Abstract and Applied Analysis

Abstract and Applied Analysis / 2014 / Article

Research Article | Open Access

Volume 2014 |Article ID 702718 | 9 pages | https://doi.org/10.1155/2014/702718

A Double Inequality for the Trigamma Function and Its Applications

Academic Editor: Wing-Sum Cheung
Received19 Mar 2014
Accepted26 May 2014
Published16 Jul 2014

Abstract

We prove that and are the best possible parameters in the interval such that the double inequality holds for . As applications, some new approximation algorithms for the circumference ratio and Catalan constant are given. Here, is the trigamma function.

1. Introduction

For real and positive values of , the classical Euler’s gamma function and its logarithmic derivative , the so-called psi function, are defined as For extension of these functions to complex variables and for basic properties, see [1]. The derivatives are known as polygamma functions (see [2]). In particular, is called trigamma function.

Recently, the bounds for the trigamma function using exponential functions have attracted the attention of many researchers. For example, Elezović et al. [3] proved that the inequality holds for all . In [4, Theorem 2.7], Batir proved that and are the best possible constants such that the double inequality holds for all , where is Euler’s constant. Batir [4] also showed that for all . In [5, (1.11)], Guo and Qi established that if and , . They [6, Lemma 2] found a very simple upper bound for trigamma function in terms of exponential function as follows: for all . The inequality (6) was generalized in [7, Theorem 3.1], [8, Theorem 1.1], and [9, Theorem 1.1] to a complete monotonicity which reads that the difference is completely monotonic on . Many other new results involving the psi and trigamma functions can be found in the literature [10, 11].

Suppose that and and are the real functions defined on . is said to be the best possible constant in such that the inequality holds for all if on , or and are not comparable on , and for any satisfies on the interval .

The main purpose of this paper is to find the best possible constants such that the double inequality or equivalently holds for all , where Our main result is the following Theorem 1.

Theorem 1. and are the best possible constants in the interval such that the double inequality (7) or (8) holds for all .

From Theorem 1, we clearly see the following.

Corollary 2. The double inequality holds for all .

2. Lemmas

Lemma 3. Let the function be defined on by (9). Then the function is strictly decreasing with respect to on and strictly increasing on .

Proof. It follows from (9) that If , then ; that is, is strictly increasing with respect to . Therefore, which implies that .
If , then ; that is, is strictly decreasing with respect to , which leads to the conclusion that

Lemma 4. Let the function be defined on by (9) and Then the equation has two roots such that for and for .

Proof. Differentiation yields which reveals that the function is strictly increasing on and strictly decreasing on . Therefore, Lemma 4 follows from the piecewise monotonicity of the function and the numerical computations results:

Lemma 5. Let , , and and let be defined as in Lemma 4. Then the following statements are true:(i)if the inequality holds for all , then ;(ii)if the inequality holds for all , then .

Proof. It follows from the series formulas that and we get (i)If inequality holds for all , then, from and Lemma 4, we clearly see that . (ii)If inequality holds for all , then and Lemma 4 lead to the conclusion that

Lemma 6. Let the function be defined on by (9). Then and are not comparable for all if .

Proof. For and , let Then simple computation leads to From (20) and (21), we have Differentiation yields which shows that is strictly decreasing on . Therefore, if . On the other hand, we clearly see that for .

Lemma 7. Let and with and let be the polynomial of degree defined by where and for with . Then, there exists such that and for and for .

Proof. Differentiating gives Note that for .
From , we clearly see that is strictly increasing on . Then and lead to the conclusion that there exists such that and for and for . Therefore, is strictly decreasing on and strictly increasing on .
It follows from the piecewise monotonicity of and that for , and there exists such that and for and for . Therefore, is strictly decreasing on and strictly increasing on .
After repeating the same steps as above times, we deduce that there exists such that and for and for .

Lemma 8. Let the function be defined on by (9). Then there exists such that and are not comparable for all if , and for all if .

Proof. For and , let Then simple computation leads to
(i) We prove that there exists such that and are not comparable for all if . For this end, it suffices to prove that there exists such that and if .
Indeed, it follows from that the function is strictly increasing on . Numerical computations show that
Therefore, there exists such that and for and for .
On the other hand, it follows from together with (20) and (21) that
(ii)  We prove that for all if . From Lemma 3, we know that is strictly decreasing on , so it suffices to prove that for all .
Let Then Differentiation yields where Note that where the second equation in (46) follows from (38).
Differentiating leads to where We assert that there exists a unique such that for and for , which leads to the conclusion that is strictly decreasing on and strictly increasing on . To this end, it is enough to verify that the coefficients of satisfy the conditions of Lemma 7. In fact, it follows from that we have From the piecewise monotonicity of together with (46) and (47), we clearly see that that is, for . Then (43) and (45) lead to the conclusion that for , which implies that is strictly increasing on and for .
Therefore, follows easily from (40) and .

Lemma 9 (see [12, pp. 258–260]). Let and . Then

From the proof of [4, Theorem 2.6], we get the following.

Lemma 10. The inequality holds for .

The following lemma can be derived immediately from the proof of [4, Theorem 2.1].

Lemma 11 (see [4, Theorem 2.1]). Let be the function defined on by Then for .

The well-known Hermite-Hadamard inequality for convex function can be stated as follows.

Lemma 12 (see [13]). Let be an interval, with , and let be a convex function. Then

3. Proofs of Theorem 1

Proof of Theorem 1. We divide the proof into four parts.
(I) We prove the first inequality in (7); that is, where is defined by (14).
It follows from Lemma 10 that where
We clearly see that it is enough to prove that for .
Differentiating gives where denotes the logarithmic mean of positive numbers and , and Differentiating leads to for , which means that is strictly increasing on and . It in turn implies that is strictly decreasing on . Therefore, for .
(II) We prove that is the best possible constant such that for all .
From Lemma 5, we know that if for all , where . It follows from Lemma 6 that and are not comparable for all if ; that is to say, is not a better lower bound of than even if there exists such that .
For any , (22) leads to
It follows from (19), (20), and (21) that we get
(III) We prove the second inequality (7); that is, for , where is defined by (14).
Lemma 11 implies that the function is strictly convex on . Then, making use of Lemma 12, we get for . That is,
Therefore, inequality (64) follows easily from (52) and the last inequality above.
(IV) We prove that is the best possible constant such that for all .
From Lemma 5, we know that if for all , where .
It follows from Lemma 8 that there exists such that and are not comparable for all if and for all if . Lemma 3 leads to the conclusion that for all if .
If there exists such that for all , then, from (19), (20), (21), and (22), we get

From the above proof and Lemma 3 we get the following.

Corollary 13. Let the function be defined on by (9) and let be the root of (38) on . Then the inequalities or equivalently hold for all .

4. Remarks

Remark 14. It follows from (67) and the facts that that we clearly see that the upper bound in Theorem 1 for the trigamma function is better than the upper bounds given in (2), (3), (4), (5), and (6) if is large enough.

Lemma 15. One has

Proof. Differentiation leads to If , then, from , we get , which leads to If , then the first inequality in (71) still holds by making a change of variable . If , since , we see that for and for . Hence,

Remark 16. Using inequality (71), one has for , which shows that the upper bound in (8) is better than the upper bound in (6).

Remark 17. The conclusion that the difference is completely monotonic on given in [6, Lemma 2] implies that or It is easy to check that the lower bound and for given in (10) are not comparable due to

Remark 18. Guo et al. [14] proved that for if . In particular, if , one has
We clearly see that the upper bound given in (69) is better than that in (80) for the trigamma function .

Finally, we give remarks on two mathematical constants and (Catalan constant).

Remark 19. It is well known that Let , and then . From Theorem 1, we clearly see that the double inequality holds for all .
Let