Two Formulas for Successive Derivatives and Their Applications

1 2 Journal of Integer Sequences, Vol. 12 (2009), 3 Article 09.8.2 47 6 23 11

Grzegorz Rz¸adkowski Faculty of Mathematics and Natural Sciences Cardinal Stefan Wyszy´nski University in Warsaw Dewajtis 5 01 - 815 Warsaw Poland [email protected]

Abstract We recall two formulas, due to C. Jordan, for the successive derivatives of functions with an exponential or logarithmic inner function. We apply them to get addition formulas for the Stirling numbers of the second kind and for the Stirling numbers of the ﬁrst kind. Then we show how one can obtain, in a simple way, explicit formulas for the generalized Euler polynomials, generalized Euler numbers, generalized Bernoulli polynomials and the Bell polynomials.

1 Introduction

n By k we mean the Stirling number of the second kind (the number of ways of partitioning a set of n elements into k nonempty subsets; see Graham et al. [4] and sequence A008277 n 0 of Sloane’s On-line Encyclopedia [10]). As usual, we set 0 = 0 if n > 0, 0 = 1, and n = 0 for k > n or k < 0. Let us recall that the Stirling numbers satisfy the identities k k k n 1 k 1 k = (−1)k−j jn = (−1)j (k − j)n, (1) k k! j k! j j=0 j=0 X X n + 1 n n = k + , (2) k k k − 1 and appear in the Taylor expansion ∞ (ew − 1)k n wn = . (3) k! k n! n k X= 1 Peregrino [9] proved the following addition formula for the Stirling numbers of the second kind v v u + v v v − i u = ki(−1)v+i+j . (4) k i j k − j j=0 i=0 X X n By k we denote the Stirling number of the first kind (number of ways of partitioning a set of n elements into k nonempty cycles, see [4], sequence A008275 in [10]). Similarly n = 0 0 if n> 0, 0 = 1, n = 0 for k > n or k < 0. The Stirling numbers of the first kind fulfil the 0 k recurrence formula n + 1 n n = n + . (5) k k k − 1 We use common notation for the falling factorial

(x)k = x(x − 1) ··· (x − k + 1) and for the rising factorial (Pochhammer’s symbol)

x(k) = x(x + 1) ··· (x + k − 1).

The paper is organized as follows. We recall and prove two formulas, due to C. Jordan [6], in section 2. The formulas involve Stirling numbers of both kinds. Since the original Peregrino’s proof of (4), by induction on v, is long we give shorter and more direct proof of his formula and similar formulas in section 3. We prove addition formulas for Stirling numbers of the ﬁrst kind in section 4. Sections 5,6,7 are devoted to show explicit formulas respectively for generalized Euler polynomials, generalized Bernoulli polynomials and the Bell polynomials. All proofs, in the last three sections, are more simple and direct than the proofs which exist in the literature. In the case of generalized Bernoulli polynomials (and generalized Bernoulli numbers) our results seem to be new.

2 Formulas for successive derivatives

We have the following formulas for successive derivatives of composite functions with the exponential, or the logarithmic, inner function.

Lemma 1. If f ∈ C∞(R) then the following formulas for the nth order (n = 1, 2, 3,...) derivatives hold

dn n n (f(et)) = f (k)(et)ekt, (6) dtn k k=1 X n dn 1 n (f(log t)) = (−1)n−k f (k)(log t). (7) dtn tn k k X=1

2 Proof. To prove formula (6), we proceed by induction with respect to n. Denote g(t)= f(et). For n = 1, (6) is obviously true. Let us suppose that for an integer n formula (6) holds. Then using (2) we have

d n n g(n+1)(t)= f (k)(et)ekt dt k k ! X=1 n n = (f (k+1)(et)e(k+1)t + kf (k)(et)ekt) k k X=1 = f ′(et)et + f (n+1)(et)e(n+1)t n n n + k + f (k)(et)ekt k k − 1 k=2 Xn +1 n + 1 = f (k)(et)ekt, k k X=1 which ends the proof of (6). Analogously, using (5), formula (7) can be shown. Formulas (6), (7) are known and can be found, with proofs based on ﬁnite diﬀerences, in the C. Jordan’s book [6, pp. 205–206]. The formulas are given, as exercises without proofs and without direct referencing to [6], also in the L. Comtet’s book [2, Ex. 6, p. 157]. It is easy to see that formula (6) holds, with the same proof, for complex variable t and a holomorphic function f. Formula (7) holds for complex t, a branch of logarithm and a holomorphic function f. For example if f(x)= xm, g(t)= emt then using (6) we get

n n g(n)(t)= m(m − 1) ··· (m − k + 1)e(m−k)tekt. (8) k k X=1 From the other side g(n)(t)= mnemt, (9) and comparing (8) with (9) we obtain the well-known generating function for the Stirling numbers of the second kind n n (m) = mn. k k k X=1 3 Addition formulas for Stirling numbers of the second kind

Let us substitute f(x) = exp(x) in (6). We have

u+v t u + v t (ee )(u+v) = ee · ekt. (10) k k X=1 3 From the other side u t u t (ee )(u) = ee · ent n n=1 X and

u t t u t (ee )(u+v) = [(ee )(u)](v) = (ee · ent)(v) n n=1 u v Xm u v m t = ( ee · eit) · nv−ment n m i n=1 m=0 i=0 Xu Xv Xm u v m t = ( ee · e(i+n)tnv−m). (11) n m i n=1 m=0 i=0 X X X Theorem 2. The following addition formula for the Stirling numbers of the second kind holds. k v u+v u v m = nv−m. (12) k n m k − n n=1 m k−n X X= t Proof. Formula (12) follows by comparing the coeﬃcients of ee ekt in (10) and (11) for i + n = k. Denoting in (12) m = v − i, (i = 0, 1, 2,...,v − k + n) we have

k v−k n u+v u + v v − i = ni, k n i k − n n=1 i=0 X X and then letting n = k − j, (j = 0, 1, 2,...,k − 1) we obtain

k− v−j u+v 1 v v − i u = (k − j)i . (13) k i j k − j j=0 i=0 X X Formula (4) follows easily from (13). To see this let us observe that the range of the variables i, j in (13) can be changed to the same range as in (4) i.e., i, j = 0, 1, 2,...,v. This is allowed u v−i because the number k−j is zero if j ≥ k and j equals zero for j>v −i i.e., if i+j>v. − We rearrange (13) into (4) using formulas (1), for the symbol v i , respectively in the start j

4 and end of the following calculation:

v v u+v v v − i u = (k − j)i k i j k − j j=0 i=0 X X v v j u v 1 j = (k − j)i (−1)j−l lv−i k − j i j! l j=0 i=0 l X X X=0 v j v u 1 j v = (−1)j−l (k − j)ilv−i k − j j! l i j=0 l i=0 X X=0 X v j u 1 j = (−1)j−l (k − j + l)v k − j j! l j=0 l X X=0 v j v u 1 j v = (−1)j−l (−1)v−i ki(j − l)v−i k − j j! l i j=0 l i=0 X X=0 X v v j u v 1 j = ki(−1)v+i+j (−1)l (j − l)v−i k − j i j! l j=0 i=0 l=0 Xv v X X v v − i u = ki(−1)v+i+j . i j k − j j=0 i=0 X X

4 Addition formulas for Stirling numbers of the ﬁrst kind

Let us substitute f(x) = log x in (7). We have (t> 1)

u v (−1)u+v+1 + u + v (k − 1)! (log log t)(u+v) = . (14) tu+v k k k log t X=1 From the other side u (−1)u+1 u (n − 1)! (log log t)(u) = tu n logn t n=1 X

5 and by using the Leibniz formula and formula (7) for g(t) = (log t)−n we get (log log t)(u+v) = [(log log t)(u)](v) u u 1 1 (v) =(−1)u+1 (n − 1)! · n logn t tu n=1 Xu v − u v 1 (m) 1 (v m) =(−1)u+1 (n − 1)! n m logn t tu n=1 m=1 Xu Xv u v (−u) − =(−1)u+1 (n − 1)! v m n m tu+v−m n=1 m=1 m X X 1 m (−n) × (−1)m−i i . (15) tm i (log t)n+i i=0 X Theorem 3. The following addition formula for the Stirling numbers of the ﬁrst kind holds.

v v u + v v v − i u = u(i) . (16) k i j k − j j=0 i=0 X X Proof. By comparing the coeﬃcients of (tu+v logk t)−1 in (14) and (15) for i+n = k, i = k−n we get k u + v v u (k − 1)! = (−1) (n − 1)! (−n) − k n k n n=1 v X v m−k+n m × (−u) − (−1) m v m k − n m k−n X= and then using the identities

(n − 1)!(−n)k−n k−n v−m (v−m) =(−1) , (−u) − (−1) = u (k − 1)! v m we obtain the formula k v u + v u v m = u(v−m) . (17) k n m k − n n=1 m k−n X X= By the same manner as for the Stirling numbers of the second kind, formula (17) can be rearranged to the form (16).

5 Generalized Euler polynomials

µ The generalized Euler polynomials En (z) of degree n = 0, 1, 2,..., complex order µ and complex argument z (see Nörlund [8]) can be defined by the generating function ∞ Eµ(z) 2µewz n wn = , |w| < π. n! (ew + 1)µ n=0 X 6 The generalized Euler polynomials play an important role in the calculus of finite differences. µ We will show, in a very easy way, an explicit formula for the Euler polynomial En (z). By (6) we get m dm 1 m ekw = (−µ) dwm (ew + 1)µ k k (ew + 1)µ+k k X=1 and then by the Leibniz formula

n m dn 2µewz znewz n m ekwewzzn−m = 2µ + (−µ) . dwn (ew + 1)µ (ew + 1)µ m k k (ew + 1)µ+k m=1 k ! X X=1 Thus n m dn 2µewz n m (−µ) Eµ(z)= = zn + zn−m k . (18) n dwn (ew + 1)µ m k 2k w=0 m=1 k=1 X X

Another approach to formula (18) is presented by Howard [5]. Putting in (18) z = 0 we µ obtain the following explicit formula (see Todorov [12]) for the generalized Euler number En

n n (−µ) Eµ = Eµ(0) = k . (19) n n k 2k k X=1 Luo [7] deals with the Apostol–Euler polynomials, which are a further generalization of µ the polynomials {En (z)}. Formula (19) coincides, as a particular case, with formula (32) of this paper.

6 Generalized Bernoulli polynomials

µ The generalized Bernoulli polynomials Bn (z) of degree n = 0, 1, 2,..., complex order µ and complex argument z (see N¨orlund [8] ) can be deﬁned by the generating function

∞ Bµ(z) wµewz n wn = , |w| < 2π. n! (ew − 1)µ n=0 X We will show that by applying formula (6) one can obtain explicit formulas for the polyno- µ µ mials and then for the generalized Bernoulli numbers Bn = Bn (0). Our approach can be seen as consistent with the spirit of the paper of Gould [3]. Using (6) and the Leibniz formula we get successively

j dj 1 j ekw = (−µ) , dwj (ew − 1)µ k k (ew − 1)µ+k k X=0

m µ m j kw d w m j (−µ)ke µ−m+j = (µ) − w (20) dwm (ew − 1)µ j k (ew − 1)µ+k m j j=0 k X X=0

7 and dn wµewz wµznewz n n dm wµ = + zn−mewz. (21) dwn (ew − 1)µ (ew − 1)µ m dwm (ew − 1)µ m=1 X Rewriting the right hand side of (20) into the form

µ+m m j w 1 m j kw w m−k j (µ) − (−µ) e (e − 1) w , (ew − 1)µ+m w2m j m j k k j=0 k ! X X=0 using (3) in the expression ekw 1 k wjekw(ew − 1)m−k = wj(ew − 1)m = wj(ew − 1)m 1+ (ew − 1)k ew −1 k k k 1 k = wj(ew − 1)m = wj (ew − 1)m−l l (ew − 1)l l l=0 l=0 X∞ X k k i wi = wj (m − l)! , l m − l i! l i m−l X=0 =X and grouping terms of power w2m we get j dm wµ m m j lim = (µ)m−j (−µ)k w→0 dwm (ew − 1)µ j k j=0 k X X=0 k k 2m − j (m − l)! × . l m − l (2m − j)! l X=0 Thus by (21) we obtain the following explicit formula for the generalized Bernoulli polynomials n µ wz n m µ d w e n n n−m m B (z) = | = z + z (µ) − n dwn (ew − 1)µ w=0 m j m j m=1 j=0 X X j k j k 2m − j (m − l)! × (−µ) . (22) k k l m − l (2m − j)! k l X=0 X=0 Comparing it with the formula given by Srivastava and Todorov [11, Eq. (3), p. 510], we see that formula (22) does not involve any hypergeometric function. In particular, for µ = 1 we obtain the common Bernoulli polynomials n 1 n B (z) = zn − zn−1 + zn−m n 2 m m=2 m− X k 1 m − 1 k m + 1 (m − l)! × m (−1)kk! k l m − l (m + 1)! k=1 l=0 mX k X m k m (m − l)! + (−1)kk! k l m − l m! k l ! X=1 X=0 8 and putting z = 0, the common Bernoulli numbers

n− k 1 n − 1 k n + 1 (n − l)! B = n (−1)kk! n k l n − l (n + 1)! k=1 l=0 Xn k X n k n (n − l)! + (−1)kk! . (23) k l n − l n! k l X=1 X=0 Applying, to the expression (23), the two identities

k n−k k n + 1 n − k (n − l)! = (−1)j(n − j)n+1, l n − l j l j=0 X=0 X k n−k k n n − k (n − l)! = (−1)j(n − j)n, l n − l j l j=0 X=0 X and then the next two

n− j 1 n−1 n−k n+1 (−1)kk! =(−1)j+n+1 (−1)k(j+1−k)n−1, k j k k k X=1 X=0

n j n n − k n + 1 (−1)kk! = (−1)j+n (−1)k(j + 1 − k)n k j k k k X=1 X=0 n = (−1)j+n , j

n where j are Eulerian numbers (see [4], and A008292 in [10]), we obtain the following formulaD forE the nth Bernoulli number

n− j (−1)n 1 n + 1 B = n (n − j)n+1 (−1)k+1 (j + 1 − k)n−1 n (n + 1)! k " j=0 k=0 X − X n 1 n +(n + 1) (n − j)n . j j=0 # X 7 Bell polynomials

The Bell polynomials Bn(z) can be deﬁned by the generating function (see Bell [1])

∞ B (z) w n wn = e(e −1)z. (24) n! n=0 X

9 Using (6) we compute the nth derivative of the right hand side of (24)

n n d w n w e(e −1)z = e(e −1)zekwzk. (25) dwn k k X=1 The well known explicit formula for Bn(z)

n n B (z)= zk, n k k X=1 follows immediately from (25) by putting w = 0.

8 Acknowledgements

I would like to thank M. Skwarczy´nski, J. A. Rempa la and K. Jezuita for valuable discussions during preparation of the paper and the anonymous referee for his/her helpful comments.

References

[1] E. T. Bell, Exponential polynomials, Ann. Math. 35 (1934), 258–277.

[2] L. Comtet, Advanced Combinatorics, D. Reidel Publishing Co., 1974.

[3] H. W. Gould, Explicit formula for Bernoulli numbers, Amer. Math. Monthly 78 (1972), 44–51.

[4] R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. A Foundation for Computer Science, Addison-Wesley Company, Inc., 1994.

[5] F. T. Howard, A theorem relating Potential and Bell polynomials, Discrete Math. 39 (1982), 129–143.

[6] C. Jordan, Calculus of Finite Diﬀerences, Chelsea P. Company, 1950.

[7] Qiu-Ming Luo, Apostol-Euler polynomials of higher order and Gaussian hypergeometric functions, Taiwanese J. Math. 10 (2006), 917–925.

[8] N. E. Nörlund, Vorlesungen über Differenzrechnung, Springer, 1924.

[9] R. S. Peregrino, The Lucas congruence for Stirling numbers of the second kind, Acta Arith. 94 (2000), 41–52.

[10] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, 2009. http://www.research.att.com/~njas/sequences/.

[11] H. M. Srivastava, P. G. Todorov, An explicit formula for the generalized Bernoulli polynomials, J. Math. Anal. Appl., 130 (1988), 509–513.

10 [12] P. G. Todorov, Explicit and recurrence formulas for generalized Euler numbers, Funct. Approx. Comment. Math. 22 (1993), 13–17 (1994).

2010 Mathematics Subject Classiﬁcation: Primary 11B73; Secondary 11B68. Keywords: Stirling numbers, addition formulas, generalized Euler polynomials, generalized Bernoulli polynomials.

(Concerned with sequences A008275, A008277 and A008292.)

Received May 8 2009; revised version received November 12 2009. Published in Journal of Integer Sequences, November 16 2009.

Return to Journal of Integer Sequences home page.