ZEROS OF FEKETE POLYNOMIALS
Brian Conrey, Andrew Granville, Bjorn Poonen and K. Soundararajan
1. Introduction
Dirichlet noted that, from the formula
?(s) = ns
Z1
0
xs?1e?nx dx = ns
Z1
0
(? log t)s?1 tn?1 dt;
we may obtain the identity
(1.1)
X n n?1
X ?np Z 1
s
?
1
t dt
?(s)L(s; ( p )) = ?(s)
s = 0 (? log t)
n1 p
n1 n
Z 1 (? log t)s?1 fp(t)
=
t
1 ? tp dt:
0
?
Here p is the Legendre symbol and
(1.2)
fp(t) :=
p?1 X
a a
a=0
p t:
?
Equation (1.1) allowed Dirichlet to dene L(s; p ) as a regular function for all complex
?
s. Fekete observed that if fp (t) has no real zeros t with 0 < t < 1, then L(s; p ) has
no real?zeros
s > 0; and the fp(t) are thus now known as Fekete polynomials. Indeed,
if L(s; p ) = 0 then by (1.1) and the mean value theorem there is a t in (0; 1) with
(? log t)s?1 fp (t) = 0, and so f (t) = 0 here.
p
t
1?tp
Among small primes p, there are only a few for which the Fekete polynomial fp (t) has
a real zero t in the range 0 < t < 1. In fact, we may verify computationally that there
are just 23 primes up to 1000?for
which fp has a zero in (0; 1). This implies that there are
no positive real zeros of L(s; p ) for most such primes p, and in particular no Siegel zeros
(that is, real zeros \especially close to 1"). It is interesting to note that for thosepprimes
p 3 mod 4 for which fp(t) does have a zero in (0; 1), the class number of Q( ?p) is
surprisingly small (for example p = 43; 67; 163; : : : ). Unfortunately this trend does not
persist: Indeed Baker and Montgomery [1] proved that fp(t) has a large number of zeros
in (0; 1) for almost all primes p (that is, the number of such zeros ! 1 as p ! 1, and it
seems likely that there are, in fact, log log p such zeros).
The rst three authors are all supported, in part, by grants from the N.S.F. The rst and fourth
authors are supported by the American Institute of Mathematics. The second author is a Presidential
Faculty Fellow. The third author is an Alfred P. Sloan Research Fellow and a Packard Fellow.
Typeset by A S-T X
2
CONREY ET. AL.
In this paper we shall study the complex zeros of fp (t). Using zero locating software one
nds that, for primes p up to 1000, about half of the zeros lie on the unit circle; leading
one to expect this to be the general phenomenon. It turns out to be fairly easy to prove
that at least half of the zeros of fp (t) are on the unit circle (that is jtj = 1): First note
that
?1
(pX
?1)=2 a
?
p=
2
a
?
p=
2
p=
2
?
a
Fp (z) := z fp (z) =
+ p z
p z
a=1
by combining the a and p ? a terms1. Taking z = e2it we have
8 P(p?1)=2 a < 2 a=1 p cos((2a ? p)t) if p 1 mod 4
?
2
it
(1.3)
Fp e
= : P(p?1)=2 a sin((2a ? p)t) if p 3 mod 4:
2i a=1
p
Dene Hp(t) = Fp (e2it) if p 1 (mod 4), and Hp (t) = ?iFp (e2it ) if p 3 (mod 4). By
(1.3) we see that Hp(t) is a periodic, continuous real-valued function when t is real.
Now if p =pe2i=p then, for all k not divisible by p, fp (p
pk ) is a Gauss sum and has
absolute value p (see section 2 of [2]); therefore jFp (pk )j = p. Moreover
k X
k p?1 p?1 X
a
ak
k
k
?
p=
2
ak
k
ak
k
Fp (p ) = (p )
p = (?1) p
p = (?1) p Fp (p):
a=1 p
a=1 p
? ? then H (k=p) and H ((k + 1)=p) have dierent signs. Since H (t)
Therefore if kp = k+1
p
p
p
p
is real-valued and continuous, it must have a zero in-between k=p and (k + 1)=p, by the
intermediate value theorem. Thus the number of zeros of Hp (t) in [0; 1) (and so of Fp (z)
on the unit circle) is
k k + 1 p ? 3
= 2 ;
# k : 1 k p ? 2 and p = p
as we shall see in Lemma 2.
Other than possible zeros at z = ?1 and at z = 1, this accounts for all the zeros on the
unit circle for each prime p < 500. So the question is, is this all, for all p? The answer is
\no" and indeed one nds more zeros when p = 661. In general one has the following:
Theorem 1. There exists a constant 0 , 1 > 0 > 21 such that
#fz : jzj = 1 and fp (z) = 0g 0 p as p ! 1:
We determine 0 in terms of another constant 1 dened as follows:
Theorem 2. Let FJ be the set of rational functions
X j
g(x) = x1 + 1 ?1 x +
jj j<J x + j
j 6=0;?1
where we allow each j to take value +1 or ?1. There exists a constant 1 ; 21 > 1 > 0,
such that
#fg 2 FJ : g(x) = 0 for some x 2 (0; 1)g 1 #fg 2 FJ g
as J ! 1.
The constants 0 and 1 are related as follows:
1 Here z = e2it with 0 t < 1, so that there is no ambiguity in the meaning of z ?p=2 .
ZEROS OF FEKETE POLYNOMIALS
3
Theorem 1 12 . In fact 0 = 12 + 1 .
It is still an open question to determine the
p value of 0 . It is known that a \random"
trigonometricppolynomial of degree p has p= 3 zeros in [0; 1) (see [7]), so one might guess
that 0 = 1= 3 0:5773 : : : . However this is not the case. We will show
0:500813 > 0 > 0:500668:
While it is theoretically easy to nd the value of 0 , we do not know a good practical way
of achieving this.
As well as determining precisely the proportion, 0 , of the zeros of fp(t) which lie on
the unit circle, we would also like to understand the distribution of the set of zeros in the
complex plane. There are several easy remarks to make: By (1.2) we have
?1 p
t fp (1=t) = p fp (t)
and so the zeros of fp (t), other than t = 0, are symmetric about the unit circle (i.e. they
come in pairs other than at t = 0; 1). We also note that, for jtj > 1,
X
p?2
p?1 X
1 > 1? 1 :
1
a
p
?
1
1
?
jfp (t)=t j = p
?
1
?
a
p
?
1?a
jtj ? 1
a=0 jtj
a=0 p t
However if jtj 2 then 1 ? 1=(jtj? 1) 0, and so fp (t) has no zeros in jtj 2. By symmetry
it has no zeros in jtj 21 except 0. Thus
Proposition 1. The zeros of fp (t), other than at 0; 1 and ?1 come in pairs ; 1=.
Moreover, other than 0, they all lie in the annulus fr 2 : 21 < jrj < 2g.
As for the distribution of the arguments of the roots of fp(t) we can use a beautiful
result of Erd}os and Turan (Theorem 1 of [3]), which immediately implies that, for any
0 < < 1,
p
(1.4)
#f 2 : fp ( ) = 0; < arg( )=2 < g = ( ? )p + O( p log p):
The arguments above, and those used in proving Theorems 1 and 2, focus on determining
which arcs (pK ; pK +1) of the unit circle contain a zero of fp (t). Evidently (1.4) cannot be
used so precisely. However we can show that there are zeros of fp(t) near to such an arc,
so long as fp (t) gets \small" on that arc.
Theorem 3. Suppose that > 0 is a suciently small constant. If p is a suciently large
prime and K an integer such thatpthere exists a value of t on the unit circle in the arc
from pK to pK +1 with jfp(t)j < p, then there exists = rpK + with fp ( ) = 0 where
0 < < 1 and 1 ? 1=3 =p < r < 1.
Remark. Applying Proposition 1 we also have fp ((1=r)?pK + ) = 0.
P
As we have already discussed, Gauss sums ap?=11 ap pak (and many generalizations)
have the surprising property that they have absolute value exactly equal to pp. It is, we
think, of interest to ask what happens when we replace the primitive pth root of unity
pk in the expression for a Gauss sum above, by some primitive 2pth root of unity. These
may be written as pk+1=2 or 22pk+1, or ?pk ; so we must consider the values of fp(?pk ). Do
these all take on the same absolute value ? The answer we now see is \no", as we evaluate
the distribution of these absolute values:
C
C
4
CONREY ET. AL.
Theorem 4. For any xed real number # k : 1 k p such that Hp
as p ! 1 where
k + 1=2 p
p
< p c p
Z1
Y 2 2x dx
1
1
c = 2 + sin(x)
cos n x :
x=0
n1
n odd
Moreover c? and 1 ? c = exp(? exp(=2 + O(1))) for positive .
After proving this in section 6, we indicate how our proof may be modied to establish several related results. First, to show that maxjzj=1 jfp (z)j pp log log p, so
re-establishing a result of Montgomery [5]. Second to understand the distribution of the
values of the Fekete polynomial at (p ? 1)st roots of unity.
Acknowledgements: We thank Je Lagarias for facilitating this joint endeavour; and
Peter Borwein, Neil Dummigan, Hugh Montgomery, Pieter Moree, Mike Mossingho, Bob
Vaughan and Trevor Wooley for some helpful remarks.
2. First results
Let be any character (mod p) and let k be an integer not divisible by p. Note that
(2.1)
p?1
X
a=1
(a)pak = (k)
?
p?1
X
a=1
(ak)pak = (k)
p?1
X
b=1
(b)pb:
In particular we see that fp (pk ) = kp fp (p), whereas in contrast fp (1) = 0. Recall that
Pp?1 (a) a. Thus f ( )
for a non-principal character
(mod
q
),
the
Gauss
sum
(
)
is
p p
p
a=1
?
is the Gauss sum ( p ). It is easy to determine the magnitude of jfp (p)j: Note that
p?1 p?1 X
p?1
X
X
ab (a+b)k
k
2
2
(p ? 1)fp (p) = fp (p ) =
p
k=0 a;b=0 p
k=0
?1
p?1 X
p?1
p?1 X
X
ab
ab
(
a
+
b
)
k
=
= p p (p ? 1):
p
=p
p
p
a;b=1
k=0
b=a=1
p?a
? Hence we have fp(p )2 = ?p1 p, and so jfp(p )j = pp. Gauss showed more and determined
that
pp if p 1 (mod 4);
fp (p) = p
i p if p 3 (mod 4):
?
Since fp (pk ) = kp fp (p), for 1 k p ? 1, and fp (1) = 0, we get by Lagrangian
interpolation
p?1
pY
?1 z ? j X
p :
k
fp (z) = fp (p )
j
k
j =0 p ? p
k=0
j 6=k
ZEROS OF FEKETE POLYNOMIALS
Note that
pY
?1
j =1
j 6=k
(z ? pj ) =
zp ? 1 ;
z ? pk
5
pY
?1
pY
?1
k
j
k
(
p
?
1)
(1 ? pj ) = pp?k :
(p ? p ) = p
j =1
j =1
j 6=k
and that
Hence
p?1 k
? p2 fp (z ) X
k
p
f
(
z
)
z
p
p :
p
=
(2.2)
p =
p
p
?
fp (p) z ? 1 fp (p) z 2 ? z 2 k=1 p z ? pk
? If jzj = 1 then note that z p2 ? z? p2 2 i , and from (1.3) and fp (p)2 = ?p1 p we have
z? p2 fp (z)=fp(p ) 2 . Thus the right side of (2.2) 2 i for all jzj = 1. To facilitate studying
fp (z) as z goes around the unit circle from pK to pK +1 , we write z = pK +x = pK e2ix=p
and then let
R
R
(2.3)
R
K p f (z) gp;K (x) := i p f ( ) zpp ? 1 p p
z=
K K+(X?2 1 ) k
=i
p
K +x
p
p k=K ?( p?1 ) p
2
1
:
pK ?k+x ? 1
Thus gp;K (x) is a real valued function of x 2 [0; 1].
? ? Proposition 2. If 0 K p ? 1 is an integer with Kp = Kp+1 then gp;K (x) has exactly
one zero in (0; 1). Equivalently,
fp(z) has exactly one zero on the arc of the unit circle
?
?
K
K
+1
K
K
+1
from p to p . If p = ? p then gp;K has either no zeros, or exactly two zeros in
(0; 1). Equivalently, fp (z) has exactly 0 or 2 zeros on the arc from pK to pK +1 .
Remark. In the above Proposition, and henceforth,
P we count zeros with multiplicity.
Before proving the Proposition, we evaluate kp?=11 sin21( kp ) .
Lemma 1. For all integers p 2,
p?1
X
1 = p2 ? 1 :
2 k
3
k=1 sin ( p )
Q
Proof. Put A(z) = kp?=11 (z ? pk ). Logarithmic dierentiation shows that
(
)
p?1 k
p?1
0 (z )0 A0 (z ) X
X
A
1 :
z A(z) + A(z) = ? (1 ?p k )2 = 14
2
k
p
z=1
k=1
k=1 sin ( p )
However, A(z) = zzp??11 = zp?1 + zp?2 + : : : + 1 and using this to evaluate the left side
above, we get the lemma.
Proof of Proposition
2. Note
that with g = gp;K , we have limx!0+ g(x) = 1, and
?
?
K
+1
K
limx!1? g(x) = ? p p 1. Further observe that
K X k K?k+x
2
p
0
g (x) = p p
K
?
p
(p k+x ? 1)2
jk?K j< p2
K X k 1
:
= ? 2p p
2
jk?K j< p p sin ( p (K ? k + x))
2
6
CONREY ET. AL.
? ? If Kp = Kp+1 then, by Lemma 1,
1
X
1
1
0
jg (x)j 2p sin2 ( x) + sin2 ( (1 ? x)) ?
2 p
p
j 6=0;1 sin ( p (x ? j ))
jj j<p=2
2
2 ? 1
p
2p sin2 ( ) ? 3 > 0;
2p
(2.4)
since the sum of the rst two terms is minimized when x = 21 . Hence g0(x) 6= 0 for all
x 2 (0; 1), so that g is monotone decreasing in [0; 1] going from 1 to ?1. Thus g has
exactly one zero in this interval.
Moreover
2 K X K cos( p (K ? k + x))
00
g (x) = p2 p
:
3 jk?K j<p=2 p sin ( p (K ? k + x))
? ? Now if Kp = ? Kp+1 then
2 cos( p x)
X cos( p (j ? x)) cos( p (1 ? x))
00
jg (x)j p2 sin3 ( x) + sin3 ( (1 ? x)) ?
3 :
p
p
jj j<p=2 j sin( p (j ? x))j
j=
6 0;1
Let be the minimum of cot( p t) over t = x, 1 ? x. Since cot t decreases rapidly as t goes
from 0 to we see that the above is
2
2
X
1
1
?
> 0;
p2 sin21( x) + sin2 ( (1
2
p
p ? x)) j 6=0;1 sin ( p (x ? j ))
jj j<p=2
as in (2.4). Thus g0 (x) is monotone increasing in (0; 1) going from ?1 to +1. Thus there
is a unique x0 in (0; 1) with g0(x0 ) = 0, and the minimum value of g(x) is attained at x0.
Plainly g has 0 or 2 zeros depending on whether g(x0) > 0, or g(x0) 0. This proves the
proposition.
>From Proposition 2 we know
that? fp (z) has at least as many zeros on jzj = 1, as there
?
K
are values 1 K p ? 1 with p = Kp+1 . We next determine the number of such values
K.
Lemma 2 (Gauss). For any non-principal character (mod p), we have
(2.5)
Hence
and
p?1
X
b=1
p ? 1 if pjk
(b)(b + k) =
?1
if p k:
-
b b + 1 p ? 3
= 2 ;
# b (mod p) : p = p
b b + 1 p ? 1
# b (mod p) : p = ?
p
= 2 :
ZEROS OF FEKETE POLYNOMIALS
7
P
?1 j(b)j2 = p ? 1. Suppose now that p - k,
Proof. If pjk then the right side of (2.5) is bp=1
and let c = (b + k)=b = 1 + k=b. As b runs over the non-zero residue classes (mod q), note
that c runs over all residue classes except the residue class 1 (mod p). Hence the right side
of (2.5) is
X
(c) = ?1;
c (mod p)
c61 (mod p)
as desired.
? ? If Kp = ? Kp+1 then we need to determine (in the notation of the? proof of Proposition
2) whether g(x0) > 0 or 0. This depends heavily on the values of kp for k neighbouring
K . The following Lemma shows that these neighbouring values behave like independent
random variables.
Lemma 3 (Weil). Fix j 2 f?1; 1g for each j , jj j < J . Uniformly for all positive integers
J,
p
x ? j # x (mod p) : p = j for all jj j < J = 22J ?1 + O(J pp):
Proof. The above equals
p
X
1 Y 1 + x ? j + O(J )
j
2J ?1
p
x=1 2
jj j<J
1 X X
p Q (x ? j )
p
j
2
S
=
+J :
+O
22J ?1
22J ?3 Sfjjj<J g x=1
p
S 6=;
By Weil's Theorem [8], if f (x) is a squarefree polynomial (mod p) then
X
p f (x) (degree f )pp:
x=1 p Hence the above is
!
pp 2X
J ?3
p
2J ? 3 m + J ;
= 22J ?3 + O 22J ?3
m
m=1
and the result follows.
We conclude this section by determining the order of the zeros of fp (z) at 1. In fact
p+1
p?1
we shall determine the number of zeros of fp (z) on the arcs p 2 to p 2 (which contains
?1), and p?1 to p (which contains 1).
Lemmap?14. If pp+1 1 (mod 4) then fp (z) has only a simple zero at z = ?1, on the arc
from p 2 to p 2 , and fp (z) has only a double zero at z = 1, on the arc from p?1 to p.
8
CONREY ET. AL.
?1
p+1
p
If p 3 (mod 4) then there are no zeros of fp(z) on the arc from p 2 to p 2 , and fp (z)
has only a simple zero at z = 1 on the arc from p?1 to p .
? Proof. We make free use of the fact that ?p1 = 1, or ?1 depending on whether p 1
p+1
p?1
2 to p 2 . We take K = p?1
(mod 4), or 3 (mod 4). Let's? begin
with
the
arc
from
p
2
? ? ? in Proposition 2. Note that Kp = Kp+1 if p 1 (mod 4), and Kp = ? Kp+1 if p 3
(mod 4). In the rst case, Proposition
2? tells
usPthat there's? exactly
one
(simple) zero
?
P
p
?
1
p
?
1
p
?
a
a
a
1
a
a
on this arc. Since fp(?1) = a=1 (?1) p = 2 a=1 (?1) ( p ? p ) = 0 for p 1
(mod 4), this simple zero is at ?1. Now suppose p 3 (mod 4). By Proposition 2, we
know that there are 0 or 2 zeros on this arc, depending on whether minx gp;K (x) > 0 or
not. We now show that this minimum is attained at x = 21 , and the minimum value is
positive. Putting j = K ? k in (2.3) we have
K X K ? j 1
gp;K (x) = i
p jjj p?1
2
p
pj+x ? 1
?1 K X
2 K ? j 1
=i
p
p j=0
p
pj+x ? 1
1
;
? ?j?1+x
p
?1
since K + j + 1 ?(K ? j ) (mod q). Evidently gp;K (1 ? x) = gp;K (x), so gp;K (1 ? x) =
gp;K (x) since gp;K (x) is real-valued. However we see that the minimum of gp;K (x) is
obtained at a unique point in (0; 1), so that must be at x = 21 . Now
fp (?1) =
p?1
X
a=1
(?1)a
p?1 X
p?1 X
a
p
?
b
p = a=1 p ? b=1 p
a
a even
b even
where a = p ? b is odd in the second sum,
=2
(pX
?1)=2 d=1
?1)=2 d 2 2d = 2 2 (pX
p
p d=1 p = 2 2 p ? 1 h(?p);
where h(?pp) is the class number of (p ?p) (see section 2 of [2]). By (2.3), and since
fp (p) = i p by Gauss, we have
Q
2K K p
p
2 fp (?1) = p ?2 p + p h(?p)
K p
= p 2 + p h(?p) > 0:
gp;K ( 12 ) = ?
K pp
?1
p
p+1
This shows that fp (z) has no zeros on the arc from p 2 to p 2 when p 3 (mod 4).
Now let's consider the arc from p?1 to p . Take K = p ? 1, and consider gp;K
(x) as
?
K
in (2.3). Usually gp;K (x) would have a discontinuity at 1, but here since p+1 =
?dened
0 = 0 we do not have this problem. Thus gp;K is a continuous function on (0; 2), and
p
we may study fp (z) on the arc from p?1 to p by studying gp;K (x) on (0; 2). Note that
ZEROS OF FEKETE POLYNOMIALS
P ?
9
for any p, fp (1) = ap?=11 ap = 0, so that there is at least a simple zero at z = 1. Also
fp0 (1) = ?i(?1=p)fp (p)gp;p?1(1) by (2.3). Since fp (z) = (?1=p)zp fp (z), we deduce that
gp;p?1(x) = ?(?1=p)gp;p?1(2 ? x).
If p 1 (mod 4) then gp;p?1(1) = 0 and so fp0 (1) = 0. Now, as in the proof of (2.4), the
0 (x)j > 0 for all x 2 (0; 2). Therefore
rst part of the proof of Proposition 2, we have jgp;K
g has only a simple zero at x = 1, and thus fp has a double zero at 1.
00 (x)j > 0
If p 3 (mod 4) then, as in the second part of the proof of Proposition 2, jgp;K
for x 2 (0; 2). Thus there is a unique minimum of gp;K (x) on (0; 2), but since gp;p?1p(x) =
gp;p?1(2 ? x) this must be attained at x = 1. However, by (2.3), and as fp(p ) = i p by
Gauss,
p?1 X
fp0 (1)
1
gp;K (1) = ? pp = ? pp a ap = pph(?p) > 0;
a=1
(see [2], section 2), and so gp;K (x) > 0 and thus has no zeros in (0; 2). Therefore fp has
only a simple zero at z = 1 on this arc.
3. Functions with random coefficients
If g 2 FJ then, for any x 2 (0; 1), we have
(3.1)
j
1 g00 (x) = 1 + 1 + X
3
3
2
x (1 ? x) jjjJ (x + j )3
j 6=0;?1
X 1
1 ? 2 (3) > 0:
x13 + (1 ?1 x)3 ?
>
2
1
3
( 2 )3
jj jJ (x + j )
j 6=0;?1
Since t!
lim0+ g0(t) = ?1 and t!
lim1? g0(t) = 1 we deduce that g0(x) has exactly one zero in
(0; 1), call it x0. Note that g(x) attains its minimum value at x0. If 0 t < 1 then
1
1
1
1
2 > 0:
?
2
+
+
+
:
:
:
=
?
2
2
2
2
2
t
(1=2) (3=2) (5=2)
t
?g0 (t) 1
Similarly if 1 ? 1 < t 1 then g0(t) > 0. Thus
(3.2)
x0 2 [ 1 ; 1 ? 1 ]:
We now show that few g are small in absolute value, at their minimum x0 .
Proposition 3. We have jg(x0)j > J ? 14 for almost all g 2 FJ , where g0(x0) = 0, uniformly
as J ! 1.
Proof. Consider the subset S of FJ with all the j xed given values, except when j 2
[I; I + I 21 ] where I = J 41 . Let f 2 S with j = ?1 for all j 2 [I; I + I 21 ]. Suppose that
f 0(x1 ) = 0 and let
X j
=
x +j
jj jJ 1 1
j 2= [I;I +I 2 ]
10
CONREY ET. AL.
where 0 = 1, ?1 = ?1. Let g be any element of S with g0(x0) = 0.
By (3.1) note that
Z x1
00
jx1 ? x0j f (t)dt = jf 0(x0 ) ? f 0 (x1)j = jf 0(x0 )j
x0
X
1
1
0
0
= jf (x0) ? g (x0)j 2
(3.3)
Hence, keeping in mind x0 , x1 2 [ 1 ; 1 ? 1 ],
g(x0) ? =
X
j 2[I;I +I 21 ]
j + O
x0 + j
(x0 + j )
j 2[I;I +I 21 ]
2
I:
X 1
1
x0 + j ? x1 + j jj jJ
j 2= [I;I +I 21 ]
X 1 1 ?
j + O
x0 + j + jx1 ? x0 j
12
12 I
j 2[I;I +I ]
j 2[I;I +I ]
X
1
1
= I1
=I
X
j 2[I;I +I 12 ]
j + O I ;
since each j1=I ? 1=(x0 + j )j 1=I 23 and there are I 12 such terms. Therefore if jg(x0)j I1
then
X
(3.4)
j = ?I + O(1):
j 2[I;I +I 21 ]
Now, the j are independent binomial random variables, so the distribution of their sum
tends towards the normal distribution. Therefore the maximum probability
for (3.4) to
1
?
4
occur happens when = 0; and so (3.4) holds with probability O(I ), for any , implying
Proposition 3.
4. Proof of Theorem 2
Suppose that g 2 FJ and f 2 FK , with J < K , such that the j are the same in each for
jj j < J . Select x0; x1 2 (0; 1) so that g0(x0) = 0 and f 0 (x1) = 0. Now
X jx ? x j
X 1
1
1
0 jx ? x j;
jf (x ) ? f (x )j ?
1
0
jj j<K x1 + j
x0 + j 2
jj j<K j + 1
1
0
since x0 , x1 2 [1=; 1 ? 1=]. Arguing exactly as in (3.3), we see that jx0 ? x1 j J1 , and
so we have
(4.1)
jf (x1) ? f (x0)j J1 :
We next consider the mean-square of
X j :
jf (x0) ? g(x0)j = J jj j<K x0 + j
ZEROS OF FEKETE POLYNOMIALS
This is
11
X X j 2
22K ?2J 2f?1;1gJ jjj<K x0 + j J jij<K
X
1
i
=
=
X
1
1
j1 j2
2K ?2J
J jj1 j;jj2 j<K (x0 + j1 )(x0 + j2 ) 2
i 2f?1;1g
X
1 1:
2 J
J jj j<K (x0 + j )
Thus if J ! 1 as J ! 1 then
(4.2)
J jij<K
X j < J;
J 12
x
+
J jj j<K 0 j
for almost all choices of the j .
Combining (4.1) and (4.2), we see that for almost all choices of j (J jj j < K ) we
have
(4.3)
jf (x1) ? g(x0)j jf (x1) ? f (x0)j + jf (x0) ? g(x0)j < 2 21J :
J
Taking J = J 14 =2, and combining this with Proposition 3 we see that for almost all
g 2 FJ , and almost all extensions f of g to FK , f (x1) has the same sign as g(x0). Thus
Theorem 2 follows.
5. Proofs of Theorems 1 and 1 21
? ? Let 1 K p ? 1 be an integer. If Kp = Kp+1 then by Proposition 2 there is exactly
one zero of fp (z) on the arc
from ?pK to pK +1 ; by Lemma 2 this happens for p2 values
?
of K . Suppose now that Kp = ? Kp+1 so that fp (z) has either 0 or 2 zeros on the arc
from pK to pK +1 depending on whether minx2(0;1) gp;K (x) is positive or not. To decide
this question we need the following proposition:
Proposition 4. Suppose J pp, and J ! 1 as p ! 1. For almost all 1 K p ? 1
we have
K X K ? j 1
p
p
+O 1 ;
gp;K (x) = 2 p
p
j
+
x
J3
jj j<J
uniformly for all x 2 (0; 1).
Proof. Note that for J jj j < p2 ,
1
x?1
1
p
px p2 ;
j+x ? j = j+x
j
p ? 1 p ? 1
(p ? 1)(p ? 1) j (j + x) j
and, for jj j < J ,
1
p
1
=
+ O(1):
pj+x ? 1 2i (j + x)
12
CONREY ET. AL.
Hence, putting j = K ? k in (2.3), we have
K X K ? j 1
gp;K (x) = i p
p pj+x ? 1
jj j< p2
K X K ? j 1 K X K ? j 1
p
p
= 2 p
+
O
J+J :
+
i
j ?1
p
j
+
x
p
p
p
p
jj j<J
J jj j<
2
We now show that the mean-square of the second term above is small, which proves the
Proposition. By Lemma 2,
p X X
K ? j 1 2
p pj ? 1 K =1 J jj j< p2
p
X
K ? j1
1
=
j
?
j
1
2
p
J jj1 j; jj2 j< p2 (p ? 1)(p ? 1) K =1
X 1 2
X
1
? =p
j
j
2
J jj j< p2 p ? 1
J jj j< p2 jp ? 1j
X
K ? j p
2
X p 2 X p 2 p3 2 2
p
+
J + p log p:
j
j
J jj j<p=2
J jj j<p=2
This proves the Proposition.
? ? By Proposition 4 we know that for almost all K with Kp = ? Kp+1 the minimum value
? P ? of 2p gp;K (x) equals the minimum of Kp jjj<J Kp?j j+1 x + O(J ? 31 ).? For
such K? the min
P
imum value of gp;K (x) is non-positive if and only if the minimum of Kp jjj<J Kp?j j+1 x
is non-positive, unless
K X K ? j 1
1:
(5.1)
p jjj<J p j + x J 13
Now choose J = [ log10p ]. Given any choice of j 2 f?1; 1g, 0 <? jj?j < Jwith 0 = 1, and
?1 = ?1, by Lemma 3 there are p=22J ?2 values of K with Kp Kp?j = j for each j .
Therefore (5.1) fails, for almost all K , by Proposition
2. Appealing
now to Theorem 2 we
?
?
K
+1
K
have proved that for 1 p=2 values of K with p = ? p , the minimum of gp;K (x)
is < 0. For such K , fp(z) has two zeros on the arc from pK to pK +1 , so that the total
number of such zeros is 1 p. Theorems 1 and 1 21 follow.
6. Pseudo-Gauss Sums: Proof of the first part of Theorem 4
1
In this section,pwe wish to study the distribution of fp (pK + 2 ). By (2.3) and Proposition
4 we have (if ( p >)J ! 1 as p ! 1) for almost all 1 K p ? 1,
X K ? j 1
1 if
(
)
K
+ 21
p
p
fp(p ) = 1 + O J 31
p
j
+
K ? j 1 2 1 pp Xjjj<J
(6.1)
;
= 1 + O J 31
p
j
+
2
jj j<J
ZEROS OF FEKETE POLYNOMIALS
13
where = 1 or i is xed. Thus, by Lemma 3, we have that for any xed real number K + 1 p 1
2 < p
lim # K : 1 K p and Hp
p!1 p
p
exists and equals
X j
(6.2)
lim Prob j 2 f?1; 1g : jj j < J :
1 < :
J !1
j
+
2
jj j<J
R
One may obtain an expression for this probability as follows: Recall that 01 siny y dy = 2 ,
and so for any k 6= 0
2 Z 1 sin(kx) dx = sgn(k) 2 Z 1 sin(jkjx) dx = sgn(k) 2 Z 1 sin y dy = sgn(k);
0
x
0
x
0 y
where sgn(k) is the sign of k (= 1 if k > 0 and ?1 if k < 0). Hence the probability (6.2)
equals
dx X 1 1 Z 1 X j
1
22J ?1 i 2f?1;1g 2 ? 0 sin jJ j<J j + 12 ? x x
jij<J
P
Z
ix
1
X
1
e jj
= 12 ? 1
22J ?1
j
j <J j +1=2
0
i 2f?1;1g
jij<J
?
? e?ix
P
2i
jjj<J j+1=2 ?
j
dx
x
Z 1 Y e j+1=2 + e? j+1=2 e?ix ? eix dx
1
1
=2?
2
2i
x
0 jj j<J
Z1
Y 2x dx
1
1
=2+
sin(x)
cos 2j + 1 x :
x=0
jj j<J
ix
Letting J ! 1, we get
Z1
1
1
sin(x)C (x) dx
c = 2 + x
0
ix
where
C (x) :=
Y
n1
n odd
cos2
2x n ;
and thus Theorem 4 is proved. Note that this integral does converge: For any x > 0 we
have
C (x) 13x
2
since this estimate is trivial for x 1, and otherwise we note that j cos( 2nx )j < 21 if
3x= < n < 6x=. Thus the part of the integral with x 1 is easily bounded. Since
sin(x) x, the portion of the integral from 0 to 1 is also easily bounded.
1
Remark 1. We use the above to study the multiplicative average size of fp (pk+ 2 ). Due to
the symmetry of c we have that
?1 f (pk+ 21 ) Z 1
1 log pY
pp
1 ):
=
2
log
d(
c
?
2
p?1
p
0
k=1
14
CONREY ET. AL.
Using our expression for c one can show that this is
= + log ?
Z 1 C (x) ? 1
Z 1 C (x)
x dx ? 1 x dx:
All of these integrals converge, though we do not know their exact values.
Remark 2. The expansion given in (6.1) for fp , and the general technique involved, is very
similar to that used by Montgomery [5] in showing that,
1) jfp(z)j pp log p for all jzj = 1.
2) If p is suciently
large then there exists some value of z with jzj = 1 for which
p
2
jfp(z)j > p log log p.
Indeed to prove a result like that in (2) we note that we may select each j equal to the sign
of j for jj j < J = " log p. By Lemma 3 there are many such K and we proceed as before
with the expansion in (6.1), but now taking a little more care over the set of excluded K .
Remark 3. Fix t 2 (0; 1). By the argument above, we have, for any xed real number ,
0
1
K + t < pp
c;t : = plim
!1 p # K : 1 K p and Hp
p
X j
< sin(t)
= Jlim
Prob j 2 f?1; 1g : jj j < J :
!1
j
+
t
jj j<J
dx
Z
1
Y
1
x
1
x
sin sin(t)
=2+
cos j + t x :
x=0
j 2Z
Remark 4. We can also use these techniques to investigate the distribution of values of
Hp (t) at t = a=(p ? 1) for 1 a p ? 1. We note that if K p then pK?1 = pK +f1 +
we
can get an expression similar to (6.1) for almost all Fp (pK?1), but now
o( p1 )g. Therefore
?
P
with jjj<J Kp?j j+1 replacing the sum in (6.1), and multiplying the whole expression
through by sin(). Thus the density of those K , for which Hp (K=(p ? 1)) pp, is
1 + 1 Z 1 Z 1 sin x Y cos
dx d:
x
2 =0 x=0
sin() m2Z
m+ x
We cannot see how to obtain a simpler expression.
It is not hard to modify this technique to determine the distribution of values of the
Fekete polynomial (or, in fact, Hp (t)) at any \reasonably" distributed set of values.
2 FJ as J ! 1.
We now look at the limiting distribution of g( 21 ) ? 4 for g 2 FJ as J ! 1. Dene, for
N 1,
X
7. The distribution of g ( 21 ) for g
SN () =
j ;
1
jj + 1 j>N j + 2
2
where each j = 1 or ?1 with probability 12 . We will prove that the distribution function
of S1 () decays double exponentially.
ZEROS OF FEKETE POLYNOMIALS
15
Theorem 5. As x ! 1, we have Prob(jS1()j > x) = exp(?e x2 +O(1) ).
Proof ofx second part of Theorem 4. Note that Prob(S1() > x) =Prob(S1 () < ?x) =
exp(?e 2 +O(1) ), by symmetry. Taking x = , the result follows from (6.2).
To prove Theorem 5 we study the 2k-th moment of SN (), call it MN (k), that is, the
expectation of SN ()2k . For example
MN (1) =
X
1 :
1 2
jj + 1 j>N (j + 2 )
2
Our aim is to determine the asymptotic behaviour of M1(k) for large k.
Proposition 5. For large k,
M1(k) = (2 log k ? 2 log log k + O(1))2k :
Proof. To establish the lower bound, consider such that j = 1 for all 1 jj + 12 j k= log k; and such that Sk= log k () > 0. The probability of this happening is 1=22k= log k ,
and S1 () 2 log k ? 2 log log k + O(1) for such . Hence
M1(k) 22k=1log k (2 log k ? 2 log log k + O(1))2k = (2 log k ? 2 log log k + O(1))2k :
Now
MN (k) =
X
j1 ;j2 ;:::j2k
E
j1
j2 : : : j2k ;
j1 + 12 j2 + 12 j2k + 21
where stands for the expectation. Observe that a summand above is non-zero only if each
value of j appears an even Q
number of times amongst
j1 ; j2; : : :j2k . In particular j` = j1
Q
for some ` > 1, and then ( 1i2k ji ) = ( 1i2k; i6=1;` ji ). Summing over all 2k ? 1
possibilities for ` in the above, we deduce that
E
E
(7.1)
MN (k) (2k ? 1)
E
X
1 M (k ? 1);
1 )2 N
(
j
+
1
2
jj + j>N
2
for all k 1 and all N 1. Iterating this inequality, we obtain
X
k
1
MN (k) (2k ? 1) (2k ? 3) 3 1 1 2
jj + 1 j>N (j + 2 )
(7.2)
2
k
(2k!2k)!k N ?2 1 = k!(N(2k?)!1 )k :
2
2
Now
jS1 () ? SN ()j 2N ; where N :=
X
1 = log N + O(1):
1
N j + 1 1 j + 2
2
16
CONREY ET. AL.
Evidently the odd moments of SN () are zero. Therefore, by the binomial theorem and
(7.2),
M1 (k) =
k 2k
X
2k?2j )
2j MN (j ) (jS1 () ? SN ()j
E
j =0
k 2k
X
j =0
(2j )! (2 )2k?2j
2j j !(N ? 21 )j N
X
(2N )2k j1!
j =0
k
k2
(N ? 12 )2N
j
(2N
)2k exp
k2
(N ? 21 )2N
:
Taking N = k= log k we obtain the upper bound of the Proposition.
Proof of Theorem 5. Take k = c1 xex=2 + O(1) for some c1 > 0, and then
Prob(jS1()j > x) x?2k M1(k) exp(?c2 ex=2 ) for some constant c2 > 0, if c1 is suciently small, by Proposition 5.
The lower bound is more involved. Select integer k so that 2 log k ? 2 log log k is as
close as possible to x. The contribution to M1 (k) of those with jS1()j < x ? c3 is
(x ? c3 )2k M1(k)=4 if Rc3 is suciently large. The contribution
to M1 (k) of those R
with jS1 ()j > x + c3 is t>x+c3 Prob(jS1()j > t)t2k dt t>x+c3 exp(?c2 et=2 )t2k dt M1(k)=4 if c3 is suciently large, using the upper bound from the paragraph above. Thus
M1(k)=2 Prob(x ? c3 jS1 ()j x + c3)(x + c3 )k which implies that Prob(jS1()j x ? c3 ) M1 (k)=2(x + c3 )k exp(?c4 ex=2 ) for some constant c4 > 0, by Proposition 5.
Replacing x ? c3 by x gives the lower bound and thus our result.
Remark. We follow up on remark 3 of section 6. The arguments above (Theorem 5 and
Proposition 5) hold just as well with \1=2" replaced by any xed t 2 (0; 1). Thus 1 ? c;t
and c?;t = exp(? exp(=2 sin(t) + O(1))) for > 0.
8. Bounds on 1
Applying the method of section 6, we note that for any real ,
(8.1)
: = Jlim
Probfg 2 FJ : g(1=2) < 4g
!1 Z
1
Y 2 x dx
1
1
sin((1 ? )x)
cos 2n x :
=2?
0
n3
n odd
We can use this to obtain numerical bounds on 1 using the following result.
Proposition 6. We have :013496::: 1 0.
Using Simpson's rule to compute the integrals in (8.1) we obtain :000813 > :013496::: 1 0 > :000668, from which we deduce the bounds on 0 in the introduction.
Proof. Again selecting x0 so that g(x0) is minimal, we have, by denition, that
1 = Jlim
Probfg 2 FJ : g(x0) 0g:
!1
ZEROS OF FEKETE POLYNOMIALS
17
Since g(x0) g(1=2) we deduce the lower bound on 1 above.
To get the upper bound, write x0 = 12 + so that j j < 21 . If g(x0) 0 then
g( 21 ) g( 21 ) ? g(x0) = 4 ? x1
0
2
? 1 4? 2 +
1
X
X
j (x0 ? 21 )
? 1?x +
1
0
jj jJ (j + 2 )(j + x0 )
1
j j
1
1
2 )(j + 2
j 6=0;?1
?2
X
+
j j
1
1
2 )(j + 2
+ ) j=?1 (j +
+ )
2
(2j j + 4 2 ) + tan(jvj):
= ? 1 4 2 +
=
?
1 2
1 ? 2
2
4 ?
4
j =1 (j + 2 ) ? 4
j =1 (j +
1
X
2j j
Using Maple to compute the max , we obtain
g( 21 ) max1
j j
2
2)
(2
j
j
+
4
= 0:053986 : : :
tan(jvj) ? 1 2
4 ?
Remark. One can rene the above to get better bounds for 1 . First note that g(x) =
1=x + 1=(1 ? x) is the only element in F1, and in this case x0 = 1=2; thus \1=2" appears in
the denition of . More generally, let J be some positive integer. For each 2 FJ select
0 so that (0) is minimal. We again have g(x0) g(0), so if g(0) 0 then g(x0) 0.
On the other hand, if g(x0) 0 then we can again get an explicit upper bound on g(0)
and proceed as above. This can be used to give another proof that 1 exists.
9. Zeros off the unit circle
Proof of Theorem 3. Theorem 3 holds trivially if there is a zero of fp (t) on the unit circle
in the arc from pK to pK +1 . Thus
we shall henceforth assume that there is no such
zero. Let h(x) := Hp ((K + x)=p) Hp (K=p), so that jh(x)j = jfp (pK +x)=ppj, and h(x)
is a continuous real-valued function. Now the hypothesis implies that h(y) < for some
y 2 (0; 1) (in fact, t = pK +y ), while our assumption above implies that h(x) 6= 0 for all
x 2 (0; 1). By (2.3) we have, uniformly for jxj 2=3,
0
h(x) = sin(x) @
K X
(k=p)
p
sin((x + K ? k)=p)
1jK ?kj<p=2
K X (k=p)
:
= 1 ? (C + O(1))x; where C := ? p
K
?
k
1jK ?kj<p=2
1
sin(x=p) + p
1
A
So if h(y) < for some suciently small y then h(2y) = 2h(y) ? 1+ O(y) < 0, contradicting
our assumption. Therefore we
may assume that y 1, and also 1?y 1 by the symmetric
p
argument. Thus gp;K (y) pjfp (t)j= sin(y) p by (2.3), so that
gp;K (x0 ) gp;K (y) p
where x0 is dened as in section 3.
18
CONREY ET. AL.
Let x1 = x0 ? 1=2 , and x2 = x0 + 1=2 , and then j = pxj for j = 1; 2. Let R = 1 ? 13 =p.
We shall consider the variation in argument of
K p f (z) K X k 1
;
G(z) := i p f ( ) zpp ? 1 = i p
?k
p p
jK ?kj< 2 p zp ? 1
p
as z goes around (in the anti-clockwise direction) the box bounded by the four curves, C1 ,
the arc of the unit circle from 1 to 2, then C2 , the straight line segment from 2 to R2,
then C3 , the arc of the circle of radius R, from R2 to R1, then nally C4 , the straight
line segment from R1 back to 1 .
We know that G(z) is real valued and positive on the arc C1 . We shall show that G(z)
has positive imaginary part on C2 , that G(z) has negative real part on C3 , and that G(z)
has negative imaginary part on C4 , This shows that the change in argument of G(z) is 2
as we go around our box, so that there is exactly one zero in our box. This implies a little
more than Theorem 3.
To estimate H (r; x) := G(rp(K +x)=p ) when R r 1, for a value of x 2 [x1 ; x2], we
calculate the Taylor series expansion around r = 1, which is
2 p
3
2 p 2
(1
?
r
)
1
?
r
(1
?
r
)
00
0
4 :
g
(
x
)
+
i
g
(
x
)
+
O
p
H (r; x) = gp;K (x) ? 2r
p;K
p;K
2
2r 2
r
From the proof of Proposition 2 we have, since x is bounded away from 0 and 1,
0 (x) (x ? x0 )p; and g 00 (x) p:
gp;K (x) = gp;K (x0 ) + O((x ? x0 )2p); gp;K
p;K
Therefore
Im(G(z)) = Im(H (r; x)) 12 p2 (1 ? r) + O((1 ? r) 32 p2 ) > 0 on C2 ;
Im(G(z)) = Im(H (r; x)) ? 12 p2 (1 ? r) + O((1 ? r) 32 p2 ) < 0 on C4 ;
Re(G(z)) = Re(H (r; x)) ? 23 p + O(p) < 0 on C3 ;
as required.
References
1. R.C. Baker and H.L. Montgomery, Oscillations of Quadratic L-functions, Analytic Number Theory
(ed. B.C. Berndt et.al.), Birkhauser, Boston, 1990, p. 23{40.
2. H. Davenport, Multiplicative Number Theory (2nd ed.), Springer-Verlag, New York, 1980.
3. P. Erd}os and P. Turan, On the distribution of roots of polynomials, Ann. of Math. 51 (1950), 105{119.
4. M. Fekete, Palermo Rend 34 (1912), 89.
5. H.L. Montgomery, An exponential polynomial formed with the Legendre symbol, Acta Arithmetica 37
(1980), 375{380.
6. G. Polya, Verschiedene Bemerkung zur Zahlentheorie, Jber. deutsch Math. Verein 28 (1919), 31{40.
7. M. Sambandham and V. Thangaraj, On the average number of real zeros of a random trigonometric
polynomial, J. Indian Math. Soc 47 (1983), 139{150.
8. A. Weil, Sur les fonctions algebriques a corps de constantes ni, C.R. Acad. Sci., Paris 210 (1940),
592{594.
(Conrey) American Institute of Mathematics. 360 Portage Ave, Palo Alto, California
94306, USA
ZEROS OF FEKETE POLYNOMIALS
19
E-mail address : [email protected]
(Conrey) Department of Mathematics, Oklahoma State University, Stillwater, Oklahoma 74078, USA
E-mail address : [email protected]
(Granville) Department of Mathematics, The University of Georgia, Athens, Georgia
30602-7403, USA
E-mail address : [email protected]
(Poonen) Department of Mathematics, The University of California, Berkeley, California 94720-3840, USA
E-mail address : [email protected]
(Soundararajan) Department of Mathematics, Princeton University, Princeton, New
Jersey 08544, USA
E-mail address : [email protected]