Arab. J. Math. (2017) 6:95–108 DOI 10.1007/s40065-017-0166-4 Arabian Journal of Mathematics Feng Qi · Xiao-Ting Shi · Fang-Fang Liu Expansions of the exponential and the logarithm of power series and applications Received: 12 June 2016 / Accepted: 2 April 2017 / Published online: 17 April 2017 © The Author(s) 2017. This article is an open access publication Abstract In the paper, the authors establish explicit formulas for asymptotic and power series expansions of the exponential and the logarithm of asymptotic and power series expansions. The explicit formulas for the power series expansions of the exponential and the logarithm of a power series expansion are applied to find explicit formulas for the Bell numbers and logarithmic polynomials in combinatorics and number theory. Mathematics Subject Classification 34E05 · 11B73 · 11B75 · 11B83 · 30B10 ﺍﻟﻤﻠﺨﺺ ﻓﻲ ﻫﺬﻩ ﺍﻟﻮﺭﻗﺔ ﻳﻜﺮﺱ ﺍﻟﻤﺆﻟﻔﻮﻥ ﺻﻴﻐﺎ ﺻﺮﻳﺤﺔ ﻟﺘﻤﺪﻳﺪﺍﺕ ّﻣﻘﺎﺭﺑﻴ ﺔ ﻭﻣﺘﺴﻠﺴﻼﺕ ﻗﻮﻯ ﻷﺱ ﻭﻟﻮﻏﺎﺭﻳﺘﻢ ﺗﻤﺪﻳﺪﺍﺕ ﻣﻘﺎﺭﺑﻴﺔ ﻭﻣﺘﺴﻠﺴﻼﺕ ﻗﻮﻯ. ﺗﻢ ﺗﻄﺒﻴﻖ ﺍﻟﺼﻴﻎ ﺍﻟﺼﺮﻳﺤﺔ ﻟﺘﻤﺪﻳﺪﺍﺕ ﻣﺘﺴﻠﺴﻼﺕ ﺍﻟﻘﻮﻯ ﻷﺱ ﻭﻟﻮﻏﺎﺭﻳﺘﻢ ﺗﻤﺪﻳﺪ ﻣﺘﺴﻠﺴﻼﺕ ﻗﻮﻯ ﻹﻳﺠﺎﺩ ﺻﻴﻎ ﺻﺮﻳﺤﺔ ﻷﻋﺪﺍﺩ ﺑﻞ ﻭﻛﺜﻴﺮﺍﺕ ﺍﻟﺤﺪﻭﺩ ﺍﻟﻠﻮﻏﺎﺭﻳﺘﻤﻴﺔ ﻓﻲ ﻧﻈﺮﻳﺔ ﺍﻟﺘﺮﻛﻴﺒﺎﺕ ﻭﻧﻈﺮﻳﺔ ﺍﻷﻋﺪﺍﺩ. 1 Introduction Throughout this paper, we understand an empty sum to be 0 and regard an empty product as 1. Let us recall definitions for an asymptotic expansion. ∞ an Definition 1.1 [11, p. 31, Definition 2.1] Let F be a function of a real or complex variable z;let n=0 zn denote a convergent or divergent formal power series, of which the sum of the first n terms is denoted by Sn(z); and let Rn(z) = F(z) − Sn(z). F. Qi (B) Institute of Mathematics, Henan Polytechnic University, Jiaozuo 454010, Henan, China E-mail: [email protected]; [email protected]; [email protected] URL: https://qifeng618.wordpress.com F. Qi College of Mathematics, Inner Mongolia University for Nationalities, Tongliao 028043, Inner Mongolia Autonomous Region, China X.-T. Shi · F.-F. Liu Department of Mathematics, College of Science, Tianjin Polytechnic University, Tianjin 300387, China E-mail: [email protected]; [email protected] F.-F. Liu E-mail: [email protected] 123 96 Arab. J. Math. (2017) 6:95–108 In other words, − n 1 a F(z) = k + R (z) = S (z) + R (z), n ≥ 0, zk n n n k=0 where we assume that when n = 0wehaveF(z) = R0(z). Next, assume that for each n ≥ 0 the relation 1 R (z) = O , z →∞ n zn ∞ an ( ) holds in some unbounded domain . Then, n=0 zn is called an asymptotic expansion of the function F z and we denote this by ∞ a F(z) ∼ n , z →∞, z ∈ . zn n=0 Definition 1.2 [12, p. 151] A divergent series ∞ An , zn n=0 in which the sum of the first n + 1termsisSn(z), is said to be an asymptotic expansion of a function f (z) for a given range of values of arg z, if the expression n Rn(z) = z [ f (z) − Sn(z)] satisfies the condition lim Rn(z) = 0 (n fixed), |z|→∞ even though lim |Rn(z)|=∞ (z fixed). n→∞ We denote the fact that the series is the asymptotic expansion of f (z) by writing ∞ A f (z) ∼ n . zn n=0 On the asymptotic expansion of the composite of a function and an asymptotic expansion, we recite the following four results. ( ) = ∞ α n , ( ) Theorem 1.3 [6, p. 540, Theorem 2] If g z n=0 n z is a power series with positive radius r if F x possesses the asymptotic representation ∞ a F(x) ∼ k , xk k=0 and if |a0| < r, then the function (x) = g(F(x)) ∞ , an also possesses an asymptotic representation and this is again calculated exactly as if n=0 xn were convergent, where, since F(x) → a0 as x →∞, and since |a0| < r, the function (x) is obviously defined for every sufficiently large x. 123 Arab. J. Math. (2017) 6:95–108 97 In [6, p. 541], it was stated that, when taking g(z) = ez in Theorem 1.3, we obtain, without any restrictions, 2 ( ) a1 a /2 + a2 eF x ∼ ea0 1 + + 1 +··· . x x2 Theorem 1.4 [9, Remark 3.2] If ∞ a f (x) ∼ k , x →∞, xk k=0 then ∞ ( ) α e f x ∼ k , x →∞, xk k=0 a where α0 = e 0 and k j 1 a0 αk = e ai (1.1) j! j=1 j = , =1 =1 i k 0≤i≤k,1≤≤ j, i∈{0}∪N Theorem 1.5 [1, Lemma 4] Let ∞ a A(x) = n (1.2) xn n=1 be a given asymptotic expansion. Then, the composition B(x) = eA(x) has the asymptotic expansion ∞ b B(x) = n , xn n=0 where b0 = 1 and n 1 b = ka b − , n ∈ N. (1.3) n n k n k k=1 Theorem 1.6 [1, Lemma 5] Let c0 = 0 and ∞ c C(x) = n xn n=0 be a given asymptotic expansion. Then, the composition A(x) = ln C(x) has the expansion ∞ a A(x) = n , xn n=1 where n− 1 1 1 a = c − ka c − , n ∈ N. (1.4) n c n n k n k 0 k=1 123 98 Arab. J. Math. (2017) 6:95–108 In [3], many useful conclusions on asymptotic expansions were obtained. We note that no accurate and explicit references were given in [9] to cite the formula (1.1) in Theorem 1.4. We observe that the constant term did not appear in (1.2) in Theorem 1.5 and that the formulas (1.3)and (1.4) are recurrences. In Sect. 2 of this paper, we will add a constant term into (1.2) and acquire a modified version of Theorem 1.5. In Sect. 3, we will derive from the recurrences (1.3)and(1.4) explicit formulas for the above an and bn.In Sect. 4, we will simply confirm explicit formulas for power series expansions of the exponential and the logarithm of a power series expansion. In Sect. 5, we will apply the explicit formulas for the power series expansions of the exponential and the logarithm of a power series expansion to find explicit formulas for the Bell numbers and logarithmic polynomials extensively studied in combinatorics and number theory. 2 A slightly modified version of Theorem 1.5 Now we are in a position to give a slightly modified version of Theorem 1.5. Theorem 2.1 Let ∞ d D(x) = k xk k=0 be an asymptotic expansion. Then, the function E(x) = eD(x) has the asymptotic expansion ∞ e E(x) = k , xk k=0 where d0 e0 = e (2.1) and k k− 1 1 1 e = de − = (k − )d −e, k ∈ N. (2.2) k k k k k =1 =0 First proof Differentiation yields ( ) E (x) = eD x D (x) = E(x)D (x) which can be written as: ∞ ∞ ∞ e e d (−k) k = k (−k) k xk+1 xk xk+1 k=1 k=0 k=1 ∞ ∞ ∞ k 1 ek dk+1 1 1 = (−k − 1) = (− − 1)d+ e − , x2 xk xk x2 1 k xk k=0 k=0 k=0 =0 that is, ∞ ∞ k−1 ∞ k ek 1 1 k = ( + 1)d+ e −− = de − . xk+1 1 k 1 xk+1 k xk+1 k=1 k=1 =0 k=1 =1 1 Equating coefficients of xk+1 results in (2.2). Taking the limit x →∞on both sides of E(x) = eD(x), we arrive at (2.1). The proof of Theorem 2.1 is complete. 123 Arab. J. Math. (2017) 6:95–108 99 ∞ dk Second proof Set D(x) = d + D (x),whereD (x) = = . Then, by virtue of Theorem 1.5,wehave 0 1 1 k 1 xk ∞ ( ) ( ) βk E(x) = eD x = ed0 eD1 x = ed0 , xk k=0 where β0 = 1and k 1 β = dβ −, k ∈ N. k k k =1 d This means that the Eq. (2.1) is valid and that ek = e 0 βk for k ∈ N. Hence, the sequence ek for k ∈ N satisfies k ek 1 ek− = d , k ∈ N. ed0 k ed0 =1 Consequently, the recurrence relation (2.2) follows. The proof of Theorem 2.1 is complete. 3 Explicit formulas for an and en In this section, we derive explicit formulas for an and en from recurrence relations (1.4)and(2.2). Theorem 3.1 Under the conditions of Theorem 1.6, we have n−1 j cn 1 j cmi an = + (−1) m j , n ∈ N. (3.1) c0 n c0 j=1 j = , i=0 i=0 mi n mi ≥1,0≤i≤ j Proof From the recurrence relation (1.4) and by induction, it follows that n−1 cn 1 1 an = − akkcn−k c0 c0 n k=1 ⎛ ⎞ n−1 k−1 cn 1 1 ⎝ck 1 1 ⎠ = − − a pc − kc − c c n c c k p k p n k 0 0 k=1 0 0 p=1 n−1 n−1 k−1 cn 1 1 1 1 = − c kc − + a pc − c − 2 k n k 2 p k p n k c0 c0 n = c0 n = = k 1 k 1 p ⎛1 ⎞ n−1 n−1 k−1 p−1 cn 1 1 1 1 ⎝cp 1 1 ⎠ = − c kc − + − a qc − pc − c − c c 2 n k n k c 2 n c c p q p q k p n k 0 0 k=1 0 k=1 p=1 0 0 q=1 n−1 n−1 k−1 cn 1 1 1 1 = − c kc − + c pc − c − c c 2 n k n k c 3 n p k p n k 0 0 k=1 0 k=1 p=1 n− k− p−1 1 1 1 1 − a qc − c − c − c 3 n q p q k p n k 0 k=1 p=1 q=1 n−1 n−1 k−1 cn 1 1 1 1 = − c kc − + c pc − c − c c 2 n k n k c 3 n p k p n k 0 0 k=1 0 k=1 p=1 123 100 Arab.