Table Of ContentBounding the Growth Rate of Approximation
Coefficients using the Symmetry of Jager Pairs
Avraham Bourla
2
1 Department of Mathematics
0
2 Saint Mary’s College of Maryland
l Saint Mary’s City, MD, 20686
u
J [email protected]
6
1
July 17, 2012
]
T
N
h. Abstract
t
a
Symmetrical subdivisions in the space of Jager Pairs will provide us with bounds on
m
the rate of growth for the sequence of approximation coefficients corresponding to
[
continued fractions like expansions. Results will also apply to the classical regular
3
andbackwardscontinuedfractionsexpansions,whicharerealizedasspecialcases.
v
7
6
0 1 Introduction
1
.
7
Given a real number r and a rational number, written as the unique quotient p of the rel-
0 q
2 atively prime integers p and q > 0, our fundamental object of interest from diophantine
1 (cid:12) (cid:12)
: approximationistheapproximationcoefficientθ(r, p) := q2(cid:12)r− p(cid:12). Smallapproximation
v q (cid:12) q(cid:12)
i coefficients correspond to high quality approximations, combining accuracy with sim-
X
plicity. For instance, the error in approximating π using 355 = 3.1415920353982 is smaller
r 113
a than the error of its decimal expansion to the fifth digit 3.14159 = 314159. Since the former
100000
rational also has a much smaller denominator, it is of far greater quality than the latter.
(cid:0) (cid:1) (cid:0) (cid:1)
Indeedθ π, 355 < 0.0341whereasθ π, 314159 > 26,535.
113 100000
Adding any integer to a fraction does not change its denominator, hence we may restrict
ourattentiontotheunitinterval. Morespecifically,wehaveθ(r, p) = θ(x, p −(cid:98)r(cid:99)),where
q q
(cid:98)r(cid:99) is the largest integer smaller than or equal to r (a.k.a. the the floor of r) and x :=
r −(cid:98)r(cid:99) ∈ [0,1). Expanding an irrational initial seed x ∈ (0,1)−Q as an infinite regular
0
continuedfraction
1
x = [a ,a ,a ,...] :=
0 1 2 3 0 a + 1
1 a2+a3+1...
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AvrahamBourla
provides us with the unique symbolic representation via the sequence {a }∞ of positive
n 1
integers,knownasthepartialquotientsordigitsofexpansion. Foralln ≥ 1,welabelthe
approximationcoefficientassociatedwiththenth convergentofx , pn := [a .a ,...,a ] by
0 qn 1 2 n 0
θ andrefertothesequence{θ }∞ asthesequenceofapproximationcoefficients.
n n 1
(cid:12) (cid:12)
For all n ≥ 1, it is well known [3, Theorem 4.6] that (cid:12)x − pn(cid:12) < 1 < 1 , hence
(cid:12) 0 qn(cid:12) qnqn+1 qn2
θ < 1. Conversely, Legendre [3, Theorem 5.12] proved that if θ(x , p) < 1 then p is
n 0 q 2 q
a convergent of x . In 1891, Hurwitz [5, Theorem 5.1.4] showed there exist infinitely
0
(cid:16) (cid:17)
many pairs of integers p and q such that the inequality θ x , p < √1 is sharp. Conclude
0 q 5
that all irrational numbers possess infinitely many high quality approximations using
rational numbers, whose associated approximation coefficients are less than √1 , all of
5
which must belong to the sequence of regular continued fraction convergents for x . In
0
fact, the value of liminf{θ (x )} measures how well can x be approximated by rational
n 0 0
n→∞
numbers,leadingtotheconstructionoftheLagrangeSpectrum(formoredetails,referto
[4]). In1978,JurkatandPeyerimhoff[8]showedthattheSpaceofJagerPairs
(cid:8)(θ (x ),θ (x )) ⊂ R2, x ∈ (0,1)−Q, n ≥ 1(cid:9)
n 0 n+1 0 0
isadensesubsetoftheinteriorofthetriangleinthecartesianplanewithvertices(0,0), (0,1)
and(1,0).
Our goal is two folded: we will first show that the symmetry in this space discovered in
2002byDajaniandKraaikamp[5],extendstothetwooneparameterfamiliesofcontinued
fractions-like expansions, first introduced in 2003 by Haas and Molnar [6]. In particular,
theresultswillapplytotheSpaceofJagerPairsassociatedwiththebackwardscontinued
fractionsexpansion
1
x = 1− .
0 a +1− 1
1 a2+1−a3+1...
Next, a simple plane geometry argument will provide us with upper and lower bounds
for the growth rate of the sequence of approximation coefficients, given some a-priori
knowledge of the digits of expansion. For instance, if a = a = 1 and a = 3 are the
2 3 4
digits of the classical regular continued fraction expansion, then corollary 4.4 will imply
√ √ √
that|θ −θ | < 2 and 2 2 < |θ −θ | < 3 2.
2 1 3 7 3 2 5
2 Preliminaries
This section is a paraphrased summery of excerpts from [6, 7], given for sake of com-
pleteness. In general, the fractional part of Mo¨bius transformation which map [0,1] onto
[0,∞]leadstoexpansionofrealnumbersascontinuedfractions. Tocharacterizeallthese
transformations,werecallthatMo¨biustransformationsareuniquelydeterminedbytheir
valuesforthreedistinctpoints. Thus,wewillneedtointroduceaparameterfortheimage
of an additional point besides 0 and 1, which we will naturally take to be ∞. Since our
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AvrahamBourla
mapsfixtherealline,theimageof∞,denotedby−k,cantakeanyvaluewithinthesetof
negativerealnumbers. Weletm ∈ {0,1}equalzerofororientationreversingtransforma-
tions, i.e. 0 (cid:55)→ ∞, 1 (cid:55)→ 0 and let m equal one for orientation preserving transformations,
i.e. 0 (cid:55)→ 0, 1 (cid:55)→ ∞. Conclude that all such transformations must be the extension of the
maps x (cid:55)→ k(1−m−x), k > 0, mapping (0,1) homeomorphically to (0,∞), to the extended
x−m
complexplane. ThemapsT : [0,1) → [0,1), 0 (cid:55)→ 0,
(m,k)
(cid:22) (cid:23)
k(1−m−x) k(1−m−x)
T (x) = − , x > 0
(m,k)
x−m x−m
arecalledGauss-likeandRenyi-likeform = 0andm = 1respectively.
We expand the initial seed x ∈ (0,1) as an (m,k)-continued fraction using the following
0
iterationprocess:
1. Setn := 1.
2. Ifx = 0,writethe(m,k)-expansionofx as[a ,...,a ] andexit.
n−1 0 1 n−1 (m,k)
3. Set the digit and future of x at time n to be a := (cid:4)k(1−m−xn−1)(cid:5) ∈ Z+ := Z∩[0,∞)
0 n xn−1−m
andx := k(1−m−xn−1) −a ∈ [0,1). Increasenbyoneandgotostep2.
n xn−1−m n
Foralln ≥ 0,wethushave
k(1−m−x )
n
x = T (x ) = −a
n+1 (m,k) n n
x −m
n
sothat
k(1−2m)
x = m+ .
n
a +k +x
n+1 n+1
Therefore,thisiterationschemeleadstotheexpansionoftheinitialseedx as
0
k(1−2m) k(1−2m)
x = m+ = m+ = ...
0 a1 +k +x1 a1 +k +m+ ak2(+1−k2+mx2)
Remark 2.1. The classical k = 1 cases lead to the regular and backwards continued frac-
tions expansions for m = 0 and m = 1 respectively, bit with digits which are smaller by
onethantheirclassicalrepresentation. Forinstance,
k k2 +4k +2
[0,1,2] = =
(0,k) k k2 +5k +4
0+k +
k
1+k +
2+k
and
k k +4
[0,1,2] = 1− =
(1,k) k k3 +3k2 +5k +4
0+k +1−
k
1+k +1−
2+k
willyield,afterpluggingk = 1,thefractions[1,2,3] = 7 and[1,2,3] = 5 .
0 10 1 13
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AvrahamBourla
Since for the classical regular and backwards continued fraction expansions, this itera-
tionprocesseventuallyterminatespreciselywhentheinitialseedx isarationalnumber.
0
Analogously,wedenotethecountablesetofallnumbersintheintervalwithfinite(m,k)-
expansionbyQ . Foralla ∈ Z+,thecylinderset
(m,k)
(cid:18) (cid:19)
(1−m)k +ma (1−m)k +m(a+1)
∆ := , (1)
a
a+k +1−m a+k +m
isdefinedsuchthatx ∈ ∆ iffa = aor,moregenerally,
0 a 1
x ∈ ∆ ⇐⇒ a = a, n ≥ 0. (2)
n a n+1
The restriction of the map T to ∆ is a homeomorphism onto (0,1). We further define
(m,k) a
thepastofx attimen ≥ 2tobe
0
Y (x ) = Y := m−k −a −[a ,...,a ] ∈ (m−k −a −1,m−k −a ). (3)
n 0 n n n−1 1 (m,k) n n
The set (cid:8)(x ,Y ) : x ∈ (0,1)−Q , n ≥ 2(cid:9) is a subset of the space Ω := (0,1) ×
n n 0 (m,k) (m,k)
(−∞,m−k).
The sequence of approximation coefficients {θ (x )}∞ for the (m,k)-expansion is defined
(cid:12) n 0 (cid:12)1
just like the classical object, i.e. θ (x ) := q2(cid:12)x − pn(cid:12), where pn = [a ,...,a ] ∈ Q
n 0 n(cid:12) 0 qn(cid:12) qn 1 n (m,k) (m,k)
are the appropriate convergents. The sequence of approximation coefficients relates to
thefutureandpastsequencesofx usingtheidentity
0
1
θ (x ) = ,
n 0
x −Y
n+1 n+1
first proven for the classical regular continued fraction case in 1921 by Perron [9]). The
equicontinuousfamilyofcontinuousmaps
(cid:18) (cid:19)
1 (m−x)(m−y)
Ψ : Ω → R2, (x,y) (cid:55)→ , , (4)
(m,k) (m,k)
x−y (2m−1)k(x−y)
(θ ,θ ) = Ψ (x ,Y ).
n n+1 (m,k) n+1 n+1
Each map is a homeomorphism onto its image Ψ (Ω ), which we will denote by
(m,k) (m,k)
Γ . thecorrespondingSpaceofJagerPairs
(m,k)
{(θ (x ),θ (x )) : x ∈ (0,1)−Q , n ≥ 1}
n 0 n+1 0 0 (m,k)
is a dense subset of Γ . In our excursion, we will require the parameter k to be grater
(m,k)
than one, treating the classical regular and backwards continued fraction expansions as
thelimitofthe(m,k)-expansions. Ask → 1+. Thechoiceoftakingthelimitfromtheright
is not arbitrary, for the treatment of the 0 < k < 1 cases will exhibit certain pathologies
(forinstance,see[2,section3.3]).
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AvrahamBourla
3 The finer structure for the Space of Jager Pairs
Fix m ∈ {0,1}, k ∈ (1,∞), and an initial seed x ∈ (0,1) − Q . In order to ease the
0 (m,k)
notation,wewillomitthesubscripts(cid:3) fromnowon. Foralla ∈ Z+,definedefinethe
(m,k)
regions
P = P := (0,1)×(m−a−k −1,m−a−k)
(m,k,a) a
and
F = F := ∆ ×(−∞,m−k),
(m,k,a) a a
where ∆ is the cylinder set (1). Using the identity (2) and the definition (3) of Y , we see
a n
thatthatforalln ≥ 1wehave
(x ,y ) ∈ P ∩F ⇐⇒ a = aanda = b.
n+1 n+1 a b n+1 n+2
We label the images of each of these subsets of Ω under Ψ by P# and F#. Using formula
a a
(4)andthefactthatΨisahomeomorphism,weseethatforalln ≥ 1,wehave
(θ ,θ ) ∈ P# ∩F# ⇐⇒ a = aanda = b. (5)
n n+1 a b n+1 n+2
Next,letp = p betheopenhorizontallinesegment(0,1)×{m−a−k},letf = f
(m,k,a) a (m,k,a) a
(cid:110) (cid:111)
betheopenverticalray (1−m)k+m ×(−∞,m−k)andletp# andf# betheirimageunder
a+k a a
Ψ. Since both the collections {Pa ∪pa} Z+ and {Fa ∪fa} Z+ partition Ω, the image of
a∈ a∈
(cid:110) (cid:111)
their intersections under Ψ, (P# ∪p#)∩(F# ∪f#) will partition Γ. We will call
a a b b Z+
a,b∈
eachmemberofthisrefinedpartitionasubdivision.
Proposition 3.1. For all k ∈ R > 1 and a,b ∈ Z+ the region P# ∩F# is the open interior
a b
2
ofthequadrangleinIE withvertices
(cid:18) (cid:19)
b+k a+k
, ,
(1−2m)k +(a+k)(b+k) (1−2m)k +(a+k)(b+k)
(cid:18) (cid:19)
b+k a+k +1
, ,
(1−2m)k +(a+k +1)(b+k) (1−2m)k +(a+k +1)(b+k)
(cid:18) (cid:19)
b+k +1 a+k
,
(1−2m)k +(a+k)(b+k +1) (1−2m)k +(a+k)(b+k +1)
and
(cid:18) (cid:19)
b+k +1 a+k +1
, .
(1−2m)k +(a+k +1)(b+k +1) (1−2m)k +(a+k +1)(b+k +1)
Proof. Fix a ∈ Z+ and for every (x,y ) ∈ p i.e. x ∈ (0,1), y := m − k − a, set (u,v) :=
0 a 0
Ψ(x,y ). Usingthedefinition(4)ofΨ,wewrite
0
1 1 1 1
u = = : → asx : 0 → 1.
x−y x+a+k −m a+k −m a+k +1−m
0
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AvrahamBourla
Sincem−x = m−y − 1,weexpressv intermsofuas
u
(m−x)(m−y ) (m−x)(m−y )u a+k
0 0
v = = = ((a+k)u−1)
(2m−1)k(x−y ) (2m−1)k (2m−1)k
0
sothat
m(a+k) (1−m)(k +a)
v : → asx : 0 → 1.
k(a+k −m) k(k +a+1−m)
Since Ψ : Ω → Γ is a homeomorphism, we see that p# is an open segment of the line (a+
a
(cid:16) (cid:17) (cid:16) (cid:17)
k)2u+(1−2m)kv = a+kbetweenthepoints 1 , m(a+k) and 1 , (1−m)(k+a) .
a+k−m k(a+k−m) a+k+1−m k(a+k+1−m)
1 1
k(cid:45)1,k(cid:45)1
1
k
1 k(cid:43)3
k(cid:43)4,k(cid:72)k(cid:43)4(cid:76)
k(cid:43)13,k(cid:72)kk(cid:43)(cid:43)23(cid:76) 1k,kk(cid:43)21
k(cid:43)12,k(cid:72)kk(cid:43)(cid:43)12(cid:76) k(cid:43)11,k(cid:43)11 k(cid:43)12,k(cid:72)kk(cid:43)(cid:43)32(cid:76) k(cid:43)11,k(cid:72)kk(cid:43)(cid:43)21(cid:76)
P2(cid:240) P1(cid:240) P0(cid:240)
P2(cid:240) P1(cid:240) P0(cid:240)
1 1 1 1 1 1 1 1
k(cid:43)3 k(cid:43)2 k(cid:43)1 k k(cid:43)3 k(cid:43)2 k(cid:43)1 k
Γ Γ
(0,k) (1,k)
Next, fix b ∈ Z+ and set x := (1−m)k+mb. We use the definition (4) of Ψ to first write
0 b+k
m−y = m−x+ 1 andthenexpressv intermsofuas
u
(cid:18) (cid:19) (cid:18) (cid:19)
u 1 1 (2m−1)k
v = (m−x ) m−x + = u+1 . (6)
0 0
(2m−1)k u b+k b+k
Since Ψ is a homeomorphism, this allows us to conclude that Ψ maps f to an open
b
segment of a line in the uv plane, which is the reflection of the line p# along the diag-
b
onal u − v = 0. In particular, P# ∩ F# is the region interior to the quadrangle with
a b
vertices p# ∩ f#, p# ∩ f#, p# ∩ f# and p# ∩ f# . But (x ,y ) = p ∩ f , so that
a b a+1 b a b+1 a+1 b+1 0 0 a b
(u ,v ) := Ψ(x ,y ) = p# ∩f#. Thedefinition(4)ofΨandformula(6)nowyield
0 0 0 0 a b
1 b+k
u = =
0
x −y (1−2m)k +(a+k)(b+k)
0 0
and
a+k
v =
0
(1−2m)k +(a+k)(b+k)
asdesired.
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AvrahamBourla
Whenm = 0,weobtainthedepictedSpacesofJagerPairs:
Γ revisited Γ
(0,k) (0,1)
This result was first proved for the classical regular continued fractions corresponding
to k = 1 by Dajani and Kraaikamp [5, exercise 5.3.4]. For this case, the region P# ∩ F#
0 0
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
degenerates to the interior of the triangle with vertices 1, 2 , 2, 1 and 2, 2 . When
3 3 3 3 5 5
m = 1,weobtainthedepictedSpacesofJagerPairs:
Γ revisited Γ
(1,k) (1,1)
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AvrahamBourla
The picture for the classical backwards continued fractions k = 1, is obtained after we
let k → 1+ and use the equicontinuity of the family {Ψ } . For this case, the region
(1,k) k>1
P# ∩ F# expands to the unbounded region in the uv plane, which is the intersection of
0 0
theregionsu−v > 1, v −u < 1, 4u−v > 2and4v −u > 2.
4 Results
This symmetry exhibited by the subdivisions in the Space of Jager Pairs will allow us to
provide uniform bounds for the growth rate sequence for {θ }∞, which is the sequence
n 1
whose member at time n is {|θ −θ |}. But first, we already know that Γ is the
n+1 n (1,1)
unbounded region in the first quadrant of the uv plane bounded by the lines u − v < 1
andv −u < 1,henceweobtainatoncethat
Theorem 4.1. In general, the sequence of approximation coefficients associated with the
classical backwards continued fraction expansion m = k = 1 has no uniform upper
bound. Howeveritsgrowthratesequenceisuniformlyboundedbyone.
Ourmainresultis:
Theorem 4.2. Given m ∈ {0,1}, k ∈ R > 1, x ∈ (0,1) − Q and N ≥ 1, write l :=
0 (m,k)
min{a ,a ,a } and L := max{a ,a ,a }, where a is the digit at time n in
N+1 N+2 N+3 N+1 N+2 N+3 n
the(m,k)-expansionforx . Thentheinequality
0
(cid:18) L−l+1 (cid:19)2
(θ −θ )2 +(θ −θ )2 < 2
N+1 N N+2 N+1
(1−2m)k +(l+k)(L+k +1)
issharp. Furthermore,ifL−l > 1,thentheinequality
(cid:18) L−l (cid:19)2
(θ −θ )2 +(θ −θ )2 > 2
N+1 N N+2 N+1
(1−2m)k +(l+k +1)(L+k)
issharp.
Toprovethistheorem,wewillfirstprove:
Lemma 4.3. Assuming the hypothesis of the theorem, write a := min{a ,a }, A :=
N+1 N+2
max{a ,a }, b := min{a ,a }andB := max{a ,a }. Thentheinequality
N+1 N+2 N+2 N+3 N+2 N+3
(θ −θ )2 +(θ −θ )2
N+1 N N+2 N+1
(cid:18) B +k +1 b+k (cid:19)2
< −
(1−2m)k +(a+k)(B +k +1) (1−2m)k +(b+k)(A+k +1)
(cid:18) A+k +1 a+k (cid:19)2
+ −
(1−2m)k +(b+k)(A+k +1) (1−2m)k +(a+k)(B +k +1)
issharp. Furthermore,ifmax{A−a,B −b > 1},thentheinequality
(θ −θ )2 +(θ −θ )2
N+1 N N+2 N+1
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AvrahamBourla
(cid:18) B +k b+k +1 (cid:19)2
> −
(1−2m)k +(a+k +1)(B +k) (1−2m)k +(b+k +1)(A+k)
(cid:18) A+k a+k +1 (cid:19)2
+ −
(1−2m)k +(b+k +1)(A+k) (1−2m)k +(a+k +1)(B +k)
issharp.
Proof. LetR = R betheset
(m,k,a,b,A,B)
(cid:91)
(P# ∩F#),
i j
a≤i≤A,b≤j≤B
so that, from formula (5), we obtain that (θ ,θ ) ∈ P# ∩ F# ⊂ R as well as
N N+1 aN+1 aN+2
(θ ,θ ) ∈ P# ∩F# ⊂ R. Fromproposition3.1,weseethatRistheinteriorofthe
N+1 N+2 aN+2 aN+3
bounded convex quadrangle, consisting of sections from the line segments p#, p# , f#
a A+1 b
andf# . Unlessmax{A−a,B −b} ≤ 1,wealsoletr = r bethenon-emptyset
B+1 (m,k,a,b,A,B)
(cid:91)
(P# ∩F#),
i j
a+1≤i≤A−1,b+1≤j≤B−1
sothat
{(θ ,θ ),(θ ,θ )} ⊂ R−r. (7)
N N+1 N+1 N+2
From the same proposition, we see that r ⊂ R is the interior of the convex quadrangle,
consistingofsectionsfromthesegmentsp# , p#, f# andf#,thatis,thequadrangler is
a+1 A b+1 B
obtainedfrompeelingtheouterlayerofR’ssubdivisions.
Whenm = 0,let(u ,v ) := p# ∩f# and(U ,V ) := p# ∩f#. Then
0 0 A b+1 0 0 a+1 B
(u,v) ∈ P# ∩F# =⇒ u < u andv > v
A b 0 0
and
(u,v) ∈ P# ∩F# =⇒ u > U andv < V .
a B 0 0
Whenm = 1,let(u ,v ) := p# ∩f# and(U ,V ) := p# ∩f# . Then
1 1 A B 1 1 a+1 b+1
(u,v) ∈ P# ∩F# =⇒ u < u andv < v
a b 0 0
and
(u,v) ∈ P# ∩F# =⇒ u > U andv > V .
A B 0 0
Intandemwithformula(7),thisallowsustoconcludethatforeithercasewehave
diam(r)2 < d((θ ,θ ),(θ ,θ )) = (θ −θ )2 +(θ −θ )2 < diam(R)2, (8)
N N+1 N+1 N+2 N N+1 N+1 N+2
wheredisthefamiliardistanceformulabetweentwopointsinIE2 anddiam(R)standsfor
theeuclideandiameteroftheregionR.
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AvrahamBourla
2
The diameter of the interior of a convex quadrangle in IE with opposite acute angles is
the length of the diagonal connecting the vertices corresponding to these acute angles.
To verify this simple observation from plane geometry, observe that from the continuity
of the euclidean distance formula, the diameter of a convex polygon is either a side or
a diagonal. The diagonal connecting the two acute angles will lie opposite to an obtuse
angle in either one of the induced triangles obtained. Since for any triangle, the length of
asideisproportionaltothemeasureofitsoppositeangle,weconcludethatthisdiagonal
is longer than all of the four sides in this quadrangle. To verify it is longer than the other
diagonal, consider one of the four induced triangles obtained after drawing both diago-
nalsandusethesameoppositeangleargument.
When m = 0 and c ∈ Z+, the slope of the line containing p# lie in (−∞,−1) whereas the
c
slope of the line containing f# lie in (−1,0). Conclude that both R and r have opposite
c
acuteangles,sothattheirdiametersarethelengthofthediagonalconnectingthevertices
p# ∩f# withp# ∩f# andp# ∩f# withp# ∩f# respectively. Hence
a B+1 A+1 b a+1 B A b+1
diam(R) = d(p# ∩f# ,p# ∩f#)
a B+1 A+1 b
and
diam(r) = d(p# ∩f#,p# ∩f# ).
a+1 B A b+1
When m = 1, the slope of the line containing p# lie in (1,∞) whereas the slope of the
c
line containing f# lie in (0,1). Conclude that both R and r has opposite acute angles, so
c
that their diameters are the length of the diagonal connecting the vertices p# ∩ f# with
a b
p# ∩f# andp# ∩f# withp# ∩f# respectively. Hence
A+1 B+1 a+1 b+1 A B
diam(R) = d(p# ∩f#,p# ∩f# )
a b A+1 B+1
and
diam(r) = d(p# ∩f# ,p# ∩f#).
a+1 b+1 A B
Foreithercase,wehave
(cid:18) B +k +1 b+k (cid:19)2
diam(R)2 = −
(1−2m)k +(A+k +1)(B +k +1) (1−2m)k +(a+k)(b+k)
(cid:18) A+k +1 a+k (cid:19)2
+ −
(1−2m)k +(A+k +1)(B +k +1) (1−2m)k +(a+k)(b+k)
and
(cid:18) B +k b+k +1 (cid:19)2
diam(r)2 = −
(1−2m)k +(A+k)(B +k) (1−2m)k +(a+k +1)(b+k +1)
(cid:18) A+k a+k +1 (cid:19)2
+ − .
(1−2m)k +(A+k)(B +k) (1−2m)k +(a+k +1)(b+k +1)
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