Borwein's algorithm (others)

• Start out by setting

  <math>a_0 = \frac{1}{3}</math>

  <math>s_0 = \frac{\sqrt{3} - 1}{2}</math>

• Then iterate

  <math>r_{k+1} = \frac{3}{1 + 2(1-s_k^3)^{1/3}}</math>

  <math>s_{k+1} = \frac{r_{k+1} - 1}{2}</math>

  <math>a_{k+1} = r_{k+1} a_k - 3^k(r_{k+1}^2-1)</math>

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

• Start out by setting

  <math>a_0 = \sqrt{2}</math>

  <math>b_0 = 0</math>

  <math>p_0 = 2 + \sqrt{2}</math>

• Then iterate

  <math>a_{n+1} = \frac{\sqrt{a_n} + 1/\sqrt{a_n}}{2}</math>

  <math>b_{n+1} = \frac{\sqrt{a_n} (1 + b_n)}{a_n + b_n}</math>

  <math>p_{n+1} = \frac{p_n b_{n+1} (1 + a_{n+1})}{1 + b_{n+1}}</math>

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.

• Start out by setting

  <math>a_0 = \frac{1}{2}</math>

  <math>s_0 = 5(\sqrt{5} - 2)</math>

• Then iterate

  <math>x_{n+1} = \frac{5}{s_n} - 1</math>

  <math>y_{n+1} = (x_{n+1} - 1)^2 + 7</math>

  <math>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}</math>

  <math>s_{m+1} = \frac{25}{(z_{n+1} + x_{n+1}/z_{n+1} + 1)^2 s_n}</math>

  <math>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)</math>

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

<math>0 < a_n - \frac{1}{\pi} < 16\cdot 5^n\cdot e^{-5^n}\pi</math>

• Start out by setting

  <math>a_0 = \frac{1}{3}</math>

  <math>r_0 = \frac{\sqrt{3} - 1}{2}</math>

  <math>s_0 = (1 - r_0^3)^{1/3}</math>

• Then iterate

  <math>t_{n+1} = 1 + 2r_k</math>

  <math>u_{n+1} = (9r_k (1 + r_n + r_n^2))^{1/3}</math>

  <math>v_{n+1} = t_{n+1}^2 + t_{n+1}u_{n+1} + u_{n+1}^2</math>

  <math>w_{n+1} = \frac{27 (1 + s_n + s_n^2)}{v_{n+1}}</math>

  <math>a_{n+1} = w_{n+1}a_n + 3^{2n-1}(1-w_{n+1})</math>

  <math>s_{n+1} = \frac{(1 - r_n)^3}{t_{n+1} + 2u_{n+1})v_{n+1}}</math>

  <math>r_{n+1} = (1 - s_n^3)^{1/3}</math>

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