To: heureka and others..........

If you calculate the value of yk+1 in (1) and feed it into (2) and repeat the iterations at least 2-3 times, what does the result converge to and what is the rate of convegence, i.e., is it: quadratic, cubic, quartic.....etc. Also, what is the source of this algorithm? Thanks and have fun!.

Set a0 = 6−4√2 and y0 = √2−1 and k=0, 1, 2, 3................... Iterate:

yk+1 = 1−(1−y^4 k)^1/4 / 1 + (1−y^4 k)^1/4 ..............................(1)

and ak+1 = ak(1 + yk+1)^4 − 2^(2k+3)yk+1(1 + yk+1 + y^2 k+1)..(2)

Guest Feb 11, 2016

#1**+10 **

**To: heureka and others..........**

**If you calculate the value of yk+1 in (1) and feed it into (2) and repeat the iterations at least 2-3 times, what does the result converge to and what is the rate of convegence, i.e., is it: quadratic, cubic, quartic.....etc. Also, what is the source of this algorithm? Thanks and have fun!.**

**Set a0 = 6−4√2 and y0 = √2−1 and k=0, 1, 2, 3................... Iterate:**

**yk+1 = 1−(1−y^4 k)^1/4 / 1 + (1−y^4 k)^1/4 ..............................(1)**

**and ak+1 = ak(1 + yk+1)^4 − 2^(2k+3)yk+1(1 + yk+1 + y^2 k+1)..(2)**

\(\begin{array}{rcll} \text{Borwein's algorithm:}\\\\ \end{array} \\ \begin{array}{rcll} \text{Quartic algorithm (1985)}\\\\ \end{array} \\ \text{Start out by setting[1]}\\ \begin{array}{rcll} a_0 & =& 2\big(\sqrt{2}-1\big)^2 = 6-4\sqrt{2}\\ y_0 & =& \sqrt{2}-1 \end{array} \\ \begin{array}{rcll} \\ \text{Then iterate}\\ \end{array} \\ \\ \begin{array}{rcll} y_{k+1} & =& \frac{1-(1-y_k^4)^{1/4}}{1+(1-y_k^4)^{1/4}} \\ a_{k+1} & =& a_k(1+y_{k+1})^4 - 2^{2k+3} y_{k+1} (1 + y_{k+1} + y_{k+1}^2) \end{array} \\ \\ \text{Then }~a_k \text{ converges quartically against }~ 1/\pi; \\ \text{ that is, each iteration approximately quadruples the number of correct digits.} \)

see: https://en.wikipedia.org/wiki/Borwein's_algorithm

heureka Feb 11, 2016

#3**0 **

heureka: Brilliant as usual!!. Thanks for that. It was discovered by two Canadian mathematicians, who are brothers, Jonathan Borwein and Peter Borwein. However, the algorithm is derived from modular equations of the great Indian mathematician, S. Ramanujan, even though, on the surface, it seems to have no resemblance to Chudnowsky-like formula, which is widely used in most modern calculations of Pi to trillions of digits.

Guest Feb 11, 2016