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Pi

\begin{figure}\begin{center}\BoxedEPSF{Circle.epsf}\end{center}\end{figure}

A Real Number denoted $\pi$ which is defined as the ratio of a Circle's Circumference $C$ to its Diameter $d=2r$,

\begin{displaymath}
\pi \equiv {C\over d}={C\over 2r}
\end{displaymath} (1)

It is equal to
\begin{displaymath}
\pi=3.141 592 653 589 793 238 462 643 383 279 502 884 197\ldots
\end{displaymath} (2)

(Sloane's A000796). $\pi$ has recently (August 1997) been computed to a world record $51,539,600,000 \approx 3\cdot 2^{34}$ Decimal Digits by Y. Kanada. This calculation was done using Borwein's fourth-order convergent algorithm and required 29 hours on a massively parallel 1024-processor Hitachi SR2201 supercomputer. It was checked in 37 hours using the Brent-Salamin Formula on the same machine.


The Simple Continued Fraction for $\pi$, which gives the ``best'' approximation of a given order, is [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, ...] (Sloane's A001203). The very large term 292 means that the Convergent

\begin{displaymath}[3, 7, 15, 1]= [3, 7, 16] = {\textstyle{355\over 113}} = 3.14159292\ldots
\end{displaymath} (3)

is an extremely good approximation. The first few Convergents are 22/7, 333/106, 355/113, 103993/33102, 104348/33215, ... (Sloane's A002485 and A002486). The first occurrences of $n$ in the Continued Fraction are 4, 9, 1, 30, 40, 32, 2, 44, 130, 100, ... (Sloane's A032523).


Gosper has computed 17,001,303 terms of $\pi$'s Continued Fraction (Gosper 1977, Ball and Coxeter 1987), although the computer on which the numbers are stored may no longer be functional (Gosper, pers. comm., 1998). According to Gosper, a typical Continued Fraction term carries only slightly more significance than a decimal Digit. The sequence of increasing terms in the Continued Fraction is 3, 7, 15, 292, 436, 20776, ... (Sloane's A033089), occurring at positions 1, 2, 3, 5, 308, 432, ... (Sloane's A033090). In the first 26,491 terms of the Continued Fraction (counting 3 as the 0th), the only five-Digit terms are 20,776 (the 431st), 19,055 (15,543rd), and 19,308 (23,398th) (Beeler et al. 1972, Item 140). The first 6-Digit term is 528,210 (the 267,314th), and the first 8-Digit term is 12,996,958 (453,294th). The term having the largest known value is the whopping 9-Digit 87,878,3625 (the 11,504,931st term).


The Simple Continued Fraction for $\pi$ does not show any obvious patterns, but clear patterns do emerge in the beautiful non-simple Continued Fractions

\begin{displaymath}
{4\over\pi} = 1+{\strut\displaystyle 1^2\over\strut\displays...
...+{\strut\displaystyle 7^2\over\strut\displaystyle 2+\ldots}}}}
\end{displaymath} (4)

(Brouckner), giving convergents 1, 3/2, 15/13, 105/76, 315/263, ... (Sloane's A025547 and A007509) and
\begin{displaymath}
{\pi\over 2} = 1-{1\over\strut\displaystyle 3-{\strut\displa...
...\displaystyle 5\cdot 6\over\strut\displaystyle 3-\ldots}}}}}}}
\end{displaymath} (5)

(Stern 1833), giving convergents 1, 2/3, 4/3, 16/15, 64/45, 128/105, ... (Sloane's A001901 and A046126).


$\pi$ crops up in all sorts of unexpected places in mathematics besides Circles and Spheres. For example, it occurs in the normalization of the Gaussian Distribution, in the distribution of Primes, in the construction of numbers which are very close to Integers (the Ramanujan Constant), and in the probability that a pin dropped on a set of Parallel lines intersects a line (Buffon's Needle Problem). Pi also appears as the average ratio of the actual length and the direct distance between source and mouth in a meandering river (Støllum 1996, Singh 1997).


A brief history of Notation for pi is given by Castellanos (1988). $\pi$ is sometimes known as Ludolph's Constant after Ludolph van Ceulen (1539-1610), a Dutch $\pi$ calculator. The symbol $\pi$ was first used by William Jones in 1706, and subsequently adopted by Euler. In Measurement of a Circle, Archimedes (ca. 225 BC ) obtained the first rigorous approximation by Inscribing and Circumscribing $6\cdot 2^n$-gons on a Circle using the Archimedes Algorithm. Using $n=4$ (a 96-gon), Archimedes obtained

\begin{displaymath}
3 +{\textstyle{10\over 71}} < \pi < 3 +{\textstyle{1\over 7}}
\end{displaymath} (6)

(Shanks 1993, p. 140).


The Bible contains two references (I Kings 7:23 and Chronicles 4:2) which give a value of 3 for $\pi$. It should be mentioned, however, that both instances refer to a value obtained from physical measurements and, as such, are probably well within the bounds of experimental uncertainty. I Kings 7:23 states, ``Also he made a molten sea of ten Cubits from brim to brim, round in compass, and five cubits in height thereof; and a line thirty cubits did compass it round about.'' This implies $\pi=C/d=30/10=3$. The Babylonians gave an estimate of $\pi$ as $3+1/8 =
3.125$. The Egyptians did better still, obtaining $2^8/3^4=3.1605\ldots$ in the Rhind papyrus, and 22/7 elsewhere. The Chinese geometers, however, did best of all, rigorously deriving $\pi$ to 6 decimal places.


A method similar to Archimedes' can be used to estimate $\pi$ by starting with an $n$-gon and then relating the Area of subsequent $2n$-gons. Let $\beta$ be the Angle from the center of one of the Polygon's segments,

\begin{displaymath}
\beta = {\textstyle{1\over 4}}(n-3)\pi.
\end{displaymath} (7)

Then
\begin{displaymath}
\pi={{\textstyle{1\over 2}}n\sin(2\beta)\over \cos\beta\cos\...
...\beta\over 2^2}\right)\cos\left({\beta\over 2^3}\right)\cdots}
\end{displaymath} (8)

(Beckmann 1989, pp. 92-94). Viète (1593) was the first to give an exact expression for $\pi$ by taking $n=4$ in the above expression, giving
\begin{displaymath}
\cos\beta=\sin\beta={1\over \sqrt{2}}={\textstyle{1\over 2}}\sqrt{2},
\end{displaymath} (9)

which leads to an Infinite Product of Continued Square Roots,
\begin{displaymath}
{2\over \pi} = \sqrt{{\textstyle{1\over 2}}}\sqrt{{\textstyl...
...}+{\textstyle{1\over 2}}\sqrt{{\textstyle{1\over 2}}}}} \cdots
\end{displaymath} (10)

(Beckmann 1989, p. 95). However, this expression was not rigorously proved to converge until Rudio (1892). Another exact Formula is Machin's Formula, which is
\begin{displaymath}
{\pi\over 4}=4\tan^{-1}({\textstyle{1\over 5}})-\tan^{-1}({\textstyle{1\over 239}}).
\end{displaymath} (11)

There are three other Machin-Like Formulas, as well as other Formulas with more terms. An interesting Infinite Product formula due to Euler which relates $\pi$ and the $n$th Prime $p_n$ is
$\displaystyle \pi$ $\textstyle =$ $\displaystyle {2\over \prod_{i=n}^\infty \left[{1+{\sin({\textstyle{1\over 2}}\pi p_n)\over p_n}}\right]}$ (12)
  $\textstyle =$ $\displaystyle {2\over \prod_{i=n}^\infty \left[{1+{(-1)^{(p_n-1)/2}\over p_n}}\right]}$ (13)

(Blatner 1997, p. 119), plotted below as a function of the number of terms in the product.

\begin{figure}\begin{center}\BoxedEPSF{PiProduct.epsf}\end{center}\end{figure}


The Area and Circumference of the Unit Circle are given by

$\displaystyle A$ $\textstyle =$ $\displaystyle \pi=4\int_0^1 \sqrt{1-x^2}\,dx$ (14)
  $\textstyle =$ $\displaystyle \lim_{n\to\infty} {4\over n^2} \sum_{k=0}^n \sqrt{n^2-k^2}$ (15)

and
$\displaystyle C$ $\textstyle =$ $\displaystyle 2\pi = 4\int_0^1 {dx\over\sqrt{1-x^2}}$ (16)
  $\textstyle =$ $\displaystyle 4\int_0^1 \sqrt{1+\left({{d\over dx}\sqrt{1-x^2}\,}\right)^2}\,dx.$ (17)

The Surface Area and Volume of the unit Sphere are
$\displaystyle S$ $\textstyle =$ $\displaystyle 4\pi$ (18)
$\displaystyle V$ $\textstyle =$ $\displaystyle {\textstyle{4\over 3}}\pi.$ (19)


$\pi$ is known to be Irrational (Lambert 1761, Legendre 1794) and even Transcendental (Lindemann 1882). Incidentally, Lindemann's proof of the transcendence of $\pi$ also proved that the Geometric Problem of Antiquity known as Circle Squaring is impossible. A simplified, but still difficult, version of Lindemann's proof is given by Klein (1955).


It is also known that $\pi$ is not a Liouville Number (Mahler 1953). In 1974, M. Mignotte showed that

\begin{displaymath}
\left\vert{\pi-{p\over q}}\right\vert\leq q^{-20}
\end{displaymath} (20)

has only a finite number of solutions in Integers (Le Lionnais 1983, p. 50). This result was subsequently improved by Chudnovsky and Chudnovsky (1984) who showed that
\begin{displaymath}
\left\vert{\pi-{p\over q}}\right\vert> q^{-14.65},
\end{displaymath} (21)

although it is likely that the exponent can be reduced to $2+\epsilon$, where $\epsilon$ is an infinitesimally small number (Borwein et al. 1989). It is not known if $\pi$ is Normal (Wagon 1985), although the first 30 million Digits are very Uniformly Distributed (Bailey 1988). The following distribution is found for the first $n$ Digits of $\pi-3$. It shows no statistically Significant departure from a Uniform Distribution (technically, in the Chi-Squared Test, it has a value of ${\chi_s}^2=5.60$ for the first $5\times 10^{10}$ terms).


digit $1\times 10^5$ $1\times 10^6$ $6\times 10^9$ $5\times 10^{10}$
0 9,999 99,959 599,963,005 5,000,012,647
1 10,137 99,758 600,033,260 4,999,986,263
2 9,908 100,026 599,999,169 5,000,020,237
3 10,025 100,229 600,000,243 4,999,914,405
4 9,971 100,230 599,957,439 5,000,023,598
5 10,026 100,359 600,017,176 4,999,991,499
6 10,029 99,548 600,016,588 4,999,928,368
7 10,025 99,800 600,009,044 5,000,014,860
8 9,978 99,985 599,987,038 5,000,117,637
9 9,902 100,106 600,017,038 4,999,990,486


The digits of $1/\pi$ are also very uniformly distributed ( ${\chi_s}^2=7.04$), as shown in the following table.

digit $5\times 10^{10}$
0 4,999,969,955
1 5,000,113,699
2 4,999,987,893
3 5,000,040,906
4 4,999,985,863
5 4,999,977,583
6 4,999,990,916
7 4,999,985,552
8 4,999,881,183
9 5,000,066,450


It is not known if $\pi+e$, $\pi/e$, or $\ln\pi$ are Irrational. However, it is known that they cannot satisfy any Polynomial equation of degree $\leq 8$ with Integer Coefficients of average size 109 (Bailey 1988, Borwein et al. 1989).


$\pi$ satisfies the Inequality

\begin{displaymath}
\left({1+{1\over\pi}}\right)^{\pi+1}\approx 3.14097 < \pi.
\end{displaymath} (22)


Beginning with any Positive Integer $n$, round up to the nearest multiple of $n-1$, then up to the nearest multiple of $n-2$, and so on, up to the nearest multiple of 1. Let $f(n)$ denote the result. Then the ratio

\begin{displaymath}
\lim_{n\to\infty} {n^2\over f(n)} = \pi
\end{displaymath} (23)

(Brown). David (1957) credits this result to Jabotinski and Erdös and gives the more precise asymptotic result
\begin{displaymath}
f(n) = {n^2\over \pi} + {\mathcal O}(n^{4/3}).
\end{displaymath} (24)

The first few numbers in the sequence $\{f(n)\}$ are 1, 2, 4, 6, 10, 12, 18, 22, 30, 34, ... (Sloane's A002491).


A particular case of the Wallis Formula gives

\begin{displaymath}
{\pi\over 2} = \prod_{n=1}^\infty \left[{(2n)^2\over(2n-1)(2...
... \,{4\cdot 4\over 3\cdot 5} \,{6\cdot 6\over 5\cdot 7} \cdots.
\end{displaymath} (25)

This formula can also be written
\begin{displaymath}
\lim_{n\to\infty} {2^{4n}\over n{2n\choose n}^2}=\pi \lim_{n...
...{n[\Gamma(n)]^2\over[\Gamma({\textstyle{1\over 2}}+n)]^2}=\pi,
\end{displaymath} (26)

where ${n\choose k}$ denotes a Binomial Coefficient and $\Gamma(x)$ is the Gamma Function (Knopp 1990). Euler obtained
\begin{displaymath}
\pi = \sqrt{6\left({1+{1\over 2^2}+{1\over 3^2}+{1\over 4^2}+\ldots}\right)},
\end{displaymath} (27)

which follows from the special value of the Riemann Zeta Function $\zeta(2)=\pi^2/6$. Similar Formulas follow from $\zeta(2n)$ for all Positive Integers $n$. Gregory and Leibniz found
\begin{displaymath}
{\pi\over 4}=1-{1\over 3}+{1\over 5}+\ldots,
\end{displaymath} (28)

which is sometimes known as Gregory's Formula. The error after the $n$th term of this series in Gregory's Formula is larger than $(2n)^{-1}$ so this sum converges so slowly that 300 terms are not sufficient to calculate $\pi$ correctly to two decimal places! However, it can be transformed to
\begin{displaymath}
\pi=\sum_{k=1}^\infty {3^k-1\over 4^k} \zeta(k+1),
\end{displaymath} (29)

where $\zeta(z)$ is the Riemann Zeta Function (Vardi 1991, pp. 157-158; Flajolet and Vardi 1996), so that the error after $k$ terms is $\approx (3/4)^k$. Newton used
$\displaystyle \pi$ $\textstyle =$ $\displaystyle {\textstyle{3\over 4}}\sqrt{3}+24\int_0^{1/4}\sqrt{x-x^2}\,dx$ (30)
  $\textstyle =$ $\displaystyle {3\sqrt{3}\over 4}+24\left({{1\over 12}-{1\over 5\cdot 2^5}-{1\over 28\cdot 2^7}-{1\over 72\cdot 2^9} -\ldots}\right)$  
      (31)

(Borwein et al. 1989). Using Euler's Convergence Improvement transformation gives
$\displaystyle {\pi\over 2}$ $\textstyle =$ $\displaystyle {1\over 2} \sum_{n=0}^\infty {(n!)^2 2^{n+1}\over (2n+1)!} = \sum_{n=0}^\infty {n!\over (2n+1)!!}$  
  $\textstyle =$ $\displaystyle 1+{1\over 3}+{1\cdot 2\over 3\cdot 5}+{1\cdot 2\cdot 3\over 3\cdot 5\cdot 7}+\ldots$ (32)
  $\textstyle =$ $\displaystyle 1+{1\over 3}\left({1+{2\over 5}\left({1+{3\over 7}\left({1+{4\over 9}\left({1+\ldots}\right)}\right)}\right)}\right)$ (33)

(Beeler et al. 1972, Item 120). This corresponds to plugging $x=1/\sqrt{2}$ into the Power Series for the Hypergeometric Function ${}_2F_1(a,b;c;x)$,
\begin{displaymath}
{\sin^{-1}x\over\sqrt{1-x^2}}=\sum_{i=0}^\infty {(2x)^{2i+1}...
...\over 2(2i+1)!} = {}_2F_1(1, 1; {\textstyle{3\over 2}}; x^2)x.
\end{displaymath} (34)

Despite the convergence improvement, series (33) converges at only one bit/term. At the cost of a Square Root, Gosper has noted that $x = 1/2$ gives 2 bits/term,
\begin{displaymath}
{\textstyle{1\over 9}}\sqrt{3}\pi={1\over 2}\sum_{i=1}^\infty {(i!)^2\over(2i+1)!},
\end{displaymath} (35)

and $x = \sin(\pi/10)$ gives almost 3.39 bits/term,
\begin{displaymath}
{\pi\over 5\sqrt{\phi+2}}={1\over 2}\sum_{i=0}^\infty {(i!)^2\over\phi^{2i+1}(2i+1)!},
\end{displaymath} (36)

where $\phi$ is the Golden Ratio. Gosper also obtained


\begin{displaymath}
\pi = 3+{1\over 60}\left({8+{2\cdot 3\over 7\cdot 8\cdot 3}\...
... 7\over 13\cdot 14\cdot 3}(23+\ldots)}\right)}\right)}\right).
\end{displaymath} (37)

An infinite sum due to Ramanujan is
\begin{displaymath}
{1\over \pi}=\sum_{n=0}^\infty {2n \choose n}^3 {42n+5\over 2^{12n+4}}
\end{displaymath} (38)

(Borwein et al. 1989). Further sums are given in Ramanujan (1913-14),
${4\over \pi}=\sum_{n=0}^\infty {(-1)^n(1123+21460n)(2n-1)!!(4n-1)!!\over 882^{2n+1} 32^n (n!)^3}$

(39)
and


\begin{displaymath}
{1\over\pi} =\sqrt{8}\sum_{n=0}^\infty {(1103+26390n)(2n-1)!...
...1} \sum_{n=0}^\infty {(4n)!(1103+26390n)\over (n!)^4 396^{4n}}
\end{displaymath} (40)

(Beeler et al. 1972, Item 139; Borwein et al. 1989). Equation (40) is derived from a modular identity of order 58, although a first derivation was not presented prior to Borwein and Borwein (1987). The above series both give
\begin{displaymath}
\pi\approx {9801\over 2206\sqrt{2}} = 3.14159273001\ldots
\end{displaymath} (41)

as the first approximation and provide, respectively, about 6 and 8 decimal places per term. Such series exist because of the rationality of various modular invariants. The general form of the series is
\begin{displaymath}
\sum_{n=0}^\infty [a(t)+nb(t)] {(6n)!\over(3n)!(n!)^3}{1\over[j(t)]^n}={\sqrt{-j(t)}\over\pi},
\end{displaymath} (42)

where $t$ is a Quadratic Form Discriminant, $j(t)$ is the j-Function,
$\displaystyle b(t)$ $\textstyle =$ $\displaystyle \sqrt{t[1728-j(t)]}$ (43)
$\displaystyle a(t)$ $\textstyle =$ $\displaystyle {b(t)\over 6}\left\{{1-{E_4(t)\over E_6(t)}\left[{E_2(t)-{6\over\pi\sqrt{t}}}\right]}\right\},$ (44)

and the $E_i$ are Ramanujan-Eisenstein Series. A Class Number $p$ field involves $p$th degree Algebraic Integers of the constants $A=a(t)$, $B=b(t)$, and $C=c(t)$. The fastest converging series that uses only Integer terms corresponds to the largest Class Number 1 discriminant of $d=-163$ and was formulated by the Chudnovsky brothers (1987). The 163 appearing here is the same one appearing in the fact that $e^{\pi\sqrt{163}}$ (the Ramanujan Constant) is very nearly an Integer. The series is given by


$\displaystyle {1\over \pi}$ $\textstyle =$ $\displaystyle 12\sum_{n=0}^\infty {(-1)^n(6n)!(13591409+545140134n)\over (n!)^3(3n)!(640320^3)^{n+1/2}}$  
  $\textstyle =$ $\displaystyle {163\cdot 8\cdot 27\cdot 7\cdot 11\cdot 19\cdot 127\over 640320^{...
...1\cdot 19\cdot 127}+n}\right){(6n)!\over (3n)!(n!)^3} {(-1)^n\over 640320^{3n}}$ (45)

(Borwein and Borwein 1993). This series gives 14 digits accurately per term. The same equation in another form was given by the Chudnovsky brothers (1987) and is used by Mathematica ${}^{\scriptstyle\circledRsymbol}$ (Wolfram Research, Champaign, IL) to calculate $\pi$ (Vardi 1991),
\begin{displaymath}
\pi={426880\sqrt{10005}\over A
[\,{}_3F_2({\textstyle{1\ove...
..., {\textstyle{3\over 2}}, {\textstyle{11\over 6}}; 2, 2; B)]},
\end{displaymath} (46)

where
$\displaystyle A$ $\textstyle \equiv$ $\displaystyle 13591409$ (47)
$\displaystyle B$ $\textstyle \equiv$ $\displaystyle -{\textstyle{1\over 151931373056000}}$ (48)
$\displaystyle C$ $\textstyle \equiv$ $\displaystyle {\textstyle{30285563\over 1651969144908540723200}}.$ (49)


The best formula for Class Number 2 (largest discriminant $-427$) is

\begin{displaymath}
{1\over \pi} = 12\sum_{n=0}^\infty {(-1)^n(6n)!(A+Bn)\over (n!)^3(3n)!C^{n+1/2}},
\end{displaymath} (50)

where
$\displaystyle A$ $\textstyle \equiv$ $\displaystyle 212175710912\sqrt{61}+1657145277365$ (51)
$\displaystyle B$ $\textstyle \equiv$ $\displaystyle 13773980892672\sqrt{61}+107578229802750$ (52)
$\displaystyle C$ $\textstyle \equiv$ $\displaystyle [5280(236674+30303\sqrt{61}\,)]^3$ (53)

(Borwein and Borwein 1993). This series adds about 25 digits for each additional term. The fastest converging series for Class Number 3 corresponds to $d=-907$ and gives 37-38 digits per term. The fastest converging Class Number 4 series corresponds to $d=-1555$ and is
\begin{displaymath}
{\sqrt{-C^3}\over\pi}=\sum_{n=0}^\infty {(6n)!\over(3n)!(n!)^3}{A+nB\over C^{3n}},
\end{displaymath} (54)

where


$\displaystyle A$ $\textstyle =$ $\displaystyle 63365028312971999585426220$  
  $\textstyle \phantom{=}$ $\displaystyle +28337702140800842046825600\sqrt{5}$  
  $\textstyle \phantom{=}$ $\displaystyle +384\sqrt{5}(108917285511711782004674\cdots$  
  $\textstyle \phantom{=}$ $\displaystyle \cdots 36212395209160385656017+487902908657881022\cdots$  
  $\textstyle \phantom{=}$ $\displaystyle \cdots 5077338534541688721351255040\sqrt{5}\,)^{1/2}$ (55)
$\displaystyle B$ $\textstyle =$ $\displaystyle 7849910453496627210289749000$  
  $\textstyle \phantom{=}$ $\displaystyle +3510586678260932028965606400\sqrt{5}$  
  $\textstyle \phantom{=}$ $\displaystyle +2515968\sqrt{3110}(62602083237890016\cdots$  
  $\textstyle \phantom{=}$ $\displaystyle \cdots 36993322654444020882161+2799650273060444296\cdots$  
  $\textstyle \phantom{=}$ $\displaystyle \cdots 577206890718825190235\sqrt{5}\,)^{1/2}$ (56)
$\displaystyle C$ $\textstyle =$ $\displaystyle -214772995063512240-96049403338648032\sqrt{5}$  
  $\textstyle \phantom{=}$ $\displaystyle -1296\sqrt{5}(10985234579463550323713318473$  
  $\textstyle \phantom{=}$ $\displaystyle +4912746253692362754607395912\sqrt{5}\,)^{1/2}.$ (57)

This gives 50 digits per term. Borwein and Borwein (1993) have developed a general Algorithm for generating such series for arbitrary Class Number. Bellard gives the exotic formula

\begin{displaymath}
\pi={1\over 740025}\left[{\sum_{n=1}^\infty {3P(n)\over {7n\choose 2n}2^{n-1}}-20379280}\right],
\end{displaymath} (58)

where
$P(n)\equiv -885673181n^5+3125347237n^4-2942969225n^3$
$+1031962795n^2-196882274n+10996648.\quad$ (59)


A complete listing of Ramanujan's series for $1/\pi$ found in his second and third notebooks is given by Berndt (1994, pp. 352-354),

$\quad {4\over\pi}=\sum_{n=0}^\infty {(6n+1){({\textstyle{1\over 2}})_n}^3\over 4^n(n!)^3}$ (60)
$\quad {16\over\pi}=\sum_{n=0}^\infty {(42n+5){({\textstyle{1\over 2}})_n}^3\over (64)^n(n!)^3}$ (61)
$\quad {32\over\pi}=\sum_{n=0}^\infty {(42\sqrt{5}\,n+5\sqrt{5}+30n-1){({\textstyle{1\over 2}})_n}^3\over (64)^n(n!)^3}\left({\sqrt{5}-1\over 2}\right)^{8n}$ (62)
$\quad {27\over 4\pi}=\sum_{n=0}^\infty {(15n+2)({\textstyle{1\over 2}})_n({\tex...
...{1\over 3}})_n({\textstyle{2\over 3}})_n\over(n!)^3} \left({2\over 27}\right)^n$ (63)
$\quad {15\sqrt{3}\over 2\pi}=\sum_{n=0}^\infty {(33n+4)({\textstyle{1\over 2}})...
...1\over 3}})_n({\textstyle{2\over 3}})_n\over (n!)^3}\left({4\over 125}\right)^n$ (64)
$\quad {5\sqrt{5}\over 2\pi\sqrt{3}}=\sum_{n=0}^\infty {(11n+1)({\textstyle{1\ov...
...{1\over 6}})_n({\textstyle{5\over 6}})_n\over(n!)^3}\left({4\over 125}\right)^n$ (65)
$\quad {85\sqrt{85}\over 18\pi\sqrt{3}}=\sum_{n=0}^\infty {(133n+8)({\textstyle{...
...e{1\over 6}})_n({\textstyle{5\over 6}})_n\over(n!)^3}\left({4\over 85}\right)^n$ (66)
$\quad {4\over\pi}=\sum_{n=0}^\infty{(-1)^n(20n+3)({\textstyle{1\over 2}})_n({\textstyle{1\over 4}})_n({\textstyle{3\over 4}})_n\over(n!)^3 2^{2n+1}}$ (67)
$\quad {4\over\pi\sqrt{3}}=\sum_{n=0}^\infty {(-1)^n(28n+3)({\textstyle{1\over 2}})_n({\textstyle{1\over 4}})_n({\textstyle{3\over 4}})_n\over(n!)^3 3^n 4^{n+1}}$ (68)
$\quad {4\over\pi}=\sum_{n=0}^\infty {(-1)^n(260n+23)({\textstyle{1\over 2}})_n({\textstyle{1\over 4}})_n({\textstyle{3\over 4}})_n\over(n!)^3(18)^{2n+1}}$ (69)
$\quad {4\over\pi\sqrt{5}}=\sum_{n=0}^\infty {(-1)^n(644n+41)({\textstyle{1\over...
...n({\textstyle{1\over 4}})_n({\textstyle{3\over 4}})_n\over(n!)^35^n(72)^{2n+1}}$ (70)
$\quad {4\over\pi}=\sum_{n=0}^\infty {(-1)^n(21460n+1123)({\textstyle{1\over 2}})_n({\textstyle{1\over 4}})_n({\textstyle{3\over 4}})_n\over(n!)^3(882)^{2n+1}}$ (71)
$\quad {2\sqrt{3}\over\pi}=\sum_{n=0}^\infty {(8n+1)({\textstyle{1\over 2}})_n({\textstyle{1\over 4}})_n({\textstyle{3\over 4}})_n\over(n!)^3 9^n}$ (72)
$\quad {1\over 2\pi\sqrt{2}}=\sum_{n=0}^\infty {(10n+1)({\textstyle{1\over 2}})_n({\textstyle{1\over 4}})_n({\textstyle{3\over 4}})_n\over(n!)^3 9^{2n+1}}$ (73)
$\quad {1\over 3\pi\sqrt{3}}=\sum_{n=0}^\infty {(40n+3)({\textstyle{1\over 2}})_n({\textstyle{1\over 4}})_n({\textstyle{3\over 4}})_n\over(n!)^3(49)^{2n+1}}$ (74)
$\quad {2\over\pi\sqrt{11}}=\sum_{n=0}^\infty{(280n+19)({\textstyle{1\over 2}})_n({\textstyle{1\over 4}})_n({\textstyle{3\over 4}})_n\over(n!)^3(99)^{2n+1}}$ (75)
$\quad {1\over 2\pi\sqrt{2}}=\sum_{n=0}^\infty{(26390n+1103)({\textstyle{1\over 2}})_n({\textstyle{1\over 4}})_n({\textstyle{3\over 4}})_n\over(n!)^3(99)^{4n+2}}.$ (76)
These equations were first proved by Borwein and Borwein (1987, pp. 177-187). Borwein and Borwein (1987b, 1988, 1993) proved other equations of this type, and Chudnovsky and Chudnovsky (1987) found similar equations for other transcendental constants.


A Spigot Algorithm for $\pi$ is given by Rabinowitz and Wagon (1995). Amazingly, a closed form expression giving a digit extraction algorithm which produces digits of $\pi$ (or $\pi^2$) in base-16 was recently discovered by Bailey et al. (Bailey et al. 1995, Adamchik and Wagon 1997),


\begin{displaymath}
\pi=\sum_{n=0}^\infty \left({{4\over 8n+1}-{2\over 8n+4}-{1\over 8n+5}-{1\over 8n+6}}\right)\left({1\over 16}\right)^n,
\end{displaymath} (77)

which can also be written using the shorthand notation
\begin{displaymath}
\pi=\sum_{i=1}^\infty {p_i\over 16^{\left\lfloor{i/8}\right\...
...\qquad
\{p_i\}=\{\,\overline{4, 0, 0, -2, -1, -1, 0, 0}\,\},
\end{displaymath} (78)

where $\{p_i\}$ is given by the periodic sequence obtained by appending copies of $\{4, 0, 0, -2, -1, -1, 0, 0\}$ (in other words, $p_i\equiv p_{[(i-1) {\rm\ (mod\ 8})]+1}$ for $i>8$) and $\left\lfloor{x}\right\rfloor $ is the Floor Function. This expression was discovered using the PSLQ Algorithm and is equivalent to
\begin{displaymath}
\pi=\int_0^1 {16y-16\over y^4-2y^3+4y-4}\,dy.
\end{displaymath} (79)

A similar formula was subsequently discovered by Ferguson, leading to a 2-D lattice of such formulas which can be generated by these two formulas. A related integral is
\begin{displaymath}
\pi={\textstyle{22\over 7}}-\int_0^1 {x^4(1-x)^4\over 1+x^2}\,dx
\end{displaymath} (80)

(Le Lionnais 1983, p. 22). F. Bellard found the more rapidly converging digit-extraction algorithm (in Hexadecimal)


\begin{displaymath}
\pi={1\over 2^6}\sum_{n=0}^\infty {(-1)^n\over 2^{10n}}\left...
...n+3}-{2^2\over 10n+5}-{2^2\over 10n+7}+{1\over 10n+9}}\right).
\end{displaymath} (81)

More amazingly still, S. Plouffe has devised an algorithm to compute the $n$th Digit of $\pi$ in any base in ${\mathcal O}(n^3(\log n)^3)$ steps.


Another identity is

\begin{displaymath}
\pi^2=36\mathop{\rm Li}\nolimits _2({\textstyle{1\over 2}})-...
...er 8}})+6\mathop{\rm Li}\nolimits _2({\textstyle{1\over 64}}),
\end{displaymath} (82)

where $L_n$ is the Polylogarithm. (82) is equivalent to
\begin{displaymath}
{\pi^2\over 36}=\sum_{i=1}^\infty {a_i\over 2^i i^2}\qquad \{a_i\}=[\,\overline{1, -3, -2, -3, 1, 0}\,]
\end{displaymath} (83)

and
\begin{displaymath}
\pi^2=12L_2({\textstyle{1\over 2}})+6(\ln 2)^2
\end{displaymath} (84)

(Bailey et al. 1995). Furthermore


\begin{displaymath}
\pi^2={1\over 8}\sum_{k=0}^\infty {1\over 64^k}\left[{{144\o...
...}-{72\over(6k+3)^2}-{54\over(6k+4)^2}+{9\over(6k+5)^2}}\right]
\end{displaymath} (85)

and


\begin{displaymath}
\pi^2=\sum_{k=0}^\infty {1\over 16^k}\left[{{16\over(8k+1)^2...
...^2}-{4\over(8k+5)^2}-{4\over(8k+6)^2}+{2\over(8k+7)^2}}\right]
\end{displaymath} (86)

(Bailey et al. 1995, Bailey and Plouffe).


A slew of additional identities due to Ramanujan , Catalan, and Newton are given by Castellanos (1988, pp. 86-88), including several involving sums of Fibonacci Numbers.


Gasper quotes the result

\begin{displaymath}
\pi={16\over 3}\left[{\,\lim_{x\to\infty} x\,{}_1F_2({\textstyle{1\over 2}}; 2, 3; -x^2)\,}\right]^{-1},
\end{displaymath} (87)

where ${}_1F_2$ is a Generalized Hypergeometric Function, and transforms it to
\begin{displaymath}
\pi=\lim_{x\to\infty} 4x\,{}_1F_2({\textstyle{1\over 2}}; {\textstyle{3\over 2}}, {\textstyle{3\over 2}}; -x^2).
\end{displaymath} (88)

Fascinating results due to Gosper include
\begin{displaymath}
\lim_{n\to\infty} \prod_{i=n}^{2n} {\pi\over 2\tan^{-1} i}=4^{1/\pi} = 1.554682275\ldots
\end{displaymath} (89)

and


\begin{displaymath}
\sum_{n=1}^\infty {1\over n^2}\cos\left({9\over n\pi+\sqrt{n^2\pi^2-9}}\right)= -{\pi^2\over 12e^3} = -0.040948222\ldots.
\end{displaymath} (90)

Gosper also gives the curious identity

${1\over e}\prod_{n=1}^\infty \left({{1\over 3n}+1}\right)^{3n+1/2}={3\cdot 3^{1...
...({\textstyle{1\over 3}})\over 12\pi}-{2\zeta'(2)\over\pi^2}-1}\right]\pi^{5/6}}$
$ = 1.012378552722912\ldots.\quad$ (91)
Another curious fact is the Almost Integer

\begin{displaymath}
e^\pi-\pi=19.999099979\ldots,
\end{displaymath} (92)

which can also be written as
\begin{displaymath}
(\pi+20)^i = -0.9999999992-0.0000388927i \approx -1
\end{displaymath} (93)


\begin{displaymath}
\cos(\ln(\pi + 20)) \approx -0.9999999992.
\end{displaymath} (94)

Applying Cosine a few more times gives


\begin{displaymath}
\cos(\pi\cos(\pi\cos(\ln(\pi+20))))\approx -1+3.9321609261\times 10^{-35}.
\end{displaymath} (95)


$\pi$ may also be computed using iterative Algorithms. A quadratically converging Algorithm due to Borwein is

$\displaystyle x_0$ $\textstyle =$ $\displaystyle \sqrt{2}$ (96)
$\displaystyle \pi_0$ $\textstyle =$ $\displaystyle 2+\sqrt{2}$ (97)
$\displaystyle y_1$ $\textstyle =$ $\displaystyle 2^{1/4}$ (98)

and
$\displaystyle x_{n+1}$ $\textstyle =$ $\displaystyle {1\over 2}\left({\sqrt{x_n}+{1\over\sqrt{x_n}}}\right)$ (99)
$\displaystyle y_{n+1}$ $\textstyle =$ $\displaystyle {y_n\sqrt{x_n}+{1\over\sqrt{x_n}}\over y_n+1}$ (100)
$\displaystyle \pi_n$ $\textstyle =$ $\displaystyle \pi_{n-1}{x_n+1\over y_n+1}.$ (101)

$\pi_n$ decreases monotonically to $\pi$ with
\begin{displaymath}
\pi_n-\pi < 10^{-2^{n+1}}
\end{displaymath} (102)

for $n\geq 2$. The Brent-Salamin Formula is another quadratically converging algorithm which can be used to calculate $\pi$. A quadratically convergent algorithm for $\pi/\ln 2$ based on an observation by Salamin is given by defining
\begin{displaymath}
f(k)=k 2^{-k/4}\left[{\sum_{n=1}^\infty 2^{-k{n\choose 2}}}\right]^2,
\end{displaymath} (103)

then writing
\begin{displaymath}
g_0\equiv {f(n)\over f(2n)}.
\end{displaymath} (104)

Now iterate
\begin{displaymath}
g_k=\sqrt{{1\over 2}\left({g_{k-1}+{1\over g_{k+1}}}\right)}
\end{displaymath} (105)

to obtain
\begin{displaymath}
\pi=2(\ln 2)f(n)\prod_{k=1}^\infty g_k.
\end{displaymath} (106)


A cubically converging Algorithm which converges to the nearest multiple of $\pi$ to $f_0$ is the simple iteration

\begin{displaymath}
f_n=f_{n-1}+\sin(f_{n-1})
\end{displaymath} (107)

(Beeler et al. 1972). For example, applying to 23 gives the sequence
\begin{displaymath}
\{23, 22.1537796, 21.99186453, 21.99114858, \dots\},
\end{displaymath} (108)

which converges to $7\pi\approx 21.99114858$.


A quartically converging Algorithm is obtained by letting

$\displaystyle y_0$ $\textstyle =$ $\displaystyle \sqrt{2}-1$ (109)
$\displaystyle \alpha$ $\textstyle =$ $\displaystyle 6-4\sqrt{2},$ (110)

then defining
\begin{displaymath}
y_{n+1}={1-(1-{y_n}^4)^{1/4}\over 1+(1-{y_n}^4)^{1/4}}
\end{displaymath} (111)


\begin{displaymath}
\alpha_{n+1} =(1+y_{n+1})^4\alpha _n-2^{2n+3}y_{n+1}(1+y_{n+1}+{y_{n+1}}^2).
\end{displaymath} (112)

Then
\begin{displaymath}
\pi=\lim_{n\to\infty} {1\over \alpha_n}
\end{displaymath} (113)

and $\alpha_n$ converges to $1/\pi$ quartically with
\begin{displaymath}
\alpha_n-{1\over \pi} < 16\cdot 4^n e^{-2\pi\cdot 4^n}
\end{displaymath} (114)

(Borwein and Borwein 1987, Bailey 1988, Borwein et al. 1989). This Algorithm rests on a Modular Equation identity of order 4.


A quintically converging Algorithm is obtained by letting

$\displaystyle s_0$ $\textstyle =$ $\displaystyle 5(\sqrt{5}-2)$ (115)
$\displaystyle \alpha_0$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}.$ (116)

Then let
\begin{displaymath}
s_{n+1}={25\over (z+{x\over z}+1)^2s_n},
\end{displaymath} (117)

where
$\displaystyle x$ $\textstyle =$ $\displaystyle {5\over s_n}-1$ (118)
$\displaystyle y$ $\textstyle =$ $\displaystyle (x-1)^2+7$ (119)
$\displaystyle z$ $\textstyle =$ $\displaystyle [{\textstyle{1\over 2}}x(y+\sqrt{y^2-4x^3}\,)]^{1/5}.$ (120)

Finally, let
\begin{displaymath}
\alpha_{n+1}={s_n}^2\alpha_n-5^n[{\textstyle{1\over 2}}({s_n}^2-5)+\sqrt{s_n({s_n}^2-2s_n+5)}\,],
\end{displaymath} (121)

then
\begin{displaymath}
0<\alpha_n-{1\over\pi}<16\cdot 5^n e^{-\pi 5^n}
\end{displaymath} (122)

(Borwein et al. 1989). This Algorithm rests on a Modular Equation identity of order 5.


Another Algorithm is due to Woon (1995). Define $a(0)\equiv 1$ and

\begin{displaymath}
a(n)=\sqrt{1+\left[{\,\sum_{k=0}^{n-1} a(k)}\right]^2}.
\end{displaymath} (123)

It can be proved by induction that
\begin{displaymath}
a(n)=\csc\left({\pi\over 2^{n+1}}\right).
\end{displaymath} (124)

For $n=0$, the identity holds. If it holds for $n\leq t$, then
\begin{displaymath}
a(t+1)=\sqrt{1+\left[{\,\sum_{k=0}^t \csc\left({\pi\over 2^{k+1}}\right)}\right]^2},
\end{displaymath} (125)

but
\begin{displaymath}
\csc\left({\pi\over 2^{k+1}}\right)=\cot\left({\pi\over 2^{k+2}}\right)-\cot\left({\pi\over 2^{k+1}}\right),
\end{displaymath} (126)

so
\begin{displaymath}
\sum_{k=0}^t \csc\left({\pi\over 2^{k+1}}\right)=\cot\left({\pi\over 2^{t+2}}\right).
\end{displaymath} (127)

Therefore,
\begin{displaymath}
a(t+1)=\csc\left({\pi\over 2^{t+2}}\right),
\end{displaymath} (128)

so the identity holds for $n=t+1$ and, by induction, for all Nonnegative $n$, and
$\displaystyle \lim_{n\to\infty} {2^{n+1}\over a(n)}$ $\textstyle =$ $\displaystyle \lim_{n\to \infty} 2^{n+1}\sin\left({\pi\over 2^{n+1}}\right)$  
  $\textstyle =$ $\displaystyle \lim_{n\to\infty} 2^{n+1} {\pi\over 2^{n+1}} {\sin\left({\pi\over 2^{n+1}}\right)\over {\pi\over 2^{n+1}}}$  
  $\textstyle =$ $\displaystyle \pi \lim_{\theta\to 0} {\sin\theta\over \theta} = \pi.$ (129)


Other iterative Algorithms are the Archimedes Algorithm, which was derived by Pfaff in 1800, and the Brent-Salamin Formula. Borwein et al. (1989) discuss $p$th order iterative algorithms.


Kochansky's Approximation is the Root of

\begin{displaymath}
9x^4-240x^2+1492.
\end{displaymath} (130)

given by
\begin{displaymath}
\pi\approx \sqrt{{\textstyle{40\over 3}}-\sqrt{12}} \approx 3.141533.
\end{displaymath} (131)

An approximation involving the Golden Mean is
\begin{displaymath}
\pi\approx {\textstyle{6\over 5}}\phi^2 = {6\over 5}\left({\...
...ght)^2 = {\textstyle{3\over 5}}(3+\sqrt{5}\,) = 3.14164\ldots.
\end{displaymath} (132)


Some approximations due to Ramanujan


$\displaystyle \pi$ $\textstyle \approx$ $\displaystyle {19\sqrt{7}\over 16}$ (133)
  $\textstyle \approx$ $\displaystyle {\textstyle{7\over 3}}(1+{\textstyle{1\over 5}}\sqrt{3}\,)$ (134)
  $\textstyle \approx$ $\displaystyle {\textstyle{9\over 5}}+\sqrt{{\textstyle{9\over 5}}}$ (135)
  $\textstyle \approx$ $\displaystyle \left({9^2+{19^2\over 22}}\right)^{1/4}=\left({102-{2222\over 22^2}}\right)^{1/4}$ (136)
  $\textstyle \approx$ $\displaystyle (97+{\textstyle{1\over 2}}-{\textstyle{1\over 11}})^{1/4} =(97+{\textstyle{9\over 22}})^{1/4}$ (137)
  $\textstyle \approx$ $\displaystyle {63\over 25}\left({17+15\sqrt{5}\over 7+15\sqrt{5}}\right)$ (138)
  $\textstyle \approx$ $\displaystyle {355\over 113}\left({1-{0.0003\over 3533}}\right)$ (139)
  $\textstyle \approx$ $\displaystyle {12\over\sqrt{130}}\ln\left[{(3+\sqrt{13}\,)(\sqrt{8}+\sqrt{10}\,)\over 2}\right]$ (140)
  $\textstyle \approx$ $\displaystyle {24\over\sqrt{142}}\ln\left[{\sqrt{10+11\sqrt{2}}+\sqrt{10+7\sqrt{2}}\over 2}\right]$ (141)
  $\textstyle \approx$ $\displaystyle {12\over\sqrt{190}}\ln[(3+\sqrt{10}\,)(\sqrt{8}+\sqrt{10}\,)]$ (142)
  $\textstyle \approx$ $\displaystyle {12\over \sqrt{310}}\ln\left[{{\textstyle{1\over 4}}(3+\sqrt{5}\,)(2+\sqrt{2}\,)\left({5+2\sqrt{10}+\sqrt{61+20\sqrt{10}}\,}\right)}\right]$ (143)
  $\textstyle \approx$ $\displaystyle {4\over\sqrt{522}} \ln\left[{\left({5+\sqrt{29}\over\sqrt{2}}\rig...
...eft({\sqrt{9+3\sqrt{6}\over 4}+\sqrt{5+3\sqrt{6}\over 4}\,}\right)^6\,}\right],$ (144)

which are accurate to 3, 4, 4, 8, 8, 9, 14, 15, 15, 18, 23, 31 digits, respectively (Ramanujan 1913-1914; Hardy 1952, p. 70; Berndt 1994, pp. 48-49 and 88-89).


Castellanos (1988) gives a slew of curious formulas:

$\displaystyle \pi$ $\textstyle \approx$ $\displaystyle (2e^3+e^8)^{1/7}$ (145)
  $\textstyle \approx$ $\displaystyle \left({553\over 311+1}\right)^2$ (146)
  $\textstyle \approx$ $\displaystyle ({\textstyle{3\over 14}})^4({\textstyle{193\over 5}})^2$ (147)
  $\textstyle \approx$ $\displaystyle ({\textstyle{296\over 167}})^2$ (148)
  $\textstyle \approx$ $\displaystyle \left({66^3+86^2\over 55^3}\right)^2$ (149)
  $\textstyle \approx$ $\displaystyle 1.09999901\cdot 1.19999911\cdot 1.39999931\cdot 1.69999961$  
      (150)
  $\textstyle \approx$ $\displaystyle {47^3+20^3\over 30^3}-1$ (151)
  $\textstyle \approx$ $\displaystyle 2+\sqrt{1+({\textstyle{413\over 750}})^2}$ (152)
  $\textstyle \approx$ $\displaystyle ({\textstyle{77729\over 254}})^{1/5}$ (153)
  $\textstyle \approx$ $\displaystyle \left({31+{62^2+14\over 28^4}}\right)^{1/3}$ (154)
  $\textstyle \approx$ $\displaystyle {1700^3+82^3-10^3-9^3-6^3-3^3\over 69^5}$ (155)
  $\textstyle \approx$ $\displaystyle \left({95+{93^4+34^4+17^4+88\over 75^4}}\right)^{1/4}$ (156)
  $\textstyle \approx$ $\displaystyle \left({100-{2125^3+214^3+30^3+37^2\over 82^5}}\right)^{1/4},$ (157)

which are accurate to 3, 4, 4, 5, 6, 7, 7, 8, 9, 10, 11, 12, and 13 digits, respectively. An extremely accurate approximation due to Shanks (1982) is
\begin{displaymath}
\pi\approx {6\over\sqrt{3502}}\ln(2u)+7.37\times 10^{-82},
\end{displaymath} (158)

where $u$ is the product of four simple quartic units. A sequence of approximations due to Plouffe includes
$\displaystyle \pi$ $\textstyle \approx$ $\displaystyle 43^{7/23}$ (159)
  $\textstyle \approx$ $\displaystyle {\ln 2198\over\sqrt{6}}$ (160)
  $\textstyle \approx$ $\displaystyle ({\textstyle{13\over 4}})^{1181/1216}$ (161)
  $\textstyle \approx$ $\displaystyle {689\over 396\ln\left({689\over 396}\right)}$ (162)
  $\textstyle \approx$ $\displaystyle ({\textstyle{2143\over 22}})^{1/4}$ (163)
  $\textstyle \approx$ $\displaystyle \sqrt{9\over 67}\ln 5280$ (164)
  $\textstyle \approx$ $\displaystyle ({\textstyle{63023\over 30510}})^{1/3}+{\textstyle{1\over 4}}+{\textstyle{1\over 2}}(\sqrt{5}+1)$ (165)
  $\textstyle \approx$ $\displaystyle {\textstyle{48\over 23}} \ln\left({60318\over 13387}\right)$ (166)
  $\textstyle \approx$ $\displaystyle (228+{\textstyle{16\over 1329}})^{1/41}+2$ (167)
  $\textstyle \approx$ $\displaystyle {\textstyle{125\over 123}}\ln\left({28102\over 1277}\right)$ (168)
  $\textstyle \approx$ $\displaystyle {\textstyle{276694819753963\over 226588}}^{1/158}+2$ (169)
  $\textstyle \approx$ $\displaystyle {\ln 262537412640768744\over\sqrt{163}},$ (170)

which are accurate to 4, 5, 7, 7, 8, 9, 10, 11, 11, 11, 23, and 30 digits, respectively.


Ramanujan (1913-14) and Olds (1963) give geometric constructions for 355/113. Gardner (1966, pp. 92-93) gives a geometric construction for $3+16/113=3.1415929\ldots$. Dixon (1991) gives constructions for $6/5(1+\phi)=3.141640\ldots$ and $\sqrt{4+[3-\tan(30^\circ)]^2}=3.141533\ldots$. Constructions for approximations of $\pi$ are approximations to Circle Squaring (which is itself impossible).


A short mnemonic for remembering the first eight Decimal Digits of $\pi$ is ``May I have a large container of coffee?'' giving 3.1415926 (Gardner 1959; Gardner 1966, p. 92; Eves 1990, p. 122, Davis 1993, p. 9). A more substantial mnemonic giving 15 digits (3.14159265358979) is ``How I want a drink, alcoholic of course, after the heavy lectures involving quantum mechanics,'' originally due to Sir James Jeans (Gardner 1966, p. 92; Castellanos 1988, p. 152; Eves 1990, p. 122; Davis 1993, p. 9; Blatner 1997, p. 112). A slight extension of this adds the phrase ``All of thy geometry, Herr Planck, is fairly hard,'' giving 24 digits in all (3.14159265358979323846264).


An even more extensive rhyming mnemonic giving 31 digits is ``Now I will a rhyme construct, By chosen words the young instruct. Cunningly devised endeavour, Con it and remember ever. Widths in circle here you see, Sketched out in strange obscurity.'' (Note that the British spelling of ``endeavour'' is required here.)


The following stanzas are the first part of a poem written by M. Keith based on Edgar Allen Poe's ``The Raven.'' The entire poem gives 740 digits; the fragment below gives only the first 80 (Blatner 1997, p. 113). Words with ten letters represent the digit 0, and those with 11 or more digits are taken to represent two digits.


Poe, E.: Near a Raven.


Midnights so dreary, tired and weary.

Silently pondering volumes extolling all by-now obsolete lore.

During my rather long nap-the weirdest tap!

An ominous vibrating sound disturbing my chamber's antedoor.

`This,' I whispered quietly, `I ignore.'

Perfectly, the intellect remembers: the ghostly fires, a glittering ember.

Inflamed by lightning's outbursts, windows cast penumbras upon this floor.

Sorrowful, as one mistreated, unhappy thoughts I heeded:

That inimitable lesson in elegance--Lenore--

Is delighting, exciting... nevermore.


An extensive collection of $\pi$ mnemonics in many languages is maintained by A. P. Hatzipolakis. Other mnemonics in various languages are given by Castellanos (1988) and Blatner (1997, pp. 112-118).


In the following, the word ``digit'' refers to decimal digit after the decimal point. J. H. Conway has shown that there is a sequence of fewer than 40 Fractions $F_1$, $F_2$, ... with the property that if you start with $2^n$ and repeatedly multiply by the first of the $F_i$ that gives an integral answer, then the next Power of 2 to occur will be the $2^n$th decimal digit of $\pi$.


The first occurrence of $n$ 0s appear at digits 32, 307, 601, 13390, 17534, .... The sequence 9999998 occurs at decimal 762 (which is sometimes called the Feynman Point). This is the largest value of any seven digits in the first million decimals. The first time the Beast Number 666 appears is decimal 2440. The digits 314159 appear at least six times in the first 10 million decimal places of $\pi$ (Pickover 1995). In the following, ``digit'' means digit of $\pi-3$. The sequence 0123456789 occurs beginning at digits 17,387,594,880, 26,852,899,245, 30,243,957,439, 34,549,153,953, 41,952,536,161, and 43,289,964,000. The sequence 9876543210 occurs beginning at digits 21,981,157,633, 29,832,636,867, 39,232,573,648, 42,140,457,481, and 43,065,796,214. The sequence 27182818284 (the digits of e) occur beginning at digit 45,111,908,393. There are also interesting patterns for $1/\pi$. 0123456789 occurs at 6,214,876,462, 9876543210 occurs at 15,603,388,145 and 51,507,034,812, and 999999999999 occurs at 12,479,021,132 of $1/\pi$.


Scanning the decimal expansion of $\pi$ until all $n$-digit numbers have occurred, the last 1-, 2-, ... digit numbers appearing are 0, 68, 483, 6716, 33394, 569540, ... (Sloane's A032510). These end at digits 32, 606, 8555, 99849, 1369564, 14118312, ... (Sloane's A036903).

See also Almost Integer, Archimedes Algorithm, Brent-Salamin Formula, Buffon-Laplace Needle Problem, Buffon's Needle Problem, Circle, Dirichlet Beta Function, Dirichlet Eta Function, Dirichlet Lambda Function, e, Euler-Mascheroni Constant, Gaussian Distribution, Maclaurin Series, Machin's Formula, Machin-Like Formulas, Relatively Prime, Riemann Zeta Function, Sphere, Trigonometry


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Almkvist, G. ``Many Correct Digits of $\pi$, Revisited.'' Amer. Math. Monthly 104, 351-353, 1997.

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Assmus, E. F. ``Pi.'' Amer. Math. Monthly 92, 213-214, 1985.

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Bailey, D. H. ``The Computation of $\pi$ to 29,360,000 Decimal Digit using Borwein's' Quartically Convergent Algorithm.'' Math. Comput. 50, 283-296, 1988b.

Bailey, D.; Borwein, P.; and Plouffe, S. ``On the Rapid Computation of Various Polylogarithmic Constants.'' http://www.cecm.sfu.ca/~pborwein/PAPERS/P123.ps.

Ball, W. W. R. and Coxeter, H. S. M. Mathematical Recreations and Essays, 13th ed. New York: Dover, p. 55 and 274, 1987.

Beckmann, P. A History of Pi, 3rd ed. New York: Dorset Press, 1989.

Beeler, M.; Gosper, R. W.; and Schroeppel, R. HAKMEM. Cambridge, MA: MIT Artificial Intelligence Laboratory, Memo AIM-239, Feb. 1972.

Berggren, L.; Borwein, J.; and Borwein, P. Pi: A Source Book. New York: Springer-Verlag, 1997.

Bellard, F. ``Fabrice Bellard's Pi Page.'' http://www-stud.enst.fr/~bellard/pi/.

Berndt, B. C. Ramanujan's Notebooks, Part IV. New York: Springer-Verlag, 1994.

Blatner, D. The Joy of Pi. New York: Walker, 1997.

Blatner, D. ``The Joy of Pi.'' http://www.joyofpi.com.

Borwein, P. B. ``Pi and Other Constants.'' http://www.cecm.sfu.ca/~pborwein/PISTUFF/Apistuff.html.

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