Approximations of π

1. In August 2009, a Japanese supercomputer called the T2K Open Supercomputer more than doubled the previous record by calculating π to roughly 2.6 trillion digits in approximately 73 hours and 36 minutes.
2. In December 2009, Fabrice Bellard used a home computer to compute 2.7 trillion decimal digits of π. Calculations were performed in base 2 (binary), then the result was converted to base 10 (decimal). The calculation, conversion, and verification steps took a total of 131 days.[40]
3. In August 2010, Shigeru Kondo used Alexander Yee's y-cruncher to calculate 5 trillion digits of π. This was the world record for any type of calculation, but significantly it was performed on a home computer built by Kondo.[41] The calculation was done between 4 May and 3 August, with the primary and secondary verifications taking 64 and 66 hours respectively.[42]
4. In October 2011, Shigeru Kondo broke his own record by computing ten trillion (10¹³) and fifty digits using the same method but with better hardware.[43][44]
5. In December 2013, Kondo broke his own record for a second time when he computed 12.1 trillion digits of π.[45]
6. In October 2014, Sandon Van Ness, going by the pseudonym "houkouonchi" used y-cruncher to calculate 13.3 trillion digits of π.[46]
7. In November 2016, Peter Trueb and his sponsors computed on y-cruncher and fully verified 22.4 trillion digits of π (22,459,157,718,361 (π^e × 10¹²)).[47] The computation took (with three interruptions) 105 days to complete,[46] the limitation of further expansion being primarily storage space.[45]
8. In March 2019, Emma Haruka Iwao, an employee at Google, computed 31.4 (approximately 10π) trillion digits of pi using y-cruncher and Google Cloud machines. This took 121 days to complete.[48]
9. In January 2020, Timothy Mullican announced the computation of 50 trillion digits over 303 days.[49][50]
10. On 14 August 2021, a team (DAViS) at the University of Applied Sciences of the Grisons announced completion of the computation of π to 62.8 (approximately 20π) trillion digits.[51][52]
11. On 8 June 2022, Emma Haruka Iwao announced on the Google Cloud Blog the computation of 100 trillion (10¹⁴) digits of π over 158 days using Alexander Yee's y-cruncher.[2]

The so-called "Indiana Pi Bill" from 1897 has often been characterized as an attempt to "legislate the value of Pi". Rather, the bill dealt with a purported solution to the problem of geometrically "squaring the circle".[56]

The bill was nearly passed by the Indiana General Assembly in the U.S., and has been claimed to imply a number of different values for π, although the closest it comes to explicitly asserting one is the wording "the ratio of the diameter and circumference is as five-fourths to four", which would make π = 16⁄5 = 3.2, a discrepancy of nearly 2 percent. A mathematics professor who happened to be present the day the bill was brought up for consideration in the Senate, after it had passed in the House, helped to stop the passage of the bill on its second reading, after which the assembly thoroughly ridiculed it before postponing it indefinitely.

It is sometimes claimed that the Hebrew Bible implies that "π equals three", based on a passage in 1 Kings 7:23 and 2 Chronicles 4:2 giving measurements for the round basin located in front of the Temple in Jerusalem as having a diameter of 10 cubits and a circumference of 30 cubits.

The issue is discussed in the Talmud and in Rabbinic literature.[57] Among the many explanations and comments are these:

• Rabbi Nehemiah explained this in his Mishnat ha-Middot (the earliest known Hebrew text on geometry, ca. 150 CE) by saying that the diameter was measured from the outside rim while the circumference was measured along the inner rim. This interpretation implies a brim about 0.225 cubit (or, assuming an 18-inch "cubit", some 4 inches), or one and a third "handbreadths," thick (cf. NKJV and NKJV).
• Maimonides states (ca. 1168 CE) that π can only be known approximately, so the value 3 was given as accurate enough for religious purposes. This is taken by some[58] as the earliest assertion that π is irrational.

There is still some debate on this passage in biblical scholarship.[59][60] Many reconstructions of the basin show a wider brim (or flared lip) extending outward from the bowl itself by several inches to match the description given in NKJV[61] In the succeeding verses, the rim is described as "a handbreadth thick; and the brim thereof was wrought like the brim of a cup, like the flower of a lily: it received and held three thousand baths" NKJV, which suggests a shape that can be encompassed with a string shorter than the total length of the brim, e.g., a Lilium flower or a Teacup.

Archimedes, in his Measurement of a Circle, created the first algorithm for the calculation of π based on the idea that the perimeter of any (convex) polygon inscribed in a circle is less than the circumference of the circle, which, in turn, is less than the perimeter of any circumscribed polygon. He started with inscribed and circumscribed regular hexagons, whose perimeters are readily determined. He then shows how to calculate the perimeters of regular polygons of twice as many sides that are inscribed and circumscribed about the same circle. This is a recursive procedure which would be described today as follows: Let p_k and P_k denote the perimeters of regular polygons of k sides that are inscribed and circumscribed about the same circle, respectively. Then,

${\displaystyle P_{2n}={\frac {2p_{n}P_{n}}{p_{n}+P_{n}}},\quad \quad p_{2n}={\sqrt {p_{n}P_{2n}}}.}$

Archimedes uses this to successively compute P₁₂, p₁₂, P₂₄, p₂₄, P₄₈, p₄₈, P₉₆ and p₉₆.[62] Using these last values he obtains

${\displaystyle 3{\frac {10}{71}}<\pi <3{\frac {1}{7}}.}$

It is not known why Archimedes stopped at a 96-sided polygon; it only takes patience to extend the computations. Heron reports in his Metrica (about 60 CE) that Archimedes continued the computation in a now lost book, but then attributes an incorrect value to him.[63]

Archimedes uses no trigonometry in this computation and the difficulty in applying the method lies in obtaining good approximations for the square roots that are involved. Trigonometry, in the form of a table of chord lengths in a circle, was probably used by Claudius Ptolemy of Alexandria to obtain the value of π given in the Almagest (circa 150 CE).[64]

Advances in the approximation of π (when the methods are known) were made by increasing the number of sides of the polygons used in the computation. A trigonometric improvement by Willebrord Snell (1621) obtains better bounds from a pair of bounds obtained from the polygon method. Thus, more accurate results were obtained from polygons with fewer sides.[65] Viète's formula, published by François Viète in 1593, was derived by Viète using a closely related polygonal method, but with areas rather than perimeters of polygons whose numbers of sides are powers of two.[66]

The last major attempt to compute π by this method was carried out by Grienberger in 1630 who calculated 39 decimal places of π using Snell's refinement.[65]

For fast calculations, one may use formulae such as Machin's:

${\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}}$

together with the Taylor series expansion of the function arctan(x). This formula is most easily verified using polar coordinates of complex numbers, producing:

${\displaystyle (5+i)^{4}\cdot (239-i)=2^{2}\cdot 13^{4}(1+i).}$

({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,[38] 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:[38]

${\displaystyle {\frac {\pi }{4}}=6\arctan {\frac {1}{8}}+2\arctan {\frac {1}{57}}+\arctan {\frac {1}{239}}}$

and they used another Machin-like formula,

${\displaystyle {\frac {\pi }{4}}=12\arctan {\frac {1}{18}}+8\arctan {\frac {1}{57}}-5\arctan {\frac {1}{239}}}$

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:

${\displaystyle {\frac {\pi }{4}}=12\arctan {\frac {1}{49}}+32\arctan {\frac {1}{57}}-5\arctan {\frac {1}{239}}+12\arctan {\frac {1}{110443}}}$

K. Takano (1982).

${\displaystyle {\frac {\pi }{4}}=44\arctan {\frac {1}{57}}+7\arctan {\frac {1}{239}}-12\arctan {\frac {1}{682}}+24\arctan {\frac {1}{12943}}}$

F. C. M. Størmer (1896).

Other formulae that have been used to compute estimates of π include:

Liu Hui (see also Viète's formula):

${\displaystyle {\begin{aligned}\pi &\approx 768{\sqrt {2-{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+1}}}}}}}}}}}}}}}}}}\\&\approx 3.141590463236763.\end{aligned}}}$

Madhava:

${\displaystyle \pi ={\sqrt {12}}\sum _{k=0}^{\infty }{\frac {(-3)^{-k}}{2k+1}}={\sqrt {12}}\sum _{k=0}^{\infty }{\frac {(-{\frac {1}{3}})^{k}}{2k+1}}={\sqrt {12}}\left({1 \over 1\cdot 3^{0}}-{1 \over 3\cdot 3^{1}}+{1 \over 5\cdot 3^{2}}-{1 \over 7\cdot 3^{3}}+\cdots \right)}$

Newton / Euler Convergence Transformation:[67]

${\displaystyle {\begin{aligned}\arctan x&={\frac {x}{1+x^{2}}}\sum _{k=0}^{\infty }{\frac {(2k)!!\,x^{2k}}{(2k+1)!!\,(1+x^{2})^{k}}}={\frac {x}{1+x^{2}}}+{\frac {2}{3}}{\frac {x^{3}}{(1+x^{2})^{2}}}+{\frac {2\cdot 4}{3\cdot 5}}{\frac {x^{5}}{(1+x^{2})^{3}}}+\cdots \\[10mu]{\frac {\pi }{2}}&=\sum _{k=0}^{\infty }{\frac {k!}{(2k+1)!!}}=\sum _{k=0}^{\infty }{\cfrac {2^{k}k!^{2}}{(2k+1)!}}=1+{\frac {1}{3}}\left(1+{\frac {2}{5}}\left(1+{\frac {3}{7}}\left(1+\cdots \right)\right)\right)\end{aligned}}}$

where m!! is the double factorial, the product of the positive integers up to m with the same parity.

Euler:

${\displaystyle {\pi }=20\arctan {\frac {1}{7}}+8\arctan {\frac {3}{79}}}$

(Evaluated using the preceding series for arctan.)

Ramanujan:

${\displaystyle {\frac {1}{\pi }}={\frac {2{\sqrt {2}}}{9801}}\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}}$

David Chudnovsky and Gregory Chudnovsky:

${\displaystyle {\frac {1}{\pi }}=12\sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k+3/2}}}}$

Ramanujan's work is the basis for the Chudnovsky algorithm, the fastest algorithms used, as of the turn of the millennium, to calculate π.

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 ${\displaystyle y_{0}={\sqrt {2}}-1,\ a_{0}=6-4{\sqrt {2}}}$ and

${\displaystyle y_{k+1}=(1-f(y_{k}))/(1+f(y_{k}))~,~a_{k+1}=a_{k}(1+y_{k+1})^{4}-2^{2k+3}y_{k+1}(1+y_{k+1}+y_{k+1}^{2})}$

where ${\displaystyle f(y)=(1-y^{4})^{1/4}}$, the sequence ${\displaystyle 1/a_{k}}$ converges quartically to π, giving about 100 digits in three steps and over a trillion digits after 20 steps. The Gauss–Legendre algorithm (with time complexity ${\displaystyle O(n\log ^{2}n)}$, using Harvey–Hoeven multiplication algorithm) is asymptotically faster than the Chudnovsky algorithm (with time complexity ${\displaystyle O(n\log ^{3}n)}$) – but which of these algorithms is faster in practice for "small enough" ${\displaystyle n}$ depends on technological factors such as memory sizes and access times.[68] For breaking world records, the iterative algorithms are used less commonly than the Chudnovsky algorithm since they are memory-intensive.

The first one million digits of π and 1⁄π are available from Project Gutenberg.[69][70] A former calculation record (December 2002) by Yasumasa Kanada of Tokyo University stood at 1.24 trillion digits, which were computed in September 2002 on a 64-node Hitachi supercomputer with 1 terabyte of main memory, which carries out 2 trillion operations per second, nearly twice as many as the computer used for the previous record (206 billion digits). The following Machin-like formulae were used for this:

${\displaystyle {\frac {\pi }{4}}=12\arctan {\frac {1}{49}}+32\arctan {\frac {1}{57}}-5\arctan {\frac {1}{239}}+12\arctan {\frac {1}{110443}}}$ (Kikuo Takano (1982))

${\displaystyle {\frac {\pi }{4}}=44\arctan {\frac {1}{57}}+7\arctan {\frac {1}{239}}-12\arctan {\frac {1}{682}}+24\arctan {\frac {1}{12943}}}$ (F. C. M. Størmer (1896)).

These approximations have so many digits that they are no longer of any practical use, except for testing new supercomputers.[71] Properties like the potential normality of π will always depend on the infinite string of digits on the end, not on any finite computation.

Historically, base 60 was used for calculations. In this base, π can be approximated to eight (decimal) significant figures with the number 3;8,29,44₆₀, which is

${\displaystyle 3+{\frac {8}{60}}+{\frac {29}{60^{2}}}+{\frac {44}{60^{3}}}=3.14159\ 259^{+}}$

(The next sexagesimal digit is 0, causing truncation here to yield a relatively good approximation.)

In addition, the following expressions can be used to estimate π:

• accurate to three digits:

${\displaystyle {\frac {22}{7}}=3.143^{+}}$

• accurate to three digits:

${\displaystyle {\sqrt {2}}+{\sqrt {3}}=3.146^{+}}$

Karl Popper conjectured that Plato knew this expression, that he believed it to be exactly π, and that this is responsible for some of Plato's confidence in the omnicompetence of mathematical geometry—and Plato's repeated discussion of special right triangles that are either isosceles or halves of equilateral triangles.

${\displaystyle {\sqrt {15}}-{\sqrt {3}}+1=3.1409^{+}}$

• accurate to four digits:

${\displaystyle {\sqrt[{3}]{31}}=3.1413^{+}}$[72]

${\displaystyle {\sqrt {63}}-{\sqrt {23}}=3.14142\ 2^{+}}$

${\displaystyle {\sqrt {51}}-{\sqrt {16}}=3.14142\ 8^{+}}$

• accurate to four digits (or five significant figures):

${\displaystyle {\sqrt {7+{\sqrt {6+{\sqrt {5}}}}}}=3.1416^{+}}$[73]

• an approximation by Ramanujan, accurate to 4 digits (or five significant figures):

${\displaystyle {\frac {9}{5}}+{\sqrt {\frac {9}{5}}}=3.1416^{+}}$

• accurate to five digits:

${\displaystyle {\frac {7^{7}}{4^{9}}}=3.14156^{+}}$

${\displaystyle {\sqrt[{5}]{306}}=3.14155^{+}}$[74]

• accurate to six digits:

${\displaystyle \left(2-{\frac {\sqrt {2{\sqrt {2}}-2}}{2^{2}}}\right)^{2}=3.14159\ 6^{+}}$

${\displaystyle {\sqrt {2}}+{\sqrt {3}}+{\frac {{\sqrt {2}}-{\sqrt {3}}}{68}}=3.14159\ 0^{+}}$

${\displaystyle 4\left(1-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+{\frac {1}{9}}\right)-{\frac {2}{10+{\frac {1^{2}}{10+{\frac {2^{2}}{10+{\frac {3^{2}}{10}}}}}}}}=3.14159\ 3^{+}}$[75][76]

• accurate to seven digits:

${\displaystyle {\frac {355}{113}}=3.14159\ 29^{+}}$

${\displaystyle {\frac {{\sqrt {883}}-{\sqrt {21}}}{8}}=3.14159\ 25^{+}}$

${\displaystyle {\frac {9801}{2206{\sqrt {2}}}}}$ - inverse of first term of Ramanujan series.

${\displaystyle \left({\frac {2\cdot 4\cdot 6}{3\cdot 5\cdot 7}}\right)^{2}\left(15+{\frac {1^{2}}{30+{\frac {3^{2}}{30+{\frac {5^{2}}{30}}}}}}\right)=3.14159\ 27^{+}}$[77]

• accurate to eight digits:

${\displaystyle {\frac {{\sqrt {2669}}-{\sqrt {547}}}{9}}=3.14159\ 269^{+}}$

${\displaystyle {\frac {16}{5{\sqrt[{38}]{2}}{\sqrt[{3846}]{2}}}}=3.14159\ 260^{+}}$

${\displaystyle {\frac {99733}{31746}}=3.14159\ 264^{+}}$

${\displaystyle \left({\frac {\sqrt {58}}{4}}-{\frac {37{\sqrt {2}}}{33}}\right)^{-1}={\frac {66{\sqrt {2}}}{33{\sqrt {29}}-148}}=3.14159\ 263^{+}}$[78]

This is the case that cannot be obtained from Ramanujan's approximation (22).[79]

• accurate to nine digits:

${\displaystyle {\sqrt[{4}]{3^{4}+2^{4}+{\frac {1}{2+({\frac {2}{3}})^{2}}}}}={\sqrt[{4}]{\frac {2143}{22}}}=3.14159\ 2652^{+}}$

This is from Ramanujan, who claimed the Goddess of Namagiri appeared to him in a dream and told him the true value of π.[79]

• accurate to ten digits:

${\displaystyle {\sqrt[{15}]{28658146}}}$

• accurate to ten digits:

${\displaystyle {\frac {63}{25}}\times {\frac {17+15{\sqrt {5}}}{7+15{\sqrt {5}}}}=3.14159\ 26538^{+}}$

• accurate to ten digits (or eleven significant figures):

${\displaystyle {\sqrt[{193}]{\frac {10^{100}}{11222.11122}}}=3.14159\ 26536^{+}}$

This curious approximation follows the observation that the 193rd power of 1/π yields the sequence 1122211125... Replacing 5 by 2 completes the symmetry without reducing the correct digits of π, while inserting a central decimal point remarkably fixes the accompanying magnitude at 10¹⁰⁰.[80]

• accurate to eleven digits:

${\displaystyle {\sqrt[{18}]{888582403}}}$

• accurate to twelve digits:

${\displaystyle {\sqrt[{20}]{8769956796}}}$

• accurate to 12 decimal places:

${\displaystyle \left({\frac {\sqrt {163}}{6}}-{\frac {181}{\sqrt {10005}}}\right)^{-1}}$

This is obtained from the Chudnovsky series (truncate the series (1.4)[81] at first term and let E₆(τ₁₆₃)²/E₄(τ₁₆₃)³ = 151931373056001/151931373056000 ≈ 1).

• accurate to 16 digits:

${\displaystyle {\frac {2510613731736{\sqrt {2}}}{1130173253125}}}$ - inverse of sum of first two terms of Ramanujan series.

${\displaystyle {\frac {165707065}{52746197}}=3.14159\ 26535\ 897\ 934^{+}}$

• accurate to 18 digits:

${\displaystyle {\frac {80{\sqrt {15}}(5^{4}+53{\sqrt {89}})^{\frac {3}{2}}}{3308(5^{4}+53{\sqrt {89}})-3{\sqrt {89}}}}}$[82]

This is based on the fundamental discriminant d = 3(89) = 267 which has class number h(-d) = 2 explaining the algebraic numbers of degree 2. The core radical ${\displaystyle \scriptstyle 5^{4}+53{\sqrt {89}}}$ is 5³ more than the fundamental unit ${\displaystyle \scriptstyle U_{89}=500+53{\sqrt {89}}}$ which gives the smallest solution { x, y} = {500, 53} to the Pell equation x² − 89y² = −1.

• accurate to 18 decimal places:

${\displaystyle \left({\frac {\sqrt {253}}{4}}-{\frac {643{\sqrt {11}}}{903}}-{\frac {223}{172}}\right)^{-1}}$

This is the approximation (22) in Ramanujan's paper[79] with n = 253.

• accurate to 24 digits:

${\displaystyle {\frac {2286635172367940241408{\sqrt {2}}}{1029347477390786609545}}}$ - inverse of sum of first three terms of Ramanujan series.

• accurate to 25 decimal places:

${\displaystyle {\frac {1}{10}}\ln \left({\frac {2^{21}}{({\sqrt[{4}]{5}}-1)^{24}}}+24\right)}$

This is derived from Ramanujan's class invariant g₁₀₀ = 2^5/8/(5^1/4−1).[79]

• accurate to 30 decimal places:

${\displaystyle {\frac {\ln(640320^{3}+744)}{\sqrt {163}}}=3.14159\ 26535\ 89793\ 23846\ 26433\ 83279^{+}}$

Derived from the closeness of Ramanujan constant to the integer 640320³+744. This does not admit obvious generalizations in the integers, because there are only finitely many Heegner numbers and negative discriminants d with class number h(−d) = 1, and d = 163 is the largest one in absolute value.

• accurate to 52 decimal places:

${\displaystyle {\frac {\ln(5280^{3}(236674+30303{\sqrt {61}})^{3}+744)}{\sqrt {427}}}}$

Like the one above, a consequence of the j-invariant. Among negative discriminants with class number 2, this d the largest in absolute value.

• accurate to 52 decimal places:

${\displaystyle {\frac {\ln(2^{-30}((3+{\sqrt {5}})({\sqrt {5}}+{\sqrt {7}})({\sqrt {7}}+{\sqrt {11}})({\sqrt {11}}+3))^{12}-24)}{{\sqrt {5}}{\sqrt {7}}{\sqrt {11}}}}}$

This is derived from Ramanujan's class invariant G₃₈₅.[79]

• accurate to 161 decimal places:

${\displaystyle {\frac {\ln {\big (}(2u)^{6}+24{\big )}}{\sqrt {3502}}}}$

where u is a product of four simple quartic units,

${\displaystyle u=(a+{\sqrt {a^{2}-1}})^{2}(b+{\sqrt {b^{2}-1}})^{2}(c+{\sqrt {c^{2}-1}})(d+{\sqrt {d^{2}-1}})}$

and,

${\displaystyle {\begin{aligned}a&={\tfrac {1}{2}}(23+4{\sqrt {34}})\\b&={\tfrac {1}{2}}(19{\sqrt {2}}+7{\sqrt {17}})\\c&=(429+304{\sqrt {2}})\\d&={\tfrac {1}{2}}(627+442{\sqrt {2}})\end{aligned}}}$

Based on one found by Daniel Shanks. Similar to the previous two, but this time is a quotient of a modular form, namely the Dedekind eta function, and where the argument involves ${\displaystyle \tau ={\sqrt {-3502}}}$. The discriminant d = 3502 has h(−d) = 16.

• The continued fraction representation of π can be used to generate successive best rational approximations. These approximations are the best possible rational approximations of π relative to the size of their denominators. Here is a list of the first thirteen of these:[83][84]

${\displaystyle {\frac {3}{1}},{\frac {22}{7}},{\frac {333}{106}},{\frac {355}{113}},{\frac {103993}{33102}},{\frac {104348}{33215}},{\frac {208341}{66317}},{\frac {312689}{99532}},{\frac {833719}{265381}},{\frac {1146408}{364913}},{\frac {4272943}{1360120}},{\frac {5419351}{1725033}}}$

Of these, ${\displaystyle {\frac {355}{113}}}$ is the only fraction in this sequence that gives more exact digits of π (i.e. 7) than the number of digits needed to approximate it (i.e. 6). The accuracy can be improved by using other fractions with larger numerators and denominators, but, for most such fractions, more digits are required in the approximation than correct significant figures achieved in the result.[85]

Numerical approximation of π: as points are randomly scattered inside the unit square, some fall within the unit circle. The fraction of points inside the circle approaches π/4 as points are added.

Pi can be obtained from a circle if its radius and area are known using the relationship:

${\displaystyle A=\pi r^{2}.}$

If a circle with radius r is drawn with its center at the point (0, 0), any point whose distance from the origin is less than r will fall inside the circle. The Pythagorean theorem gives the distance from any point (x, y) to the center:

${\displaystyle d={\sqrt {x^{2}+y^{2}}}.}$

Mathematical "graph paper" is formed by imagining a 1×1 square centered around each cell (x, y), where x and y are integers between −r and r. Squares whose center resides inside or exactly on the border of the circle can then be counted by testing whether, for each cell (x, y),

${\displaystyle {\sqrt {x^{2}+y^{2}}}\leq r.}$

The total number of cells satisfying that condition thus approximates the area of the circle, which then can be used to calculate an approximation of π. Closer approximations can be produced by using larger values of r.

Mathematically, this formula can be written:

${\displaystyle \pi =\lim _{r\to \infty }{\frac {1}{r^{2}}}\sum _{x=-r}^{r}\;\sum _{y=-r}^{r}{\begin{cases}1&{\text{if }}{\sqrt {x^{2}+y^{2}}}\leq r\\0&{\text{if }}{\sqrt {x^{2}+y^{2}}}>r.\end{cases}}}$

In other words, begin by choosing a value for r. Consider all cells (x, y) in which both x and y are integers between −r and r. Starting at 0, add 1 for each cell whose distance to the origin (0,0) is less than or equal to r. When finished, divide the sum, representing the area of a circle of radius r, by r² to find the approximation of π. For example, if r is 5, then the cells considered are:

(−5,5)	(−4,5)	(−3,5)	(−2,5)	(−1,5)	(0,5)	(1,5)	(2,5)	(3,5)	(4,5)	(5,5)
(−5,4)	(−4,4)	(−3,4)	(−2,4)	(−1,4)	(0,4)	(1,4)	(2,4)	(3,4)	(4,4)	(5,4)
(−5,3)	(−4,3)	(−3,3)	(−2,3)	(−1,3)	(0,3)	(1,3)	(2,3)	(3,3)	(4,3)	(5,3)
(−5,2)	(−4,2)	(−3,2)	(−2,2)	(−1,2)	(0,2)	(1,2)	(2,2)	(3,2)	(4,2)	(5,2)
(−5,1)	(−4,1)	(−3,1)	(−2,1)	(−1,1)	(0,1)	(1,1)	(2,1)	(3,1)	(4,1)	(5,1)
(−5,0)	(−4,0)	(−3,0)	(−2,0)	(−1,0)	(0,0)	(1,0)	(2,0)	(3,0)	(4,0)	(5,0)
(−5,−1)	(−4,−1)	(−3,−1)	(−2,−1)	(−1,−1)	(0,−1)	(1,−1)	(2,−1)	(3,−1)	(4,−1)	(5,−1)
(−5,−2)	(−4,−2)	(−3,−2)	(−2,−2)	(−1,−2)	(0,−2)	(1,−2)	(2,−2)	(3,−2)	(4,−2)	(5,−2)
(−5,−3)	(−4,−3)	(−3,−3)	(−2,−3)	(−1,−3)	(0,−3)	(1,−3)	(2,−3)	(3,−3)	(4,−3)	(5,−3)
(−5,−4)	(−4,−4)	(−3,−4)	(−2,−4)	(−1,−4)	(0,−4)	(1,−4)	(2,−4)	(3,−4)	(4,−4)	(5,−4)
(−5,−5)	(−4,−5)	(−3,−5)	(−2,−5)	(−1,−5)	(0,−5)	(1,−5)	(2,−5)	(3,−5)	(4,−5)	(5,−5)

This circle as it would be drawn on a Cartesian coordinate graph. The cells (±3, ±4) and (±4, ±3) are labeled.

The 12 cells (0, ±5), (±5, 0), (±3, ±4), (±4, ±3) are exactly on the circle, and 69 cells are completely inside, so the approximate area is 81, and π is calculated to be approximately 3.24 because 81⁄5² = 3.24. Results for some values of r are shown in the table below:

For related results see The circle problem: number of points (x,y) in square lattice with x^2 + y^2 <= n.

Similarly, the more complex approximations of π given below involve repeated calculations of some sort, yielding closer and closer approximations with increasing numbers of calculations.

Besides its simple continued fraction representation [3; 7, 15, 1, 292, 1, 1, ...], which displays no discernible pattern, π has many generalized continued fraction representations generated by a simple rule, including these two.

${\displaystyle \pi ={3+{\cfrac {1^{2}}{6+{\cfrac {3^{2}}{6+{\cfrac {5^{2}}{6+\ddots \,}}}}}}}}$

${\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{3+{\cfrac {2^{2}}{5+{\cfrac {3^{2}}{7+{\cfrac {4^{2}}{9+\ddots }}}}}}}}}}=3+{\cfrac {1^{2}}{5+{\cfrac {4^{2}}{7+{\cfrac {3^{2}}{9+{\cfrac {6^{2}}{11+{\cfrac {5^{2}}{13+\ddots }}}}}}}}}}}$

The remainder of the Madhava–Leibniz series can be expressed as generalized continued fraction as follows.[75]

${\displaystyle \pi =4\sum _{n=1}^{m}{\frac {(-1)^{n-1}}{2n-1}}+{\cfrac {2(-1)^{m}}{2m+{\cfrac {1^{2}}{2m+{\cfrac {2^{2}}{2m+{\cfrac {3^{2}}{2m+\ddots }}}}}}}}\qquad (m=1,2,3,\ldots )}$

Note that Madhava's correction term is

${\displaystyle {\frac {2}{2m+{\frac {1^{2}}{2m+{\frac {2^{2}}{2m}}}}}}=4{\frac {m^{2}+1}{4m^{3}+5m}}}$.

The well-known values 22⁄7 and 355⁄113 are respectively the second and fourth continued fraction approximations to π. (Other representations are available at The Wolfram Functions Site.)

The Gregory–Leibniz series

${\displaystyle \pi =4\sum _{n=0}^{\infty }{\cfrac {(-1)^{n}}{2n+1}}=4\left({\frac {1}{1}}-{\frac {1}{3}}+{\frac {1}{5}}-{\frac {1}{7}}+-\cdots \right)}$

is the power series for arctan(x) specialized to x = 1. It converges too slowly to be of practical interest. However, the power series converges much faster for smaller values of ${\displaystyle x}$, which leads to formulae where ${\displaystyle \pi }$ arises as the sum of small angles with rational tangents, known as Machin-like formulae.

Knowing that 4 arctan 1 = π, the formula can be simplified to get:

${\displaystyle {\begin{aligned}\pi &=2\left(1+{\cfrac {1}{3}}+{\cfrac {1\cdot 2}{3\cdot 5}}+{\cfrac {1\cdot 2\cdot 3}{3\cdot 5\cdot 7}}+{\cfrac {1\cdot 2\cdot 3\cdot 4}{3\cdot 5\cdot 7\cdot 9}}+{\cfrac {1\cdot 2\cdot 3\cdot 4\cdot 5}{3\cdot 5\cdot 7\cdot 9\cdot 11}}+\cdots \right)\\&=2\sum _{n=0}^{\infty }{\cfrac {n!}{(2n+1)!!}}=\sum _{n=0}^{\infty }{\cfrac {2^{n+1}n!^{2}}{(2n+1)!}}=\sum _{n=0}^{\infty }{\cfrac {2^{n+1}}{{\binom {2n}{n}}(2n+1)}}\\&=2+{\frac {2}{3}}+{\frac {4}{15}}+{\frac {4}{35}}+{\frac {16}{315}}+{\frac {16}{693}}+{\frac {32}{3003}}+{\frac {32}{6435}}+{\frac {256}{109395}}+{\frac {256}{230945}}+\cdots \end{aligned}}}$

with a convergence such that each additional 10 terms yields at least three more digits.

Another formula for ${\displaystyle \pi }$ involving arctangent function is given by

${\displaystyle {\frac {\pi }{2^{k+1}}}=\arctan {\frac {\sqrt {2-a_{k-1}}}{a_{k}}},\qquad \qquad k\geq 2,}$

where ${\displaystyle a_{k}={\sqrt {2+a_{k-1}}}}$ such that ${\displaystyle a_{1}={\sqrt {2}}}$. Approximations can be made by using, for example, the rapidly convergent Euler formula[86]

${\displaystyle \arctan(x)=\sum _{n=0}^{\infty }{\frac {2^{2n}(n!)^{2}}{(2n+1)!}}\;{\frac {x^{2n+1}}{(1+x^{2})^{n+1}}}.}$

Alternatively, the following simple expansion series of the arctangent function can be used

${\displaystyle \arctan(x)=2\sum _{n=1}^{\infty }{{\frac {1}{2n-1}}{\frac {{{a}_{n}}\left(x\right)}{a_{n}^{2}\left(x\right)+b_{n}^{2}\left(x\right)}}},}$

where

${\displaystyle {\begin{aligned}&a_{1}(x)=2/x,\\&b_{1}(x)=1,\\&a_{n}(x)=a_{n-1}(x)\,\left(1-4/x^{2}\right)+4b_{n-1}(x)/x,\\&b_{n}(x)=b_{n-1}(x)\,\left(1-4/x^{2}\right)-4a_{n-1}(x)/x,\end{aligned}}}$

to approximate ${\displaystyle \pi }$ with even more rapid convergence. Convergence in this arctangent formula for ${\displaystyle \pi }$ improves as integer ${\displaystyle k}$ increases.

The constant ${\displaystyle \pi }$ can also be expressed by infinite sum of arctangent functions as

${\displaystyle {\frac {\pi }{2}}=\sum _{n=0}^{\infty }\arctan {\frac {1}{F_{2n+1}}}=\arctan {\frac {1}{1}}+\arctan {\frac {1}{2}}+\arctan {\frac {1}{5}}+\arctan {\frac {1}{13}}+\cdots }$

and

${\displaystyle {\frac {\pi }{4}}=\sum _{k\geq 2}\arctan {\frac {\sqrt {2-a_{k-1}}}{a_{k}}},}$

where ${\displaystyle F_{n}}$ is the n-th Fibonacci number. However, these two formulae for ${\displaystyle \pi }$ are much slower in convergence because of set of arctangent functions that are involved in computation.

Observing an equilateral triangle and noting that

${\displaystyle \sin \left({\frac {\pi }{6}}\right)={\frac {1}{2}}}$

yields

${\displaystyle {\begin{aligned}\pi &=6\sin ^{-1}\left({\frac {1}{2}}\right)=6\left({\frac {1}{2}}+{\frac {1}{2\cdot 3\cdot 2^{3}}}+{\frac {1\cdot 3}{2\cdot 4\cdot 5\cdot 2^{5}}}+{\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 7\cdot 2^{7}}}+\cdots \!\right)\\&={\frac {3}{16^{0}\cdot 1}}+{\frac {6}{16^{1}\cdot 3}}+{\frac {18}{16^{2}\cdot 5}}+{\frac {60}{16^{3}\cdot 7}}+\cdots \!=\sum _{n=0}^{\infty }{\frac {3\cdot {\binom {2n}{n}}}{16^{n}(2n+1)}}\\&=3+{\frac {1}{8}}+{\frac {9}{640}}+{\frac {15}{7168}}+{\frac {35}{98304}}+{\frac {189}{2883584}}+{\frac {693}{54525952}}+{\frac {429}{167772160}}+\cdots \end{aligned}}}$

with a convergence such that each additional five terms yields at least three more digits.

Pi Hex was a project to compute three specific binary digits of π using a distributed network of several hundred computers. In 2000, after two years, the project finished computing the five trillionth (5*10¹²), the forty trillionth, and the quadrillionth (10¹⁵) bits. All three of them turned out to be 0.

Most computer algebra systems can calculate π and other common mathematical constants to any desired precision.

Functions for calculating π are also included in many general libraries for arbitrary-precision arithmetic, for instance Class Library for Numbers, MPFR and SymPy.

Programs designed for calculating π may have better performance than general-purpose mathematical software. They typically implement checkpointing and efficient disk swapping to facilitate extremely long-running and memory-expensive computations.

• TachusPi by Fabrice Bellard[92] is the program used by himself to compute world record number of digits of pi in 2009.
• y-cruncher by Alexander Yee[46] is the program which every world record holder since Shigeru Kondo in 2010 has used to compute world record numbers of digits. y-cruncher can also be used to calculate other constants and holds world records for several of them.
• PiFast by Xavier Gourdon was the fastest program for Microsoft Windows in 2003. According to its author, it can compute one million digits in 3.5 seconds on a 2.4 GHz Pentium 4.[93] PiFast can also compute other irrational numbers like e and √2. It can also work at lesser efficiency with very little memory (down to a few tens of megabytes to compute well over a billion (10⁹) digits). This tool is a popular benchmark in the overclocking community. PiFast 4.4 is available from Stu's Pi page. PiFast 4.3 is available from Gourdon's page.
• QuickPi by Steve Pagliarulo for Windows is faster than PiFast for runs of under 400 million digits. Version 4.5 is available on Stu's Pi Page below. Like PiFast, QuickPi can also compute other irrational numbers like e, √2, and √3. The software may be obtained from the Pi-Hacks Yahoo! forum, or from Stu's Pi page.
• Super PI by Kanada Laboratory[94] in the University of Tokyo is the program for Microsoft Windows for runs from 16,000 to 33,550,000 digits. It can compute one million digits in 40 minutes, two million digits in 90 minutes and four million digits in 220 minutes on a Pentium 90 MHz. Super PI version 1.9 is available from Super PI 1.9 page.