James Wan MATRIX 17/1/17...James Wan MATRIX 17/1/17 1/24 Ramanujan seriesRecent development Proofs...

Ramanujan series Recent development

Ramanujan-type series for 1/π

James Wan

MATRIX

17/1/17

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Ramanujan series Recent development Proofs

Ramanujan series for 1/π

In 1914, Ramanujan found a class of truly innovative series for 1/πusing algebraic (sometimes rational) summands:

∞∑n=0

(s)n(12)n(1− s)nn!3

(a+ bn) zn0 =1

π,

where a, b, z0 are algebraic, and s ∈ {1/2, 1/3, 1/4, 1/6}.(These are the values of s for which the Pochhammer combinationcan be written in terms of binomial coefficients.)

S. Ramanujan, Modular equations and approximations to π,Quart. J. Math. (Oxford) 45 (1914), 350–372

In terms of hypergeometric series:

4F3

(s, 12 , 1− s, 1 + a

b

1, 1, ab

∣∣∣∣z0) =1

a π.

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Ramanujan series for 1/π – examples

∞∑n=0

( 12 )3nn!3

(1 + 4n) (−1)n =2

π(1)

∞∑n=0

( 12 )3nn!3

(1 + 6n)(−1

8

)n=

2√

2

π(2)

∞∑n=0

( 13 )n( 1

2 )n( 23 )n

n!3(2 + 15n)

(2

27

)n=

27

4π(3)

∞∑n=0

( 14 )n( 1

2 )n( 34 )n

n!3(1 + 7n)

(32

81

)n=

9

2π(4)

∞∑n=0

( 16 )n( 1

2 )n( 56 )n

n!3

(13591409

+ 545140134n

)( −1

533603

)n=

6403203/2

12π(5)

Equation (1) was due to Bauer (1859). Equation (3) was due toRamanujan.

Equation (4) was due to Berndt, Chan, Liaw (2001). Equation (5) was

use by the Chudnovskys to compute over 2 billion digits of π, then a

world record. 3 / 24


Proof technique 1: Wilf-Zeilberger

For Bauer’s series (1), consider

F (m,n) :=(12)2n(−m)n

n!2(m+ 32)n

(1 + 4n) (−1)n.

Construct G(m,n) =2n2

(1 + 4n)(1 +m− n)F (m,n), which

satisfies

G(m,n+ 1)−G(m,n) = F (m,n)− 2m+ 2

2m+ 3F (m+ 1, n).

Sum in n:∑n

F (m,n) =2m+ 2

2m+ 3

∑n

F (m+ 1, n).

It follows that∑n

F (m,n) =Γ(m+ 3/2)

Γ(3/2)Γ(m+ 1).

Let m→ −1/2, which is justified by Carlson’s theorem. �4 / 24


Proof technique 2: hypergeometric evaluations

For Bauer’s series (1), take a = c = d = 1/2 in Dougall’s formulafor a 4F3 at −1:

4F3

(a, a/2 + 1, c, d

a/2, a− c+ 1, a− d+ 1

∣∣∣∣− 1

)=

Γ(a− c+ 1)Γ(a− d+ 1)

Γ(a+ 1)Γ(a− c− d+ 1).

�

Remark : Dougall’s 5F4 formula can be used to prove some seriesfor 1/π2. However, in general, 1/π2 series do not appear to bemodular, and seem to be connected with Calabi-Yau differentialequations.

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Proof technique 2: hypergeometric evaluations

For the series (2), we use the Clausen-type formula,

1√1− x 3F2

( 12 ,

12 ,

12

1, 1

∣∣∣∣ −x24(1− x)

)︸︷︷︸

f(x)

= 2F1

( 12 ,

12

1

∣∣∣∣x)2

.

Observe that the series is simply (1/√

2) f ′(1/2).

We can evaluate this from the RHS expression, using Gauss’second summation theorem for a 2F1 at 1/2 (and its contiguousversions). �

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General idea

Clausen’s formula, 1828

3F2

(s, 12 , 1− s

1, 1

∣∣∣∣4x(1− x)

)= 2F1

(s, 1− s

1

∣∣∣∣x)2

. (6)

A 1/π series is just a suitable linear combination of the above andits derivative, at an appropriate x. We demonstrate the generalproof for s = 1/2.

Complete elliptic integral of the first kind

K(k) :=π

22F1

( 12 ,

12

1

∣∣∣∣k2) ,k′ :=

√1− k2, K ′(k) := K(k′).

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Properties of K and k

Let τ be in the upper half-plane. The Jacobi theta functions are

θ2(τ) =∑n

eπi(n+1/2)2τ , θ3(τ) =∑n

eπin2τ , θ4(τ) =

∑n

(−1)neπin2τ .

Incredibly, with τ = iK ′(k)

K(k), we have

k(τ) =θ22(τ)

θ23(τ), k′(τ) =

θ24(τ)

θ23(τ), K(k(τ)) =

π

2θ23(τ).

This makes k(τ)2 a modular function on the congruence subgroupΓ0(4).

Therefore, when τ is a quadratic irrationality, k(τ) is a computablealgebraic number. We now flesh out some details.

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Modular details

For p ∈ N, k(pτ) is an algebraic function of k(τ), i. e. there is acomputable polynomial Pp with integer coefficients such that

Pp(k(τ), k(pτ)

)= 0.

Pp is known as the pth order modular equation.

At τp :=√−p/p, k(pτp) = k′(τp), so k(τp) is computable and

algebraic.

Moreover, Mp(τ) :=K(k(pτ))

K(k(τ))is also a computable algebraic

function of k(τ), called the multiplier of order p.

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Eisenstein series

Note that

η(τ) =

(θ2(τ)θ3(τ)θ4(τ)

2

)1/3

,

E2(τ) = 1− 24∑n≥1

ne2nπiτ

1− e2nπiτ=

12

πi

d

dqlog η(τ).

It follows that

E2(τ) =4

π2

[(1− 2k2)K(k)2 +

3

2k(1− k2) d

dkK(k)2

]. (7)

At τp =√−p/p, the terms in black are computable.

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Multiplier

Also,

Rp(τ) :=π2

4

pE2(pτ)− E2(τ)

K(k(pτ))K(k(τ)),

is a computable algebraic function of k(τ). (Proof: use (7) andthe multiplier Mp.)

On the other hand,

E2(−1/τ) = τ2E2(τ) +6τ

πi,

so by letting τ = τp =√−p/p, E2(pτp) = −E2(τp)/p+ 6τp/(πi).

Substitute this into the equation for Rp:

E2(τp) =3√p

π− 2

π2Rp(τp)Mp(τp)K(k(τp))

2. (8)

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Completing the proof

We combine equations (7) and (8) to eliminate τp:

(2− 4k(τp)

2 +Rp(τp)Mp(τp)) ( 2

πK(k(τp))

)2+ 3k(τp)

(1− k(τp)

2) d

dk

( 2

πK(k(τp))

)2=

6√p

π.

The brown terms correspond to the RHS of Clausen’s (6). �

Extending the proof to other quadratic irrationalities, and to others, use similar ideas.

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References

J. M. Borwein and P. B. Borwein, Pi and the AGM: A study inanalytic number theory and computational complexity, Wiley,New York, 1987

N. D. Baruah, B. C. Berndt, H. H. Chan, Ramanujan’s Seriesfor 1/π: A Survey, Amer. Math. Monthly 116 (2009),567–587

M. D. Rogers and A. Straub, A solution of Sun’s $520challenge concerning 520

π , IJNT 9 (2013), 1273–1288

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Ramanujan series Recent development Legendre polynomials Hypergeometric ODE

Generalizations

The term(s)n(12)n(1− s)n

n!3can be replaced by any sequence as

long as its generating function can be related to 3F2

(s, 1

2,1−s1,1

∣∣∣∣z).

This includes Apery-like sequences.

Examples:

∞∑n=0

n∑k=0

(n

k

)2(n+ k

k

)2

(10− 3√

5 + 20n)(√5− 1

2

)12n=

√15

6(4√

5− 9)π

∞∑n=0

n∑k=0

(2n− 2k

n− k

)2(2k

k

)2

n( 1

22

)n=

2

π

∞∑n=0

(2n

n

) n∑k=0

(n

k

)3

(2 + 9n)( 1

50

)n=

25

2π

∞∑n=0

n∑k=0

(n

k

)4

(1 + 4n)( 1

36

)n=

18√15π

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Legendre polynomials

Let Pn(x) = 2F1

(−n, n+1

1

∣∣∣∣1−x2 ) denote the Legendre polynomials.

In 2011, Sun conjectured dozens of 1/π series involving Pn(x).They can all be explained by

Brafman’s formula, 1951

∞∑n=0

(s)n(1− s)nn!2

Pn(x)zn

= 2F1

(s, 1− s

1

∣∣∣∣ 1− ρ− z2

)2F1

(s, 1− s

1

∣∣∣∣ 1− ρ+ z

2

),

where ρ = (1− 2xz + z2)1/2.

We pick x, z such that 1−ρ−z2 = k(τ0) and 1−ρ+z

2 = k(pτ0) forsome p and quadratic irrational τ0. Then, use the multiplier Mp

and Clausen’s formula.15 / 24


Legendre polynomials – examples

For instance,∞∑n=0

( 13 )n( 2

3 )n

n!2Pn

(3√

3

5

)(2 + 15n)

(5

6√

3

)n=

45√

3

4π,

and is in fact equivalent to (3).

Along these lines, other examples are produced, such as∞∑n=0

( 12 )2nn!2

P2n

(3√

3

5

)(2 + 15n)

(2√

2

5

)2n

=15

π,

∞∑n=0

{ n∑k=0

k∑j=0

(n

k

)(−1

8

)k(k

j

)3}Pn

(5

3√

3

)n

(4

3√

3

)n=

9√

3

2π.

H. H. Chan, J. G. Wan and W. Zudilin, Legendre polynomials andRamanujan-type series for 1/π, Israel J. Math. 194 (2013), 183–207

J. G. Wan and W. Zudilin, Generating functions of Legendrepolynomials: a tribute to Fred Brafman, J. Approx. Theory 164(2012), 488–503

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Supercongruences

As was first observed by Van Hamme, Ramanujan-type seriessatisfy supercongruences, such as

p−1∑n=0

(12)3nn!3

(1 + 4n) (−1)n ≡(−1

p

)p mod p3

p−1∑n=0

(14)n(12)n(34)n

n!3(3 + 20n)

(−1

4

)n≡ 3

(−1

p

)p mod p3

It seems that (almost?) all known Ramanujan-type series satisfy asupercongruence; conversely, supercongruences can be used todiscover Ramanujan-type series.

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Supercongruences

Question 1

Find and prove supercongruences for the 1/π series involving Legendrepolynomials. For instance, show thatp−1∑n=0

( 13 )n( 2

3 )n

n!2Pn

(3√

3

5

)(2 + 15n)

(5

6√

3

)n?≡[5

(−3

p

)− 1

]p

2mod p2

p−1∑n=0

( 12 )2nn!2

Pn

(1

2

)(3 + 14n)

(3

8

)n?≡ 3

(−2

p

)p mod p2

The infinite version of the 2nd sum evaluates to 8√2

π but is not modular,and does not fit into the general theory for Legendre polynomials.

J. G. Wan, Series for 1/π using Legendre’s relation,Integr. Transf. Spec. F. 25 (2014), 1–14

The idea is to use Legendre’s relation and ‘translations’ to avoid modularmachinery. For similar ideas, see for instance

J. Guillera and W. Zudilin, Ramanujan-type formulae for 1/π: Theart of translation, Ramanujan Math. Soc. Lecture Notes Series, 20(2013), 181–195 18 / 24


More challenging series

There were many other conjectures, due to Sun, for instance

∞∑n=0

[(2n

n

) n∑k=0

(2n− 2k

n− k

)(2k

k

)2

6−2k](1 + 12n)

( 3

20

)2n=

75

8π.

These conjectures all have building blocks of the form

G(x, z) =

∞∑n=0

[ n∑k=0

F (n, k)xk]zn.

It is routine to find an ODE in z (with coefficients depending on x)for G, however, in all these conjectures the ODEs have orders ≥ 4.

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Order reduction

Guess, based on numerical evidence, that x and z are connected bysome rational function ra,b, namely z = ra,b(x). One possibleguess is

G(x,

x

(a+ bx)2

)=

∞∑n=0

n∑k=0

F (n, k)xk+n

(a+ bx)2n.

Compute enough coefficients in its x-expansion and try to find aand b such that the coefficients satisfy an order 3 recurrence,which corresponds to a order 3 ODE for G.

Then, solve the ODE. In our case,∞∑n=0

(2n

n

) n∑k=0

(2n− 2k

n− k

)(2k

k

)2xk+n

(1 + 4x)2n

= (1 + 4x)

∞∑n=0

( 13 )n( 1

2 )n( 23 )n

n!3(108x2(1− 4x)

)n.

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Satellite identity

The conjecture uses x = 6−2 and z = x/(1 + 4x)2 = (3/20)2; thisx corresponds to Ramanujan’s (3).

However, x-differentiation (to introduce the n term) alsointroduces a k term.

We eliminate the extra k term using a vanishing ‘satellite identity’,

∞∑n=0

(2n

n

) n∑k=0

(2n− 2k

n− k

)(2k

k

)2xk+n

(1 + 4x)2n[4x+2k(4x+1)+n(4x−1)

]= 0.

The coefficients of the satellite identity was first guessed, thenproven using Wilf-Zeilberger.

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More results

This technique has been used successfully to prove many of theconjectures, including, for instance,

∞∑n=0

[(2n

n

) n∑k=0

(n

k

)(2n− 2k

n− k

)(2k

k

)2−6k

](19 + 140n)

( 2

17

)2n=

289

3π

∞∑n=0

[(2n

n

) n∑k=0

(n

k

)2(2k

k

)142k

](541 + 3920n)

( 1

198

)2n=

42471

8√

7π

Much more information (such as modular parametrizations) can befound at

S. Cooper, J. G. Wan and W. Zudilin, Holonomic alchemy andseries for 1/π, arXiv:1512.04608v2 (Aug 2016)

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Open questions

Question 2

Is there a systematic way to guess z = ra,b(x) which leads to anorder reduction?

Question 3

Suppose the generating function∑∞

n=0An(x) zn satisfies a 2ndorder linear ODE in z. What can we say about

∞∑n=0

(2n

n

)An(x) zn?

Again, x and z may be related.

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Open questions

Question 4

Find a closed form for

∞∑n=0

Pn(x)3 zn,

where x and z may be related in some way.

Thank you.

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James Wan MATRIX 17/1/17...James Wan MATRIX 17/1/17 1/24 Ramanujan seriesRecent development Proofs...

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