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Ramanujan-type series for 1? with quadratic
irrationals
A thesis presented in partial fulfillment
of the requirements for the degree of
Master of Science
in
Mathematics
at Massey University, Albany,
New Zealand.
Antesar Mohammed Aldawoud
2012
Abstract
In 1914, Ramanujan discovered 17 series for 1/?, 16 are rational and one is irrational.
They are classified into four groups depending on a variable  called the level, where
 = 1, 2, 3 and 4. Since then, a total of 36 rational series have been found for these
levels. In addition, 57 series have been found for other levels. Moreover, 14 irrational
series for 1/? were found. This thesis will classify the series that involve quadratic
irrationals for the levels  ? {1, 2, 3, 4}. A total of 90 series are given, 76 of which are
believed to be new. These series were discovered by numerical experimentations using
the mathematical software tool ?Maple? and they will be listed in tables at the end of
this thesis.
Acknowledgements
I am heartily thankful to my supervisor Dr. Shaun Cooper, whose encouragement,
guidance and support from the initial to the final level enabled me to develop an un-
derstanding of the subject. I am grateful to all my family and friends, most importantly
to my husband Saud, whose love and guidance is with me in whatever I pursue. Lastly,
I offer my regards and blessings to all of those who supported me in any respect during
the completion of the project.
1
Contents
1 Introduction 4
1.1 Literature Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 Series for 1/? that involve quadratic irrationals . . . . . . . . . . . . . 7
2 Background theory 8
2.1 Preliminary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.2 The Level . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.3 The Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.3.1 Level 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.3.2 Level 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.3.3 Level 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.3.4 Level 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3 Methodology: Maple 16
3.1 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.2 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
3.3 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
2
List of Tables
2.1 Modular forms and solutions to recurrence relations . . . . . . . . . . . 12
3.1 Irrational Series for level l = 1 . . . . . . . . . . . . . . . . . . . . . . . 22
3.2 Irrational Series for level l = 1 . . . . . . . . . . . . . . . . . . . . . . . 23
3.3 Irrational Series for level l = 2 . . . . . . . . . . . . . . . . . . . . . . . 24
3.4 Irrational Series for level l = 2 . . . . . . . . . . . . . . . . . . . . . . . 25
3.5 Irrational Series for level l = 2 . . . . . . . . . . . . . . . . . . . . . . . 26
3.6 Irrational Series for level l = 3 . . . . . . . . . . . . . . . . . . . . . . . 27
3.7 Irrational Series for level l = 3 . . . . . . . . . . . . . . . . . . . . . . . 28
3.8 Irrational Series for level l = 3 . . . . . . . . . . . . . . . . . . . . . . . 29
3.9 Irrational Series for level l = 4 . . . . . . . . . . . . . . . . . . . . . . . 30
3.10 Irrational Series for level l = 4 . . . . . . . . . . . . . . . . . . . . . . . 31
3
Chapter 1
Introduction
In 1914, Ramanujan [15] gave 17 extraordinary series for 1/?. Two of his best known
examples are
1
?
=
2
?
2
9801
?
?
n=0
(4n)!
(n!)4
1
3964n
(1103 + 26390n), (1.1)
16
?
=
?
?
n=0
(2n)!3
n!6
1
212n
(42n+ 5). (1.2)
These series have excellent convergence properties. They are relatively scarce, and
the numbers and the coefficients are striking. They are widely famous, nearly every
mathematician will recognize (1.1). The proofs of these series require sophisticated
number theory, such as modular equations and class invariants. The series (1.1) con-
verges very fast: each term contributes 8 digits of ?.
The ratio test for the series (1.1) shows: if
tn =
(4n)!
(n!)4
(1103 + 26390n)
3964n
,
Then :
tn+1
tn
=
(4n+ 4)!
((n+ 1)!)4
(1103 + 26390(n+ 1))
3964n+4
?
(n!)4
(4n)!
3964n
(1103 + 26390n)
,
=
(4n+ 4)(4n+ 3)(4n+ 2)(4n+ 1)
(n+ 1)4
1
3964
(1103 + 26390(n+ 1))
(1103 + 26390n)
.
Taking the limit:
lim
n??
tn+1
tn
= lim
n??
44
3964
,
=
1
994
,
 10?8.
4
In 1985, Gosper programmed (1.1) on a computer and got a world record of more than
17 million digits of ? [14].
Baruah, Berndt and Chan (2009) wrote in the survey [1]: ?The series (1.2) appeared
in the ?Walt Disney? film High School Musical, starring Vanessa Anne Hudgens, who
plays an exceptionally bright high school student named Gabriella Montez. Gabriella
points out to her teacher that she had incorrectly written the left-hand side of (1.2) as
8/? instead of 16/? on the blackboard, After first claiming that Gabriella is wrong, her
teacher checks (possibly Ramanujan?s Collected Papers? ) and admits that Gabriella is
correct. Formula (1.2) was correctly recorded on the blackboard? (p. 568).
In the paper [6], the authors derived a general statement that encapsulates all known
rational analogues of Ramanujan?s series for 1/?.
In this thesis, I will classify new series for 1/? that involve quadratic irrationals. An
example of one of our series for 1/? from level 3 is given by:
1
?
= 2
?
1? 108x
?
?
k=0
(3k)!(2k)!
(k!)5
(k + ?)xk,
with x = ?1
6
+ 7
72
?
3 and ? = 5
22
?
1
22
?
3.
I include a list of all the known quadratic irrational series for 1/?. Of the 90 series
listed in Tables 3.1?3.10, 14 were known previously and 76 are believed to be new.
1.1 Literature Review
In 1914, Ramanujan published a paper entitled?Modular equations and approximations
to ?? in England [15]. In the paper, he gave 17 series for 1/?. The first three are (1.2)
and :
4
?
=
?
?
n=0
(2n)!3
n!6
1
28n
(6n+ 1), (1.3)
32
?
=
?
?
n=0
(2n)!3
n!6
(
3?
?
5
16
)
4n
(
(42
?
5 + 30)n+ (5
?
5? 1)
)
. (1.4)
He classified the 17 series into four types based on four functions:
q
1
= exp
(
?2? 2
F
1
(1
6
, 5
6
; 1; 1? x)
2
F
1
(1
6
, 5
6
; 1; x)
)
, q
2
= exp
(
?
2?
?
2
2
F
1
(1
4
, 3
4
; 1; 1? x)
2
F
1
(1
4
, 3
4
; 1; x)
)
q
3
= exp
(
?
2?
?
3
2
F
1
(1
3
, 2
3
; 1; 1? x)
2
F
1
(1
3
, 2
3
; 1; x)
)
, q
4
= exp
(
??
2
F
1
(1
2
, 1
2
; 1; 1? x)
2
F
1
(1
2
, 1
2
; 1; x)
)
5
where
2
F
1
is the hypergeometric function defined by:
2
F
1
(a, b; c; x) =
?
?
n=0
(a)n(b)n
(c)nn!
xn.
All of Ramanujan?s examples above have the property a+ b = c = 1. The function q
4
is the classical theory of Jacobi. While, q
1
, q
2
and q
3
are new theories of Ramanujan.
Of the 17 series, two are based on q
1
, ten are based on q
2
, two based on q
3
and the
remaining three series are given by (1.2)?(1.4) above, are based on q
4
.
We define a series as a ?rational? series for 1/?, if C/? can be expressed as a series of
rational numbers for some algebraic number C.
The main contributions to Ramanujan series for 1/?
Many books and survey papers have been written to discuss Ramanujan?s prominent
work. One example is the valuable survey done by Nayandeep D. Baruah, Bruce C.
Berndt and Heng Huat Chan [1].
Here is a summary of some of the main contributions to Ramanujan?s series for 1/?,
and the information in this section comes mainly from [6] and [1].
Fourteen years after Ramanujan?s work, Sarvadaman Chowla [8] proved the series (1.3)
of Ramanujan?s series for 1/?. In 1985, R. William Gosper programmed Ramanujan?s
series (1.1) which is based on q
2
on a computer. He calculated 17,526,100 digits of ?
which was at that time a world record. The problem with Gosper?s calculation was
the series had not been proved yet [1]. In 1987, David and Gregory Chudnovsky [9]
developed a theory and derived new series representations for 1/? and used one of
them, the following series which is based on q
1
:
1
?
= 12
?
?
n=0
(?1)n
(6n)!
(n!)3(3n)!
1
(640320)3n+3/2
(13591409 + 545140134n),
to calculate 2,260,331,336 digits of ? which was also a world record in 1989. The above
series yields 14 digits of ? per term and is the fastest convergent rational series for 1/?.
In 1987, Jonathan and Peter Borwein [3] proved all 17 of Ramanujan?s series for 1/?
successfully, they were the first to give complete proofs. In [4] they listed all the series
for 1/? which are based on the function q
1
that were discovered by the Chudnovskys.
They also gave one that is new. In 2001, Chan, Wen-Chin Liaw and Victor Tan [7]
found new identities helped them to prove the following series:
6
1
?
=
1
1500
?
3
?
?
n=0
(3n)!(2n)!
(n!)5
(?1)n
3003n
(14151n+ 827).
This series had not been discovered by Ramanujan and it corresponds to q
3
. In 2001,
Berndt and Chan [2] determined a series for 1/? that corresponds to q
1
that is not
rational which yields about 73 or 74 digits per term that appears to be one of the fastest
known convergent series for 1/?. In 2012, Chan and Shaun Cooper [6] introduced
classification by level and stated a general theorem that is satisfied by 93 rational
series for 1/?, 40 of which were discovered by them. Also Cooper in [11], derived three
new series for 1/?. Some mathematicians, namely Takeshi Sato and Matthew Rogers
found series for 1/? based on different functions, these series can be found in [6], [16]
and [17].
1.2 Series for 1/? that involve quadratic irrationals
A series is a ?quadratic-irrational? series for 1/?, if C/? can be expressed as a series
of quadratic irrational numbers for some algebraic number C. The first non-rational
series for 1/? is (1.4), it was given by Ramanujan in [15]. Also, the Borweins gave 14
non-rational series that involve quadratic irrationals which correspond to q
1
and q
2
[4].
These series are listed in Tables 3.1?3.5 denoted by ?*?.
7
Chapter 2
Background theory
2.1 Preliminary Results
In this section we present the definitions that will be used to find the series for 1/?.
Let q be a complex number that satisfies |q| < 1.
Ramanujan?s Eisenstein series and Dedekind?s eta function
Ramanujan?s Eisenstein series are defined by:
P = P (q) = 1? 24
?
?
j=1
jqj
1? qj
, (2.1)
Q = Q(q) = 1 + 240
?
?
j=1
j3qj
1? qj
, (2.2)
R = R(q) = 1? 504
?
?
j=1
j5qj
1? qj
. (2.3)
Let
?n = ?n(q) = qn/24
?
?
j=1
(1? qnj). (2.4)
Another formula for the eta function given by Euler is [3]:
?n = ?n(q) =
?
?
j=??
(?1)j q(6j?1)
2/24. (2.5)
This function, when n = 1, is Dedekind?s eta-function. The eta functions that will be
encountered in this thesis are:
(
?
2
?
1
)
24
,
(
?
3
?
1
)
12
and
(
?
4
?
1
)
8
.
8
All these functions have leading term q, so there will be no ambiguity in the branch
of qn/24 in (2.4).
2.2 The Level
The function x defined in Theorem 1 in the next section satisfies an involution of the
form:
x
(
e?2?
?
t/
)
= x
(
e?2?/
?
t
)
, t > 0.
In this thesis, the positive integer  will be called the level. As stated before, Ramanu-
jan?s series are based on the four functions q,  ? {1, 2, 3, 4}. For reasons to use the
level see [10].
2.3 The Main Result
In this section, we introduce the next result for each level which gives 90 quadratic-
irrational series for 1/? of 4 different types given by (2.7).
The results in this section are from [6].
Theorem 2.3.1 Let  ? {1, 2, 3, 4}. Let ? = ?(q) and (a, b) ? Z2 be as in Table 2.1.
Let s(k) be the sequence defined by the recurrence relation:
(k + 1)2s(k + 1) = (ak2 + ak + b)s(k)
and initial conditions
s(?1) = 0, s(0) = 1.
Let
x = x(q) = ?(1? a?). (2.6)
Let N be a positive integer called the degree. Either: let ? and q take the values:
? = 2?
?
N/ and q = exp (??);
Or
? =
?
?
?
?
?
?
?
?
?
2?
?
N/4 if  ? 1 (mod 2),
2?
?
N/2 if  ? 2 (mod 4), and q = ? exp(??).
2?
?
N/ if  ? 0 (mod 4)
9
Then the identity:
?
1? 4ax
?
?
k=0
(
2k
k
)
s(k)(k + ?)xk =
1
?
(2.7)
holds, provided the series converges.
In Tables 3.1?3.10 we give 90 sets of values of ,N, x and ?. All these values form
series for 1/? that involve quadratic irrationals.
The parameter ? is given by the following result of Chan, Chan and Liu [5]. First we
need the following definition [10]:
Definition 2.3.2 For 1 ?  ? 4, define z = z(q) by
z = z(q) =
?
?
?
(Q(q))1/4 if  = 1,
(
P (q)?P (q)
?1
)
1/2
if  = 2, 3 or 4.
Theorem 2.3.3 Let Z, x, u and h(k) be defined by:
Z = Z(q) = z2 ,
x = x(q) = ?(1? a?),
u = u(q) =
?
1? 4ax
and
h(k) =
(
2k
k
)
s(k).
Suppose t > 0. Suppose x = x(q), Z = Z(q) and u = u(q) satisfy the properties
tZ
(
exp?2?
?
t
)
= Z
(
exp?2?
?
t/
)
,
Z(q) =
?
?
k=0
= h(k)xk(q),
and
q
d
dq
log x(q) = u(q)Z(q).
For any integer N ? 2, let
M(q) =
Z(q)
Z(qN)
.
Let ?, X and U be defined by
? =
x
2N
dM
dx
?
?
?
?
q=e?2?/
?
N
,
10
X = x(e?2?
?
N/),
U = u(e?2?
?
N/).
Then
?

N
1
2?
= U
?
?
k=0
h(k)(k + ?)Xk.
The identity (2.7) can be used to prove Ramanujan?s series for 1/?, provided that x
is defined by (2.6) can be evaluated for specific values of q for various N . The Table 2.1
contains the definitions of the modular forms in terms of the results 2.1?2.4, for each
level as well as the recurrence relations and their solutions in terms of the binomial
coefficients.
11
 (a, b) ?(q) s(k) z = z(q) z = z(x)
1 (432,60) 1
864
(
1? R(q)
Q(q)3/2
)
(
6k
3k
)(
3k
k
)
(Q(q))1/4
2
F
1
(
1
6
, 5
6
; 1; x
)
2 (64,12) ?
24
2
?24
1
+64?24
2
(
4k
2k
)(
2k
k
)
(2P (q2)? P (q))1/2
2
F
1
(
1
4
, 3
4
; 1; x
)
3 (27,6) ?
12
3
?12
1
+27?12
3
(
3k
k
)(
2k
k
)
(
3P (q3)?P (q)
2
)
1/2
2
F
1
(
1
3
, 2
3
; 1; x
)
4 (16,4) ?
8
4
?8
1
+16?8
4
(
2k
k
)
2
(
4P (q4)?P (q)
3
)
1/2
2
F
1
(
1
2
, 1
2
; 1; x
)
Table 2.1: for the level
Here is a summary for the results and definitions for each level:
2.3.1 Level 1
Let z
1
be defined as in Table 2.1;
z
1
= z
1
(q) = (Q(q))1/4,
then the branch of the root is determined by requiring z
1
= 1 when q = 0.
Define the modular function ? to be:
? = ?(q) =
1
864
(
1?
R(q)
Q(q)3/2
)
.
By substituting the values of a = 432 and ? into (2.6) we define x for level 1 to be:
x = x(q) = ?(1? 432?) =
1
1728
(
Q(q)3 ?R(q)2
Q(q)3
)
.
Theorem 4.8 in [10] gives:
z
1
= z
1
(x) =
2
F
1
(
1
6
,
5
6
; 1; x
)
.
The series (2.7) for  = 1 becomes:
?
1? 1728x
?
?
k=0
(6k)!
(k!)3(3k)!
(k + ?)xk =
?
1/N
2?
, (2.8)
when q = exp (?2?
?
N). And when q = ? exp (??
?
N) it becomes:
?
1? 1728x
?
?
k=0
(6k)!
(k!)3(3k)!
(k + ?)xk =
?
1/N
?
. (2.9)
Both series (2.8) and (2.9) converge for |x| < 1
1728
. The series (2.8) holds for 7 values
of x and ? given in Table 3.1, while there are 4 series for 1/? satisfied by the series
(2.9) given in Table 3.2.
12
2.3.2 Level 2
Let z
2
be defined as in Table 2.1;
z
2
= z
2
(q) = (2P (q2)? P (q))1/2,
then the branch of the root is determined by requiring z
2
= 1 when q = 0.
Define the modular function ? to be:
? = ?(q) =
?24
2
?24
1
+ 64?24
2
.
By substituting the values of a = 64 and ? into (2.6) we define x for level 2 to be:
x = x(q) = ?(1? 64?) =
?24
1
?24
2
(?24
1
+ 64?24
2
)2
.
Theorem 4.8 in [10] gives:
z
2
= z
2
(x) =
2
F
1
(
1
4
,
3
4
; 1; x
)
.
The series (2.7) for  = 2 becomes:
?
1? 256x
?
?
k=0
(4k)!
(k!)4
(k + ?)xk =
?
2/N
2?
, (2.10)
when q = exp (?2?
?
N/2). And when q = ? exp (??
?
N), it becomes:
?
1? 256x
?
?
k=0
(4k)!
(k!)4
(k + ?)xk =
?
1/N
?
. (2.11)
Both series (2.10) and (2.11) converge for |x| < 1
256
. The series (2.10) holds for 9 values
of x and ? given in Table 3.3 and Table 3.4, while there are 7 series for 1/? satisfied
by (2.11) given in Table 3.5.
2.3.3 Level 3
Let z
3
be defined as in Table 2.1;
z
3
= z
3
(q) =
(
3P (q3)? P (q)
2
)
1/2
,
then the branch of the root is determined by requiring z
3
= 1 when q = 0.
Define the modular function ? to be:
? = ?(q) =
?12
3
?12
1
+ 27?12
3
.
13
By substituting the values of a = 27 and ? into (2.6) we define x for level 3 to be:
x = x(q) = ?(1? 27?) =
?12
1
?12
3
(?12
1
+ 27?12
3
)2
.
Theorem 4.8 in [10] gives:
z
3
= z
3
(x) =
2
F
1
(
1
3
,
2
3
; 1; x
)
.
The series (2.7) for  = 3 becomes:
?
1? 108x
?
?
k=0
(3k)!(2k)!
(k!)5
(k + ?)xk =
?
3/N
2?
, (2.12)
when q = exp (?2?
?
N/3). And when q = ? exp (??
?
N/3), it becomes:
?
1? 108x
?
?
k=0
(3k)!(2k)!
(k!)5
(k + ?)xk =
?
3/N
?
. (2.13)
Both series (2.12) and (2.13) converge for |x| < 1
108
. The series (2.12) holds for 12
values of x and ? given in Table 3.6, while there are 20 series for 1/? satisfied by (2.13)
given in Tables 3.7 and 3.8.
2.3.4 Level 4
Let z
4
be defined as in Table 2.1;
z
4
= z
4
(q) =
(
4P (q4)? P (q)
3
)
1/2
,
then the branch of the root is determined by requiring z
4
= 1 when q = 0. Define the
modular function ? to be:
? = ?(q) =
?8
4
?8
1
+ 16?8
4
.
By substituting the values of a = 16 and ? into (2.6) we define x for level 4 to be:
x = x(q) = ?(1? 16?) =
?8
1
?8
4
(?8
1
+ 16?8
4
)2
. (2.14)
Theorem 4.8 in [10] gives:
z
4
= z
4
(x) =
2
F
1
(
1
2
,
1
2
; 1; x
)
.
The series (2.7) for  = 4 becomes:
?
1? 64x
?
?
k=0
(2k)!3
(k!)6
(k + ?)xk =
?
1/N
?
, (2.15)
14
when q = exp (??
?
N). And when q = ? exp (??
?
N), it becomes:
?
1? 64x
?
?
k=0
(2k)!3
(k!)6
(k + ?)xk =
?
1/N
?
. (2.16)
Both series (2.15) and (2.16) converge for |x| < 1
64
. The series (2.15) holds for 8 values
of x and ? given in Table 3.9, while there are 9 series for 1/? satisfied by (2.16) given
in Table 3.10.Series for 1/? of other levels are discussed in the conclusion.
15
Chapter 3
Methodology: Maple
In our search for series for 1/?, we performed numerical experiments using Maple to
compute the values of x and ?. For both cases q < 0 and q > 0, and for  = 1, 2, 3, 4,
we examined every degree up to N = 1000. Let x be defined by (2.6). As N is a
positive integer, it is known that x is an algebraic number [6]. Also, using the identity
(2.7), and by Theorem 2.3.1 ? is an algebraic number.
Example 3.0.4
If  = 4, q = e??
?
N , and N = 5,
then x =
9
64
?
1
16
?
5, (3.1)
and ? =
1
4
?
1
20
?
5. (3.2)
To determine x and ? in Maple, we use the command identify to give the corre-
sponding algebraic numbers. Simply, if the input is a floating-point constant, then
the identify command searches for an exact expression for the number. For example, if
x = 1.732050808, using identify command determines that this is the constant x =
?
3.
> x := 1.732050808;
> identify(x);
(1/2)
3
For the identify command to succeed, enough digits must be provided to approximate
the number.
For finding x in Example 3.0.4: we define the eta function by (2.5) and x for level 4 is
16
given by (2.14). Then we convert the x as series of q into polynomial and substitute
q = ??
?
5 into x. For this example, 46 digits were needed to determine x, at less than
46 the identify command does not give the quadratic irrational number (3.1).
In Maple, we write the code:
> e := proc (n) local i;
sum((-1)^i*q^((1/24)*(6*i-1)^2*n), i = -20 .. 20) end:
> x := convert(series
(e(1)^8*e(4)^8/(e(1)^8+16*e(4)^8)^2, q, 200), polynom);
> Digits := 46;
> evalf(subs(q = exp(-Pi*sqrt(5)), x));
0.0008707514062631439744266457042952352849613525244
> identify(%);
9 1 (1/2)
-- - -- 5
64 16
Although there is a theoretical formula for ?, it?s too complicated?indeed, impractical?
to use in practice. Therefore, it is computed last, by summing the series. From (2.7),
we can write ? as
? =
(
?
/N
2?
?
1?4ax
)
? s
1
s
2
,
where
s
1
=
?
?
k=0
(
2k
k
)
s(k) k xk,
and
s
2
=
?
?
k=0
(
2k
k
)
s(k) xk.
For finding ? in Example 3.0.4, we only need 15 digits of x to determine ?. Using less
than 15 digits, does not identify the quadratic irrational expression for ? in (3.2). The
following code is used in Maple to compute ?:
17
> Digits := 15;
> l := 4;
> n := 5;
> x := 0.000870751406263149;
> s1 := 0;
> s2 := 0;
> k := 0;
> t := 1;
> while abs(t) > 10^(-15) do
s1 := evalf(s1+t*k):
s2 := evalf(s2+t):
t := t*x*(2*k+1)^3*2^3/(k+1)^3:
k := k+1:
do:
> k;
11
> evalf((sqrt(l/n)/(2*Pi*sqrt(1-64*x))-s1)/s2);
0.138196601125012
> identify(%);
1 1 (1/2)
- - -- 5
4 20
Numbers of digits and the terms in the series are modified depending on the speed
of the convergence of the series. Some values of x and ? are easily computed using
identify with a small number of digits, for instance, Example 3.0.4; we need 46 digits
to identify x and 15 digits to identify ?. However some series need bigger numbers of
digits to be identified. These series have much slower convergence.
For example: The series of 1/? of level 4 and q = ?e??
?
N that corresponds to N = 7
needed only 30 digits to identify x and ?. While the series of level 3 and q = ?e?2?
?
N/3
that corresponds to N = 13 needed 200 digits to identify x and ?.
In some cases the command identify does not work for finding the algebraic number
of ?, and the command PSLQ can be used. Simply, when given a list (or a Vector) v
18
of n real numbers, the PSLQ(v) command outputs a list (or a vector) u of n integers
such that
?n
i=1 uivi is minimized. Thus the PSLQ function finds an integer relation
between a vector of linearly dependent real numbers if the input has enough precision.
To find ? we take the numerical value of ? and include it in a vector with 1, and
the same square root in x. This PSLQ command produces a minimal polynomial for
? that can be solved to give a quadratic irrational value of ?.
Example 3.0.5 The series of 1/? of level 3 and q = e?2?
?
N/3 that corresponds to
N = 3 was found by using PSLQ.
We used this code in Maple:
> Digits := 20:
> lambda := .14854314511050557756:
> with(IntegerRelations);
[LLL, LinearDependency, PSLQ]
> v := [lambda, 1, sqrt(3)]:
> u := PSLQ(v);
[22, -5, 1]
The PSLQ produces the vector u, and
?
ui vi ? i = 1, 2, 3 is minimized. If ? is a
relation linear combination of 1 and
?
3, then since PSLQ produces the minimum, we
have:
22?? 5 +
?
3 = 0, solving for ? gives ? = 5
22
?
1
22
?
3.
Another way to determine the values of x and ?, is to use the modular equations of
degree N that is satisfied by x. This method is illustrated in [6]. However, this method
was only practical for small degrees of N .
Example 3.0.6 The values of x and ? for the series of level 2 and degree N = 3 is
found by using the modular equations
Consider the level  = 2 and degree N = 3. Then x = x(q) and v = x(q3) satisfy the
modular equation:
x4 + v4 + 49152(x4v + xv4) + 905969664(x4v2 + x2v4) + 7421703487488
(x4v3 + x3v4) + 22799473113563136x4v4 ? 19332(x3v + xv3) + 526860288
(x3v2 + x2v3)? 918519021568x3v3 + 312(x2v + xv2) + 362266x2v2 ? xv = 0.
19
For the value q = e?2?
?
3/2 ,  = 2 and N = 3 we have:
v = x(q3) = x
(
e?2?
?
3/2
)
= x
(
e?2?/
?
6
)
= x,
and the modular equation simplifies and factorizes to:
x2(144x+ 1)(2304x? 1)(1024x+ 1)2(?1 + 256x)2 = 0. (3.3)
We deduce that:
x ? {0,
?1
44
,
1
2304
,
?1
1024
,
1
256
}.
By using numerical approximation to x to determine which root of (3.3) to select, we
have:
x
(
e?2?
?
3/2
)
= x
(
e?2?/
?
6
)
=
1
2304
.
Using Theorem 2.3.2 we have:
M(q) =
Z(q)
Z(q3)
=
u(q3)
u(q)
q d
dq
log x(q)
q d
dq
log x(q3)
,
=3
?
1? 256v
?
1? 256x
v
x
dx
dv
.
The derivative dx/dv can be calculated by differentiating the modular equation im-
plicitly and therefore M(q) becomes an algebraic function of x and v. Differentiating
M(q) with respect to x and substituting q gives:
dM
dx
?
?
?
?
q=(?2?/sqrt6)
= 1728.
Using the formula for ? from Theorem 2.3.2 gives:
? =
1/2304
6
? 1728 =
1
8
.
20
3.1 Results
In this section we first present tables of all the values of x and ? for series of 1/?. A
total of 90 series are included, 76 of which are believed to be new. In the discussion
section, we analyze these results and observe the important points and patterns.
Tables
This section contains tables of values of the parameters x and ?. Tables 3.1?3.10
with values of x and ? that give series for 1/? are organized according to the level
 ? {1, 2, 3, 4} and according to q > 0 or q < 0. Entries within each table are organized
according to the degree N . An asterisk ?*? next to the degree indicates that the series
was already given in [4]; there are 14 of these series and they belong to the levels
 ? {1, 2}. The other 76 series are believed to be new.
21
q N x ?
6? 1399
8489664
?
247
2122416
?
2 25
276
?
5
276
?
2
8 209
97336
?
18473
12167000
?
2 78
713
?
375
9982
?
2
10? 4927
210720960
?
51
4077880
?
5 83
1116
?
1
93
?
5
12 389
1706850
?
20213
161716500
?
3 2938
35673
?
250
11891
?
3
13 ? 3193
525614400
+ 41
24334000
?
13 103
1548
?
125
20124
?
13
e?2?
?
N 15 10968319
90769370526
?
122629507
2269234263150
?
5 1052
13629
?
20
1239
?
5
16 ? 190338695
34106789907
?
19939227
5052857764
?
2 10
99
?
10
231
?
2
22? 2914279
14627903313600
?
95403
677217746000
?
2 1013
19908
?
125
10428
?
2
28 8145698488
79267303539275
?
570145303
14679130284125
?
7 90352
1274007
?
59000
3822021
?
7
37 ? 9180598102
36565937573779944000
144743
3687948
?
2054375
955178532
?
37
+ 27940649
67714699210703600
?
37
58? 1399837865393267
68847549597949709007000000
6117973
195168708
?
4314395
2223529209
?
29
?
1203441508269
318738655546063467625000
?
29
Table 3.1: Irrational Series of level l = 1 for q = e?2?
?
N
22
q N x ?
35 9
20480
?
161
819200
?
5 873
8246
?
96
4123
?
5
75 ? 36983
588791808
+ 18377
654213120
?
5 6965
92158
?
800
46079
?
5
91 2927165
2173353984
?
60137
160989184
?
13 15
154
?
160
9009
?
13
-e??
?
N 99 9487199
28334096384
?
18166603
311675060224
?
33 2085
22078
?
120
11039
?
33
235? ? 2796157855
81160240398336
+ 578925837
37574185369600
?
5 1425191
22391082
?
214720
11195541
?
5
267? ? 1423834769537
76885078511663972352
2012743871
51659265306
?
33230537000
16091861142819
?
89
+ 18865772964857
9610634813957996544000
?
89
427? ? 26514389807073851
526643727429777408000
1594143977
25145035062
?
14138201440
2300770708173
?
61
+ 251468129201653
39010646476279808000
?
61
Table 3.2: Irrational Series of level l = 1 for q = ?e??
?
N
23
q N x ?
4 ? 457
19208
+ 325
19208
?
2 6
35
?
3
70
?
2
6 373
263538
?
425
527076
?
3 14
115
?
2
115
?
3
7 249
12544
?
11
784
?
2 9
56
?
3
56
?
2
8 ? 221
194481
+ 209
259308
?
2 626
4991
?
162
4991
?
2
e?2?
?
N/2 14 102376
855036681
?
38675
855036081
?
7 4528
50995
?
552
50995
?
7
15 6449
5531904
+ 95
115248
?
2 29
280
?
9
280
?
2
17 2177
20736
?
11
432
?
17 31
208
?
81
3536
?
17
21? 150889
300259584
?
1925
9383112
?
6 13
140
?
1
56
?
6
Table 3.3: Irrational Series of level l = 2 for q = e?2?
?
N/2
24
q N x ?
35? 18287
11757312
?
1265
2571912
?
10 1543
16120
?
147
8060
?
10
39? 243407089
5108829513984
?
3585725
106433948208
?
2 5983
83720
?
2097
83720
?
2
41 54600721
5802782976
?
88825
60445656
?
41 79
592
?
1863
121360
?
41
e?2?
?
N/2 51 ? 13062107489
1026527766038784
?
33000275
10692997562904
?
17 10735
171080
?
2763
363545
?
17
65? 86801836241
242945341583616
+ 112150445
2530680641496
?
65 9887
123830
?
5184
804895
?
65
95? 427925331521
5029625975060736
+ 6303935495
104783874480432
?
2 50572309
741263320
+ 23655969
741263320
?
2
Table 3.4: Irrational Series of level l = 2 for q = e?2?
?
N/2
25
q N x ?
17 103
32768
?
25
32768
?
17 3
20
?
3
170
?
17
21 ?97
9216
+ 7
1152
?
3 53
364
?
4
91
?
3
33 ? 1867
1179648
+ 325
1179648
?
33 73
620
?
37
3410
?
33
49 85
36864
?
253
290304
?
7 773
5828
?
48
1457
?
7
?e??
?
N 57 ? 542267
2832334848
+ 71825
2832334848
?
57 157
1820
?
101
17290
?
57
73 152107
4718592
?
160225
42467328
?
73 4511
32660
?
14073
1192090
?
73
85 ? 22861961
10809345024
+ 231035
450389376
?
17 1531
13780
?
7614
409955
?
17
177? ? 38480821035067
86466866929622581248
3562829
74444860
?
3950637
2196123370
?
177
+ 964131876175
28822288976540860416
?
177
253? - 12633605109401
680849241375744
+ 158715635975
28368718390656
?
11 7152017
57366140
?
995652
31551377
?
11
Table 3.5: Irrational Series of level l = 2 for q = ?e??
?
N
26
q N x ?
3 ?1
6
+ 7
72
?
3 5
22
?
1
22
?
3
6 463
125000
?
91
62500
?
6 33
230
?
3
230
?
6
7 ? 17
2916
+ 13
5832
?
7 1
6
?
1
42
?
7
8 ?265
108
+ 17
12
?
3 29
138
?
3
46
?
3
e?2?
?
N/3 10 223
157464
?
35
78732
?
10 11
90
?
1
90
?
10
11 ? 97
71879
+ 25
31944
?
3 43
330
?
3
110
?
3
13 17743
39366
?
4921
39366
?
13 10
51
?
22
663
?
13
19 ? 4261
22781250
+ 3913
91125000
?
19 49
510
?
73
9690
?
19
20 205694
47832147
?
13195
5314683
?
3 128
1065
?
12
355
?
3
31 ? 684197
33215062500
+ 245791
66430125000
?
31 1007
13530
?
1877
419430
?
31
34 3555313
278957081304
?
152425
69739270326
?
34 2033
28710
?
197
48807
?
34
59 ? 730612447
1641786993734274
+ 187475575
729683108326344
?
3 5472653
101652870
?
351123
33884290
?
3
Table 3.6: Irrational Series of level l = 3 for q = e?2?
?
N/3
27
q N x ?
2 ?265
108
?
17
12
?
3 the series diverges
3 ?1
6
?
7
72
?
3 the series diverges
4 ? 18551
421875
?
561
31250
?
6 the series diverges
7 ? 17
2916
?
13
5832
?
7 the series diverges
11 ? 97
71874
?
25
31944
?
3 43
330
+ 3
110
?
3
?e??
?
N/3 13 23
1458
?
7
1458
?
13 2
9
?
2
117
?
13
19 ? 4261
22781250
?
3913
91125000
?
19 49
510
+ 73
9690
?
19
31 ? 684197
33215062500
?
245791
66430125000
?
31 1007
13530
+ 1877
419430
?
31
33 ? 523
8192
+ 91
8192
?
33 96
493
?
93
5423
?
33
59 ? 730612447
1641786993734274
?
187475575
729683108326344
?
3 5472653
101652870
+ 351123
33884290
?
3
65 ? 649
1728
+ 5
48
?
13 83
435
?
64
1885
?
13
Table 3.7: Irrational Series of level l = 3 for q = ?e??
?
N/3
28
q N x ?
73 1555
46656
?
91
23328
?
73 73
207
?
202
15111
?
73
97 33161
5971968
?
3367
5971968
?
97 16
99
?
103
9603
?
97
121 24323
221184
?
1725
90112
?
33 8
45
?
1
45
?
33
145 ? 1769693
161243136
+ 146965
161243136
?
145 314
2115
?
523
62335
?
145
?e??
?
N/3 169 14339
1259712
?
12925
4094064
?
13 941
5865
?
64
1955
?
13
185 ? 14983669
3061257408
+ 68425
85034928
?
37 12079
91635
?
3392
226033
?
37
241 489978007
729000000
?
15781129
364500000
?
241 22051
124785
?
273526
30073185
?
241
265 ? 30814033
20155392000
?
1892891
20155392000
?
265 53662
498249
?
119771
26407197
?
265
409 77088425801
5971968000000
?
3811777333
5971968000000
?
409 1094279
7115265
?
18207572
2910143385
?
409
Table 3.8: Irrational Series of level l = 3 for q = ?e??
?
N/3
29
q N x ?
2 ?7
8
+ 5
8
?
2 2
7
?
1
14
?
2
4 ?35 + 99
4
?
2 2
7
?
2
21
?
2
5 9
64
?
1
16
?
5 1
4
?
1
20
?
5
e??
?
N 9 97
64
?
7
8
?
3 1
4
?
1
12
?
3
13 649
64
?
45
16
?
13 1
4
?
7
156
?
13
15 47
8192
?
21
8192
?
5 3
22
?
4
165
?
5
25 51841
64
?
1449
4
?
5 1
4
?
1
12
?
5
37 1555849
64
?
63945
16
?
37 1
4
?
101
3108
?
37
Table 3.9: Irrational Series of level l = 4 for q = e??
?
N
30
q N x ?
6 ?17
64
+ 3
16
?
2 1
4
?
1
12
?
2
7 ?2024 + 765
?
7 16
57
?
8
133
?
7
10 ?161
64
+ 9
8
?
5 1
4
?
1
15
?
5
12 ? 13
512
+ 15
1024
?
3 13
66
?
2
33
?
3
?e??
?
N 16 35
512
?
99
2048
?
2 3
14
?
2
21
?
2
18 ?4801
64
+ 245
8
?
6 1
4
?
1
14
?
6
22 ?19601
64
+ 3465
16
?
2 1
4
?
17
132
?
2
28 ?253
512
+ 765
4096
?
7 25
114
?
8
133
?
7
58 ?192119201
64
+ 4459455
8
?
29 1
4
?
37
957
?
29
Table 3.10: Irrational Series of level l = 4 for q = ?e??
?
N
31
3.2 Discussion
In Tables 3.1-3.10 the values of x and ? were presented to give a total of 90 series for
1/?. These series were obtained from the identity (2.7). It is obvious that the values of
x and ? in the quadratic irrational forms share the same square roots for each degree.
In most cases, it was a square root of the level, the degree or a divisor of the degree.
There are 4 examples where the conjugate series (i.e., the series obtained by replacing
each quadratic irrational with its conjugate) also gives a series for 1/?, but the con-
vergence is much slower. These series are from level 3 and correspond to the degrees
N = {11, 19, 31, 59}, where these values of N satisfy the condition N ? 3 (mod 4). As
for the convergence, all series found are convergent except for 4 series. These series are
from level 3 and correspond to N = {2, 3, 4, 7}. They diverge because x > 1
108
; and 1
108
is the radius of convergence for  = 3.
3.3 Conclusion
We have listed 90 Ramanujan-type series for 1/? that involve quadratic irrationals of
the levels  ? {1, 2, 3, 4}. Of the 90 series, 76 are believed to be new and 14 were
previously known. There are series of higher levels;  ? {5, 6, 7, 8, 9, 10}. The main
difference is that, instead of the 2-term recurrence relations in Theorem 2.3.1, the
underlying sequences satisfy three-term relation. Heun functions appears instead of
hypergeometric functions. Only six examples are known, and these are for levels 5,6 (3
cases), 8, and 9. For more details on rational series for 1/? of the levels  ? {5, 6, 8, 9}
see [6]. In addition, there are series of level  = 7, 10, 18, not investigated in this thesis
but details are in [12] and [13]. It is not known if there other series for 1/? that can
be obtained from the identity (2.7). Future work could involve rational and irrational
series for 1/? of other levels.
32
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