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This article was stimulated by the recent SIAM ''100 DigitChallenge'' of Nick Trefethen, beautifully described in a recent book. Indeed, these ten numeric challenge problems are also listed in a recent book by two of present authors, where they are followed by the ten symbolic/numeric challenge problems that are discussed in this article. Our intent was to present ten problems that are characteristic of the sorts of problems that commonly arise in ''experimental mathematics''. The challenge in each case is to obtain a high precision numeric evaluation of the quantity, and then, if possible, to obtain a symbolic answer, ideally one with proof. Our goal in this article is to provide solutions to these ten problems, and in the process present a concise account of how one combines symbolic and numeric computation, which may be termed ''hybrid computation'', in the process of mathematical discovery.
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Ten Problems in Experimental Mathematics
David H. Bailey, Jonathan M. Borwein, Vishaal Kapoor, and Eric W. Weisstein
January 25, 2006
1. INTRODUCTION. This article was stimulated by the recent SIAM “100 Digit
Challenge” of Nick Trefethen, beautifully described in  (see also ). Indeed, these
ten numeric challenge problems are also listed in [15, pp. 22–26], where they are followed
by the ten symbolic/numeric challenge problems that are discussed in this article. Our
intent in  was to present ten problems that are characteristic of the sorts of problems
that commonly arise in “experimental mathematics” . The challenge in each case
is to obtain a high precision numeric evaluation of the quantity, and then, if possible, to
solutions to these ten problems, and at the same time, to present a concise account of
how one combines symbolic and numeric computation, which may be termed “hybrid
computation,” in the process of mathematical discovery.
The passage from object αto answer Ω often relies on being able to compute the
object to suﬃciently high precision, for example, to determine numerically whether αis
algebraic or is a rational combination of known constants. While some of this is now
automated in mathematical computing software such as Maple and Mathematica, in most
cases intelligence is needed, say in choosing the search space and in deciding the degree of
polynomial to hunt for. In a similar sense, using symbolic computing tools such as those
incorporated in Maple and Mathematica often requires signiﬁcant human interaction to
produce material results. Such matters are discussed in greater detail in  and .
Integer relation detection. Several of these solutions involve the usage of integer
relation detection schemes to ﬁnd experimentally a likely relationship. For a given real
vector (x1, x2,·· ·, xn) an integer relation algorithm is a computational scheme that either
ﬁnds the n-tuple of integers (a1, a2,···, an), not all zero, such that a1x1+a2x2+···anxn=
0 or else establishes that there is no such integer vector within a ball of some radius about
the origin, where the metric is the Euclidean norm (a2
1+a2
2+· ·· +a2
n)1/2.
At the present time, the best known integer relation algorithm is the PSLQ algorithm
 of Helaman Ferguson, who is well known in the community for his mathematical
sculptures. Simple formulations of the PSLQ algorithm and several variants are given in
. Another widely used integer relation detection scheme involves the Lenstra-Lenstra-
Lovasz (LLL) algorithm. The PSLQ algorithm, together with related lattice reduction
schemes such as LLL, was recently named one of ten “algorithms of the century” by the
publication Computing in Science and Engineering .
1
Perhaps the best-known application of PSLQ is the 1995 discovery, by means of a
PSLQ computation, of the “BBP” formula for π:
π=
X
k=0
1
16kµ4
8k+ 1 2
8k+ 4 1
8k+ 5 1
8k+ 6.
This formula permits one to calculate directly binary or hexadecimal digits beginning at
the nth digit, without the need to calculate any of the ﬁrst n1 digits . This result has,
in turn, led to more recent results that suggest a possible route to a proof that πand some
other mathematical constants are 2-normal (i.e., that every m-long binary string occurs
in the binary expansion with limiting frequency bm). The BBP formula even has
some practical applications: it is used, for example, in the g95 compiler for transcendental
function evaluations .
All integer relation schemes require very high precision arithmetic, both in the input
data and in the operation of the algorithms. Simple reckoning shows that if an integer
relation solution vector (ai, a2,···, an) has Euclidean norm 10d, then the input data must
be speciﬁed to at least dn digits, lest the true solution be lost in a sea of numerical arti-
facts. In some cases, including one mentioned at the end of the next section, thousands
of digits are required before a solution can be found with these methods. This is the
principal reason for the great interest in high-precision numerical evaluations in experi-
mental mathematics research. It is the also the motivation behind this set of ten challenge
problems.
2. THE BIFURCATION POINT B3.
Problem 1. Compute the value of rfor which the chaotic iteration xn+1 =rxn(1 xn),
starting with some x0in (0,1), exhibits a bifurcation between four-way periodicity and
eight-way periodicity. Extra credit: This constant is an algebraic number of degree not
exceeding twenty. Find the minimal polynomial with integer coeﬃcients that it satisﬁes.
History and context. The chaotic iteration xn+1 =rxn(1 xn) has been studied since
the early days of chaos theory in the 1950s. It is often called the “logistic iteration,” since
it mimics the behavior of an ecological population that, if its growth one year outstrips
its food supply, often falls back in numbers for the following year, thus continuing to vary
in a highly irregular fashion. When ris less than one iterates of the logistic iteration
converge to zero. For rin the range 1 < r < B1= 3 iterates converge to some nonzero
limit. If B1< r < B2= 1 + 6 = 3.449489 . . ., the limiting behavior bifurcates—every
other iterate converges to a distinct limit point. For rwith B2< r < B3iterates hop
between a set of four distinct limit points; when B3< r < B4, they select between a set of
eight distinct limit points; this pattern repeats until r > B= 3.569945672 . . ., when the
iteration is completely chaotic (see Figure 1). The limiting ratio limn(BnBn1)/(Bn+1
Bn) = 4.669201 . . . is known as Feigenbaum’s delta constant.
A very readable description of the logistic iteration and its role in modern chaos
theory are given in Gleick’s book . Indeed, John von Neumann had suggested using
the logistic map as a random number generator in the late 1940s. Work by W. Ricker in
2
Figure 1: Bifurcation in the logistic iteration.
1954 and detailed analytic studies of logistic maps beginning in the 1950s with Paul Stein
and Stanislaw Ulam showed the existence of complicated properties of this type of map
beyond simple oscillatory behavior [35, pp. 918-919].
Solution. We ﬁrst describe how to obtain a highly accurate numerical value of B3using
a relatively straightforward search scheme. Other schemes could be used to ﬁnd B3;
we present this one to underscore the fact that computational results suﬃcient for the
purposes of experimental mathematics can often be obtained without resorting to highly
sophisticated techniques.
Let f8(r, x) be the eight-times iterated evaluation of rx(1 x), and let g8(r, x) =
f8(r, x)x. Imagine a three-dimensional graph, where rranges from left to right and x
ranges from bottom to top (as in Figure 1), and where g8(r, x) is plotted in the vertical
(out-of-plane) dimension. Given some initial rslightly less than B3, we compute a “comb”
of function values at nevenly spaced xvalues (with spacing hx) near the limit of the
iteration xn+1 =f8(r, xn). In our implementation, we use n= 12, and we start with
r= 3.544, x = 0.364, hr= 104, and hx= 5 ×104. With this construction, the comb has
n/2 negative function values, followed by n/2 positive function values. We then increment
rby hrand reevaluate the “comb,” continuing in this fashion until two sign changes are
observed among the nfunction values of the “comb.” This means that a bifurcation
occurred just prior to the current value of r, so we restore rto its previous value (by
subtracting hr), reduce hr, say by a factor of four, and also reduce the hxroughly by
a factor of 2.5. We continue in this fashion, moving the value of rand its associated
3
“comb” back and forth near the bifurcation point with progressively smaller intervals hr.
The center of the comb in the x-direction must be adjusted periodically to ensure that
n/2 negative function values are followed by n/2 positive function values, and the spacing
parameter hxmust be adjusted as well to ensure that two sign changes are disclosed
when this occurs. We quit when the smallest of the nfunction values is within two or
three orders of magnitude of the “epsilon” of the arithmetic (e.g., for 2000-digit working
precision, “epsilon” is 102000). The ﬁnal value of ris then the desired value B3, accurate
to within a tolerance given by the ﬁnal value of rh. With 2000-digit working precision,
our implementation of this scheme ﬁnds B3to 1330-digit accuracy in about ﬁve minutes
on a 2004-era computer. The ﬁrst hundred digits are as follows:
B3= 3.544090359551922853615965986604804540583099845444573675457812530
3058429428588630122562585664248917999626 . . .
With even a moderately accurate value of rin hand (at least two hundred digits or
so), one can use a PSLQ program (such as the PSLQ programs available at the URL
http://crd.lbl.gov/~dhbailey/mpdist) to check whether ris an algebraic constant. This
is done by computing the vector (1, r, r2,··· , rn) for various n, beginning with a small
value such as two or three, and then searching for integer relations among these n+ 1 real
numbers. When n12, the relation
0 = r12 12r11 + 48r10 40r9193r8+ 392r7+ 44r6+ 8r5977r4
604r3+ 2108r2+ 4913 (1)
can be recovered.
A symbolic solution that explicitly produces the polynomial (1) can be obtained as
follows. We seek a sequence x1, x2, .., x4that satisﬁes the equations
x2=rx1(1 x1), x3=rx2(1 x2), x4=rx3(1 x3), x1=rx4(1 x4),
and
1 = ¯¯¯¯¯
4
Y
i=1
r(1 2xi)¯¯¯¯¯.
The ﬁrst four conditions represent a period-4 sequence in the logistic equation xn+1 =
rxn(1 xn), and the last condition represents the stability of the cycle, which must be 1
or 1 for a bifurcation point (see  for details).
First, we deal with the system corresponding to 1 + Q4
i=1 r(1 2xi) = 0.We compute
the lexicographic Groebner basis in Maple:
with(Groebner):
L := [x2 - r*x1*(1-x1),x3 - r*x2*(1-x2),x4 - r*x3*(1-x3),
x1 - r*x4*(1-x4),r^4*(1-2*x1)*(1-2*x2)*(1-2*x3)*(1-2*x4)+1];
gbasis(L,plex(x1,x2,x3,x4,r));
4
After a cup of coﬀee, we discover the univariate element
(r4+ 1)(r48r3+ 24r232r+ 17) ×(r44r34r2+ 16r+ 17) ×
(r12 12r11 + 48r10 40r9193r8+ 392r7+ 44r6+ 8r5977r4
604r3+ 2108r2+ 4913)
in the Groebner basis, in which the monomial ordering is lexicographical with rlast.
The ﬁrst three of these polynomials have no real roots, and the fourth has four real
roots. Using trial and error, it is easy to determine that B3is the root of the minimal
polynomial
r12 12r11 + 48r10 40r9193r8+ 392r7+ 44r6+ 8r5977r4
604r3+ 2108r2+ 4913,
which has the numerical value stated earlier. The corresponding Mathematica code reads:
GroebnerBasis[{x2 - r x1(1 - x1), x3 - r x2(1 - x2),
x4 - r x3(1 - x3), x1 - r x4(1 - x4),
r^4(1 - 2x1)(1 - 2x2)(1 - 2x3)(1 - 2x4) + 1},
r,
{x1, x2, x3, x4}, MonomialOrder -> EliminationOrder] // Timing
This requires only 1.2 seconds on a 3 GHz computer. These computations can also be
recreated very quickly in Magma, an algebraic package available at
http://magma.maths.usyd.edu.au/magma:
Q := RationalField(); P<x,y,z,w,r> := PolynomialRing(Q,5);
I:= ideal< P| y - r*x*(1-x), z - r*y*(1-y), w - r*z*(1-z),
x - r*w*(1-w), r^4*(1-2*x)*(1-2*y)*(1-2*z)*(1-2*w)+1>;
time B := GroebnerBasis(I);
This took 0.050 seconds on a 2.4Ghz Pentium 4.
The signiﬁcantly more challenging problem of computing and analyzing the constant
B4= 3.564407266095 ··· is discussed in . In this study, conjectural reasoning suggested
that B4might satisfy a 240-degree polynomial, and, in addition, that α=B4(B42)
might satisfy a 120-degree polynomial. The constant αwas then computed to over
10,000-digit accuracy, and an advanced three-level multi-pair PSLQ program was em-
ployed, running on a parallel computer system, to ﬁnd an integer relation for the vector
(1, α, α2,··· , α120). A numerically signiﬁcant solution was obtained, with integer coeﬃ-
cients descending monotonically from 25730, which is a 73-digit integer, to the ﬁnal value,
which is one (a striking result that is exceedingly unlikely to be a numerical artifact).
This experimentally discovered polynomial was recently conﬁrmed in a large symbolic
computation .
Additional information on the Logistic Map is available at
http://mathworld.wolfram.com/LogisticMap.html.
5
Problem 2. Evaluate
X
(m,n,p)6=0
(1)m+n+p
m2+n2+p2,(2)
where convergence means the limit of sums over the integer lattice points enclosed in in-
creasingly large cubes surrounding the origin. Extra credit: Usefully identify this constant.
History and context. Highly conditionally convergent sums like this are very common
in physical chemistry, where they are usually written down with no thought of convergence.
The sum in question arises as an idealization of the electrochemical stability of NaCl. One
computes the total potential at the origin when placing positive and negative charges at
each nonzero point of the cubic lattice [16, chap. 4].
Solution. It is important to realize that this sum must be viewed as the limit of the sum
in successively larger cubes. The sum diverges when spheres are used instead. To clarify
this consider, for complex s, the series
b2(s) = X
(m,n)6=0
(1)m+n
(m2+n2)s/2, b3(s) = X
(m,n,p)6=0
(1)m+n+p
(m2+n2+p2)s/2.(3)
These converges in two and three dimensions, respectively, over increasing “cubes,” pro-
vided that Re s > 0. When s= 1, one may sum over circles in the plane but not spheres
in three-space, and one may not sum over diamonds in dimension two. Many chemists do
not know that b3(1) 6=Pn(1)nr3(n)/n, a series that arises by summing over increasing
spheres but that diverges. Indeed, the number r3(n) of representations of nas a sum of
three squares is quite irregular—no number of the form 8n+7 has such a representation—
and is not O³n1/2´.This matter is somewhat neglected in the discussion of Madelung’s
constant in Julian Havil’s deservedly popular recent book Gamma: Exploring Euler’s
Constant, , which contains a wealth of information related to each of our problems in
Straightforward methods to compute (3) are extremely unproductive. Such techniques
produce at most three digits—indeed, the physical model should have a solar-system sized
salt crystal to justify ignoring the boundary. Thus, we are led to using more sophisticated
methods. We note
b3(s) = X0(1)i+j+k
(i2+j2+k2)s/2,
where P0signiﬁes a sum over Z3\{(0,0,0)}, and let Ms(f) denote the Mellin transform
Ms(f) = Z
0f(x)xs1dx.
6
The quantity that we wish to compute is b3(1). It follows by symmetry that
b3(1) = X0(1)i+j+k(i2+j2+k2)
(i2+j2+k2)3/2
= 3 X0(1)i(i2)(1)j+k
(i2+j2+k2)3/2.(4)
We note that Ms(et) = Γ(s), so
M3/2µqn2+j2+k2= Γµ3
2(n2+j2+k2)3/2,
where n, j, and kare arbitrary integers and q=et.Continuing, we rewrite equation (4)
as
Γµ3
2b3(1) = 3M3/2·
X
n=−∞
(1)nn2qn2θ2
4(x)¸,
where θ4(x) = P
−∞(1)nxn2is the usual Jacobi theta-function. Since the theta trans-
form—a form of Poisson summation—yields θ4(eπ/s) = 2(e ), it follows that
Γµ3
2b3(1) = 3
X
n=−∞
n2M3/2µX(1)nn2qn2π
xθ2
2Ãπ2
x!.
Also, Γµ3
2=π/2, so
b3(1) = 12π
X
n=1
(1)nn2X
(j,k) odd Z
0[en2x(π2/4x)(j2+k2)]x1/2dx.
The integral is evaluated in [19, Exercise 4, sec. 2.2] and is (π/n2)1/2eπ nj2+k2,
whence
b3(1) = 48π
X
k=0
X
j=0
X
n=1
(1)nneπn((2j+1)2+(2k+1)2.
Finally, when a > 0,
4
X
n=1
(1)n+1nean =4ea
(1 + ea)2= sech2µa
2,
from which we obtain
b3(1) = 12πX
m,n1
m,n odd
sech2µπ
2(m2+n2)1/2.(5)
Summing mand nfrom 1 up to 81 in (5) gives
b3(1) = 1.74756459463318219063621203554439740348516143662474175
8152825350765040623532761179890758362694607891 . . . .
7
It is possible to accelerate the convergence further still. Details can be found in ,.
There are closed forms for sums with an even number of variables, up to 24 and
beyond. For example, b2(2s) = 4α(s)β(s), where
α(s) = X
n0
(1)n/(n+ 1)s
and
β(s) = X
n0
(1)n/(2n+ 1)s.
In particular, b2(2) = πlog 2. No such closed form for b3is known, while much work
has been expended looking for one. The formula for b2is due to Lorenz (1879). It was
rediscovered by G. H. Hardy and is equivalent to Jacobi’s Lambert series formula for
θ2
3(q):
θ2
3(q)1 = 4 X
n0
(1)nq2n+1
1q2n+1 .
This, in turn, is equivalent to the formula for the number r2(n) of representations of nas
a sum of two squares, counting order and sign,
r2(n) = 4 (d1(n)d3(n)) ,
where dkis the number of divisors of ncongruent to kmodulo four. The analysis of three
squares is notoriously harder.
http://mathworld.wolfram.com/LatticeSum.html.
4. DOUBLE EULER SUMS.
Problem 3. Evaluate the sum
X
k=1 µ11
2+· ·· + (1)k+1 1
k2
(k+ 1)3.(6)
Extra credit: Evaluate this constant as a multiterm expression involving well-known math-
ematical constants. This expression has seven terms and involves π, log 2, ζ(3), and
Li5(1/2), where Lin(x) = Pk>0xn/nkis the nth polylogarithm. (Hint: The expression
is “homogenous,” in the sense that each term has the same total “degree.” The degrees of
πand log 2 are each 1, the degree of ζ(3) is 3, the degree of Li5(1/2) is 5, and the degree
of αnis ntimes the degree of α.)
History and context. In April 1993, Enrico Au-Yeung, an undergraduate at the Uni-
versity of Waterloo, brought to the attention of one of us (Borwein) the curious result
X
k=1 µ1 + 1
2+· ·· +1
k2
k2= 4.59987 . . . 17
4ζ(4) = 17π4
360 .(7)
8
The function ζ(s) in (7) is the classical Riemann zeta-function:
ζ(s) =
X
n=1
1
ns.
Euler had solved Bernoulli’s Basel problem when he showed that, for each positive integer
n,ζ(2n) is an explicit rational multiple of π2n[16, sec. 3.2].
Au-Yeung had computed the sum in (7) to 500,000 terms, giving an accuracy of
ﬁve or six decimal digits. Suspecting that his discovery was merely a modest numerical
coincidence, Borwein sought to compute the sum to a higher level of precision. Using
Fourier analysis and Parseval’s equation, he obtained
1
2πZπ
0(πt)2log2(2 sin t
2)dt =
X
n=1
(Pn
k=1 1
k)2
(n+ 1)2.(8)
The idea here is that the series on the right of (8) permits one to evaluate (7), while
the integral on the left can be computed using the numerical quadrature facility of Math-
ematica or Maple. When he did this, Borwein was surprised to ﬁnd that the conjectured
identity holds to more than thirty digits. We should add here that, by good fortune,
17/360 = 0.047222 . . . has period one and thus can plausibly be recognized from its ﬁrst
six digits, so that Au-Yeung’s numerical discovery was not entirely far-fetched.
Solution. We deﬁne the multivariate zeta-function by
ζ(s1, s2,··· , sk) = X
n1>n2>···>nk>0
k
Y
j=1
n−|sj|
jσnj
j,
where the s1, s2, . . . , skare nonzero integers and σj= signum(sj). A fast method for
computing such sums based on H¨older convolution is discussed in  and implemented
in the EZFace+ interface, which is available as an online tool at the URL
http://www.cecm.sfu.ca/projects/ezface+. Expanding the squared term in (6), we have
X
0<i,j<k
k>0
(1)i+j+1
ijk3=2ζ(3,1,1) + ζ(3,2).(9)
Evaluating this in EZFace+ we quickly obtain
C= 0.156166933381176915881035909687988193685776709840303872957529354
497075037440295791455205653709358147578 . . . .
Given this numerical value, PSLQ or some other integer-relation-ﬁnding tool can be
used to see if this constant satisﬁes a rational linear relation with the following constants
(as suggested in the hint): π5, π4log(2), π3log2(2), π2log3(2), π log4(2),log5(2),
π2ζ(3), π log(2)ζ(3),log2(2)ζ(3), ζ(5),Li5(1/2). The result is quickly found to be
C= 4 Li5µ1
21
30 log5(2) 17
32ζ(5) 11
720π4log(2) + 7
4ζ(3) log2(2)
+1
18π2log3(2) 1
8π2ζ(3).
9
This result has been proved in various ways, both analytic and algebraic. Indeed, all
evaluations of sums of the form ζ(±a1,±a2,··· ,±am) with weight w=Pkam,(k < 8),
as in (9) have been established.
Further history and context. What Borwein did not know at the time was that Au-
Yeung’s suspected identity follows directly from a related result proved by De Doelder in
1991. In fact, it had cropped up even earlier as a problem in this Monthly, but the
story goes back further still. Some historical research showed that Euler considered these
summations. In response to a letter from Goldbach, he examined sums that are equivalent
to
X
k=1 µ1 + 1
2m+···+1
km(k+ 1)n.(10)
The great Swiss mathematician was able to give explicit values for certain of these sums
in terms of the Riemann zeta-function.
Starting from where we left oﬀ in the previous section provides some insight into
evaluating related sums. Recall that the Taylor expansion of f(x) = 1
2log(1 x) log (1 +
x) takes the form
f(x) =
X
k=0 µ11
2+1
3− · ·· +1
2k1x2k
2k.
Applying Parseval’s identity to f(eit), we have an eﬀective way of computing
X
k=0 ³11
2+1
3− · ·· +1
2k1´2
(2k)2
in terms of an integral that can be rapidly evaluated in Maple or Mathematica.
Alternatively, we may compute
X
k=0 ³1 + 1
2+1
3+· ·· +1
k´2
k2.
The Fourier expansions of (πt)/2 and log |2 sin(t/2)|are
X
n=1
sin(nt)
n=πt
2,(0 < t < 2π)
and
X
n=1
cos(nt)
n=log |2 sin(t/2)|,(0 < t < 2π),(11)
respectively. Multiplying these together, simplifying, and doing a partial fraction decom-
position gives
log |2 sin(t/2)| · πt
2=
X
n=1
1
n
n1
X
k=1
1
ksin(nt)
10
on (0,2π). Applying Parseval’s identity results in
1
4πZ2π
0(πt)2log2(2 sin(t/2))dt =
X
n=1 ³1 + 1
2+1
3+· ·· +1
n´2
(n+ 1)2.
The integral may be computed numerically in Maple or Mathematica, delivering an ap-
proximation to the sum.
The Clausen functions deﬁned by
Cl2(θ) =
X
n=1
sin()
n2,Cl3(θ) =
X
n=1
cos()
n3,Cl4(θ) =
X
n=1
sin()
n4,···
arise as repeated antiderivatives of (11). They are useful throughout harmonic analysis
and elsewhere. For example, with α= 2 arctan 7, one discovers with the aid of PSLQ
that
6Cl2(α)6Cl2(2α) + 2Cl2(3α)?
= 7Cl2µ2π
7+ 7Cl2µ4π
77Cl2µ6π
7,(12)
(here the question mark is used because no proof is yet known) or, in what can be shown
to be equivalent, that
24
77Zπ/2
π/3log Ã¯¯¯¯¯
tan (t) + 7
tan (t)7¯¯¯¯¯!dt ?
=L7(2) = 1.151925470 . . . . (13)
This arises from the volume of an ideal tetrahedron in hyperbolic space [15, pp. 90–91].
(Here L7(s) = Pn>0χ7(n)nsis the primitive L-series modulo seven, whose character
pattern is 1,1,1,1,1,1,0, which is given by
χ7(k) = 2(sin( ) + sin(2 )sin(3 ))/7
with τ= 2π/7.)
Although (13) has been checked to twenty thousand decimal digits, by using a nu-
merical integration scheme we shall describe in section 8, and although it is known for
K-theoretic reasons that the ratio of the left- and right-hand sides of (12) is rational
, to the best of our knowledge there is no proof of either (12) or (13). We might add
that recently two additional conjectured identities related to (13) have been discovered
by PSLQ computations. Let Inbe the deﬁnite integral of (13), except with limits nπ/24
and (n+ 1)π/24. Then
2I22I32I42I5+I8+I9I10 I11
?
= 0,
I2+ 3I3+ 3I4+ 3I5+ 2I6+ 2I73I8I9
?
= 0.(14)
Readers who attempt to calculate numerical values for either the integral in (13) or the
integral I9in (14) should note that the integrand has a nasty singularity at t= arctan 7.
In retrospect, perhaps it was for the better that Borwein had not known of De Doelder’s
and Euler’s results, because Au-Yeung’s intriguing numerical discovery launched a fruitful
11
line of research by a number of researchers that has continued until the present day.
Sums of this general form are known nowadays as “Euler sums” or “Euler-Zagier sums.”
Euler sums can be studied through a profusion of methods: combinatorial, analytic, and
algebraic. The reader is referred to [16, chap. 3] for an overview of Euler sums and their
applications. We take up the story again in Problem 9.
Additional information on Euler sums is available at
http://mathworld.wolfram.com/EulerSum.html.
5. KHINTCHINE’S CONSTANT.
Problem 4. Evaluate
K0=
Y
k=1 "1 + 1
k(k+ 2)#log2k
=
Y
k=1
k[log2(1+ 1
k(k+2) )].(15)
Extra credit: Evaluate this constant in terms of a less-well-known mathematical constant.
History and context. Given some particular continued fraction expansion α= [a0, a1,···],
consider forming the limit
K0(α) = lim
n→∞ (a0a1···an)1/n.
Based on the Gauss-Kuzmin distribution, which establishes that the digit distribution of
a random continued fraction satisﬁes Prob (ak=n) = log2(1 + 1/k/(k+ 2)), Khintchine
showed that the limit exists for almost all continued fractions and is a certain constant,
which we now denote K0. This circle of ideas is accessibly developed in . As such a
constant has an interesting interpretation, computation seems like the next step.
Taking logarithms of both sides of (15) and simplifying , we have
log 2 ·log K0=
X
n=1
log n·logµ1 + 1
n(n+ 2).
Such a series converges extremely slowly. Computing the sum of the ﬁrst 10000 terms
gives only two digits of log 2 ·log K0.Thus, direct computation again proves to be quite
diﬃcult.
Solution. Rewriting log nas the telescoping sum
log n= (log nlog(n1)) + ·· · + (log 2 log 1) =
n
X
k=2
logµk
k1,
we see that
log 2 ·log K0=
X
n=2
n
X
k=2
log k
k1·log (n+ 1)2
n(n+ 2).
We interchange the order of summation to obtain
log 2 ·log K0=
X
k=2
X
n=k
log (n+ 1)2
n(n+ 2) log k
k1.(16)
12
But
X
n=k
log (n+ 1)2
n(n+ 2) = log (k+ 1)
k= logµ1 + 1
k,
so (16) transforms into
log 2 ·log K0=
X
k=2
logµ11
klogµ1 + 1
k.(17)
The Maclaurin series for log(1 x) log(1 + x) is
X
k=1 µ11
2+1
3− · ·· +1
2k1x2k
k.
This allows us to rewrite log 2 ·log K0as
log 2 log K0=
X
k=1 µ11
2+1
3− · ·· +1
2k11
k
X
n=2
n2k
=
X
k=1 µ11
2+1
3− · ·· +1
2k11
k(ζ(2k)1).
Appealing to either Maple or Mathematica, we can easily compute this sum. Taking the
ﬁrst 161 terms, we obtain one hundred digits of K0:
K0= 2.68545200106530644530971483548179569382038229399446295
3051152345557218859537152002801141174931847709 . . . .
However, faster convergence is possible, and the constant has now been computed to
more than seven thousand places. Moreover, the harmonic and other averages are sim-
ilarly treated. It appears to satisfy its own predicted behavior (for details, see ,).
Correspondingly, using 108terms one can obtain the approximation K0(π)2.675 . . ..
Note however that K0(e) = = limn→∞ 3n
q(2n)!, since eis a member of the measure
zero set of exceptions not having K0(α) = K0, as a result of the non-Gauss-Kuzmin
distribution of terms in the continued fraction e= [2,1,2,1,1,4,1,1,6, . . .].
We emphasize that while it is known that almost all numbers αhave limits K0(α)
that equal K0, this has not been exhibited for any explicit number α, excluding artiﬁcial
examples constructed using their continued fractions .
6. RAMANUJAN’S AGM CONTINUED FRACTION.
Problem 5. For positive real numbers a, b, and ηdeﬁne Rη(a, b)by
Rη(a, b) = a
η+b2
η+4a2
η+9b2
η+...
.
13
Calculate R1(2,2). Extra credit: Evaluate this constant as a two-term expression involving
a well-known mathematical constant.
History and context. This continued fraction arises in Ramanujan’s Notebooks. He
discovered the beautiful fact that
Rη(a, b) + Rη(b, a)
2=RηÃa+b
2,ab!.
The authors wished to record this in  and to check the identity computationally.
A ﬁrst attempt to ﬁnd R1(1,1) by direct numerical computation failed miserably, and
with some eﬀort only three reliable digits were obtained: 0.693 . . . . With hindsight, it was
realized that the slowest convergence of the fraction occurs in the mathematically simplest
case, namely, when a=b. Indeed, R1(1,1) = log 2 as the ﬁrst primitive numerics had
tantalizingly suggested.
Solution. Attempting a direct computation of R1(2,2) using a depth of twenty thousand
gives only two digits. Thus we must seek more sophisticated methods. From [16, (1.11.70)]
we learn that when 0 < b < a,
R1(a, b) = π
2X
nZ
aK(k)
K2(k) + a2n2π2sech ÃK(k0)
K(k)!,(18)
where k=b/a =θ2
22
3and k0=1k2. Here θ2and θ3are Jacobian theta-functions,
and Kis a complete elliptic integral of the ﬁrst kind.
Writing (18) as a Riemann sum, we ﬁnd that
R(a) = R1(a, a) = Z
0
sech(πx/(2a))
1 + x2dx = 2a
X
k=1
(1)k+1
1 + (2k1)a,(19)
where the ﬁnal equality follows from the Cauchy-Lindel¨of theorem. This sum may also
be written as
R(a) = 2a
1 + aFµ1
2a+1
2,1; 1
2a+3
2;1,
where F(·) denotes the hypergeometric function [1, p. 556]. The latter form is what we
use in Maple or Mathematica to determine
R(2) = 0.97499098879872209671990033452921084400592021999471060574526825
1285877387455708594352325320911129362 . . . .
This constant, as written, is a bit diﬃcult to recognize, but if one ﬁrst divides by
2 and exploits the Inverse Symbolic Calculator, an online tool available at the URL
http://www.cecm.sfu.ca/projects/ISC/ISCmain.html, it becomes apparent that the quo-
tient is π/2log(1 + 2). Thus we conclude, experimentally, that
R(2) = 2[π/2log(1 + 2)].
14
Indeed, it follows (see ) that
R(a)=2Z1
0
t1/a
1 + t2dt.
Note that R(1) = log 2. No nontrivial closed-form expression is known for R(a, b) when
a6=b, although
R1Ã1
4πβµ1
4,1
4,2
8πβµ1
4,1
4!=1
2X
nZ
sech()
1 + n2
is almost closed. It would be pleasant to ﬁnd a direct proof of (19). Further details are
to be found in ,, and .
7. EXPECTED DISTANCE ON A UNIT SQUARE.
Problem 6. Calculate the expected distance E2between two random points on diﬀerent
sides of the unit square:
E2=2
3Z1
0Z1
0qx2+y2dx dy +1
3Z1
0Z1
0q1+(uv)2du dv. (20)
Extra credit: Express this constant as a three-term expression involving algebraic constants
and an evaluation of the natural logarithm with an algebraic argument.
History and context. This evaluation and the next were discovered, in slightly more
complicated form, by James D. Klein [16, p. 66]. He computed the numerical integral
and compared it with a the result of a Monte Carlo simulation. Indeed, a straightfor-
ward approach to a quick numerical value for an arbitrary iterated integral is to use a
Monte-Carlo simulation, which entails approximating the integral by a sum of function
values taken at pseudo-randomly generated points within the region. It is important to
use a good pseudo-random number generator for this purpose. We tried doing a Monte
Carlo evaluation for this problem, using a pseudo-random number generator based on the
recently discovered class of provably normal numbers ,[15, pp. 169–70]. The results we
obtained for the two integrals in question, with 108pseudo-random pairs, are 0.765203 . . .
and 1.076643 . . ., respectively, yielding an expected distance of 0.869017 . . . . Unfortu-
nately, none of these three values immediately suggests a closed form, and they are not
suﬃciently accurate (because of statistical limitations) to be suitable for PSLQ or other
constant recognition tools. More digits are needed.
Solution. It is possible to calculate high-precision numerical values for these two integrals
using a two-dimensional quadrature (numerical integration) program. In our program, we
employed a two-dimensional version of the “tanh-sinh” quadrature algorithm, which we
will discuss in more detail in Problem 8. Two-dimensional quadrature is usually much
more expensive than one-dimensional quadrature, at a given precision level, because many
more function evaluations must be performed. Often a highly parallel computer system
must be used to obtain a high-precision result in reasonable run time . Nonetheless,
15
in this case we were able to evaluate the ﬁrst of the two integrals to 108-digit accuracy
in twenty-one minutes runtime on a 2004-era computer, and the second to 118-digit ac-
curacy in just twenty seconds. The ﬁrst is more diﬃcult due to nondiﬀerentiability of the
integrand at the origin.
Indeed, in this case both Maple and Mathematica are able to evaluate each of these
integrals numerically, as is, to over one hundred decimal digit accuracy in just a few
minutes run time. This is because these software packages are able to integrate the inner
integrals symbolically, leaving only the outer integrals to be evaluated numerically. Maple,
Mathematica, and the two-dimensional quadrature program all agreed on the following
numerical value for the expected distance:
α= 0.86900905527453446388497059434540662485671927963168056
9660350864584179822174693053113213554875435754 . . .
Using PSLQ, with the basis elements α, 2,log(2 + 1), and 1, we obtain
α=1
92 + 5
9log(2 + 1) + 2
9.(21)
An alternate solution is to attempt to evaluate the integrals symbolically! In fact, in
this case Version 5.1 of Mathematica can do both the integrals “out of the box,” whereas
in the ﬁrst case Maple appears to need coaxing, for instance, by converting to polar
coordinates:
2Zπ/4
0Zsec θ
0r2drdθ =2
3Zπ/4
0sec3θdθ =1
321
6log(2) + 1
3log(2 + 2),
since the radius for a given θis 1/cos θ. As for the second integral, Maple and Mathematica
both give
1
321
2log(21) + 1
2log(1 + 2) + 2
3.
To obtain the second integral analytically, write it as 2 R1
0Ru
0q1 + (uv)2dvdu. Now
change variables (set t=uv) to obtain 1/2R1
0nu1 + u2+ arcsinh uodu. Thus, the
expected distance is
1
921
9log(2) + 2
9log(2 + 2) 1
6log(21) + 1
6log(1 + 2) + 2
9,
which can be simpliﬁed to the formula (21) above.
Additional information on the problem is available at
http://mathworld.wolfram.com/SquareLinePicking.html.
16
8. EXPECTED DISTANCE ON A UNIT CUBE.
Problem 7. Calculate the expected distance between two random points on diﬀerent faces
of the unit cube. (Hint: This can be expressed in terms of integrals as
E3: = 4
5Z1
0Z1
0Z1
0Z1
0qx2+y2+ (zw)2dw dx dy dz
+1
5Z1
0Z1
0Z1
0Z1
0q1+(yu)2+ (zw)2du dw dy dz .)
Extra credit: Express this constant as a six-term expression involving algebraic constants
and two evaluations of the natural logarithm with algebraic arguments.
History and context. As we noted earlier, this evaluation was discovered, in essentially
the same form, by Klein [16, p. 66]. As with Problem 6, a Monte Carlo integration
scheme can be used to obtain a quick approximation to the integrals. The values we
obtained were 0.870792 . . . and 1.148859 . . ., respectively, yielding an expected distance
of 0.926406 . . . . Once again, however, these numerical values do not immediately suggest
a closed-form evaluation, yet the accuracy is too low to apply PSLQ or other constant
recognition schemes. What’s more, in this case, unlike Problem 6, neither Maple nor
Mathematica are able to evaluate these four-fold integrals directly—though Mathematica
comes close. As in most cases “help” is needed, in the form of mathematical manipulation
to render these integrals in a form where mathematical computing software can evaluate
them–numerically or symbolically.
Solution. Let 2F1(·) denote the hypergeometric function [1, p. 556]. One may show that
the ﬁrst integral evaluates to
2π
5
X
n=2
2F1(1/2,n+ 2; 3/2; 1/2)
(2 n+ 1) Γ (n+ 2) Γ (5/2n)+4
15 2 + 2
5log ³2 + 1´1
75 π
and the second formally evaluates to
π
10
X
n=0
4F3(1,1/2,1/2n, n1; 2,1/2n, 3/2; 1)
(2 n+ 1) Γ (n+ 2) Γ (3/2n)
2
25 +2
50 +1
10 log ³2+1´.
(Although the second diverges as a Riemann sum, both Maple and Mathematica can
handle it, with some human help, producing numerical values of the corresponding Borel
sum). Both expressions are consequences of the binomial theorem, modulo an initial
integration with respect to zin the ﬁrst case. These expansions allow one to compute the
expectation to high precision numerically and to express both of the individual integrals
in terms of the same set of constants. The numerical value of the desired expectation is
0.926390055174046729218163586547779014444960190107335046732521921271418
504594036683829313473075349968212 . . .
17
An integer relation search in the span of {1, π, 2,3,log(1 + 2),log(2 + 3)}produces
4
75 +17
75 22
25 37
75 π+7
25 log ³1 + 2´+7
25 log ³7+43´.
With substantial eﬀort we were able to nurse the symbolic integral out of Maple. We
started, as in the previous problem, by integrating with respect to wover [0, z], doubling,
and continuing in this fashion until we reduced the problem to showing that
3Z1
0
(x2+ 1) ln ³2 + x21´+ ln ³21´
x2(x2+ 1) dx
Z1
0³2x3+ 6 x2+ 3´ln ³2 + x21´dx =
5
3π+7
62 + 7
2ln ³1 + 2´3
2ln (2) + ln³1 + 3´+37
24 +3
4ln ³1 + 2´π,
which we leave to the reader to establish.
4/5 Integrate[Sqrt[x^2 + y^2 + (z - w)^2], {x, 0, 1}, {y, 0, 1},
{w, 0, 1}, {z, 0, 1}] // Timing
{52.483021*Second, (168*Sqrt - 24*Sqrt - 44*Pi + 72*ArcSinh +
162*ArcSinh[1/Sqrt] + 24*Log - 240*Log[-1 + Sqrt] +
192*Log[1 + Sqrt] + 20*Log[26 + 15*Sqrt] + 3*Log[70226 +
40545*Sqrt])/900}
This form is what the shipping version of Mathematica 5.1 returns on a 3.0 GHz Pentium
4. It evaluates the ﬁrst integral directly, while the second one can be done with a little
help. The combined outcomes can then be simpliﬁed symbolically to the result shown.
There is also an ingenious method due to Michael Trott using a Laplace transform
to reduce the four-dimensional integrals to integrals over one-dimensional integrands. It
proceeds by eliminating the square roots (which cause most of the diﬃculty in symbolic
evaluation of the multiple integrals) at the expense of introducing one additional (but
“easy”) integral. The original problem can then be written in terms of the single integral
Z
0"14
25ez2πerf2(z) + 28e2z2erf(z)
25z+7ez2erf(z)
25z12e3z2
25π+68e2z2
75π+8ez2
75π#dz,
which can be evaluated directly in Mathematica to produce the symbolic expression for
E3.
Nonetheless, we must emphasize that (i) one needs to proceed with conﬁdence, since
such symbolic computations can take several minutes, and (ii) phrases like “Maple can
not” or “Mathematica can” are release-speciﬁc and may also depend on the skill of the
human user to make use of expert knowledge in mathematics, symbolic computation, or
both, in order to produce a form of the problem that is most amenable to computation
in a given software system. This explains our desire to illustrate various solution paths
here and elsewhere.
18
Additional information on this problem is available at
Laplace transform trick applied to the related problem of expected distance in a unit
hypercube, see http://mathworld.wolfram.com/HypercubeLinePicking.html.
9. AN INFINITE COSINE PRODUCT.
Problem 8. Calculate
π2=Z
0cos(2x)
Y
n=1
cos µx
ndx.
History and context. The challenge of showing that π2< π/8 was posed by Bernard
Mares, Jr., along with the problem of demonstrating that
π1=Z
0
Y
n=1
cos µx
ndx < π
4.
This is indeed true, although the error is remarkably small, as we shall see.
Solution. The computation of a high-precision numerical value for this integral is rather
challenging, owing in part to the oscillatory behavior of Qn1cos(x/n) (see Figure 2) but
mostly because of the diﬃculty of computing high-precision evaluations of the integrand.
Note that evaluating thousands of terms of the inﬁnite product would produce only a few
correct digits. Thus it is necessary to rewrite the integrand in a form more suitable for
computation.
Let f(x) signify the integrand. We can express f(x) as
f(x) = cos(2x)"m
Y
1
cos(x/k)#exp(fm(x)),(22)
where we choose mgreater than xand where
fm(x) =
X
k=m+1
log cos µx
k.(23)
The kth summand can be expanded in a Taylor series [1, p. 75], as follows:
log cos µx
k=
X
j=1
(1)j22j1(22j1)B2j
j(2j)! µx
k2j
,
in which B2jare Bernoulli numbers. Observe that since k > m > x in (23), this series
converges. We can then write
fm(x) =
X
k=m+1
X
j=1
(1)j22j1(22j1)B2j
j(2j)! µx
k2j
.(24)
19
n=2
n=5
n=10
C(x)
–0.2
0
0.2
0.4
0.6
0.8
1
12 3 4
x
Figure 2: Approximations to Qn1cos(x/n).
After applying the identity [1, p. 807]
B2j=(1)j+12(2j)!ζ(2j)
(2π)2j
and interchanging the sums, we obtain
fm(x) =
X
j=1
(22j1)ζ(2j)
2j
X
k=m+1
1
k2j
x2j.
Note that the inner sum can also be written in terms of the zeta-function, as follows:
fm(x) =
X
j=1
(22j1)ζ(2j)
2j"ζ(2j)
m
X
k=1
1
k2j#x2j.
This can now be reduced to a compact form for purposes of computation as
fm(x) =
X
j=1
ajbj,mx2j,(25)
where
aj=(22j1)ζ(2j)
2j,(26)
bj,m =ζ(2j)
m
X
k=1
1/k2j.(27)
20
We remark that ζ(2j), aj, and bj,m can all be precomputed, say for jup to some
speciﬁed limit and for a variety of m. In our program, which computes this integral to
120-digit accuracy, we precompute bj,m for m= 1,2,4,8,16, ..., 256 and for jup to 300.
During the quadrature computation, the function evaluation program picks mto be the
ﬁrst power of two greater than the argument x, and then applies formulas (22) and (25).
It is not necessary to compute f(x) for xlarger than 200, since for these large arguments
|f(x)|<10120 and thus may be presumed to be zero.
The computation of values of the Riemann zeta-function can be done using a simple
algorithm due to Peter Borwein  or, since what we really require is the entire set of
values {ζ(2j):1jn}for some n, by a convolution scheme described in . It is
important to note that the computation of both the zeta values and the bj,m must be
done with a much higher working precision (in our program, we use 1600-digit precision)
than the 120-digit precision required for the quadrature results, since the two terms being
subtracted in formula (27) are very nearly equal. These values need to be calculated to a
relative precision of 120 digits.
With this evaluation scheme for f(x) in hand, the integral (22) can be computed
using, for instance, the tanh-sinh quadrature algorithm, which can be implemented fairly
easily on a personal computer or workstation and is also well suited to highly parallel
processing ,,[16, p. 312]. This algorithm approximates an integral f(x) on [1,1]
by transforming it to an integral on (−∞,) via the change of variable x=g(t), where
g(t) = tanh(π/2·sinh t):
Z1
1f(x)dx =Z
−∞ f(g(t))g0(t)dt =h
X
j=−∞
wjf(xj) + E(h).(28)
Here xj=g(hj) and wj=g0(hj) are abscissas and weights for the tanh-sinh quadrature
scheme (which can be precomputed), and E(h) is the error in this approximation.
The function g0(t) = π/2·cosh t·sech2(π/2·sinh t) and its derivatives tend to zero
very rapidly for large |t|. Thus, even if the function f(t) has an inﬁnite derivative,
a blow-up discontinuity, or oscillatory behavior at an endpoint, the product function
f(g(t))g0(t) is in many cases quite well behaved, going rapidly to zero (together with all
of its derivatives) for large |t|. In such cases, the Euler-Maclaurin summation formula
[2, p. 180] can be invoked to conclude that the error E(h) in the approximation (28)
decreases very rapidly—faster than any power of h. In many applications, the tanh-sinh
algorithm achieves quadratic convergence (i.e., reducing the size hof the interval in half
produces twice as many correct digits in the result).
The tanh-sinh quadrature algorithm is designed for a ﬁnite integration interval. In
this problem, where the interval of integration is [0,), it is necessary to convert the
integral to a problem on a ﬁnite interval. This can be done with the simple substitution
s= 1/(x+ 1), which yields an integral from 0 to 1.
In spite of the substantial computation required to construct the zeta- and b-arrays,
as well as the abscissas xjand weights wjneeded for tanh-sinh quadrature, the entire
calculation requires only about one minute on a 2004-era computer, using the ARPREC
21
arbitrary precision software package available at http://crd.lbl.gov/~dhbailey/mpdist.
The ﬁrst hundred digits of the result are the following:
0.392699081698724154807830422909937860524645434187231595926812285162
093247139938546179016512747455366777....
AMathematica program capable of producing 100 digits of this constant is available on
Michael Trott’s website:
Using the Inverse Symbolic Calculator, for instance, one ﬁnds that this constant is
likely to be π/8. But a careful comparison with a high-precision value of π/8, namely,
0.392699081698724154807830422909937860524646174921888227621868074038
477050785776124828504353167764633497...,
reveals that they are not equal—the two values diﬀer by approximately 7.407 ×1043.
Indeed, these two values are provably distinct. This follows from the fact that
55
X
n=1
1/(2n+ 1) >2>
54
X
n=1
1/(2n+ 1).
See [16, chap. 2] for additional details. We do not know a concise closed-form expression
for this constant.
Further history and context. Recall the sinc function
sinc x=sin x
x,
and consider, the seven highly oscillatory integrals:
I1=Z
0sinc x dx =π
2,
I2=Z
0sinc xsinc µx
3dx =π
2,
I3=Z
0sinc xsinc µx
3sinc µx
5dx =π
2,
...
I6=Z
0sinc xsinc µx
3···sinc µx
11dx =π
2,
I7=Z
0sinc xsinc µx
3···sinc µx
13dx =π
2.
It comes as something of a surprise, therefore, that
I8=Z
0sinc xsinc µx
3···sinc µx
15dx
=467807924713440738696537864469
935615849440640907310521750000π0.499999999992646π.
22
When this was ﬁrst discovered by a researcher, using a well-known computer algebra
package, both he and the software vendor concluded there was a “bug” in the software.
Not so! It is fairly easy to see that the limit of the sequence of such integrals is 2π1. Our
analysis, via Parseval’s theorem, links the integral
IN=Z
0sinc(a1x) sinc (a2x)··· sinc (aNx)dx
with the volume of the polyhedron PNdescribed by
PN={x:|
N
X
k=2
akxk| ≤ a1,|xk| ≤ 1,2kN},
for x= (x2, x3,··· , xN). If we let
CN={(x2, x3,···, xN) : 1xk1,2kN},
then
IN=π
2a1
Vol(PN)
Vol(CN).
Thus, the value drops precisely when the constraint PN
k=2 akxka1becomes active
and bites the hypercube CN. That occurs when PN
k=2 ak> a1. In the foregoing,
1
3+1
5+· ·· +1
13 <1,
but on addition of the term 1/15, the sum exceeds 1, the volume drops, and IN=π/2
no longer holds. A similar analysis applies to π2. Moreover, it is fortunate that we began
with π1or the falsehood of π2= 1/8 would have been much harder to see.
Additional information on this problem is available at
http://mathworld.wolfram.com/InfiniteCosineProductIntegral.html and
http://mathworld.wolfram.com/BorweinIntegrals.html.
10. A MULTIVARIATE ZETA-FUNCTION.
Problem 9. Calculate
X
i>j>k>l>0
1
i3jk3l.
Extra credit: Express this constant as a single-term expression involving a well-known
mathematical constant.
History and context. We resume the discussion from Problem 3. In the notation
introduced there, we ask for the value of ζ(3,1,3,1). The study of such sums in two
variables, as we noted, originates with Euler. These investigations were apparently due
to a serendipitous mistake. Euler wrote to Goldbach [15, pp. 99–100]:
23
When I recently considered further the indicated sums of the last two series
in my previous letter, I realized immediately that the same series arose due to
a mere writing error, from which indeed the saying goes, “Had one not erred,
one would have achieved less. [Si non errasset, fecerat ille minus].”
Euler’s reduction formula is
ζ(s, 1) = s
2ζ(s+ 1) 1
2
s2
X
k=1
ζ(k+ 1)ζ(s+ 1 k),
which reduces the given double Euler sums to a sum of products of classical ζ-values.
Euler also noted the ﬁrst reﬂection formulas
ζ(a, b) + ζ(b, a) = ζ(a)ζ(b)ζ(a+b),
certainly valid when a > 1 and b > 1. This is an easy algebraic consequence of adding
the double sums. Another marvelous fact is the sum formula
X
Σai=n,ai0
ζ(a1+ 2, a2+ 1,··· , ar+ 1) = ζ(n+r+ 1) (29)
for nonnegative integers nand r. This, as David Bradley observes, is equivalent to the
generating function identity
X
n>0
1
nr(nx)=X
k1>k2>···kr>0
r
Y
j=1
1
kjx.
The ﬁrst three nontrivial cases of (29) are ζ(3) = ζ(2,1), ζ(4) = ζ(3,1) + ζ(2,2), and
ζ(2,1,1) = ζ(4).
Solution. We notice that such a function is a generalization of the zeta-function. Similar
to the deﬁnition in section 4, we deﬁne
ζ(s1, s2,··· , sk;x) = X
n1>n2>···>nk>0
xn
1
ns1
1ns2
2···nsr
r
,(30)
for s1, s2, . . . , sknonnegative integers. We see that we are asked to compute ζ(3,1,3,1; 1).
Such a sum can be evaluated directly using the EZFace+ interface at
http://www.cecm.sfu.ca/projects/ezface+, which employs the H¨older convolution, giv-
ing us the numerical value
0.005229569563530960100930652283899231589890420784634635522547448
97214886954466015007497545432485610401627 . . . . (31)
Alternatively, we may proceed using diﬀerential equations. It is fairly easy to see [16, sec.
3.7] that
d
dxζ(n1, n2,···, nr;x) = 1
xζ(n11, n2,···, nr;x),(n1>1),(32)
d
dxζ(n1, n2,···, nr;x) = 1
1xζ(n2,···, nr;x),(n1= 1),(33)
24
with initial conditions ζ(n1; 0) = ζ(n1, n2; 0) = ···=ζ(n1,··· , nr; 0) = 0,and ζ(·;x)1.
Solving
> dsys1 =
> diff(y3131(x),x) = y2131(x)/x,
> diff(y2131(x),x) = y1131(x)/x,
> diff(y1131(x),x) = 1/(1-x)*y131(x),
> diff(y131(x),x) = 1/(1-x)*y31(x),
> diff(y31(x),x) = y21(x)/x,
> diff(y21(x),x) = y11(x)/x,
> diff(y11(x),x) = y1(x)/(1-x),
> diff(y1(x),x) = 1/(1-x);
> init1 = y3131(0) = 0,y2131(0) = 0, y1131(0) = 0,
> y131(0)=0,y31(0)=0,y21(0)=0,y11(0)=0,y1(0)=0;
in Maple, we obtain 0.005229569563518039612830536519667669502942 (this is valid to
thirteen decimal places). Maple’s identify command is unable to identify portions of
this number, and the inverse symbolic calculator does not return a result. It should be
mentioned that both Maple and the ISC identiﬁed the constant ζ(3,1) (see the remark
under the “history and context” heading). From the hint for this question, we know this
is a single-term expression. Suspecting a form similar to ζ(3,1),we search for a constants
cand dsuch that ζ(3,1,3,1) = d. This leads to c= 1/81440 = 2/10! and d= 8.
Further history and context. We start with the simpler value, ζ(3,1). Notice that
log(1 x) = x+1
2x2+1
3x3+···,
so
f(x) = log(1 x)/(1 x) = x+ (1 + 1
2)x2+ (1 + 1
2+1
3)x3+···
=X
nm>0
xn
m.
As noted in the section on double Euler sums,
(1)m+1
Γ(m)Z1
0xnlogm1x dx =1
(n+ 1)m,
so integrating fusing this transform for m= 3, we obtain
ζ(3,1) = (1)
2Z1
0f(x) log2x dx
= 0.270580808427784547879000924 . . . .
The corresponding generating function is
X
n0
ζ({3,1}n)) x4n=cosh(πx)cos(πx)
π2x2,
25
equivalent to Zagier’s conjectured identity
ζ({3,1}n) = 2π4n
(4n+ 2).
Here {3,1}ndenotes n-fold concatenation of {3,1}.
The proof of this identity (see [16, p. 160]) derives from a remarkable factorization of
the generating function in terms of hypergeometric functions:
X
n0
ζ({3,1}n)x4n=2F1Ãx(1 + i)
2,x(1 + i)
2; 1; 1!2F1Ãx(1 i)
2,x(1 i)
2; 1; 1!.
Finally, it can be shown in various ways that
ζ({3}n) = ζ({2,1}n)
for all n, while a proof of the numerically-conﬁrmed conjecture
ζ({2,1}n)?
= 23nζ({−2,1}n) (34)
remains elusive. Only the ﬁrst case of (34), namely,
X
n=1
1
n2
n1
X
m=1
1
m= 8
X
n=1
(1)n
n2
n1
X
m=1
1
m(= ζ(3))
has a self-contained proof . Indeed, the only other established case is
X
n=1
1
n2
n1
X
m=1
1
m
m1
X
p=1
1
p2
p1
X
q=1
1
q= 64
X
n=1
(1)n
n2
n1
X
m=1
1
m
m1
X
p=1
(1)p
p2
p1
X
q=1
1
q(= ζ(3,3)).
This is an outcome of a complete set of equations for multivariate zeta functions of depth
four.
There has been abundant evidence amassed to support identity (34) since it was found
in 1996. For example, very recently Petr Lisonek checked the ﬁrst eighty-ﬁve cases to one
thousand places in about forty-one hours with only the expected roundoﬀ error. And he
checked n= 163 in ten hours. This is the only identiﬁcation of its type of an Euler sum
with a distinct multivariate zeta-function.
11. A WATSON INTEGRAL.
Problem 10. Evaluate
W=1
π3Zπ
0Zπ
0Zπ
0
1
3cos xcos ycos zdx dy dz. (35)
26
History and context. The integral arises in Gaussian and spherical models of ferro-
magnetism and in the theory of random walks. It leads to one of the most impressive
closed-form evaluations of an equivalent multiple integral due to G. N. Watson:
c
W=Zπ
πZπ
πZπ
π
1
3cos xcos ycos zdx dy dz
=1
96 (31) Γ2µ1
24Γ2µ11
24(36)
= 4 π³18 + 12 210 376´K2(k6),
where k6=³23´³32´is the sixth singular value. The most self-contained
derivation of this very subtle result is due to Joyce and Zucker in  and , where
more background can also be found.
Solution. In , it is shown that a simpliﬁcation can be obtained by applying the
formula
1
λ=Z
0eλt dt (Reλ > 0) (37)
to W3. The three-dimension integral is then reducible to a single integral by using the
identity
1
πZ
0exp(tcos θ)=I0(t),(38)
in which I0(t) is the modiﬁed Bessel function of the ﬁrst kind. It follows from this that
W=R
0exp(3t)I3
0(t)dt. This integral can be evaluated to one hundred digits in Maple,
giving
W3= 0.50546201971732600605200405322714025998512901481742089
21889934878860287734511738168005372470698960380 . . . . (39)
Finally, an integer relation hunt to express log Win terms of log π, log 2,log Γ(k/24), and
log(31) will produce (36).
We may also write W3as a product solely of values of the gamma function. This is
what our Mathematician’s ToolKit returned:
0= -1.* log[w3] + -1.* log[gamma[1/24]] + 4.*log[gamma[3/24]] +
-8.*log[gamma[5/24]] + 1.* log[gamma[7/24]] + 14.*log[gamma[9/24]] +
-6.*log[gamma[11/24]] + -9.*log[gamma[13/24]] + 18.*log[gamma[15/24]] +
-2.*log[gamma[17/24]] + -7.*log[gamma[19/24]]
Proving this is achieved by comparing the result with (36) and establishing the implicit
gamma representation of (31)2/96.
27
Similar searches suggest there is no similar four-dimensional closed form—the relevant
Bessel integral is W4=R
0exp(4t)I4
0(t)dt. (N.B. R
0exp(2t)I2
0(t)dt =.) In this
case it is necessary to compute exp(t)I0(t) carefully, using a combination of the formula
exp(t)I0(t) = exp(t)
X
n=0
t2n
22n(n!)2
for tup to roughly 1.2·d, where dis the number of signiﬁcant digits desired for the result,
and
exp(t)I0(t)1
2πt
N
X
n=0 Qn
k=1(2k1)2
(8t)nn!
for large t, where the upper limit Nof the summation is chosen to be the ﬁrst index nsuch
that the summand is less than 10d(since this is an asymptotic expansion, taking more
terms than Nmay increase, not decrease the error). We have implemented this as ‘bessel-
exp’ in our Mathematician’s ToolKit, available at http://crd.lbl.gov/~dhbailey/mpdist.
Using this software, which includes a PSLQ facility, we found that W4is not expressible as a
product of powers of Γ(k/120) (0 < k < 120) with coeﬃcients having fewer than 80 digits. This
result does not, of course, rule out the possibility of a larger relation, but it does cast some
doubt, in an experimental sense, that such a relation exists. Enough to stop looking.
Additional information on this problem is available at
http://mathworld.wolfram.com/WatsonsTripleIntegrals.html.
12. CONCLUSION. While all the problems described herein were studied with a great deal
of experimental computation, clean proofs are known for the ﬁnal results given (except for
Problem 7), and in most cases a lot more has by now been proved. Nonetheless, in each case
the underlying object suggests plausible generalizations that are still open.
The “hybrid computations” involved in these solutions are quite typical of modern experi-
mental mathematics. Numerical computations by themselves produce no insight, and symbolic
computations frequently fail to produce full-ﬂedged, closed-form solutions. But when used to-
gether, with signiﬁcant human interaction, they are often successful in discovering new facts of
mathematics and in suggesting routes to formal proof.
ACKNOWLEDGMENTS. Bailey’s work was supported by the Director, Oﬃce of Compu-
tational and Technology Research, Division of Mathematical, Information, and Computational
Sciences of the U.S. Department of Energy, under contract number DE-AC02-05CH11231; also
by the NSF, under Grant DMS-0342255. Borwein’s work was supported in part by NSERC
and the Canada Research Chair Programme. Kapoor’s work was performed as part of an inde-
pendent study project at the University of British Columbia. Weisstein’s work on MathWorld
is supported by Wolfram Research as a free service to the world mathematics and Internet
communities.
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30
DAVID H. BAILEY received his B.S. at Brigham Young University and received his Ph.D.
(1976) from Stanford University. He worked for the Department of Defense and for SRI In-
ternational, before spending fourteen years at NASA’s Ames Research Center in California.
Since 1997 he has been the chief technologist of the Computational Research Department at
the Lawrence Berkeley National Laboratory. In 1993, he was a corecipient of the Chauvenet
Prize from the MAA, and in that year was also awarded the Sidney Fernbach Award from the
IEEE Computer Society. His research spans computational mathematics and high-performance
computing. He is the author (with Jonathan Borwein and, for volume two, Roland Girgensohn)
of two recent books on experimental mathematics.
Lawrence Berkeley National Laboratory, Berkeley, CA 94720
dhbailey@lbl.gov
JONATHAN M. BORWEIN received his DPhil from Oxford (1974) as a Rhodes Scholar.
He taught at Dalhousie, Carnegie-Mellon, and Waterloo, before becoming the Shrum Professor
of Science and a Canada Research Chair in Information Technology at Simon Fraser University.
At SFU he was founding director of the Centre for Experimental and Constructive Mathematics.
In 2004, he rejoined Dalhousie University in the Faculty of Computer Science. Jonathan has
received several awards, including the 1993 Chauvenet Prize of the MAA, Fellowship in the
Royal Society of Canada, and Fellowship in the American Association for the Advancement of
Science. His research spans computational number theory and optimization theory, as well as
numerous topics in computer science. He is the author of ten books, and is cofounder of Math
Resources, Inc., an educational software ﬁrm.
Faculty of Computer Science, Dalhousie University, Halifax, NS, B3H 2W5
jmborwein@cs.dal.ca
VISHAAL KAPOOR graduated from Simon Fraser University with a B.S. degree in math-
ematics and is currently completing his M.S. at the University of British Columbia under the
supervision of Greg Martin.
Department of Mathematics, University of BC, Vancouver, BC, V6T 1Z2
vkapoor@math.ubc.ca
ERIC W. WEISSTEIN graduated from Cornell University, with a B.A. degree in physics, and
from the California Institute of Technology (M.S., 1993; Ph.D., 1996) with degrees in planetary
astronomy. Upon completion of his doctoral thesis, Weisstein became a research scientist in
the Department of Astronomy at the University of Virginia in Charlottesville. Since 1999 he
has been a member of the Scientiﬁc Information group at Wolfram Research, where he holds
the oﬃcial title “Encyclopedist.” Eric is best known as the author of the MathWorld website
(http://mathword.wolfram.com), an online compendium of mathematical knowledge, as well as
the author of the CRC Concise Encyclopedia of Mathematics.
Wolfram Research Inc., Champaign, IL 61820
eww@wolfram.com
31