APPROXIMATIONS TO , DEDEKIND’S ETA FUNCTION AND …sequence ffing1 n=1 recursively for which lim n!1 fin = 1 …: Let p be an integer ‚ 1. We say a sequence ffing1 n=1 converges

APPROXIMATIONS TO π, DEDEKIND’S ETA FUNCTIONAND MODULAR EQUATIONS

By

AMITAVA GHOSH

A THESIS PRESENTED TO THE GRADUATE SCHOOLOF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT

OF THE REQUIREMENTS FOR THE DEGREE OFMASTER OF SCIENCE

UNIVERSITY OF FLORIDA

2007

1

c© 2007 Amitava Ghosh

2

To my most loving parents, Pranab Kumar Ghosh and Mita Ghosh

3

TABLE OF CONTENTS

page

LIST OF TABLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

ABSTRACT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

CHAPTER

1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1 Algorithms For Computing π . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 αp(r) and p-th Order Iteration Construction . . . . . . . . . . . . . . . . . 91.3 The Goal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2 DEDEKIND’S ETA-FUNCTION . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.1 The Function αp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.2 Modular Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.3 Transformation Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.4 Finding Values of αp(r) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3 MAPLE PROGRAMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.1 An Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.2 MAPLE Steps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.3 MAPLE Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4 EVALUATIONS OF εN,P (1/N) . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

BIOGRAPHICAL SKETCH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4

LIST OF TABLES

Table page

1-1 Iterations of the Borwein quartic algorithm . . . . . . . . . . . . . . . . . . . . . 8

1-2 Iterations of the Borwein quadratic algorithm . . . . . . . . . . . . . . . . . . . 11

4-1 Initial values αp(r0) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

4-2 Values of εN,p(1/N) for N = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37





5

Abstract of Thesis Presented to the Graduate Schoolof the University of Florida in Partial Fulfillment of the

Requirements for the Degree of Master of Science

APPROXIMATIONS TO π, DEDEKIND’S ETA FUNCTIONAND MODULAR EQUATIONS

By

Amitava Ghosh

August 2007

Chair: F.G. GarvanMajor: Mathematics

In 1989, Bailey, Borwein and Borwein gave a method for constructing series and

algorithms that converge to π to high order using Jacobi’s theta-functions. In 1997,

Borwein and Garvan developed a family of functions αp(r), defined in terms of Dedekind’s

eta-function, for constructing pth order algorithms that converge to π. Their method

involved finding initial values αp(r0) and finding certain modular equations. Their paper

included some experimental values of αp(r0). In this thesis a method for finding and

proving the validity of such evaluations is given. We confirm Borwein and Garvan’s

evaluations as well as giving some new ones.

6

CHAPTER 1INTRODUCTION

1.1 Algorithms For Computing π

In Bailey, Borwein and Borwein [3] an overview is given for a method for constructing

series and algorithms for the rapid computation of π. These methods involve defining a

sequence {αn}∞n=1 recursively for which

limn→∞

αn =1

π.

Let p be an integer ≥ 1. We say a sequence {αn}∞n=1 converges to α∞ to pth order or has

pth order convergence if

|αn+1 − α∞| ≤ c|αn − α∞|p

for some constant c > 0. For example, we give Borwein and Borwein’s [2, p.700] quartic

algorithm. Define three sequences {αn}, {sn}, {s∗n} by

α0 :=1

3

s0 =√

2− 1

s∗n = (1− s4n)1/4

sn =1− s∗n−1

1 + s∗n−1

αn = (1 + sn)4αn−1 +4n

3(1− (1 + sn)4).

Then αn converges quartically (4th order) to 1π. We illustrate this by MAPLE. We write

and execute the following program where we create a MAPLE function called bb4.

bb4:=proc(n,D)local a,m,i,diff1:Digits:=D:a:=1/3:s:=sqrt(2)-1:for i from 1 to n doss:=(1-s∧4)∧(1/4):s:=(1-ss)/(1+ss):a:=evalf((1+s)∧4*a+4∧i/3*(1-(1+s)∧4)):diff1:=evalf(abs(1/a-Pi)):

7

lprint(i,evalf(1/a),evalf(diff1,6));od:RETURN():end:

> read bb4:> it2(2,50);1, 3.1415926487741770159263140171966648531299770424765, .481562e-82, 3.1415926535897932384626433832795028841971331019203, .362975e-40

Table 1-1. Iterations of the Borwein quartic algorithm

n∣∣∣ 1αn− π

∣∣∣

1 10−8

2 10−40

3 10−170

4 10−694

5 10−2789

6 10−11171

We see that the digit accuracy quadruples with each iteration. This illustrates the 4th

order convergence. Computation of 1α6

gives π correctly to 11, 170 digits. It turns out that

in the algorithm above αn = α(16n.4) where α(r) is a certain function defined in terms of

the classical theta functions.

θ2(q) :=∞∑

n=−∞q(n+ 1

2)2

θ3(q) :=∞∑

n=−∞qn2

θ4(q) :=∞∑

n=−∞(−1)n.qn2

Then

α(r) :=

1π− 4

√rq θ4(q)

θ4(q)

θ43(q)

(1–1)

where q = exp(−π√

r) and θ4(q) = ddq

θ4(q). In [3], Borwein, Borwein and Bailey were able

to express α(p2r) in terms of α(r) and various theta functions. Utilizing p-order modular

8

equations for the theta functions, they were able to construct p-th order iterations that

converge to 1π.

1.2 αp(r) and p-th Order Iteration Construction

In [5], Borwein and Garvan defined a family of functions αp(r) for each p ≥ 2, and

showed how to construct p-th order iterations converging to 1π. The definition of αp(r) is

given in the next chapter. And it turns out that

αp(p2r) = αp(r).mp,p(r) +

√rp

3(1−mp,p(r))

where mp,p(r) is defined in (2-14). For a fixed initial r0 and p ≥ 2 define the sequence

{αn} by

αn := αp(r0p2n)

Then αn converges to 1π

to p-th order. To construct the corresponding p-th order iteration,

we need 3 ingredients :

(1) Initial value αp(r0)

(2) Initial value mp,p(r0)

(3) a modular equation giving mp,p(p2r) in terms of mp,p(r).

(1) We can always take r0 = 1p

and so α0 := αp(1p) = 1

3. See (2-20). (2) Also,

mp,p(1p) = p. See (2-19). (3) It turns out that there is always an algebraic relation between

mp,p(p2r) and mp,p(r).

EXAMPLE (p = 2)

m(q2) =4

1 +√

(4−m(q))(2 + m(q))(1–2)

Here, m(q) = m2,2(r), where q = exp(−2π√

r/√

2). We define two sequences {αn},{mn} by

{αn} := α2(22n−1),

9

and

{mn} := m2,2(22n).

Then by (1-2), we have

mn =4

1 +√

(4−mn−1)(2 + mn−1)(1–3)

and by (1-2), we have

αn = mn−1αn−1 +2n−1

3(1−mn−1) (1–4)

Equations (1-2), (1-3) give a quadratic iteration such that

limn→∞

αn =1

π

We illustrate this by MAPLE.

quadit:=proc(n,D)local a,m,i,diff1:Digits:=D:a:=1/3:m:=2:for i from 1 to n doa:=evalf(m*a + 2∧(i-1)/3*(1-m)):m:=4/(1 + sqrt( (4-m)*(2+m) )):diff1:=evalf(abs(1/a-Pi)):lprint(i,evalf(1/a),evalf(diff1,6));od:RETURN():end:

> read quadit;> quadit(5,45);1, 3.00000000000000000000000000000000000000000000, .1415932, 3.14075448203408147040144747789402692448829019, .838172e-33, 3.14159264682484879608649690272348368662752472, .676494e-84, 3.14159265358979323828694721365612084950195671, .175696e-185, 3.14159265358979323846264338327950288419711582, .5358e-40

We see that the digit accuracy at least doubles with each iteration. This illustrates

the quadratic convergence. Computation of 1α10

gives π to 1391 digits. A table of the first

10 iterations is given below in Table (1-2).

1.3 The Goal

In this thesis, we concentrate on the evaluation of αp(r0) which is the first step in

iteration construction discussed in the previous section. When N is a positive integer, we

10

Table 1-2. Iterations of the Borwein quadratic algorithm

n∣∣∣ 1αn− π

∣∣∣

1 < 10−1

2 < 10−3

3 < 10−8

4 < 10−18

5 < 10−40

6 < 10−83

7 < 10−170

8 < 10−344

9 < 10−693

10 < 10−1392

prove that αp(N) is algebraic. See Theorem 1.4.1. We also give an algorithm for finding

the minimal polynomial of αp(N) (see Chapter 2). In [5], some evaluations of αp(N) were

given without proof. We rigorously verify these evaluations and find new evaluations using

MAPLE. Tables of these new evaluations are given in Chapter 4.

11

CHAPTER 2DEDEKIND’S ETA-FUNCTION

2.1 The Function αp

In this section we describe Borwein and Garvan’s [5] infinite family of functions αp,

where p is any integer greater than 1. Theoretically it is possible to find an equation

relating αp(N2r) with αp(r). This relation is particularly nice when N = p. This will give

rise to p-th order iterations with a nice form. The functions αp are constructed from the

Dedekind eta function instead of the theta functions.

Let q := exp(2πiτ) (with =τ > 0). As usual the Dedekind eta function is defined as

η(τ) := exp(πiτ/12)∞∏

n=1

(1− exp(2πinτ)) (2–1)

= q1/24

∞∏n=1

(1− qn) (2–2)

Then

η(−1/τ) =

√τ

iη(τ). (2–3)

See [1, p.121] for a proof. Now for p > 1 (a positive integer) we define

Bp(r) :=ηp(τ)

η(pτ), Cp(r) :=

ηp(pτ)

η(τ). (2–4)

where τ = i√

r/√

p and q = exp(−2π√

r/√

p). It should be noted that the functions B3

and C3 occured naturally in the Borwein-Borwein cubic iteration [2], [4]. Define

αp(r) :=

(1π− q 8

√r

(p−1)√

pBB

)

Ap(r), (2–5)

where

Ap(r) := q

(24

p2 − 1

) {C

C− B

B

}. (2–6)

Here B = dBdq

. From (2–2) we have

Ap(r) = 1 + O(q), (2–7)

12

and

Ap(1/r) = rAp(r), (2–8)

which follows from (2–3). The definition of αp was chosen so that it had a form analogous

to that of (1–1) and that it satisfied a transformation like (1–1) below. Using (2–3) and

(2–8) it is not hard to show that

αp(1/r) =

(p+1)3√

p

√r − αp(r)

r. (2–9)

Substituting r = 1 gives

αp(1) =p + 1

6√

p, (2–10)

Since q → 0 as r →∞ we see that

limr→∞

αp(r) =1

π. (2–11)

The following theorem is given in [5, p.93].

Theorem 2.1.1. Let N, p ≥ 1 be fixed. We have

αp(N2r) = αp(r)mN,p(r) +

√rεN,p(r), (2–12)

where

εN,p =p + 1

3√

p

{q B

B−NqN B

B(qN)

qN CC(qN)− qN B

B(qN)

}, (2–13)

and

mN,p =Ap(r)

Ap(N2r). (2–14)

Further

Ap =1

p− 1[pP (qp)− P (q)] , (2–15)

where

P (q) := 1− 24∞∑

n=1

nqn

1− qn= 24q

η

η. (2–16)

13

Proof. The statement (2–12) follows easily from (2–5) and (2–6). Equation (2–15) follows

easily from the product expansion (2–2) and the definition in (2–4).

When N = p, the function εN,p has a nice form

εp,p(r) =

√p

3(1−mp,p(r)) , (2–17)

so that

αp(p2r) = αp(r)mp,p(r) +

√rp

3(1−mp,p(r)) . (2–18)

The proof of (2–17) follows easily from (2–2), (2–4) and (2–15). From (2–8) and (2–14) we

have

mp,p(1/p) = p. (2–19)

By using (2–18) and (2–19) we find that (2–9) and (2–18) give rise to two equations

involving αp(p) and αp(1/p). These equations may be solved easily to yield

αp(p) = αp(1/p) =1

3(Not a bad starting point for 1/π). (2–20)

We know that

η(−1/τ) =

√τ

iη(τ). (2–21)

Now let’s prove

Ap(1/r) = rAp(r), (2–22)

PROOF : (Step1)

Let’s find Bp(−1/pτ) in terms of Cp(τ).

Now,

Bp

(− 1

pτ

)=

ηp(− 1pτ

)

η(p · −1pτ

)(2–23)

=ηp(− 1

pτ)

η(− 1τ)

=ηp(− 1

pτ)√

τiη(τ)

14

=(√

pτi)pηp(pτ)√τiη(τ)

=(√

pτi)p

√τi

Cp(τ)

Step 2: Let’s find Cp(−1/pτ) in terms of Bp(τ).

Now,

Cp

(− 1

pτ

)=

ηp(p.−1pτ

)

η(−1pτ

)(2–24)

=ηp(− 1

τ)

η(− 1pτ

)

=(√

τi)p ηp(τ)

η(− 1pτ

)

=(√

τi)p ηp(τ)√

pτi

η(pτ)

=(√

τi)p

√pτi

Bp(τ)

Step 3: Let’s findB′pBp

(− 1pτ

) in terms of τ, Cp(τ) and C ′p(τ). (please note that the sign ′

means ddτ

). If we calculate Bp

Bp(− 1

pτ), we will see that:

B′p

Bp

(− 1

pτ

)=

p(p− 1)τ

2+ pτ 2

C ′p

Cp

(τ) (2–25)

Now it’s easy to see that as r goes to 1r, p is mapped to − 1

pτ.

Step 4: Let’s find q Bp

Bp(1

r) in terms of r and Cp(r).

qBp

Bp

(1

r) = p(−r

p)q

Cp

Cp

(r) +p(p− 1)

2.2πi.i√

r√p

= −r(qCp

Cp

) +

√p(p− 1)

√r

4π

Step 5:

Let’s find q Cp

Cp(1

r) in terms of r and Bp(r).

We know that :

15

C(− 1

pτ) = τ

p−12 i

1−p2 p−

12 Bp(τ)

C ′(− 1

pτ) = i

1−p2 p−

12 [

p− 1

2τ

p−32 Bp(τ) + τ

p−12 B′

p(τ)]

B′p

Bp

(− 1

pτ

)= p(p− 1)τ + pτ 2

C ′p

Cp

(τ)

qBp

Bp

(1

r) = −r(q

Cp

Cp

) +

√p(p− 1)

√r

4π

So,

Ap(1

r) =

24

p2 − 1.q(

C

C(1

r)− B

B(1

r))

=24

p2 − 1r(q

C

C− q

B

B)

= r(24

p2 − 1q(

C

C− B

B))

So, Ap(1r) = rAp(r).

We can show that

αp(1

r) =

p+13√

p

√r − αp(r)

r.

When r = 1, we have

αp(1) =p + 1

6√

p.

Since q → 0 as r →∞, we see that :

limr→∞

αp(r) =1

π.

Now, we define

mN,p =Ap(r)

Ap(N2r)

16

When N = p and r = 1p,

mp,p =Ap(

1p)

Ap(p2.1p)

=Ap(

1p)

Ap(p)=

p.Ap(p)

Ap(p)= p

Also, we have :

αp(p2r) = αp(r)mp,p(r) +

√rp

3(1−mp,p(r))

When r = 1p, we get :

αp(p2.1

p) = αp(

1p)mp,p(

1p) +

√1p.p

3(1−mp,p(

1p))

or, αp(p) = αp(1p) + 1

3(1− p)

or, (3p)αp(1p) = 3αp(p) + (p− 1)

Also, we know that :

αp(1

r) =

p+13√

p.√

r − αp(r)

r

Putting r = p, we have :

αp(1

p) =

p+13− αp(p)

p

Simplifying the expression we get :

3p.αp(1

p) = p + 1− 3αp(p)

From the two equations, we have :

αp(p) =1

3= αp(

1

p)

Now, let’s prove

P (q) := 1− 24∞∑

n=1

nqn

1− qn= 24q

η

η

PROOF: We know that

η = q1/24.

∞∏n=1

(1− qn)

17

log η =1

24log(q) +

∞∑n=1

log(1− qn)

Now, differentiating both sides w.r.t q,

η

η=

1

24q+

∞∑n=1

−nqn−1

1− qn, where η =

dη

dq

or,

qη

η=

1

24− q

∞∑n=1

nqn−1

1− qn

or,

24η

η= 1− 24

∞∑n=1

nqn

1− qn

Also let’s prove :

Ap =1

p− 1[pP (qp)− P (q)]

If we simplify the right hand side, we get :

p

p− 1(1− 24

∞∑n=1

nqpn

1− qpn)− 1

p− 1(1− 24

∞∑n=1

nqn

1− qn)

= 1− 24p

p− 1

∞∑n=1

nqpn

1− qpn+

24

p− 1

∞∑n=1

nqn

1− qn

Now’s let’s evaluate the left hand side (LHS) i.e. Ap. We know that

Ap(r) =24q

p2 − 1(C

C− B

B)

Now,

Bp(τ) =ηp(τ)

η(pτ)=

qp/24∏∞

n=1(1− qn)p

qp/24∏∞

n=1(1− qpn)

=

∏∞n=1(1− qn)p

∏∞n=1(1− qpn)

So, log Bp = p∑∞

n=1 log(1− qn)−∑∞n=1 log(1− qpn)

18

Differentiating both sides w.r.t. q, we get the following :

Bp

Bp

= p∞∑

n=1

−nqn−1

1− qn−

∞∑n=1

−pnqpn−1

1− qpn

= −p

∞∑n=1

nqn−1

1− qn+ p

∞∑n=1

nqpn−1

1− qpn

Now,

Cp(τ) =ηp(pτ)

η(τ)

=qp2/24

∏∞n=1(1− qpn)p

q1/24∏∞

n=1(1− qn)

So,

log(Cp) = (p2 − 1

24) log(q) + p

∞∑n=1

log(1− qpn)−∞∑

n=1

log(1− qn)

Differentiating both sides w.r.t. q, we get :

Cp

Cp

=p2 − 1

24q+ p

∞∑n=1

−pn.qpn−1

1− qpn+

∞∑n=1

n.qn−1

1− qn

So, now we have :

Ap(r) =24q

p2 − 1(C

C− B

B) =

24

p2 − 1(q

C

C− q

B

B)

=24

p2 − 1[p2 − 1

24− p2

∞∑n=1

n.qpn

1− qpn+

∞∑n=1

n.qn

1− qn+ p

∞∑n=1

n.qn

1− qn− p

∞∑n=1

n.qpn

1− qpn]

=24

p2 − 1[p2 − 1

24− p(p + 1)

∞∑n=1

n.qpn

1− qpn+ (p + 1)

∞∑n=1

n.qn

1− qn

= 1− 24p

p− 1

∞∑n=1

n.qpn

1− qpn+

24

p− 1

∞∑n=1

n.qn

1− qn

= Right Hand Side (hence proved)

2.2 Modular Forms

Let H denote the complex upper half plane. SL2(Z) acts transitively on H by linear

fractional transformations

19

Aτ =aτ + b

cτ + d, where A =

a b

c d

∈ SL2(Z),

and τ ∈ H. Let N be a positive integer. We define

Γ(N) :=

a b

c d

∈ SL2(Z) : a ≡ d ≡ 1 (mod N) and b ≡ c ≡ 0 (mod N)

,

so that Γ(1) = SL2(Z). A subgroup of Γ(1) is called a congruence subgroup of level N if it

contains Γ(N). We will be concerned with the following congruence subgroup

Γ0(N) :=

a b

c d

∈ Γ(1) : c ≡ 0 (mod N)

Let k be a non-negative integer and let Γ′ ⊂ Γ(1) be a congruence subgroup of level N

with [Γ(1) : Γ′] < ∞. Also, Γ(N) ⊂ Γ′. A function f : H → C is a modular form of weight

k for Γ′ if

(i) f is holomorphic on H,

(ii) f(Aτ) = (cτ + d)kf(τ) for all A =

a b

c d

∈ Γ′,

(iii) f is holomorphic at the cusps A(∞) for A ∈ Γ(1)

i.e. there is an expansion

(cτ + d)−kf(Aτ) =∞∑

n=0

ane2πinτ/N

for Im(τ) > δ.

We let Mk(Γ′) be set of such modular forms. It turns out that this is a finite

dimensional C -vector space. We will need the valence formula

∑

ζ∈F∗ord(f, ζ, Γ′) =

k[Γ(1) : Γ′]12

20

if f 6= 0, f ∈ Mk(Γ′) and F∗ is a proper fundamental region for Γ′. See Theorem 4.14

(p.98) of Rankin’s book [9]. We will need

[Γ(1) : Γ0(N)] = N∏

p|N(1 +

1

p)

See Theorem 4.2.5 (p.106) of Miyake’s book [8].

2.3 Transformation Formulas

We are going to assume throughout that p is prime. We define the following

Fp(τ) := pη′(pτ)

η(pτ)− η′(τ)

η(τ)

Here, f ′(τ) = ddτ

f(τ). If f(τ) = F (q) where q = e2πiτ then

f ′(τ) = 2πiqdF

dqand

qdF

dq=

1

2πif ′(τ)

We note that

Fp(τ) =1

p + 1

(C ′

p

Cp

(τ)− B′p

Bp

(τ)

)

Let A =

a b

c d

∈ Γ(1), so that

Aτ =aτ + b

cτ + d

and

d(Aτ)

dτ=

1

(cτ + d)2

We need the following

η(Aτ) = ε(a, b, c, d)√−i(cτ + d)η(τ)

See theorem 3.4 (p.52) of Tom Apostol’s book on Modular Functions and Dirichlet

Series [1].

21

Proposition 2.3.1. For p prime, Fp(τ) is a modular form of weight 2 on Γ0(p).

Proof. Let A =

a b

c d

∈ Γ(1). Then we get the following

η′(Aτ)dA

dτ= ε

√−i(cτ + d)η′(τ)

−icεη(τ)

2√−i(cτ + d)

so that

η′(Aτ)

η(Aτ).dA

dτ=

η′(τ)

η(τ)+

c/2

cτ + d

and

η′(Aτ)

η(Aτ)= (cτ + d)2.

η′(τ)

η(τ)+

1

2c(cτ + d) (2–26)

Now let

B =

α β

γ δ

∈ Γ0(p)

so that

B∗ =

α pβ

γ/p δ

∈ Γ(1),

p 0

0 1

B = B∗

p 0

0 1

,

pB(τ) = B∗(pτ)

We have the following

η′(Bτ)

η(Bτ)= (γτ + δ)2.

η′(τ)

η(τ)+

1

2γ(γτ + δ)

Now,

pη′(pBτ)

η(pBτ)= p

η′(B∗pτ)

η(B∗pτ)

= p[(γτ + δ)2η′(pτ)

η(pτ)+

1

2.γ

p(γτ + δ)]

22

= p(γτ + δ)2η′(pτ)

η(pτ)+

γ

2(γτ + δ)

Hence, Fp(Bτ) = p(γτ + δ)2 η′(pτ)η(pτ)

− (γτ + δ)2 η′(τ)η(τ)

i.e. Fp(Bτ) = (γτ + δ)2Fp(τ)

So, Fp(τ) is a modular form of weight 2 on Γ0(p).

Proposition 2.3.2. For p prime and N any integer ≥ 2, define

Gp,N(τ) :=B′

p(τ)

Bp(τ)−N

B′p(Nτ)

Bp(Nτ)

Then, Gp,N(τ) is a modular form of weight 2 on Γ0(pN)

Proof. Let A =

a b

c d

∈ Γ0(p)

Then we have

Bp(Aτ) = χ(d)(cτ + d)p−12 Bp(τ),

where χ(d) = (dp) (Legendre symbol mod p). See Corollary (2.2.12) in [6].

So,

B′p(Aτ)

dA

dτ= χ(d)(cτ + d)

p−12 B′

p(τ) + χ(d)(p− 1

2)(cτ + d)

p−32 Bp(τ)

andB′

p(Aτ)

Bp(Aτ)= (cτ + d)2

B′p(τ)

Bp(τ)+

p− 1

2c(cτ + d)v

Let A =

α β

γ δ

∈ Γ0(pN) so that

NAτ =Nατ + Nβ

γτ + δ=

α(Nτ) + NβγN

(Nτ) + δ

and

A∗ =

α Nβ

γ/N δ

∈ Γ0(p)

23

So,B′

p(Aτ)

Bp(Aτ)= (γτ + δ)2

B′p(τ)

Bp(τ)+

p− 1

2γ(γτ + δ)

NB′

p(NAτ)

Bp(NAτ)= N(γτ + δ)2

B′p(Nτ)

Bp(Nτ)+

p− 1

2γ(γτ + δ)

So, Gp,N(Aτ) = (γτ + δ)2Gp,N(τ)

Therefore, Gp,N(τ) is a modular form of weight 2 on Γ0(pN).

Proposition 2.3.3. Let p be prime and N ≥ 1. Then Fp(− 1pτ

) = −pτ 2Fp(τ),

Gp,N(− 1

pNτ) = pNτ 2

(N

C ′p(Nτ)

Cp(Nτ)− C ′

p(τ)

Cp(τ)

)

Proof. By (2–26),

η′(− 1τ)

η(− 1τ)

= τ 2.η′(τ)

η(τ)+

τ

2

So,

Fp

(− 1

pτ

)= p

η′(− 1τ)

η(− 1τ)−

η′(− 1pτ

)

η(− 1pτ

)

= p [ τ 2.η′(τ)

η(τ)+

τ

2]− [ p2τ 2.

η′(pτ)

η(pτ)+

pτ

2]

= −p [ pτ 2.η′(pτ)

η(pτ)− τ 2.

η′(τ)

η(τ)]

= −pτ 2Fp(τ)

By (2–25),

Gp,N(− 1

pNτ) =

B′p(− 1

pNτ)

Bp(− 1pNτ

)−N

B′p(− 1

pτ)

Bp(− 1pτ

)

= [pN2τ 2(NC ′

p(Nτ)

Cp(Nτ)+

p(p− 1)

2Nτ ] − N [ pτ 2

C ′p(τ)

Cp(τ)+

p(p− 1)τ

2]

24

= pNτ 2[NC ′

p(Nτ)

Cp(Nτ)− C ′

p(τ)

Cp(τ)].

Proposition 2.3.4. Let p be prime and N ≥ 1. Then Fp(Nτ) is a modular form of weight

2 on Γ0(pN).

Proof. Let A =

α β

γpN δ

∈ Γ0(pN), so that A∗ =

α Nβ

γp δ

∈ Γ0(p)

and

NAτ =N(ατ + β)

γpNτ + δ=

α(Nτ) + Nβ

γp(Nτ) + δ

So,

Fp(NAτ) = (γp(Nτ) + δ)2Fp(Nτ)

which implies that Fp(Nτ) is a modular form of weight 2 on Γ0(pN).

2.4 Finding Values of αp(r)

The next proposition shows that computing αp(N) reduces to computing εN,p(1N

).

Proposition 2.4.1.

αp(N) =1

2√

N

((p + 1)N

3√

p+ εN,p(

1

N)

)

Proof. From (I-8), (I-14) we have

mN,p(1

N) =

Ap(1N

)

Ap(N)= N

From (I-9) and (I-12), we have

αp(N) = αp(1

N)mN,p(

1

N) +

1√N

εN,p(1

N)

= Nαp(1

N) +

1√N

εN,p(1

N)

25

=(p + 1)

√N

3√

p− αp(N) +

1√N

εN,p(1

N)

and the result follows.

εN,p =p + 1

3√

pεN,p

where

εN,p(r) =q B

B(q)−NqN B

B(qN)

qN CC(qN)− qN B

B(qN)

=Gp,N(τ)

(p + 1)Fp(Nτ)

Here using q = e2πiτ , τ = i√

r√p

Now,

εN,p

(1

N2r

)=

Gp,N(− 1pNτ

)

(p + 1)Fp(− 1pτ

)

= N

Cp

Cp(τ)−N Cp

Cp(Nτ)

Cp

Cp(τ)− Bp

Bp(τ)

= N

q Cp

Cp(q)−NqN Cp

Cp(qN)

q Cp

Cp(q)− q Bp

Bp(q)

We note that

εN,p(r) = εN,p(1

N2r)

when r = 1N

. We let

Gp,N(τ) = NC ′

p

Cp

(Nτ)− C ′p

Cp

(τ).

Then Gp,N(τ) is a modular form of weight 2 on Γ0(pN).

Theorem 2.4.2. Let p be prime and N ≥ 1. For sufficiently large `, there is a polynomial

P (x) ∈ C[x] of degree 2` such that

P (εN,p(1/N)) = 0

26

Proof. Let p, N be fixed. For ` ≥ 1, consider the set of monomials

Gp,N(τ)i((p + 1)Fp(Nτ))`−i(NGp,N(τ))j(−(p + 1)Fp(τ))`−j (2–27)

for 0 ≤ i, j ≤ `. The number of such monomials is (` + 1)2. Each monomial is a modular

form of weight 4` on Γ0(pN). The dimension of this space of modular forms is ∼ `. So for

sufficiently large `, the monomials in (2–27) are linearly dependent. Since the coefficients

of all q-expansions are rational, there exist aij ∈ Q (not all zero) such that

∑

0≤i,j≤`

ai,jGp,N(τ)i((p + 1)Fp(Nτ))`−i(NGp,N(τ))j(−(p + 1)Fp(τ))`−j = 0

and

P

(εN,p(r), εN,p(

1

N2r)

)= 0. (2–28)

where

P (x, y) =∑

0≤i,j≤`

aijxiyj (2–29)

Therefore, x = εN,p(1N

) satisfies the polynomial

P (x) = P (x, x) =∑

0≤i,j≤`

aijxi+j = 0.

From the proof of the theorem, we see that we need to find a polynomial

Q(x1, x2, y2, y2) =∑

0≤i,j≤`

(ai,j)(xi1)(x

`−i2 )(yj

1)(y`−j2 ) (2–30)

with rational coefficients such that

Q(Gp,N(τ), (p + 1)Fp(Nτ), NGp,N(τ),−(p + 1)Fp(τ)

)= 0

27

Because of the transformation formulas, we assume that aij = aji for all i, j. So, now

we consider monomials


for 0 ≤ i, j ≤ ` and


+ Gp,N(τ)j((p + 1)Fp(Nτ))`−j(NGp,N(τ))i(−(p + 1)Fp(τ))`−i

= Gp,N(τ)i((p + 1)Fp(Nτ))`−j(NGp,N(τ))i(−(p + 1)Fp(τ))l−j

(((p + 1)Fp(Nτ))j−i(NGp,N(τ))j−i + Gp,N(τ)j−i(−(p + 1)Fp(τ))j−i)

for 0 ≤ i < j ≤ `.

We can use MAPLE to find the polynomial in (2–30). We compute the q-expansion

up to qT for each monomial hk(τ) in (2–31) and (2–32).

hk(τ) =T∑

j=0

bj,k qj + O(qT+1) (2–33)

for 1 ≤ k ≤ `(`+1)2

. Here, T > `(`+1)2

.

Let

B = (bj,k)0≤j≤T, 1≤k≤ `(`+1)2

. (2–34)

We choose l large enough so that B has a non-trivial nullspace. We use MAPLE to

find a vector ~x ∈ Qk with ~x ∈ N(B). This gives a linear relation between the monomials

hk(τ) and the polynomial in (2–30). If the functions hk(τ) are linearly independent then

some linear combination

h(τ) =∑

βkhk(τ) 6= 0. (2–35)

satisfies vi∞(h) ≥ `(`+1)2

.

28

By the valence formula,

`(` + 1)

2≤ vi∞(h) ≤

∑

ζ∈F∗ord(h, ζ, Γ′) =

4l[Γ(1) : Γ′]12

(2–36)

=`Np

3

∏

t|Np, t prime

(1 +

1

t

), (2–37)

where Γ′ = Γ0(pN).

So,

` + 1 ≤ 2Np

3

∏

t|Np, t prime

(1 +

1

t

)(2–38)

Hence if

` ≥ 2Np

3

∏

t|Np, t prime

(1 +

1

t

)(2–39)

the functions hk(τ) will be linearly dependent. Now for a given `, T with T > `(`+1)2

, we

assume that we have used MAPLE to find ~x ∈ N(B) and hence a candidate polynomial

Q(x1, x2, y1, y2).

Let L(τ) = Q(Gp,N(τ), (p + 1)Fp(Nτ), NGp,N(τ),−(p + 1)Fp(τ)

).

Then,

L

(− 1

pNτ

)= p2`N2`τ 4`L(τ) (2–40)

since we have assumed that the coefficients aij of the polynomial Q are symmetric in i and

j. It follows that

vi∞(L) = v0(L). (2–41)

Hence to show that L = 0, we need only show that

vi∞(L) >`[Γ : Γ0(pN)]

6=

`Np

6

∏

t|Np, t prime

(1 +

1

t

). (2–42)

29

CHAPTER 3MAPLE PROGRAMS

3.1 An Example

As an example we show how to find and prove the evaluation

ε2,11(1/2) = −275√

2

102. (3–1)

First we compute the first 100 decimal places of ε2,11(1/2) and use the minpoly

MAPLE function to find a quadratic polynomial with small coefficients that “fits” this

evaluation. The minpoly function utilizes the Lenstra-Lenstra-Lovasz [7] lattice reduction

algorithm.

> x11:=evalf(epsilon0(2,11,1/2));

x11 := −3.8128306828686386119653372466437938392809780953790\

26667878303215171092466441955251802515411177465014

> minpoly(x11,2);

−75625 + 5202X2

> solve(%,2);

275

102

√2, −275

102

√2

From this calculation we suspect the evaluation given in (3–1). The code for all MAPLE

functions used is given in section 3.3. Our MAPLE function epsilon0(N, p, r) computes

εN,p(r) to a 100 decimal places.

First we compute q-expansions of

Gp,N(τ), Fp(Nτ), Gp,N(τ), Fp(τ)

for N = 2, p = 11.

> dop1(2,11,100):

30

Here, dop1(N, p, T ) computes the q-expansions of these functions up to qT . Next we

find ` large enough so that the matrix B in (1-31) has non-trivial nullspace.

> newdop3(5,2,11,101,60):

No polynomial relation found

Note that newdop3(`,N, p, t, T ) computes the nullspace of B modulo t for a large prime

t. Here, ` = 5 gives a zero nullspace and hence no polynomial relation mod t. By trial and

error, we use newdop3(`,N, p, t, T ) to find the correct candidate for `. Now, let’s do the

next step.

> newdop3(6,2,11,101,60):

This gives a polynomial relation mod t for ` = 6. We need to recalculate over Z. We

also need to find the minimum number of terms to prove the result.

> numterms(6,2,11);

36

Here, numterms(`,N, p) computes ` [Γ: Γ0(pN)]6

which is the right hand side of (1-41).

So we must take T > 36.

> P:= newdop2a(6,2,11,50);

We successfully found P (X,Y ) in this case.

> f:= factor(subs(Y=X,P));

f :=(−75625 + 5202X2)(18X2 − 1925)2(X − 10)6

1685448

The MAPLE function newdop2a(`, N, p, T ) finds the polynomial P (x, y) in (2–29) and

checks that (2–28) holds up to qT . If T > numterms(`,N, p), this calculation proves

that (2–28) holds. In the calculation above, we have proved that the minimal polynomial

of ε2,11

(12

)must divide the polynomial f . A little checking shows that the minimal

31

polynomial is

P (x) = 5202x2 − 75625

which proves the evaluation ε2,11

(12

)= −275

√2

102.

3.2 MAPLE Steps

To find and prove an evaluation of εN,p

(1N

)we perform the following steps.

STEP 1: Find numterms(`,N, p).

STEP 2: For T > numterms(`,N, p) use dop1(N, p, T ) to calculate the q-expansions

of

Gp,N(τ), Fp(Nτ), Gp,N(τ), Fp(τ)

upto qT .

STEP 3: Choose a large prime t and use newdop3(`,N, p, t, T ) for different values of

` until we find B with non-trivial nullspace mod t.

STEP 4: Use newdop2a(`,N, p, T ) to find and verify the polynomial P (x, y) in

(2–29).

STEP 5: Factor the polynomial P (x, x) = P (x). εN,p

(1N

)will be a root of one of the

factors. Use the MAPLE command evalf to find the correct factor.

STEP 6: If possible, try to factorize in order to identify εN,p

(1N

)in terms of radicals.

3.3 MAPLE Code

epsilon0:=proc(N,p,r)local BB,CC,qq;qq:=exp(-2*Pi*sqrt(r)/sqrt(p)):BB:=qlogdiff(B(p)):CC:=qlogdiff(C(p)):(BB(q)-N*BB(q∧N))/(CC(q∧N)-BB(q∧N))/(24/(p∧2-1)):subs(q=qq,%):end:

dop1:=proc(N,p,T)localEPL;globalX1,X2,Y1,Y2:

32

EPL:=NEWEPFUNCS(N,p,T):X1:=EPL[1]:X2:=EPL[2]:Y1:=EPL[3]:Y2:=EPL[4]:RETURN():end:

Digits:=100:with(polytools):etaq:=proc(q,i,trunk)local k,x,z1,z,w:z1:=(i + sqrt( i*i + 24*trunk*i ) )/(6*i):z:=1+trunc( evalf(z1) ):x:=0:for k from -z to z dow:=i*k*(3*k-1)/2:if w<=trunk thenx:=x+ q∧( w )*(-1)∧k:fi:od:RETURN(x):end:

findpoly2:=proc() #This proc looks for a polynomiallocalx,y,q,deg1,deg2,ARGLIST,TYPELIST,CORRECTL,check,dim1,dim2,num,k,j,B,qq,l,POLY,kkk,i,POLYg,POLYfunc,ss,numtyp,tA,opkk,POLYSET,INDLIST,COFMAT,indd,ind1,ind2;global A,X,Y,kk,AQ,COFMATSET:lprint(‘WARNING: X,Y are global.‘);## This version assumes coeff matrix is symmetric#relation between x,y of degree deg1 in x, and degree deg2 in y.#The relation is checked to order O(q∧check).if nargs<5 thenERROR(‘ number of arguments must be 5 or 6.‘);fi:x:=args[1]:y:=args[2]:q:=args[3]:deg1:=args[4]:deg2:=args[5]:ARGLIST:=[q,deg1,deg2];TYPELIST:=map(whattype,ARGLIST);POLYSET:=:INDLIST:=[]:COFMAT:=array(1..(deg1+1),1..(deg2+1)):COFMATSET:=:CORRECTL:=[symbol,integer,integer]:if TYPELIST=CORRECTL thenif nargs=6 thencheck:=args[6]:fi:if nargs>6 thenERROR(‘ findpoly can at most 6 arguments‘);fi:dim1:=(deg1+1)*(deg1+2)/2:dim2:=dim1+10:print(‘ dims ‘, dim1, dim2);A:=array(1..dim1,1..dim2):AQ:=array(1..(deg1+1),1..(deg2+1)):

33

num:=0:for k from 0 to deg1 dofor j from k to deg2 donum:=num+1:if i=j thenB[num]:=X∧k*Y∧j:qq:=series(x∧k*y∧j,q,dim2+10):elseB[num]:=X∧k*Y∧j+X∧j*Y∧k:qq:=series(x∧k*y∧j+x∧j*y∧k,q,dim2+10):fi:INDLIST:=[op(INDLIST),[k+1,j+1]]:AQ[k+1,j+1]:=degree(convert(qq,polynom),q):for l from 0 to (dim2-1) doA[num,l+1]:=coeff(qq,q,l):od:od:od:tA:=linalg[transpose](A):kk:=linalg[kernel](tA):if kk= thenlprint(‘ NO polynomial relation found. ‘);elsefor opkk in kk doPOLY:=0;##opkk:=op(kk);kkk:=opkk;for i from 1 to num doPOLY:=POLY+B[i]*kkk[i]:indd:=INDLIST[i]:ind1:=indd[1]:ind2:=indd[2]:COFMAT[ind1,ind2]:=kkk[i]:od:COFMATSET:=COFMATSET union COFMAT:POLYg:=0;for j from 0 to deg2 doPOLYg:=POLYg+factor(coeff(POLY,Y,deg2-j))*Y∧(deg2-j):od:if nargs=6 thenprint(‘The polynomial is‘);print(POLYg);POLYfunc:=unapply(POLYg,X,Y):ss:=series(POLYfunc(x,y),q,check+1):print(‘Checking to order‘,check);print(ss);##RETURN(POLYg):elseprint(‘The polynomial is‘);fi:POLYSET:=POLYSET union POLYg:od:RETURN(POLYSET):fi:elsenumtyp:=nops(TYPELIST):ERROR(‘ Wrong type of argument. ARGLIST has type ‘,seq(TYPELIST[i],i=1..numtyp), ‘It should have type‘,seq(CORRECTL[i],i=1..numtyp)); fi:end:

dop2:=proc(l,N,p,T)

34

local dd;dd:=dimpoly(l,p,N):print("dd=",dd);findpoly2(L1,L2,q,l,l,T);RETURN(%):end:newnewdop2:=proc()local dd,l,N,p,T;l:=args[1]:N:=args[2]:p:=args[3]:LP:=args[4]:if nargs=5 thenT:=args[5]:fi:dd:=dimpoly(l,p,N):print("dd=",dd);NL1:=modp(L1,LP):NL2:=modp(L2,LP):if nargs= 5 thennewfindpoly3(NL1,NL2,q,l,l,LP,T);elsenewfindpoly3(NL1,NL2,q,l,l,LP);fi:RETURN(%):end:

dimpoly:=proc(l,p,N)local S,x,t;S:=numtheory[factorset](p*N):x:=1:for t in S dox:=x*(1+1/t):od:RETURN(x*p*N*l/3);end:

cosetdim:=proc(N)local S,x,t;S:=numtheory[factorset](N):x:=1:for t in S dox:=x*(1+1/t):od:RETURN(x*N);end:

numterms:=proc(l,N,p)l*cosetdim(N*p)/6;end:

newdop2:=proc()local dd,l,N,T;global p:l:=args[1]:N:=args[2]:

35

p:=args[3]:if nargs=4 thenT:=args[4]:fi:dd:=dimpoly(l,p,N):print("dd=",dd);if nargs= 4 thennewfindpoly2([X1,X2],[Y1,Y2],q,l,l,T);elsenewfindpoly2([X1,X2],[Y1,Y2],q,l,l);fi:RETURN(%):end:

newdop2a:=proc()local dd,l,N,T;global p:l:=args[1]:N:=args[2]:p:=args[3]:if nargs=4 thenT:=args[4]:fi:dd:=dimpoly(l,p,N):print("dd=",dd);if nargs= 4 thennewfindpoly2a([X1,X2],[Y1,Y2],q,l,l,T);elsenewfindpoly2a([X1,X2],[Y1,Y2],q,l,l);fi:RETURN(%):end:

36

CHAPTER 4EVALUATIONS OF εN,P (1/N)

In this chapter we tabulate the results for evaluations of εN,p(1/N) using the methods

of Chapter 2.

Table 4-1. Initial values αp(r0)

p r0 αp(r0) p r0 αp(r0)

2 3 (3√

6− 2√

3)/12 4 2 (3√

2− 2)/6

2 4 (3√

2− 2)/7 4 3 (10√

3− 9)/24

2 5 (3√

10− 4√

2)/12 5 2 (6√

10− 5√

2)/30

2 6 (13√

3− 12)/33 5 3 (21√

15− 25√

3)/105

2 6 (13√

3− 12)/33 6 2 (21√

3− 20)/39

3 2 (2√

6−√3)/9 6 3 (90√

3− 17√

2)/348

3 4 (4√

3− 4)/9 7 3 (12√

21− 14√

7 + 7)/63

3 5 (10√

15− 14√

3)/45 7 4 (200√

7− 336)/525

3 6 (28√

2− 9√

3)/75 9 2 (45√

2− 14√

3)/81

3 7 (2√

21− 2√

7− 1)/9 9 3√

49/27 + (1/3) 3√

16− (14/9) 3√

4

Table 4-2. Values of εN,p(1/N) for N = 2

p ε2,p(1/2) ` numterms [Γ : Γ0(pN)]

3 −√

26

2 4 12

5 −√

53

2 4 12

7 −√

2√

8√

2−10

24 24 24

11 −275√

2102

6 36 36

13 −√

39√

29−γ−1/3−γ1/3

3where γ =

√5617 + 636

√78 6 42 28

17 −4√

17√

6√

17−22

38 72 45

29 −539√

29111

14 210 90


p ε3,p(1/3) ` numterms [Γ : Γ0(pN)]3 −1

22 4 12

5 −5√

57

4 16 24

7 −14√

3+√

216

6 32 32

11 −675√

33+200√

11386

8 64 48

13 −√

741+546√

13

38 75 56

37



2 −√

56

2 6 18

3 −7√

515

4 16 245 −10

32 10 30



2 −7+8√

344

4 16 24

3 −44+9√

650

6 36 36



2 −√

10+8√

2

84 16 24

3 −14√

3+√

2118

4 22 32

38

REFERENCES

[1] Tom M. Apostol, Modular Functions and Dirichlet Series in Number Theory,Springer-Verlag, 2nd edition, 1990.

[2] J.M. Borwein and P.B. Borwein, A cubic counterpart of Jacobi’s identity and theAGM, Trans. Amer. Math. Soc., 323 1991, 691–701.

[3] J.M. Borwein, P.B. Borwein and D.H. Bailey, Ramanujan, modular equations,and approximations to pi or how to compute one billion digits of pi, Amer. Math.Monthly, 96 (1989), 201–219.

[4] J.M. Borwein, P.B. Borwein and F.G. Garvan, Some cubic modular identities ofRamanujan, Trans. Amer. Math. Soc., 343 1994, 35–47.

[5] J.M. Borwein and F.G. Garvan, Approximations to π via the Dedekind eta function.Organic Mathematics (Burnaby, BC, 1995), CMS Conf. Proc., 20, Amer. Math.Soc., Providence, RI, 1997, 89–115.

[6] F.G. Garvan, Some congruences for partitions that are p-cores, Proc. London Math.Soc.(3) 66, no. 3, 1993, 449–478.

[7] A.K.Lenstra, H.W.Lenstra and L.Lovasz, Factoring polynomials with rationalcoefficients. Math. Ann. 261, no.4, 1982, 515–534.

[8] Toshitsune Miyake, Modular forms, Translated from the 1976 Japanese original byYoshitaka Maeda. Reprint of the first 1989 English edition. Springer Monographs inMathematics. Springer-Verlag, Berlin, 2006.

[9] Robert A. Rankin, Modular Forms and Functions, Cambridge University Press,Cambridge-New York-Melbourne, 1977.

39

BIOGRAPHICAL SKETCH

Amitava Ghosh was born on December 10, 1979, in Durgapur (West Bengal), India.

He received his high school education at St.Xavier’s School, Durgapur. He then went to

DAV Model School for eleventh and twelfth grade. He participated in the International

Mathematics Olympiad training camp at Mumbai in 1995 and 1996. He has independent

proofs of the Bertrand’s conjecture regarding primes and the Gauss’s reciprocity laws.

He also has worked on the construction of Abelian Galois groups of any arbitrary order.

In 2002, he graduated from Regional Engineering College (University of Burdwan)

with a bachelor’s degree in metallurgical engineering. After completing his diploma, he

went to the Institute of Mathematical Sciences–Chennai, India as a visiting student in

mathematics. In the fall of 2004, he entered the University of Florida–Gainesville to

undertake a Master of Science degree in mathematics by research.

Currently, he finished his master’s degree in mathematics from the University of

Florida–Gainesville via the thesis option.

40

Documents

APPROXIMATIONS TO , DEDEKIND’S ETA FUNCTION AND …sequence ffing1 n=1 recursively for which lim n!1 fin = 1 …: Let p be an integer ‚ 1. We say a sequence ffing1 n=1 converges