SOME INEQUALITIES FOR THE q-DIGAMMA
FUNCTION
TOUFIK MANSOUR
ARMEND SH. SHABANI
Department of Mathematics
University of Haifa
31905 Haifa, Israel
Department of Mathematics
University of Prishtina
Avenue "Mother Theresa"
5 Prishtine 10000, Republic of Kosova
EMail: armend_shabani@hotmail.com
EMail: toufik@math.haifa.ac.il
Some Inequalities for the
q-Digamma Function
Toufik Mansour and
Armend Sh. Shabani
vol. 10, iss. 1, art. 12, 2009
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Contents
Received:
17 June, 2008
Accepted:
21 February, 2009
Communicated by:
A. Laforgia
2000 AMS Sub. Class.:
33D05
Key words:
q-digamma function, inequalities.
Abstract:
For the q-digamma function and it’s derivatives are established the functional
inequalities of the types:
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2
f (x · y) ≶ f (x) · f (y),
f (x + y) ≶ f (x) + f (y).
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Contents
1
Introduction
3
2
Inequalities of the type f 2 (x · y) ≶ f (x) · f (y)
6
3
Inequalities of the Type f (x + y) ≶ f (x) + f (y)
11
Some Inequalities for the
q-Digamma Function
Toufik Mansour and
Armend Sh. Shabani
vol. 10, iss. 1, art. 12, 2009
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1.
Introduction
The Euler gamma function Γ(x) is defined for x > 0 by
Z ∞
Γ(x) =
tx−1 e−t dt.
0
The digamma (or psi) function is defined for positive real numbers x as the logarithmic derivative of Euler’s gamma function, ψ(x) = Γ0 (x)/Γ(x). The following
integral and series representations are valid (see [1]):
Z ∞ −t
1 X
x
e − e−xt
(1.1)
ψ(x) = −γ +
dt
=
−γ
−
+
,
−t
1−e
x n≥1 n(n + x)
0
where γ = 0.57721 . . . denotes Euler’s constant. Another interesting series representation for ψ, which is “more rapidly convergent" than the one given in (1.1), was
discovered by Ramanujan [3, page 374].
Jackson (see [5, 6, 7, 8]) defined the q-analogue of the gamma function as
(q; q)∞
Γq (x) = x
(1 − q)1−x ,
(q ; q)∞
(1.2)
0 < q < 1,
and
vol. 10, iss. 1, art. 12, 2009
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−1
(1.3)
Some Inequalities for the
q-Digamma Function
Toufik Mansour and
Armend Sh. Shabani
Γq (x) =
Q
−1
(q ; q )∞
1−x (x2 )
(q
−
1)
q ,
(q −x ; q −1 )∞
j
q > 1,
where (a; q)∞ = j≥0 (1 − aq ).
The q-analogue of the psi function is defined for 0 < q < 1 as the logarithmic
derivative of the q-gamma function, that is,
ψq (x) =
d
log Γq (x).
dx
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Many properties of the q-gamma function were derived by Askey [2]. It is well
known that Γq (x) → Γ(x) and ψq (x) → ψ(x) as q → 1− . From (1.2), for 0 < q < 1
and x > 0 we get
X q n+x
(1.4)
ψq (x) = − log(1 − q) + log q
1 − q n+x
n≥0
X q nx
= − log(1 − q) + log q
1 − qn
n≥1
and from (1.3) for q > 1 and x > 0 we obtain
Some Inequalities for the
q-Digamma Function
Toufik Mansour and
Armend Sh. Shabani
vol. 10, iss. 1, art. 12, 2009
!
(1.5)
1 X q −n−x
−
2 n≥0 1 − q −n−x
!
1 X q −nx
= − log(q − 1) + log q x − −
.
2 n≥1 1 − q −n
ψq (x) = − log(q − 1) + log q x −
A Stieltjes integral representation for ψq (x) with 0 < q < 1 is given in [4]. It is
well-known that ψ 0 is strictly completely monotonic on (0, ∞), that is,
n
0
(n)
(−1) (ψ (x))
> 0 for x > 0 and n ≥ 0,
ψq0
see [1, Page 260]. From (1.4) and (1.5) we conclude that
has the same property
for any q > 0
(−1)n (ψq0 (x))(n) > 0 for x > 0 and n ≥ 0.
If q ∈ (0, 1), using the second representation of ψq (x) given in (1.4), it can be shown
that
X nk · q nx
(1.6)
ψq(k) (x) = logk+1 q
1 − qn
n≥1
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(k)
and hence (−1)k−1 ψq (x) > 0 with x > 1, for all k ≥ 1. If q > 1, from the second
representation of ψq (x) given in (1.5), we obtain
!
X nq −nx
(1.7)
ψq0 (x) = log q 1 +
1 − q −nx
n≥1
and for k ≥ 2,
(1.8)
ψq(k) (x) = (−1)k−1 logk+1 q
X nk q −nx
1 − q −nx
n≥1
Some Inequalities for the
q-Digamma Function
Toufik Mansour and
Armend Sh. Shabani
vol. 10, iss. 1, art. 12, 2009
(k)
and hence (−1)k−1 ψq (x) > 0 with x > 0, for all q > 1.
In this paper we derive several inequalities for ψ (k) (x), where k ≥ 0.
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2.
Inequalities of the type f 2 (x · y) ≶ f (x) · f (y)
We start with the following lemma.
Lemma 2.1. For 0 < q <
1
2
and 0 < x < 1 we have that ψq (x) < 0.
Proof. At first let us prove that ψq (x) < 0 for all x > 0. From (1.4) we get that
X q nx
qx
ψq (x) =
log q − log(1 − q) + log q
.
n
1−q
1
−
q
n≥2
Some Inequalities for the
q-Digamma Function
Toufik Mansour and
Armend Sh. Shabani
vol. 10, iss. 1, art. 12, 2009
In order to see that ψq (x) < 0, we need to show that the function
g(x) =
qx
log q − log(1 − q)
1−q
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qx
1−q
is a negative for all 0 < x < 1 and 0 < q < 12 . Indeed g 0 (x) =
implies that g(x) is an increasing function on 0 < x < 1, hence
log2 q > 0, which
q
g(x) < g(1) =
log q − log(1 − q)
1−q
1
qq
=
log
< 0,
1−q
(1 − q)1−q
for all 0 < q < 21 .
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Theorem 2.2. Let 0 < q <
1
2
and 0 < x, y < 1. Let k ≥ 0 be an integer. Then
ψq(k) (x)ψq(k) (y) < (ψq(k) (xy))2 .
Proof. We will consider two different cases: (1) k = 0 and (2) k ≥ 1.
(1) Let f (x) = ψq2 (x) defined on 0 < x < 1. By Lemma 2.1 we have that
f 0 (x) = 2ψq (x)ψq0 (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
ψq2 (xy) > ψq2 (x)
and
Some Inequalities for the
q-Digamma Function
Toufik Mansour and
Armend Sh. Shabani
ψq2 (xy) > ψq2 (y),
which gives that
ψq4 (xy) > ψq2 (x)ψq2 (y).
Since ψq (x)ψq (y) > 0 for all 0 < x, y < 1, see Lemma 2.1, we obtain that
vol. 10, iss. 1, art. 12, 2009
ψq2 (xy) > ψq (x)ψq (y),
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as claimed.
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(2) From (1.6) we have that
ψq(k) (x)ψq(k) (y) − (ψq(k) (xy))2
!
!
X nk q nx
X nk q ny
= logk+1 q
logk+1 q
−
n
1
−
q
1 − qn
n≥1
n≥1
k nx
= (logk+1 q)2
k my
logk+1 q
X nk q nxy
n≥1
1−
!2
qn
k (n+m)xy
X n q
X (nm) q
m q
k+1
2
·
−
(log
q)
1 − qn 1 − qm
(1 − q n )(1 − q m )
n,m≥1
n,m≥1
X (nm)k (q nx+my − q (n+m)xy )
= (logk+1 q)2
.
n )(1 − q m )
(1
−
q
n,m≥1
For 0 < x, y < 1 , q nx+my − q (n+m)xy < 0 and for x, y > 1, q nx+my − q (n+m)xy > 0
and the results follow.
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Note that the above theorem for k ≥ 1 remains true also for q ∈
x, y > 1, k ≥ 1 and 0 < q < 1 then
2
ψq(k) (x)ψq(k) (y) > ψq(k) (xy) .
1
,
1
. Also, if
2
Now we extend Lemma 2.1 to the case q > 1. In order to do that we denote the
q−3
zero of the function f (q) = 2(q−1)
log(q) − log(q − 1), q > 1, by q ∗ . The numerical
solution shows that q ∗ ≈ 1.56683201 . . . as shown on Figure 1.
Lemma 2.3. For q > q ∗ and 0 < x < 1 we have that ψq (x) < 0.
vol. 10, iss. 1, art. 12, 2009
Proof. From (1.5) we get that
X q −nx
q −x
1
ψq (x) = −
log
q
−
log(q
−
1)
+
log
q
x
−
−
log
q
.
1 − q −1
2
1 − q −n
n≥2
In order to show our claim, we need to prove that
−x
g(x) = −
q
1
log q − log(q − 1) + log q x −
−1
1−q
2
<0
−x
q
2
on 0 < x < 1. Since g 0 (x) = 1−q
−1 log q + log q > 0, it implies that g(x) is an
increasing function on 0 < x < 1. Hence
g(x) < g(1) =
Some Inequalities for the
q-Digamma Function
Toufik Mansour and
Armend Sh. Shabani
q−3
log q − log(q − 1) < 0,
2(q − 1)
for all q > q ∗ , see Figure 1.
Theorem 2.4. Let q > 2 and 0 < x, y < 1. Let k ≥ 0 be an integer. Then
2
ψq(k) (x)ψq(k) (y) < ψq(k) (xy) .
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1
Some Inequalities for the
q-Digamma Function
Toufik Mansour and
Armend Sh. Shabani
0.5
vol. 10, iss. 1, art. 12, 2009
0
1.5
2
2.5
3
3.5
4
4.5
5
q
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–0.5
Contents
–1
Figure 1: Graph of the function
q−3
2(q−1)
log q − log(q − 1).
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Proof. As in the previous theorem we will consider two different cases: (1) k = 0
and (2) k ≥ 1.
(1) As shown in the introduction the function ψq0 (x) is an increasing function on
0 < x < 1. Therefore, for all 0 < x, y < 1 we have that
ψq (xy) < ψq (x)
and
ψq (xy) < ψq (y).
Hence, Lemma 2.3 gives that ψq2 (xy) > ψq (x)ψq (y), as claimed.
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(2) Analogous to the second case of Theorem 2.2.
Note that Theorem 2.4 for k ≥ 1 remains true also for q > 1. Also, if x, y > 1,
k ≥ 1 and q > 1 then
2
ψq(k) (x)ψq(k) (y) > ψq(k) (xy) .
Some Inequalities for the
q-Digamma Function
Toufik Mansour and
Armend Sh. Shabani
vol. 10, iss. 1, art. 12, 2009
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3.
Inequalities of the Type f (x + y) ≶ f (x) + f (y)
The main goal of this section is to show that ψq (x + y) ≥ ψq (x) + ψq (y), for all
0 < x, y < 1 and 0 < q < 1. In order to do that we define
ρ(q) = log(1 − q) + log q
X q j (q j − 2)
.
1 − qj
j≥1
Lemma 3.1. For all 0 < q < 1, ρ(q) > 0.
P
q j (q j −2)
Proof. Let 0 < q < 1 and let gm (q) = c + m−1
with constant c > 0 for
j=1
1−q j
m ≥ 2. Then gm (0) = c, limq→1− gm (q) < 0 and gm (q) is a decreasing function
since
m−1
X jq j−1 (1 + (1 − q j )2 )
0
gm (q) = −
< 0.
(1 − q j )2
j=1
On the other hand
q m (q m − 2)
<0
1 − qm
for all 0 < q < 1. Hence, for all m ≥ 2 we have that
gm+1 (q) − gm (q) =
gm+1 (q) < gm (q),
0 < q < 1.
Thus, if bm is the positive zero of the function gm (q) (because gn (q) is decreasing)
on 0 < q < 1 (by Maple or any mathematical programming we can see that b1 =
0.38196601 . . ., b2 = 0.3184588966 and b3 = 0.3055970874), then gm (q) > 0 for all
0 < q < bm and gm (q) < 0 for all bm < q < 1. Furthermore, the sequence {bm }m≥0
is a strictly decreasing sequence of positive real numbers, that is 0 < bm+1 < bm ,
Some Inequalities for the
q-Digamma Function
Toufik Mansour and
Armend Sh. Shabani
vol. 10, iss. 1, art. 12, 2009
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and bounded by zero, which implies that
lim gm (q) = c +
m→∞
X q j (q j − 2)
< 0,
j
1
−
q
j≥1
(c is positive since 0 < q < 1),
for all 0 < q < 1. Hence, if we choose c = 2 log(1−q)
log q
then we have that
X q j (q j − 2)
2 log(1 − q)
<
−
,
j
1
−
q
log
q
j≥1
Some Inequalities for the
q-Digamma Function
Toufik Mansour and
Armend Sh. Shabani
vol. 10, iss. 1, art. 12, 2009
which implies that
ρ(q) = log(1 − q) + log q
X q j (q j − 2)
1 − qj
j≥1
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> −2 log(1 − q) + log(1 − q)
= − log(1 − q) > 0,
as requested.
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Theorem 3.2. For all 0 < q < 1 and 0 < x, y < 1,
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ψq (x + y) > ψq (x) + ψq (y).
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Proof. From the definitions we have that
ψq (x + y) − ψq (x) − ψq (y) = log(1 − q) + log q
Close
X q n(x+y) − q nx − q ny
n≥1
1 − qn
.
Since 0 < x, y, q < 1, we have that
q n(x+y) − q nx − q ny = (1 − q nx )(1 − q ny ) − 1
< (1 − q n )2 − 1
= q n (q n − 2).
Hence, by Lemma 3.1
ψq (x + y) − ψq (x) − ψq (y) > ρ(q) > 0,
which completes the proof.
Some Inequalities for the
q-Digamma Function
Toufik Mansour and
Armend Sh. Shabani
vol. 10, iss. 1, art. 12, 2009
The above theorem is not true for x, y > 1, for example
ψ1/10 (4) = 0.1051046497 . . . ,
ψ1/10 (9) = 0.1053605131 . . . .
ψ1/10 (5) = 0.1053349312 . . . ,
Theorem 3.3. For all q > 1 and 0 < x, y < 1,
ψq (x + y) > ψq (x) + ψq (y).
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Proof. From the definitions we have that
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X Qn(x+y) − Qnx − Qny
1
ψq (x+y)−ψq (x)−ψq (y) = log(q −1)+ log q +log Q
,
n
2
1
−
Q
n≥1
where Q = 1/q. Thus
ψq (x + y) − ψq (x) − ψq (y)
1
= log(q − 1) + log q + ψQ (x + y) − ψQ (x) − ψQ (y) − log(1 − Q).
2
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Using Theorem 3.2 we get that
ψq (x + y) − ψq (x) − ψq (y) > log(q − 1) +
1
log q − log(q − 1) + log q > 0,
2
which completes the proof.
Note that the above theorem holds for q > 2 and x, y > 1, since
ψq (x + y) − ψq (x) − ψq (y)
= log(q − 1) +
X q −nx (1 − q −ny ) + q −ny
1
log q + log q
> 0.
−n
2
1
−
q
n≥1
The above theorem is not true for x, y > 1 when 1 < q < 2, for example
ψ3/2 (4) = 1.83813910 . . . ,
ψ3/2 (9) = 4.10745515 . . . .
Some Inequalities for the
q-Digamma Function
Toufik Mansour and
Armend Sh. Shabani
vol. 10, iss. 1, art. 12, 2009
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ψ3/2 (5) = 2.34341101 . . . ,
Theorem 3.4. Let q ∈ (0, 1). Let k ≥ 1 be an integer.
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(1) If k is even then
ψq(k) (x + y) ≥ ψq(k) (x) + ψq(k) (y).
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(2) If k is odd then
ψq(k) (x + y) ≤ ψq(k) (x) + ψq(k) (y).
Proof. From (1.6) we have
ψqk (x + y) − ψqk (x) − ψqk (x) = logk+1 q
X
n≥1
nk
(q n(x+y) − q nx − q ny ).
1 − qn
Close
Since the function f (z) = q nz is convex from
x+y
1
f
≤ (f (x) + f (y)),
2
2
we obtain that
(3.1)
2 · qn
x+y
2
≤ q nx + q ny .
On the other hand it is clear that
(3.2)
2 · qn
x+y
2
> q n(x+y) .
From (3.1) and (3.2) we have that
Some Inequalities for the
q-Digamma Function
Toufik Mansour and
Armend Sh. Shabani
vol. 10, iss. 1, art. 12, 2009
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q n(x+y) − q nx − q ny < 0.
(1) Since for q ∈ (0, 1) and k even we have logk+1 q < 0, hence
ψq(k) (x + y) − ψq(k) (x) − ψq(k) (x) ≥ 0.
(2) The other case can be proved in a similar manner.
Using a similar approach one may prove analogue results for q > 1.
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References
[1] M. ABRAMOWITZ AND I.A. STEGUN, Handbook of Mathematical Functions
with Formulas and Mathematical Tables, Dover, NewYork, 1965.
[2] R. ASKEY, The q-gamma and q-beta functions, Applicable Anal., 8(2) (1978/79)
125–141.
[3] B.C. BERNDT, Ramanujan’s Notebook, Part IV. Springer, New York 1994.
[4] M.E.H. ISMAIL AND M.E. MULDOON, Inequalities and monotonicity properties for gamma and q-gamma functions, in: R.V.M. Zahar (Ed.), Approximation and Computation, International Series of Numerical Mathematics, vol. 119,
Birkhäuser, Boston, MA, 1994, pp. 309–323.
[5] T. KIM, On a q-analogue of the p-adic log gamma functions and related integrals,
J. Number Theory, 76 (1999), 320–329.
[6] T. KIM, A note on the q-multiple zeta functions, Advan. Stud. Contemp. Math.,
8 (2004), 111–113.
Some Inequalities for the
q-Digamma Function
Toufik Mansour and
Armend Sh. Shabani
vol. 10, iss. 1, art. 12, 2009
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[7] T. KIM AND S.H. RIM, A note on the q-integral and q-series, Advanced Stud.
Contemp. Math., 2 (2000), 37–45.
[8] H.M. SRIVASTAVA, T. KIM AND Y. SIMSEK, q-Bernoulli numbers and polynomials associated with multiple q-zeta functions and basic L-series, Russian J.
Math. Phys., 12 (2005), 241–268.
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