New bounds for the exponential function with cotangent.

In this paper, new bounds for the exponential function with cotangent are found by using the recurrence relation between coefficients in the expansion of power series of the function [Formula: see text] and a new criterion for the monotonicity of the quotient of two power series.

Introduction

In 1978, Becker and Stark [1] proved the double inequality or holds for all . Since then, many inequalities for cotangent function were established by different ideas and methods; see e.g. [2-18]. Very recently, Lv, Yang, Luo, and Zheng [19] gave a new type of bounds for the function . More precisely, they proved the following results.

Theorem A

Let , be the unique zero of the function on , where if and . Then the double inequality holds for all if and only if and .

Theorem B

For , the double inequality holds.

Throughout the full text, we suppose that where is the unique zero of the function on .

Now considering the asymptotic expansion of , we have It is interesting that the power series above has not item of , which also remind us to establish a more accurate estimate for . The first aim of this paper is to determine the best parameters p and q such that the double inequality holds for all . The main conclusions of this paper are proved by the recursive method and a new criterion for the monotonicity of the quotient of two power series. The following result is a theorem on the recurrence relation of coefficients in the series expansion of the function .

Theorem 1

Let and . Then the function can be expressed in the form of a series, where and, for , Moreover, for all .

Our main results are contained in the following theorems.

Theorem 2

Let and .

If , then the function is strictly increasing on , and therefore the double inequality holds, where

If , then there is an such that the function is strictly decreasing on and strictly increasing on . Consequently, the inequality holds, where . In particular, we have

If , then the function is strictly decreasing on , and therefore the double inequality (1.8) is reversed.

As a consequence of Theorem 2, we immediately get the following.

Theorem 3

Let and . Then the double inequality holds for all with the best coefficients and . In particular, we have for all .

The second aim of this paper is to refine some known results presented in [19], we shall state it carefully in the fifth section.

Lemmas

In this paper, we will use some methods, such as the monotone form of l’Hospital’s rule, an important criterion for the monotonicity of the quotient of two power series, and the latest promotion of the latter.

Lemma 1

([20, 21])

For , let be continuous functions that are differentiable on , with or . Assume that for each x in . If is increasing (decreasing) on , then so is .

Lemma 2

([22-24])

Let and be two real power series converging on () with for all k. If the sequence is increasing (decreasing) for all k, then the function is also increasing (decreasing) on .

Now, we will introduce a useful auxiliary function . For , let f and g be differentiable on and on . Then the function is defined by The function has some good properties [25, Property 1] and plays an important role in the proof of a monotonicity criterion for the quotient of power series (see [26]).

Lemma 3

([26, Theorem 1])

Let and be two real power series converging on and for all k. Suppose that for certain , the non-constant sequence is increasing (resp. decreasing) for and decreasing (resp. increasing) for . Then the function is strictly increasing (resp. decreasing) on if and only if (resp. ≤) 0. Moreover, if (resp. >) 0, then there exists such that the function is strictly increasing (resp. decreasing) on and strictly decreasing (resp. increasing) on .

Lemma 4

([27])

For , the Bernoulli numbers satisfy

Lemma 5

For , let be defined by (1.4). Then has a unique zero such that for and for .

Proof

Differentiation yields which together with the facts that reveals that there is a unique such that for and for . Numerically, the equation for p on has the solution . This completes the proof. □

Proof of Theorem 1

Since and , we see that which shows that holds for all . It remains to determine the coefficients . Differentiation for the two sides of (3.1) gives which is equivalent to Comparing coefficients gives the recurrence formulas (1.7) and (1.6).

From the second equality of (3.1) we easily find that for all , which completes the proof. □

Proofs of Theorems 2 and 3

Proof of Theorem 2

Using the expansion the function can be expressed as by Theorem 1, where . We now observe the monotonicity of the sequence . Since for all , it suffices to determine the sign of . Direct computations yield We claim that for . In fact, by means of the recurrence formula (1.7), we have Clearly, if we prove for , then it follows that . Using the right hand side of (2.2) we have for . This together with yields for .

If , then for , that is, the sequence is strictly increasing, so is on by Lemma 2. Therefore, we conclude that which implies (1.8).

If , then , and for , which indicates that the sequence is decreasing for and increasing for . By Lemma 3, to determine the monotonicity of on , we have to observe the sign of . A simple computation leads us to

Subcase 2.1: For . By Lemma 5 we see that . It follows from Lemma 3 that there is an such that the function is strictly decreasing on and strictly increasing on , where . Consequently, we obtain which implies (1.9).

In particular, if , that is, , then the inequality (1.10) holds. If , that is, , then the inequality (1.11) holds.

Subcase 2.2: For . By Lemma 5 we see that . From Lemma 3 it is deduced that is strictly decreasing on , and so the inequalities (1.8) reverse. This completes the proof. □

Proof of Theorems 3

Let Since we find that the function is decreasing with respect to p on . Then by the left hand side of (1.8) and by (1.10) we can complete the proof of Theorem 3. □

Consequences and remarks

Remark 1

One can obtain the double inequality (1.12) using the key theorem of Wu and Debnath [28].

Let in (iii) of Theorem 2. Then we have the following.

Corollary 1

The function is strictly decreasing on , and therefore, the inequalities hold for , where are the best constants.

It suffices to show the first inequality in (5.1). Consider the monotonicity of the function on . Since holds for all , we see that the function is decreasing on . So which completes the proof of Corollary 1. □

Theorem 4

Let . Then the function is strictly decreasing on if and only if . And therefore, for , the double inequality holds for , where .

The necessity follows from To prove the sufficiency, we note that where is positive and decreasing on by Corollary 1, it thus suffices to prove the function is positive and decreasing on . A simple computation gives which indicates that is strictly decreasing on by Lemma 1. Meanwhile is obviously positive for , which proves the sufficiency.

Inequalities (5.3) follow from the decreasing property of the function .

The proof is finished. □

Remark 2

It is easy to check that, for and , where , . In fact, we have Due to , the first factor is positive. And, since is decreasing in p, it follows that, for , These imply that the inequality (5.4) holds. It thus can be seen that the above theorem partly refines Lv et al.’s result [19].

Remark 3

We claim that the lower bound in the double inequality (5.3) is strictly increasing with respect to the parameter p. In fact, put with , then differentiation yields which indicates that is increasing in p by Lemma 1.

Remark 4

Taking , , and in Theorem 4. Using the monotonicity of the lower and upper bounds in (5.3), we can obtain where . This shows that our double inequality (5.3) is a generalization and refinement of the one (1.2) (see [19]).

The following theorem gives a sufficient condition for the function to be increasing on .

Theorem 5

The function defined by (5.2) is strictly increasing on if . And therefore, for , the double inequality (5.3) is reversed.

We have If we prove for , then by Lemma 2 is increasing on , and the reverse of double inequality (5.3) follows. Using the right hand side of (2.2) we have for . This together with and yields for .

This completes the proof. □

Remark 5

Likewise, taking , , in the above theorem, we have where , .

Finally, we consider the monotonicity of the function on . In this case, we have to assume that .

Theorem 6

Let . Then the function defined by (5.2) is strictly decreasing on . And therefore, the inequality holds for all .

From Lemma 2 and the proof of the previous theorem, it suffices to prove for and . Using the left hand side of (2.2) we get for , which, by Lemma 2, proves the decreasing property of . □

Remark 6

Let in Theorem 3, we can obtain which is sharper than the right hand side of (1.2):

Remark 7

Let in Theorem 2 or Theorem 3, we can obtain Comparing the inequality above with the left hand side of (1.2), we find that they are not included in each other.

Conclusions

In the present study, we first obtain some new bounds for the exponential function with cotangent by using the recurrence relation between coefficients in the expansion of power series of the function and a new criterion for the monotonicity of the quotient of two power series. Then we refine some well-known results presented in [19].