Τρίτη, 4 Απριλίου 2017

Pi Formulas -- from Wolfram MathWorld - Προσεγγίζοντας το π

Pi Formulas -- from Wolfram MathWorld



Pi Formulas

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There are many formulas of pi of many types. Among others, these include
series, products, geometric constructions, limits, special values, and pi
iterations
.


pi is intimately related to the properties of circles
and spheres. For a circle of radius r, the circumference
and area are given by


C=2pir
(1)
A=pir^2.
(2)
Similarly, for a sphere of radius r, the surface area
and volume enclosed are


S=4pir^2
(3)
V=4/3pir^3.
(4)
An exact formula for pi in terms of the inverse
tangents
of unit fractions is Machin's
formula



 1/4pi=4tan^(-1)(1/5)-tan^(-1)(1/(239)).
(5)
There are three other Machin-like formulas,
as well as thousands of other similar formulas having more terms.


GregorySeries

Gregory and Leibniz found


pi/4=sum_(k=1)^(infty)((-1)^(k+1))/(2k-1)
(6)
=1-1/3+1/5-...
(7)
(Wells 1986, p. 50), which is known as the Gregory series and may be obtained by plugging x=1 into the Leibniz series for tan^(-1)x. The error
after the nth term of this series in the Gregory
series
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


 pi=sum_(k=1)^infty(3^k-1)/(4^k)zeta(k+1),
(8)
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.


An infinite sum series to Abraham Sharp (ca. 1717) is given by


 pi=sum_(k=0)^infty(2(-1)^k3^(1/2-k))/(2k+1)
(9)
(Smith 1953, p. 311). Additional simple series in which pi appears are


1/4pisqrt(2)=sum_(k=1)^(infty)[((-1)^(k+1))/(4k-1)+((-1)^(k+1))/(4k-3)]
(10)
=1+1/3-1/5-1/7+1/9+1/(11)-...
(11)
1/4(pi-3)=sum_(k=1)^(infty)((-1)^(k+1))/(2k(2k+1)(2k+2))
(12)
=1/(2·3·4)-1/(4·5·6)+1/(6·7·8)-...
(13)
1/6pi^2=sum_(k=1)^(infty)1/(k^2)
(14)
=1+1/4+1/9+1/(16)+1/(25)+...
(15)
1/8pi^2=sum_(k=1)^(infty)1/((2k-1)^2)
(16)
=1+1/(3^2)+1/(5^2)+1/(7^2)+...
(17)
(Wells 1986, p. 53).


In 1666, Newton used a geometric construction to derive the formula


pi=3/4sqrt(3)+24int_0^(1/4)sqrt(x-x^2)dx
(18)
=(3sqrt(3))/4+24(1/(12)-1/(5·2^5)-1/(28·2^7)-1/(72·2^9)-...),
(19)
which he used to compute pi (Wells 1986, p. 50; Borwein et
al.
1989; Borwein and Bailey 2003, pp. 105-106). The coefficients can be
found from the integral


I(x)=intsqrt(x-x^2)dx
(20)
=1/4(2x-1)sqrt(x-x^2)-1/8sin^(-1)(1-2x)
(21)
by taking the series expansion of I(x)-I(0) about
0, obtaining


 I(x)=2/3x^(3/2)-1/5x^(5/2)-1/(28)x^(7/2)-1/(72)x^(9/2)-5/(704)x^(11/2)+...
(22)
(OEIS A054387 and A054388). Using Euler's convergence improvement
transformation gives


pi/2=1/2sum_(n=0)^(infty)((n!)^22^(n+1))/((2n+1)!)=sum_(n=0)^(infty)(n!)/((2n+1)!!)
(23)
=1+1/3+(1·2)/(3·5)+(1·2·3)/(3·5·7)+...
(24)
=1+1/3(1+2/5(1+3/7(1+4/9(1+...))))
(25)
(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),


 (sin^(-1)x)/(sqrt(1-x^2))=sum_(i=0)^infty((2x)^(2i+1)i!^2)/(2(2i+1)!)=_2F_1(1,1;3/2;x^2)x.
(26)
Despite the convergence improvement, series (◇) converges at only one bit/term. At the cost of a square root, Gosper has noted that
x=1/2 gives 2 bits/term,


 1/9sqrt(3)pi=1/2sum_(i=0)^infty((i!)^2)/((2i+1)!),
(27)
and x=sin(pi/10) gives almost 3.39 bits/term,


 pi/(5sqrt(phi+2))=1/2sum_(i=0)^infty((i!)^2)/(phi^(2i+1)(2i+1)!),
(28)
where phi is the golden
ratio
. Gosper also obtained


 pi=3+1/(60)(8+(2·3)/(7·8·3)(13+(3·5)/(10·11·3)(18+(4·7)/(13·14·3)(23+...)))).
(29)
A spigot algorithm for pi is given by Rabinowitz
and Wagon (1995; Borwein and Bailey 2003, pp. 141-142).


More amazingly still, a closed form expression giving a digit-extraction algorithm which produces digits of pi (or pi^2) in base-16
was discovered by Bailey et al. (Bailey et al. 1997, Adamchik and Wagon
1997),


 pi=sum_(n=0)^infty(4/(8n+1)-2/(8n+4)-1/(8n+5)-1/(8n+6))(1/(16))^n.
(30)
This formula, known as the BBP formula, was discovered using the PSLQ algorithm (Ferguson et al. 1999)
and is equivalent to


 pi=int_0^1(16y-16)/(y^4-2y^3+4y-4)dy.
(31)
There is a series of BBP-type formulas for pi in powers of (-1)^k, the first
few independent formulas of which are


pi=4sum_(k=0)^(infty)((-1)^k)/(2k+1)
(32)
=3sum_(k=0)^(infty)(-1)^k[1/(6k+1)+1/(6k+5)]
(33)
=4sum_(k=0)^(infty)(-1)^k[1/(10k+1)-1/(10k+3)+1/(10k+5)-1/(10k+7)+1/(10k+9)]
(34)
=sum_(k=0)^(infty)(-1)^k[3/(14k+1)-3/(14k+3)+3/(14k+5)+4/(14k+7)+4/(14k+9)-4/(14k+11)+4/(14k+13)]
(35)
=sum_(k=0)^(infty)(-1)^k[2/(18k+1)+3/(18k+3)+2/(18k+5)-2/(18k+7)-2/(18k+11)+2/(18k+13)+3/(18k+15)+2/(18k+17)]
(36)
=sum_(k=0)^(infty)(-1)^k[3/(22k+1)-3/(22k+3)+3/(22k+5)-3/(22k+7)+3/(22k+9)+8/(22k+11)+3/(22k+13)-3/(22k+15)+3/(22k+17)-3/(22k+19)+1/(22k+21)].
(37)
Similarly, there are a series of BBP-type formulas for pi in powers of 2^k, the first few
independent formulas of which are


pi=sum_(k=0)^(infty)1/(16^k)[4/(8k+1)-2/(8k+4)-1/(8k+5)-1/(8k+6)]
(38)
=1/2sum_(k=0)^(infty)1/(16^k)[8/(8k+2)+4/(8k+3)+4/(8k+4)-1/(8k+7)]
(39)
=1/(16)sum_(k=0)^(infty)1/(256^k)[(64)/(16k+1)-(32)/(16k+4)-(16)/(16k+5)-(16)/(16k+6)+4/(16k+9)-2/(16k+12)-1/(16k+13)-1/(16k+14)]
(40)
=1/(32)sum_(k=0)^(infty)1/(256^k)[(128)/(16k+2)+(64)/(16k+3)+(64)/(16k+4)-(16)/(16k+7)+8/(16k+10)+4/(16k+11)+4/(16k+12)-1/(16k+15)]
(41)
=1/(32)sum_(k=0)^(infty)1/(4096^k)[(256)/(24k+2)+(192)/(24k+3)-(256)/(24k+4)-(96)/(24k+6)-(96)/(24k+8)+(16)/(24k+10)-4/(24k+12)-3/(24k+15)-6/(24k+16)-2/(24k+18)-1/(24k+20)]
(42)
=1/(64)sum_(k=0)^(infty)1/(4096^k)[(256)/(24k+1)+(256)/(24k+2)-(384)/(24k+3)-(256)/(24k+4)-(64)/(24k+5)+(96)/(24k+8)+(64)/(24k+9)+(16)/(24k+10)+8/(24k+12)-4/(24k+13)+6/(24k+15)+6/(24k+16)+1/(24k+17)+1/(24k+18)-1/(24k+20)-1/(24k+21)]
(43)
=1/(96)sum_(k=0)^(infty)1/(4096^k)[(256)/(24k+2)+(64)/(24k+3)+(128)/(24k+5)+(352)/(24k+6)+(64)/(24k+7)+(288)/(24k+8)+(128)/(24k+9)+(80)/(24k+10)+(20)/(24k+12)-(16)/(24k+14)-1/(24k+15)+6/(24k+16)-2/(24k+17)-1/(24k+19)+1/(24k+20)-2/(24k+21)]
(44)
=1/(96)sum_(k=0)^(infty)1/(4096^k)[(256)/(24k+1)+(320)/(24k+3)+(256)/(24k+4)-(192)/(24k+5)-(224)/(24k+6)-(64)/(24k+7)-(192)/(24k+8)-(64)/(24k+9)-(64)/(24k+10)-(28)/(24k+12)-4/(24k+13)-5/(24k+15)+3/(24k+17)+1/(24k+18)+1/(24k+19)+1/(24k+21)-1/(24k+22)]
(45)
=1/(96)sum_(k=0)^(infty)1/(4096^k)[(512)/(24k+1)-(256)/(24k+2)+(64)/(24k+3)-(512)/(24k+4)-(32)/(24k+6)+(64)/(24k+7)+(96)/(24k+8)+(64)/(24k+9)+(48)/(24k+10)-(12)/(24k+12)-8/(24k+13)-(16)/(24k+14)-1/(24k+15)-6/(24k+16)-2/(24k+18)-1/(24k+19)-1/(24k+20)-1/(24k+21)]
(46)
=1/(4096)sum_(k=0)^(infty)1/(65536^k)[(16384)/(32k+1)-(8192)/(32k+4)-(4096)/(32k+5)-(4096)/(32k+6)+(1024)/(32k+9)-(512)/(32k+12)-(256)/(32k+13)-(256)/(32k+14)+(64)/(32k+17)-(32)/(32k+20)-(16)/(32k+21)-(16)/(32k+22)+4/(32k+25)-2/(32k+28)-1/(32k+29)-1/(32k+30)]
(47)
=1/(4096)sum_(k=0)^(infty)1/(65536^k)[(32768)/(32k+2)+(16384)/(32k+3)+(16384)/(32k+4)-(4096)/(32k+7)+(2048)/(32k+10)+(1024)/(32k+11)+(1024)/(32k+12)-(256)/(32k+15)+(128)/(32k+18)+(64)/(32k+19)+(64)/(32k+20)-(16)/(32k+23)+8/(32k+26)+4/(32k+27)+4/(32k+28)-1/(32k+31)].
(48)
F. Bellard found the rapidly converging BBP-type
formula



 pi=1/(2^6)sum_(n=0)^infty((-1)^n)/(2^(10n))(-(2^5)/(4n+1)-1/(4n+3)+(2^8)/(10n+1)-(2^6)/(10n+3)-(2^2)/(10n+5)-(2^2)/(10n+7)+1/(10n+9)).
(49)
A related integral is


 pi=(22)/7-int_0^1(x^4(1-x)^4)/(1+x^2)dx
(50)
(Dalzell 1944, 1971; Le Lionnais 1983, p. 22; Borwein, Bailey, and
Girgensohn 2004, p. 3; Boros and Moll 2004, p. 125; Lucas 2005; Borwein et al.
2007, p. 14). This integral was known by K. Mahler in the mid-1960s
and appears in an exam at the University of Sydney in November 1960 (Borwein, Bailey,
and Girgensohn, p. 3). Beukers (2000) and Boros and Moll (2004, p. 126)
state that it is not clear if these exists a natural choice of rational polynomial
whose integral between 0 and 1 produces pi-333/106, where
333/106 is the next convergent. However, an integral exists for the fourth
convergent, namely


 pi=(355)/(113)-1/(3164)int_0^1(x^8(1-x)^8(25+816x^2))/(1+x^2)dx.
(51)
(Lucas 2005; Bailey et al. 2007, p. 219). In fact, Lucas (2005) gives
a few other such integrals.


Backhouse (1995) used the identity


I_(m,n)=int_0^1(x^m(1-x)^n)/(1+x^2)dx
(52)
=2^(-(m+n+1))sqrt(pi)Gamma(m+1)Gamma(n+1)×_3F_2(1,(m+1)/2,(m+2)/2;(m+n+2)/2,(m+n+3)/2;-1)
(53)
=a+bpi+cln2
(54)
for positive integer m and n and where a, b, and c are rational constant
to generate a number of formulas for pi. In particular,
if 2m-n=0 (mod 4), then c=0 (Lucas 2005).


A similar formula was subsequently discovered by Ferguson, leading to a
two-dimensional lattice of such formulas which can be generated by these
two formulas given by


 pi=sum_(k=0)^infty((4+8r)/(8k+1)-(8r)/(8k+2)-(4r)/(8k+3)-(2+8r)/(8k+4)-(1+2r)/(8k+5)-(1+2r)/(8k+6)+r/(8k+7))(1/(16))^k
(55)
for any complex value of r (Adamchik and Wagon), giving the BBP
formula
as the special case r=0.


PiFormulasWagonIdentity

An even more general identity due to Wagon is given by


 pi+4tan^(-1)z+2ln((1-2z-z^2)/(z^2+1))=sum_(k=0)^infty1/(16^k)[(4(z+1)^(8k+1))/(8k+1)-(2(z+1)^(8k+4))/(8k+4)-((z+1)^(8k+5))/(8k+5)-((z+1)^(8k+6))/(8k+6)]
(56)
(Borwein and Bailey 2003, p. 141), which holds over a region of the complex plane excluding two triangular portions symmetrically placed about the real
axis
, as illustrated above.


A perhaps even stranger general class of identities is given by


 pi=4sum_(j=1)^n((-1)^(j+1))/(2j-1)+((-1)^n(2n-1)!)/4sum_(k=0)^infty1/(16^k)[8/((8k+1)_(2n))-4/((8k+3)_(2n))-4/((8k+4)_(2n))-2/((8k+5)_(2n))+1/((8k+7)_(2n))+1/((8k+8)_(2n))]
(57)
which holds for any positive integer n, where (x)_n is a Pochhammer
symbol
(B. Cloitre, pers. comm., Jan. 23, 2005). Even more amazingly,
there is a closely analogous formula for the natural
logarithm of 2
.


Following the discovery of the base-16 digit BBP formula and related formulas, similar formulas in other bases were investigated. Borwein,
Bailey, and Girgensohn (2004) have recently shown that pi has no Machin-type
BBP arctangent formula that is not binary, although this does not rule out a completely
different scheme for digit-extraction algorithms
in other bases.


S. Plouffe has devised an algorithm to compute the nth digit
of pi in any base in O(n^3(logn)^3) steps.


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


 sum_(k=0)^infty((-1)^k(4k+1)[(2k-1)!!]^3)/([(2k)!!]^3)=sum_(k=0)^infty((-1)^k(4k+1)[Gamma(k+1/2)]^3)/(pi^(3/2)[Gamma(k+1)]^3)=2/pi
(58)
(Hardy 1923, 1924, 1999, p. 7).


Plouffe (2006) found the beautiful formula


 pi=72sum_(n=1)^infty1/(n(e^(npi)-1))-96sum_(n=1)^infty1/(n(e^(2npi)-1)) 
 +24sum_(n=1)^infty1/(n(e^(4npi)-1)).
(59)
PiBlatnerProduct

An interesting infinite product formula due to Euler which relates pi and the nth prime p_n is


pi=2/(product_(n=1)^(infty)[1+(sin(1/2pip_n))/(p_n)])
(60)
=2/(product_(n=2)^(infty)[1+((-1)^((p_n-1)/2))/(p_n)])
(61)
(Blatner 1997, p. 119), plotted above as a function of the number of terms in the product.


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,


 beta=1/4(n-3)pi,
(62)
then


 pi=(2sin(2beta))/((n-3)product_(k=0)^(infty)cos(2^(-k)beta))
(63)
(Beckmann 1989, pp. 92-94).


Vieta (1593) was the first to give an exact expression for pi by taking n=4 in the above expression, giving


 cosbeta=sinbeta=1/(sqrt(2))=1/2sqrt(2),
(64)
which leads to an infinite product of nested
radicals
,


 2/pi=sqrt(1/2)sqrt(1/2+1/2sqrt(1/2))sqrt(1/2+1/2sqrt(1/2+1/2sqrt(1/2)))...
(65)
(Wells 1986, p. 50; Beckmann 1989, p. 95). However, this expression was not rigorously proved to converge until Rudio in 1892.


A related formula is given by


 pi=lim_(n->infty)2^(n+1)sqrt(2-sqrt(2+sqrt(2+sqrt(2+...+sqrt(2))))_()_(n)),
(66)
which can be written


 pi=lim_(n->infty)2^(n+1)pi_n,
(67)
where pi_n is defined using the iteration


 pi_n=sqrt((1/2pi_(n-1))^2+[1-sqrt(1-(1/2pi_(n-1))^2)]^2)
(68)
with pi_0=sqrt(2) (J. Munkhammar, pers.
comm., April 27, 2000). The formula


 pi=2lim_(m->infty)sum_(n=1)^msqrt([sqrt(1-((n-1)/m)^2)-sqrt(1-(n/m)^2)]^2+1/(m^2))
(69)
is also closely related.


A pretty formula for pi is given by


 pi=(product_(n=1)^(infty)(1+1/(4n^2-1)))/(sum_(n=1)^(infty)1/(4n^2-1)),
(70)
where the numerator is a form of the Wallis formula for pi/2 and the denominator is a telescoping
sum
with sum 1/2 since


 1/(4n^2-1)=1/2(1/(2n-1)-1/(2n+1))
(71)
(Sondow 1997).


A particular case of the Wallis formula gives


 pi/2=product_(n=1)^infty[((2n)^2)/((2n-1)(2n+1))]=(2·2)/(1·3)(4·4)/(3·5)(6·6)/(5·7)...
(72)
(Wells 1986, p. 50). This formula can also be written


 lim_(n->infty)(2^(4n))/(n(2n; n)^2)=pilim_(n->infty)(n[Gamma(n)]^2)/([Gamma(1/2+n)]^2)=pi,
(73)
where (n; k) denotes a binomial
coefficient
and Gamma(x) is the gamma
function
(Knopp 1990). Euler obtained


 pi=sqrt(6(1+1/(2^2)+1/(3^2)+1/(4^2)+...)),
(74)
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.


An infinite sum due to Ramanujan is


 1/pi=sum_(n=0)^infty(2n; n)^3(42n+5)/(2^(12n+4))
(75)
(Borwein et al. 1989; Borwein and Bailey 2003, p. 109; Bailey et al.
2007, p. 44). Further sums are given in Ramanujan (1913-14),


 4/pi=sum_(n=0)^infty((-1)^n(1123+21460n)(2n-1)!!(4n-1)!!)/(882^(2n+1)32^n(n!)^3)
(76)
and


1/pi=sqrt(8)sum_(n=0)^(infty)((1103+26390n)(2n-1)!!(4n-1)!!)/(99^(4n+2)32^n(n!)^3)
(77)
=(sqrt(8))/(9801)sum_(n=0)^(infty)((4n)!(1103+26390n))/((n!)^4396^(4n))
(78)
(Beeler et al. 1972, Item 139; Borwein et al. 1989; Borwein and Bailey 2003, p. 108; Bailey et al. 2007, p. 44). Equation (78)
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


 pi approx (9801)/(2206sqrt(2))=3.14159273001...
(79)
(Wells 1986, p. 54) 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


 sum_(n=0)^infty[a(t)+nb(t)]((6n)!)/((3n)!(n!)^3)1/([j(t)]^n)=(sqrt(-j(t)))/pi,
(80)
where t is a binary
quadratic form discriminant
, j(t) is the j-function,


b(t)=sqrt(t[1728-j(t)])
(81)
a(t)=(b(t))/6{1-(E_4(t))/(E_6(t))[E_2(t)-6/(pisqrt(t))]},
(82)
and the E_i are Eisenstein
series
. A class number p field involves
pth degree algebraic
integers
of the constants A=a(t), B=b(t), and C=c(t). Of all series consisting of only integer
terms, the one gives the most numeric digits in the shortest period of time 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^(pisqrt(163))
(the Ramanujan constant) is very nearly an
integer. Similarly, the factor 640320^3 comes from
the j-function identity for j(1/2(1+isqrt(163))).
The series is given by


1/pi=12sum_(n=0)^(infty)((-1)^n(6n)!(13591409+545140134n))/((n!)^3(3n)!(640320^3)^(n+1/2))
(83)
=(163·8·27·7·11·19·127)/(640320^(3/2))sum_(n=0)^(infty)((13591409)/(163·2·9·7·11·19·127)+n)((6n)!)/((3n)!(n!)^3)((-1)^n)/(640320^(3n))
(84)
(Borwein and Borwein 1993; Beck and Trott; Bailey et al. 2007, p. 44). 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 the Wolfram
Language
to calculate pi (Vardi 1991; Wolfram Research),


 pi=(426880sqrt(10005))/(A[_3F_2(1/6,1/2,5/6;1,1;B)-C_3F_2(7/6,3/2,(11)/6;2,2;B)]),
(85)
where


A=13591409
(86)
B=-1/(151931373056000)
(87)
C=(30285563)/(1651969144908540723200).
(88)
The best formula for class number 2 (largest discriminant -427) is


 1/pi=12sum_(n=0)^infty((-1)^n(6n)!(A+Bn))/((n!)^3(3n)!C^(n+1/2)),
(89)
where


A=212175710912sqrt(61)+1657145277365
(90)
B=13773980892672sqrt(61)+107578229802750
(91)
C=[5280(236674+30303sqrt(61))]^3
(92)
(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


 (sqrt(-C^3))/pi=sum_(n=0)^infty((6n)!)/((3n)!(n!)^3)(A+nB)/(C^(3n)),
(93)
where


A=63365028312971999585426220+28337702140800842046825600sqrt(5)+384sqrt(5)(10891728551171178200467436212395209160385656017+4870929086578810225077338534541688721351255040sqrt(5))^(1/2)
(94)
B=7849910453496627210289749000+3510586678260932028965606400sqrt(5)+2515968sqrt(3110)(6260208323789001636993322654444020882161+2799650273060444296577206890718825190235sqrt(5))^(1/2)
(95)
C=-214772995063512240-96049403338648032sqrt(5)-1296sqrt(5)(10985234579463550323713318473+4912746253692362754607395912sqrt(5))^(1/2).
(96)
This gives 50 digits per term. Borwein and Borwein (1993) have developed a general algorithm for generating such series for arbitrary
class number.


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),


4/pi=sum_(n=0)^(infty)((6n+1)(1/2)_n^3)/(4^n(n!)^3)
(97)
(16)/pi=sum_(n=0)^(infty)((42n+5)(1/2)_n^3)/(64^n(n!)^3)
(98)
(32)/pi=sum_(n=0)^(infty)((42sqrt(5)n+5sqrt(5)+30n-1)(1/2)_n^3)/(64^n(n!)^3)((sqrt(5)-1)/2)^(8n)
(99)
(27)/(4pi)=sum_(n=0)^(infty)((15n+2)(1/2)_n(1/3)_n(2/3)_n)/((n!)^3)(2/(27))^n
(100)
(15sqrt(3))/(2pi)=sum_(n=0)^(infty)((33n+4)(1/2)_n(1/3)_n(2/3)_n)/((n!)^3)(4/(125))^n
(101)
(5sqrt(5))/(2pisqrt(3))=sum_(n=0)^(infty)((11n+1)(1/2)_n(1/6)_n(5/6)_n)/((n!)^3)(4/(125))^n
(102)
(85sqrt(85))/(18pisqrt(3))=sum_(n=0)^(infty)((133n+8)(1/2)_n(1/6)_n(5/6)_n)/((n!)^3)(4/(85))^(3n)
(103)
4/pi=sum_(n=0)^(infty)((-1)^n(20n+3)(1/2)_n(1/4)_n(3/4)_n)/((n!)^32^(2n+1))
(104)
4/(pisqrt(3))=sum_(n=0)^(infty)((-1)^n(28n+3)(1/2)_n(1/4)_n(3/4)_n)/((n!)^33^n4^(2n+1))
(105)
4/pi=sum_(n=0)^(infty)((-1)^n(260n+23)(1/2)_n(1/4)_n(3/4)_n)/((n!)^3(18)^(2n+1))
(106)
4/(pisqrt(5))=sum_(n=0)^(infty)((-1)^n(644n+41)(1/2)_n(1/4)_n(3/4)_n)/((n!)^35^n(72)^(2n+1))
(107)
4/pi=sum_(n=0)^(infty)((-1)^n(21460n+1123)(1/2)_n(1/4)_n(3/4)_n)/((n!)^3(882)^(2n+1))
(108)
(2sqrt(3))/pi=sum_(n=0)^(infty)((8n+1)(1/2)_n(1/4)_n(3/4)_n)/((n!)^39^n)
(109)
1/(2pisqrt(2))=sum_(n=0)^(infty)((10n+1)(1/2)_n(1/4)_n(3/4)_n)/((n!)^39^(2n+1))
(110)
1/(3pisqrt(3))=sum_(n=0)^(infty)((40n+3)(1/2)_n(1/4)_n(3/4)_n)/((n!)^3(49)^(2n+1))
(111)
2/(pisqrt(11))=sum_(n=0)^(infty)((280n+19)(1/2)_n(1/4)_n(3/4)_n)/((n!)^3(99)^(2n+1))
(112)
1/(2pisqrt(2))=sum_(n=0)^(infty)((26390n+1103)(1/2)_n(1/4)_n(3/4)_n)/((n!)^3(99)^(4n+2)).
(113)
These equations were first proved by Borwein and Borwein (1987a,
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 (Bailey et al. 2007, pp. 44-45).


A complete list of independent known equations of this type is given by


4/pi=sum_(n=0)^(infty)((6n+1)(1/2)_n^3)/(4^n(n!)^3)
(114)
(16)/pi=sum_(n=0)^(infty)((42n+5)(1/2)_n^3)/(64^n(n!)^3)
(115)
(32)/pi=sum_(n=0)^(infty)((42sqrt(5)n+5sqrt(5)+30n-1)(1/2)_n^3)/(64^n(n!)^3)((sqrt(5)-1)/2)^(8n)
(116)
(5^(1/4))/pi=sum_(n=0)^(infty)((540sqrt(5)n-1200n-525+235sqrt(5))(1/2)_n^3(sqrt(5)-2)^(8n))/((n!)^3)
(117)
(12^(1/4))/pi=sum_(n=0)^(infty)((24sqrt(3)n-36n-15+9sqrt(3))(1/2)_n^3(2-sqrt(3))^(4n))/((n!)^3)
(118)
for m=1 with nonalternating signs,


2/pi=sum_(n=0)^(infty)((-1)^n(1/2)_n^3(12sqrt(2)n-12n-5+4sqrt(2))(sqrt(2)-1)^(4n))/((n!)^3)
(119)
2/pi=sum_(n=0)^(infty)((-1)^n(1/2)_n^3(60n-24sqrt(5)n+23-10sqrt(5))(sqrt(5)-2)^(4n))/((n!)^3)
(120)
2/pi=sum_(n=0)^(infty)((-1)^n(1/2)_n^3(420n-168sqrt(6)n+177-72sqrt(6)))/((n!)^3)
(121)
(2sqrt(2))/pi=sum_(n=0)^(infty)((-1)^n(1/2)_n^3(2sqrt(2))^(2n))/((n!)^3)
(122)
for m=1 with alternating signs,


(128)/(pi^2)=sum_(n=0)^(infty)((-1)^n(1/2)_n^5(820n^2+180n+13))/(32^(2n)(n!)^5)
(123)
(32)/(pi^2)=sum_(n=0)^(infty)((-1)^n(1/2)_n^5(20n^2+8n+1))/(2^(2n)(n!)^5)
(124)
for m=2 (Guillera 2002, 2003, 2006),


 (32)/(pi^3)=sum_(n=0)^infty((1/2)_n^7(168n^3+76n^2+14n+1))/(32^(2n)(n!)^5)
(125)
for m=3 (Guillera 2002, 2003, 2006), and no others for
m>3 are known (Bailey et al. 2007, pp. 45-48).


Bellard gives the exotic formula


 pi=1/(740025)[sum_(n=1)^infty(3P(n))/((7n; 2n)2^(n-1))-20379280],
(126)
where


 P(n)=-885673181n^5+3125347237n^4-2942969225n^3+1031962795n^2-196882274n+10996648.
(127)
Gasper quotes the result


 pi=(16)/3[lim_(x->infty)x_1F_2(1/2;2,3;-x^2)]^(-1),
(128)
where _1F_2 is a generalized
hypergeometric function
, and transforms it to


 pi=lim_(x->infty)4x_1F_2(1/2;3/2,3/2;-x^2).
(129)
A fascinating result due to Gosper is given by


 lim_(n->infty)product_(i=n)^(2n)pi/(2tan^(-1)i)=4^(1/pi)=1.554682275....
(130)
pi satisfies the inequality


 (1+1/pi)^(pi+1) approx 3.14097<pi.
(131)
D. Terr (pers. comm.) noted the curious identity


 (3,1,4)=(1,5,9)+(2,6,5) (mod 10)
(132)
involving the first 9 digits of pi.


SEE ALSO: BBP Formula, Digit-Extraction Algorithm, Pi, Pi Approximations,
Pi Continued Fraction, Pi
Digits
, Pi Iterations, Pi
Squared
, Spigot Algorithm








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CITE THIS AS:


Weisstein, Eric W. "Pi Formulas." From
MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/PiFormulas.html

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