Generalized Continued Logarithms and Related Continued Fractions

Transcription

1 Generalized Continued Logarithms and Related Continued Fractions Jonathan M Borwein Kevin G Hare Jason G Lynch June Abstract We study continued arithms as introduced by Bill Gosper and studied by J Borwein et al After providing an overview of the type I and type II generalizations of binary continued arithms introduced by Borwein et al we focus on a new generalization to an arbitrary integer base b We show that all of our so-called type III continued arithms converge and all rational numbers have finite type III continued arithms As with simple continued fractions we show that the continued arithm terms for almost every real number follow a specific distribution We also generalize Khinchine s constant from simple continued fractions to continued arithms and show that these arithmic Khinchine constants have an elementary closed form Finally we show that simple continued fractions are the limiting case of our continued arithms and briefly consider how we could generalize past continued arithms Introduction Continued fractions especially simple continued fractions have been well studied throughout history Continued binary arithms however appear to have first been introduced by Bill Gosper in his appendix on Continued Fraction Arithmetic [4] More recently in [2] J Borwein et al proved some basic results about binary continued arithms and applied experimental methods to determine the term distribution of binary continued arithms They conjectured and indicated a proof that like in the case of continued fractions almost every real number has continued arithm terms that follow a specific distribution They then introduced two different generalizations of binary continued arithms to arbitrary bases The Structure of This Paper Section introduces some basic definitions and results for continued fractions briefly describes binary continued arithms as introduced by Gosper and provides an overview jonathanborwein@newcastleeduau CARMA University of Newcastle Callaghan NSW 2308 Australia Research of J M Borwein was supported by CARMA University of Newcastle kghare@uwaterlooca Department of Pure Mathematics Univsersity of Waterloo Waterloo Ontario N2L 3G Canada Research of K G Hare was supported by NSERC Grant RGPIN j4lynch@uwaterlooca Research of J G Lynch was supported by CARMA University of Newcastle

2 of results relating to the Khinchine constant for continued fractions Sections 2 and 3 then provide an overview of the type I and type II continued arithms introduced by Borwein et al Further details on these can be found in [2] Section 4 comprises the main body of the paper In Section 4 we define type III continued arithms and extend to them the standard continued fraction recurrences Section 42 then proves that type III continued arithms are guaranteed to converge to the correct value and that every rational number has a finite type III continued arithm These are two desirable properties of continue fractions and binary continued arithms that a complete generalization should have In Section 43 we describe how measure theory can be used to investigate the distribution of continued arithm terms This is then applied in Section 44 to determine the distribution and Section 45 to determine the arithmic Khinchine constant The main proofs of these sections are quite technical and are separated out into Appendices A and B respectively Finally Section 46 derives some relationships between simple continued fractions and the limiting case of type III continued arithms Finally we close the paper in Section 5 by briefly introducing one way to generalize past continued arithms 2 Continued Fractions The material in this section can be found in many places including [3] Definition A continued fraction is an expression of the form or y α 0 + y 2 α 0 + α + α + α 2 + β β 2 α 2 + β 3 β β 2 β 3 + β n α n For the sake of simplicity we will sometimes denote the above as or y α 0 + β α + β 2 α 2 + y 2 α 0 + β α + β 2 α β n α n respectively The terms α 0 α are called denominator terms and the terms β β 2 are called numerator terms Definition 2 Two continued fractions y α 0 + β α + β 2 α 2 + and y α 0 + β α 2 + β 2 α 2 +

3 are called equivalent if there is a sequence d n n0 with d 0 such that α n d n α n for all n 0 and β n d n d n β n for all n The c n terms can be thought of as constants that are multiplied by both numerators and denominators of successive terms Definition 3 The nth convergent of the continued fraction y α 0 + β α + β 2 α 2 + is given by x n α 0 + β α + β 2 α β n α n Definition 4 The nth remainder term of the continued fraction y α 0 + β α + β 2 α 2 + is given by r n α n + β n+ α n+ + β n+2 α n+2 + The following results will be useful for generalizing to continued arithms Fact Suppose x α 0 + β + β 2 + where α α α n β n > 0 for all n Then the 2 convergents are given by x n p n q n where p q 0 p 0 α 0 q 0 p n α n p n + β n p n 2 n q n α n q n + β n q n 2 n Fact 2 Suppose x α 0 + β + β 2 + where α α α n β n > 0 for all n Then the continued 2 fraction for x converges to x if α nα n+ n β n+ Remark Throughout this paper we will use MA or just MA to denote the Lebesgue measure of a set A R 3 Binary Continued Logarithms Let α R Let y 0 α and recursively define a n 2 y n If y n 2 an terminate Otherwise set y n+ 2an y n 2 an 0 then 3

4 and recurse This produces the binary base 2 continued arithm for y 0 : y 0 2 a 0 + 2a 0 2 a + 2a 2 a 2 + 2a2 2 a 3 + These binary continued arithms were introduced explicitly by Gosper in his appendix on Continued Fraction Arithmetic [4] Borwein et al studied binary continued arithms further in [2] extending classical continued fraction recurrences for binary continued s and investigating the distribution of aperiodic binary continued arithm terms for quadratic irrationalities such as can not occur for simple continued fractions Remark 2 Jeffrey Shallit [7] proved some limits on the length of a finite binary continued arithm Specifically the binary continued arithm for a rational number p/q has at most 2 2 p + O terms Furthermore this bound is tight as can be seen by considering the continued fraction for 2 n Moreover the sum of the terms of the continued arithm of p/q is bounded by 2 p2 2 p Khinchine s Constant In [5] Khinchine proved that for almost every α 0 where α a + a 2 + a 3 + the denominator terms a a 2 a 3 follow a specific limiting distribution That is let P α k lim N {n N : a N n k} This is the limiting ratio of the denominator terms that equal k if this limit exists Then for almost every α 0 + kk+2 P α k 2 for every k N It then follows for almost every α 0 that the limiting geometric average of the denominator terms is given by lim n n a a 2 a n k 2 r rr + 2 This constant is now known as Khinchine s constant K 2 Type I Continued Logarithms 2 Type I Definition and Preliminaries Fix an integer base b 2 We define type I continued arithms as follows Definition 5 Let α The base b continued arithm of type I for α is b a 0 + b ba 0 b a + b ba b a 2 + b ba2 b a 3 + [b a 0 b a b a 2 ] cl b 4

5 where the terms a 0 a a 2 are determined by the recursive process below terminating at the term b an if at any point y n b an y 0 α a n b y n n 0 y n+ b ban y n b an n 0 The numerator terms b b an are defined as such to ensure that y n for all n Indeed notice that for each n we must have b an y n < b an+ Thus 0 y n b an < b b an If y n b an 0 then we terminate otherwise we get 0 < y n b an < b b an so y n+ b ban y n b an Borwein et al proved that the type I continued fraction of α will converge to α [2 Theorem 5] Additionally numbers with finite type I continued arithms must be rational However for b 3 rationals need not have finite continued arithms For example the type I ternary continued arithm for 2 is [ ] cl 3 22 Distribution of Type I Continued Logarithm Terms and Type I Logarithmic Khinchine Constant We now look at the limiting distribution of the type I continued arithm terms Consider α [b a 0 b a b a 2 ] cl b Assume that the continued arithm for α is infinite Furthermore assume without loss of generality that a 0 0 so that α b Definition 6 For n N let D n k {α b : a n k} denote the set of α b for which the nth continued arithm term is b k Definition 7 Let x [ b a b a 2 ] cl b b The nth remainder term of x is r n r n x [b an b a n+ ] cl b as in Definition 4 Define and wherever this limit exists z n z n x r n b an [ ba n+ b a n+2 ] cl b b M n x {α b : z n α < x} b m n x b MM nx 0 mx lim n m n x Notice that since < z n α < b for all n N and α b we must have m n 0 and m n b for all n N We can now derive a recursion for the functions m n Theorem The sequence of functions m n is given by the recursive relationship m 0 x x b m n x m n + b m n + x b n 2 for x b k0 5

6 The proof of this is similar to that of Theorem 6 We next derive a formula for D n k in terms of the function m n Theorem 2 b MD n+k m n + b m n + b + The proof of this theorem is similar to that of Theorem 9 Thus if the limiting distribution mx exists it immediately follows that lim n b MD nk m + b m + b Experimentally Determining the Type I Distribution Now suppose b > is an arbitrary integer Let µ b denote the limiting distribution function m for the base b assuming it exists We may investigate the form of µ b x by iterating the recurrence relation of Theorem at points evenly spaced over the interval [ b] starting with m 0 x x At each b iteration we fit a spline to these points evaluating each infinite sum to 00 terms and breaking the interval [ b] into 00 pieces This is practicable since the continued arithm converges much more rapidly than the simple continued fraction We find good convergence of µ b x after around 0 iterations We use the 0 data points from this process to seek the best fit to a function of the form µ b x C b αx + β γx + δ We set γ to eliminate any common factor between the numerator and denominator To meet the boundary condition µ b 0 we must have δ α + β and to meet the boundary condition µ b b we must have C leaving the functional form to be fit as µ b x b b bα+β b α+β+b αx+β x+α+β bα+β α+β+b 4 We sought this superposition form when the simpler structure for simple continued fractions failed Fitting our data to the model suggests candidate values of α b and β from b b which we get µ b x bx x+b 5 When we then apply 3 we get lim n b MD nk b2 2b + bk b 3 b k+ +b 2 b2 2b A proof of this distribution and of the type I Khinchine constant for each integer base b using ergodic theory can be found in [6] Additionally it is likely that the proofs in Appendices A and B for the type III continued arithm distribution and arithmic Khinchine constant could be appropriately adjusted to prove these results 6

7 If a type I base b Khinchine constant KL I b exists ie almost every α has the same limiting geometric mean of denominator terms and if a limiting distribution Dk lim n D n k of denominator terms exists then KL I b k0 b k MDk b b k0 k MDk b This is because the limiting distribution of denominator terms if it exists is essentially the average distribution over all numbers α b If we then assume that almost every α b has the same limiting geometric mean of denominator terms then this limiting geometric mean the arithmic type I Khinchine constant must equal the limiting geometric mean of the average distribution Thus if we assume KL I b exists and that the distribution in 3 is correct then we must have KL I b b A where A k0 k MDk b k[µ b + b µ b + b + ] k0 b b2 2b by Theorem 2 and a lengthy but straightforward algebraic manipulation These conjectured type I arithmic Khinchine constants for 2 b 0 are given in Figure b KL I b Figure : Type I arithmic Khinchine constants for 2 b 0 These conjectured values of the type I arithmic Khinchine constants were supported by empirical evidence as the numerically computed limiting geometric means of denominator terms for various irrational constants give the expected values Notice that the type I arithmic Khinchine constants have a simple closed form which is noteworthy as no simple closed form has been found for the Khinchine constant for simple continued fractions 3 Type II Continued Logarithms 3 Type II Definition and Preliminaries Fix an integer base b 2 We define type II continued arithms as follows 7

8 Definition 8 Let α R The base b continued arithm for α is c 0 b a 0 + c 0b a 0 c b a + c b a c 2 b a 2 + c 2b a2 c 3 b a 3 + [c 0 b a 0 c b a c 2 b a 2 ] cl2 b where the terms a 0 a a 2 and c 0 c c 2 are determined by the recursive process below terminating at the term c n b an if at any point y n c n b an y 0 α a n b y n n 0 yn c n n 0 b an y n+ c nb an n 0 y n c n b an Remark 3 The numerator terms c n b an are defined to match the corresponding denominator terms Recall that in the type I case the term y n+ could take any value in regardless of the value of a n This is no longer true since y n c n b an 0 b an so y n+ c n We will see later that this results in type II continued arithms having a more complicated distribution for which we could not find a closed form This issue was the inspiration for the definition of type III continued arithms where the numerator terms are b an instead of c n b an Borwein et al proved that the type II continued fraction of α will converge to α and that α has a finite continued arithm if and only if α Q [2 Theorems 9 and 20] unlike the situation for type I 32 Distribution of Type II Continued Logarithm Terms and Type II Logarithmic Khinchine Constant We now look at the limiting distribution of the type II continued arithm terms Consider α [c 0 b a 0 c b a c 2 b a 2 ] cl2 b Assume that α / Q so that the continued arithm for α is infinite Furthermore assume without loss of generality a 0 0 and c 0 so that α 2 Definition 9 Let n N Let D n k l {α 2 : a n k c n l} denote the α 2 for which the nth continued arithm term is lb k Definition 0 Let x [ c b a c 2 b a 2 ] cl2 b 2 with nth remainder term r n r n x [c n b an c n+ b a n+ ] cl2 b as in Definition 4 Define and z n z n x wherever this limit exists r n c n b an [ c n+b a n+ c n+2 b a n+2 ] cl2 b 2 M n x {α 2 : z n α < x} 2 m n x MM n x 0 mx lim n m n x 8

9 Notice that since z n α 2 for all n N and α 2 we must have m n 0 and m n 2 for all n N We may now derive a recursion relation for the functions m n Theorem 3 The sequence of functions m n is given by the recursive relationship m 0 x x 6 b m n x m n + l m n max{ + l x + l + } n k0 l for x 2 We can now derive a formula for D n k l in terms of the function m n Theorem 4 MD n+ k l m n + l m n + l + Thus if the limiting distribution mx exists it immediately follows that lim MD nk l m + l m + l + 8 n 33 Experimentally Determining the Type II Distribution Again suppose b is arbitrary Let µ b denote the limiting distribution function m for the base b assuming it exists We may again investigate the form of µ b x by iterating the recurrence relation of Theorem 3 at points evenly spaced over the interval [ 2] starting with m 0 x x At each iteration we fit a spline to these points evaluating each infinite sum to 00 terms and breaking the interval [ 2] into 00 pieces We find good convergence of µ b x after around 0 iterations However we have been unable to find a closed form for µ b for b > 2 It appears that µ b is a continuous non-monotonic function that is smooth on 2 except at x j+ for j 2 b j If a arithmic Khinchine constant KL II b exists ie almost every α has the same limiting geometric mean of denominator terms and if a limiting distribution Dk l lim n D n k l of denominator terms exists then KL II b b lb k MDk l k0 l This is because the limiting distribution of denominator terms if it exists is essentially the average distribution over all numbers α 2 If we then assume that almost every α 2 has the same limiting geometric mean of denominator terms then this limiting geometric mean the arithmic Khinchine constant must equal the limiting geometric mean of the average distribution However since we do not know the limiting distribution we can only approximate the arithmic Khinchine constants This conjectured values of the type II arithmic Khinchine constants are supported by empirical evidence as the limiting geometric means of denominator terms for various irrational constants give the conjectured values 9 7

10 b KL II b Figure 2: Experimental type II arithmic Khinchine constants for 2 b 0 4 Type III Continued Logarithms Fix an integer base b 2 In this section we will introduce our third generalization of base 2 continued arithms This appears to be the best of the three generalizations as we will show that type III continued arithms have guaranteed convergence rational finiteness and closed forms for the limiting distribution and arithmic Khinchine constant Additionally type III continued arithms converge to simple continued fractions if one looks at limiting behaviour as b 4 Type III Definitions and Recurrences We start with some definitions notation and lemmas related to continued arithm recurrences Definition Let α R The type III base b continued arithm for α is c 0 b a 0 + ba 0 c b a + ba c 2 b a 2 + ba2 c 3 b a 3 + [c 0 b a 0 c b a c 2 b a 2 ] cl3 b where the terms a 0 a a 2 and c 0 c c 2 are determined by the recursive process below terminating at the term c n b an if at any point y n c n b an y 0 α a n b y n n 0 yn c n n 0 b an b an y n+ n 0 y n c n b an Remark 4 We can and often will think of the a n and c n as functions a 0 a a 2 : Z 0 and c 0 c c 2 : { 2 b } since the terms a 0 c 0 a c a 2 c 2 are uniquely determined by α Conversely given the complete sequences a 0 a a 2 and c 0 c c 2 one can recover the value of α Remark 5 Let α [c 0 b a 0 c b a c 2 b a 2 ] cl3 b Based on Definitions 3 and 4 the nth convergent and nth remainder term of α are given by x n α c 0 b a 0 + ba 0 c b a + ba c 2 b a 2 + ba2 c 3 b a ban c n b an 0

11 and r n α c n b an + b an c n+ b a n+ + b a n+ c n+2 b a n+2 + respectively Note that the terms r n are the same as the terms y n from Definition Lemma 5 The nth convergent of α [c 0 b a 0 c b a c 2 b a 2 ] cl3 b is given by x n p n q n where and for n p q 0 p 0 c 0 b a 0 q 0 p n c n b an p n + b a n p n 2 q n c n b an q n + b a n q n 2 Proof This follows from Fact where for continued arithms we have α n c n b an β n b a n and Lemma 6 We have the following lower bounds on the denominators q n : q n 2 n /2 > 2 2n/2 for n 0 q n b a + +a n for n 0 Proof For the first bound note that q n c n b an q n + b a n q n 2 c n b an + b a n q n 2 2q n 2 A simple inductive argument then gives q n 2 n/2 q 0 2 n/2 > 2 n /2 for even n and q n 2 n /2 q 2 n /2 for odd n For the second bound note that q n c n b an q n + b a n q n 2 b an q n from which another simple inductive argument gives q n b an+a n + +a q 0 b a + +a n Lemma 7 For n 0 Proof For n 0 we have p n q n q n p n n b a 0+ +a n p 0 q q 0 p c 0 b a 0 0 b 0 Now suppose that the statement is true for some n 0 Then by Lemma 5 p n+ q n q n+ p n c n+ b a n+ p n + b an p n q n c n+ b a n+ q n + b an q n p n b an p n q n q n p n b an n b a 0+ +a n n b a 0+ +a n so the result follows by induction The following lemma is equivalent to Lemma 5 and will be used to prove Theorem 9

12 Lemma 8 Let a 0 Then for all n 0 pn p n n cj b a j b a j 0 Proof For n 0 we have 0 j0 q n cj b a j b a j 0 j0 q n j0 c0 b a 0 b a 0 c0 b a 0 0 Now suppose for induction that n cj b a j pn p b a n 2 j 0 Then by Lemma 5 n cj b a j pn p b a n 2 cn b an j 0 b a n 0 j0 as asserted q n q n 2 Theorem 9 For arbitrary k n q n q n 2 p0 p q 0 q cn b an p n + b a n p n 2 p n pn p c n b an q n + b a n n q n 2 q n q n q n [c 0 b a 0 c b a c n b an ] cl3 b p k r k + p k 2 b a k q k r k + q k 2 b a k Proof First notice that r k [c k b a k cn b an ] cl3 b p k where q k p k q k ck b a k n 0 jk+ cj b a j b a j 0 Also note that ck b a k 0 ck b b a a k k 0 0 b a k 0 Then Thus pn q n 0 n cj b a k j cj b b a a j ck b a k n cj b a j j 0 0 b a j 0 b a k 0 b a j 0 0 j0 j0 jk+ pk p k 2 0 ck b a k n cj b a j q k q k 2 0 b a k 0 b a j 0 0 jk+ pk p k 2 b a k p k pk p q k q k 2 b a k q k k + p k 2b a k q k q k p k + q k 2b a k q k [c 0 b a 0 c n b an ] cl3 b p n q n p k p k + p k 2b ak q k as required q k p k + q p p k k + p q k 2 b a k k k 2b a k q k + q k 2 b a k 2 q k p k q k p k r k + p k 2 b a k q k r k + q k 2 b a k

13 42 Convergence and Rational Finiteness of Type III Continued Logarithms Theorem 0 The type III continued arithm for a number x converges to x Proof Suppose that the continued arithm for x [c 0 b a 0 c b a c n b an ] cl3 b is finite From the construction we have x y 0 where y k c k b a k + ba k for 0 k n From Definition since the continued arithm terminates we have y n c n b an at which point we simply have y k+ x c 0 b a 0 + ba 0 c b a + ba c 2 b a 2 + ba2 c 3 b a ban c n b an This shows convergence in the case of finite termination If the continued arithm for x does not terminate then convergence follows from Fact 2 since n α n α n+ β n+ n while all terms are positive as required Lemma If c n b an c n+ b a n+ b an c n c n+ b a n+ n y c 0 b a 0 + ba 0 c b a + ba c 2 b a 2 + ba2 c 3 b a 3 + y c 0 b a 0 + c b a 0 a + c c 2 b a 2 + c 2 c 3 b a 3 + then y and y are equivalent The form y is called the denominator-reduced continued arithm for y Proof Take d 0 and d n c n b an for n to satisfy the conditions of Definition 2 Theorem 2 The type III continued arithm for a number x will terminate finitely if and only if x Q Proof Clearly if the continued arithm for x terminates finitely then x Q Conversely suppose x c 0 b a 0 + ba 0 c b a + ba c 2 b a 2 + is rational By Lemma we can write x c 0 b a 0 + c 0 c b a + c c 2 b a 2 Let y n denote the nth tail of the continued arithm that is + y n + c n c n+b a n+ + c n+c n+2b a n+2 3 +

14 Notice that so y n + c n y n+ c n c n+b a n+ y n+ c n+b a n+ y n Since each y n is rational write y n un v n for positive relatively prime integers u n and v n Hence or equivalently u n+ y n+ c n v n+ c n+b a n+ u n v n v n v n c n c n+ b a n+ un v n c n c n+ b a n+ u n v n u n+ v n v n+ Notice that since y n for all n u n v n 0 so each multiplicative term in the above equation is a nonnegative integer Since u n+ and v n+ are relatively prime we must have u n+ v n so u n+ v n u n If at any point we have u n+ v n u n then y n un v n and the continued arithm terminates Otherwise u n+ < u n so u n is a strictly decreasing sequence of nonnegative integers so the process must terminate again giving a finite continued arithm 43 Using Measure Theory to Study the Type III Continued Logarithm Terms We now look at the relative frequency of the continued arithm terms Specifically the main theorem of this section places bounds on the measure of the set { } a x 2 : k a 2 k 2 a n k n a n+ k c l c 2 l 2 c n l n c n+ l in terms of the measure of the set { a x 2 : k a 2 k 2 a n k n c l c 2 l 2 c n l n } and the value of k and l From that we can get preliminary bounds on the measure of {x 2 : a n k c n l} in terms of k and l Consider α [c 0 b a 0 c b a c 2 b a 2 ] cl3 b Assume that α / Q so that the continued arithm for α is infinite Furthermore assume a 0 and c 0 so that α 2 Notice that in order to have a k and c l we must have + l + < α + l Thus we can partition 2 into countably many intervals J J J 2 J b J such that a 2 k and c l for k all α J This gives in general l k J + l l + b + ] k l b k We call these intervals the intervals of first rank 4

15 k Now fix some interval of first rank J and consider the values of a l 2 and c 2 for k α J One can show that we have a k c l a 2 k 2 and c 2 l 2 on the interval l [ k k J l l 2 l b k + b k l 2 b k 2 + l b k + b k l 2 +b k 2 These are the intervals of second rank We may repeat this process indefinitely to get the intervals of nth rank noting that each interval of rank n is just a subinterval of an interval of rank n Definition 2 Let n N The intervals of nth rank are the intervals of the form { } k k J 2 k n a n α 2 : k a 2 k 2 a n k n l l 2 l n c l c 2 l 2 c n l n where k k 2 k n Z 0 and l l 2 l n { 2 b } Remark 6 The intervals of nth rank will be half-open intervals that are open on the left if n is odd and open on the right if n is even However for simplicity we will ignore what happens at the endpoints and treat these intervals as open intervals This will not affect the main theorems of this paper as the set of endpoints is a set of measure zero Definition 3 Suppose m n N with m n Let a n+ a m Z 0 and c n+ c m { b } Let f be a function that maps intervals of rank m to real numbers Then we define n a a f J 2 a m m c c 2 c m and similarly we define n a a 2 a m Jm c c 2 c m Definition 4 Let n N Let b a c b a c b a n c n b a n c n D n k l {α 2 : a n k c n l} f a a J 2 a m m c c 2 c m a a J 2 a m m c c 2 c m denote the set of points where the nth continued arithm term is lb k Remark 7 D n k l is a countable union of intervals of rank n specifically D n k l a a 2 a n k Jn c c 2 c n l n a a Lemma 3 Let J 2 a n n be an interval of rank n The endpoints of J c c 2 c n n are p n p n + p n b an and q n q n + q n b an 5

16 a a Proof Let α J 2 a n n be arbitrary Note that α [ c c c 2 c b a c n b an r n+ ] cl3 b n where r n+ can take any real value in [ From Theorem 9 we have Notice that α p nr n+ + p n b an q n r n+ + q n b an α p n p nr n+ + p n b an q n q n r n+ + q n b an a a and on J 2 a n n c c 2 c n q np n p n q n b an q n q n r n+ + q n b an all of p n q n p n q n a n are fixed Thus α is a monotonic a a function of r n+ so the extreme values of α on J 2 a n n will occur at the c c 2 c n extreme values of r n+ Taking r n+ gives α pn+p n b an q n+q n and letting r b an n+ gives a a α pn q n Thus the endpoints of J 2 a n n are c c 2 c n p n q n and p n + p n b an q n + q n b an as claimed Theorem 4 Suppose n N a a 2 a n k Z 0 and c c 2 c n l { b } Let a a a n and c c c n Then a a k 4ll + b MJ 2 a k n MJ c n+ c l ll + b MJ k n c a Proof From Lemma 3 we know that the endpoints of J n are c p n p n + p n b an and q n q n + q n b an a k Now in order to be in J n+ we must have a c l n+ k and c n+ l so lb k a k r n+ l + b k Thus the endpoints of J n+ will be c l p n lb k + p n b an q n lb k + q n b an and p n l + b k + p n b an q n l + b k + q n b an Thus a MJ n p n c p n + p n b an q n q n + q n b p n q n b an p n q n b an an q n q n + q n b an b a + +a n q n q n + q n b an b a+ +an qn 2 + q n b an q n 6

17 and a k MJ n+ p n lb k + p n b an c l q n lb k + q n b p nl + b k + p n b an an q n l + b k + q n b an p n q n lb an+k + p n q n l + b an+k p n q n l + b an+k p n q n lb an+k q n lb k + q n b an q n l + b k + q n b an b a + +a n+k ll + b 2k qn 2 + q n b an q nlb + q n b an k q nl+b k so ll + b k q 2 n a k MJ n+ c l a MJ n c b a + +a n + q n b an q nlb k + q n b an q nl+b k + q n b an q n ll + b k + q n b an q nlb + q n b an k q nl+b k Now notice that q n c n b an q n +b a n q n 2 b an q n so 0 q n b an and 0 q n b an q nl+b k and thus q n 0 q n b an q nlb k q n b an q n q n b an q nlb + q n b an s q nl+b k Therefore 4ll + b k MJ n a c a k MJ n+ c l 2 ll + b k MJ n a c and we are done Corollary 5 Let n N k Z 0 and l { b } Then 4ll + b k MD n+k l 2 ll + b k Proof Note that any two distinct intervals of rank n are disjoint Thus we can add up the above inequality over all intervals of rank n noting that so and that n a a n Jn 2 c c n M 2 c c n n a a n MJn n a a n k Jn+ D c c n l n+ k l 7

18 so This gives as needed MD c c n l n+ k l n a a n k MJn+ 4ll + b k MD n+ k l 2 ll + b k 44 Distribution of Type III Continued Logarithm Terms Definition 5 Let x [ c b a c 2 b a 2 ] cl3 b 2 and r n r n x [c n b an c n+ b a n+ ] cl3 b as per Remark 5 Define and z n z n x r n b an c n + [ c n+ b a n+ c n+2 b a n+2 ] cl3 b 2 wherever this limit exists M n x {α 2 : z n α < x} 2 m n x MM n x 0 mx lim n m n x We now get a recursion relation for the sequence of functions m n Theorem 6 The sequence of functions m n is given by the recursive relationship m 0 x x 9 b m n x m n + l m n + x + l n 0 for x 2 k0 l Proof Notice that r 0 α α a 0 0 and c 0 so z 0 α r 0 b a 0 c 0 + α and thus M 0 x {α 2 : z 0 α < x} {α 2 : α < x} x so m 0 x x Now fix n Since a n Z 0 and c n { b } we have m n x M{α 2 : z n < x} M Fix x 2 and let b {α 2 : z n < x a n k c n l} k0 l A kl {α 2 : z n < x a n k c n l} for k Z 0 and l { b } By Definition 5 z n < x if and only if r n b an c n + < x 8

19 Notice that z n [ c n b an c n+ b a n+ ] cl3 b [ r n ] cl3 b + r n so z n < x if and only if or equivalently b an z n c n + < x z n > + x + c n b an + x + l Additionally in order to have a n k and c n l we must have lb k r n < l + b k or equivalently + l + < z n + l 2 Now notice that since x < 2 + l + < + x + l and thus the left hand inequality in 2 is implied by Therefore z n < x with a n k and c n l if and only if Thus + x + l < z n + l 3 A kl {α 2 : + x + l < z n + l } 4 Now suppose k k 2 Z and l l 2 { b } with k l k 2 l 2 We claim that A k l and A k2 l 2 are disjoint Consider two cases: Case : k k 2 Suppose without loss of generality that k 2 < k so k 2 k Also note that l 2 b and x < 2 so l 2 + x < b Then we have +l +l b k 2 k 2 +l b 2 +b 2 < +l 2 +x 2 5 Case 2: k k 2 l l 2 Suppose without loss of generality that l > l 2 so indeed l l 2 + Then since x < + l + l 2 + l < + l 2 + x 2 6 Now suppose a A k l and a 2 A k2 l 2 By 4 and either 5 or 6 a + l < + l 2 + x 2 a 2 so a a 2 and thus A k l and A k2 l 2 must be disjoint Therefore m n x M b k0 l A kl b MA kl 7 k0 l Finally since m n x M{α 2 : z n < x} by 4 and 7 we can conclude m n x b mn + l m n + x + l k0 l which proves the recursion 0 and completes the proof of the theorem 9

20 Theorem 7 There exist constants A λ > 0 such that m nx bx x+b < Ae λ for all n 0 and x 2 b+ The proof of this theorem which is based on the proof in Section 5 of [5] is lengthy and somewhat technical It is provided in appendix A and the following corollary immediately follows n Corollary 8 We have for all x 2 Theorem 9 We have mx bx x + b 2b b + MD n+ k l m n + l + m n + l Proof Suppose that α D n+ k l Then a n+ k and c n+ l so z n [ lb k r n+2 ] + lb k + where r n+2 can take any value in Clearly z n is a monotonic function of r n+2 for fixed k l so the extreme values of z n on D n+ k l will occur at the extreme values of r n+2 Letting r n gives z n + + l + and letting r lb k +b k n gives z n + lb k +0 l Thus bk r n+2 D n+ k l {α 2 : + l + < z n α + l } M n + l \ M n + l + so MD n+ k l m n + l m n + l + Theorem 20 There exist constants A λ > 0 such that MD nk l lb k +l+b k+ + n < Ae λ ll + b k lb k+ +l+b k + b+ for all k Z 0 l { 2 b } and n Z 0 We then immediately get the following limiting distribution Notice that like with type II continued fractions the distribution is non-monotonic This is due to the gaps in possible denominator terms For example for base 4 the possible denominator terms are The jump from 4 to 8 causes a spike in the limiting distribution 20

21 Corollary 2 We have lim MD nk l n for k Z 0 and l { 2 b } lbk+l+bk++ lb k+ +l+b k + b+ 45 Type III Logarithmic Khinchine Constant We now extend the Khinchine constant to type III continued arithms Note that we only gave an overview for type I and type II but here we will be much more rigourous Definition 6 Let α have type III continued arithm [c 0 b a 0 c b a c 2 b a 2 ] cl3 b Let k Z 0 and l { 2 b } We define {n N : a n k c n l} P α k l lim N N to be the limiting proportion of continued arithm terms of α that have a n k and c n l if this limit exists Note that for the theorems which follow we will restrict our study to 2 instead of The results can be easily extended to by noting that every α corresponds to an α 2 in the sense that the continued arithm of α is just the continued arithm of α with the first term replace by Since we are looking at limiting behaviour over all terms changing the first term will have no impact The following two theorems are proved in Appendix B The proofs are based on the anaous proofs for simple continued fractions that are presented in Sections 5 and 6 of [5] Theorem 22 For almost every α 2 with continued arithm [ c b a c 2 b a 2 ] cl3 b we have +l b kb+l+ b k b+l P α k l +l+ b+ for all k Z 0 and l { 2 b } Theorem 23 For almost every α 2 with continued arithm [ c b a c 2 b a 2 ] cl3 b we have N /N c n b an b A b where A b lim N b b+ 2b n b l2 + l l The values of the Khinchine constant given by the above formula for 2 b 0 are shown in Figure 3 2

22 b KL III b Figure 3: Type III arithmic Khinchine constants for 2 b 0 Remark 8 Notice that Theorem 22 is similar to Corollary 2 However Corollary 2 is about the limiting proportion of numbers α 2 that have a n k and c n l whereas Theorem 22 is about the limiting proportion of terms of a number α 2 for which a n k and c n l The fact that these two limits are the same is not a coincidence: one can show that Corollary 2 is a consequence of Theorem 22 Based on Theorem 23 we denote where A b is as in Theorem 23 KL III b b A b 46 Type III Continued Logarithms and Simple Continued Fractions Now suppose b is no longer fixed Let µ b denote the limiting distribution for a given base b as shown in Corollary 8 That is x+b b+ µ b x bx Furthermore let KL III b denote the base b arithmic Khinchine constant as in Remark 8 and let K denote the Khinchine constant for simple continued fractions as in Section 4 We now have an interesting relationship between these arithmic Khinchine constants and the Khinchine constant for simple continued fractions based on the following lemma Lemma 24 [] Lemma c l2 + K 2 l l Theorem 25 lim b KLIII b K 22

23 Proof We will show that lim b KL III b K from which the desired limit immediately follows lim b KLIII b lim b A b lim ba b b b b lim b b b+ 2b lim b b+ 2b b k2 b k2 lim b 2 2 k2 b+ b k k2 k k + k + k + k k + K k Furthermore as b the distribution function µ b approaches the appropriately shifted continued fraction distribution µ cl The continued fraction distribution function is given by µ cl x 2 + x x 0 See Section 34 of [3] Since the continued fraction for a number will be unchanged except for the first term when adding an integer we can shift this distribution to the right and think of it as a distribution over 2 instead of 0 in order to compare it to µ b We define the shifted continued fraction distribution We then have lim µ bx lim b x+b bx b b+ 2b µ clx µ cl x 2 x x 2 lim b + b x bx x b+ 2 b 2 x 2 2x µ cfx This shows that in some sense as we let b for type III continued arithms we get in the limit simple continued fractions 5 Generalizing Beyond Continued Logarithms A natural question that arises is how one can define something more general than continued arithms Consider the following definition of generalized continued fractions Definition 7 Let c n n0 be an increasing sequence of natural numbers with c 0 Let α The generalized continued fraction for α determined by c n n0 is a 0 + b 0 a + b a 2 + b 2 a 3 + [a 0 a a 2 ] gcf 23

24 where the the terms a 0 a and b 0 b are determined by the following recursive process terminating at the term a n if y n a n y 0 α j n max{j : c j y n } n 0 a n c jn n 0 b n c jn+ c jn n 0 y n+ b n c j n+ c jn y n a n y n c jn n 0 Remark 9 This is a generalization of simple continued fractions and of type I and type III continued arithms Indeed for simple continued fractions the term sequence c n n0 consists of the natural numbers For type I continued arithms the term sequence consists of the powers b 0 b b 2 For type III continued arithms the term sequence consists of terms of the form lb k where k Z 0 and l { b } Recall from Remark 3 that type II continued arithms did not have the property that y n+ could take any value in regardless of the values of a n c n This is a desirable property to have since it uniquely determines the numerator terms based on the corresponding denominator terms We have defined generalized continued arithms so that they have this property and for that reason they are not a generalization of type II continued arithms Remark 0 As per Definitions 3 and 4 the nth convergent and nth remainder term are given by x n [a 0 a a n ] gcf and r n [a n a n+ a n+2 ] gcf respectively Note that the remainder terms r n and the terms y n from Definition 7 are in fact the same We can derive various results for generalized continued fractions that are similar to those for continued arithms Most notably we get the following sufficient criteria for guaranteed convergence and rational finiteness Theorem 26 Suppose there is a constant M > 0 such that c j+ c j < Mc j for all j Then every infinite continued fraction with term sequence c n n0 will converge Theorem 27 Suppose c n+ c n c n for all n Then for every α > the continued fraction of α is finite if and only if α Q We are also able to extend some of the measure-theoretic results to generalized continued fractions though details are not provided here We conjecture that the main results that we derived for the distribution and Khinchine constant of continued arithms would extend likely with some additional restrictions on the sequence c n n0 to our generalized continued fractions Acknowledgements We would like to thank Andrew Mattingly for his input and assistance This research was initiated at and supported by the Priority Research Centre for Computer-Assisted Research Mathematics and its Applications at the University of Newcastle 24

25 References [] David H Bailey Jonathan M Borwein Richard E Crandall On the Khinchine Constant Mathematics of Computation pp [2] Jonathan M Borwein Neil J Calkin Scott B Lindstrom Andrew Mattingly Continued Logarithms and Associated Continued Fractions To appear in Experimental Mathematics [3] Jonathan M Borwein Alf van der Poorten Jeff Shallit and Wadim Zudilin Neverending Fractions Australia Mathematical Society Lecture Series Cambridge University Press 204 [4] Bill Gosper Continued Fraction Arithmetic Perl Paraphernalia Accessed October [5] Alexsander YA Khinchin Continued Fractions Third Edition University of Chicago Press 964 [6] Dan Lascu A Gauss-Kuzmin Theorem for Continued Fractions Associated with Nonpositive Integer Powers of an Integer M 2 The Scientific World Journal [7] Jeffrey Shallit Length of the continued arithm algorithm on rational inputs Preprint Appendix A: Proof of the Type III Continued Logarithm Distribution This appendix is devoted to proving Theorems 7 and 20 restated below: Theorem 7 Restated There exist constants A λ > 0 such that m nx bx x+b < n Ae λ for all n 0 and x 2 Theorem 20 Restated There exist constants A λ > 0 such that MD nk l lb k +l+b k+ + n < Ae λ ll + b k b+ lb k+ +l+b k + b+ for all k Z 0 l { 2 b } and n Z 0 These proofs are based extensively on the proof presented in Section 5 of [5] which proves similar statements for simple continued fractions 25

26 Lemma 28 For x > Proof b k0 l x + l 2 + x + l b + x + l xx + b b k0 l x + l 2 + x + l b + x + l b k0 l b b b b b k b k x + l + b k+ x + l + b l k0 b [ l b l [ b k b k x + l + b k+ x + l lim k [ x + l x + b x b k b k x + l + ] x + l ] b [ b k+ x + l + ] ] b xx + b xx + b Theorem 29 The sequence of functions m nx relationship d m dx nx is given by the recursive m 0x 8 b m nx x + l 2 m n + x + l n 9 for x 2 k0 l Proof Equation 8 follows immediately from 9 Notice that 9 is the result of differentiating both sides of 0 In general if m n+ is bounded and continuous for some n then the series on the right hand side of 9 will converge uniformly on 2 Thus the sum of the series will be bounded and continuous and will equal m n+ so 9 follows by induction since m 0 is clearly bounded and continuous We will now prove a number of lemmas and theorems about the following classes of sequences of functions to which m n n0 belongs Definition 8 Let f 0 f be a sequence of functions on 2 We will say f n n0 A if for all x 2 and n 0 f n+ x b k0 l x + l f 2 n + 20 x + l Furthermore we say that f n n0 A if f n n0 A and there exist constants M µ > 0 such that for all x 2 we have 0 < f 0 x < M and f 0x < µ 26

27 Lemma 30 n b a 0 + +a n q n q n + b an q n Proof Since the intervals of rank n are disjoint and we have that n a a n Jn 2 c c n M 2 c c n n a a n MJn Now notice that by Lemma 3 and Lemma 7 a a MJ n n p n c c n p n + b an p n q n q n + b an q n b an p n q n q n p n q n q n + b an q n n b a 0+ +a n q n q n + b an q n b a 0+ +a n q n q n + b an q n and thus n b a 0 + +a n q n q n + b an q n c c n n a a n MJn Lemma 3 If f n n0 A then for n 0 n pn + b an p n x b n j0 a j f n x f0 q n + b an q n x q n + b an q n x 2 2 Proof For n 0 we just have a single interval so 0 p0 + b a 0 p x f0 q 0 + b a 0 q x q 0 + b a 0 q x 2 + x f 0 + 0x + 0x f 0x 2 b a 0 27

28 Now suppose 2 holds for n Then f n+ x b k0 l b k0 l n b k0 l n b k0 l x + l f 2 n + x + l 2 x + l n pn + b an p n + f0 q n + b an q n + f 0 pn b k x + l + b an p n q n b k x + l + b an q n lb k p n + b an p n + b k p n x f 0 lb k q n + b an q n + b k q n x b k x+l b k x+l n+ cn+ b a n+ p n + b an p n + b a n+ p n x f0 c n+ b a n+ qn + b an q n + b a n+ qn x n+ pn+ + b a n+ p n x f0 q n+ + b a n+ qn x so the result follows by induction Lemma 32 If f n n0 A then for n 0 b n j0 a j q n + b an q n + b k x+l 2 b n j0 a j b k q n b k x + l + b an q n 2 b n+ j0 a j q n+ + b a n+ qn x 2 f nx 3µ + 4M 2n/2 b n j0 a j b k lb k q n + b an q n + b k q n x 2 Proof Differentiate 2 termwise letting u pn+ban p n x q n+b an q n to get x f nx n f 0 u n b 2 n j0 a j q n + b an q n x 4 2 n f0 u b n j0 a j b a n+ c n+ b a n+ qn + b an q n + b a n+ qn x 2 b an q n b n j0 a j q n + b an q n x 3 22 The validity of termwise differentiation follows from the uniform convergence of both sums on the right hand side for x 2 Notice that n b 2 n j0 a j q n + b an q n x 4 q n j0 nb a j 2b n j0 a j 23 2 n /2 q n q n + b an q n by Lemma 6 Lemma 5 and the fact that q n + b a n q n 2q n Additionally q 4 n b an q n b n j0 a j q n + b an q n x ban q n b 3 q 3 n n j0 a j 2b n j0 a j q n q n + b an q n since b an q n q n and q n + b an q n 2q n Since f n n0 A we have by Definition 8 that f 0 x < M and f 0x < µ for all x 2 Thus we have by and Lemma 30 n f nx f n b 2 n j0 a j n 0 u q n + b an q n x b an q n b n j0 a j f0 u q n + b an q n x 3 2µ 2 n /2 n b n j0 a j n q n q n + b an q n + 4M 2 2µ + 4M + 4M < 3µ + 4M n /2 2 n/2 2n/2 2µ 2 28 b n j0 a j q n q n + b an q n 24

29 Lemma 33 If f n n0 A and for some constants T > t > 0 t xx + b < f nx < T xx + b x 2 then t xx + b < f n+x < Proof By 20 and Lemma 28 we have f n+ x > b k0 l b k0 l t xx + b x + l f 2 n + T xx + b x + l x 2 t x + l 2 + x + l b + x + l and a similar derivation shows f n+ x < T xx + b from which the result follows Lemma 34 If f n n0 A then for all n 0 Proof Notice that f n z dz f 0 z dz 2] b + l + + l ] k0 l where the intervals are pairwise disjoint We then have by 20 that f n z dz b z + l f 2 n + k0 l b +l+ k0 l b +l k0 l +l f n u du +l+ f n u du f n u du from which the result follows by induction 29 dz z + l

30 Lemma 35 Suppose f n n0 A and there are constants g G > 0 such that for all x 2 g xx + b < f G 0x < xx + b 25 Then there exist n N and g G > 0 such that and where δ 2b 4b b+ Proof First define ϕ n x f n x g xx + b < f nx < G xx + b 26 g < g < G < G 27 G g < G gδ + 2 n/2 µ + G 28 g xx + b ψ nx G xx + b f nx 29 which are both positive functions by 25 and Lemma 33 Notice that for the functions G hx and Hx we have by Lemma 28 that hx g xx+b g xx + b b k0 l k0 l xx+b b x + l 2 h + g x + l 2 + x + l b + x + l x + l and similarly Hx Thus for n we have b k0 l x + l H + 2 x + l ϕ n+ x f n+ x hx b and similarly k0 l b k0 l x + l 2 f n x + l 2 ϕ n ψ n+ x + + b k0 l x + l x + l b k0 l x + l ψ 2 n + x + l h + 2 x + l Thus ϕ n n0 ψ n n0 A so by Lemma 3 setting u pn+ban p n x q n+b an q n we get x ϕ n x n ϕ0 u b n j0 a j q n + b an q n x n n ϕ0 u b j0 a j q 2 n x + l 30

31 and similarly since ψ n x 4 n n ψ0 u b j0 a j q n + b an q n x 2q n x 2 On the other hand the mean value theorem gives and 4 4 where for each interval and the length of the interval is ϕ n x 4 ϕ 0 z dz 4 q 2 n 3 n b n j0 a j ϕ0 u q n q n + b an q n 32 ψ 0 z dz n b n j0 a j ψ0 u 2 4 q n q n + b an q n 33 p n q n pn+ban p n q n+b an q n of rank n u and u 2 are points in the interval nj0 a b j q nq n+b an q n From 30 and 32 we then get and from 3 and 33 we get ψ n x 4 ϕ 0 z dz 4 ψ 0 z dz 4 n b n j0 a j [ϕ0 u ϕ 0 u ] q n q n + b an q n 34 n b n j0 a j [ψ0 u ψ 0 u 2 ] q n q n + b an q n 35 Now for x 2 we have ϕ 0x f 0x +g µ+g and ψ 0x f 0x +G µ+g so it follows by Lemma 6 that b n j0 a j ϕ 0 u ϕ 0 u µ + g u u µ + g q n q n + b an q n µ + g q n 2 µ + g 36 2n/2 and similarly ψ 0 u 2 ψ 0 u 2 µ + G 37 2n/2 Then by Lemma and 36 give where l 4 ϕ n x > 4 l 4 l µ + g 2 2 n/2 ϕ 0 z dz 4 n b n j0 a j [ϕ0 u ϕ 0 u] q n q n + b an q n n b n j0 a j ϕ0 u ϕ 0 u q n q n + b an q n n ϕ 0z dz Similarly 35 and 37 give n b j0 a j q n q n + b an q n l µ + g l µ + g 2 2 n/2 2 n/2+ ψ n x L G + µ 2 n/2+ 3

32 where L 4 ψ 0z dz Now by 29 we have f n x g xx + b + ϕ nx > > g + l 2 n/2 µ + g xx + b where g g + l 2 n/2 µ + g and f n x G xx + b ψ nx < < G l + 2 n/2 µ + G xx + b g xx + b + l µ + g 2 n/2+ g xx + b 38 G xx + b L + µ + G 2 n/2+ G xx + b 39 where G G L + 2 n/2 µ + G Now since l L > 0 we can choose n sufficiently large so that 2 n/2 µ + g < l and 2 n/2 µ + G < L so that we get g < g < G < G 40 Thus by and 40 we have found g G and n that satisfy 26 and 27 Notice that we also have G g G g L + l + 2 n/2 2µ + g + G < G g L + l + 2 n/2 µ + G 4 Now since l + L 4 G g xx + b dx G g 2b 4b b + 4 becomes 2b G g < G g + 2 n/2 µ + G δg g + 2 n/2 µ + G 4b b + so we see that g G and n also satisfy 28 completing the proof Remark Notice that the value of n chosen depends only on the values of µ and G and that if we make 0 < µ < µ and 0 < G < G the value of n chosen for µ and G will also work for µ and G In other words we can make µ and G smaller without having to increase n This will be useful in the proof of the Theorem 36 Theorem 36 Suppose f n n0 A Then there exist constants λ A > 0 such that for all n 0 and x 2 f a nx xx + b < Ae λ n where a b b+ f 0 z dz 32

33 Proof By assumption f 0 is differentiable and continuous on [ 2] so there is some constant m > 0 such that m < f 0 x < M for all x [ 2] Then since < < 2b+ xx+b b for all x 2 we have bm xx + b < f 0x < 2b + M xx + b x 2 Thus let g bm and G 2b + M and apply Lemma 35 to f 0 g and G to get g G and n such that and g xx + b < f nx < G xx + b g < g < G < G G g < δg g + 2 n/2 µ + G x 2 By Lemma 32 f nx < µ 3µ 2 n/2 + 4M and we can arrange to have µ < µ by making µ and n sufficiently large By Remark the results above are still valid for the new values of µ and n We can then apply Lemma 35 again with f n g and G instead of f 0 g and G This gives us new constants g 2 and G 2 such that again due to Remark and g 2 xx + b < f 2nx < G 2 xx + b g < g < g 2 < G 2 < G < G G 2 g 2 < δg g + 2 2n/2 µ + G x 2 Repeating this in a similar fashion gives in general constants g r G r such that and g r xx + b < f nrx < G r xx + b x 2 g < g < < g r < g r < G r < G r < < G < G G r g r < δg r g r + 2 rn/2 µ r + G r where µ r is a constant such that f nr x < µ r for all x 2 By Lemma 32 we can take µ r 3µ + 4M and then can choose r 2 nr/2 0 N such that µ r < 5M for all r r 0 Then since G r < G 2b + M we have G r g r < δg r g r + 2b + 7M2 nr/2 δg r g r + M 2 nr/2 42 for all r r 0 where M 2b + 7M We now claim that for all k 0 For k 0 from 42 we have G r0 +k g r0 +k < δ k G g + δ k M 2 nr 0/2 k 2 nj/2 δ j 43 G r0 g r0 < δg r0 g r0 + M 2 nr 0/2 < G g + M 2 nr 0/2 δ 0 G g + M δ 0 2 nr 0/2 j0 0 2 nj/2 δ j j0 33

34 Now suppose 43 holds for k Notice that so by 42 M 2 nr 0+k+/2 M δ k+ 2 nr 0/2 2 nk+/2 δ k+ G r0 +k+ g r0 +k+ < δg r0 +k g r0 +k + M 2 nr 0+k+/2 k < δ δ k G g + M δ k 2 nr 0/2 2 nj/2 δ j + M 2 nr 0+k+/2 so 43 follows by induction Now notice that for k > 0 j0 k δ k+ G g + M δ k+ 2 nr 0/2 2 nj/2 δ j + 2 nk+/2 δ k+ j0 k+ δ k+ G g + M δ k+ 2 nr 0/2 2 nj/2 δ j j0 k 2 nj/2 δ j < j0 2 n/2 δ j j0 2 /2 δ j γ < j0 since 2 /2 δ 2 2b > 2 2 > 43 then becomes 4b b+ 4 G r0 +k g r0 +k < δ k G g + M 2 nr 0/2 γ δ k c where c > 0 is a constant Then for r r 0 we have G r g r < δ r r 0 c δ r δ r 0 c δ r d where again d > 0 is a constant Finally since δ < we can choose B λ > 0 such that G r g r < Be λr Thus there is clearly some common limit a lim r g r lim r G r and we have setting r n that f n 2x a xx + b < Be λn x 2 44 Thus we have lim n f n 2z dz so by Lemma 34 f 0z dz a 2b b b+ a b b+ a xx + b and thus 34 f 0 z dz a b 2b b +

35 Now for arbitrary N r 2 0 we can choose n r 0 such that n 2 N < n + 2 We then have by 44 a 2b + Be λn xx + b < a xx + b Be λn < f n 2x < a xx + b +Be λn < a + 2b + Be λn xx + b for all x 2 Then by Lemma 33 so f a Nx xx + b a 2b + Be λn xx + b < 2b + Be λn xx + b < f N x < a + 2b + e λn xx + b < 2b+Be λn 2b+Be λ e λn+ < A e λ N where A 2b+Be λ is a constant Now for 0 N < r0 2 note that each f N is continuous since f 0 is differentiable and thus continuous and f N+ is an absolutely convergent sum of continuous transformations of f N Thus we can choose A 0 A A r 2 0 such that for 0 N r0 2 f a Nx xx + b < A ne λ N x 2 N {0 r0 2 } for all x 2 Finally take A max{a 0 A A r0 A } so we have f a Nx xx + b < Ae λ N x 2 N Z 0 proving the theorem Corollary 37 There exist constants λ A > 0 such that for all n 0 and x 2 a m nx xx + b < Ae λ n where a b b+ Proof By Theorem 29 m n n0 A Then Theorem 36 gives constants A λ > 0 such that m a nx xx + b < Ae λ n where proving the corollary a b b+ m 0z dz b b+ dz b b+ Our main goals Theorems 7 and 20 follow easily from Corollary 37 35

36 Theorem 7 Restated There exist constants A λ > 0 such that m nx bx x+b < n Ae λ for all n 0 and x 2 b+ Proof First note that since m n 0 for all n so by the Fundamental Theorem of Calculus Thus x m nz dz b b+ m n x m n + x Then by Theorem 36 we have m nx bx x+b b+ x m nz dz x m nz dz zz + b dz m nx b bx x+b b b+ x x m nz b b+ m nz b b+ x Ae λ n < Ae λ n zz + b dz zz + b dz < x+b b+ m n x bx x Ae λ n 45 Theorem 20 Restated There exist constants A λ > 0 such that MD nk l lb k +l+b k+ + n < Ae λ ll + b k lb k+ +l+b k + b+ for all k Z 0 l { 2 b } and n Z 0 Proof By Theorem 9 MD n k l m n + l m n + l + +l +l+ m n z dz Then by Corollary 37 it follows that there is are constants A λ > 0 such that +l m n z dz b +l +l+ b+ +l+ b zz + b dz k +l m n z b zz + b dz < +l+ +l +l+ Ae λ b+ n dz l l + Ae λ n Ae λ n ll + b k 36