Applied Mathematics E-Notes, 7(2007), 237-246 c
Available free at mirror sites of http://www.math.nthu.edu.tw/∼amen/
ISSN 1607-2510
Construction Of Binomial Sums For π And
Polylogarithmic Constants Inspired By BBP Formulas∗
Boris Gourévitch†, Jesús Guillera Goyanes‡
Received 28 October 2006
Abstract
We present new sums involving binomial coefficients for π and various logarithms and polylogarithms constants. These sums are a generalization of BBP
formulas first introduced by D. Bailey, P. Borwein and S. Plouffe in 1995. In this
paper, we describe how to find and prove such sums using the Beta function at
integer and rational arguments.
1
Introduction
In 1995, David Bailey, Peter Borwein and Simon Plouffe discovered a new seminal
formula for π
∞
X
4
1
2
1
1
π=
−
−
−
.
(1)
16n 8n + 1 8n + 4 8n + 5 8n + 6
n=0
This amazing formula allows extracting the n-th binary digit of π without knowledge
of previous digits [1]. Such sums are called BBP (Bailey-Borwein-Plouffe) formulas.
Many new BBP formulas for π, G (Catalan’s constant) or ζ(3) (Riemann Zeta function)
were subsequently discovered by Adamchik [2], Bellard [3], Broadhurst [4], Lupas [5]
and Huvent [6]. The main tool for finding BBP formulas is an algorithm designed
to detect linear combinations on the set of natural numbers N like PSLQ (from its
use of a partial-sum-of-squares (PS) vector and lower-diagonal-orthogonal (LQ) matrix
factorization) or LLL (Lenstra–Lenstra–Lovász) [7], [8]. To then prove such P
formulas,
∞
zn
an integral is evaluated giving rise to logarithms or polylogarithms (Lip (z) = n=1 n
p,
[9]).
In October 2001, Almkvist and colleagues presented new sums for π involving binomial coefficients such as
∞
X
50n − 6
π=
(2)
3n n .
2
n
n=0
∗ Mathematics
Subject Classifications: 40C10
rue Montera, 75012 Paris, email: boris@pi314.net
‡ Av. Cesareo Alierta, 31 esc. izda 4oVA, Zaragoza
† 17
237
238
Binomial Sums for π and Polylogarithmic Constants
They proved this formula using the so-called Beta method which is based on use of
R1
Beta function B(r, s) = 0 xr−1 (1 − x)s−1dx for positive integer values r and s [10].
In this paper, using an adapted Beta method, we show how to prove some more
general sums of the form
T =
∞
X
n=0
1
qn
mn
pn
b1
b2
bk
bm−1
+
+···+
+ ···+
mn + 1 mn + 2
mn + k
mn + (m − 1)
(3)
where T is a constant often related to π and the natural logarithm ln(z). These sums
involve binomial coefficients and they are similar to BBP formulas. We therefore call
them “BBP binomial formulas”. One example of such formulas is
∞
π=
1 X 1
16807 n=0 2n 7n
2n
59296
10326
3200
1352
792
552
−
−
−
−
+
7n + 1 7n + 2 7n + 3 7n + 4 7n + 5 7n + 6
(4)
that we discovered with LLL algorithm. An explicit proof of this formula is detailed in
this paper. Then we show how to generate an infinite set of similar formulas for π and
the natural logarithm ln(z). We also present outlines for proofs of formulas showing
that Beta method is not restrained to integer parameters.
2
Proofs of BBP Binomial Formulas
Consider the Eq. 3. Proof of such formula is reduced to the following proposition:
PROPOSITION 1 (The Beta Method). There exists a polynomial P (X) ∈ Qm−2 [X]
(polynomial of degree m − 2 with coefficients in the set of rational numbers Q) such
that
Z 1
qP (x)
T =
dx
(5)
p
(m−p)
0 q − x (1 − x)
PROOF. The Beta function B(r, s) is
B(r, s) =
Γ(r)Γ(s)
Γ(r + s)
(6)
R∞
where Γ(z) = 0 tz−1e−t dt is the Gamma function. With parameters m, p, n ∈ N, we
use Γ(n) = (n − 1)! to obtain
1
mn = (mn + 1)
pn
Z
1
xpn (1 − x)(m−p)n dx.
0
(7)
B. Gourévitch and J. Guillera Goyanes
Summing Eq. (7) on n with
∞
X
n=0
1
qn
mn
pn
1
(mn+1)( mn
pn )
239
on the left side, we obtain
Z 1
∞
X
1
1
xpn(1 − x)(m−p)n dx
=
mn + 1 n=0 qn 0
n
Z 1X
∞ p
x (1 − x)(m−p)
=
dx
q
0 n=0
Z 1
q
=
dx.
p (1 − x)(m−p)
q
−
x
0
(8)
We can evaluate this integral by partial fraction decomposition. More generally, to
1
obtain the elements
in Eq. (3) we write
mn + k
Z 1
1
1 (pn + 1)(pn + 2)...(pn + k − 1)
xpn+k−1(1−x)(m−p)n dx = mn+k = mn
. (9)
(mn + 1)(mn + 2)...(mn + k)
0
pn
pn+k−1
For a given k, the partial fraction decomposition of
(ak,j )j=1,..,k such that
Z
1
xk−1xpn (1 − x)(m−p)n dx =
0
1
mn
pn
(pn+1)(pn+2)...(pn+k−1)
(mn+1)(mn+2)...(mn+k)
in n gives
ak,1
ak,2
ak,k
+
+ ... +
mn + 1 mn + 2
mn + k
(10)
with ak,j ∈ Q. For values k ranging from 1 to m − 1, we linearly combine (ak,j )j=1,..,k
in order to find the m − 1 values bk of Eq. (3). This is equivalent to solving the system




a1,1 a2,1 ... am−1,1
b1
 0


a2,2 ... am−1,2 

 V =  b2  .
(11)
 ....


0
...
...
.... 
0
0
... am−1,k
bm−1


1
 X 

Vector V contains the coefficients of the polynomial P (X) = V t . 
 ....  of degree
X m−2
m − 2, such that
∞
X
1
b1
b2
bm−1
+
+
·
·
·
+
mn + 1 mn + 2
mn + (m − 1)
qn mn
pn
n=0
Z
∞
X 1
xpn(1 − x)(m−p)n
=
P (x)
dx
qn
n=0 0
Z 1
qP (x)
=
dx
p (1 − x)(m−p)
q
−
x
0
when we use Eq. (10).
240
Binomial Sums for π and Polylogarithmic Constants
R1
1
= (3n + 1) 0 x2n(1 − x)n dx
(3n
n)
in order to prove the formula in Eq. (2). In general, they consider formulas of the form
REMARK 1. Almkvist et al. [10] use the relation
π=
∞
X
S(n)
mn n
q
n=0 pn
(12)
where S ∈ Qm−2 [X] and q ∈ N. However,
forms of sums lead to the intricate
such
n
P∞
xp (1−x)m−p
evaluation of f(x) = n=0 (mn + 1)S(n)
. f(x) is indeed an integral
q
whose denominator is a power of the one considered in Eq. (8). Our method uses
1
on the left side of Eq. (7), which produces formulas of the general form of
(mn+1)(mn
pn )
Eq. 3.
3
Proof of Eq. 4
We write Eq. (4) as
∞
X
Un = π
n=0
Using Proposition 1, we are looking for the polynomial P (x) such that
Z 1
Z 1
∞
∞
X
X
x2n(1 − x)5n
1
Un =
P (x)
dx
=
P (x)
dx
2
n
2
x (1 − x)5
0
0
n=0
n=0
1−
2
Z 1
2
=
P (x) 2
dx.
(x − 2x + 2) (x5 − 3x4 + 2x3 + x + 1)
0
R1
4
Because 0 2
dx = π, it is likely that P (x) = 2 x5 − 3x4 + 2x3 + x + 1 .
(x − 2x + 2)
Let us verify this conjecture.
To obtain the equivalent linear system in (11), and therefore vector V , we calculate
all coefficients (ak,i)i (columns of left matrix in Eq. (11)) with help from the computation of the integral in Eq. (10). For example, for k = 2 (2nd column of matrix), we
have
Z 1
(2n + 1)
1
x.x2n(1 − x)5n dx = B(5n + 1, 2n + 2) = 7n
(7n + 1)(7n + 2)
0
2n
1
5
1
3
1
= 7n
−
.
7 (7n + 1) 7 (7n + 2)
2n
Consequently, a2,1 = 57 and a2,2 = − 37 . The whole linear system is


 59296
30
190
1235
8151
1 57
49
343
2401
16 807
16807
3
30
255
2040
15
810
 0 −
 − 10326
− 49 − 343 − 2401 − 16 807 
7
16807



4
60
660
6380
 0 0

 − 3200

49
343
2401
16 807  V = 
16807
13
260
3510
 0 0

 − 1352
0


343
2401
16 807 
16807
99
2475 
 0 0
 − 792
0
0
− 2401
− 16
807
16807
552
276
0 0
0
0
0
16 807
16807




.



B. Gourévitch and J. Guillera Goyanes
Its solution is V =
1 1
0 2 −3
241
1 . Finally, we have


1
 X 
3
4
5

P (X) = V t 
 ....  = 2 1 + X + 2X − 3X + X
X5
which is the expected polynomial.
4
Construction and Prediction of BBP Binomial Formulas
qP (x)
Using proposition 1, we choose a suitable P (x) that simplifies q−x(m−p)
. Know(1−x)p
ing P then gives rise to a BBP formula by using Eq. (10), after combination and
simplification.
qP (x)
As an example, we find formulas for π choosing P (X) such that q−x(m−p)
=
(1−x)p
R
1
1
π
1
2
due to 0 1+x2 dx = 4 . To find m and p, we have to know when 1 + X divides
1+x2
q − X (m−p) (1 − X)p i.e. when i and −i are roots of q − X (m−p) (1 − X)p . Case m = p
in Eq. (3) is excluded since it only provides classical BBP formulas without binomial
coefficients.
Let us choose p = 2, m = 3, then i(1 − i)2 = i(1 + i)2 = 2. Now there exists a
formula for π with 3n
, p = 2 and q = 2. Moreover, i and −i are fourth roots of 1, so
2n
we can multiply x(1 − x)2 by x4k and keep roots i and −i. Thus there exists formulas
2n
for π involving C(3+4k)n
, k ∈ N.
More generally, if K(X) = q − X (m−p) (1 − X)p ,
n
o
{i and − i are roots of K, p < m} ⇔ q − (±i)(m−p) (1 ∓ i)p = 0, p < m
⇔ {(1 ± i)p = (−1)m+p q(±i)m , p < m}
⇔ {(1 + i)p = (−1)m+p qim , p < m} by complex conjugate
n √ p
o
⇔
2 = q and p π4 ≡ (m + p)π + m π2 [2π] , p < m with modulus and argument, where [x] designates modulo x
p
⇔ 2 2 = q, p < m and − p ≡ 2m [8]
⇔ p = 2k, k ∈ N, 2k = q, p < m and 7p ≡ 2m [8] .
R1 1
This idea also leads to other constants using the root −1 ( ln(2) = 0 1+x
dx), or
√
R
9 1
1
2
the polynomial x − x + 1 (π 3 = 2 0 1−x+x2 dx).
P∞
bm−1
b1
b2
For non-alternating sums n=0 qn 1mn mn+1
with equiv+ mn+2
+ ... + mn+(m−1)
( pn )
R1
qP (x)
alent integral representation 0 q−xp (1−x)
(m−p) dx, we obtain the following table for
242
Binomial Sums for π and Polylogarithmic Constants
p < m:
Polynomial
1 + x2
1+x
2+x
z−1+x
Constant
π
ln(2)
√
π 3
ln 32
ln z+1
z
1 + (z − 1)x
ln (z)
1 − x + x2
Polynomial
1 + x2
1+x
(−2)
(3)p = q
(m−p)
(1 − z)
(z)p = q
(m−p) p
1
1
− z−1
1 + z−1
=q
q = 1, m = 2p + 6k
m = p + 2k, q = 22k 3p
m = p + 2k, q = (1 − z)2k z p
m = p + 2k, q =
P∞
2+x
z−1+x
Constant
π
ln(2)
√
π 3
ln 32
ln z+1
z
1 + (z − 1)x
ln (z)
1 − x + x2
5
(m−p)
Existence of a sum for p < m
p = 2k, 2k = q, 7p ≡ 2m [8]
m = p + 2k, q = 2p
bm−1
(−1)m
b1
b2
with
n=0 q n (mn ) mn+1 + mn+2 + ... + mn+(m−1)
pn
R1
qP (x)
representation 0 q+xp (1−x)(m−p) dx, we get the following table:
For alternating sums
lent integral
Condition given by roots
(m−p)
(i)
(1 − i)p = q
m−p p
(−1)
2 =q
√ p+m
p 1+i 3
(−1)
=q
2
Condition given by roots
(m−p)
(i)
(1 − i)p = −q
m−p+1 p
(−1)
2 =q
√ p+m
p+1 1+i 3
(−1)
=q
2
(m−p)
(−2)
(3)p = −q
(m−p)
(1 − z)
(z)p = −q
(m−p) p
1
1
− z−1
1 + z−1
=q
zp
(z−1)p+2k
equiva-
Existence of a sum for p < m
p = 2k, 2k = q, 7p ≡ 2m + 4 [8]
m = p + 2k + 1, q = 2p
q = 1, m = 2p + 3 + 6k
m = p + 2k + 1, q = 22k 3p
m = p + 2k + 1, q = (1 − z)2k z p
m = p + 2k + 1, q =
zp
(z−1)p+2k
Generalized Beta Function Method
Albeit more intricate, some similar formulas can also be found using fractional parameters in Beta function. Considering Beta function and ab , dc ∈ Q, we have
Z
1
x
pn+ a
b
(1 − x)
mn+ dc
dx =
0
=
Γ pn + ab + 1 Γ mn + dc + 1
Γ (p + m)n + ab + dc + 2
a
c
a
c
b pn+1 d mn+1 Γ b Γ d
a
c
a
c
b + d (p+m)n+2 Γ b + d
(13)
Γ (x + n)
= x(x+1) · · ·(x+n−1) is Pochhammer symbol. Γ ab is often
Γ (x)
known for small integers values of a et b. All following formulas stem from summing
this relation on n for particular values of ab , dc . We now emphasize three particular
cases of interest.
where (x)n =
B. Gourévitch and J. Guillera Goyanes
5.1
Case 1 :
a
b
= − 12 ,
c
d
243
=0
We have
Z
1
0
Z
∞
1 X xpn (1 − x)mn
√
dx
x n=0
qn
0
∞ Z 1
X
1
=
q−n xpn− 2 (1 − x)mn dx
1
q
√
dx =
p
x q − x (1 − x)m
1
=
n=0 0
∞
X
−n Γ
=
n=0
∞
X
q
pn + 12 Γ (mn + 1)
Γ (p + m)n + 12 + 1
(−1)n
n=0
(2pn)!(mn + pn)!(mn)!
.
qn (2mn + 2pn)!(pn)!
(14)
EXAMPLE 1. Case p = 3, m = 1. We have
Z 1
Z 1
1
1
2
3
2
√
√ x − 2x + 2x − 2
−π = −2
dx =
dx
x(x + 1)
x
2 − x3(1 − x)
0
0
Z 1
∞
X
1
(−1)n x3n(1 − x)n
√ x3 − 2x2 + 2x − 2
=
dx
x
2n
0
n=0
Z
∞
X
x3n
(−1)n 1 3
=
x − 2x2 + 2x − 2 √ (1 − x)ndx
(15)
n
2
x
0
n=0
So now using Eq. (14), we have
Z 1
x3n
−
x3 − 2x2 + 2x − 2 √ (1 − x)ndx
x
0
Z 1
5 3 1
1 T
= −[−1, 2, −2, 2] ×
x3n+[ 2 , 2 , 2 ,− 2 ] (1 − x)ndx
0
T Γ 3n + 72 , 52 , 32 , 12
Γ(n + 1)
= [−1, 2, −2, 2] ×
9 7 5 3 T Γ 4n + 2 , 2 , 2 , 2
(6n)!(4n)!n!4n 1
1885
−965
363
−51
=
+
+
+
(3n)!(8n)! 512 8n + 1 8n + 3 8n + 5 8n + 7
(16)
in simplified notations. Finally with Eq. (15) and Eq. (16), we have
∞
6n
X
1885
−965
363
−51
9
n n
3n
2 π=
(−1) 2 8n 4n
+
+
+
.
8n + 1 8n + 3 8n + 5 8n + 7
4n
n
n=0
5.2
Case 2 :
a
b
+
c
d
∈ Z (positive and negative integers)
This case allows us to obtain special values of Γ ab Γ dc (equal to π multiplied by
an algebraic number). In some cases, π is absent and we obtain amazing formulas for
algebraic numbers.
244
Binomial Sums for π and Polylogarithmic Constants
EXAMPLE 2. The following relations are based on Beta values for
or = − 13 , dc = − 23 or ab = − 53 , dc = − 13
a
b
√ Z 1
3
4
2
xn− 3 (1 − x)n− 3 dx −
=
27n(3n − 1)
4π 0
√ Z 1
3n
3
5
1
n
xn− 3 (1 − x)n− 3 dx −
=
n
27 (3n − 2)
8π 0
3n
n
a
b
= − 43 , dc = − 23
√ Z 1
3
1
2
xn− 3 (1 − x)n− 3 dx
2π 0
√ Z 1
3
1
2
xn− 3 (1 − x)n− 3 dx
4π 0
which gives the formula
∞
3n √
−1
8 X (−1)n n
1
3
+
= 2.
n
3 n=0
54
3n − 1 3n − 2
5.3
Case 3 :
a
b
+
c
d
∈ Q\N (rationals but not naturals)
Γ ab Γ dc
in the outcome.
We obtain Pochhammer symbols in sums then factors
Γ ab + dc
EXAMPLE 3. Considering
Z 1
Γ n + 34 Γ n + 12
n− 14
n− 12
x
(1 − x)
dx =
Γ 2n + 54
0
Z 1
Γ n + 74 Γ n + 12
n+ 34
n− 12
x
(1 − x)
dx =
Γ 2n + 94
0
and the following integral
Z
1
(x − 2)
0
Z 1
∞
X
(−1)n n− 1
1
1
4 (1 − x)n− 2 dx = −2
√
x
dx = −π
√
n
4
2
x 1 − x(x + 1)
0
n=0
we obtain finally
1
∞
X
(−1)n
8n
n=0
where K(k) =
1
2 n
1
4 n
R π/2
0
2 2n
1
4 2n
11
1
+
8n + 1 8n + 5
dθ
1−k 2 sin 2 θ
√
2
1
√ Γ 4
√
1
= 2 √
= 4 2K √
π
2
is the complete elliptic integral of the first kind, thus
making the connection with singular values of the elliptic integral.
6
Higher Order Constants
One point of interest is to generalize this method to higher order constants. A constant
is of order p if it is equal to a linear combination in N of p-th order polylogarithms. For
example, π or ln(2) are of order 1 whereas π ln(2) or G are of order 2 and ζ(3) or π3
B. Gourévitch and J. Guillera Goyanes
245
are of order 3. A BBP binomial sum involves a constant of order p, as for BBP sums,
if the polynomial in n in the denominator is of degree p. For example, we have
∞
X
n=0
(−1)n
2
2n n
n 2 (n + 1)
3
(2n + 1)
= −2 ln3 (2) + 3ζ(3).
(17)
However, the straightforward application of integral representation does not completely help since, for order 2 for instance, we have
∞
X
n mn
n=0 q
pn
1
(mn + 1)
2
=
1Z 1
Z
0
0
q
dxdy.
q − ym xm (1 − x)(m−p)
The evaluation is not often easy, there is no known proof of Eq. (17) to our knowledge. Nevertheless, there is evidence for existence of BBP binomial formulas for higher
order constants.
Acknowledgements. The authors would like to thank Typhenn Brichieri-Colombi,
Professor Sui Sun Cheng and the anonymous reviewer for their helpful comments and
corrections.
References
[1] D. Bailey, P. Borwein and S. Plouffe, On The Rapid Computation of Various
Polylogarithmic Constants, Math. Comp., 66(1997), 903–913.
[2] V. Adamchik, π2 : A 2000-Year Search Changes Direction, Education and Research, 5(1)(1996), 11–19
[3] F. Bellard, Computation of the n’th digit of π in any base in O n2 , preprint
(1997), http://fabrice.bellard.free.fr/pi/pi n2.ps
[4] D. J. Broadhurst,
Polylogarithmic ladders,
hypergeometric series
and the ten millionth digits of ζ (3) and ζ (5) , preprint (1998),
http://front.math.ucdavis.edu/math.CA/9803067
[5] A. Lupas, Some BBP functions, preprint, Sibiu University, Romania (2000).
[6] G. Huvent, Formules BBP, Seminar ANO, IRMA, Lille University (2001),
http://ano.univ-lille1.fr/seminaries/expo huvent01.pdf
[7] A. K. Lenstra, H. W. Jr. Lenstra, L. Lovász, Factoring polynomials with rational
coefficients, Mathematische Annalen 261(4)(1982), 515–534.
[8] Pari-GP, software pour PC, MacOS et Unix, http://www.parigp-home.de/, documentation at http://www.gn-50uma.de/ftp/pari-2.1/manuals/users.ps.gz.
[9] L. Lewin, Polylogarithms and Associated Functions, North-Holland, New York,
1981.
246
Binomial Sums for π and Polylogarithmic Constants
[10] G. Almkvist, C. Krattenthaler and J. Petersson, Some new formulas for π, preprint, Matematiska Institutionen, Lunds Universitet, Sweden. and Institut fur Mathematik der Universitat Wien, Austria (2001),
http://www.arxiv.org/pdf/math.NT/0110238