A congruence sum and rational approximations

0.6 MB·

by  Sandro Bettin

A congruence sum and rational approximations
Sandro Bettin
7 January 25, 2017
1
0
2
Abstract
n
a We give a reciprocity formula for a two-variable sum where the variables satisfy a linear
J congruence condition. We also prove that such sum is a measure of how well a rational is
3 approximablefrombelowandshowthatthereciprocityformulaisasimpleconsequenceofthis
2 fact.

]
T 1 A congruence sum
N
. Sumsovertheintegerssubjecttosomecongruenceconditionsareubiquitousinnumbertheoryand
h
t often appear in several other areas of mathematics. They are related to many classical arithmetic
a
problems (e.g. that of primes in arithmetic progression, the Dirichlet’s divisor problem, etc.) and
m
evenwhentheyseemparticularlyinnocuousthereisactuallyalotofarithmeticinformationhidden
[ in them. This is the case of the sum

1
v S(a/q):= 1,
14 ammn≡<nq(cid:88),(mn,omd>q0),
6
6 where a,q Z, q > 0, and (a,q) = 1, which counts the number of points (m,n) in (Z/qZ)2 which
∈
0 belong to the line am n (mod q) and are contained in the hyperbolic region mn < q (here of
≡
. course we mean the representatives with 0 < m,n q). Notice that S can be interpreted as a
1
≤
0 1-periodic function defined over the rational numbers.
7 Two “visual” examples of such function are given in Figure 1 below. Notice that in both cases
1 S( 226,307)=S(307,226)=7andthetwographslookoverallverysimilar. Thisisactuallynota
: −
v coincidence: indeed one always have that S( a,q) is close to S(q,a) (for q =307 their difference is
−
i always 3 and is typically either 0 or 1, whereas the maximum value of S(a,307) is 17). A similar
X ≤
relation holds for S(a,q) and S( q,a), with the difference that in this case one also has a main
r −
a term. These two facts are expressed in the following theorem.
Theorem 1. Let 1 a<q. Then,
≤
S(a/q) S( q/a)= q/a+E (a,q), (1.1)
+
− −
S( a/q) S(q/a)=E (a,q), (1.2)
(cid:112)
− − −
with E (a,q) 3k +3wherek isthenumberofstepsintheEucliddivisionq/a(inparticular
| ± |≤ 2 a/q a/q
E (a,q) log(2+q)).
± (cid:28)
2010Mathematics subject classification. 11A55(primary),11A07,11A15,11N37,11J70(secondary)
Key words and phrases. Congruencesum;rationalapproximation,continuedfractions.
1
n n
(0,307) (0,226)
m m
(0,0) S( 226/307)=7 (307,0) (0,0) S(307/226)=7 (226,0)
�
FigFuigruer1e:1A: AvivsuisaulaclocmompuptuattaitoionnooffSS((aa//qq))iinntthheeccaassee((aa,,qq)) == (( 222266,,330077))aanndd((aa,,qq))==3(0370,72,2262).6).
ThTehpeopinotinstrserperperseesnetntththeecocouuppleless((mm,,nn)) 00<< rr <<qq 22ssaattiissffyyini�−nggaamm nn(m(mododq)q,)t,htehreedredonoenses
bebinegintghtehoenoensesbebleolwowththeehhyyppeerrbboolalammnn==∈2qq{{.. }} ⌘≡
The partoftheseinequalitieswasalreadyobtainedbyYoung[You]inhisextensionofConrey’s
The partoftheseinequalitieswasalreadyobtainedbyYoung[You]inhisextensionofConrey’s

recipro≤city formula for the twisted second moment of Dirichlet L-functions [Con]. Young also
reciprocity formula for the twisted second moment of Dirichlet L-functions [Con]. Young also
obtained a similar version of these formulas in the case when the sharp cut-o↵ mn q is replaced
obtained a similar version of these formulas in the case when the sharp cut-off mn q is replaced
byasmoothone,thatisbyinsertingafactoroff(mn/q)wheref(x)isasmoothfunc≤tiongoingto
by a smooth one, that is by inserting a factor of f(mn/q) where f(x) is a smooth function going
zero faster than any polynomial at + (see also [Bet] for an alternative treatment the smoothed
to zero faster than any polynomial at1+ (see also [Bet15] for an alternative treatment of the
case). The method used in this paper can∞[You]
smoothed case). We remark that the same method used to prove Theorem 1 can be used to obtain
Theorem 1 suggests that one can obtain a formula for S(a/q) in terms of the coe�cients of the
a simpler proof of the Conrey-Young reciprocity formula for the twisted second moment.
continued fraction expansion [0;b , ,b ] of a/q. Indeed one can repeatedly alternate the use of
1 k
Theorem 1 suggests that one can··o·btain a formula for S(a/q) in terms of the coefficients of the
one among (1.1) or (1.2) and the reduction modulo the denominator and obtain
continued fraction expansion [0;b , ,b ] of a/q. Indeed one can repeatedly alternate the use of
1 k
···
one among (1.1) or (1.2) and the reductio[(nkm1)o/d2]ulo the denominator and obtain
�
S(a/q)= b +O(k2 ).
2j+1 a/q
[(k 1)/2]
S(a/q)= −Xj=0 pb +O(k2 ).
2j+1 a/q
However, in fact one can prove directly a stronger form of this result, with O(k2 ) replaced by
(cid:88)j=0 (cid:112) a/q
O(k ), and then deduce Theorem 1 from it.
a/q
However, in fact one can prove directly a stronger form of this result, with O(k2 ) replaced by
a/q
O(Tfkraahc/eqtoi)o,rneamnexdp2at.hnesLinoentdeaod,fquac2.eTTZhheweniotrhemq >1 f0r,oma 6=it.0 and (a,q) = 1. Let [0;b1,··· ,bk] be the continued
q
Theorem 2. Let a,q Z with q > 0, a = 0 and (a,q) = 1. Let [0;b1, ,bk] be the continued
∈ (cid:54) ···
fraction expansion of aq. Then [(k�1)/2]
S(a/q)= [ b ]+E,
2j+1
[(k−Xj=10)/2]p
S(a/q)= [ b ]+E,
where 0 E 3k+1 log(2+q) and [x] denotes the2jin+t1eger part of x.
  2 ⌧ j=0
(cid:88) (cid:112)
Theorem2doesn’tprovideausefulasymptoticformulaformostvaluesofa. Indeed,astandard
wh(ebruet0somEewha3tkle+ng1thy)locogm(2p+utqa)tioanndw[ixth] dDeinroicthelsett’hsehiynpteergbeorlapamrtetohfoxd.shows that the average
≤ ≤ 2 (cid:28)
Theorem2doesn’tprovideausefulasymptoticformulaformostvaluesofa. Indeed,astandard
(but somewhat lengthy) computation with Dirichlet’s hyperbola method shows that the average
2
2
value of S(a/q) is
1 ϕ(q)σ (q)
1
S(a/q)= − (logq+O(loglogq)) logq,
ϕ(q) q (cid:16)
0<a<q,
(cid:88)
(a,q)=1
as q , where σ (q) = d 1 and ϕ is Euler’s totient function.1 However, Theorem 2 is
→ ∞ −1 d|q −
still useful as it determines exactly all the large values of S(a/q). Indeed, the Theorem shows that
(cid:80)
S(a/q)is“large”ifandonlyifthecontinuedfractionexpansionofa/q hasa“large”oddcoefficient.
Theorem 2 can also be used to prove some density results for S(a/q), following the method
introduced by Hickerson [Hic] for the Dedekind sum. For example, we can prove the following.
Corollary 1. Let κ>1. Then the set aq,loSgκ(a(2/+q)q) | aq ∈Q is dense in R×R>0.
Itwouldbeinterestingtounderstand(cid:110)(cid:16)howtheset (cid:17)(a,S(a/q(cid:111))) a Q isdistributedasq
{ q | q ∈ } →∞
(see[Var]and[Bet15]forthecomputationofthedistributionoftwosomewhatsimilarsumsviathe
use of reciprocity formulas close to (1.1) and (1.2)).
2 Rational approximations
Theorem 1 and 2 actually admit a rather simple explanation, once one realizes that S(a/q) can be
defined also in another simple and apparently unrelated way. Indeed, S(a/q) coincides with the
number of ways a/q can be “well approximated” from below by fractions of smaller denominator.
Thetheoryofcontinuedfractionstellsusthat“goodapproximations”ofarealnumberareobtained
byitsconvergents,whichthenexplainswhythecoefficientsofthecontinuedfractionexpansionarise
in Theorem 2.
Before stating the precise result, we give an example for the case of S( 226/307) considered
−
in Figure 1. The 7 points (m,n) appearing in the sum defining S( 226/307) are (235,1), (91,3),
−
(19,4), (38,8), (4,17), (8,34), (1,81). Now, compare them with the fractions r/s with s < 307
whichsatisfy0< 226 r < 1 . Thesearetheevenconvergents 1, 3, 14,the“double”ofthe
−307−s s2 −1 −4 −19
last two 6, 28, and two of the semi-convergents 67 and 173. Notice that the denominators
−8 −38 −91 −235
ofsuchfractionsarepreciselythem-coordinatesofthe7points! Thefollowingsimpletheoremtells
us that this is no coincidence.
Theorem 3. Let a,q Z, with q >0 and (a,q)=1. Then,
∈
S(a/q)=# (c,d) Z2 d>0, 0< a c < 1 . (2.1)
{ ∈ | q − d d2}
Proof of Theorem 3. Let0<n,m<q witham n(mod q). Letr bethelargestintegersuchthat
≡
r < a; clearly 0< a r < 1. Also,
m q q − m m
a r a r a r
am=q m=q + m q m (mod q). (2.2)
q m q − m ≡ q − m
(cid:18) (cid:19) (cid:18) (cid:19)
Thus, since 0<q(a r)m<q, we must have q(a r)m=n. We then have that mn<q if and
q − m q − m
only if q(a r)m2 <q and thus if and only if (a r)< 1 . Thus, we obtain the side of (2.2);
q −m q −m m2 ≤
it is clear that repeating the same argument in the opposite direction yields the other inequality
and so the proof is completed.
1Notice that if q is prime, then the average of S(a/q) reduces exactly to the Dirichlet’s divisor problem:
ϕ(1q)(cid:80)0<a<q,S(a/q)=(cid:80)n<qd(n),whered(n)isthenumberofdivisorsofn.
(a,q)=1
3
Notice that the set on the right hand side of (2.1) doesn’t have the condition (c,d) = 1, that
is we are not requiring the fractions c to be in reduced form. If one wants to count only reduced
d
fractions, then one immediately sees that S(a/q) corresponds to a weighted sum, where the weight
takes into account how good the approximation is:
1
S(a/q)= [ε−2],
c/d
(2.3)
(c,d)(cid:88)=1,d>0
0<a c 1
q−d≤d2
where ε = d2 a c . The theory of continued fractions then helps us recover Theorem 2
c/d |q − d|
from (2.3) and Theorem 1 will then follow easily.
Before proving the various results we make one last remark. If in S(a/q) we add the condition
(m,n)=1andincludealsothesolutionsofam n(mod q),thenwehavethefollowinganalogue
≡−
of Theorem 3:
1=# c Q , (c,d)=1, 0= a c < 1
{d ∈ | (cid:54) q − d d2}
ammn≡<±qn(cid:88), n(m,mod>0q,), (cid:12)(cid:12) (cid:12)(cid:12) (2.4)
(m,n)=1 (cid:12) (cid:12)
validforqprime(ifqisnotprimeonehastoaddthecondition(q,d)=1inthesetontherighthand
side). Now, using M¨obius inversion formula and the asymptotic for Dirichlet’s divisor problem, we
have that for q prime the average value of the left hand side is
1 12
1= logq+O(1),
ϕ(q) π2
a((am(cid:88),qo)d=1q),ammn≡<±qn(cid:88), n(m,mod>0q,),
(m,n)=1
as q . Also, it is known (cf. the next Section) that all the convergents to a (different from
→ ∞ q
a) are contained in the set of the right hand side of (2.4) and that, on average over a, there are
q
asymptotically 12log2logq convergents of a as q goes to infinity among primes [Hei]. In particular,
π2 q
on average over a and as q among primes, we have that log2 69.3% of the solutions to
→ ∞ ≈
a c < 1 are partial quotients of a, whereas 30.7% are not. (In Lemma 2 below we will see
|q − d| d2 q ≈
that these other solutions are certain semi-convergents of a/q).
We conclude this section by observing that the second moment of S(a/q) is very closely related
to the 4-th moment of Dirchtlet L-functions at the central point. In particular, Theorem 3 opens
analternativeapproachtothisproblemviamethodsofDiophantineapproximation(seealso[CK]).
3 Proofs of Theorems 2 and 1 and of Corollary 1
Proof of Theorem 2. Itiswellknown(see[Khi], Chapter1)thatallconvergentsh /k ofa/q (with
j j
hj = a) satisfy
kj (cid:54) q
1 h a 1 1
j
< < <
k (k +k ) k − q k k k2
j j+1 j (cid:12) j (cid:12) j j+1 j
(cid:12) (cid:12)
and that one has hj < a if and only if j is (cid:12)(cid:12)even, so(cid:12)(cid:12)that all the even convergents appear in the
kj q
sum (2.3). Also, the above inequalities give
k k
j j
<ε <
(k +k ) hj/kj k
j+1 j j+1
4
andso,sincekj+1 =bj+1kj+kj−1 and0≤kj−1 <kj forj ≥0,weobtainbj+1 ≤ε−hj1/kj ≤bj+1+2.
1 1 1
By the inequality [√x+2] [√x] 1 for x 1 we then have [b2 ] [ε−2 ] [b2 ]+1.
− ≤ ≥ j+1 ≤ hj/kj ≤ j+1
The solutions c/d to 0 < c a < 1 are not all convergents of a/q, however this is not far
|d − q| d2
for being true. Indeed, all solutions c/d with (c,d)=1 of the stricter inequality 0< c a < 1
|d − q| 2d2
are convergents (see [Khi], Theorem 19), so that ε 1 2 if c/d is not a convergent. In
c/d −
| | ≤
particular, (2.3) gives
k
1
S(a/q)= [bj2]+S∗(a/q)+Ea,q
j=1,
(cid:88)
j odd
where 0 (k +1)/2 and S (a/q) is the number of reduced rationals c/d satisfying 0 <
a,q ∗
≤ E ≤
a c < 1 which are not convergents of a/q. By Lemma 2 below S (a/q) is bounded by twice
q − d d2 ∗
the number of even convergents of a (different from a/q), so that S (a/q) k and the proof of
q ∗ ≤
Theorem 2 is completed.
Lemma 2. Let aq ∈ Q with continued fraction expansion [b0;b1,...,bk] and convergents hkjj for
1 j k. Then, every c Q which satisfies 0 = a c < 1 with (c,d) = 1 is either a
− ≤ ≤ d ∈ (cid:54) |q − d| d2
convergent hj of a (with 1 j k 1) or a semi-convergent of the form hj+ghj+1 with g =1 or
kj q − ≤ ≤ − kj+gkj+1
g =b 1 (and 1 j k 2). Moreover, in both cases c/d<a/q if and only if j is even.
j+2
− − ≤ ≤ −
Proof. We assume a < c, the proof being essentially identical otherwise. Now, if 0 < a c < 1
q d q − d d2
with (c,d)=1, then c is a “best rational approximation from below” for a/q, that is c is closer to
d d
a than any rational which is less than a/q and have a smaller denominator. Indeed, if 0 < d < d
q (cid:48)
and 0< aq − dc(cid:48)(cid:48) ≤ aq − dc for some c(cid:48) ∈Z, then
1 c c a c a c 1
(cid:48) (cid:48)
<
dd ≤ d − d ≤ q − d − q − d d2
(cid:48) (cid:48) (cid:18) (cid:19) (cid:18) (cid:48)(cid:19)
which gives a contradiction. Now, all the best rational approximation from above or below for a
q
which are not convergents are semi-convergents of the form hj+ghj+1 for g an integer satisfying
kj+gkj+1
1 g < b (Theorem 15 of [Khi] proves this for best rational approximations, but the proof
j+2
≤
carries over also for best approximations from below or from above); also this fraction is smaller
than a if and only if j is even. We will now show that among these only the values g = 1 and
q
g =b 1 might be such that the inequality hj+ghj+1 a < 1 is satisfied.
j+2− |kj+gkj+1 − q| (kj+gkj+1)2
Clearly we can assume b 4, since otherwise there is nothing to prove. For j even we have
j+2
≥
k h k h =1 and so
j+1 j j j+1
−
h h +gh b g
j+2 j j+1 j+2
= − .
k − k +gk (k +b k )(k +gk )
j+2 j j+1 j j+2 j+1 j j+1
Since hj+ghj+1 < hj+2 a, then it follows that in order to have a hj+ghj+1 < 1 it is
kj+gkj+1 kj+2 ≤ q q − kj+gkj+1 (kj+gkj+1)2
necessary that
(b g)(k +gk )<(k +b k ).
j+2 j j+1 j j+2 j+1
−
5
and a fortiori one must have (b g)g <b , since b g 1. Solving for g we obtain
j+2 j+2 j+2
− − ≥
b b2 4b b + b2 4b
j+2− j+2− j+2 j+2 j+2− j+2
g < or g >
(cid:113) 2 (cid:113) 2
and thus, reminding that g is a positive integer greater than or equal to 2, the only possibilities are
g =1 and g =b 1, as desired.
j+2
−
Proof of Theorem 1. A simple computation shows that if a/q = [b ;b ,b , b ] with b 0 and
0 1 2 k 0
a=0, then the continued fraction expansion of q is ··· ≥
(cid:54) −a
q [ b 1;1,b 1,b ,b ,...,b ] if b =1,
1 2 3 4 k 2
= − − − (3.1)
−a (cid:40)[ b1 1;b3+1,b4,b5,...,bk] if b2 >1,
− −
if b =0 and k 2 (if k =1 then q = b ) and
0 ≥ −a − 1
q [ 1;1,b 1,b ,b ,...,b ] if b =1,
0 1 2 k 0
= − − (3.2)
−a (cid:40)[ 1;b1+1,b2,b3...,bk] if b0 >1,
−
ifb >0(ifk =0andb =1then q = 1). Equation(1.1)thenfollowsfrom(3.1)andTheorem2
0 0 −a −
upon observing that b q/a<b +1 and thus [√b ] q/a<[√b ]+1. Equation (1.2) follows
1 1 1 1
≤ ≤
in the same way from (3.2) exchanging the roles of a and q.
(cid:112)
Proof of Corollary 1. Let (x,y) R R>0. For every fixed ε > 0 we need to find a/q such
that a/q x < ε and S(a/q)/∈logκ(2×+q) y < ε. Since S(a/q) is 1-periodic, we can assume
| − | | − |
0<x<1. Let b1, ,b2r be such that x [0;b1, b2r] ε/2 for some r ε 1. For m,n N, let
··· | − ··· |≤ (cid:28) ∈
a =[0;b , b ,m,n] so that q =(dm+d)n+d for some d,d 1 (considering b , b ,x,y,r
q 1 ··· 2r (cid:48) (cid:48) (cid:28) 1 ··· 2r
as fixed). Thus, by Theorem 2, we have
S(a/q) √m
= +o(1).
logκ(2+q) logκ(2+mn)
as m,n . We take n=[em21κy−κ1] and the Corollary follows by taking m large enough.
→∞
Acknowledgement
ThisnotewasmostlywrittenwhiletheauthorwasapostdoctoralfellowattheCentrederecherches
math´ematiques in Montr´eal.
References
[Bet15] Bettin, S. On the distribution of a cotangent sum. Int Math Res Notices (2015) 2015 (21):
11419–11432.
[Bet16] Bettin, S. On the reciprocity law for Dirichlet L-functions. Trans. Amer. Math. Soc. 368
(2016), 6887–6914.
[Con] Conrey, J.B. The mean-square of Dirichlet L-functions. arxiv math.NT/0708.2699.
6
[CK] Conrey, J.B.; Keating, J.P.Moments of Zeta and Correlations of Divisor-Sums: II.Advances
in the Theory of Numbers. Volume 77 of the series Fields Institute Communications pp. 75–85.
[Hei] Heilbronn, H. On the average length of a class of finite continued fractions. 1969 Number
Theory and Analysis (Papers in Honor of Edmund Landau) pp. 87–96 Plenum, New York
[Hic] Hickerson, D. Continued fractions and density results for Dedekind sums. J. Reine Angew.
Math. 290 (1977), 113–116.
[Khi] Khinchin, A.Y. Continued fractions. The University of Chicago Press, Chicago, Ill.-London
1964.
[Var] Vardi, I. Dedekind sums have a limiting distribution. Int Math Res Notices (1993) 1993 (1):
1–12.
[You] Young, M.P. The reciprocity law for the twisted second moment of Dirichlet L-functions.
Forum Math. 23 (2011), no. 6, 1323–1337.
S.Bettin,DipartimentodiMatematica,Universita`diGenova;viaDodecaneso35;16146Genova;Italy.
E-mail address: [email protected]
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