LEONHARD EULER and a Q-ANALOGUE of the LOGARITHM

LEONHARD EULER and a Q-ANALOGUE of the LOGARITHM

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 137, Number 5, May 2009, Pages 1663–1676 S 0002-9939(08)09374-X Article electronically published on December 12, 2008 LEONHARD EULER AND A q-ANALOGUE OF THE LOGARITHM ERIK KOELINK AND WALTER VAN ASSCHE (Communicated by Peter A. Clarkson) On the 300th anniversary of Euler’s birth Abstract. We study a q-logarithm which was introduced by Euler and give some of its properties. This q-logarithm has not received much attention in the recent literature. We derive basic properties, some of which were already given by Euler in a 1751 paper and in a 1734 letter to Daniel Bernoulli. The corresponding q-analogue of the dilogarithm is introduced. The relation to the valuesat1and2ofaq-analogue of the zeta function is given. We briefly describe some other q-logarithms that have appeared in the recent literature. 1. Introduction In a paper from 1751, Leonhard Euler (1707–1783) introduced the series [8, §6] ∞ (1 − x)(1 − x/a) ···(1 − x/ak−1) (1.1) s = . 1 − ak k=1 We will take q =1/a. Then this series is convergent for |q| < 1andx ∈ C.Inthis paper we will assume 0 <q<1. Then this becomes ∞ qk (1.2) S (x)=− (x; q) , q 1 − qk k k=1 k−1 where (x; q)0 =1,(x; q)k =(1− x)(1 − xq) ···(1 − xq ). This can be written as a basic hypergeometric series q(1 − x) q, q, qx S (x)=− φ ; q, q . q 1 − q 3 2 q2, 0 Euler had come across this series much earlier in an attempt to interpolate the logarithm at powers ak (or q−k); see, e.g., Gautschi’s comment [11] discussing Euler’s letter to Daniel Bernoulli where Euler introduced the function for a = 10. Euler was aware that this interpolation did not work very well; see [11, §§3-4]. The function in (1.2) does not seem to appear in the recent literature, even though it has some nice properties. We will prove some of its properties, some already obtained Received by the editors March 6, 2007. 2000 Mathematics Subject Classification. Primary 33B30, 33E30. The second author was supported by research grant OT/04/21 of Katholieke Universiteit Leu- ven, research project G.0455.04 of FWO-Vlaanderen, and INTAS research network 03-51-6637. c 2008 American Mathematical Society 1663 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1664 ERIK KOELINK AND WALTER VAN ASSCHE by Euler [8], and indicate why this should be called a q-analogue of the logarithm. Afirstreasonisthatfor0<q<1, ∞ ∞ − − k k 1 q (1 x) lim(1 − q)Sq(x)=− lim q (x; q)k = − =logx, q→1 q→1 1 − qk k k=1 k=1 which is only a formal limit transition, since interchanging limit and sum seems hard to justify. In Sections 2–3 we study this q-analogue of the logarithm more closely. In par- ticular, we reprove some of Euler’s results. Then we go on to extend the definition in Section 4. Finally, we study the corresponding q-analogue of the dilogarithm in Section 5. It involves also the values at 1 and 2 of a q-analogue of the ζ-function. We give a (incomplete) list of some other q-analogues of the logarithm appearing in the literature in Section 6. The purpose of this note is to draw attention to the q-analogues of the logarithm, dilogarithm and ζ-function for which we expect many interesting results remain to be discovered. Many results in this paper use the q-binomial theorem [10, §1.3], [1, §10.2] ∞ (ax; q)∞ (a; q) (1.3) = j xj, |x| < 1. (x; q)∞ (q; q) j=0 j We also use the q-exponential functions [10, p. 9], [1, p. 492] ∞ 1 zn eq(z)= = , |z| < 1, (z; q)∞ (q; q) n=0 n ∞ n(n−1)/2 q n E (z)=(−z; q)∞ = z . q (q; q) n=0 n 2. The q-logarithm as an entire function First of all we will show that the function Sq in (1.2) is an entire function, and as such it is a nicer function than the logarithm, which has a cut along the negative real axis. Property 2.1. The function Sq defined in (1.2) is an entire function of order zero. Proof. For k ∈ N the q-Pochhammer (z; q)k is a polynomial of degree k with zeros at 1, 1/q,...,1/qk−1.For|z|≤r we have the simple bound k−1 |(z; q)k|≤(1 + r)(1 + r|q|) ···(1 + r|q| )=(−r; |q|)k < (−r; |q|)∞, and hence the partial sums are uniformly bounded on the ball |z|≤r: ∞ n qk |q|k − (z; q) ≤ (−r; |q|)∞ . 1 − qk k 1 −|q|k k=1 k=1 The partial sums therefore are a normal family and are uniformly convergent on every compact subset of the complex plane. The limit of these partial sums is Sq(z) and is therefore an entire function of the complex variable z. Let M(r)=max|z|≤r |Sq(z)|.Then ∞ |q|k M(r) ≤ (−r; |q|)∞ 1 −|q|k k=1 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use EULER’S q-LOGARITHM 1665 and (−r; |q|)∞ = E|q|(r)isthemaximumofE|q|(z) on the ball {|z|≤r}. The func- tion Eq is an entire function of order zero, which can be seen from the coefficients an of its Taylor series and the formula [2, Theorem 2.2.2] n log n (2.1) lim sup n→∞ log(1/|an|) ∞ n for the order of n=0 anz . Hence also Sq has order zero. Observe that for 0 <q<1wehave ∞ qk M(r)=max|Sq(z)| = (−r; q)k |z|≤r 1 − qk k=1 and some simple bounds give ∞ ∞ qk qk ∞ − ≤ ≤ − ∞ (q; q) ( r; q)k M(r) ( r; q) k . (q; q)k 1 − q k=1 k=1 For the lower bound we can use the q-binomial theorem (1.3) to find ∞ qk (−rq; q)∞ − (q; q)∞ ≤ M(r) ≤ (−r; q)∞ , 1 − qk k=1 which shows that M(r) behaves like Eq(qr)−C1 ≤ M(r) ≤ C2Eq(r), where C1 and C2 are constants (which depend on q). Euler [8, §§14-15] essentially also stated the following Taylor expansion. Property 2.2. The q-logarithm (1.2) has the following Taylor series around x =0: ∞ k − k − q k(k−1)/2 ( x) Sq(x)= k 1+q . 1 − q (q; q)k k=1 Proof. Use the q-binomial theorem (1.3) with x = zqk and a = q−k to find k k j(j−1)/2 j k (q; q)k (2.2) (z; q)k = q (−z) , = . j j (q; q) (q; q) − j=0 j k j Use this in (1.2), and change the order of summation to find ∞ ∞ ∞ k k − q − j(j−1)/2 − j q (q; q)k Sq(x)= k q ( x) k . 1 − q 1 − q (q; q)j(q; q)k−j k=1 j=1 k=j With a new summation index k = j + this becomes ∞ ∞ ∞ k j j − q − q j(j−1)/2 − j (q ; q) Sq(x)= k j q ( x) q . 1 − q 1 − q (q; q) k=1 j=1 =0 Now use the q-binomial theorem (1.3) to sum over to find ∞ ∞ k j − j − q − q j(j−1)/2 ( x) Sq(x)= k j q . 1 − q 1 − q (q; q)j k=1 j=1 If we combine both series, the required expansion follows. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1666 ERIK KOELINK AND WALTER VAN ASSCHE This result can be written in terms of basic hypergeometric series as q q, q qx q, q S (x)=− φ ; q, q − φ ; q, q2x . q 1 − q 2 1 q2 (1 − q)2 2 2 q2,q2 The growth of the coefficients in this Taylor series again shows thatSq is an entire ∞ n function of order zero if we use the formula (2.1) for the order of n=0 anz ;see also [11, §4]. Next we mention the following q-integral representation, where we use Jackson’s q-integral (see [10, §1.11]) a ∞ k k (2.3) f(t) dqt =(1− q)a f(aq ) q , 0 k=0 defined for functions f whenever the right-hand side converges. Property 2.3. For every x ∈ C we have − 1 −q(1 x) Sq(x)= − Gq(qx, qt) dqt, 1 q 0 with ∞ x, q 1 q G (x, t)= tk(x; q) = φ ; q, t = φ ; q, xt . q k 2 1 0 1 − t 1 1 qt k=0 a → a → → − − Since 0 f(t) dqt 0 f(t) dt when q 1andGq(x, t) 1/(1 t(1 x)) when q → 1forx>0, we see (at least formally) that Property 2.3 is a q-analogue of the integral representation 1 − − 1 x ∈ −∞ log(x)= − − dt, x / ( , 0] 0 1 t(1 x) for the logarithm. Proof. Observe that ∞ 1 − q 1 =(1− q) q(k+1)p = tk d t. 1 − qk+1 q p=0 0 Inserting this in the definition (1.2) of Sq and interchanging summations, which is justified by the absolute convergence of the double sum, give the result.

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    14 Page
  • File Size
    -

Download

Channel Download Status
Express Download Enable

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

  • Not to be reproduced or distributed without explicit permission.
  • Not used for commercial purposes outside of approved use cases.
  • Not used to infringe on the rights of the original creators.
  • If you believe any content infringes your copyright, please contact us immediately.

Support

For help with questions, suggestions, or problems, please contact us