On Qi’s Normalized Remainder of Maclaurin Power Series Expansion of Logarithm of Secant Function ^†

Abstract

In the study, the authors introduce Qi’s normalized remainder of the Maclaurin power series expansion of the function $\ln \sec x=-\ln \cos x$; in view of a monotonicity rule for the ratio of two Maclaurin power series and by virtue of the logarithmic convexity of the function $(2^x-1)\zeta \left(x\right)$ on $(1,\infty )$, they prove the logarithmic convexity of Qi’s normalized remainder; with the aid of a monotonicity rule for the ratio of two Maclaurin power series, the authors present the monotonic property of the ratio between two Qi’s normalized remainders.

1. Definition of Qi’s Normalized Remainders

According to [1] (Fact 13.3), for $z\in \mathbb{C}$ such that $\mathcal{R}\left(z\right)>1$, the Riemann zeta function $\zeta \left(z\right)$ can be defined by

$$\zeta \left(z\right)=\sum _{k=1}^{\infty }{\displaystyle \frac{1}{k^z}}={\displaystyle \frac{1}{1-2^{1-z}}}\sum _{k=1}^{\infty }{\displaystyle \frac{{(-1)}^{k-1}}{k^z}}={\displaystyle \frac{1}{1-2^{1-z}}}\eta \left(z\right),$$

(1)

where $\eta \left(z\right)$ is called the Dirichlet eta function.

In [2] (Section 3.5, pp. 57–58), the Riemann zeta function $\zeta \left(z\right)$ is analytically extended from $z\in \mathbb{C}$ such that $\mathcal{R}\left(z\right)>1$ to the punctured complex plane $\mathbb{C}\setminus \left\{1\right\}$, such that the only singularity $z=1$ is a simple pole with residue 1. In other words, the Riemann zeta function $\zeta \left(z\right)$ is meromorphic with a simple pole at $z=1$. Consequently, by virtue of the relation (1), the Dirichlet eta function

$$\eta \left(z\right)=\left(1-2^{1-z}\right)\zeta \left(z\right)$$

(2)

can be continued as an entire function in $z\in \mathbb{C}$. See also [3] (Chapter 6) and [4].

The Stirling numbers of the second kind $S(n,k)$ for $n\ge k\ge 0$ can be analytically generated [5] (pp. 131–132) by

$${\left({\displaystyle \frac{{\mathrm{e}}^x-1}{x}}\right)}^k=\sum _{n=0}^{\infty }{\displaystyle \frac{S(n+k,k)}{\left(\genfrac{}{}{0pt}{}{n+k}{k}\right)}}{\displaystyle \frac{x^n}{n!}},k\ge 0.$$

To answer the question posed by Gottfried Helms (Germany) in February 2013 on the website https://math.stackexchange.com/q/307274 (accessed on 3 July 2024), Qi and his coauthors discussed and established in [6] several forms of the Maclaurin power series expansion of the function $\ln (1+{\mathrm{e}}^x)$. Among other things, two main results in [6] are the following theorems.

Theorem 1

([6] (Theorem 3)). For $\left|x\right|<\pi$, we have

$$\ln (1+{\mathrm{e}}^x)=\ln 2+{\displaystyle \frac{x}{2}}+\sum _{n=1}^{\infty }\eta (1-2n){\displaystyle \frac{x^{2n}}{\left(2n\right)!}}.$$

(3)

Theorem 2

([6] (Theorem 4)). For $\left|x\right|<\pi$, we have

$$\ln (1+{\mathrm{e}}^x)=\ln 2+{\displaystyle \frac{x}{2}}+\sum _{n=1}^{\infty }\left[\sum _{k=1}^{2n}{(-1)}^{k-1}{\displaystyle \frac{(k-1)!}{2^k}}S(2n,k)\right]{\displaystyle \frac{x^{2n}}{\left(2n\right)!}}.$$

(4)

The Maclaurin power series expansion (4) is a simplification of [6] (Theorem 4) by virtue of the fact that the function $\ln (1+{\mathrm{e}}^x)-\left(\ln 2+{\displaystyle \frac{x}{2}}\right)$ is even on $(-\infty ,\infty )\supset (-\pi ,\pi )$ or with the help of the identity

$$\sum _{k=1}^{2n+1}{(-1)}^k{\displaystyle \frac{(k-1)!}{2^k}}S(2n+1,k)=0,n\in \mathbb{N}$$

derived in [6] (Remark 2).

In [7] (p. 807, Entries 23.2.14 and 23.2.15), the identity

$$\zeta (1-2n)=-{\displaystyle \frac{B_{2n}}{2n}},n\in \mathbb{N}$$

(5)

is collected, where $B_n$ denotes the Bernoulli numbers, which can be generated by

$${\displaystyle \frac{z}{{\mathrm{e}}^z-1}}=\sum _{n=0}^{\infty }{B}_n{\displaystyle \frac{z^n}{n!}}=1-{\displaystyle \frac{z}{2}}+\sum _{n=1}^{\infty }{B}_{2n}{\displaystyle \frac{z^{2n}}{\left(2n\right)!}},\left|z\right|<2\pi .$$

(6)

Combining (2) with (5) results in

$$\eta (1-2n)=\left(2^{2n}-1\right){\displaystyle \frac{B_{2n}}{2n}},n\in \mathbb{N}.$$

(7)

Substituting (7) into (3) yields

$$\ln (1+{\mathrm{e}}^x)=\ln 2+{\displaystyle \frac{x}{2}}+\sum _{n=1}^{\infty }\left(2^{2n}-1\right){\displaystyle \frac{B_{2n}}{2n}}{\displaystyle \frac{x^{2n}}{\left(2n\right)!}},\left|x\right|<\pi .$$

(8)

Comparing (4) with (8) deduces

$$B_{2n}={\displaystyle \frac{2n}{2^{2n}-1}}\sum _{k=1}^{2n}{(-1)}^{k-1}{\displaystyle \frac{(k-1)!}{2^k}}S(2n,k),n\in \mathbb{N}.$$

This identity has been established in [8] (Theorem 1.1).

We observe that, from the definition $\cosh z={\displaystyle \frac{{\mathrm{e}}^z+{\mathrm{e}}^{-z}}{2}}$, the Maclaurin power series expansions (3), (4), and (8) can be reformulated as

$$\begin{array}{cc}\hfill \ln \cosh {\displaystyle \frac{x}{2}}& =\sum _{n=1}^{\infty }\eta (1-2n){\displaystyle \frac{x^{2n}}{\left(2n\right)!}},\left|x\right|<\pi ,\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill \ln \cosh {\displaystyle \frac{x}{2}}& =\sum _{n=1}^{\infty }\left[\sum _{k=1}^{2n}{(-1)}^{k-1}{\displaystyle \frac{(k-1)!}{2^k}}S(2n,k)\right]{\displaystyle \frac{x^{2n}}{\left(2n\right)!}},\left|x\right|<\pi ,\hfill \end{array}$$

(10)

and

$$\begin{array}{cc}\hfill \ln \cosh {\displaystyle \frac{x}{2}}& =\sum _{n=1}^{\infty }\left(2^{2n}-1\right){\displaystyle \frac{B_{2n}}{2n}}{\displaystyle \frac{x^{2n}}{\left(2n\right)!}},\left|x\right|<\pi .\hfill \end{array}$$

(11)

In [9] (p. 55), we find the Maclaurin power series expansion

$$\ln \cos x=-\sum _{n=1}^{\infty }{\displaystyle \frac{2^{2n-1}(2^{2n}-1)}{n\left(2n\right)!}}|B_{2n}|x^{2n},\left|x\right|<{\displaystyle \frac{\pi }{2}}.$$

(12)

Since

$$\cosh x=\cos (x \mathrm{i})\mathrm{and}\cos x=\cosh (x \mathrm{i}),$$

(13)

the Maclaurin power series expansions in (3), (4), (8), (9), (10), (11), and (12) are equivalent to each other. This observation was announced at https://math.stackexchange.com/a/4940352 (accessed on 1 July 2024) and in its first comment on 2 July 2024.

In the papers [10,11,12,13,14,15,16] (Remark 7), Qi and his coauthors invented and investigated the normalized remainders or the normalized tails

$${\mathrm{SinR}}_n\left(x\right)=\{\begin{array}{cc}{(-1)}^n{\displaystyle \frac{(2n+1)!}{x^{2n+1}}}\left[\sin x-{\displaystyle \sum _{k=0}^{n-1}}{(-1)}^k{\displaystyle \frac{x^{2k+1}}{(2k+1)!}}\right],\hfill & x\ne 0\hfill \\ 1,\hfill & x=0\hfill \end{array}$$

(14)

and

$${\mathrm{CosR}}_n\left(x\right)=\{\begin{array}{cc}{(-1)}^n{\displaystyle \frac{\left(2n\right)!}{x^{2n}}}\left[\cos x-{\displaystyle \sum _{k=0}^{n-1}}{(-1)}^k{\displaystyle \frac{x^{2k}}{\left(2k\right)!}}\right],\hfill & x\ne 0\hfill \\ 1,\hfill & x=0\hfill \end{array}$$

(15)

for $n\in {\mathbb{N}}_0$ of the Maclaurin power series expansions

$$\sin x=\sum _{k=0}^{\infty }{(-1)}^k{\displaystyle \frac{x^{2k+1}}{(2k+1)!}}\mathrm{and}\cos x=\sum _{k=0}^{\infty }{(-1)}^k{\displaystyle \frac{x^{2k}}{\left(2k\right)!}}$$

for $x\in \mathbb{R}$. In the articles [17,18], Qi and his coauthors considered the normalized remainders of the Maclaurin power series expansion of the tangent function $\tan x$ and its square ${\tan }^{2}x$. In [19,20], Qi and his coauthors invented and discussed the normalized remainder of the Maclaurin power series expansion of the exponential function ${\mathrm{e}}^x$. In the work [21], basing on the power series expansion (6), Qi and his coauthors created and studied the normalized remainders

$$T_n\left(x\right)=\{\begin{array}{cc}{\displaystyle \frac{(2n+2)!}{B_{2n+2}}}{\displaystyle \frac{1}{x^{2n+2}}}\left[{\displaystyle \frac{x}{{\mathrm{e}}^x-1}}-1+{\displaystyle \frac{x}{2}}-{\displaystyle \sum _{k=1}^n}{B}_{2k}{\displaystyle \frac{x^{2k}}{\left(2k\right)!}}\right],\hfill & x\ne 0\\ 1,\hfill & x=0\end{array}$$

(16)

for $n\in \mathbb{N}$. Since the function ${\displaystyle \frac{x}{{\mathrm{e}}^x-1}}-1+{\displaystyle \frac{x}{2}}$ is even in $x\in (-\infty ,\infty )$, the normalized remainder $T_n\left(x\right)$ is an even function of $x\in (-\infty ,\infty )$.

Basing on the Maclaurin power series expansion (12), motivated by the ideas in the sequence of the papers [16] (Remark 7), [18] (Section 1), [19] (Remark 5), and [10,11,12,13,14,15,17,21], and similar to Qi’s normalized remainders (14), (15), and (16), we now introduce Qi’s normalized remainder associated with $\ln \sec x=-\ln \cos x$, as follows.

Definition 1.

For $n\in {\mathbb{N}}_0=\left\{0\right\}\cup \mathbb{N}$ and $x\in \left(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}\right)$, define Qi’s normalized remainder associated with $\ln \sec x=-\ln \cos x$ by

$${\mathrm{QiR}}_n\left(x\right)=\{\begin{array}{cc}{\displaystyle \frac{(n+1)(2n+2)!\left[\ln \sec x-{\sum }_{k=1}^n\frac{2^{2k-1}(2^{2k}-1)}{k\left(2k\right)!}\left|B_{2k}\right|x^{2k}\right]}{2^{2n+1}(2^{2n+2}-1)\left|B_{2n+2}\right|x^{2n+2}}},\hfill & x\ne 0;\\ 1,\hfill & x=0,\end{array}$$

(17)

where an empty sum is understood to be 0.

It is easy to see that Qi’s normalized remainder ${\mathrm{QiR}}_n\left(x\right)$ associated with $\ln \sec x=-\ln \cos x$ can be rewritten as

$${\mathrm{QiR}}_n\left(x\right)={\displaystyle \frac{(n+1)(2n+2)!}{(2^{2n+2}-1)\left|B_{2n+2}\right|}}\sum _{k=0}^{\infty }{\displaystyle \frac{2^{2k}\left(2^{2n+2k+2}-1\right)\left|B_{2n+2k+2}\right|}{(n+k+1)(2n+2k+2)!}}{x}^{2k}$$

(18)

for $n\in {\mathbb{N}}_0$ and $x\in \left(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}\right)$. Therefore, Qi’s normalized remainder ${\mathrm{QiR}}_n\left(x\right)$ for $n\in {\mathbb{N}}_0$ is even and positive for $x\in \left(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}\right)$, absolutely monotonic for $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$, and completely monotonic for $x\in \left(-{\displaystyle \frac{\pi }{2}},0\right)$.

In this study, we will prove that

• Qi’s normalized remainder ${\mathrm{QiR}}_n\left(x\right)$ for $n\in {\mathbb{N}}_0$ is a logarithmically convex function of $x\in \left(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}\right)$;
• The ratio ${\displaystyle \frac{{\mathrm{QiR}}_{n+1}\left(x\right)}{{\mathrm{QiR}}_n\left(x\right)}}$ for $n\in {\mathbb{N}}_0$ is an increasing function of $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$ and a decreasing function of $x\in \left(-{\displaystyle \frac{\pi }{2}},0\right)$.

2. Two Lemmas

For verifying our main results, we need the following two lemmas.

Lemma 1

(Monotonicity rule for the ratio of two Maclaurin power series [22]). Let ${\alpha }_k$ and ${\beta }_k$ for $k\in {\mathbb{N}}_0$ be two real sequences and let the Maclaurin power series

$$U\left(x\right)=\sum _{k=0}^{\infty }{\alpha }_k{x}^k\mathrm{and}V\left(x\right)=\sum _{k=0}^{\infty }{\beta }_k{x}^k$$

converge on $(-r,r)$ for some $r>0$. If ${\beta }_k>0$ and the sequence ratio ${\displaystyle \frac{{\alpha }_k}{{\beta }_k}}$ increases in $k\ge 0$, then the function ratio ${\displaystyle \frac{U\left(x\right)}{V\left(x\right)}}$ increases in $x\in (0,r)$.

Lemma 2

([18] (Lemma 1)). The function $\varphi \left(x\right)=(2^x-1)\zeta \left(x\right)$ is logarithmically convex on $(1,\infty )$.

3. Logarithmic Convexity of Qi’s Normalized Remainder

We now start off to verify the logarithmic convexity of Qi’s normalized remainder ${\mathrm{QiR}}_n\left(x\right)$ defined for $n\in {\mathbb{N}}_0$ in $x\in \left(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}\right)$ by (17).

Theorem 3.

Qi’s normalized remainder ${\mathrm{QiR}}_n\left(x\right)$ for given $n\in {\mathbb{N}}_0$ is a logarithmically convex function in $x\in \left(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}\right)$.

Proof. 

Taking the logarithm on both sides of (18) and differentiating give

$$\begin{array}{cc}\hfill {\left[\ln {\mathrm{QiR}}_n\left(x\right)\right]}^{\prime }& ={\displaystyle \frac{{\displaystyle \sum _{k=1}^{\infty }\frac{2^{2k}(2^{2n+2k+2}-1)\left|B_{2n+2k+2}\right|}{(n+k+1)(2n+2k+2)!}\left(2k\right)x^{2k-1}}}{{\displaystyle \sum _{k=0}^{\infty }\frac{2^{2k}(2^{2n+2k+2}-1)\left|B_{2n+2k+2}\right|}{(n+k+1)(2n+2k+2)!}{x}^{2k}}}}\hfill \\ \hfill & =2x{\displaystyle \frac{{\displaystyle \sum _{k=0}^{\infty }\frac{2^{2k+2}(2^{2n+2k+4}-1)\left|B_{2n+2k+4}\right|}{(n+k+2)(2n+2k+4)!}(k+1)x^{2k}}}{{\displaystyle \sum _{k=0}^{\infty }\frac{2^{2k}(2^{2n+2k+2}-1)\left|B_{2n+2k+2}\right|}{(n+k+1)(2n+2k+2)!}{x}^{2k}}}}\hfill \end{array}$$

for $n\in {\mathbb{N}}_0$ and $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$.

In order to prove ${\left[\ln {\mathrm{QiR}}_n\left(x\right)\right]}^{″}>0$ for $n\in {\mathbb{N}}_0$ and $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$, in view of Lemma 1, it is sufficient to show that the ratio

$$\begin{array}{c}{\displaystyle \frac{{\displaystyle \frac{2^{2k+2}(2^{2n+2k+4}-1)\left|B_{2n+2k+4}\right|}{(n+k+2)(2n+2k+4)!}(k+1)}}{{\displaystyle \frac{2^{2k}(2^{2n+2k+2}-1)\left|B_{2n+2k+2}\right|}{(n+k+1)(2n+2k+2)!}}}}\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}={\displaystyle \frac{2(k+1)(n+k+1)}{{(n+k+2)}^2(2n+2k+3)}}{\displaystyle \frac{2^{2n+2k+4}-1}{2^{2n+2k+2}-1}}\left|{\displaystyle \frac{B_{2n+2k+4}}{B_{2n+2k+2}}}\right|\hfill \end{array}$$

(19)

is an increasing sequence in $k\in {\mathbb{N}}_0$ for all given $n\in {\mathbb{N}}_0$. In light of the relation

$$B_{2n}={(-1)}^{n+1}{\displaystyle \frac{2\left(2n\right)!}{{\left(2\pi \right)}^{2n}}}\zeta \left(2n\right),n\in \mathbb{N},$$

(20)

see [2] (p. 5, Equation (1.14)) and [7] (pp. 807–808, Section 23.2), the right-hand side of the equality (19) becomes

$${\displaystyle \frac{k+1}{{\pi }^2}}{\displaystyle \frac{n+k+1}{n+k+2}}{\displaystyle \frac{2^{2n+2k+4}-1}{2^{2n+2k+2}-1}}{\displaystyle \frac{\zeta (2n+2k+4)}{\zeta (2n+2k+2)}},k,n\in {\mathbb{N}}_0.$$

By Lemma 2, we can easily derive that the sequence

$${\displaystyle \frac{2^{2n+2k+4}-1}{2^{2n+2k+2}-1}}{\displaystyle \frac{\zeta (2n+2k+4)}{\zeta (2n+2k+2)}}$$

(21)

is increasing in $k\in {\mathbb{N}}_0$ for given $n\in {\mathbb{N}}_0$. On the other hand, it is easier to see that the sequence ${\displaystyle \frac{k+1}{{\pi }^2}}{\displaystyle \frac{n+k+1}{n+k+2}}$ is also increasing in $k\in {\mathbb{N}}_0$ for given $n\in {\mathbb{N}}_0$. Accordingly, the product of these two sequences is increasing in $k\in {\mathbb{N}}_0$ for given $n\in {\mathbb{N}}_0$. Hence, the sequence in (19) is increasing in $k\in {\mathbb{N}}_0$ for given $n\in {\mathbb{N}}_0$. Consequently, the first derivative ${\left[\ln {\mathrm{QiR}}_n\left(x\right)\right]}^{\prime }$ is increasing in $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$ for given $n\in {\mathbb{N}}_0$. Equivalently speaking, Qi’s normalized remainder ${\mathrm{QiR}}_n\left(x\right)$ for $n\in {\mathbb{N}}_0$ is a logarithmically convex function for $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$.

Since Qi’s normalized remainder ${\mathrm{QiR}}_n\left(x\right)$ for $n\in {\mathbb{N}}_0$ is an even function for $x\in \left(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}\right)$, then ${\mathrm{QiR}}_n\left(x\right)$ for $n\in {\mathbb{N}}_0$ is also a logarithmically convex function for $x\in \left(-{\displaystyle \frac{\pi }{2}},0\right)$. The proof of Theorem 3 is thus completed. □

Theorem 4.

For $n\in {\mathbb{N}}_0$ and $x\in \left(-{\displaystyle \frac{\pi }{2}},0\right)\cup \left(0,{\displaystyle \frac{\pi }{2}}\right)$, we have

$$\begin{array}{c}{\displaystyle \frac{\tan x}{x}}+(2n+2){\displaystyle \frac{\ln \cos x}{x^2}}\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}>{\displaystyle \frac{2^{2n+3}\left(2^{2n+4}-1\right)\left|B_{2n+4}\right|}{{(n+2)}^2(2n+3)!}}{x}^{2n+2}-\sum _{k=0}^{n-1}{\displaystyle \frac{(n-k)2^{2k+2}\left(2^{2k+2}-1\right)\left|B_{2k+2}\right|}{(k+1)(2k+2)!}}{x}^{2k},\hfill \end{array}$$

(22)

where an empty sum is conventionally regarded as 0. In particular, we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{\tan x}{x}}+{\displaystyle \frac{2\ln \cos x}{x^2}}& >{\displaystyle \frac{x^2}{6}}\hfill \end{array}$$

(23)

and

$${\displaystyle \frac{\tan x}{x}}+{\displaystyle \frac{4\ln \cos x}{x^2}}>{\displaystyle \frac{2x^4}{45}}-1$$

(24)

for $x\in \left(-{\displaystyle \frac{\pi }{2}},0\right)\cup \left(0,{\displaystyle \frac{\pi }{2}}\right)$.

Proof. 

From the proof of Theorem 3, we conclude that the function

$$\begin{array}{cc}\hfill {\displaystyle \frac{{\left[\ln {\mathrm{QiR}}_n\left(x\right)\right]}^{\prime }}{2x}}& ={\displaystyle \frac{{\displaystyle \sum _{k=0}^{\infty }\frac{2^{2k+2}(2^{2n+2k+4}-1)\left|B_{2n+2k+4}\right|}{(n+k+2)(2n+2k+4)!}(k+1)x^{2k}}}{{\displaystyle \sum _{k=0}^{\infty }\frac{2^{2k}(2^{2n+2k+2}-1)\left|B_{2n+2k+2}\right|}{(n+k+1)(2n+2k+2)!}{x}^{2k}}}},n\in {\mathbb{N}}_0\hfill \end{array}$$

is increasing in $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$ and the limit as $x\to 0^{+}$ is equal to

$${\displaystyle \frac{2(n+1)}{{(n+2)}^2(2n+3)}}{\displaystyle \frac{2^{2n+4}-1}{2^{2n+2}-1}}\left|{\displaystyle \frac{B_{2n+4}}{B_{2n+2}}}\right|,n\in {\mathbb{N}}_0.$$

This means that

$${\displaystyle \frac{{\left[\ln {\mathrm{QiR}}_n\left(x\right)\right]}^{\prime }}{2x}}>{\displaystyle \frac{2(n+1)}{{(n+2)}^2(2n+3)}}{\displaystyle \frac{2^{2n+4}-1}{2^{2n+2}-1}}\left|{\displaystyle \frac{B_{2n+4}}{B_{2n+2}}}\right|$$

for $n\in {\mathbb{N}}_0$ and $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$.

A direct computation gives

$$\begin{array}{cc}\hfill {\left[{\mathrm{QiR}}_n\left(x\right)\right]}^{\prime }& ={\displaystyle \frac{(n+1)(2n+2)!}{2^{2n+1}(2^{2n+2}-1)\left|B_{2n+2}\right|}}{\left[{\displaystyle \frac{\ln \sec x-{\sum }_{k=1}^n\frac{2^{2k-1}(2^{2k}-1)}{k\left(2k\right)!}\left|B_{2k}\right|x^{2k}}{x^{2n+2}}}\right]}^{\prime }\hfill \\ \hfill & =(n+1)(2n+2)!{\displaystyle \frac{x\tan x+(2n+2)\ln \cos x+{\sum }_{k=1}^n\frac{(n-k+1)2^{2k}(2^{2k}-1)\left|B_{2k}\right|}{k\left(2k\right)!}{x}^{2k}}{2^{2n+1}(2^{2n+2}-1)\left|B_{2n+2}\right|x^{2n+3}}}\hfill \end{array}$$

for $n\in {\mathbb{N}}_0$ and $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$. As a result, we arrive at

$$\begin{array}{c}(n+1)(2n+2)!{\displaystyle \frac{x\tan x+(2n+2)\ln \cos x+{\sum }_{k=1}^n\frac{(n-k+1)2^{2k}(2^{2k}-1)\left|B_{2k}\right|}{k\left(2k\right)!}{x}^{2k}}{2^{2n+2}(2^{2n+2}-1)\left|B_{2n+2}\right|x^{2n+4}}}\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}>{\displaystyle \frac{2(n+1)}{{(n+2)}^2(2n+3)}}{\displaystyle \frac{2^{2n+4}-1}{2^{2n+2}-1}}\left|{\displaystyle \frac{B_{2n+4}}{B_{2n+2}}}\right|\hfill \end{array}$$

for $n\in {\mathbb{N}}_0$ and $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$. This inequality can be simplified as (22) for $n\in {\mathbb{N}}_0$ and $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$.

The inequalities (23) and (24) for $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$ are special cases $n=0,1$ of the inequality (22) for $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$.

Considering the evenness, the inequalities (22) for $n\in {\mathbb{N}}_0$, (23), and (24) are valid for $x\in \left(-{\displaystyle \frac{\pi }{2}},0\right)$. The proof of Theorem 4 is thus completed. □

Remark 1.

Comparing (5) with (20) results in the known identity

$$\zeta (1-2n)={(-1)}^n{\displaystyle \frac{2(2n-1)!}{{\left(2\pi \right)}^{2n}}}\zeta \left(2n\right),n\in \mathbb{N}.$$

From the Maclaurin power series expansion (9), it follows that

$$\ln \cosh x=\sum _{n=1}^{\infty }{2}^{2n}\eta (1-2n){\displaystyle \frac{x^{2n}}{\left(2n\right)!}},\left|x\right|<{\displaystyle \frac{\pi }{2}}.$$

Replacing x by $x \mathrm{i}$ and utilizing the second relation in (13) lead to

$$\ln \cos x=\sum _{n=1}^{\infty }{(-1)}^n{2}^{2n}\eta (1-2n){\displaystyle \frac{x^{2n}}{\left(2n\right)!}},\left|x\right|<{\displaystyle \frac{\pi }{2}}.$$

Comparing this series with (12) deduces the known relation

$$B_{2n}={\displaystyle \frac{2n}{2^{2n}-1}}\eta (1-2n),n\in \mathbb{N}.$$

4. Monotonic Property of Ratio Between Qi’s Normalized Remainders

We now verify the monotonic property of the ratio ${\displaystyle \frac{{\mathrm{QiR}}_{n+1}\left(x\right)}{{\mathrm{QiR}}_n\left(x\right)}}$ for $n\in {\mathbb{N}}_0$ in $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$.

Theorem 5.

The ratio ${\displaystyle \frac{{\mathrm{QiR}}_{n+1}\left(x\right)}{{\mathrm{QiR}}_n\left(x\right)}}$ for $n\in {\mathbb{N}}_0$ is an increasing function of $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$ and a decreasing function of $x\in \left(-{\displaystyle \frac{\pi }{2}},0\right)$. Consequently, the inequality

$${\mathrm{QiR}}_{n+1}\left(x\right)\ge {\mathrm{QiR}}_n\left(x\right),n\in {\mathbb{N}}_0$$

(25)

is sound and the equality in (25) is valid only if $x=0$.

Proof. 

Making use of the expression (18), we obtain

$${\displaystyle \frac{{\mathrm{QiR}}_{n+1}\left(x\right)}{{\mathrm{QiR}}_n\left(x\right)}}={\displaystyle \frac{{\displaystyle \frac{(n+2)(2n+4)!}{(2^{2n+4}-1)\left|B_{2n+4}\right|}\sum _{k=0}^{\infty }\frac{2^{2k}\left(2^{2n+2k+4}-1\right)\left|B_{2n+2k+4}\right|}{(n+k+2)(2n+2k+4)!}{x}^{2k}}}{{\displaystyle \frac{(n+1)(2n+2)!}{(2^{2n+2}-1)\left|B_{2n+2}\right|}\sum _{k=0}^{\infty }\frac{2^{2k}\left(2^{2n+2k+2}-1\right)\left|B_{2n+2k+2}\right|}{(n+k+1)(2n+2k+2)!}{x}^{2k}}}}$$

for $n\in {\mathbb{N}}_0$ and $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$. In order to verify the increasing property of the ratio ${\displaystyle \frac{{\mathrm{QiR}}_{n+1}\left(x\right)}{{\mathrm{QiR}}_n\left(x\right)}}$ in $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$ for $n\in {\mathbb{N}}_0$, it suffices to show that the ratio

$${\displaystyle \frac{{\displaystyle \frac{2^{2k}\left(2^{2n+2k+4}-1\right)\left|B_{2n+2k+4}\right|}{(n+k+2)(2n+2k+4)!}}}{{\displaystyle \frac{2^{2k}\left(2^{2n+2k+2}-1\right)\left|B_{2n+2k+2}\right|}{(n+k+1)(2n+2k+2)!}}}}={\displaystyle \frac{(n+k+1)}{2{(n+k+2)}^2(2n+2k+3)}}{\displaystyle \frac{2^{2n+2k+4}-1}{2^{2n+2k+2}-1}}\left|{\displaystyle \frac{B_{2n+2k+4}}{B_{2n+2k+2}}}\right|$$

(26)

is increasing in $k\in {\mathbb{N}}_0$ for given $n\in {\mathbb{N}}_0$.

Employing the expression (20), we can rewrite the right-hand side of (26) as

$${\displaystyle \frac{1}{4{\pi }^2}}{\displaystyle \frac{{(n+k+1)}^2}{{(n+k+2)}^2}}{\displaystyle \frac{2^{2n+2k+4}-1}{2^{2n+2k+2}-1}}{\displaystyle \frac{\zeta (2n+2k+4)}{\zeta (2n+2k+2)}},k,n\in {\mathbb{N}}_0.$$

(27)

Since the ratio (21) and the fraction ${\displaystyle \frac{{(n+k+1)}^2}{{(n+k+2)}^2}}$ are increasing sequences in $k\in {\mathbb{N}}_0$ for given $n\in {\mathbb{N}}_0$, the product of these sequences, the sequence (27), is increasing in $k\in {\mathbb{N}}_0$ for given $n\in {\mathbb{N}}_0$. With the help of Lemma 1, we derive that the ratio ${\displaystyle \frac{{\mathrm{QiR}}_{n+1}\left(x\right)}{{\mathrm{QiR}}_n\left(x\right)}}$ is increasing in $x\in \left(0,{\displaystyle \frac{\pi }{2}}\right)$ for given $n\in {\mathbb{N}}_0$.

Since Qi’s normalized remainder ${\mathrm{QiR}}_n\left(x\right)$ is even on $\left(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}\right)$, we see easily that the ratio ${\displaystyle \frac{{\mathrm{QiR}}_{n+1}\left(x\right)}{{\mathrm{QiR}}_n\left(x\right)}}$ is also even on $\left(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}\right)$. Accordingly, from the increasing property of the ratio ${\displaystyle \frac{{\mathrm{QiR}}_{n+1}\left(x\right)}{{\mathrm{QiR}}_n\left(x\right)}}$ on $\left(0,{\displaystyle \frac{\pi }{2}}\right)$, we deduce the decreasing property on $\left(-{\displaystyle \frac{\pi }{2}},0\right)$.

The inequality (25) follows from the trivial property ${\mathrm{QiR}}_n\left(0\right)=1$ for $n\in {\mathbb{N}}_0$. The proof of Theorem 5 is thus completed. □

Remark 2.

We pose one problem now: Prove that the ratio ${\displaystyle \frac{{\mathrm{QiR}}_{n+1}\left(x\right)}{{\mathrm{QiR}}_n\left(x\right)}}$ for $n\in {\mathbb{N}}_0$ is a logarithmically convex function of $x\in \left(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}\right)$. For example, prove that the function

$${\displaystyle \frac{{\mathrm{QiR}}_1\left(x\right)}{{\mathrm{QiR}}_0\left(x\right)}}=\{\begin{array}{cc}3\left({\displaystyle \frac{2}{x^2}}+{\displaystyle \frac{1}{\ln \cos x}}\right),\hfill & x\ne 0\\ 1,\hfill & x=0\end{array}$$

is a logarithmically convex function of $x\in \left(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}\right)$.

Remark 3.

It is common knowledge that

$$\begin{array}{cc}\hfill \ln {\mathrm{QiR}}_n\left(x\right)& =\sum _{k=0}^{\infty }{\left[\ln {\mathrm{QiR}}_n\left(x\right)\right]}_{x=0}^{\left(k\right)}{\displaystyle \frac{x^k}{k!}}\hfill \\ \hfill & =\sum _{k=1}^{\infty }{\left[{\displaystyle \frac{{\mathrm{QiR}}_n^{\prime }\left(x\right)}{{\mathrm{QiR}}_n\left(x\right)}}\right]}_{x=0}^{(k-1)}{\displaystyle \frac{x^k}{k!}}\hfill \\ \hfill & =\sum _{k=1}^{\infty }{\left[{\displaystyle \frac{{\mathrm{QiR}}_n^{\prime }\left(x\right)}{{\mathrm{QiR}}_n\left(x\right)}}\right]}_{x=0}^{(2k-1)}{\displaystyle \frac{x^{2k}}{\left(2k\right)!}},\hfill \end{array}$$

where we used the evenness of Qi’s normalized remainder ${\mathrm{QiR}}_n\left(x\right)$ defined in (17) for $n\in {\mathbb{N}}_0$ and $x\in \left(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}\right)$. In order to expand the logarithm $\ln {\mathrm{QiR}}_n\left(x\right)$ into a Maclaurin power series, it is sufficient to compute the derivatives ${\left[{\displaystyle \frac{{\mathrm{QiR}}_n^{\prime }\left(x\right)}{{\mathrm{QiR}}_n\left(x\right)}}\right]}_{x=0}^{(2k-1)}$ for $k\in \mathbb{N}$ and $n\in {\mathbb{N}}_0$. This derivative computation can be carried out with the aid of the expression (18) and the following general derivative formula (28), which is a reformulation of [23] (p. 40, Exercise 5).

• Let $u\left(x\right)$ and $v\left(x\right)\ne 0$ be two n-time differentiable functions on an interval I for a given integer $n\ge 0$. Then the nth derivative of the ratio ${\displaystyle \frac{u\left(x\right)}{v\left(x\right)}}$ is

  $${\displaystyle \frac{{\mathrm{d}}^n}{\mathrm{d}{x}^n}}\left[{\displaystyle \frac{u\left(x\right)}{v\left(x\right)}}\right]={(-1)}^n{\displaystyle \frac{\left|W_{(n+1)\times (n+1)}\left(x\right)\right|}{v^{n+1}\left(x\right)}},n\ge 0,$$

  (28)

  where the matrix

  $$W_{(n+1)\times (n+1)}\left(x\right)={\left(\begin{array}{cc}{U}_{(n+1)\times 1}\left(x\right)& V_{(n+1)\times n}\left(x\right)\end{array}\right)}_{(n+1)\times (n+1)},$$

  the matrix $U_{(n+1)\times 1}\left(x\right)$ is an $(n+1)\times 1$ matrix whose elements satisfy $u_{k,1}\left(x\right)=u^{(k-1)}\left(x\right)$ for $1\le k\le n+1$, the matrix $V_{(n+1)\times n}\left(x\right)$ is an $(n+1)\times n$ matrix whose elements are

  $$v_{\mathcal{l},j}\left(x\right)=\left\{\begin{array}{cc}\left({\displaystyle \genfrac{}{}{0pt}{}{\mathcal{l}-1}{j-1}}\right)v^{(\mathcal{l}-j)}\left(x\right),\hfill & \mathcal{l}-j\ge 0\hfill \\ 0,\hfill & \mathcal{l}-j<0\hfill \end{array}\right.$$

  for $1\le \mathcal{l}\le n+1$ and $1\le j\le n$, and the notation $|W_{(n+1)\times (n+1)}\left(x\right)|$ denotes the determinant of the $(n+1)\times (n+1)$ matrix $W_{(n+1)\times (n+1)}\left(x\right)$.

Because the computation is straightforward, we omit the details.

The general derivative formula (28) was effectively, successfully, and significantly applied by Qi and his coauthors since the papers [24,25] in 2015; see also [10,11,13,14,17] and closely-related references therein. A list of papers applied the derivative formula (28) is at the website https://qifeng618.wordpress.com/2020/03/22/some-papers-authored-by-dr-prof-feng-qi-and-utilizing-a-general-derivative-formula-for-the-ratio-of-two-differentiable-functions/ (accessed on 3 July 2024).

We guess that all the coefficients in the Maclaurin power series expansion of the function $\ln {\mathrm{QiR}}_n\left(x\right)$ are positive, that is, all the derivatives ${\left[{\displaystyle \frac{{\mathrm{QiR}}_n^{\prime }\left(x\right)}{{\mathrm{QiR}}_n\left(x\right)}}\right]}_{x=0}^{(2k-1)}$ are positive for all $k\in \mathbb{N}$ and $n\in {\mathbb{N}}_0$.

Remark 4.

When $n=0$ in the inequality (25), we deduce the inequality

$$\cos x\le \exp \left({\displaystyle \frac{3x^2}{x^2-6}}\right),x\in \left(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}\right).$$

Remark 5.

In this paper, we essentially considered Qi’s normalized remainder for the Maclaurin power series expansion of the logarithm $\ln {\mathrm{CosR}}_0\left(x\right)=\ln \cos x=-\ln \sec x$, where Qi’s normalized remainder ${\mathrm{CosR}}_n\left(x\right)$ for $n\in {\mathbb{N}}_0$ is defined by (15) and an empty sum is conventionally understood to be 0. In [26], a subsequent work of the article [10], the authors investigated Qi’s normalized remainder for the Maclaurin power series expansion of the logarithm

$$\ln {\mathrm{CosR}}_1\left(x\right)=\{\begin{array}{cc}\ln {\displaystyle \frac{2(1-\cos x)}{x^2}},\hfill & 0<\left|x\right|<2\pi ;\hfill \\ 0,\hfill & x=0.\hfill \end{array}$$

5. Conclusions

In this paper, after retrospecting the backgrounds, motivated by previous ideas and concepts of Qi’s normalized remainders, see [20] (Section 1), [18] (Section 1), and [19] (Section 5), we introduced a family of Qi’s normalized remainders of the Maclaurin power series expansion of the function $\ln \sec x=-\ln \cos x$ for $x\in \left(-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}\right)$. By virtue of the logarithmic convexity of the function $(2^x-1)\zeta \left(x\right)$ on $(0,\infty )$, we proved the logarithmic convexity of the newly-introduced Qi’s normalized remainder and presented the monotonic property of the ratio between two Qi’s normalized remainders associated with the function $\ln \sec x=-\ln \cos x$. As by-products, lower bounds of the functions ${\displaystyle \frac{\tan x}{x}}+(2n+2){\displaystyle \frac{\ln \cos x}{x^2}}$ for $n\in {\mathbb{N}}_0$ and $0<\left|x\right|<{\displaystyle \frac{\pi }{2}}$ were obtained in terms of polynomials of degree $2n+2$.

Basing on those results of Qi’s normalized remainders in the papers [10,11,12,13,14,15,16,17,18,19,20,21] (Remark 7) since 2023, we believe that the exploration of Qi’s normalized remainders will become more and more extensive, active, and prosperous.

On 12 September 2024, an academic editor commented on the paper [20] that “I found this paper to be a delightful and engaging read, and it was truly a pleasure to invest time in understanding its contents. I believe that numerous other readers will also find this article intriguing and will approach it with a keen interest”.

In the paper [27], there is a detailed academic bibliography of Professor Dr. Feng Qi.