円周率.jp (http://xn--w6q13e505b.jp/formula/borwein.html)

Borwein 兄弟は AGM 系の公式に類似した公式として， n 次収束する公式が任意の n について存在する証明をし，そのいくつかについて具体的な方法を示している． 以下でそのいくつかを紹介する．

\[ x_0 = \sqrt{2}, \pi_0 = 2 + \sqrt{2}, y_1 = 2^{1/4} \] \[ \left\{ \begin{eqnarray} x_{n+1} &=& \dfrac{1}{2} \left( \sqrt{x_n} + \dfrac{1}{\sqrt{x_n}} \right) & {\rm for\ }n\geq 0\\ y_{n+1} &=& \dfrac{y_n\sqrt{x_n}+1/\sqrt{x_n}}{y_n+1} & {\rm for\ }n\geq 1\\ \pi_n &=& \pi_{n-1} \dfrac{x_n+1}{y_n+1} & {\rm for\ }n \geq 1 \end{eqnarray} \right. \] \[ {\rm 収束：} |\pi_n-\pi| \lt 10^{-2^{n+1}} \]

\[ a_0 = \frac{1}{3}, s_0 = \frac{\sqrt{3}-1}{2} \] \[ \left\{ \begin{eqnarray} r_{n+1} &=& \dfrac{3}{1+2(1-s_n^3)^{1/3}}\\ s_{n+1} &=& \dfrac{r_{n+1}-1}{2}\\ a_{n+1} &=& r_{n+1}^2 a_n - 3^n(r_{n+1}^2-1) \end{eqnarray} \right. \] \[ \frac{1}{a_n} \rightarrow \pi\quad (n\rightarrow\infty) \]

\[ a_0 = 6 - 4\sqrt{2}, y_0 = \sqrt{2} - 1 \] \[ \left\{ \begin{eqnarray} y_{n+1} &=& \frac{1-(1-y_n^4)^{1/4}}{1+(1-y_n^4)^{1/4}}\\ a_{n+1} &=& a_n(1+y_{n+1})^4 - 2^{2n+3}y_{n+1}(1+y_{n+1}+y_{n+1}^2) \end{eqnarray}\right. \] \[ {\rm 収束: }\ |a_n - 1/\pi| \lt 16\cdot4^n\cdot2e^{-4^n\cdot2\pi} \]

\[ s_0 = 5(\sqrt{5} - 2), a_0 = 1/2 \] \[ \left\{ \begin{eqnarray} x &=& \frac{5}{s_n} - 1\\ y &=& (x-1)^2 + 7\\ z &=& \left(\frac{x}{2}\left(y + \sqrt{y^2-4x^3}\right)\right)^{1/5}\\ s_{n+1} &=& \frac{25}{s_n(z+x/z+1)^2}\\ a_{n+1} &=& s_n^2 a_n - 5^n \left(\frac{s_n^2-5}{2} + \sqrt{s_n(s_n^2-2s_n+5)} \right) \end{eqnarray}\right. \] \[ {\rm 収束: }\ |a_n - 1/\pi| \lt 16\cdot 5^n e^{-\pi5^n} \]

\[ a_0 = 1/3, r_0 = (\sqrt{3}-1)/2, s_0 = (1-r_0^3)^{1/3} \] \[ \left\{ \begin{eqnarray} t &=& 1 + 2r_n\\ u &=& (9r_n(1 + r_n + r_n^2))^{1/3}\\ y &=& t^2 + tu + u^2\\ m &=& \dfrac{27(1+s_n+s_n^2)}{v}\\ a_{n+1} &=& ma_n + 3^{2n-1}(1-m)\\ s_{n+1} &=& \dfrac{(1-r_n)^3}{(t+2u)v}\\ r_{n+1} &=& (1-s_n^3)^{1/3} \end{eqnarray}\right. \] \[ \frac{1}{a_n} \rightarrow \pi \quad (n\rightarrow \infty) \]