Abstract
In this paper, we obtain some improved exponential approximation inequalities for the functions \((\sin x)/x\) and \(\sec(x)\), and we prove them by using the properties of Bernoulli numbers and new criteria for the monotonicity of quotient of two power series.
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1 Introduction
The following result is known as the Mitrinovic–Adamovic inequality [1, 2]:
Nishizawa [3] gave the upper bound of the function \(( ( \sin x ) /x ) ^{3}\) in the form of the above inequality (1.1) and obtained the following power exponential inequality:
Chen and Sándor [4] looked into the bounds for the function secx and obtain the following result for \(0< x<\pi /2\):
Nishizawa [3] obtained the following inequality with power exponential functions derived from the right-hand inequality side of (1.3):
The purpose of this article is to establish some exponential approximation inequalities which improve the ones of (1.1)–(1.4). We prove these results for circular functions by using the properties of Bernoulli numbers and new criteria for the monotonicity of quotient of two power series.
Theorem 1.1
Let \(0< x<\pi /2\), \(a=2/15\thickapprox 0.13333\) and \(b=4/\pi^{2}\thickapprox 0.40528\). Then we have
where a and b are the best constants in (1.5).
Theorem 1.2
Let \(0< x<\pi /2\), \(c=19/945\thickapprox 0.02011\) and \(d=8(30-\pi^{2})/(15 \pi^{4})\thickapprox 0.11022\). Then we have
where c and d are the best constants in (1.6).
Theorem 1.3
Let \(0< x<\pi /2\), \(b=4/\pi^{2}\thickapprox 0.40528\) and \(p=1/(2\ln (4/ \pi ))\thickapprox 2.0698\). Then we have
where b and p are the best constants in (1.7).
Theorem 1.4
Let \(0< x<\pi /2\),
Then we have
where α and β are the best constants in (1.8).
We note that the right-hand side of the inequality (1.5) is stronger than that one in (1.2) due to
while the double inequality (1.6) and (1.8) are sharper than the (1.5) and (1.7), respectively.
2 Lemmas
Lemma 2.1
Let \(B_{2n}\) be the even-indexed Bernoulli numbers, \(n=1,2,\ldots \) . Then
Lemma 2.2
Let \(B_{2n}\) be the even-indexed Bernoulli numbers. Then the following power series expansion:
and
hold.
Proof
The following power series expansions can be found in [9, 1.3.1.4(2)(3)]:
and
□
Lemma 2.3
([10])
Let \(a_{n}\) and \(b_{n}\) (\(n=0,1,2,\ldots \)) be real numbers, and let the power series \(A(t)=\sum_{n=0}^{\infty }a_{n}t^{n}\) and \(B(t)=\sum_{n=0} ^{\infty }b_{n}t^{n}\) be convergent for \(\vert t \vert < R\) (\(R\leq +\infty \)). If \(b_{n}>0\) for \(n=0,1,2,\ldots \) , and if \(\varepsilon_{n}=a_{n}/b_{n}\) is strictly increasing (or decreasing) for \(n=0,1,2,\ldots \) , then the function \(A(t)/B(t)\) is strictly increasing (or decreasing) on \((0,R)\) (\(R\leq +\infty \)).
In order to prove Theorem 1.4, we need the following lemma. We 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 and plays an important role in the proof of a monotonicity criterion for the quotient of power series.
Lemma 2.4
([11])
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 \(m\in N\), 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 \(H_{A,B} ( r^{-} ) \geq \textit{(resp. }{\leq} \textit{) }0\). Moreover, if \(H_{A,B} ( r^{-} ) <\textit{(resp. }{>}\textit{) }0\), 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 \(( t_{0},r ) \).
3 Proof of Theorem 1.1
Let
where
and
by Lemma 2.2. Let
where
We now show that \(\{ e_{n} \} \) is increasing for \(n\geq 1\). Since
by Lemma 2.1, the proof of \(e_{n-1}< e_{n}\) for \(n\geq 2\) can be completed when proving
In fact,
for \(n\geq 2\). So \(\{ a_{n}/b_{n} \} _{n\geq 1}\) is decreasing, and \(F_{1}(x)\) is decreasing on \((0,\pi /2)\) by Lemma 2.3. In view of \(F_{1}(0^{+})=-2/15\), and \(F_{1}((\pi /2)^{-})=-4/ \pi^{2}\), the proof of Theorem 1.1 is complete.
4 Proof of Theorem 1.2
(i) We first prove the left-hand side inequality of (1.6). Let
Then by Lemma 2.2 we have
where
By Lemma 2.1, we have
with
due to \(\pi^{2}<79/8\), where
It is not difficult to verify
and
for \(n\geq 3\). So \(i_{n}>0\) for \(n\geq 3\), and \(F_{2}(x)>0\) for \(x \in (0,\pi /2)\).
(ii) Then we prove the right-hand side inequality of (1.6). Let
Then by Lemma 2.2 we have
where
By Lemma 2.1 we have
that is,
where
for \(n\geq 2\). So \(l_{n}<0\) for \(n\geq 2\) and \(F_{3}(x)<0\) for \(x \in (0,\pi /2)\).
(iii) Let
Then
This complete the proof of Theorem 1.2.
5 Proof of Theorem 1.3
(1) Let
Then we get
where
We now show
for \(n\geq 1\), that is,
or
holds for \(n\geq 1\). In fact, by Lemma 2.1 we have
so (5.1) holds as long as we can prove that
that is,
which is equivalent to
for \(n\geq 1\). So \(k_{n}<0\) for \(n\geq 1\), which leads to \(G_{1}^{ \prime \prime \prime }(x)=\sum_{n=2}^{\infty }2n(2n-1)(2n-2)k_{n}x ^{2n-3}<0\), and \(G_{1}^{\prime \prime }(x)\) is decreasing on \((0,\pi /2)\). We can compute
which give
Then there exists an unique real number \(x_{1}\in (0,\pi /2)\) such that \(G_{1}^{\prime \prime }(x)>0\) on \((0,x_{1})\) and \(G_{1}^{\prime \prime }(x)<0\) on \((x_{1},\pi /2)\). So \(G_{1}^{\prime }(x)\) is increasing on \((0,x_{1})\) and decreasing on \((x_{1},\pi /2)\). Since
there exists an unique real number \(x_{2}\in (x_{1},\pi /2)\) such that \(G_{1}^{\prime }(x)>0\) on \((0,x_{2})\) and \(G_{1}^{\prime }(x)<0\) on \((x_{2},\pi /2)\). So \(G_{1}(x)\) is increasing on \((0,x_{2})\) and decreasing on \((x_{2},\pi /2)\). In view of \(G_{1}(0^{+})=0=G_{1}(( \pi /2)^{-})\), the proof of the left-hand side inequality of (1.7) is complete.
(2) Let
Then we get
where
We now show \(w_{n}>0\) for \(n\geq 1\), that is,
holds for \(n\geq 1\). In fact, by Lemma 2.1 we have
so (5.2) holds as long as we can prove that
which is true for \(n\geq 1\). So \(G_{2}^{\prime }(x)>0\), and \(G_{2}(x)\) is increasing on \((0,\pi /2)\). We can compute \(G_{2}(0^{+})=0\) and \(G_{2}((\pi /2)^{-})=+\infty \), the proof of the right-hand side inequality of (1.7) is complete.
(3) Let
Then
this completes the proof of Theorem 1.3.
6 Proof of Theorem 1.4
Let
where
and
with
Since
and
we can obtain
but
for \(n\geq 2\). The inequality (6.1) is equivalent to
By Lemma 2.1, we have
and
So (6.1) holds when we prove
or
which is ensured for \(n\geq 2\).
So
Since
we see that \(G_{4}(x)\) is increasing on \((0,\pi /2)\) by Lemma 2.4. In view of
the proof of Theorem 1.4 is complete.
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The first author was supported by the National Natural Science Foundation of China (no. 11471285 and no. 61772025).
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Zhu, L., Nenezić, M. New approximation inequalities for circular functions. J Inequal Appl 2018, 313 (2018). https://doi.org/10.1186/s13660-018-1910-9
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DOI: https://doi.org/10.1186/s13660-018-1910-9