A convenient way to represent $\pi$ is with an infinite series. When the terms of an infinite series are added together they can converge to a single value or they can diverge by increasing to infinity or by oscillating within a range of values. For the purposes of representing $\pi$ we need to use a convergent infinite sequence.
The Nilakantha Series is easy to understand and easy to calculate but converges to $\pi$ slowly. \[ \pi = 3 +\frac{4}{2\times3\times4} -\frac{4}{4\times5\times6} +\frac{4}{6\times7\times8} -\frac{4}{8\times9\times10}\cdots \] The Nilakantha Series can be expressed in sigma notation as: \[\pi = 3+\displaystyle\sum_{k=1}^{\infty} \frac{4(-1)^{k-1}}{2k\times (2k+1) \times (2k+2)} \] \[\pi = 3+4\displaystyle\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{8k^{3}+12k^{2}+4k} \] The Chudnovsky Series is more complex but converges to $\pi$ quickly. \[\frac{1}{\pi} = 12\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k}640320^{\frac{3}{2}}}\] This blog will demonstrate how to calculate the Chudnovsky series in C++ and will attempt to do so with sufficient detail such that any motivated novice can understand the process. The Chudnovsky series can be reformulated to allow for a more convenient calculation.
\[\frac{1}{\pi} = 12\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k}640320^{\frac{3}{2}}}\] \[\frac{1}{\pi} = \frac{12}{640320^{\frac{3}{2}}}\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k}}\] \[\frac{1}{\pi} = \frac{12}{640320^{1}\times640320^{\frac{1}{2}}}\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k}}\] \[\frac{1}{\pi} = \frac{12}{640320\times\sqrt{640320}}\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k}}\] \[\frac{1}{\pi} = \frac{12}{12\times53360\times\sqrt{64\times10005}}\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k}}\] \[\frac{1}{\pi} = \frac{1}{53360\times8\times\sqrt{10005}}\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k}}\] \[\frac{1}{\pi} = \frac{1}{426880\sqrt{10005}}\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k}}\] \[\frac{1}{\pi} = \frac{1}{426880\sqrt{10005}}\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k}(6k)!13591409}{(3k)!(k!)^{3}640320^{3k}}+\frac{(-1)^{k}(6k)!545140134k}{(3k)!(k!)^{3}640320^{3k}}\] \[a = \frac{(-1)^{k}(6k)!}{(3k)!(k!)^{3}640320^{3k}}\] \[b = \frac{(-1)^{k}(6k)!k}{(3k)!(k!)^{3}640320^{3k}}\] \[b = a\times k\] \[\frac{1}{\pi} = \frac{1}{426880\sqrt{10005}}\displaystyle\sum_{k=1}^{\infty}13591409a+545140134b\] \[\frac{1}{\pi} = \displaystyle\sum_{k=1}^{\infty}\frac{13591409a+545140134b}{426880\sqrt{10005}}\] \[\pi = \displaystyle\sum_{k=1}^{\infty}\frac{426880\sqrt{10005}}{13591409a+545140134b}\] At this point the series has a favorable form where knowing a allows b to be quickly calculated. This is further improved by finding a multiple such that ak can be calculated quickly from a(k-1). \[\frac{a_k}{a_{(k-1)}}=\frac{(-1)^{k}(6k)!}{(3k)!(k!)^{3}640320^{3k}}\div\frac{(-1)^{k-1}(6(k-1))!}{(3(k-1))!((k-1)!)^{3}640320^{3(k-1)}}\] \[\frac{a_k}{a_{(k-1)}}=\frac{(-1)^{k}(6k)!}{(3k)!(k!)^{3}640320^{3k}}\times\frac{(3(k-1))!((k-1)!)^{3}640320^{3(k-1)}}{(-1)^{k-1}(6(k-1))!}\] \[\frac{a_k}{a_{(k-1)}}=\frac{(-1)^{k}(6k)!(3(k-1))!((k-1)!)^{3}640320^{3(k-1)}}{(-1)^{k-1}(6(k-1))!(3k)!(k!)^{3}640320^{3k}}\] \[\frac{a_k}{a_{(k-1)}}=\frac{(-1)^{k}(6k)!(3k-3)!((k-1)!)^{3}640320^{3k-3}}{(-1)^{k}(-1)^{-1}(6k-6)!(3k)!(k!)^{3}640320^{3k}}\] \[\frac{a_k}{a_{(k-1)}}=\frac{(6k)!(3k-3)!((k-1)!)^{3}640320^{3k}640320^{-3}}{(-1)^{-1}(6k-6)!(3k)!(k!)^{3}640320^{3k}}\] \[\frac{a_k}{a_{(k-1)}}=\frac{-1(6k)!(3k-3)!((k-1)!)^{3}640320^{-3}}{(6k-6)!(3k)!(k\times(k-1)!)^{3}}\] \[\frac{a_k}{a_{(k-1)}}=\frac{-1(6k)(6k-1)(6k-2)(6k-3)(6k-4)(6k-5)(6k-6)!(3k-3)!((k-1)!)^{3}}{(6k-6)!(3k)(3k-1)(3k-2)(3k-3)!k^{3}((k-1)!)^{3}640320^{3}}\] \[\frac{a_k}{a_{(k-1)}}=\frac{-6k(6k-1)(6k-2)(6k-3)(6k-4)(6k-5)}{3k(3k-1)(3k-2)k^{3}640320^{3}}\] \[\frac{a_k}{a_{(k-1)}}=\frac{-2(6k-1)(2)(3k-1)(3)(2k-1)(2)(3k-2)(6k-5)}{(3k-1)(3k-2)k^{3}640320^{3}}\] \[\frac{a_k}{a_{(k-1)}}=\frac{-24(6k-1)(2k-1)(6k-5)}{k^{3}640320^{3}}\] Here is a basic implementation of the algorithm in C++. The code is free to use on an as is basis. No attribution is required if you use it. No warranty of any kind is implied.
void chudnovsky (void)
{
}
