THREE ADVENTURES:

SYMBOLICALLY DISCOVERED

IDENTITIES FOR ZETA(4N+3) AND

LIKE MATTERS

Jonathan Borwein, Simon Fraser University

C E C M

Centre for Experimental &Constructive Mathematics

July 14, 1997 Vienna (FPSAC'97)

9th International Conference on: Formal

Power Series and Algebraic Combinatorics

Joint work: in parts with David Broadhurst (OU),David Bradley (CECM), Roland Girgensohn (Munich),Petr Lisonek (CECM), John Macdonald (Melbourne)

Papers: www.cecm.sfu.ca/preprints/

Talks: www.cecm.sfu.ca/~ jborwein/talks.html/

WebPage: www.cecm.sfu.ca/~ jborwein/fpsac.html/

1

ABSTRACT.

I will describe three sets of results in which

computer search and computer algebra played

a large role in the discovery/and or proof of

results - each ultimately relating to hypergeo-

metric functions.

1. Ap�ery: A generating function for �(4n+3).

2. Ramanujan: Integrals, means and transfor-

mations of the hypergeometric function

2F1(1=3;2=3;1; 1� x3):

3. Euler: Multivalued �{function values and a

product of hypergeometric 2F1's.

y In each case, my emphasis is on the role of

disciplined experimentation and computation.

2

�

�

OUTLINE

Part 0. Preliminaries: Integer Relation Algorithms

Part I. Ramanujan: Cubic Arithmetic{Geometric

Mean

a. Functional Equations: Power series

b. Integral Representations: Substitutions

Part II. Ap�ery: Evaluations of �(4n+3)

Part III. Euler: Sums

a. Reducibility: Dimensional conjectures

b. Duality: Evaluations and computations

c. Proof: A conjecture due to Zagier

3

COMMENTS and PRELIMINARIES

� Extensive use of \Integer Relation Algorithms":PSLQ/LLL { or lattice basis reduction.

� Exclusion bounds are especially useful.

�Many proofs are out of reach { understanding

is not.

� Great test bed for \experimental methodol-

ogy":�

�

EZ{FACE

the remote interface we are building (Web,

Maple, C).

� Special thanks to Open University { and

Memorial { computers.

4

PART 0: Integer Relation Detection

� Let x = (x1; x2; � � � ; xn) be a vector of real

numbers. Then x is said to possess an integer

relation if there exist integers ai not all zero

such that

a1x1+ a2x2+ � � �+ anxn = 0

Problem: Find the integers ai if they exist. If

they do not, obtain a sequence of bounds on

the size of any possible integers ai.

� Euclid's algorithm gives solution for n = 2.

� Euler, Jacobi, Poincare, Minkowski, Perron,

Brun, Bernstein, and others tried to �nd a gen-

eral algorithm for n > 2.

� First general algorithm was discovered in 1977

by Ferguson and Forcade.

� Algorithms discovered since 1977: LLL, HJLS,PSOS, PSLQ (1991).

5

�

�

Application to Number Theory

Suppose � can be computed to high precision.

Then form the n-long vector x = (1; �; �2; � � � ; �n�1)and apply an integer relation algorithm.

� If a relation is found, the solution integers

ai are precisely the coe�cients of a poly-

nomial satis�ed by �.

� If no relation is found, bounds are obtained

within which no polynomial can exist.

It is also possible to explore whether � satis�es

an identify of the form

�p = 2a 3b 5c 7d �k el m � � �by simply taking logarithms.

� g46 is a root of 1�5292x�7450x2�5292x3+x4 = 0.

6

AN EXEMPLARY APPLICATION

� Thanks to Ap�ery it is well known that

�(2) = 31Xk=1

1

k2�2kk

��(3) =

5

2

1Xk=1

(�1)k�1k3�2kk

��(4) =

36

17

1Xk=1

1

k4�2kk

�

� These results suggest that

Z5 = �(5)=1Xk=1

(�1)k�1k5�2kk

�might also be a simple rational or algebraic

number.

Integer relation result: If Z5 satis�es a poly-

nomial of degree 25 or less, then the Euclidean

norm of the coe�cients must exceed 2�1037.7

Part I. Ramanujan: The cubic AGM �

ABSTRACT. We use Maple's gfun library to

study the limit formulae for a two-term re-

currence (iteration) AGN , which for N = 2

becomes the well-known Arithmetic-Geometric

Mean iteration of Gauss. Our main aim is to

rediscover and prove the limit formulae for two

classical cases (N = 2;3) in a completely auto-

mated manner and to open the way for study-

ing the remaining cases (N > 3).

� We also present a new, self-contained proof

of the integral representation of the limit for

N = 3, which avoids associating this limit with

hypergeometric functions and/or di�erential

equations.

�Jonathan M. Borwein, Petr Lison�ek, and JohnA. Macdonald, \Arithmetic-geometric means revis-

ited," MapleTech, Special Issue on Maple in the Math-

ematical Sciences. 4 (1997) 20{27.

8

The Generalized AGM Iteration

� Our �rst goal is to identify this limit as a

power series.

Let a and b be real numbers, a > b > 0, and let

N be an integer greater than 1. The iteration

AGN is the following two-term recursion:

a0 = a

b0 = b

and, for any k � 0,

ak+1 =ak + (N � 1)bk

N

bk+1 =

Nq(ak+ (N � 1)bk)

N � (ak � bk)N

N:

� In the case N = 2 we get the famous Arithmetic-

Geometric Mean (AGM) iteration. The case

N = 3 was studied in detail in the '80's (Bor-

wein et al.).

9

�

�

Convergence

We have

aNk+1 � bNk+1 =

�ak � bk

N

�N; (1)

for any k � 0, and N > 1.

� This shows both the global convergence and

local Nth-order convergence of the iteration.

� So there is a common limit of (ak) and (bk),

MN(a0; b0) := limk!1

ak = limk!1

bk:

10

�

�

Representation

� Let � be the involution on [0;1] de�ned by

x� := Nq1� xN :

When we use the function symbol �, the valueof N is clear in context.

� From the homogeneity

MN(�a; �b) = �MN(a; b) for � > 0 (2)

it is enough to investigate

AN(xN) :=

1

MN(1; x�)

for 0 < x < 1.

Fix a0 = 1 and b0 = z 2 D where D = ft 2C j jt � 1j < 1

2g. From (1) we can argue that

aNk (z) and bNk (z) are analytic in D and converge

uniformly to MN(1; z) therein. Thus AN(xN)

is analytic near zero and AN(0) = 1.

11

The Functional Equation for AN

For any N > 1 and any 0 < x < 1, applying one

step of the AGN iteration,

MN(1+ (N � 1)x;1� x) =MN(1; x�): (3)

Further, by (2) we have

MN(1+ (N � 1)x;1� x) =

(1+ (N � 1)x)MN

1;

1� x

1+ (N � 1)x

!(4)

Combining right-hand sides of (4) and (3) we

arrive at the FE for AN :

(1 + (N � 1)x) �AN(xN)

= AN

0@1� 1� x

1+ (N � 1)x

!N1A (5)

Notice that (5) uniquely determines the Tay-

lor series for AN at 0 and hence has a unique

analytic solution

AN(xN) =

1

MN(1; x�):

12

�

�

The Quadratic Case

� Gauss can be replaced by routine computer

algebra in 1996 (not in 1986)

Power Series Expansion

For N = 2, the equation (5) specializes to

(1+ x) �A2(x2) = A2

4x

(1+ x)2

!(6)

whose solution was found by Gauss in the form

A2(x2) =

1Xi=0

�2ii

�2 �x4

�2i= 2F1

�1=2; 1=2

1j x2

�:

(7)

� Gauss discovered this closed form after hav-

ing been inspired by a great number of exper-

imental numerical calculations - and Stirling's

tables.

� Nowadays the expansion (7) can be discov-

ered as well as proved in a completely mechan-

ical way.

13

Integral Representation

Another way of expressing the AG2 limit is by

means of the following de�nite integral:

I2(a0; b0) =

Z 10

dtq(t2+ a20)(t

2+ b20)(8)

Then it is straightforward to prove that

M2(a0; b0) =I2(1;1)

I2(a0; b0)=

�

I2(a0; b0):

The core step of this proof, namely showing

the AG2-invariance

I2(a; b) = I2(a+ b

2;pab)

follows fairly easily on substituting u = 12(t� ab=t)

(a nice CAS exercise).

� Yet another way of proving the AG2 limit

formula is by identifying M2(1; x) as a solution

of a second-order linear di�erential equation.

14

�

�

The Cubic Case

� The closed form for M3(1; x�) was identi�ed

and proved in several ways in the '80s.

Power Series Expansion

Again, the discovery of the closed form for

A3(x3) as a formal power series again is now

routine task. We obtain

1

M3(1; x�) = A3(x

3) = 2F1

�1=3; 2=3

1j x3

�:

(9)

� The ways of proving (9) using computer alge-bra are discussed in Appendix A of our paper.

15

Integral Representation

The cubic counterpart of (8) is

I3(a0; b0) =

Z 10

t dt

3q(t3+ a30)(t

3+ b30)2: (10)

We have

M3(a0; b0) =I3(1;1)

I3(a0; b0)=

2�p27

� 1

I3(a0; b0):

(11)

Again, the crucial part of the proof of (11)

is to show that the integral is invariant with

respect to the AG3 iteration, that is,

I3(a; b) = I3

0@a+2b

3;3

sba2+ ab+ b2

3

1A (12)

for all a > b > 0.

16

� The proof of (12) was proposed by the JMB

as Part (a) of the American Mathematical Monthly

Problem #10281. The published solutions are

concluded with the editorial comment which

points out that:

\There is still no self-contained proof

that avoids exploiting the identi�cation

with a hypergeometric function."

Recently, J.A. Macdonald found a proof of (12)

which consists of a chain of variable substitu-

tions together with a split of the integration

range at the point 1.

� Here we present a variation of this proof, in

which all integral substitutions have been sim-

pli�ed to the extent that they can be checked

by a computer. The proof was announced in

the \Revivals" section of the Monthly. Maple

code verifying the core steps of the proof can

be found in Appendix B of the paper.

17

The Solution to AMM 10281

Much the proof of (12) is encapsulated in the

following lemma.

Lemma. For any 2 ]0;1],

I3(1;3q1� 3)

=

Z 11

dxq(x� 1)((x+3)x2 � 4 3)

: (13)

Proof. We �rst note that

I3(1;3q1� 3) =

Z 10

uduq(u3 � 1)2+4(1� 3)u3

(14)

using the substitution

u3 = t3(1+ t3)=((1� 3) + t3).

18

Further, the equalityZ 10

uduq(u3 � 1)2+4(1� 3)u3

=

Z 11

(u+1) duq(u3 � 1)2+4(1� 3)u3

(15)

follows fromZ 10

uduq(u3 � 1)2+4(1� 3)u3

=

Z 10

duq(u3 � 1)2+4(1� 3)u3

and Z 1

0

(u+ 1) duq(u3 � 1)2+4(1� 3)u3

=

Z 11

(u+1) duq(u3 � 1)2+4(1� 3)u3

which both easily verify on substitution of u= 1=u .

19

Now use the substitution x = u+1=u� 1 to

complete the Lemma:Z 11

dxq(x� 1)((x+3)x2 � 4 3)

=

Z 11

(u+1) duq(u3 � 1)2+4(1� 3)u3

(16)

�

�

The remains of the proof

� For any y; z > 0 we have

I3(y; z) =1

y� I3

1;z

y

!: (17)

Let c := b=a. Use (17) to rewrite (12) as

2c+1

3� I3(1; c) = I3(1; c

^) (18)

where x 7! x^ is de�ned by

x^ := 3

vuut9x(1+ x+ x2)

(1 + 2x)3

for any x 2 [0;1].20

� Denote function composition in the obvious

way: by c�^ we mean�c��^ and note that, for

any c 2 [0;1],

(2c+1)(2c^�+1) = 3 (19)

or, in other words,

c^� = 1� c

1+ 2c: (20)

Recall that � is an involution on [0;1]. From

(20) it follows that ^� is also an involution on

[0;1]. Thus,

^�^ = � (21)

as functions on [0;1].

By inspection of the function ^�, we see that

8c 2 ]0;1]2c+1

3� I3(1; c) = I3(1; c

^)

is equivalent to

21

8C 2 [0;1[ 2C^�+1

3�I3(1; C^�) = I3(1; C

^�^):

� Using (21), (19) and swapping sides, the laststatement is equivalent to

8C 2 [0;1[ (2C+1) � I3(1; C�) = I3(1; C^�):(22)

Using (13), the equality (22) translates to

(2C +1) �Z 11

dxq(x� 1)((x+3)x2 � 4C3)

=

Z 11

dzs(z � 1)

�(z+3)z2 � 36

C(1+C+C2)(2C+1)3

�which can be proven by the substitution

z =(x�C)3(x�1)

(x3�C3)(2C+1)+1:

This completes the proof of (18) and thus also

the proof of (12).

22

�

�

Comments

� While the proof of (12) involves not only

syntactic steps (substitutions) but also seman-

tic steps, numerical evidence suggests that the

composition of all substitutions involved in the

proof is a reasonable candidate for a single sub-

stitution step on the original integral.

� Thus, we hope the proof presented above

can be shortened considerably. Let t and t0

be the integration variables in (18). To �nd a

direct substitution in t and t0 we rewrote all in-tegral substitutions as multivariate polynomial

equations, and used resultants to eliminate in-

termediate integration variables.

23

� In the end, however, we had an irreducible

polynomial of degree 36 in t and of degree 12

in t0. This particular relation between t and t0is quite non-trivial even for c = 1, when the

sides of (18) are syntactically equal.

� We also looked at parameterizing the result-

ing t; t0-relation rationally. Sadly, even in for

c = 1 the curve is of genus 2 and thus has no

rational parameterization.

Open Problems

Question 1. Find a purely syntactic transfor-

mation for the cubic integral.

� For any N one can compute many terms of

AN(xN) in a very short time. So far we were

unable to identify the functions AN for N > 3.

Question 2. Find a power series expansion

for AN which allows one to \read o�" the hy-

pergeometric series when N = 2 or N = 3.

24

Part II. Ap�ery Evaluations of ZETA

SEARCHING SYMBOLICALLY FOR

AP�ERY-LIKE FORMULAE�

ABSTRACT. We discuss the search for iden-

tities using computer algebra and symbolic meth-

ods. To keep the discussion as concrete as

possible, we focus on so-called Ap�ery-like for-

mulae for special values of the Riemann Zeta

function. Many of these results are new, and

much more work needs to be done before they

can all be formally proved and properly classi-

�ed.

�J.M. Borwein and D.M. Bradley, \Empirically deter-

mined Ap�ery{like formulae for zeta(4n+3)," Experi-

mental Mathematics, in press [CECM Research Report

96:069]

25

Introduction. Again, the Riemann Zeta func-

tion is

�(s) =1Xk=1

1

ks; <(s) > 1:

In view of the \Ap�ery-like" formulae

�(2) = 31Xk=1

1

k2�2kk

�; �(3) =5

2

1Xk=1

(�1)k+1

k3�2kk

� ;

(23)

and

�(4) =36

17

1Xk=1

1

k4�2kk

�;one is tempted to speculate there is an analo-

gous formula for �(5), �(6), �(7) and so on.

� The key word here is analogous.

26

� As we saw, extensive computation has ruled

out the possibility of formulae of the form

�(5) =a

b

1Xk=1

(�1)k+1

k5�2kk

� ; �(6) =c

d

1Xk=1

1

k6�2kk

�;where a, b, c, d are moderately sized integers.

� Such negative results are useful, as they tell

us it would be a waste of time to search for

interesting formulae of a given form. Thus, it

would seem there are no corresponding Ap�ery-

like formulae for higher zeta values.

� End of story?�

�We hunted for Ramanujan-like formulae for Catalan's

constant in the wrong haystack:

G =�

8log

10 +

p50� 22

p5

10�p50 � 22

p5

!+5

8

Xm�0

�2m+1 � �2m+1

(2m+ 1)2�2m

m

�;where � := (

p5 + 1)=2, � := (

p5� 1)=2 (Bradley).

27

Consider however, the following result of Koecher:

�(5) = 21Xk=1

(�1)k+1

k5�2kk

� � 5

2

1Xk=1

(�1)k+1

k3�2kk

� k�1Xj=1

1

j2:

(24)

� This highlights a key issue with symbolic

searching. Negative results need to be inter-

preted carefully, lest they be given more weight

than they deserve and unnecessarily discourage

further investigation. Also, symbolic searching

is much limited by the need to know fairly pre-

cisely the form one is searching for in advance.

Koecher's formula (24) suggests that one might

pro�t by searching for a formula of the form

�(7) = r1

1Xk=1

(�1)k+1

k7�2kk

� + r2

1Xk=1

(�1)k+1

k5�2kk

� k�1Xj=1

1

j2

+ r3

1Xk=1

(�1)k+1

k3�2kk

� k�1Xj=1

1

j4;

where r1, r2, r3 are rational numbers.

28

The next (empirical) formula for �(7) was then

found using high precision arithmetic and Maple's

integer relations algorithms:

�(7) =5

2

1Xk=1

(�1)k+1

k7�2kk

� +25

2

1Xk=1

(�1)k+1

k3�2kk

� k�1Xj=1

1

j4:

(25)

� Ultimately, we were led (deus ex machina)

to:

5

2

1Xk=1

(�1)k+1

k3�2kk

� 1

1� z4=k4

k�1Yj=1

j4+4z4

j4 � z4

=1Xk=1

1

k3(1� z4=k4); (26)

for z 2 C.

� This is now proven elegantly by Almquist and

Granville from a �nite dual \Gosper re ection

formula" that we had subsequently obtained:

5

2

nXk=1

�2kk

� k2

4n4+ k4

k�1Yj=1

n4 � j4

4n4+ j4=

1

n2; (27)

for n 2 Z+.29

� Equivalent to (27), we have the following,

apparently new, strange evaluation of a termi-

nating 6F5:

6F5

�2; 3=2; 1 + n; 1� n; 1 + in; 1� in

1; 2 + n+ in; 2 + n� in; 2� n+ in; 2� n� inj � 4

�=

4n4+1

5n2;

for n 2 Z+.

� We can also rewrite (26) as a formula for a

non-terminating 6F5:

6F5

�2;2;1 + z+ iz;1 + z � iz;1� z+ iz;1� z � iz

3=2; 2 + z; 2� z; 2 + iz; 2� izj � 1

4

�= (1� z4)� 4

5

1Xk=1

1

k3(1� z4=k4);

for z 2 C.

� Observe the remarkable dual nature of thesetwo results (top to bottom, �4 to �1

4).

30

y Note the constant coe�cient in (26) gives

the formula for �(3) in (23). The coe�cient

of z4 in (26) gives (25). We arrived at (26) by

extensive use of boot-strapped� use of Maple's

lattice algorithms, combined with a good deal

of insightful guessing.Here Maple's convert(series,

ratpoly) played a really signi�cant role.

� Comparing our generating function formula

(26) with Koecher's

1Xk=1

(�1)k+1

k3�2kk

� (1

2+

2

1� z2=k2)k�1Yj=1

(1� z2=j2)

=1Xk=1

1

k3(1� z2=k2)

raises some interesting issues related to for-

mula redundancy, which remain unresolved.

�To obtain enough structure one either has to work

with huge data sets, and ultra high precision, or tomake inferences along the way.

31

�

�

Redundancy Relations.

De�ne l, � by

l(m;nY

j=1

Prj) :=1Xk=1

(�1)k+1

km�2kk

� nYj=1

Prj(k);

�(m;nY

j=1

Prj) :=1Xk=1

1

km�2kk

� nYj=1

Prj(k):

using the power sum symmetric functions

Pr(k) :=k�1Xj=1

j�r; r � 1 ; P0 = 1

Thus, (23) becomes

�(2) = 3�(2; P0); �(3) =5

2l(3; P0);

�(4) =36

17�(4; P0); (28)

while (24) and (25) become

�(5) = 2l(5; P0)�5

2l(3; P2);

�(7) =5

2l(7; P0) +

25

2l(3; P4): (29)

32

� To illustrate the issue of redundancy, con-

sider Koecher's formula: �(7) =

2l(7; P0)� 2l(5; P2) +5

4l(3; P2

2 )�5

4l(3; P4)(30)

In view of the second formula in (29), the mid-

dle two terms of (28) must be redundant. In-

deed, lattice-based reduction shows that

10l(3; P22 )� 8l(5; P2) = 55l(3; P4) + 2l(7; P0)

� While we have little understanding why such

interrelations between l sums hold, we decided

to limit our search to identities in which no

such interrelations exist.� This was carried

out by starting with a \full set" of l sums and

checking that a relation holds with the relevant

Zeta value.

�Of course, we cannot prove that (25) contains no re-

dundancy, since, for example, we cannot even provethat �(7) is irrational.

33

Now recurse, using the following scheme.

� From any found relation, toss out the Zeta

value. If no relation is found amongst the re-

maining sums, output the relation that held

when the Zeta value was included, and report

it as non-redundant. Otherwise, systematically

discard the various l sums from the list, until

a non-redundant relation remains.

y The aforementioned procedure yields formu-

lae on the next slide, like

17�(4)�36�(4; P0) = 5�(4)� 108�(2; P2) = 0;

which evidently exhaust the list of non-redundant

formulae for each given Zeta value.

� No additional formulae other than the for-

mulae given above were found for �(2), �(3)

and �(5).

34

7�(6) + 1944�(4; P4)� 1944�(2; P 22 )

= 215�(6) � 2592�(4; P2)� 3888�(2; P4)

= 229�(6) � 2592�(4; P2)� 3888�(2; P 22 )

= 1481�(6) � 2592�(6; P0)� 3888�(2; P 22 )

= 313�(6) � 648�(6; P0) + 648�(4; P2)

= 163�(6) � 288�(6; P0)� 432�(2; P4)

= 0;

2�(7)� 5l(7; P0)� 25l(3; P4)

= 4�(7)� 25l(3; P 22 ) + 40l(5; P2) + 225l(3; P4)

= 22�(7) � 25l(3; P 22 ) + 40l(5; P2)� 45l(7; P0)

= 0;

72�(9) + 135l(7; P2)� 147l(9; P0)� 60l(5; P 22 )

�85l(3; P6) + 25l(3; P 32 )

= 36�(9) � 540l(5; P4)� 96l(9; P0) + 60l(5; P 22 )

�1130l(3; P6) + 675l(3; P4P2)� 25l(3; P 32 )

= 4�(9) + 196l(5; P4) + 32l(7; P2)� 36l(5; P 22 )

+390l(3; P6)� 245l(3; P4P2) + 15l(3; P 32 )

= 4�(9)� 20l(5; P4) + 5l(7; P2)� 9l(9; P0)

�45l(3; P6) + 25l(3; P4P2)

= 116�(9) + 68l(5; P4) + 226l(7; P2)� 234l(9; P0)

�108l(5; P 22 )� 85l(3; P4P2) + 45l(3; P 3

2 )

= 0;

35

Uniqueness and �(4n+3)

� Extending the list above, makes it apparent

that �(4n+ 3) has a unique representation in

terms of l sums of the form l(m;Pr) in which

4jr. We exploited this vigorously� to arrive at

our generating function formula (26). Unfor-

tunately, there seems to be no analogous se-

lection for �(4n+1).

� Our Maple code producing all possible non-

redundant formulae for �(13) ran for over two

months before it was killed. The resulting in-

complete �le is over three thousand lines long

and contains hundreds and hundreds of inde-

pendent formulae.

� If a generating function identity for �(4n+1)

is found, it is unlikely that it will be discovered

by hunting for the appropriate representatives

from the identities for �(9), �(13), etc. and

looking for a pattern.

�We could use \partitions of n" terms (rather than 2n)and so compute as far as �(4 � 11 + 3).

36

The case of 4N+1

Recall Ramanujan's formulae:

2�(4n+3) = �41Xk=1

k�4n�3

e2�k � 1+

(2�)4n+32n+2Xk=0

(�1)k+1 B2k

(2k)!

B4n+4�2k(4n+4� 2k)!

and

2�(4n+1) = ��n

1Xk=1

k�4n

sinh2(�k)� 4

1Xk=1

k�4n�1

e2�k � 1+

(2�)4n+1 1

2n

2n+1Xk=0

(�1)k+1(2k � 1)B2k

(2k)!

B4n+2�2k(4n+2� 2k)!

� Here, the additional complexity in the 4n+1

case arises from taking the derivative of the

appropriate modular transformation formula.

QUESTION. Is there an analogous phenomenon

operating in the case of these Ap�ery-like iden-

tities as well.

37

Part IIIa. Reducibility: Dimensional conjectures �

�

�

EULER SUMS

�(i1; i2; : : : ; ik ; �1; �2; : : : ; �k) :=

Xn1>n2>:::>nk>0

�n11 �

n22 � � ��nkk

ni11 n

i22 � � �n

ikk

� �i 2 f1;�1g de�ne Euler sums. General �iyield Eulerian polylogarithms.

�

�

MZVs (MULTIPLE ZETA VALUES)

�(i1; i2; : : : ; ik) :=X

n1>n2>:::>nk>0

1

ni11 n

i22 � � �n

ikk

� That is: �i � 1.

�J.M. Borwein, D.M. Bradley and D.J. Broadhurst,\Evaluations of k�fold Euler/Zagier sums: a com-pendium of results for arbitrary k," Electronic Journal

of Combinatorics, 4 (1997), R5 (21 pages).

38

A REDUCTION of EULER'S

� As a �rst taste:

�(a; b) + �(b; a) = �(a)�(b)� �(a+ b)

reduces �(a; a).

� Maple can \prove" Euler's result: generat-

ingfunctionology yields

�(m;1) =(�1)m(m� 1)!

Z 1

0

lnm�1(t) ln(1� t)

1� tdt

=(�1)m

2(m� 1)!

Z 1

0

(m� 1) lnm�2(t) ln2(1� t)

tdt

which yields a �-function derivative

�(m;1) =(�1)m

2(m� 2)!B(m�2)1 (0) (31)

where B1(y) :=@2

@x2�(x; y)

����x=1

.

39

Since

@2

@x2�(x; y) =

�(x; y)�((x)�(x+ y))2+ (0(x)�0(x+ y))

�;

we have a digamma representation via

B1(y) =1

y

�(� �(y+1))2+ (�(2)�0(y+1))

�:

� Indeed, we may implement (31) in Maple or

Mathematica very painlessly and discover its

Riemann �-function expression:

�(n;1) =1Xk=1

�1+

1

2+ � � �+ 1

k

�(k+1)�n

=n�(n+1)

2� 1

2

n�2Xk=1

�(n� k)�(k+1):

40

A KEY PROBLEM

� Find the dimension of aminimal generating

set for a (Q;+; �)-algebra that contains

� all Euler sums of weight n and depth k,

generated by Euler sums ... En;k

� all MZVs of weight n and depth k,

generated by Euler sums ... Mn;k

� all MZVs of weight n and depth k,

generated by MZVs ... Dn;k

41

CONJECTURED GENERATING FUNCTIONS

(Broadhurst{Kreimer, BBK, Zagier)

Yn�3

Yk�1

(1� xnyk)En;k?= 1� x3y

(1� x2)(1� xy)

Yn�3

Yk�1

(1� xnyk)Mn;k?= 1� x3y

1� x2

Yn�3

Yk�1

(1� xnyk)Dn;k?= 1� x3y

1� x2

+x12y2(1� y2)

(1� x4)(1� x6)

y For k = 2, n odd and k = 3, n even, the re-

sult for Dn;k is proven by \elementary meth-

ods" by Borwein & Girgensohn.

� Dn;k has a disconcertingly complicated ratio-

nal generating function.

42

�

�

EXAMPLES of Reductions

MZV via MZVs

�(4;1;3) =

��(5;3)+ 7136�(8)� 5

2�(5)�(3)+ 1

2�(3)2�(2)

MZV via MZVs and Euler sums�

�(4;2;4;2) =

�102427

�(�9;�3)� 2679915528

�(12)� 104027

�(9;3)

�763�(9)�(3)� 160

9�(7)�(5)+ 2�(6)�(3)2

+14�(5;3)�(4) + 70�(5)�(4)�(3)� 16�(3)4

� �(5;3); �(�9;�3) are irreducible

� Functional equations assist:�(a; b; c) + �(a; c; b) + �(c; a; b) =

�(c)�(a; b)� �(a; b+ c)� �(a+ c; b)

�We write �si to denote alternation in the i-th position.

43

En;k

k 1 2 3 4 5 6

n

3 1

4 1

5 1 1

6 1 1

7 1 2 1

8 2 2 1

9 1 3 3

10 2 5 3

11 1 5 7

12 3 8 9

13 1 7 14

14 3 14 20

15 1 9 25

16 4 20 42

17 1 12 42

18 4 30 75

19 1 15 66

20 5 40 132

44

Mn;k

k 1 2 3 4 5 6

n

3 1

4

5 1

6

7 1

8 1

9 1

10 1

11 1 1

12 2

13 1 2

14 2 1

15 1 3

16 3 2

17 1 5 1

18 3 5

19 1 7 3

20 4 8 1

45

Dn;k

k 1 2 3 4 5 6

n

3 1

4

5 1

6

7 1

8 1

9 1

10 1

11 1 1

12 1 1

13 1 2

14 2 1

15 1 2 1

16 2 3

17 1 4 2

18 2 5 1

19 1 5 5

20 3 7 3

46

� Generating functions con�rmed numerically�in the following ranges:

En;k: (with REDUCE and PSLQ)

k = 2 and n � 44 ... k = 7 and n � 8

Mn;k: (with REDUCE and PSLQ)

k = 2 and n � 17 ... k = 7 and n � 20

Dn;k: (modulo a big prime)

�With REDUCE and FORTRAN at OU: (DEC�,

256Mb, 333Mhz)

k = 3 and n � 141 ... k = 7 and n � 21

� With FORTRAN at MUN:(DEC�, 4 � 1Gb,

400Mhz)

k = 3 and n � 161 ... k = 7 and n � 23

�A sizable subset is proven symbolically.Tools: partial fractions/functional equations/shu�es

47

ETINGHOF: \Let VN � C be the Q-vector

space generated by MZV's of weight N , and

WN be the subspace of VN of all elements rep-

resentable as rational polynomials of MZVs of

lower weights, with all terms have weight N .

Let DN = dimQ(VN=WN). You conjectured�

that DN = D�N where D�N are given by the for-

mula

1YN=1

(1� xN)D�

N = 1� x2 � x3:

This conjecture consists of 2 parts: the upper

bound DN � D�N , and the lower bound DN �D�N . The upper bound involves proving that

�(2n+1) is irrational and is well beyond the range

of accessibility for today's number theory. On

the other hand, the lower bound reduces to a

purely algebraic problem, which can be solved

modulo:"

�A very special sub{case: DN =P

kDN;k.

48

Drinfeld (1991)-Deligne Conjecture.

The graded Lie algebra of Grothendieck-Teichmuller

has no more than one generator in odd de-

grees, and no generators in even degrees.

� Hybrid Evaluate+PSLQ ) exact reduction

code has been performed for:

1. All alternating (Euler) sums to weight 9;

2. All MZV's to weight 14.

Thus all these evaluations are established.

49

Part IIIb. Duality Evaluations and computations

For non-negative integers s1; : : : ; sk, let

�a(s1; : : : ; sk) :=X

nj>nj+1>0

a�n1kY

j=1

n�sjj ;

(32)

a special case of our multidimensional polylog-

arithm. Note that

�a(s) =Xn>0

1

anns= Lis(a

�1)

is the usual polylogarithm for s 2 N and jaj > 1.

� Put� := �1 � := �2;

!a :=dx

x� a;

�(�1; : : : ; �k) :=X

nj>nj+1>0

kYj=1

�njj

nj;

a Unit Euler sum.

50

� s := s1 + � � � + sk and r := r1 + � � � + rk,

denote weights of strings in N. Iterated integral

representations:�

�a(s1; : : : ; sk) = (�1)kZ 1

0

kYj=1

!sj�10 !a;

and dually

�a(s1; : : : ; sk) = (�1)s+kZ 1

0

1Yj=k

!1�a!sj�11 ;

follow on changing x 7! 1�x at each level. So:(�1)k�a(s1+2; f1gr1; : : : ; sk +2; f1grk)

= (�1)rZ 1

0

kYj=1

!sj+1

0 wrj+1a ;

and dually

(�1)k�a(s1+2; f1gr1; : : : ; sk +2; f1grk)

= (�1)sZ 1

0

1Yj=k

!rj+1

1�a wsj+1

0 :

�Integrated over 0 � x1 � x2 � : : : � xs � 1.

51

� a := 1 gives the \duality for MZVs":

�(s1+2; f1gr1; : : : ; sk +2; f1grk)

= �(rk+2; f1gsk; : : : ; r1+ 2; f1gs1):� a := 2 gives a corresponding \kappa-to-unit-

Euler" duality:

�(s1+2; f1gr1; : : : ; sk+2; f1grk)

= (�1)r+k�(1; f1grk;1; f1gsk; : : : ;1; f1gr11; f1gs1):� A more general, less convenient, \kappa-to-

unit-Euler" duality similarly derivable is

�(s1; : : : ; sk) = (�1)k�(�1; �2=�1; �3=�2; : : : ; �s=�s�1);where

[�1; : : : ; ��] = [�1; f1gsk�1; : : : ;�1; f1gs1�1]:

52

SOME �$ � DUALITY EXAMPLES

�(1) =Xn�1

1

n2n= � log(1=2)

=Xn�1

(�1)n+1

n= ��(1);

�(2) =Xn�1

1

n22n= Li2(1=2)

=Xn�1

(�1)n+1

n

n�1Xk=1

(�1)kk

= ��(1;1);

�(r+2) =Xn�1

1

nr+22n= Lir+2(1=2)

= ��(1;1; f1gr); (r � 0)

53

SOME MORE DUALITY EXAMPLES

�(f1gn) = (�1)n�(�1; f1gn�1) =(ln 2)n

n!

�(2; f1gn) = (�1)n+1�(�1; f1gn;�1)

�(f1gm+1;2; f1gn) = (�1)m+n

�(�1; f1gn; f�1g2; f1gm)

� �(1; n+2) = �(�1;�1; f1gn;�1)

and

�(1; n) =

Z 12

0

Lin(z)

1� zdz:

In particular,

�(1;2) =5

7Li2(

1

2)Li1(

1

2)�2

7Li3(

1

2)+

5

21Li1(

1

2)3

and

�(1;3) = Li3(1

2)Li1(

1

2)� 1

2Li2(

1

2)2:

54

TWO �-REDUCTIONS

� Using integral ideas similar to below:

GOOD. Every MZV of depth N is a sum of

2N �'s of depth N . (Hence easily computed.)

BETTER: !0 = dx=x, !1 = �dx=(1� x),

�(s1; : : : ; sk) :=X

nj>nj+1>0

kYj=1

n�sjj

again has representation

�(s1; : : : ; sk) = (�1)kZ 1

0!s1�10 !1 : : : !

sk�10 !1:

� Domain, 1 > xj > xj+1 > 0, in n =Pj sj

variables, splits into n+1 parts: each being a

product of regions 1 > xj > xj+1 > �, for �rst

r variables, and � > xj > xj+1 > 0, for rest.

� xj ! 1�xj replaces integral of the former by

the latter type, with � replaced by �� := 1� �.

Thence:

55

� Let S(!0; !1) be the n{string !s1�10 !1 : : : !sk�10 !1

specifying a MZV. Let Tr denote the substring

of the �rst r letters and Un�r the complemen-tary substring, on the last n� r letters, so that

S = TrUn�rfor n � r � 0. Then

�(s1; : : : ; sk) =

Z 1

0S =

nXr=0

�Z ��

0

eTr Z �

0Un�r

where e indicates reversal of letter order.� The alternate polylogarithm integral

�z(s1; : : : ; sk) =X

nj>nj+1>0

z�n1kY

j=1

n�sjj

=

Z z

0!s1�10 !1 : : : !

sk�10 !1

produces the MZV as the scalar product of two

vectors, composed of �z{values with z := p and

z := q, for any 0 < p < 1 and

1

p+1

q= 1:

56

�

�

ILLUSTRATION

� Usually set p = q = 2 (�2 = �)

� Thus, for any 1p +

1q = 1

�(2;1;2;1;1;1) = �p(2;1;2;1;1;1)

+�p(1;1;2;1;1;1)�q(1) + �p(1;2;1;1;1)�q(2)

+�p(2;1;1;1)�q(3) + �p(1;1;1;1)�q(1;3)

+�p(1;1;1)�q(2;3)+ �p(1;1)�q(3;3)

+�p(1)�q(4;3)+ �q(5;3) = �(5;3)

� Uses 2n �'s (no higher depth, same weight)

� This is also a duality result (q !1)!

� Di�erentiation ) �(0; f1gn) = �(f1gn).

� Applied to �(n + 2) this provides a lovely

closed form for �(2; f1gn).57

Seq := proc(s, t) local k, n;

if 1 < s[1] then

[s[1] - 1, seq(s[k], k = 2 .. nops(s))], [1, op(t)]

else [seq(s[k], k = 2 .. nops(s))],

[t[1] + 1, seq(t[k], k = 2 .. nops(t))]

fi end

SEQ := proc(a) local w, k, s;

w := convert(a, `+`); s := a, [];

for k to w do s := Seq(s); print(s, k) od;

s[2] end

>SEQ([5,3]);

[4, 3], [1], 1

[3, 3], [1, 1], 2

[2, 3], [1, 1, 1], 3

[1, 3], [1, 1, 1, 1], 4

[3], [2, 1, 1, 1], 5

[2], [1, 2, 1, 1, 1], 6

[1], [1, 1, 2, 1, 1, 1], 7

[], [2, 1, 2, 1, 1, 1], 8

[2, 1, 2, 1, 1, 1]

58

� Time for D digits for MZV �(s1; : : : ; sk), of

weight n, is roughly c(n)D precision D multi-

plications (with c(n) / n, for large n, whatever

the depth, k).

� Idea extends appropriately to all Euler sums.

� 100 digits of �(5;3) takes only 0:06 CPU sec-

onds and 1;000 digits only 8:53 CPU seconds

on Manyjars (CECM's 194 MHz R10000 SGI).

� Bailey's superb MPFUN took 47 minutes to

get 20;000 digits, on Tempus (400 MHz Dec

at MUN).

� 5;000 digit PSLQ computations now feasi-

ble/in train { this allows for 145-term relation-

ships to be found at 5,000 digits (256-terms

with 2;000 digits).

59

Part IIIc. Proof of a conjecture due to Zagier

� For r � 1 and n1; : : : ; nr � 1, again specialize

the polylogarithm to one variable�

L(n1; : : : ; nr;x) :=X

0<mr<:::<m1

xm1

mn11 : : :mnr

r:

� Thus

L(n;x) =x

1n+

x2

2n+

x3

3n+ � � �

is the classical polylogarithm, while

L(n;m; x) =1

1m

x2

2n+ (

1

1m+

1

2m)x3

3n

+(1

1m+

1

2m+

1

3m)x4

4n+ � � � ;

L(n;m; l; x) =1

1l

1

2m

x3

3n

+(1

1l

1

2m+

1

1l

1

3m+

1

2l

1

3m)x4

4n+ � � �

�L(n1; : : : ; nr; x) = �x�1(n1; : : : ; nr) to highlight depen-dence on x.

60

� Series converge absolutely for jxj < 1 (condi-

tionally on jxj= 1 unless n1 = 1 and x = 1).

� These polylogarithms

L(nr; : : : ; n1; x) =X

0<m1<:::<mr

xmr

mnrr : : :m

n11

;

are determined uniquely by

d

dxL(nr; : : : ; n1; x) =

1

xL(nr � 1; : : : ; n2; n1; x)

if nr � 2; while for nr = 1

d

dxL(nr; : : : ; n1; x) =

1

1� xL(nr�1; : : : ; n1;x)

with the initial conditions

L(nr; : : : ; n1; 0) = 0

for r � 1 and

L(;;x) � 1:

61

� Thus, if s := (s1; s2; : : : ; sN) and w :=Psi,

every periodic polylogarithm leads to a func-

tion

Ls(x; t) :=XnL(fsgn;x)twn

which solves an algebraic ODE in x, and leads

to nice recurrence relations.

� In the simplest case, with N = 1, the ODE

is DsF = tsF where

Ds :=

�(1� x)

d

dx

�1 �xd

dx

�s�1and the solution (by series) is a generalized

hypergeometric function:

Ls(x; t) := 1+Xn�1

xnts

ns

n�1Yk=1

1+

ts

ks

!;

as follows from considering Ds(xn).

62

� For N = 1 and negative integers , we similarly

obtain

L�s(x; t) := 1+Xn�1

(�x)n ts

ns

n�1Yk=1

1+ (�1)k t

s

ks

!;

and, L�1(2x � 1; t) solves the hypergeometric

ODE.

� IndeedL�1(1; t) =

1

�(1+ t=2;1=2� t=2):

� We correspondingly obtain ODEs for even-

tually periodic Euler sums. Thus L�2;1(x; t) isa solution of

t6F = x2(x� 1)2(x+1)2D6F

+ x(x� 1)(x+1)(15x2 � 6x� 7)D5F

+ (x� 1)(65x3+14x2 � 41x� 8)D4F

+ (x� 1)(90x2 � 11x� 27)D3F

+ (x� 1)(31x� 10)D2F

+ (x� 1)DF

63

� Again, let F(a; b; c;x) denote the hypergeo-

metric function. Then:

THEOREM (B3GL).

1Pn=0

L(3;1;3;1; : : : ;3;1| {z }n�fold

; x) t4n =

F

t(1 + i)

2;�t(1 + i)

2; 1;x

!F

t(1� i)

2;�t(1� i)

2; 1; x

!

Proof. Both sides of the putative identity

start

1 +t4

8x2+

t4

18x3+

t8+44t4

1536x4+ � � �

and are annihilated by the operator

D31 :=

�(1� x)

d

dx

�2 �xd

dx

�2� t4

� Once discovered this can be checked vari-

ously in Maple!

64

THE MAPLE CODE

deq:=proc(F) D(D(F))+A*D(F)+B*F; end;

eqns:=

{(deq(H))(x),D((deq(H)))(x),D(D((deq(H))))(x)};

for p from 0 to 4 do

eqns:=subs((`@@`(D,p))(H)(x)=y[p],eqns); od:

yi_sol:=solve(eqns,{seq(y[i], i=0..4)});

id:=x->x; T:=x->t;

# The annihilator to be checked for any product

A31:=proc(F)

(1-id)*D((1-id)*D(id*D(id*D(F)))) - T^4*F; end;

Z:=expand(A31(F1*F2)(x));

for p from 0 to 4 do

Z:=subs((`@@`(D,p))(F1)(x)=y[1,p],

(`@@`(D,p))(F2)(x)=y[2,p],Z);

od:

65

# The annihilator to be checked for this product

a[1]:= x -> 1/x;

b[1]:= x -> I*t^2/2*1/(x*(1-x));

a[2]:= x -> 1/x;

b[2]:= x -> -I*t^2/2*1/(x*(1-x));

for i from 1 to 2 do

for o from 2 to 4 do

y[i,o]:=

subs(subs(A=a[i],B=b[i],

y[0]=y[i,0],y[1]=y[i,1],yi_sol),y[o]):

od:

od:

normal(Z);

# IS ZERO SHOWING THE 2_F_1 PRODUCT

IS ANNIHILATED

66

Corollary (Zagier Conjecture).

�(3;1;3;1; : : : ;3;1| {z }n�fold

) =2�4n

(4n+2)!

Proof. We have

F(a;�a; 1; 1) = 1

�(1� a)�(1 + a)=

sin�a

�a

and hence, setting x = 1,

F

t(1 + i)

2;�t(1 + i)

2; 1; 1

!F

t(1� i)

2;�t(1� i)

2; 1; 1

!

=2

�2t2sin

�1+ i

2�t

�sin

�1� i

2�t

�

=cosh�t� cos�t

�2t2=

1Xn=0

2�4nt4n

(4n+2)!

� This proof is Zagier's adaptation of Broad-

hurst's based on extensive empirical work by

B3GL.

67

Generalizations of the Zagier identity

� Broadhurst{Lisonek have an alternative proof

of Zagier: no DEs, just combinatorial ma-

nipulations of the iterated integral repre-

sentations:

\pure shu�e identity = integral iden-

tity independent of the di�erential forms

that appear in the identity."

� By the same technique, Broadhurst{Lisonek

obtain a \Zagier dressed with 2" identity:

Xs�(s) =

�4n+2

(4n+3)!(33)

where s runs over all 2n+1 possible inser-

tions of the number 2 in the string f3;1gn.

� Also (33) is just the beginning of a large

family of conjectured identities we are cur-

rently investigating intensively using PSLQ.

68

� Compare the much easier:

1Xn=0

L(f2gn;x) t2n = F (it;�it; 1;x)

and, more generally,P1n=0L(fpgn; x) tpn

= pFp�1(�!t;�!3t; : : : ;�!2p�1t; 1; : : : ;1; x)

where !p = �1:

� The amazing factorizations in the result for

�(f3;1gn) and

�(f3;1gn) = 4�n�(f4gn) = 1

2n+1�(f2;2gn)

beg the

QUESTION. \What other deep Clausen{like

hypergeometric factorizations lurk within?"

y And why is

8n�(f�2;1gn) = �(f2;1gn)?69