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---
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.
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