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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   n −n n+1 k k (q ; q)k(q ; q)k q x = , 0

n (q; q) = n , (5) k (q; q) (q; q)  q k n−k and the formulas

−n k −nk+k(k−1)/2 (q; q)n n+1 (q; q)n+k (q ; q)k =( 1) q , (q ; q)k = , (6) − (q; q)n−k (q; q)n then (3) reduces to

n n n + k −nk+k(k+1)/2 k Pn(x q)= q ( x) , (7) | k q k q − Xk=0     and since n n lim = , q↑1 k k  q   we see that we indeed find the Legendre polynomials on [0, 1] by letting q tend to 1

n n n + k k lim Pn(x q)= ( x) = Pn(x). q↑1 | k k − Xk=0    Let p =1/q so that p> 1 whenever 0

n n (q; q) =( 1)kp−k(k+1)/2(p; p) , = p−k(n−k) , (8) k − k k k  q  p and we can rewrite the little q-Legendre polynomial as

n n n + k −kn+k(k−1)/2 k Pn(x q)= p ( x) . (9) | k p k p − Xk=0    

3 There is a Rodrigues formula for these little q-Legendre polynomials in terms of the q- difference operator Dq for which

f(z) f(qz) − if z =0, Dqf(z)= (1 q)z 6 (10)  ′ − f (0) if z =0, namely 

n(n−1)/2 n q (1 q) n n Pn(x q)= − Dp [(qx; q)nx ], (11) | (q; q)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, k = 0, 1, 2,...,n . We will need to use some q-series for this purpose. Recall {the q-analog of Newton’s binomium} formula

n n k(k−1)/2 k (x; q)n = q ( x) , (12) k q − Xk=0   and its dual

n n n k −nk+k(k+1)/2 x = ( 1) q (x; q)k. (13) k q − Xk=0   A more general result is the q-binomial series

∞ (a; q) (ax; q) n xn = ∞ , q < 1, x < 1. (14) (q; q) (x; q) | | | | n=0 n ∞ X Use (13) with argument qn+1x in the Rodrigues formula (11), then we have

n(n+1) n n p (1 p) n k −nk+k(k+1)/2 n Pn(x q)= − ( 1) q Dp (qx; q)n+k. | (p; p)n k q − Xk=0   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 n + k k (n−k)(n−k+1)/2 Pn(x q)=( 1) ( 1) p (qx; q)k. (16) | − k p k p − Xk=0    

4 3 The q-harmonic series

Orthogonal polynomials naturally arise in Pad´eapproximation 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 Pn(x) (n =0, 1, 2,... ) are orthogonal polynomials for µ, i.e., Pn is 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) Q (z)= n − n dµ(x), (18) n z x Z − then it is easy to see that P (x) P (z)f(z) Q (z)= n 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 − Xk=0 − 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 n−1 ∞ approximation so that Qn(z)/Pn(z) is the [ n ] Pad´eapproximant for f(z) near z = . 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 Xk=0 The Stieltjes function for this measure is

∞ qk ∞ 1 f(z)= = . (20) z qk zpk 1 Xk=0 − Xk=0 − We will need this function at pn, where it gives

∞ 1 n−1 1 f(pn)= = h (1) . (21) pn+k 1 p − pk 1 Xk=0 − Xk=1 −

5 Hence if p > 1 is an integer, then f(pn) gives h (1) up to n−1 1/(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 ∞ k n 1 n Pn(q q) k Pn(p q) hp(1) k Qn(p q)= n | k q . (22) | − p 1! − | p q Xk=1 − Xk=0 − Observe that (9) gives

n n n n + k k k(k−1)/2 Pn(p q)= ( 1) p , (23) | k p n p − Xk=0     n+1 which is nearly the bn found in [1, p. 277] (their bn corresponds to Pn(p q)). Observe also n+1 | that Borwein’s construction [5, Lemma 2] uses Pn−1(cp q). The p-binomial numbers n | k p are polynomials in p with integer coefficients, which follows easily from the q-version of Pascal’s triangle identities   n n 1 n 1 n 1 n 1 = − + pk − = − + pn−k − , k k 1 k k k 1  p  − p  p  p  − p n n+k hence if p> 1 is an integer then k p and k p are integers. This means that (23) implies n n Pn(p q) to be an integer. Furthermore, since p > 1 and all the zeros of Pn(x q) are in |   n  n | [0, 1], we also may conclude that ( 1) Pn(p q) is positive for all n. − | n The second important quantity in (22) is Qn(p q). This associated little q-Legendre 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 − k qj. n | − k k − x qj p p j=0 Xk=0     X − Now use

(qx; q) (qy; q) k k − k = qℓ(qy; q) (qℓ+1x; q) , x y − ℓ−1 k−ℓ − Xℓ=1 which one can easily prove by induction, then this gives

n n+1 n n + k k (n−k)(n−k+1)/2 Qn(x q)=( 1) ( 1) p | − k p k p − Xk=0     k ∞ qℓ(qℓ+1x; q) qj(qj+1; q) . × k−ℓ ℓ−1 j=0 Xℓ=1 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=0 j=0 j X X (q; q) (qℓ+1; q) = ℓ−1 ∞ (q; q)∞ 1 = , 1 qℓ − so that

n k ℓ+1 n+1 n n + k k (n−k)(n−k+1)/2 (q x; q)k−ℓ Qn(x q)=( 1) ( 1) p ℓ . (24) | − k p k p − p 1 Xk=0     Xℓ=1 − Evaluating at x = pn, and using

ℓ+1 n n−k (q p ; q)k−ℓ =(p ; p)k−ℓ, then gives

n k n−k n n+1 n n + k k (n−k)(n−k+1)/2 (p ; p)k−ℓ Qn(p q)=( 1) ( 1) p ℓ . (25) | − k p k p − p 1 Xk=0     Xℓ=1 − 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

dn(p)= Φk(p), (26) kY=1 where

n Φ (x)= (x ωk), ω = e2πi/n, (27) n − n n k=1 gcd(Yk,n)=1 are the cyclotomic polynomials [15, 4.8]. Each cyclotomic polynomial is monic, has § integer coefficients, and the degree of Φn is φ(n) (Euler’s totient function). It is known that

xn 1= Φ (x), (28) − d Yd|n and that every cyclotomic polynomial is irreducible over Q[x]. Hence dn(p) is a multiple 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 dn(p) be given by (26). Then

1/n2 3/π2 lim dn(p) = p . (29) n→∞ n Proof: The degree of dn(p) as a monic polynomial in p is k=1 ϕ(k) and an old result of Mertens (1874) shows that for n this grows like [12, Theorem 330] →∞ P n 3 ϕ(k)= n2 + (n log n). (30) π2 O Xk=1 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 − Yd|n where µ is the M¨obius function. Taking logarithms in (26) gives

n log d (p)= µ(k/d) log(pd 1). n − Xk=1 Xd|k Changing the order of summation gives

n ⌊n/ℓ⌋ log d (p)= µ(ℓ) log(pd 1). n − Xℓ=1 Xd=1 Now m m(m + 1) log(pd 1) = log pm(m+1)/2(q; q) = log p + log(q; q) , − m 2 m Xd=1 so that n2 n µ(ℓ) n 1 log dn(p) = log p 2 + n 2 ℓ O ℓ ! Xℓ=1 Xℓ=1 n2 log p = + (n log n). 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 Xk=1 − are integers and ∞ P (qk q) b h (1) a = d (p) n | qk. (33) n p − n n pn qk Xk=0 −

8 We now want to show that b h (1) a = 0 for all n and n p − n 6

lim (bnhp(1) an)=0, n→∞ − so that Lemma 1 implies the irrationality of hp(1). First observe that

∞ k ∞ k n Pn(q q) k 1 Pn(q q)Pn(p q) k n | k q = n |n k | q . p q Pn(p q) p q Xk=0 − | Xk=0 − If we add and subtract P (qk q) in the sum, then we have n | ∞ k ∞ n k ∞ 2 k Pn(q q) k 1 k Pn(p q) Pn(q q) k 1 Pn (q q) k n | k q = n Pn(q q) | n − k | q + n n | k q . p q Pn(p q) | p q Pn(p q) p q Xk=0 − | Xk=0 − | Xk=0 − The first sum on the right hand side vanishes because of the orthogonality, so that

∞ 2 k dn(p) Pn (q q) k bnhp(1) an = n n | k q . (34) − Pn(p q) p q | Xk=0 − 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 ∞ ∞ P 2(qk q) 1 ∞ P 2(qk q)qk n | qk P 2(qk q)qk. pn n | ≤ pn qk ≤ pn 1 n | Xk=0 Xk=0 − − Xk=0 Now we can use the norm of the little q-Legendre polynomial (4) to find

p ∞ P 2(qk q) pn+1 n | qk . (35) p2n+1 1 ≤ pn qk ≤ (pn 1)(p2n+1 1) − Xk=0 − − − 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 xj,n (1 j n) of Pn are such that xj,n M, with M independent of n. Then we have for x >≤1 and≤ every c C | | ≤ | | ∈ n 1/n2 lim Pn(cx ) = x . (36) n→∞ | | | |

Proof: Factoring the polynomial Pn gives

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 2 M x c P (cxn) 1/n x c + , | | | | − x n ≤| n | ≤| | | | x n  | |   | |  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

1/n2   lim κn = √p, n→∞ | | and hence Lemma 3 gives for x > 1 and c C | | ∈ n 1/n2 lim Pn(cx q) = √p x . (37) n→∞ | | | | |

Theorem 1 Suppose p > 1 is an integer. Let an and bn be given by (31)–(32), then a Z, ( 1)nb N, and ( 1)n(b h (1) a ) > 0 for n> 1. Furthermore n ∈ − n ∈ − n p − n π2− 2 3( 2) 1/n − π2 lim bnhp(1) an = p 2 < 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

n 1/n2 3/2 lim Pn(p q) = p , n→∞ | | | and combining with (29) we have for the integer bn in (31)

π2 2 3( +2) 1/n π2 lim bn = p 2 . n→∞ | | Observe that b has the same sign as P (pn q) which is ( 1)n. Furthermore (34) and (35) n n | − show that

2 1/n π2− 1/n2 limn→∞ dn(p) − 3( 2) 2π2 lim bnhp(1) an = 2 = p < 1. n→∞ | − | lim P (pn q) 1/n n→∞ | n | |

10 The irrationality now follows from Lemma 1. Observe that this gives rational approxi- mants an/bn for hp(1) satisfying

2 −3(π −2) 2 ( π2 +ǫ)n an p 2 hp(1) = | − bn | O  bn 

3n2(π2+2)/(2π2)+o(n2)  for every ǫ> 0. Now bn = p , hence

an 1 hp(1) = π2− 1+ 2 −ǫ | − bn | O  π2+2  bn   2 2 for every ǫ> 0, which gives for the measure of irrationality the bound r 1+ π +2 = 2π . ≤ π2−2 π2−2

The upper bound for the measure of irrationality is better than the upper bound 4.8 obtained in [1], but the same as the (earlier) upper bound of Bundschuh and V¨a¨an¨anen [8].

4 The q-analog of the logarithm of 2

Next we show that a very similar analysis also proves the irrationality of lnp(2) for every integer p> 1. First of all we rewrite lnp(2) using the geometric series ∞ qk ∞ ∞ ln (2) = ( 1)k = ( 1)kqk qjk. p − 1 qk − j=0 Xk=1 − Xk=1 X Fubini’s theorem allows us to change the order of the sums whenever 0

n n n n + k k(k−1)/2 Pn( p q)= p , − | k p k p Xk=0     is a positive integer, and if we use (24) and ( pnqℓ+1; q) =( pn−k; p) then − k−ℓ − k−ℓ n k n−k n n+1 n n + k k (n−k)(n−k+1)/2 ( p ; p)k−ℓ Qn( p q)=( 1) ( 1) p − ℓ . − | − k p k p − p 1 Xk=0     Xℓ=1 − This quantity has the numbers pℓ 1 (ℓ = 1, 2,...,n) in the denominator, so that we need to multiply it by a multiple of− all pℓ 1 with 1 ℓ n. Now we have an additional − ≤ ≤ quantity

n−1 1 pk +1 Xk=1 and in order to make this an integer we need to multiply it by a multiple of all pk +1 for k =1, 2,...,n 1. If we choose − n ˆ 2 dn(p)= Φk(p ), kY=1 2k k k ˆ k then because of p 1=(p 1)(p + 1) we see that dn(p) is a multiple of all p + 1 and k − − 2 all p 1 for k = 1,...,n. Note that dˆn(p) = dn(p ), where dn is given by (26), so that Lemma− 2 gives the growth

2 1/n2 6/π2 lim dn(p ) = p . (38) n→∞ So if we choose

b = d (p2)P ( pn q), (39) n n n − | n−1 1 a = d (p2)Q ( pn q) b , (40) n n n − | − n pk +1 Xk=1 then an and bn are integers and

∞ P (qk q) b ln (2) a = d (p2) n | qk. (41) n p − n n pn qj Xk=0 − −

Theorem 2 Suppose p > 1 is an integer. Let an and bn be given by (39)–(40), then a Z, b N, and b ln (2) a < 0. Furthermore n ∈ n ∈ n p − n π2− 2 3( 4) 1/n − π2 lim bn lnp(2) an = p 2 < 1, n→∞ | − |

2π2 which implies that lnp(2) is irrational. Its measure of irrationality satisfies r π2−4 = 3.36295 ... . ≤

12 Proof: Use c = 1 and x = p in (37), then together with (38) we find − π2 2 3( +4) 2 1/n π2 n lim bn = p 2 . n→∞ Furthermore, we have

∞ k ∞ 2 k Pn(q q) k 1 Pn (q q) k n | j q = n n | j q p q Pn( p q) p q Xk=0 − − − | Xk=0 − − and pn pn + qj pn + 1 for every j, combined with (41) implies ≤ ≤ d (p2) pn+1 d (p2) p n (b ln (2) a ) n . P ( pn q) (pn + 1)(p2n+1 1) ≤ − n p − n ≤ P ( pn q) p2n+1 1 n − | − n − | − From this one easily finds the required asymptotics and the irrationality then follows from 3n2(π2+4)/(2π2)+o(n2) Lemma 1. Observe that bn = p and

an 1 lnp(2) = π2− 1+ 4 −ǫ | − bn | O  π2+4  bn

  2 2 for every ǫ> 0, which gives for the measure of irrationality the bound r 1+ π +4 = 2π . ≤ π2−4 π2−4

The upper bound is better than the upper bounds 4.8 (obtained in [1]), 4.311 (obtained in [8]), and 3.9461 (obtained in [14]).

5 Extensions

The construction of rational approximants for hp(1) and lnp(2) can be extended with little effort to series of the form ∞ 1 L = , cpk 1 Xk=1 − where c = a/b is a rational number and cpk = 1 for every k 1. Indeed, these series can 6 ≥ be obtained by evaluating the Stieltjes function f in (20) at cpn, giving

∞ 1 n−1 1 f(cpn)= . cpk 1 − cpk 1 Xk=1 − Xk=1 − We then get

n−1 ∞ k n 1 n Pn(q q) k Pn(cp q) L k Qn(cp q)= n | k q , | − cp 1! − | cp q Xk=1 − Xk=0 − where n n n n + k k(k−1)/2 k Pn(cp q)= p ( c) , | k p k p − Xk=0     13 and

n k n−k n n+1 n n + k k (n−k)(n−k+1)/2 (cp ; p)k−ℓ Qn(cp q)=( 1) ( 1) p ℓ . | − k p k p − p 1 Xk=0     Xℓ=1 − n n In order to have integers, the quantities Pn(cp q) and Q(cp q) now need to be multiplied by bn and by a multiple of all pℓ 1 for 1 ℓ| n and of all| apk b for 1 k n 1. 2n − ≤ ≤ − ≤ ≤ − A possible factor is b dn(p)(c; p)n. This factor grows like

2 3 1 2n 1/n π2 + lim b dn(p)(c; p)n = p 2 . n→∞ | | In a way similar to the proof of Theorems 1 and 2 we can then prove:

Theorem 3 Suppose p > 1 is an integer and c = a/b is rational but cpk = 1 for every 6 k =1, 2,... Let an and bn be given by

b = b2nd (p)(c; p) P (cpn q), (42) n n n n | n−1 1 a = b2nd (p)(c; p) Q (cpn q)+ b . (43) n n n n | n cpk 1 Xk=1 − Then a , b Z, and n n ∈ 2 2 π −3 1/n − 2 lim bnL an = p π < 1, n→∞ | − | which implies that the infinite sum L is irrational. Its measure of irrationality satisfies 2 r 3π =4.310119 ... . ≤ π2−3 The upper bound for the measure of irrationality corresponds to the upper bound given by Bundschuh and V¨a¨an¨anen [8, p. 178]. For the cases c = 1 and c = 1, which we handled in Theorems 1 and 2, one can find better upper bounds. Note that− the results in [8] and [14] are also valid for p and c in other number fields. Our main purpose in this paper, however, was to emphasise the use of little q-Legendre polynomials in the construction of rational approximants for certain important Lambert series.

Acknowledgments

This research was carried out while visiting Georgia Institute of Technology. The author wishes to thank the School of Mathematics for its hospitality. Also thanks to a referee for pointing out references [3], [8], [11], and [14], and to Roberto Costas for a useful conversation that led to Lemma 2. This research is partially funded by FWO research project G.0278.97 and INTAS 2000-272.

References

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14 [2] R. Ap´ery, Irrationalit´ede ζ(2) et ζ(3), Ast´erisque 61 (1979), 11–13. [3] J-P. B´ezivin, Ind´ependance lin´eaire des valeurs des solutions transcendantes de cer- taines ´equations fonctionelles, Manuscripta Math. 61 (1988), 103–129.

1 [4] P. Borwein, On the irrationality of qn+r , J. Number Theory 37 (1991), 253–259. [5] P. Borwein, On the irrationality ofP certain series, Proc. Cambridge Philos. Soc. 112 (1992), 141–146. [6] J. M. Borwein, P. B. Borwein, and the AGM — A Study in Analytic Number Theory and Computational Complexity, Wiley, New York, 1987. [7] P. Borwein, T. Erd´elyi, Polynomials and polynomial inequalities, Graduate Texts in Mathematics 161, Springer-Verlag, New York, 1995. [8] P. Bundschuh, K. V¨a¨an¨anen, Arithmetical investigations of a certain infinite product, Compositio Math. 91 (1994), 175–199. [9] P. Erd˝os, On arithmetical properties of Lambert series, J. Indiana Math. Soc. 12 (1948), 63–66. [10] G. Gasper, M. Rahman, Basic Hypergeometric Series, Encyclopedia of Mathematics and its Applications 35, Cambridge University Press, Cambridge, 1990. [11] A. O. Gel’fond, Functions which take on integral values, Mat. Zam. 1 No. 5 (1967), 509–513; translated in Math. Notes 1 (1967), 337–340. [12] G. H. Hardy, E. M. Wright, An Introduction to the Theory of Numbers, Oxford University Press, 1938 (Fifth Edition, 1979). [13] R. Koekoek, R. F. Swarttouw, The Askey-scheme of hypergeometric orthogonal poly- nomials and its q-analogue, Faculty of Technical Mathematics and Informatics Report 98-17, Technical University Delft, 1998. Available at ftp://ftp.twi.tudelft.nl/TWI/publications/tech-reports/1998/DUT-TWI-98-17.ps.gz [14] T. Matala-aho, K. V¨a¨an¨anen, On approximation measures of q-logarithms, Bull. Aus- tralian Math. Soc. 58 (1998), 15–31. [15] J. Stillwell, Elements of Algebra, Undergraduate Texts in Mathematics, Springer, Berlin, 1994. [16] W. Van Assche, Multiple orthogonal polynomials, irrationality and transcendence, Contemporary Mathematics [17] A. van der Poorten, A proof that Euler missed ... , Ap´ery’s proof of the irrationality of ζ(3), Math. Intell. 1 (1979), 195–203. Department of Mathematics Katholieke Universiteit Leuven Celestijnenlaan 200 B B-3001 Leuven BELGIUM [email protected]

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