\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{amscd} \usepackage{graphicx} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics} \usepackage{latexsym} \usepackage{epsf} \usepackage{breakurl} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{https://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf Log-Concavity of Recursively Defined \\ \vskip .1in Polynomials} \vskip 1cm \large Bernhard Heim and Markus Neuhauser \\ Department of Mathematics and Science\\ Faculty of Science\\ German University of Technology in Oman (GUtech)\\ PO Box 1816\\ Athaibah PC 130\\ Sultanate of Oman\\ \href{mailto:bernhard.heim@gutech.edu.om}{\tt bernhard.heim@gutech.edu.om} \\ \href{mailto:markus.neuhauser@gutech.edu.om}{\tt markus.neuhauser@gutech.edu.om} \end{center} \vskip .2 in \begin{abstract} Fourier coefficients of powers of the Dedekind eta function can be studied by polynomials introduced by M. Newman. We generalize the defining recurrence relations in this paper. From this we derive new families of polynomials, which approximate these polynomials from below and above. Although these families are recursively defined, we are able to determine explicit closed formulas for both approximating polynomials. (For the original polynomials closed formulas are not yet known.) Furthermore, we obtain that both approximating families and the coefficients involved are log-concave and unimodal. \end{abstract} \section{Introduction} Let $\eta(\tau)$ be the Dedekind eta function \cite{On03}. Put $q := e^{2 \pi \, i \, \tau}$, where $\tau$ is in the upper half-space $\mathbb{H}$. Let further $z \in \mathbb{C}$. We consider the Fourier expansion \begin{equation} \label{start} \left( q^{-\frac{1}{24}} \eta(\tau)\right)^{-z} = \prod_{n=1}^{\infty} \left( 1 - q^n \right)^{-z} = \sum_{n=0}^{\infty} P_n(z) \, q^n. \end{equation} We recover several famous sequences \cite[Introduction]{HLN18}. For example, the partition numbers $p(n)$ and the Ramanujan numbers $\tau(n)$ are directly linked to $z=1$ and $z=-24$, \begin{eqnarray*} \left( P_n(1)\right) _{n=0}^{\infty} = \left( p(n)\right) _{n=0}^{\infty}&:& 1,1,2,3,5,\ldots \quad (\seqnum{A000041});\\ \left( P_{n-1}(-24)\right) _{n=1}^{\infty} = \left( \tau(n)\right) _{n=1}^{\infty} &:& 1,-24,252, \ldots \quad (\seqnum{A000594}). \end{eqnarray*} The integer-valued polynomials $P_n(X)$ have degree $n$ and positive integral coefficients $A_k^n$ for $0 \leq k \leq n-1$: \begin{equation} P_n(X) = \frac{X}{n!} \,\, \sum_{k=0}^{n-1} A_k^{n} \, \, X^k. \end{equation} Newman \cite{Ne55} calculated the coefficients recursively for $ 0 \leq k \leq n-1$. For example $A_0^n = (n-1)! \sigma(n)$, $A_{n-1}^n = 1$, and $A_{n-2}^n = 3 n (n-1)/2$. Here $\sigma(n)$ is the sum of the divisors of $n$. Further Newman determined the first ten polynomials and their integral zeros and found a recursion formula for the involved polynomials. Apparently there is yet no closed explicit formula for these polynomials defined by \begin{equation} \label{recursion} P_n(X) = \frac{X}{n} \,\, \sum_{k=1}^{n} \sigma(k) P_{n-k}(X) \end{equation} and the involved coefficients. Numerical calculations indicate that the sequence of the coefficients $A_k^n$ seems to be close to unimodal and stable (Hurwitz polynomial). Here we normalize $P_n(X)$ by $n!/X$. A direct approach seems to be out of reach at the moment, since there is still a lack of understanding the properties of the arithmetic function $\sigma(n)$ completely. Nevertheless the value distribution of the polynomials $P_n(X)$, especially the zeros, have significant applications. For $r$ even and positive, the sequences $\left( P_n(-r)\right) _{n}$ are lacunary if and only if $r= 2,4,6,8,10,14,26$. This result is due to Serre \cite{Se85} and is in relation to modular forms with complex multiplication (CM-forms). The non-existence of $-24$ as a zero is equivalent to the Lehmer conjecture \cite{Le47}. In this paper we generalize the recurrence relation (\ref{recursion}), studying closed formulas for the polynomials associated with the functions $g(n)=n$ and $g(n)=n^2$. They are integer valued and approximate $\sigma(n)$ from below and above. \begin{definition} Let $g(n)$ be an arithmetic function. We define a family of polynomials $P_n^g(X)$ associated with $g$. Let $P_0^g( X):=1$ and \begin{eqnarray}\label{generalrecursion} P_n^g(X) &:=& \frac{X}{n} \sum_{k=1}^{n} g(k) \,\, P_{n-k}^g(X),\\ &=& \frac{X}{n!} \sum_{k=0}^{n-1} A_k^{n} (g) \, X^k. \end{eqnarray} \end{definition} Let $\varphi_1(n) = n $ and $\varphi_2(n)=n^2$. In what follows, we study the properties of the associated polynomials $P_n^{\varphi_1}(X)$ and $P_n^{\varphi_2}(X)$. Their properties are related to $P_n(X) = P_n^{\sigma}(X)$, since $\varphi_1(n) < \sigma(n) < \varphi_2(n)$ for $n >1$. We obtain $A_{n-1}^n(\varphi_1) = A_{n-1}^n = A_{n-1}^n\left( \varphi _{2}\right) =1$. Further, let $ 0 \leq k \leq n-2$. Then \begin{equation} A_k^n (\varphi_1) < A_k^n < A_k^n (\varphi_2). \end{equation} \begin{theorem}\label{varphi1} Let $\varphi_1(n) =n$. Then the coefficients of $P_n^{\varphi_1}(X)$ are given by \begin{equation} A_k^n(\varphi_1) = \frac{n!}{(k+1)!} \, \binom{n-1}{k}. \label{eq:linear} \end{equation} \end{theorem} Although the binomial coefficients are twisted by $1/(k+1)!$, they remain log-concave. \begin{corollary}\label{uni1} The sequence of the coefficients of $P_n^{\varphi_1}(X)$ $$\left( A_k^n(\varphi_1)\right) _{k=0}^{n-1}$$ is strongly log-concave and hence unimodal. \end{corollary} Further we can determine the index, and therefore the size of the largest coefficient. \begin{corollary} \label{max1} The index $K_1$ of the maximal coefficient $A_k^n(\varphi_1)$ is given by \begin{equation} \sqrt{n+1} -2 \leq K_1 < \sqrt{n+1} -1. \end{equation} In general, $K_1$ is not always unique. \end{corollary} Numerical calculations indicate that the zeros of $P_n^{\varphi_1}(X)$ are simple and that the polynomials are stable in the sense of Hurwitz (neglecting $X=0$). It is also possible to get similar results for $P_n^{\varphi_{2}}(X)$, although the involved coefficients $A_k^n(\varphi_2)$ are more complicated. We also studied the polynomials $P^{\varphi}_n(X)$ associated with $\varphi(n) := n \, \mbox{ln}(n)$ and $\varphi(n) := n \, \sqrt{n}$. But a closed formula for the coefficients is very difficult to obtain. \begin{theorem}\label{varphi2} Let $\varphi_2(n) =n^2$. Then the coefficients of $P_n^{\varphi_2}(X)$ are given by \begin{equation} A_k^n(\varphi_2) = \frac{n!}{(k+1)!} \, \binom{n+k}{2k+1}\ . \label{eq:quadratisch} \end{equation} \end{theorem} \begin{corollary}\label{uni2} The sequence of the coefficients of $P_n^{\varphi_2}(X)$ $$\left( A_k^n(\varphi_{2})\right) _{k=0}^{n-1}$$ is strongly log-concave and hence unimodal. \end{corollary} \begin{corollary} \label{max2} The coefficients $A_k^n(\varphi_2)$ assume their maximum at the index $K_2$, such that $K_{2}\leq k 1$ exists, such that \begin{equation} \frac{ A_k^n (\varphi_1)^2 }{ A_{k+1}^n(\varphi_1) \, A_{k-1}^n(\varphi_1) } \geq \alpha. \end{equation} This quotient results in \begin{eqnarray*} \frac{\left( \frac{1}{\left( k+1\right) !}\binom{n-1}{k}\right) ^{2}}{\frac{1}{k!}\binom{n-1}{k-1}\frac{1}{\left( k+2\right) !}\binom{n-1}{\left( k+1\right) }} &=&\frac{\frac{1}{k+1 }\left( \frac{\left( n-1\right) !}{k!\left( n-k-1\right) !}\right) ^{2}}{\frac{\left( n-1\right) !}{\left( k-1\right) !\left( n-k\right) !}\frac{1}{k+2}\frac{\left( n-1\right) !}{\left( k+1\right) !\left( n-k-2\right) !}} \\ & = & \frac{\frac{1}{k+1}\frac{1}{k\left( n-k-1\right) }}{\frac{1}{n-k}\frac{1}{k+2}\frac{1}{k+1}} \\ &=&\frac{\left( n-k\right) \left( k+2\right) }{k\left( n-k-1\right) } \geq \alpha >1. \end{eqnarray*} Hence the sequence $A_k^n(\varphi_1)$ is strongly log-concave. This implies unimodality. We obtain $\alpha =\alpha _{n}\geq 1+\left( 3n+1+2\sqrt{2n\left( n+1\right) }\right) \left( n-1\right) ^{-2}$. Depending on~$n$ $\alpha $ decreases to $1$. To show that $\alpha $ decreases to $1$, consider the denominator of the derivative of $\frac{\left( n-k\right) \left( k+2\right) }{k\left( n-1-k\right) }$ with respect to~$k$. The denominator yields $2n\left( n+1\right) -\left( k-2n\right) ^{2}$. Hence the minimum of the denominator for $1\leq k\leq n-2$ is obtained at $k=2n-\sqrt{2n\left( n+1\right) }$. This yields the claimed value. \subsection{Proof of Corollary \ref{uni2}} Let $n \geq 2$ be given. We consider the sequence \begin{equation*} a_k :=A_k^n(\varphi_2) = \frac{n!}{(k+1)!} \, \binom{ n + k }{2k +1} \end{equation*} for $k=0, \ldots, n-1$. Hence $a_0,a_1,\ldots, a_m$ with $m=n-1$. We show that for each pair $(n,k)$ with $0 \leq k \leq n-2$ a $\beta >1$ exists such that \begin{equation} \frac{ A_k^n (\varphi_{2})^2 }{ A_{k+1}^n(\varphi_{2}) \, A_{k-1}^n(\varphi_{2}) } > \beta. \end{equation} This quotient results in \begin{eqnarray*} \frac{\left( \frac{1}{\left( k+1\right) !}\binom{n+k}{2k+1}\right) ^{2}}{\frac{1}{k!}\binom{n+k-1}{2k-1}\frac{1}{\left( k+2\right) !}\binom{n+k+1}{2k+3}} & = & \frac{\frac{1}{k+1}\left( \frac{\left( n+k\right) !}{\left( 2k+1\right) !\left( n-k-1\right) !}\right) ^{2}}{\frac{\left( n+k-1\right) !}{\left( 2k-1\right) !\left( n-k\right) !}\frac{1}{k+2}\frac{\left( n+k+1\right) !}{\left( 2k+3\right) !\left( n-k-2\right) !}} \\ &=& \frac{\frac{1}{k+1}\frac{n+k}{\left( 2k+1\right) 2k\left( n-k-1\right) }}{\frac{1}{n-k}\frac{1}{k+2}\frac{n+k+1}{\left( 2k+3\right) \left( 2k+2\right) }} \\ & = & \frac{\left( n+k\right) \left( n-k\right) \left( k+2\right) \left( 2k+3\right) \left( 2k+2\right) }{\left( k+1\right) \left( n+k+1\right) \left( 2k+1\right) 2k\left( n-k-1\right) } \\ &>& \frac{n^{2}-k^{2}}{n^{2}-\left( k+1\right) ^{2}}\geq \beta >1. \end{eqnarray*} Hence the sequence $A_k^n(\varphi_2)$ is strongly log-concave. This implies unimodality. We have $\frac{n^{2}-k^{2}}{n^{2}-\left( k+1\right) ^{2}}=1+\frac{2k+1}{n^{2}\left( k+1\right) ^{2}}$. This is increasing as a function of~$k$ for $1\leq k \leq n-2$. So the minimum is attained at $k=1$ and results in $1+\frac{3}{n^{2}-4}$. For $n\rightarrow \infty $ $1+\frac{3}{n^{2}-4}$ again decreases to $1$. Even if we could determine the minimum of the expression before the `strictly less' sign, we would obtain the same limiting behavior. \subsection{Proof of Corollary \ref{max1}} The idea is simple. Let $n \geq 2$. Let $a_k = A_k^n(\varphi_1)$, $0 \leq k \leq n-1$. We analyze the possible sign changes of \begin{equation} \Delta_k (\varphi_1) := a_{k+1} - a_k = A_{k+1}^n (\varphi_1) - A_{k}^n (\varphi_1), \quad 0 \leq k \leq n-2. \end{equation} We obtain \begin{equation} \Delta_k (\varphi_1) = \gamma(k,n) \left( \frac{1}{\left( k+2\right) \left( k+1\right) }-\frac{1}{n-k-1}\right) , \end{equation} where $\gamma(k,n)$ is a positive rational number. Hence $\Delta_k(\varphi_1)$ has the same sign as $$ n+1 - (k+2)^2.$$ Hence the coefficients increase for $k<\sqrt{n+1}-2$, and decrease for $k>\sqrt{n+1}-2$. In the case of equality we have two maxima. \subsection{Proof of Corollary \ref{max2}} Let $n \geq 2$. Let $a_k = A_k^n(\varphi_2)$, $0 \leq k \leq n-1$. We analyze the possible sign changes of \begin{equation} \Delta_k (\varphi_2) := a_{k+1} - a_k = A_{k+1}^n (\varphi_2) - A_{k}^n (\varphi_2), \quad 0 \leq k \leq n-2. \end{equation} We obtain \begin{equation} \Delta_k (\varphi_2) = \gamma'(k,n) \, \left( \frac{n+k+1}{\left( k+2\right) \left( 2k+3\right) \left( 2k+2\right) }-\frac{1}{n-k-1}\right) . \end{equation} where $\gamma'(k,n)$ is a positive rational number. Hence $\Delta_k(\varphi_2)$ has the same sign as \begin{equation*} -4k^{3}-19k^{2}-28k-13+n^{2}. \end{equation*} As a function of $k \geq 0$, the expression $\Delta_k(\varphi_2)$ is decreasing. Hence there exists exactly one $0 \leq K \leq n-1$ such that $$ A_{K-1}^n (\varphi_2) < A_K^n (\varphi_2) \geq A_{K+1}^n (\varphi_2).$$ We used the computer algebra system \texttt{Maple} to calculate the algebraic expression defining $K$. Nevertheless, it is obvious that $K \approx 2^{-2/3} \,\, n^{\frac{2}{3}}$ for large $n$. \section{Explicit formulas for $P_n^{\varphi_1}(X)$} Let us start with a list of the polynomials for $n=1,2,\ldots ,7$. Note that they are no longer integer-valued polynomials, although the modified coefficients are integral. \begin{eqnarray*} P_{1}^{\varphi_1} (X) & = & X \\ P_{2}^{\varphi_1} (X) & = & \frac{1}{2}X\left( X+2\right) \\ P_{3}^{\varphi_1} (X) & = & \frac{1}{6} X\left( X^{2}+6X+6\right) \\ P_{4}^{\varphi_1} (X) & = & \frac{1}{24}X\left( X^{3}+12X^{2}+36X+24\right) \\ P_{5}^{\varphi_1} (X) & = & \frac{1}{120}X\left( X^{4}+20X^{3}+120X^{2}+240X+120\right) \\ \end{eqnarray*} \begin{eqnarray*} P_{6}^{\varphi_1} (X) & = & \frac{1}{720}X\left(X^{5}+30X^{4}+300X^{3}+1200X^{2}+1800X+720\right) \\ P_{7}^{\varphi_1} (X) & = & \allowbreak \frac{1}{5040}X\left( X^{6}+42X^{5}+630X^{4}+4200X^{3}+12600X^{2} +15120X+5040\right) . \end{eqnarray*} \begin{remark} The first polynomials $\frac{n!}{X} P_{n}^{\varphi_1}(X)$ are irreducible. \end{remark} \begin{proof}[Proof of Theorem \ref{varphi1}] We start with the identity \begin{equation} \label{qn} \sum_{n=0}^{\infty }P_{n}^{\varphi_1} \left( X \right) \,\, q^{n}=\exp \left( X \,\, \sum_{n=1}^{\infty }q^{n}\right). \end{equation} It is useful to substitute $\left( \sum_{n=1}^{\infty}q^{n}\right) ^{k}$ by the multi-index sum $$\sum_{m_{1},\ldots ,m_{k}=1}^{\infty }q^{m_{1}+\cdots + m_{k}}.$$ Comparing the coefficients of $q^n$ in (\ref{qn}) leads to \begin{eqnarray*} P_{n}\left( X\right) &=&\sum_{k=1}^{n}\frac{1}{k!}\left( \sum_{m_{1}+\cdots +m_{k}=n}1\right) X^{k} \\ &=&\frac{X}{n!}\sum_{k=0}^{n-1}\frac{n!}{\left( k+1\right) !}\left( \sum_{m_{1}+\cdots +m_{k+1}=n}1\right) X^{k}. \end{eqnarray*} Note that \begin{equation} \sum_{m_{1}+\cdots +m_{k}=n}1=\binom{n-1}{k-1}. \label{eq:linearcoeff} \end{equation} We prove (\ref{eq:linearcoeff}) by induction. For $k=1$ we have $1=\binom{n-1}{0}=\binom{n-1}{k-1}$. Suppose the formula holds for~$k$ then \begin{eqnarray*} \sum_{m_{1}+\cdots +m_{k+1}=n}1 &=&\sum_{m_{k+1}=1}^{n-k}\sum_{m_{1}+\cdots +m_{k}=n-m_{k+1}}1=\sum_{m_{k+1}=1}^{n-k}\binom{n-m_{k+1}-1}{k-1} \\ &=&\sum_{m_{k+1}=k}^{n-1}\binom{m_{k+1}-1}{k-1}=\binom{n-1}{k}. \end{eqnarray*} The proof (again by induction) of the identity $\sum_{m=k}^{n-1}\binom{m-1}{k-1}=\allowbreak \binom{n-1}{k}$ we leave to the reader. \end{proof} \section{Explicit formulas for $P_n^{\varphi_2}(X)$} Let us start with a list of the polynomials for $n=1,2,\ldots 7$. Note that they are no longer integer-valued polynomials, although the modified coefficients are integral. \begin{eqnarray*} P_{1}^{\varphi_2} (X) & = & X \\ P_{2}^{\varphi_2} (X) & = & 1/2\,X \left( 4+X \right) \\ P_{3}^{\varphi_2} (X) & = & 1/6\,X \left( {X}^{2}+12\,X+18 \right) \\ P_{4}^{\varphi_2} (X) & = & 1/24\,X \left( {X}^{3}+24\,{X}^{2}+120\,X+96 \right) \\ P_{5}^{\varphi_2} (X) & = & \frac{1}{120}X \left( {X}^{4}+40\,{X}^{3}+420\,{X}^{2}+1200\,X+600 \right) \\ P_{6}^{\varphi_2} (X) & = & \frac{1}{720}X \left( {X}^{5}+60\,{X}^{4}+1080\,{X}^{3}+6720\,{X}^{2}+12600 \,X+4320 \right) \\ P_{7}^{\varphi_2} (X) & = & \frac{1}{5040}X \left( {X}^{6}+84\,{X}^{5}+2310\,{X}^{4}+25200\,{X}^{3}+ 105840\,{X}^{2} +141120\,X+35280 \right) \end{eqnarray*} Before we prove Theorem \ref{varphi2} we show the following useful property. \begin{lemma}\label{lemma} For $n\geq 1$ and $1\leq k\leq n$ we obtain \begin{equation} \sum_{m_{1}+\cdots +m_{k}=n}m_{1} m_2 \cdots m_{k}=\binom{n+k-1}{2k-1}. \label{eq:quadraticcoeff} \end{equation} \end{lemma} \begin{proof} The proof is by induction on~$n$ and $k$. For $k=1$ we have $n=\binom{n}{1}=\binom{n+k-1}{2k-1}$. Let now $N\geq K\geq 2$ be fixed. We assume that formula (\ref{eq:quadraticcoeff}) holds for $k