--- title: "Approximations of pi" chunk: 9/10 source: "https://en.wikipedia.org/wiki/Approximations_of_pi" category: "reference" tags: "science, encyclopedia" date_saved: "2026-05-05T16:19:48.727542+00:00" instance: "kb-cron" --- π 2 = ∑ n = 0 ∞ arctan ⁡ 1 F 2 n + 1 = arctan ⁡ 1 1 + arctan ⁡ 1 2 + arctan ⁡ 1 5 + arctan ⁡ 1 13 + ⋯ {\displaystyle {\frac {\pi }{2}}=\sum _{n=0}^{\infty }\arctan {\frac {1}{F_{2n+1}}}=\arctan {\frac {1}{1}}+\arctan {\frac {1}{2}}+\arctan {\frac {1}{5}}+\arctan {\frac {1}{13}}+\cdots } and π 4 = ∑ k ≥ 2 arctan ⁡ 2 − a k − 1 a k , {\displaystyle {\frac {\pi }{4}}=\sum _{k\geq 2}\arctan {\frac {\sqrt {2-a_{k-1}}}{a_{k}}},} where F n {\displaystyle F_{n}} is the n-th Fibonacci number. However, these two formulae for π {\displaystyle \pi } are much slower in convergence because of set of arctangent functions that are involved in computation. ==== Arcsine ==== Observing an equilateral triangle and noting that sin ⁡ ( π 6 ) = 1 2 {\displaystyle \sin \left({\frac {\pi }{6}}\right)={\frac {1}{2}}} yields π = 6 sin − 1 ⁡ ( 1 2 ) = 6 ( 1 2 + 1 2 ⋅ 3 ⋅ 2 3 + 1 ⋅ 3 2 ⋅ 4 ⋅ 5 ⋅ 2 5 + 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ⋅ 7 ⋅ 2 7 + ⋯ ) = 3 16 0 ⋅ 1 + 6 16 1 ⋅ 3 + 18 16 2 ⋅ 5 + 60 16 3 ⋅ 7 + ⋯ = ∑ n = 0 ∞ 3 ⋅ ( 2 n n ) 16 n ( 2 n + 1 ) = 3 + 1 8 + 9 640 + 15 7168 + 35 98304 + 189 2883584 + 693 54525952 + 429 167772160 + ⋯ {\displaystyle {\begin{aligned}\pi &=6\sin ^{-1}\left({\frac {1}{2}}\right)=6\left({\frac {1}{2}}+{\frac {1}{2\cdot 3\cdot 2^{3}}}+{\frac {1\cdot 3}{2\cdot 4\cdot 5\cdot 2^{5}}}+{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7\cdot 2^{7}}}+\cdots \!\right)\\&={\frac {3}{16^{0}\cdot 1}}+{\frac {6}{16^{1}\cdot 3}}+{\frac {18}{16^{2}\cdot 5}}+{\frac {60}{16^{3}\cdot 7}}+\cdots \!=\sum _{n=0}^{\infty }{\frac {3\cdot {\binom {2n}{n}}}{16^{n}(2n+1)}}\\&=3+{\frac {1}{8}}+{\frac {9}{640}}+{\frac {15}{7168}}+{\frac {35}{98304}}+{\frac {189}{2883584}}+{\frac {693}{54525952}}+{\frac {429}{167772160}}+\cdots \end{aligned}}} with a convergence such that each additional five terms yields at least three more digits. == Digit extraction methods == The Bailey–Borwein–Plouffe formula (BBP) for calculating π was discovered in 1995 by Simon Plouffe. Using a spigot algorithm, the formula can compute any particular base 16 digit of π—returning the hexadecimal value of the digit—without computing the intervening digits. π = ∑ n = 0 ∞ ( 4 8 n + 1 − 2 8 n + 4 − 1 8 n + 5 − 1 8 n + 6 ) ( 1 16 ) n {\displaystyle \pi =\sum _{n=0}^{\infty }\left({\frac {4}{8n+1}}-{\frac {2}{8n+4}}-{\frac {1}{8n+5}}-{\frac {1}{8n+6}}\right)\left({\frac {1}{16}}\right)^{n}} In 1996, Plouffe derived an algorithm to extract the nth decimal digit of π (using base 10 math to extract a base 10 digit), and which can do so with an improved speed of O(n3(log n)3) time. The algorithm does not require memory for storage of a full n-digit result, so the one-millionth digit of π could in principle be computed using a pocket calculator. (However, it would be quite tedious and impractical to do so.) π + 3 = ∑ n = 1 ∞ n 2 n n ! 2 ( 2 n ) ! {\displaystyle \pi +3=\sum _{n=1}^{\infty }{\frac {n2^{n}n!^{2}}{(2n)!}}} The calculation speed of Plouffe's formula was improved to O(n2) by Fabrice Bellard, who derived an alternative formula (albeit only in base 2 math) for computing π. π = 1 2 6 ∑ n = 0 ∞ ( − 1 ) n 2 10 n ( − 2 5 4 n + 1 − 1 4 n + 3 + 2 8 10 n + 1 − 2 6 10 n + 3 − 2 2 10 n + 5 − 2 2 10 n + 7 + 1 10 n + 9 ) {\displaystyle \pi ={\frac {1}{2^{6}}}\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{2^{10n}}}\left(-{\frac {2^{5}}{4n+1}}-{\frac {1}{4n+3}}+{\frac {2^{8}}{10n+1}}-{\frac {2^{6}}{10n+3}}-{\frac {2^{2}}{10n+5}}-{\frac {2^{2}}{10n+7}}+{\frac {1}{10n+9}}\right)} == Efficient methods == Many other expressions for π were developed and published by Indian mathematician Srinivasa Ramanujan. He worked with mathematician Godfrey Harold Hardy in England for a number of years. Extremely long decimal expansions of π are typically computed with the Gauss–Legendre algorithm and Borwein's algorithm; the Salamin–Brent algorithm, which was invented in 1976, has also been used. In 1997, David H. Bailey, Peter Borwein and Simon Plouffe published a paper (Bailey, 1997) on a new formula for π as an infinite series: π = ∑ k = 0 ∞ 1 16 k ( 4 8 k + 1 − 2 8 k + 4 − 1 8 k + 5 − 1 8 k + 6 ) . {\displaystyle \pi =\sum _{k=0}^{\infty }{\frac {1}{16^{k}}}\left({\frac {4}{8k+1}}-{\frac {2}{8k+4}}-{\frac {1}{8k+5}}-{\frac {1}{8k+6}}\right).} This formula permits one to fairly readily compute the kth binary or hexadecimal digit of π, without having to compute the preceding k − 1 digits. Bailey's website contains the derivation as well as implementations in various programming languages. The PiHex project computed 64 bits around the quadrillionth bit of π (which turns out to be 0). Fabrice Bellard further improved on BBP with his formula: π = 1 2 6 ∑ n = 0 ∞ ( − 1 ) n 2 10 n ( − 2 5 4 n + 1 − 1 4 n + 3 + 2 8 10 n + 1 − 2 6 10 n + 3 − 2 2 10 n + 5 − 2 2 10 n + 7 + 1 10 n + 9 ) {\displaystyle \pi ={\frac {1}{2^{6}}}\sum _{n=0}^{\infty }{\frac {{(-1)}^{n}}{2^{10n}}}\left(-{\frac {2^{5}}{4n+1}}-{\frac {1}{4n+3}}+{\frac {2^{8}}{10n+1}}-{\frac {2^{6}}{10n+3}}-{\frac {2^{2}}{10n+5}}-{\frac {2^{2}}{10n+7}}+{\frac {1}{10n+9}}\right)} Other formulae that have been used to compute estimates of π include: π 2 = ∑ k = 0 ∞ k ! ( 2 k + 1 ) ! ! = ∑ k = 0 ∞ 2 k k ! 2 ( 2 k + 1 ) ! = 1 + 1 3 ( 1 + 2 5 ( 1 + 3 7 ( 1 + ⋯ ) ) ) {\displaystyle {\frac {\pi }{2}}=\sum _{k=0}^{\infty }{\frac {k!}{(2k+1)!!}}=\sum _{k=0}^{\infty }{\frac {2^{k}k!^{2}}{(2k+1)!}}=1+{\frac {1}{3}}\left(1+{\frac {2}{5}}\left(1+{\frac {3}{7}}\left(1+\cdots \right)\right)\right)} Newton.