Borwein's Algorithm (Others)

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

     $a\_0\; =\; \backslash frac\{1\}\{3\}$

     $s\_0\; =\; \backslash frac\{\backslash sqrt\{3\}\; -\; 1\}\{2\}$

   • Then iterate

     $r\_\{k+1\}\; =\; \backslash frac\{3\}\{1\; +\; 2(1-s\_k^3)^\{1/3\}\}$

     $s\_\{k+1\}\; =\; \backslash frac\{r\_\{k+1\}\; -\; 1\}\{2\}$

     $a\_\{k+1\}\; =\; r\_\{k+1\}\; a\_k\; -\; 3^k(r\_\{k+1\}^2-1)$

   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

     $a\_0\; =\; \backslash sqrt\{2\}$

     $b\_0\; =\; 0$

     $p\_0\; =\; 2\; +\; \backslash sqrt\{2\}$

   • Then iterate

     $a\_\{n+1\}\; =\; \backslash frac\{\backslash sqrt\{a\_n\}\; +\; 1/\backslash sqrt\{a\_n\}\}\{2\}$

     $b\_\{n+1\}\; =\; \backslash frac\{\backslash sqrt\{a\_n\}\; (1\; +\; b\_n)\}\{a\_n\; +\; b\_n\}$

     $p\_\{n+1\}\; =\; \backslash frac\{p\_n\; b\_\{n+1\}\; (1\; +\; a\_\{n+1\})\}\{1\; +\; b\_\{n+1\}\}$

   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

     $a\_0\; =\; \backslash frac\{1\}\{2\}$

     $s\_0\; =\; 5(\backslash sqrt\{5\}\; -\; 2)$

   • Then iterate

     $x\_\{n+1\}\; =\; \backslash frac\{5\}\{s\_n\}\; -\; 1$

     $y\_\{n+1\}\; =\; (x\_\{n+1\}\; -\; 1)^2\; +\; 7$

     $z\_\{n+1\}\; =\; \backslash left(\backslash frac\{1\}\{2\}\; x\_\{n+1\}\backslash left(y\_\{n+1\}\; +\; \backslash sqrt\{y\_\{n+1\}^2\; -\; 4x\_\{n+1\}^3\}\backslash right)\backslash right)^\{1/5\}$

     $s\_\{m+1\}\; =\; \backslash frac\{25\}\{(z\_\{n+1\}\; +\; x\_\{n+1\}/z\_\{n+1\}\; +\; 1)^2\; s\_n\}$

     $a\_\{n+1\}\; =\; s\_n^2\; a\_n\; -\; 5^n\backslash left(\backslash frac\{s\_n^2\; -\; 5\}\{2\}\; +\; \backslash sqrt\{s\_n(s\_n^2\; -\; 2s\_n\; +\; 5)\}\backslash right)$

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

4. Nonical convergence:
   • Start out by setting

     $a\_0\; =\; \backslash frac\{1\}\{3\}$

     $r\_0\; =\; \backslash frac\{\backslash sqrt\{3\}\; -\; 1\}\{2\}$

     $s\_0\; =\; (1\; -\; r\_0^3)^\{1/3\}$

   • Then iterate

     $t\_\{n+1\}\; =\; 1\; +\; 2r\_n$

     $u\_\{n+1\}\; =\; (9r\_n\; (1\; +\; r\_n\; +\; r\_n^2))^\{1/3\}$

     $v\_\{n+1\}\; =\; t\_\{n+1\}^2\; +\; t\_\{n+1\}u\_\{n+1\}\; +\; u\_\{n+1\}^2$

     $w\_\{n+1\}\; =\; \backslash frac\{27\; (1\; +\; s\_n\; +\; s\_n^2)\}\{v\_\{n+1\}\}$

     $a\_\{n+1\}\; =\; w\_\{n+1\}a\_n\; +\; 3^\{2n-1\}(1-w\_\{n+1\})$

     $s\_\{n+1\}\; =\; \backslash frac\{(1\; -\; r\_n)^3\}\{(t\_\{n+1\}\; +\; 2u\_\{n+1\})v\_\{n+1\}\}$

     $r\_\{n+1\}\; =\; (1\; -\; s\_\{n+1\}^3)^\{1/3\}$

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

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