1 Introduction
In 1978, Becker and Stark [1] proved the double inequality
or
holds for all \(x\in (0,\pi /2)\). 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 \(\exp (x\cot x-1)\). More precisely, they proved the following results.
$$\frac{8}{\pi^{2}-4x^{2}}< \frac{\tan x}{x}< \frac{\pi^{2}}{\pi^{2}-4x ^{2}} $$
$$-\frac{4x^{2}}{\pi^{2}}< x\cot x-1< \frac{ ( \pi^{2}-8 ) -4x ^{2}}{8} $$
Theorem A
Let
\(p,q\in ( -\infty,4/\pi^{2} ] \), \(p^{\ast }\approx 0.13484\)
be the unique zero of the function
\(\alpha_{p} ( \pi /2 ) -1\)
on
\(( -\infty,4/\pi^{2} ) \), where
\(\alpha_{p} ( x ) =\exp ( x\cot x-1 ) / ( 1-px^{2} ) ^{1/ ( 3p ) }\)
if
\(p\neq 0\)
and
\(\alpha_{0} ( x ) =\exp ( x\cot x-1+x^{2}/3 ) \). Then the double inequality
holds for all
\(x\in ( 0,\pi /2 ) \)
if and only if
\(p\geq p^{\ast }\)
and
\(q\leq 2/15\approx 0.13333\).
$$\begin{aligned} \bigl( 1-px^{2} \bigr) ^{{1/ ( 3p ) }} < &e^{x\cot x-1} \\ < & \bigl( 1-qx^{2} \bigr) ^{{1/ ( 3q ) }} \end{aligned}$$
(1.1)
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Theorem B
For
\(x\in (0,\pi /2)\), the double inequality
holds.
$$ \biggl( 1-\frac{4}{3\pi^{2}}x^{2} \biggr) ^{\pi^{2}/4}< e^{x\cot x-1}< \biggl( 1-\frac{2}{15}x^{2} \biggr) ^{5/2} $$
(1.2)
Throughout the full text, we suppose that
where \(p_{0}\) is the unique zero of the function
on \(( 0,p_{2} ) \).
$$\begin{aligned}& \begin{aligned}&p_{3} =\frac{32 ( 30-\pi^{2} ) }{15\pi^{6}}\approx 0.04467, \\ &p_{2} =\frac{4}{70\text{,}875}\approx 5. 643 7 \times 10^{-5}, \\ &p_{1} =\frac{64 ( 1-\pi^{2}/30-e^{-2/5} ) }{\pi^{6}}\approx 4.6143\times 10^{-5}, \\ &p_{0} \approx 3.799533\times 10^{-5}, \end{aligned} \end{aligned}$$
(1.3)
$$ h ( p ) =\ln \biggl( 1-\frac{\pi^{2}}{30}-\frac{\pi^{6}}{64}p \biggr) - \frac{8(45\pi^{4}p+32)}{15 \pi^{6}p+32\pi^{2}-960} $$
(1.4)
Now considering the asymptotic expansion of \(( e^{x\cot x-1} ) ^{2/5}\), we have
It is interesting that the power series above has not item of \(x^{4}\), which also remind us to establish a more accurate estimate for \(\exp ( x\cot x-1 ) \). The first aim of this paper is to determine the best parameters p and q such that the double inequality
holds for all \(x\in ( 0,\pi /2 ) \). 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 \(\ln (1-2x^{2}/15-px^{6})\).
$$\bigl( e^{x\cot x-1} \bigr) ^{2/5}=1-\frac{2}{15}x^{2}- \frac{4}{70\text{,}875}x^{6}+\frac{2}{1\text{,}063\text{,}125}x^{8}+O \bigl( x^{10} \bigr). $$
$$\biggl( 1-\frac{2}{15}x^{2}-px^{6} \biggr) ^{5/2}< e^{x\cot x-1}< \biggl( 1- \frac{2}{15}x^{2}-qx^{6} \biggr) ^{5/2} $$
Theorem 1
Let
\(0< p< p_{3}\)
and
\(x\in ( 0,\pi /2 ) \). Then the function
\(f ( x ) =\ln ( 1-2x^{2}/15-px^{6} ) \)
can be expressed in the form of a series,
where
and, for
\(n\geq 3\),
Moreover, \(a_{n}>0\)
for all
\(n\geq 1\).
$$ \ln \biggl( 1-\frac{2}{15}x^{2}-px^{6} \biggr) =-\sum _{n=1}^{\infty }a _{n}x^{2n}, $$
(1.5)
$$ a_{1}=\frac{2}{15},\qquad a_{2}=\frac{2}{225},\qquad a_{3}=p+ \frac{8}{10\text{,} 125} , $$
(1.6)
$$ a_{n+1}=\frac{2n}{15 ( n+1 ) }a_{n}+p\frac{n-2}{n+1}a_{n-2}. $$
(1.7)
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Our main results are contained in the following theorems.
Theorem 2
Let
\(0< p< p_{3}\)
and
\(x\in ( 0,\pi /2 ) \).
(i)
If
\(p_{2}\leq p< p_{3}\), then the function
is strictly increasing on
\(( 0,\pi /2 ) \), and therefore the double inequality
holds, where
$$x\mapsto \frac{\ln ( 1-2x^{2}/15-px^{6} ) }{x\cot x-1}:=\frac{f ( x ) }{g ( x ) } $$
$$ \biggl( 1-\frac{2}{15}x^{2}-px^{6} \biggr) ^{5/2}< e^{x\cot x-1}< \biggl( 1- \frac{2}{15}x^{2}-px^{6} \biggr) ^{1/\lambda_{p}} $$
(1.8)
$$\lambda_{p}=-\ln \biggl( 1-\frac{\pi^{2}}{30}-\frac{\pi^{6}}{64}p \biggr). $$
(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
$$ e^{x\cot x-1}< \biggl( 1-\frac{2}{15}x^{2}-px^{6} \biggr) ^{1/\theta_{p}} $$
(1.9)
$$\begin{aligned}& e^{x\cot x-1} < \biggl( 1-\frac{2}{15}x^{2}-px^{6} \biggr) ^{5/2} \quad \textit{for }p_{0}< p\leq p_{1}, \end{aligned}$$
(1.10)
$$\begin{aligned}& e^{x\cot x-1} < \biggl( 1-\frac{2}{15}x^{2}-px^{6} \biggr) ^{1/\lambda _{p}}\quad \textit{for }p_{1}< p< p_{2}. \end{aligned}$$
(1.11)
(iii)
If
\(0< p< p_{0}\), then the function
\(f/g\)
is strictly decreasing on
\(( 0,\pi /2 ) \), and therefore the double inequality (1.8) is reversed.
As a consequence of Theorem 2, we immediately get the following.
Theorem 3
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 particular, we have
for all
\(x\in (0,\pi /2)\).
$$ \biggl( 1-\frac{2}{15}x^{2}-px^{6} \biggr) ^{5/2}< e^{x\cot x-1}< \biggl( 1- \frac{2}{15}x^{2}-qx^{6} \biggr) ^{5/2} $$
(1.12)
$$\biggl( 1-\frac{2}{15}x^{2}-\frac{4}{70\text{,}875}x^{6} \biggr) ^{5/2}< e^{x \cot x-1}< \biggl( 1-\frac{2}{15}x^{2}- \frac{64 ( 1-\pi^{2}/30-e ^{-2/5} ) }{\pi^{6}}x^{6} \biggr) ^{5/2} $$
The second aim of this paper is to refine some known results presented in [19], we shall state it carefully in the fifth section.
2 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
For
\(-\infty < a< b<\infty \), let
\(f,g:[a,b]\rightarrow \mathbb{R}\)
be continuous functions that are differentiable on
\(( a,b ) \), with
\(f ( a ) =g ( a ) =0\)
or
\(f ( b ) =g ( b ) =0\). Assume that
\(g^{\prime }(x)\neq 0\)
for each
x
in
\((a,b)\). If
\(f^{\prime }/g^{\prime }\)
is increasing (decreasing) on
\((a,b)\), then so is
\(f/g\).
Lemma 2
Let
\(A ( t ) =\sum_{k=0}^{\infty }a_{k}t^{k}\)
and
\(B ( t ) =\sum_{k=0}^{\infty }b_{k}t^{k}\)
be two real power series converging on
\(( -r,r ) \) (\(r>0\)) with
\(b_{k}>0\)
for all
k. If the sequence
\(\{a_{k}/b_{k}\} \)
is increasing (decreasing) for all
k, then the function
\(t\mapsto A ( t ) /B ( t ) \)
is also increasing (decreasing) on
\(( 0,r ) \).
Now, we will introduce a useful auxiliary function \(H_{f,g}\). For \(-\infty \leq a< b\leq \infty \), let f and g be differentiable on \((a,b)\) and \(g^{\prime }\neq 0\) on \((a,b)\). Then the function \(H_{f,g}\) is defined by
The function \(H_{f,g}\) 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]).
$$ H_{f,g}:=\frac{f^{\prime }}{g^{\prime }}g-f. $$
(2.1)
Lemma 3
([26, Theorem 1])
Let
\(A ( t ) =\sum_{k=0}^{\infty }a_{k}t^{k}\)
and
\(B ( t ) =\sum_{k=0}^{\infty }b_{k}t^{k}\)
be two real power series converging on
\(( -r,r ) \)
and
\(b_{k}>0\)
for all
k. Suppose that for certain
\(m\in \mathbb{N} \), the non-constant sequence
\(\{ a_{k}/b_{k} \} \)
is increasing (resp. decreasing) for
\(0\leq k\leq m\)
and decreasing (resp. increasing) for
\(k\geq m\). Then the function
\(A/B\)
is strictly increasing (resp. decreasing) on
\(( 0,r ) \)
if and only if
\(H_{A,B} ( r^{-} ) \geq \) (resp. ≤) 0. Moreover, if
\(H_{A,B} ( r^{-} ) <\) (resp. >) 0, then there exists
\(t_{0}\in ( 0,r ) \)
such that the function
\(A/B\)
is strictly increasing (resp. decreasing) on
\(( 0,t_{0} ) \)
and strictly decreasing (resp. increasing) on
\(( t_{0},r ) \).
Lemma 4
([27])
For
\(n\in N\), the Bernoulli numbers satisfy
$$ \frac{1}{ ( 2\pi ) ^{2}}\frac{2n ( 2n-1 ) ( 2^{2n-3}-1 ) }{2^{2n-3}}< \frac{\vert B_{2n}\vert }{ \vert B_{2n-2}\vert }< \frac{1}{ ( 2\pi ) ^{2}} \frac{2n ( 2n-1 ) 2^{2n-1}}{2^{2n-1}-1}. $$
(2.2)
Lemma 5
For
\(0< p< p_{2}=4/70\text{,}875\), let
\(h(p)\)
be defined by (1.4). Then
\(h ( p ) \)
has a unique zero
\(p_{0}\approx 3.799533\times 10^{-5}\)
such that
\(h ( p ) <0\)
for
\(p\in ( 0,p_{0} ) \)
and
\(h ( p ) >0\)
for
\(p\in ( p_{0},p_{2} ) \).
Proof
Differentiation yields
which together with the facts that
reveals that there is a unique \(p_{0}\in ( 0,p_{2} ) \) such that \(h ( p ) <0\) for \(p\in ( 0,p_{0} ) \) and \(h ( p ) >0\) for \(p\in ( p_{0},p_{2} ) \). Numerically, the equation \(h ( p ) =0\) for p on \(( 0,p_{2} ) \) has the solution \(p_{0}\approx 3.799533\times 10^{-5}\). This completes the proof. □
$$h^{\prime } ( p ) =15\pi^{4}\frac{15\pi^{8}p+32\pi^{4}-1472 \pi^{2}+23\text{,}040}{ ( 15\pi^{6}p+32\pi^{2}-960 ) ^{2}}>0, $$
$$\begin{aligned} h \bigl( 0^{+} \bigr) =&\ln \biggl( 1-\frac{\pi^{2}}{30} \biggr) - \frac{8}{ \pi^{2}-30}\approx -0.001557 5< 0, \\ h ( p_{2} ) =&h \biggl( \frac{4}{70\text{,}875} \biggr) =\ln \biggl( 1- \frac{ \pi^{2}}{30}-\frac{\pi^{6}}{11\text{,}340\text{,}000} \biggr) -\frac{24\pi^{4}+3\text{,}024\text{,} 000}{378\text{,}000\pi^{2}+\pi^{6}-11\text{,}340\text{,}000} \\ \approx &0.0007572>0, \end{aligned}$$
3 Proof of Theorem 1
Proof
Since \(0< p< p_{3}\) and \(x\in ( 0,\pi /2 ) \), we see that
which shows that
holds for all \(x\in ( 0,\pi /2 ) \). It remains to determine the coefficients \(a_{n}\). Differentiation for the two sides of (3.1) gives
which is equivalent to
Comparing coefficients gives the recurrence formulas (1.7) and (1.6).
$$0< \frac{2}{15}x^{2}+px^{6}< 1, $$
$$ \ln \biggl( 1-\frac{2}{15}x^{2}-px^{6} \biggr) =-\sum _{n=1}^{\infty } \frac{1}{n} \biggl( \frac{2}{15}x^{2}+px^{6} \biggr) ^{n}:=-\sum _{n=1} ^{\infty }a_{n}x^{2n} $$
(3.1)
$$\frac{90px^{5}+4x}{15px^{6}+2x^{2}-15}=-\sum_{n=1}^{\infty }2na_{n}x ^{2n-1}, $$
$$\begin{aligned} 45px^{4}+2 =&- \bigl( 15px^{6}+2x^{2}-15 \bigr) \sum_{n=1}^{\infty }na _{n}x^{2n-2} \\ =&15a_{1}+ ( 30a_{2}-2a_{1} ) x^{2}+ ( 45a_{3}-4a_{2} ) x^{4} \\ &{}+\sum_{n=3}^{\infty } \bigl[ 15 ( n+1 ) a_{n+1}-15p ( n-1 ) a_{n-2}-2na_{n} \bigr] x^{2n}. \end{aligned}$$
From the second equality of (3.1) we easily find that \(a_{n}>0\) for all \(n\geq 1\), which completes the proof. □
4 Proofs of Theorems 2 and 3
Proof of Theorem 2
Using the expansion
the function \(f/g\) can be expressed as
by Theorem 1, where \(x^{2}=t\). We now observe the monotonicity of the sequence \(\{a_{n}/b_{n}\}_{n\geq 1}\). Since \(b_{n}>0\) for all \(n\geq 1\), it suffices to determine the sign of \(c_{n}:=a_{n+1}- ( b_{n+1}/b_{n} ) a_{n}\). Direct computations yield
We claim that \(c_{n}>0\) for \(n\geq 3\). In fact, by means of the recurrence formula (1.7), we have
Clearly, if we prove
for \(n\geq 3\), then it follows that \(c_{n}>0\). Using the right hand side of (2.2) we have
for \(n\geq 4\). This together with \(d_{3}=0\) yields \(d_{n}\geq 0\) for \(n\geq 3 \).
$$g ( x ) =x\cot x-1=-\sum_{n=1}^{\infty } \frac{2^{2n}}{ ( 2n )!}\vert B_{2n}\vert x^{2n},\quad \vert x \vert < \pi, $$
$$\frac{f ( x ) }{g ( x ) }=\frac{\ln ( 1-2x^{2}/15-px ^{6} ) }{x\cot x-1}=\frac{-\sum_{n=1}^{\infty }a_{n}x^{2n}}{- \sum_{n=1}^{\infty }\frac{2^{2n}}{ ( 2n )!}\vert B_{2n}\vert x ^{2n}}:=\frac{\sum_{n=1}^{\infty }a_{n}t^{n}}{\sum_{n=1}^{\infty }b _{n}t^{n}} $$
$$\begin{aligned}& c_{1} =a_{2}-\frac{b_{2}}{b_{1}}a_{1}=0, \\ & c_{2} =a_{3}-\frac{b_{3}}{b_{2}}a_{2}=p- \frac{4}{70\text{,}875}. \end{aligned}$$
$$\begin{aligned} c_{n} =&a_{n+1}-\frac{b_{n+1}}{b_{n}}a_{n} \\ =&a_{n+1}- \frac{2}{ ( n+1 ) ( 2n+1 ) }\frac{\vert B_{2n+2}\vert }{ \vert B_{2n}\vert }a_{n} \\ =&\frac{2}{n+1} \biggl( \frac{n}{15}-\frac{1}{2n+1} \frac{\vert B _{2n+2}\vert }{\vert B_{2n}\vert } \biggr) a_{n}+p \frac{n-2}{n+1}a_{n-2}. \end{aligned}$$
$$d_{n}=\frac{n}{15}-\frac{1}{2n+1}\frac{\vert B_{2n+2}\vert }{ \vert B_{2n}\vert }>0 $$
$$\begin{aligned} d_{n} >&\frac{n}{15}-\frac{1}{2n+1} \frac{1}{ ( 2\pi ) ^{2}} \frac{2 ( n+1 ) ( 2n+1 ) 2^{2n+1}}{2^{2n+1}-1} \\ >&\frac{n}{15}-\frac{1}{2n+1}\frac{ ( n+1 ) ( 2n+1 ) }{4\times 19/2}\frac{2^{2n}-1}{2^{2n+1}-1} \\ =&\frac{1}{285}\frac{ ( 8n-30 ) 2^{2n}-19n}{2^{2n+1}-1}>0 \end{aligned}$$
(i)
If \(p_{2}\leq p< p_{3}\), then \(c_{n}=a_{n+1}- ( b_{n+1}/b_{n} ) a_{n}>0\) for \(n\geq 1\), that is, the sequence \(\{a_{n}/b_{n}\}_{n\geq 1}\) is strictly increasing, so is \(f/g\) on \(( 0,\pi /2 ) \) by Lemma 2. Therefore, we conclude that
which implies (1.8).
$$\frac{2}{5}=\lim_{x\rightarrow 0^{+}}\frac{f ( x ) }{g ( x ) }< \frac{f ( x ) }{g ( x ) }< \lim_{x\rightarrow ( \pi /2 ) ^{-}}\frac{f ( x ) }{g ( x ) }=-\ln \biggl( 1- \frac{\pi^{2}}{30}-\frac{\pi^{6}}{64}p \biggr) =\lambda_{p}, $$
(ii)
If \(0< p< p_{2}\), then \(c_{1}=0\), \(c_{2}<0\) and \(c_{n}>0\) for \(n\geq 3\), which indicates that the sequence \(\{a_{n}/b_{n}\}\) is decreasing for \(n=1,2,3\) and increasing for \(n\geq 3\). By Lemma 3, to determine the monotonicity of \(f/g\) on \(( 0, \pi /2 ) \), we have to observe the sign of \(H_{-f ( \sqrt{t} ),-g ( \sqrt{t} ) } ( ( \pi^{2}/4 ) ^{-} ) \). A simple computation leads us to
$$\begin{aligned} H_{-f ( \sqrt{t} ),-g ( \sqrt{t} ) } \biggl( \biggl( \frac{\pi^{2}}{4} \biggr) ^{-} \biggr) =& \lim_{t\rightarrow ( \pi^{2}/4 ) ^{-}} \biggl[ \frac{-f^{ \prime } ( \sqrt{t} ) }{-g^{\prime } ( \sqrt{t} ) } \bigl( -g ( \sqrt{t} ) \bigr) +f ( \sqrt{t} ) \biggr] \\ =&\ln \biggl( 1-\frac{\pi^{2}}{30}-\frac{\pi^{6}}{64}p \biggr) - \frac{8(45 \pi^{4}p+32)}{15\pi^{6}p+32\pi^{2}-960} \\ =&h ( p ). \end{aligned}$$
Subcase 2.1: For \(p_{0}< p< p_{1}\). By Lemma 5 we see that \(H_{-f ( \sqrt{t} ),-g ( \sqrt{t} ) } ( ( \pi^{2}/4 ) ^{-} ) >0\). It follows from Lemma 3 that there is an \(t_{0}\in ( 0,\pi^{2}/4 ) \) such that the function \(-f ( \sqrt{t} ) / ( -g ( \sqrt{t} ) ) \) is strictly decreasing on \(( 0,t_{0} ) \) and strictly increasing on \(( t_{0},\pi /2 ) \), where \(t=x^{2}\). Consequently, we obtain
which implies (1.9).
$$\frac{f ( x ) }{g ( x ) }< \max \biggl( \lim_{x\rightarrow 0^{+}}\frac{f ( x ) }{g ( x ) }, \lim_{x\rightarrow ( \pi /2 ) ^{-}}\frac{f ( x ) }{g ( x ) } \biggr) =\max \biggl( \frac{2}{5},\lambda_{p} \biggr), $$
Proof of Theorems 3
Let
Since
we find that the function \(H(p,x)\) is decreasing with respect to p on \((0,p_{3})\). Then by the left hand side of (1.8) and by (1.10) we can complete the proof of Theorem 3. □
$$H(p,x)= \biggl( 1-\frac{2}{15}x^{2}-px^{6} \biggr) ^{5/2},\quad 0< x< \frac{ \pi }{2},0< p< p_{3}. $$
$$\frac{\partial }{\partial p}H(p,x)= -\frac{5}{2}x^{6} \biggl( 1-\frac{2}{15}x^{2}-px^{6} \biggr) ^{3/2}< 0, $$
5 Consequences and remarks
Let \(p\rightarrow 0^{+}\) in (iii) of Theorem 2. Then we have the following.
Corollary 1
The function
is strictly decreasing on
\(( 0,\pi /2 ) \), and therefore, the inequalities
hold for
\(x\in ( 0,\pi /2 ) \), where
are the best constants.
$$x\mapsto F_{2/15} ( x ) =\frac{\ln ( 1-2x^{2}/15 ) }{x\cot x-1} $$
$$ \alpha \biggl( 1-\frac{2}{15}x^{2} \biggr) ^{5/2}< \biggl( 1-\frac{2}{15}x ^{2} \biggr) ^{1/\lambda_{0}}< \exp ( x\cot x-1 ) < \biggl( 1- \frac{2}{15}x^{2} \biggr) ^{5/2} $$
(5.1)
$$\begin{aligned} \lambda_{0} =&-\ln \biggl( 1-\frac{\pi^{2}}{30} \biggr) \approx 0.39897, \\ \alpha =&e^{-1} \bigl( 1-\pi^{2}/30 \bigr) ^{-5/2} \approx 0.997 42, \end{aligned}$$
Proof
It suffices to show the first inequality in (5.1). Consider the monotonicity of the function
on \(( 0,\pi /2 ) \). Since
holds for all \(x\in ( 0,\pi /2 ) \), we see that the function \(K(x)\) is decreasing on \(( 0,\pi /2 ) \). So
which completes the proof of Corollary 1. □
$$K(x)= \biggl( 1-\frac{2}{15}x^{2} \biggr) ^{1/\lambda_{0}-5/2} $$
$$K^{\prime }(x)= \biggl( \frac{1}{\lambda_{0}}-\frac{5}{2} \biggr) \biggl( -\frac{4}{15}x \biggr) \biggl( 1-\frac{2}{15}x^{2} \biggr) ^{1/\lambda_{0}-7/2}< 0 $$
$$K(x)>K\bigl( ( \pi /2 ) ^{-}\bigr)=e^{-1} \bigl( 1- \pi^{2}/30 \bigr) ^{-5/2}=\alpha, $$
Theorem 4
Let
\(0< p\leq 4/\pi^{2}\). Then the function
is strictly decreasing on
\(( 0,\pi /2 ) \)
if and only if
\(0< p\leq 2/15\). And therefore, for
\(0< p\leq 2/15\), the double inequality
holds for
\(x\in ( 0,\pi /2 ) \), where
\(\beta_{p}=-\ln ( 1-p\pi^{2}/4 ) \).
$$ x\mapsto F_{p} ( x ) =\frac{\ln ( 1-px^{2} ) }{x \cot x-1} $$
(5.2)
$$ \bigl( 1-px^{2} \bigr) ^{1/\beta_{p}}< e^{x\cot x-1}< \bigl( 1-px^{2} \bigr) ^{1/ ( 3p ) } $$
(5.3)
Proof
The necessity follows from
To prove the sufficiency, we note that
where \(f_{1}\) is positive and decreasing on \(( 0,\pi /2 ) \) by Corollary 1, it thus suffices to prove the function \(f_{2}\) is positive and decreasing on \(( 0,\pi /2 ) \). A simple computation gives
which indicates that \(f_{2}\) is strictly decreasing on \(( 0, \pi /2 ) \) by Lemma 1. Meanwhile \(f_{2} ( x ) \) is obviously positive for \(p\in (0,2/15]\), which proves the sufficiency.
$$\lim_{x\rightarrow 0+}\frac{F_{p}^{\prime } ( x ) }{x}= \frac{1}{5}p ( 15p-2 ) \leq 0. $$
$$\frac{\ln ( 1-px^{2} ) }{x\cot x-1}=\frac{\ln ( 1-2x ^{2}/15 ) }{x\cot x-1}\times \frac{\ln ( 1-px^{2} ) }{ \ln ( 1-2x^{2}/15 ) }:=f_{1} ( x ) \times f_{2} ( x ), $$
$$\begin{aligned}& \frac{ [ \ln ( 1-px^{2} ) ] ^{\prime }}{ [ \ln ( 1-2x^{2}/15 ) ] ^{\prime }} =\frac{1}{2}p\frac{15-2x ^{2}}{1-px^{2}}, \\& \biggl( \frac{1}{2}p\frac{15-2x^{2}}{1-px^{2}} \biggr) ^{\prime } =15px \frac{p-2/15}{ ( px^{2}-1 ) ^{2}}< 0, \end{aligned}$$
Inequalities (5.3) follow from the decreasing property of the function \(F_{p} ( x ) \).
The proof is finished. □
Remark 2
It is easy to check that, for \(0< p\leq 2/15\) and \(x\in ( 0, \pi /2 ) \),
where \(\alpha_{p}=e^{-1} ( 1-\pi^{2}p/4 ) ^{-1/ ( 3p ) }\), \(\beta_{p}=-\ln ( 1-p\pi^{2}/4 ) \). In fact, we have
Due to \(x\in ( 0,\pi /2 ) \), the first factor is positive. And, since \(p\mapsto 1+ ( \ln ( 1-p\pi^{2}/4 ) ) / ( 3p ) \) is decreasing in p, it follows that, for \(0< p\leq 2/15\),
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].
$$ \bigl( 1-px^{2} \bigr) ^{1/\beta_{p}}>\alpha_{p} \bigl( 1-px^{2} \bigr) ^{1/ ( 3p ) }, $$
(5.4)
$$\begin{aligned}& \frac{\ln ( 1-px^{2} ) }{\beta_{p}}-\ln \alpha_{p}- \frac{1}{3p}\ln \bigl( 1-px^{2} \bigr) \\& \quad =\frac{\ln ( 1-px^{2} ) }{-\ln ( 1-p\pi^{2}/4 ) }+1+\frac{1}{3p}\ln \bigl( 1-p\pi^{2}/4 \bigr) -\frac{1}{3p}\ln \bigl( 1-px^{2} \bigr) \\& \quad =\frac{1}{\ln ( 1-p\pi^{2}/4 ) }\ln \frac{1-p\pi^{2}/4}{1-px ^{2}}+\frac{1}{3p}\ln \frac{1-p\pi^{2}/4}{1-px^{2}} \\& \quad = \biggl[ 1-\frac{\ln ( 1-px^{2} ) }{\ln ( 1-p\pi^{2}/4 ) } \biggr] \biggl[ 1+\frac{\ln ( 1-p\pi^{2}/4 ) }{3p} \biggr] . \end{aligned}$$
$$\biggl( 1+\frac{\ln ( 1-p\pi^{2}/4 ) }{3p} \biggr) > \frac{5}{2}\ln \biggl( 1- \frac{\pi^{2}}{30} \biggr) +1\approx 0.002583 8>0. $$
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 \(h_{1} ( 0^{+} ) =h_{2} ( 0^{+} ) =0\), then differentiation yields
which indicates that \(h_{1}/h_{2}\) is increasing in p by Lemma 1.
$$\ln \bigl( 1-px^{2} \bigr) ^{1/\beta_{p}}=\frac{\ln ( 1-px^{2} ) }{-\ln ( 1-p\pi^{2}/4 ) }:= \frac{h_{1} ( p ) }{h_{2} ( p ) } $$
$$\begin{aligned}& \frac{h_{1}^{\prime } ( p ) }{h_{2}^{\prime } ( p ) } =\frac{\frac{d}{dp}\ln ( 1-px^{2} ) }{-\frac{d}{dp} \ln ( 1-p\pi^{2}/4 ) }=-\frac{x^{2}}{\pi^{2}}\frac{4-\pi ^{2}p}{1-px^{2}}, \\& \biggl( \frac{h_{1}^{\prime } ( p ) }{h_{2}^{\prime } ( p ) } \biggr) ^{\prime } =\frac{x^{2}}{\pi^{2}} \frac{\pi^{2}-4x ^{2}}{ ( 1-px^{2} ) ^{2}}>0, \end{aligned}$$
Remark 4
Taking \(p=0^{+}\), \(1/\pi^{2}\), and \(2/15\) in Theorem 4. Using the monotonicity of the lower and upper bounds in (5.3), we can obtain
where \(\beta_{2/15}=-\ln ( 1-\pi^{2}/30 ) \approx 0.398 97\). This shows that our double inequality (5.3) is a generalization and refinement of the one (1.2) (see [19]).
$$\begin{aligned} e^{-4x^{2}/\pi^{2}} < & \biggl( 1-\frac{x^{2}}{\pi^{2}} \biggr) ^{1/ \ln ( 4/3 ) }< \biggl( 1-\frac{2}{15}x^{2} \biggr) ^{1/\beta_{2/15}}< e ^{x\cot x-1} \\ < & \biggl( 1-\frac{2}{15}x^{2} \biggr) ^{5/2}< \biggl( 1-\frac{x^{2}}{\pi ^{2}} \biggr) ^{\pi^{2}/3}< e^{-x^{2}/3}, \end{aligned}$$
The following theorem gives a sufficient condition for the function \(F_{p} ( x ) \) to be increasing on \(( 0,\pi /2 ) \).
Theorem 5
Proof
We have
If we prove \(( p-u_{n} ) \geq 0\) for \(n\geq 1\), then by Lemma 2
\(F_{p}\) is increasing on \(( 0,\pi /2 ) \), and the reverse of double inequality (5.3) follows. Using the right hand side of (2.2) we have
for \(n\geq 3\). This together with \(p-u_{1}=p-2/15>0\) and \(p-u_{2}=p-1/7 \geq 0 \) yields \(( p-u_{n} ) \geq 0\) for \(n\geq 1\).
$$\begin{aligned}& F_{p} ( x ) =\frac{-\ln ( 1-px^{2} ) }{- ( x \cot x-1 ) }=\frac{\sum_{n=1}^{\infty }\frac{p^{n}}{n}x^{2n}}{ \sum_{n=1}^{\infty }\frac{2^{2n}}{ ( 2n )!}\vert B_{2n}\vert x ^{2n}}:= \frac{\sum_{n=1}^{\infty }a_{n}^{\prime }x^{2n}}{\sum_{n=1} ^{\infty }b_{n}^{\prime }x^{2n}}, \\& c_{n}^{\prime } =a_{n+1}^{\prime }- \frac{b_{n+1}^{\prime }}{b_{n} ^{\prime }}a_{n}^{\prime } \\& \hphantom{c_{n}^{\prime }}=\frac{p^{n+1}}{n+1}- \frac{2}{ ( n+1 ) ( 2n+1 ) }\frac{\vert B_{2n+2}\vert }{ \vert B_{2n}\vert }\frac{p^{n}}{n} \\& \hphantom{c_{n}^{\prime }} =\frac{p^{n}}{n+1} \biggl( p-\frac{2}{n ( 2n+1 ) }\frac{ \vert B_{2n+2}\vert }{\vert B_{2n}\vert } \biggr):=\frac{p^{n}}{n+1} ( p-u_{n} ). \end{aligned}$$
$$\begin{aligned} p-u_{n} \geq &\frac{1}{7}-\frac{2}{n ( 2n+1 ) } \frac{1}{ ( 2\pi ) ^{2}} \frac{2 ( n+1 ) ( 2n+1 ) 2^{2n+1}}{2^{2n+1}-1} \\ =&\frac{1}{7\pi^{2}}\frac{ ( ( \pi^{2}-7 ) n-7 ) 2^{2n+1}- \pi^{2}n}{n ( 2^{2n+1}-1 ) }>0 \end{aligned}$$
This completes the proof. □
Remark 5
Likewise, taking \(p=1/7\), \(1/3\), \(4/\pi^{2}\) in the above theorem, we have
where \(\beta_{1/7}=-\ln ( 1-\pi^{2}/28 ) \approx 0.434 61\), \(\beta_{1/3}=-\ln ( 1-\pi^{2}/12 ) \approx 1.728 6\).
$$\begin{aligned} \biggl( 1-\frac{4x^{2}}{\pi^{2}} \biggr) ^{\pi^{2}/12} < &1- \frac{x^{2}}{3}< \biggl( 1-\frac{x^{2}}{7} \biggr) ^{7/3}< e^{x\cot x-1} \\ < & \biggl( 1-\frac{x^{2}}{7} \biggr) ^{1/\beta_{1/7}}< \biggl( 1- \frac{x ^{2}}{3} \biggr) ^{1/\beta_{1/3}}, \end{aligned}$$
Finally, we consider the monotonicity of the function \(F_{p} ( x ) \) on \(( 0,\pi ) \). In this case, we have to assume that \(0< p\leq 1/\pi^{2}\).
Theorem 6
Let
\(0< p\leq 1/\pi^{2}\). Then the function
\(F_{p} ( x ) \)
defined by (5.2) is strictly decreasing on
\(( 0,\pi ) \). And therefore, the inequality
holds for all
\(x\in ( 0,\pi ) \).
$$e^{x\cot x-1}< \bigl( 1-px^{2} \bigr) ^{1/ ( 3p ) } $$
Proof
From Lemma 2 and the proof of the previous theorem, it suffices to prove \(p-u_{n}\leq 0\) for \(n\geq 1\) and \(0< p\leq 1/\pi ^{2}\). Using the left hand side of (2.2) we get
for \(n\geq 1\), which, by Lemma 2, proves the decreasing property of \(F_{p}\). □
$$\begin{aligned} p-u_{n} =&p-\frac{2}{n ( 2n+1 ) }\frac{\vert B_{2n+2}\vert }{ \vert B_{2n}\vert } \\ \leq &\frac{1}{\pi^{2}}-\frac{2}{n ( 2n+1 ) }\frac{1}{ ( 2\pi ) ^{2}} \frac{2 ( n+1 ) ( 2n+1 ) ( 2^{2n-1}-1 ) }{2^{2n-1}} \\ =&-\frac{2^{2n}-2n-2}{2^{2n}\pi^{2}n}\leq 0 \end{aligned}$$
Remark 6
Let \(p=p_{1}=32 ( 30-\pi^{2}-30e^{-2/5} ) / ( 15\pi^{6} ) \approx 4.614 3\times 10^{-5}\) in Theorem 3, we can obtain
which is sharper than the right hand side of (1.2):
$$e^{x\cot x-1}< \biggl( 1-\frac{2}{15}x^{2}-p_{1}x^{6} \biggr) ^{5/2},\quad 0< x< \frac{ \pi }{2}, $$
$$e^{x\cot x-1}< \biggl( 1-\frac{2}{15}x^{2} \biggr) ^{5/2},\quad 0< x< \frac{ \pi }{2}. $$
Remark 7
Let \(p=p_{2}=4/70\text{,}875\) 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.
$$\biggl( 1-\frac{2}{15}x^{2}-\frac{4}{70\text{,}875}x^{6} \biggr) ^{5/2}< e^{x \cot x-1},\quad 0< x< \frac{\pi }{2}. $$
6 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 \(\ln (1-2x^{2}/15-px^{6})\) and a new criterion for the monotonicity of the quotient of two power series. Then we refine some well-known results presented in [19].
Acknowledgements
The author is thankful to anonymous referees for their careful corrections to and valuable comments on the original version of this paper.
Competing interests
The author declares that he has no competing interests.
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