Chan H H, Chan S H, Liu Z G - Domb's Numbers and Ramanujan-Sato Type Series for Inv(Pi)

8/9/2019 Chan H H, Chan S H, Liu Z G - Domb's Numbers and Ramanujan-Sato Type Series for Inv(Pi)

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DOMBS NUMBERS AND RAMANUJAN-SATO TYPESERIES FOR 1/

HENG HUAT CHAN, SONG HENG CHAN, AND ZHIGUO LIU

Abstract. In this article, we construct a general series for 1

. We

indicate that Ramanujans 1

series are all special cases of this general

series and we end the paper with a new class of 1

series. Our work ismotivated by series recently discovered by Takeshi Sato.

1. Introduction

In [14], S. Ramanujan recorded a total of 17 series for 1/, one of whichis

(1.1)4

=

k=0

(6k + 1)

(k!)3

12

3k

4k,

where

(a)k = (a)(a + 1) (a + k 1).Ramanujan did not indicate how he arrived at these series but instead hintedthat some of these series belonged to what are now known, after the workof B.C. Berndt, S. Bhargava and F.G. Garvan [2], as the theories of ellipticfunctions to alternative bases.

Ramanujans series for 1/ were not extensively studied until around1987. The Chudnovskys succeeded in extending Ramanujans list of seriesand derived the following series which they used to compute over a billiondigits of :

(1.2)6403203/2

= 12

k=0

(1)k (6k)!(k!)3(3k)!

(13591409 + 545140134k)

6403203k.

For a recent discussion of the Chudnovskys series, see [3].In [6], the Borweins provided rigorous proofs of Ramanujans series for

the first time and derived many new series for 1/. Both the Borweins and

Keywords : Ramanujan-type series, theta functions, hypergeometric series, Clausensidentity, modular equations, Dombs numbers, differential equations.

2000 Mathematics Subject Classifications : 11F11, 11F27, 11Y60, 33C20, 33C05, 05A10.

The first and third authors are funded by National University of Singapore AcademicResearch Fund, Project Number R146000027112.

1


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2 HENG HUAT CHAN, SONG HENG CHAN, AND ZHIGUO LIU

the Chudnovskys admitted that Clausens identity, namely, [6, p. 188]

2F1

a, b; a + b +

1

2 ; z2

= 3F2

2a, 2b, a + b; a + b +

1

2 , 2a + 2b; z

,

plays an important role in their derivations of such series.In this article, we will derive a general series for 1/ without using

Clausens identity. We will show that all the existing series for 1/ arespecial cases of this general series. We will then specialize our series furtherto derive new classes of series for 1/. In particular, we will show that

(1.3)83

=k=0

kj=0

kj

22(k j)

k j

2jj

1

64

k(1 + 5k).

Our work is motivated by the following series discovered recently by TakeshiSato [12] 1:

1

15120(4

5 9)

=k=0

kj=0

kj

2k +j

j

212 3

20

5 + k

5 1

2

12k.

The above series shows that Clausens formula is probably not needed inthe derivation of Ramanujans series since there are no Clausen-type trans-formation formula for series such as

k=0

k

j=0

kj

2

k +j

j 2

xk.

2. A general series for 1/

Suppose the function Z(q) satisfies the relation

(2.1) rZ(e2r/s) = Z(e2/

rs).

By differentiating (2.1) with respect to r and simplifying, we find that

(2.2)

rs

= e2/

rs Z

(e2/rs)

Z(e2/rs)

+ re2r/s Z

(e2r/s)

Z(e2r/s)

,

where

Z(q) =dZ(q)

dq .Next, set

(2.3) MN(q) =Z(q)

Z(qN).

1The authors are grateful to S. Kanemitsu for sending the abstract of Satos talk whichcontains several new series for 1/.


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DOMBS NUMBERS AND RAMANUJAN-SATO TYPE SERIES FOR 1/ 3

By differentiating (2.3) with respect to q, we find that

MN(q)

MN(q) =

Z(q)

Z(q) NqN

1 Z(qN)

Z(qN) ,

which implies that

(2.4) qMN(q)MN(q)

= qZ(q)Z(q)

N qNZ(qN)

Z(qN).

Substituting q = e2/Ns into (2.4) and using (2.1), we find that

(2.5)

e2/NsMN(e

2/Ns)

N= e2/

Ns Z

(e2/Ns)

Z(e2/Ns)

N e2N/s Z

(e2N/s)

Z(e2N/s)

.

Solving (2.5) and (2.2) with r = N, we conclude that

(2.6)

Ns

= 2Ne2

N/s Z

(e2N/s)

Z(e2N/s)

+e2/

NsMN(e

2/Ns)

N.

At this stage, we suppose that there exist functions X(q) and U(q) suchthat

(2.7) Z(q) =k=0

AkXk(q), Ak Q

and

(2.8) qdX(q)

dq

= U(q)X(q)Z(q).

Relation (2.3) and (2.7) also imply that MN may be expressed as a functionof two variables, namely,

(2.9) MN = MN(X(q), X(qN)).

We may also assume that X(q) and X(qN) satisfies a polynomial relationand hence,using (2.8), (2.1) and (2.7),we deduce that

e2/NsMN(e

2/Ns)

N=

1

Nq

dX(q)

dq

dMN(X(q), X(qN))

dX(q)

q=e

2/Ns

(2.10)

= 1N

U(e2/Ns)X(e2/Ns)Z(e2/Ns) dMN(X(q), X(qN

))dX(q)

q=e2/

Ns

= U(e2/Ns)X(e2/

Ns)Z(e2

N/s)

dMN(X(q), X(qN))

dX(q)

q=e2/

Ns

= U(e2/Ns)X(e2/

Ns)

dMN(X(q), X(qN))

dX(q)

q=e2/

Ns

k=0

AkXk(e2

N/s).


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This simplifies the second term on the right hand side of (2.6). Next, by(2.7) and (2.8), we find that

qZ(q) =k=0

kAkXk1(q)qX(q)

=k=0

kAkXk1(q)U(q)X(q)Z(q).

Hence,

(2.11) qZ(q)Z(q)

= U(q)k=0

kAkXk(q).

Substituting q = e2N/s into (2.11) and combining with (2.10) and (2.6),

we deduce the following result:

Theorem 2.1. Suppose Z(q) satisfies (2.1) and let MN(q), X(q), Ak andU(q) be defined as in (2.3), (2.7) and (2.8). Then

(2.12)

s

N

1

2=

k=0

(bNk + aN) AkXkN,

where

aN = U(e2/

Ns)

X(e2/Ns)

2N

dMNdX(q)

(X(q), X(qN))

q=e2/

Ns

,

bN = U(e2

N/s),

and XN = X(e2N/s).

We end this section by listing s, x, Z, U and Ak which correspond toRamanujans series for 1/. In Table 1, x is given by the relation

X = 4x(1 x),

f(q) =k=1

(1 qk),

and

j(q) = 1728Q3

qf24

(q),

where

Q := 1 + 240k=1

k3qk

1 qk .

In the next section, we will illustrate how we derive the fourth class of series

without using Clausens identity. We will then give a new proof of (1.1).


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Table 1

s x Z U Ak

11

2

1

1 1728

j(q)

Q 1 2x

12

k

16

k

56

k

(k!)3

2

1 +

1

64q

f(q)

f(q2)241 f8(q)

f4(q2)

1 + 64q

f(q2)f(q)

241 2x

12

k

14

k

34

k

(k!)3

3

1 + 127q

f(q)

f(q3)

121

m,n=qm

2+mn+n22

1 2x

12k

13k

23k

(k!)3

4

1 +

1

16q

f(q)

f(q4)81

m,n=qm

2+n2

21 2x

12

3k

(k!)3

3. Ramanujans series for 1/ associated with the Jacobiantheta functions

Let Z(q) = 4(q) where

(q) =

k=qk

2

.

Note that Z(q) satisfies the transformation formula [1, p. 43, Entry 27(i)]

(3.1) NZ(eN) = Z(e/

N),

and hence, s = 4.It is known that if 2

(3.2) x = x(q) = 16q4(q2)

4(q) ,

with

(q) =k=0

qk(k+1)/2,

2Our x given here is equal to the x given in the Table in Section 2.


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then z = z(q) = 2(q) and x satisfy the differential equation [15, p.54,(3.83)]

(3.3) x(1 x) d2zdx2

+ (1 2x) dzdx

z4

= 0.

Substituting z =

Z and X = 4x(1 x) into (3.3), we deduce that3

(3.4) 3X(Z) = X

X +

1

2

3(Z),

where the operator

X = Xd

dX.

If

Z =

k=0

AkXk,

then (3.4) yields a recurrence relation satisfied by Ak and we deduce imme-diately that

Ak =

12

3k

(k!)3.

In other words, we find that

(3.5) Z = 3F2

1

2,

1

2,

1

2; 1, 1; X

.

It is also known that [1, p. 120, Entry 9 (i)]

(3.6) qdX

dq= ZX(1

2x),

which means thatU(q) = 1 2x(q).

By (3.6), we find that if Y = X(qN) and y = x(qN), then

(3.7)dX

dY=

1

NMN

X

Y

1 2x1 2y .

Squaring (3.7), we deduce that

(3.8) M2N = N2 Y

2(1 Y)X2(1 X)

dX

dY

2.

From the above relation,dMN

dX can be computed explicitly once we computethe modular relation satisfied by X and Y.

We now give an explicit series for 1/. First, from (3.1) and [1, p. 43,Entry 27 (ii)], we find that

4(e) = 4(e),

3The calculations are tedious but straightforward.


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and

16e4(e2) = 4(

e),

if = 2. By (3.2) and the above relations,

(3.9) x(e/N) =

4(eN)

4(eN)

.

Using the famous Jacobi identity [1, p. 40, Entry 25 (vii)],

4(q) 4(q) = 16q4(q2),

we conclude that

(3.10) x(e/N

) = 1 x(eN

).

Now, let N = 3, then from Ramanujans modular equation [1, p. 230, Entry]

(x(q)x(q3))1/4 + ((1 x(q))(1 x(q3)))1/4 = 1,

and (3.10), we find that

(3.11) X(e/

3) =1

4.

Solving (3.11) for x, we deduce that

(3.12) x(e3) = 12 3

4.

Next, one can show that X = X(q) and Y = X(q3) satisfy the relation 4

4096XY + 4608X2Y + 4608XY2 900X3Y + 28422X2Y2 900XY3 + 4608X3Y2 + 4608X2Y3 4096X3Y3 + X4 + Y4 = 0.

Differentiating the above relation with respect to Y and substituting theresult into (3.8), we obtain an expression of M3 in terms of X and Y.Differentiating the resulting relation with respect to X and using (3.11) and(3.12), we conclude that

a3 =

3

12and b3 =

3

2.

This gives Ramanujans series (1.1).

4This modular equation can be proved using [1, p. 231, Entry 5(xii)]


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4. A new class of series for 1/

From the last section, we find that in order to derive a new class of series

different from that of Ramanujan, one needs to seek new pair of functionsX and Z. In this section, we will construct a new class of series for 1/and derive (1.3). Series similar to (1.3) where all the coefficients in the sumare rational are very rare. It turns out that Ramanujan nearly exhaustedall these series in [14]. The Chudnovskys 5 and the Borweins 6 constructedseveral such series for 1/ associated with the first and second type of seriesin Table 1, respectively. Recently, Chan, W.C. Liaw and V. Tan added atotal of six such series [8] associated essentially to the third type of series inTable 1.

Let

Z =(f(q)f(q3))4

(f(

q2)f(

q6))2

and

X = q

f(q2)f(q6)f(q)f(q3)

6.

From our discussion in Section 3, we need to determine U(q) and Ak. Itfollows from [4, Theorem 10.6]

(4.1) qdX

dq= ZX

(4X+ 1)(16X+ 1)

that

(4.2) U(q) =

(4X+ 1)(16X+ 1).To determine Ak, we need to establish the differential equation satisfied

by Z and X.Let

(4.3)1

t= 1 +

1

27q

f12(q)f12(q3) .

The differential equation satisfied by

u = 4t(1 t) and f = m,n=

qm2+mn+n2

2

is

(4.4) 3uf = u

u +

1

2

u +

1

3

u +

2

3

(f).

5The Chudnovskys series (1.2) yields the fastest convergent series for 1/ which in-volves only rational coefficients.

6See [5, p. 145] for the two series missed by Ramanujan.


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The proof of (4.4) is similar to that of (3.4) but instead of using (3.3), weuse the differential equation [7, p. 200]

t(1 t) d2zdt2

+ (1 2t) dzdt

29

z = 0,

where z2 = f. Incidently (4.4) is the differential equation needed to deter-mine Ak in the third type of series in Table 1. Our method of deriving adifferential equation satisfied by Z and X is to first establish the identities

(4.5) f = Z(1 + 16X)

and

(4.6) u =108X

(1 + 16X)3

.

Using (4.5) and (4.6), we conclude that

uf =(1 + 16X)2XZ+ 16X(1 + 16X)Z

1 32X ,(4.7)

2uf =1 + 16X

(1 32X)3

(4.8)

{(1 768X2 8192X3)2XZ+ (80X+ 256X2 16384X3)XZ+ (16X+ 512X2

8192X3)Z

}and

3uf =1 + 16X

(1 32X)5

(4.9)

{(1 16X 1280X2 + 4096X3 + 524288X4 + 4194304X5)3XZ+ (192X 1536X2 147456X3 393216X4 + 12582912X5)2XZ+ (96X+ 8448X2 12288X3 1179648X4 + 12582912X5)XZ+ (16X+ 2560X2 + 24576X3 524288X4 + 4194304X5)Z}.

Substituting (4.7)(4.9) into (4.4) and simplifying, we immediately obtainthe differential equation

(1 + 20X+ 64X2)3XZ+ (192X2 + 30X)2XZ(4.10)

+ (192X2 + 18X)XZ+ (64X2 + 4X)Z = 0,

where X = Xd

dX. For an alternative proof of (4.10), see [16].


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It remains to establish (4.5) and (4.6). From the definition (4.3), we findthat

1

u = 2 +1

27 P6 +27

P6 ,

where

P = q1/6f2(q)

f2(q3) .Identity (4.6) then follows immediately from [4, Lemma 10.3, (10.4)]

P3 +27

P3

2=

(1 + 16X)3

X.

To prove (4.5), recall from [7, p. 200] that

(4.11) qdu

dq= f u

1 u.

This is the analogue of (3.6). From (4.6), we find that

(4.12)du

dX= 108

1 32X(1 + 16X)4

.

Combining (4.12), (4.11), (4.6) and (4.1), we conclude that

f = Z(1 + 16X),

which is (4.5).Now, from (4.10), we conclude that if

Z =

k=0

AkXk,

then Ak satisfies the difference equation

(4.13) k3Ak + 2(2k 1)(5k2 5k + 2)Ak1 + 64(k 1)3Ak2 = 0.A check with N. Sloanes Online sequences [13] shows that

(4.14) Ak = (1)kk

j=0

kj

22(k j)

k j

2jj

.

This sequence appears to be first discovered by C. Domb [9]. The fact thatthis sequence satisfies the recurrence (4.13) can be verified using the methodof creative telescoping implemented in D. Zeilbergers MAPLE programme

EKHAD [11, Chapter 7]. More precisely, Zeilbergers programme producesthe function

R(k, j) =

(12k3 + 62k2 + 104k + 56 26k2j 89kj 74j + 18kj2 + 30j2 4j3)j3

4(k + 1)2(2k 2j + 1)

(k + 1 j)3(k + 2 j)3 ,


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which satisfies the relation

(k + 2)3f(k + 2, j) + 2(2(k + 2) 1)(5(k + 2)2 5(k + 2) + 2)f(k + 1, j)(4.15)

+ 64(k + 1)3f(k, j) = R(k, j + 1)f(k, j + 1) R(k, j)f(k, j),where

f(k, j) = (1)k

kj

22(k j)

k j

2jj

.

Summing (4.15) for j from 0 to k + 2 completes the proof that Ak satisfies(4.13).

We now turn to the proof of (1.3). First, we let Z = Z(q). Then fromthe transformation formulas [1, Entry 27(iii), (iv)]

e/12

f(

e2) = e/12f(e2)(4.16)

and

e/24

f(e) = e/24

f(e),(4.17)

where = 2, we deduce that

(4.18) Z(e/

3N) = NZ(eN/3).

This shows that s = 12. We may now apply Theorem 2.1 with the functionsZ and X = X(q). Note that

Z =

k=0

Ak(X)k,

and hence our corresponding

Ak = (1)kAk =k

j=0

kj

22(k j)

k j

2jj

.

From (4.2), we have

U(q) =

(1 4X)(1 16X).In addition, if

MN = Z(q)Z(qN)

,

then the analogue of (3.8) is

(4.19) M2N = N2 Y

2

X2(1 4Y)(1 16Y)(1 4X)(1 16X)

dX

dY

2,

where Y = X(qN).


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Now let N = 5. We have 7

(X, Y) = X6 + Y6 + 30XY2 + 30X2Y 285XY3 285X3Y + 970X4Y(4.20)

+ 970XY4 990XY5 990X5Y 930X2Y3 930X3Y2 25665X2Y4 25665X4Y2 + 62080X2Y5 + 62080X5Y2 + 92860X3Y3 59520X3Y4 59520X4Y3 1167360X3Y5 1167360X5Y3 + 7864320X4Y5+ 7864320X5Y4 16777216X5Y5 XY + 815X2Y2 + 3338240X4Y4 = 0.

When q = e/

15, then from (4.16) and (4.17), we find that

X(e/

15) = X(e

5/3) = Y(e/

15).

This shows that X(e/

15

) is a root of the polynomial (T, T) where isgiven by (4.20). This polynomial factors as

T2(T 1)(64T 1)(64T2 11T + 1)(64T2 + 1)(1 + 8T)2,and a quick check shows that

(4.21) X(e/

15) =1

64.

Next, from (4.20), we can computedX

dYand substitute the result into

(4.19). Differentiating the result with respect to X and using (4.21) and(4.18), we conclude that

(4.22)dMN

dX

q=e/

15

= 128.

With these values, we conclude that

an =3

8

5and bn =

3

5

8.

Substituting these values into Theorem 2.1 yields (1.3).

Acknowledgements. The authors wish to thank Professor D. Zeilberger forhis wonderful program which allows us to verify that the Dombs sequence

satisfies the difference equation (4.13). The first author wishes to thank P.Hammond from the University of Sussex for bringing his attention to [10].This work is partly motivated by this excellent article by M. Kontsevich andD. Zagier.

7This can be verified using Ramanujans modular equation of degree 5 in the theory ofsignature 3 [2].


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References

[1] B. C. Berndt, Ramanujans Notebooks, Part III, Springer-Verlag, New York, 1991.

[2] B.C. Berndt, S. Bhargava and F.G. Garvan, Ramanujans theories of elliptic functionsto alternative bases, Trans. Amer. Math. Soc., 347 (1995), 4163-4244.

[3] B.C. Berndt and H.H. Chan, Eisenstein series and approximations to , Illinois J.Math., 45, no. 1 (2001), 75-90.

[4] B.C. Berndt, H.H. Chan and S.S. Huang, Incomplete elliptic integrals in RamanujansLost Notebook, Contemp. Math., 254 (2000), 79126.

[5] B. C. Berndt, H. H. Chan and W. C. Liaw, On Ramanujans quartic theory of ellipticfunctions, J. Number Theory, 88 (2001), 129-156.

[6] J.M. Borwein and P.B. Borwein, Pi and the AGM, Wiley, New York, 1987.[7] H.H. Chan, On Ramanujans cubic transformation formula for 2F1(

1

3, 23

; 1; z), Math.Proc. Camb. Phil. Soc., 124, (1998), 193-204.

[8] H.H. Chan, W.C. Liaw and V. Tan, Ramanujans class invariantn and a new classof series for 1/, J. Lond. Math. Soc. (2), 64, (2001), 93-106.

[9] C. Domb, On the theory of cooperative phenomena in crystals, Advances in Phys., 9

(1960), 149-361.[10] M. Kontsevich, D. Zagier, Periods, Math. Unlimited 2001 and beyond, 771-808,

Springer, Berlin, 2001.[11] M. Petkovsek, H.S. Wilf and D. Zeilberger, A = B, A K Peters, U.S.A., 1996.[12] T. Sato, Apery numbers and Ramanujans series for 1/, Abstract of a talk presented

at the Annual meeting of the Mathematical Society of Japan, March 28-31, 2002.[13] N.J.A. Sloane, http://www.research-att.com/ njas/sequences/index.html.[14] S. Ramanujan, Modular equations and approximations to , Quart. J. Math. (Oxford),

45 (1914), 350-372.[15] K. Venkatachaliengar, Development of elliptic functions according to Ramanujan

(1988) Madurai Kamaraj University, Madurai.[16] Y.F. Yang, On differential equations satisfied by modular forms, preprint.

National University of Singapore, Department of Mathematics, 2 Science

Drive 2, Singapore 117543E-mail address: [email protected]

University of Illinois at Urbana-Champaign, Department of Mathematics,

1409 West Green Street, IL61801, U.S.A.

E-mail address: [email protected]

East China Normal University, Department of Mathematics, Shanghai 200062,

P.R. China

E-mail address: [email protected]

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Chan H H, Chan S H, Liu Z G - Domb's Numbers and Ramanujan-Sato Type Series for Inv(Pi)