Post on 09-Aug-2020
Ramanujan’s series for 1/π
Shaun Cooper
Massey University, Auckland, New Zealand
Srinivasa Ramanujan, 1887 – 1920
Jeremy Irons and Dev Patel as G. H. Hardy and S. Ramanujan in“The Man who knew Infinity”, based on a book by Robert Kanigel.
Goals of today’s lecture:
1. To outline a general theory that accounts for all of these series.
2. To sketch the ideas of a proof, with specific reference to theseries
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
3. To indicate some directions for further research.
Goals of today’s lecture:
1. To outline a general theory that accounts for all of these series.
2. To sketch the ideas of a proof, with specific reference to theseries
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
3. To indicate some directions for further research.
Goals of today’s lecture:
1. To outline a general theory that accounts for all of these series.
2. To sketch the ideas of a proof, with specific reference to theseries
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
3. To indicate some directions for further research.
Goals of today’s lecture:
1. To outline a general theory that accounts for all of these series.
2. To sketch the ideas of a proof, with specific reference to theseries
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
3. To indicate some directions for further research.
Ramanujan’s series rely on four hypotheses.
Hypothesis 1. There is a positive integer `, called the level, suchthat the transformation formulas
t Z(e−2π√
t/`)
= Z(e−2π/
√t`)
andX(e−2π√
t/`)
= X(e−2π/
√t`)
hold for all values of a positive real variable t.
For the next slide, recall that
P(q) = 1− 24∞∑j=1
jqj
1− qjand ηm = qm/24
∞∏j=1
(1− qmj).
Ramanujan’s series rely on four hypotheses.
Hypothesis 1. There is a positive integer `, called the level, suchthat the transformation formulas
t Z(e−2π√
t/`)
= Z(e−2π/
√t`)
andX(e−2π√
t/`)
= X(e−2π/
√t`)
hold for all values of a positive real variable t.
For the next slide, recall that
P(q) = 1− 24∞∑j=1
jqj
1− qjand ηm = qm/24
∞∏j=1
(1− qmj).
Ramanujan’s series rely on four hypotheses.
Hypothesis 1. There is a positive integer `, called the level, suchthat the transformation formulas
t Z(e−2π√
t/`)
= Z(e−2π/
√t`)
andX(e−2π√
t/`)
= X(e−2π/
√t`)
hold for all values of a positive real variable t.
For the next slide, recall that
P(q) = 1− 24∞∑j=1
jqj
1− qjand ηm = qm/24
∞∏j=1
(1− qmj).
Z4 = 13
(4P(q4)− P(q)
), X4 =
(η41η
44
η42 Z4
)2
, ` = 4
Z3 = 12
(3P(q3)− P(q)
), X3
(η21η
23
Z3
)3
, ` = 3
Z2 = 2P(q2)− P(q), X2 =
(η21η
22
Z2
)4
, ` = 2
Z1 = Q(q)1/2, X1 =
(η41Z1
)6
, ` = 1
Z10 =1
12
(10P(q10 + 5P(q5)− 2P(q2)− P(q)
)X10 =
(η1η2η5η10
Z10
)4/3
, ` = 10
Z4 = 13
(4P(q4)− P(q)
), X4 =
(η41η
44
η42 Z4
)2
, ` = 4
Z3 = 12
(3P(q3)− P(q)
), X3
(η21η
23
Z3
)3
, ` = 3
Z2 = 2P(q2)− P(q), X2 =
(η21η
22
Z2
)4
, ` = 2
Z1 = Q(q)1/2, X1 =
(η41Z1
)6
, ` = 1
Z10 =1
12
(10P(q10 + 5P(q5)− 2P(q2)− P(q)
)X10 =
(η1η2η5η10
Z10
)4/3
, ` = 10
Z4 = 13
(4P(q4)− P(q)
), X4 =
(η41η
44
η42 Z4
)2
, ` = 4
Z3 = 12
(3P(q3)− P(q)
), X3
(η21η
23
Z3
)3
, ` = 3
Z2 = 2P(q2)− P(q), X2 =
(η21η
22
Z2
)4
, ` = 2
Z1 = Q(q)1/2, X1 =
(η41Z1
)6
, ` = 1
Z10 =1
12
(10P(q10 + 5P(q5)− 2P(q2)− P(q)
)X10 =
(η1η2η5η10
Z10
)4/3
, ` = 10
Z4 = 13
(4P(q4)− P(q)
), X4 =
(η41η
44
η42 Z4
)2
, ` = 4
Z3 = 12
(3P(q3)− P(q)
), X3
(η21η
23
Z3
)3
, ` = 3
Z2 = 2P(q2)− P(q), X2 =
(η21η
22
Z2
)4
, ` = 2
Z1 = Q(q)1/2, X1 =
(η41Z1
)6
, ` = 1
Z10 =1
12
(10P(q10 + 5P(q5)− 2P(q2)− P(q)
)X10 =
(η1η2η5η10
Z10
)4/3
, ` = 10
Z4 = 13
(4P(q4)− P(q)
), X4 =
(η41η
44
η42 Z4
)2
, ` = 4
Z3 = 12
(3P(q3)− P(q)
), X3
(η21η
23
Z3
)3
, ` = 3
Z2 = 2P(q2)− P(q), X2 =
(η21η
22
Z2
)4
, ` = 2
Z1 = Q(q)1/2, X1 =
(η41Z1
)6
, ` = 1
Z10 =1
12
(10P(q10 + 5P(q5)− 2P(q2)− P(q)
)X10 =
(η1η2η5η10
Z10
)4/3
, ` = 10
Hypothesis 2. There is a power series expansion
Z =∞∑j=0
h(j)X j
that converges in a neighborhood of X = 0.
Z4 =1
3
(4P(q4)− P(q)
)= 1+8q+24q2+32q3+24q4+48q5+· · ·
X4 =
(η41η
44
η42 Z4
)2
= q − 24q2 + 296q3 − 2528q4 + 16928q5 + · · ·
Z4 = 1 + 8X4 + 216X 24 + 8032X 3
4 + · · ·
In fact,
Z4 =∞∑j=0
(2j
j
)3
X j4, so h(j) =
(2j
j
)3
.
Hypothesis 2. There is a power series expansion
Z =∞∑j=0
h(j)X j
that converges in a neighborhood of X = 0.
Z4 =1
3
(4P(q4)− P(q)
)= 1+8q+24q2+32q3+24q4+48q5+· · ·
X4 =
(η41η
44
η42 Z4
)2
= q − 24q2 + 296q3 − 2528q4 + 16928q5 + · · ·
Z4 = 1 + 8X4 + 216X 24 + 8032X 3
4 + · · ·
In fact,
Z4 =∞∑j=0
(2j
j
)3
X j4, so h(j) =
(2j
j
)3
.
Hypothesis 2. There is a power series expansion
Z =∞∑j=0
h(j)X j
that converges in a neighborhood of X = 0.
Z4 =1
3
(4P(q4)− P(q)
)= 1+8q+24q2+32q3+24q4+48q5+· · ·
X4 =
(η41η
44
η42 Z4
)2
= q − 24q2 + 296q3 − 2528q4 + 16928q5 + · · ·
Z4 = 1 + 8X4 + 216X 24 + 8032X 3
4 + · · ·
In fact,
Z4 =∞∑j=0
(2j
j
)3
X j4, so h(j) =
(2j
j
)3
.
Hypothesis 2. There is a power series expansion
Z =∞∑j=0
h(j)X j
that converges in a neighborhood of X = 0.
Z4 =1
3
(4P(q4)− P(q)
)= 1+8q+24q2+32q3+24q4+48q5+· · ·
X4 =
(η41η
44
η42 Z4
)2
= q − 24q2 + 296q3 − 2528q4 + 16928q5 + · · ·
Z4 = 1 + 8X4 + 216X 24 + 8032X 3
4 + · · ·
In fact,
Z4 =∞∑j=0
(2j
j
)3
X j4, so h(j) =
(2j
j
)3
.
Hypothesis 2. Further examples:
Z4 =∞∑j=0
(2j
j
)3
X j4,
Z3 =∞∑j=0
(4j
2j
)(2j
j
)2
X j3,
Z2 =∞∑j=0
(3j
j
)(2j
j
)2
X j2,
Z1 =∞∑j=0
(6j
3j
)(3j
j
)(2j
j
)X j1,
Z10 =∞∑j=0
h(j)X j10, h(j) =
j∑i=0
(j
i
)4
Hypothesis 2. Further examples:
Z4 =∞∑j=0
(2j
j
)3
X j4,
Z3 =∞∑j=0
(4j
2j
)(2j
j
)2
X j3,
Z2 =∞∑j=0
(3j
j
)(2j
j
)2
X j2,
Z1 =∞∑j=0
(6j
3j
)(3j
j
)(2j
j
)X j1,
Z10 =∞∑j=0
h(j)X j10, h(j) =
j∑i=0
(j
i
)4
Hypothesis 2. Further examples:
Z4 =∞∑j=0
(2j
j
)3
X j4,
Z3 =∞∑j=0
(4j
2j
)(2j
j
)2
X j3,
Z2 =∞∑j=0
(3j
j
)(2j
j
)2
X j2,
Z1 =∞∑j=0
(6j
3j
)(3j
j
)(2j
j
)X j1,
Z10 =∞∑j=0
h(j)X j10, h(j) =
j∑i=0
(j
i
)4
Hypothesis 2. Further examples:
Z4 =∞∑j=0
(2j
j
)3
X j4,
Z3 =∞∑j=0
(4j
2j
)(2j
j
)2
X j3,
Z2 =∞∑j=0
(3j
j
)(2j
j
)2
X j2,
Z1 =∞∑j=0
(6j
3j
)(3j
j
)(2j
j
)X j1,
Z10 =∞∑j=0
h(j)X j10, h(j) =
j∑i=0
(j
i
)4
Hypothesis 2. Further examples:
Z4 =∞∑j=0
(2j
j
)3
X j4,
Z3 =∞∑j=0
(4j
2j
)(2j
j
)2
X j3,
Z2 =∞∑j=0
(3j
j
)(2j
j
)2
X j2,
Z1 =∞∑j=0
(6j
3j
)(3j
j
)(2j
j
)X j1,
Z10 =∞∑j=0
h(j)X j10, h(j) =
j∑i=0
(j
i
)4
Hypothesis 3. There is an algebraic function B = B(X ) such thatthe differentiation formula
qd
dqlogX = Z B(X )
holds.
Example: In Lecture 3 we proved (for the level 4 theory)
qdx
dq= z2x(1− x).
Under the change of variables Z4 = z2 and X4 = x(1− x)/16 thisbecomes
qdX
dq= Z√
1− 64X , so B(X ) =√
1− 64X .
Hypothesis 3. There is an algebraic function B = B(X ) such thatthe differentiation formula
qd
dqlogX = Z B(X )
holds.
Example: In Lecture 3 we proved (for the level 4 theory)
qdx
dq= z2x(1− x).
Under the change of variables Z4 = z2 and X4 = x(1− x)/16 thisbecomes
qdX
dq= Z√
1− 64X , so B(X ) =√
1− 64X .
Hypothesis 3. There is an algebraic function B = B(X ) such thatthe differentiation formula
qd
dqlogX = Z B(X )
holds.
Example: In Lecture 3 we proved (for the level 4 theory)
qdx
dq= z2x(1− x).
Under the change of variables Z4 = z2 and X4 = x(1− x)/16 thisbecomes
qdX
dq= Z√
1− 64X , so B(X ) =√
1− 64X .
Hypothesis 3.
qd
dqlogX = Z B(X )
` = 4 B =√
1− 64X
` = 3 B =√
1− 108X
` = 2 B =√
1− 256X
` = 1 B =√
1− 1728X
` = 10 B =√
(1− 16X )(1 + 4X )
Hypothesis 3.
qd
dqlogX = Z B(X )
` = 4 B =√
1− 64X
` = 3 B =√
1− 108X
` = 2 B =√
1− 256X
` = 1 B =√
1− 1728X
` = 10 B =√
(1− 16X )(1 + 4X )
Hypothesis 3.
qd
dqlogX = Z B(X )
` = 4 B =√
1− 64X
` = 3 B =√
1− 108X
` = 2 B =√
1− 256X
` = 1 B =√
1− 1728X
` = 10 B =√
(1− 16X )(1 + 4X )
Hypothesis 3.
qd
dqlogX = Z B(X )
` = 4 B =√
1− 64X
` = 3 B =√
1− 108X
` = 2 B =√
1− 256X
` = 1 B =√
1− 1728X
` = 10 B =√
(1− 16X )(1 + 4X )
Hypothesis 3.
qd
dqlogX = Z B(X )
` = 4 B =√
1− 64X
` = 3 B =√
1− 108X
` = 2 B =√
1− 256X
` = 1 B =√
1− 1728X
` = 10 B =√
(1− 16X )(1 + 4X )
Hypothesis 4.
Fix an integer N ≥ 2, called the degree. Let Y = Y (q) = X (qN).
The functions X and Y are algebraically dependent. That is, thereexists an irreducible polynomial g(X ,Y ), not identically zero, suchthat g(X ,Y ) = 0. The equation g(X ,Y ) = 0 is said to be amodular equation of degree N.
Example: level ` = 4, degree N = 3
X := X4 =
(η41η
44
η42 Z4
)2
= q−24q2+296q3−2528q4+16928q5+· · ·
Y := Y4 = X4(q3) = q3−24q6+296q9−2528q12+16928q15+ · · ·
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0
Hypothesis 4.
Fix an integer N ≥ 2, called the degree. Let Y = Y (q) = X (qN).
The functions X and Y are algebraically dependent.
That is, thereexists an irreducible polynomial g(X ,Y ), not identically zero, suchthat g(X ,Y ) = 0. The equation g(X ,Y ) = 0 is said to be amodular equation of degree N.
Example: level ` = 4, degree N = 3
X := X4 =
(η41η
44
η42 Z4
)2
= q−24q2+296q3−2528q4+16928q5+· · ·
Y := Y4 = X4(q3) = q3−24q6+296q9−2528q12+16928q15+ · · ·
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0
Hypothesis 4.
Fix an integer N ≥ 2, called the degree. Let Y = Y (q) = X (qN).
The functions X and Y are algebraically dependent. That is, thereexists an irreducible polynomial g(X ,Y ), not identically zero, suchthat g(X ,Y ) = 0.
The equation g(X ,Y ) = 0 is said to be amodular equation of degree N.
Example: level ` = 4, degree N = 3
X := X4 =
(η41η
44
η42 Z4
)2
= q−24q2+296q3−2528q4+16928q5+· · ·
Y := Y4 = X4(q3) = q3−24q6+296q9−2528q12+16928q15+ · · ·
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0
Hypothesis 4.
Fix an integer N ≥ 2, called the degree. Let Y = Y (q) = X (qN).
The functions X and Y are algebraically dependent. That is, thereexists an irreducible polynomial g(X ,Y ), not identically zero, suchthat g(X ,Y ) = 0. The equation g(X ,Y ) = 0 is said to be amodular equation of degree N.
Example: level ` = 4, degree N = 3
X := X4 =
(η41η
44
η42 Z4
)2
= q−24q2+296q3−2528q4+16928q5+· · ·
Y := Y4 = X4(q3) = q3−24q6+296q9−2528q12+16928q15+ · · ·
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0
Hypothesis 4.
Fix an integer N ≥ 2, called the degree. Let Y = Y (q) = X (qN).
The functions X and Y are algebraically dependent. That is, thereexists an irreducible polynomial g(X ,Y ), not identically zero, suchthat g(X ,Y ) = 0. The equation g(X ,Y ) = 0 is said to be amodular equation of degree N.
Example: level ` = 4, degree N = 3
X := X4 =
(η41η
44
η42 Z4
)2
= q−24q2+296q3−2528q4+16928q5+· · ·
Y := Y4 = X4(q3) = q3−24q6+296q9−2528q12+16928q15+ · · ·
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0
Hypothesis 4.
Fix an integer N ≥ 2, called the degree. Let Y = Y (q) = X (qN).
The functions X and Y are algebraically dependent. That is, thereexists an irreducible polynomial g(X ,Y ), not identically zero, suchthat g(X ,Y ) = 0. The equation g(X ,Y ) = 0 is said to be amodular equation of degree N.
Example: level ` = 4, degree N = 3
X := X4 =
(η41η
44
η42 Z4
)2
= q−24q2+296q3−2528q4+16928q5+· · ·
Y := Y4 = X4(q3) = q3−24q6+296q9−2528q12+16928q15+ · · ·
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0
Hypothesis 4.
Fix an integer N ≥ 2, called the degree. Let Y = Y (q) = X (qN).
The functions X and Y are algebraically dependent. That is, thereexists an irreducible polynomial g(X ,Y ), not identically zero, suchthat g(X ,Y ) = 0. The equation g(X ,Y ) = 0 is said to be amodular equation of degree N.
Example: level ` = 4, degree N = 3
X := X4 =
(η41η
44
η42 Z4
)2
= q−24q2+296q3−2528q4+16928q5+· · ·
Y := Y4 = X4(q3) = q3−24q6+296q9−2528q12+16928q15+ · · ·
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0
Hypothesis 4.
Fix an integer N ≥ 2, called the degree. Let Y = Y (q) = X (qN).
The functions X and Y are algebraically dependent. That is, thereexists an irreducible polynomial g(X ,Y ), not identically zero, suchthat g(X ,Y ) = 0. The equation g(X ,Y ) = 0 is said to be amodular equation of degree N.
Example: level ` = 4, degree N = 3
X := X4 =
(η41η
44
η42 Z4
)2
= q−24q2+296q3−2528q4+16928q5+· · ·
Y := Y4 = X4(q3) = q3−24q6+296q9−2528q12+16928q15+ · · ·
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0
Let X , Y , Z , B, h(j), ` and N be as for Hypotheses 1–4.
Let λ = λ(q) be defined by
λ =X
2
d
dX
(Y B(Y )
X B(X )÷ dY
dX
),
where the derivatives may be computed from the modular equationby implicit differentiation.Let X`,N , B`,N and λ`,N be defined by
X`,N = X(e−2π√
N/`), B`,N = B (X`,N)
andλ`,N = λ
(e−2π/
√N`).
Then
∞∑j=0
h(j) (j + λ`,N) (X`,N)j =1
2π× 1
B`,N×√`
N.
Let X , Y , Z , B, h(j), ` and N be as for Hypotheses 1–4.Let λ = λ(q) be defined by
λ =X
2
d
dX
(Y B(Y )
X B(X )÷ dY
dX
),
where the derivatives may be computed from the modular equationby implicit differentiation.
Let X`,N , B`,N and λ`,N be defined by
X`,N = X(e−2π√
N/`), B`,N = B (X`,N)
andλ`,N = λ
(e−2π/
√N`).
Then
∞∑j=0
h(j) (j + λ`,N) (X`,N)j =1
2π× 1
B`,N×√`
N.
Let X , Y , Z , B, h(j), ` and N be as for Hypotheses 1–4.Let λ = λ(q) be defined by
λ =X
2
d
dX
(Y B(Y )
X B(X )÷ dY
dX
),
where the derivatives may be computed from the modular equationby implicit differentiation.Let X`,N , B`,N and λ`,N be defined by
X`,N = X(e−2π√
N/`), B`,N = B (X`,N)
andλ`,N = λ
(e−2π/
√N`).
Then
∞∑j=0
h(j) (j + λ`,N) (X`,N)j =1
2π× 1
B`,N×√`
N.
Let X , Y , Z , B, h(j), ` and N be as for Hypotheses 1–4.Let λ = λ(q) be defined by
λ =X
2
d
dX
(Y B(Y )
X B(X )÷ dY
dX
),
where the derivatives may be computed from the modular equationby implicit differentiation.Let X`,N , B`,N and λ`,N be defined by
X`,N = X(e−2π√
N/`), B`,N = B (X`,N)
andλ`,N = λ
(e−2π/
√N`).
Then
∞∑j=0
h(j) (j + λ`,N) (X`,N)j =1
2π× 1
B`,N×√`
N.
Example:
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
Here, the level is 4 and degree is 3.
By Hypothesis 2 with ` = 4, we have
Z4 =1
3
(4P(q4)− P(q)
), X4 =
(η41η
44
η42 Z4
)2
Z4 =∞∑j=0
(2j
j
)3
X j4, so h(j) =
(2j
j
)3
.
Example:
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
Here, the level is 4 and degree is 3.
By Hypothesis 2 with ` = 4, we have
Z4 =1
3
(4P(q4)− P(q)
), X4 =
(η41η
44
η42 Z4
)2
Z4 =∞∑j=0
(2j
j
)3
X j4, so h(j) =
(2j
j
)3
.
Example:
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
Here, the level is 4 and degree is 3.
By Hypothesis 2 with ` = 4, we have
Z4 =1
3
(4P(q4)− P(q)
), X4 =
(η41η
44
η42 Z4
)2
Z4 =∞∑j=0
(2j
j
)3
X j4, so h(j) =
(2j
j
)3
.
The degree is N = 3. With Y (q) = X (q3) we have (Hypothesis 4)
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0
where
X = q∞∏j=1
(1− qj)24(1− q4j)24
(1− q2j)48.
The degree is N = 3. With Y (q) = X (q3) we have (Hypothesis 4)
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0
When q = e−2π√
N/` = e−π/√3, Hypothesis 1 implies X = Y .
The modular equation reduces to
X 2(1− 256X )(1− 16X )(1 + 64X 2) = 0.
A numerical approximation determines that X (e−π/√3) = 1
256 .
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
The degree is N = 3. With Y (q) = X (q3) we have (Hypothesis 4)
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0
When q = e−2π√
N/` = e−π/√3, Hypothesis 1 implies X = Y .
The modular equation reduces to
X 2(1− 256X )(1− 16X )(1 + 64X 2) = 0.
A numerical approximation determines that X (e−π/√3) = 1
256 .
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
The degree is N = 3. With Y (q) = X (q3) we have (Hypothesis 4)
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0
When q = e−2π√
N/` = e−π/√3, Hypothesis 1 implies X = Y .
The modular equation reduces to
X 2(1− 256X )(1− 16X )(1 + 64X 2) = 0.
A numerical approximation determines that X (e−π/√3) = 1
256 .
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
The degree is N = 3. With Y (q) = X (q3) we have (Hypothesis 4)
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0
When q = e−2π√
N/` = e−π/√3, Hypothesis 1 implies X = Y .
The modular equation reduces to
X 2(1− 256X )(1− 16X )(1 + 64X 2) = 0.
A numerical approximation determines that X (e−π/√3) = 1
256 .
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
The degree is N = 3. With Y (q) = X (q3) we have (Hypothesis 4)
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0
When q = e−2π√
N/` = e−π/√3, Hypothesis 1 implies X = Y .
The modular equation reduces to
X 2(1− 256X )(1− 16X )(1 + 64X 2) = 0.
A numerical approximation determines that X (e−π/√3) = 1
256 .
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
∞∑j=0
h(j) (j + λ`,N) (X`,N)j =1
2π× 1
B`,N×√`
N.
` = 4, N = 3, X`,N =1
256
B =√
1− 64X , B4,3 =√
1− 64/256 =
√3
2, by Hypothesis 3.
1
2π× 1
B`,N×√`
N=
1
2π× 2√
3×√
4
3=
2
3× 1
π.
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
∞∑j=0
h(j) (j + λ`,N) (X`,N)j =1
2π× 1
B`,N×√`
N.
` = 4, N = 3, X`,N =1
256
B =√
1− 64X , B4,3 =√
1− 64/256 =
√3
2, by Hypothesis 3.
1
2π× 1
B`,N×√`
N=
1
2π× 2√
3×√
4
3=
2
3× 1
π.
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
∞∑j=0
h(j) (j + λ`,N) (X`,N)j =1
2π× 1
B`,N×√`
N.
` = 4, N = 3, X`,N =1
256
B =√
1− 64X ,
B4,3 =√
1− 64/256 =
√3
2, by Hypothesis 3.
1
2π× 1
B`,N×√`
N=
1
2π× 2√
3×√
4
3=
2
3× 1
π.
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
∞∑j=0
h(j) (j + λ`,N) (X`,N)j =1
2π× 1
B`,N×√`
N.
` = 4, N = 3, X`,N =1
256
B =√
1− 64X , B4,3 =√
1− 64/256 =
√3
2, by Hypothesis 3.
1
2π× 1
B`,N×√`
N=
1
2π× 2√
3×√
4
3=
2
3× 1
π.
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
∞∑j=0
h(j) (j + λ`,N) (X`,N)j =1
2π× 1
B`,N×√`
N.
` = 4, N = 3, X`,N =1
256
B =√
1− 64X , B4,3 =√
1− 64/256 =
√3
2, by Hypothesis 3.
1
2π× 1
B`,N×√`
N=
1
2π× 2√
3×√
4
3=
2
3× 1
π.
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
∞∑j=0
h(j) (j + λ`,N) (X`,N)j =1
2π× 1
B`,N×√`
N.
` = 4, N = 3, X`,N =1
256
B =√
1− 64X , B4,3 =√
1− 64/256 =
√3
2, by Hypothesis 3.
1
2π× 1
B`,N×√`
N=
1
2π× 2√
3×√
4
3=
2
3× 1
π.
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
λ =X
2
d
dX
(Y B(Y )
X B(X )÷ dY
dX
),
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0,
B(X ) =√
1− 64X .
Then, put X = Y = 1/256 to get λ = 1/6.
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
This is Ramanujan’s formula (28).
λ =X
2
d
dX
(Y B(Y )
X B(X )÷ dY
dX
),
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0,
B(X ) =√
1− 64X .
Then, put X = Y = 1/256 to get λ = 1/6.
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
This is Ramanujan’s formula (28).
λ =X
2
d
dX
(Y B(Y )
X B(X )÷ dY
dX
),
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0,
B(X ) =√
1− 64X .
Then, put X = Y = 1/256 to get λ = 1/6.
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
This is Ramanujan’s formula (28).
λ =X
2
d
dX
(Y B(Y )
X B(X )÷ dY
dX
),
X 4+Y 4−16777216X 3Y 3+294912 (X 3Y 2+X 2Y 3)−900(X 3Y+XY 3)
+28422X 2Y 2 + 72(X 2Y + XY 2)− XY = 0,
B(X ) =√
1− 64X .
Then, put X = Y = 1/256 to get λ = 1/6.
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π.
This is Ramanujan’s formula (28).
Other examples
∞∑j=0
(2j
j
)3 (42j + 5)
212j=
16
π
Ramanujan–Disney series
Other examples
∞∑j=0
(2j
j
)3 (42j + 5)
212j=
16
π
Ramanujan–Disney series
Other examples
∞∑j=0
(2j
j
)3 (42j + 5)
212j=
16
π
Ramanujan–Disney series
Other examples:
∞∑n=0
(4n
2n
)(2n
n
)2(n +
1103
26390
)(1
396
)4n
=9801
√2
105560× 1
π.
It is Ramanujan’s formula (44), his last example.
Each term adds 8 decimal digits.
It comes from the level 2 theory, with degree 29.
Used by W. Gosper in 1985 to compute over 17,500,000 decimaldigits of π.
The value 1103 hadn’t been proved when Gosper set the record.
There are now three proofs of this result, all within the last twoyears.
Other examples:
∞∑n=0
(4n
2n
)(2n
n
)2(n +
1103
26390
)(1
396
)4n
=9801
√2
105560× 1
π.
It is Ramanujan’s formula (44), his last example.
Each term adds 8 decimal digits.
It comes from the level 2 theory, with degree 29.
Used by W. Gosper in 1985 to compute over 17,500,000 decimaldigits of π.
The value 1103 hadn’t been proved when Gosper set the record.
There are now three proofs of this result, all within the last twoyears.
Other examples:
∞∑n=0
(4n
2n
)(2n
n
)2(n +
1103
26390
)(1
396
)4n
=9801
√2
105560× 1
π.
It is Ramanujan’s formula (44), his last example.
Each term adds 8 decimal digits.
It comes from the level 2 theory, with degree 29.
Used by W. Gosper in 1985 to compute over 17,500,000 decimaldigits of π.
The value 1103 hadn’t been proved when Gosper set the record.
There are now three proofs of this result, all within the last twoyears.
Other examples:
∞∑n=0
(4n
2n
)(2n
n
)2(n +
1103
26390
)(1
396
)4n
=9801
√2
105560× 1
π.
It is Ramanujan’s formula (44), his last example.
Each term adds 8 decimal digits.
It comes from the level 2 theory, with degree 29.
Used by W. Gosper in 1985 to compute over 17,500,000 decimaldigits of π.
The value 1103 hadn’t been proved when Gosper set the record.
There are now three proofs of this result, all within the last twoyears.
Other examples:
∞∑n=0
(4n
2n
)(2n
n
)2(n +
1103
26390
)(1
396
)4n
=9801
√2
105560× 1
π.
It is Ramanujan’s formula (44), his last example.
Each term adds 8 decimal digits.
It comes from the level 2 theory, with degree 29.
Used by W. Gosper in 1985 to compute over 17,500,000 decimaldigits of π.
The value 1103 hadn’t been proved when Gosper set the record.
There are now three proofs of this result, all within the last twoyears.
Other examples:
∞∑n=0
(4n
2n
)(2n
n
)2(n +
1103
26390
)(1
396
)4n
=9801
√2
105560× 1
π.
It is Ramanujan’s formula (44), his last example.
Each term adds 8 decimal digits.
It comes from the level 2 theory, with degree 29.
Used by W. Gosper in 1985 to compute over 17,500,000 decimaldigits of π.
The value 1103 hadn’t been proved when Gosper set the record.
There are now three proofs of this result, all within the last twoyears.
Other examples:
∞∑n=0
(4n
2n
)(2n
n
)2(n +
1103
26390
)(1
396
)4n
=9801
√2
105560× 1
π.
It is Ramanujan’s formula (44), his last example.
Each term adds 8 decimal digits.
It comes from the level 2 theory, with degree 29.
Used by W. Gosper in 1985 to compute over 17,500,000 decimaldigits of π.
The value 1103 hadn’t been proved when Gosper set the record.
There are now three proofs of this result, all within the last twoyears.
∞∑n=0
(6n
3n
)(3n
n
)(2n
n
)(n +
13591409
545140134
)(−1
640320
)3n
=26 × 53/2 × 233/2 × 293/2
33/2 × 7× 11× 19× 127× 163× 1
π
David and Gregory Chudnovsky (1987).
Level 1, degree 163.
Each term contributes roughly 14 decimal digits per term.
Used by the Chudnovskys (and others) in world-record calculationsof π.
∞∑n=0
(6n
3n
)(3n
n
)(2n
n
)(n +
13591409
545140134
)(−1
640320
)3n
=26 × 53/2 × 233/2 × 293/2
33/2 × 7× 11× 19× 127× 163× 1
π
David and Gregory Chudnovsky (1987).
Level 1, degree 163.
Each term contributes roughly 14 decimal digits per term.
Used by the Chudnovskys (and others) in world-record calculationsof π.
∞∑n=0
(6n
3n
)(3n
n
)(2n
n
)(n +
13591409
545140134
)(−1
640320
)3n
=26 × 53/2 × 233/2 × 293/2
33/2 × 7× 11× 19× 127× 163× 1
π
David and Gregory Chudnovsky (1987).
Level 1, degree 163.
Each term contributes roughly 14 decimal digits per term.
Used by the Chudnovskys (and others) in world-record calculationsof π.
∞∑n=0
(6n
3n
)(3n
n
)(2n
n
)(n +
13591409
545140134
)(−1
640320
)3n
=26 × 53/2 × 233/2 × 293/2
33/2 × 7× 11× 19× 127× 163× 1
π
David and Gregory Chudnovsky (1987).
Level 1, degree 163.
Each term contributes roughly 14 decimal digits per term.
Used by the Chudnovskys (and others) in world-record calculationsof π.
Data for h(j) =(
6j3j
)(3jj
)(2jj
)q, ` N x λ Ramanujan
2 1203
328
q = e−2π√N 3 4
603111 (33)
` = 1 4 1663
563
7 12553
8133 (34)
7 −1153
863
11 −1323
15154
q = −e−π√N 19 −1
96325342
` = 4 27 −94803
31506
43 −19603
2635418
67 −152803
10177261702
163 −16403203
13591409545140134
Data for h(j) =(
4j2j
)(2jj
)2
q, ` N x λ Ramanujan
2 264
17
3 1482
18 (40)
q = e−2π√
N/2 5 1124
110 (41)
` = 2 9 1284
340 (42)
11 115842
19280 (43)
29 13964
110326390 (44)
5 −1210
320 (35)
7 −1632
865
q = −e−π√N 9 −3
1922328 (36)
` = 4 13 −13242
23260 (37)
25 −53602
41644 (38)
37 −16722
112321460 (39)
Data for h(j) =(
3jj
)(2jj
)2
q, ` N x λ Ramanujan
2 163
16
e−2π√
N/3 4 4183
215 (31)
5 1153
433 (32)
9 −1192
15
17 −1123
751
−e−π√
N/3 25 −25603
19
41 −1483
53615
49 −492523
13165
89 −13003
82714151
Data for h(j) =(
2jj
)3
q = e−π√N 3 1
25616 (28)
` = 4 7 14096
542 (29)
q = −e−π√N 2 −1
6414
` = 4 4 −1512
16
Directions for further research
Sporadic sequences
f (4) :=4P(q4)− P(q)
3=∞∑j=0
(2j
j
)3( η41η44
η42 f (4)
)2j
f (6) :=30P(q6)− 3P(q3) + 2P(q2)− 5P(q)
24
=∞∑j=0
{j∑
k=0
(j
k
)2(j + k
k
)2}(
η1η2η3η6f (6)
)2j
(j+1)3tj+1 = (2j+1)(17j2+17j+5)tj− j3tj−1 : Apery, ζ(3) 6∈ Q
Not obvious from the recurrence relation that tj is always aninteger.
Sporadic sequences
f (4) :=4P(q4)− P(q)
3=∞∑j=0
(2j
j
)3( η41η44
η42 f (4)
)2j
f (6) :=30P(q6)− 3P(q3) + 2P(q2)− 5P(q)
24
=∞∑j=0
{j∑
k=0
(j
k
)2(j + k
k
)2}(
η1η2η3η6f (6)
)2j
(j+1)3tj+1 = (2j+1)(17j2+17j+5)tj− j3tj−1 : Apery, ζ(3) 6∈ Q
Not obvious from the recurrence relation that tj is always aninteger.
Sporadic sequences
f (4) :=4P(q4)− P(q)
3=∞∑j=0
(2j
j
)3( η41η44
η42 f (4)
)2j
f (6) :=30P(q6)− 3P(q3) + 2P(q2)− 5P(q)
24
=∞∑j=0
{j∑
k=0
(j
k
)2(j + k
k
)2}(
η1η2η3η6f (6)
)2j
(j+1)3tj+1 = (2j+1)(17j2+17j+5)tj− j3tj−1 : Apery, ζ(3) 6∈ Q
Not obvious from the recurrence relation that tj is always aninteger.
Apery, 1978
(k + 1)2sk+1 = (11k2 + 11k + 3)sk + k2sk−1, s0 = 1
sk =k∑
j=0
(k
j
)2(k + j
j
)
Franel, 1894
(k + 1)2sk+1 = (7k2 + 7k + 2)sk + 8k2sk−1, s0 = 1
sk =k∑
j=0
(k
j
)3
Zagier, 1998, 2009
(k + 1)2sk+1 = (ak2 + ak + b)sk + ck2sk−1, s0 = 1
Apery, 1978
(k + 1)2sk+1 = (11k2 + 11k + 3)sk + k2sk−1, s0 = 1
sk =k∑
j=0
(k
j
)2(k + j
j
)
Franel, 1894
(k + 1)2sk+1 = (7k2 + 7k + 2)sk + 8k2sk−1, s0 = 1
sk =k∑
j=0
(k
j
)3
Zagier, 1998, 2009
(k + 1)2sk+1 = (ak2 + ak + b)sk + ck2sk−1, s0 = 1
(k + 1)2sk+1 = (ak2 + ak + b)sk + ck2sk−1, s0 = 1
(a, b, c) s(k)
(11, 3, 1)∑j
(k
j
)2(k + j
j
)(−17,−6,−72)
∑j ,`
(−8)k−j(k
j
)(j
`
)3
(10, 3,−9)∑j
(k
j
)2(2j
j
)(7, 2, 8)
∑j
(k
j
)3
(12, 4,−32)∑j
4k−2j(k
2j
)(2j
j
)2
(−9,−3,−27)∑j
(−3)k−3j(k
j
)(k − j
j
)(k − 2j
j
)
Analogue of Beukers’ result
(k + 1)2sk+1 = (7k2 + 7k + 2)sk + 8k2sk−1, s0 = 1
z =∞∑k=0
skxk =
∞∑k=0
k∑
j=0
(k
j
)3 xk
x = q∞∏j=1
(1− qj)3(1− q6j)9
(1− q2j)3(1− q3j)9= rc(q)3
z =∞∏j=1
(1− q2j)(1− q3j)6
(1− qj)2(1− q6j)3
1. rc(q) is Ramanujan’s cubic continued fraction2. Similar results hold for Zagier’s other examples
Analogue of Beukers’ result
(k + 1)2sk+1 = (7k2 + 7k + 2)sk + 8k2sk−1, s0 = 1
z =∞∑k=0
skxk =
∞∑k=0
k∑
j=0
(k
j
)3 xk
x = q∞∏j=1
(1− qj)3(1− q6j)9
(1− q2j)3(1− q3j)9= rc(q)3
z =∞∏j=1
(1− q2j)(1− q3j)6
(1− qj)2(1− q6j)3
1. rc(q) is Ramanujan’s cubic continued fraction2. Similar results hold for Zagier’s other examples
(k + 1)2sk+1 = (ak2 + ak + b)sk + ck2sk−1, s0 = 1
(a, b, c) Level
(11, 3, 1) 5
(−17,−6,−72) 6
(10, 3,−9) 6
(7, 2, 8) 6
(12, 4,−32) 8
(−9,−3,−27) 9
The sequences satisfy several congruence properties.
Another computer search (C., 2012)
(n + 1)3vn+1 = (2n + 1)(an2 + an + b)vn + n(cn2 + d)vn−1
where v0 = 1, v−1 = 0.
(a, b, c , d) = (13, 4, 27,−3): appears to be integer valued.
What are the parameterizing modular forms?
z =∞∑n=0
vn xn, z =?, x =?
z =7P(q7)− P(q)
6, x =
x
1 + 13y + 49y2,
where
y := q∞∏j=1
(1− q7j)4
(1− qj)4
Another computer search (C., 2012)
(n + 1)3vn+1 = (2n + 1)(an2 + an + b)vn + n(cn2 + d)vn−1
where v0 = 1, v−1 = 0.
(a, b, c , d) = (13, 4, 27,−3): appears to be integer valued.
What are the parameterizing modular forms?
z =∞∑n=0
vn xn, z =?, x =?
z =7P(q7)− P(q)
6, x =
x
1 + 13y + 49y2,
where
y := q∞∏j=1
(1− q7j)4
(1− qj)4
Another computer search (C., 2012)
(n + 1)3vn+1 = (2n + 1)(an2 + an + b)vn + n(cn2 + d)vn−1
where v0 = 1, v−1 = 0.
(a, b, c , d) = (13, 4, 27,−3): appears to be integer valued.
What are the parameterizing modular forms?
z =∞∑n=0
vn xn, z =?, x =?
z =7P(q7)− P(q)
6, x =
x
1 + 13y + 49y2,
where
y := q∞∏j=1
(1− q7j)4
(1− qj)4
Another computer search (C., 2012)
(n + 1)3vn+1 = (2n + 1)(an2 + an + b)vn + n(cn2 + d)vn−1
where v0 = 1, v−1 = 0.
(a, b, c , d) = (13, 4, 27,−3): appears to be integer valued.
What are the parameterizing modular forms?
z =∞∑n=0
vn xn, z =?, x =?
z =7P(q7)− P(q)
6, x =
x
1 + 13y + 49y2,
where
y := q∞∏j=1
(1− q7j)4
(1− qj)4
Another computer search (C., 2012)
(n + 1)3vn+1 = (2n + 1)(an2 + an + b)vn + n(cn2 + d)vn−1
where v0 = 1, v−1 = 0.
(a, b, c , d) = (13, 4, 27,−3): appears to be integer valued.
What are the parameterizing modular forms?
z =∞∑n=0
vn xn, z =?, x =?
z =7P(q7)− P(q)
6, x =
x
1 + 13y + 49y2,
where
y := q∞∏j=1
(1− q7j)4
(1− qj)4
(n + 1)3vn+1 = (2n + 1)(an2 + an + b)vn + n(cn2 + d)vn−1
where v0 = 1, v−1 = 0.
(a, b, c , d) = (13, 4, 27,−3): appears to be integer valued.
In fact, W. Zudilin found that
vn =n∑
j=0
(n
j
)2(2j
n
)(n + j
j
).
Series for 1/π can be obtained, as before, e.g.,
1
π=√
7∞∑n=0
(−1)nvn(11895n + 1286)
223n+3.
Level 7, degree 61.
Other sporadic sequences, e.g., complex? Four term relations?
(n + 1)3vn+1 = (2n + 1)(an2 + an + b)vn + n(cn2 + d)vn−1
where v0 = 1, v−1 = 0.
(a, b, c , d) = (13, 4, 27,−3): appears to be integer valued.
In fact, W. Zudilin found that
vn =n∑
j=0
(n
j
)2(2j
n
)(n + j
j
).
Series for 1/π can be obtained, as before, e.g.,
1
π=√
7∞∑n=0
(−1)nvn(11895n + 1286)
223n+3.
Level 7, degree 61.
Other sporadic sequences, e.g., complex? Four term relations?
(n + 1)3vn+1 = (2n + 1)(an2 + an + b)vn + n(cn2 + d)vn−1
where v0 = 1, v−1 = 0.
(a, b, c , d) = (13, 4, 27,−3): appears to be integer valued.
In fact, W. Zudilin found that
vn =n∑
j=0
(n
j
)2(2j
n
)(n + j
j
).
Series for 1/π can be obtained, as before, e.g.,
1
π=√
7∞∑n=0
(−1)nvn(11895n + 1286)
223n+3.
Level 7, degree 61.
Other sporadic sequences, e.g., complex? Four term relations?
2. Clausen-type formulas
Chan, Tanigawa, Yang and Zudilin (2011)
(1 + cw2)
( ∞∑k=0
skwk
)2
=∞∑k=0
(2k
k
)sk
(w(1− aw − cw2)
(1 + cw2)2
)k
Almkvist, van Straten and Zudilin (2011)
(1− aw − cw2)
( ∞∑k=0
skwk
)2
=∞∑k=0
tk
(w
1− aw − cw2
)k
(k + 1)2sk+1 = (ak2 + ak + b)sk + ck2sk−1
(k + 1)3tk+1 = −(2k + 1)(ak2 + ak + a− 2b)tk − (4c + a2)k3tk−1
s0 = t0 = 1 s−1 = t−1 = 0
2. Clausen-type formulas
Chan, Tanigawa, Yang and Zudilin (2011)
(1 + cw2)
( ∞∑k=0
skwk
)2
=∞∑k=0
(2k
k
)sk
(w(1− aw − cw2)
(1 + cw2)2
)k
Almkvist, van Straten and Zudilin (2011)
(1− aw − cw2)
( ∞∑k=0
skwk
)2
=∞∑k=0
tk
(w
1− aw − cw2
)k
(k + 1)2sk+1 = (ak2 + ak + b)sk + ck2sk−1
(k + 1)3tk+1 = −(2k + 1)(ak2 + ak + a− 2b)tk − (4c + a2)k3tk−1
s0 = t0 = 1 s−1 = t−1 = 0
2. Clausen-type formulas
Chan, Tanigawa, Yang and Zudilin (2011)
(1 + cw2)
( ∞∑k=0
skwk
)2
=∞∑k=0
(2k
k
)sk
(w(1− aw − cw2)
(1 + cw2)2
)k
Almkvist, van Straten and Zudilin (2011)
(1− aw − cw2)
( ∞∑k=0
skwk
)2
=∞∑k=0
tk
(w
1− aw − cw2
)k
(k + 1)2sk+1 = (ak2 + ak + b)sk + ck2sk−1
(k + 1)3tk+1 = −(2k + 1)(ak2 + ak + a− 2b)tk − (4c + a2)k3tk−1
s0 = t0 = 1 s−1 = t−1 = 0
2. Clausen-type formulas
Chan, Tanigawa, Yang and Zudilin (2011)
(1 + cw2)
( ∞∑k=0
skwk
)2
=∞∑k=0
(2k
k
)sk
(w(1− aw − cw2)
(1 + cw2)2
)k
Almkvist, van Straten and Zudilin (2011)
(1− aw − cw2)
( ∞∑k=0
skwk
)2
=∞∑k=0
tk
(w
1− aw − cw2
)k
(k + 1)2sk+1 = (ak2 + ak + b)sk + ck2sk−1
(k + 1)3tk+1 = −(2k + 1)(ak2 + ak + a− 2b)tk − (4c + a2)k3tk−1
s0 = t0 = 1 s−1 = t−1 = 0
3. q-analogues
(x + y)n =n∑
j=0
(n
j
)xn−jyk
(n
j
)=
n!
j!(n − j)!
Exercise: find and prove a formula for the coefficients c(n, j , q) inthe expansion of
(x + y)(x + qy)(x + q2y) · · · (x + qn−1y) =n∑
j=0
c(n, j , q)xn−jy j .
Called: the q-binomial theorem. (See: Polya and Alexanderson)
3. q-analogues
(x + y)n =n∑
j=0
(n
j
)xn−jyk
(n
j
)=
n!
j!(n − j)!
Exercise: find and prove a formula for the coefficients c(n, j , q) inthe expansion of
(x + y)(x + qy)(x + q2y) · · · (x + qn−1y) =n∑
j=0
c(n, j , q)xn−jy j .
Called: the q-binomial theorem. (See: Polya and Alexanderson)
3. q-analogues
(x + y)n =n∑
j=0
(n
j
)xn−jyk
(n
j
)=
n!
j!(n − j)!
Exercise: find and prove a formula for the coefficients c(n, j , q) inthe expansion of
(x + y)(x + qy)(x + q2y) · · · (x + qn−1y) =n∑
j=0
c(n, j , q)xn−jy j .
Called: the q-binomial theorem. (See: Polya and Alexanderson)
3. q-analogues
(x + y)n =n∑
j=0
(n
j
)xn−jyk
(n
j
)=
n!
j!(n − j)!
Exercise: find and prove a formula for the coefficients c(n, j , q) inthe expansion of
(x + y)(x + qy)(x + q2y) · · · (x + qn−1y) =n∑
j=0
c(n, j , q)xn−jy j .
Called: the q-binomial theorem. (See: Polya and Alexanderson)
q-analogues
n!q = 1× (1 + q)× (1 + q + q2)× · · · × (1 + q + q2 + · · · qn−1)
=(1− q)
(1− q)× (1− q2)
(1− q)× (1− q3)
(1− q)× · · · × (1− qn)
(1− q)
→ 1× 2× 3× · · · × n = n! as q → 1.
Define the q-integer [n]q by
[n]q = 1 + q + q2 + · · ·+ qn−1 =1− qn
1− q.
Then n!q = [1]q × [2]q × · · · × [n]q.
q-analogues
n!q = 1× (1 + q)× (1 + q + q2)× · · · × (1 + q + q2 + · · · qn−1)
=(1− q)
(1− q)× (1− q2)
(1− q)× (1− q3)
(1− q)× · · · × (1− qn)
(1− q)
→ 1× 2× 3× · · · × n = n! as q → 1.
Define the q-integer [n]q by
[n]q = 1 + q + q2 + · · ·+ qn−1 =1− qn
1− q.
Then n!q = [1]q × [2]q × · · · × [n]q.
q-analogues
n!q = 1× (1 + q)× (1 + q + q2)× · · · × (1 + q + q2 + · · · qn−1)
=(1− q)
(1− q)× (1− q2)
(1− q)× (1− q3)
(1− q)× · · · × (1− qn)
(1− q)
→ 1× 2× 3× · · · × n = n! as q → 1.
Define the q-integer [n]q by
[n]q = 1 + q + q2 + · · ·+ qn−1 =1− qn
1− q.
Then n!q = [1]q × [2]q × · · · × [n]q.
q-analogues
n!q = 1× (1 + q)× (1 + q + q2)× · · · × (1 + q + q2 + · · · qn−1)
=(1− q)
(1− q)× (1− q2)
(1− q)× (1− q3)
(1− q)× · · · × (1− qn)
(1− q)
→ 1× 2× 3× · · · × n = n! as q → 1.
Define the q-integer [n]q by
[n]q = 1 + q + q2 + · · ·+ qn−1 =1− qn
1− q.
Then n!q = [1]q × [2]q × · · · × [n]q.
q-analogues
n!q = 1× (1 + q)× (1 + q + q2)× · · · × (1 + q + q2 + · · · qn−1)
=(1− q)
(1− q)× (1− q2)
(1− q)× (1− q3)
(1− q)× · · · × (1− qn)
(1− q)
→ 1× 2× 3× · · · × n = n! as q → 1.
Define the q-integer [n]q by
[n]q = 1 + q + q2 + · · ·+ qn−1 =1− qn
1− q.
Then n!q = [1]q × [2]q × · · · × [n]q.
Ramanujan’s formula
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π
has the q-analogue
∞∑j=0
qj2[6j + 1]q
(q; q2)2k(q2; q4)k(q4; q4)3k
=(1 + q)(q2; q4)∞(q6; q4)∞
(q4; q4)2∞.
Notation:
(x ; q)∞ = (1− x)(1− qx)(1− q2x)(1− q3x) · · · , |q| < 1.
Victor J. W. Guo and Ji-Cai Liu.
Ramanujan’s formula
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π
has the q-analogue
∞∑j=0
qj2[6j + 1]q
(q; q2)2k(q2; q4)k(q4; q4)3k
=(1 + q)(q2; q4)∞(q6; q4)∞
(q4; q4)2∞.
Notation:
(x ; q)∞ = (1− x)(1− qx)(1− q2x)(1− q3x) · · · , |q| < 1.
Victor J. W. Guo and Ji-Cai Liu.
Ramanujan’s formula
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π
has the q-analogue
∞∑j=0
qj2[6j + 1]q
(q; q2)2k(q2; q4)k(q4; q4)3k
=(1 + q)(q2; q4)∞(q6; q4)∞
(q4; q4)2∞.
Notation:
(x ; q)∞ = (1− x)(1− qx)(1− q2x)(1− q3x) · · · , |q| < 1.
Victor J. W. Guo and Ji-Cai Liu.
Ramanujan’s formula
∞∑j=0
(2j
j
)3(j +
1
6
)(1
256
)j
=2
3× 1
π
has the q-analogue
∞∑j=0
qj2[6j + 1]q
(q; q2)2k(q2; q4)k(q4; q4)3k
=(1 + q)(q2; q4)∞(q6; q4)∞
(q4; q4)2∞.
Notation:
(x ; q)∞ = (1− x)(1− qx)(1− q2x)(1− q3x) · · · , |q| < 1.
Victor J. W. Guo and Ji-Cai Liu.
The last slide.
1. Elliptic functions
2. Modular functions
3. Jacobi’s inversion formula
4. Parametrization of hypergeometric and Heun functions bymodular forms
5. All of the above lead to Ramanujan-type series for 1/π.
6. Many ideas are involved. That leads to fruitful researchquestions.
The end.
The last slide.
1. Elliptic functions
2. Modular functions
3. Jacobi’s inversion formula
4. Parametrization of hypergeometric and Heun functions bymodular forms
5. All of the above lead to Ramanujan-type series for 1/π.
6. Many ideas are involved. That leads to fruitful researchquestions.
The end.
The last slide.
1. Elliptic functions
2. Modular functions
3. Jacobi’s inversion formula
4. Parametrization of hypergeometric and Heun functions bymodular forms
5. All of the above lead to Ramanujan-type series for 1/π.
6. Many ideas are involved. That leads to fruitful researchquestions.
The end.
The last slide.
1. Elliptic functions
2. Modular functions
3. Jacobi’s inversion formula
4. Parametrization of hypergeometric and Heun functions bymodular forms
5. All of the above lead to Ramanujan-type series for 1/π.
6. Many ideas are involved. That leads to fruitful researchquestions.
The end.
The last slide.
1. Elliptic functions
2. Modular functions
3. Jacobi’s inversion formula
4. Parametrization of hypergeometric and Heun functions bymodular forms
5. All of the above lead to Ramanujan-type series for 1/π.
6. Many ideas are involved. That leads to fruitful researchquestions.
The end.
The last slide.
1. Elliptic functions
2. Modular functions
3. Jacobi’s inversion formula
4. Parametrization of hypergeometric and Heun functions bymodular forms
5. All of the above lead to Ramanujan-type series for 1/π.
6. Many ideas are involved. That leads to fruitful researchquestions.
The end.
The last slide.
1. Elliptic functions
2. Modular functions
3. Jacobi’s inversion formula
4. Parametrization of hypergeometric and Heun functions bymodular forms
5. All of the above lead to Ramanujan-type series for 1/π.
6. Many ideas are involved. That leads to fruitful researchquestions.
The end.
The last slide.
1. Elliptic functions
2. Modular functions
3. Jacobi’s inversion formula
4. Parametrization of hypergeometric and Heun functions bymodular forms
5. All of the above lead to Ramanujan-type series for 1/π.
6. Many ideas are involved. That leads to fruitful researchquestions.
The end.