% TeX file for a 5 page paper \input amssym.def \input amssym \magnification = \magstephalf \parskip=12pt \nopagenumbers \font\smcp=cmcsc8 \font\smcpp=cmcsc10 \font\lbf=cmbx12 \headline={\ifnum\pageno>1 {\smcp the electronic journal of combinatorics 2 (1995), \#R1\hfill\folio} \fi} \def\half{{1\over 2}} \def\thalf{{\textstyle{\half}}} \def\intl{\int\limits} \def\({\left(} \def\){\right)} \centerline{\lbf Counting distinct zeros of the Riemann zeta-function} \vskip .25in \centerline{David W.~Farmer} \vskip .15in \centerline{Submitted: December 1, 1994; Accepted: December 13, 1994.} \footnote{}{1991 Mathematical Subject Classification: 05A20, 11M26} \vskip .4in \vbox{\narrower\noindent{\smcpp Abstract.} Bounds on the number of simple zeros of the derivatives of a function are used to give bounds on the number of distinct zeros of the function. } The Riemann $\xi$-function is defined by $\xi(s)=H(s)\zeta(s)$, where $H(s)=\half s(s-1)\pi^{-\half s} \Gamma(\half s)$ and $\zeta(s)$ is the Riemann $\zeta$-function. The zeros of $\xi(s)$ and its derivatives are all located in the critical strip $0<\sigma<1$, where $s=\sigma+it$. Since $H(s)$ is regular and nonzero for $\sigma>0$, the nontrivial zeros of $\zeta(s)$ exactly correspond to those of $\xi(s)$. Let $\rho^{(j)}=\beta+i\gamma$ denote a zero of the $j^{\rm th}$ derivative $\xi^{(j)}(s)$, and denote its multiplicity by $m(\gamma)$. Define the following counting functions: $$\eqalign{ N^{(j)}(T)&= \sum_{\rho^{(j)}=\beta+i\gamma}^{\phantom{o}}1\phantom{XXXXXXX} \hbox{zeros of $\xi^{(j)}(\sigma+it)$ with $00.40219$, $\alpha_1>0.79874$, $\alpha_2> 0.93469$, $\alpha_3>0.9673$, $\alpha_4>0.98006$, and $\alpha_5>0.9863$. These bounds were obtained by combining Theorem~2 of~[C2] with the methods of~[C1]. Trivially, $\beta_j\ge\alpha_j$. Let $N_d(T)$ be the number of distinct zeros of $\xi(s)$ in the region $0 \kappa\, N(T)$, with $\kappa = 0.40219$. We will use the bounds on $\beta_j$ to obtain the following \proclaim Theorem. For $T$ sufficiently large, $$ N_d(T) > k\, N(T), $$ with $k=0.63952\ldots\,$. Furthermore, given the bounds on $\beta_j$, this result is best possible. We present two methods for determining lower bounds for $N_d(T)$. These methods employ combinatorial arguments involving the $\beta_j$. We note that the added information that $\alpha_j$ detects zeros on the critical line is of no use in improving our result. Everything below is phrased in terms of the Riemann $\xi$-function, but the manipulations work equally well for any function such that it and all of its derivatives have the same number of zeros. We write $f(T)\gtrsim g(T)$ for $f(T)\ge g(T)+o(N(T))$ as $T\to \infty$. For example, $N^{(j)}_s(T) \gtrsim \beta_j N(T)$ means $N^{(j)}_s(T)\ge\(\beta_j +o(1)\) N(T)$ as $T\to\infty$. The first method starts with the following inequality of Conrey, Ghosh, and Gonek [CGG]. A simple counting argument yields $$ N_d(T) \ge \sum_{r=1}^R {M_{\le r}(T)\over r(r+1)} + {M_{\le R+1}(T)\over R+1}. \eqno (2) $$ To obtain lower bounds for $M_{\le r}(T)$ we note that if $\rho$ is a zero of $\xi(s)$ of order $m \ge n+2$ then $\rho$ is a zero of order $m-n\ge 2m/(n+2)\ge 2$ for $\xi^{(n)}(s)$. Thus, $$ N^{(n)}_s(T) \le N(T)-{2\over n+2} \bigl(N(T) -M_{\le n+1}(T)\bigr), $$ which gives $$ M_{\le n}(T) \gtrsim \({\beta_{n-1} (n+1) -n+1\over 2}\) N(T). \eqno(3) $$ The bounds for $\alpha_j$ now give: $M_{\le 1}(T) \gtrsim 0.40219 N(T)$, $M_{\le 2}(T) \gtrsim 0.69812 N(T)$, $M_{\le 3}(T) \gtrsim 0.86938 N(T)$, $M_{\le 4}(T) \gtrsim 0.91825 N(T)$, $M_{\le 5}(T) \gtrsim 0.94019 N(T)$, and $M_{\le 6}(T) \gtrsim 0.9520 N(T)$. Inserting these bounds into inequality (2) with $R=5$ gives $N_d(T) \gtrsim 0.62583 N(T)$. We note that the lower bounds for $M_{\le n}(T)$ are best possible in the sense that, for each $n$ separately, equality could hold in~(3). However, inequality (3) is not simultaneously sharp for all $n$, and this possibility imparts some weakness to the result. A lower bound for $N_d(T)$ was calculated in [CGG] in a spirit similar to the above computation, but it was mistakenly assumed that $M_{\le n}(T)\gtrsim \beta_{n-1}N(T)$, rendering their bound invalid. Our second method eliminates the loss inherent in the first method. We start with this \proclaim Lemma. In the notation above, $$ N^{(n)}_s(T) \le \sum_{j=1}^{n+1} M_j(T) + n\sum_{j=n+2}^\infty {M_j(T)\over j}. $$ Proof. Suppose $\rho$ is a zero of order $j$ for $\xi(s)$. If $j\ge n+2$ then $\rho$ is a zero of order $j-n$ for $\xi^{(n)}(s)$, so $\xi^{(n)}(s)$ has at least $\displaystyle{\sum_{j=n+2}^\infty {(j-n)M_j(T)\over j}}\ $ zeros of order $\ge 2$. Thus, $$\eqalign{ N^{(n)}_s(T) &\le N^{(n)}(T) - \sum_{j=n+2}^\infty {(j-n)M_j(T)\over j}\cr &= \sum_{j=0}^\infty M_j(T) - \sum_{j=n+2}^\infty {(j-n)M_j(T)\over j}\cr &=\sum_{j=1}^{n+1} M_j(T) + n\sum_{j=n+2}^\infty {M_j(T)\over j}, }$$ as claimed. Combining the Lemma with (1) we get $$ N^{(n)}_s(T)\le n N_d(T) + n\sum_{j=1}^{n+1} \({1\over n}-{1\over j}\)\! M_j(T). \eqno (4) $$ Let $I_n$ denote the inequality (4). Then, in the obvious notation, a straightforward calculation finds that the inequality $$ I_J \ +\ \sum_{n=1}^{J-1} 2^{J-n-1} I_n $$ is equivalent to $$ \bigl(2^J -1\bigr)N_d(T) + \sum_{n=1}^{J+1}{M_n(T)\over n} \ \ge\ 2^{J-1}M_1(T)+N^{(J)}_s(T) + \sum_{n=1}^{J-1} 2^{J-n-1} N^{(n)}_s(T). \eqno (5) $$ This implies $$\eqalignno{ N_d(T) &\ge 2^{-J}\(2^{J-1} N^{(0)}_s(T)+ N^{(J)}_s(T) + \sum_{n=1}^{J-1} 2^{J-n-1} N^{(n)}_s(T)\)\cr \cr &\gtrsim 2^{-J}\(2^{J-1} \beta_0+ \beta_J + \sum_{n=1}^{J-1} 2^{J-n-1} \beta_n\)\!N(T). &(6) }$$ Choose $J=5$ and use the trivial inequality $\beta_j\ge\alpha_j$ and the bounds for $\alpha_j$ to obtain the Theorem. Finally, we show that our result is best possible. In other words, if our lower bounds for the $\beta_j$ were actually equalities, then the lower bound given by (6) is sharp. We will accomplish this by showing that the $M_n(T)$, the number of zeros of $\xi(s)$ with multiplicity exactly $n$, can be assigned values which achieve the bounds on $\beta_j$, and which yield a value of $N_d(T)$ which is arbitrarily close to the lower bound given by (6). Suppose we have lower bounds for $\beta_j$, for $0\le j\le J$, and let $K\ge J+2$. Suppose we had the following four equalities: $$\displaylines{ M_1(T) = \beta_0 N(T),\cr \cr M_K(T)={\textstyle{K\over K-J}} (1-\beta_J)N(T), \cr\cr M_{J+1}(T) = {J+1\over 2}\(\beta_J-\beta_{J-1} -{1-\beta_J\over K-J}\)N(T),\cr }$$ and for $2\le n\le J$, $$ M_n(T) = {n\over 2}\({3\beta_{n-1}\over 2} -\beta_{n-2} -2^{n-J-1}\beta_J -{1-\beta_J\over 2^{J-n+1}(K-J)} -\sum_{j=n}^{J-1}2^{n-j-2}\beta_j\)\! N(T) $$ and $M_j(T)=0$ otherwise. Then $\displaystyle{\sum_{j=1}^\infty M_j(T) = N(T)}$ and for $0\le n\le J$ we have $$ \sum_{j=1}^{n+1} M_j(T) + n\sum_{j=n+2}^\infty {M_j(T)\over j}= \beta_n N(T), \eqno(7) $$ and $$ \sum_{n=1}^\infty {M_n(T)\over n}= 2^{-J}\!\(2^{J-1} \beta_0+ \beta_J + \sum_{n=1}^{J-1} 2^{J-n-1} \beta_n\)\!N(T) + {(1-\beta_J)2^{-J}\over K-J}N(T). \eqno(8) $$ Since the left side of (8) is $N_d(T)$ and the second term on the right side can be made arbitrarily small by choosing $K$ large, we conclude that (6) is sharp. There are two things left to check. The given values of $M_n(T)$ must be positive when $K$ is large. It is easy to check this for $J=5$ and our lower bounds for $\beta_j$. And since we supposed that our bounds for $\beta_j$ are sharp, we must show that $N_s^{(j)}(T) = \beta_j N(T)$. To see this, note that, generically, the left side of (7) equals $N_s^{(j)}(T)$. In other words, the zeros of the derivatives of a generic function are all simple, except for those which are ``tied up'' in high-order zeros of the original function. By computing further values of $\alpha_j$, enabling us to take a larger value of $J$ in~(6), we could improve the result slightly: this is due to a decrease in the loss in passing from~(5) to~(6). The bound $M_{\le 6}(T) \gtrsim 0.952 N(T)$ implies that this improvement could increase the lower bound we obtained by at most $0.00021 N(T)$. \noindent{\bf References} {\parskip =8pt \item{[C1]} {\sl J.~B.~Conrey}, Zeros of derivatives of Riemann's $\xi$-function on the critical line, II, J.~Number Theory {\bf 17} (1983), 71-75. \item{[C2]} {\sl J.~B.~Conrey}, More than two fifths of the zeros of the Riemann zeta function are on the critical line, J.~reine angew.~Math. {\bf 399} (1989), 1-26. \item{[CGG]} {\sl J.~B.~Conrey, A.~Ghosh}, and {\sl S.~M.~Gonek}, Mean values of the Riemann zeta-function with application to distribution of zeros, Number Theory, Trace Formulas and Discrete Groups, (1989), 185-199. \item{[L]} {\sl N.~Levinson}, More than one-third of the zeros of Riemann's zeta-function are on $\sigma=\half$, Adv.~in Math., {\bf 13} (1974), 383-436. } \vskip .3in {\parskip 0pt \parindent 0pt Mathematical Sciences Research Institute 1000 Centennial Drive Berkeley, CA \hskip .1in 94720 farmer@msri.org \vfill} \bye