Little q-Legendre polynomials and irrationality of certain Lambert series

15 Pages·0.17 MB·English

by  Walter Van Assche

Little q-Legendre polynomials and irrationality of
certain Lambert series
1
0 Walter Van Assche ∗
0
2 Katholieke Universiteit Leuven and Georgia Institute of Technology

n
a February 1, 2008
J

3
2

Abstract
]
A
Certain q-analogs h (1) of the harmonic series, with p = 1/q an integer greater
C p
than one, were shown to be irrational by Erd˝os [9]. In 1991–1992 Peter Borwein [4]
.
h [5]usedPad´eapproximation andcomplex analysis toprovetheirrationality ofthese
t
a q-harmonic series and of q-analogs ln (2) of the natural logarithm of 2. Recently
p
m
Amdeberhan and Zeilberger [1] used the qEKHAD symbolic package to find q-WZ
[
pairs that provide a proof of irrationality similar to Ap´ery’s proof of irrationality of

1 ζ(2)andζ(3). Theyalsoobtainanupperboundforthemeasureofirrationality, but
v
betterupperboundswereearliergivenbyBundschuhandV¨aa¨na¨nen[8]andrecently
7
8 also by Matala-aho and V¨a¨ana¨nen [14] (for lnp(2)). In this paper we show how
1 one can obtain rational approximants for h (1) and ln (2) (and many other similar
p p
1
quantities) by Pad´e approximation usinglittle q-Legendrepolynomials andwe show
0
1 that properties of these orthogonal polynomials indeed prove the irrationality, with
0 anupperboundof themeasureof irrationality whichis as sharpas theupperbound
/
h given by Bundschuh and V¨a¨ana¨nen for h (1) and a better upper bound as the one
p
at given by Matala-aho and V¨a¨ana¨nen for lnp(2).
m
:
v 1 Introduction
i
X
r Most important special functions, in particular hypergeometric functions, have q-exten-
a
sions, usually obtained by replacing Pochhammer symbols (a) = a(a+1) (a+n 1)
n
··· −
by their q-analog (a;q) = (1 a)(1 aq)(1 aq2) (1 aqn−1). Since
n
− − − ··· −
(qa;q)
n
lim = (a) ,
q→1 (1 q)n n
−
one usually retrieves the original special function from its q-extension by letting q 1. A
→
good source for q-extensions of hypergeometric series (basic hypergeometric series) is the
book [10] by Gasper and Rahman. Our interest in this paper is the following q-extension
of the harmonic series
∞ 1 ∞ qk 1
h (1) = = , 0 < q = < 1, (1)
p pk 1 1 qk p
k=1 − k=1 −
X X
∗
WVA is a Research Director of the Belgian Fund for Scientific Research (FWO).
1
and of the natural logarithm of 2
∞ ( 1)k ∞ ( q)k 1
ln (2) = − = − , 0 < q = < 1. (2)
p pk 1 1 qk p
k=1 − k=1 −
X X
In 1948, Paul Erd˝os [9] proved that h (1) is irrational. Peter Borwein [4] [5] showed that
2
h (1) (and other similar numbers) are irrational for every integer p > 1 and also proved
p
the irrationality of ln (2) for every integer p > 1. If we denote E (z) = (qz;q) , with
p p ∞
q = 1/p, then B´ezivin [3] had earlier shown that E (α),E′(α),... ,E(k)(α) are linearly
p p p
independent over Q for every k N and α Q with αpj = 1 for every integer j. The
∈ ∈ 6 −
case k = 1 corresponds to irrationality of
∞
1
.
α+pj
j=1
X
Borwein used Pad´e approximation techniques and complex analysis to obtain good ratio-
nal approximants to h (1) and ln (2). Indeed, one can use the following lemma to prove
p p
irrationality [16, Lemma 5.1]:
Lemma 1 Let x be a real number, and suppose there exist integers a ,b (n N) such
n n
∈
that
1. x = a /b for every n N,
n n
6 ∈
2. lim (b x a ) = 0,
n→∞ n n
−
then x will be irrational.
This lemma expresses the fact that the order of approximation of a rational number by
rational numbers is one and not higher [12, Theorem 186]. Furthermore, if x a /b =
n n
| − |
(1/b1+s) with 0 < s < 1 and b < b < b1+o(1), then the measure of irrationality r(x)
O n n n+1 n
(Liouville-Roth number, order of approximation)
r(x) = inf r R : x a/b < 1/br has at most finitely many integer solutions (a,b)
{ ∈ | − | }
satisfies 2 r(x) 1 + 1/s (see, e.g., [6, exercise 3 on p. 376] for the upper bound;
≤ ≤
the lower bound follows since every irrational number is approximable to order 2 [12,
Theorem 187]). Recently Amdeberhan and Zeilberger [1] found q-WZ pairs to obtain
rational approximants for h (1) and ln (2), improving the upper bound for the measure
p p
of irrationality to 4.8 = 24/5. However, four years earlier Bundschuh and V¨a¨an¨anen [8,
p.178]hadestablished better upperbounds: r(h (1)) 1+(π2+2)/(π2 2) = 2.50828...
p
≤ −
and r(ln (2)) 1+(2π2+3)/(π2 3) = 4.310119.... The upper bound for r(ln (2)) was
p p
≤ −
improved by Matala-aho and V¨a¨an¨anen [14] to 3.9461.
Inthispaperweshowthatonecanfindrationalapproximantswhicharerelatedtolittle
q-Legendre polynomials and hence return to Pad´e approximation. We can then use some
results for little q-Legendre polynomials to prove the irrationality once more, and with
the aid of some elementary number theory we obtain the same bound for the measure of
irrationalityastheoneobtainedbyBundschuh andV¨a¨an¨anenforh (1)andabetterupper
p
bound as the one given by Matala-aho and V¨a¨an¨anen for ln (2). The connection with
p
little q-Legendre polynomials opens the way for proving the irrationality of q-extensions
of ζ(2) and ζ(3) in Ap´ery’s spirit [2], using multiple orthogonal q-polynomials [16].
2
2 Little q-Legendre polynomials
The little q-Legendre polynomials are defined by
q−n,qn+1
P (x q) = φ q;qx
n | 2 1 q
(cid:18) (cid:12) (cid:19)
n (q−n;q)k(qn+(cid:12)(cid:12)1;q)k qkxk
= (cid:12) , 0 < q < 1, (3)
(q;q) (q;q)
k k
k=0
X
and they are orthogonal polynomials on the exponential lattice qk,k = 0,1,2,... :
{ }
∞ qn
qkP (qk q)P (qk q) = δ . (4)
m | n | 1 q2n+1 n,m
k=0 −
X
If we use the q-binomial coefficients
n (q;q)
n
= , (5)
k (q;q) (q;q)
(cid:20) (cid:21)q k n−k
and the formulas
(q;q) (q;q)
(q−n;q) = ( 1)kq−nk+k(k−1)/2 n , (qn+1;q) = n+k, (6)
k k
− (q;q) (q;q)
n−k n
then (3) reduces to
n
n n+k
P (x q) = q−nk+k(k+1)/2( x)k, (7)
n
| k k −
k=0(cid:20) (cid:21)q(cid:20) (cid:21)q
X
and since
n n
lim = ,
q↑1 k k
(cid:20) (cid:21)q (cid:18) (cid:19)
we see that we indeed find the Legendre polynomials on [0,1] by letting q tend to 1
n
n n+k
limP (x q) = ( x)k = P (x).
n n
q↑1 | k k −
k=0(cid:18) (cid:19)(cid:18) (cid:19)
X
Let p = 1/q so that p > 1 whenever 0 < q < 1, then
n n
(q;q) = ( 1)kp−k(k+1)/2(p;p) , = p−k(n−k) , (8)
k k
− k k
(cid:20) (cid:21)q (cid:20) (cid:21)p
and we can rewrite the little q-Legendre polynomial as
n
n n+k
P (x q) = p−kn+k(k−1)/2( x)k. (9)
n
| k k −
k=0(cid:20) (cid:21)p(cid:20) (cid:21)p
X
3
There is a Rodrigues formula for these little q-Legendre polynomials in terms of the q-
difference operator D for which
q
f(z) f(qz)
− if z = 0,
D f(z) = (1 q)z 6 (10)
q
 −
f′(0) if z = 0,

namely 
qn(n−1)/2(1 q)n
P (x q) = − Dn[(qx;q) xn], (11)
n | (q;q) p n
n
which will be useful later. We refer to [13] for more information and references for little
q-Legendre polynomials.
Equation (9) expresses the little q-Legendre polynomials in the basis 1,x,x2,... ,xn
{ }
of monomials. Sometimes it is more convenient to use another basis of polynomials, and
for orthogonal polynomials on qk, k = 0,1,2,... a convenient set of basis functions is
{ }
(qx;q) , k = 0,1,2,... ,n . We will need to use some q-series for this purpose. Recall
k
{ }
the q-analog of Newton’s binomium formula
n
n
(x;q) = qk(k−1)/2( x)k, (12)
n
k −
k=0(cid:20) (cid:21)q
X
and its dual
n
n
xn = ( 1)kq−nk+k(k+1)/2(x;q) . (13)
k
k −
k=0(cid:20) (cid:21)q
X
A more general result is the q-binomial series
∞
(a;q) (ax;q)
nxn = ∞, q < 1, x < 1. (14)
(q;q) (x;q) | | | |
n ∞
n=0
X
Use (13) with argument qn+1x in the Rodrigues formula (11), then we have
pn(n+1)(1 p)n n n
P (x q) = − ( 1)kq−nk+k(k+1)/2Dn(qx;q) .
n | (p;p) k − p n+k
n k=0(cid:20) (cid:21)q
X
One finds easily that
(q;q)
Dk(qx;q) = n (qx;q) , (15)
p n (q;q) (1 p)k n−k
n−k
−
so that, using (8) we find the required expansion
n
n n+k
P (x q) = ( 1)n ( 1)kp(n−k)(n−k+1)/2(qx;q) . (16)
n k
| − k k −
k=0(cid:20) (cid:21)p(cid:20) (cid:21)p
X
4
3 The q-harmonic series
Orthogonal polynomials naturally arise in Pad´e approximation of a Stieltjes function
dµ(x)
f(z) = , z / supp(µ). (17)
z x ∈
Z −
Suppose µ is a positive measure on the real line with infinite support and for which all
the moments exist. If P (x) (n = 0,1,2,...) are orthogonal polynomials for µ, i.e., P is
n n
of degree n and
P (x)P (x)dµ(x) = 0, m = n,
n m
6
Z
and if Q are the polynomials of degree n 1 given by
n
−
P (z) P (x)
n n
Q (z) = − dµ(x), (18)
n
z x
Z −
then it is easy to see that
P (x)
n
P (z)f(z) Q (z) = dµ(x), z / supp(µ). (19)
n n
− z x ∈
Z −
If we expand 1/(z x) around z = as
− ∞
1 n−1 xk xn 1
= + ,
z x zk+1 zn z x
− k=0 −
X
then by orthogonality we have
1 P (x)xn
P (z)f(z) Q (z) = n dµ(x) = (1/zn+1).
n n
− zn z x O
Z −
This is precisely the (linearized) form of the interpolation conditions near z = for Pad´e
∞
approximation so that Q (z)/P (z) is the [n−1] Pad´e approximant for f(z) near z = .
n n n ∞
For little q-Legendre polynomials the measure µ is supported on qk,k = 0,1,2,... ,
{ }
which is a bounded set in [0,1] with one accumulation point at 0. The measure is given
by
∞
g(x)dµ(x) = g(qk)qk.
Z k=0
X
The Stieltjes function for this measure is
∞ qk ∞ 1
f(z) = = . (20)
z qk zpk 1
k=0 − k=0 −
X X
We will need this function at pn, where it gives
∞ n−1
1 1
f(pn) = = h (1) . (21)
pn+k 1 p − pk 1
k=0 − k=1 −
X X
5
Hence if p > 1 is an integer, then f(pn) gives h (1) up to n−11/(pk 1), which is a
p k=1 −
rational number. Now use (19) for little q-Legendre polynomials at z = pn to find
P
n−1 1 ∞ P (qk q)
P (pn q) h (1) Q (pn q) = n | qk. (22)
n | p − pk 1 − n | pn qk
!
k=1 − k=0 −
X X
Observe that (9) gives
n
n n+k
P (pn q) = ( 1)kpk(k−1)/2, (23)
n
| k n −
k=0(cid:20) (cid:21)p(cid:20) (cid:21)p
X
whichisnearlytheb foundin[1,p.277](theirb correspondstoP (pn+1 q)). Observealso
n n n
|
that Borwein’s construction [5, Lemma 2] uses P (cpn+1 q). The p-binomial numbers
n−1
|
n are polynomials in p with integer coefficients, which follows easily from the q-version
k p
of Pascal’s triangle identities
(cid:2) (cid:3)
n n 1 n 1 n 1 n 1
= − +pk − = − +pn−k − ,
k k 1 k k k 1
(cid:20) (cid:21)p (cid:20) − (cid:21)p (cid:20) (cid:21)p (cid:20) (cid:21)p (cid:20) − (cid:21)p
hence if p > 1 is an integer then n and n+k are integers. This means that (23) implies
k p k p
P (pn q) to be an integer. Furthermore, since pn > 1 and all the zeros of P (x q) are in
n | (cid:2) (cid:3) (cid:2) (cid:3) n |
[0,1], we also may conclude that ( 1)nP (pn q) is positive for all n.
n
− |
The second important quantity in (22) is Q (pn q). This associated little q-Legendre
n
|
polynomial can be computed explicitly using (18) and is given by
∞ P (x q) P (qj q)
Q (x q) = n | − n | qj.
n
| x qj
j=0 −
X
Use (16) to write this as
n n n+k ∞ (qx;q) (qj+1;q)
Q (x q) = ( 1)n ( 1)kp(n−k)(n−k+1)/2 k − kqj.
n
| − k k − x qj
k=0(cid:20) (cid:21)p(cid:20) (cid:21)p j=0 −
X X
Now use
k
(qx;q) (qy;q)
k − k = qℓ(qy;q) (qℓ+1x;q) ,
ℓ−1 k−ℓ
x y −
− ℓ=1
X
which one can easily prove by induction, then this gives
n
n n+k
Q (x q) = ( 1)n+1 ( 1)kp(n−k)(n−k+1)/2
n
| − k k −
k=0(cid:20) (cid:21)p(cid:20) (cid:21)p
X
k ∞
qℓ(qℓ+1x;q) qj(qj+1;q) .
k−ℓ ℓ−1
×
ℓ=1 j=0
X X
6
Using the q-binomial series (14) we can calculate the modified moments
∞ ∞ (qℓ;q)
qj(qj+1;q) = (q;q) qj j
ℓ−1 ℓ−1
(q;q)
j
j=0 j=0
X X
(q;q) (qℓ+1;q)
ℓ−1 ∞
=
(q;q)
∞
1
= ,
1 qℓ
−
so that
n n n+k k (qℓ+1x;q)
Q (x q) = ( 1)n+1 ( 1)kp(n−k)(n−k+1)/2 k−ℓ. (24)
n | − k k − pℓ 1
k=0(cid:20) (cid:21)p(cid:20) (cid:21)p ℓ=1 −
X X
Evaluating at x = pn, and using
(qℓ+1pn;q) = (pn−k;p) ,
k−ℓ k−ℓ
then gives
n n n+k k (pn−k;p)
Q (pn q) = ( 1)n+1 ( 1)kp(n−k)(n−k+1)/2 k−ℓ. (25)
n
| − k k − pℓ 1
k=0(cid:20) (cid:21)p(cid:20) (cid:21)p ℓ=1 −
X X
All the terms in the sum for Q (pn q) are now integers, except for the pℓ 1 in the
n
| −
denominators. In order to obtain an integer we therefore need to multiply everything by
a multiple of all pℓ 1 for ℓ = 1,2,... ,n. We will choose
−
n
d (p) = Φ (p), (26)
n k
k=1
Y
where
n
Φ (x) = (x ωk), ω = e2πi/n, (27)
n − n n
k=1
Y
gcd(k,n)=1
are the cyclotomic polynomials [15, 4.8]. Each cyclotomic polynomial is monic, has
§
integer coefficients, and the degree of Φ is φ(n) (Euler’s totient function). It is known
n
that
xn 1 = Φ (x), (28)
d
−
d|n
Y
and that every cyclotomic polynomial is irreducible over Q[x]. Hence d (p) is a multiple
n
of all pℓ 1 for ℓ = 1,2,... ,n. The growth of this sequence is given by the following
−
lemma, which was essentially given by A. O. Gel’fond, who obtained the upper bound in
[11, Equation (7)]. We give a proof to make this paper self-contained.
7
Lemma 2 Suppose p is an integer greater than one and let d (p) be given by (26). Then
n
lim d (p)1/n2 = p3/π2. (29)
n
n→∞
Proof: The degree of d (p) as a monic polynomial in p is n ϕ(k) and an old result of
n k=1
Mertens (1874) shows that for n this grows like [12, Theorem 330]
→ ∞ P
n
3
ϕ(k) = n2 + (nlogn). (30)
π2 O
k=1
X
We will adapt the classical proof of (30) to prove our lemma. For this we use M¨obius
inversion of (28) to find the representation
Φ (x) = (xd 1)µ(n/d),
n
−
d|n
Y
where µ is the M¨obius function. Taking logarithms in (26) gives
n
logd (p) = µ(k/d)log(pd 1).
n
−
k=1 d|k
XX
Changing the order of summation gives
n ⌊n/ℓ⌋
logd (p) = µ(ℓ) log(pd 1).
n
−
ℓ=1 d=1
X X
Now
m
m(m+1)
log(pd 1) = logpm(m+1)/2(q;q) = logp+log(q;q) ,
m m
− 2
d=1
X
so that
n2 n µ(ℓ) n 1
logd (p) = logp + n
n 2 ℓ2 O ℓ
!
ℓ=1 ℓ=1
X X
n2logp
= + (nlogn).
2ζ(2) O
The lemma now follows by using ζ(2) = π2/6.
So far we have found that the numbers
b = d (p)P (pn q), (31)
n n n
|
n−1
1
a = d (p)Q (pn q)+b . (32)
n n n | n pk 1
k=1 −
X
are integers and
∞ P (qk q)
b h (1) a = d (p) n | qk. (33)
n p − n n pn qk
k=0 −
X
8
We now want to show that b h (1) a = 0 for all n and
n p n
− 6
lim(b h (1) a ) = 0,
n p n
n→∞ −
so that Lemma 1 implies the irrationality of h (1). First observe that
p
∞ P (qk q) 1 ∞ P (qk q)P (pn q)
n | qk = n | n | qk.
pn qk P (pn q) pn qk
n
k=0 − | k=0 −
X X
If we add and subtract P (qk q) in the sum, then we have
n
|
∞ P (qk q) 1 ∞ P (pn q) P (qk q) 1 ∞ P2(qk q)
n | qk = P (qk q) n | − n | qk + n | qk.
n
pn qk P (pn q) | pn qk P (pn q) pn qk
n n
k=0 − | k=0 − | k=0 −
X X X
The first sum on the right hand side vanishes because of the orthogonality, so that
d (p) ∞ P2(qk q)
b h (1) a = n n | qk. (34)
n p − n P (pn q) pn qk
n
| k=0 −
X
All the terms in the sum are now positive, and ( 1)nP (pn q) is positive for all n, hence
n
− |
we may conclude that
( 1)n(b h (1) a ) > 0, n = 1,2,...
n p n
− −
Next we show that this quantity converges to zero. Clearly pn 1 pn qk pn so that
− ≤ − ≤
1 ∞ ∞ P2(qk q) 1 ∞
P2(qk q)qk n | qk P2(qk q)qk.
pn n | ≤ pn qk ≤ pn 1 n |
k=0 k=0 − − k=0
X X X
Now we can use the norm of the little q-Legendre polynomial (4) to find
p ∞ P2(qk q) pn+1
n | qk . (35)
p2n+1 1 ≤ pn qk ≤ (pn 1)(p2n+1 1)
− k=0 − − −
X
What remains is to find the asymptotic behavior of P (pn q) as n . For this we can
n
| → ∞
use a very general theorem for sequences of polynomials with uniformly bounded zeros.
Lemma 3 Suppose P (n N) is a sequence of monic polynomials of degree n and that
n
∈
the zeros x (1 j n) of P are such that x M, with M independent of n. Then
j,n n j,n
≤ ≤ | | ≤
we have for x > 1 and every c C
| | ∈
lim P (cxn) 1/n2 = x . (36)
n
n→∞| | | |
Proof: Factoring the polynomial P gives
n
n
P (x) = x x .
n j,n
| | | − |
j=1
Y
9
We have the obvious bounds
x M x x x +M,
j,n
| |− ≤ | − | ≤ | |
hence when cxn > M
| |
( cxn M)n P (cxn) ( cxn +M)n.
n
| |− ≤ | | ≤ | |
For n large enough this easily gives
1/n 1/n
M M
x c P (cxn) 1/n2 x c + ,
| | | |− x n ≤ | n | ≤ | | | | x n
(cid:18) | | (cid:19) (cid:18) | | (cid:19)
which gives the desired result.
Observe that we may allow M to grow with n subexponentially. For little q-Legendre
polynomials the zeros are all in [0,1] so that we can use the Lemma with M = 1. The
leading coefficient κ of P (x q) is, by (9), equal to κ = ( 1)n 2n p−n(n+1)/2, giving
n n | n − n p
(cid:2) (cid:3)
lim κ 1/n2 = √p,
n
n→∞| |
and hence Lemma 3 gives for x > 1 and c C
| | ∈
lim P (cxn q) 1/n2 = √p x . (37)
n
n→∞| | | | |
Theorem 1 Suppose p > 1 is an integer. Let a and b be given by (31)–(32), then
n n
a Z, ( 1)nb N, and ( 1)n(b h (1) a ) > 0 for n > 1. Furthermore
n n n p n
∈ − ∈ − −
lim bnhp(1) an 1/n2 = p−3(π22π−22) < 1,
n→∞| − |
which implies that h (1) is irrational with measure of irrationality r 2π2 = 2.50828...
p ≤ π2−2
Proof: If we take x = p and c = 1 in (37) then we have
lim P (pn q) 1/n2 = p3/2,
n
n→∞| | |
and combining with (29) we have for the integer b in (31)
n
lim bn 1/n2 = p3(π22π+22).
n→∞| |
Observe that b has the same sign as P (pn q) which is ( 1)n. Furthermore (34) and (35)
n n
| −
show that
nl→im∞|bnhp(1)−an|1/n2 = limlinm→n∞→∞Pnd(np(np)q1)/n12/n2 = p−3(π22π−22) < 1.
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10