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%  Dec. 5-7 2006
%  Ibaraki National College of Technology
%  Nakane 866, Hitachinaka-shi, Ibaraki 312-8508, Japan
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%���M�E�C�����鏑����(2007.6.7)�A��{�E�g��
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\begin{document}%
\keywords{Golden Ratio appears to
 discret time optimization problem}
\mathclass{Primary 90C39; Secondary 11A99.}
%\keywords{These must be included}

\abbrevauthors{S.~Iwamoto and M.~Yasuda}
\abbrevtitle{Golden Ratio}

\title{Golden Ratio appears in 
 Discrete-time \\ 
 Dynamic Optimization Processes}
%

\author{Akifumi KIRA} 
\address{
   Kyushu University,
   Fukuoka 812-8581, Japan\\ 
   E-mail: {\tt \RED{ec207004@s.kyushu-u.ac.jp}}
   %
   % �g�� �m���iAkifumi Kira)
   % ��B��w�o�ϊw����{�[�~�S��
   % �O�X�O�|�U�W�X�O�|�U�X�P�S
   % ec103049@s.kyushu-u.ac.jp(�p�\�R��)
   % tel.09068906914@docomo.ne.jp(�g��)
}
\author{Masami YASUDA}
\address{%
   %Department of Mathematical and Informatics,
   %Faculty of Science, \\
   Chiba University,
   Chiba 263-8522, Japan \\
   E-mail: {\tt yasuda@math.s.chiba-u.ac.jp}
   %Tel\&fax. +81(43)290-3662 }
}
\author{Seiichi IWAMOTO} 
\address{
   %Department of Economic Engineering,
   %Graduate School of Economics, \\
   Kyushu University,
   Fukuoka 812-8581, Japan\\ 
   E-mail: {\tt iwamoto@en.kyushu-u.ac.jp}
   %Tel\&fax. +81(92)642-2488} 
}
%
%\date{\today}
%
\maketitleMTE
% \maketitle 
%
\begin{abstract}%
  Since antiquity, the Golden Ratio ($\phi$) has been keeping 
  to show up in many aspects of\author{Seiichi IWAMOTO} 
\address{
   %Department of Economic Engineering,
   %Graduate School of Economics, \\
   Kyushu University,
   Fukuoka 812-8581, Japan\\ 
   E-mail: {\tt iwamoto@en.kyushu-u.ac.jp}
   %Tel\&fax. +81(92)642-2488} 
}
  geometry, architecture, music, art and even philosophy, and 
  it appears also in newer domains. Here a few source of information 
  about this will be listed as follows:
  A gambling in the coin tossing gives a amount of bet and 
  observes a martingale property when it is adapted a probability 
  of win called as the Golden probability(2-$\phi$). 
  We will also show a typical Discrete-time Dynamic Optimization Processes for 
  Allocation problem, Linear-Quadratic control problem and 
  Multi-variate stopping problem, of which the ratio has an interest rule.   
\end{abstract}

%
\section{Introductholds from (\ref{eqn:2.4}). Soion.}                            \label{sec:1}
%***************************************************************

The Golden Ratio ($\phi = 1.61803\cdots $) has been a profound 
influence since ancient times such as the Parthenon at Athens. 
The beauty appears even in the ingredient of nature creatures.
The most influential mathematics textbook by Euclid of Alexandria 
defines the proportion. 
These information presents a broad sampling of $\phi-$related 
topics in an engaging and easy-to-understand format.

The Fibonacci sequence ($1,1,2,3,5,8,13,$ $\cdots$ ) is closely 
related to the Golden Ratio, which is a limiting ratio of 
 its two adjacent numbers. It is also known
that the diagonal summation produces the Fibonacci sequence 
in the Pascal's triangle. These repeated procedure or iteration
have something in common.

The principle of Dynamic Programming is said to \lq divide and conquer.'
In fact, if it is not possible to work out directly, divide up
a problem into smaller ones. The basic idea 
is aiming to provide the original problem an effective family of sub-problems.
The Bellman's curse of dimensionality conquers the computational
explosion with the problem dimension through the use of parametric
representations. The more it's in a complex, the more it is divided.
When a problem is in a multi-stage decision form,
we should consider the problem repeatedly. 
If this reduction procedure gives a self-similar one, 
the methodology turns out to be effective.  
The Golden Ratio is created repeatedly by its own in a quite same form.
A recurrence relation is ubiquitous. Let us say that a beautiful 
continued fraction represents the Golden number. 
It is interesting that this quite introductory problem of Dynamic
programming produces the basic mathematical aspects. 

  In this way the Golden ratio is no isolated phenomenon, 
  but, in many cases, the first and indeed, a simple example 
  in the context of a sequence of generalizations of a common idea. 
  Therefore the Golden ratio has a pedagogical significance as well.
  
  A theme of this note is presented originally at the conference MTE2006
  Dec. 2006
  and also Research Meeting RIMS at Feb. 2007 as  
  "Golden ratio in optimal value function" by
   Seiichi Iwamoto and Masami Yasuda.
   Then adding the next section, this note is prepared.


\section{Gambling}
holds from (\ref{eqn:2.4}). So
A coin tossing is the game of win or loss in the typical gambling
for the probability theory\RED{.}
%and a deep development has been developmented
\RED{The gambling theory has been developed} in the related various fields. 
%>From the Wikipedia, The free Encyclopedia, 
\RED{Martingale is referred to as a popular class of betting strategies
 in 18th century France.} 
The simplest of these strategies was designed for a game in which 
the gambler wins his stake if a coin comes up heads and loses 
it if the coin comes up tails.  The strategy had the gambler double 
his bet after every loss, so that the first win would recover 
all previous losses\RED{. plus yield} a profit equal to the original stake. 
Since as a gambler's wealth and available time jointly approach 
infinity\RED{.} his probability of \RED{eventual} flipping heads 
approaches 1\RED{.} 
\RED{Thus} the martingale betting strategy was seen as a sure thing by those 
who practiced it. Of course in reality the exponential growth of 
the bets would eventually bankrupt those foolish enough to use 
the martingale for a long time. 
  The concept of martingale in probability theory 
was introduced by Paul Pierre Levy(1986-1971), and much of the original 
development of the theory was done by Joseph Leo Doob(1910-2004). 
Part of the motivation for that work was to show the impossibility of 
successful betting strategies.

The following gambling with a toin tossing is considered.
Assume that there are no ties and \RED{the} result of any game is 
determined randomly. 
Let $p$ be the probability of \RED{winning} and $q=1-p$ losing. 
If $y$ is the amount of the commencing bet and
let $x^{\RED{n-1}}$ be the finite number of bets you can afford in the
$n$-th play of time.
Then $y - x^{\RED{n-1}}$ is the amount when one losses, however, 
one can get $x$-time more if one wins in the betting,
$x^{\RED{n}}$ when one wins. 
That is, at $n$-th time play($n=\RED{1}, 2,\cdots $) 
\begin{equation}                   \label{eqn:2.1}
 \begin{array}{ll}
   y + x ^{\RED{n}}  & : p \\
   y - x^{\RED{n-1}} & : q = 1-p
 \end{array} 
\end{equation}
When you counitue \RED{to lose} all your bets, the amount of money you lose is
\(  1 + x + x^2 + \cdots + x^{n-1} \).
\RED{Let $W_n$ denote} the amount of money in total until $n$-th time play. 
So the dynamics of the amount is expressed as follows:
\begin{equation}                  \label{eqn:2.2}
  W_{n+1} = W_n + A_{n\RED{+1}} , \quad n=\RED{0,1,} \dots
\end{equation}
where the gain of $n$-th time is $A_n$:
\begin{equation}                  \label{eqn:2.2.1}
  A_n = x^n \epsilon_n \RED{-} x^{n-1} (1-\epsilon_n)
\end{equation}
and
\begin{equation}                  \label{eqn:2.3}
   \epsilon_n = \left\{ \begin{array}{ll}
                 +1 & \mbox{with probability } p, \\ 
                 0 & \mbox{with probability } q=1-p.
                \end{array} \right. 
\end{equation}

Alternatively, if we change the gain to $B_n$ from $A_n$:
\begin{equation}                  \label{eqn:2.4}
  B_n = (x^{n \RED{-1}} + x^{n \RED{-2}}) \epsilon_n \RED{-}
 x^{n-1} (1-\epsilon_n)
\end{equation}

Because, if the gambler fails to win at \RED{($n-1$)} and wants \RED{to}
continue the next play \RED{$n$}, the regret is \RED{the loss} $x^{n
\RED{-2}}$ \RED{at $(n-1)$}. \RED{And $x^{n-1}$ is a bet for the next
play $n$.} To recover the loss, one \RED{can} get 
\RED{not $x$-time but $x^{n-2}$ more if one wins in the betting.}  
%the gain two amount $x^{n} + x^{n-1}$ which is
%the last and the second last of loss.   
And then
\begin{equation}                  \label{eqn:2.5}
  V_{n+1} = V_n + B_{n \RED{+1}} , \quad n=\RED{0,1}, \dots
\end{equation}
provided that $V_0 = 0$.

\begin{dfn}(\cite{Nev75,Wil91} )                \label{dfn:matingale}
  A process $(X_n, {\cal B}_n)$ is called a martingale if 
  {\rm (i)} it is adapted, that is, for each $n$, $X_n$ is 
     ${\cal B}_n)$-measurable,
  {\rm (ii)} $E(\vert X_n \vert ) < \infty$ ,  $\forall n$, and 
  {\rm (iii)} $E(\vert X_{n \RED{+1}} \, \vert \, {\cal B}_n ) = X_n$, a.s. $ (n \geq 1)$.
\end{dfn}

%If the information(natural filtration) of $\sigma$-algebra denotes by
%${\cal B}_n = \sigma (e_k; k \leq n)$, 
\RED{Let ${\cal B}_n = \sigma (e_k; k \leq n)$ denote the information (natural
 filtration)}, then we have the following theorem.
\begin{thm}                      \label{thm:2.1}
 In order to ensure each process a martingale, the relation between
 the probability and the reward is given as 
  \begin{itemize}
    \item[\rm(i)]                \label{thm:2.1(i)}
     $(W_n, {\cal B}_n; n \geq 1) $ constitutes 
     a martingale if and only if $p= 1/(1+x), x >0$,
    \item[\rm{ (ii)}]            \label{thm:2.1(ii)}
     $(V_n, {\cal B}_n; n \geq 1) $ constitutes 
     a martingale if and only if $p= 1/(2+ 1/x), x>0$.
  \end{itemize}
\end{thm}
{\it Proof.}
\RED{The adaptation} is clear and \RED{it holds that} $E\vert \epsilon_n
\vert  < \infty$ for each $n$\RED{.} \RED{The} condition 
makes zero-mean random variables, \RED{and} so the sum is a martingale.

\begin{dfn}(\cite{Nev75,Wil91} )                        \label{dfn:UI}
The class $(X_n, n \geq 1)$ of random variables %is called 
%uniformly integrable or 
\RED{is said to satisfy} the uniform integrability 
condition
if given $\epsilon > 0$, there exists $K> 0$ such that
\(
  E(\vert X_n \vert ; \vert X_n \vert > K) < \epsilon, \forall n. 
\)
\end{dfn}

\begin{thm}                      \label{thm:2.2}
 The boundedness of each process is given as follows:
  \begin{itemize}
    \item[\rm(i)]                \label{thm:2.2(i)}
     $(W_n, {\cal B}_n; n \geq 1) $ satisfies 
     the uniform integrability condition 
     if $\RED{x^2 - x - 1 =0}$. \RED{Thus the Golden ratio $x = \phi$ appears.}
    \item[\rm{ (ii)}]            \label{thm:2.2(ii)}
     $(V_n, {\cal B}_n; n \geq 1) $ satisfies 
     the uniform integrability condition 
     if $x^2 - x - 1 =0 $. \RED{Thus the Golden ratio $x = \phi$ appears.}
%This case appears the Golden probability $x= 2 - \phi$. 
 \end{itemize}
\end{thm}
{\it Proof.}
(i)
%Since
\RED{From (\ref{eqn:2.3}), it holds that}
\[ 
\begin{array}{ll}
  & P(\epsilon_1 = \epsilon_2 = \cdots = \epsilon_{n-1} =0, \epsilon_n =1  )\\ 
= & P(W_{n+1} = -1  -x  -x^2 - \cdots - x^{n-1} + x^{n \RED{+1}})\\ 
= & P\Big(W_{n+1} = \dfrac{(\RED{x^2-x-1})x^n +1}{x-1}\Big) = p q^n
\end{array}
\] 
%holds from (\ref{eqn:2.3}), this imply
\RED{Substituting $x = \phi $} impl\RED{ies} that
$\sup_n P(\vert W_n \vert >a ) \to 0$ as $ a \uparrow  \infty  $\RED{.}
%inserting $x=2$ in the equation. 
Hence the condition \RED{is satisfied}.

(ii) Similarly
\[ 
\begin{array}{ll}
  & P(\epsilon_1 = \epsilon_2 = \cdots = \epsilon_{n-1} =0, \epsilon_n =1  )\\ 
= & P(V_{n+1} = -1  -x  -x^2 - \cdots - x^{n-1} + x^{n-1} + x^n)\\ 
= & P\Big(V_{n+1} = \dfrac{(x^2 - x - 1)x^{n-1} +1}{x-1}\Big) = p q^n
\end{array}
\] 
holds from (\ref{eqn:2.4}). So 
$(V_n, {\cal B}_n; n \geq 1) $ satisfies 
     the uniform integrability condition 
     if $x^2 - x - 1 =0 $.

\begin{cor}
 Let $x=\phi $. \RED{That is, by} letting \RED{an} amount $x$ of  bet \RED{equal} the Golden 
 ratio $\phi$, two problems (\ref{eqn:2.2}) and (\ref{eqn:2.5}) are equal:
 $ A_n = B_n$ and $ V_n = W_n$ for all $n$. \RED{And then, both
 processes $(W_n, {\cal A}_n; n \geq 1) $ and $(V_n, {\cal B}_n; n \geq
 1) $ constitute a martingale if and only if $p=2-\phi$ which is the Golden
 probability.}
\end{cor}
{\it Proof.} 
\RED{It suffices to note that the Golden ratio $\phi$ satisfies
$\phi^{n+2} = \phi^{n+1} + \phi^n, \; 1/(1+\phi) = 2-\phi$ 
and $1/\phi = \phi-1$.}
%The property of $\phi^n = a_n \phi + a_{n-1}$ imply the
%assertion where $\{a_n\} $ is the fibonacci sequence by L.Euler. 
%>From $\phi^{n+2} - \phi^{n+1} - \phi^n 
%   = a_{n+2} \phi + a_{n+1}
%         -a_{n+1} \phi + a_{n} - a_n \phi - a_{n-1}
%   = a_{n+2} \phi -a_{n+1} \phi - a_n \phi
%     + a_{n+1} + a_{n}  - a_{n-1}
%   = 0 \cdot \phi + 0 = 0 $, 
%therefore $  \phi^{n+2} =  \phi^{n+1} +  \phi^n$.
%This imply the two problem is equal.



\section{Discrete-time Dynamic Optimization}          \label{sec:3}
 
In the following sections, we treat typical dynamic programming problems;
Allocation problem and Linear-Quadratic Control problem. 
However problems are in a simple fashion, it figures out 
the essence of Dynamic Programming.

Let us consider a typical type of criterion in a deterministic 
optimization. 
We minimize the next quadratic criteria :
\begin{equation}                                  \label{QC:11}
%\begin{eqnarray*}
  I(x) = %\eql 
    \dsum_{n=0}^{\infty} \left[x_{n}^{2} + (x_{n} - x_{n+1})^{2} \right], %\\
\end{equation}
and
\begin{equation}                                  \label{QC:12}
  J(x) = %\eql 
     \dsum_{n=0}^{\infty}\left[(x_{n} - x_{n+1})^{2} + x_{n+1}^{2} \right].
%\end{eqnarray*}
\end{equation}
%
%*******************************************************************
Let $ R^{\infty} $ be the set of all sequences of real values : 
$$ 
  R^{\infty} = \{ x = (x_{0}, x_{1}, \ldots , x_{n}, \ldots ) 
    \, | \,x_{n} \in R^{1}~~n = 0, 1, \ldots \, \}.  
$$
%
First we take the quadratic criterion (\ref{QC:11}).
%\begin{eqnarray*}
%  I(x) = %\eql 
%    \dsum_{n=0}^{\infty}\left[x_{n}^{2} + (x_{n} - x_{n+1})^{2} \right]
%     %~~~\mbox{on}~~ R^{\infty}. 
%\end{eqnarray*}
%

Now we consider a mathematical programming problem for any given 
initial value $ c $ :
%
\begin{eqnarray*}
  \underline{{\rm MP_{1}(c):}}\hspace{0mm}
  & &
    {\rm ~minimize} \hspace{2ex} I(x) \hspace{3mm} 
    {\rm subject~ to}\hspace{2ex} {\rm (i)}~ x \in R^{\infty},
  ~~{\rm (ii)}~x_{0} = c.
\end{eqnarray*}
%
%
Let us evaluate a few special trajectories :
\begin{exm}                                   \label{ex:1}
   %\item 
   First all but on and after nothing $\, 
     y = \{ c,~ 0,~ 0,~ \ldots ,~ 0,~ \ldots \} \,$ yields 
       $\, I(y) \dql 2c^{2}.  $
\end{exm}
%
\begin{exm}                                  \label{ex:2}
  %\item  
  Always all constant $\, z = \{ c,~ c,~ \ldots ,~ c,~ \ldots \} \,$ 
     yields $\, I(z) \dql \infty.  $
\end{exm}
%
\begin{exm}                                  \label{ex:3}
  %\item  
  A proportional (geometrical) $\, w = \{ c,~ \rho c,~ \ldots ,
       ~ \rho^{n} c,~ \ldots \} \,$ yields 
    \begin{equation}
    \begin{array}{rl}
      I(w) = & %\eql 
       \left\{ c^{2} +  (1 - \rho)^{2} c^{2} \right\}
       \left( 1 + \rho^{2} + \cdots + \rho^{2n} + \cdots \right) \\
       = & %\eql 
       \dfrac{\,1 +  (1 - \rho)^{2} \,}{\, 1 - \rho^{2} \,}c^{2}
       ~~~~~~(0 < \rho < 1).
    \end{array}
    \end{equation}
    Since 
      \( %\begin{equation}
      \min_{0 \le \rho < 1} \dfrac{\,1 +  (1 - \rho)^{2} \,}
      {\, 1 - \rho^{2} \,} 
      %\nonumber 
      \) %\end{equation}
    is attained at $ \hat{\rho} = 2 - \phi $, we have the minimum value 
    \begin{equation}
      \dfrac{\,1 +  \left\{1 - (2 - \phi)\right\}^{2} \,}
      {\, 1 - (2 - \phi)^{2} \,} \dql \phi. \nonumber 
    \end{equation}
\end{exm}
%
\begin{exm}                                   \label{ex:4}
  %\item  
  The propotional 
  $\, \hat{u} = \{ c,~ (2 - \phi) c,~ \ldots ,
     ~ (2 - \phi)^{n} c,~ \ldots \} \,$, with ratio $(2 - \phi)$, yields 
  \begin{equation}
    I(\hat{u}) = %\eql 
       \phi c^{2}.
  \end{equation}
%
%
Thus $ \hat{u} = (\hat{u}_{n}) $ gives a Golden optimal trajectory, because $\hat{u}_{n+1} $ is always a Golden section point of interval $ [0, \hat{u}_{n}]. $ 
\end{exm}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\subsection{Control process I; quadratic in current state} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Next let us now consider a control process with an additive transition 
$ T(x, u) = x + u$. 
\begin{center}
\begin{tabular}{lll}
  &   {\rm minimize}  &
        $\dsum_{n=0}^{\infty} \left( x_{n}^{2} + u_{n}^{2} \right)$ \\
  & {\rm subject  to}  &  (i)  
         $ x_{n+1} = x_{n} + u_{n},  \quad  n \ge 0 $ \\ 
  & &  (ii)  
         $ - \infty < u_{n} < \infty $ \\ 
  & &  (iii) 
         $ x_{0} = c$. 
\end{tabular}
\end{center}
%
Then the value function $ v $ satisfies Bellman equation : 
\begin{equation}                                            \label{eqn:cp1} 
   v(x) \dql \min_{-\infty < u < \infty}
    \left[x^{2} + u^{2} + v(x + u) \right]. 
\end{equation}
Eq. (\ref{eqn:cp1}) has a quadratic form $ v(x) = \phi  x^{2} $.  

Second we take the following quadratic criterion 
\begin{eqnarray*}
   J(x) = %\eql 
       \dsum_{n=0}^{\infty}\left[(x_{n} - x_{n+1})^{2} + x_{n+1}^{2} 
\right].
\end{eqnarray*}
%
We consider a problem of form:
%
\begin{eqnarray*}
 \underline{{\rm MP_{2}(c):}}\hspace{0mm}
  & &{\rm ~minimize} \hspace{2ex} J(x) \hspace{3mm} 
  {\rm subject~ to}\hspace{2ex} 
  {\rm (i)}~ x \in R^{\infty},~~{\rm (ii)}~x_{0} = c.
\end{eqnarray*}
%
%
Since 
%\begin{eqnarray*}
$  J(x) = %\eql 
         I(x) - c^{2}$,  
%\end{eqnarray*}
%
%$ {\rm MP_{2}(c)} $ has 
%
the minimum value is 
%\begin{eqnarray*}
$  J(\hat{u}) = %\eql 
               (\phi - 1) c^{2}$
%\end{eqnarray*} 
at the Golden trajectory 
$$
  \, \hat{u} = 
     \{ c,~ \mu c,~ \ldots ,~ \mu^{n} c,~ \ldots \} \, 
     ;~~\mu = 2 - \phi. 
$$
%

%\vskip 3mm
\noindent
In fact, a proportional $\, w = \{ c,~ \rho c,~ \ldots ,~ \rho^{n} c,~ \ldots \} \,$ yields 
%\begin{eqnarray*}
$$
  \begin{array}{ll}
  J(w) & = %\eql 
     \left\{ \rho^{2} c^{2} +  (1 - \rho)^{2} c^{2} \right\}
      \left( 1 + \rho^{2} + \cdots + \rho^{2n}  + \cdots \right)  \\
      & = %\eql 
     \dfrac{\,\rho^{2} +  (1 - \rho)^{2} \,}{\, 1 - \rho^{2} \,}c^{2}~~~~~~(0 < \rho < 1).
  \end{array}
$$
%\end{eqnarray*}
%
%
Figure 1 in the Appendix shows that 
\( %\begin{equation}
  \min_{0 \le x < 1} \dfrac{\,x^{2} +  (1 - x)^{2} \,}{\, 1 - x^{2} \,} 
  %\nonumber
\) %\end{equation}
is attained at $ \hat{x} = 2 - \phi $ with the minimum value 
\begin{equation}
  \dfrac{\,(2 - \phi)^{2} +  \left\{1 - (2 - \phi)\right\}^{2} \,}{\, 1 - (2 - \phi)^{2} \,} 
   \dql \phi - 1. 
  \nonumber 
\end{equation}



%******************************************************************
\section{An Illustrative Graph. }                     \label{sec:2}
%******************************************************************
Let us now describe a graph which has dual Golden extremum points
in the previous section.
The graph is $ x = f(u) = \dfrac{u^2 + (1-u)^2}{1-u^2}$. 
See Figure 1 in Appendix.
%   For this function, it is seen that two inequalities:
For this function, it is seen that two equalities:
\begin{equation}                        \label{eq:2.1}
    \min_{0<u<1}f(u) 
  = \min_{0<u<1}\dfrac{u^2 + (1-u)^2}{1-u^2}
  = -1 + \phi %\quad \mbox{at} \quad u = 2 -\phi, 
\end{equation} 
and 
$   \max_{1<u<\infty}f(u) 
  = \max_{1<u<\infty}\dfrac{u^2 + (1-u)^2}{1-u^2}
  = -\phi$ hold.   %$\quad \mbox{at} \quad u = 1 + \phi$,
%   Equivalently, the latter inequality is that
Equivalently, the latter equality is that
\begin{equation}                        \label{eq:2.2}
    \min_{1<u<\infty}\{-f(u)\} 
  = \min_{1<u<\infty}\dfrac{u^2 + (1-u)^2}{u^2-1}
  = \phi  %\quad \mbox{at} \quad u = 1 + \phi.
\end{equation} 
\noindent
%  The first equality (\ref{eq:2.1}) attains iff $ \hat{u} = 2 - \phi $,
The minimum in (\ref{eq:2.1}) attains iff $ \hat{u} = 2 - \phi $, 
%  and the second equality (\ref{eq:2.2}) attains iff $ u^{*} = 1 + \phi. $
and the minimum in (\ref{eq:2.2}) attains iff $ u^{*} = 1 + \phi. $ 
Thus we have the inequality
%\begin{eqnarray*}
% f(u) \lf -2 ~~~\mbox{on}~~(-\infty, -1) \nonumber \\
\[
 f(u) \geq %\nge 
   -1 +\phi ~~~\mbox{on}~~(-1, 1) \quad \mbox{and}  \quad  %\nonumber 
 f(u) \leq %\nle 
   -\phi ~~~~\mbox{on}~~(-\infty, -1) \cup (1, \infty).   %\nonumber 
\]
%\end{eqnarray*}
%
Refer to the shape for this graph in the Figure 1 of Appendix.


%\section{Allocation processes}

\section{Dynamic Programming of the discrete-time system.}     
                                                        \label{sec:DP}

The conceptual cluster of Dynamic Programming are investigated 
%                                                 ^^^^^^^^^^^^
throughout the mathematics. Not only the analytical aspect of
optimization method, but also the investigate problem
with a repeated structure.
%                ^^^^^^^^
To give a useful explanation and an interesting implication,
we show some explicitly solvable problems.  
%                               ^^^^^^^^^ 
%   ��L�̕��������������Ă��Ă��܂�
%   
%


First the general setting of Dynamic Programming problem 
are illustrated. It is composed as 
$(S,A,r,T)$.
Let $S$ be a state space in the Euclidean space
$R$ and $A=(A_x), A_x \subset R, x \in S$ means a feasible action
space depending on a current state $x \in S$.
The immediate reward is a function of 
$r=r(x,a,t), x \in S, a \in A_x, t=0,1,2,\cdots  $.
And the terminal reward $K=K(x), x \in S$ is given.
The transition law from the current $x$ to the new $y = m(x,a,t)$
by the action or decision $ a \in A_x$ at a time $t$.
If the transition law $m(x,a,t)$ does not depend $t$,
it is called a stationary $m(x,a,t)$ and we treat it in this paper.
Here we consider additive costs and the optimal value of $a_t$
will be depend on the decision history.
Assume its value at time $t$ denoted by $x_t$, which enjoy the
following properties:
 \begin{itemize}
    \item [(a)] The value of $x_t$ is observable at time $t$.
%    \item [(b)] The sequence $\{ x_t \}$ follow a recurrence in time:
    \item [(b)] The sequence $\{ x_t \}$ follows a recurrence in time:
   \begin{equation}\label{eq:21}
     x_{t+1} = m(x_t, a_t, t). 
   % (2.1)
   \end{equation}
     It is termed that the function 
     $y = m(x,\pi(x,t),t)$ means a move from the current $x$ to the 
     next $y$ at $t$ so 
     the law of motion or the plant equation by adapting a policy
     $a= \pi(x,t)$.
    \item [(c)] The set of $a_t$ may adopt depends on $x_t$ and $t$.
    \item [(d)] The cost function $C_\pi(x,t) $ starting a state $x$
    at time-to-go $T_t = T-t$ to optimize 
    over all policies $\pi$ has the additive form;
   \begin{equation}\label{eq:22}
      C_\pi(x_0,t_0 ) = \sum_{t=t_0}^{T-1} r(x_t,a_t,t) + K(x_T)
   % (2.2)
   \end{equation}
   with $x_0 = x_{t_0}$ and 
      $$
       x_{t+1} = m(x_t,\pi(x_t,t),t ), \quad a_t = \pi(x_t,t)
      $$
   where $T$ is a given finite planning horizon.
   % If $T < \infty$, the problem is a finite horizon of planning 
   % problem and alternatively if it is $T=\infty$ and 
   %\begin{equation}\label{eq:23}
   %   C = \sum_{t=0}^{\infty} g(x_t,a_t,t), 
   %% (2.3)
   %\end{equation}
   % provided it is convergent,
   % is an infinity horizon of planning problem.
 \end{itemize}

Let 
$$
 F(x,t) = \inf_{\pi} C_{\pi }(x,t).
$$
%  It is well known that the optimization with state structure of 
%  recursively satisfies the optimality equation(DP equation):
It is well known that the sequence $ \{F(\cdot,t)\} $ 
satisfies the optimality equation(DP equation):
 \begin{equation}\label{eq:24}
    F(x,t+1) = \inf_{a \in A_x} [r(x, a, t) 
    + F(m(x,a,T_t), t)]
 % (2.4)
 \end{equation}
with the boundary $F(x,T) = K(x)$
where $T_t = T-t$ for $x \in S$, $0 \leq t < T $.

All of these are referred from text books by Bertsekas\cite{Ber76}, 
Whittle\cite{Whi82}, Sniedovich\cite{Sni92}, etc.

The relation between The Golden ratio formula and Fibonacci sequence 
is known as \cite{Dun97} etc.
To produce the Fibonacci sequence, it is a good example in
a recursive programming \cite{Kno97}.
Also the Fibonacci sequence are related with continued fraction.
For the notation of continued fraction, 
we adopt ourself to the following notations:
\[  b_0 + \cfrac{c_1}{b_1 +
           \cfrac{c_2}{b_2 +
            \cfrac{c_3}{b_3 +
              \cfrac{c_4}{b_4+ \cdots }}}}
= b_0 + \dfrac{c_1}{b_1 +} \dfrac{c_2}{b_2 +} 
         \dfrac{c_3}{b_3 + } \dfrac{c_4}{b_4 + } \cdots. 
\] 

%   Note that the Golden number satisfy the quadratic equation:
Note that the Golden number satisfies 
$\phi^2 = 1 + \phi$. 
By using this relation repeatedly,  
$\phi = 1 + \dfrac{1}{\phi} = 1+ \dfrac{1}{ 1 + \dfrac{1}{\phi}} 
= 1 + \dfrac{1}{1+}{\dfrac{1}{\phi }}
= 1 + \dfrac{1}{1+}\dfrac{1}{1+}{\dfrac{1}{\phi }}
= 1 + \dfrac{1}{1+}\dfrac{1}{1+}\dfrac{1}{1+}\cdots   $.
Similarly the reciprocal 
(or sometimes called as a dual Golden number) is denoted
$\phi^{-1}= 0.618\cdots = \phi - 1 = \dfrac{1}{\phi } 
= \dfrac{1}{1+}{\dfrac{1}{\phi }}
= \dfrac{1}{1+}\dfrac{1}{1+}{\dfrac{1}{\phi }}
= \dfrac{1}{1+}\dfrac{1}{1+}\dfrac{1}{1+}\cdots   $.

This reproductive property suggests our fundamental claim for 
%    the following a typical example of Dynamic Programming.
the following typical example of Dynamic Programming.
%   Before we solve the problem, let us induce sequences
Before we solve the problem, let us induce a sequence
$ \{\phi_n\}$  as
\begin{equation}                                            \label{eq:2-5}
  \phi_{n+1} = 1 + \dfrac{1}{1 + {1}/{\phi_n} } 
    = 1 + \dfrac{1}{1+}\dfrac{1}{\phi_n} \quad (n \ge 1),\ 
  \phi_1 = 1.
\end{equation}
Also let $\{\hat{\phi_n}\}$ as
\begin{equation}                                            \label{eq:2-6}
  \begin{array}{l}
  \hat\phi_{n+1} 
    = \dfrac{1}{1+}\dfrac{1}{1+}{\hat\phi_n} \quad (n \ge 1), \ 
  \hat\phi_1 = 1, \\
  i.e. \\ 
  \hat\phi_{n+1}^{-1} = \dfrac{1}{\hat\phi_{n+1}}
    =  1+ \dfrac{1}{1 + \hat\phi_n } 
    = 1 + \dfrac{1}{1+}{\hat\phi_n} \quad (n \ge 1). 
  \end{array}
\end{equation}

\noindent
The sequence $\{ \phi_{n} \} $ of (\ref{eq:2-5})  satisfies
\[
 \begin{array}{ll}
  \phi_{n+1} % & = 1 + \cfrac{1}{1+\cfrac{1}{\phi_n}} 
             %   = 1 + \dfrac{1}{1+}\dfrac{1}{\phi_n} \\[2ex] 
          & = 1 + 
               \dfrac{1}{1+}\dfrac{1}{1+}\dfrac{1}{1+}
               \dfrac{1}{\phi_{n-1}} \\
          & = 1 + 
               \dfrac{1}{1+}\dfrac{1}{1+}\dfrac{1}{1+}
               \dfrac{1}{1+}\dfrac{1}{1+}
               \dfrac{1}{\phi_{n-2}} \\
 \end{array}
\]


\noindent
Similarly $\{ \hat\phi_{n} \} $ of (\ref{eq:2-6}) 
satisfies
\[
 \begin{array}{ll}
  \dfrac{1}{\hat\phi_{n+1}} 
          % & = 1 + \dfrac{1}{1 + \hat\phi_{n}}
          %   = 1 + \dfrac{1}{1+}\hat\phi_{n} \\ 
          & = 1 + 
               \dfrac{1}{1+}\dfrac{1}{1+}\dfrac{1}{1+}
               {\hat\phi_{n-1}} \\
          & = 1 + 
               \dfrac{1}{1+}\dfrac{1}{1+}\dfrac{1}{1+}
               \dfrac{1}{1+}{\hat\phi_{n-2}} \\
 \end{array}
\]


\noindent
>From this definition, it is seen easily that
\begin{equation}
   \lim_{n \rightarrow \infty } \phi_n = \phi  = (\sqrt{5}+1)/2
\end{equation}
\begin{equation}
   \lim_{n \rightarrow \infty } \hat\phi_n  
   =  1/\phi = (\sqrt{5}-1)/2
\end{equation}





%=== section 4 ================================================
\section{Linear-Quadratic Control problem.}                \label{LQ}
%*******************************************************************

%  The Linear Quadratic(LQ) control problem is to consider
%  minimizing a control for the linear system
%  over the quadratic cost function.
The Linear Quadratic (LQ) control problem is to minimize 
the quadratic cost function over the linear system. 
If the state of system $\{ x_t \}$ moves on
\begin{equation}
 %x_{n+1} =  x_n +  u_n + \epsilon_n , \ n=1,2,\cdots
 x_{t+1} =  x_t +  a_t, \ t=0,1,2,\cdots
\end{equation}
with $x_0 =1$ by an input control $\{ a_t, -\infty < a_t < \infty \}$
%with a random noize $\epsilon_n$ such that $E\epsilon_n =0$
. 
The cost incured is
\begin{equation}
  %\sum_{n=1,2,\cdots,n} E(x_n^2 + u_n^2).
  \sum_{t=0}^{T-1} (x_t^2 + a_t^2) + x_T^2.
\end{equation}



%  LQ of the DP equation
Then DP equation of LQ is 
\begin{equation}                                            \label{eq:LQ} 
   v_{t+1}(x) 
 = \min_{a \in A_x} \left\{ r(a,x) + v_{t}(a+x)\right\}
\end{equation}
where
\[ 
  \begin{array}{l}
    r(a,x) = a^2 + x^2, \\
    a \in A_x = (-\infty, \infty ),
    x \in (-\infty, \infty ) 
  \end{array}
\] 




\begin{thm}
 The solution of (\ref{eq:LQ}) is given by 
 \[\left\{
  \begin{array}{l}
   v_0(x) = \phi_T x^2 \\[1.2ex]
   v_t(x) = \phi_{T-t} x^2, t= 1,2,\cdots 
  \end{array} \right.
 \] 
using the Golden number related sequence $\{ \phi_n \} $ by
(\ref{eq:2-5}).
\end{thm}

\noindent
{\it (Proof)}
%   The proof is immediately obtained by the elementary quadratic
The proof is immediately obtained by an elementary quadratic
minimization and then the mathematical induction.
$\Box$

%==== section 5 =======================================
\section{Allocation problem.}                             \label{Alloc}
%*******************************************************************

Allocation problem or sometime called as partition problem 
%  is the problem of the form
is of the form
\begin{equation}                                          \label{eq:AL} 
   v_{t+1}(x) 
 = \min_{a \in A_x} \left\{ r(a,x) + v_{t}(a)\right\}
\end{equation} 
for $ t= 0,1,2,\cdots, T,$ where
\[ 
  \begin{array}{l}
    r(a,x) = a^2 + (x - a)^2, \\
    a \in A_x = [0, x],
    x \in (-\infty, \infty ) .
  \end{array}
\] 



\begin{thm}
The solution of (\ref{eq:AL}) is given by using the dual
golden number as
\[
\left\{
  \begin{array}{l}
   v_0(x) = \hat\phi_T x^2 \\[1.2ex]
   v_t(x) = \hat\phi_{T-t} x^2, t= 1,2,\cdots 
  \end{array} \right.
\] 
\end{thm}

\noindent
{\it (Proof)}
Using the Schwartz inequality, the following holds
immediately:
For given positive constants $A$ and $B$ with a fixed $x$,
$$
 \min_{0 \le a \le 1} \{ A a^2 + B(x-a)^2 \} = \dfrac{x^2}{1/A + 1/B}. 
$$ 
So the proof could be done inductively. $\Box$ 


\vspace{2ex}
\noindent
{\bf Remark 1 :} We note here that the number $\phi^{-1}= 0.618\cdots$ of reciprocal of 
the Golden number is called sometimes 
{\it Dual Golden number}. The above two problems are closely related.

\vspace{2ex}
\noindent
{\bf Remark 2 :} It is seen that the same quadratic function
%   of the form; $ v(x) = C x^2$ where $c$ is a constant, 
  of the form; $ v(x) = c x^2$ where $c$ is a constant, 
 becomes a solution if
 the DP equation is, for Allocation and LQ,
\begin{equation}
  \label{eq:AL2} 
   v_{t+1}(x) 
 = \min_{a \in A_x} 
   \left\{ r(a,x) + 2 \int_{0}^{a}v_{t}(y)/y \, dy\right\}
\end{equation} 
\begin{equation}
  \label{eq:LQ2} 
   v_{t+1}(x) 
 = \min_{a \in A_x} 
   \left\{ r(a,x) + 2 \int_{0}^{a+x}v_{t}(y)/y \, dy \right\}
\end{equation}
respectively. Refer to \cite{Iwa05}.



\section{Monotone Stopping Game.}           \label{sec:MonoStop}

A monotone rule is introduced to sum up individual declarirations in 
a multi-variate stopping problem\cite{YNK82}. The rule is defined 
by a monotone logical function and is equivalent to the winning class 
of Kadane\cite{Kad78}.
There given $p$-dimensional random process $\{X_n; n=1,2,\cdots \}$ 
and a stopping rule $\pi$ by which the group decision determined
from the declararion of $p$ players at each stage.
The stopping rule is $p$-variate $\{0,1\}$-valued monotone 
logical function.  We consider two cases of rules with $p = 3$ 
as follows:
\begin{equation}                                 \label{rule(a)}
  \pi(x_1,x_2,x_3) = x_1 + x_2
\end{equation}
and
\begin{equation}                                 \label{rule(b)}
  \pi(x_1,x_2,x_3) = x_1 x_2 + x_1 x_3.
\end{equation}
  That is, in case of the case (\ref{rule(a)}), 
if either of player 1 and 2 declares
stop, then the system stops neglecting of $x_3$. 
In case of (\ref{rule(a)}), we have the system stops when 
either of player 1 and 2 declares stop accompanying with player 1.
%
\begin{figure}
%  \begin{center}
  \begin{minipage}{.5\textwidth}
  \begin{center}
  \begin{tabular}[t]{|ll|c|} \hline               \label{fig:1}
  $x_1$ &  $x_2$ &  $\pi(x_1,x_2,x_3)$ \\ \hline
    0   &    0   &   0 \\
    0   &    1   &   1 \\
    1   &    0   &   1 \\
    1   &    1   &   1 \\ \hline
  \end{tabular}
  \caption{$\pi(x_1,x_2,x_3) = x_1+x_2$ for any $ x_3$}
  \end{center}
  \end{minipage}
%  �󔒍s�Ȃ�
  \begin{minipage}{.5\textwidth}
  \begin{center}
  \begin{tabular}[t]{|lll|c|} \hline               \label{fig:2}
  $x_1$ &  $x_2$ & $x_3$ &  $\pi(x_1,x_2,x_3)$ \\ \hline
    0   &    0   &   0   &    0 \\
    0   &    0   &   1   &    0 \\
    0   &    1   &   0   &    0 \\
    0   &    1   &   1   &    0 \\
    1   &    0   &   0   &    0 \\
    1   &    0   &   1   &    1 \\
    1   &    1   &   0   &    1 \\
    1   &    1   &   1   &    1 \\ \hline
  \end{tabular}
  \caption{$\pi(x_1,x_2,x_3) = x_1x_2 + x_1x_3$}
  \end{center}
  \end{minipage}
%  \end{center}
\end{figure}
%
Without loss of generality, we can assume that each $X_n$ takes
the uniformly distribution on $[0,1]$. 
Then equilibrium expected values for each player is given as
\begin{center}
 {\doublerulesep=\arrayrulewidth  % �r���̑�����2�{�� \hline*2
 \begin{tabular}{l|ccc} \hline\hline \hline \\[-2ex]    
      & Player 1 & Player 2 & Player 3 \\[1.5ex] \hline \hline\hline \\[-1.5ex]
   $\pi(x_1,x_2,x_3) = x_1 + x_2$ & $\dfrac{\sqrt{5}-1}{2}$ & 
       $\dfrac{\sqrt{5}-1}{2}$ & $0.5$     \\[2ex] \hline \\[-1.5ex]
   $\pi(x_1,x_2,x_3) = x_1x_2 + x_1x_3$ &  $1$   &  $\dfrac{\sqrt{5}-1}{2}$    
       & $\dfrac{\sqrt{5}-1}{2}$    \\[2ex] \hline\hline 
 \end{tabular}
 } %doublerule
 \\[1.5ex] 
 {Table 1. The equilibrium expected value for each players.}
\end{center}

\vspace{2ex}
In order to derive the value $ \phi^{-1} = \dfrac{\sqrt{5}-1}{2}$ , 
we consider an equilibrium stopping strategy of threshold type 
in the form $\{X_n > a\}$ for some $a$.
Bellman type equation for this game version will be given as 
\cite{YNK82}.
That is, each player declares  ^^ ^^ stop" or ^^ ^^ continue" 
if the observed value exceeds some $a$ or not. 
The event of the occurence is denoted by 
$D_n^{i} = \{ \mbox {Player $i$ declares stop}\}$.
Two trivial cases are the whole event $\Omega $ and the empty event 
$\emptyset$. 

In generally a logical function is assumed ^^ ^^ monotone" so
its function can be written as
$$
   \pi(x^1,\cdots, x^p ) = x^i \cdot \pi(x^1,\cdots , \stackrel{i}{1} , 
       \cdots x^p) + \overline{ x^i} \cdot \pi(x^1,\cdots , \stackrel{i}{0} , 
       \cdots x^p)  \quad x^i \in \{ 0, 1\} \,\, \,  \forall i
$$ 
where $\overline{ x^i} = 1-  x^i$.  Corresponding to this expression,
$$
   \Pi(D^1,\cdots, D^p ) = D^i \cdot \Pi(D^1,\cdots , \stackrel{i}{\Omega } , 
       \cdots D^p) + \overline{ D^i} \cdot 
       \Pi(D^1,\cdots , \stackrel{i}{\emptyset } , \cdots D^p) 
$$ 
where $\overline{ D^i} $ is the complement of the event $D^i$.
The general equation for the expected for player $i$ equals
\begin{equation}                                    \label{eq:101}
   E\left[(X_n^i - v^i)^{+} 
      {\bf 1}_{\Pi(D_n^1,\cdots , \stackrel{i}{\Omega}, \cdots D_n^p)}\right]
   + E\left[(X_n^i - v^i)^{-} 
      {\bf 1}_{\Pi(D_n^1,\cdots , \stackrel{i}{\emptyset}, 
      \cdots D_n^p)}\right]
\end{equation}
where $D_n^i = \{ X_n^i \geq v^i\}$ and 
$(x)^+ = \max\{x,0\}, (x)^- = \min\{x,0\}$. 
If we assume an independence case 
between player's random variable $X_n^i$ for each $i$. The equation
(\ref{eq:101}) becomes as
\begin{equation}                                    \label{eq:102}
   \beta_n^{\Pi(i) } E\left[(X_n^i - v^i)^{+}\right]
   - \alpha_n^{\Pi(i) } E\left[(X_n^i - v^i)^{-} \right]
\end{equation}
where
{rule(a)}Our objection is to find an equilibrium strategy and values of playes 
for a given monotone rule as the rule (\ref{rule(a)}) and (\ref{rule(b)}).
A sequence of expected value (a net gain) under the situation formulated 
in the section is obtained as 
\begin{equation}                                    \label{eq:103}
   v_{n+1}^i = v_n^i + \beta_n^{\Pi(i) } E\left[(X_n^i - v_n^i)^{+}\right]
   - \alpha_n^{\Pi(i) } E\left[(X_n^i - v_n^i)^{-} \right]
\end{equation}
for player $ i = 1, \cdots, p$ and $n$  denotes a time-to-go.
The details refer to Theorem 2.1 in YKN\cite{YNK82}.
%
Under these derivation, now we are able to calculate the optimal (equilibrium) 
value $ v^i = \lim_{n } v_n^i$ for player $i= 1,2,3$ for the rule (\ref{rule(a)}) 
and (\ref{rule(b)}) in the table .



\newpage
%*****************************
%\subsection{$ x = f(u) $}
%*****************************
%
\clearpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                  �} 1-1~~ �Ȑ� $ x = f(u) $  
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\vspace*{70mm}\hspace*{00mm}  % \vspace*{60mm}\hspace*{00mm}
%
%
\begin{center}
\unitlength=2mm
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\bezier{500}(10.5,  -40)(11.5,  -24)(15,  -20)   % C1  ��������  %
%\curve(15,  -20,  26.18,-16.18,  40,-19)   % C2   �E������  %
\bezier{500}(15,  -20)(23,-13)(40,-19)   %CC2   �E������  %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  �� %%%%%%%%%%%%%%%%%%%%
\put(-30,38){\makebox(0,0){{\bf 7. Apppendix.}}}  %<<<<<<<<<<<<<<
%
\put(-30,30){\makebox(0,0){$
f(u) \dql \dfrac{u^{2} + (1 - u)^{2}}{1 - u^{2}}
$}}
%
\put(-29,22){\makebox(0,0){$
f^{\prime}(u) \dql (-2)\dfrac{u^{2} - 3u + 1}{(u^{2} - 1)^{2}}
$}}
%
\put(-28,14){\makebox(0,0){$
f^{\prime\prime}(u) \dql 2\dfrac{2u^{3} - 9u^{2} + 6u - 3}{(u^{2} - 1)^{3}}
$}}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  �� %%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  ������ %%%%%%%%%%%%%%%%%%%%
\put(28,30){\makebox(0,0){The golden ratio}}
%
\put(28,22){\makebox(0,0){$
\dfrac{1 + \phi}{-(-\phi)} \dql \dfrac{-1 + \phi}{2 - \phi} \dql 
\dfrac{\phi}{1}
$}}
%
\put(28,14){\makebox(0,0){$
\dfrac{-(-\phi)}{1} \dql \dfrac{1}{-1 + \phi} \dql \dfrac{\phi}{1}
$}}
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  ������ %%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\put(42.4,00){\makebox(0,0){$u$}}
\put(00,37.4){\makebox(0,0){$x$}}
\put(40,-00.13){\makebox(0,0){$\rightarrow$}}
\put(00,35){\makebox(0,0){$\uparrow$}}
%
%\put(85,20){\makebox(0,0){$t$}}
%\put(37,55){\makebox(0,0){$y$}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\put(00,00){\makebox(0,0){{$\bullet$}}}   %  $\bullet$   ���_
\put(-02,-02){\makebox(0,0){$O$}}   %  {$O$}} ���_
%\curve(-40,0,40,0)   %      �� u ��
\put(00,00){\line(1,0){40}}   %      �c x �� �E��
\put(00,00){\line(-1,0){40}}   %      �c x �� ����
%\curve(0,35,0,-45)   %      �c x ��
\put(00,00){\line(0,1){35}}   %      �c x �� �㔼
\put(00,00){\line(0,-1){45}}   %      �c x �� �㔼
%%%\curve(-40,0,40,0)   %      �� u ��
%%%\curve(0,35,0,-45)   %      �c x ��
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  u = 1  �Q�ߐ�
\begin{dottedjoin}{0.8}         %   epic  is needed
\thicklines
\jput(10,33){\makebox(0,0){$\cdot$}}
\jput(10,-43){\makebox(0,0){$\cdot$}}
\end{dottedjoin}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  u = -1  �Q�ߐ�
\begin{dottedjoin}{0.8}         %   epic  is needed
\thicklines
\jput(-10,33){\makebox(0,0){$\cdot$}}
\jput(-10,-43){\makebox(0,0){$\cdot$}}
\end{dottedjoin}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  x = -2  �Q�ߐ�
\begin{dottedjoin}{0.8}         %   epic  is needed
\thicklines
\jput(-40,-20){\makebox(0,0){$\cdot$}}
\jput(40,-20){\makebox(0,0){$\cdot$}}
\end{dottedjoin}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  x = -1 + p  �Q�ߐ�
\begin{dottedjoin}{0.4}         %   epic  is needed
\thicklines
\jput(00,6.18){\makebox(0,0){$\cdot$}}
\jput(10,6.18){\makebox(0,0){$\cdot$}}
\end{dottedjoin}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\put(3.82,00){\makebox(0,0){$\bullet$}}   % u����_ (2-p, 0)
\put(3.82,-2){\makebox(0,0){$2 - \phi$}}   % u����_ (2-p, 0)  2 - \phi
\put(-10,0){\makebox(0,0){$+$}}   % u����_ (-1, 0)  +
\put(-12,2){\makebox(0,0){$-1$}}   % u����_ (-1, 0)  -1
\put(10,0){\makebox(0,0){$+$}}   % u����_ (1, 0)  +
\put(9,2){\makebox(0,0){$1$}}   % u����_ (1, 0)  1
\put(15,0){\makebox(0,0){$+$}}   % u����_ (3/2, 0)  +
\put(15,2){\makebox(0,0){$3/2$}}   % u����_ (3/2, 0)  3/2
\put(20,0){\makebox(0,0){$+$}}   % u����_ (2, 0)  +
\put(20,2){\makebox(0,0){$2$}}   % u����_ (2, 0)  2
\put(30,0){\makebox(0,0){$+$}}   % u����_ (3, 0)  +
\put(30,2){\makebox(0,0){$3$}}   % u����_ (3, 0)  3
\put(3.82,6.18){\makebox(0,0){$\bigstar$}}   % �ɏ��_ (2-p, p-1)
\put(26.18,00){\makebox(0,0){$\bullet$}}   % u����_ (1+p, 0)
\put(26.18,2){\makebox(0,0){$1 + \phi$}}   % u����_ (1+p, 0)  1 + \phi
\put(26.18,-16.18){\makebox(0,0){$\bigstar$}}   % �ɑ�_(1+p, -p)
\put(00,6.18){\makebox(0,0){$\bullet$}}   % �c���ؕ�(0, p-1)
\put(-4.5,6.18){\makebox(0,0){$-1 + \phi$}}   % �c���ؕ�(0, -1+p)  -1 + \phi
\put(-2,10){\makebox(0,0){$1$}}   % �c���ؕ�(0, 1)  1 
\put(-2,20){\makebox(0,0){$2$}}   % �c���ؕ�(0, 2)  2 
\put(00,-16.18){\makebox(0,0){$\bullet$}}   % �c���ؕ�(0, -p)
\put(00,-20){\makebox(0,0){$\bullet$}}   % �c���ؕ�(0, -2)
\put(00,10){\makebox(0,0){$\bullet$}}   % �c���ؕ� (0, 1)
\put(00,20){\makebox(0,0){$+$}}   % �c���ؕ� (0, 2)
\put(15,-20){\makebox(0,0){$\bullet$}}   % x = -2�Q�ߐ��ؕ� (1.5, -2)
\put(-3,-21.3){\makebox(0,0){$-2$}}   % x����_ (0, -p)  -2
\put(-3,-16.18){\makebox(0,0){$-\phi$}}   % x����_ (0, -p)  -\phi
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\thicklines    
\drawline(0,0) (26.18, 00) (26.18,-16.18) (00,-16.18)  % epic
\drawline(0,0) (3.82, 00) (3.82,6.18) (00, 6.18)  % epic
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{picture}
\end{center}
\vspace*{88mm}
\begin{center}
%�} 1~~ �Ȑ� $ x = f(u) $;~~~~Dual extremum points $\bigstar$ are golden 
%,~~ f(u) = \dfrac{u^{2} + (1 - u)^{2}}{1 - u^{2}} $ 
Figure 1.~~ The curve $ x = f(u) $ has dual golden extremum points with a marked $\bigstar$. 
%,~~ f(u) = \dfrac{u^{2} + (1 - u)^{2}}{1 - u^{2}} $
\end{center}
\vspace*{6mm}


%\clearpage


\newpage
%*******************************************************************
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  %�Q�O�O�T�N�V���A�����D�Q�V�\�S�R�D

%\bibitem{Iwa87}
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%, Yokohama,2005, pp.173--180???

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%�����W��u�s�m�����̉��ł̈ӎv����Ɛ������f�v�A���吔�����u���^�A�Q�O�O�U�N�Aunder submission. 
%�P�H�H�H�A�Q�O�O�U�N�H���A�����D�R�H�H�\�S�H�H�D


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\end{thebibliography}


\end{document}

