Sharper bounds and new proofs of the exponential function with cotangent

1.   Introduction

In 1978, Becker and Stark ^[1] proved the double inequality

holds for all $x \in (0, \pi/2)$. It is clear that the product of $x \cot(x)$ and $\dfrac{\tan(x)}{x}$ is equal to 1, while $F(0) = 0$. In ^[16], it proves that

Since then, many inequalities for cotangent function were established by different ideas and methods; see e.g. ^{[2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]}, including the following inequality

Recently, a new type of bounds for

is proposed in ^[7], see more details in Theorems 1 and 2.

Theorem 1. ^[7] Let $p, q \in (-\infty, 4/\pi^2]$, $p^\star \approx 0.13484$ be the unique zero of the function $\alpha_p(\pi/2)-1$ on $(-\infty, 4/\pi^2)$, where $\alpha_p(x) = G(x)/(1-p x^2)^{1/(3p)}$ if $p \neq 0$ and $\alpha_0(x) = e^{x \cot(x)-1+x^2/3}$. Then the double inequality

holds for all $x \in (0, \pi/2)$ if and only if $p \geq p^\star$ and $q \leq 2/15 \approx 0.13333$.

Theorem 2. ^[7] For $x \in (0, \pi/2)$, one has that the double inequality

Very recently, Zhu ^[17] proposed the following results in Theorems 3 and 4, where $h(p) = \ln (1-\dfrac{\pi^2}{30}-\dfrac{\pi^6}{64}p)-\dfrac{8(45 \pi^4 p + 32)}{15 \pi^6 p + 32 \pi^2 -960}$,

and $p_0$ is the unique zero of the function $h(p)$.

Theorem 3. ^[17] Let $0 < p < p_3$ and $x \in (0, \pi/2)$.

(ⅰ) If $p_2 \leq p < p_3$, then the function $x \rightarrow \dfrac {\ln(1 - 2 x^2/15 - p x^6)}{x \cot x - 1} : = \dfrac{f(x)}{g(x)}$ is strictly increasing on $(0, \pi/2)$, and therefore the double inequality

holds, where $\lambda_p = -\ln(1- \dfrac{\pi^2}{ 30} - \dfrac{\pi^6}{64} p)$.

(ⅱ) If $p_0 < p < p_2$, then there is an $x_0 \in (0, \pi/2)$ such that the function $f/g$ is strictly decreasing on $(0, x_0)$ and strictly increasing on $(x_0, \pi/2)$. Consequently, the inequality

holds, where $\theta_p = \max(2/5, \lambda_p)$. In particular, we have

(ⅲ) If $0 < p < p_0$, then the function $f/g$ is strictly decreasing on $(0, \pi/2)$, and therefore the double inequality (1.9) is reversed.

Theorem 4. ^[17] Let $p_2 \leq p < p_3$ and $p_0 < q \leq p_1$. Then the double inequality

holds for all $x \in (0, \pi/2)$ with the best coefficients $p = p_2$ and $q = p_1$. In particularly, we have

for all $x \in (0, \pi/2)$.

In this paper, we present a new method for finding new bounds of both $F(x)$ and $G(x)$, and also provide a new method for the proofs. The first thing is to find four polynomials such that $0 \leq l_1(x) \leq F(x) \leq l_2(x)$ and $0 \leq l_3(x)^{\rho} \leq G(x) \leq l_4(x)^{\rho}$, where $x \in (0, \pi/2)$ and $\rho = 8/3$ in this paper. Our method for finding the bounds is as follows. Suppose that $B(x) = \sum\limits_{i=0}^{n} b_i x^i$ is a bounding polynomial of degree $n$ to be found. Let $h(x) = (x \cot(x)-1-B(x))\cdot \sin(x) = x \cos(x) - (1+B(x)) \sin(x)$, where $h(0) = 0$. By selecting a suitable $k \in (0, n)$, the unknown parameter of $b_i$ can be determined by the following constraints

which are linear in $b_i$. On the other hand, let $H(x) = G(x)^{1/\rho}-B(x)$. By selecting a suitable $l \in (0, n)$, the unknown parameter of $b_i$ can be determined by the following constraints

which are linear in $b_i$. Through the above way, together with the Maple software, one can find better bounds for both $F(x)$ and $G(x)$.

The main results are as follows, see also the details of Theorems 5 and 6. We also present a new method for proving the new bounds, see more details in the proofs.

Theorem 5. For all $x \in (0, \pi/2)$, we have that

Theorem 6. For all $x \in (0, \pi/2)$, we have that

where

Remark 1. This paper uses Maple software to deduce and verify the formulae and inequalities.

2.   Lemmas

We introduce Theorem 3.5.1 in Page 67, Chapter 3.5 of ^[19] as follows.

Theorem 7. ^[19] Let $w_0$, $w_1$, $\cdots$, $w_r$ be $r$+1 distinct points in $[a, b]$, and $n_0$, $\cdots$, $n_r$ be $r$+1 integers $\geq 0$. Let $N = n_0+ \cdots + n_r + r$. Suppose that $\gamma(t)$ is a polynomial of degree $N$ such that

Then there exists $\xi_1(t) \in [a, b]$ such that

Lemma 1. For all $x \in (0, \pi/2)$, we have that

where

Proof. By using Maple software, it can be verified that

Combining Eq (2.3) with Theorem 7, for all $x \in (0, \pi/2)$, there exists $\xi_j(x) \in (0, \pi/2)$ such that

From Eq (2.4), we have completed the proof.

Lemma 2. For all $x \in (0, \pi/2)$, we have that

where

Proof. By using Maple software, it can be verified that

Combining Eq (2.6) with Theorem 7, for all $x \in (0, \pi/2)$, there exists $\xi_j(x) \in (0, \pi/2)$ such that

From Eq (2.7), we have completed the proof.

Lemma 3. For all $x \in (0, \pi/2)$, we have that

where $D_1(x) = \ln(l_3(x))$ and $D_2(x) = \ln(l_4(x))$.

Proof. By using Maple software, it can be verified that $D_i'(x) = \dfrac{l_{i+2}'(x)}{l_{i+2}(x)}$ is a rational polynomial,

and for all $x \in (0, \pi/2]$,

Combining Eq (2.12) with (2.17), we have that

and complete the proof.

Lemma 4. For all $x \in (0, \pi/2)$, we have that

where

Proof. By using Maple software, it can be verified that

Combing Eq (2.19) with Theorem 7, for all $x \in (0, \pi/2)$, there exists $\xi_i(x) \in (0, \pi/2)$ such that

Combining Eq (2.20) with Lemma 3, we have that

which is equivalent to Eq (2.18), and we complete the proof.

3.   Proof of Theorem 5

Prove that $F(x)-l_1(x) > 0$ and $F(x)-l_2(x) < 0$, for all $x \in (0, \pi/2)$.

Let $E_{i+2}(x) = (F(x)-l_i(x)) \cdot \sin(x) = x \cos(x) - (1+l_i(x)) \sin(x)$, $i = 1, 2$. It is equivalent to prove

For $\forall x \in (0, \pi/2)$, note that $1+l_i(x) > 0$, $i = 1, 2$, combining with Lemma 1, we have that

Combining with Eq (3.2) and Eq (3.3), we have that $E_3(x) > 0$ and $E_4(x) < 0$, $\forall x \in (0, \pi/2)$, and complete the proof of Eq (3.1).

4.   Proof of Theorem 6

It is equivalent to prove that for all $x \in (0, \pi/2)$

Let $E_{i+4}(x) = (x \cot(x)-1 - 8/3 \cdot D_i(x)) \cdot \sin(x) = x \cos(x) - (1+8/3 \cdot D_i(x)) \cdot \sin(x)$, $i = 1, 2$. Eq (4.1) is equivalent to

By using Maple software, it can be verified that $l_i(x) \geq e^{-3/8}$ and $1+8/3 \cdot \ln(l_i(x)) > 0$, $i = 3, 4$. For all $x \in (0, \pi/2)$, combining with Lemmas 2 and 4, we have that

For all $\forall x \in (0, \pi/2)$, $i = 0, 1, \cdots, 15$ and $j = 0, 1, \cdots, 19$, we have that

From Eqs (4.3–4.13), we have that

and complete the proof.

5.   Discussions

It can be verified that

where $\bar{G}(x) = (G(x))^{2/5}$. From the constraints in Eq (5.1) and Eq (5.2), we can recover the resulting bounds $l_{zhu}(x)$ and $u_{zhu}(x)$ in Eq (1.13). In principle, one can find much better bounds by adding other more constraints. Figure 1 shows the error plots of $L_C(x)-L_{Zhu}(x)$ and $U_C(x)-U_{Zhu}(x)$. It shows that $L_{Zhu}(x) < L_C(x) < G(x) < U_C(x) < U_{Zhu}(x)$, for all $x \in (0, \pi/2)$.

6.   Conclusions

This paper provides a new method to find better bounds for the exponential function with cotangent, by using the interpolation constraints at two end-points of the parametric interval. Many previous results can be recovered by using the new method in this paper. Usually, more constraints, much tighter bounds. In principle, it can be directly extended to more cases with other functions. Moreover, it also presents a new method for proving the corresponding bounds, instead of prevailing methods based on the monotonicity of some special functions.

Acknowledgments

We would like to thank you for following the instructions above very closely in advance. It will definitely save us lot of time and expedite the process of your paper's publication. This research work was partially supported by Zhejiang Key Research and Development Project of China (LY19F020041, LY18F020016, 2018C01030), the National Natural Science Foundation of China (61972120, 61672009, 61972122).

Conflict of interest

The authors declare that they have no conflict interests.