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

Chapter 5 Corollary 5.1.7

$0 \lt x \lt 1$を満たす$x$について $$F\left(\frac{1}{2},\frac{1}{2};1;1-\frac{\left(x-1\right)^2}{\left(x+1\right)^2}\right)=(1+x)\,F\left(\frac{1}{2},\frac{1}{2};1;x^2\right)$$ が成り立つ。

証明の方針は単純です。$1-\frac{\left(x-1\right)^2}{\left(x+1\right)^2}=\frac{4\,x}{\left(x+1\right)^2}$を示せば、 $$F\left(\frac{1}{2},\frac{1}{2};1;\frac{4\,x}{(1+x)^2}\right)=(1+x)\,F\left(\frac{1}{2},\frac{1}{2};1;x^2\right)$$ より明らかです。

In [5]:
1-((x-1)/(x+1))^2;
Out[5]:
\[\tag{${\it \%o}_{5}$}1-\frac{\left(x-1\right)^2}{\left(x+1\right)^2}\]
In [6]:
xthru(%);
Out[6]:
\[\tag{${\it \%o}_{6}$}\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x+1\right)^2}\]
In [7]:
factor(expand(%));
Out[7]:
\[\tag{${\it \%o}_{7}$}\frac{4\,x}{\left(x+1\right)^2}\]
In [9]:
hypergeometric([1/2,1/2],[1],1-((x-1)/(x+1))^2)=(1+x)*hypergeometric([1/2,1/2],[1],x^2);
Out[9]:
\[\tag{${\it \%o}_{8}$}F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,1-\frac{\left(x-1\right)^2}{\left(x+1\right)^2}\right)=F\left( \left. \begin{array}{c}\frac{1}{2},\;\frac{1}{2}\\1\end{array} \right |,x^2\right)\,\left(x+1\right)\] 
SB-KERNEL:REDEFINITION-WITH-DEFUN: redefining MAXIMA::SIMP-HYPERGEOMETRIC in DEFUN
In [ ]: