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Padé approximant related to remarkable inequalities involving trigonometric functions
Journal of Inequalities and Applications volume 2016, Article number: 99 (2016)
Abstract
In this paper we, respectively, give simple proofs of some remarkable trigonometric inequalities, based on the Padé approximation method. We also obtain rational refinements of these inequalities. We are convinced that the Padé approximation method offers a general framework for solving many other similar inequalities.
1 Introduction and motivation
The starting point of this paper is the following famous inequalities.
The Cusa-Huygens inequalities state that for \(x\in ( 0,\frac{\pi }{2} ) \),
The Wilker inequality asserts that
The Huygens inequality for sine and tangent functions states that for \(x\in ( 0,\frac{\pi }{2} ) \),
The inequality
was established by Adamović and Mitrinović (see, e.g. [1], p.238).
Another inequality which is of interest to us is the following:
In [2], the following inequality was given:
The classical Jordan inequality states that
This inequality has been further refined in the past year. For example, in [3], Redheffer proposed the inequality
A similar result, which involves the rational approximation, has been obtained in [4] as follows:
These inequalities were of great interest throughout the research, since they were extended in different forms in the recent past. We refer to [1–15] and closely related references therein. Some of the recent improvements are nice through their symmetric form, but these inequalities have also practical importance, because they provide some bounds for a given expression. We noticed that many of these new inequalities were obtained using Taylor’s expansions of some trigonometric functions. That is why in our paper we propose a new approach based on the Padé approximation. It is known that a Padé approximant is the ‘best’ approximation of a function by a rational function of given order. This method has the somewhat magical property of converting a poorly converging power series into a usually much better behaved rational polynomial. The rational approximation is particularly good for series with alternating terms and poor polynomial convergence. For these reasons, longer versions of optimized rational polynomials are frequently used in computer calculations.
Following the techniques developed and the results obtained in recent work [2, 6, 10, 11], we expect that the Padé approximation method will be useful in solving and refining others problems concerning inequalities.
The Padé approximant [L/M] corresponds to the Taylor series. When it exists, the [L/M] Padé approximant to any power series \(A ( x ) =\sum_{j=0}^{\infty }a_{j}x^{j}\) is unique. If \(A ( x ) \) is a transcendental function, then the terms are given by the Taylor series about \(x_{0}\), \(a_{n}=\frac{1}{n!}A^{ ( n ) } ( x_{0} ) \).
The coefficients are found by setting \(A ( x ) =\frac{ p_{0}+p_{1}x+\cdots+p_{L}x^{L}}{1+q_{1}x+\cdots+q_{M}x^{M}}\). These give the set of equations
For example, we consider the Taylor series for sin: \(\sin x=x-\frac{1}{6}x^{3}+\frac{1}{120}x^{5}+O ( x^{7} ) \) and its associated polynomial: \(x-\frac{1}{6}x^{3}+\frac{1}{120}x^{5}\).
The Padé approximant \(\sin _{[3/3]}(x)=\frac{p_{0}+p_{1}x+p_{2}x^{2}+p_{3}x^{3}}{1+q_{1}x+q_{2}x^{2}+q_{3}x^{3}}\) satisfies
We find
Therefore
The first order versions of a few trigonometric functions which we will use are in Table 1.
2 Some lemmas
In order to attain our aim, we first prove several lemmas.
Lemma 2.1
For every \(x\in ( 0,\frac{\pi }{2} ) \), one has
and also
Proof
We introduce the function
Easy computation yields
Evidently \(r^{ ( 5 ) }>0\) on \(( 0,\frac{\pi }{2} ) \). Then \(r^{ ( 4 ) }\) is strictly increasing on \(( 0,\frac{\pi }{2} ) \). As \(r^{ ( 4 ) } ( 0 ) =0\), we get \(r^{ ( 4 ) }>0\) on \(( 0,\frac{\pi }{2} ) \). Continuing the algorithm, finally we obtain \(r ( x ) >0\) for all \(x\in ( 0,\frac{\pi }{2} ) \).
Let
Then
The function \(s^{ ( 7 ) }<0\) for all \(x\in ( 0,\frac{\pi }{2} ) \), therefore \(s^{ ( 6 ) }\) is strictly decreasing on \(( 0,\frac{\pi }{2} ) \). As \(s^{ ( 6 ) } ( 0 ) =0 \), we get \(s^{ ( 6 ) }<0\) on \(( 0,\frac{\pi }{2} ) \). Using the same arguments, finally we have \(s ( x ) <0\) for every \(x\in ( 0,\frac{\pi }{2} ) \).
For proving the last part of the lemma, we consider the function
and its derivatives
We have \(p^{ ( 3 ) }<0\) on \(( 0,\frac{\pi }{2} ) \), hence \(p^{ ( 2 ) }\) is strictly decreasing on \(( 0,\frac{\pi }{2} ) \). As \(p^{ ( 2 ) } ( 0 ) =0\), it follows that \(p^{ ( 2 ) }<0\) on \(( 0,\frac{\pi }{2} ) \). As above, finally we deduce that \(p<0\) on \(( 0,\frac{\pi }{2} ) \).
This completes the proof. □
Lemma 2.2
For every \(x\in ( 0,\frac{\pi }{2} ) \), one has
and also
Proof
Let
We get
Evidently \(f^{ ( 6 ) }<0\) on \(( 0,\frac{\pi }{2} ) \). It follows that \(f^{ ( 5 ) }\) is strictly decreasing on \(( 0,\frac{\pi }{2} ) \). As \(f^{ ( 5 ) } ( 0 ) =0\), we get \(f^{ ( 5 ) }<0\) on \(( 0,\frac{\pi }{2} ) \). Using the same algorithm, we finally obtain \(f ( x ) <0\) for \(x\in ( 0,\frac{\pi }{2} ) \).
If we consider
then we have
The positivity of \(g^{ ( 6 ) }\) on \(( 0,\frac{\pi }{2} ) \) shows that \(g^{ ( 5 ) }\) is strictly increasing on \(( 0,\frac{\pi }{2} ) \). As \(g^{ ( 5 ) } ( 0 ) =0\), we get \(g^{ ( 5 ) }>0\) on \(( 0,\frac{\pi }{2} ) \). Similar arguments lead us to the positivity of g on \(( 0,\frac{\pi }{2} ) \).
Next, we get
Elementary calculations reveal that
We remark that \(h^{ ( 4 ) }>0\) on \(( 0,\frac{\pi }{2} ) \).
Using a similar algorithm to above, we find \(h>0\) on \(( 0,\frac{\pi }{2} ) \).
This completes the proof of the second lemma. □
Lemma 2.3
For every \(x\in ( 0,\frac{\pi }{2} ) \), one has
Remark 2.1
The denominator \(-6x^{2}+15\) is positive for all \(x\in ( 0,\sqrt{\frac{5}{2}} ) = ( 0, 1.58 ) \). Since \(x\in ( 0,\frac{\pi }{2} ) = ( 0, 1.57075 ) \), the denominator of the function from the right-hand side is positive.
Proof of Lemma 2.3
We introduce the function
The derivative is \(q^{\prime } ( x ) =x [ ( 3-x^{2} ) \sin x-3x\cos x ] \). Then we consider the function \(u ( x ) = ( 3-x^{2} ) \sin x-3x\cos x\) and its derivative \(u^{\prime } ( x ) =x ( \sin x-x\cos x ) \). Again we consider the function \(v ( x ) =\sin x-x\cos x\) and its derivative \(v^{\prime } ( x ) =x\sin x\). We see that \(v^{\prime }>0\) on \(( 0,\frac{\pi }{2} ) \). As \(v ( 0 ) =0\), it follows that \(v>0\) on \(( 0,\frac{\pi }{2} ) \), therefore \(u^{\prime } ( x ) >0\) for every \(x\in ( 0,\frac{\pi }{2} ) \).
Due to similar arguments, finally it follows that \(q ( x ) >0\) for every \(x\in ( 0,\frac{\pi }{2} ) \).
The proof of Lemma 2.3 is complete. □
3 Results
In this section we will formulate and prove the announced rational inequalities, using as our main tool the estimates from the above lemmas.
First, we establish new inequalities of the Cusa-Huygens type as follows.
Theorem 3.1
For every \(0< x<\frac{\pi }{2}\), the rational inequalities chain
holds.
Proof
The inequalities
and
are easy consequences of Lemmas 2.1 and 2.2.
We will calculate the difference
The function
has the real roots
and
We obtain \(P ( x ) <0\) for all \(x\in ( 0,\frac{\pi }{2} ) \subset ( -\frac{2}{3}\sqrt{5 ( \sqrt{31}-2 ) },\frac{2}{3}\sqrt{5 ( \sqrt{31}-2 ) } ) \) and therefore \(E ( x ) >0\) for every \(x\in ( 0,\frac{\pi }{2} ) \).
We will calculate the difference
The function
has the real roots
and
We have \(Q ( x ) <0\) for all \(x\in ( 0,\frac{\pi }{2} ) \subset ( -\sqrt{\frac{120+6\sqrt{2{,}710}}{11}},\sqrt{\frac{120+6\sqrt{2{,}710}}{11}} ) \) and therefore \(F ( x ) >0\) for all \(x\in ( 0,\frac{\pi }{2} ) \). □
Using the Padé approximation method we improve the celebrated Wilker inequality as follows.
Theorem 3.2
Let \(0< x<\frac{\pi }{2}\). Then
holds.
Proof
Using the results from Lemma 2.1, respectively Lemma 2.3, we have
The inequality
has the equivalent form
This inequality is obviously true for all \(x\in ( 0,\frac{\pi }{2} ) \). □
We refine the Huygens inequality for the sine and tangent functions (1.3) as follows.
Theorem 3.3
For \(0< x<\frac{\pi }{2}\), we have
Proof
Using again the results from Lemma 2.1, respectively Lemma 2.3, we deduce
The inequality
has the equivalent form
which is true for all \(x\in ( 0,\frac{\pi }{2} ) \). □
In the following we will refine the inequality (1.4) as follows.
Theorem 3.4
For every \(0< x<\frac{\pi }{2}\), we have
Proof
In the light of the results of the above lemmas, we have only to prove the inequality
for every \(x\in ( 0,\frac{\pi }{2} ) \).
The difference
can be re-written as
The polynomial function \(R ( x ) =-81x^{6}-4{,}720x^{4}-15{,}600x^{2}+1{,}728{,}000\) has the real roots \(x_{1}\approx -3.9658\) and \(x_{2}\approx 3.9658\). Therefore \(R ( x ) >0\) for all \(x\in ( 0,\frac{\pi }{2} ) \) and also the above difference is positive for all \(x\in ( 0,\frac{\pi }{2} ) \). □
Using the Padé approximation method, we will sharpen the inequality (1.5) as follows.
Theorem 3.5
For \(0< x<\frac{\pi }{2}\), the rational inequality
holds.
Proof
The inequalities from Lemma 2.1, respectively Lemma 2.3, give us the estimate
which is greater than 2, since \(75x^{4}+150x^{2}>0\) for all \(x\in ( 0,\frac{\pi }{2} ) \). □
Using the Padé approximation for the sine and cosine functions, we obtain an improved version of the inequality (1.6).
Theorem 3.6
For every \(x\in ( 0,\frac{\pi }{2} ) \), the rational inequality
holds.
Proof
The estimates from the sine and cosine functions obtained in Lemma 2.1, respectively Lemma 2.2, lead to the inequality
which is obviously greater than 4 for all \(x\in ( 0,\frac{\pi }{2} ) \). □
In the following we will determine a lower rational bound for the sine function.
The sine function
appears in a variety of applications.
Using the lower rational bound founded in Lemma 2.2, we will provide a refined version of the classical Jordan inequality (1.7):
Theorem 3.7
The inequality
is true for all \(x\in ( 0,\sqrt{\frac{60\pi -120}{7\pi +6}} ) \).
Proof
We only have to find x such that \(\frac{60-7x^{2}}{60+3x^{2}}>\frac{2}{\pi }\). Elementary calculations lead to the solution \(0< x<\sqrt{\frac{60\pi -120}{7\pi +6}}\). □
We remark that \(\sqrt{\frac{60\pi -120}{7\pi +6}}\approx 1.5642\) and \(\frac{\pi }{2}\approx 1.5707\), therefore our rational refinement of Jordan’s inequality gives good results near the origin.
We also remark that \(\frac{60-7x^{2}}{60+3x^{2}}>\frac{\pi ^{2}-x^{2}}{\pi ^{2}+x^{2}}\) for \(x\in ( 0,\sqrt{\frac{60-5\pi ^{2}}{2}} ) = ( 0, 2.3081 ) \), so we improved the Redheffer inequality (1.8) near the origin.
Finally, we see that \(\frac{60-7x^{2}}{60+3x^{2}}>\frac{53}{53+9x^{2}}\) for \(x\in ( 0,\sqrt{\frac{10}{63}} ) \supset ( 0, \frac{1}{3} ) \), so our lower bound for the function sinc offers a better approximation near the origin than the fractional function \(\frac{53}{53+9x^{2}}\).
4 Final remarks
We are convinced that the Padé approximation method is suitable for proving and refining many other inequalities. We also mention that these refinements are of rational type, so these can easily be used in computer calculations.
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Bercu, G. Padé approximant related to remarkable inequalities involving trigonometric functions. J Inequal Appl 2016, 99 (2016). https://doi.org/10.1186/s13660-016-1044-x
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DOI: https://doi.org/10.1186/s13660-016-1044-x