

Here, we have not classified the  quantum special cases, i.e. the quantum $\hbar$-systems that are special cases of classical superintegrable systems.
\medskip

 {\bf The TTW solution}:

\medskip

The TTW method applied to the system $S7$ with horocyclic coordinates leads to an infinite family of higher order superintegrable systems, and exactly one  system  is 3rd order superintegrable:
\begin{equation}
U(u)=U_0+U_1 u_1+U_2 u_1^2, \quad V(v)=U_2 u_2^4.
\end{equation}
Thus $A=\partial_{u_1u_1}^2+U_0+U_1 u_1+U_2 u_1^2$.
We can verify that the Hamiltonian with this potential is third order superintegrable with
\begin{align}
B=&\frac{2u_2U_2}{U_1}\partial_{u_1 u_1}^2\partial_{u_2}+\frac{(2U_2u_1+U_1)}{U_1}\partial_{u_1}^3
-\left(\frac{2U_2^2u_1^2}{U_1}+2U_2u_1+\frac12 U_1\right)u_2\partial_{u_2}\\
+&\frac{3U_1^2u_1+(4U_2u_2^2+6U_2u^2-U_1u_1+2U _0)U_1+ 8u_1(U_2u_2^2+\frac12 U_2u_1^2+\frac12 U_0)U_2}{2U_1}\partial_{u_1} \nonumber \\
+&\left(\frac{2u_2^2U_2^2}{U_1}+\frac{2U_2^2u_1^2}{U_1}+2U_2u_1\right)\nonumber
\end{align}

\medskip

{\bf The structure equations}:

\medskip

Then, with $C=[A,B]$ we have
\begin{eqnarray}{[A,C]} &=&-16\ U_2 B-\frac{32\ U_2^2}{U_1} A+\frac{4U_2(4U_0U_2-3U_1^2)}{U_1},\nonumber\\
{ [B,C]} &=&a_1 A^3+a_2A^2+a_3HA+a_4B+a_5A+a_6H+a_7,\nonumber\end{eqnarray}

with
\begin{eqnarray} a_1&&=\frac{ -32U_2^2}{U_1^2},\ a_2 = \frac{12U_2(4 U_0U_2-U_1^2)}{U_1^2},\ a_3 = \frac{-64U_2^3}{U_1^2},\  a_4 = \frac{32U_2^2}{U_1},\nonumber \\
 a_5&&= -\frac{(16U_0^2U_2^2-8U_0U_1^2U_2+U_1^4-160U_2^3)}{U_1^2 },\   a_6  = \frac{16U_2^2(4U_0U_2-U_1^2)}{U_1^2 }    \nonumber           \\
 a_7&&= \frac{-4U_2^2(20U_0U_2-9U_1^2)}{U_1^2}.\nonumber\end{eqnarray}
 \begin{eqnarray}C^2&&=b_1\{A,B^2\} +b_2A^4+b_3BAB
 +b_{4}B^2+b_{5}A^3+b_{6}HA^2+b_{7}A^2\nonumber
 \\
 &&+b_{8}HA+b_{9}A+b_{10}H+b_{11}\{A,B\}+b_{12},  \nonumber
 \end{eqnarray} where we have omitted the constants $b_j$.

The eigenvalues of $A$ are of the form $\lambda=-4\sqrt{U_2}\ n+{\rm constant}$, linear in $n$.
Note that this system is St\"ackel equivalent to a 3rd order Euclidean space superintegrable system in Cartesian coordinates.

\medskip

{\bf The remaining 3rd order nonlinear superintegrable systems.}

\medskip

These are the  systems treated in (\ref{horocyclicnonlinear}).

\section{Discussion and conclusions} \label{section9} We have developed a new approach to the classification of  higher order superintegrable systems on any 2D manifold that admit a separation of variables, not just Euclidean space. As a check we have  reproduced the striking Tremblay and Winternitz result that for Euclidean space and polar coordinates one of the 3rd order superintegrable  potentials can be expressed in terms of Painlev\'e VI.  As new results we show that  Painlev\'e VI also appears for 3rd order superintegrable systems on the 2-sphere and the 2-hyperboloid,  this occurrence is common for all constant curvature spaces.  We  derived the possible separable systems on the 2-sphere in spherical coordinates and the two sheet 2-hyperboloid in horocyclic coordinates that admit a third order symmetry operator and provide information about their symmetry algebras. Our aim is  to develop a convenient tool that can be used study higher order superintegrability on spheres, hyperboloids and more general Riemannian spaces, and to compare results between spaces.

\section*{Acknowledgments}
We thank Pavel Winternitz for helpful discussions and Adrian Escobar for pointing out the
relevance of the paper \cite{KKM2010} to classification of 3rd order superintegrable systems.
W.M. was partially supported by a grant from the Simons Foundation (\# 412351 to Willard Miller, Jr.).
B.K.B.\ acknowledges support from the G\"oran Gustafsson Foundation and the European Research Council, Grant Agreement No. 682537. \ I.M. was supported by the Australian Research Council Discovery Grant DP160101376 and Future Fellowship FT180100099.

config: arxiv


TITLE: How is Bayesian inference better than classical inference on small samples?
QUESTION [9 upvotes]: I have heard many times that Bayesian inference is better than classical inference for small samples.
Why is it the case? In what kind of problems can I see this difference?

REPLY [5 votes]: This is necessarily an opinion-based question. There is vast variety opinion
among statisticians about the value of the Baysian approach to statistical
analysis--ranging from dismissively hostile to mindlessly evangelical.
I am mainly an applied statistician, so I am willing to use whatever works,
and in some settings I feel the Bayesian approach works better.
More directly to your question, the assertion that Bayesian inference works
better than classical frequentist inference probably arises from the fact
that Bayesian inference allows prior experience and expert opinion to be
used in formulating a prior distribution. Both the
prior distribution and the data are used to get the final result. The prior information may be especially important when there is not much data.
Almost all experiments and surveys arise in a context where there is some
prior knowledge. Otherwise, planning the details of the study would be
difficult.

If items (specimens or subjects) are to be weighed, we usually know whether we will
use a laboratory balance, a postal scale, a bathroom scale, or a truck scale
to do the weighing.
A preliminary power computation to plan how many items
to use requires educated guesses (a) of variability and (b) of how large an effect
needs to be, in order to have practical importance.
Also, one may have some
idea in advance whether data will be normal, and that is important information.

So even frequentist
statisticians make some use of prior experience when planning an experiment.
The main distinction is that a Bayesian study will begin with a formal
statement of some of that prior knowledge in terms of a prior probability distribution.

Here is a specific (overly simple)  example. Suppose we are seeking a preliminary guess whether
a particular ballot proposition will win in an upcoming election. At this early
juncture, suppose we have been able to poll only $n = 20$ randomly chosen people.
Suppose 4 are in favor and 16 are opposed. A traditional 95% frequentist confidence interval based on that information would be something like
$(.05, .40)$ (with slight differences, depending whether a Wald or Agresti CI is used).
However, very few US elections result in such a one-sided outcome. Propositions do not get on the ballot without many signatures on petitions, and thus they must have at
least minimal support. Even
with limited knowledge of the specific circumstances, one might feel comfortable
using the prior distribution $\mathsf{Beta}(3,3)$ for the proportion $\theta$
in favor. This distribution puts about 95% of its probability between $.15$ and $.85.$
This prior distribution together with the likelihood function for the
data (4 in favor out of 20), gives a 95% Bayesian probability interval
of $(0.12, 0.45).$ There is so little information in the "survey" of 20
people, that the prior distribution has had a major effect on the resulting
interval estimate. I would feel more comfortable with the more optimistic
Bayesian interval estimate than with the frequentist one. I have about as much
 faith in the general knowledge about US elections as in the
information from 20 subjects.
Note: There is an important difference in interpretation between frequentist CI's and
Bayesian PI's. The former 'confidence' intervals speak only of the process used to derive them, without directly addressing the survey at hand. The
Bayesian 'Posterior probability' intervals directly address the current survey, partly because the prior distribution is considered relevant. However, this
difference between frequentist and Bayesian inference is quite general and
is not restricted to the small sample sizes of your question.


TITLE: Maximise the earnings
QUESTION [0 upvotes]: Imagine you have two boxes, of which both are empty. Also, let's say you have a 100 turns. In each turn, you have two options:

You decide to place a coin :- A box will be randomly chosen, and you have to place a coin in that box. Assume choosing any box is equally likely.
You decide to choose a box:- A box will be randomly chosen, and you will take all the coins from the box and add them to your earnings. Again, both boxes have equal probability of being chosen.

Devise a strategy to maximise the earnings and estimate the number of coins won.
I tried solving the question by writing an expression for the expected value of the amount of coins won but couldn't make much progress beyond that.

REPLY [3 votes]: First, if you choose to put coins before pull coins, the expectation should be improved if you swap these two actions. Therefore, the best strategy must be first putting $100-k$ coins and then pulling $k$ times.
Next, let $X_1$ and $X_2$ be the number of coins in boxes, we have $X_1+X_2=100-k$. As we pull $k$ times, we have $1/2^k$ probability that getting only $X_1$, $1/2^k$ probability that getting only $X_2$, and $1-2/2^k$ probability that getting $X_1+X_2$, the expectation is hence $(1-1/2^k)(100-k)$.
Finally, we search for the best $k$ (or you can take the derivative and then try the integer nearby), and you should have $k=6$ with the maximum expectation $2961/32$.


TITLE: Solve $x^2+y^2=199^2$
QUESTION [1 upvotes]: Question :

Find the number of integral points on the circle with centre $(199,0)$ and radius $199$.(Integral point is a point $(x,y)$  with both $x,y \in \mathbb{Z}$.

My Approach : Since no. of integral coordinates won't change on shifting of origin : I shifted origin to $(199,0)$. Now the equation of circle becomes $x^2+y^2=199^2$.
But how do I solve this equation.I can conclude that one of $x$ and $y$ is odd and other is even, since some of their squares is odd.
Trivial solutions are $(\pm 199,0)$ and $(0,\pm199)$ .
But how do I prove that the these are the only solutions, or others also exist, and if more points exist, how to find them.
Thanks.

REPLY [1 votes]: Without using any theory of Pythagorean triples here is an approach from first principle:
Note that $199$ is a prime number, hence gcd$(x,y)=1$. One among $x,y$ should be even and the other odd. Assume $x$ is odd. Now given condition means
$ x^2= (199+y)(199-y)$. So for any prime $p$ dividing $x$ we have either  $p^2$ divides $(199+y)$ or $p^2$ divides $(199-y)$, but not both. (otherwise, it will divide their difference $2y$, and being odd will divide $y$, contradicting the fact that $x,y$ are relatively prime).
SO we can conclude that $x^2=(199+y)(199-y)$ is a factorization of $x^2$ into two squares which are equidistant from 199. This can be verified to be false proceeding as below:
Look at all perfect squares less than 199 (e.g., 169, 144 etc, these are the candidates for $199-y$) Check that corresponding $199+y$ are not squares, e.g. $199+30, 199+55$ are not squares.
The advantage now is instead of looking at squares upto $199^2$, we reduced to checking with squares upto $199$.


TITLE: How to distinguish trial and experiment?
QUESTION [4 upvotes]: How to distinguish trial and experiment in probability? I have checked Ross's definition and wikipedia's intro on the definition of them for a while, but not quite get it till now. What is the difference of them? And when it comes to Bernoulli trail or Bernoulli experiment, do we call Bernoulli trial, or Bernoulli trials, or Bernoulli experiment, or Bernoulli experiments?

REPLY [0 votes]: Experiment is the some activity whose outcomes are sometimes known to us and sometimes not. For example, "After many experiments Edison was able to invent an electric bulb. This was an experiment in which were not having any idea about what will be the outcome. Experiment in which we know that our outcome will from some possible out comes that are known to us, for example, when we roll a dice, we know that, the outcome will be either from "1,2,3,4,5,6". There will be know outcome other than these six numbers. We call such experiments as random experiments.

Trial- When we repeat a experiment many times, then each of that performed experiment is known as trail.

Use following link for reference-

https://www.cuemath.com/data/terms-of-probability/


TITLE: Show that if $G$ is $2-$connected, then for any two vertices $u$ and $v$ there exists a cycle $C$ such that $u, v \in V (C)$
QUESTION [1 upvotes]: Show that if a graph $G$ is $2-$connected, then for any two vertices $u$ and $v$ there exists a cycle $C$ such that $u, v \in V (C)$
I tried to use the fact that a graph $G$ with at least $3$ vertices is $2-$connected, if and only if for
any $u, v \in V$, with $u \neq v$, there are at least $2
(u, v)-$ trajectories that do not have internal vertices in common.
Any suggestions would be great!

REPLY [2 votes]: This is for an undirected graph $G$:
We show that there is a cycle by using induction on $d_G(u,v)$ i.e., the number of edges in the shortest path from $u$ to $v$, without using Menger's Theorem.
We first show that if $d_G(u,v)$ is $1$ i.e., $uy$ is an edge in $G$, then there is a cycle $C$ containing both $u$ and $v$. To this end, we note the following: First of all, if $G$ is $2$-vertex-connected, then $G$ is $2$-edge-connected. In particular, $G \setminus \{uv\}$ is connected. So letting $P$ be a path from $u$ to $v$ in $G \setminus \{uv\}$, note that $C=P+\{uv\}$ is a cycle.
So now let $k$ be a positive integer. we now assume the following:
(IH) Let $u$ and $v$ be two vertices in $G$ such that $d_G(u,v) \le k$. Then there is a cycle $C$ containing $u$ and $v$.
We now use this to show the following:
THM 1 Let $v'$ be a vertex such that $d_G(u,v')=k+1$. If (IH) is true, then there is a cycle $C'$ containing $u$ and $v'$.
Then from Thm 1 the result will follow.
Proof of THM 1: There is, as $d_G(u,v')=k$, a path $P=uv_1v_2 \ldots v_kv'$ with $k+1$ edges. Now $d_G(u,v_k)=k$ [and also, $v_kv'$ is an edge in $G$; we will use this for Cases 1 and 2 below], so if (IH) is true, there is a cycle $C$ containing both $u$ and $v_k$. We may also assume that $v' \not \in C$ lest we are already done. Write $C=uy_1y_2 \ldots y_rv_ky_{r+1} \ldots y_su$ for some integers $r$ and $s$, and

write $uy_1y_2 \ldots v_k$ the left-side of $C$ and

write $v_ky_{r+1} \ldots y_su$ the right-side of $C$.

Then, by the fact that $G$ is $2$-connected, there is a path $P'$ from $u$ to $v'$ that does not contain $v_k$. We consider 2 cases:
Case 1: $P'$ intersects $C$ at a point besides $u$. Then writing $P'=v'u'_1u'_2u'_3 \ldots u'_qu$ [for some integer $q$], let $i$ be the smallest integer such that $u'_i$ is in $C$. Then none of $u'_1,\ldots, u'_{i-1}$ is in $C$. Just as crucially, $u'_i$ is distinct from $v_k$.

Then if on the one hand $u'_i$ is in $\{y_1,y_2, \ldots, y_r\}$ say $u'_i=y_j$; $j \in \{1,2,\ldots, r\}$, then let $C'$ be the cycle

$$C' = \underbrace{uy_1 \ldots y_j}_{{\substack{\text{from $u$} \\ \text{up the left-side} \\ \text{of $C$ to} \\ \text{$y_j=u'_i \in P'\cap C$}}}} \ \underbrace{u'_{i-1} \ldots u'_1v'}_{\substack{\text{then $P'$ from $u'_iu'_{i-1}$} \\ \text{to $v'$}}}\ \underbrace{v_ky_{r+1} \ldots y_su}_{\substack{\text{then} \\ \text{$v'v_k$ then down} \\ \text{the right-side of $C$}\\ \text{back to $u$}}}.$$

If on the other hand $u'_i$ is in $\{y_{r+1},\ldots, y_s\}$ say $u'_i=y_j$ for some $j \in \{r+1,\ldots, s\}$, then let $C'$ be the cycle

$$C' = \underbrace{uy_s \ldots y_j}_{{\substack{\text{from $u$} \\ \text{up the right-side} \\ \text{of $C$ to}\\  \text{$y_j=u'_i \in C\cap P'$}}}} \ \underbrace{u'_{i-1} \ldots u'_1v'}_{\substack{\text{then $P'$ from $u'_iu'_{i-1}$} \\ \text{to $v'$}}}\ \underbrace{v_ky_{r}y_{r-1} \ldots y_1u}_{\substack{\text{then} \\ \text{$v'v_k$ then down} \\ \text{the left-side of $C$}\\ \text{back to $u$}}}.$$
Note that $C'$ indeed is a cycle containing both $u$ and $v'$, using the fact that $u_i$ and $v_k$ are distinct vertices, and $u_1,\ldots, u_{i-1}$ are not in $C$.
Case 2: $P'$ does not intersect $C$ at a point besides $u$. Then set
$$C' = P'+v'v_kv_{r+1} \ldots v_su,$$ i.e., $C'$ is the cycle formed by taking $P'$ from $u$ to $v'$, and then from $v'$ take the edge $v'v_k$ and then go back down the right-side of $C$ back to $u$. Note that $C'$ for this subcase as well indeed is a cycle containing both $u$ and $v'$.
As Cases 1 and 2 exhaust the possibilities, THM 1 follows and thus the intended result.

It is natural to ask if an analog to THM 1 exists for directed graphs. Interestingly, the answer to this is NO as there is a  graph $G$ such that both:

Between any $2$ vertices $u,v$ there are both (a) $2$ dipaths from $u$ to $v$ that are vertex-disjoint except for the endpoints $u$ and $v$ themselves, and (b) $2$ dipaths from $v$ to $u$ that are vertex-disjoint except for the endpoints $v$ and $u$ themselves.

There are also $2$ vertices $u$ and $v$ such that there is no directed cycle containing both $u$ and $v$ simultaneously.

Let $G$ be the graph on $\{v_1,v_2,v_3,v_4,v_5,v_6\}$ where
the arcs in $G$ are $v_1v_4, v_4v_1, v_4v_5,v_5v_6$, $v_5v_1$,
$v_1v_2,v_2v_3,v_3v_6, v_6v_3, v_6v_2$, $v_3v_5$. $v_2v_4$.
Then draw the graph out, and check that

Between any $2$ vertices $u,v \in \{v_1,v_2,v_3,v_4,v_5,v_6\}$ there are both (a) $2$ dipaths from $u$ to $v$ that are vertex-disjoint except for the endpoints $u$ and $v$ themselves, and (b) $2$ dipaths from $v$ to $u$ that are vertex-disjoint except for the endpoints $v$ and $u$ themselves.

There is no directed cycle containing both $v_1$ and $v_6$ simultaneously.


TITLE: Interpretation of the Boltzmann factor and partition function
QUESTION [5 upvotes]: $$p_i = \frac{ \exp\left(-\frac{\epsilon _i}{k_BT} \right)}{Z} $$
$$ Z= \sum_{i} \exp\left(-\frac{\epsilon _i}{k_BT} \right)$$
A) Is $p_i$ the probability of the system having an energy equal to $\epsilon_i$? (Probability to be in any of the many microstates that have energy $\epsilon_i$).
B) Or is $p_i$ the probability of the system being in one particular microstate which happens to have energy $\epsilon_i$? (This microstate is not the only microstate with the same energy).
If A) is correct then:
$$ Z= \sum_{\epsilon_i} \exp\left(-\frac{\epsilon _i}{k_BT} \right)$$
If B) is correct then:
$$ Z= \sum_{\epsilon_i} \Omega_i\exp\left(-\frac{\epsilon _i}{k_BT} \right),$$
where $\Omega_i$ is the multiplicity of the macrostate of energy $\epsilon_i$.
From the derivation of the Boltzmann distribution I am inclined to understand it as B). But I have never seen the multiplicity in the partition function.
What is the correct interpretation of the Boltzmann distribution?

REPLY [10 votes]: To the first question, the answer is B: $p_i$ is the probability of being in the $i$-th microstate, which happens to have an energy $\varepsilon_i$. However, microstates other than the $i$-th one may also have an energy $\varepsilon_i$.
The reason you never see the multiplicity in the partition function is because you are probably looking at summations done over the microstates: $$Z=\sum_i e^{-\frac{\varepsilon_i}{k_BT}}$$ instead of over the internal energies as you’ve written above.


TITLE: Why are the tangential derivatives in this diffraction problem zero?
QUESTION [1 upvotes]: I'm considering the diffraction problem described in section 3.16 of "Linear and quasilinear elliptic equations" of Ladyzhenskaya and Uraltseva (1968). Let $\Omega$ be an open bounded subset in $\mathbb{R}^n$, and suppose the surface $\Gamma$ partition $\Omega$ into sub-regions $\Omega_1$ and $\Omega_2$ where $\Omega = \Omega_1 \cup \Omega_2$. For simplicity, I assume $\Gamma$ to be the surface $\{x_n=0\}$. Now suppose $u(x)$ satisfies the following condition in $\Omega_1$ and $\Omega_2$
\begin{equation}
Lu \equiv \partial_{x_i} (a^{ij}(x) \partial_{x_j}u) + b^i(x) \partial_{x_i}u + c(x) u = f(x) \quad (16.1)
\end{equation}
and the conditions
\begin{equation}
u|_{\partial\Omega} = 0, \quad [u]|_\Gamma = 0, \quad \left.\left[ \frac{\partial u}{\partial N} \right]\right|_\Gamma = 0 \quad (16.3)
\end{equation}
on $\partial\Omega$ and $\Gamma$.
Here the symbol $[v]|_\Gamma$ denotes the jump in the function $v$ as it crosses $\Gamma$. Also, $\displaystyle \left.\frac{\partial u}{\partial N} \right|_\Gamma = a^{ij} (\partial_{x_j}u) \cos(n, x_i)$, where $n$ is the normal to $\Gamma$.
$u \in H^1_0(\Omega)$ is a weak solution of the problem (16.1) and (16.3) if $u$ satisfies
\begin{equation}
\int_\Omega \left[ a^{ij}(\partial_{x_j}u) (\partial_{x_i}v) -b^i (\partial_{x_i}u) v - cuv \right] \,dx = - \int_\Omega fv \,dx \quad (16.5)
\end{equation}
for all $v \in H^1_0(\Omega)$. Assume boundedness of the coefficients $a^{ij}, \partial_{x_k}a^{ij}, b^i, c$, the uniformly elliptic condition, and $f \in L^{q/2}(\Omega)$ for $q > n$.
Suppose we already showed $u \in H^2(\Omega_i)$. In (16.5), take $v(x) = \partial_{x_s} \eta$ for $s \le n-1$, where $\eta$ is a sufficiently smooth function with compact support in $K_\rho$, where $K_\rho$ is a sphere inside $\Omega$ with center on $\Gamma$. Let $\Gamma_1 := K_\rho \cap \Gamma$.
Substituting into (16.5) and integrating by parts we get
\begin{equation}
\int_\Omega \left[ a^{ij} (\partial_{x_s}\partial_{x_j}u) (\partial_{x_i}\eta) + \partial_{x_s} a^{ij} (\partial_{x_j}u) (\partial_{x_i}\eta) + b^i (\partial_{x_i}u) (\partial_{x_s}\eta) + cu(\partial_{x_s}\eta) \right] \,dx = \int_\Omega f (\partial_{x_s} \eta) \,dx. \quad (16.13)
\end{equation}
The author stated "On the basis of the conditions (16.13), we have $\left.[\partial_{x_s}u] \right|_{\Gamma_1} = 0$ for $s \le n-1$.
I don't understand why it is the case. I'm new to this type and really want to understand it.
Thank you.
EDIT: Following John Jiang's comment, I did the computation out. Substituting $v = \partial_{x_s} \eta$ into (16.5) and integrating by parts the first term, we get (justifying derivatives of $u$ makes sense)
\begin{equation}
\int_{K_\rho} \left[ -\partial_{x_s}(a^{ij}\partial_{x_j}u)(\partial_{x_i}\eta) - b^i(\partial_{x_i}u)(\partial_{x_s}\eta) - cu(\partial_{x_s}\eta) \right] \,dx + \int_{\Gamma_1} \big[a^{ij}(\partial_{x_j}u)(\partial_{x_i}\eta)\big]\Big|_\Gamma \,dS = -\int_{K_\rho} f(\partial_{x_s}\eta) dx.
\end{equation}
The integral over $\Gamma_1$ vanishes by the hypothesis that $\displaystyle \left.\left[ \frac{\partial u}{\partial N} \right]\right|_\Gamma = 0$. However, I haven't seen why the tangential derivatives of $u$ across $\Gamma_1$, $[\partial_{x_s}u]|_{\Gamma_1}$ must be 0.

REPLY [2 votes]: You can exchange the partial derivatives on $\nu = \partial_{x_s} \eta$. Taking one partial of $u$ is allowed in the whole $\Omega$ since $u \in H^1_0(\Omega) \subset W^1_0(\Omega)$. So $\partial_{x_s} u$ makes sense there. Mixed second partials of $u$ are justified by equation 16.10, which you didn't mention; it's established by assuming $u$ to be a solution of the original equation 16.1 satisfying condition 16.3. I believe 16.10 alone tells you there is no gap of $u$ across $\Gamma$. This is of course in the almost sure sense since $u$ is an equivalence class of functions. Any gap of positive measure in $\Gamma$ would show up as a boundary term upon integration by parts. For the simplest situation refer to this. I encourage you to write down the details carefully.


text:
\begin{document}
\bibliographystyle{plainnat}
\setcitestyle{numbers}
\title{On the functional limits for sums of a function of partial sums}
\author{Kamil Marcin Kosi\'nski}
\thanks{This research was done while the author stayed at the Vrije Universiteit, Amsterdam.}
\email{k.m.kosinski@uva.nl}
\address{Wydzia{\l} Matematyki, Informatyki i Mechaniki, Uniwersytet Warszawski, Warszawa, Poland}
\curraddr{Korteweg-de Vries Institute for Mathematics, University of Amsterdam, P.O. Box 94248,
1090 GE Amsterdam, The Netherlands}

\date{March 6, 2009}
\subjclass[2000]{Primary G0F05}
\keywords{Limit distributions, Product of sums, Stable laws}

\begin{abstract}
We derive a functional central limit theorem (fclt) for normalised sums of a function of the partial sums of independent and
identically distributed random variables. In particular, we show, using a technique presented in
 Huang and Zhang (Electron. Comm. Probab. 12 (2007), 51--56), that the result from
 Qi (Statist. Probab. Lett. 62 (2003), 93--100), for normalised products of partial sums, can be generalised in this fashion to a fclt.
\end{abstract}

\maketitle

\section{Introduction}
While considering limiting properties of sums of records, \citet{Arnold} obtained the following
version of the central limit theorem (clt) for a sequence $(X_n)$ of independent, identically
distributed (iid) exponential random variables (rv's) with the mean equal one:
\[
  \left(\prod_{k=1}^n\frac{S_k}{k}\right)^{1/\sqrt n}\dto e^{\sqrt 2\cn}\as n,
\]
where $S_n=\sum_{k=1}^n X_k$ and $\cn$ is a standard normal random variable.
\par Later \citet{Remiweso} extended such a clt to general iid positive rv's $(X_n)$. Namely,
provided that $\ee X_1^2<\infty$,
\beq
\label{res1}
  \left(\prod_{k=1}^n\frac{S_k}{k\mu}\right)^{\gamma/\sqrt n}\dto e^{\sqrt 2\cn}\as n,
\eeq
where $\mu=\ee X_1$ and $\gamma=\mu/\sigma$ with $\sigma^2=\Var X_1>0$.
\par This result was generalised by \citet{Qi} by assuming that the underlying distribution of $X_1$
is in the domain of attraction of a stable law with index $\alpha\in(1,2]$.
In this case
\beq
\label{res2}
  \left(\prod_{k=1}^n\frac{S_k}{k\mu}\right)^{\mu/a_n}\dto e^{(\Gamma(\alpha+1))^{1/\alpha}\cl}\as n,
\eeq
where $\Gamma(\alpha+1)=\int_0^\infty x^\alpha e^{-x}\, dx$ and the sequence $a_n$ is taken such that
\[
  \frac{S_n-n\mu}{a_n}\dto\cl,
\]
where $\cl$ is one of the stable distributions with index $\alpha\in(1,2]$. \citet{LuiQi} obtained a similar
result in the case $\alpha=1$ with $\ee |X_1|<\infty$. In a paper by \citet{Huang} it is shown that \eqref{res1}
follows from the weak invariance principle and the whole result can be reformulated to a functional theorem.
\par Let $(S_n)$ be \textit{any} nondecreasing sequence of positive rv's (which do not have
to be a sequence of partial sums). Suppose there exists a standard
Wiener process $(W(t))_{t\ge0}$ and two positive constants $\mu$ and $\sigma$ such that
\beq
\label{cond1}
  \frac{S_{[nt]}-[nt]\mu}{\sigma\sqrt{n}}\dto W(t)\quad\text{in}\quad D[0,1]\as n,
\eeq
and
\beq
\label{cond2}
  \sup_n \frac{\ee |S_n-n\mu|}{\sqrt n}<\infty.
\eeq
Then the result from \citet{Huang} states that
\beq
\label{res3}
  \left(\prod_{k=1}^{[nt]}\frac{S_k}{k\mu}\right)^{\gamma/\sqrt n}\dto\exp\left\{\int_0^t\frac{W(x)}{x}\,dx\right\}
\quad\text{in}\quad D[0,1]\as n,
\eeq
where $\gamma=\mu/\sigma$.
\par For example, if $(X_n)$ are iid positive rv's with mean $\mu$ and variance
$\sigma^2$ and $S_n$ are the partial sums, then \eqref{cond1} is satisfied by the invariance principle, cf.
\cite[Theorem 14.1]{Billingsley}, and \eqref{cond2} follows from the Cauchy-Schwarz inequality. Then, one can
check that \eqref{res3} implies \eqref{res1}.
\par The purpose of this paper is to show that the technique used by \citet{Huang} can be utilised to obtain a similar
result for the rv's in the domain of attraction of a stable law with index $\alpha\in(1,2]$. Furthermore, we will
set our discussion in a more general setting. It is straightforward that \eqref{res1}, and analogously \eqref{res2}, is
a simple corollary from
\[
  \frac{\sum_{k=1}^n f(S_k/k) - b_n}{a_n}\dto\cn,
\]
if one sets $f(x)=\log x$ and chooses the sequences $a_n$, $b_n$ properly.
\section{Preliminaries}
The following theorem is well known and can be easily found in the literature, see e.g., \cite{Bingham}.
\bt[Stability Theorem]
\label{general:stab}
  The general stable law with index $\alpha\in(0,2]$ is given by a characteristic function of one of the following forms:
\begin{enumerate}
  \item $\phi(t)=\exp(-\sigma^\alpha|t|^\alpha(1-i\beta(\sgn t)\tan\frac{1}{2}\pi\alpha)+i\mu t)$, $\alpha\ne1$,
  \item $\phi(t)=\exp(-\sigma|t|(1+i\beta(\sgn t)\frac{2}{\pi}\log|t|)+i\mu t)$,  $\alpha=1$,
  \item $\phi(t)=\exp(-\sigma^2t^2/2+i\mu t)$, $\alpha=2$
\end{enumerate}
with $\beta\in[-1,1]$, $\mu\in\rr$ and $\sigma>0$.
\et
From the above theorem one can see that every stable law with index $\alpha\in(0,2)$ can be parametrized by four parameters and written
as $\gs$. We distinguish the case $\alpha=2$ because otherwise $S_2(\sigma,\beta,\mu)\de\cn(\mu,2\sigma^2)$ and $\beta$
plays no role, moreover one would like to think of $S_2(1,\beta,0)$ as $\cn$ not $\cn(0,2)$.
\par Let $(X_n)$ be a sequence of iid rv's, set $S_n=\sum_{k=1}^n X_k$ and assume $X_1$ is in the
domain of attraction of a stable law with index $\alpha\in(1,2]$. Note that for such $X_1$ we have $\ee |X_1|<\infty$.
Recall that a sequence of iid rv's $(X_n)$ is said to be in the domain of attraction of a stable law $\gs$, if
there exists constants $a_n>0$ and $b_n\in\rr$ such that
\beq
\label{def}
  \frac{S_n-b_n}{a_n}\dto\gs.
\eeq
Clearly by scaling it suffices to let $\sigma=1$ and $\mu=0$, hence only the parameters $\alpha$ and $\beta$ are
unaltered by scaling. In case $\alpha=2$ as mentioned above, we understand $S_2(1,\beta,0)$ as $\cn$.
Moreover, in case $\alpha\in(1,2]$, the sequence $b_n$ can be taken equal to $\mu n$.
On how to choose $a_n$ we refer to \cite[Theorem 4.5.1]{Whitt}. The choice is irrelevant for our discussion,
the only fact that plays the crucial role is the fact that if \eqref{def} holds, then $a_n=n^{1/\alpha}L(n)$, where
$L$ is slowly varying.
\par Furthermore, in addition to stable clt \eqref{def}, there is convergence in distribution
\beq
\label{stable}
  S_n(t):=\frac{S_{[nt]}-[nt]\mu }{a_n}\dto \cl(t)\quad\text{in}\quad D[0,1],
\eeq
where $\cl$ is a standard $(\alpha,\beta)$-stable L\'evy motion, with
\[
  \cl(t)\de t^{1/\alpha}S_\alpha(1,\beta,0)\de S_\alpha(t,\beta,0)
\]
and as before, for $\alpha=2$, $\cl$ is a standard Wiener process, cf. \cite[Theorem 4.5.3]{Whitt}.
\section{Main Results}
\bt
  \label{result}
Let $(X_n)$ be a sequence of iid rv's in the domain of attraction of
the stable law $S_\alpha(1,\beta,0)$ with $\alpha\in(1,2]$, so that \eqref{def} holds
for some sequence $a_n$ and $b_n=n\mu$, where $\mu=\ee X_1$. Let $f$ be a real function defined on an interval $I$
such that $\pp(X_1\in I)=1$ and $f'(\mu)$ exists. Then, as $n\toi$
\beq
  \label{tw3:teza}
\frac{1}{a_n}\sum_{k=1}^{[nt]}\left(f(S_k/k)-f(\mu)\right)\dto f'(\mu)\int_0^t\frac{\cl(x)}{x}\,dx\quad\text{in}\quad D[0,1],
\eeq
where $S_k$ denotes the $k$-th partial sum.
\et
Because $\cl(x)$ is c\'adl\'ag, it has at most countably many discontinuity points, so the integral
on the right hand side of \eqref{tw3:teza} exists and is finite almost surely if
\[
  \int_0^1\frac{|\cl(x)|}{x}\,dx<\infty\quad\text{a.s}.
\]
To ensure this, note that for a positive nondecreasing function $h$ we have
\[
  \int_0^t\frac{|\cl(x)|}{x}\,dx\le\sup_{0\le s\le t}\left\{\frac{|\cl(s)|}{h(s)}\right\}\int_0^t\frac{h(x)}{x}\, dx.
\]
Setting $h(x)=x^\gamma$ with $\gamma\in(0,1/\alpha)$ we get $\int_0^1\frac{h(x)}{x}\, dx<\infty$ and
\[
  \sup_{0\le s\le t}\left\{\frac{|\cl(s)|}{h(s)}\right\}\to 0\,\text{ a.s. as}\quad t\to 0,
\]
by Khintchine's Theorem, see e.g., \cite[Theorem 2.1]{Nielsen}. This guaranties the existence of the integral in
\eqref{tw3:teza} as well as implies that
\[
  \sup_{0\le s\le t}\left|\int_0^t\frac{\cl(x)}{x}\,dx\right|=0\,\text{ a.s. as}\quad t\to 0,
\]
a fact that is going to be used later in the proof.
\begin{rmk}
\label{remark:calc}
Observe that
\begin{align*}
  \int_0^t\frac{\cl(x)}{x}\,dx &=\lim_{n\toi}\sum_{k=1}^n\frac{t}{n}\frac{n}{tk}\cl\left(\frac{tk}{n}\right)
=\lim_{n\toi}\sum_{k=1}^n\sum_{i=1}^k\frac{1}{k}\left(\cl\left(\frac{ti}{n}\right)-\cl\left(\frac{t(i-1)}{n}\right)\right)\\
&=\lim_{n\toi}\sum_{i=1}^n\left(\cl\left(\frac{ti}{n}\right)-\cl\left(\frac{t(i-1)}{n}\right)\right)\sum_{k=i}^n\frac{1}{k}\\
&\de \lim_{n\toi}\left(\sum_{i=1}^n\left(\sum_{k=i}^n\frac{1}{k}\right)^\alpha\right)^{1/\alpha}S_\alpha\left(\frac{t}{n},\beta,0\right)\\
&\de  S_\alpha\left(t,\beta,0\right) \lim_{n\toi}\left(\sum_{i=1}^n\frac{1}{n}\left(\sum_{k=i}^n\frac{1}{k}\right)^\alpha\right)^{1/\alpha}\\
&=S_\alpha\left(t,\beta,0\right)\left(\int_0^1(-\log x)^\alpha\,dx\right)^{1/\alpha}\\
&=S_\alpha\left(t,\beta,0\right)\left(\Gamma(\alpha+1)\right)^{1/\alpha}.
\end{align*}
If $X_1$ is a positive rv, then the limiting stable law has $\beta=1$. Setting
$f(x)=\mu\log(x/\mu)$, Theorem \ref{result} yields
\[
  \left(\prod_{k=1}^n\frac{S_k}{k\mu}\right)^{\mu/a_n}\dto

 \exp\left(S_\alpha\left(\Gamma(\alpha+1),1,0\right)\right),\as n.
\]
which is the result \eqref{res2} obtained by \citet{Qi}.
\end{rmk}
\begin{rmk}
If $\ee X_1^2<\infty$, then $X_1$ is in the domain of attraction of normal distribution $\cn$
and $a_n\sim\sigma\sqrt n$, where $\sigma^2=\Var(X_1)>0$. If furthermore $X_1$ is positive, then setting $\gamma=\mu/\sigma$
\[
  \left(\prod_{k=1}^{[nt]}\frac{S_k}{k\mu}\right)^{\gamma/\sqrt n}\dto \exp\left(\int_0^t\frac{W(x)}{x}\,dx\right),\quad\text{in}\quad D[0,1]\as n,
\]
which coincides with the result \eqref{res3} by \citet{Huang}.
\end{rmk}
Before proceeding to the proof of the main theorem, we need a technical lemma.
\bl Under the assumptions of Theorem \ref{result}
\label{lemma:sb}
\[
  \sum_{k=1}^n\frac{\ee|S_k-k\mu|}{k}=O(a_n).
\]
\el
\begin{proof}
Note that
\[
  \sum_{k=1}^n\frac{\ee|S_k-k\mu|}{k} \le
\sup_{k\le n}\left\{\ee\frac{|S_k-k\mu|}{a_k}\right\}\sum_{k=1}^n\frac{a_k}{k}.
\]
By Theorem 6.1 in \citet{De}
\beq
\label{cond:sim2}
  \ee\frac{|S_n-n\mu|}{a_n}=O(1).
\eeq
Now, for a regularly varying function $A>0$ with index $\gamma>-1$, its easy to see that
\[
  \sum_{k\le x} A(k)\sim\int_1^x A(t)dt\sim\frac{1}{1+\gamma}xA(x)\as x\quad\text{if }\gamma>-1,
\]
where the last asymptotic equivalence follows from the Karamata's Theorem, cf. \cite[Theorem 1.5.8]{Bingham}.
Recall that $a_n$ is slowly varying with index $1/\alpha>0$, this implies
\[
  \sum_{k=1}^n\frac{a_k}{k}=O(a_n),
\]
and proves the Lemma.
\end{proof}
Now we may proceed to the proof of the main theorem. The proof follows the steps of the proof of \eqref{res3} in
\citet{Huang}.
\begin{proof}[Proof of Theorem \ref{result}]
 Expand $f$ in the neighbourhood of $\mu$, then
\beq
\label{expansion}
\frac{1}{a_n}\sum_{k=1}^{[nt]}(f(S_k/k)-f(\mu))=\frac{f'(\mu)}{a_n}\sum_{k=1}^{[nt]}(S_k/k-\mu)+\frac{1}{a_n}\sum_{k=1}^{[nt]}(S_k/k-\mu)r(S_k/k),
\eeq
where $r(x)\to 0$ as $x\to\mu$. Note that $\ee |X_1|<\infty$ so by the SLLN $r(S_k/k)\to 0$ a.s..
It now follows from Lemma \ref{lemma:sb} that
\[
  \sup_{0\le t\le 1}\left|\frac{1}{a_n}\sum_{k=1}^{[nt]}(S_k/k-\mu)r(S_k/k)\right|
\le \frac{1}{a_n}\sum_{k=1}^{n}\frac{|S_k-k\mu|}{k}|r(S_k/k)|=o_\pp(1).
\]
So, according to \eqref{expansion} it suffices to show that, as $n\toi$
\beq
\label{cond}
  Y_n(t):=\frac{1}{a_n}\sum_{k=1}^{[nt]}\frac{S_k-k\mu}{k}\dto\int_0^t\frac{\cl(x)}{x}\,dx,\quad\text{in}\quad D[0,1].
\eeq
Let
\[
  H_\varepsilon(f)(t)=\left\{
 \begin{array}{rl}
 \int_\varepsilon^t\frac{f(x)}{x}\,dx,& \varepsilon<t\le 1\\
 0,& 0\le t\le\varepsilon
 \end{array}\right.
\]
and
\[
  Y_{n,\varepsilon}(t)=\left\{
 \begin{array}{rl}
 \frac{1}{a_n}\sum_{k=[n\varepsilon]+1}^{[nt]}\frac{S_k-k\mu}{k},& \varepsilon<t\le 1\\
 0,& 0\le t\le\varepsilon,
 \end{array}\right.
\]
It is obvious that
\beq
\label{eq:1}
  \sup_{0\le t\le 1}\left|\int_{0}^t\frac{\cl(x)}{x}\,dx-H_\varepsilon(\cl)(t)\right|=
\sup_{0\le t\le \varepsilon}\left|\int_{0}^t\frac{\cl(x)}{x}\,dx\right|\to0\,\text{ a.s. as }\,\varepsilon\to 0
\eeq
and
\beq
\label{eq:2}
  \ee\max_{0\le t\le 1}\left|Y_n(t)-Y_{n,\varepsilon}(t)\right|\le
\frac{1}{a_n}\sum_{k=1}^{[n\varepsilon]}\frac{\ee|S_k-k\mu|}{k}\le C \varepsilon^{1/\alpha}
\eeq
by the same argumentation as in the proof of Lemma \ref{lemma:sb}.
\par On the other hand, it is
easily seen that, for $n$ large enough such that $n\varepsilon\ge 1$,
\begin{align*}
\sup_{\varepsilon\le t\le 1}\Bigg|\sum_{k=[n\varepsilon]+1}^{[nt]}\frac{S_k-k\mu}{k}&-
\int_{n\varepsilon}^{nt}\frac{S_{[x]}-[x]\mu}{x}\,dx\Bigg|
=\sup_{\varepsilon\le t\le 1}\left|\int_{[n\varepsilon]+1}^{[nt]+1}\frac{S_{[x]}-[x]\mu}{[x]}\,dx-
\int_{n\varepsilon}^{nt}\frac{S_{[x]}-[x]\mu}{x}\,dx\right|\\
&\le\left |\int_{n\varepsilon}^{[n\varepsilon]+1}\frac{S_{[x]}-[x]\mu}{x}\,dx \right|
+\sup_{\varepsilon\le t\le 1}\left|\int_{nt}^{[nt]+1}\frac{S_{[x]}-[x]\mu}{x}\,dx\right|\\
&\quad+\sup_{\varepsilon\le t\le 1}\left|\int_{[n\varepsilon]+1}^{[nt]+1}(S_{[x]}-[x]\mu)\left(\frac{1}{x}-\frac{1}{[x]}\right)\,dx\right|\\
&\le\max_{k\le n}|S_k-k\mu|\sup_{\varepsilon\le t\le 1}\left(\frac{2}{n\varepsilon}+\frac{2}{nt}+\frac{1}{n\varepsilon}\right)\\
&\le 5\max_{k\le n}|S_k-k\mu|/(n\varepsilon)=O_\pp(a_n/n)=o_\pp(1),
\end{align*}
by noticing that $\max_{k\le n} |S_k-k\mu|/a_n\dto \sup_{0\le t\le 1} |\cl(t)|$ according to \eqref{stable}.
So
\[
  \frac{1}{a_n}\sum_{k=[n\varepsilon]+1}^{[nt]}\frac{S_k-k\mu}{k}=\frac{1}{a_n}\int_{n\varepsilon}^{nt}\frac{S_{[x]}-[x]\mu}{x}\,dx+o_\pp(1)
=\int_{\varepsilon}^{t}\frac{S_{n}(x)}{x}\,dx+o_\pp(1)
\]
uniformly in $t\in[\varepsilon,1]$. Notice that $H_\varepsilon(\cdot)$ is a continuous mapping on the space $D[0,1]$. Using
the continuous mapping theorem (cf. \cite[Theorem 2.7]{Billingsley}) it follows that
\beq
\label{eq:3}
  Y_{n,\varepsilon}(t)=H_\varepsilon(S_n)(t)+o_\pp(1)\dto H_\varepsilon(\cl)(t)\quad\text{in}\quad D[0,1]\as n.
\eeq
Combining \eqref{eq:1}-\eqref{eq:3} yields \eqref{cond} by \cite[Theorem 3.2]{Billingsley}.
\end{proof}
\section{Extensions}
To prove Lemma \ref{lemma:sb}, we have only used the property \eqref{cond:sim2} (which is in fact
the condition \eqref{cond2}) and the fact that $a_n$ varies regularly with a positive index. The proof
of Theorem \ref{result} was based on the convergence \eqref{stable} and the fact that $S_k/k\to\mu$ a.s.. All
those conditions are satisfied when $S_k$ is defined to be the partial sum of a sequence of iid rv's in the
domain of attraction of a stable law with index greater than one. However, we do not need to assume anything
about $S_k$ and only require that it satisfies the aforementioned conditions. This leads to
\bt
\label{result2}
Let $(S_k)$ be a sequence of random variables. Suppose there exists an $(\alpha,\beta)$-stable L\'evy process $(\cl(t))_{t\ge 0}$,
a constant $\mu$ and a sequence $a_n$ such that as $n\toi$
\[
  \frac{S_{[nt]}-[nt]\mu}{a_n}\dto\cl(t)\quad\text{in}\quad D[0,1],
\]
where $a_n$ can be written as $a_n=n^{1/\alpha}L(n)$ with $\alpha\in(1,2]$ and $L$ slowly varying. In addition,
suppose that
\beq
\label{cond2:again}
  \sup_n \frac{\ee |S_n-n\mu|}{a_n}=O(1),
\eeq
and $S_n/n\to\mu$ a.s., then, as $n\toi$
\[
\frac{1}{a_n}\sum_{k=1}^{[nt]}\left(f(S_k/k)-f(\mu)\right)\dto f'(\mu)\int_0^t\frac{\cl(x)}{x}\,dx\quad\text{in}\quad D[0,1],
\]
for any real function $f$ defined on an interval $I$ such that $\pp(S_k/k\in I)=1$ for all $k$, provided that $f'(\mu)$ exists.
\et
In their paper, Huang and Zhang showed that if $(S_k)$ is a nondecreasing (in fact we only need monotonicity) sequence
satisfying \eqref{cond2:again}, then $S_k/k\to\mu$ a.s.. Thus, Theorem \ref{result2} is an extension of the result \eqref{res3}
from \citet{Huang}.
\section{Acknowledgments}
The author is grateful to dr. R. Lata{\l}a for supervision of his work and would also like to thank the referees for
pointing out some subtleties in the previous version of this paper.


TITLE: Charged plasma and ion grid in interaction in ion thrusters
QUESTION [1 upvotes]: I was just wondering .....
In this Image of an ion thruster, when the positively charged particles pass through the grids, wouldn't they just bombard the negatively charged grid(a fraction of them). This means that there must be a constant adjustment to maintain the potential difference between the grids. Is this the reason for the high energy consumption of these engines (along with ionization..)?

REPLY [0 votes]: You say "constant adjustment" as if you expect a motor with a PID controller to be spinning cat's fur on the grid to keep its charge ;P
The grid keeps it charge with 1.) a battery and 2.) by connecting both electron guns' filament to the positive grid. An electron gun is a special filament that when heated (by passing current through it) expels electrons.
As the gas ions are passing by the negative grid, the electrons are jumping off and neutralizing the gas as it exits.  This is just part of the circuit under normal operating conditions - the Thevenin equivalent voltage of the circuit would be such that it's adequately charged at all times.
Energy consumed by the system is 1.) work done to accelerate the ions and 2.) the work done to heat the electron gun filament.
Here's the circuit:


TITLE: Formula for the nth prime number.
QUESTION [2 upvotes]: I need to see how to make a certain development, and did not find any reference to do, the only quote I found, so is the following, I know (have shown)
$$ \pi(m)=-1+\sum_{j=1}^{m} F(j) $$ with $$F(j)=\left[ \cos^2 \pi\frac{(j-1)!+1}{j}\right]$$ and also I have to
$$ \pi(m)=\sum_{j=1}^{m} H(j) $$ for all $ m\ge2 $, at where $$ H(j)=\dfrac{\sin^2\pi\dfrac{((j-1)!)^2}{j}}{\sin^2\dfrac{\pi}{j}} $$
this second part also fails to prove if someone wants to put some demonstration would be very happy, but the focus is the same demonstration that the nth prime number is given by
$$ p_n=1+\sum_{m=1}^{2^n} \left[\left(\frac{n}{\sum_{j=1}^{m} F(j)} \right )^{1/n} \right ]$$

REPLY [4 votes]: The key fact in the formulae above is Wilson's theorem:

If $n$ is prime, then $(n-1)!+1$ is divisible by $n$.

And the slightly more obvious:

If $n$ is composite and $n\neq4,$ then $(n-1)!$ is divisible by $n$.

Let's state upfront, though: These formula are essentially "nonsense," in that they add zero knowledge beyond our knowledge from Wilson's theorem, and are certainly not computationally useful.
Formula 1
$$\theta_j = \pi\frac{(j-1)!+1}{j}$$
will be a multiple of $\pi$ if $j$ is prime, and something else if $j$ is not prime. So $\cos^2\theta_j = 1$ if $j$ is prime and otherwise it is some value in $[0,1)$ so
$$\left\lfloor \cos^2\theta_j \right\rfloor = \begin{cases}1&\text{if $j$ is prime or $j=1$}\\
0&\text{if $j$ composite}
\end{cases}$$
The case $j=1$ is why there is a $-1$ added to the sum above.
Formula 2
The second case has $H(j)=F(j)$ for $j>1$. It's not actually defined when $j=1$, so you'd have to deal with that to fix it.
If $j$ is composite, then $\pi\frac{(j-1)!^2}{j}$ is a multiple of $\pi$, so the numerator is zero, so $H(j)=0$.
When $j$ is prime, then $\pi\frac{(j-1)!^2}{j} = \frac{\pi}{j}+M\pi$ for some integer $M$, so the sine of this angle is $\pm\sin \frac{\pi}{j}$, and therefore the square of the quotient is $1$.
Formula 3
The last formula can be rewritten as:
$$p_n=1+\sum_{m=1}^{2^n} \left\lfloor\left(\frac{n}{1+\pi(m)} \right )^{1/n} \right\rfloor$$
But it is easy to show that $n^{1/n}<2$ for all $n$, and thus
$$\left\lfloor\left(\frac{n}{1+\pi(m)} \right )^{1/n}\right\rfloor=\begin{cases}1&\text{if }\pi(m)<n\\
0&\text{otherwise}
\end{cases}$$
Since the first $n$ primes are known to be less than $2^n$, this sum counts all $m$ with $\pi(m)<n$, which is is $p_n-1$. Adding $1$ gives $p_n$.