The Hurwitz continued fraction expansion as applied to real numbers

0.14 MB·

by  David Simmons

THE HURWITZ CONTINUED FRACTION EXPANSION AS APPLIED TO REAL
NUMBERS
DAVIDSIMMONS
6
1
0 Abstract. Hurwitz (1887) defined a continued fraction algorithm for complex numbers which is better
2 behaved in many respects than a more “natural” extension of the classical continued fraction algorithm
to the complex plane would be. Although the Hurwitz complex continued fraction algorithm is not
n
“reducible”toanother complexcontinued fractionalgorithm,overtherealsthestoryisdifferent. Inthis
u
note we make clear the relation between the restriction of Hurwitz’s algorithm to the real numbers and
J
theclassicalcontinuedfractionalgorithm. AsanapplicationwereprovethemainresultofChoudhuriand
9
Dani(2015).
2

]
T
N 1. Hurwitz’s algorithm
.
h Let x be a complex number such that x∈/ Q(i). The (positive) Hurwitz continued fraction expansion of
t x (see [2, 3, 4]) is defined to be the expression
a
m
1
(1.1) a + ,
[ 0
1

a +
3 1 ..
v a2+ .
8 where the Gaussian integers (a )∞ (the partial quotients) and the complex numbers (x )∞ are chosen
3 n 0 n 0
recursively according to the Hurwitz algorithm:
8
7 • x =x.
0
0 • If x is defined, then a is the Gaussian integer closest to x , which we denote by [x ].1
n n n n
.
1 • If xn and an are both defined, then xn+1 =1/(xn−an).
0
It is not hard to see that the Hurwitz continued fraction expansionof x alwaysconvergesto x, and in fact
6
the corresponding partial quotients are in some sense the “best approximations possible” [2, Theorem 1].
1
: Someauthors[1,5]alsoconsiderthenegativeHurwitzcontinuedfraction expansion ofanumberx,which
v
is the expression
i
X 1 1
a − =a − ,
r 0 1 0 1
a a − −a −
1 1
e a −... a −...
2 2
where (a )∞ are defined in the sameeway as in the positive Hurwitz continued fraction expansion, and
n 0
an =(−1)nan. Note that by the identity e
1 1
(1.2) x+ =x− ,
e y+z −y−z
theconvergentsofthenegativeHurwitzcontinuedfractionexpansionarethesameastheconvergentsofthe
positive Hurwitz continued fraction expansion. Thus for many purposes, it is not necessary to distinguish
between the positive and negative Hurwitz expansions.
Inthis note,weconsiderthe restrictionofthe Hurwitz algorithmtothe realline. Inthis case,itis clear
that the numbers (x )∞ and (a )∞ will all be real. Moreover, unlike the case of the complex Hurwitz
n 0 n 0
1Thetiebreakingmechanismisnotrelevantforthepurposesofthispaper,butforthesakeofdefinitenessletus(agreeing
with[2])set[x]=[Rex]+i[Imx],where[t]denotes theintegernearesttot∈R,roundeddowninthecaseofatie.
1
2 DAVIDSIMMONS
expansion,it is possible to say exactly when a sequence (a )∞ is the sequence of partialquotients of some
n 0
real number:
Proposition 1.1. 2 For a sequence of integers (a )∞, the following are equivalent:
n 0
(A) The expression (1.1) is the Hurwitz continued fraction expansion of some (irrational) real number.
(B) For all n≥1, we have |a |≥2, with a a >0 if equality holds.
n n n+1
(C) For all n≥1, we have |a |≥2, with a a <0 if equality holds.
n n n+1
Obviously, (B) and (C) are reformulations of each other, so we prove (A) ⇔(B):
e e e
Proof of (A) ⇒(B). By definition, for all n ≥ 0 we have |x − a | ≤ 1/2 and thus |x | ≥ 2 and
n n n+1
|a |≥2. Ifequalityholds,then a hasthe same signasx −a , whichin turnhasthe same sign
n+1 n+1 n+1 n+1
as a . (cid:3)
n+2
Proof of (B) ⇒(A). For each n,N with n≤N, let
1
x =a + ·
n,N n
.. 1
.+
a
N
Reverse induction on n shows that whenever n≥1, we have |x |≥2 and a =[x ]. It follows that
n,N n n,N
1 1
|x −x |≤ |x −x | and |x −x |≤ ,
n,M n,N n+1,M n+1,N N,M N,N
4 2
which implies that |x −x |≤(1/4)min(M,N)−n, and thus for each n the limit
n,M n,N
1
x = lim x =a +
n n,N n
N→∞ 1
a +
n+1
..
.
exists. We have a = [x ] and x = 1/(x −a ), and thus (1.1) is the Hurwitz continued fraction
n n n+1 n n
expansion of x . (cid:3)
0
2. Relation to the classical algorithm
We nowshowthatthe restrictionofthe Hurwitz algorithmto the realline isinsomesense“equivalent”
to the classical continued fraction algorithm:
Theorem 2.1. The sequence of convergents of the Hurwitz continued fraction expansion of a real number
x is a subsequence of the sequence of convergents of the classical continued fraction expansion of x. This
sequence has the property that it omits no two consecutive convergents, and it also contains all rational
approximants p/q that satisfy the inequality |x−p/q|≤1/(3q2).
Proof. The key to the proof is the identity
1 1
(2.1) =1− ,
1 n+1+y
1+
n+y
which is easily verified for all n and y. Now let us denote the classical continued fraction expansion of a
real number x by
1
(2.2) b + ,
0
1
b +
1
b +...
2
so that (b )∞ is a sequence of integers and b ≥ 1 for all n ≥ 1. Let S = {n ≥ 1 : b = 1}, and let S′ be
n 0 n n
the unique subset of S with the following property:
2Thispropositionisnotoriginal;thewordingof[1]seemstosuggestthatitwasproveninthedifficult-to-find[5].
THE REAL HURWITZ CONTINUED FRACTION EXPANSION 3
• For all n∈S, we have either n∈S′ or n−1∈S′, but not both.
The set S′ can be constructed by taking each “block” of S and selecting “every other element”, starting
from the first element of that block; for example, if S = {1,4,5,6,9,10}, then S′ = {1,4,6,9}, since the
“blocks” are {1}, {4,5,6},and {9,10}.
For each n∈S′, in the expression (2.2) we replace
1 1
by 1−
1 ..
bn+ (bn+1+1)+ .
b +...
n+1
according to (2.1); this is possible since b =1. This results in an expression of the form
n
1
(2.3) c +ε +(−1)ε0 ,
0 0
1
c1+ε1+(−1)ε1
1
c2+ε2+(−1)ε2
...
where ε ,ε ,···∈{0,1}, and c ≥2 for all n≥1. Here, we have used the facts that 1 =0+(−1)01 and
0 1 n x x
1− 1 = 1+(−1)11 to represent the expressions 1 and 1− 1 in a uniform manner as ε+(−1)ε1, where
x x x x x
ε∈{0,1}. Repeatedly applying the identity (1.2) yields the Hurwitz expansionof x, so the convergentsof
(2.3) are the same as the convergentsof the Hurwitz expansion. But these are precisely those convergents
p /q of the classical expansion (2.2) such that n ∈/ S′. So the sequence of partial convergents of
n−1 n−1
the Hurwitz expansion is a subsequence of the sequence of convergents of the classical expansion, which
omits no two consecutive convergents (by the definition of S′). The omitted convergents are of the form
p /q , where n∈S′, and these convergents satisfy
n−1 n−1
p 1 1 1 1
n−1
x− > = > =
q q (q +q ) q (b q +q +q ) (b +2)q2 3q2
(cid:12) n−1(cid:12) n−1 n n−1 n−1 n n−1 n−2 n−1 n n−1 n−1
(cid:12) (cid:12)
(cf. [6, T(cid:12)heorem 1(cid:12)3]). Here we have used the fact that bn = 1 for all n ∈ S′. On the other hand,
(cid:12) (cid:12)
approximants that are not convergents of the classical expansion satisfy |x−p/q| ≥ 1/(2q2) [6, Theorem
19]. SoallapproximantsthatarenotconvergentsoftheHurwitzexpansionsatisfy|x−p/q|>1/(3q2). (cid:3)
Aside from the relation between the sequences of convergents described in Theorem 2.1, the Hurwitz
continuedfractionexpansionalsosharesthefollowingformalsimilaritywiththeclassicalcontinuedfraction
expansion:
Proposition 2.2. Let (a )∞ be the sequenceof Hurwitz (resp. classical) partial quotients of a real number
n 0
x. If the sequences (p )∞ and (q )∞ are defined recursively via the formulas
n −2 n −2
(2.4) p =1, p =0, q =0, q =1,
−1 −2 −1 −2
(2.5) p =a p +p , q =a q +q ,
n n n−1 n−2 n n n−1 n−2
then (p /q )∞ is precisely the sequence of Hurwitz (resp. classical) convergents of x.3
n n 0
Proof. The proof of [6, Theorem 1] is valid for both the classical and Hurwitz setups, since both use the
same formal expressions for the convergents and partial quotients. (cid:3)
However, there are differences from the classical algorithm as well. For example, while the error terms
p /q −x corresponding to the classical convergents always alternate in sign [6, Theorem 4], the error
n n
terms corresponding to the Hurwitz convergents can be described as follows:
3Note that in the Hurwitz case there is some ambiguity as to how to represent each convergent as a fraction (p/q vs.
(−p)/(−q)), and this proposition gives a way to resolve this ambiguity (namely to take the sequences (pn)∞0 and (qn)∞0
definedbytherecursiverelations). Theambiguitywouldberesolvedinthesamewayifonetooktheexpressiondefiningthe
convergent andsimplifieditrepeatedlyaccordingtotherules(p/q)−1=q/pandn+p/q=(nq+p)/q.
4 DAVIDSIMMONS
Proposition 2.3. If p /q is the nth convergent of the Hurwitz algorithm, then the sign of the error term
n n
p /q −x is the same as the sign of the nth partial quotient a , i.e. (−1)n+1 times the sign of the nth
n n n+1
partial quotient a .
n+1
Proof. Since x and a share the same sign, comparing e
n+1 n+1
p 1 1
n
=a + vs. x=a +
0 0
qn ...+ 1 ...+ 1
an+0 1
a +
n
x
n+1
yields the desired conclusion. (cid:3)
Another difference between the Hurwitz and classical expansions is that the Hurwitz expansion yields
a faster rate of exponential growth for the denominators of the convergents. In the classical setup, the
sequence(q )∞ alwayssatisfiesliminf q /q ≥2>1,4butitis possiblethatliminf q /q =
n 0 n→∞ n+2 n n→∞ n+1 n
1. By contrast:
Proposition 2.4. If p /q denotes the nth convergent of the Hurwitz algorithm, then for all n≥1,
n n
(2.6) (|a |−(2−φ))|q |<|q |<(|a |+(φ−1))|q |,
n n−1 n n n−1
where φ denotes the golden ratio. In particular,
(2.7) |q |>φ|q |.
n n−1
Proof. For each n ≥ 1 let y = q /q . Then by (2.4) and (2.5), we have y = 0, and for all n ≥ 1 we
n n−1 n 0
have
1
y = ·
n
a +y
n n−1
By induction, |y |<1 for all n≥1, so y shares the same sign as a . We will prove by induction that
n−1 n n
(2.8) −(2−φ)<y sgn(a )<φ−1
n−1 n
for all n. The base case n=1 is trivial, so suppose that (2.8) holds for some n≥1. Then
|y |= 1 < 1 ≤ φ1 |an|=2 = φ−1 |an|=2·
n |an|+yn−1sgn(an) |an|−(2−φ) (1+1φ |an|≥3 (2−φ |an|≥3
To complete the inductive step, we need to show that if |a | =2, then y sgn(a )>0. But this follows
n n n+1
from Proposition 1.1, since sgn(y )=sgn(a ).
n n
Combining (2.8) with the formula
|q |=(|a |+y sgn(a ))|q |
n n n−1 n n−1
demonstrates (2.6). Finally, the inequality |a |≥2 gives (2.7). (cid:3)
n
3. Relation with Diophantine approximation
Although the connection between the classicalcontinued fraction expansionof a realnumber x and the
Diophantine properties of x has been dealt with extensively in a number of places, the connection with
the Hurwitz algorithm has not been stated precisely before. Many results can be proven simply from the
identification of the Hurwitz convergentsequence with a subsequence of the classicalconvergentsequence,
i.e. Theorem 2.1. For brevity we do not list these here. One place where a difference does appear is in the
basic estimates for the accuracy of the approximationof a convergent. In the classical setting, we have
1 p 1
n−1
< x− <
(b +2)q2 q b q2
n n−1 (cid:12) n−1(cid:12) n n−1
(cid:12) (cid:12)
(e.g. this follows from [6, Theorems 9 and 13]). By(cid:12) contrast,(cid:12)in the Hurwitz setup we have:
(cid:12) (cid:12)
4Ingeneral, qn+k/qn is always atleast the (k+1)st Fibonacci number. Thisisbecause ifFk denotes the kthFibonacci
number,thenaninductionargumentshowsthatqn+k=Fk+1qn+Fkqn−1.
THE REAL HURWITZ CONTINUED FRACTION EXPANSION 5
Proposition 3.1. If p /q denotes the nth convergent of the Hurwitz expansion of x, then
n n
1 p 1
n−1
< x− < ·
(|a |+(φ−0.5))q2 q (|a |−(2.5−φ))q2
n n−1 (cid:12) n−1(cid:12) n n−1
(cid:12) (cid:12)
Proof. By [6, Theorem 5], we have (cid:12) (cid:12)
(cid:12) (cid:12)
x p +p
n n−1 n−2
x= ,
x q +q
n n−1 n−2
where x is as in the definition of the Hurwitz algorithm, i.e.
n
1
x =a + ·
n n
a +...
n+1
Thus
p (q x p +q p )−(p x q +p q )
q2 x− n−1 =q n−1 n n−1 n−1 n−2 n−1 n n−1 n−1 n−2
n−1 q n−1 x q +q
(cid:12) n−1(cid:12) (cid:12) n n−1 n−2 (cid:12)
(cid:12) (cid:12) (cid:12) 1 1 1 (cid:12)
(cid:12) (cid:12)=q (cid:12) = = (cid:12) ·
(cid:12) (cid:12) n−1|(cid:12)x q +q | |x +q /q | |x −a +q /q(cid:12) |
n n−1 n−2 n n−2 n−1 n n n n−1
Since |x −a |≤1/2, combining with (2.6) completes the proof. (cid:3)
n n
4. Comparison with Choudhuri and Dani (2015)
Inthis sectionweshowthatbycombiningtheresultsofprevioussectionsinanappropriateway,wecan
strengthena resultof ChoudhuriandDani [1]. We state andproveour theorembelow andthen show that
it implies the main result of [1].
Theorem 4.1. Let(a )∞ be theHurwitzpartial quotient sequenceof arealnumber x, and fix0<δ ≤1/3.
n 0
For each ρ>0, let
#{(p,q)∈Z2 primitive:0<q ≤ρ, |q(qx−p)|≤δ}
(4.1) X = ·
ρ
log(ρ)
Then
# j =1,...,n:|a |≥δ−1+(2.5−φ)
j+1
(4.2) liminfX ≥liminf
ρ→∞ ρ n→∞ (cid:8) nj=1log(|aj|+(φ−1)) (cid:9)
# j =1P,...,n:|aj+1|≥δ−1−(φ−0.5)
(4.3) limsupX ≤limsup
ρ n
ρ→∞ n→∞ (cid:8) j=1log(|aj|−(2−φ)) (cid:9)
Proof. By Theorem 2.1, the condition δ ≤ 1/3 impPlies that the set appearing in (4.1) contains only pairs
(p,q)suchthatp/q is aconvergentofthe Hurwitz expansionofx. Thus the numeratorof (4.1)is constant
with respect to ρ along intervals of the form (|q |,|q |), and increases by at most 1 from |q |−o(1) to
n−1 n n
|q |+o(1). It follows that liminf X =liminf X , andsimilarly for the limsup. Now, applying
n ρ→∞ ρ n→∞ |qn|
Theorem 2.1 again, we have
#{j =1,...,n:|q (q x−p )|≤δ}
j j j
X = ·
|qn| log|q |
n
To finish the proof, we have to bound this expression between the corresponding expressions in the right
hand sides of (4.2) and (4.3). And indeed, by Proposition 3.1 we have
|a |≥δ−1+(2.5−φ) ⇒ |q (q x−p )|≤δ ⇒ |a |≥δ−1−(φ−0.5)
j+1 j j j j+1
and thus
#{j =1,...,n:|a |≥δ−1+(2.5−φ)}≤#{j =1,...,n:|q (q x−p )|≤δ}
j+1 j j j
≤#{j =1,...,n:|a |≥δ−1−(φ−0.5)}.
j+1
6 DAVIDSIMMONS
On the other hand, iterating (2.6) and taking logarithms gives
n n
log(|a |−(2−φ))≤log|q |≤ log(|a |+(φ−1))
j n j
j=1 j=1
X X
and dividing these two pairs of inequalities completes the proof. (cid:3)
We now show that Theorem4.1 implies the main resultof[1]. Since the statementofthe main theorem
of that paper contains a few inaccuracies, we state a corrected version here, which is equivalent to the
version that appears in the authors’ erratum (currently unpublished, but available from the authors upon
request).
Theorem 4.2 (Corrected version of [1, Theorem 1.1]). Let Q(p,q) = (aq +bp)(cq+dp) be a quadratic
form, where a,b,c,d∈R, ad−bc=1, b6=0, and a is irrational. Let (a )∞ be the Hurwitz partial quotient
b n 0
sequence of a. Let
b
n n
1
α− =liminf log|a |, α+ =limsup log|a |.
j j
n→∞ n n→∞
j=1 j=1
X X
For each A>0 let
1
D−(A)=liminf #{j =1,...,n:|a |≥A}
j+1
n→∞ n
1
D+(A)=limsup #{j =1,...,n:|a |≥A}.
j+1
n
n→∞
Fix 0<δ < 1, and let e(δ)=D−(δ−1+1) and f(δ)=D+(δ−1−3). Let κ>0 be fixed and for each ρ>0
π 2
let
G(ρ)={(p,q)∈Z2 primitive:0<|Q(p,q)|<δ, cq+dp>κ, k(p,q)k≤ρ}.
Then we have the following:
(i) if α+ <∞ then there exists ρ such that for all ρ≥ρ we have
0 0
e(δ)
#G(ρ)≥ log(ρ);
α++3
(ii) Let M = max(1log(9), 1α−) if α− < ∞, and let M < ∞ be arbitrary if α− = ∞. Then for any
4 5 8
m>f(δ), there exists ρ such that for all ρ≥ρ we have
0 0
m
#G(ρ)≤ log(ρ).
M
Proof using Theorem 4.1. Let x=−a and y =bd. Since ad−bc=1, we have
b
Q(p,q)=(qx−p)(q+y(qx−p))
and thus
|Q(p,q)| |Q(p,q)|
(4.4) lim = lim =1.
(p,q)∈G(∞)|q(qx−p)| (p,q)∈Z2 |q(qx−p)|
k(p,q)k→∞ |q(qx−p)|≤1
q→∞
Moreover,the Hurwitz partialquotientsequence of x is the same as the Hurwitz partialquotient sequence
of a except for minus signs.
b
Fix 0<δ < 1. We prove (i) and (ii):
π
(i) Since 1 > 2.5−φ, there exists 0 < δ < δ < 1/π < 1/3 such that δ−1+1 ≥ δ−1 +(2.5−φ). It
follows that
e(δ)=D−(δ−e1+1)≤D−(δ−1+(2.5−φ)). e
Now by (4.4), we have
e
X (δ)≤#G(ρ)+C
ρ
e
THE REAL HURWITZ CONTINUED FRACTION EXPANSION 7
for some constant C depending on δ and δ. Thus if α+ <∞, then
#G(ρ) D−(δ−1+(2.5−φ)) e(δ)
liρm→i∞nf log(ρ) ≥liρm→i∞nfXρ(δ)e(4≥.2) α++(φ−1) > α++3,
e
which implies (i). e
(ii) Since 3 > φ−0.5, there exists 0 < δ < δ < 1/π < 1/3 such that δ−1− 3 ≤ δ−1−(φ−0.5). It
2 2
follows that
f(δ)=D+(δ−1− 3e)≥D+(δ−1−(φ−0.5)). e
2
Now by (4.4), we have
e
X (δ)≥#G(ρ)−C
ρ
for some constant C depending on δ and δ. Thus
e
#G(ρ) D+(δ−1−(φ−0.5)) f(δ)
liρm→i∞nf log(ρ) ≤liρm→i∞nfXρ(δ)(4≤.3) e α−−(2−φ) ≤ max(14log(95),81α−),
e
which implies (ii). In the last ineequality, we have used the bound
α−−(2−φ)≥max 1log(9),1α− ,
4 5 8
which follows from the fact that α− ≥ log(2)(cid:0)(cf. Propositio(cid:1)n 1.1) together with the numerical
bound
log(2)−(2−φ)>max 1log(9),1log(2) . (cid:3)
4 5 8
Acknowledgements. The author was supported(cid:0)by the EPSRC P(cid:1)rogramme Grant EP/J018260/1.
The author thanks the anonymous referee for helpful comments.
References
1. Manoj Choudhuri and S. G. Dani, On values of binary quadratic forms at integer points, Math. Res. Lett. 22 (2015),
no.4,1023–1045. MR3391875
2. D. Hensley, The Hurwitz complex continued fraction, http://www.math.tamu.edu/~doug.hensley/SanAntonioShort.pdf,
preprint2006.
3. Doug Hensley, Continued fractions, World Scientific Publishing Co. Pte. Ltd., Hackensack, NJ, 2006. MR 2351741
(2009a:11019)
4. A. Hurwitz, U¨ber die Entwicklung complexer Gr¨ossen in Kettenbru¨che, Acta Math. 11 (1887), no. 1-4, 187–200.
MR1554754
5. SvetlanaKatokandIlieUgarcovici,Arithmetic coding of geodesics on the modular surface via continued fractions,Euro-
peanwomeninmathematics—Marseille2003,CWITract,vol.135,CentrumWisk.Inform.,Amsterdam,2005,pp.59–77.
MR2223106 (2006k:37075)
6. A.Y.Khinchin,Continued fractions,TheUniversityofChicagoPress,Chicago,Ill.-London,1964.
University of York,DepartmentofMathematics,Heslington,York YO105DD, UK
E-mail address: [email protected]
URL:https://sites.google.com/site/davidsimmonsmath/