The natural algorithmic approach of mixed trigonometric-polynomial problems

TL;DR

This paper introduces a new algorithm that simplifies mixed trigonometric-polynomial inequalities to polynomial inequalities, enabling automated proofs and applications in approximations and inequality improvements.

Contribution

The paper presents a novel algorithm for proving mixed trigonometric-polynomial inequalities by reduction to polynomial inequalities, enhancing automated proof capabilities.

Findings

Successfully applied to rational approximations of cos^2(x)

Improved a class of inequalities by Z.-H. Yang

Algorithm suitable for automated proof systems

Abstract

The aim of this paper is to present a new algorithm for proving mixed trigonometric-polynomial inequalities by reducing to polynomial inequalities. Finally, we show the great applicability of this algorithm and as examples, we use it to analyze some new rational (Pade) approximations of the function $\cos^2(x)$, and to improve a class of inequalities by Z.-H. Yang. The results of our analysis could be implemented by means of an automated proof assistant, so our work is a contribution to the library of automatic support tools for proving various analytic inequalities.

Equations

$$\sum _{i=1}^n{\alpha }_i{x}^{p_i}{\cos }^{q_i}x{\sin }^{r_i}x>0,$$

$$\displaystyle\begin{array}[]{ccccc}\mathop{\mbox{$\sum$}}\limits_{i=0}^{2s+1}\displaystyle\frac{(-1)^{i}x^{2i+1}}{(2i+1)!}\!\!&\!\!<\!\!&\!\!\sin x\!\!&\!\!<\!\!&\!\!\mathop{\mbox{$\sum$}}\limits_{i=0}^{2s}\displaystyle\frac{(-1)^{i}x^{2i+1}}{(2i+1)!}\\[10.76385pt] \displaystyle\mathop{\mbox{$\sum$}}\limits_{i=0}^{2k+1}\frac{(-1)^{i}x^{2i}}{(2i)!}\!\!&\!\!<\!\!&\!\!\cos x\!\!&\!\!<\!\!&\!\!\displaystyle\mathop{\mbox{$\sum$}}\limits_{i=0}^{2k}\frac{(-1)^{i}x^{2i}}{(2i)!},\end{array}$$

$$\displaystyle\cos x-\left(\frac{\sin x}{x}\right)^{\!3}>\,-\frac{x^{4}}{15},\;\;\;\;x\!\in\!\left(0,\mbox{\small$\displaystyle\frac{\pi}{2}$}\right)\!,$$

$$\displaystyle\cos x\,-\left(\frac{\sin x}{x}\right)^{\!3}\!+\,\frac{x^{4}}{15}>1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}\,-\!\left(\!\frac{x-\mbox{\small$\displaystyle\frac{x^{3}}{3!}$}+\mbox{\small$\displaystyle\frac{x^{5}}{5!}$}}{x}\!\right)^{\!\!3\!}\!+\,\frac{x^{4}}{15}=\frac{x^{6}R\left(x^{2}\right)}{1728000},$$

$$f(x)=\sum _{i=1}^n{\alpha }_i{x}^{p_i}{\cos }^{q_i}x{\sin }^{r_i}x,x\in ({\delta }_1,{\delta }_2),$$

$$T_n^{\varphi ,a}(x)=\sum _{k=0}^n\frac{{\varphi }^{(k)}(a)}{k!}(x-a{)}^k$$

$$f(x)>P(x),$$

$$T_n(x)\ge T_{n+4}(x)\ge \sin x,\text{mbox}forevery0\le x\le \sqrt{(n+3)(n+4)};$$

$$T_n(x)\le T_{n+4}(x)\le \sin x,\text{mbox}forevery-\sqrt{(n+3)(n+4)}\le x\le 0.$$

$$T_n(x)\le T_{n+4}(x)\le \sin x,\text{mbox}forevery0\le x\le \sqrt{(n+3)(n+4)};$$

$$T_n(x)\ge T_{n+4}(x)\ge \sin x,\text{mbox}forevery-\sqrt{(n+3)(n+4)}\le x\le 0.$$

$$T_n(x)\ge T_{n+4}(x)\ge \cos x,\text{mbox}forevery-\sqrt{(n+3)(n+4)}\le x\le \sqrt{(n+3)(n+4)}.$$

$$T_n(x)\le T_{n+4}(x)\le \cos x,\text{mbox}forevery-\sqrt{(n+3)(n+4)}\le x\le \sqrt{(n+3)(n+4)}.$$

$$T_n^{\sin ,0}(x)\ge 0.$$

$${\left({\underline{T}}_{4s+3}^{\sin ,0}(x)\right)}^p\le {\sin }^{p}x\le {\left({\overline{T}}_{4s+1}^{\sin ,0}(x)\right)}^p.$$

$${\cos }^{p}x\le {\left({\overline{T}}_{4k}^{\cos ,0}(x)\right)}^p.$$

$${\cos }^{2}x\ge {\left({\underline{T}}_{4k+2}^{\cos ,0}(x)\right)}^2.$$

$$\left(c_k\in \left(\delta ,\frac{\pi }{2}\right)\wedge {\underline{T}}_{4k+2}^{\cos ,0}(c_k)=0\right)⟺{\underline{T}}_{4k+2}^{\cos ,0}(\delta )\ge 0.$$

$$\cos^{2p}{\!x}\,=\,\left(1-\sin^{2}{\!x}\right)^{p}\,=\,\mathop{\displaystyle\sum}_{i=0}^{p}(-1)^{i}\mbox{\footnotesize$\left(\!\!\begin{array}[]{c}\mbox{\footnotesize$\!p\!$}\\[1.72221pt] \mbox{\footnotesize$\!i\!$}\end{array}\!\!\right)$}\sin^{2i}{\!x}.$$

$$f(x)=\sum _{i=1}^{\text{mbox}\mathsf{m}}{\alpha }_i{x}^{p_i}{\cos }^{2k_i}x{\sin }^{r_i}x+f_1(x)$$

$$k_1+k_2+\dots +k_{\text{mbox}\mathsf{m}};$$

$$f(x)>{\mathcal{T}\mathcal{P}}^f(x).$$

$${\mathcal{T}\mathcal{P}}^f(x)>0,$$

$${\left({\cos }^{2}x\right)}_{[6/4]}=\frac{-59x^6+962x^4-3675x^2+4095}{17x^4+420x^2+4095}$$

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

$${\left({\cos }^{2}x\right)}_{[6/4]}<{\cos }^{2}x<{\left({\cos }^{2}x\right)}_{[4/4]}$$

$$G_1(x)\ge 0\text{mbox}foreveryx\in (0,\delta ]$$

$$G_{1}(x)<0\;\;\;\mbox{for every}\;\;\;x\in\left(\delta,\mbox{\small$\displaystyle\frac{\pi}{2}$}\right).$$

$$G_1(x)<{\cos }^{2}x,x\in \left(0,\delta \right].$$

$$G_1(x)<{\left({\underline{T}}_{4k+2}^{\cos ,0}(x)\right)}^2<{\cos }^{2}x.$$

Full text

The natural algorithmic approach of mixed trigonometric-polynomial problems

Tatjana Lutovac{}^{\mbox{\tiny,1)}}, Branko Malešević{}^{\mbox{\tiny,1),\ast}}, Cristinel Mortici{}^{\mbox{\tiny,2)}} ††$\!\!\!\!\!\!\!\!\!\!\!\!{}^{\ast}\,$Corresponding author, Telephone: +381113218321, Fax: +381113248681 ††E-mails: Tatjana Lutovac$\,<$tatjana.lutovac@etf.rs$>$, Branko Malešević$\,<$malesevic@etf.rs$>$, Cristinel Mortici$\,<$cristinel.mortici@valahia.ro$>$

*1)**Faculty of Electrical Engineering, University of Belgrade,

Bulevar kralja Aleksandra 73, 11000 Belgrade, Serbia

2)Valahia University of Târgovişte, Bd. Unirii 18, 130082 Târgovişte, Romania;

Academy of Romanian Scientists, Splaiul Independenţei 54, 050094 Bucharest, Romania;

University Politehnica of Bucharest, Splaiul Independenţei 313, 060042 Bucharest, Romania*

**Abstract. ** The aim of this paper is to present a new algorithm for proving mixed trigonometric-polynomial inequalities of the form

[TABLE]

by reducing to polynomial inequalities. Finally, we show the great applicability of this algorithm and as examples, we use it to analyze some new rational (Pad$\acute{\mbox{e}}$) approximations of the function $\cos^{2}{\!x}$, and to improve a class of inequalities by Z.-H. Yang. The results of our analysis could be implemented by means of an automated proof assistant, so our work is a contribution to the library of automatic support tools for proving various analytic inequalities.

MSC 2010: 41A10; 26D05; 68T15; 12L05 41A58

Keywords: mixed trigonometric-polynomial functions; Taylor series; approximations; inequalities; algorithms; automated theorem proving

1 Introduction and Motivation

In this paper, we propose a general computational method for reducing some inequalities involving trigonometric functions to the corresponding polynomial inequalities. Our work has been motivated by many papers [10], [15], [17], [18], [22], [23], [26] - [32] recently published in this area. As an example, we mention the work of Mortici [17] who extended Wilker-Cusa-Huygens inequalities, using a method, he called the natural approach method. This method consists in comparing and replacing $\sin x$ and $\cos x$ by their corresponding Taylor polynomials, as follows:

[TABLE]

for every integers $s,k\!\in\!{\mathbb{N}}_{0}$ and $x\!\in\!\left(0,\pi/2\right)$.

In this way, complicated trigonometric expressions can be reduced to polynomial, or rational expressions, which can be, at least theoretically, easier studied (this can be done using some softwares for symbolic computation, such as Maple).

For example, Mortici in [17] (Theorem 1), proved the next inequality:

[TABLE]

by intercalating the following Taylor polynomials, as follows:

[TABLE]

where $\displaystyle R\left(t\right)=20000-1560t+60t^{2}-t^{3}.$

Although transformation based on the natural approach method has been made by several researchers in their isolated studies, a unified approach has not been given yet. Moreover, it is interesting to note that just trigonometric expressions involving odd powers of $\cos x$ were studied only, as the natural approach method cannot be directly applicable for the function $\displaystyle\cos^{2}\!x$ in the entire over interval $(0,\pi/2)$.

The aim of this paper is to extend and formalize the ideas of the natural approach method for a wider class of trigonometric inequalities, including also those containing even powers of $\cos x$, with no further restrictions.

Let $\delta_{1}\!\leq\!0\!\leq\!\delta_{2},$ with $\delta_{1}\!<\!\delta_{2}.$ Recall that a function defined by the formula

[TABLE]

is named a mixed trigonometric-polynomial function, denoted in the sequel by MTP function [20], [27]. Here, $\alpha_{i}\!\in\!\mathbb{R}\!\setminus\!\{0\}$, $p_{i},q_{i},r_{i}\!\in\!\mathbb{N}_{0}$, $n\!\in\!\mathbb{N}$. Moreover, an inequality of the form $f(x)\!>\!0$ is called a mixed trigonometric-polynomial inequality (MTP inequality).

MTP functions currently appear in the monographs on the theory of analytical inequalities [3], [7] and [22], while concrete MTP inequalities are employed in numerous engineering problems (see e.g. [14], [19]). A large class of inequalities arising from different branches of science, can be reduced to MTP inequalities. Notwithstanding, the development of formal methods and procedures for automated generation of proofs of analytical inequalities remains a challenging and important task of artificial intelligence and automated reasoning [6], [9].

Notice the logical-hardness general problem under consideration. According to Wang [4], for every function $G$ defined by arithmetic operations and a composition over polynomials and sine functions of the form $\sin\pi x$, there is a real number $r$ such that the problem $G(r)=0$ is undecidable (see [21]). In 2003, M. Laczkovich [8] proved that this result can be derived if the function $G$ is defined in terms of the functions $x,\sin x$ and $\sin(x\sin x^{n})$, $n=1,2,\ldots$ (without involving $\pi$). On the other hand, several algorithms [1], [11] and [25] have been developed to determine the sign and the real zeroes of a given polynomial, so that such problems can be considered as decidable (see also [5], [21]).

Let us denote by

[TABLE]

the Taylor polynomial of $n$-th degree associated to the function $\phi$ at a point $a$. Here, $\overline{T}_{n}^{\phi,a}(x)$ and $\underline{T}_{n}^{\phi,a}(x)$ represent the Taylor polynomial of $n$-th degree associated to the function $\phi$ at a point $a$, in case $\displaystyle T_{n}^{\phi,a}(x)\geq\phi(x)$, respective $\displaystyle T_{n}^{\phi,a}(x)\leq\phi(x)$, for every $x\in\left(a,b\right).$ We will call $\overline{T}_{n}^{\phi,a}(x)$ and $\underline{T}_{n}^{\phi,a}(x)$ an upward, respective a downward approximation of $\phi,$ on $\left(a,b\right).$

We present a new algorithm for approximating a given MTP function $f(x)$ by a polynomial function $P(x)$ such that

[TABLE]

using the upward and downward Taylor approximations $\underline{T}_{\,n}^{\,\sin,0\!}{(x)}$, $\overline{T}_{n}^{\,\sin,0\!}{(x)}$, $\underline{T}_{\,n}^{\,\cos,0\!}{(x)}$, $\overline{T}_{n}^{\,\cos,0\!}{(x)}$.

2 The natural approach method and the associated algorithm

The following two lemmas [27] related to the Taylor polynomials associated to sine and cosine functions will be of great help in our study.

Lemma 1

*Let T_{n}(x)\!=\!\!\mathop{\mbox{\displaystyle\sum}}\limits_{i=0}^{(n-1)/2}\displaystyle\frac{(-1)^{i}x^{2i+1}}{(2i+1)!}.

$(i)$ If $n=4s+1$, with $s\in{\mathbb{N}}_{0},$ then:*

[TABLE]

and

[TABLE]

$(ii)$* If $n=4s+3$, with $s\in{\mathbb{N}}_{0},$ then:*

[TABLE]

and

[TABLE]

Lemma 2

*Let T_{n}(x)=\!\mathop{\mbox{\displaystyle\sum}}\limits_{i=0}^{n/2}\displaystyle\frac{(-1)^{i}x^{2i}}{(2i)!}.

$(i)$ If $n=4k$, with $k\in\mathbb{N}_{0}$, then:*

[TABLE]

$(ii)$* If $n=4k+2$, with $k\in\mathbb{N}_{0}$, then:*

[TABLE]

According to Lemmas 1-2, the upper bounds of the approximation intervals of the functions $\sin x$ and $\cos x$ are $\varepsilon_{1}=\sqrt{(n_{1}+3)(n_{1}+4)}$ and $\varepsilon_{2}=\sqrt{(n_{2}+3)(n_{2}+4)}$, respectively. As \varepsilon_{1}>\mbox{\small\displaystyle\frac{\pi}{2}} and \varepsilon_{2}>\mbox{\small\displaystyle\frac{\pi}{2}}, the results of these lemmas are valid in particular, in the entire interval \left(0,\mbox{\small\displaystyle\frac{\pi}{2}}\right).

Lemma 3

1. Let $n\in{\mathbb{N}}$ and x\in\left(0,\mbox{\small\displaystyle\frac{\pi}{2}}\right). Then:*

[TABLE]

2) Let $s\in{\mathbb{N}}_{0}$, $\,p\in{\mathbb{N}}$ and x\in\left(0,\mbox{\small\displaystyle\frac{\pi}{2}}\right). Then:

[TABLE]

Lemma 4

Let $k\in\mathbb{N}_{0}$, $p\in\mathbb{N}$ and x\in\left(0,\mbox{\small\displaystyle\frac{\pi}{2}}\right). Then:*

[TABLE]

In contrast to the function $\sin x$ and its downward Taylor approximations, in the interval \left(0,\mbox{\small\displaystyle\frac{\pi}{2}}\right) the function $\cos x$ and the downward Taylor approximations \underline{T}_{\,4k+2}^{\,\cos,0}(x)=\mathop{\mbox{\displaystyle\sum}}_{i=0}^{2k+1}{\mbox{\small\displaystyle\frac{(-1)^{i}x^{2i}}{(2i)!}}}, $\,k\in{\mathbb{N}}_{0}$, require special attention as there is no downward Taylor approximation $\displaystyle\underline{T}_{\,4k+2}^{\;\cos,0}(x)$, such that $\cos^{2}{\!x}\,\geq\,\left(\underline{T}_{\,4k+2}^{\;\cos,0}(x)\right)^{2},$ for every x\in\left(0,\mbox{\small\displaystyle\frac{\pi}{2}}\right).

We present the following results related to the problem with downward Taylor approximations of the cosine function.

Proposition 5

1. For every $k\in\mathbb{N}_{0}\,$, the downward Taylor approximation $\displaystyle\underline{T}_{\,4k+2}^{\;\cos,0}(x)$ is a strictly decreasing function on \left(0,\mbox{\small\displaystyle\frac{\pi}{2}}\right).

2. For every $k\in\mathbb{N}_{0}\,$, there exists an unique c_{k}\in\left(0,\mbox{\small\displaystyle\frac{\pi}{2}}\right) such that $\,\displaystyle\underline{T}_{\,4k+2}^{\;\cos,0}(c_{k})=0$.

3. The sequence $\displaystyle\left(c_{k}\right)_{k\in\mathbb{N}_{0}},$ with $c_{0}=\sqrt{2}$, is strictly increasing and $\displaystyle\,\lim_{k\rightarrow+\infty}{\!c_{k}}=\frac{\pi}{2}$.*

4) For every $k\in{\mathbb{N}}_{0}$, there exists d_{k}\in\left(c_{k},\mbox{\small\displaystyle\frac{\pi}{2}}\right) such that $\cos{d_{k}}\,=\,\left|\,\underline{T}_{\,4k+2}^{\;\cos{},0}(d_{k})\,\right|$.

5) The sequence $\displaystyle\left(d_{k}\right)_{k\in\mathbb{N}_{0}}$ is strictly increasing and $\displaystyle\,\lim_{k\rightarrow+\infty}{\!d_{k}}=\frac{\pi}{2}$.

Proof. 1) The function $\displaystyle\underline{T}_{\,4k+2}^{\;\cos,0}(x)$ is strictly decreasing on \left(0,\mbox{\small\displaystyle\frac{\pi}{2}}\right), since, according to Lemma 1, $\,\,\displaystyle\left(\,\underline{T}_{\,4k+2}^{\;\cos,0}(x)\,\right)^{\prime}\,=\,-\overline{T}_{4k+1}^{\,\sin,0}\,(x)\,\leq\,0.$

2) The existence of $c_{k}$ follows from the fact that $\displaystyle\underline{T}_{\,4k+2}^{\;\cos,0}(0)\!=\!1\,>\,0$ and \displaystyle\,\underline{T}_{\,4k+2}^{\;\cos,0}\left(\mbox{\small\displaystyle\frac{\pi}{2}}\right)\,<\,\cos{\left(\mbox{\small\displaystyle\frac{\pi}{2}}\right)}\!=\!0.

3) The monotonicity of the sequence \displaystyle{\big{(}}c_{k}{\big{)}}_{k\in\mathbb{N}_{0}} is a result of the monotonicity of $\displaystyle\,\underline{T}_{\,4k+2}^{\;\cos,0}(x)$ and Lemma 2 (ii).

The convergence of the sequence \displaystyle{\big{(}}T_{n}^{\;\cos,0}(x){\big{)}}_{n\in\mathbb{N}} implies the convergence of the sequence \displaystyle{\big{(}}c_{k}{\big{)}}_{k\in{\mathbb{N}}_{0}} to $\displaystyle\,\frac{\pi}{2}$.

4) The function $\left|\,\underline{T}^{\,\cos{},0}_{4k+2}(x)\,\right|$ is decreasing on $\displaystyle(0,c_{k})$ and increasing on \displaystyle\left(c_{k},\mbox{\small\displaystyle\frac{\pi}{2}}\right). Based on Lemma 2 (ii), it follows that there exists d_{k}\in\left(c_{k},\mbox{\small\displaystyle\frac{\pi}{2}}\right) such that $\cos{d_{k}}\,=\,\left|\,\underline{T}^{\;\cos{},0}_{\,4k+2}(d_{k})\,\right|$.

5) This statement is a consequence of the monotonicity of the sequence \displaystyle{\big{(}}c_{k}{\big{)}}_{k\in\mathbb{N}_{0}} and the increasing monotonicity of the function $\left|\,\underline{T}_{4k+2}^{\,\cos{},0}(x)\,\right|$ on \displaystyle\left(c_{k},\mbox{\small\displaystyle\frac{\pi}{2}}\right). $\blacksquare$

Corollary 6

Let $\,k\in{\mathbb{N}}_{0}$ and $\,p\in{\mathbb{N}}$. Then:*

1) $\displaystyle\cos^{2p}{\!x}>\left(\underline{T}_{4k+2}^{\,\cos,0}(x)\right)^{2p}\!\!\!,\,$ for every \,\displaystyle x\in{\big{(}}0,d_{k}{\big{)}};

2) $\displaystyle\cos^{2p}{\!x}<\left(\underline{T}_{4k+2}^{\,\cos,0}(x)\right)^{2p}\!\!\!,\,$ for every x\in{\big{(}}d_{k},\mbox{\small\displaystyle\frac{\pi}{2}}{\big{)}}.

Based on the above results, we have:

Corollary 7

Let $\,k\in{\mathbb{N}}_{0}$ and $\,p\in{\mathbb{N}}$. Then $\underline{T}_{\,4k+2}^{\;\cos{},0}(x)$ is not a downward approximation of the MTP function \cos^{2p}{\!x}\on \displaystyle\left(d_{k},\mbox{\small\displaystyle\frac{\pi}{2}}\right).*

In order to ensure the correctness of the algorithm ( [5], [12]) we will develop next in the sequel, the following problem needs to be considered:

**Problem. **

For a given \delta\!\in\!\left(0,\mbox{\small\displaystyle\frac{\pi}{2}}\right) and \,\mathcal{I}\!\subseteq\!\left(0,\mbox{\small\displaystyle\frac{\pi}{2}}\right), find $\displaystyle\widehat{k}\!\in\!\mathbb{N}_{0}$ such that for all $\,k\!\in\!\mathbb{N}_{0}$, $k\!\geq\!\widehat{k}$ and $x\!\in\!\mathcal{I}\,:$

[TABLE]

Remark. If $\cos x$ appears in odd powers only in the given MTP function $f(x)$, we take $\widehat{k}=0$.

One of the method to solve the problem of downward approximation of the function $\cos^{2p}{\!x},\,p\in\mathbb{N}$ is the method of multiple angles developed in [27]. All degrees of the functions $\sin x$ and $\cos x$ are eliminated from the given MTP function $f(x)$, through conversion into multiple-angle expressions. This removes all even degrees of the function $\cos x$, but then sine and cosine functions appear in the form \sin$$\kappa$$x or \cos$$\kappa$$x where \mbox{\boldmath\kappa}\,x\!\in\!\left(0,\mbox{\boldmath\kappa}\,\mbox{\small\displaystyle\frac{\pi}{2}}\right) and \mbox{\boldmath\kappa}\!\in\!\mathbb{N}. In this case, in order to use the results of Lemmas 1-2, we are forced to choose large enough values of $k\!\in\!\mathbb{N}_{0}$, such that \sqrt{(k+3)(k+4)}\!>\!\mbox{\boldmath\kappa}\,\mbox{\small\displaystyle\frac{\pi}{2}}. Note that higher value of $k$ implies a higher degree of the downward Taylor approximations and of the polynomial $P(x)$ in (2) (for instance, see [29] and [31]).

Several more ideas to solve the above problem are proposed and considered below, under the names of Method A-D. In the following, the numbers $c_{k}$ and $d_{k}$ are those defined in Proposition 5.

$\underline{\mbox{Method A}}$

If \displaystyle\delta\,<\,\mbox{\small\displaystyle\frac{\pi}{2}}, find the smallest $k\in\mathbb{N}_{0}$ such that \displaystyle d_{k}\in\left(\delta,\mbox{\small\displaystyle\frac{\pi}{2}}\right). Then $\widehat{k}=k$.

Note that Method A assumes the solving of a transcendental equation of the form $\displaystyle\cos{x}=\underline{T}_{\,4k+2}^{\;\cos,0}(x),$ that requires numerical methods.

$\underline{\mbox{Method B}}$

If \displaystyle\delta\,<\,\mbox{\small\displaystyle\frac{\pi}{2}}, find the smallest $k\in\mathbb{N}_{0}$ such that \displaystyle c_{k}\in\left(\delta,\mbox{\small\displaystyle\frac{\pi}{2}}\right). Then $\widehat{k}=k$.

$\underline{\mbox{Method C}}$

If \displaystyle\,\delta\,<\,\mbox{\small\displaystyle\frac{\pi}{2}}, find the smallest $k\in\mathbb{N}_{0}$ such that $\displaystyle\,\underline{T}_{\,4k+2}^{\;\cos,0}(\delta)\,\geq\,0$. Then $\widehat{k}=k$.

Note that Method B and Method C return the same output as for a given $\delta$ and for every $k\in{\mathbb{N}}_{0}$ the following equivalence holds true:

[TABLE]

As Method B assumes the determining the root $c_{k}$ of the downward Taylor approximation $\displaystyle\underline{T}_{\,4k+2}^{\;\cos,0}(x)$ and Method C assumes the checking the sign of the downward Taylor approximation at point the $x=\delta$, it is notable that Method C presents a faster and simpler procedure.

$\underline{\mbox{Method D}}$

Eliminate all even degrees of the function $\cos x$ using the transformation

[TABLE]

Then $\widehat{k}=0$.

Note that Method D can be applied for any $0<\delta\leq\pi/2$. Hence, if a MTP function $f(x)$ is considered in the whole interval \left(0,\mbox{\small\displaystyle\frac{\pi}{2}}\right), then Method D is applicable only (apart from the multiple-angle method). However, Method D implies an increasement of the number of terms needed to be estimated. Let us represent a given MTP function $f$ in the following form:

[TABLE]

where there are no terms of the form $\cos^{2J}{\!x},\,\,J\!\in\!\mathbb{N},$ in $f_{1}(x)$. The elimination of all terms of the form $\cos^{2k_{i}}{\!x}$ from (11) using the transformation (10), will increase the number of addends in (11), in the general case with

[TABLE]

consequently, it will increase the number of terms of the form $\sin^{j}{\!x}$, $j\in\mathbb{N}$, in (11) needed to be estimated.

2.1 An algorithm based on the natural approach method

Let $f$ be a MTP function and $\mathcal{I}\subseteq\left(0,\pi/2\right)$. We concentrate to find a polynomial $\mathcal{TP}^{f}(x)$ such that for every $x\in\mathcal{I},$

[TABLE]

In this case, the associated MTP inequality $\,f(x)>0$ can be proved if we show that for every $x\in\mathcal{I},$

[TABLE]

which is a decidable problem according to Tarski [1], [21].

The following algorithm describes the method for finding such a polynomial $\mathcal{TP}^{f}(x)$.

$\underline{\mathbf{ALGORITHM~{}\emph{Natural Approach}}}$

INPUT:   function $f$,   $\delta\in\left(0,\frac{\pi}{2}\right).$

OUTPUT:   polynomial $\mathcal{TP}^{f}(x).$

1. $/\ast$ Solve a problem involving downward approximations depending on $\cos^{2p}{\!x},$ $\ast/$

$/\ast$ i.e. determining $\widehat{k}\in{\mathbb{N}}_{0}$, such that for all $k\in{\mathbb{N}}_{0},$ $k\geq\widehat{k},$ it holds: $\ast/$

$/\ast$ $\,\,\,\displaystyle\cos^{2}x\,\geq\,\left(\,\underline{T}_{\,4k+2}^{\;\cos,0}(x)\,\right)^{2},$  for every  $x\in\left(0,\delta\right].\,\,$  $\ast/$

If   $\delta>\sqrt{2}$ and there are even degrees of the function $\cos{x}$ then

If   \displaystyle\delta<\mbox{\small\displaystyle\frac{\pi}{2}}  then    use $\underline{\mbox{Method C}}$ or $\underline{\mbox{Method D}}$

else   use $\underline{\mbox{Method D}}$

else   $\widehat{k}:=0$

2. $/\ast$ In the procedure Estimation (described below), for a given MTP function $\ast/$

$/\ast$ $f(x),$ each addend $a_{i}(x)$ in the function $f(x)$ is estimated.  $\ast/$

PROCEDURE Estimation$\,\left(\,f(x)\,\right)$

END /$\ast$ Algorithm $\ast$/

$\underline{\mathbf{PROCEDURE~{}Estimation}}$

INPUT:   the function $f(x)=\sum_{i=1}^{n}a_{i}(x),\,$ where $a_{i}(x)\,=\,\alpha_{i}x^{p_{i}}\!\cos^{q_{i}}{\!x}\sin^{r_{i}}{\!x}$.

OUTPUT:

the polynomial ${\cal TP}^{f}(x)$ and array $\left(\,\left(s_{i},k_{i}\right)\,\right),\,i=1,...,n$ where $s_{i}$ and $k_{i}$ represent the number that determines the degree of the Taylor approximation of the function $\sin{x}$, respective $\cos{x}$ in the addend $a_{i}(x)$.

Estimate each addend $a_{i}(x)$ with $q_{i}^{2}+r_{i}^{2}\neq 0$, as follows:

I   If $\alpha_{i}>0$, then:

/$\ast$ First select the degrees of the downward approximations $\ast$/

Select $s_{i}\geq 0$ and $k_{i}\geq\widehat{k}$.

Estimate: $a_{i}(x)\geq\alpha_{i}x^{p_{i}}\left(\underline{T}_{\,4s_{i}+3}^{\;\sin,0}(x)\right)^{q_{i}}\left(\underline{T}_{\,4k_{i}+2}^{\;\cos,0}(x)\right)^{r_{i}};$

II   If $\alpha_{i}<0$, (i.e. $\alpha_{i}=-\beta_{i},$ with $\beta_{i}>0$)

/$\ast$ First select the degrees of the downward approximations $\ast$/

Select $s_{i}\geq 0$ and $k_{i}\geq 0$.

Estimate:

$~{}a_{i}(x)=-\beta_{i}x^{p_{i}}(\sin x)^{q_{i}}(\cos x)^{r_{i}}\geq-\beta_{i}x^{p_{i}}\left(\overline{T}^{\,\sin,0}_{4s_{i}+1}(x)\right)^{q_{i}}\left(\overline{T}^{\,\cos,0}_{4k_{i}}(x)\right)^{r_{i}}$;

$/\ast$ Estimation of each addend $a_{i}(x)$ in function $f(x)$ yields a polynomial $\ast/$

$/\ast$ of the form: $\ast/$

$/\ast$ \;P(x)=\mathop{\mbox{\displaystyle\sum}}_{i=1}^{n}\alpha_{i}x^{p_{i}}\left(T_{\,n_{i}}^{\;\sin,0}(x)\right)^{q_{i}}\left(T_{\,m_{i}}^{\;\cos,0}(x)\right)^{r_{i}}, where $T\in\{\underline{T},\overline{T}\}$.   $\ast/$

Return: the polynomial $\,\mathcal{TP}^{f}(x)\,\,\,$ and     array $\left(\,\left(s_{i},k_{i}\right)\,\right)$.

END /$\ast$ Procedure $\ast$/

Comment on step II of the Procedure Estimation: in the general case, the addend $\,a_{i}(x)=\,\,-\beta_{i}x^{p_{i}}(\sin x)^{q_{i}}(\cos x)^{r_{i}}$ can be estimated in one of the following three ways:

$\!\!\!\!\;\,(i)~{}~{}a_{i}(x)=-\beta_{i}x^{p_{i}}(\sin x)^{q_{i}}(\cos x)^{r_{i}}\geq\beta_{i}x^{p_{i}}\left(\underline{T}^{\,\sin,0}_{4s_{i}+3}(x)\right)^{\!q_{i}}\!\!\left(-\overline{T}^{\,\cos,0}_{4k_{i}}(x)\right)^{\!r_{i}}\!\!\!\!,$

$\!\!\!\!\,(ii)~{}~{}a_{i}(x)=-\beta_{i}x^{p_{i}}(\sin x)^{q_{i}}(\cos x)^{r_{i}}\geq\beta_{i}x^{p_{i}}\left(-\overline{T}^{\,\sin,0}_{4s_{i}+1}(x)\right)^{\!q_{i}}\!\!\left(\underline{T}^{\,\cos,0}_{4k_{i}+2}(x)\right)^{\!r_{i}}\!\!\!\!,$

$\!\!\!\!(iii)~{}~{}a_{i}(x)=-\beta_{i}x^{p_{i}}(\sin x)^{q_{i}}(\cos x)^{r_{i}}\geq-\beta_{i}x^{p_{i}}\left(\overline{T}^{\,\sin,0}_{4s_{i}+1}(x)\right)^{\!q_{i}}\!\!\left(\overline{T}^{\,\cos,0}_{4k_{i}}(x)\right)^{\!r_{i}}\!\!\!\!.$

Note that for fixed $s_{i},k_{i},q_{i}$ and $r_{i}$, the method $(iii)$ generates polynomials of the smallest degree.

We present the following characteristic ([2], [12]) for the Natural Approach algorithm.

Theorem 8

The Natural Approach algorithm is correct.

Proof. Every step in the algorithm is based on the results obtained from Lemmas 1-4 and Proposition 5. Hence, for every input instance (i.e. for any MTP function $f(x)$ over a given interval ${\cal I}\subseteq\left(0,\pi/2\right)$), the algorithm halts with the correct output (i.e. the algorithm returns the corresponding polynomial). $\blacksquare$

3 Some applications of the algorithm

We present an application of the Natural Approach algorithm in the proof (Application 1 - Theorem 9) of certain new rational (Pad$\acute{\mbox{e}}$) approximations of the function $\cos^{2}{\!x}$, as well as in the improvement of a class of inequalities (20) by Z. H. Yang (Application 2, Theorem 10).

Application 1

Bercu [26] used the Pad$\acute{\mbox{e}}$ approximations to prove certain inequalities for trigonometric functions. Let us denote by $\left(f(x)\right)_{[m/n]}$ the Pad$\acute{\mbox{e}}$ approximant $\left[m/n\right]$ of the function $f(x)$.

In this example we introduce a constraint of the function $\cos^{2}{\!x}$ by the following Pad$\acute{\mbox{e}}$ approximations:

[TABLE]

and

[TABLE]

Theorem 9

The following inequalities hold true, for every \displaystyle x\!\in\!\left(0,\mbox{\small\displaystyle\frac{\pi}{2}}\right)\!:

[TABLE]

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

[TABLE]

and

[TABLE]

Moreover, $G_{1}(x)<\cos^{2}{\!x}$, for every x\in\left(\delta,\,\mbox{\small\displaystyle\frac{\pi}{2}}\right). We prove now that

[TABLE]

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

[TABLE]

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

[TABLE]

We can use Method C, or Method D from the Natural Approach algorithm, since \delta<\mbox{\small\displaystyle\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 $k_{1}=2.$(1)(1)(1)For the selection $s_{1}=0$ and $k_{1}=1$, the output of the Natural Approach algorithm is the polynomial:

$\mathcal{TP}(x)=\underline{T}_{6}^{\,\cos,0}(x)=1-\displaystyle\frac{x^{2}}{2!}+\displaystyle\frac{x^{4}}{4!}-\displaystyle\frac{x^{6}}{6!}$

such that \mathcal{TP}(x)\!\!\begin{array}[]{c}\mbox{\scriptsize<}\\[-5.38193pt] \mbox{\scriptsize>}\end{array}\!\!G_{1}(x) holds for some $x\in(0,\delta]$. As a result of this selection, the output of the Natural Approach algorithm is the polynomial:

[TABLE]

We prove that

[TABLE]

This is true, since

[TABLE]

where

[TABLE]

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

[TABLE]

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

[TABLE]

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

[TABLE]

This is true, as

[TABLE]

where

[TABLE]

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

[TABLE]

$\blacksquare$

Note: Using Pad$\acute{\mbox{e}}$ approximations, Bercu [26], [32] recently refined certain trigonometric inequalities over various intervals \mathcal{I}=(0,\delta)\subseteq(0,\mbox{\small\displaystyle\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.

Z.-H. Jang [23] proved the following inequalities, for every $x\in\left(0,\pi\right):$

[TABLE]

Previously, Kl$\acute{\mbox{\rm e}}$n, Visuri, and Vuorinen [15] proved the above inequality on {\big{(}}0,\sqrt{27/5}\,{\big{)}} only.

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

Theorem 10

The following inequalities hold true, for every $x\in\left(0,\pi\right)$ and a\in\displaystyle\left(1,\mbox{\small\displaystyle\frac{3}{2}}\right):

[TABLE]

Proof. As $\displaystyle a>1$ and $\displaystyle 0<\displaystyle\frac{\sin{x}}{x}<1$, we have:

[TABLE]

We prove now the following inequality:

[TABLE]

for every $x\in\left(0,\pi\right)$ and a\in\displaystyle\left(1,\mbox{\small\displaystyle\frac{3}{2}}\right). It suffices to show that the following mixed logarithmic-trigonometric-polynomial function [30]

[TABLE]

is positive, for every $x\in\left(0,\pi\right)$ and a\in\left(1,\mbox{\small\displaystyle\frac{3}{2}}\right). Given that

[TABLE]

based on the ideas from [30], we connect the function $\,F(x)$ to the analysis of its derivative:

[TABLE]

where

[TABLE]

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

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

[TABLE]

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

[TABLE]

with $\widehat{k}\,=\,0$. In the Estimation procedure only(2)(2)(2)Because for every fixed $a\in\left(1,\frac{3}{2}\right)$: $\alpha_{1}=4(1-a)<0$ and $\alpha_{2}=-2a<0$. the step II can be applied to the first and second addends in (26), where $\displaystyle s_{i}\geq 0$ and $\displaystyle k_{i}\geq 0$, $i=1,2$ should be selected. Let us, for example, select $\displaystyle s_{1}=k_{1}=s_{2}=k_{2}=1$. As a result of this selection, the Natural Approach algorithm yields the polynomial

[TABLE]

for which $f(t)>\mathcal{TP}(t)$, for every t\!\in\!\left(0,\mbox{\small\displaystyle\frac{\pi}{2}}\right) and a\!\in\!\left(1,\mbox{\small\displaystyle\frac{3}{2}}\right). The inequality $f(t)>0$ is reduced to a decidable problem:

[TABLE]

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

[TABLE]

where

[TABLE]

and

[TABLE]

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

[TABLE]

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

[TABLE]

But

[TABLE]

so the inequality (27) is true and consequently, $F^{\prime\!}(x)>0$ on $(0,\pi),$ for every a\in\left(1,\mbox{\small\displaystyle\frac{3}{2}}\right). But $\displaystyle\lim_{x\rightarrow 0}{F(x)}=0$, so $F(x)>0$ on $(0,\pi)$, for every a\in\left(1,\mbox{\small\displaystyle\frac{3}{2}}\right).

$\,\hfill\blacksquare$

Remark on Theorem 10.

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

[TABLE]

($B_{i}$ are the Bernoulli numbers, see e.g. [24]). 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 [27]. Let us consider the real analytical function

[TABLE]

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

[TABLE]

i.e.

[TABLE]

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

[TABLE]

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

[TABLE]

[TABLE]

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 the inequality (20) are possible:

Corollary 11

Let a\in\left[\mbox{\small\displaystyle\frac{3}{2}},\,+\infty\right). There exists $\delta>0$ such that for every $x\in\left(0,\delta\right)$, it holds:

[TABLE]

Corollary 12

Let $a\in\displaystyle\left(-\infty,1\right)$. There exists $\delta>0$ such that for every $x\in\left(0,\delta\right)$, it holds:

[TABLE]

4 Conclusions and Future Work

The results of our analysis could be implemented by means of an automated proof assistant [13], so our work is a contribution to the library of automatic support tools [16] for proving various analytic inequalities.

Our general algorithm associated to the natural approach method can be successfully applied to prove a wide category of classical MTP inequalities. For example, the Natural Approach algorithm has recently been used to prove some several open problems that involve MTP inequalities (see e.g. [27] - [31]).

It is our contention that the Natural Approach algorithm can be used to introduce and solve other new similar results. Chen [18] used a similar method to prove the following inequalities, for every $x\in\left(0,1\right)$:

[TABLE]

and

[TABLE]

then he proposed the following inequalities as a conjecture:

[TABLE]

and

[TABLE]

Very recently, Malešević et al. [31] solved this open problem using the same procedure - the natural approach method - associated to upwards and downwards approximations of the inverse trigonometric functions.

Finally, we present other ways for approximating the function $\cos^{2n}{\!x}$, $n\in{\mathbb{N}}$. It is well known that the power series of the function $\cos^{2n}{\!x}$ converges to the function everywhere on ${\mathbb{R}}$. The power series of the function $\cos^{2n}{\!x}$ is an alternating sign series. For example, for $n=1$ and $x\in\mathbb{R}$, we have:

[TABLE]

Therefore, for the above power (Taylor) series it is not hard to determine (depending on $m$) which partial sums (i.e. Taylor polynomials) \displaystyle T_{m}^{\,\cos^{2}\!x,\mbox{\scriptsize0}}(x) become good downward or upward approximations of the function $\cos^{2}{\!x}$ in a given interval ${\cal I}$. Assuming the following representation of the function $\cos^{2n}{\!x}$ in power (Taylor) series

[TABLE]

with $a_{j}^{(2n)}>0$ $\left(j=0,2,4,6,\ldots\right)$, the power (Taylor) series of function $\cos^{2n+2}{\!x}$ will be an alternating sign series as follows:

[TABLE]

with $a_{j}^{(2n+2)}>0$ {\big{(}}j=0,2,4,6,\ldots{\big{)}}.

Therefore, in general, for the function $\cos^{2n}{\!x}$ it is possible to determine, depending on the form of the real natural number $m$, the upward (downward) Taylor approximations \overline{T}^{\,\cos^{2n}{\!x},\mbox{\scriptsize0}}_{m}(x) (\underline{T}^{\,\cos^{2n}\!x,\mbox{\scriptsize0}}_{m}(x)) that are all above (below) the considered function in a given interval $\mathcal{I}$. Such estimation of the function $\cos^{2n}{\!x}$ and the use of corresponding Taylor approximations will be the object of future research.

Acknowledgements. The first and the second author was supported in part by the Serbian Ministry of Education, Science and Technological Development, Projects ON 174032, III 44006 and TR 32023. The third author was supported by a Grant of the Romanian National Authority for Scientific Research, CNCS-UEFISCDI, with the Project Number PN-II-ID-PCE-2011-3-0087.