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Some Inequalities for the Q-Digamma Function 1 Introduction and Preliminaries

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Some Inequalities for the Q-Digamma Function 1 Introduction and Preliminaries Mathematica Aeterna, Vol. 4, 2014, no. 5, 515 - 519 Some Inequalities for the q-Digamma Function Kwara Nantomah Department of Mathematics, University for Development Studies, Navrongo Campus, P. O. Box 24, Navrongo, UE/R, Ghana. [email protected], [email protected] Edward Prempeh Department of Mathematics, Kwame Nkrumah University of Science and Technology, Kumasi, Ghana. [email protected] Abstract Some inequalities involving the q-digamma function are presented. These results are the q-analogues of some recent results. Mathematics Subject Classification: 33B15, 26A48. Keywords: digamma function, q-digamma function, Inequality. 1 Introduction and Preliminaries The classical Euler’s Gamma function Γ(t) and the digamma function ψ(t) are commonly defined as ∞ d Γ′(t) Γ(t)= e−xxt−1 dx, and ψ(t)= ln(Γ(t)) = , t> 0. dt Γ(t) Z0 Similarly the q-Gamma and q-digamma functions are defined as (see [1]) ∞ 1 − qn Γ (t)=(1 − q)1−t , q ∈ (0, 1), t> 0. q 1 − qt+n n=1 Y and ′ d Γq(t) ψq(t)= ln(Γq(t)) = dt Γq(t) 516 Kwara Nantomah and Edward Prempeh The functions ψ(t) and ψq(t) as defined above exhibit the following series representations. ∞ 1 ψ(t)= −γ +(t − 1) , t> 0. (1 + n)(n + t) n=0 X ∞ qnt ψ (t)= − ln(1 − q) + (ln q) , q ∈ (0, 1), t> 0. q 1 − qn n=1 X where γ is the Euler-Mascheroni’s constant. By taking the m-th derivative of the above functions, we arrive at the following statements for m ∈ N. ∞ 1 ψ(m)(t)=(−1)m+1m! , t> 0. (n + t)m+1 n=0 ∞X nmqnt ψ(m)(t) = (ln q)m+1 , q ∈ (0, 1), t> 0. q 1 − qn n=1 X In 2011, Sulaiman [3] presented the following results. ψ(t + s) ≥ ψ(t)+ ψ(s) (1) where t> 0 and 0 <s< 1. ψ(m)(t + s) ≤ ψ(m)(t)+ ψ(m)(s) (2) where m is a positive odd integer and t,s > 0. ψ(m)(t + s) ≥ ψ(m)(t)+ ψ(m)(s) (3) where m is a positive even integer and t,s > 0. In a recent paper, Sroysang [2] presented the following geralizations of the above inequalities. α α ψ(t + βisi) ≥ ψ(t)+ βiψ(si) (4) i=1 i=1 X X where t> 0, βi > 0 and 0 <si < 1 for all i ∈ Nα. α α (m) (m) (m) ψ (t + βisi) ≤ ψ (t)+ βiψ (si) (5) i=1 i=1 X X Some Inequalities for the q-Digamma Function 517 where m is a positive odd integer, t> 0, βi > 0 and si > 0 for all i ∈ Nα. α α (m) (m) (m) ψ (t + βisi) ≥ ψ (t)+ βiψ (si) (6) i=1 i=1 X X where m is a positive even integer, t> 0, βi > 0 and si > 0 for all i ∈ Nα. The objective of this paper is to establish that the inequalities (4), (5) and (6) still hold true for the q-digamma function. 2 Main Results We now present our results. Theorem 2.1. Let q ∈ (0, 1), t > 0, βi > 0 and 0 < si < 1 for all i ∈ Nα. Then the following inequality is valid. α α ψq(t + βisi) ≥ ψq(t)+ βiψq(si) (7) i=1 i=1 Xα Xα Proof. Let u(t)= ψq(t + i=1 βisi) − ψq(t) − i=1 βiψq(si). Then fixing si for each i we have, P P α ′ ′ ′ u (t)= ψq(t + βisi) − ψq(t) i=1 X∞ α nqn(t+Pi=1 βisi) nqnt = (ln q)2 − 1 − qn 1 − qn n=1 X∞ α nqnt(qn Pi=1 βisi − 1) = (ln q)2 ≤ 0. 1 − qn n=1 X That implies u is non-increasing. Moreover, α α lim u(t) = lim ψq(t + βisi) − ψq(t) − βiψq(si) t→∞ t→∞ " i=1 i=1 # α X X = ln(1 − q) βi i=1 X∞ α α qn(t+Pi=1 βisi) qnt β qnsi + (ln q) lim − − i t→∞ 1 − qn 1 − qn 1 − qn n=1 " i=1 # α X ∞ α X β qnsi = ln(1 − q) β − (ln q) i ≥ 0. i 1 − qn i=1 n=1 i=1 X X X Therefore u(t) ≥ 0 concluding the proof. 518 Kwara Nantomah and Edward Prempeh Theorem 2.2. Let q ∈ (0, 1), t> 0, βi > 0 and si > 0 for all i ∈ Nα. Suppose that m is a positive odd integer, then the following inequality is valid. α α (m) (m) (m) ψq (t + βisi) ≤ ψq (t)+ βiψq (si) (8) i=1 i=1 X X (m) α (m) α (m) Proof. Let v(t)= ψq (t + i=1 βisi) − ψq (t) − i=1 βiψq (si). Then fixing si for each i we have, P P α ′ (m+1) (m+1) v (t)= ψq (t + βisi) − ψq (t) i=1 ∞X α nm+1qn(t+Pi=1 βisi) nm+1qnt = (ln q)m+2 − 1 − qn 1 − qn n=1 X∞ α nm+1qnt(qn Pi=1 βisi − 1) = (ln q)m+2 ≥ 0. (since m is odd) 1 − qn n=1 X That implies v is non-decreasing. Moreover, α α (m) (m) (m) lim v(t) = lim ψq (t + βisi) − ψq (t) − βiψq (si) t→∞ t→∞ " i=1 i=1 # ∞X α X α m n(t+Pi=1 βisi) m nt m nsi m+1 n q n q n q = (ln q) lim − − βi t→∞ 1 − qn 1 − qn 1 − qn n=1 " i=1 # ∞Xα X nmqnsi = −(ln q)m+1 β ≤ 0. (since m is odd) i 1 − qn n=1 i=1 X X Therefore v(t) ≤ 0 concluding the proof. Theorem 2.3. Let q ∈ (0, 1), t> 0, βi > 0 and si > 0 for all i ∈ Nα. Suppose that m is a positive even integer, then the following inequality is valid. α α (m) (m) (m) ψq (t + βisi) ≥ ψq (t)+ βiψq (si) (9) i=1 i=1 X X (m) α (m) α (m) Proof. Let w(t)= ψq (t+ i=1 βisi)−ψq (t)− i=1 βiψq (si). Then fixing P P Some Inequalities for the q-Digamma Function 519 si for each i we have, α ′ (m+1) (m+1) w (t)= ψq (t + βisi) − ψq (t) i=1 ∞X α nm+1qn(t+Pi=1 βisi) nm+1qnt = (ln q)m+2 − 1 − qn 1 − qn n=1 X∞ α nm+1qnt(qn Pi=1 βisi − 1) = (ln q)m+2 ≤ 0. (since m is even) 1 − qn n=1 X That implies w is non-increasing. Moreover, α α (m) (m) (m) lim w(t) = lim ψq (t + βisi) − ψq (t) − βiψq (si) t→∞ t→∞ " i=1 i=1 # ∞X α X α m n(t+Pi=1 βisi) m nt m nsi m+1 n q n q n q = (ln q) lim − − βi t→∞ 1 − qn 1 − qn 1 − qn n=1 " i=1 # ∞Xα X nmqnsi = −(ln q)m+1 β ≥ 0. (since m is even) i 1 − qn n=1 i=1 X X Therefore w(t) ≥ 0 concluding the proof. Remark 2.4. If we let q → 1− in inequalities (7), (8) and (9) then we repectively recover the inequalities (4), (5) and (6). References [1] T. Mansour, Some inequalities for the q-Gamma Function, J. Ineq. Pure Appl. Math. 9(1)(2008), Art. 18. [2] B. Sroysang, More on some inequalities for the digamma function, Math. Aeterna, 4(2)(2014), 123-126. [3] W. T. Sulaiman, Turan inequalites for the digamma and polygamma func- tions, South Asian J. Math. 1(2)(2011), 49-55. Received: May, 2014.
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