Journal of Inequalities in Pure and
Applied Mathematics
π AND SOME OTHER CONSTANTS
ANTHONY SOFO
School of Computer Science and Mathematics
Victoria University
PO Box 14428, MCMC 8001
VIC, Australia.
volume 6, issue 5, article 138,
2005.
Received 18 March, 2005;
accepted 26 October, 2005.
Communicated by: J. Sándor
EMail: anthony.sofo@vu.edu.au
URL: http://rgmia.vu.edu.au/sofo/
Abstract
Contents
JJ
J
II
I
Home Page
Go Back
Close
c
2000
Victoria University
ISSN (electronic): 1443-5756
123-05
Quit
Abstract
We consider a particular definite integral and reduce it to hypergeometric form.
Then we develop identities for some numerical constants and the number π.
2000 Mathematics Subject Classification: Primary 33B15; Secondary 33C05.
Key words: Hypergeometric summation, Definite integration, Binomial type series.
This paper is based on the talk given by the author within the “International
Conference of Mathematical Inequalities and their Applications, I”, December 0608, 2004, Victoria University, Melbourne, Australia [http://rgmia.vu.edu.au/
conference]
π and Some Other Constants
Anthony Sofo
Title Page
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
The Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3
Illustrative Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4
Some Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
References
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 2 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
1.
Introduction
π, is a real number and defined as the ratio of the circumference of a circle to
its diameter. π 0s digits have many interesting properties and π has a rich history
dating back to the time of the Babylonians and Egyptians, circa 2000 B.C. The
Bible has two references, I Kings 7:23 and Chronicles 4:2, to Pi and gives it
an estimate of about 3. The Babylonians gave an estimate of π as 3 18 and the
Egyptians also obtained 3 13
as an estimate. We know that 3 18 < π < 3 13
.
81
81
Many researchers have increasingly calculated the number of decimal places
for the value of π. Apparently in September 2002, Dr. Kanada and his team,
from the University of Tokyo, calculated π to 1.2411 trillion digits, indeed a
world record. Many, many formulae also exist for the representation of π, and
a collection of these formulae is listed below.
Vieta (~1593), see [7], gave an infinite product of nested radicals for the
reciprocal of π, namely
v
s
r s
r u
r
u
2
1 1 1 1 t1 1 1 1 1
=
+
+
+
··· .
π
2 2 2 2 2 2 2 2 2
π and Some Other Constants
Anthony Sofo
Title Page
Contents
JJ
J
II
I
Go Back
Wallis (~1650), see [7], gave
∞ π Y
1
=
1−
.
4 r=1
(2r + 1)2
Leibnitz (~1670), see [7], gave the very slow converging series
∞
π X (−1)r
=
.
4
2r + 1
r=0
Close
Quit
Page 3 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
Newton (~1666) admitted being ashamed at having computed π to fifteen
decimal places, by the formula
√
Z 1√
4
3 3
π=
x − x2 dx
+ 24
4
0
√
3 1
1
1
5
3 3
+2−
+
+
+
+ ···
=
4
4 5 7 · 24 9 · 27 11 · 212
√
∞ 1
3 3
3 X 2k
=
+2−
k
k
4
4 k=0
16 (k + 1) (2k + 5)
Euler (~1750) gave many representations of π including:
"
#
n
X
1
1
1
π
= lim
+
+ 4n
.
2 + j2
2 n→∞ n 6n2
n
j=1
Ramanujan (~1914) has also given many representations of π and its reciprocal, including:
√
∞
2 2 X (4r)! (1103 + (2 · 5 · 7 · 13 · 29) r)
1
= 4
.
π
3 · 112 r=0
(r!)4 (22 · 32 · 11)4r
For a fuller account of Ramanujan’s work the interested reader is referred to the
books of Berndt [4].
Comtet (1974) gave
∞
23 · 34 · 5 X 1
.
π =
17
r4 2rr
r=1
4
π and Some Other Constants
Anthony Sofo
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 4 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
D. and G. Chudnovsky (1989) gave
∞
X
1
(−1)r (6r)! 13 · 1045493 + 2 · 32 · 7 · 11 · 19 · 127 · 163r
= 12
·
.
1
π
(r!)3 (3r)!
(218 · 33 · 53 · 233 · 293 )r+ 2
r=0
Bailey, Borwein and Plouffe (1996) gave
∞
X
4
2
1
1
1
(1.1)
π=
−
−
−
.
16r 8r + 1 8r + 4 8r + 5 8r + 6
r=0
π and Some Other Constants
Anthony Sofo
Bellard (1997) gave
"∞
#
X 3P (r)
1
4
π= 2 2
− 2 · 5 · 254741 ,
3 · 5 · 11 · 13 · 23 r=1 7r
2r−1
2r
where
P (r) = −13 · 29 · 2351653r5 + 193 · 16193509r4 − 52 · 7 · 79 · 212873r3
+ 5 · 206392559r2 − 2 · 98441137r + 23 · 13 · 43 · 2459.
Title Page
Contents
JJ
J
II
I
Go Back
Close
Lupas [10], gave
π =4+
∞
X
k=1
k
(−16)
2k
k
(40k 2 + 16k + 1)
.
2
4k 2
2k
(4k
+
1)
2k
The original Lupas formula contained a minor misprint which has been corrected here.
Quit
Page 5 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
Borwein and Girgensohn (2003), wrote
π = ln 4 + 10
∞
X
r=1
1
2r r
3r
r
.
Sofo [16] has given
∞ √
X
4r
1
π √
= 2 + ln
2−1 +
r
2
2r 16 (2r + 1) (4r + 1)
r=0
and
∞
1308 12 X
π =
+
135
5 r=1 r2
2
4r
.
2r
(r + 1) (2r + 1) (2r + 3)
r
Many other representations of π exist, including the famous Machin-type
formulae such as
π
1
1
= 4 arctan
− arctan
.
4
5
239
There are also many other connections of π and other mathematical constants,
including:
eiπ + 1 = 0,
∞
X
(−1)r
π 3 + 8π = 56 − 8
3,
r
(r
+
1)
(2r
+
1)
r=1
∞
X (−1)r (r!)2
π2
2
= 3 ln φ +
, where φ is the Golden ratio
6
(2r)! (2r + 1)2
r=0
π and Some Other Constants
Anthony Sofo
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 6 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
and
π 2 = −12e3
∞
X
r=1

1
cos 
r2

9
rπ +
q
2
(rπ) −
.
32
A selection of some series expansion representations of π including some of
the above is given by Sebah and Gourdon [15].
There are other nice articles and books relating to π including [2, 3, 6, 7, 8,
11, 12].
The aim of this paper is to derive representations of π, as well as some other
constants, by the consideration of a particular definite integral. The following
integral will now be investigated.
π and Some Other Constants
Anthony Sofo
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 7 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
2.
The Integral
Theorem 2.1. For k, m and α real positive numbers and a ≥ 1, then
Z 1
a
xm
(2.1) I (a, k, m, α) =
dx
k α
0 (1 − x )
∞
X
(α)r
(2.2)
=
r! (rk + m + 1) ark+m+1
r=0
m+1
1
,
α
k
(2.3)
= T0 2 F1
m+1+k
ak
k
m+1
m+1
1
−k
(2.4)
=
B 1−α,
−B 1−a ; 1−α,
,
k
k
k
where
(2.5)
π and Some Other Constants
Anthony Sofo
Title Page
Contents
1
T0 =
,
(m + 1) am+1
(b)s is Pochhammer’s symbol defined by

 (b)0 = 1
(2.6)

(b)s = b (b + 1) · · · (b + s − 1) =
JJ
J
Go Back
Close
Γ(b+s)
,
Γ(b)
Γ (b) is the classical Gamma function, 2 F1 [· ·] is the Gauss Hypergeometric
function, B (s, t) is the classical Beta function and
Z z
B (z; s, t) =
us−1 (1 − u)t−1 du
0
II
I
Quit
Page 8 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
is the incomplete Beta function.
Proof.
Z
1
a
I (a, k, m, α) =
0
Z
1
a
=
0
xm
dx
(1 − xk )α
∞
X
−α
r
(−1)
xkr+m .
r
r=0
π and Some Other Constants
where we have utilised
1
(1 + z)β
and from
=
∞ X
−β
r
r=0
Anthony Sofo
zr
Title Page
−β
r
= (−1)r
β+r−1
r
we have
Z
I(a, k, m, α) =
0
1
a
∞
X
r=0
r
=
(−1) (β)r
r!
(α)r kr+m
x
.
r!
Reversing the order of integration and summation and substituting the integration limits we obtain the result (2.2). The result (2.4) is obtained by the use of
the substitution u = 1 − xk .
Binomial sums are intrinsically associated with generalised hypergeometric
functions and if from (2.2) we let
(2.7)
(α)r
Tr =
,
r! (rk + m + 1) ark+m+1
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 9 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
then we get the ratio
(2.8)
(α + r) r +
Tr+1
= k
Tr
a (r + 1) r +
m+1
k
m+1+k
k
where T0 is given by (2.5). From (2.5) and (2.2) we can write
m+1
1
,
α
k
I(a, k, m, α) = T0 2 F1
m+1+k
ak
k
m+1
m+1
1
−k
=
B 1 − α,
− B 1 − a ; 1 − α,
,
k
k
k
which is the result (2.3). We can now match (2.2) and (2.3) so that
∞
X
(α)r
r! (rk + m + 1) ark+m+1
r=0
m+1
1
,
α
k
= T0 2 F1
m+1+k
ak
k
1
m+1
m+1
−k
=
B 1 − α,
− B 1 − a ; 1 − α,
,
k
k
k
and the infinite series converges for a−k < 1.
π and Some Other Constants
Anthony Sofo
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 10 of 31
In a previous paper, Sofo [17] has utilised (2.1) for the case a = 1 and
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
developed identities for π and other constants, such as
15p!
π=
p
√
√ ∞
7
30 + 6 5 + 1 − 5 X
30 r
,
7
r!
(30r
+
30p
+
7)
4 30
p
r=0
for p = 0, 1, 2, 3, . . .
Remark
√ 1. Bailey, Borwein, Borwein and Plouffe see [7] utilised (2.1) for
a = 2, α = 1, k = 8 and m = β − 1, β < 8 to prove the new formula
(1.1). Subsequently Hirschhorn [9] has shown that (1.1) can be obtained from
standard integration procedures.
The following lemma will be useful in the consideration of the integral (2.1).
Lemma 2.2. For p = 0, 1, 2, . . . the following well-known identities are given
by Beyer [5]
p
(−1) X
2p + 1
j
2p+1
sin
x = 2p
(−1)
sin ((2p + 1 − 2j) x)
2
p
j=0
p
(2.9)
and
sin2p+2 x
"
#
p
X
1
2p + 1
2p
+
2
= 2p+1
− (−1)p
(−1)j
cos ((2p + 2 − 2j) x) .
2
p
p
j=0
(2.10)
π and Some Other Constants
Anthony Sofo
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 11 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
The integral (2.1) can be simplified as follows. Consider the case k = 2;
then from (2.1)
Z 1
a
xm
(2.11)
I (a, m, 2, α) =
dx.
2 α
0 (1 − x )
The following lemma concerns the integral (2.11).
Lemma 2.3.
(i) For m = 2p + 1, 2α = 2q + 1; p = 0, 1, 2, . . . , and q = 0, 1, 2, . . .
(2.12) I (a, m, 2, α)
Z 1
Z θ∗
a
x2p+1
sin2p+1 θ
=
dθ
2q+1 dx =
cos2q θ
0 (1 − x2 ) 2
0
1
1 − 2q
−2 1 − 2q
=
B
,p + 1 − B 1 − a ;
,p + 1
2
2
2
q−1
j
j 1 2p+2 Y
1 2p+2
X
(−1)
1
2 (p − q + i) − 1
a
a
=
·
+
2q−1
∗
2(q−j)−1
∗
2q − 1 cos
θ
cos
θ i=1 2 (q − i) + 1
j=1
q
Y
2 (p − q + i) − 1 (−1)p
q
+ (−1)
2 (q − i) + 1
22p
i=1
#
p
X
2p + 1
(−1)s
{1 − cos ((2p − 2s + 1) θ∗ )} .
×
2p
−
2s
+
1
s
s=0
where x = sin θ and θ∗ = arcsin a1 .
Note, the middle term in the right hand side of (2.12) is identically zero for
q = 1 and only the last sum applies for q = 0.
π and Some Other Constants
Anthony Sofo
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 12 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
(ii) For m = 2p + 2, 2α = 2q + 1; p = 0, 1, 2, . . . , and q = 0, 1, 2, . . .
(2.13)
I (a, m, 2, α)
Z 1
Z θ∗
a
x2p+2
sin2p+2 θ
=
dθ
2q+1 dx =
cos2q θ
0 (1 − x2 ) 2
0
1 − 2q
3
1
3
−2 1 − 2q
=
B
,p +
−B 1−a ;
,p +
2
2
2
2
2
2p+3
2p+3
q−1
j
1
X (−1)j 1
Y 2 (p − q + i + 1)
1
a
=
· a2q−1 ∗ +
θ
cos2(q−j)−1 θ∗ i=1 2 (q − i) + 1
2q − 1 cos
j=1
q
Y
1
2p + 1 ∗
2 (p − q + i + 1)
q
θ
+ (−1)
2p+1
2
(q
−
i)
+
1
2
p
i=1
#)
p
s
X
(−1)
2p
+
2
− (−1)p
sin ((2p − 2s + 2) θ∗ )
.
2p
−
2s
+
2
s
s=0
where x = sin θ and θ∗ = arcsin a1 .
Note, the middle term in the right hand side of (2.13) is identically zero for
q = 1 and only the last two terms apply for q = 0.
Proof.
π and Some Other Constants
Anthony Sofo
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
(i) From
Z
I (a, m, 2, α) =
0
θ∗
2p+1
sin
θ
dθ,
cos2q θ
Page 13 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
integrating by parts once leads to
"
#
θ∗
Z θ∗
1
sin2p+2 θ
sin2p+1 θ
I (a, m, 2, α) =
− (2p + 3 − 2q)
dθ
2q − 1 cos2q−1 θ 0
cos2q−2 θ
0
Z ∗
1 2p+2
2p − 2q + 3 θ sin2p+1 θ
a
=
−
dθ,
(2q − 1) cos2q−1 θ∗
2q − 1
cos2q−2 θ
0
and repeated integration by parts gives us
1 2p+2
(2.14) I (a, m, 2, α) =
π and Some Other Constants
Anthony Sofo
a
(2q − 1) cos2q−1 θ∗
2p+2 j
q−1
X
Y 2 (p − q + i) + 1
(−1)j a1
+
cos2(q−j)−1 θ∗ i=1 2 (q − i) + 1
j=1
Z ∗
q
Y
2 (p − q + i) + 1 θ
q
+ (−1)
sin2p+1 θdθ.
2 (q − i) + 1
0
i=1
Substituting (2.9), from Lemma 2.2, into (2.14) and integrating, results in
(2.12) hence part (i) of the lemma is proved.
(ii) The proof of part (ii) of the lemma follows the same footsteps as part (i).
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 14 of 31
The following lemma is given and will be useful in the simplification of the
left hand side of (2.17) and (2.18).
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
Lemma 2.4. For r = 0, 1, 2, . . . and q = 1, 2, 3, . . . then
q−1
q + 21 r
1 2r Y 2r + 2ρ + 1
(2.15)
= r
,
r!
4
r ρ=0 2ρ + 1
and for q = 0,
1
2 r
1
= r
r!
4
(2.16)
2r
r
.
π and Some Other Constants
Proof. For q = 0, then
Anthony Sofo
1
2 r
Γ r + 12
1 2r
= r
.
=
4
r
r!
r!Γ 12
Title Page
This result is well known and is also given by Wilf [18].
For q ≥ 1, let
q−1
q + 21 r
1 2r Y 2r + 2ρ + 1
= r
P (q) :=
r!
4
r ρ=0 2ρ + 1
Contents
JJ
J
II
I
Go Back
Close
then
3
2 r
(2r + 1) Γ
P (1) =
=
r!
2
Γ
1
2
3
2
and from (2.16)
2r + 1
P (1) =
4r
2r
r
1
2 r
Quit
r!
,
Page 15 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
,
http://jipam.vu.edu.au
which satisfies the right hand of (2.15) for q = 1.
Consider
q + 23 r
Γ q + r + 32
P (q + 1) =
=
r!
r!Γ q + 32
2q + 2r + 1 Γ q + r + 12
=
2q + 1
r!Γ q + 12
q + 12 r
2q + 2r + 1
=
2q + 1
r!
π and Some Other Constants
Anthony Sofo
and from (2.15)
q−1
2q + 2r + 1 1 2r Y 2r + 2ρ + 1
P (q + 1) =
2q + 1
4r r ρ=0 2ρ + 1
q
1 2r Y 2r + 2ρ + 1
= r
4
r ρ=0 2ρ + 1
hence the lemma is proved.
For k = 2 the following theorem now applies.
Theorem 2.5.
(i) For m = 2p + 1, 2α = 2q + 1; p = 0, 1, 2, . . . , q = 0, 1, 2, . . . ,
∞
X
q + 21 r
(2.17)
r! (2r + 2p + 2) a2r+2p+2
r=0
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 16 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
Qq−1 2r+2ρ+1
∞ X
2r
ρ=0 2ρ+1
=
r
r
4 (2r + 2p + 2) a2r+2p+2
r=0
p + 1, 21 (2q + 1) 1
∗
= T0 2 F1
a2
p+2
1
1 − 2q
−2 1 − 2q
=
B
,p + 1 − B 1 − a ;
,p + 1
2
2
2
2p+2 j
q−1
1 2p+2
Y 2 (p − q + i) + 1
X
(−1)j a1
a
+
=
(2q − 1) cos2q−1 θ∗ j=1 cos2(q−j)−1 θ∗ i=1 2 (q − i) + 1
q
Y
2 (p − q + i) + 1
q
+ (−1)
2 (q − i) + 1
i=1
#
"
p
2p + 1
(−1)p X (−1)s
∗
×
{1 − cos ((2p − 2s + 1) θ )}
22p s=0 2p − 2s + 1
s
where
T0∗ =
and θ∗ = arcsin
1
a
1
(2p + 2) a2p+2
.
(ii) For m = 2p + 2, 2α = 2q + 1; p = 0, 1, 2, . . . , q = 0, 1, 2, . . . ,
∞
X
q + 12 r
(2.18)
r! (2r + 2p + 3) a2r+2p+3
r=0
π and Some Other Constants
Anthony Sofo
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 17 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
Qq−1 2r+2ρ+1
∞ X
2r
ρ=0 2ρ+1
=
r
r
4 (2r + 2p + 3) a2r+2p+3
r=0
1
(2p + 3) , 12 (2q + 1) 1
∇
2
= T0 2 F1
1
a2
(2p + 5)
2
1
1 − 2q
3
3
−2 1 − 2q
=
B
,p +
−B 1−a ;
,p +
2
2
2
2
2
2p+3 j
q−1
1 2p+3
Y 2 (p − q + i + 1)
X
(−1)j a1
a
+
=
(2q − 1) cos2q−1 θ∗ j=1 cos2(q−j)−1 θ∗ i=1 2 (q − i) + 1
q
Y
1
2p + 1 ∗
2 (p − q + i + 1)
q
+ (−1)
θ
2 (q − i) + 1
22p+1
p
i=1
#)
p
X
2p + 2
(−1)s
p
∗
− (−1)
sin ((2p − 2s + 2) θ )
,
2p − 2s + 2
s
s=0
where
T0∇ =
and θ∗ = arcsin
1
a
1
(2p + 3) a2p+3
.
The proof of Theorem 2.5 follows directly from Lemma 2.3 , (2.14) and
Lemma 2.4.
Some examples will now be given expressing π and other constants in terms
of an infinite series.
π and Some Other Constants
Anthony Sofo
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 18 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
3.
Illustrative Examples
Example 3.1. From (2.17) with q = 2, a = 2 and θ∗ = π6 , we have
2
√
3
8
− 2 (2p − 1) + 8 (−1)p (2p − 1) (2p + 1)
3
p
X
2p + 1 n
πo
(−1)s
1 − cos (2p − 2s + 1)
×
2p − 2s + 1
s
6
s=0
∞
X 2r (2r + 1) (2r + 3)
=
.
r (r + p + 1)
r
16
r=0
Hence, for p = 7,
√
3=
π and Some Other Constants
Anthony Sofo
Title Page
3 · 222
32 · 11
− 6
7 · 163 · 6367 2 · 7 · 163 · 6376
Example 3.2. From (2.18) with q = 2, a =
16p (p + 1)
8 − 4p + √ 2p+3
3
p
X
p
− (−1)
s=0
2p + 1
p
(−1)s
2p − 2s + 2
π
3
∞ X
r=0
√2
3
2r
r
∗
and θ =
(2r + 1) (2r + 3)
.
16r (r + 8)
π
,
3
we have
#
2p + 2
π
sin (2p − 2s + 2)
s
3
r
∞
X
2r (2r + 1) (2r + 3) 3
=
r
(2r + 2p + 3)
16
r=0
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 19 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
or rearranging,
3
3p+ 2
π
=
3
16p(p + 1) 2p+1
p
r
∞ X
2r (2r + 1) (2r + 3) 3
4p − 8 +
r
(2r + 2p + 3)
16
r=0
(
p
(−1)p X
+
2p+1
p
s=0
(−1)s
2p − 2s + 2
2p + 2
s
)
π
sin (2p − 2s + 2) ,
3
and for p = 2
√
r
∞ 1 X 2r (2r + 1) (2r + 3) 3
20 3π
.
=1+
81
16 r=0 r
2r + 7
16
√
For p = 2, a = 2
r
∞ 8
1 X 2r (2r + 1) (2r + 3) 1
.
π= + √
3 3 2 r=0 r
2r + 5
8
√
Example 3.3. For q = 3, p = 2 and a = 2,
r
∞ 52
1 X 2r (2r + 1) (2r + 3) (2r + 5) 1
− √
.
π=
15 30 2 r=0 r
2r + 7
8
Example 3.4. For q = 4, p = 3 and a = 2,
√
∞ 1712 3
1 X 2r (2r + 1) (2r + 3) (2r + 5) (2r + 7)
+
.
π=
945
8960 r=0 r
(2r + 9) 16r
π and Some Other Constants
Anthony Sofo
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 20 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
Example 3.5. For q = 5, p = 0 and a =
√
∞
X
211
5=
3 · 5 · 193 · 2731 r=0
2r
r
√
5,
(2r + 1) (2r + 5) (2r + 7) (2r + 9) (2r + 11)
.
20r
Example 3.6. For q = 6, p = 5 and a =
√
∞
X
23 · 1289
1
√
π=
+
5 · 7 · 9 · 11 5 · 16 · 829 · 2 r=0
2,
2r
r
(2r + 1) (2r + 3) (2r + 5) (2r + 7) (2r + 9) (2r + 11)
×
.
(2r + 13) 8r
Example 3.7. For q = 4 α = 92 , p = 59 (m = 120) and a = 2,
π=
Ω1
1
√ +
Ω2 3 Ω3
∞ X
r=0
2r
r
(2r + 1) (2r + 3) (2r + 5) (2r + 7)
,
(2r + 121) 16r
where
Ω1 = 15604102274295581508678435968572864501995513795052733
= (46042305118509401202197) (338907929004245243145594887689) ,
π and Some Other Constants
Anthony Sofo
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 21 of 31
Ω2 = 26 · 35 · 52 · 72 · 11 · 13 · 17 · 19 · 23 · 29 · 31 · 37 · 41 · 43 · 47 · 53 · 59
· 61 · 67 · 71 · 73 · 79 · 83 · 89 · 97 · 101 · 103 · 107 · 109 · 113
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
and
Ω3 = 211 · 33 · 5 · 7 · 13 · 17 · 19 · 23 · 29 · 31 · 37 · 59
· 61 · 67 · 71 · 73 · 79 · 83 · 89 · 97 · 101 · 103 · 107 · 109 · 113.
In this case the first term of the sum gives π accurate to over forty decimal
places.
Other particular values of constants may be obtained from (2.13).
Example 3.8. For a = 2, m = 10, k = 1 and α = 7
∞ X
27947
1
r+6
1
ln 2 = 7 2
+ 12
.
2 · 3 · 5 · 7 2 · 3 · 5 · 7 r=0
r
(r + 11) 2r
It is of some interest to note that from (2.18) for m = 0 and α = w > 1, integer,
we may eventually write, after integration by parts, and using (2.2)
"∞ w
a+1
2 (w − 1)! X r + w − 1
1
ln
=
a−1
(2w − 3)!! r=0
r
(2r + 1) a2r+1
#
2 w−j Y
j
w−1
X
1
1
a
2w − 2ν + 1
−
,
a (2w − 1) j=1 2j a2 − 1
w−ν
ν=1
where (2w − 3)!! = (2w − 3) (2w − 5) · · · 5 · 3 · 1, and a > 1.
For a = 11 and w = 7, we have
∞ X
211
r+6
1
11 · 179 · 17047711
6
=
−
.
ln
5
3 · 7 · 112 r=0
r
(2r + 1) 121r
29 · 38 · 56
π and Some Other Constants
Anthony Sofo
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 22 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
Remark 2. In the degenerative case of w = 1 then we obtain the well-known
formula as listed in Abramowitz and Stegun [1], namely
ln
a+1
a−1
=2
∞
X
r=0
1
.
(2r + 1) a2r+1
π and Some Other Constants
Anthony Sofo
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 23 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
4.
Some Estimates
It is useful to be able to obtain some estimates of the representation of the series
(2.2). This is done in the following theorems.
Theorem 4.1. Given that
(4.1)
S (a, k, α, m) =
∞
X
r=0
(α)r
,
r! (rk + m + 1) ark+m+1
π and Some Other Constants
then
(4.2)
Anthony Sofo
1
(m + 1) am+1
< S (a, k, α, m)

p1
1
1
1
−k

,
B
1
−
αp,
−
B
1
−
a
;
1
−
αp,
1 1

k
k
 ((mq+1)amq+1 ) q k p
≤
k α



a
1
,
a > 1,
(m+1)am+1
ak −1
for real numbers p and q where p > 1, p1 + 1q = 1, B (s, t) is the classical Beta
function and B (z; s, t) is the incomplete Beta function as described in Theorem
2.1.
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 24 of 31
and g (x) = xm . Since |f (x)|p and |g (x)|q are
integrable functions defined on x ∈ 0, a1 , then by Hölder’s integral inequality
Proof. Let f (x) =
1
α
(1−xk )
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
[14]
Z
S (a, k, α, m) ≤
1
a
! 1q
xmq dx
0
Z
0
1
a
dx
(1 − xk )αp
! p1
.
Now
Z
0
1
a
dx
1
1
1
−k
=
B 1 − αp,
− B 1 − a ; 1 − αp,
(1 − xk )αp
k
k
k
by the substitution u = 1 − xk , and hence
π and Some Other Constants
Anthony Sofo
1
S (a, k, α, m) ≤
1
1
((mq + 1) amq+1 ) q k p
1
1 p
1
−k
.
× B 1 − αp,
− B 1 − a ; 1 − αp,
k
k
1
The lower bound on S (a, k, α, m) is (m+1)a
m+1 since the sum (2.2) is one of
positive terms.
The second part of the inequality (4.2) is obtained from
Z x1
Z x1
|f (x) g (x)| dx ≤ ess sup |f (x)|
|g (x)| dx.
x∈[x0 ,x1 ]
x0
x0
Since f (x) is monotonic on x ∈ 0, a1 ,
k α
1
a
ess sup
=
α
k
k
(1 − x )
a −1
x∈[0, 1 ]
a
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 25 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
and
Z
0
1
a
xm dx =
1
,
(m + 1) am+1
hence
1
< S (a, k, α, m) ≤
(m + 1) am+1
ak
ak − 1
α
1
.
(m + 1) am+1
The result (4.2) follows and the theorem is proved.
The next theorem develops an inequality of (4.1) based on the pre-Grüss
result.
Theorem 4.2. For a > 1,
1
(4.3) S (a, k, α, m) −
k (m + 1) am
1
1
−k
− B 1 − a ; 1 − α,
× B 1 − α,
k
k
k α
m
a
√
≤
−1 .
ak − 1
2 (m + 1) am+1 2m + 1
Proof. Define
T (g, f ) :=
π and Some Other Constants
Anthony Sofo
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
1
x1 − x0
Z
x1
x0
f (x) g (x) dx
Z x1
Z x1
1
1
−
f (x) dx ·
g (x) dx
x 1 − x 0 x0
x 1 − x 0 x0
Page 26 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
for f (x)
and
g (x) integrable functions, as given in Theorem 4.1, and defined
1
on x ∈ 0, a , then the pre-Grüss inequality [13] states that
|T (g, f )| ≤
1
Γ−γ
[T (g, g)] 2
2
for γ ≤ f (x) ≤ Γ.
Now, for x ∈ 0, a1
1
γ = 1 ≤ f (x) =
≤
(1 − xk )α
Z
T (g, f ) = a
0
1
a
xm
dx − a2
(1 − xk )α
Z
0
2
1
a
ak
ak − 1
α
dx
(1 − xk )α
π and Some Other Constants
= Γ,
Z
1
a
xm dx
0
a
= aS (a, k, α, m) −
(m + 1) am+1 k
1
1
−k
× B 1 − α,
− B 1 − a ; 1 − α,
.
k
k
Anthony Sofo
Title Page
Contents
JJ
J
II
I
Go Back
Close
In a similar fashion
1
[T (g, g)] 2 =
m
√
.
(m + 1) am 2m + 1
Quit
Page 27 of 31
1
Combining T (g, f ) and [T (g, g)] 2 gives us the result (4.3) after a little algebraic simplification.
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
Open Problem 1. From Example 3.2, we have that
(
r )
3
∞ X
3p+ 2
2r (2r + 1) (2r + 3) 3
4p − 8 +
U∞ =
r
(2r + 2p + 3)
16
16p(p + 1) 2p+1
r=0
p
p
(−1)p X
+
2p+1
p
s=0
(−1)s
2p − 2s + 2
2p + 2
s
sin (2p − 2s + 2)
π
π
= .
3
3
π and Some Other Constants
Now let us consider the following. For a finite positive integer R let
3
3p+ 2
UR =
16p(p + 1) 2p+1
p
(
r
R X
2r (2r + 1) (2r + 3) 3
4p − 8 +
r
(2r + 2p + 3)
16
r=0
p
(−1)p X
+
2p+1
p
s=0
(−1)s
2p − 2s + 2
2p + 2
s
Anthony Sofo
)
π
sin (2p − 2s + 2) ,
3
Title Page
Contents
JJ
J
II
I
Go Back
UR = V + W,
Close
π
.
3
For a fixed positive integer R, it can be shown, by standard analysis methods,
Quit
in fact
UR <
Page 28 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
that
3
3p+ 2
lim V = lim
p−>∞
p−>∞
16p(p + 1) 2p+1
p
(
r )
R X
2r (2r + 1) (2r + 3) 3
× 4p − 8 +
=0
r
(2r + 2p + 3)
16
r=0
and as p− > ∞ uniformly
p
(−1)p X
W =
2p+1
p
s=0
(−1)s
2p − 2s + 2
π and Some Other Constants
2p + 2
s
sin (2p − 2s + 2)
π
π
u .
3
3
Title Page
This implies that for p− > ∞, W u U∞ and
p
(−1)p X
W =
2p+1
p
Anthony Sofo
s=0
(−1)s
2p − 2s + 2
2p + 2
s
Contents
sin (2p − 2s + 2)
π
π
u .
3
3
JJ
J
II
I
Go Back
An open problem is to prove, or provide a contradiction to,


p
s
 (−1)p X
(−1)
2p + 2
π
lim W = lim
sin (2p − 2s + 2)
p−>∞
p−>∞  2p+1
2p − 2s + 2
s
3
s=0
Close
Quit
Page 29 of 31
p
=
π
.
3
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
References
[1] M. ABRAMOWITZ AND I. STEGUN, Handbook of Mathematical functions, Dover Publications Inc. New York, 1970.
[2] J. ARNDT
2001.
AND
C. HAENEL, Pi-Unleashed, Springer-Verlag, Germany,
[3] P. BECKMANN, A History of Pi, 3rd Edition, New York, 1989.
[4] B. BERNDT, Ramanujans Notebooks, Springer Verlag, New York. Part I,
1985. Part II, 1989. Part III, 1991. Part IV, 1994. Part V, 1998.
π and Some Other Constants
Anthony Sofo
[5] W.H. BEYER, CRC Standard Mathematical Tables, 28th Ed., Boca Raton
FL., CRC Press, 1987.
Title Page
[6] D. BLATNER, The Joy of Pi. Walker, New York, 1997.
Contents
[7] J. BORWEIN AND P. BORWEIN, Pi and the AGM: A Study in Analytic
Number Theory and Computational Complexity, Wiley, New York, 1987,
reprinted 1998.
[8] S.R. FINCH, Mathematical Constants. Encyclopedia of Mathematics and
its Applications, 94. Cambridge University Press, USA, 2003.
[9] M. HIRSCHHORN, A new formula for π. Australian Math. Soc. Gazette,
25 (1999), 82–83.
[10] A. LUPAŞ, Formulae for some classical constants, Schriftenreihe des
fachbereichs Mathematik, Gerhard Mercator Universitat Dusiburg, 2000,
70–76.
JJ
J
II
I
Go Back
Close
Quit
Page 30 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au
[11] Pi, Wolfram Research, 2003. http://mathworld.wolfram.com/
Pi
[12] Pi Formulas, Wolfram Research,
wolfram.com/PiFormulas
2003. http://mathworld.
[13] M. MATIĆ, J.E. PEČARIĆ AND N. UJEVIĆ, On new estimation of the
remainder in generalised Taylor’s formula, Math. Ineq. and App., 2 (1999),
343–361.
[14] D.S. MITRINOVIĆ, J.E. PEC̆ARIĆ AND A.M. FINK, Classical and New
Inequalities in Analysis, Kluwer Academic Publishers, Boston, 1993.
[15] P. SEBAH AND X. GOURDON, Collection of Series for π,
2002. http://numbers.computation.free.fr/constants/
Pi/pi.html
[16] A. SOFO, Computational Techniques for the Summation of Series, Kluwer
Publishing Co., New York, 2003.
[17] A. SOFO, Some Representations of π, Australian Math. Soc. Gazette, 31
(2004), 184–189.
π and Some Other Constants
Anthony Sofo
Title Page
Contents
JJ
J
II
I
Go Back
Close
[18] H.S. WILF, Generatingfunctionology, Academic Press Inc., New York,
1994.
Quit
Page 31 of 31
J. Ineq. Pure and Appl. Math. 6(5) Art. 138, 2005
http://jipam.vu.edu.au