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\headline={\ifnum\pageno>1 {\smcp the electronic journal of
combinatorics 2 (1995), \#R1\hfill\folio} \fi}


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\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:
$$\eqalign{
N^{(j)}(T)&=
\sum_{\rho^{(j)}=\beta+i\gamma}^{\phantom{o}}1\phantom{XXXXXXX}
\hbox{zeros of $\xi^{(j)}(\sigma+it)$ with $0<t<T$}\cr
\phantom{\sum_{\gamma}}	N(T)&=N^{(0)}(T) \phantom{XXXXXXXX}
\hbox{zeros of $\xi(\sigma+it)$ with $0<t<T$}	\cr
        N^{(j)}_s(T)&= 
		\sum_{\rho^{(j)}=\beta+i\gamma\atop m(\gamma)=1}1
\phantom{XXXXXXX}\,
\hbox{simple zeros of $\xi^{(j)}(\sigma+it)$ with $0<t<T$}       \cr
        N^{(j)}_{s,\half}(T)&= 
	\sum_{\rho^{(j)}=\half+i\gamma\atop m(\gamma)=1}^{\phantom{o}}1
\phantom{XXXXXXX}
\hbox{simple zeros of $\xi^{(j)}(\thalf+it)$ with $0<t<T$}
\cr
        M_r(T)&= 
	\sum_{\rho^{(0)}=\beta+i\gamma\atop m(\gamma)=r}^{\phantom{o}}1
\phantom{XXXXXXX}\,
\hbox{zeros of $\xi(\sigma+it)$ of multiplicity $r$ with $0<t<T$}
\cr
        M_{\le r}(T)&= 
	\sum_{\rho^{(0)}=\beta+i\gamma\atop m(\gamma)\le r}^{\phantom{o}}1
\phantom{XXXXXXX}\,
\hbox{zeros of $\xi(\sigma+it)$ of multiplicity $\le r$ with $0<t<T$}
}$$
where all sums are over $0<\gamma<T$, and zeros are counted according 
to their multiplicity.
It is well known that $N^{(j)}(T) \sim {1\over 2\pi} T\log T$.  Let
$$
        \alpha_j =\liminf_{T\to\infty}{N^{(j)}_{s,\half}(T)\over N^{(j)}(T)}.
\phantom{XXXXXXXXXXXXX}
        \beta_j =\liminf_{T\to\infty}{N^{(j)}_{s}(T)\over N^{(j)}(T)}.
$$
Thus, $\beta_j$ is the proportion of zeros of $\xi^{(j)}(s)$ 
which are simple,
and $\alpha_j$ is the proportion 
which are simple
and on the critical line.  
The best currently available bounds are
$\alpha_0>0.40219$, $\alpha_1>0.79874$,
$\alpha_2> 0.93469$, $\alpha_3>0.9673$, 
$\alpha_4>0.98006$, 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 $0<t<T$. That is,
$$
	N_d(T) = \sum_{n=1}^\infty {M_n(T)\over n}.
\eqno (1)
$$
It is conjectured that all of the zeros of $\xi(s)$ are distinct: 
$N_d(T)=N(T)$, or equivalently, all of the zeros are simple:
$N_s^{(0)}(T)=N(T)$.  From the bound on $\alpha_0$ we have
$N_s^{(0)}(T) > \kappa\, N(T)$, 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.

}

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Mathematical Sciences Research Institute

1000 Centennial Drive

Berkeley, CA  \hskip .1in  94720

farmer@msri.org

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