I know that, as a hobbyist mathematician, this is generally a term we can use to express pi
\begin{equation*} \frac{\pi}{4} = 1 – \frac{1}{3} + \frac{1}{5} – \frac{1}{7} + \frac{1}{9} – \frac{1}{11} + \frac{1}{13} – \frac{1}{15} + \frac{1}{17} – \cdots \end{equation*}
This is great representation , and it works just fine
However, I've been introduced to some iterative series recently (they're great for finding the roots to an equation by the way) and I was wondering, if there was a iterative series for pi, could we get a more justified value?
I've looked online, and there only seems to be geometric expressions using sine and cosine.
I was wondering if a numerical formula could be derived.
Edit: An iterative series is a series much like an algorithm, for example (N+1) = root(N+ 2/N) the idea being this series will converge on a value
Best Answer
Well, there are iterative algorithms. There are two beautiful ones by the Borwein brothers, based on work by Ramanujan. Algorithm 1 involves the silver ratio, and Algorithm 2 involves the cube of the golden ratio.
I. Algorithm 1. Start with seed values:
$$y_0 = -1+\sqrt{2}$$
$$a_0 = 2(-1+\sqrt{2})^2$$
and two iterative rules,
$$y_{n+1} = \frac{1-(1-{y_n}^4)^{1/4}}{1+(1-{y_n}^4)^{1/4}}\tag1$$
$$a_{n+1} = a_n(y_{n+1}+1)^4-2^{2n+3}\,y_{n+1}\big(y_{n+1}^2+y_{n+1}+1\big)\tag2$$
Then,
$$\quad\quad\quad\lim_{n\to\infty} \frac{1}{a_n} = \pi\quad\text{(very fast)}$$
The difference grows quartically,
$$\quad\quad\quad\quad\frac{1}{a_n} - \pi \approx 4^{n+2} q^{4^n},\quad \text{where}\;q = e^{-2\pi}$$
Thus for $n=1,2,3$, the difference is about $10^{-10},\,10^{-42},\,10^{-172},$ or more than the fourth power of the previous. It's that fast.