SOME INEQUALITIES FOR THE q-digamma FUNCTION

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Volume 10 (009), Issue 1, Article 1, 8 pp SOME INEQUALITIES FOR THE -DIGAMMA FUNCTION TOUFIK MANSOUR AND ARMEND SH SHABANI DEPARTMENT OF MATHEMATICS UNIVERSITY OF HAIFA 31905 HAIFA, ISRAEL

1 Volume 10 (009), Issue 1, Article 1, 8 pp SOME INEQUALITIES FOR THE -DIGAMMA FUNCTION TOUFIK MANSOUR AND ARMEND SH SHABANI DEPARTMENT OF MATHEMATICS UNIVERSITY OF HAIFA HAIFA, ISRAEL toufik@mathhaifaacil DEPARTMENT OF MATHEMATICS UNIVERSITY OF PRISHTINA AVENUE "MOTHER THERESA" 5 PRISHTINE 10000, REPUBLIC OF KOSOVA armend_shabani@hotmailcom Received 17 June, 008; accepted 1 February, 009 Communicated by A Laforgia ABSTRACT For the -digamma function and it s derivatives are established the functional ineualities of the types: f (x y) f(x) f(y), f(x + y) f(x) + f(y) Key words and phrases: -digamma function, ineualities 000 Mathematics Subject Classification 33D05 1 INTRODUCTION The Euler gamma function Γ(x) is defined for x > 0 by Γ(x) = 0 t x 1 e t dt The digamma (or psi) function is defined for positive real numbers x as the logarithmic derivative of Euler s gamma function, ψ(x) = Γ (x)/γ(x) The following integral and series representations are valid (see [1]): (11) ψ(x) = γ + 0 e t e xt dt = γ 1 1 e t x + x n(n + x), where γ = denotes Euler s constant Another interesting series representation for ψ, which is more rapidly convergent" than the one given in (11), was discovered by Ramanujan [3, page 374]

2 TOUFIK MANSOUR AND ARMEND SH SHABANI Jackson (see [5, 6, 7, 8]) defined the -analogue of the gamma function as (1) Γ (x) = (; ) (1 ) 1 x, ( x ; ) 0 < < 1, and (13) Γ (x) = ( 1 ; 1 ) ( x ; 1 ) ( 1) 1 x (x ), > 1, where (a; ) = j 0 (1 aj ) The -analogue of the psi function is defined for 0 < < 1 as the logarithmic derivative of the -gamma function, that is, ψ (x) = d dx log Γ (x) Many properties of the -gamma function were derived by Askey [] It is well known that Γ (x) Γ(x) and ψ (x) ψ(x) as 1 From (1), for 0 < < 1 and x > 0 we get (14) ψ (x) = log(1 ) + log n 0 and from (13) for > 1 and x > 0 we obtain (15) = log(1 ) + log ( n+x 1 n+x nx 1 n n x ψ (x) = log( 1) + log x 1 1 n x n 0 ( = log( 1) + log x 1 ) nx 1 n A Stieltjes integral representation for ψ (x) with 0 < < 1 is given in [4] It is well-known that ψ is strictly completely monotonic on (0, ), that is, ( 1) n (ψ (x)) (n) > 0 for x > 0 and n 0, see [1, Page 60] From (14) and (15) we conclude that ψ has the same property for any > 0 ( 1) n (ψ (x)) (n) > 0 for x > 0 and n 0 If (0, 1), using the second representation of ψ (x) given in (14), it can be shown that (16) ψ (k) (x) = log k+1 n k nx 1 n and hence ( 1) k 1 ψ (k) (x) > 0 with x > 1, for all k 1 If > 1, from the second representation of ψ (x) given in (15), we obtain ( (17) ψ (x) = log 1 + ) n nx 1 nx and for k, (18) ψ (k) (x) = ( 1) k 1 log k+1 n k nx 1 nx and hence ( 1) k 1 (x) > 0 with x > 0, for all > 1 ) J Ineual Pure and Appl Math, 10(1) (009), Art 1, 8 pp

3 SOME INEQUALITIES FOR THE -DIGAMMA FUNCTION 3 In this paper we derive several ineualities for (x), where k 0 INEQUALITIES OF THE TYPE f (x y) f(x) f(y) We start with the following lemma Lemma 1 For 0 < < 1 and 0 < x < 1 we have that ψ (x) < 0 Proof At first let us prove that ψ (x) < 0 for all x > 0 From (14) we get that ψ (x) = x 1 log log(1 ) + log n nx 1 n In order to see that ψ (x) < 0, we need to show that the function g(x) = x log log(1 ) 1 is a negative for all 0 < x < 1 and 0 < < 1 Indeed g (x) = x 1 log > 0, which implies that g(x) is an increasing function on 0 < x < 1, hence g(x) < g(1) = log log(1 ) 1 = 1 1 log < 0, (1 ) 1 for all 0 < < 1 Theorem Let 0 < < 1 and 0 < x, y < 1 Let k 0 be an integer Then (x)ψ (k) (y) < (ψ (k) (xy)) Proof We will consider two different cases: (1) k = 0 and () k 1 (1) Let f(x) = ψ (x) defined on 0 < x < 1 By Lemma 1 we have that f (x) = ψ (x)ψ (x) < 0 for all 0 < x < 1, which gives that f(x) is a decreasing function on 0 < x < 1 Hence, for all 0 < x, y < 1 we have ψ (xy) > ψ (x) and ψ (xy) > ψ (y), which gives that ψ 4 (xy) > ψ (x)ψ (y) Since ψ (x)ψ (y) > 0 for all 0 < x, y < 1, see Lemma 1, we obtain that as claimed ψ (xy) > ψ (x)ψ (y), J Ineual Pure and Appl Math, 10(1) (009), Art 1, 8 pp

4 4 TOUFIK MANSOUR AND ARMEND SH SHABANI () From (16) we have that (x)ψ (k) (y) (ψ (k) (xy)) ( ) ( n k nx 1 n = log k+1 = (log k+1 ) n,m 1 = (log k+1 ) n,m 1 ) ( n k ny log k+1 1 n log k+1 n k nx 1 mk my n 1 m (logk+1 ) (nm) k ( nx+my (n+m)xy ) (1 n )(1 m ) n,m 1 ) n k nxy 1 n (nm) k (n+m)xy (1 n )(1 m ) For 0 < x, y < 1, nx+my (n+m)xy < 0 and for x, y > 1, nx+my (n+m)xy > 0 and the results follow Note that the above theorem for k 1 remains true also for [ 1, 1] Also, if x, y > 1, k 1 and 0 < < 1 then ψ (k) (x)ψ (k) (y) > ( ψ (k) (xy) ) Now we extend Lemma 1 to the case > 1 In order to do that we denote the zero of the function f() = 3 log() log( 1), > 1, by ( 1) The numerical solution shows that as shown on Figure Figure 1: Graph of the function 3 ( 1) log log( 1) Lemma 3 For > and 0 < x < 1 we have that ψ (x) < 0 Proof From (15) we get that x ψ (x) = log log( 1) + log 1 1 ( x 1 ) log n nx 1 n J Ineual Pure and Appl Math, 10(1) (009), Art 1, 8 pp

5 SOME INEQUALITIES FOR THE -DIGAMMA FUNCTION 5 In order to show our claim, we need to prove that x on 0 < x < 1 Since g (x) = function on 0 < x < 1 Hence for all >, see Figure 1 g(x) = log log( 1) + log 1 1 ( x 1 ) < 0 x log + log > 0, it implies that g(x) is an increasing 1 1 g(x) < g(1) = 3 log log( 1) < 0, ( 1) Theorem 4 Let > and 0 < x, y < 1 Let k 0 be an integer Then (x) (y) < ( (xy) ) Proof As in the previous theorem we will consider two different cases: (1) k = 0 and () k 1 (1) As shown in the introduction the function ψ (x) is an increasing function on 0 < x < 1 Therefore, for all 0 < x, y < 1 we have that ψ (xy) < ψ (x) and ψ (xy) < ψ (y) Hence, Lemma 3 gives that ψ (xy) > ψ (x)ψ (y), as claimed () Analogous to the second case of Theorem Note that Theorem 4 for k 1 remains true also for > 1 Also, if x, y > 1, k 1 and > 1 then (x)ψ (k) (y) > ( ψ (k) (xy) ) 3 INEQUALITIES OF THE TYPE f(x + y) f(x) + f(y) The main goal of this section is to show that ψ (x + y) ψ (x) + ψ (y), for all 0 < x, y < 1 and 0 < < 1 In order to do that we define ρ() = log(1 ) + log j 1 Lemma 31 For all 0 < < 1, ρ() > 0 j ( j ) 1 j Proof Let 0 < < 1 and let g m () = c + m 1 j ( j ) j=1 with constant c > 0 for m Then 1 j g m (0) = c, lim 1 g m () < 0 and g m () is a decreasing function since m 1 g m() = j=1 j j 1 (1 + (1 j ) ) (1 j ) < 0 On the other hand g m+1 () g m () = m ( m ) < 0 1 m for all 0 < < 1 Hence, for all m we have that g m+1 () < g m (), 0 < < 1 Thus, if b m is the positive zero of the function g m () (because g n () is decreasing) on 0 < < 1 (by Maple or any mathematical programming we can see that b 1 = , b = and b 3 = ), then g m () > 0 for all 0 < < b m and g m () < 0 J Ineual Pure and Appl Math, 10(1) (009), Art 1, 8 pp

6 6 TOUFIK MANSOUR AND ARMEND SH SHABANI for all b m < < 1 Furthermore, the seuence {b m } m 0 is a strictly decreasing seuence of positive real numbers, that is 0 < b m+1 < b m, and bounded by zero, which implies that lim g m() = c + j ( j ) < 0, m 1 j j 1 for all 0 < < 1 Hence, if we choose c = log(1 ) log have that which implies that as reuested j 1 j ( j ) 1 j < (c is positive since 0 < < 1), then we log(1 ), log ρ() = log(1 ) + log j ( j ) 1 j j 1 > log(1 ) + log(1 ) = log(1 ) > 0, Theorem 3 For all 0 < < 1 and 0 < x, y < 1, Proof From the definitions we have that ψ (x + y) > ψ (x) + ψ (y) ψ (x + y) ψ (x) ψ (y) = log(1 ) + log Since 0 < x, y, < 1, we have that n(x+y) nx ny 1 n n(x+y) nx ny = (1 nx )(1 ny ) 1 < (1 n ) 1 = n ( n ) Hence, by Lemma 31 ψ (x + y) ψ (x) ψ (y) > ρ() > 0, which completes the proof The above theorem is not true for x, y > 1, for example ψ 1/10 (4) = , ψ 1/10 (5) = , ψ 1/10 (9) = Theorem 33 For all > 1 and 0 < x, y < 1, Proof From the definitions we have that ψ (x + y) > ψ (x) + ψ (y) ψ (x + y) ψ (x) ψ (y) = log( 1) + 1 log + log Q Q n(x+y) Q nx Q ny 1 Q n, J Ineual Pure and Appl Math, 10(1) (009), Art 1, 8 pp

7 SOME INEQUALITIES FOR THE -DIGAMMA FUNCTION 7 where Q = 1/ Thus ψ (x + y) ψ (x) ψ (y) = log( 1) + 1 log + ψ Q(x + y) ψ Q (x) ψ Q (y) log(1 Q) Using Theorem 3 we get that ψ (x + y) ψ (x) ψ (y) > log( 1) + 1 log log( 1) + log > 0, which completes the proof Note that the above theorem holds for > and x, y > 1, since ψ (x + y) ψ (x) ψ (y) = log( 1) + 1 log + log The above theorem is not true for x, y > 1 when 1 < <, for example ψ 3/ (4) = , ψ 3/ (5) = , ψ 3/ (9) = Theorem 34 Let (0, 1) Let k 1 be an integer (1) If k is even then (x + y) (x) + (y) () If k is odd then Proof From (16) we have (x + y) (x) + (y) ψ k (x + y) ψ k (x) ψ k (x) = log k+1 Since the function f(z) = nz is convex from ( ) x + y f 1 (f(x) + f(y)), we obtain that (31) n x+y nx + ny On the other hand it is clear that (3) n x+y > n(x+y) From (31) and (3) we have that n(x+y) nx ny < 0 (1) Since for (0, 1) and k even we have log k+1 < 0, hence (x + y) (x) ψ (k) (x) 0 nx (1 ny ) + ny 1 n > 0 n k 1 n (n(x+y) nx ny ) () The other case can be proved in a similar manner Using a similar approach one may prove analogue results for > 1 J Ineual Pure and Appl Math, 10(1) (009), Art 1, 8 pp

8 8 TOUFIK MANSOUR AND ARMEND SH SHABANI REFERENCES [1] M ABRAMOWITZ AND IA STEGUN, Handbook of Mathematical Functions with Formulas and Mathematical Tables, Dover, NewYork, 1965 [] R ASKEY, The -gamma and -beta functions, Applicable Anal, 8() (1978/79) [3] BC BERNDT, Ramanujan s Notebook, Part IV Springer, New York 1994 [4] MEH ISMAIL AND ME MULDOON, Ineualities and monotonicity properties for gamma and -gamma functions, in: RVM Zahar (Ed), Approximation and Computation, International Series of Numerical Mathematics, vol 119, Birkhäuser, Boston, MA, 1994, pp [5] T KIM, On a -analogue of the p-adic log gamma functions and related integrals, J Number Theory, 76 (1999), [6] T KIM, A note on the -multiple zeta functions, Advan Stud Contemp Math, 8 (004), [7] T KIM AND SH RIM, A note on the -integral and -series, Advanced Stud Contemp Math, (000), [8] HM SRIVASTAVA, T KIM AND Y SIMSEK, -Bernoulli numbers and polynomials associated with multiple -zeta functions and basic L-series, Russian J Math Phys, 1 (005), J Ineual Pure and Appl Math, 10(1) (009), Art 1, 8 pp

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