Expansions of the Exponential and the Logarithm of Power Series and Applications

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. J. Math. (2017) 6:95–108

⎛ ⎞ n−1 k−1 p−1 q−1 1 1 ⎝cq 1 1 ⎠ − − a tc − qc − c − c − c 3 n c c q t q t p q k p n k 0 k=1 p=1 q=1 0 0 t=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 n− k− p−1 q−1 1 1 1 1  1 1 1 1   − c qc − c − c − + a tc − c − c − c − c 4 n q p q k p n k c 4 n t q t p q k p n k 0 k=1 p=1 q=1 0 k=1 p=1 q=1 t=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=2 p=1 n− k− p−1 n− k− p−1 q−1 1 1 1 1  1 1 1 1   − c qc − c − c − + a tc − c − c − c − c 4 n q p q k p n k c 4 n t q t p q k p n k 0 k=3 p=2 q=1 0 k=4 p=3 q=2 t=1

 −  − − − n−2 j n−1 1 1 j1 1 j 1 cn 1 (−1) = + ··· c  c − c − j+1 j j n 1 i i+1 c0 n c0 j=1 1= j 2= j−1  j =1 i=1 n−  −1  − −1 n− 1 (−1)n−1 1 1 n2 2 + ··· a  − c − c − n−1 n−1 n 1 n 1 i i+1 n c0 1=n−1 2=n−2 n−1=1 i=1  −  − − − n−2 j n−1 1 1 j1 1 j 1 cn 1 (−1) = + ··· c  c − c − j+1 j j n 1 i i+1 c0 n c0 j=1 1= j 2= j−1  j =1 i=1 n−  −1  − −1 n− (−1)n−1 1 1 1 n2 2 + ··· a  − c − c − n−1 1 n 1 n 1 i i+1 c0 n 1=n−1 2=n−2 n−1=1 i=1  −  − − − n−2 j n−1 1 1 j1 1 j 1 cn 1 (−1) = + ··· c  c − c − j+1 j j n 1 i i+1 c0 n c0 j=1 1= j 2= j−1  j =1 i=1 n−  −1  − −1 n− (−1)n−1 1 1 1 n2 2 + ···  − −  − n c1 n 1cn 1 c i i+1 c0 n 1=n−1 2=n−2 n−1=1 i=1  −  − − − n−2 j n−1 1 1 j1 1 j 1 cn 1 (−1) = + ··· c  c − c − j+1 j j n 1 i i+1 c0 n c0 j=1 1= j 2= j−1  j =1 i=1 n−  −1  − −1 n− (−1)n−1 1 1 1 n2 2 + ···   − −  − n c n−1 n 1cn 1 c i i+1 c0 n 1=n−1 2=n−2 n−1=1 i=1  −  − − − n−1 j n−1 1 1 j1 1 j 1 cn 1 (−1) = + ··· c  c − c − j+1 j j n 1 i i+1 c0 n c0 j=1 1= j 2= j−1  j =1 i=1 − n−1 j  j 1 cn 1 (−1) = +  c c − c − . j+1 j j n 1 i i+1 c0 n c0 j=1 j≤1≤n−1,2≤k≤ j, i=1 j−k+1≤k ≤k−1−1

123 Arab. J. Math. (2017) 6:95–108 101

Let m0 = n − 1, m j =  j ,andmi = i − i+1 for 1 ≤ i ≤ j − 1. Then

j k−1 mi = n,k = n − mi , 1 ≤ k ≤ j, i=0 i=0 and

n− j c 1 1 (−1) j   a = n + m c n j+1 j mi c0 n c0  j=1 j = , i=0 i=0 mi n mi ≥1,0≤i≤ j n−1  j cn 1 j cmi = + (−1) m j . c0 n  c0 j=1 j = , i=0 i=0 mi n mi ≥1,0≤i≤ j The proof of Theorem 3.1 is complete. Theorem 3.2 Under the conditions of Theorem 2.1, we have ⎛ ⎞

⎜ n−1  j ⎟ ⎜ mi dm ⎟ = d0 ⎜ +  i ⎟ , ∈ N. en e dn i− n (3.2) ⎝  n − 1 m ⎠ j=1 j = , i=0 q=0 q i=0 mi n mi ≥1,0≤i≤ j

Proof From the recurrence relation (2.2) and by induction, it follows that

n− 1 1 e − e d = e (n − k)d − n 0 n n k n k k=1 ⎡ ⎤ n−1 k−1 1 ⎣ 1 ⎦ = e d + e (k − p)d − (n − k)d − n 0 k k p k p n k k=1 p=1 n−1 n−1 k−1 e0 1 1 = dk(n − k)dn−k + ep(k − p)dk−p(n − k)dn−k n = n = k = k 1 k 1 p 1⎡ ⎤ n−1 n−1 k−1 p−1 e0 1 1 ⎣ 1 ⎦ = d (n − k)d − + e d + e (p − q)d − (k − p)d − (n−k)d − n k n k n k 0 p p q p q k p n k k=1 k=1 p=1 q=1 n−1 n−1 k−1 e0 e0 1 = d (n − k)d − + d (k − p)d − (n − k)d − n k n k n k p k p n k k=1 k=1 p=1 n− k− p−1 1 1 1 1 1  + e (p − q)d − (k − p)d − (n − k)d − n k p q p q k p n k k=1 p=1 q=1 n−1 n−1 k−1 e0 e0 1 = dk(n − k)dn−k + dp(k − p)dk−p(n − k)dn−k n = n = k = k 1 ⎡ k 1 p 1 ⎤ n−1 k−1 p−1 q−1 1 1 1 ⎣ 1 ⎦ + e d + e (q − t)d − n k p 0 q q t q t k=1 p=1 q=1 t=1

× (p − q)dp−q (k − p)dk−p(n − k)dn−k 123 102 Arab. J. Math. (2017) 6:95–108

n−1 n−1 k−1 e0 e0 1 = d (n − k)d − + d (k − p)d − (n − k)d − n k n k n k p k p n k k=1 k=1 p=1 n−1 k−1 p−1 e0 1 1 + d (p − q)d − (k − p)d − (n − k)d − n k p q p q k p n k k=1 p=1 q=1 n− k− p−1 q−1 1 1 1 1 1  1  + e (q − t)d − (p − q)d − (k − p)d − (n − k)d − n k p q t q t p q k p n k k=1 p=1 q=1 t=1 n−1 n−1 k−1 e0 e0 1 = d (n − k)d − + d (k − p)d − (n − k)d − n k n k n k p k p n k k=1 k=2 p=1 n−1 k−1 p−1 e0 1 1 + d (p − q)d − (k − p)d − (n − k)d − n k p q p q k p n k k=3 p=2 q=1 n− k− p−1 q−1 1 1 1 1 1  1  + e (q − t)d − (p − q)d − (k − p)d − (n − k)d − n k p q t q t p q k p n k k=4 p=3 q=2 t=1  −  − n−2 n−1 1−1 j−2 1 j−1 1 = e0 1 1 ··· 1 ( −  ) d j n 1 dn−1 n 1 2  j−1 j=1 1= j 2= j−1  j−1=2  j =1 − j 1 n−1 1−1 × ( −  ) + 1 1 1 i i+1 di −i+1 n 1 2 i=1 1=n−1 2=n−2  −  − n−3 1 n−2 1 n−2 ··· 1 ( −  ) ( −  ) en−1 n 1 dn−1 i i+1 di −i+1 n−2 n−2=2 n−1=1 i=1  −  − n−2 n−1 1−1 j−2 1 j−1 1 = e0 1 1 ··· 1 ( −  ) d j n 1 dn−1 n 1 2  j−1 j=1 1= j 2= j−1  j−1=2  j =1 − j 1 n−1 1−1 × ( −  ) + 1 1 1 i i+1 di −i+1 n 1 2 i=1 1=n−1 2=n−2  −  − n−3 1 n−2 1 n−2 ··· 1 ( −  ) ( −  ) e1 n 1 dn−1 i i+1 di −i+1 n−2 n−2=2 n−1=1 i=1  −  − n−2 n−1 1−1 j−2 1 j−1 1 = e0 1 1 ··· 1 ( −  ) d j n 1 dn−1 n 1 2  j−1 j=1 1= j 2= j−1  j−1=2  j =1 − j 1 n−1 1−1 × ( −  ) + 1 1 1 i i+1 di −i+1 n 1 2 i=1 1=n−1 2=n−2  −  − n−3 1 n−2 1 n−2 ··· 1 ( −  ) ( −  ) e0d1 n 1 dn−1 i i+1 di −i+1 n−2 n−2=2 n−1=1 i=1  −  − n−2 n−1 1−1 j−2 1 j−1 1 = e0 1 1 ··· 1 ( −  ) d j n 1 dn−1 n 1 2  j−1 j=1 1= j 2= j−1  j−1=2  j =1

123 Arab. J. Math. (2017) 6:95–108 103

− j 1 n−1 1−1 × ( −  ) + e0 1 1 i i+1 di −i+1 n 1 2 i=1 1=n−1 2=n−2  −  − n−3 1 n−2 1 n−2 ··· 1 ( −  ) ( −  ) dn−1 n 1 dn−1 i i+1 di −i+1 n−2 n−2=2 n−1=1 i=1  −  − − n−1 n−1 1−1 j−2 1 j−1 1 j 1 = e0 1 1 ··· 1 ( −  ) ( −  ) d j n 1 dn−1 i i+1 di −i+1 n 1 2  j−1 j=1  = j  = j−1  − =2  =1 i=1 1 2 ⎛ j 1 ⎞ j − j j−1 d0 n 1    = e ⎝ 1 ⎠  ( −  ) ( −  ) . j d j n 1 dn−1 i i+1 di −i+1 n i j=1 j≤1≤n−1,2≤k≤ j, i=1 i=1 j−k+1≤k ≤k−1−1

Let m0 = n − 1, m j =  j ,andmi = i − i+1 for 1 ≤ i ≤ j − 1. Then

j k−1 mi = n,k = n − mi , 1 ≤ k ≤ j, i=0 i=0 and

n−1 j j ed0    1  = d0 +  en e dn i−1 mi dmi n = − ≥ ≥ , ≤ ≤ , = n − = mq = j 1 n j m0 1 2 k j i 1 q 0 i 0 k−2 n− j+k−1− m ≥m − ≥1,  i=0 i k 1 j = i=0 mi n n−1  j mi dm = d0 + d0  i e dn e i−1 = ≤ ≤ − , ≤ ≤ , = n − = mq j 1 1 m0 n j 2 kj i 0 q 0 k−2 1≤m − ≤n− j+k−1− m , k 1  i=0 i j m =n ⎛ i=0 i ⎞

⎜ n−1  j ⎟ ⎜ mi dm ⎟ = d0 ⎜ +  i ⎟ . e dn i− ⎝  n − 1 m ⎠ j=1 j = , i=0 q=0 q i=0 mi n mi ≥1,0≤i≤ j The proof of Theorem 3.2 is complete.

4 Expansions of the exponential and logarithm of power series

By similar arguments as in the proofs of Theorems 3.1 and 3.2, we can obtain the following power series expansions of the exponential and the logarithm of a power series expansion. For simplicity, we do not write down their proofs in details. Theorem 4.1 Let ∞ k D(x) = dk x k=0 be a power series expansion. Then, the function E(x) = eD(x) has the power series expansion ∞ k E(x) = ek x , k=0 123 104 Arab. J. Math. (2017) 6:95–108

where the coefficients ek for k ∈{0}∪N satisfy the formulas (2.1), (2.2), (3.2), and

k j  1   d0 ek = e di , k ∈ N. (4.1) j!  j=1 j = , =1 =1 i k i≥1,1≤≤ j

Theorem 4.2 Let c0 = 0 and ∞ n C(x) = cn x n=0 be a given power series expansion. Then, the composition A(x) = ln C(x) has the expansion ∞ n A(x) = an x , n=1 where the coefficients an for n ∈ N satisfy (1.4) and (3.1).

5 Applications of Theorems 4.1 and 4.2

In this section, we will apply Theorems 4.1 and 4.2, respectively, to the Bell numbers and logarithmic poly- nomials which are extensively studied in combinatorics and number theory.

5.1 An application of Theorems 4.1 to the Bell numbers

In combinatorics, Bell numbers, usually denoted by Bn for n ∈{0}∪N, count the number of ways a set with n elements can be partitioned into disjoint and nonempty subsets. Every Bell number Bn can be generated by

∞ k x − x ee 1 = B . k k! k=0

The first few Bell numbers Bn are

B0 = 1, B1 = 1, B2 = 2, B3 = 5, B4 = 15, (5.1) B5 = 52, B6 = 203, B7 = 877, B8 = 4140, B9 = 21147. For more detailed information, refer to [2, pp. 210–212, Section 5.4].

Theorem 5.1 The Bell numbers Bk for k ∈ N can be computed by   k−1 − = k 1 , Bk  B (5.2) =0 k− 1  1 = + !     , Bk 1 k j − (5.3)  (m − )! k − i 1 m j=1 j = , i=0 i 1 q=0 q i=0 mi k mi ≥1,0≤i≤ j and

k  1  1 B = k!  . (5.4) k ! j j  i! j=1 j = , =1 =1 i k i≥1,1≤≤ j

123 Arab. J. Math. (2017) 6:95–108 105

 ( ) = x − = ∞ xk Proof Applying Theorem 4.1 to D x e 1 k=1 k! yields

1 B d = 0, d = , and e = k 0 k k! k k! for k ∈ N.From(2.2) in Theorem 4.1, it follows that

k k−1 B 1 1 B − 1 1 B k = k = , k ∈ N. k! k ( − 1)! (k − )! k (k −  − 1)! ! =1 =0

This is equivalent to

k k−1 1 Bk− 1 B Bk = (k − 1)! = (k − 1)! ( − 1)! (k − )! (k −  − 1)! ! =1 =0 k   k−    k − 1 1 k − 1 = Bk− = B  − 1  =1 =0 for k ∈ N. d From (2.1)and(3.2) in Theorem 4.1, it follows that B0 = 0!e0 = 0!e 0 = 1and

= ! Bk k e⎧k ⎫ ⎪ ⎪ ⎨⎪ k− ⎬⎪ 1 1  1 = k! +      ⎪ j i− ⎪ ⎪k!  (m − 1)! k − 1 m ⎪ ⎩⎪ j=1 j = , i=0 i q=0 q ⎭⎪ i=0 mi k mi ≥1,0≤i≤ j k− 1  1 = + !      1 k j −  (m − )! k − i 1 m j=1 j = , i=0 i 1 q=0 q i=0 mi k mi ≥1,0≤i≤ j for k ∈ N. The formula (5.3) follows. From (4.1) in Theorem 4.1, it follows that

k j k B  1   1  1  1 k = e0 =  , k ∈ N. ! ! ! ! j k j  i j  i! j=1 j = =1 j=1 j = =1 =1 i k =1 i k

This can be easily rewritten as (5.4). The required proof is complete.

5.2 An application of Theorem 4.2 to logarithmic polynomials

According to the monograph [2, pp. 140–141], the logarithmic polynomials Ln can be defined by   ∞ ∞  tn  tn ln g = L , (5.5) n n! n n! n=0 n=1

(n) where g0 = G(a) = 1, gn = G (a) for n ∈ N,andG(x) is an infinitely differentiable function at x = a, and they are expressions for the nth derivative of ln G(x) at the point x = a. 123 106 Arab. J. Math. (2017) 6:95–108

Theorem 5.2 For n ∈ N, the logarithmic polynomials Ln can be computed by

n−   1 n − 1 Ln = gn − gn−kLk (5.6) k − 1 k=1 and n−1  j j gmi Ln = gn + (n − 1)! (−1) m j . (5.7)  mi ! j=1 j = , i=0 i=0 mi n mi ≥1,0≤i≤ j Proof Applying Theorem 4.2 to the Eq. (5.5)gives g L c = g = 1, c = n , a = n , n ∈ N. 0 0 n n! n n! Substituting these quantities into (1.4)and(3.1) produces n−1 n−1 Ln gn 1 Lk gn−k 1 (n − 1)! = − k = gn − gn−kLk n! n! n k! (n − k)! n! (k − 1)!(n − k)! k=1 k=1 and n−1  j Ln gn 1 j gmi = + (−1) m j . n! n! n  mi ! j=1 j = , i=0 i=0 mi n mi ≥1,0≤i≤ j Hence, the formulas (5.6)and(5.7) follow immediately. The proof of Theorem 5.2 is complete.

6 Remarks

Remark 6.1 When d0 = 0, Theorem 2.1 becomes [1, Lemma 4]. Therefore, our Theorem 2.1 is a slight generalization of [1, Lemma 4]. The constant term d0 does not appear in the recursion formula (2.2), but it has a relation with the constant term e0. On the other hand, the second proof of Theorem 2.1 reveals that Theorems 1.5 and 2.1 are equivalent to each other. However, it is clear that Theorem 2.1 can be used more conveniently. Remark 6.2 The formulas (3.2)and(4.1) are not that different. The difference is that in (4.1)wehavethe ··· products di1 di2 di whereas in (3.2) we have these products with weight factors. But if we group terms which are obtained by permutation of the indices, then these weights add up to the expected value. For example, if we want to find the coefficient of z6 in   ∞ 3 k dk z , k=1 then d1d2d3 appears six times which makes the total contribution 6d1d2d3. In the formula (3.2), these six terms are written as:

 2 m d  i mi , 6 2 i=0 q=i mq where the sum is over the six permutations of (1, 2, 3) for (m0, m1, m2). The total contribution of d1d2d3 is the same as before. Therefore, one could say that the formula (3.2) is not that different from (4.1). Since the proofs of the formulas (5.3)and(5.4) based on the formulas (3.2)and(4.1), the formula (5.3)is not that different from (5.4)yet.

123 Arab. J. Math. (2017) 6:95–108 107

Remark 6.3 Letting k = 3, 4in(5.4), respectively, gives

3 1  1 B3 = 3!  j! j ! =  j = i j 1 i=3 1 ⎛ =1 ⎞ ⎜ 1  1 1  1 1  1 ⎟ = 3! ⎝  +  +  ⎠ 1!  1 ! 2!  2 ! 3!  3 ! 1 = i 2 = i 3 = i i=3 1 i=3 1 i=3 1 ! =1   =1 " =1 1 1 1 1 1 1 = 3! + + + 3! 2! 1!2! 2!1! 3! 1!1!1! = 5 and 4 1  1 B4 = 4!  j! j ! =  j = i j 1 i=4 1 ⎛ =1 ⎜ 1  1 1  1 = 4! ⎝  +  1!  1 ! 2!  2 ! 1 = i 2 = i i=4 1 i=4 1 =1 =1 ⎞ 1  1 1  1 ⎟ +  +  ⎠ 3!  3 ! 4!  4 ! 3 = i 4 = i i=4 1 i=4 1 ! =1  =1   " 1 1 1 1 1 1 1 1 1 1 1 = 4! + + + + + + + 4! 2! 1!3! 2!2! 3!1! 3! 1!1!2! 1!2!1! 2!1!1! 4! 1!1!1!1! = 15. These two concrete results are the same as corresponding ones listed in (5.1). Remark 6.4 Finding various expressions for a mathematical quantity is meaningful and significant in mathe- matics. For example, mathematicians find several infinite product expressions for π. From this point of view, the formulas (5.3)and(5.4) are all useful somewhat.

Remark 6.5 In [2, pp. 140–141, Theorem A], it was given that the logarithmic polynomials Ln equal n k−1 Ln = Ln(g1, g2,...,gn) = (−1) (k − 1)!Bn,k(g1, g2,...,gn−k+1), k=1 where Bn,k denotes the Bell polynomials of the second kind, which are defined by − +  n k 1# $ n! xi i B , (x , x ,...,x − + ) =  n k 1 2 n k 1 n−k+1 ! = i ! i 1≤i≤n,i ∈{0}∪N, i 1 i=1 n  = , i=1 i i n n  = i=1 i k for n ≥ k ≥ 0. This is different from the formulas (5.6)and(5.7). Remark 6.6 The recurrence (5.2) is a recovery of a known result listed in [4, p. 373, Excercise 15]. However, the explicit formula (5.3)isnew.

Remark 6.7 The Bell numbers Bk are connected with the Lah numbers L(n, k) and the Stirling numbers of the second kind S(n, k). See, for example, the papers [5,7,8] and closely related references therein. Importantly, some inequalities for determinants and products of the Bell numbers Bk were established in [8]. Remark 6.8 This paper is a revised version of the preprint [10].

123 108 Arab. J. Math. (2017) 6:95–108

Acknowledgements The authors appreciate the anonymous referees for their careful corrections and valuable comments on the original version of this paper. Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http:// creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

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