Directory UMM :Data Elmu:jurnal:J-a:Journal of Computational And Applied Mathematics:Vol99.Issue1-2.1998:
Journal of Computational and Applied Mathematics 99 (1998) 167–175
Uni ed treatment of Gautschi–Kershaw type inequalities
for the gamma function1
C. Giordano a; ∗ , A. Laforgia b , J. Pecaric c
a
b
Dipartimento di Matematica – Universita, Via Carlo Alberto, 10, 10123 Torino, Italy
Dipartimento di Matematica, Universita di Roma Tre, Largo S. Leonardo Murialdo, 1, Roma, Italy
c
Faculty 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 uni ed 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
c 1998 Elsevier Science B.V. All rights reserved.
of this type.
AMS classi cation: 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 + 1)1−s ;
(x + 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 + 21 s]1−s ¡
(x + 1)
¡[x − 21 + (s + 14 )1=2 ]1−s
(x + s)
1
(1.2)
This work was supported by the Consiglio Nazionale delle Ricerche of Italy and by the Ministero dell’Universita e
della Ricerca Scienti ca e Tecnologica of Italy.
∗
Corresponding author. E-mail: giordano@alpha01.dm.unito.it
c 1998 Elsevier Science B.V. All rights reserved.
0377-0427/98/$ – see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 8 ) 0 0 1 5 4 - X
168
C. Giordano et al. / Journal of Computational and Applied Mathematics 99 (1998) 167–175
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:
(x + 1)
¿[x + 21 s]1−s
(x + s)
for 0¡s¡1 or s¿2;
(1.3)
(x + 1)
¡[x + 21 s]1−s
(x + s)
for 1¡s¡2:
(1.4)
Lorch [10] successfully used (1.3) to prove a sharpened inequality for the ultraspherical polynomials
of degree n and parameter , 0¡¡1, Pn() (x),
(n + )1− (sin ) |Pn() (cos )|¡21− = ();
0 6 6 ;
(1.5)
n = 0; 1; 2; : : :, 0¡¡1, which re nes the customary Bernstein’s inequality [12, p. 171]
n1− (sin ) |Pn() (cos )|¡21− = ();
066
For the important special case of Legendre polynomials, where = 12 , (1.5) gives [2]
1
|Pn (cos )|¡ q
;
(=2)(n + 21 ) sin
0¡¡; n = 1; 2; : : :
which improves the Bernstein inequality [12, p. 165]
1
;
|Pn (cos )|¡ p
(=2)n sin
0¡¡; n = 1; 2; : : : :
Once more the upper bound for the gamma function [5]
s
4
1 4
4
1+
n + 21
16
(n=2 + 12 )
¡1; n = 1; 2; : : :
2 (n=2 + 1)
2
was the key of the proof for the following inequality for the Legendre polynomials:
|Pn (cos )|¡ q
4
1+
1
4
(n
16
+ 12 )4 sin4
;
0¡¡; n = 1; 2; : : :
due to Elbert and Laforgia [5].
In this paper we shall give a uni ed 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
Theorem 2.1. Let x¿0. The inequalities (1.2) hold for 0¡s¡1 or s¿2; while the reverse inequalities hold for 1¡s¡2.
C. Giordano et al. / Journal of Computational and Applied Mathematics 99 (1998) 167–175
Proof. We shall use the same idea of the proofs in [8–10]. From the relation [1, p. 257]
(x + a)
= 1;
(x + b)
lim xb−a
x→∞
a¿0; b¿0
we see that
lim f(x) = 1;
x→∞
where
f(x) =
(x + 1)
(x + )s−1
(x + s)
for all x¿0 and s; ¿0. Now let
F(x) =
x+s
f(x)
=
f(x + 1) x + 1
x+ +1
x+
1−s
then
F ′ (x) (1 − s)[( 2 + − s) + (2 − s)x]
=
:
F(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; 21 s) = 12 s(1 − s)( 21 s − 1);
Ax (s; 12 s)¿0;
1¡s¡2; x¿0;
Ax (s; 21 s)¡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 + 41 )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
Ax (s; s0 )¿0;
0¡s¡1 or s¿2; x¿0;
or
169
170
C. Giordano et al. / Journal of Computational and Applied Mathematics 99 (1998) 167–175
i.e.
F ′ (x)¡0;
F decreases if 1¡s¡2; x¿0;
F ′ (x)¿0;
F increases if 0¡s¡1 or s¿2; x¿0:
So if F strictly decreases, since F(x) → 1 as x → ∞, it follows that F(x)¿1 for x¿0. Hence
f(x)¿f(x + n). Take the limit as n → ∞ to give the result that f(x)¿1. If F is strictly increasing
we have reverse results, i.e. f(x)¡1. So the proof of Theorem 2.1 is complete.
Remark 2.1. In [9] Laforgia proved the inequality
(x + s)
s 1
¡ x+ +
(x + 1)
2 8
s−1
;
1¡s¡2; x ¿ 0:
(2.1)
Note that
− 12 + (s + 41 )1=2 6 12 s + 81 ;
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).
Remark 2.2. Instead of the ratio (x + 1)= (x + s) we can consider the ratio (x + a)= (x + b) for
x; a; b¿0. Namely we shall use only substitutions: x → x + a − 1, s → b − a + 1 in (1.2). So the
inequalities
a+b−1
x+
2
a−b
¡
(x + a)
¡[x + a − 23 + (b − a + 54 )1=2 ]a−b
(x + b)
are valid if x +a¿1, 0¡a−b¡1 or b−a¿1, while the reverse inequalities are valid if 0¡b−a¡1.
3. Inequalities valid for x ¿ x0 ¿ 0
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:
(x + s)
¿(x + 32 s)s−1 ;
(x + 1)
0¡s¡1; x ¿ 1;
√
(x + s)
¿(x + 21 (−11 + 121 + 24s))s−1 ;
(x + 1)
(x + s)
¡(x + 21 s +
(x + 1)
1 s−1
) ;
10
(3.1)
0¡s¡1; x ¿ 5;
1¡s¡2; x ¿ 1:
Here we shall formulate more general results.
(3.2)
(3.3)
171
C. Giordano et al. / Journal of Computational and Applied Mathematics 99 (1998) 167–175
Theorem 3.1. Let x¿x0 ¿0. If either 0¡s¡1 or s¿2 then
(x + s)
¿(x + )s−1 ;
(x + 1)
(3.4)
q
where = 21 (−2x0 −1+ 4x02 + 4x0 (1 + s) + 4s + 1). If 1¡s¡2 we have reverse inequality in (3.4).
Proof. As in the proof of the Theorem 2.1 we consider the function Ax (s; ) to deduce the sign of
F ′ (x). Note that Ax (s; ) = 0 if x = (s − 2 − )=(2 − s). Now we take this value as x0 , i.e.
x0 =
s − 2 −
2 − s
(3.5)
q
and we can deduce the corresponding = 21 (−2x0 − 1 + 4x02 + 4x0 (1 + s) + 4s + 1).
For our consideration the interest is in the sign of Ax (s; ) for x¿x0 . Note that we can write
Ax (s; ) = (1 − s)(2 − s)(x − x0 ):
(3.6)
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 21 s¡ ¡ − 21 + (s + 14 )1=2 ;
(2) 1¡s¡2 when 21 s¡ ¡ − 21 + (s + 14 )1=2 ;
(3) s¿2 when − 21 + (s + 14 )1=2 ¡ ¡ 12 s:
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 de ned 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. For x0 = 5 and x¿0 we get (3.2) from (3.4), but not only for 0¡s¡1. This inequality
is also valid for s¿2, while for 1¡s¡2 we have reverse inequality. Similarly taking x0 = 1 we have
for x¿1
√
(x + s)
¿[x + 21 (−3 + 9 + 8s)]s−1 if 0¡s¡1 or s¿2
(3.8)
(x + 1)
√
while for 1¡s¡2 we have reverse inequality in (3.8). Note that 21 (−3 + 9 + 8s)¡ 23 s, so (3.8) is
√
also more stringent than (3.1). Also for s ∈ (1; 2) we have 21 (−3 + 9 + 8s)¡ 12 s + 101 ; so (3.8) is
better than (3.3).
172
C. Giordano et al. / Journal of Computational and Applied Mathematics 99 (1998) 167–175
Remark 3.2. The results of Theorems 2.1 and 3.1 in fact give us a complete picture about inequalities of the form
(x + 1)
(x + )1−s ¡
(3.9)
(x + s)
or
(x + 1)
¡(x + )1−s :
(3.9′ )
(x + s)
For example, let 0¡s¡1. From the rst inequality in (1.2) we have that (3.4) is valid for all 6 21 s
and x¿0. For values ∈ ( 21 s; − 21 + (s + 41 )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 ¿ − 21 + (s + 41 )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 Pn( ; ) (x) for − 12 6 ; 6 21 and 0 6 6
sin
2
+ 1
2
cos
2
+ 1
2
|Pn( ; ) (cos )| 6
(q + 1)
( 12 )
1
n+q
N −q− 2 ;
n
(4.1)
where q = max( ; ) and N = n + 21 ( + + 1). If = , the Jacobi polynomials reduces to the
ultraspherical polynomials Pn() (x), = + 12 and (4.1) gives [4]
(sin ) |Pn() (cos )| 6
21− (n + 2)
(n + )− :
() (n + 1)
(4.2)
In [4] it is pointed out that Lorch’s result (1.5) follows from the inequality [10]
(n + 2)
¡(n + )2−1 ;
(n + 1)
provided that 0¡2¡1. In the case 1¡2¡2, by using the Lorch’s inequality [10]
(n + 2)
¡(n + 2)2−1
(n + 1)
there follows the only slightly weaker result
(sin ) |Pn() (cos )| 6
21− (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 + 41 )1=2 ]1−s ¡
(x + 1)
(x + s)
C. Giordano et al. / Journal of Computational and Applied Mathematics 99 (1998) 167–175
173
we deduce from (4.2)
(sin )
|Pn() (cos )| 6
"
21−
1
1
n − + 2 +
()
2
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)
¡ exp[(1 − s) (x + 1)];
(x + s)
(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)
¡ exp[(1 − s) (x + 12 (s + 1))]:
(x + s)
(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. De ne the function g(x) by
(x + 1)
exp[(s − 1) (x + )]
(x + s)
g(x) =
for x¿0 and s¿0. We have again that
lim g(x) = 1:
x→∞
As in [3] set
G(x) =
x+s
1−s
g(x)
=
exp
:
g(x + 1) x + 1
x+
Then
G ′ (x)
( 2 − s) + (2 − s − 1)x
= (1 − s)
:
G(x)
(x + 1)(x + s)(x + )2
So the sign of G ′ (x) depends on the sign of the expression
Bx (s; ) = (1 − s)[( 2 − s) + (2 − s − 1)x]:
174
C. Giordano et al. / Journal of Computational and Applied Mathematics 99 (1998) 167–175
If = s1=2 , we have
√
Bx (s; s1=2 ) = − (1 − s)( s − 1)2 x:
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 = 21 (1 + s), we have
Bx (s; 21 (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 substitution 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)
¡ exp[(a − b) (x + 12 (a + b))]
(x + 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)
¡ exp[(s − 1) (x + )];
(x + 1)
where = − x0 +
(5.3)
q
x02 + 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
q
and derive = − x0 + x02 + x0 (1 + s) + s.
√
Since x0 ¿0, we have that s¡ ¡ 21 (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;
and as in the previous proofs we have the results of our theorem, assuming = .
C. Giordano et al. / Journal of Computational and Applied Mathematics 99 (1998) 167–175
175
Example. For x¿1 we have
p
(x + s)
¡ exp[(s − 1) (x − 1 + 2(1 + s))]
(x + 1)
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)
¿ exp[(1 − s) (x + )];
(x + s)
(5.4)
(x + 1)
¡ exp[(1 − s) (x + )];
(x + s)
(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 and x¿0. For values ∈ ( s; 12 (1 + s)) we have inequality (5; 4′ ), if x¿( 2 − s)=(s + 1 − 2 ).
Finally, for ¿ 21 (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.
References
[1] M. Abramowitz, I.A. Stegun (Eds.), Handbook of Mathematical Functions, Appl. Math. Series 55, National Bureau
of Standards, Washington, DC, 1964.
[2] V.A. Antonov, K.V. Holsevnikov, An estimate of the remainder in the expansion of the generating function for the
Legendre polynomials (Generalization and improvement of Bernstein’s inequality), Vestnik Leningrad Univ. Math.
13 (1981) 163 –166.
[3] J. Bustoz, M.E.H. Ismail, On gamma function inequalities, Math. Comp. 47 (176) (1986) 659 – 667.
[4] Y. Chow, L. Gatteschi, R. Wong, A Bernstein-type inequality for the Jacobi polynomial, Proc. Amer. Math. Soc.
121 (3) (1994) 703 –709.
[5] A. Elbert, A. Laforgia, An inequality for Legendre polynomials, J. Math. Phys. 35 (3) (1994) 1348 –1360 .
[6] W. Gautschi, Some elementary inequalities relating to the gamma and incomplete gamma function, J. Math. Phys.
38 (1959) 77– 81.
[7] W. Gautschi, The incomplete gamma functions since Tricomi, in: Proc. Internat. Meeting on Tricomi’s Ideas and
Contemporary Applied Mathematics, Atti Convegni Lincei (1998) Spec. Vol. (in print).
[8] D. Kershaw, Some extensions of W. Gautschi’s inequalities for the gamma function, Math. Comp. 41 (164) (1983)
607– 611.
[9] A. Laforgia, Further inequalities for the gamma function, Math. Comp. 42 (166) (1984) 597– 600.
[10] L. Lorch, Inequalities for ultraspherical polynomials and the gamma function, J. Approx. Theory 40 (1984) 115 –120.
[11] B. Palumbo, A generalization of some inequalities for the gamma function, J. Comput. Appl. Math. 88 (2) (1998)
255 –268.
[12] G. Szego, Orthogonal Polynomials, 4th ed., Colloq. Publ., vol. 23, Amer. Math. Soc., Providence, RI, 1975.
Uni ed treatment of Gautschi–Kershaw type inequalities
for the gamma function1
C. Giordano a; ∗ , A. Laforgia b , J. Pecaric c
a
b
Dipartimento di Matematica – Universita, Via Carlo Alberto, 10, 10123 Torino, Italy
Dipartimento di Matematica, Universita di Roma Tre, Largo S. Leonardo Murialdo, 1, Roma, Italy
c
Faculty 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 uni ed 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
c 1998 Elsevier Science B.V. All rights reserved.
of this type.
AMS classi cation: 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 + 1)1−s ;
(x + 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 + 21 s]1−s ¡
(x + 1)
¡[x − 21 + (s + 14 )1=2 ]1−s
(x + s)
1
(1.2)
This work was supported by the Consiglio Nazionale delle Ricerche of Italy and by the Ministero dell’Universita e
della Ricerca Scienti ca e Tecnologica of Italy.
∗
Corresponding author. E-mail: giordano@alpha01.dm.unito.it
c 1998 Elsevier Science B.V. All rights reserved.
0377-0427/98/$ – see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 8 ) 0 0 1 5 4 - X
168
C. Giordano et al. / Journal of Computational and Applied Mathematics 99 (1998) 167–175
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:
(x + 1)
¿[x + 21 s]1−s
(x + s)
for 0¡s¡1 or s¿2;
(1.3)
(x + 1)
¡[x + 21 s]1−s
(x + s)
for 1¡s¡2:
(1.4)
Lorch [10] successfully used (1.3) to prove a sharpened inequality for the ultraspherical polynomials
of degree n and parameter , 0¡¡1, Pn() (x),
(n + )1− (sin ) |Pn() (cos )|¡21− = ();
0 6 6 ;
(1.5)
n = 0; 1; 2; : : :, 0¡¡1, which re nes the customary Bernstein’s inequality [12, p. 171]
n1− (sin ) |Pn() (cos )|¡21− = ();
066
For the important special case of Legendre polynomials, where = 12 , (1.5) gives [2]
1
|Pn (cos )|¡ q
;
(=2)(n + 21 ) sin
0¡¡; n = 1; 2; : : :
which improves the Bernstein inequality [12, p. 165]
1
;
|Pn (cos )|¡ p
(=2)n sin
0¡¡; n = 1; 2; : : : :
Once more the upper bound for the gamma function [5]
s
4
1 4
4
1+
n + 21
16
(n=2 + 12 )
¡1; n = 1; 2; : : :
2 (n=2 + 1)
2
was the key of the proof for the following inequality for the Legendre polynomials:
|Pn (cos )|¡ q
4
1+
1
4
(n
16
+ 12 )4 sin4
;
0¡¡; n = 1; 2; : : :
due to Elbert and Laforgia [5].
In this paper we shall give a uni ed 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
Theorem 2.1. Let x¿0. The inequalities (1.2) hold for 0¡s¡1 or s¿2; while the reverse inequalities hold for 1¡s¡2.
C. Giordano et al. / Journal of Computational and Applied Mathematics 99 (1998) 167–175
Proof. We shall use the same idea of the proofs in [8–10]. From the relation [1, p. 257]
(x + a)
= 1;
(x + b)
lim xb−a
x→∞
a¿0; b¿0
we see that
lim f(x) = 1;
x→∞
where
f(x) =
(x + 1)
(x + )s−1
(x + s)
for all x¿0 and s; ¿0. Now let
F(x) =
x+s
f(x)
=
f(x + 1) x + 1
x+ +1
x+
1−s
then
F ′ (x) (1 − s)[( 2 + − s) + (2 − s)x]
=
:
F(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; 21 s) = 12 s(1 − s)( 21 s − 1);
Ax (s; 12 s)¿0;
1¡s¡2; x¿0;
Ax (s; 21 s)¡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 + 41 )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
Ax (s; s0 )¿0;
0¡s¡1 or s¿2; x¿0;
or
169
170
C. Giordano et al. / Journal of Computational and Applied Mathematics 99 (1998) 167–175
i.e.
F ′ (x)¡0;
F decreases if 1¡s¡2; x¿0;
F ′ (x)¿0;
F increases if 0¡s¡1 or s¿2; x¿0:
So if F strictly decreases, since F(x) → 1 as x → ∞, it follows that F(x)¿1 for x¿0. Hence
f(x)¿f(x + n). Take the limit as n → ∞ to give the result that f(x)¿1. If F is strictly increasing
we have reverse results, i.e. f(x)¡1. So the proof of Theorem 2.1 is complete.
Remark 2.1. In [9] Laforgia proved the inequality
(x + s)
s 1
¡ x+ +
(x + 1)
2 8
s−1
;
1¡s¡2; x ¿ 0:
(2.1)
Note that
− 12 + (s + 41 )1=2 6 12 s + 81 ;
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).
Remark 2.2. Instead of the ratio (x + 1)= (x + s) we can consider the ratio (x + a)= (x + b) for
x; a; b¿0. Namely we shall use only substitutions: x → x + a − 1, s → b − a + 1 in (1.2). So the
inequalities
a+b−1
x+
2
a−b
¡
(x + a)
¡[x + a − 23 + (b − a + 54 )1=2 ]a−b
(x + b)
are valid if x +a¿1, 0¡a−b¡1 or b−a¿1, while the reverse inequalities are valid if 0¡b−a¡1.
3. Inequalities valid for x ¿ x0 ¿ 0
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:
(x + s)
¿(x + 32 s)s−1 ;
(x + 1)
0¡s¡1; x ¿ 1;
√
(x + s)
¿(x + 21 (−11 + 121 + 24s))s−1 ;
(x + 1)
(x + s)
¡(x + 21 s +
(x + 1)
1 s−1
) ;
10
(3.1)
0¡s¡1; x ¿ 5;
1¡s¡2; x ¿ 1:
Here we shall formulate more general results.
(3.2)
(3.3)
171
C. Giordano et al. / Journal of Computational and Applied Mathematics 99 (1998) 167–175
Theorem 3.1. Let x¿x0 ¿0. If either 0¡s¡1 or s¿2 then
(x + s)
¿(x + )s−1 ;
(x + 1)
(3.4)
q
where = 21 (−2x0 −1+ 4x02 + 4x0 (1 + s) + 4s + 1). If 1¡s¡2 we have reverse inequality in (3.4).
Proof. As in the proof of the Theorem 2.1 we consider the function Ax (s; ) to deduce the sign of
F ′ (x). Note that Ax (s; ) = 0 if x = (s − 2 − )=(2 − s). Now we take this value as x0 , i.e.
x0 =
s − 2 −
2 − s
(3.5)
q
and we can deduce the corresponding = 21 (−2x0 − 1 + 4x02 + 4x0 (1 + s) + 4s + 1).
For our consideration the interest is in the sign of Ax (s; ) for x¿x0 . Note that we can write
Ax (s; ) = (1 − s)(2 − s)(x − x0 ):
(3.6)
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 21 s¡ ¡ − 21 + (s + 14 )1=2 ;
(2) 1¡s¡2 when 21 s¡ ¡ − 21 + (s + 14 )1=2 ;
(3) s¿2 when − 21 + (s + 14 )1=2 ¡ ¡ 12 s:
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 de ned 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. For x0 = 5 and x¿0 we get (3.2) from (3.4), but not only for 0¡s¡1. This inequality
is also valid for s¿2, while for 1¡s¡2 we have reverse inequality. Similarly taking x0 = 1 we have
for x¿1
√
(x + s)
¿[x + 21 (−3 + 9 + 8s)]s−1 if 0¡s¡1 or s¿2
(3.8)
(x + 1)
√
while for 1¡s¡2 we have reverse inequality in (3.8). Note that 21 (−3 + 9 + 8s)¡ 23 s, so (3.8) is
√
also more stringent than (3.1). Also for s ∈ (1; 2) we have 21 (−3 + 9 + 8s)¡ 12 s + 101 ; so (3.8) is
better than (3.3).
172
C. Giordano et al. / Journal of Computational and Applied Mathematics 99 (1998) 167–175
Remark 3.2. The results of Theorems 2.1 and 3.1 in fact give us a complete picture about inequalities of the form
(x + 1)
(x + )1−s ¡
(3.9)
(x + s)
or
(x + 1)
¡(x + )1−s :
(3.9′ )
(x + s)
For example, let 0¡s¡1. From the rst inequality in (1.2) we have that (3.4) is valid for all 6 21 s
and x¿0. For values ∈ ( 21 s; − 21 + (s + 41 )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 ¿ − 21 + (s + 41 )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 Pn( ; ) (x) for − 12 6 ; 6 21 and 0 6 6
sin
2
+ 1
2
cos
2
+ 1
2
|Pn( ; ) (cos )| 6
(q + 1)
( 12 )
1
n+q
N −q− 2 ;
n
(4.1)
where q = max( ; ) and N = n + 21 ( + + 1). If = , the Jacobi polynomials reduces to the
ultraspherical polynomials Pn() (x), = + 12 and (4.1) gives [4]
(sin ) |Pn() (cos )| 6
21− (n + 2)
(n + )− :
() (n + 1)
(4.2)
In [4] it is pointed out that Lorch’s result (1.5) follows from the inequality [10]
(n + 2)
¡(n + )2−1 ;
(n + 1)
provided that 0¡2¡1. In the case 1¡2¡2, by using the Lorch’s inequality [10]
(n + 2)
¡(n + 2)2−1
(n + 1)
there follows the only slightly weaker result
(sin ) |Pn() (cos )| 6
21− (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 + 41 )1=2 ]1−s ¡
(x + 1)
(x + s)
C. Giordano et al. / Journal of Computational and Applied Mathematics 99 (1998) 167–175
173
we deduce from (4.2)
(sin )
|Pn() (cos )| 6
"
21−
1
1
n − + 2 +
()
2
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)
¡ exp[(1 − s) (x + 1)];
(x + s)
(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)
¡ exp[(1 − s) (x + 12 (s + 1))]:
(x + s)
(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. De ne the function g(x) by
(x + 1)
exp[(s − 1) (x + )]
(x + s)
g(x) =
for x¿0 and s¿0. We have again that
lim g(x) = 1:
x→∞
As in [3] set
G(x) =
x+s
1−s
g(x)
=
exp
:
g(x + 1) x + 1
x+
Then
G ′ (x)
( 2 − s) + (2 − s − 1)x
= (1 − s)
:
G(x)
(x + 1)(x + s)(x + )2
So the sign of G ′ (x) depends on the sign of the expression
Bx (s; ) = (1 − s)[( 2 − s) + (2 − s − 1)x]:
174
C. Giordano et al. / Journal of Computational and Applied Mathematics 99 (1998) 167–175
If = s1=2 , we have
√
Bx (s; s1=2 ) = − (1 − s)( s − 1)2 x:
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 = 21 (1 + s), we have
Bx (s; 21 (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 substitution 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)
¡ exp[(a − b) (x + 12 (a + b))]
(x + 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)
¡ exp[(s − 1) (x + )];
(x + 1)
where = − x0 +
(5.3)
q
x02 + 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
q
and derive = − x0 + x02 + x0 (1 + s) + s.
√
Since x0 ¿0, we have that s¡ ¡ 21 (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;
and as in the previous proofs we have the results of our theorem, assuming = .
C. Giordano et al. / Journal of Computational and Applied Mathematics 99 (1998) 167–175
175
Example. For x¿1 we have
p
(x + s)
¡ exp[(s − 1) (x − 1 + 2(1 + s))]
(x + 1)
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)
¿ exp[(1 − s) (x + )];
(x + s)
(5.4)
(x + 1)
¡ exp[(1 − s) (x + )];
(x + s)
(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 and x¿0. For values ∈ ( s; 12 (1 + s)) we have inequality (5; 4′ ), if x¿( 2 − s)=(s + 1 − 2 ).
Finally, for ¿ 21 (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.
References
[1] M. Abramowitz, I.A. Stegun (Eds.), Handbook of Mathematical Functions, Appl. Math. Series 55, National Bureau
of Standards, Washington, DC, 1964.
[2] V.A. Antonov, K.V. Holsevnikov, An estimate of the remainder in the expansion of the generating function for the
Legendre polynomials (Generalization and improvement of Bernstein’s inequality), Vestnik Leningrad Univ. Math.
13 (1981) 163 –166.
[3] J. Bustoz, M.E.H. Ismail, On gamma function inequalities, Math. Comp. 47 (176) (1986) 659 – 667.
[4] Y. Chow, L. Gatteschi, R. Wong, A Bernstein-type inequality for the Jacobi polynomial, Proc. Amer. Math. Soc.
121 (3) (1994) 703 –709.
[5] A. Elbert, A. Laforgia, An inequality for Legendre polynomials, J. Math. Phys. 35 (3) (1994) 1348 –1360 .
[6] W. Gautschi, Some elementary inequalities relating to the gamma and incomplete gamma function, J. Math. Phys.
38 (1959) 77– 81.
[7] W. Gautschi, The incomplete gamma functions since Tricomi, in: Proc. Internat. Meeting on Tricomi’s Ideas and
Contemporary Applied Mathematics, Atti Convegni Lincei (1998) Spec. Vol. (in print).
[8] D. Kershaw, Some extensions of W. Gautschi’s inequalities for the gamma function, Math. Comp. 41 (164) (1983)
607– 611.
[9] A. Laforgia, Further inequalities for the gamma function, Math. Comp. 42 (166) (1984) 597– 600.
[10] L. Lorch, Inequalities for ultraspherical polynomials and the gamma function, J. Approx. Theory 40 (1984) 115 –120.
[11] B. Palumbo, A generalization of some inequalities for the gamma function, J. Comput. Appl. Math. 88 (2) (1998)
255 –268.
[12] G. Szego, Orthogonal Polynomials, 4th ed., Colloq. Publ., vol. 23, Amer. Math. Soc., Providence, RI, 1975.