Applied Mathematical Sciences, vol. 8, 2014, no. 149, 7423 - 7430 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ams.2014.49727 On New Polynomials Related to Bernoulli Polynomials and Stirling Polynomials C. S. Ryoo Department of Mathematics, Hannam University, Daejeon 306-791, Korea Copyright c 2014 C. S. Ryoo. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract In this paper, we observe an interesting phenomenon of `scattering' of the zeros of the polynomials SRn(x) in complex plane Mathematics Subject Classification: 11B68, 11S40, 11S80 Keywords: Bernoulli numbers, Bernoulli polynomials, Stirling polynomi- als, complex roots 1 Introduction Bernoulli numbers and polynomials, Euler numbers and polynomials, Genocchi numbers and polynomials, Tangent numbers and polynomials, and Stirling numbers and polynomials possess many interesting properties and arising in many areas of mathematics, mathematical physics and statistical physics. It is the aim of this paper to investigate the roots of the polynomials SRn(x). We also carried out computer experiments for doing demonstrate a remarkably regular structure of the complex roots of the Stirling polynomials Sn(x). Throughout this paper, we always make use of the following notations:N = f1; 2; 3; · · · g denotes the set of natural numbers, N0 = f0; 1; 2; 3; · · · g denotes the set of nonnegative integers, Z denotes the set of integers, R denotes the set of real numbers, C denotes the set of complex numbers. 7424 C. S. Ryoo For a real or complex parameter x, the Stirling polynomials are defined by the following generating function 1 x+1 X tn t S (x) = (jtj < 2π; 1x := 1); see [1]: n n! 1 − e−t n=0 The polynomials SRn(x) and numbers SRn are defined by the following generating functions 1 X tn t SR (x) = ext; jtj < 2π; (1:1) n n! 1 − e−t n=0 1 X tn t SR = ; jtj < 2π; (1:2) n n! 1 − e−t n=0 respectively. Clearly, we obtain SRn(x) = Bn(x + 1) and SRn = Sn(0) = Bn(1) for n 2 N0; in terms of the classical Bernoulli polynomials Bn(x) and Bernoulli numbers Bn. For n 2 N, numbers SRn meet SR2n+1 = 0. By using computer, the numbers SRn can be determined explicitly. The first few BRn are listed in Table 1. Table 1. The first few numbers SRn degree n 0 1 2 3 4 5 6 7 8 9 10 1 1 1 1 1 5 SR 1 0 − 0 0 − 0 n 2 6 30 42 30 66 By the above definition, we obtain 1 1 1 X tl t X tn X tm BR (x) = ext = SR xm l l! 1 − e−t n n! m! l=0 n=0 m=0 1 l ! 1 l ! X X tn tl−n X X l tl = SR xl−n = SR xl−n : n n! (l − n)! n n l! l=0 n=0 l=0 n=0 tl By using comparing coefficients , we have the following theorem. l! Theorem 1.1 For n 2 N0, one has n X n SR (x) = SR xn−l: n l l l=0 Polynomials related to Bernoulli polynomials and Stirling polynomials 7425 By Theorem 1.1, after some elementary calculations, we have n Z b X n Z b SR (x)dx = SR xn−ldx n l l a l=0 a n n−l+1 b X n x = SRl l n − l + 1 l=0 a n+1 1 X n + 1 n−l+1 b = SRl x : n + 1 l a l=0 Hence we get Z b SRn+1(b) − SRn+1(a) SRn(x)dx = : (1:3) a n + 1 Since SRn(0) = SRn, by (1.3), we have the following theorem. Theorem 1.2 For n 2 N, one has Z x SRn(x) = SRn + n SRn−1(t)dt: 0 Then, it is easy to deduce that SRn(x) are polynomials of degree n. By using computer, the polynomials SRn(x) can be determined explicitly. Here is the list of the first Stirling SRn(x)'s polynomials. 1 SR (x) = x + ; 1 2 1 SR (x) = x2 + x + ; 2 6 3x2 x SR (x) = x3 + + ; 3 2 2 1 SR (x) = x4 + 2x3 + x2 − ; 4 30 5x4 5x3 x SR (x) = x5 + + − ; 5 2 3 6 5x4 x2 1 SR (x) = x6 + 3x5 + − + ; 6 2 2 42 7x6 7x5 7x3 x SR (x) = x7 + + − + ; 7 2 2 6 6 14x6 7x4 2x2 1 SR (x) = x8 + 4x7 + − + − : 8 3 3 3 30 7426 C. S. Ryoo 2 The zeros of the polynomials SRn(x) In this section, an interesting phenomenon of scattering of zeros of SRn(x) is observed. We investigate the beautiful zeros of the SRn(x) by using a computer. We plot the zeros of SRn(x) for x 2 C(Figure 1). In Figure 1(top- left), we choose n = 25. In Figure 1(top-right), we choose n = 40. In Figure 1(bottom-left), we choose n = 55: In Figure 1(bottom-right), we choose n = 70. 10 10 7.5 7.5 5 5 2.5 2.5 ImHwL 0 ImHwL 0 -2.5 -2.5 -5 -5 -7.5 -7.5 -6 -4 -2 0 2 4 6 -6 -4 -2 0 2 4 6 ReHwL ReHwL 10 10 7.5 7.5 5 5 2.5 2.5 ImHwL 0 ImHwL 0 -2.5 -2.5 -5 -5 -7.5 -7.5 -6 -4 -2 0 2 4 6 -6 -4 -2 0 2 4 6 ReHwL ReHwL Figure 1: Zeros of SRn(x) Our numerical results for numbers of real and complex zeros of SRn(x) are displayed in Table 2. Polynomials related to Bernoulli polynomials and Stirling polynomials 7427 Table 2. Numbers of real and complex zeros of SRn(x) degree n real zeros complex zeros degree n real zeros complex zeros 1 1 0 13 5 8 2 2 0 14 6 8 3 3 0 15 7 8 4 4 0 16 8 8 5 5 0 17 5 12 6 2 4 18 6 12 7 3 4 19 7 12 8 4 4 20 8 12 9 5 4 21 9 12 10 6 4 22 6 16 11 7 4 23 7 16 12 4 8 24 8 16 We calculated an approximate solution satisfying SRn(x); x 2 R. The results are given in Table 3. Table 3. Approximate solutions of SRn(x) = 0 degree n x 1 −0:5 2 −0:788675; −0:211325 3 −1; −0:5; 0 4 −1:1577; −0:759665; −0:240335; 0:157704 5 −1:26376; −1; −0:5; 0; 0:263763 6 −0:752459; −0:247541 7 −1; −0:5; 0 8 −1:24722; −0:75062; −0:24938; 0:247215 9 −1:44911; −1; −0:5; 0; 0:449106 Stacks of zeros of SRn(x) for 1 ≤ n ≤ 50 forming a 3D structure are presented(Figure 2). In Figure 2(top-right), we draw y and z axes but no x axis in three dimensions. In Figure 2(bottom-left), we draw x and y axes but no z axis in three dimensions. In Figure 2(bottom-right), we draw x and z axes but no y axis in three dimensions. Plot of real zeros of SRn(x) for 1 ≤ n ≤ 50 structure are presented(Figure 3). 7428 C. S. Ryoo 5 ImHwL 2.5 0 -2.5 40 -5 40 s n 20 20 0 -4 -2 0 0 5 2.5 0 -2.5 -5 Re w H L 2 ImHwL ReHwL ReHHLwL -4 -2 0 2 -4 -2 0 2 5 40 2.5 ImHwL 0 s 20 -2.5 -5 0 Figure 2: Stacks of zeros of SRn(x) for 1 ≤ n ≤ 50 By (1.1), we obtain 1 1 X (−1)ntn t X tn SR (−1 − x) = ext = SR (x) : n n! 1 − e−t n n! n=0 n=0 Hence we have the following theorem. Theorem 2.1 For n 2 N0, one has n SRn(x) = (−1) SRn(−1 − x): (2:1) By (2.1), we also have the following theorem. 1 Theorem 2.2 For n 2 , if n ≡ 1 (mod 2), then SR − = 0: N0 n 2 Since 1 1 X tn X tn (SR (x + 1) − SR (x)) = m(x + 1)m−1 ; n n n! n! n=0 n=0 we have the following theorem. Theorem 2.3 For n 2 N0, if n ≡ 1 (mod 2), then SRn(0) = SRn(−1) = 0: Polynomials related to Bernoulli polynomials and Stirling polynomials 7429 40 n 20 0 -4 -3 -2 -1 0 1 2 3 ReHxL Figure 3: Real zeros of SRn(x) for 1 ≤ n ≤ 50 We made a series of the following conjectures: 1 Conjecture 1. Prove that SRn(x); x 2 C; has Re(x) = − 2 reflection symmetry in addition to the usual Im(x) = 0 reflection symmetry analytic complex functions. The obvious corollary is that the zeros of SRn(x) will also inherit these symmetries. ∗ ∗ If SRn(x0) = 0; then SRn(1 − x0) = SRn(x0) = SRn(1 − x0) = 0 ∗ denotes complex conjugation (see Figure 1, Figure 2, Figure 3). Conjecture 2. Prove that SRn(x) = 0 has n distinct solutions. Since n is the degree of the polynomial SRn(x), the number of real zeros RSRn(x) lying on the real plane Im(x) = 0 is then RSRn(x) = n−CSRn(x), where CSRn(x) denotes complex zeros.