\documentclass[reqno]{amsart} \usepackage{graphicx} \usepackage{hyperref} \AtBeginDocument{{\noindent\small Sixth Mississippi State Conference on Differential Equations and Computational Simulations, {\em Electronic Journal of Differential Equations}, Conference 15 (2007), pp. 229--238.\newline ISSN: 1072-6691. URL: http://ejde.mathMississippi State University\.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \setcounter{page}{229} \title[\hfilneg EJDE-2006/Conf/15\hfil Global attractivity] {Global attractivity in a nonlinear difference equation} \author[C. Qian, Y. Sun\hfil EJDE/Conf/15 \hfilneg] {Chuanxi Qian, Yijun Sun} % in alphabetical order \address{Chuanxi Qian \newline Department of Mathematics and Statistics\\ Mississippi State University\\ Mississippi State, MS 39762, USA} \email{qian@math.msstate.edu} \address{Yijun Sun \newline Department of Mathematics and Statistics\\ Mississippi State University\\ Mississippi State, MS 39762, USA} \email{ys101@msstate.edu} \thanks{Published February 28, 2007.} \subjclass[2000]{39A10} \keywords{Nonlinear difference equation; global attractor; \hfill\break\indent unimodal function; positive equilibrium} \begin{abstract} In this paper, we study the asymptotic behavior of positive solutions of the nonlinear difference equation $$x_{n+1}=x_n f(x_{n-k}),$$ where $f:[0,\infty)\to(0, \infty)$ is a unimodal function, and $k$ is a nonnegative integer. Sufficient conditions for the positive equilibrium to be a global attractor of all positive solutions are established. Our results can be applied to to some difference equations derived from mathematical biology. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \section{Introduction} Our aim in this paper is to study the global attractivity of the difference equation $$x_{n+1}=x_nf(x_{n-k}), \quad n=0,1,\dots \label{e1.1}$$ where $k\in \{0,1,2,\dots\}$, and $f:[0,\infty)\to (0,\infty)$ is a unimodal function; i.e, $f$ is first increasing, and then decreasing. Global attractivity of \eqref{e1.1} under different assumptions on $f$ has been studied by several authors, see, for example, Kocic and Ladas \cite{k1}, Qian \cite{q1} and Graef and Qian \cite{g2}. However, few results can be found under the assumption that $f$ is a unimodal function. Besides its theoretic interest, our motivation to study \eqref{e1.1} comes from the following observation. Consider the delay logistic equation $$x'(t)=x(t)[a+bx(t-\tau)-cx^2(t-\tau)], \quad t\geq 0, \label{e1.2}$$ where $$a,c,\tau\in(0,\infty) \quad \mbox{and}\quad b\in(-\infty,\infty).\label{e1.3}$$ This equation is a model of single species with a quadratic per-capita growth rate (see \cite{g1} for details). One may see \cite{h1} and \cite{l1} also for some extensions of \eqref{e1.2}. The following difference equation with piecewise constant arguments $$x'(t)=x(t)[a+bx([t-k])-cx^2([t-k])], \quad t\geq 0, \label{e1.4}$$ where $[\cdot]$ denotes the greatest integer function, may be viewed as a semidiscretization of \eqref{e1.2}. By an argument similar to that in \cite[Section 8.2]{g3}, one can see that \eqref{e1.4} leads to the following equation $$x_{n+1}=x_n e^{a+bx_{n-k}-cx_{n-k}^{2}},\quad n=0,1,\dots \label{e1.5}$$ whose oscillatory and stability properties completely characterize the same properties for \eqref{e1.4}, and so this leads us to study \eqref{e1.5}, which is a special case of \eqref{e1.1}. In the following, we only consider the positive solutions of \eqref{e1.1}. If we let $$x_{-k}, x_{-k+1}, \dots , x_0 \label{e1.6}$$ be $k+1$ given nonnegative numbers with $x_0>0$, and $\bar{x}$ be the unique solution of $f(x)=1$, then \eqref{e1.1} has a unique positive solution with initial condition \eqref{e1.6}. Clearly, $\bar{x}$ is the only positive equilibrium. In the following section, we establish some sufficient conditions such that $\bar{x}$ attracts all the positive solutions of \eqref{e1.1}. Then, in Section 3, we apply our results about \eqref{e1.1} to \eqref{e1.5} to establish some sufficient conditions for the global attractivity of the positive equilibrium of \eqref{e1.5}. \section{Global attractivity of \eqref{e1.1}} Consider the difference equation $$x_{n+1}=g(x_n),\quad n=1,2,\dots ,\label{e2.1}$$ where $g\in C[R,R]$. Let $G$ be any set in $R$. We say that $V$ is a Liapunov function for \eqref{e2.1} on $G$ if \begin{itemize} \item[(i)] $V$ is continuous on $G$, and \item[(ii)] $\dot{V}(x)=V(g(x))-V(x)\leq 0$ for all $x\in G$. \end{itemize} The following lemma on the asymptotic behavior of \eqref{e2.1} is taken from \cite{l2} and will be needed later. \begin{lemma} \label{lem1} Let $G$ be a bounded open positively invariant set. If \begin{itemize} \item[(i)] $V$ is a Liapunov function for \eqref{e2.1} on $G$, \item[(ii)] $M\subset G$, where $M$ is the largest invariant set of $E=\{x:\dot{V}(x)=0, x\in \bar{G}\}$, \item[(iii)] $V$ is constant on $M$. \end{itemize} Then $M$ is globally asymptotically stable relative to $G$. \end{lemma} The following theorem is our main result in this section. \begin{theorem} \label{thm2.1} Let $f:[0,\infty)\to(0,\infty)$ satisfy the following assumptions: \begin{itemize} \item[(i)] $f$ is a unimodal function obtaining its maximum at $x=x^*$; \item[(ii)] $\bar{x}>x^*$ is the unique equilibrium point; \item[(iii)] $[\ln f(x)]''\leq 0$ on $(x^*, m_0)$; \item[(iv)] $\bar{x}f(m_0)^{k+1}\geq x^*$; \item[(v)] $f(m_0)\geq \frac{1}{f(x^*)}$, \end{itemize} where $m_0=\bar{x}f(x^*)^{k+1}$. Then $\bar{x}$ is the global attractor of all positive solutions of \eqref{e1.1}. \end{theorem} \begin{proof} First, we show that every non-oscillatory (about $\bar{x}$) solution $\{x_n\}$ tends to $\bar{x}$. We assume that $x_n\geq \bar{x}$ eventually. The proof for the case that $x_n\leq \bar{x}$ eventually is similar, and so is omitted. By \eqref{e1.1}, $$\frac{x_{n+1}}{x_n}=f(x_{n-k}).$$ Since for large $n$, $x_{n-k}\geq \bar{x}$, and $\bar{x}>x^*$, we have $\frac{x_{n+1}}{x_n}\leq f(\bar{x})=1$. Hence, $\{x_n\}$ is non-increasing and $\lim_{n\to\infty}x_n=l$ exists. Obviously, $l=\bar{x}$. Next, we show that every oscillatory solution tends to $\bar{x}$. Suppose that $\{x_n\}$ is an oscillatory solution of \eqref{e1.1}. We say that $x_s$ is a local maximum of $\{x_n\}$, if $$x_s\geq \bar{x},\quad x_s\geq x_{s-1}, \quad x_s\geq x_{s+1}.$$ Similarly, we say that $x_r$ is a local minimum of $\{x_n\}$, if $$x_r\leq \bar{x}, \quad x_r\leq x_{r-1}, \quad x_r\leq x_{r+1}.$$ By \eqref{e1.1}, $$x_s=x_{s-1}f(x_{s-k-1})\geq x_{s-1},$$ so we have $f(x_{s-k-1})\geq 1$, and hence $x_{s-k-1}\leq\bar{x}$. Thus, $$x_s=x_{s-k-1}\prod_{i=s-k}^s f(x_{i-k-1})\leq\bar{x}[f(x^*)]^{k+1}=m_0.$$ So, $m_0$ is an upper bound of $\{x_n\}$. By a similar argument, we have $$x_r\geq \bar{x}[f(m_0)]^{k+1}=m_1.$$ Thus, there exists some $N_0>0$, such that $$m_1\leq x_n\leq m_0, \mbox{ for } n\geq N_0.$$ Notice that under assumption (iv), $m_1=\bar{x}f(m_0)^{k+1}\geq x^*$, by induction, we can prove that $$m_{2s+1}\leq x_n\leq m_{2s}, \quad \mbox{for } n\geq N_s$$ where $\{m_s\}$ is defined by $$\begin{gathered} m_{s+1}=\bar{x}[f(m_s)]^{k+1}\\ m_0=\bar{x}[f(x^*)]^{k+1}. \end{gathered}\label{e2.2}$$ To prove $x_n$ tends to $\bar{x}$, it suffices to show that $m_s$ tends to $\bar{x}$. Let $G=(0, m_0)$. Clearly, $G$ is a positively invariant set of IVP \eqref{e2.2}. Define $$V(x)=\big(\ln\frac{x}{\bar{x}}\big)^2, \quad x\in G.$$ Then $$\dot{V}(x) =[(k+1)\ln f(x)]^2-\big(\ln\frac{x}{\bar{x}}\big)^2.$$ Let $$g(x)= (k+1)\ln f(x).$$ To get $\dot{V}(x)<0$ on $G$ for $x\neq\bar{x}$, we need $|g(x)|<|\ln\frac{x}{\bar{x}}|$, which is equivalent to $$\begin{gathered} g(x)<-\ln\frac{x}{\bar{x}}=\ln\frac{\bar{x}}{x} \quad \mbox{for } x< \bar{x};\\ g(x)>-\ln\frac{x}{\bar{x}}=\ln\frac{\bar{x}}{x}\quad \mbox{for } x> \bar{x}. \end{gathered}\label{e2.3}$$ Let $h(x)=\ln\frac{\bar{x}}{x}$. Observe that $$g'(x)=\frac{(k+1)f'(x)}{f(x)}<0, \quad g''(x)=(k+1)(\ln f(x))''\leq 0 \quad \mbox{on } (x^*, m_0); \label{e2.4}$$ and $$\big(h(x)\big)'=-\frac{1}{x}<0, \quad \big(h(x)\big)''=\frac{1}{x^2}>0 \quad \mbox{for } x>0. \label{e2.5}$$ So, $g$ and $h$ look as in the following graph \begin{figure}[ht] \begin{center} \includegraphics[width=0.7\textwidth]{fig1} \end{center} \caption{$g(x^*)= \max g(x),\ \ g(\bar{x})=h(\bar{x})$, by the concavity of $f$ and $g$, $h(x)>g(x) \ \ \mbox{for } x<\bar{x},\ \ h(x)\bar{x}$} \end{figure} We first show that \eqref{e2.3} holds on $(\bar{x}, m_0)$. Since $g(\bar{x})=\ln\frac{\bar{x}}{\bar{x}}=0$, by the monotonicity and concavity of $g$ and $\ln\frac{\bar{x}}{x}$, it's enough to show that $g(m_0)\geq \ln(\bar{x}/m_0)$; i.e, $$[f(m_0)]^{k+1}\geq \frac{\bar{x}}{m_0}=\frac{\bar{x}}{\bar{x}f(x^*)^{k+1}} =\frac{1}{f(x^*)^{k+1}}.$$ From (v), we can see that \eqref{e2.3} holds immediately on $(\bar{x}, m_0)$. Furthermore, we have $g'(x)>\big(\ln\frac{\bar{x}}{x}\big)'$ at $x=\bar{x}$. So, for $x\in(\bar{x}, x^*)$, again by \eqref{e2.4} and \eqref{e2.5}, we know that $g(x)>\ln\frac{\bar{x}}{x}$. For $x\leq x^*$, since $f$ is increasing and $\ln\frac{\bar{x}}{x}$ is decreasing, \eqref{e2.3} is satisfied automatically. Thus, we have $$\dot{V}(x)<0 \quad\mbox{for } x\in G \quad\mbox{and} \quad x\neq\bar{x},$$ and $$E=\{x: \dot{V}(x)=0,\; x\in G\}=\{\bar{x}\}.$$ Hence, by Lemma \ref{lem1}, $\bar{x}$ is a global attractor relative to $G$, and so every solution $\{m_s\}$ of IVP \eqref{e2.2} tends to $\bar{x}$. Then, it follows that $\{x_n\}$ tends to $\bar{x}$. The proof of Theorem \ref{thm2.1} is complete. \end{proof} We can get a linearized stability result by using the following lemma. \begin{lemma}[\cite{l3}] \label{lem2} Assume that $q\in {\bf R}$ and $k\in\{0, 1, 2,\dots\}$. Then the delay difference equation $$x_{n+1}-x_n+qx_{n-k}=0,\quad n=0,1,\dots\label{e2.6}$$ is asymptotically stable if and only if 00$since$f'(\bar{x})<0$. By \eqref{e2.7}, we have that \eqref{e2.10} is stable if \eqref{e2.8} holds. Hence, \eqref{e1.1} is locally stable. Then combining with the global attractivity from Theorem \ref{thm2.1}, we get the global stability result. \end{proof} Although we can not prove it now, we believe that if the conditions in Theorem \ref{thm2.1} hold, then \eqref{e2.8} holds, and so the conditions in Theorem \ref{thm2.1} imply the global stability of \eqref{e1.1}. In Section 3, we will show that this is true for \eqref{e1.5}. \section{Global attractivity of \eqref{e1.5}} In this section, we apply our results in Section 2 to establish some sufficient conditions for the global attarctivity of \eqref{e1.5}. Two cases of \eqref{e1.5} with$b\leq 0$and$b>0$are considered. For \eqref{e1.5},$\bar{x}=\frac{b+\sqrt{b^2+4ac}}{2c}$is the only positive equilibrium. \begin{theorem} \label{thm3.1} Assume that$b\leq 0$, and $$(k+1)a \leq \ln\frac{b+\sqrt{b^2+8ac}}{b+\sqrt{b^2+4ac}}.\label{e3.1}$$ Then$\bar{x}$is a global attractor of all positive solutions of \eqref{e1.5}. \end{theorem} \begin{proof} To apply Theorem \ref{thm2.1}, we need to show that all assumptions of Theorem \ref{thm2.1} are satisfied. Here,$f(x)=\exp\big(a+bx-cx^2\big)$, and clearly,$f(x)$is decreasing on$(0, \infty)$, obtaining its maximum at$x^*=0$. Note that $$m_0=\bar{x}f(0)^{k+1}=\bar{x}e^{(k+1)a},\label{e3.2}$$ and it is easy to see that assumption (ii), (iii) and (iv) of Theorem \ref{thm2.1} are satisfied. Now, wee only need to check (v), which is $$e^{a+bm_0-cm_0^2}\geq \frac{1}{e^{a}}=e^{-a},$$ that is, $$a+bm_0-cm_0^2\geq -a.\label{e3.3}$$ Let$g(x)=cx^2-bx-2a$. Then$x_1=\frac{b+\sqrt{b^2+8ac}}{2c}$is the only positive solution of$g(x)=0$. Since$g$is increasing on$(0,\infty)$, \eqref{e3.3} is equivalent to$m_0\leq x_1$, which is \eqref{e3.1}. Thus, all assumptions of Theorem \ref{thm2.1} are satisfied, and so$\bar{x}$is a global attractor of \eqref{e1.5}. \end{proof} The following result is a consequence of the above theorem and Corollary \ref{coro2.1}. \begin{corollary} \label{coro3.1} If$b\leq 0$and \eqref{e3.1} hold, then \eqref{e1.5} is globally asymptotically stable. \end{corollary} \begin{proof} By \eqref{e2.8} in Corollary \ref{coro2.1}, we need $$-f'(\bar{x})\bar{x}=\frac{\sqrt{b^2+4ac}(b+\sqrt{b^2+4ac})}{2c} <2\cos\frac{k\pi}{2k+1}.\label{e3.4}$$ Now, we claim that \eqref{e3.1} implies \eqref{e3.4}. First, we want to simplify the expression. Let$A=(k+1)a$and$C=c/(k+1)$. From \eqref{e3.1}, it is easy to see that$ A\leq \ln2$, and \eqref{e3.4} can be written as $$\frac{\sqrt{b^2+4AC}(b+\sqrt{b^2+4AC})}{2C}<2(k+1)\cos \frac{k\pi}{2k+1}. \label{e3.5}$$ Next, we let $$B=\frac{|b|}{\sqrt C}=-\frac{b}{\sqrt C}>0.$$ Then \eqref{e3.5} can be written as $$\frac{\sqrt{B^2+4A}(-B+\sqrt{B^2+4A})}{2}<2(k+1)\cos \frac{k\pi}{2k+1}.$$ After simplification, this becomes $$\frac{\sqrt{B^2+4A}}{B+\sqrt{B^2+4A}}A<(k+1)\cos \frac{k\pi}{2k+1}. \label{e3.6}$$ Since$A\le \ln 2$, the left hand side of Inequality \eqref{e3.6} is less than$\ln 2$. On the other hand, the right hand side of \eqref{e3.6} can be written as $(k+1)\sin\frac{\pi}{4(k+\frac 12)}.$ If we use$s=k+\frac 12$, then the right hand side of Inequality \eqref{e3.6} is $$g(s)=(s+\frac 12)\sin\frac{\pi}{4s},\quad s\ge \frac 12.$$ We claim that$g$is a decreasing function for$s\ge \frac 12$. Observe that $$g'(s)=\sin\frac{\pi}{4s}-\left(\frac{\pi}{4s}+\frac{\pi}{8s^2}\right) \cos\frac{\pi}{4s},$$ and notice that$\frac{\pi}{4s}\geq \sin\frac{\pi}{4s}$for$s\geq\frac{1}{2}, \begin{align*} g''(s) &=\frac{\pi}{4s^3}\cos\frac{\pi}{4s} -\Big(\frac{\pi^2}{16s^3}+\frac{\pi^2}{32s^4}\Big)\sin\frac{\pi}{4s}\\ &\geq\frac{\pi}{4s^3}\Big[1-\left(\frac{\pi}{4s}\right)^2\Big] -\Big(\frac{\pi^2}{16s^3}+\frac{\pi^2}{32s^4}\Big)\frac{\pi}{4s}\\ &= \frac{\pi}{4s^3}\Big(1-\frac{\pi^2}{16s^2}-\frac{\pi^2}{16s} -\frac{\pi^2}{32s^2}\Big)\\ &\geq \frac{\pi}{4s^3}\Big(1-\frac{\pi^2}{12}\Big)>0. \end{align*} Sog'$is increasing,$\max_{s\geq\frac12}g'(s)=\lim_{s\to\infty}g'(s)=0$, and$g$is decreasing for$s\geq \frac12$. We get $$\min_{s\geq\frac12}g(s)=\lim_{s\to\infty}(s+\frac 12)\sin\frac{\pi}{4s} =\frac{\pi}4,$$ which is greater than$\ln 2$. Thus \eqref{e3.6} holds, and therefore \eqref{e3.4} holds. The proof is complete. \end{proof} \begin{example} \label{exa1} \rm Consider the difference equation $$x_{n+1}=x_ne^{0.1-x_{n-1}-x_{n-1}^2}.$$ Here$k=1$,$a=0.1$,$b=-1$,$c=1$, and $$(k+1)a=.2<\ln\frac{b+\sqrt{b^2+8ac}}{b+\sqrt{b^2+4ac}}\approx 0.623.$$ Hence, by Corollary \ref{coro3.1},$\bar{x}$is globally asymptotically stable. \end{example} \begin{theorem} \label{thm3.2} Assume$b> 0$and $$\frac{(k+1)D}{4c}\leq \ln\frac{b+\sqrt{2D}}{b+\sqrt{D}}, \label{e3.7}$$ where$D=b^2+4ac$. Then all positive solutions of \eqref{e1.5} tend to$\bar{x}$. \end{theorem} \begin{proof} Let$f(x)=\exp\big(a+bx-cx^2\big)$. We show that$f$satisfies all the conditions assumed in Theorem \ref{thm2.1}. Clearly,$f$is increasing on$(0, x^*)$and decreasing on$(x^*, \infty)$, where$x^*=\frac{b}{2c}$, and so assumption (i) is satisfied. (ii) and (iii) are also easy to check. We see that to have$\bar{x}f(m_0)^{k+1}\geq x^*$, we need $$(a+bm_0-cm_0^2)\geq \frac{1}{k+1}\ln\frac{b}{b+\sqrt{D}}.\label{e3.8}$$ By a direct but tedious calculation, \eqref{e3.8} is equivalent to $$\frac{(k+1)D}{4c}\leq \ln\frac{b+\sqrt{D+\frac{4c}{k+1}\ln\frac{b+\sqrt{D}}{b}}}{b+\sqrt{D}}. \label{e3.9}$$ We claim that \eqref{e3.7} implies \eqref{e3.9}. To prove our claim, it suffices to show $$\frac{b+\sqrt{2D}}{b+\sqrt{D}}\leq \frac{b+\sqrt{D+\frac{4c}{k+1}\ln\frac{b+\sqrt{D}}{b}}}{b+\sqrt{D}},$$ which is equivalent to $$\sqrt{2D}\leq \sqrt{D+\frac{4c}{k+1}\ln\frac{b+\sqrt{D}}{b}},$$ that is $$\frac{(k+1)D}{4c}\leq \ln\frac{b+\sqrt{D}}{b}.$$ By \eqref{e3.7}, it is sufficient to show that $$\ln\frac{b+\sqrt{2D}}{b+\sqrt{D}}\leq \ln\frac{b+\sqrt{D}}{b}. \label{e3.10}$$ It is not difficult to see that \eqref{e3.10} is equivalent to $$\frac{\sqrt{2}-1}{b+\sqrt{D}}\leq \frac{1}{b},$$ which is obviously true. Thus, (iv) is satisfied. To check (v), we need$\exp\big(a+bm_0-cm_0^2)\geq \exp\big(-(a+bx^*-cx^{*2})\big)$; i.e, $$a+bm_0-cm_0^2\geq -\frac{D}{4c}.\label{e3.11}$$ Let $$h(x)=a+bx-cx^2+\frac{D}{4c}.\label{e3.12}$$ We see that the positive solution of \eqref{e3.12} is$x_2=\frac{b+\sqrt{2D}}{2c}$. Since both$m_0$and$x_2$are larger than$x^*$, and on$(x^*,\infty)$,$h$is decreasing, \eqref{e3.11} is equivalent to$m_0\leq x_2$, which is satisfied by \eqref{e3.7}. Hence, it follows by Theorem \ref{thm2.1} that$\{x_n\}$tends to$\bar{x}$. \end{proof} \begin{corollary} \label{coro3.2} Assume that \eqref{e3.7} holds. Then \eqref{e1.5} is globally asymptotically stable. \end{corollary} \begin{proof} By condition \eqref{e2.8} in Corollary \ref{coro2.1}, we need $$-f'(\bar{x})\bar{x}=\frac{\sqrt{D}(b+\sqrt{D})}{2c} <2\cos\frac{k\pi}{2k+1},$$ which is equivalent to $$\frac{D}{4c}<\cos\frac{k\pi}{2k+1}-\frac{b\sqrt{D}}{4c}.$$ Then, combining this with Theorem \ref{thm3.2}, we know that if $$\frac{D}{4c}\leq \min\{\frac{1}{k+1}\ln\frac{b+\sqrt{2D}}{b+\sqrt{D}}, \cos\frac{k\pi}{2k+1}-\frac{b\sqrt{D}}{4c}\}\label{e3.13}$$ holds, then \eqref{e1.5} is globally asymptotically stable. Now, we show that $$\frac{1}{k+1}\ln \frac{b+\sqrt {2D}}{b+\sqrt D}\le \cos \frac{k\pi}{2k+1}-\frac{b\sqrt D}{4c}, \label{e3.14}$$ under the assumption \eqref{e3.7}. Observe that $$\frac{b\sqrt D}{4c}=\frac{b}{\sqrt D}\frac{D}{4c}\le \frac{b}{\sqrt D}\frac{1}{k+1}\ln \frac{b+\sqrt {2D}}{b+\sqrt D}$$ by \eqref{e3.7}. So it suffices to show that $$(\frac{b}{\sqrt D}+1)\frac 1{k+1}\ln \frac{b+\sqrt {2D}}{b+\sqrt D} \le \cos \frac{k\pi}{2k+1}.$$ After simplification, this becomes $$(\frac{b}{\sqrt D}+1)\ln \frac{b+\sqrt {2D}}{b+\sqrt D} \le (k+1)\sin \frac{\pi}{4k+2} .$$ Let$\frac b{\sqrt D}=t$($00$,$f$is increasing for$0