I have an equation for the value of PI from the Madhava-Gregory series (but with an ability to make it converge faster). It is quite an obvious equation. But I couldn't find these algorithms on the internet.
There are 2 of them one is the Madhava-Gregory series itself (modified):
$${\pi} = {K} \sum\limits_{n = 0}^\infty {\frac{{\left( -1 \right)^{n} {(\tan {\pi \over K})^{2n + 1}}}}{{2n + 1}}} \quad \small{(\text{where}\: K \ge 4)}$$
Here we have to calculate the value K and T (ie. $\tan {\pi \over K}$) externally. They are interlinked constants and are calculated by an algorithm which, (for the lack of a better word) I call TAN-DIP. The whole algorithm looks like this:
\begin{alignat*}{3} & \small{\text{Start TAN-DIP iteration with initial value of 1/2}} \nonumber\\ &\nonumber\\ v_1 &= ({1 \over 2}) \nonumber\\ v_2 &= \sqrt{{1 + v_1} \over 2} \nonumber\\ v_3 &= \sqrt{{1 + v_2} \over 2} \nonumber\\ &. \nonumber\\ &. \nonumber\\ &. \nonumber\\ v_{R} &= \sqrt{{1 + v_{R-1}} \over 2} \label{ref19}\\ &\nonumber\\ &\small{\text{After the TAN-DIP iterations, get the value of tan}} \nonumber\\ &\nonumber\\ T &= \sqrt {{1 - v_{R}} \over {1 + v_{R}}} \label{ref20}\\ &\nonumber\\ &\small{\text{Determine K}} \nonumber\\ \nonumber\\ K &= 3 \times 2^{R+1} \label{ref21}\\ &\nonumber\\ &\small{\text{Use K and T in the series and run L iterations}} \nonumber\\ &\nonumber\\ {P} &= {K} \sum\limits_{n = 0}^L {\frac{{\left( -1 \right)^{n} {(T)^{2n + 1}}}}{{2n + 1}}} \label{ref22}\\ \end{alignat*}
The second is a very basic one - without any geometry etc. In the algorithm above Just start from 0.5 (or your choice of $cos({\pi \over x})$ value $x \ge 4$; you may have to change the $K$ equation) and do just a very high TAN-DIP iteration with very high precision. Ideally it is infinite TAN-DIP with infinite precision. And then just multiply T with K to get PI.
I have made a quick video - sorry a bit monotonus and long. https://www.youtube.com/watch?v=c1mRnbQUfH8
And I have written a self-published paper (non-academic style) available on the internet. https://eplogica.com/teplics/mgl-series/icmgl-01/
It will take a lot of effort for me to explore it beyond this. But if you are good at numbers, trigonometry, calculus etc., it would be nice if you can make some comments. However, it would be nice to have your thoughts as well.
Thanks in advance.