22 KiB
| title | chunk | source | category | tags | date_saved | instance |
|---|---|---|---|---|---|---|
| Approximations of pi | 5/10 | https://en.wikipedia.org/wiki/Approximations_of_pi | reference | science, encyclopedia | 2026-05-05T16:19:48.727542+00:00 | kb-cron |
((x),(y) = {239, 132} is a solution to the Pell equation x2 − 2y2 = −1.) Formulae of this kind are known as Machin-like formulae. Machin's particular formula was used well into the computer era for calculating record numbers of digits of π, but more recently other similar formulae have been used as well. For instance, Shanks and his team used the following Machin-like formula in 1961 to compute the first 100,000 digits of π:
π
4
=
6
arctan
1
8
+
2
arctan
1
57
+
arctan
1
239
{\displaystyle {\frac {\pi }{4}}=6\arctan {\frac {1}{8}}+2\arctan {\frac {1}{57}}+\arctan {\frac {1}{239}}}
and they used another Machin-like formula,
π
4
=
12
arctan
1
18
+
8
arctan
1
57
−
5
arctan
1
239
{\displaystyle {\frac {\pi }{4}}=12\arctan {\frac {1}{18}}+8\arctan {\frac {1}{57}}-5\arctan {\frac {1}{239}}}
as a check. The record as of December 2002 by Yasumasa Kanada of Tokyo University stood at 1,241,100,000,000 digits. The following Machin-like formulae were used for this:
π
4
=
12
arctan
1
49
+
32
arctan
1
57
−
5
arctan
1
239
+
12
arctan
1
110443
{\displaystyle {\frac {\pi }{4}}=12\arctan {\frac {1}{49}}+32\arctan {\frac {1}{57}}-5\arctan {\frac {1}{239}}+12\arctan {\frac {1}{110443}}}
K. Takano (1982).
π
4
=
44
arctan
1
57
+
7
arctan
1
239
−
12
arctan
1
682
+
24
arctan
1
12943
{\displaystyle {\frac {\pi }{4}}=44\arctan {\frac {1}{57}}+7\arctan {\frac {1}{239}}-12\arctan {\frac {1}{682}}+24\arctan {\frac {1}{12943}}}
F. C. M. Størmer (1896).
=== Other classical formulae === Other formulae that have been used to compute estimates of π include: Liu Hui (see also Viète's formula):
π
≈
768
2
−
2
+
2
+
2
+
2
+
2
+
2
+
2
+
2
+
1
≈
3.141590463236763.
{\displaystyle {\begin{aligned}\pi &\approx 768{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+1}}}}}}}}}}}}}}}}}}\\&\approx 3.141590463236763.\end{aligned}}}
Madhava:
π
=
12
∑
k
=
0
∞
(
−
3
)
−
k
2
k
+
1
=
12
∑
k
=
0
∞
(
−
1
3
)
k
2
k
+
1
=
12
(
1
1
⋅
3
0
−
1
3
⋅
3
1
+
1
5
⋅
3
2
−
1
7
⋅
3
3
+
⋯
)
{\displaystyle \pi ={\sqrt {12}}\sum _{k=0}^{\infty }{\frac {(-3)^{-k}}{2k+1}}={\sqrt {12}}\sum _{k=0}^{\infty }{\frac {(-{\frac {1}{3}})^{k}}{2k+1}}={\sqrt {12}}\left({1 \over 1\cdot 3^{0}}-{1 \over 3\cdot 3^{1}}+{1 \over 5\cdot 3^{2}}-{1 \over 7\cdot 3^{3}}+\cdots \right)}
Newton / Euler Convergence Transformation:
arctan
x
=
x
1
+
x
2
∑
k
=
0
∞
(
2
k
)
!
!
x
2
k
(
2
k
+
1
)
!
!
(
1
+
x
2
)
k
=
x
1
+
x
2
+
2
3
x
3
(
1
+
x
2
)
2
+
2
⋅
4
3
⋅
5
x
5
(
1
+
x
2
)
3
+
⋯
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 {\begin{aligned}\arctan x&={\frac {x}{1+x^{2}}}\sum _{k=0}^{\infty }{\frac {(2k)!!\,x^{2k}}{(2k+1)!!\,(1+x^{2})^{k}}}={\frac {x}{1+x^{2}}}+{\frac {2}{3}}{\frac {x^{3}}{(1+x^{2})^{2}}}+{\frac {2\cdot 4}{3\cdot 5}}{\frac {x^{5}}{(1+x^{2})^{3}}}+\cdots \\[10mu]{\frac {\pi }{2}}&=\sum _{k=0}^{\infty }{\frac {k!}{(2k+1)!!}}=\sum _{k=0}^{\infty }{\cfrac {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)\end{aligned}}}
where m!! is the double factorial, the product of the positive integers up to m with the same parity. Euler:
π
=
20
arctan
1
7
+
8
arctan
3
79
{\displaystyle {\pi }=20\arctan {\frac {1}{7}}+8\arctan {\frac {3}{79}}}
(Evaluated using the preceding series for arctan.) Ramanujan:
1
π
=
2
2
9801
∑
k
=
0
∞
(
4
k
)
!
(
1103
+
26390
k
)
(
k
!
)
4
396
4
k
{\displaystyle {\frac {1}{\pi }}={\frac {2{\sqrt {2}}}{9801}}\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}}
David Chudnovsky and Gregory Chudnovsky:
1
π
=
12
∑
k
=
0
∞
(
−
1
)
k
(
6
k
)
!
(
13591409
+
545140134
k
)
(
3
k
)
!
(
k
!
)
3
640320
3
k
+
3
/
2
{\displaystyle {\frac {1}{\pi }}=12\sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k+3/2}}}}
Ramanujan's work is the basis for the Chudnovsky algorithm, the fastest algorithms used, as of the turn of the millennium, to calculate π.
=== Modern algorithms === Extremely long decimal expansions of π are typically computed with iterative formulae like the Gauss–Legendre algorithm and Borwein's algorithm. The latter, found in 1985 by Jonathan and Peter Borwein, converges extremely quickly: For
y
0
=
2
−
1
,
a
0
=
6
−
4
2
{\displaystyle y_{0}={\sqrt {2}}-1,\ a_{0}=6-4{\sqrt {2}}}
and
y
k
+
1
=
(
1
−
f
(
y
k
)
)
/
(
1
+
f
(
y
k
)
)
,
a
k
+
1
=
a
k
(
1
+
y
k
+
1
)
4
−
2
2
k
+
3
y
k
+
1
(
1
+
y
k
+
1
+
y
k
+
1
2
)
{\displaystyle y_{k+1}=(1-f(y_{k}))/(1+f(y_{k}))~,~a_{k+1}=a_{k}(1+y_{k+1})^{4}-2^{2k+3}y_{k+1}(1+y_{k+1}+y_{k+1}^{2})}