Borwein's algorithm (others)

Jonathan and Peter Borwein devised various algorithms to calculate the value of π. The most prominent and oft-used one is explained under Borwein's algorithm. Other algorithms found by them include the following:

1. Cubical convergence, 1991:
   • Start out by setting

     $a_{0}={\frac {1}{3}}$

     $s_{0}={\frac {{\sqrt {3}}-1}{2}}$

   • Then iterate

     $r_{k+1}={\frac {3}{1+2(1-s_{k}^{3})^{1/3}}}$

     $s_{k+1}={\frac {r_{k+1}-1}{2}}$

     $a_{k+1}=r_{k+1}a_{k}-3^{k}(r_{k+1}^{2}-1)$

   Then a_k converges cubically against 1/π; that is, each iteration approximately triples the number of correct digits.

2. Quartical convergence, 1984:
   • Start out by setting

     $a_{0}={\sqrt {2}}$

     $b_{0}=0$

     $p_{0}=2+{\sqrt {2}}$

   • Then iterate

     $a_{n+1}={\frac {{\sqrt {a_{n}}}+1/{\sqrt {a_{n}}}}{2}}$

     $b_{n+1}={\frac {{\sqrt {a_{n}}}(1+b_{n})}{a_{n}+b_{n}}}$

     $p_{n+1}={\frac {p_{n}b_{n+1}(1+a_{n+1})}{1+b_{n+1}}}$

   Then p_k converges quartically against π; that is, each iteration approximately quadruples the number of correct digits. The algorithm is not self-correcting; each iteration must be performed with the desired number of correct digits of π.

3. Quintical convergence:
   • Start out by setting

     $a_{0}={\frac {1}{2}}$

     $s_{0}=5({\sqrt {5}}-2)$

   • Then iterate

     $x_{n+1}={\frac {5}{s_{n}}}-1$

     $y_{n+1}=(x_{n+1}-1)^{2}+7$

     $z_{n+1}=\left({\frac {1}{2}}x_{n+1}\left(y_{n+1}+{\sqrt {y_{n+1}^{2}-4x_{n+1}^{3}}}\right)\right)^{1/5}$

     $s_{m+1}={\frac {25}{(z_{n+1}+x_{n+1}/z_{n+1}+1)^{2}s_{n}}}$

     $a_{n+1}=s_{n}^{2}a_{n}-5^{n}\left({\frac {s_{n}^{2}-5}{2}}+{\sqrt {s_{n}(s_{n}^{2}-2s_{n}+5)}}\right)$

   Then a_k converges quintically against 1/π (that is, each iteration approximately quintuples the number of correct digits), and the following condition holds:

   $0<a_{n}-{\frac {1}{\pi }}<16\cdot 5^{n}\cdot e^{-5^{n}}\pi$

4. Nonical convergence:
   • Start out by setting

     $a_{0}={\frac {1}{3}}$

     $r_{0}={\frac {{\sqrt {3}}-1}{2}}$

     $s_{0}=(1-r_{0}^{3})^{1/3}$

   • Then iterate

     $t_{n+1}=1+2r_{n}$

     $u_{n+1}=(9r_{n}(1+r_{n}+r_{n}^{2}))^{1/3}$

     $v_{n+1}=t_{n+1}^{2}+t_{n+1}u_{n+1}+u_{n+1}^{2}$

     $w_{n+1}={\frac {27(1+s_{n}+s_{n}^{2})}{v_{n+1}}}$

     $a_{n+1}=w_{n+1}a_{n}+3^{2n-1}(1-w_{n+1})$

     $s_{n+1}={\frac {(1-r_{n})^{3}}{(t_{n+1}+2u_{n+1})v_{n+1}}}$

     $r_{n+1}=(1-s_{n+1}^{3})^{1/3}$

   Then a_k converges nonically against 1/π; that is, each iteration approximately multiplies the number of correct digits by nine.