Modified Degenerate Tangent Numbers and Polynomials

Modified Degenerate Tangent Numbers and Polynomials

Global Journal of Pure and Applied Mathematics. ISSN 0973-1768 Volume 12, Number 2 (2016), pp. 1567-1574 © Research India Publications http://www.ripublication.com/gjpam.htm Modified degenerate tangent numbers and polynomials Cheon Seoung Ryoo Department of Mathematics, Hannam University, Daejeon 306-791, Korea. Abstract In this paper, we introduce the modified degenerate tangent numbers and poly- nomials. We also obtain some explicit formulas for modified degenerate tangent numbers and polynomials. AMS subject classification: 11B68, 11S40, 11S80. Keywords: Tangent numbers and polynomials, degenerate tangent numbers and polynomials, modified degenerate tangent numbers and polynomials. 1. Introduction Recently, many mathematicians have studied in the area of degenerate Bernoulli poly- nomials and Euler polynomials (see [1, 2, 3, 4, 6]). For example, in [2], L. Carlitz introduced the degenerate Bernoulli polynomials. In [3], Feng Qi et al. studied the par- tially degenerate Bernoull polynomials of the first kind in p-adic field. In this paper, we introduce the modified degenerate tangent numbers and polynomials. We also establish some interesting properties for modified degenerate tangent numbers and polynomials. Throughout this paper, we always make use of the following notations: N denotes the set of natural numbers and Z+ = N ∪{0}, C denotes the set of complex numbers. The Tangent numbers are defined by means of the following generating function: ∞ tn 2 π T = |t| < . (1.1) n n! e2t + 1 2 n=0 In [5], we defined Tangent polynomials by multiplying ext on the right side of the Eq. (1.1) as follows: ∞ tn 2 π T (x) = ext |t| < . (1.2) n n! e2t + 1 2 n=0 1568 Cheon Seoung Ryoo For more theoretical properties of the tangent numbers and polynomials, the readers may refer to [4, 6, 7]. We recall that the classical Stirling numbers of the first kind S1(n, k) and S2(n, k) are defined by the relations (see [7]) n n k n (x)n = S1(n, k)x and x = S2(n, k)(x)k, k=0 k=0 respectively. Here (x)n = x(x−1) ···(x−n+1) denotes the falling factorial polynomial of order n. The numbers S2(n, m) also admit a representation in terms of a generating function ∞ tn (et − 1)m S (n, m) = .(1.3) 2 n! m! n=m We also have ∞ tn (log(1 + t))m S (n, m) = .(1.4) 1 n! m! n=m The generalized falling factorial (x|λ)n with increment λ is defined by n−1 (x|λ)n = (x − λk) (1.5) k=0 for positive integer n, with the convention (x|λ)0 = 1; we may also write n n−k k (x|λ)n = S1(n, k)λ x .(1.6) k=0 Note that (x|λ) is a homogeneous polynomials in λ and x of degree n,soifλ = 0 then n −1 n (x|λ)n = λ (λ x|1)n. Clearly (x|0)n = x . We also need the binomial theorem: for a variable x, ∞ tn (1 + λt)x/λ = (x|λ) .(1.7) n n! n=0 For a variable t, we consider the degenerate tangent polynomials which are given by the generating function to be ∞ 2 tn (1 + λt)x/λ = T (x, λ) .(1.8) (1 + λt)2/λ + 1 n n! n=0 When x = 0, Tn(0,λ)= Tn(λ) are called the degenerate tangent numbers(see [6]). Note that (1 + λt)1/λ tends to et as λ → 0. In [6], we investigated some properties which are related to degenerate tangent numbers Tn(λ) and polynomials Tn(x, λ). Modified degenerate tangent numbers and polynomials 1569 2. On modified degenerate tangent numbers and polynomials In this section, we define the modified degenerate tangent numbers and polynomials, and we obtain explicit formulas for them. For a variable t, we consider the modified degenerate tangent polynomials which are given by the generating function to be ∞ 2 tn (1 + λ)xt/λ = T (x) .(2.1) (1 + λ)2t/λ + 1 n,λ n! n=0 When x = 0, Tn,λ(0) = Tn,λ are called the modified degenerate tangent numbers. Note that (1 + λ)t/λ tends to et as λ → 0. From (2.1) and (1.2), we note that ∞ n t 2 xt/λ lim Tn,λ(x) = lim (1 + λ) λ→0 n! λ→0 (1 + λ)2t/λ + 1 n=0 ∞ tn = T (x) . n n! n=0 Thus, we have lim Tn,λ(x) = Tn(x), (n ≥ 0). λ→0 From (2.1) and (1.5), we have ∞ tn 2 T (x) = (1 + λ)xt/λ n,λ n! (1 + λ)2t/λ + 1 n=0 ∞ ∞ tm log(1 + λ) n tn = T xn (2.2) m,λ m! λ n! m=0 n=0 ∞ n + l n n log(1 λ) l t = T − x . l n l,λ λ n! n=0 l=0 Therefore, by (2.2), we obtain the following theorem. Theorem 2.1. For n ≥ 0, we have n + l n log(1 λ) l T (x) = T − x . n,λ l n l,λ λ l=0 1570 Cheon Seoung Ryoo By (1.4), (2.1), and using Cauchy product, we see that ∞ tn 2 T (x) = (1 + λ)xt/λ n,λ n! (1 + λ)2t/λ + 1 n=0 ∞ ∞ tk tx λn = T xm k,λ k! λ n! k=0 m=0 m ∞ ∞ tk m tx l λm = T S (m, l) k,λ k! 1 λ m! k=0 m=0 l=0 ∞ ∞ ∞ (2.3) tk tx l λm = T S (m, l) k,λ k! 1 λ m! k=0 l=0 m=l ∞ ∞ ∞ tk x l λm tl = T S (m, l) l! k,λ k! 1 λ m! l! k=0 l=0 m=l ∞ n ∞ n x l λm tn = S (m, l) l! T − . l 1 λ m! n l,λ n! n=0 l=0 m=l tl Comparing the coefficients of in (2.3), we obtain the following theorem. l! Theorem 2.2. For n ≥ 0, we have n ∞ n x l λm T (x) = T − S (m, l) l! . n,λ l n l,λ 1 λ m! l=0 m=l By theorem 2.1 and Theorem 2.2, we have the following corollary. Corollary 2.3. For n ≥ 0, we have n ∞ n l m + l n x λ n log(1 λ) l T − S (m, l) l! = T − x . l n l,λ 1 λ m! l n l,λ λ l=0 m=l l=0 Modified degenerate tangent numbers and polynomials 1571 From (2.1), we can derive the following recurrence relation: ∞ tn 2 = ((1 + λ)2t/λ + 1) T n,λ n! n=0 ∞ ∞ tn tn = (1 + λ)2t/λ T + T n,λ n! n,λ n! n=0 n=0 ∞ ∞ ∞ (2.4) log(1 + λ) l tl tm tn = 2l T + T λ l! m,λ m! n,λ n! l=0 m=0 n=0 ∞ n + l n n l log(1 λ) t = 2 T − + T . l λ n l,λ n,λ n! n=0 l=0 tn By comparing of the coefficients on the both sides of (2.4), we have the following n! theorem. Theorem 2.4. For n ∈ Z+,wehave n + l n l log(1 λ) 2, if n = 0, 2 T − + T = l λ n l,λ n,λ 0, if n = 0. l=0 Also, we can obtain similar recurrence relation: n ∞ − n 2ll!λm l 2, if n = 0, T − S (m, l) + T = l n l,λ 1 m! n,λ 0, if n = 0. l=0 m=l By (2.1), we have ∞ ∞ tn tn T (x + 2) + T n,λ n! n,λ n! n=0 n=0 2 + 2 = (1 + λ)(x 2)t/λ + (1 + λ)xt/λ (1 + λ)2t/λ + 1 (1 + λ)2t/λ + 1 (2.5) = 2(1 + λ)xt/λ ∞ log(1 + λ) n tn = 2 xn . λ n! n=0 tn By comparing of the coefficients on the both sides of (2.4), we obtain the following n! theorem. 1572 Cheon Seoung Ryoo Theorem 2.5. For n ∈ Z+,wehave log(1 + λ) n T (x + 2) + T (x) = 2 xn. n,λ n,λ λ By (2.1), we get ∞ n t 2 − T (x) = (1 + λ)(xt 2t)/λ n,λ n! (1 + λ)−2t/λ + 1 n=0 2 − = (1 + λ)(2 x)t/λ (2.6) (1 + λ)−2t/λ + 1 ∞ tn = (−1)nT (2 − x) . n,λ n! n=0 tn By comparing of the coefficients on the both sides of (2.6), we have the following n! theorem. Theorem 2.6. For n ∈ Z+,wehave n Tn,λ(x) = (−1) Tn,λ(2 − x), In particular, n Tn,λ = (−1) Tn,λ(2). For d ∈ N with d ≡ 1(mod 2), we have ∞ tn 2 T (x) = (1 + λ)xt/λ n,λ n! (1 + λ)2t/λ + 1 n=0 − 2 d 1 = (1 + λ)xt/λ (−1)l(1 + λ)2lt/λ + 2dt/λ + (1 λ) 1 = l 0 ∞ − d 1 2l + x tn = dn (−1)lT . n,λ d n! n=0 l=0 tm By comparing coefficients of in the above equation, we have the following theorem: m! Theorem 2.7. For d ∈ N with d ≡ 1(mod 2) and n ∈ Z+,wehave − d 1 2l + x T (x) = dn (−1)lT . n,λ n,λ d l=0 Modified degenerate tangent numbers and polynomials 1573 In particular, − d 1 2l T = dn (−1)lT .

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    8 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