Explicit zero density theorems for Dedekind zeta functions

0.28 MB·English

by  Habiba Kadiri

EXPLICIT ZERO DENSITY THEOREMS FOR DEDEKIND ZETA
FUNCTIONS
2
1 HABIBAKADIRIANDNATHANNG
0
2

n
a
J

8
1 1. Introduction

] This article concernsthe zerosofDedekind zetafunctions. We provebounds for
T
thenumberofzerosofDedekindzetafunctionsinboxesandweprovezerorepulsion
N
theoremsfor these zeros. The formula forzerosin boxesis classicalanddates back
.
h to Riemann and von Mangoldt. The zero repulsion property is commonly referred
t to as the Deuring-Heilbronn phenomenon. It asserts that if an L-function has a
a
m zero very close to s=1 then the other zeros of this L-function are pushed further
awayfroms=1. Theoremsofthis type alreadyexistin the literature,but they do
[
notgiveexplicitconstants. Ourtheoremswill determine suchconstantsandthis is

1
importantinourrelatedworkwhichconcernsanexplicitboundfor theleastprime
v
ideal in Chebotarev’s density theorem. Let K be a number field and K a Galois
2 0
3 extension of K0 with ring of integers K. Its degree is denoted nK =[K :Q] and
O
9 itsabsolutediscriminantisdK. LetGbetheGaloisgroupofK/K0,andletC G
⊂
3 be a conjugacy class. We show in [6] that there exists an unramified prime ideal p
1. ofdegreeonesuchthatits Frobenius σp =C andits normNp≤dCK0 foranexplicit
0 constantC >0. Thistheoremmakesuseofvariousresultsconcerningthelocation
0
2 and number of zeros of the Dedekind zeta function of K.
1
: Wenowstateourresults. Throughoutthisarticleweshallencounterthequantity
v
i logdK. From this point on, we shall employ the abbreviation
X
L =logd . (1)
r K
a
The Dedekind zeta function of K is
1
ζ (s)=
K (Na)s
a⊂XOK
where a ranges through non-zero ideals. We now define a function which counts
the zerosofζ (s) inboxes. Throughoutthis articlewe shalldenote the non-trivial
K
zeros of ζ (s) as ̺=β+iγ where β,γ R. We set for T 0
K
∈ ≥
N (T)=# ̺ ζ (̺)=0, 0<β <1, γ T .
K K
{ | | |≤ }
Our first result is
1991 Mathematics Subject Classification. Primary: 11M41;Secondary: 11R42.
1
2 HABIBAKADIRIANDNATHANNG
Theorem 1. Let T 1 and 0<η 1 then
≥ ≤ 2
T
|NK(T)− π log((2Tπe)nKdK)|≤c1(η)(L +nKlogT)+c2(η)nK +7.6227
where
1+2η
c (η)= ,
1
πlog2
2 ζ(1+η)2 2
c (η)=0.2675 0.2680η+ log + logζ(3 +2η).
2 − log2 ζ(2+2η) π 2
These results were proven by following arguments of Backlund [1], Rosser [17],
and McCurley [12] who obtained analogous results for the Riemann zeta function
andDirichletL-functions. Theirworkisimportantinargumentswhichgiveexplicit
zero-free regions for L-functions and explicit bounds for prime counting functions.
Our next result is an inequality for the real partof the logarithmic derivative of
ζ (s). We establish
K
Theorem 2. Let 0<ǫ 10 2, s=σ+it, σ >1, t 1, and
−
≤ | |≤
1 1
φ= − √5 =0.276393... (2)
2
We define a multiset of non-trivial zeros of ζ (s) by
K
= ̺ ζ (̺)=0, 1 ǫ β <1, γ t 1 . (3)
ǫ,t K
R { | − ≤ | − |≤ }
For 1<Re(s) 1+ǫ, we have
≤
ζ 1 1
Re K′ (s) Re Re +φL 0.0354n
K
− ζ ≤ s 1 − s ̺ −
(cid:18) K (cid:19) (cid:16) − (cid:17) ̺∈XRǫ,t (cid:16) − (cid:17)
+5ǫ( +1)+0.1216. (4)
ǫ,t
|R |
It is well known that there exists an explicit constant C such that n C L
0 K 0
≤
for K =Q. This follows from an inequality due to Minkowski:
6
d > π nK nnKK 2 for K =Q.
K
4 n ! 6
K
(cid:16) (cid:17) (cid:16) (cid:17)
This combined with Theorem 1 implies there exists an explicit constant C such
1
that
N (2) C L
ǫ,t K 1
|R |≤ ≤
for all t 1.
| |≤
Corollary 2.1. Let 0 < ǫ 10 2, s = σ + it, σ > 1, and t 1. If d is
− K
≤ | | ≤
sufficiently large, then there exists a positive constant C such that
2
ζ 1 1
Re K′ (s) Re Re +(φ+C2ǫ)L. (5)
− ζ ≤ s 1 − s ̺
(cid:18) K (cid:19) (cid:16) − (cid:17) ̺∈XRǫ,t (cid:16) − (cid:17)
The inequality given in Corollary 2.1 will play an important role in obtaining
zero-free regions and zero-repulsion theorems. Versions of this result have been
proveninGraham[2]forDirichletL-functionsandimplicitlyinStechkin[19]forthe
Riemann zeta function. Our proof, as in [19], uses the global method, namely the
classical explicit formula for Re(ζ (s)/ζ (s)) in conjunction with the Stechkin
− K′ K
EXPLICIT ZERO DENSITY THEOREMS FOR DEDEKIND ZETA FUNCTIONS 3
differencing trick. In order to obtain a negative contribution in n , we need to
K
improveonalemmaofMcCurleyon Re(Γ(s)/Γ(s)). In[3],Heath-Brownemploys
′
−
a local method using a Jensen type formula which produces much better values
of φ (φ 1). Recently, Li [10, Lemma 4] applied this method to ζ (s) and
≤ 6 K
was able to obtain an inequality like (5) with φ = 1 and an extra term of size
4
2n log( L )+O(n ). If n =o(L), then Li’s result is superior to ours. On the
othKer hanndK, for thosKe fields KK with n L this error term becomes weaker than
K
≫
(5).
In order to obtain good zero-free regions with nice constants we will develop a
smooth variant of the above theorem. Such results have already been proven by
Heath-Brown in the case of Dirichlet L-functions. We shall follow closely his ap-
proach,thoughthereareseveraldifferencesintheargument. Letf beacontinuous
function from [0, ) to R and supportedin [0,x ). In addition, f is twice differen-
0
∞
tiable on (0,x ) with a bounded and continuous second derivative. Its associated
0
Laplace transform is F(z)= 0∞f(t)e−ztdt.
Theorem 3. Let 0<ǫ≤10−R2, 0<δ <1, s=σ+it with σ >1−(1−δx)0(LlogL), and
t 1. Suppose f(0) 0. If d is sufficiently large, then there exists a positive
K
| | ≤ ≥
constant C such that
3
Λ(a)
Re f(L−1logNa) LReF((s 1)L)
(Na)s ≤ −
(cid:16)a⊂OXK\{0} (cid:17)
L Re(F((s ̺)L))+f(0)(φ+C ǫ)L.
3
− −
̺∈XRǫ,t
An advantage of the inequality in Theorem 3 over the one in Corollary 2.1 is
that it allows for a wide variety of functions f and it allows σ to be chosen inside
the critical strip.
We now state our zero-repulsion theorems. It is known that ζ (s) possesses at
K
most one real zero in a region close to one. For example, it was proven in [8] that
there exists a positive constant R such that ζ (s) does not vanish in
K
1
Re(s) 1 and Im(s) 1 (6)
≥ − RL | |≤
with the exception of possibly one real zero β . Recently Kadiri [5] proved that
1
R = 12.74 is a valid constant. We now examine the consequences of the existence
of this possible exceptional zero. Let
β =1 λ L 1 where λ >0.
1 1 − 1
−
Let ̺ =β +iγ be another zero satisfying
′ ′ ′
β′ =1 λ′L−1 where λ′ >0 and γ′ 1.
− | |≤
We shall prove
Theorem 4. Let β be an exceptional zero of ζ (s) satisfying λ < R 1. Let
1 K 1 −
̺ =β +iγ be another zero of ζ (s) satisfying γ 1, λ < 1 (logL), and β
′ ′ ′ K | ′|≤ ′ 13.85 ′
is maximal with respect to these conditions. If d is sufficiently large, then
K
λ′ ≥0.6546log(λ−11).
4 HABIBAKADIRIANDNATHANNG
The firstproofof this type, inthe case of DirichletL-functions, is due to Linnik
[11]. Hisproofwascomplicated;itmadeuseofBrun’ssieveandconvexitytheorems
forentirefunctions. ItshouldbenotedthatLinnik’sresultplaysanimportantrole
in the proof that the least prime in an arithmetic progression modulo q is qC4
≪
for some positive constant C . Knapowski [7] simplified the argument by applying
4
Turan’s power sum method. Later, Motohashi [14] and Jutila [4] independently
showedthat an argumentrelated to Selberg’s sieve led to better numerical results.
Finally, Heath-Brown[3] made significantnumericalimprovementsby employing a
smoothed version of the explicit formula. His corresponding theorem for Dirichlet
L-functions has a 2 in place of our 0.6546 (see Lemma 8.1 and Table 2 of [3]). He
makes use of an explicit formula for Dirichlet L-functions like our Theorem 3. It
turnsoutthatthecoefficientoflog(λ−11)dependsonhowsmallφis. Thiswasoneof
many ingredients in his proof that C =5.5 is valid. In [8], Lagarias,Montgomery,
4
and Odlyzko proved an inexplicit version of Theorem 4. They used a smoothing
throughdifferentiationinconjunctionwithavariantofTuran’spowersummethod.
Instead, we shall prove the above theorem by following Heath-Brown’s method.
Conventions and Notation. We shall use extensively big notation and Linnik’s
O
notation. For a complex number A and a real number B we shall use the notation
A= (B) and A B or B A to mean there exists M >0 suchthat A MB
O ≪ ≫ | |≤
for A sufficiently large.
2. Properties of the Dedekind zeta function
The Dedekind zeta function of K possesses the Euler product
ζK(s)= (1 (Np)−s)−1
−
p
Y
where p ranges over all prime ideals in and Re(s) > 1. It is convenient to
K
O
consider the completed zeta function
ξ (s)=s(s 1)(d )s/2γ (s)ζ (s), (7)
K K K K
−
γK(s)=(πnK22r2)−s/2Γ(s/2)r1Γ(s)r2, (8)
where r and r are the number of real and complex places in K. The benefit of
1 2
working with ξ is that it is entire of order 1, it satisfies the functional equation
K
ξ (s)=ξ (1 s), (9)
K K
−
and its zeros are the non-trivial zeros of ζ (s).
K
3. Proof of Theorem 1 and Theorem 2
Proof of Theorem 1. Let 0<η 1 and define
≤ 2
3
σ = +2η.
1
2
Throughout this proof we shall let θ , for j =1,...,4, denote real numbers which
j
satisfy θ 1. We followthe argumentofMcCurley[12] whichgeneralizedearlier
j
| |≤
arguments of Backlund and Rosser. Assume that T does not coincide with the
±
ordinate of a zero. We consider the rectangle with vertices σ iT,σ +iT,1
1 1
R − −
EXPLICIT ZERO DENSITY THEOREMS FOR DEDEKIND ZETA FUNCTIONS 5
σ +iT,and1 σ iT whereσ >1. Sinceξ (s)isentire,theargumentprinciple
1 1 1 K
− −
yields
1
N (T)= ∆ argξ (s).
K K
2π R
Let be the part of the contour in Re(s) 1 and the part of the contour
C R ≥ 2 C0
in Re(s) 1 and Im(s) 0. By the functional equation and by the formula
≥ 2 ≥
ξ (s)=ξ(s) it follows that
K
∆ argξ (s)=2∆ argξ (s)=4∆ argξ (s)
R K C K C0 K
and therefore
2
N (T)= ∆ argξ (s). (10)
K π C0 K
We write ξK(s)=sBs2Γ(2s)r1Γ(s)r2(s−1)ζK(s) where B = πnKdK22r2. Hence
∆C0argξK(s)=∆C0args+∆C0argBs2 +r1∆C0argΓ(2s)+r2∆C0argΓ(s)
+∆ arg((s 1)ζ (s)).
C0 − K
A straightforwardcalculation yields
∆ args=arctan(2T),
C0
T T d
s K
∆ argB2 = logB = log .
C0 2 2 πnK22r2
(cid:16) (cid:17)
Tocompute ∆ argΓ(s)=∆ (ImlogΓ(s)), we use Stirling’s formulaasgivenby
C0 C0
Olver [15, p. 294]
1 log2π θ
logΓ(z)=(z )logz z+ + (11)
− 2 − 2 6z
| |
with argz π and θ 1. It follows, as in p. 268 of [12], that
| |≤ 2 | |≤
T T T T 1 1 1
ImlogΓ 1 +i log log 1+ + arctan(2T)+ ,
4 2 − 2 2e ≤ 4 4T2 4 3 1 +T2
(cid:12) (cid:16) (cid:17) (cid:16) (cid:17)(cid:12) (cid:16) (cid:17) 4
(cid:12) (cid:12)
(cid:12) T (cid:12) T 1 1 q
ImlogΓ(1 +iT) Tlog log 1+ + .
2 − e ≤ 2 4T2 6 1 +T2
(cid:12) (cid:16) (cid:17)(cid:12) (cid:16) (cid:17) 4
(cid:12) (cid:12)
As(cid:12)both functions on the right ar(cid:12)e decreasing for T 1, itqfollows that
≥
T T T
ImlogΓ 1 +i log 0.630716,
4 2 − 2 2e ≤
(cid:12) (cid:16) (cid:17) (cid:16) T(cid:17)(cid:12)
(cid:12) ImlogΓ 1 +iT T log (cid:12) 0.260643.
(cid:12) 2 − e (cid:12)≤
(cid:12) (cid:16) (cid:17) (cid:16) (cid:17)(cid:12)
Therefore (cid:12) (cid:12)
(cid:12) (cid:12)
T T
∆ argΓ(s)= log +0.630716θ ,
C0 2 2 2e 1
(cid:16)T (cid:17)
∆ argΓ(s)=T log +0.260643θ .
C0 e 2
(cid:16) (cid:17)
Combining these facts, we obtain
T T T T
∆ argξ (s)=arctan(2T)+ log(B)+r log +r T log
C0 K 2 12 2e 2 e
(cid:18) (cid:16) (cid:17) (cid:16) (cid:17)(cid:19)
+0.630716r θ +0.260643r θ +∆ arg((s 1)ζ (s)).
1 1 2 2 C0 − K
6 HABIBAKADIRIANDNATHANNG
Since r +2r =n , we have
1 2 K
T T T T T T nK
log(B)+r log +r Tlog = log d .
1 2 K
2 2 2e e 2 2πe
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:16) (cid:17) (cid:17)
Combining (10) with the last two equations yields
T T nK 2 1.261431nK
N (T)= log d + ∆ arg((s 1)ζ (s))+ θ +θ .
K π K 2πe π C0 − K π 3 4
(cid:16) (cid:16) (cid:17) (cid:17) (12)
In order to complete the argument we must bound 2∆ arg((s 1)ζ (s)). We
π C0 − K
divide into the contours and as follows:
0 1 2
C C C
1
:σ to σ +iT and :σ +iT to +iT.
1 1 1 2 1
C C 2
We begin with the argument change on . If σ >1 then
1
C
argζ (s) logζ (s) logζ (σ) n logζ(σ)
K K K K
| |≤| |≤ ≤
and therefore, since σ = 3 +2η,
1 2
∆ argζ (s) n logζ(3 +2η).
| C1 K |≤ K 2
In addition, ∆ arg(s 1)=arctan( T )=arctan( T ) and we deduce that
C1 − σ1−1 2η+12
2 2n
∆ arg(s 1)ζ (s) K logζ(3 +2η)+1. (13)
π| C1 − K |≤ π 2
We nowboundthe argumentchangeon . Leta(w)=(w 1)ζ (w) andconsider
2 K
C −
1
f(w)= (a(w+iT)N +a(w iT)N), where N N. (14)
2 − ∈
Note that
f(σ)=Rea(σ+iT)N if σ R.
∈
Suppose f(σ) has n real zeros in the interval 1 σ σ . These zeros partition
2 ≤ ≤ 1
the interval into n+1 subintervals. On each of these subintervals arga(σ+iT)N
can change by at most π, since Rea(σ+iT)N is nonzero on the interior of each
subinterval. It follows that
1 (n+1)π
∆ arga(s) = ∆ arga(s)N . (15)
| C2 | N| C2 |≤ N
We now provide an upper bound for n. Let 0<η < 1 and
2
σ =1+η.
0
Jensen’s theorem asserts that
1+2η n(r)dr 1 32π
log f(σ ) + = log f(σ +(1+2η)eiθ dθ,
0 0
| | Z0 r 2π Z−π2 | |
wheren(r)denotesthenumberofzerosoff(z)inthecirclecenteredatσ ofradius
0
r. Observe that n(r) n for r 1 +η and thus
≥ ≥ 2
3π
1 2
nlog2 log f(σ +(1+2η)eiθ dθ log f(σ ). (16)
0 0
≤ 2π π | | − | |
Z−2
EXPLICIT ZERO DENSITY THEOREMS FOR DEDEKIND ZETA FUNCTIONS 7
The next step is to provide an upper bound for the integral. Rademacher proved
an explicit version of the Phragm´en-Lindel¨of Theorem. Theorem 4 of [16] states
that
|ζK(w)|≤3|11+ww|(dK(|w2+π1|)nK)1+η−2Re(w)ζ(1+η)nK
| − |
uniformly for η Re(w) 1+η. An examination of the proof reveals that the
− ≤ ≤
slightly stronger bound
|ζK(w)|≤3|11+ww|(dK(|w2+π1|)nK)1+η−2Re(w)ζK(1+η), (17)
| − |
holds for η Re(w) 1+η. Bound (7.1) in [16] is ζK(1+η+it) ζ(1+η)nK
− ≤ ≤ | |≤
for η >0. However, this may be replaced by
ζ (1+η+it) ζ (1+η)
K K
| |≤
for η > 0 and this change in the argument immediately leads to (17). It follows
that, for w=σ +(1+2η)eiθ with θ [π,3π],
0 ∈ 2 2
|a(w±iT)|≤3|1+w±iT|(dK(|1+w2π±iT|)nK)−12(1+2η)cosθζK(1+η). (18)
Since T 1 and 0<η 1,
≥ ≤ 2
1+w iT 1+σ iT +1+2η= T2+(2+η)2+1+2η T2+(5)2+2 (19)
| ± |≤| 0± | ≤ 2
and thus p q
log 1+w iT log(b T), (20)
1
| ± |≤
where
b = (1+(5)2)+2=4.692582.... (21)
1 2
q
Putting together (14), (18), (19), taking logarithms, and then applying (20) gives
N
log f(σ +(1+2η)eiθ) (1+2η)(cosθ)(L +n log(b1T))
| 0 |≤− 2 K 2π
+N(log(3b T)+logζ (1+η)),
1 K
valid for θ [π,3π]. Applying this bound on the left-hand side of the contour in
∈ 2 2
(16) and employing the integrals 1 32π(cosθ)dθ = 1 and 1 32π dθ = 1, we
−2π π2 π 2π π2 2
find that
R R
3π
1 2 log f(σ +(1+2η)eiθ)dθ N (1+2η)(L +n log(b1T))
2π π | 0 | ≤ 2π K 2π
Z2
N N
+ log(3b T)+ logζ (1+η). (22)
1 K
2 2
For the right part of the contour in (16), we shall make use of the bound
f(σ +(1+2η)eiθ) (1+3η+T)Nζ (1+η)N
0 K
| |≤
valid for θ [ π/2,π/2]. This implies that
∈ −
π
1 2 N N
log f(σ +(1+2η)eiθ dθ log(1+3η+T)+ logζ (1+η). (23)
0 K
2π π | | ≤ 2 2
Z−2
8 HABIBAKADIRIANDNATHANNG
Together with (16), (22), and (23), we obtain
N N N
nlog2 (1+2η)(L +n log(b1T))+ log(3b T)+ log(1+3η+T)
≤ 2π K 2π 2 1 2
+Nlogζ (1+η) log f(1+η). (24)
K
− | |
Tocomplete our boundfor n,we requirea lowerbound forlog f(1+η). We write
| |
a(1+η +iT) = reiφ and then choose (by Dirichlet’s approximation theorem) a
sequence of N’s tending to infinity such that Nφ tends to 0 modulo 2π. It follows
that
f(1+η)
lim =1. (25)
N a(1+η+iT)N
→∞| |
Note that, for σ >1, we have
1 ζ (2σ)
ζ (s) = 1 N(p) s 1 (1+ ) 1 = K (26)
| K | | − − |− ≥ N(p)σ − ζ (σ)
p p K
Y Y
and 1+η+iT 1 = η2+T2 so that
| − |
p ζ (2+2η)
a(1+η+iT) η2+T2 K . (27)
| |≥ ζ (1+η)
K
p
Thus we derive from (25), (26), and (27) that
ζ (2+2η)
log f(1+η) Nlog η2+T2 K +o(1),
| |≥ ζ (1+η)
(cid:18) K (cid:19)
p
where the term o(1) 0 as N . Equation (24) becomes
→ →∞
N N
nlog2 (1+2η)(L +n log(b1T))+ (log(3b T)+log(5 +T))
≤ 2π K 2π 2 1 2
ζ (2+2η)
+Nlogζ (1+η) Nlog K Nlog η2+T2+o(1).
K
− ζ (1+η) −
K
p
We combine the third and fourth terms and then use the inequality ζK(σ)2
ζK(2σ) ≤
ζ(σ)2 nK to obtain
ζ(2σ)
(cid:16) (cid:17)
N N
nlog2 (1+2η)(L +n log(b1T))+ (log(3b T)+log(5 +T))
≤ 2π K 2π 2 1 2
ζ(1+η)2
+Nn log Nlog η2+T2+o(1).
k
ζ(2+2η) −
p
By the last inequality and by (15), we have
2 1+2η 1
∆ arga(s) L +n log b1T + (log(3b T)+log(5 +T))
π| C2 |≤ πlog2 K 2π log2 1 2
(cid:0) 2n (cid:0) ζ((cid:1)1(cid:1)+η)2 log(η2+T2)
K
+ log +o(1) (28)
log2 ζ(2+2η) − log2
EXPLICIT ZERO DENSITY THEOREMS FOR DEDEKIND ZETA FUNCTIONS 9
where o(1) 0 as N . We let N and combine the results obtained for
→ → ∞ → ∞
in (13) and for in (28):
1 2
C C
2 1+2η log(3b T)+log(5 +T)
∆ arg(s 1)ζ (s) L +n log b1T + 1 2
π| C0 − K |≤ πlog2 K 2π log2
2n ζ(1+η(cid:0))2 log(η2+(cid:0) T2)(cid:1)(cid:1) 2n
+ K log + K logζ(3 +2η)+1.
log2 ζ(2+2η) − log2 π 2
Inserting this in (12) yields
N (T) T log d T nK c (η)(L +n logT)+c (η)n +g(T),
K K 1 K 2 K
| − π 2πe |≤
where (cid:16) (cid:16) (cid:17) (cid:17)
1+2η
c (η)= , (29)
1
πlog2
1+2η b 2 ζ(1+η)2 2 1.261431
c (η)= log 1 + log + logζ(3 +2η)+ ,
2 πlog2 2π log2 ζ(2+2η) π 2 π
(cid:16) (cid:17) (30)
1
g(T)= log(3b T)+log(5 +T) log(η2+T2) +2. (31)
log2 1 2 −
Observe that (cid:0) (cid:1)
2 ζ(1+η)2 2
c (η)=b b η+ log + logζ(3 +2η)
2 2− 3 log2 ζ(2+2η) π 2
where b is defined in (21), and
1
log(b1) 1.261431
b = 2π + =0.267481..., (32)
2
π(log2) π
2log(b1)
b = 2π =0.268089... . (33)
3
− πlog2
Since 0<η 1, we have that, for T 1,
≤ 2 ≥
1 log(3b )
g(T) log(1+ 5 )+2+ 1 =7.622699....
≤ log2 2T log2
(cid:3)
WenowmoveontotheproofofTheorem2. First,werequireacoupleoflemmas
on some real-valued functions. For a,b,c,x R, define
∈
a b c
g(a,b,c;x):=κ + , (34)
a2+x2 b2+x2 − c2+x2
(cid:16) (cid:17)
with κ= 1 .
√5
Lemma 5. Let a = √5 1,b = 1+√5, and c =1.
0 2− 0 2 0
(i) The inequality
0.121585... g(a ,b ,c ;x) 0
0 0 0
− ≤ ≤
is valid for all x R.
∈
(ii) Let 0 < ǫ 10 2 and let a,b,c R. If a a < 2ǫ, b b < 2ǫ, and
− 0 0
≤ ∈ | − | | − |
c c <2ǫ, then
0
| − |
0.121585... 5ǫ g(a,b,c;x) 5ǫ.
− − ≤ ≤
10 HABIBAKADIRIANDNATHANNG
Proof. (i) Differentiating, we find that
g (a ,b ,c ;x)
′ 0 0 0
κa (b2+x2)2(c2+x2)2 κb (a2+x2)2(c2+x2)2+c (a2+x2)2(b2+x2)2
=2x− 0 0 0 − 0 0 0 0 0 0 .
(a2+x2)2(b2+x2)2(c2+x2)2
0 0 0
The polynomial in the numerator is of the form Ax8+Bx6+Cx4+Dx2+E, and
it may be checked that A = D = 0, B = 2, C = 4, and E = 1. Observe that
−
the polynomial2x6+4x4 1 has one positive realrootβ =0.672016.... Itfollows
−
from calculus that
0 g(a ,b ,c ;x) g(a ,b ,c ;β)= 0.121585....
0 0 0 0 0 0
≥ ≥ −
(ii) We begin by considering the difference
g(a,b,c;x) g(a ,b ,c ;x)
0 0 0
−
a a b b c c
0 0 0
=κ + + .
a2+x2 − a2+x2 b2+x2 − b2+x2 − c2+x2 c2+x2
0 0 0
(cid:16) (cid:17)
For real numbers u and u , we have that
0
u u (u u )(x2 uu ) u u
0 0 0 0
= − − | − | .
u2+x2 − u2+x2 (u2+x2)(u2+x2) ≤ min(u, u )2
(cid:12) 0 (cid:12) (cid:12) 0 (cid:12) | | | 0|
Using this(cid:12)bound, the triangle(cid:12)ine(cid:12)quality implies that(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
2ǫ 2ǫ 2ǫ
g(a,b,c;x) g(a ,b ,c ;x) κ + +
| − 0 0 0 |≤ (a 2ǫ)2 (b 2ǫ)2 (c 2ǫ)2
0 0 0
(cid:16) − − (cid:17) −
κ κ 1
2ǫ + + <5ǫ.
≤ (a 2 10 2)2 (b 2 10 2)2 (c 2 10 2)2
(cid:16) 0− · − 0− · − 0− · − (cid:17)
The above combined with (i) yields (ii). (cid:3)
Lets=σ+it ands =σ +itwhereσ = 1(1+√1+4σ2). Fora=0,1,define
1 1 1 2
the function
1 Γ s+a 1 Γ s +a
f (σ,t)= Re ′ ′ 1 .
a
2 Γ 2 − √5 Γ 2
(cid:18) (cid:18) (cid:19) (cid:18) (cid:19)(cid:19)
In order to abbreviate notation, we set ψ(z)= Γ′(z). We shall prove
Γ
Lemma 6. Let ǫ>0, σ [1,1+ǫ], and t [0,1]. Then
∈ | |∈
f (σ,t) C (ǫ),
a a
≤
where
1 1+a 1 1 1+ǫ 1+√5 +a 1
C (ǫ)=f (1,1)+ǫ S , + S 2 , (35)
a a
4 2 2 2√5 1+4(1+ǫ)2 2 2
(cid:16) (cid:16) (cid:17) (cid:16) (cid:17)(cid:17)
and p
∞ 1
S(x,y)= . (36)
(x+n)2+y2
n=0
X