We present an application of the *Natural Approach* algorithm in the proof (Application 1 - Theorem 9) of certain new rational (Padé) approximations of the function cos^{2}
*x*, as well as in the improvement of a class of inequalities (20) by Yang (Application 2, Theorem 10).

### Application 1

Bercu [7] used the Padé approximations to prove certain inequalities for trigonometric functions. Let us denote by \(( f(x) ) _{[m/n]}\) the Padé approximant \([ m/n ] \) of the function \(f(x)\).

In this example we introduce a constraint of the function cos^{2}
*x* by the following Padé approximations:

$$\bigl(\cos^{2}{ x} \bigr)_{[6/4]} = \frac{-59 x^{6} + 962 x^{4} - 3{,}675 x^{2} + 4{,}095}{17 x^{4} + 420 x^{2}+ 4{,}095} $$

and

$$\bigl(\cos^{2}{ x} \bigr)_{[4/4]} = \frac{163 x^{4} - 780 x^{2} + 945}{13 x^{4} + 165 x^{2} + 945}. $$

### Theorem 9

*The following inequalities hold true for every*
\(x \in ( 0, \frac{\pi}{2} )\):

$$ \bigl(\cos^{2}{ x} \bigr)_{[6/4]} < \cos ^{2}{ x} < \bigl(\cos^{2}{ x} \bigr)_{[4/4]}. $$

(12)

### Proof

We first prove the left-hand side inequality (11). Using the computer software for symbolic computations, we can conclude that the function \(G_{1}(x) = ( \cos ^{2}{ x} ) _{[6/4]}\) has exactly one zero \(\delta =1.551413\ldots\) in the interval \(( 0, \frac{\pi}{2} )\). As \(G_{1}(0)=1>0\) and \(G_{1} ( \frac{\pi}{2} ) =-0.000431\ldots <0\), we deduce that

$$ G_{1}(x)\geq 0 \quad \mbox{for every } x\in (0,\delta ] $$

(13)

and

$$ G_{1}(x)< 0 \quad \mbox{for every } x\in \biggl(\delta , \frac{\pi}{2} \biggr). $$

(14)

Moreover, \(G_{1}(x)<\cos ^{2}{ x}\) for every \(x\in ( \delta , \frac{\pi}{2} ) \). We prove now that

$$ G_{1}(x)< \cos ^{2}{ x},\quad x\in ( 0, \delta ]. $$

(15)

We search a downward Taylor polynomial \(\underline{T}_{ 4k+2}^{ \cos ,0}(x) \) such that for every \(x\in ( 0, \delta ] \),

$$ G_{1}(x)< \bigl( \underline{T}_{ 4k+2}^{ \cos ,0}(x) \bigr) ^{2}< \cos ^{2}{ x}. $$

(16)

We apply the *Natural Approach* algorithm to the function \(f(x)=\cos ^{2}{ x}\), \(x\in (0,\delta ]\), to determine the downward Taylor polynomial \(\underline{T}_{ 4k+2}^{ \cos ,0}(x)\) such that

$$\bigl( \underline{T}_{4k+2}^{ \cos ,0}(x) \bigr)^{2} < \cos ^{2} x,\quad x\in (0,\delta]. $$

We can use Method C or Method D from the *Natural Approach* algorithm since \(\delta < \frac{\pi}{2}\). In this proof, we choose Method C.

The smallest *k* for which \(\underline{T}_{ 4k+2}^{ \cos ,0}(\delta )>0\) is \(k=1\). Therefore \(\widehat{k}=1\). In the *Estimation* procedure only step I can be applied to the (single) addend cos^{2}
*x*. In this step, \(s_{1}\geq 0\) and \(k_{1}\geq \widehat{k}=1\) should be selected. Let us select \(s_{1}=0\) and \(k_{1}=2\).^{Footnote 1} As a result of this selection, the output of the *Natural Approach* algorithm is the polynomial

$$\mathcal{TP}(x)= \bigl( \underline{T}_{10}^{ \cos ,0}(x) \bigr) ^{2}= \biggl(1- \frac{x^{2}}{2!}+ \frac{x^{4}}{4!}- \frac{x^{6}}{6!}+ \frac{x^{8}}{8!}- \frac{x^{10}}{10!} \biggr)^{2}. $$

We prove that

$$ \bigl( \underline{T}_{10}^{ \cos ,0}(x) \bigr) ^{2}-G_{1}(x)>0,\quad x\in ( 0, \delta ] . $$

(17)

This is true since

$$\bigl( \underline{T}_{10}^{ \cos ,0}(x) \bigr) ^{2}-G_{1}(x)= \frac{x^{12}}{13{,}168{,}189{,}440{,}000 (17 x^{4}+420 x^{2}+4{,}095)} Q(x), $$

where

$$\begin{aligned} Q(x) = & 17 x^{12}+15 x^{8}\bigl(15{,}837-176 x^{2}\bigr)+8{,}100 x^{4}\bigl(64{,}519-1{,}687 x^{2}\bigr) \\ & {}+3{,}200 \bigl(50{,}205{,}015-4{,}035{,}906 x^{2}\bigr) > 0. \end{aligned}$$

Finally, we have \(G_{1}(x)<\cos ^{2}{ x}\) for every \(x\in ( 0, \delta ]\). According to (14), we have

$$G_{1}(x)< \cos ^{2}{ x} \quad \mbox{for every } x\in \biggl( 0, \frac{\pi}{2} \biggr). $$

Now we prove the right-hand side inequality (12). For \(G_{2}(x)= ( \cos ^{2}{ x} ) _{[4/4]}\), we prove the following inequalities for every \(x\in ( 0,\frac{\pi}{2} )\):

$$ \cos ^{2}{ x}< \bigl( \overline{T}_{8}^{ \cos ,0}(x) \bigr) ^{2}< G_{2}(x). $$

(18)

Based on Proposition 5, it is enough to prove that for every \(x\in (0,\frac{\pi}{2} )\),

$$ \bigl( \overline{T}_{8}^{ \cos {x},0}(x) \bigr) ^{2}< G_{2}(x). $$

(19)

This is true as

$$G_{2}(x)- \bigl( \overline{T}_{8}^{ \cos {x},0}(x) \bigr) ^{2} = \frac{x^{10} }{1{,}625{,}702{,}400 (13 x^{4}+165 x^{2}+945)} R(x), $$

where

$$R(x)=x^{8}\bigl(1{,}291-13x^{2}\bigr)+x^{4} \bigl(2{,}004{,}240-66{,}913x^{2}\bigr)+480\bigl(632{,}604-74{,}625x^{2} \bigr) > 0. $$

Since \(\cos ^{2}{ x}\leq ( \overline{T}_{4k}^{ \cos ,0}(x) ) ^{2} \), for every \(k\in \mathbb {N}_{0}\) and all \(x\in (0, \frac{\pi}{2})\), we have

$$\cos ^{2}{ x}< G_{2}(x)\quad\mbox{for every } x\in \biggl( 0, \frac{\pi}{2} \biggr) . $$

□

### Note

Using Padé approximations, Bercu [7, 13] recently refined certain trigonometric inequalities over various intervals \(\mathcal{I}=(0,\delta ) \subseteq (0, \frac{\pi}{2})\). All such inequalities can be proved in a similar way and using the *Natural Approach* algorithm as in the proof of Theorem 9.

### Application 2

Jang [6] proved the following inequalities for every \(x\in ( 0,\pi )\):

$$ \cos ^{2}{ \frac{x}{2}}\leq \frac{\sin {x}}{x} \leq \cos ^{3}{ \frac{x}{3}}\leq \frac{2+\cos {x}}{3}. $$

(20)

Previously, Klén et al. [2] proved the above inequality on \((0,\sqrt{27/5})\) only.

In this example we propose the following improvement of (20).

### Theorem 10

*The following inequalities hold true for every*
\(x \in (0, \pi )\)
*and*
\(a \in (1, \frac{3}{2} )\):

$$ \cos^{2}{ \frac{x}{2}} \leq \biggl( \frac{\sin{x}}{x} \biggr)^{a} \leq \frac{\sin{x}}{x}. $$

(21)

### Proof

As \(a>1\) and \(0< \frac{\sin {x}}{x}<1\), we have

$$\biggl( \frac{\sin {x}}{x} \biggr) ^{ a} < \frac{\sin {x}}{x}. $$

We prove now the following inequality:

$$ \cos ^{2}{ \frac{x}{2}}< \biggl( \frac{\sin {x}}{x} \biggr) ^{ a} $$

(22)

for every \(x\in ( 0,\pi ) \) and \(a\in ( 1, \frac{3}{2} ) \). It suffices to show that the following mixed logarithmic-trigonometric-polynomial function [11]

$$ F(x)=a\ln \biggl( \frac{\sin {x}}{x} \biggr) -2\ln \biggl( \cos { \frac{x}{2} } \biggr) $$

(23)

is positive for every \(x\in ( 0,\pi ) \) and \(a\in ( 1, \frac{3}{2} ) \). Given that

$$ \lim_{x\rightarrow 0}F(x)=0, $$

(24)

based on the ideas from [11], we connect the function \(F(x)\) to the analysis of its derivative

$$F^{\prime }(x)= \frac{1}{2} \frac{ f (\frac{x}{2} )}{x\sin \frac{x}{2}\cos \frac{x}{2}}, $$

where

$$ f(t)=4t(a-1)\cos ^{2}{ t}-2a\sin {t}\cos {t}-2t(a-2). $$

(25)

Let us note that \(F^{\prime }(x)\) is the quotient of two MTP functions.

The inequality \(F^{\prime }(x)>0\) is equivalent to \(f(t)>0\). The proof of the later inequality will be done using the *Natural Approach* algorithm for the function \(f(t)\) on \(( 0, \frac{\pi}{2} ) \), with \(a \in ( 1, \frac{3}{2} )\). As before, we search a polynomial \(\mathcal{TP}(t)\) such that

$$f(t)>\mathcal{TP}(t)>0. $$

In step 1 of the *Natural Approach* algorithm, we can use Method D only because \(\delta = \frac{\pi}{2}\). Then

$$\begin{aligned} f(t) = & 4 t (a-1) \bigl(1-\sin ^{2}{t}\bigr) - 2 a \sin {t} \cos {t} - 2 t (a-2) \\ = & 4 t (1-a) \sin ^{2}{t} - 2 a \sin {t} \cos {t} + 2 t a \end{aligned}$$

(26)

with \(\widehat{k} = 0\). In the *Estimation* procedure only^{Footnote 2} step II can be applied to the first and second addends in (26), where \(s_{i}\geq 0\) and \(k_{i}\geq 0\), \(i=1,2\), should be selected. Let us, for example, select \(s_{1}=k_{1}=s_{2}=k_{2}=1\). As a result of this selection, the *Natural Approach* algorithm yields the polynomial

$$\mathcal{TP}(t) = 4t(1 - a) \biggl( t - \frac{1}{6} t^{3} + \frac{1}{120} t^{5} \biggr)^{ 2} - 2a \biggl( t - \frac{1}{6} t^{3} + \frac{1}{120} t^{5} \biggr) \biggl( 1 - \frac{1}{2} t^{2} + \frac{1}{24} t^{4} \biggr) + 2 t a $$

for which \(f(t)>\mathcal{TP}(t)\), for every \(t \in ( 0, \frac{\pi}{2} )\) and \(a \in ( 1, \frac{3}{2} ) \). The inequality \(f(t)>0\) is reduced to a decidable problem

$$ \mathcal{TP}(t)>0, \quad \mbox{for every } t\in \biggl( 0, \frac{\pi}{2} \biggr) \mbox{ and } a\in \biggl( 0,\frac{3}{2} \biggr) . $$

(27)

The sign of the polynomial \(\mathcal{TP}(t)\) can be determined in several ways. For example, let us represent the polynomial \(\mathcal{TP}(t)\) as

$$ \mathcal{TP}(t) = p(t)a + q(t), $$

(28)

where

$$p(t) = - \frac{t^{3} ( 2t^{8} - 75t^{6} + 1{,}120t^{4} - 7{,}680t^{2} + 19{,}200 ) }{7{,}200} $$

and

$$q(t) = 4 t \biggl( t - \frac{1}{6} t^{3} + \frac{1}{120} t^{5}\biggr)^{ 2} . $$

For every fixed \(t\in ( 0,\frac{\pi}{2} )\), the function \(\mathcal{TP}(t)=p(t)a+q(t)\) is linear, monotonically decreasing with respect to \(a \in (1, \frac{3}{2} )\) since for every \(t \in ( 0, \frac{\pi}{2} )\),

$$p(t)=- \frac{t^{3}}{7{,}200} \bigl(2 t^{8}+5 t^{4} \bigl(224-15t^{2}\bigr)+3{,}840 \bigl(5-2t^{2}\bigr) \bigr) < 0. $$

Hence, for every fixed \(t \in ( 0, \frac{\pi}{2} )\), the value of (28) is greater than the value of the same expression for \(a= \frac{3}{2}\):

$$p(t) \frac{3}{2} + q(t)=- \frac{t^{5}}{14{,}400} \bigl(2 t^{6}-65 t^{4}+800 t^{2}-3{,}840\bigr). $$

But

$$p(t) \frac{3}{2} + q(t)= \frac{t^{5}}{14{,}400} \bigl( t^{4} \bigl(65-2 t^{2}\bigr) +160\bigl( 24-5t^{2}\bigr) \bigr)>0,$$

so inequality (27) is true; and consequently, \(F^{\prime }(x)>0\) on \((0,\pi )\) for every \(a\in ( 1, \frac{3}{2} ) \). But \(\lim_{x\rightarrow 0}{F(x)}=0\), so \(F(x)>0\) on \((0,\pi)\) for every \(a\in ( 1,\frac{3}{2} )\). □

### Remark on Theorem 10

Let us consider possible refinements of inequality (20) by a real analytical function \(\varphi _{a}(x) = ( \frac{\sin x}{x} ) ^{ a}\) for \(x \in ( 0,\delta ) \) and \(a \in \mathbb{R}\). The function \(\varphi _{a}(x)\) is real analytical as it is related to the analytical function

$$ t(x)=a\ln \biggl( \frac{\sin x}{x} \biggr) =a \sum _{k=1}^{\infty } \frac{(-1)^{k}2^{2k-1}B_{2k}}{k(2k)!} x^{2k} $$

(29)

(\(B_{i}\) are the Bernoulli numbers; see, e.g., [30]). The following consideration of the sign of the analytical function in the left and right neighborhood of zero is based on Theorem 2.5 from [8]. Let us consider the real analytical function

$$ f_{1}(x) = \biggl( \frac{\sin x}{x} \biggr) ^{ a}-\cos ^{2}{ \frac{x}{2}} = \biggl( - \frac{a}{6}+ \frac{1}{4} \biggr)x^{2} + \biggl( \frac{{a}^{2}}{72} - \frac{a}{180} - \frac{1}{48} \biggr){x}^{4} + \cdots, $$

(30)

\(x \in (0,\pi)\). The restriction

$$ f_{1}^{\prime \prime }(0)=- \frac{a}{3}+ \frac{1}{2} >0, $$

(31)

i.e.,

$$ a\in \biggl( -\infty , \frac{3}{2} \biggr), $$

(32)

is a necessary and sufficient condition for \(f_{1}(x)>0\) to hold on an interval \((0,\delta _{1}^{(a)} )\) (for some \(\delta _{1}^{(a)} > 0\)). Also, the restriction

$$ a\in \biggl( \frac{3}{2},\infty \biggr) $$

(33)

is a necessary and sufficient condition for \(f_{1}(x)<0\) to hold on an interval \((0,\delta _{2}^{(a)})\) (for some \(\delta _{2}^{(a)} > 0\)). The following equivalences hold true for every \(x \in (0,\pi )\):

$$\begin{aligned}& a\in ( 1,\infty ) \quad \Longleftrightarrow \quad \biggl( \frac{\sin x}{x} \biggr) ^{ a}< \frac{\sin x}{x}, \end{aligned}$$

(34)

$$\begin{aligned}& a \in ( -\infty ,1 ) \quad \Longleftrightarrow \quad \frac{\sin x}{x}< \biggl( \frac{\sin x}{x} \biggr) ^{ a}. \end{aligned}$$

(35)

The refinement in Theorem 10 is given based on the possible values of the parameter *a* in (33) and (34). A similar analysis shows us that only the following refinements of inequality (20) are possible.

### Corollary 11

*Let*
\(a\in [\frac{3}{2}, +\infty )\). *There exists*
\(\delta >0\)
*such that for every*
\(x\in ( 0,\delta ) \), *it holds*

$$ \biggl( \frac{\sin {x}}{x} \biggr) ^{a}\leq \cos ^{2} \frac{x}{2}. $$

(36)

### Corollary 12

*Let*
\(a\in ( -\infty ,1 ) \). *There exists*
\(\delta >0\)
*such that for every*
\(x\in ( 0,\delta ) \), *it holds*

$$ \frac{2+\cos {x}}{3}\leq \biggl( \frac{\sin {x}}{x} \biggr) ^{a}. $$

(37)