B.C.Berndt: Number Theory in the Spirit of Ramanujanより

準備として$0\lt x \lt 1$の範囲で関数$F(x)$を $$F(x)=exp\left(-\pi\,\frac{{}_2 F_1\left(\frac{1}{2},\frac{1}{2};1;1-x\right)}{{}_2 F_1\left(\frac{1}{2},\frac{1}{2};1;x\right)}\right)$$ と定義します。


Chapter 5 Corollary 5.2.1
$0 \lt x \lt 1$の時 $$F(x^2)=F\left(\frac{4\,x}{(1+x)^2}\right)^2$$

今回はこの補題を証明していきます。

In [1]:
F(x):=exp(-%pi*hypergeometric([1/2,1/2],[1],1-x)/hypergeometric([1/2,1/2],[1],x));
Out[1]:
\[\tag{${\it \%o}_{0}$}F\left(x\right):=\exp \left(\frac{-\pi\,{\it hypergeometric}\left(\left[ \frac{1}{2} , \frac{1}{2} \right] , \left[ 1 \right] , 1-x\right)}{{\it hypergeometric}\left(\left[ \frac{1}{2} , \frac{1}{2} \right] , \left[ 1 \right] , x\right)}\right)\]

証明するべき式を確認します。

In [2]:
F(x^2)=F(4*x/(1+x)^2)^2;
Out[2]:
\[\tag{${\it \%o}_{1}$}e^ {- \frac{\pi\,F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,1-x^2\right)}{F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,x^2\right)} }=e^ {- \frac{2\,\pi\,F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,1-\frac{4\,x}{\left(x+1\right)^2}\right)}{F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,\frac{4\,x}{\left(x+1\right)^2}\right)} }\] 
SB-KERNEL:REDEFINITION-WITH-DEFUN: redefining MAXIMA::SIMP-HYPERGEOMETRIC in DEFUN

Corollary 5.1.7で示した式 $$F\left(\frac{1}{2},\frac{1}{2};1;1-\frac{\left(x-1\right)^2}{\left(x+1\right)^2}\right)=(1+x)\,F\left(\frac{1}{2},\frac{1}{2};1;x^2\right)$$ において、$x$を$\frac{1-x}{1+x}$で置換します。

In [3]:
Col517:hypergeometric([1/2,1/2],[1],1-((x-1)/(x+1))^2)=(1+x)*hypergeometric([1/2,1/2],[1],x^2);
Out[3]:
\[\tag{${\it \%o}_{2}$}F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,1-\frac{\left(x-1\right)^2}{\left(x+1\right)^2}\right)=F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,x^2\right)\,\left(x+1\right)\]
In [4]:
C2:subst((1-x)/(1+x),x,Col517),ratsimp;
Out[4]:
\[\tag{${\it \%o}_{3}$}F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,1-x^2\right)=\frac{2\,F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,\frac{x^2-2\,x+1}{x^2+2\,x+1}\right)}{x+1}\]

右辺の引数を因数分解します。

In [5]:
C3:substpart(factor(part(C2,2,1,2,3)),C2,2,1,2,3);
Out[5]:
\[\tag{${\it \%o}_{4}$}F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,1-x^2\right)=\frac{2\,F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,\frac{\left(x-1\right)^2}{\left(x+1\right)^2}\right)}{x+1}\]

因数分解した部分について以下の式が成り立つので、左辺で置き換えます。

In [6]:
A=factor(A),A=1-4*x/(x+1)^2;
Out[6]:
\[\tag{${\it \%o}_{5}$}1-\frac{4\,x}{\left(x+1\right)^2}=\frac{\left(x-1\right)^2}{\left(x+1\right)^2}\]
In [7]:
C4:substpart(1-4*x/(x+1)^2,C3,2,1,2,3);
Out[7]:
\[\tag{${\it \%o}_{6}$}F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,1-x^2\right)=\frac{2\,F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,1-\frac{4\,x}{\left(x+1\right)^2}\right)}{x+1}\]

Theorem 5.1.6では楕円積分$K(k)$について$K(\frac{2\,\sqrt{x}}{x+1})=(x+1)\,K(x)$　を示しました。これを超幾何関数の言葉で書くと以下の式になるのでした。

In [8]:
C5:(x+1)*hypergeometric([1/2,1/2],[1],x^2)=hypergeometric([1/2,1/2],[1],4*x/(x+1)^2);
Out[8]:
\[\tag{${\it \%o}_{7}$}F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,x^2\right)\,\left(x+1\right)=F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,\frac{4\,x}{\left(x+1\right)^2}\right)\]

%o7の両辺を%o8の両辺で割って両辺を入れ替えると以下の式を得ます。

In [9]:
C6:rhs(C4)/rhs(C5)=lhs(C4)/lhs(C5);
Out[9]:
\[\tag{${\it \%o}_{8}$}\frac{2\,F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,1-\frac{4\,x}{\left(x+1\right)^2}\right)}{F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,\frac{4\,x}{\left(x+1\right)^2}\right)\,\left(x+1\right)}=\frac{F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,1-x^2\right)}{F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,x^2\right)\,\left(x+1\right)}\]

両辺に$-\pi\,(x+1)$を掛けてから指数関数に代入すると以下の式を得ます。これは証明するべき式そのものです！

In [10]:
exp(-%pi*lhs(C6)*(x+1))=exp(-%pi*rhs(C6)*(x+1));
Out[10]:
\[\tag{${\it \%o}_{9}$}e^ {- \frac{2\,\pi\,F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,1-\frac{4\,x}{\left(x+1\right)^2}\right)}{F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,\frac{4\,x}{\left(x+1\right)^2}\right)} }=e^ {- \frac{\pi\,F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,1-x^2\right)}{F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,x^2\right)} }\]

念のため確認しておきましょう。

In [11]:
F(4*x/(x+1)^2)^2=F(x^2);
Out[11]:
\[\tag{${\it \%o}_{10}$}e^ {- \frac{2\,\pi\,F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,1-\frac{4\,x}{\left(x+1\right)^2}\right)}{F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,\frac{4\,x}{\left(x+1\right)^2}\right)} }=e^ {- \frac{\pi\,F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,1-x^2\right)}{F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,x^2\right)} }\]