\documentstyle[amscd,amssymb,verbatim,oneside,12pt]{amsart} \textwidth=6in \textheight=8.4in \topmargin=-\headheight \hoffset-.5in %\oddsidemargin=0in %\evensidemargin=0in \def\Tilde{\char126\relax} \newtheorem{Definition}{\bf Definition}[section] \newtheorem{Theorem}[Definition]{\bf Theorem} \numberwithin{equation}{section} \pagestyle{myheadings} \markright{\sc the electronic journal of combinatorics 6 (1999), \#R10 \hfill} \thispagestyle{empty} \begin{document} \begin{center} \Large{A criterion for unimodality} \end{center} \vskip 10pt \begin{center} George Boros \\ Department of Mathematics, University of New Orleans \\ New Orleans, LA 70148 \\ \texttt{gboros@@math.uno.edu} \\ \end{center} \vskip 5pt \begin{center} Victor H. Moll\footnote{www: \texttt{}} \\ Department of Mathematics, Tulane University \\ New Orleans, LA 70118 \\ \texttt{vhm@@math.tulane.edu} \\ \vskip 5pt Submitted: January 23, 1999; Accepted February 2, 1999 \\ Classification 05, 33, 40 \end{center} \vskip 30pt \thanks{The authors wish to thank Doron Zeilberger for his comments on an earlier version of this paper.} \begin{center} {\bf Abstract} \end{center} \vskip 5pt We show that if $P(x)$ is a polynomial with nondecreasing, nonnegative coefficients, then the coefficient sequence of $P(x+1)$ is unimodal. Applications are given. \vskip 20pt \section{Introduction} \setcounter{equation}{0} A finite sequence of real numbers $\{ d_{0}, d_{1}, \cdots, d_{m} \}$ is said to be {\it unimodal} if there exists an index $0 \leq m^{*} \leq m$, called the {\it mode} of the sequence, such that $d_{j}$ increases up to $j = m^{*}$ and decreases from then on, that is, $d_{0} \leq d_{1} \leq \cdots \leq d_{m^{*}}$ and $d_{m^{*}} \geq d_{m^{*}+1} \geq \cdots \geq d_{m}$. A polynomial is said to be unimodal if its sequence of coefficients is unimodal. Unimodal polynomials arise often in combinatorics, geometry and algebra. The reader is referred to \cite{bre} and \cite{sta} for surveys of the diverse techniques employed to prove that specific families of polynomials are unimodal. A sequence of positive real numbers $\{d_{0}, d_{1}, \cdots, d_{m} \}$ is said to be {\it logarithmically concave} (or {\em log-concave} for short) if $d_{j+1}d_{j-1} \le d_{j}^{2}$ for $ 1 \leq j \leq m-1$. It is easy to see that if a sequence is log-concave then it is unimodal \cite{wilf}. A sufficient condition for log-concavity of a polynomial is given by the location of its zeros: if all the zeros of a polynomial are real and negative, then it is log-concave and therefore unimodal \cite{wilf}. A second criterion for the log-concavity of a polynomial was determined by Brenti \cite{bre}. A sequence of real numbers is said to have {\it no internal zeros} if whenever $d_{i}, d_{k} \neq 0$ and $ i < j < k$ then $d_{j} \neq 0$. Brenti's criterion states that if $P(x)$ is a log-concave polynomial with nonnegative coefficients and with no internal zeros, then $P(x+1)$ is log-concave. \section{The main result} \setcounter{equation}{0} \begin{Theorem} \label{thm1} If $P(x)$ is a polynomial with positive nondecreasing coefficients, then $P(x+1)$ is unimodal. \end{Theorem} {\bf Proof}. Observe first that $P_{m,r}(x) := (1+x)^{m+1} - (1+x)^{r}$ with $0 \leq r \leq m$ is unimodal with mode at $1 + \lfloor{\frac{m}{2}} \rfloor$. This follows by induction on $m \geq r$ using $P_{m+1,r}(x) = P_{m,r}(x) + x(1+x)^{m+1}$. For $m$ even, $P_{m+1,r}$ is the sum of two unimodal polynomials with the same mode. For $m = 2t+1$, the modes are shifted by $1$, so it suffices to check \begin{eqnarray} a_{t+1} + \left( {m+1 \atop t} \right) & \leq & a_{t+2} + \left( {m+1 \atop t+1} \right), \label{ineq} \end{eqnarray} \noindent where $a_{t+1}$ is the coefficient of $x^{t}$ in $P_{m,r}(x)$. The case $ t \geq r$ yields equality in (\ref{ineq}). If $t \leq r-2$ then (\ref{ineq}) is equivalent to $r \leq m+2$. The final case $t = r-1$ amounts to $0 = \left( {m+1 \atop r-1} \right) - \left( {m+1 \atop r+1} \right) \leq 1$, Now $P(x+1) = \frac{1}{x} \left( b_{0} P_{m,0}(x) + (b_{1} - b_{0})P_{m,1}(x) + \cdots + (b_{m} - b_{m-1}) P_{m,m}(x) \right)$, so $P(x+1)$ is a sum of unimodal polynomials with the same mode, and hence unimodal. \\ \noindent We now restate Theorem \ref{thm1} and offer an alternative proof. \begin{Theorem} \label{thm2} Let $b_{k} > 0$ be a nondecreasing sequence. Then the sequence \begin{eqnarray} c_{j} & := & \sum_{k=j}^{m} b_{k} \left( {k \atop j} \right), \; \; 0 \leq j \leq m \label{coefj1} \end{eqnarray} \noindent is unimodal with mode $m^{*} := \lfloor{\frac{m-1}{2}} \rfloor$. \end{Theorem} {\bf Proof}. For $0 \leq j \leq m-1$ we have \begin{eqnarray} (j+1) ( c_{j+1} - c_{j} ) & = & \sum_{k=j}^{m} b_{k} \left( {k \atop j} \right) \times (k - 2j -1). \label{twotwo} \end{eqnarray} \noindent Suppose first that $j \geq m^{*}$, and let $m$ be odd so that $m = 2m^{*} + 1$; the case $m$ even is treated in a similar fashion. Every term in (\ref{twotwo}) is negative because, if $j > m^{*}$, then $k-2j-1 \leq m-2j-1 = 2(m^{*} - j) < 0$, and for $j = m^{*}$, \begin{eqnarray} (m^{*} + 1)(c_{m^{*}+1} - c_{m^{*}}) & = & \sum_{k=m^{*}}^{m-1} b_{k} \left( {k \atop m^{*} } \right) \times (k-m) < 0. \end{eqnarray} \noindent Thus $c_{j+1} < c_{j}$. Now suppose $ 0 \leq j < m^{*}$ and define \begin{eqnarray} T_{1} & := & \sum_{k=j}^{2j} b_{k} \left( {k \atop j} \right) (2j+1-k) \label{five2} \end{eqnarray} \noindent and \begin{eqnarray} T_{2} & := & \sum_{k=2j+2}^{m} b_{k} \left( {k \atop j} \right) (k-2j-1) \label{five3} \end{eqnarray} \noindent so that $(j+1)(c_{j+1} - c_{j} ) = T_{2} - T_{1}$. Then \begin{eqnarray} T_{1} < b_{2j+2} \sum_{k=j}^{2j} \left( {k \atop j} \right) (2j+1-k) = b_{2j+2} \left( {2j+2 \atop j} \right) < T_{2}. \nonumber \end{eqnarray} \noindent Thus $c_{j+1} > c_{j}$. \\ \section{Examples} \setcounter{equation}{0} \noindent {\bf Example 1}. The case $P(x) = x^{n}$ in Theorem \ref{thm1} gives the unimodality of the binomial coefficients. \\ \noindent {\bf Example 2}. For $0 \leq k \leq m-1$, define \begin{eqnarray} b_{k}(m) & := & 2^{-2m+k} \left( {2m-2k \atop m-k} \right) \left( {m+k \atop m} \right) (a+1)^{k} \nonumber \end{eqnarray} \noindent for $0 \leq k \leq m-1$. Then \begin{eqnarray} \frac{b_{k+1}(m)}{b_{k}(m)} & = & \frac{(m-k)(m+k+1)}{(2m-2k-1)(k+1)} > 1 \nonumber \end{eqnarray} \noindent so the polynomial \begin{eqnarray} P_{m}(a) & := & \sum_{k=0}^{m} b_{k}(m) (a+1)^{k} \nonumber \end{eqnarray} \noindent is unimodal. We encountered $P_{m}$ in the integral formula \cite{bm1} \begin{eqnarray} \int_{0}^{\infty} \frac{dx}{(x^{4} + 2ax^{2} + 1)^{m+1} } & = & \frac{ \pi \; P_{m}(a)}{2^{m+ 3/2} (a+1)^{m+ 1/2} }. \label{n04} \end{eqnarray} \noindent This does not appear in the standard tables. \\ \noindent {\bf Example 3}. For $0 \leq k \leq m-1$, define \begin{eqnarray} b_{k}(m) & := & \frac{ (-m - \beta)_{m}}{m!} \; \frac{(-m)_{k} (m+1+ \alpha + \beta)_{k} }{(\beta + 1)_{k} \, k! \, 2^{k} }. \nonumber \end{eqnarray} \noindent Then, with $\alpha = m + \epsilon_{1}$ and $\beta = -(m + \epsilon_{2})$, we have \begin{eqnarray} \frac{b_{k+1}(m)}{b_{k}(m)} & = & \frac{m-k}{m-k+ \epsilon_{2} -1} \times \frac{k-1 + m + \epsilon_{1} - \epsilon_{2} }{2(k+1)} > 1 \nonumber \end{eqnarray} \noindent provided $0 < \epsilon_{1} \leq \epsilon_{2} < 1$. Therefore the polynomial \begin{eqnarray} P_{m}^{(\alpha, \beta)}(a) & := & \sum_{k=0}^{m} b_{k}(m) (a+1)^{k} \nonumber \end{eqnarray} \noindent is unimodal. This is a special case of the Jacobi family, where the parameters $\alpha$ and $\beta$ are not standard since they depend on $m$. These polynomials do not have real zeros, so their unimodality is not immediate. The case of Example $2$ corresponds to $\epsilon_{1} = \epsilon_{2} = \frac{1}{2}$. \\ \noindent {\bf Example 4}. Let $n, \, m \in {\mathbb{N}}$ be fixed. Then the sequences \begin{eqnarray} \alpha_{j} := \sum_{k=j}^{m} n^{k} \left( {k \atop j} \right), \; \; \beta_{j} := \sum_{k=j}^{m} k^{n} \left( {k \atop j} \right), \; \; {\rm and } \; \gamma_{j} := \sum_{k=j}^{m} k^{k} \left( {k \atop j} \right) \; \; \nonumber \end{eqnarray} \noindent are unimodal for $0 \leq j \leq m$. \\ \noindent {\bf Example 5}. Let $2 < a_{1} < \cdots < a_{p}$ and $n_{1}, \cdots, n_{p}$ be two sequences of $p$ positive integers. For $0 \leq j \leq m$, define \begin{eqnarray} c_{j} & := & \sum_{k=j}^{m} \left( { a_{1} m \atop k} \right)^{n_{1}} \, \left( { a_{2} m \atop k} \right)^{n_{2}} \cdots \left( { a_{p} m \atop k} \right)^{n_{p}} \left( {k \atop j } \right). \label{hyper} \end{eqnarray} Then $c_{j}$ is unimodal. \vskip 20pt \noindent {\bf Acknowledgement}. The authors wish to thank Doron Zeilberger for comments on an earlier version of this paper. \vskip 20pt \begin{thebibliography}{99} \bibitem{bm1} BOROS, G. - MOLL, V.: {\em An integral hidden in Gradshteyn and Rhyzik}, J. Comp. Appl. Math., to appear. \bibitem{bre} BRENTI, F.: {\em Log-concave and unimodal sequences in Algebra, Combinatorics and Geometry: an update}. Contemporary Mathematics, {\bf 178}, 71-84, 1994. \bibitem{sta} STANLEY, R.: {\em Log-concave and unimodal sequences in algebra, combinatorics and geometry}. Graph theory and its applications: East and West (Jinan, 1986), 500-535, Ann. New York Acad. Sci., {\bf 576}, New York, 1989. \bibitem{wilf} WILF, H.S.: {\em generatingfunctionology}. Academic Press, 1990. \end{thebibliography} \end{document}