Unied treatment of Gautschi–Kershaw type inequalities
for the gamma function
1C. Giordanoa;∗, A. Laforgiab, J. Pecaricc
aDipartimento di Matematica – Universita, Via Carlo Alberto, 10, 10123 Torino, Italy b
Dipartimento di Matematica, Universita di Roma Tre, Largo S. Leonardo Murialdo, 1, Roma, Italy cFaculty of Textile Thecnology, Pierottijeva 6, 10000 Zagreb, Croatia
Received 10 December 1997; received in revised form 30 April 1998
Abstract
Gautschi, Kershaw, Lorch, Laforgia and other authors gave several inequalities for the ratio (x+ 1)= (x+s) where, as usual, denotes the gamma function. In this paper we give a unied treatment of all their results and prove, among other things, new inequalities for the above ratio, which involve the psi function. Inequalities for the ratio of two gamma functions are useful, for example, to deduce Bernstein-type inequalities for ultraspherical polynomials. We give an example of this type. c1998 Elsevier Science B.V. All rights reserved.
AMS classication:33B15; 33C45
Keywords:Gamma function; Psi function; Bernstein inequality
1. Introduction
Gautschi [6] has proved the following inequalities for the Gamma function:
x1−s¡ (x+ 1)
(x+s)¡(x+ 1)
1−s; 0¡s¡1; x= 1;2; : : : : (1.1)
The proof in [6] is also valid for arbitrary x ¿0 ([7], footnote 4). Kershaw [8] has given some improvements of these inequalities such as
[x+ 1 2s]
1−s
¡ (x+ 1) (x+s)¡[x−
1
2 + (s+ 1 4)
1=2]1−s
(1.2)
1
This work was supported by the Consiglio Nazionale delle Ricerche of Italy and by the Ministero dell’Universita e della Ricerca Scientica e Tecnologica of Italy.
∗Corresponding author. E-mail: giordano@alpha01.dm.unito.it
for real x¿0 and 0¡s¡1. Many authors have studied inequalities for this function. The left-hand inequality in (1.2) was also considered by Lorch [10]. He obtained the following results for integer x¿0:
Lorch [10] successfully used (1.3) to prove a sharpened inequality for the ultraspherical polynomials of degree n and parameter , 0¡¡1, P()
For the important special case of Legendre polynomials, where =1
2, (1.5) gives [2]
which improves the Bernstein inequality [12, p. 165]
|Pn(cos)|¡
1
p
(=2)nsin
; 0¡¡; n= 1;2; : : : :
Once more the upper bound for the gamma function [5]
1
was the key of the proof for the following inequality for the Legendre polynomials:
|Pn(cos)|¡
due to Elbert and Laforgia [5].
In this paper we shall give a unied treatment and some extensions of all Gautschi–Kershaw type inequalities. A result in this direction is given in [11].
2. Inequalities valid for x ¿0
Proof. We shall use the same idea of the proofs in [8–10]. From the relation [1, p. 257] lim
x→∞x
b−a (x+a)
(x+b)= 1; a¿0; b¿0
we see that
lim
x→∞f(x) = 1;
where
f(x) = (x+ 1)
(x+s)(x+)
s−1
for all x¿0 and s; ¿0. Now let
F(x) = f(x) f(x+ 1)=
x+s x+ 1
x++ 1
x+
1−s
then
F′
(x) F(x) =
(1−s)[(2+−s) + (2−s)x]
(x+ 1)(x+s)(x+)(x++ 1):
It is obvious that the sign of F′
(x) depends on the sign of the expression
Ax(s; ) = (1−s)[(2+−s) + (2−s)x]:
If =s=2, we have [9]
Ax(s;12s) =12s(1−s)(12s−1);
Ax(s;12s)¿0; 1¡s¡2; x¿0;
Ax(s;12s)¡0; 0¡s¡1 or s¿2; x¿0;
i.e.
F′
(x)¿0; F increases if 1¡s¡2; x¿0;
F′
(x)¡0; F decreases if 0¡s¡1 or s¿2; x¿0:
If ≡s0=−12 + (s+14)1=2 we have
Ax(s; s0)≡(1−s)(2s0−s)x:
Note that 2s0−s is positive if 0¡s¡2 and negative if s¿2. Therefore, we have
Ax(s; s0)¡0; 1¡s¡2; x¿0
or
i.e.
Remark 2.1. In [9] Laforgia proved the inequality (x+s)
where equality occurs only when s=34. So the corresponding result given in Theorem 2.1 in the case 1¡s¡2 is stronger than (2.1).
Laforgia noted in [9] that some of the previous results can be improved if we consider only x¿x0¿0, for some x0. He gave the following examples:
Theorem 3.1. Let x¿x0¿0. If either 0¡s¡1 or s¿2 then
Proof. As in the proof of the Theorem 2.1 we consider the function Ax(s; ) to deduce the sign of
F′
Furthermore, the condition x0¿0 gives us two possibilities
s−2
−¿0; 2−s¿0 or s−2
−¡0; 2−s¡0: (3.7)
Therefore the following three cases are of interest: (1) 0¡s¡1 when 1
As a consequence we have
Ax(s; )¿0 if 0¡s¡1 or s¿2;
Ax(s; )¡0 if 1¡s¡2;
for values of dened in the cases 1,3,2, respectively. Hence
F′
(x)¿0 (F ր) if 0¡s¡1 or s¿2; x¿x0;
F′
(x)¡0 (F ւ) if 1¡s¡2; x¿x0:
Further as in the proof of Theorem 2.1 we have the results of our theorem, assuming =.
Remark 3.1. Forx0= 5 andx¿0 we get (3.2) from (3.4), but not only for 0¡s¡1. This inequality
is also valid fors¿2, while for 1¡s¡2 we have reverse inequality. Similarly takingx0= 1 we have
for x¿1
Remark 3.2. The results of Theorems 2.1 and 3.1 in fact give us a complete picture about
inequal-1=2) we should consider Theorem 3.1. We have that for
these values inequality (3.9) is valid if x¿(s−2−)=(2−s). Finally, for ¿−1
2 + (s+ 1 4)
1=2
we have from the second inequality in (1.2) that (3:9′
) is valid for x¿0. Similar considerations can be also given in other cases.
4. A Bernstein-type inequality for ultraspherical polynomials
Recently, Chow et al. [4] have shown a Bernstein-type inequality for the general case of the Jacobi polynomials P(;)
n (x) for −126; 6
ultraspherical polynomials P()
n (x), =+
In [4] it is pointed out that Lorch’s result (1.5) follows from the inequality [10]
(n+ 2)
there follows the only slightly weaker result
(sin)|Pn()(cos)|6 2 1−
()
(n+ 2)2−1
(n+) : (4.3)
Now, taking into account our extension of the Kershaw’s inequality in the Theorem 2.1 to the case 1¡s¡2,
[x− 12+ (s+14)1=2]1−s¡ (x+ 1)
we deduce from (4.2)
(sin)
|P()
n (cos)|6
21−
()
"
n−1 2 +
2+1 4
1=2#2
−1
(n+)−
(4.4)
for n= 1;2; : : : and 1¡2¡2. Note that inequality (4.4) improves (4.3), but not (1.5). The same conclusion is valid if we use the reverse inequalities in (3.4), for n¿n0¿0, for some n0 and
1¡2¡2.
5. Some related results involving the psi function
Gautschi [6] has also given the following inequality for 0¡s¡1 and for x¿0
(x+ 1)
(x+s)¡exp[(1−s) (x+ 1)]; (5.1)
where = ′
= , the logarithmic derivative of the gamma function. The following closer bounds were obtained by Kershaw [8], for x¿0 and 0¡s¡1
exp[(1−s) (x+s1=2)]¡ (x+ 1)
(x+s)¡exp[(1−s) (x+
1
2(s+ 1))]: (5.2)
Here we shall prove an extension of this result.
Theorem 5.1. Let x¿0. For 0¡s¡1 we have (5.2), while for s¿1 we have reverse inequalities in (5.2).
Proof. Dene the function g(x) by g(x) = (x+ 1)
(x+s)exp[(s−1) (x+)]
for x¿0 and s¿0. We have again that
lim
x→∞g(x) = 1:
As in [3] set
G(x) = g(x) g(x+ 1)=
x+s x+ 1exp
1−s
x+
:
Then
G′
(x)
G(x) = (1−s) (2
−s) + (2−s−1)x (x+ 1)(x+s)(x+)2 :
So the sign of G′
(x) depends on the sign of the expression
If =s1=2, we have
Bx(s; s1=2) =−(1−s)(
√
s−1)2x:
So
Bx(s; s1=2)¡0 if 0¡s¡1;
Bx(s; s1=2)¿0 if s¿1:
This means that for 0¡s¡1; G(x) strictly decreases, while for s¿1, G(x) strictly increases. So for 0¡s¡1 we have the rst inequality in (5.2), while for s¿1 we have reverse inequality.
If =1
2(1 +s), we have
Bx(s;12(1 +s)) =14(1−s)3:
So we have similarly that for 0¡s¡1, the second inequality in (5.2) is valid, while for s¿1 we have the reverse inequality.
Remark 5.1. Theorem 5.1 can be given in equivalent form for x; a; b¿0, if we use the same sub-stitution as in Remark 2.2. Namely, if 0¡a−b¡1, then we have
exp[(a−b) [x+a−1 + (b−a+ 1)1=2]¡ (x+a)
(x+b)¡exp[(a−b) (x+
1
2(a+b))]
while for a−b¡0 we have the reverse inequalities.
Theorem 5.2. Let x¿x0¿0. If 0¡s¡1 then
(x+s)
(x+ 1)¡exp[(s−1) (x+)]; (5.3)
where =−x0+
q
x2
0+x0(1 +s) +s; while for s¿1 we have the reverse inequality.
Proof. As in the proof of Theorem 5.1 we have that the sign of G′
(x) depends on Bx(s; ). Note
that Bx(s; ) = 0 if x= (2−s)=(s+ 1−2). Take this value to be x0 i.e.
x0=
2−s s+ 1−2
and derive =−x0+
q
x2
0+x0(1 +s) +s.
Since x0¿0, we have that √s¡¡12(s+ 1). Now we can write
Bx(s; ) = (1−s)(2−s−1)(x−x0)
and we have
Bx(s; )¡0; s¿1;
Bx(s; )¿0; 0¡s¡1;
Example. For x¿1 we have (x+s)
(x+ 1)¡exp[(s−1) (x−1 +
p
2(1 +s))]
if 0¡s¡1 and the reverse inequality if s¿1.
Remark 5.2. Previous results can be used in considerations about inequalities (x+ 1)
(x+s)¿exp[(1−s) (x+)]; (5.4)
(x+ 1)
(x+s)¡exp[(1−s) (x+)]; (5.4
′
)
in the following way: Let 0¡s¡1. From the rst inequality in (5.2) we have that (5.4) is valid for 6√s andx¿0. For values ∈(√s;1
2(1 +s)) we have inequality (5;4 ′
), ifx¿(2−s)=(s+ 1−2).
Finally, for ¿1
2(1 +s) we have from the second inequality in (5.2) that (5:4 ′
) is valid for x¿0.
Acknowledgements
The authors wish to thank Professor Luigi Gatteschi for pointing out Ref. [4] and Professor Walter Gautschi for some helpful comments.
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