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Generating Functions, Partitions, and q-SeriesModular Forms

Applications

Generating Functions, Partitions, and q-series:

An Introduction to Classical Modular Forms

Edray Herber Goins

MA 59800-614-27988: Classical Modular Forms

Department of MathematicsPurdue University

June 28, 2016

MA 59800-614-27988: Classical Modular Forms An Introduction to Classical Modular Forms


Applications

Abstract

It is natural to count the number of objects placed into a geometric shape. Forexample, “triangular numbers” are the number objects that can form an equilateraltriangle, while “pentagonal numbers” are the number of objects that can form aregular pentagon. Similarly, it is natural to ask how to partition a natural number intosmaller parts. In the 1700’s, the German mathematician Leonhard Euler found aremarkable relationship between these two sets of numbers. Euler’s PentagonalNumber Theorem paved the way for the concept of a q-series as a generating functionfor other interesting sequences of numbers.

In 1919, the English mathematician Godfrey Harold Hardy went to visit his ill friendSrinivasa Ramanujan in the hospital. Hardy told Ramanujan that had ridden intaxi-cab No. 1729, and remarked that the number seemed to be rather a dull one.Ramanujan countered the number is very interesting as it is the smallest numberexpressible as the sum of two positive cubes in two different ways. Hardy was workingwith Ramanujan to understand some of his pulchritudinous identities involving thecoefficients of a certain “modular discriminant.” This specific q-series has atransformation property which led to the theory of modular forms.

In this introductory talk, we’ll present various applications of classical modular forms toquestions which naturally arise in combinatorics, algebra, and number theory. Inparticular, we’ll discuss properties of quadratic forms (why is every positive integer thesum of four squares?), cubic forms (what are “taxi-cab” numbers?) and congruencesamong the coefficients (why is Ramanujan’s tau function so closely related to thedivisor function?).



Applications

Outline

1 Generating Functions, Partitions, and q-SeriesFigurate NumbersPartition Functionq-Series

2 Modular FormsRiemann Zeta FunctionSpecial Values of the Riemann Zeta FunctionModular Forms, Eisenstein Series, and Cusp Forms

3 ApplicationsRepresenting Integers as the Sums of SquaresRamanujan Tau FunctionElliptic Curves and the Taniyama-Shimura-Weil Conjecture



Applications

Figurate NumbersPartition Functionq-Series

Triangular Numbers

Definition

A triangular number counts the number of objects that can form an equilateraltriangle. The nth triangular number Tn is the number of dots composing atriangle with n dots on a side.

https://en.wikipedia.org/wiki/Triangular_number


https://en.wikipedia.org/wiki/Triangular_number


Applications


Applications

Handshake Problem

Tn is the total number of handshakes required if each person in a room withn + 1 people shakes hands once with each person.

Proposition

Tn is the number of edges in a complete graph Kn of order n.

There are other applications at Sloan’s Online Encyclopedia of IntegerSequences: http://oeis.org/A000217


http://oeis.org/A000217


Applications


Properties

Proposition

Let Tn denote the nth triangular number.

Tn = n + (n − 1) + · · ·+ 2 + 1 = n (n + 1)/2.

The consecutive sum Tn + Tn−1 = n2 is always a perfect square.

The consecutive difference of squares Tn2 − Tn−1

2 = n3 is a perfect cube.∞∑n=0

Tn qn =

q

(1− q)3is a generating function for Tn.

To see the last assertion, differentiate the Geometric Series twice:∞∑n=0

qn+1 =q

1− q

∞∑n=0

(n + 1) qn =1

(1− q)2

∞∑n=0

n (n + 1) qn−1 =2

(1− q)3=⇒

∞∑n=0

n (n + 1)

2qn =

q

(1− q)3



Applications


Figurate Numbers

Definition

A figurate number is a natural number that can be represented by a regulargeometrical arrangement of equally spaced points. If the arrangement forms aregular polygon, the number is called a polygonal number.

Some examples of polygonal numbers are:

triangular numbers Tn,

square numbers Sn,

pentagonal numbers Gn, and

hexagonal numbers Hn.

http://mathworld.wolfram.com/FigurateNumber.html


http://mathworld.wolfram.com/FigurateNumber.html


Applications


Square Numbers

Proposition

Let Sn denote the nth square number.

Sn = n2.∞∑n=1

Sn qn =

q (q + 1)

(1− q)3is a generating function for Sn.

To see why, consider the recursive relation:

S1 = 1

Sn+1 = (2 n + 1) + Sn

=⇒ Sn =n−1∑k=0

(2 k + 1) = n2.

http://mathworld.wolfram.com/SquareNumber.html


http://mathworld.wolfram.com/SquareNumber.html


Applications


Pentagonal Numbers

Proposition

Let Gn denote the nth pentagonal number.

Gn =n (3 n − 1)

2.

∞∑n=1

Gn qn =

q (2 q + 1)

(1− q)3is a generating function for Gn.


G1 = 1

Gn+1 = (3 n + 1) + Gn

=⇒ Gn =n−1∑k=0

(3 k + 1) =n (3 n − 1)

2.

http://mathworld.wolfram.com/PentagonalNumber.html


http://mathworld.wolfram.com/PentagonalNumber.html


Applications


Hexagonal Numbers

Proposition

Let Hn denote the nth hexagonal number.

Hn = n (2 n − 1).∞∑n=1

Hn qn =

q (3 q + 1)

(1− q)3is a generating function for Hn.


H1 = 1

Hn+1 = (4 n + 1) + Hn

=⇒ Hn =n−1∑k=0

(4 k + 1) = n (2n − 1).

http://mathworld.wolfram.com/HexagonalNumber.html


http://mathworld.wolfram.com/HexagonalNumber.html


Applications


Do You See a Pattern?



Applications


Generating Functions for Figurate Numbers

Proposition

Let Nn denote the nth figurate number associated to a regular m-gon.

Nn =n((m − 2) (n − 1) + 2

)2

.

∞∑n=1

Nn qn =

q((m − 3) q + 1

)(1− q)3

is a generating function for Nn.

To see why, consider the recursive relation Nn+1 =((m − 2) n + 1

)+ Nn:

Nn =n−1∑k=0

((m − 2) k + 1

)= (m − 2)

n (n − 1)

2+ n.

To see the last assertion, differentiate the Geometric Series twice:∞∑n=0

n qn =q

(1− q)2

∞∑n=0

n (n − 1)

2qn =

q2

(1− q)3=⇒

∞∑n=0

Nn qn =

(m − 2) q2

(1− q)3+

q

(1− q)2



Applications


Leonhard EulerApril 15, 1707 – September 18, 1783

https://en.wikipedia.org/wiki/Leonhard_Euler


https://en.wikipedia.org/wiki/Leonhard_Euler


Applications


Partition Function

Definition

A partition of a natural number n is a way of writing n as a sum of positiveintegers. The number pn is the number of partitions of n.

Here are some examples:

p1 = 1 because there is only one partition of 1

p2 = 2 because there are two partitions of 2, namely

2 = 1 + 1

p3 = 3 because there are three partitions of 3, namely

3 = 2 + 1 = 1 + 1 + 1

p4 = 5 because there are five partitions of 4, namely

4 = 3 + 1 = 2 + 2 + 2 + 1 + 1 = 1 + 1 + 1 + 1

p5 = 7 because there are seven partitions of 5, namely

5 = 4 + 1 = 3 + 2 = 3 + 1 + 1 = 2 + 2 + 1 = 2+ 1 + 1 + 1 = 1 + 1 + 1+ 1 + 1

https://en.wikipedia.org/wiki/Partition_(number_theory)


https://en.wikipedia.org/wiki/Partition_(number_theory)


Applications


Properties

Proposition (Leonhard Euler)

Let pn denote the number of partitions of n. It has the “generating function”

∞∑n=0

pn qn =

∞∏m=1

1

1− qm.

To see why, first expand out the terms in the product as Geometric Series:

∞∏m=1

1

1− qm=∞∏m=1

(∞∑e=0

qem

)=∞∑n=0

Nn qn

where Nn is the number of ways we can express n =∞∑m=1

em m where each m

appears em times. But these are just partitions of n, so Nn = pn.

Remark: We do not know a nice closed-expression for pn.



Applications


Congruences

Recall that two integers a and b are said to be “congruent modulo N” if b − ais a multiple of N. We write a ≡ b mod N.

Proposition (Srinivasa Ramanujan, 1910’s)

pn ≡ 0 mod 5 whenever n ≡ 4 mod 5.



Proposition (Arthur Oliver Lonsdale Atkin, 1960’s)

pn ≡ 0 mod 13 whenever n ≡ 237 mod 113 · 13.

Proposition (Ken Ono, 2000; Scott Ahlgren and Ken Ono, 2001)


Let N be a positive integer relatively prime to 6. Then there exist integern0 and N0 such that pn ≡ 0 mod N whenever n ≡ n0 mod N0.

https://en.wikipedia.org/wiki/Ramanujan’s_congruences


https://en.wikipedia.org/wiki/Ramanujan's_congruences


Applications


Pentagonal Number Theorem

Proposition (Leonhard Euler)

Denote Gn = n (3 n − 1)/2 as the nth pentagonal number – for all integers n.Then we have the identity

∞∏m=1

(1− qm) =

∑n∈Z

(−1)n qGn

That is, we have the expansion

(1− q) (1− q2) (1− q3) · · · = 1− q − q2 + q5 + q7 − q12 − q15 + q22 + · · · .

Corollary (∞∑n=0

pn qn

)(∑n∈Z

(−1)n qGn

)= 1.

https://en.wikipedia.org/wiki/Pentagonal_number_theorem


https://en.wikipedia.org/wiki/Pentagonal_number_theorem


Applications


Recap

Say that we have a sequence of numbers a0, a1, a2, . . . , an, . . . .

The generating function

f (q) =∞∑n=0

an qn

is also called a q-series. It is a Maclaurin series in the variable q.

As a specific example, let Nn denote the nth figurate number associated toa regular m-gon. Then we have the closed-expressions

Nn =n((m − 2) (n − 1) + 2

)2

and∞∑n=1

Nn qn =

q((m − 3) q + 1

)(1− q)3

.

Let pn denote the number of partitions of n and Gn = n (3 n − 1)/2 as thenth Pentagonal number. Then we have the relation(

∞∑n=0

pn qn

)(∑n∈Z

(−1)n qGn

)= 1.



Applications

Riemann Zeta FunctionSpecial Values of the Riemann Zeta FunctionModular Forms, Eisenstein Series, and Cusp Forms

Carl Gustav Jacob JacobiDecember 10, 1804 – February 18, 1851

https://en.wikipedia.org/wiki/Carl_Gustav_Jacob_Jacobi




Applications


Jacobi Triple Product

The Pentagonal Number Theorem follows from the following general result:

Proposition (Carl Gustav Jacob Jacobi, 1829)

∞∏m=1

(1− w 2m

)(1 + z w 2m−1

)(1 + z−1 w 2m−1

)=∑n∈Z

zn wn2

.

To see why, substitute z = −q−1/2 and w = q3/2:

∞∏m=1

(1− w 2m

)(1 + z w 2m−1

)(1 + z−1 w 2m−1

)=∞∏m=1

(1− q3m

)(1− q3m−2

)(1− q3m−1

)=∞∏m=1

(1− qm)

∑n∈Z

zn wn2

=∑n∈Z

(−1)n qGn



Applications


Jacobi Triple Product


∞∏m=1

(1− w 2m

)(1 + z w 2m−1

)(1 + z−1 w 2m−1

)=∑n∈Z

zn wn2

.

We have a transformation property:

Corollary

For z = 1 and w = e−πt2

, denote the function

Θ(t) =∞∏m=1

(1− e−2mπt2

)(1 + e(1−2m)πt2

)2

= 1 + 2∞∑n=1

e−πn2 t2

.

Θ(1/t) = t Θ(t) for all t > 0.

limt→∞Θ(t) = 1.



Applications


Motivating Question

Consider the statement

1 + 2 + 3 + 4 + · · ·+ n + · · · = − 1

12.

Is this absurd or does this actually make sense?

This is the subject of a YouTube video by Numberphile:

https://www.youtube.com/watch?feature=player_embedded&v=w-I6XTVZXww

as well as a New York Times article by Dennis Overbye featuring Ed Frenkel:

http://www.nytimes.com/2014/02/04/science/in-the-end-it-all-adds-up-to.html

“Theorem”

1 + 2 + 3 + 4 + · · ·+ n + · · · = − 1

12.

1 + 23 + 33 + 43 + · · ·+ n3 + · · · = +1

120.

1 + 22k−1 + 32k−1 + 42k−1 + · · ·+ n2k−1 + · · · = −B2k

2k.


https://www.youtube.com/watch?feature=player_embedded&v=w-I6XTVZXww

http://www.nytimes.com/2014/02/04/science/in-the-end-it-all-adds-up-to.html


Applications


Riemann Zeta Function

Consider the Riemann zeta function

ζ(s) =∞∑n=1

1

ns=

∏primes p

1

1− p−s.

Theorem (Bernhard Riemann)

For s ∈ C satisfying Re(s) > 1, define the functions

Λ(s) =1

πs/2Γ( s

2

)ζ(s), Γ(s) =

∫ ∞0

e−t ts−1 dt = (s − 1)!

Consider those s ∈ C which satisfy 0 < Re(s) < 1.

Λ(1− s) = Λ(s).

The Riemann zeta function satisfies the functional equation

ζ(1− s) =2

(2π)scos

(π s

2

)Γ(s) ζ(s).



Applications


Proof of Theorem

Θ(t) =∞∏m=1

(1− e−2mπt2

)(1 + e(1−2m)πt2

)2

= 1 + 2∞∑n=1

e−πn2 t2

, t > 0.

Lemma 1∫ ∞0

[Θ(t)− 1] ts−1 dt = Λ(s) for Re(s) > 1.

∫ ∞0

[Θ(t)− 1] ts−1 dt =∞∑n=1

2

∫ ∞0

e−πn2t2

ts−1 dt =1

πs/2Γ( s

2

) ∞∑n=1

1

ns= Λ(s).

Lemma 2

Λ(s) = Λ(1− s) when 0 < Re(s) < 1.

Using Lemmas 1 and the identity Θ(1/t) = t Θ(t) we find that

Λ(s) =1

s (1− s)+

∫ ∞1

[Θ(t)− 1][t−s + ts−1

].



Applications


Proof of Theorem (continued)

Lemma 3 (Legendre’s Duplication Formula).

For s ∈ C satisfying 0 < Re(s) < 1,

Γ(s) =2s−1

√π

Γ

(s

2

)Γ

(1 + s

2

), Γ

(1− s

2

)Γ

(1 + s

2

)= π

/cos

(π s

2

)

ζ(1− s)

2

(2π)scos

(π s

2

)Γ(s) ζ(s)

=(2π)s

2 cos

(π s

2

)Γ(s)

· π(1−s)/2 Λ(1− s)

Γ

(1− s

2

)︸︷︷︸

ζ(1−s)

/πs/2 Λ(s)

Γ

(s

2

)︸︷︷︸

ζ(s)

=

2s−1

√π

Γ

(s

2

)Γ

(1 + s

2

)/Γ(s)

Γ

(1− s

2

)Γ

(1 + s

2

)· 1

πcos

(π s

2

) · Λ(1− s)

Λ(s)

= 1.



Applications


Special Values

Theorem (Leonhard Euler)

Let Bn denote the nth Bernoulli number, that is, that rational number whichappears in the power series expansion

t

et − 1=∞∑n=1

Bntn

n!= 1 +

(−1

2

)t +

(+

1

6

)t2

2+

(− 1

30

)t4

24+ · · · .

ζ(−n) = − Bn+1

n + 1is a rational number when evaluated at any negative

integer.

ζ(−2 k) = 0 at the negative even integers.

ζ(2k) =(−1)k+1 B2k

2 (2k)!(2π)2k at the positive even integers.

The negative even integers are called the trivial zeros of the Riemann zetafunction. The Riemann Hypothesis asserts that if ζ(s) = 0 then either (1)s = −2 k is a negative even integer, or (2) Re(s) = 1/2.


https://en.wikipedia.org/wiki/Bernoulli_number


Applications


Proof

We show the second statement. Consider the function equation

ζ(1− s) =2

(2π)scos

(π s

2

)Γ(s) ζ(s), 0 < Re(s) < 1.

Any meromorphic continuation of ζ(s) to the entire complex plane must satisfythis relation. If we set s = 2 k + 1, then we have the identity

ζ(−2 k) = − 1

4k π2k+1sin(π k)

Γ(2 k + 1) ζ(2 k + 1)

= 0

for k = 1, 2, 3, . . . .

We show the third statement. Let s = 2 k be a positive even integer, so that1− s = −n for n = 2k − 1 is a negative odd integer. Then we have

−B2k

2k= ζ(1−2k) =

2

(2π)2kcos

(2 k π

2

)Γ(2k) ζ(2k) =

2 (2k − 1)! (−1)k

(2π)2kζ(2k).

Solving for ζ(2k) gives the result.



Applications


Applications

Corollary

1 + 2 + 3 + 4 + · · ·+ n + · · · = − 1

12.

1 + 23 + 33 + 43 + · · ·+ n3 + · · · = +1

120.

1 + 22k−1 + 32k−1 + 42k−1 + · · ·+ n2k−1 + · · · = −B2k

2k.

Since ζ(s) = 1 + 1/2s + 1/3s + · · ·+ 1/ns + · · · , we find that

1 + 2 + 3 + 4 + · · ·+ n + · · · = ζ(−1) = − 1

12;

1 + 23 + 33 + 43 + · · ·+ n3 + · · · = ζ(−3) = +1

120;

1 + 22k−1 + 32k−1 + 42k−1 + · · ·+ n2k−1 + · · · = ζ(1− 2k) = −B2k

2k.

(But ζ(s) = 1 + 1/2s + 1/3s + · · ·+ 1/ns + · · · only makes sense for Re(s) > 1!)



Applications


Motivating Question


Denote the function

Θ(t) =∞∏m=1

(1− e−2mt2

)(1 + e(1−2m)t2

)2

= 1 + 2∞∑n=1

e−n2 t2

.

Θ(1/t) = t Θ(t) for all t > 0.

limt→∞Θ(t) = 1.

Can this be generalized?



Applications


Modular Forms

Γ(1) = SL2(Z) acts on the upper-half plane H = {z ∈ C | Im(z) > 0} by

γ z =a z + b

c z + dwhere γ =

(a bc d

).

Definitions

Let N be a positive integer and k be a positive even integer.

Γ0(N) =

{(a bc d

)∈ SL2(Z)

∣∣∣∣ c ≡ 0 mod N

}.

A modular form of level N and weight k is a function f : H→ C satisfyingf is holomorphic: f analytic on the upper-half plane;f is cuspidal: limz→i∞ |f (z)| <∞; and

f is automorphic: f (γ z) = (c z + d)k f (z) for γ ∈ Γ0(N).

γ =

(1 10 1

)∈ Γ0(N) =⇒ f (z + 1) = f (z)

so that each modular form has a q-series expansion:

f (z) =∑n∈Z

an e2πinz =

∞∑n=0

an qn where q = e2πiz .



Applications


Eisenstein Series

Definitions

Let the divisor function σk(n) =∑

d|n dk be the sum over the kth powers

of the divisors d of a positive integer n.

Let the Bernoulli numbers Bn be defined by the power series expansion

t

et − 1=∞∑n=1

Bntn

n!= 1 +

(−1

2

)t +

(+

1

6

)t2

2+

(− 1

30

)t4

24+ · · · .

For every even integer k, denote Ek(z) = 1− 2 k

Bk

∞∑n=1

σk−1(n) qn.

Proposition

Ek is a modular form of level N = 1 and weight k. In particular,

Ek(γ z) = (c z + d)k Ek(z) for γ ∈ Γ(1) = SL2(Z).

Ek(−1/z) = zk Ek(z)

limz→i∞ Ek(z) = 1



Applications


Modular Discriminant

Proposition

Define

E4(z) = 1 + 240∞∑n=1

σ3(n) qn E12(z) = 1 +65520

691

∞∑n=1

σ11(n) qn

E6(z) = 1− 504∞∑n=1

σ5(n) qn ∆(z) = q∞∏n=1

(1− qn)24

=∞∑n=1

τ(n) qn

691E12(z) = 441E4(z)3 + 250E6(z)2.

1728 ∆(z) = E4(z)3 − E6(z)2.

∆(z) is a modular form of level N = 1 and weight k = 12. In particular,

∆(−1/z) = z12 ∆(z)limz→i∞∆(z) = 0.

∆ is called the modular discriminant. Any modular form f : H→ C satisfyinglimz→i∞ f (z) = 0 is called a cusp form. Denote the collection by Sk

(Γ0(N)

).



Applications

Representing Integers as the Sums of SquaresRamanujan Tau FunctionElliptic Curves and the Taniyama-Shimura-Weil Conjecture

Application #1:

Representing Integers

as the Sums of Squares



Applications


Adrien-Marie LegendreSeptember 18, 1752 – January 10, 1833

https://en.wikipedia.org/wiki/Adrien-Marie_Legendre


https://en.wikipedia.org/wiki/Adrien-Marie_Legendre


Applications


Legendre’s Four-Square Theorem

Theorem (Adrien-Marie Legendre, 1797-8)

Every natural number n can be written as the sum of at most four squares.

1 = 12 + 02 + 02 + 02 11 = 32 + 12 + 12 + 02 21 = 42 + 22 + 12 + 02

2 = 12 + 12 + 02 + 02 12 = 22 + 22 + 22 + 02 22 = 32 + 32 + 22 + 02

3 = 12 + 12 + 12 + 02 13 = 32 + 22 + 02 + 02 23 = 32 + 32 + 22 + 12

4 = 22 + 02 + 02 + 02 14 = 32 + 22 + 12 + 02 24 = 42 + 22 + 22 + 02

5 = 22 + 12 + 02 + 02 15 = 32 + 22 + 12 + 12 25 = 52 + 02 + 02 + 02

6 = 22 + 11 + 12 + 02 16 = 42 + 42 + 02 + 02 26 = 52 + 12 + 02 + 02

7 = 22 + 12 + 12 + 12 17 = 42 + 42 + 12 + 02 27 = 32 + 32 + 32 + 02

8 = 22 + 22 + 02 + 02 18 = 32 + 32 + 02 + 02 28 = 52 + 12 + 12 + 12

9 = 32 + 02 + 02 + 02 19 = 32 + 32 + 12 + 02 29 = 52 + 22 + 02 + 02

10 = 32 + 12 + 02 + 02 20 = 42 + 22 + 02 + 02 30 = 52 + 22 + 12 + 02

https://en.wikipedia.org/wiki/Lagrange’s_four-square_theorem


https://en.wikipedia.org/wiki/Lagrange's_four-square_theorem


Applications


Carl Gustav Jacob JacobiDecember 10, 1804 – February 18, 1851





Applications


Jacobi’s Four-Square Theorem

Recall that we considered the function

Θ(t) =∞∏m=1

(1− q2m

)(1 + q2m−1

)2

= 1 + 2∞∑n=1

qn2

, q = e−πt2

.

Theorem (Carl Gustav Jacob Jacobi, 1834)

For each natural number m, define the function

θm(q) =

(∑n∈Z

qn2

)m

=∞∑n=1

rm(n) qn.

rm(n) counts the number of ways a natural number n = n21 + n2

2 + · · ·+ n2m

can be written as the sum of m squares n2i .

θ4(q) =

(∑n∈Z

qn2

)4

= 1 + 8

(∞∑n=1

n qn

1− qn−∞∑n=1

4 n q4n

1− q4n

).

Say m ≥ 4. Then rm(n) ≥ 8(∑

d|n d −∑

4d|n d)

. In particular, rm(n) 6= 0.

https://en.wikipedia.org/wiki/Jacobi’s_four-square_theorem


https://en.wikipedia.org/wiki/Jacobi's_four-square_theorem


Applications


Application #2:

Congruences with the

Ramanujan Tau Function



Applications


Srinivasa Ramanujan IyengarDecember 22, 1887 – April 26, 1920

https://en.wikipedia.org/wiki/Srinivasa_Ramanujan


https://en.wikipedia.org/wiki/Srinivasa_Ramanujan


Applications


Ramanujan Tau Function

∆(z) = q ·∞∏n=1

(1− qn)24

=∞∑n=1

τ(n) qn where q = e2πi z .

n τ(n)1 12 −243 2524 −1472

n τ(n)5 48306 −60487 −167448 84480

n τ(n)9 −113643

10 −11592011 53461212 −370944

n τ(n)13 −57773814 40185615 121716016 987136

Proposition

τ(n) ≡ σ11(n) mod 211 whenever n ≡ 1 mod 8.

τ(n) ≡ 1217σ11(n) mod 213 whenever n ≡ 3 mod 8.



https://en.wikipedia.org/wiki/Ramanujan_tau_function


https://en.wikipedia.org/wiki/Ramanujan_tau_function


Applications


Modular Discriminant

Proposition

Define

E4(z) = 1 + 240∞∑n=1

σ3(n) qn E12(z) = 1 +65520

691

∞∑n=1

σ11(n) qn

E6(z) = 1− 504∞∑n=1

σ5(n) qn ∆(z) = q∞∏n=1

(1− qn)24

=∞∑n=1

τ(n) qn

691E12(z) = 441E4(z)3 + 250E6(z)2.

1728 ∆(z) = E4(z)3 − E6(z)2.

∆(z) is a modular form of level N = 1 and weight k = 12. In particular,

∆(−1/z) = z12 ∆(z)limz→i∞∆(z) = 0.

∆ is called the modular discriminant. Any modular form f : H→ C satisfyinglimz→i∞ f (z) = 0 is called a cusp form. Denote the collection by Sk

(Γ0(N)

).



Applications


Properties

Let the divisor function σk(n) =∑

d|n dk be the sum over the kth powers of

the divisors d of a positive integer n.

Proposition (Srinivasa Ramanujan, 1916)

τ(n) ≡ σ11(n) mod 691 for all natural numbers n.

Since p = 691 appears in the numerator of B12, this gives the congruence

691 + 65520

[∞∑n=1

σ11(n) qn

]= 691E12(z)

= 441E4(z)3 + 250E6(z)2

≡ 441 ·[E4(z)3 − E6(z)2] mod 691

≡ 65520 ∆(z) mod 691.

Conjecture

τ(n) 6= 0 for all natural numbers n.



Applications


Application #3:

Elliptic Curves and the

Taniyama-Shimura-Weil Conjecture



Applications


Godfrey Harold HardyFebruary 7, 1877 – December 1, 1947

https://en.wikipedia.org/wiki/G._H._Hardy


https://en.wikipedia.org/wiki/G._H._Hardy


Applications


Taxi Cab Number

I remember once going to see [Srinivasa Ramanujan] when he was lying illat Putney. I had ridden in taxi-cab No. 1729, and remarked that thenumber seemed to be rather a dull one, and that I hoped it was not anunfavourable omen. “No”, he replied, “it is a very interesting number; it isthe smallest number expressible as the sum of two [positive] cubes in twodifferent ways.” – Godfrey Harold Hardy, 1919

Definition

The nth taxicab number N = Taxicab(n) is defined as the least natural numberwhich can be expressed as a sum of two positive cubes in n distinct ways.

n N = Taxicab(n) Sum of Cubes

1 2 13 + 13

2 1729 13 + 123 = 93 + 103

3 87539319 1673 + 4363 = 2283 + 4233 = 2553 + 4143

4 6963472309248 24213 + 190833 = · · ·5 48988659276962496 387873 + 3657573 = · · ·6 24153319581254312065344 581623 + 289062063 = · · ·

https://en.wikipedia.org/wiki/Taxicab_number


https://en.wikipedia.org/wiki/Taxicab_number


Applications


Johann Carl Friedrich GaussApril 30, 1777 – February 23, 1855

https://en.wikipedia.org/wiki/Carl_Friedrich_Gauss


https://en.wikipedia.org/wiki/Carl_Friedrich_Gauss


Applications


Gauss’s Theorem

Fix an odd prime p, and denote the finite field of p elements

Fp = Z/p Z ={

0, 1, 2, . . . , p − 1}.

Theorem (Carl Friedrich Gauss, 1798)

Consider the Fermat curve F3 : a3 + b3 = c3.

If either p = 3 or p ≡ 2 mod 3, then #F3(Fp) = p + 1.

If p ≡ 1 mod 3, then there exist integers ap and bp such that

#F3(Fp) = p + 1− ap and 4 p = a2p + 27 b2

p.

p #F3(Fp) ap bp

7 9 −1 113 9 5 119 27 −7 131 36 −4 237 27 11 143 36 8 2

p #F3(Fp) ap bp

61 63 −1 367 63 5 373 81 −7 379 63 17 197 117 −19 1

103 117 −13 3



Applications


Proof of Gauss’s Theorem

First assume that p ≡ 2 mod 3. We have a bijection

P1(Fp) →{

(a : b : c) ∈ P2(Fp)

∣∣∣∣ a3 + b3 = c3

}(α : β) 7→

(α : β :

(α3 + β3

)(2p−1)/3)

Now assume p ≡ 1 mod 3. There exists a nontrivial cubic characterχp : F×p → C×. Upon setting χp(0) = 0, define the Gauss sum

τ(χp) =∑α∈Fp

χp(α) ζαp = −e iθp/3√p

for some angle θp. Note that for any α ∈ Fp, we have the formula

#

{a ∈ Fp

∣∣∣∣ a3 = α

}= 1 + χp(α) + χp(α)2.



Applications


Proof of Gauss’s Theorem (continued)

Now define Jacobi sum as

J(χp, χp) =∑α+β=1

χp(α)χp(β) =τ(χp)2

τ(χ2p)

= −e iθp√p.

Writing J(χp, χp) = cp + dp ζ3, we have∣∣J(χp, χp)

∣∣2 = c2p − cp dp + d2

p .

Choose the rational integers

ap = dp − 2 cp

bp = dp/3

}=⇒

4 p = a2p + 27 b2

p

ap ≡ 2 mod 3

Finally, count the number of points:

#F3(Fp) = 3 +∑α+β=1

#

{a ∈ Fp

∣∣∣∣ a3 = α

}×#

{b ∈ Fp

∣∣∣∣ b3 = β

}= p + 1 + J(χp, χp) + J(χ2

p, χ2p)

= p + 1− ap.



Applications


Application

Corollary

Consider the elliptic curve E : y 2 − y = x3 − 7. For all primes p 6= 3,(√p − 1

)2 ≤ #E(Fp) ≤(√

p + 1)2.

We have a bijection

F3 : a3 + b3 = c3 → E : y 2 − y = x3 − 7

(a : b : c) 7→ (3 c : −4 a + 5 b : a + b)

Hence

p + 1−#E(Fp) = ap =

{0 if p ≡ 2 (mod 3),

±√

4 p − 27 b2p if p ≡ 1 (mod 3).

In either case we have∣∣p + 1−#E(Fp)

∣∣ ≤ 2√p.



Applications


Elliptic Curves

Definitions

Consider an equation with coefficients in a field F , say

E : y 2 + a1 x y + a3 y = x3 + a2 x2 + a4 x + a6.

If F has characteristic different from 2 and 3, we may write

E : y 2 = x3 − 3

122c4 x −

2

123c6 and ∆(E) =

c34 − c2

6

123.

We say E is an elliptic curve defined over F if ∆(E) 6= 0.

The collection of F -rational points on E defined as the set

E(F ) =

{(x1 : x2 : x0) ∈ P2(F )

∣∣∣∣∣ x22 x0 + a1 x1 x2 x0 + a3 x2 x

20

= x31 + a2 x

21 x0 + a4 x1 x

20 + a6 x

30

}

having a specified base point O = (0 : 1 : 0).

Define the j-invariant as j(E) =c3

4

∆(E)= 123 c3

4

c34 − c2

6

.



Applications


Taxi Cab Numbers and Elliptic Curves

Definition

The nth taxicab number N = Taxicab(n) is defined as the least natural numberwhich can be expressed as a sum of two positive cubes in n distinct ways.

Proposition

Fix a nonzero integer N.

The curve a3 + b3 = N c3 is just the elliptic curve E : y 2 = x3 − 432N2

defined over F = Q.

An expression N = (a0/c0)3 + (b0/c0)3 as the sum of two positive cubescorresponds to a Q-rational point (x0, y0) on E .

If N = Taxicab(n) is a taxicab number, then E : y 2 = x3 − 432N2 hasinfinitely many Q-rational points.

We have the bijection

F3 : a3 + b3 = N c3 → E : y 2 = x3 − 432N2

(a : b : c) 7→ (12N c : 36N (a− b) : a + b)



Applications


Group Law

Definition

Let P,Q ∈ E(F ). Denote P ∗ Q as the point of intersection of E and the linethrough P and Q, and denote P ⊕ Q = (P ∗ Q) ∗ O.

-4.8 -4 -3.2 -2.4 -1.6 -0.8 0 0.8 1.6 2.4 3.2 4 4.8

-3.2

-2.4

-1.6

-0.8

0.8

1.6

2.4

3.2

PQ P*Q

P+Q

Theorem (Henri Poincare, 1901)

Consider an elliptic curve E defined over a field F . Then (E(F ), ⊕) is abeliangroup with identity O = (0 : 1 : 0) and inverses [−1]P = P ∗ O.



Applications


Taniyama-Shimura-Weil Conjecture

For an even integer k, a cusp form of level N and weight k is a functionf : H → C that satisfy

f is holomorphic: f analytic on the upper-half plane;f is cuspidal: limz→i∞ f (z) = 0; andf is automorphic: f (γ z) = (c z + d)k f (z) for γ ∈ Γ0(N).

Denote the collection of such functions by Sk (Γ0(N)).

Theorem (Christophe Breuil, Brian Conrad, Fred Diamond, Richard Taylor, andAndrew Wiles; 2001)

Let E be an elliptic curve over E . For each natural number n, define theinteger an as follows.

an =∏

p apep whenever n =∏

p pep as a product of prime powers.

When p does not divide ∆(E), define apep recursively via the relations

a1 = 1, ap = p + 1−#E(Fp), ape+1 − ap ape + p ape−1 = 0.

Then there exists a positive integer N, divisible only by the primes dividing

∆(E), such that f (z) =∞∑n=1

an qn is a cusp form of level N and weight k = 2.



Applications


Example #1

Consider the elliptic curve E : y 2 + y = x3 − x2 − 10 x − 20, and denote

#E(Fp) = 1 + #{

(x , y) ∈ Fp × Fp

∣∣ y 2 + y = x3 − x2 − 10 x − 20}.

For example, #E(F5) = 8, #E(F7) = 8, #E(F11) = 8, and #E(F13) = 16.

Proposition

Consider the following function

f (q) = q∞∏n=1

(1− qn)2(

1− q11 n)2

= q − 2 q2 − q3 + 2 q4 + q5 + 2 q6 − 2 q7 − 2 q9 − 2 q10

+ q11 − 2 q12 + 4 q13 + 4 q14 − q15 − 4 q16 − 2 q17 + 4 q18 + · · ·

This is a cusp form of level N = 11 and weight k = 2.

The coefficient of qp is ap = p + 1−#E(Fp).

http://www.lmfdb.org/EllipticCurve/Q/11/a/2




Applications


Example #2

Consider the elliptic curve E : y 2 = x3 + 5 x2 + 4 x , and denote

#E(Fp) = 1 + #{

(x , y) ∈ Fp × Fp

∣∣ y 2 = x3 + 5 x2 + 4 x}.


Proposition


f (q) = q∞∏n=1

(1− q2n

)(1− q4n

)(1− q6n

)(1− q12n

)= q − q3 − 2 q5 + q9 + 4 q11 − 2 q13 + 2 q15 + 2 q17 − 4 q19

− 8 q23 − q25 − q27 + 6 q29 + 8 q31 − 4 q33 + 6 q37 + 2 q39

− 6 q41 + 4 q43 − 2 q45 − 7 q49 − 2 q51 − 2 q53 − 8 q55 + 4 q57 + · · ·







Applications


Example #3

Consider the elliptic curve E : y 2 = x3 − x , and denote

#E(Fp) = 1 + #{

(x , y) ∈ Fp × Fp

∣∣ y 2 = x3 − x}.


Proposition


f (q) = q∞∏n=1

(1− q4n

)2 (1− q8n

)2

= q − 2 q5 − 3 q9 + 6 q13 + 2 q17 − q25 − 10 q29 − 2 q37

+ 10 q41 + 6 q45 − 7 q49 + 14 q53 − 10 q61 − 12 q65 − 6 q73 + · · ·



If p ≡ 3 mod 4 then ap = 0. If p ≡ 1 mod 4 then ap is divisible 2.





Applications


Fermat’s Last Theorem

Theorem (Pierre de Fermat, 1670; Andrew Wiles, 1993)

Let n ≥ 3 be an integer. If a, b, and c are integers satisfying an + bn = cn,then a b c = 0.

https://en.wikipedia.org/wiki/Fermats_Last_Theorem

Assume that there are nonzero integers a, b, and c such that an + bn = cn.

Say that n has no odd prime divisors. Then x = cn/4, y = bn/4, andz = an/2 are nonzero integers such that x4 − y 4 = z2. In 1670, Pierre deFermat showed no such integers exist.

Say that n has an odd prime divisor `. Then A = an/`, B = bn/`, andC = cn/` are nonzero integers such that A` + B` = C `. Gerhard Freysuggested in 1984 to consider the curve E : y 2 = x (x − A`) (x + B`) withdiscriminant ∆(E) = 16 (AB C)2`. In 1993, Andrew Wiles proved thatsuch an elliptic curve E must be modular. In 1986, Kenneth Ribet showedthat its level must be N = 2. A simple computation shows that there areno cusp forms of level N = 2 and weight k = 2.


https://en.wikipedia.org/wiki/Fermats_Last_Theorem


Applications


The Man Who Knew Infinity

Official Theatrical Release Date: April 29, 2016

https://www.youtube.com/watch?v=oXGm9Vlfx4w




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