Download Reciprocal PSLQ and the tiny Nome of Bologna David Broadhurst

Reciprocal PSLQ
and
the tiny Nome of Bologna
David Broadhurst
The Open University, UK
1
Introduction
I begin by showing how to evaluate the vacuum-diagram integrals
V (n, a, b) =
Z ∞
0
b
x2n+1 [K0 (x)]a [xK00 (x)] dx
(1)
to very high precision, using the differential equation
x2 K000 (x) + xK00 (x) = x2 K0 (x)
(2)
for the modified Bessel function
K0 (x) =
Z ∞
0
cos(xt)
√
dt
1 + t2
(3)
which is the Fourier transform of the free propagator 1/(p2 + 1) of a scalar
field with unit mass, in two Euclidean dimensions.
The key is the integration-by-parts identity
¶
d µ 2n+2
b
x
[K0 (x)]a [xK00 (x)] dx = 0
0 dx
which is valid for integers (n, a, b) with n ≥ 0 and N = a + b > 0.
Z ∞
(4)
Using the differential equation, we obtain the recurrence relation
2(n + 1)V (n, a, b) + aV (n, a − 1, b + 1) + bV (n + 1, a + 1, b − 1) = 0 (5)
which preserves the value of N = a + b.
Moreover, the small x behaviour K0 (x) = − log(x) + O(1) gives
¶
d µ
N
0
[xK0 (x)] dx = (−1)N −1
0 dx
and hence provides the rational datum
Z ∞
(−1)N −1
.
V (0, 1, N − 1) =
N
2
(6)
(7)
1
Integers at N = 1
For N = 1, we have the integer-valued solution
V (n, 1, 0) = 4n (n!)2 ,
V (n, 0, 1) = −2(n + 1)V (n, 1, 0).
3
(8)
(9)
2
Rationals at N = 2
For N = 2, we have the rational solution
22n−1 (n!)4
,
V (n, 2, 0) =
(2n + 1)!
V (n, 1, 1) = −(n + 1)V (n, 2, 0),
n+2
V (n + 1, 2, 0).
V (n, 0, 2) =
n+1
4
(10)
(11)
(12)
3
Polylogarithm at N = 3
For N = 3, the datum V (0, 1, 2) = 1/3 is insufficient; another is needed.
Recall that V (0, 3, 0) is the two-loop vacuum diagram with 3 massive propagators connecting a pair of vertices. In momentum space, we may evaluate a one-loop subdiagram from its discontinuity across the cut in the
time-like region, obtaining
1 Z ∞ dz
V (0, 3, 0) =
Π(z)
2 0 z+1
Z ∞ ds
1
q
Π(z) =
4 s + z s(s − 4)
(13)
(14)
where the one-loop propagator is given by its dispersion relation.
The transformation of variables
√
z = 2 sinh(u/2)
(15)
gives
u
2 sinh(u)
for the one-loop propagator and hence
Π(z) =
1Z∞
udu
2 0 (2 sinh(u/2))2 + 1


∞
3 X
1
1


=
−
2
2
4 n=0 (3n + 1)
(3n + 2)
√
3
1
=
Cl2 (2π/3) = √ Cl2 (π/3)
2
3
V (0, 3, 0) =
where Cln (θ) =
P
n
k>0 sin(kθ)/k
is Clausen’s polylogarithm.
5
(16)
(17)
(18)
(19)
3.1
The sixth Root of Unity
We thus arrive at a much studied dilogarithm of the sixth root of unity,
with an amazingly fast evaluation
Ã
∞
1 X
−1
V (0, 3, 0) =
9 n=0 27
!n 5
X
Vk
2
k=1 (6n + k)
(20)
where the vector V = [9, −9, −12, −3, 1] was discovered (and proven) in
the course of my investigation of 3-loop vacuum diagrams in 4 dimensions.
The discovery was inscribed by Helaman Ferguson on the bronze statue
awarded by the Clay Mathematical Institute to Andrew Wiles, to honour
the proof of Fermat’s last theorem, as was explained in the book Mathematics by Experiment: Plausible Reasoning in the 21st Century, by Jonathan
M. Borwein and David H. Bailey.
Perhaps Grigori Yakovlevich Perelman may eventually agree to accept a
similarly inscribed statue, for proving the Poincaré conjecture.
6
3.2
High Precision from Recurrence
Imagine, however, that we are encountering the polylogarithm V (0, 3, 0)
for the first time. How might we evaluate it, to high precision, in the
context of the current investigation?
The answer is very simple: use the recurrence relation!
It gives V (n, 3, 0) at large n as a linear combination of V (0, 3, 0) and 1,
with rational coefficients. We then equate this result for V (n, 3, 0) to the
asymptotic estimate, obtained from
s



m−1
X
Ak
π −x 
Am 

e 1 +
+
O
K0 (x) =
k
2x
(−x)m 
k=1 (−x)
((2k)!)2
Ak = 5k
2 (k!)3
(21)
(22)
which gives
Ã
Nπ
V (n, N, 0) =
2
!N/2

Ã
Γ(2n + 2 − N/2) 
N2
1
1−
+O 2

2n+2
N
16n
n
!


(23)
for n → ∞, with fixed N .
We can achieve a relative accuracy for V (n, N, 0) of order N 2n /(N + 2)2n ,
by taking m = b4n/(N + 2)c terms in the asymptotic expansion for K0 (x),
albeit at the expense of taking the N th power of a series with m terms.
However, I shall show that this latter refinement is not necessary.
7
3.3
Arithmetic with 48-digit Integers
To obtain 30 good digits of V (0, 3, 0), it suffices to consider
V (30, 3, 0) = (2103 × 510 × 76 × 112 × 132 )
D30 V (0, 3, 0) − N30
330
(24)
where
N30 = 561481513123014082497399430199388175954267069801 (25)
D30 = 958197498356553368770942858760632413639590720000 (26)
are 48-digit integers.
Then the first two terms in the asymptotic estimate for V (30, 3, 0) give
Ã
!
3
N30
+ 2.5922 × 10−29 1 −
V (0, 3, 0) ≈
D30
160
= 0.585976809672364722650390572218 . . .
(27)
(28)
with 30 good decimal digits, as claimed.
In general, for N > 2, we find that V (0, N, 0) may be computed to an
accuracy better than (N −2)2n /N 2n , from the leading term in the expansion
for V (n, N, 0), and to an accuracy of order (N − 2)2n /(N + 2)2n , by taking
m = b4n/(N + 2)c terms.
The former strategy is much more efficient, since it involves far fewer
operations than taking the N th power of m terms.
8
4
Euler Sum at N = 4
The situation for N = 4 is very similar.
The single (provenly) irrational datum
1Z∞
[Π(z)]2 dz
2 0
1 Z ∞ u2 du
=
4 0 sinh(u)
∞
X
1
=
3
n=0 (2n + 1)
7
= ζ(3)
8
V (0, 4, 0) =
(29)
(30)
(31)
(32)
determines all the other integrals with N = 4.
Conversely, we may evaluate ζ(3), to an accuracy better than 1/4n , by
developing the recurrence relation as far as V (n, 4, 0) and taking just the
leading term in its asymptotic expansion.
9
5
New Numbers at N = 5
In general, the integrals V (n, a, b) with N = a + b > 0 are determined by
the recurrence relation in terms of b(N + 1)/2c initial data, of which one
(and very probably only one) is rational.
Hence at N = 5 we need two new numbers, which we take to be
A = V (0, 5, 0) =
B = V (1, 5, 0) =
Z ∞
Z0∞
0
xK05 (x)dx
(33)
x3 K05 (x)dx
(34)
noting that A comes from the 4-loop vacuum diagram with 5 massive
propagators connecting two vertices in two dimensions.
Fortunately, we may evaluate both A and B, to very high precision, by
the methods developed and tested for N < 5.
I have computed V (104 , 5, 0) and V (104 + 1, 5, 0), as exact linear combinations of (A, B, 1), and thus obtained 4375 good decimal digits of A and
B, available in the file V5AB.txt via the URL
paftp.open.ac.uk/pub/staff ftp/dbroadhu/newconst/
The challenge is to determine the number-theory behind these constants.
It is a hard problem, but a clue is at hand.
10
5.1
Hope at Sunrise
To get a handle on the type of numbers that occur, we may usefully consider
the generalized sunrise integrals
S(n, a, b) =
Z ∞
0
b
x2n+1 I0 (x) [K0 (x)]a−1 [xK00 (x)] dx
(35)
with n ≥ 0 and N = a + b > 2 lines, one of which is an external line, on
the mass shell at p2 = −1. This produces the modified Bessel function
I0 (x) =
2 Z 1 cosh(xt)
√
dt
π 0
1 − t2
(36)
with the Wronskian
K0 (x)I00 (x) − K00 (x)I0 (x) =
1
x
(37)
giving the asymptotic expansion

Ã
m−1
X Bk
1 
Bm
1
+
I0 (x)K0 (x) =
+
O
2k
2x 
x2m
k=1 x
((2k)!)3
Bk = 6k
2 (k!)4
!


(38)
(39)
and the recurrence relation
2(n + 1)S(n, a, b) + aS(n, a − 1, b + 1) + bS(n + 1, a + 1, b − 1) =
−V (n, a − 2, b)
(40)
with an inhomogeneous term whose value of N has decreased by 2 and
whose homogeneous recurrence relation we already know.
11
5.2
Known Sunrises
The polylogarithms
π
√
3 3
π2
S(0, 4, 0) =
16
S(0, 3, 0) =
(41)
(42)
exemplify the relationship that we expect between the number-theory content of S(0, N, 0) and V (0, N, 0): the former involves the same type of
numbers as the latter, at a weight that is smaller by one unit.
For example, we see that S(0, 4, 0) = 83 ζ(2) is an Euler sum of weight 2,
while V (0, 4, 0) = 78 ζ(3) has weight 3.
Incidentally, the number theory with N = 4 massive lines is that of alternating Euler sums: polylogarithms of the second root of unity, as may
be seen from the 3-loop result for the anomalous magnetic moment of the
P
electron, which contains the irreducible Euler sum m>n>0 (−1)m+n /(m3 n).
So what about N = 5 lines? Perhaps we may get a handle on the weight-4
numbers in V (0, 5, 0) by studying the weight-3 numbers in S(0, 5, 0).
12
5.3
The tiny Nome of Bologna
The capital discovery of Stefano Laporta was that the finite part of the
3-loop sunrise diagram in 4 spacetime dimensions involves products of complete elliptic integrals of the first and second kind,
K(m) =
E(m) =
Z π/2 ³
0
Z π/2 ³
0
1 − m sin2 θ
1 − m sin2 θ
´−1/2
´1/2
dθ
dθ
(43)
(44)
evaluated at the 4 values of m that are roots of the quartic polynomial
P (x) = 65536x4 − 131072x3 + 77568x2 − 12032x + 1.
(45)
This wonderful polynomial is all that is needed to show that the nome of
Bologna is the tiny number
q = e−π
√
15
= 0.000005197483312 . . .
13
(46)
5.4
A quick Guide to Nome Spotting
The four Jacobian elliptic functions
θ1 (z, q) = 2
θ2 (z, q) = 2
∞
X
n=0
∞
X
2
(−1)n q (n+1/2) sin((2n + 1)z)
2
q (n+1/2) cos((2n + 1)z)
n=0
θ3 (z, q) = 1 + 2
θ4 (z, q) = 1 + 2
∞
X
n=1
∞
X
2
q n cos(2nz)
2
(−1)n q n cos(2nz)
(47)
(48)
(49)
(50)
n=1
have remarkable expansions in the nome, q. Note that the powers of the
nome increase quadratically. At z = 0, these elliptic functions have the
equally remarkable relations
θ10 (0, q) = θ2 (0, q)θ3 (0, q)θ4 (0, q)
θ34 (0, q) = θ24 (0, q) + θ44 (0, q)
(51)
(52)
with fourth powers appearing in the latter.
Consider the smallest root
√
√
16 − 7 3 − 15
m=
= 0.000083156275326 . . .
32
(53)
of the quartic P (x). The associated nome is


√
K(1 − m) 
−π 15

=e
q = exp −π
K(m)
(54)
and the nomes associated with the other roots are q 1/3 , q 1/5 and q 1/15 .
14
5.5
One new Number at N = 5
We have an algebraic result for
Ã
m
θ2 (0, q)
=
θ4 (0, q)
1−m
!1/4
√
√
8− 6
√ .
=√
5+ 3
(55)
So we obtain only one new (presumably transcendental) number for this
nome, which we may take as
θ3 (0, q) = 1 + 2
∞
X
µ
2
√
¶
exp −n π 15
(56)
n=1
= 1.00001039496662475855 . . .
(57)
This is our candidate for helping PSLQ to determine the value of
S(0, 5, 0) =
Z ∞
0
xI0 (x)K04 (x)dx
for the finite 3-loop on-shell sunrise diagram in two dimensions.
15
(58)
5.6
Conventional PSLQ
If S(0, 5, 0) were a polylogarithm, we would expect it to have weight 3.
I guessed that the best way to incorporate the tiny nome was
√ to√try√to fit
3 4
S(0, 5, 0)/(π θ3 (0, q)) as a rational linear combination of (1, 3, 5, 15).
PSLQ confirmed this conjecture, giving
1 Z∞
C = 2
xI0 (x)K04 (x)dx
0
π

Ã
!
µ
¶ 4
∞
√
X
π
1 
=
1− √
1+2
exp −n2 π 15  .
16
5
n=1
16
(59)
(60)
5.7
Reciprocal PSLQ in two Dimensions
For reasons that will be indicated later, I further conjectured that the other
datum, required to solve the recurrence relation for sunrise diagrams with
N = 5, is not a new number.
Rather, it is the reciprocal of C.
PSLQ confirmed this with the fits
Ã
! Ã
!
1 Z∞ 3
2 2
1
4
x
I
(x)K
(x)dx
=
13C
−
0
0
π2 0
15
10C
!3 Ã
!
Ã
Z
1 ∞ 5
19
4
4
43C −
x I0 (x)K0 (x)dx =
π2 0
15
40C
Ã
!5 Ã
!
Z
1 ∞ 7
9 4
1907
4
x I0 (x)K0 (x)dx =
4079C −
π2 0
4 15
40C
(61)
(62)
(63)
and thereafter I used the recurrence relation to obtain
Ã
1 Z∞ 9
4
4
x
I
(x)K
(x)dx
=
9
0
0
π2 0
15
and so on.
17
!7 Ã
959531
509483C −
160C
!
(64)
5.8
Arithmetic with 75-digit Integers
Suppose that we did not know that
√
S(0, 5, 0)
8π(1 − 1/ 5)
=C= h
i2
√ √
√
√
π2
AGM(8 2, 15 + 3 5 + 3 − 1)
(65)
is given by an arithmetic-geometric mean. How many good digits could
we obtain by pushing the recurrence relation for S(n, 5, 0) up to n = 30?
After factoring out even powers of primes less than 30 in
à !
Ã
!
16
4 30
S(30, 5, 0)
4
2
2
2 8
=
(7
×
11
×
13
×
29)
π2
5
15 
Ã
!
 4 × 17 × 19 × 23 2
G30 
× 
F30 C −
7 × 11 × 13 × 29
160C 
(66)
there remain two 75-digit integers with the non-trivial factorizations
F30 =
×
×
×
G30 =
×
×
1103
84979600677041625660283
1256342659620456814562987
1537332739493789767737751
59
61581504848316213196732415971
98426265105029157229151379953384355952518629
(67)
(68)
and then the asymptotic expansions of K0 (x) and I0 (x)K0 (x) give
Ã
!
Ã
G30 1/2
3
7 × 11 × 13 × 29
+ 8.6058 × 10−30 1 −
C ≈
4 × 17 × 19 × 23 160F30
160
= 0.1085438698336849710403527567592 . . .
!
(69)
(70)
with 31 good digits obtained from two terms in the asymptotic series.
18
5.9
Reciprocal PSLQ in four Dimensions
In D = 4 − 2ε spacetime dimensions, the Laurent expansion of the on-shell
3-loop sunrise diagram, with N = 5 lines, has the form
2
22
577
+ S1 + O(ε)
+
+
ε3 3ε2 36ε
(71)
if we use the measure dD k/(π D/2 Γ(1 + ε)) for a loop momentum k.
I conjectured that Stefano Laporta’s finite part S1 involves only π and C.
Reciprocal PSLQ confirmed this conjecture, giving
Ã
6191 π 2
7
S1 =
−
4C +
216
3
40C
!
(72)
which agrees with the expectation that no new number appears when we
move from two dimensions to four.
If we say that
Ã
π
1
C=
1− √
16
5
!
1
+2
∞
X
µ
exp −n2 π
√

¶ 4
15 
(73)
n=1
gives an elliptic twist at weight 1, then sunrise diagrams at N = 5 involve
weights 3 and 1.
Weight 2 is conspicuously absent!
19
5.10
Reciprocation demystified
My motivation for taking the reciprocal of C was Legendre’s relation
E(m)K(1 − m) + E(1 − m)K(m) = K(m)K(1 − m) +
π
2
(74)
which is valid for arbitrary m.
√
In present case, we have K(1 − m) = 15K(m) and hence find that a
linear combination of E(m) and E(1 − m), with algebraic coefficients, is
given by a linear of combination K(m) and π/K(m), also with algebraic
coefficients.
Moreover, algebraic reductions are obtained for all 4 of the elliptic integrals
of the second kind, with the pleasing pattern
√
√
√
K(m)
π
15E(m) = ( 15 + 5 + 1)
+
(75)
2
4K(m)
√
√
√
K(m)
π
5E(m3 ) = ( 15 + 5 − 1)
+
φ
(76)
2φ
4K(m)
√
√
√
K(m)
π
3E(m5 ) = ( 15 − 5 + 1)
+
φ
(77)
2φ
4K(m)
√
√
π
K(m)
+
(78)
E(m15 ) = ( 15 − 5 − 1)
2
4K(m)
√
where φ = ( 5 + 1)/2 and (m3 , m5 , m15 ) are the other three roots of P (x),
listed in increasing size and corresponding to the nomes (q 1/3 , q 1/5 , q 1/15 ).
If we take any product of a pair of elliptic integrals, of either kind, and
divide by π, the result must be a linear combination of (C, 1, 1/C) with
algebraic coefficients.
That is how I was led to the idea of reciprocal PSLQ.
20
5.11
Taylor expansions of Jacobian elliptic functions
The remarkable algebraic reductions for complete elliptic integrals of the
second kind suggested that there should be an algebraic reduction of the
second derivative of θ3 (z, q) at z = 0.
In fact, I found that
Ã
√
!
√
θ300 (0, q)
5
π 15
= 3 5+
C −1
θ3 (0, q)
2
(79)
and this is just the tip of another iceberg.
If we take any of the 4 ratios πθ1000 /θ10 or πθk00 /θk for k = 2, 3, 4, at z = 0,
and evaluate it for any of the 4 nomes (q, q 1/3 , q 1/5 , q 1/15 ), then there is a
reduction to a linear combination of C and 1, with algebraic coefficients.
For example
³
´
Ã
!
√
π θ1000 0, q 1/15
25
³
´ = 7 5+
√
C −1
2
3 15 θ10 0, q 1/15
(80)
with 14 other reductions of this kind.
It follows that the first 2 non-zero terms in the Taylor expansion of all 4
Jacobian elliptic functions, for all 4 nomes, are algebraically determined
by just two numbers: C and π.
Taking π for granted, we get 32 elliptic numbers for the price of one Feynman diagram: the simplest sunrise diagram with N = 5.
Such are the wonders of quantum field theory, revealed by:
Reciprocal PSLQ and the tiny Nome of Bologna.
21
6
Summary
Raffiniert ist die Frau Göttin, aber boshaft ist Sie nicht!
1. She was subtle, when moving up from polylogarithms, with N < 5
massive lines, to elliptic functions, with N = 5.
2. She did not use a new sunrise number in 4 spacetime dimensions.
She took the reciprocal of a diagram in 2 spacetime dimensions.
3. This adds spice to integer-relation hunting.
√
√
4. With N = 5 massive lines, the irrational numbers 5 and 3 occur
and transcendental numbers come from elliptic functions with nome
q = e−π
√
15
.
5. This nome slightly twists powers of π expected for polylogarithms.
6. The one new number that suffices for sunrise diagrams with N = 5 is
1+2
∞
X
2
qn .
n=1
7. Sunrise diagrams at N = 5 involve elliptic twists at weights 3 and 1.
8. Weight 2 is conspicuously absent.
9. The recurrence relation between 4-loop vacuum diagrams gives very
fast evaluations of V (0, 5, 0) and V (1, 5, 0).
10. By providing 4375 digits, I hope to encourage PSLQ users to hunt
for integer relations that determine these fundamental constants.
22
7
Joke
At a Zentrum für interdisziplinäre Forschung, mathematicians, computer
scientists and quantum field theorists study the function
f (n) = e−π
√
4n−1
evaluated at the integers.
1. Some folk find the value of
f (0) + 1
to be of surpassing beauty.
2. Others are worried by the 31st decimal digit of
(1/f (41) − 4! − 6!)1/3
= 0.9999999999999999999999999999990 . . .
26 × 3 × 5 × 23 × 29
where 41 comes Euler, 4! from Jacobi and 6! from Rademacher.
3. Yet others prefer to study
1+2
∞
X
[f (4)]n
2
n=1
at sunrise.
David Broadhurst
14 June 2007
PS: After my talk, Stefano Laporta pointed out to me that the AGM in C
is of a form considered by Jon Borwein in a hunt for fast evaluations of Γ
values:
√
C =
5G(1)G(2)G(4)G(8)/1200
(81)
√
1/C = 15 5G(7)G(11)G(13)G(14)
(82)
with G(n) = Γ(n/15)/Γ(1/2).
23