LXXXVI.4 (1998)
On the number of polynomials of bounded measure
by
A. Dubickas (Vilnius) and S. V. Konyagin (Moscow)
1. Introduction. The Mahler measure of the polynomial
f (x) = a
0x
d+ a
1x
d−1+ . . . + a
d= a
0(x − α
1) . . . (x − α
d), a
06= 0, is defined by
M (f ) = M (f (x)) = |a
0| Y
d j=1max(1, |α
j|).
Our purpose here is to give an upper bound for the cardinality of the set of polynomials in Z[x] of given degree and of bounded Mahler’s measure.
Since M (±x
tf (x)) = M (f (x)), let us suppose that a
0> 0 and a
d6= 0. This allows us to avoid some trivial complications. We will denote by N (d, T ) the set of polynomials f (x) ∈ Z[x] of degree d such that M (f ) ≤ T , a
0> 0, f (0) = a
d6= 0.
Firstly, notice that if T < 1 then the set N (d, T ) is empty. Secondly, if T = 1 then by Kronecker’s theorem each f (x) in N (d, 1) is a product of cyclotomic polynomials. The result of E. Dobrowolski [Do] implies that N (d, T ) is of the same structure for T < exp c
1 log log dlog d 3. For a sufficiently large d this holds with c
1= 9/4 − ε, ε > 0 [Lo]. By Lehmer’s conjecture [Le]
there exists a positive constant δ such that the set N (d, T ) has the same structure for T < 1 + δ. Therefore for small T the cardinality of N (d, T ) is equal to the number of ways to write d as a sum of the form
d = X
m≥1
d
mϕ(m)
where ϕ is the Euler totient function and d
mis a nonnegative integer. The asymptotics for this number was found by D. W. Boyd and H. L. Mont- gomery [Bo-Mo].
The structure of the set N (d, T ) for larger T is much more complicated.
Let us start by giving some trivial lower and upper bounds for card N (d, T ).
1991 Mathematics Subject Classification: 11C08, 11R04, 11R09, 12D10.
[325]
By Landau’s inequality [La],
M (f ) ≤ v u u t X
dj=0
a
2j≤ √
d + 1 max
0≤j≤d
|a
j|.
Hence all polynomials with integer coefficients such that 1 ≤ a
0≤ T / √ d + 1,
|a
j| ≤ T / √
d + 1, j = 1, . . . , d, a
d6= 0 lie in N (d, T ). If T ≥ √
d + 1, we can bound card N (d, T ) from below by
T
√ d + 1
2
T
√ d + 1
+ 1
d−12
T
√ d + 1
> 1 2
T
√ d + 1
d+1. Here and throughout the paper [ ] denotes the integral part. If 1 ≤ T <
√ d + 1, then this lower bound holds trivially. Hence we obtain the following lower bound:
(1) card N (d, T ) >
12T
d+1(d + 1)
−(d+1)/2. On the other hand, if M (f ) ≤ T then
|a
j| = |a
0| · |α
1α
2. . . α
j+ . . . + α
d−j+1. . . α
d| ≤
d j
T.
Hence card N (d, T ) is bounded from above by 2[T ]
2d−1
Y
j=1
2
d j
T
+ 1
≤ 2T
2d−1
Y
j=1
2
d j
T + 1
≤ 2 · 3
d−1T
d+1d−1
Y
j=1
d j
= 2 · 3
d−1T
d+1d!
−d−1Y
d j=1j
2j. It is an easy exercise to show by induction that
Y
d j=1j
2j≤ d
d(d+1)exp
d − d
22
.
Utilizing Stirling’s formula d! > √
2πd
d e
dexp
− 1 12d
,
we get
card N (d, T ) < T
d+1exp
d
22 + 3d
2 + d + 1
12d + (d − 1) log 3 + log 2 − d + 1
2 log(2πd)
. Thus, for sufficiently large d,
(2) card N (d, T ) < T
d+1exp(d
2/2).
Comparing these trivial estimates (1) and (2) we see that they differ by a factor depending on d only. So for “large” T we cannot expect to do much better than (2). However, for “small” T , e.g. fixed T , this bound can be substantially improved. For example, the interesting case T < 2 was considered by M. Mignotte [Mi]. He proved that the number of irreducible polynomials of degree d and of Mahler’s measure smaller than 2 is less than 2(8d)
2d+1. Combined with the results of [Bo-Mo] this implies that
(3) card N (d, T ) < d
c2dfor T < 2 and some absolute positive constant c
2.
Denote by N
1(d, T ) the set of monic (a
0= 1), irreducible polynomials f (x) ∈ Z[x] of degree d such that M (f ) ≤ T , |f (0)| = |a
d| = 1. Recently one of the authors [Ko] proved that
(4) card N
1(d, T ) < T
c3dexp(d
19/20)
provided that d is sufficiently large where c
3is an effectively computable constant.
In all what follows, let d be a sufficiently large positive integer. Through- out this paper c, c
4, c
5, c
6, . . . will be assumed to be positive effective con- stants, and ε, ε
1, ε
2will be assumed to be small positive constants. Let θ = 1.32471 . . . be the real root of the polynomial x
3− x − 1.
Theorem. (i) If 1 ≤ T < exp
94− ε
log log dlog d
3, then card N (d, T ) = c
4(log d)
−1/2d
−1exp
1 π
p 105ζ(3)d
(1 + o(1)), where c
4=
4π12(105ζ(3)e
−γ)
1/2and o(1) → 0 as d → ∞.
(ii) If exp
94− ε
log log dlog d
3≤ T < θ, then
card N (d, T ) < T
d(1/2+c log log d/log d). (iii) If θ ≤ T , then
card N (d, T ) < T
d(1+c log log d/log d).
We will prove (ii) and (iii) with c = 16. The case (i) is given here only
for completeness. As mentioned above it is a combination of the results
obtained in [Lo] and [Bo-Mo]. Combining (i)–(iii) we obtain the following general estimate for every positive ε
1:
card N (d, T ) < T
d(1+16 log log d/log d)exp(3.58 √ d) (5)
< T
(1+ε1)dexp(4 √ d).
Taking T = 2 in (5) we obtain card N (d, 2) < (2+ε
2)
d, which strengthens the inequality (3). Notice also that (5) strengthens (4), since card N (d, T ) ≥ card N
1(d, T ). However, for large T , T > exp
2c log log dd log d, the trivial inequal- ity (2) gives a stronger upper bound than (iii).
In Section 2 we give some auxiliary lemmas. Section 3 contains a sketch of proof. The proof of (iii) is given in Sections 4 and 5 where we bound the number of polynomials with small and large leading coefficients respectively.
Finally, in Section 6 we complete the proof of (ii).
The authors thank the organizers of the Number Theory Conference dedicated to Professor A. Schinzel in Zakopane (July 1997). In the pleasant atmosphere of this conference the idea to write this paper came and a part of the work was done. The research of the second named author was supported by Grants 96-01-00378 and 96-15-96072 from the Russian Foundation for Basic Research.
2. Auxiliary lemmas. Let e
1, . . . , e
dbe the natural basis of the d- dimensional Euclidean space R
d. We put
kxk = max
1≤j≤d
|x
j|
for the l
∞norm of the vector x = x
1e
1+ . . . + x
de
d. For a convex closed bounded set A ⊂ R
dand j = 1, . . . , d we put
F
j(A) = A + [−e
j/2; e
j/2] = {x + µe
j: x ∈ A, |µ| ≤ 1/2}.
Let F (A) be the 1/2-neighbourhood of A:
F (A) = {x + y : x ∈ A, kyk ≤ 1/2} = F
1(F
2. . . F
d(A) . . .).
Suppose that G ⊂ {1, . . . , d} and g = card G. We denote by O
G(A) the orthogonal projection of the set A onto the linear space spanned by the vectors e
jwith j ∈ G. Finally, we denote by Vol
gO
G(A) the volume of the g-dimensional (1 ≤ g ≤ d) convex set O
G(A), and let Vol
0O
∅(A) = 1. With this notation we have the following general lemma for d ≥ 1:
Lemma 0. We have
Vol
dF (A) = X
Vol
gO
G(A),
where the sum is taken over all subsets G of {1, . . . , d}.
P r o o f. Let b A be a convex closed bounded set in R
d, and let G ⊂ {1, . . . , d}, j ∈ G. Expressing the volumes O
G( b A) and O
G(F
j( b A)) as in- tegrals over O
G\{j}( b A) we get
Vol
gO
G(F
j( b A)) − Vol
gO
G( b A) =
12+
12Vol
g−1O
G\{j}( b A)
= Vol
g−1O
G\{j}( b A).
Therefore, we obtain the following reduction formula:
Vol
gO
G(F
j( b A)) = Vol
gO
G( b A) + Vol
g−1O
G\{j}( b A).
Note that
F (A) = O
{1,...,d}(F
1(F
2. . . F
d(A) . . .)).
Taking into account that the operators O
Gand F
jcommute for j ∈ G, we can apply the above formula when proceeding with Vol
dF (A), thus reducing one of F
j’s at each step. It is easy to see that in this way we finally get the stated formula.
The next lemma is an estimate of the 1/2-capacity [Ti, 1.1.7] of convex polytopes contained in a parallelepiped.
Lemma 1. Suppose that 0 < τ < 1/12, d > d(τ ) and let P =
Y
d j=1[−u
j/2; u
j/2]
be a parallelepiped in R
d. Let A ⊂ P be a convex polytope with at most d
−1+1/(10τ )vertices. If D is a subset of A such that the distance between any two elements of D is at least 1, then
card D < exp(d
1−9τ) Y
d j=1(1 + u
jd
6τ −1/2).
P r o o f. The unit cubes with centers at the points of D are mutually nonoverlapping. The union of these cubes is a subset of F (A), so that card D ≤ Vol
dF (A). By Lemma 0, it remains to estimate each Vol
gO
G(A) from above.
Suppose first that g ≥ d
1−10τ. Then O
G(A) is a convex polytope with at most
d
−1+1/(10τ )≤ g
(−1+1/(10τ ))/(1−10τ )= g
1/(10τ )vertices. Clearly, O
G(A) is contained in the parallelepiped
O
G(P ) = Y
j∈G
[−u
j/2; u
j/2] ⊂ R
g.
We now need an upper bound for the volume of a convex polytope A
gwith ≤ g
1/(10τ )vertices contained in a parallelepiped Π
g⊂ R
g. With- out loss of generality we can assume that Π
gis the g-dimensional cube [−1/ √
g; 1/ √
g]
ginscribed in the unit ball B
g. The volume of the ball is estimated by the volume of the cube as follows:
Vol
gB
g= π
g/2Γ (1 + g/2) < π
g/2(g/(2e))
g/2=
2πe g
g/2= (πe)
g/2Vol
gΠ
g. Using the inequality
Vol
gA
gVol
gB
g≤
10 log(g
1/(10τ )) g
g/2=
log g τ g
g/2(see [B´a–F¨ u] or [Gl]), we bound the volume of the convex polytope A
g: Vol
gA
gVol
gΠ
g<
πe log g τ g
g/2. Thus,
Vol
gO
G(A)
Vol
gO
G(P ) < (d
1/2−6τ)
−gfor d sufficiently large.
If g < d
1−10τ, then we bound Vol
gO
G(A) from above trivially:
Vol
gO
G(A) ≤ Vol
gO
G(P )
= Vol
gO
G(P )(d
1/2−6τ)
−g(d
1/2−6τ)
g< Vol
gO
G(P )(d
1/2−6τ)
−gexp(d
1−9τ).
Thus for all G ⊂ {1, . . . , d} we have the upper bound Vol
gO
G(A) < Vol
gO
G(P )(d
1/2−6τ)
−gexp(d
1−9τ)
= exp(d
1−9τ)(d
1/2−6τ)
−gY
j∈G
u
j. Now Lemma 1 follows from Lemma 0.
For a vector w = (w
1, . . . , w
d) ∈ C
ddefine S
k(w) =
X
d j=1w
kjand write kwk = max
1≤j≤d|w
j| for the l
∞norm. We denote by <z and =z the real and imaginary part of z respectively.
Lemma 2. Let W be a subset of C
dsuch that kwk ≤ exp
log d10dfor all w ∈ W and
1≤k≤d
max |<(S
k(u) − S
k(v))|/k ≥ d
−1/5for any two distinct u, v ∈ W . Then
card W < exp(d
8/9).
P r o o f. For a complex number z = a + ib we define e
z = d
−3([|a|d
3] sgn a + i[|b|d
3] sgn b).
Then |e z| ≤ |z| and
|z − e z| = d
−3({|a|d
3}
2+ {|b|d
3}
2)
1/2< √ 2d
−3.
If w = (w
1, . . . , w
d), then write e w for the vector with coordinates e w
j. For w ∈ W and 1 ≤ k ≤ d we have
|S
k( e w) − S
k(w)| =
X
d j=1( e w
kj− w
kj)
≤ d max
1≤j≤d
| e w
jk− w
jk|
< √
2d
−2k exp
(k − 1) log d 10d
< 2kd
−19/10. Hence there exists a number K = K(u, v), 1 ≤ K ≤ d, such that
|<(S
K(e u) − S
K(e v))| > |<(S
K(u) − S
K(v))| − 4Kd
−19/10≥ Kd
−1/5− 4Kd
−19/10> Kd
−1/5/2.
Let D be the set of vectors
(2d
1/5<S
1( e w), . . . , 2k
−1d
1/5<S
k( e w), . . . , 2d
−4/5<S
d( e w))
in R
dwhere w ∈ W . By the above we see that the map W → D is injective and that the distance between any two elements of D is at least 1. Let A be the convex hull of the vectors
(2d
6/5<y, . . . , 2k
−1d
6/5<y
k, . . . , 2d
1/5<y
d).
Here |y| ≤ exp
log d10dand d
3y ∈ Z[i], i.e. y runs over the set of all possible points e w
j. In particular, each vector in D is the arithmetic mean of some d of the above vectors. Hence D is a subset of A. Clearly, the number of vertices in the polytope A is bounded above by the number of Gauss’ integers in the circle |y| ≤ d
3exp
log d10d. An upper bound for these is obtained by counting the number of unit squares with center at Gauss’ integer:
π
√
2 + d
3exp
log d 10d
2< 4d
6. Moreover, A is contained in the parallelepiped
P = Y
d j=1[−u
j/2; u
j/2] with u
j= 4j
−1d
6/5exp
j log d 10d
.
Taking τ = 1/75 we now apply Lemma 1:
card D < exp(d
22/25) Y
d j=11 + 4j
−1d
39/50exp
j log d 10d
. Since 1 + x < e
x, x > 0, we bound the last product by
exp
X
dj=1
4j
−1d
39/50exp
j log d 10d
< exp(5d
22/25log d).
Thus,
card W = card D < exp(d
22/25(1 + 5 log d)) < exp(d
8/9).
Lemma 3. Let n ∈ N and let N
u∈ N, r
u∈ N ∪ {0}, b
u∈ R, b
u≥ 1, for u = 0, 1, 2, . . . If N
u≤ b
un and P
u≥0
r
u= n, then
(6) Y
u≥0
N
u− 1 + r
ur
u≤ n
nY
u≥0
b
ruuand
(7) Y
u≥0
N
u− 1 + r
ur
u≤
22
1 + B log n n
nY
u≥0
b
ruu, where B = P
u≥0
r
uu.
P r o o f. Since
N
u− 1 + r
ur
u= N
u+ r
u− 1
r
u· N
u+ r
u− 2
r
u− 1 · . . . · N
u1 ≤ N
uru, we get (6) immediately.
In order to prove (7) we assume that n ≥ 23 and B > 0. Indeed, if n ≤ 22 then inequality (7) follows from (6). If B = 0, then r
0= n and using n! > (n/e)
nwe find that
Y
u≥0
N
u− 1 + r
ur
u=
N
0− 1 + r
0r
0≤ (b
0n + n)
nn!
< (e(b
0+ 1))
n≤ (2eb
0)
n< (22b
0)
n. Put now U =
B log nn
. We have
B ≥ X
u≥U +1
r
uu ≥ (U + 1) X
u≥U +1
r
u. Thus,
X
0≤u≤U
r
u= n − X
u≥U +1
r
u≥ n − B U + 1 > n
1 − 1 log n
.
Since the function x log x is convex, we obtain X
0≤u≤U
r
ulog r
u> n
1 − 1 log n
log
n(1 − 1/log n) U + 1
. It follows that
Y
u≥0
r
uru≥ Y
0≤u≤U
r
uru>
n
1 − 1 log n
n(1−1/log n)B log n
n + 1
−n(8)
≥ n
n1 e
1 − 1 log 23
nB log n
n + 1
−n.
Utilizing the inequalities r
u≤ n and r
u! ≥ (r
u/e)
ruwe obtain
N
u− 1 + r
ur
u≤ (N
u+ n)
rur
u! ≤
e(b
u+ 1)n r
u ru≤
2eb
un r
u ru. Thus, applying (8), we see that the left hand side of (7) does not exceed
Y
u≥0
(2eb
un)
rur
u−ru<
2e
21 − 1/log 23
nB log n
n + 1
nY
u≥0
b
ruu<
22
1 + B log n n
nY
u≥0
b
ruu.
3. Sketch of proof. For each polynomial f in N (d, T ) we define a vector with nonnegative entries in the following way. Suppose first that the leading coefficient a
0is in the range 1 ≤ a
0< d
1/5/2. Then the vector has the form
(a
0, q
1, q
2, n
l, s
l, n
l+1, s
l+1, . . .).
Here l =
log d10
, n
uis the number of zeros of f lying in
K
u=
z ∈ C : exp
u d
< |z| ≤ exp
u + 1 d
, =z > 0
, s
uis the number of zeros of f lying in
L
u=
− exp
u + 1 d
; − exp
u d
∪
exp
u d
; exp
u + 1 d
, q
1= [d log |Λ
1|], q
2= [d log |Λ
2|] where Λ
1and Λ
2are the products of zeros of f lying in S
u≥l
K
uand S
u≥l
L
urespectively.
If however a
0≥ d
1/5/2, then the vector has the form (a
0, q
3, q
4, q
5, q
6, n
0, s
0, m
0, r
0, n
1, s
1, m
1, r
1, . . .).
Here a
0, n
u, s
uare as above, m
uis the number of zeros of f lying in M
u=
z ∈ C : exp
− u + 1 d
< |z| ≤ exp
− u d
, =z > 0
,
r
uis the number of zeros of f lying in R
u=
− exp
− u d
; − exp
− u + 1 d
∪
exp
− u + 1 d
; exp
− u d
, q
k= [(−1)
k+1d log |Λ
k|], k = 3, . . . , 6, where Λ
3, . . . , Λ
6are the products of the zeros of f lying in {|z| > 1, =z > 0}, {|z| ≤ 1, =z > 0}, (−∞; −1) ∪ (1; ∞) and [−1, 1] respectively.
Now we prove that the number of all different vectors which are defined as above for the polynomials in N (d, T ), T > exp((log d)
−3), is less than
(9) T
c5√d(log d)3/2
.
Indeed, notice first that the number of different vectors (a
0, q
1, . . . , q
6) is at most
T (d log T + 1)
6< T (dT )
6< T
6(log d)4+7. Put x
u= P
k≥u
n
k, u = 1, 2, . . . Then the values n
1, n
2, . . . are determined uniquely by x
1, x
2, . . . We also have x
1≥ x
2≥ . . . Since all x
kare nonneg- ative integers and
d log |Λ
3| ≥ X
u≥1
X
z∈Ku
log |z| ≥ X
u≥1
un
u= X
u≥1
x
u,
the number of different vectors (n
1, n
2, . . .) is bounded from above by
q3
X
q=0
p(q) ≤ (q
3+ 1) p(q
3)
where p(q) is the number of partitions of q. The well known asymptotic formula for the number of partitions (see, e.g., [An])
p(q) ∼ exp(π p 2q/3) 4 √
3q
implies that the number of different vectors (n
1, n
2, . . .) does not exceed c
6exp(c
7√
q
3). Similarly, the number of different vectors (s
1, s
2, . . .), (m
1, m
2, . . .), (r
1, r
2, . . .) is bounded above by c
6exp(c
7√
q
5), c
6exp(c
7√ q
4) and c
6exp(c
7√
q
6) respectively. Since the number of different (n
0, s
0, m
0, r
0) is at most d
4, bounding |Λ
1|, |Λ
2|, |Λ
3|, |Λ
4|
−1, |Λ
5|, |Λ
6|
−1from above by T , we estimate the number of all different vectors by
T
6(log d)4+7d
4c
46exp(4c
7p
d log T ) < T
c5√d(log d)3/2.
We see that the number of different vectors is less than T
c8d log log d/log d.
Therefore we only need to prove the upper estimates for the cardinality (ii)
and (iii) for the polynomials in N (d, T ), T > exp((log d)
−3), corresponding
to the fixed vector (a
0, q
1, q
2, n
l, s
l, . . .) or (a
0, q
3, q
4, q
5, q
6, n
0, s
0, m
0, r
0, . . .).
For the polynomial
f = a
0x
d+ a
1x
d−1+ . . . + a
d= a
0(x − α
1)(x − α
2) . . . (x − α
d) we define
S
k= S
k(f ) = X
d j=1α
kj. By the Newton identities we have
(10) a
0S
k+ a
1S
k−1+ . . . + a
k−1S
1+ ka
k= 0.
Suppose that f and
g = a
0x
d+ h
1x
d−1+ . . . + h
d= a
0(x − β
1)(x − β
2) . . . (x − β
d) are two distinct polynomials with the same leading coefficient a
0. There exists a positive integer k ≤ d such that h
1= a
1, . . . , h
k−1= a
k−1, but h
k6= a
k. Then S
1(f ) = S
1(g), . . . , S
k−1(f ) = S
k−1(g), but S
k(f ) 6= S
k(g).
From (10) we deduce that
(11) |S
k(f ) − S
k(g)| = |a
k− h
k| k a
0≥ k
a
0.
Consider now the polynomials in N (d, T ) corresponding to the fixed vector (a
0, q
1, q
2, n
l, s
l, n
l+1, s
l+1, . . .) where 1 ≤ a
0< d
1/5/2. Put
n = X
u≥l
n
u, s = X
u≥l
s
u.
We can assume that α
1= α
n+1, . . . , α
n= α
2nlie in S
u≥l
K
uand that α
2n+1, . . . , α
2n+s∈ S
u≥l
L
u(and similarly for the roots β
jof g). From (11) we find that either
(12)
X
d j=2n+s+1(α
kj− β
jk) ≥ k
2a
0or (13)
2n+s
X
j=1
(α
kj− β
jk) ≥ k
2a
0.
Below we will argue as follows. All K
uand L
uwill be covered by disjoint squares and intervals respectively (see Section 4). The crucial step is to estimate the number of ways to distribute n + s roots into corresponding squares and intervals. We will show that for each distribution where each square and interval contains the same number of α
j’s and β
j’s, 1 ≤ j ≤ 2n + s, the inequality opposite to (13) holds. Thus, inequality (12) holds.
We see that the l
∞norm of the vectors (0, 0, . . . , 0, α
2n+s+1, . . . , α
d) and
(0, 0, . . . , 0, β
2n+s+1, . . . , β
d) is bounded above by exp
log d10d. Since
<
X
d j=2n+s+1(α
kj− β
kj) =
X
d j=2n+s+1(α
kj− β
jk) ≥ k
2a
0> kd
−1/5, we can apply Lemma 2. For any way to distribute roots the number of distinct polynomials corresponding to this way is bounded above by (14) exp(d
8/9) < exp(d
8/9(log T )(log d)
3) < T
d9/10.
4. Polynomials with small leading coefficient. For any nonnegative integer u we cover K
uby disjoint squares with side of length
(15) exp(−(u + 1)(d − 1)/d)
7 √ 2na
0. If α
jand β
jbelong to the same square, then
|α
kj− β
jk| = |α
j− β
j| · |α
k−1j+ α
k−2jβ
j+ . . . + β
jk−1|
≤ exp(−(u + 1)(d − 1)/d) 7na
0k exp
(u + 1)(k − 1) d
≤ k
7na
0for k ≤ d. Therefore, if each square contains the same number of α
j’s and β
j’s, then
X
n j=1(α
kj− β
jk) ≤ k
7a
0. We also have
X
2n j=n+1(α
kj− β
jk) =
X
2n j=n+1(α
kj− β
kj) =
X
n j=1(α
kj− β
jk) ≤ k
7a
0. Similarly, cover L
uby disjoint intervals of length
(16) exp(−(u + 1)(d − 1)/d)
7sa
0.
If each interval contains the same number of α
j’s and β
j’s, we have
2n+s
X
j=2n+1
(α
kj− β
jk) ≤ k
7a
0. Thus,
2n+s
X
j=1
(α
kj− β
jk) ≤ 3k
7a
0, which contradicts (13).
Clearly, the number of polynomials with different distribution of roots
is bounded above by the number of ways to distribute roots into respective
squares and intervals. More precisely, this is the number of ways to put n
u, u = l, l + 1, . . . , roots into N
usquares times the number of ways to put s
u, u = l, l + 1, . . . , roots into intervals. The number of squares N
uwith side length (15) which cover K
uis bounded from above by
π
2 98n
2a
20exp
2(u + 1)(d − 1) d
exp
u + 1 d
+ exp(−(u + 1)(d − 1)/d) 7na
0 2−
exp
u d
− exp(−(u + 1)(d − 1)/d) 7na
0 2. By a short computation we see that this is bounded by c
9na
20exp(2u). The number of ways to put n
uroots into N
usquares equals
Nu+nn u−1u
. In order to estimate the product
Y
u≥l
N
u+ n
u− 1 n
uwe apply Lemma 3 with b
u= c
9a
20exp(2u) and B = q
1. Indeed, the inequal-
ity X
u≥l
un
u≤ d log |Λ
1| < q
1+ 1 implies that
X
u≥l
un
u≤ q
1.
If n ≤ q
1(log d)
−2, then by (6) we have Y
u≥l
N
u+ n
u− 1 n
u≤ n
nY
u≥l
(c
9a
20)
nuexp(2un
u) (17)
≤ c
n9a
2n0exp(q
1(log d)
−1+ 2q
1).
If n > q
1(log d)
−2, then by (7) we find Y
u≥l
N
u+ n
u− 1 n
u≤
22
1 + q
1log n n
nc
n9a
2n0exp(2q
1) (18)
< (23(log d)
3)
nc
n9a
2n0exp(2q
1)
< c
n10a
2n0exp(2q
1+ 3n log log d).
Since
q
1≥ X
u≥l
un
u≥ nl = n
log d 10
,
in both cases (17) and (18) we have
(19) Y
u≥l
N
u+ n
u− 1 n
u≤ a
2n0exp(2q
1+ 31q
1log log d/log d).
Analogously, the number of intervals of length (16) which cover L
uis bounded above by
14sa
0exp
(u + 1)(d − 1) d
exp
u + 1 d
− exp
u d
+ exp(−(u + 1)(d − 1)/d) 3sa
0< c
11a
0exp(u).
Since P
u≥l
us
u≤ q
2, the number of ways to distribute s
u, u = l, l + 1, . . . , roots into respective intervals does not exceed
(20) Y
u≥l
(c
11a
0exp(u))
su≤ c
s11a
s0exp(q
2).
Using the inequality
s ≤ q
2l ≤ 11q
2log d ,
we further bound (20) from above by a
s0exp(q
2+ q
2log log d/log d). Since 2n + s ≤ d and 2q
1+ q
2≤ d log(T /a
0), combining this inequality with (19) we estimate the number of ways to distribute roots into respective squares and intervals by
(21) a
2n+s0exp(2q
1+ q
2+ 15.5(2q
1+ q
2) log log d/log d)
< T
d(1+15.5 log log d/log d).
5. Polynomials with large leading coefficient. In this section we are left with the case a
0≥ d
1/5/2. We need an upper bound for the number of polynomials corresponding to the vector (a
0, q
3, q
4, q
5, q
6, n
0, s
0, m
0, r
0, n
1, s
1, m
1, r
1, . . .). Put now
n = X
u≥0
n
u, s = X
u≥0
s
u, m = X
u≥0
m
u, r = X
u≥0
r
u.
We enumerate the roots of two distinct polynomials f and g as follows. Let
the first n roots of f (and g) lie outside the unit circle in the upper halfplane,
then their n complex conjugates, then s real roots in (−∞; −1) ∪ (1; ∞),
then m complex roots in the unit circle in the upper halfplane, then their
m complex conjugates, and finally r real roots in [−1; 1]. As in Section 3 we
obtain
k a
0≤
X
d j=1(α
kj− β
kj) ≤ 2
X
n j=1(α
kj− β
kj) +
2n+s
X
j=2n+1
(α
kj− β
jk)
+ 2
2n+s+m
X
j=2n+s+1
(α
kj− β
jk) +
X
d j=2n+s+2m+1(α
kj− β
jk) .
It follows that one of the four sums is at least k/(7a
0). Again the crucial step is to estimate the number of ways to distribute n + s + m + r roots into corresponding squares and intervals. Suppose first that
X
n j=1(α
kj− β
jk) > k
7a
0.
As in Section 4 we cover K
uby disjoint squares with side length (15). If each square contains the same number of α
j’s and β
j’s, 1 ≤ j ≤ n, then the inequality opposite to the above holds. In a similar manner to that in Section 4 we deduce the upper estimate for the number of ways to distribute n roots into squares. Instead of (17) and (18) we have the upper bounds c
n9a
2n0exp(2q
3+ q
3(log d)
−1) and c
n10a
2n0exp(2q
3+ 3n log log d) respectively.
Combining both cases we obtain the upper estimate for the number of ways to distribute n roots:
(22) a
2n0exp(2q
3+ q
3(log d)
−1+ 3n log log d + nc
12).
Suppose now that
2n+s
X
j=2n+1
(α
kj− β
jk) > k
7a
0.
The number of ways to distribute s roots is bounded from above analogously to (20) by
(23) a
s0exp(q
5+ sc
13).
Next we suppose that the third sum is large:
2n+s+m
X
j=2n+s+1
(α
kj− β
jk) > k
7a
0. Cover M
uby disjoint squares with side length 1/(7 √
2ma
0). If α
jand β
jbelong to the same square, then
|α
kj− β
jk| ≤ |α
j− β
j|k ≤ k 7ma
0.
Thus, if each square contains the same number of α
j’s and β
j’s, then
2n+s+m
X
j=2n+s+1
(α
kj− β
jk) ≤ k
7a
0.
Therefore, we need an upper estimate for the number of ways to put m
uroots into respective squares. The number of squares S
uwhich cover M
uis bounded from above by
π
2 98m
2a
20exp
− u d
+ 1
7ma
0 2−
exp
− u + 1 d
− 1
7ma
0 2for u + 1 ≤ d log(7ma
0). This expression is less than c
14ma
20. If however u + 1 > d log(7ma
0), then
exp
− u d
< exp(1/d) 7ma
0< 1
6ma
0and M
ucan be covered by 4 squares. Thus, our upper bound S
u< c
14ma
20also holds.
If m ≤ q
4(log d)
−2, then analogously to (17), by (6) with b
u= c
14a
20, N
u= S
u, r
u= m
u, n = m we obtain the upper bound
Y
u≥0
S
u− 1 + m
um
u≤ m
mY
u≥0
(c
14a
20)
mu≤ c
m14a
2m0exp(q
4/log d).
If m > q
4(log d)
−2, then analogously to (18), by (7) with B = q
4we obtain the upper bound
Y
u≥0
S
u− 1 + m
um
u≤
22
1 + q
4log m m
mY
u≥0
(c
14a
20)
mu≤ (23(log d)
3)
mc
m14a
2m0≤ c
m15a
2m0exp(3m log log d).
Combining these two upper bounds we finally estimate the number of ways to distribute m roots by
(24) a
2m0exp(q
4(log d)
−1+ 3m log log d + mc
16).
Finally, suppose that
X
d j=2n+s+2m+1(α
kj− β
jk) > k
7a
0.
We cover R
uby disjoint intervals of length 1/(7ra
0). The number of intervals which cover R
uis bounded above by c
17a
0. Hence, the number of ways to distribute r roots is bounded above by
Y
u≥0