\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2018 (2018), No. 135, pp. 1--11.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2018 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2018/135\hfil Maclaurin series for $\sin_p$] {Maclaurin series for $\sin_p$ with $p$ an integer greater than 2} \author[L. Kotrla \hfil EJDE-2018/135\hfilneg] {Luk\'a\v{s} Kotrla} \address{Luk\'a\v{s} Kotrla \newline Department of Mathematics and NTIS, Faculty of Applied Scences, University of West Bohemia, Univerzitn\'i 22, CZ-306 14~Plze\v{n}, Czech Republic} \email{kotrla@kma.zcu.cz} \thanks{Submitted April 24, 2017. Published July 1, 2018.} \subjclass[2010]{34L10, 33E30, 33F05} \keywords{$p$-Laplacian; $p$-trigonometry; approximation; \hfill\break\indent analytic function coefficients of Maclaurin series} \begin{abstract} We find an explicit formula for the coefficients of the generalized Maclaurin series for $\sin_p$ provided $p>2$ is an integer. Our method is based on an expression of the $n$-th derivative of $\sin_p$ in the form \[ \sum_{k = 0}^{2^{n - 2} - 1} a_{k,n} \sin_p^{p - 1}(x)\cos_p^{2 - p}(x)\,, \quad x\in (0, \frac{\pi_p}{2}), \] where $\cos_p$ stands for the first derivative of $\sin_p$. The formula allows us to compute the nonzero coefficients \[ \alpha_n = \frac{\lim_{x \to 0+} \sin_p^{(np + 1)}(x)}{(np + 1)!}\,. \] \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{definition}[theorem]{Definition} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction}\label{sec:introduction} Let us consider initial value problem \begin{equation}\label{Eq:Intro:IVP} \begin{gathered} -(|u'|^{p - 2}u' )' - (p - 1)|u|^{p - 2}u = 0\,, \\ u(0) = 0\,, \quad u'(0) = 1 \,, \end{gathered} \end{equation} where $p > 1$ is a given parameter and $u \colon \mathbb{R} \to \mathbb{R}$ is a function such that $u \in C^1(\mathbb{R})$ and $|u'|^{p - 2}u' \in C^1 ( \mathbb{R} )$. It is known that the solution of \eqref{Eq:Intro:IVP} exists and is unique (see Elbert~\cite{Elbert1979}). Since the pioneering work of del~Pino, Elgueta and Man\'{a}sevich~\cite{delPinoElguetaManasevich1989}, this solution is usually denoted by $\sin_p$. Note that it generalizes the \emph{sine} function which is the unique solution of \eqref{Eq:Intro:IVP} for $p = 2$. Moreover, the function $\sin_p$ also satisfies the generalized trigonometric identity \begin{equation} \label{Eq:Intro:ptrig:id} |\sin_p(x)|^p + |\cos_p(x)|^p = 1\,, \quad x\in\mathbb{R}\,, \end{equation} where $\cos_p(x) := \frac{\rm d}{{\rm d}x} \sin_p(x)$, which resembles the classical trigonometric identity for $p = 2$. We also define \begin{equation*} \pi_p := 2 \int_0^1 \frac{1}{(1 - s^p)^{1/p}}\,\mathrm{d}s = \frac{2\pi}{p\sin(\pi/p)}. \end{equation*} \par Let us note that the function $\sin_p$ is odd, $2\pi_p$-periodic, and $\sin_p(x)=\sin_p(\pi_p-x)$ (see, e.g., \cite{Elbert1979}). These properties are frequently used when the function $\sin_p$ is evaluated numerically. In fact, any evaluation of $\sin_p$ at an arbitrary point $x\in\mathbb{R}$ can be reduced to an evaluation of $\sin_p$ at a point in the interval $[0,\pi_p/2]$. It turns out that the system of functions $\{\sin_p(k \pi_p\, x ) \}_{k = 1}^{+\infty}$ has applications in approximation theory, see Binding et al.~\cite{BindingBoultonCepickaDrabekGirg2006} for pioneering work in this direction. Indeed, there exists $p_0>1$ such that, for $p > p_0$, $\{\sin_p(k \pi_p\, x ) \}_{k = 1}^{+\infty}$ forms a Riesz basis of $L^2(0,1)$ and a Schauder basis of $L^r(0,1)$ for any $1 < r < +\infty$. The approach from \cite{BindingBoultonCepickaDrabekGirg2006} was corrected and improved by Bushell and Edmunds~\cite{BushellEdmunds2012} where the value $p_0$ was established as the solution of the transcendental equation $$ \frac{2\pi}{p_0\sin(\pi/p_0)} = \frac{2 \pi^2}{\pi^2 - 8}\,. $$ Boulton and Lord~\cite{BoultonLord2011} use the basis $\{\sin_p(k \pi_p\, x ) \}_{k = 1}^{+\infty}$ in their numerical implementation of the Galerking method for finding an approximate solution to the boundary-initial value problem \begin{equation} \label{Eq:intro:slow:fast} \begin{gathered} \frac{\partial u}{\partial t} (x, t) - \frac{\partial \phantom{u}}{\partial x} \Big( \big| \frac{\partial u}{\partial x} (x, t) \big|^{p - 2} \frac{\partial u}{\partial x} (x, t) \Big) = g(x) \\ u(x, 0) = 0\,, \quad x \in ( 0, 1 )\,, \\ u(0 ,t) = u(1, t) = 0 \quad t > 0\,, \end{gathered} \end{equation} where $g \in L^2(0, 1)$. It appears that this choice of basis leads to very accurate results using only few terms of this basis. However, a main drawback of the Galerkin method in \cite{BoultonLord2011} is the evaluation of the values of the function $\sin_p$ on $[0, \pi_p/2]$. In \cite{BoultonLord2011}, the inverse function of $\sin_p$, \begin{equation} \label{Eq:intro:arcsine} \arcsin_p(x) := \int_0^x \frac{1}{(1 - s^p)^{1/p}}\,\mathrm{d}s\,, \quad x \in [0, 1]\,, \end{equation} is used for that purpose. The function $\sin_p$ on $[0, \pi_p/2]$ is then evaluated using numerical inverse of the function $\arcsin_p$, which is a very time consuming process. Since the problem \eqref{Eq:intro:slow:fast} and its generalizations appear in various applications, see e.g. Smreker~\cite{Smreker1878} (bulding of wells), Leibenson~\cite{Leibenson1947} (extraction of oil and natural gas), Wilkins~\cite{Wilkins1955} (bulding of rock-fill dams), Aronsson et al.~\cite{AronssonEvansWu1996}, Evans et al.~\cite{EvansFeldmanGariepy1997} (sandpile growth), Kuijper~\cite{Kuijper2007} (image analysis), and Bermejo et al.~\cite{BermejoCarpioDiazTello2009} (climatology), it is important to find a more efficient numerical implementations of $\sin_p$. Last but not least, the generalized Pr\"ufer transform using $\sin_p$ and its derivative appears to be a very efficient theoretical tool in studying various initial and/or boundary value problems for quasilinears equation of the type (or some of its generalization) $$ -(|u'|^{p - 2}u' )' - q(x)|u|^{p - 2}u = f(x) $$ (under various conditions on $q$ and $f$) see, e.g., \cite{Elbert1979}, Reichel and Walter~\cite{ReichelWalter1999}, and/or Benedikt and Girg~\cite{BenediktGirg2010}. In Brown and Reichel~\cite{BrownReichel2002}, a numerical method based on the Pr\"ufer transform was proposed. Again the main drawback the method was the lack of an efficient numerical implementation of $\sin_p$. To address the issue in this paper we obtain explicit formulas for coefficients of the Maclaurin series of $\sin_p$. This is very difficult task in general and we are not able to deal with this problem for all $p > 1$. As a starting point for further research in this direction, we provide such formulas for any integer $p$ bigger than 2. Let us note that even this partial result can already be used in practical applications, since \eqref{Eq:intro:slow:fast} with $p \to +\infty$ is considered as a model for sandpile growth (see \cite{AronssonEvansWu1996} and \cite{EvansFeldmanGariepy1997} for more details). More precisely, our goal is to find Maclaurin series for $\sin_p$ provided $p$ is even and generalized Maclaurin series for $\sin_p$ provided $p$ is odd. Generalized Maclaurin series is defined as \begin{equation*} \sum_{n = 0}^{+\infty} \alpha_n x |x|^{rn}, \quad r \geq 1\,. \end{equation*} Peetre \cite{Peetre1994} conjectured that the radius of convergence of generalized Maclaurin series for $\sin_p$ is $\pi_p/2$ for any $p > 1$. Local convergence of generalized Maclaurin series was studied in Paredes and Uchiyama \cite{ParedesUchiyama2003}. Peetre's conjecture \cite{Peetre1994} was proved in Girg and Kotrla~\cite{GirgKotrla2014} for when $p > 2$ is an integer. It remains to find the coefficients of the (generalized) Maclaurin series. One can employ \eqref{Eq:intro:arcsine} and follow the ideas presented in Lang and Edmunds~\cite{LangEdmunds2016}. Since \begin{equation*} \arcsin_p(x) = \int_0^x \frac{1}{(1 - s^p)^{1/p}}\,\mathrm{d}s =x \, _2F_1(\frac{1}{p}, \frac{1}{p}, 1 + \frac{1}{p}; x^p ), \quad x\in [0,1)\,, \end{equation*} where $_2F_1(a,b,c;z)$ is Gauss's hypergeometric function, \begin{equation} \label{Eq:Intro:arcsin:series} \arcsin_p(x) = \sum_{k = 0}^{+\infty} \frac{\Gamma(k + \frac{1}{p})} {(kp + 1) \Gamma(\frac{1}{p})} \frac{x^{kp + 1}}{k!}\,, \end{equation} where $\Gamma$ stands for the gamma function. We can obtain desired coefficients using the well-known procedure for inverting power series (see, e.g., Morse and Feshbach~\cite[p. 411 - 413]{MorseFeshbach1953}). Our aim is to derive the coefficients independently of the inverse function. It was shown in Girg and Kotrla~\cite{GirgKotrla2016} that the nonzero coefficients correspond only to the monomials $x^{kp + 1}$, $k \in \mathbb{N}$. Then \begin{equation*} \sin_p(x) = \sum_{n = 0}^{+\infty} \frac{\sin_p^{(np + 1)}(0)}{(np + 1)!} x^{np + 1} \quad x\in ( -\frac{\pi_p}{2}, \frac{\pi_p}{2} )\,, \end{equation*} for p even. In addition, it was proved in \cite{GirgKotrla2016} that the series \begin{equation*} \sum_{n = 0}^{+\infty} \frac{\lim_{x \to 0+}\sin^{(np + 1)}_p(x)}{(np + 1)!} x^{np + 1} \end{equation*} coincides on $[0,\pi_p/2)$ with the series obtained by formal inversion of \eqref{Eq:Intro:arcsin:series} provided $p$ odd. Hence, by the oddness of $\sin_p$, \begin{equation*} \sin_p(x) = \sum_{n = 0}^{+\infty} \frac{\lim_{x \to 0+}\sin_p^{(np + 1)}(x)}{(np + 1)!} x|x|^{np}\,, \quad x\in ( -\frac{\pi_p}{2}, \frac{\pi_p}{2} ). \end{equation*} It remains then to find an explicit formula for \begin{equation*} \alpha_n := \frac{1}{(np + 1)!} \lim_{x \to 0+} \sin_p^{(np + 1)}(x), \quad p \in \mathbb{N},\ p > 2\,. \end{equation*} \noindent {\bf Notation:} In the presented paper, the symbol $\prod$ represents the product of a (possibly finite) sequence of terms as usual. In addition, we define $$ \prod_{i = j_1}^{j_2} b_i = 1 $$ for any sequence $b_i$ provided $j_1 = j_2 + 1$. \begin{theorem} \label{Thm:main} Let $p > 2$ be an integer and \begin{equation} \label{Eq:intro:series:sin} \sin_p(x) = \sum_{n = 0}^{+\infty} \alpha_n x|x|^{np}\,, \quad x\in (-\frac{\pi_p}{2}, \frac{\pi_p}{2}). \end{equation} Then $\alpha_0 = 1$, $ \alpha_1 = -\frac{1}{p(p + 1)}$, and for $n \geq 2$, \begin{equation} \label{Eq:Intro:sin:coeff} \begin{aligned} \alpha_n &= \frac{(-1)^n}{(np + 1)!} \sum_{\substack{i_1 = 1 \\ i_1 \neq p - 1}}^p \sum_{\substack{i_2 = i_1 + 1 \\ i_2 \neq 2p - 1}}^{2p} \dots \\ &\quad \sum_{\substack{i_{n - 1} = i_{n - 2} + 1 \\ i_{n - 1} \neq (n - 1)p - 1}}^{(n - 1)p} \Big[\prod_{m_1 = 1}^{i_1 - 1}(p - 1 - (m_1 - 1))\Big] \big(1 - (p - 1 - (i_1 - 1))\big)\\ &\quad \times\Big[\prod_{m_2 = i_1 + 1}^{i_2 - 1}(2(p - 1) - (m_2 - 2))\Big] (1 - (2(p - 1) - (i_2 - 2))) \ldots \\ & \quad\times \Big[\prod_{m_{n - 1} = i_{n - 2} + 1}^{i_{n - 1} - 1} ((n - 1)(p - 1) - (m_{n - 1} - (n - 1)))\Big] \big(1 - ((n - 1)\\ &\quad\times (p - 1) - (i_{n - 1} - (n - 1)))\big) [ n(p - 1) - (i_{n - 1} - n + 1) ]! \end{aligned} \end{equation} \end{theorem} The proof of Theorem \ref{Thm:main} is based on a method of rewriting higher derivatives of $\sin_p$ introduced in \cite{GirgKotrla2014}. The method is described again in Section \ref{Sec:Higher} for the convenience of the reader. Theorem \ref{Thm:main} is proved in Section \ref{Sec:Main}. Let us note that the above-mentioned definitions of $\sin_p$ and $\cos_p$ are not the only ones found in the literature (see, e.g., Lindqvist~\cite{Lindqvist1995}). \section{Higher order derivatives} \label{Sec:Higher} Let us state some basic notation from formal languages. \begin{definition}{\rm (Salomaa and Soittola~\cite{SalomaaSoittola1978}, I.2, p. 4, and/or Manna~\cite{Manna1974}, p. 2--3, p. 47, and p. 78) \label{def:higher:alphabet} An {\it alphabet} (denoted by $V$) is a finite nonempty set of letters. A {\it word} (denoted by $w$) over an alphabet $V$ is a finite string of zero or more letters from the alphabet $V$. The word consisting of zero letters is called the {\it empty word}. The set of all words over an alphabet $V$ is denoted by $V^{*}$ and the set of all nonempty words over an alphabet $V$ is denoted by $V^{+}$. For strings $w_1$ and $w_2$ over $V$, their juxtaposition $w_1w_2$ is called {\it catenation} of $w_1$ and $w_2$, in operator notation $\operatorname{cat}: V^{*}\times V^{*} \to V^{*}$ and $\operatorname{cat}(w_1, w_2) = w_1w_2$. We also define the length of the word $w$, in operator notation $\mathop{\mathrm{len}}: V^{*}\to\mathbb{N}\cup\{0\}$, which for a given word $w$ yields the number of letters in $w$ when each letter is counted as many times as it occurs in $w$. We also use {\it reverse function} $\operatorname{rev}:V^{*}\to V^{*}$ which reverses the order of the letters in any word $w$ (see \cite[p. 47, p. 78]{Manna1974}). } \end{definition} We consider the alphabet $V=\{0,1\}$ and the set of all nonempty words $V^{+}$. Thus words in $V^{+}$ are, e.g., $$ \text{``}{\tt 0}\text{''}, \text{``}{\tt 1}\text{''}, \text{``}{\tt 01}\text{''}, \text{``}{\tt 10}\text{''}, \text{``}{\tt 11}\text{''}\cdots\,. $$ For instance, $\operatorname{cat}(\text{``}{\tt 1110}\text{''}, \text{``}{\tt 011}\text{''}) = \text{``}{\tt 1110011}\text{''}$, and \[ \operatorname{rev}(\text{``}{\tt 010011000}\text{''}) = \text{``}{\tt 000110010}\text{''}\,, \\ \mathop{\mathrm{len}}(\text{``}{\tt 010011000}\text{''}) = 9\,. \] % Let $m \in \mathbb{N}$, $k \in \mathbb{N} \cup \{ 0 \}, 0 \leq k \leq 2^{m - 2} - 1$ and $(k)_{2,n-2}$ be the string of bits of length $m-2$ which represents binary expansion of $k$ (it means, e.g., for $k=3$ and $m=5$, $(3)_{2,5-2}=``{\tt 011}"$). The differentiability of $\sin_p(x)$ at $x = 0$ was studied in \cite{GirgKotrla2014} leading to the results in Table \ref{Tab:Diff}. \begin{table} \caption {Differentiability of $\sin_p(x)$} \label{Tab:Diff} \renewcommand{\arraystretch}{1.3} \begin{tabular}{|c|c|c|c|c|c|} \hline $p$, $k$ & $x$ in $(0,\pi_p/2)$ &$(-\pi_p/2,\pi_p/2)$ & $\mathbb{R}$ \\ \hline $p=2$& $C^{\infty}$&$C^{\infty}$&$C^{\infty}$ \\ \hline $p=2k$, $k\in\mathbb{N}\setminus\{1\}$ &$C^{\infty}$&$C^{\infty}$&$C^{1}$ \\ \hline $p=2k+1$, $k\in\mathbb{N}$ &$C^{\infty}$&$C^{p}$&$C^1$ \\ \hline $p\in\mathbb{R}\setminus\mathbb{N}$, $p>2$ &$C^{\infty}$ &$C^{\lceil p \rceil}$&$C^1$ \\ \hline $p\in(1,2)$ &$C^{\infty}$ &$C^2$ &$C^2$\\ \hline \end{tabular} \end{table} In particular, $\sin_p(\cdot) \in C^{\infty}(0, \pi_p/2 )$. Let \[ T := \{a \sin_p^q(\cdot) \cos_p^{1 - q}(\cdot)\,\colon\, a,q\in\mathbb{R}\}\,, \] and $\operatorname{D_s}\colon T \to T$ and $\operatorname{D_c}\colon T \to T$ be defined as follows: \begin{equation} \label{eq:higher:Ds} \operatorname{D_s} a \sin_p^q(\cdot) \cos_p^{1 - q}(\cdot) = \begin{cases} a q \sin_p^{q - 1}(\cdot) \cos_p^{1 - (q -1)}(\cdot)\,, & q \neq 0\,,\\ 0\,, & q = 0\,, \end{cases} \end{equation} and \begin{equation} \label{eq:higher:Dc} \operatorname{D_c} a \sin_p^q(\cdot) \cos_p^{1 - q}(\cdot) = \begin{cases} -a(1 - q) \sin_p^{q + p - 1}(\cdot) \cos_p^{1 - (q +p -1)}(\cdot)\,, & q \neq 1\,,\\ 0\,, & q = 1\,. \end{cases} \end{equation} Finally, we define $\operatorname{D}_{k,m}$ in two steps. \begin{itemize} \item[Step 1] We create an ordered $(m-2)$-tuple $d_{k,m-2}\in \{\operatorname{D_s}, \operatorname{D_c}\}^{m-2}$ (cartesian product of sets $\{\operatorname{D_s}, \operatorname{D_c}\}$ of length $m-2$) from $\operatorname{rev}((k)_{2, m-2})$ such that for $1\leq i\leq m-2$, $d_{k,m-2}$ contains $\operatorname{D_s}$ on the $i$-th position if $\operatorname{rev}((k)_{2, n-2})$ contains ``$\tt{0}$'' on the $i$-th position, and $d_{k,m}$ contains $\operatorname{D_c}$ on the $i$-th position if $\operatorname{rev}((k)_{2, m-2})$ contains ``$\tt{1}$'' on the $i$-th position (it means, e.g., for $k=3$, and $m=5$, we obtain $d_{3, 5-2}=(\operatorname{D_c}, \operatorname{D_c}, \operatorname{D_s})$). \item[Step 2] We define $\operatorname{D}_{k,m}$ as the composition of operators $\operatorname{D_s}, \operatorname{D_c}$ in the order they appear in the ordered $m - 2$-tuple $d_{k, m-2}$ (it means, e.g., for $k=3$, and $m=5$, we obtain $\operatorname{D}_{3, 5} =(\operatorname{D_c}\circ\operatorname{D_c}\circ\operatorname{D_s})$). \end{itemize} Let us point out that it is possible to recover the index $k$ from the positions of $\operatorname{D_c}$ in $\operatorname{D}_{k,m}$. We will denote by $j(k) \geq 0$ the number of $\operatorname{D_c}$ in $\operatorname{D}_{k,m}$ and, if $j(k) \neq 0$, we denote by $i_1, i_2, \ldots, i_{j(k)}$ its positions counted from back (i.e., in the order of application of $\operatorname{D_s}$ and/or $\operatorname{D_c}$). Then \begin{equation} \label{eq:high:k:dec} k = 2^{m - 2 - (i_1 - 1)} + 2^{m - 2 - (i_2 - 1)} + \ldots + 2^{m - 2 - (i_{j(k)} - 1)}\,. \end{equation} If $j(k) = 0$, $k = 0$. Definition \ref{def:higher:alphabet} and the definition of $\operatorname{D}_{k,m}$ are taken from \cite{GirgKotrla2014} in almost unchanged form for the convenience of the reader who is not familiar with our previous work. However, the rewriting diagrams in \cite{GirgKotrla2014}, where the construction of $\operatorname{D}_{k,m}$ is visualized, are not included here. It follows from the first derivative of the $p$-trigonometric identity \eqref{Eq:Intro:ptrig:id} that \begin{equation} \label{eq:higher:2der} \sin^{(2)}_p(x) = - \sin_p^{p - 1}(x)\cos_p^{2 - p}(x)\,,\quad x \in (0, \frac{\pi_p}{2})\,. \end{equation} Note that $\sin_p(x) > 0$ and $\cos_p(x) > 0$ for $x \in (0,\pi_p/2)$. Hence, we can use $\operatorname{D}_{k,n}$ to express \begin{equation}\label{eq:higher:der:main:formula} \begin{aligned} \sin^{(m)}_p(x) &= \sum_{k = 0}^{2^{m - 2} - 1} \operatorname{D}_{k,m} \sin_p^{(2)}(x) \\ & = \sum_{k = 0}^{2^{m - 2} - 1} \operatorname{D}_{k,m} (-1) \sin_p^{p - 1}(x)\cos_p^{2 - p}(x)\,, \quad x\in (0, \frac{\pi_p}{2})\,, \end{aligned} \end{equation} for $m > 2$ be a positive integer. Let us explain the procedure for $m = 3$ at first. In that case \begin{align*} &\frac{\rm d}{{\rm d} x} (-1) \sin_p^{p - 1}(x)\cos_p^{2 - p}(x)\\ & = (-1)(p - 1)\sin_p^{p - 2}(x)\cos_p^{3 - p}(x) \\ &\quad + (-1)(2 - p)\sin_p^{p - 1}(x) \cos_p^{1 - p}(x)\sin_p^{(2)}(x)\\ & = (-1)(p - 1)\sin_p^{p - 2}(x)\cos_p^{3 - p}(x)\\ &\quad + (-1)(1 - (p - 1))\sin_p^{p - 1 + p - 1}(x) \cos_p^{1 - (p - 1 + p - 1)}(x) \\ & = \operatorname{D_S}\sin_p^{(2)}(x) + \operatorname{D_c}\sin_p^{(2)}(x) \end{align*} for any $x \in (0,\pi_p/2)$ by the definition of $\operatorname{D_S}$ and $\operatorname{D_c}$. The proof of \eqref{eq:higher:der:main:formula}, which proceeds by induction, can be found in \cite[Lemma 4.5, p. 110]{GirgKotrla2014}. There are two special cases in composing the symbolic operators for $p \in \mathbb{N}, p > 2$, which can be used for reducing of terms in \eqref{eq:higher:der:main:formula}. \begin{itemize} \item[Case 1] Assume that there exists $k \in \mathbb{N} \cup \{ 0 \}$, $k \leq 2^{m - 2}-1$ such that \begin{equation} \label{eq:higher:application:q1} {\operatorname{D}}_{k,m} \sin_p^{(2)}(\cdot) = a \sin_p(\cdot) \cos_p^0(\cdot)\,. \end{equation} The further application of $\operatorname{D_c}$ is meaningless since it produce $0$ by \eqref{eq:higher:Dc}. The situation \eqref{eq:higher:application:q1} occurs, e.g., after $p - 2$ applications of $\operatorname{D_S}$ on $\sin_p^{(2)}(\cdot)$. \item[Case 2] If there exists $k \in \mathbb{N}$, $k \leq 2^{m - 2}-1$, such that \begin{equation} \label{eq:higher:application:q0} {\operatorname{D}}_{k,m} \sin_p^{(2)}(\cdot) = a \sin^0_p(\cdot) \cos_p^1(\cdot)\,, \end{equation} then the application of $\operatorname{D_s}$ produces $0$, see \eqref{eq:higher:Ds}. The situation \eqref{eq:higher:application:q0} occurs, e.g., after $p - 1$ applications of $\operatorname{D_S}$ on $\sin_p^{(2)}(\cdot)$. This is the essential argument in the proof that the exponent $q$ is always nonnegative, see \cite[Lemma 4.6, p.113]{GirgKotrla2014} for more details. \end{itemize} \section{Proof of main result} \label{Sec:Main} \begin{proof}[Proof of Theorem~\ref{Thm:main}] It follows from \cite[Theorem 6, p. 3]{GirgKotrla2016} that \begin{equation} \label{Eq:Main:coef} \alpha_n = \frac{1}{(np + 1)!} \lim_{x \to 0+} \sin_p^{(np + 1)}(x) \end{equation} for $p$ odd, and it is obvious that \eqref{Eq:Main:coef} is valid for $p$ even, since $\sin_p(\cdot)$ belongs to $C^{\infty}(-\pi_p/2, \pi_p/2)$ in this case. We obtain $\alpha_0 = \lim_{x \to 0+} \cos_p(x) = 1$ for $p \in \mathbb{N}$, $p > 2$. Let $n\in \mathbb{N}$ and $x \in (0, \pi_p/2)$. By \cite[Lemma 4.5, p. 110]{GirgKotrla2014} \begin{align*} \sin_p^{(np + 1)}(x) &= \sum_{k = 0}^{2^{np - 1} - 1} -\operatorname{D}_{k,np + 1} \sin_p^{p - 1}(x) \cos_p^{2 - p}(x) \\ & = \sum_{k = 0}^{2^{np - 1} - 1} a_{k,np + 1} \sin_p^{q_{k,np + 1}}(x) \cos_p^{1 - q_{k,np + 1}}(x)\,, \end{align*} where $a_{k,np + 1} \in \mathbb{R}$ and $q_{k,np + 1} \in \mathbb{N} \cup \{0\}$. It follows that \begin{equation} \label{Eq:Main:limita} \begin{aligned} \lim_{x \to 0+} \sin_p^{(np + 1)}(x) &= \sum_{k = 0}^{2^{np - 1} - 1} a_{k,np + 1} \lim_{x \to 0+} \sin_p^{q_{k,np + 1}}(x) \cos_p^{1 - q_{k,np + 1}}(x)\\ &= \sum_{\substack{k = 0 \\ q_{k, np + 1} = 0}}^{2^{np - 1} - 1} a_{k,np + 1}\,. \end{aligned} \end{equation} Our first aim is to describe $k \in \mathbb{N} \cup \{ 0 \}$, $0 \leq k \leq 2^{np - 1} - 1$ such that $q_{k,n} = 0$. We use the alphabet $V = \{ 0,1 \}$ introduced in Definition \ref{def:higher:alphabet} for this purpose and we employ the formula \begin{equation} \label{Eq:Main:qkn} q_{k,np + 1} = j(k)(p - 1) + (np - 1 - j(k))(-1) + p - 1\, \end{equation} proved in \cite[Lemma 4.5, p. 11)]{GirgKotrla2014}. Let us recall that $j(k)$ is the number of occurrences of $\operatorname{D_c}$ in ${\operatorname{D}}_{k,np + 1}$. It follows from the condition $q_{k,n} = 0$ that $j(k) = n - 1$. Then $k = 0$ for $n = 1$ which implies \begin{equation} \begin{aligned} \lim_{x \to 0+} \sin_p^{(p + 1)}(x) & = -\lim_{x \to 0+} {\operatorname{D}}_{0,p + 1} \sin_p^{p - 1}(x)\cos_p^{2 - p}(x) \\ \label{eq:main:n1} & = -\lim_{x \to 0+} (p - 1)!\sin_p^0(x)\cos_p^{1}(x) = -(p - 1)! \end{aligned} \end{equation} by \eqref{eq:higher:Ds}, the definition of $\operatorname{D_s}$. Substituting \eqref{eq:main:n1} into \eqref{Eq:Main:coef} we obtain $$ \alpha_1 = -\frac{1}{p (p + 1)}\,. $$ We will assume $n \geq 2$ in the rest of the proof. Then $$ k = 2^{np - 1 - (i_1 - 1)} + 2^{np - 1 - (i_2 - 1)} + \ldots + 2^{np - 1 - (i_{n - 1} - 1)} $$ by \eqref{eq:high:k:dec}. Moreover, \begin{equation} \label{Eq:main:condition:i} \forall s \in \mathbb{N},\; 1\leq s \leq n - 1 \colon\; i_s \leq sp\,. \end{equation} Indeed, let there exist $s_0 \in \mathbb{N}$, $1 \leq s_0 \leq n - 1\ \colon\ i_{s_0} > s_0 p$ and let $$ k_1 := \begin{cases} 0 & \text{for } s_0 = 1\,,\\ 2^{np - 1 - (i_1 - 1)} + 2^{np - 1 - (i_2 - 1)} + \ldots + 2^{np - 1 - (i_{s_0 - 1} - 1)} & \text{for } s_0 \geq 2\,. \end{cases} $$ The binary expansion $(k_1)_{2,i_{s_0} - 1}$ of $k_1$ defines ${\operatorname{D}}_{k_1, i_{s_0} + 1}$ by the composition of the symbolic operators $\operatorname{D_s}$ and/or $\operatorname{D_c}$ taking the first $i_{s_0} - 1$ operators from ${\operatorname{D}}_{k, np + 1}$ (in the order of its application). The exponent $q_{k_1, i_{s_0} + 1}$ in ${\operatorname{D}}_{k_1, i_{s_0} + 1} \sin_p^{(2)}(\cdot)$ satisfies \begin{equation*} q_{k_1, i_{s_0} + 1} = (s_0 - 1)(p - 1) + (i_{s_0} - 1 - s_0 + 1)(-1) + p - 1 = s_0 p - i_{s_0} < 0 \end{equation*} by \eqref{Eq:Main:qkn} and the assumption $i_{s_0} > s_0 p$. Since $q_{k, np + 1} \geq 0$ for any $n \in \mathbb{N} \cup \{ 0 \}$ and all $0 \leq k \leq 2^{np - 1} - 1$ provided $p > 1$ be an integer, we get the contradiction. Hence, \begin{equation} \label{eq:main:sums} \alpha_n = \frac{1}{(np + 1)!}\sum_{i_1 = 1}^p \sum_{i_2 = i_1 + 1}^{2p} \ldots \sum_{i_{n - 1} = i_{n - 2} + 1}^{(n - 1)p} a_{k_0,np + 1}\,, \end{equation} where $k_0 = 2^{np - 1 - (i_1 - 1)} + 2^{np - 1 - (i_2 - 1)} + \ldots + 2^{np - 1 - (i_{n - 1} - 1)}$. It remains to express $a_{k_0, np + 1}$ as the polynomial in $p$. We will apply $\operatorname{D_s}$ and/or $\operatorname{D_c}$ on $\sin_p^{(2)}(\cdot)$ recursively. Let us denote by $a_i$ the coefficient and $q_i$ the exponent obtained by $i$ steps of recursion. The base cases are $a_0 = - 1$ and $q_0 = p - 1$ by \eqref{eq:higher:2der} and inductive clauses are given by \eqref{eq:higher:Ds} and \eqref{eq:higher:Dc}, i.e., \begin{equation} \label{Eq:Main:inductive:a} a_{i + 1} = \begin{cases} q_i \cdot a_i & \text{if } \operatorname{D_s} \text{ is applied}\,,\\ -(1 - q_i)a_i & \text{if } \operatorname{D_c} \text{ is applied}\,, \end{cases} \end{equation} and \begin{equation} \label{Eq:Main:inductive:q} q_{i + 1} = \begin{cases} q_i - 1 & \text{if } \operatorname{D_s} \text{ is applied}\,,\\ q_i + p - 1 & \text{if } \operatorname{D_c} \text{ is applied}\,. \end{cases} \end{equation} It follows from the definition of $\operatorname{D}_{k_0,np + 1}$ that the operator $\operatorname{D_s}$ is applied in the first $i_1 - 1$ steps of recursion. It means that $$ a_{i_1 - 1} = -(p - 1)(p - 2)\cdots (p - 1 - (i_1 - 2)) \quad \text {and} \quad q_{i_1 - 1} = p - 1 -(i_1 - 1)\,. $$ by \eqref{eq:higher:Ds}. Applying the operator $\operatorname{D_c}$ on the next position we have \begin{gather*} a_{i_1} = -(p - 1)(p - 2) \cdots (p - 1 - (i_1 - 2)) (-1)(1 - (p - 1 -(i_1 - 1))), \\ q_{i_1} = 2(p - 1) -(i_1 - 1)\,. \end{gather*} Applying $i_2 - 1 - i_1$ times the operator $\operatorname{D_s}$ and we obtain \begin{align*} a_{i_2 - 1} &=-(p - 1)(p - 2)\cdots (p - 1 - (i_1 - 2))(-1)(1 - (p - 1 -(i_1 - 1))) \\ &\quad\times (2(p - 1) -(i_1 - 1))\cdots (2(p - 1) - (i_2 - 3)) \end{align*} and $$ q_{i_2 - 1} = 2(p - 1) - (i_2 - 2) $$ (provided $i_2 > i_1 + 1$). The application of $\operatorname{D_c}$ leads to \begin{align*} a_{i_2} &=-(p - 1)(p - 2)\cdots (p - 1 - (i_1 - 2))(-1)(1 - (p - 1 -(i_1 - 1))) \\ &\quad\times (2(p - 1) -(i_1 - 1))\cdots (2(p - 1) - (i_2 - 3))(-1)(1- (2(p - 1) - (i_2 - 2))) \end{align*} and $$ q_{i_2} = 3(p - 1) - (i_2 - 2)\,. $$ It follows by the recursive application of $\operatorname{D_s}$ and/or $\operatorname{D_c}$ that \begin{align*} a_{i_{n - 1}} &= (-1)\Big[\prod_{m_1 = 1}^{i_1 - 1}(p - 1 - (m_1 - 1))\Big] (-1)\big(1 - (p - 1 - (i_1 - 1))\big)\\ & \quad\times \Big[\prod_{m_2 = i_1 + 1}^{i_2 - 1}(2(p - 1) - (m_2 - 2))\Big] (-1)(1 - (2(p - 1) - (i_2 - 2))) \cdots \\ & \quad\times \Big[\prod_{m_{n - 1} = i_{n - 2} + 1}^{i_{n - 1} - 1}((n - 1)(p - 1) - (m_{n - 1} - (n - 1)))\Big]\\ &\quad\times (-1)(1 - ((n - 1)(p - 1) - (i_{n - 1} - (n - 1)))) \end{align*} and $$ q_{i_{n - 1}} = n(p - 1) - (i_{n - 1} - n + 1)\,, $$ where $i_{n - 1}$ is the last position of $\operatorname{D_c}$. Since the remaining symbolic operators in ${\operatorname{D}}_{k_0,np + 1}$ are $\operatorname{D_s}$ and $q_{k_0, np + 1} = 0$ by \eqref{Eq:Main:limita}, we finally get \begin{equation} \label{eq:main:koefk0} \begin{aligned} &a_{k_0,np + 1} \\ &= (-1)\Big[\prod_{m_1 = 1}^{i_1 - 1}(p - 1 - (m_1 - 1))\Big] (-1)\big(1 - (p - 1 - (i_1 - 1))\big)\\ &\quad\times \Big[\prod_{m_2 = i_1 + 1}^{i_2 - 1}(2(p - 1) - (m_2 - 2))\Big] (-1)(1 - (2(p - 1) - (i_2 - 2)))\cdots\\ &\quad \times \Big[\prod_{m_{n - 1} = i_{n - 2} + 1}^{i_{n - 1} - 1}((n - 1)(p - 1) - (m_{n - 1} - (n - 1)))\Big]\\ &\quad\times (-1)(1 - ((n - 1)(p - 1) - (i_{n - 1} - (n - 1)))) \\ &\quad\times \Big[ n(p - 1) - (i_{n - 1} - n + 1) \Big]! \end{aligned} \end{equation} Substituting \eqref{eq:main:koefk0} into \eqref{eq:main:sums} we obtain desired formula \eqref{Eq:Intro:sin:coeff}. The positions $i_s = sp - 1$ are excluded in \eqref{Eq:Intro:sin:coeff} since it produce zero due to the terms $1 - (s(p - 1) - (i_s - s))$ in product \eqref{eq:main:koefk0} (see Case 1 in Section \ref{Sec:Higher}). \end{proof} \section{Concluding remarks} \label{Sec:Remarks} \begin{remark}{\rm The proof of Theorem \ref{Thm:main} provides a procedure to generate any coefficient $\alpha_n$, $n \geq 2$ of Maclaurin series \eqref{Eq:intro:series:sin} for $\sin_p$, when $p > 2$ is an integer. It is convenient to generate all vectors $v \in \{0,1\}^{np - 1}$ with exactly $n - 1$ occurrences of ``1''s, which satisfy condition \eqref{Eq:main:condition:i}, i.e., \begin{equation*} \forall s \in \mathbb{N},\ 1\leq s \leq n - 1\ \colon\ i_s \leq sp\,. \end{equation*} Let us note that $i_s$ is the position of ``1'' in $v$. Then the recursions \eqref{Eq:Main:inductive:a} with $a_0 = -1$ and \eqref{Eq:Main:inductive:q} with $q_0 = p - 1$ can to applied by all possible vectors $v$ to obtain the coefficient $a_v \in \mathbb{R}$. Let us remind that zero and one means that $\operatorname{D_s}$ and $\operatorname{D_c}$ is applied, respectively, and the order of application $\operatorname{D_s}$ and/or $\operatorname{D_c}$ is reversed. Finally, the resulting coefficient $\alpha_n$ is given as sum of all $a_{v}$ which is divided by $(np + 1)!$.} \end{remark} \begin{remark}{\rm The coefficients $\alpha_n$, $n \geq 2$, can be also computed recursively by the formula \begin{align*} \alpha_{n + 1} &= (-1)\Big[\frac{(p - 1)!}{((n + 1)p + 1)((n + 1)p)\cdots (np + 2)}\Big] \alpha_n \\ &\quad + \frac{(-1)^{n + 1}}{((n + 1)p + 1)!} \sum_{\substack{i_1 = 1 \\ i_1 \neq p - 1}}^p \sum_{\substack{i_2 = i_1 + 1 \\ i_2 \neq 2p - 1}}^{2p} \cdots\\ &\quad \sum_{i_{n - 1} = i_{n - 2} + 1 }^{np \mathbf{ - 2}} \Big[\prod_{m_1 = 1}^{i_1 - 1}(p - 1 - (m_1 - 1))\Big] \big(1 - (p - 1 - (i_1 - 1))\big)\\ & \quad\times \Big[\prod_{m_2 = i_1 + 1}^{i_2 - 1}(2(p - 1) - (m_2 - 2))\Big] (1 - (2(p - 1) - (i_2 - 2)))\cdots\\ & \quad\times \Big[\prod_{m_{n} = i_{n - 1} + 1}^{i_{n} - 1}(n(p - 1) - (m_{n} - n)) \Big]\\ & \quad\times (1 - (n(p - 1) - (i_{n} - n))))[ n(p - 1) - (i_{n} - n) ]! \end{align*} with $\alpha_1 = - 1/(p(p + 1))$. } \end{remark} \subsection*{Acknowledgments} The author was supported by the project LO1506 of the Czech Ministry of Education, Youth and Sport. 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