Patterns of Non-Simple Continued Fractions

Transcription

1 Patterns of Non-Simple Continued Fractions Jesse Schmieg A final report written for the Uniersity of Minnesota Undergraduate Research Opportunities Program Adisor: Professor John Greene March 01

2 Contents 1 Introduction 1 11 Real Numbers and Simple Continued Fractions 1 1 Generalized Continued Fractions 1 Motiation General Results 1 Continued Fraction Algorithm and Conergents Continued Fraction Expansions 8 Periodic Results 1 1 General Periodicity 1 Period Length 1 18 Other Results and Conjectures 0 Future Work A Data Collection A1 Terminating Rationals A Periodic Rationals 8 A Continued Fraction Rational Numerators 1 B Mathematica Code 0 ii

3 1 Introduction 11 Real Numbers and Simple Continued Fractions Take a moment to reflect on the real numbers and the fact that they fall into one of two categories: rational or irrational It is known that, when written in decimal form, all rational numbers are either terminating, such as, or are non-terminating but eentually periodic, such as 10 On the other hand, irrational numbers possess neither of these properties, being non-terminating and non-periodic, such as π 11 These expansions rely on base-10 notation, which is the most common Howeer, one could just as easily write these numbers using other bases, such as base-, which would produce 00, 10 10, and π These examples show that a change of base may produce expansions that hae different properties (such as terminating) Notice that there are many different ways to write the same number, including the relatiely less common method of continued fraction expansion We will first introduce this technique and then show why it may be useful A continued fraction is an expression of the form r 0 a 0 ` a 1 ` r1 a ` r but we shall restrict ourseles to considering a constant r alue, that is, r a 0 ` a 1 ` r a ` r For a continued fraction of this form, we will occasionally use the notation a 0 ` r r a 1 ` a `, but we will more often use the compact notation ra 0, a 1, a, s r The r in this context will usually be referred to as the numerator of the continued fraction We may say that the expansion is eentually periodic (generally referred to just as periodic) with period length n if for some m, a m`i a m`n`i for i ě 0, and use the notation ra 0,, a m 1, a m,, a m`n 1 s r Moreoer, an expansion is purely periodic if m 0 in the context we just introduced The most common restriction imposed on continued fractions is to hae r 1 and then call the expression a simple continued fraction Example To demonstrate the recursie approach to expanding a number into a simple continued fraction, we will calculate the expansion for 0 Follow each step as 0 ` 6 ` 6 ` 1 1 ` 1 6, ` 1 1 ` 1 ` 1 1 This calculation demonstrates an elementary fact of simple continued fractions each rational number has exactly two expansions, differing only at the end of the expression In 1

4 general, ra 0, a 1,, a n s 1 ra 0, a 1,, a n 1, 1s 1 if a n ą 1 The method for expanding negatie numbers differs only in the first step, as seen with ` ` ` 1 ` 1 ` 1 ` 1 1` 1 1 Recall the earlier obseration that some representations of a number hae different properties (such as terminating or periodic) depending on the base being used More precisely, if x a 0 a 1 a, then 0 ď a i ď b 1 where b is the base, so the alues of each a i depend on b This is not a problem with continued fraction expansions since the entries are integers and can be altered independently Take as example that 10 r, 1s 1 with base-10 while 10 r10, 1s 1 with base- This means that one could use a different base for a continued fraction which only changes how the integers are expressed, but not what they are One of the frequently used tools for working with continued fractions is conergents Conergents are the alues formed by partial expansions of a continued fraction These hae many uses for simple continued fractions, such as finding rational approximations of an irrational number For ra 0, a 1, a, s r, the conergents are denoted as c n where c n ra 0, a 1,, a n s r Example To demonstrate the conergents of a simple continued fraction, consider the expansion of 0 r, 1, 6s 1 Then, c 0, c 1 ` 1 1, c ` 1 1 ` As mentioned, we can calculate rational approximations of irrational numbers such as π We know that the expansion of π begins as r,, 1, 1,, 1, s 1 [] By calculating the first few conergents, we find and see that 106 c 0, c 1 ` 1, c ` 1 ` , which is a decent approximation 106, There are a number of interesting properties to find from conergents, in particular by looking at the numerator and denominator parts For this reason, we often write c n pn q n 1 Generalized Continued Fractions Simple continued fractions hae many applications, particularly for number theory, and hae been studied thoroughly in the past as demonstrated in [] Howeer, the same is not true for more general forms of continued fractions While [1] looks into the option of integer numerator alues, we wish to go further The primary extension in this work is to allow for positie rational numerators That is, we will work with continued fractions of the form u{ a 0 `, a 1 ` u{ a ` u{

5 where u, P Z`, a 0 P Z, and a i P Z` for i ą 0 Additionally, we will suppose that u ě and gcdpu, q 1 For a continued fraction of this form, we will similarly use the notation a 0 ` u{ a 1 u{ ` a ` and the compact notation ra 0, a 1, a, s u{ An algorithm for computing continued fractions of this form will be shown and examples gien Section of this paper works ground up, starting with an algorithm that we can use for continued fractions with rational numerators We deelop relations that hold for conergents, which end up haing many similarities to those that hold for simple continued fractions We go on to find other theorems that hae analogues to theorems for simple continued fractions Importantly, these findings are applicable to any expansion Section of this paper has extra focus on continued fraction expansions of rational numbers that are periodic We deelop relations that hold when an expansion is periodic, which gies us a way to find and erify periodicity Furthermore, there is a special case for period length 1, to which we gie attention Section contains some potentially useful findings that do not necessarily apply to other places or hae yet to be proen Section gies future work possibilities, mostly relating to unproen ideas 1 Motiation The motiation for this work comes from a common obseration When presenting continued fractions, many authors start by giing the general form and then restricting to the simple case By allowing more generality and working out examples, it does not take long to see potential patterns and wonder what can proen In addition, one may wonder what happens to theorems that hae been found for simple continued fractions and if they carry oer This work aims to fulfill some of these wonders General Results 1 Continued Fraction Algorithm and Conergents We start by defining a recursie algorithm for computing continued fractions with rational numerators and erifying that the algorithm gies successful results Theorem 1 Gien x P R and r P Q` with r ě 1, choose sequences px n q and pa n q by x 0 x, tx n u tru ď a n ď tx n u, x n a n ď r, x n r x n 1 a n 1, with a i ą 0 for i ą 0, and terminating when x n is an integer This creates the expansion ra 0, a 1, a, s r

6 Proof There is only one potential issue with this algorithm and it is analogous to the problem discussed in [1, Theorem 1 on p] That is, an issue arises if x i ă 1 for some i ą 0, for then no choice is aailable for a i We can inductiely show that this issue cannot arise As a base case, a alid choice exists for a 0 since it is allowed to be 0 Then assume that a alid a k ą 0 is chosen for some k ą 0 Subsequently, 0 ď x k tx k u ď x k a k ď r As stated, if x k is an integer, then the algorithm terminates, otherwise, 0 ă x k tx k u ď x k a k ď r and x k`1 r x k a k ě r 1 Thus, x r k`1 ě 1, so we can make a alid choice for a k`1 By the principle of mathematical induction, a alid choice is aailable for any a n Remark A question that could be raised at this point is whether or not an infinite expansion conerges to the desired alue That is, for an infinite expansion, can we truly write x ra 0, a 1, s u{ We can refer this question to the proof of [1, Theorem 1 on p] This proof can be applied here after demonstrating that q n ě pu ` q n and q n`1 ě pu ` q n, both of which can be shown inductiely The rest of the proof follows, and we see that conergence holds With the described algorithm, one can create many different expressions for a particular alue Each expansion depends on the choices of a n, where there are as many as rrs possibilities Sometimes we will desire a certain method for the selections Definition For a choice of a n in the continued fraction algorithm, if a n is the smallest integer such that a n ě tx n u tru and x n a n ď r, then it is the min choice for a n If a n tx n u, then it is the max choice If the min choice is used for eery a n, then this is referred to as the min algorithm Similarly, we define the max algorithm Note that these two may produce identical expansions in some cases With these algorithms, it is useful to demonstrate the numerous ways to apply them Example We found earlier that 0 r, 1, 6s 1 Now instead we can make an expression that uses numerator Starting with the max algorithm, 0 ` 6 ` { ` {, 1 ` { so 0 r, 1, s { With the min algorithm, tx 0u tru t 0u t u 1, so 1 ď a 0 ď Howeer, x 0 1 ą, so the only alid (and thus smallest) choice is a 0 Continuing in this fashion, the min algorithm actually produces the same expansion as the max algorithm Suppose instead we wish to expand 0 with as a numerator While omitting the calculations, we find 0 r,,,,,, s { with the max algorithm Then, using the min algorithm, 0 1 ` { 6 1 ` { 1 ` { 6 1 ` { 1 ` { ` { 8 1 ` { 1 ` { ` { 1` { 0, so 0 r1, 1,, 1, 0s {, a different expression!

7 These examples hae shown how rational numbers may hae expansions that terminate In other cases howeer, we may find this does not happen For example, when we attempt to expand 10 with numerator and the max algorithm, then 10 ` { ` { ` ` { ` { ` { ` { ` { It appears that this pattern continues indefinitely, suggesting that 10, Later, we { will show how this periodicity can be proed It also appears that some expansions of rational numbers can continue indefinitely without becoming periodic For example, when we expand with numerator and the max 1 algorithm, the expansion begins as r1,,,,, s {1 This appears to neer terminate nor become periodic, een when checking up to one million iterations! Unfortunately, we are unable to proe this continues indefinitely; it could, for example, terminate later in the expansion Remark When we calculate the expansion of with numerator and the max algorithm, 1 one million steps can be used without terminating or becoming periodic Howeer, some expansions are quicker to expose behaior Using the max algorithm up to 100 steps, neither expansion of 10 with numerator or terminate or become periodic Howeer, using a sufficiently large number of steps, both become periodic! The moral is that using a greater number of iterations can potentially expose the true properties The conergents of continued fractions play a number of roles in these results, so they are deeloped first in the generalized setting We start by showing how we can determine the conergents, c n pn q n, by collapsing an expansion Example We wish to find the conergents of 0 when expanded with numerator and the max algorithm A series of calculations gies 0 r,,,, s { Now, for example, the calculation for c is c r, s { ` { ` { 1{ 1 We also calculate the rest of the conergents, but omit the calculations, to find ˆ pc n q 1, 1, 6, 10 8, 0 The conergents of a continued fraction hae many interesting uses, requiring a few relations The relations found here hae known analogues for simple continued fractions Theorem Suppose that x ra 0, a 1,, a n 1, a n, s u{ Let x n ra n, a n`1, s u{, so that x ra 0, a 1,, a n 1, x n s u{ For the conergents of x, pn q n, the following relations are satisfied: p 1 1 p 0 a 0 q 1 0 q 0 1

8 p n a n p n 1 ` up n p n`1 a n`1 p n ` up n 1 q n a n q n 1 ` uq n q n`1 a n`1 q n ` uq n 1 (1) p n q n 1 p n 1 q n p 1q n 1 u n () x x 0 x p n 1x n ` up n q n 1 x n ` uq n x p nx n`1 ` up n 1 q n x n`1 ` uq n 1 () Proof The proof of (1) uses a similar strategy to that of [, Theorem 1 on p1] As a base case, and p 1 a 0 ` u{ a 0a 1 ` u q 1 a 1 a 1 a 1p 0 ` up 1 a 1 q 0 ` uq 1 p a 0 ` u{ u{ a 0a 1 a ` ua 0 ` ua q a 1 ` a a 1 a ` u a p 1 ` up 0 a q 1 ` uq 0 Then, assume that (1) holds for integers k 1 and k for some k P N With this assumption, we shall show that the relations also hold for k ` 1 and k ` First, obsere that for any n, p n`1 a 0 ` u{ u{ q n`1 a 1 ` ` a n ` u{ a 0 ` u{ u{ a n`1 ` ` a n 1 ` a 1 u{, a n ` u{ a n`1 which allows the calculation of p k`1 q k`1 by replacing a k There must be some care with this obseration, again following the strategy used in [, Theorem 1 on p1] Note that if we replace a k, the alues of p k`1 and q k`1 are unaltered This is due to the manner of calculation, where the alues of p k`1 and q k`1 depend on the preceding equations and not the further a alues Additionally, the same technique is used for calculating p k` and q k` Now, starting with the assumptions for k 1 and k, we calculate and p k`1 pa u{ k ` a k`1 qp k 1 ` up k pa ka k`1 ` uqp k 1 ` up k q k`1 pa k ` u{ a k`1 qq k 1 ` uq k pa k a k`1 ` uqq k 1 ` uq k a k`1pa k p k 1 ` up k q ` up k 1 a k`1 pa k q k 1 ` uq k q ` uq k 1 a k`1p k ` up k 1 a k`1 q k ` uq k 1 p k` pa u{ k`1 ` a k` qp k ` up k 1 pa k`1a k` ` uqp k ` ua k`p k 1 q k` pa k`1 ` u{ a k` qq k ` uq k 1 pa k`1 a k` ` uqq k ` ua k` q k 1 a k`pa k`1 p k ` up k 1 q ` up k a k` pa k`1 q k ` uq k 1 q ` uq k a k`p k`1 ` up k a k` q k`1 ` uq k 6

9 By the principle of mathematical induction, (1) holds for any n P N Proof of () also follows from induction For the base case, calculations show that and p 0 q 1 p 1 q 0 a p 1q 0 1 u 0 p 1 q 0 p 0 q 1 a 0 a 1 ` u a 0 a 1 u p 1q 1 1 u 1 Now, assume that () holds for some k, k P N We shall use this assumption to show that the relation also holds for both k ` 1 and k ` Assume that Then, p k 1 q k p k q k 1 p 1q k`1 u k p k`1 q k a k`1 p k pa k q k 1 ` uq k q ` up k 1 pa k q k 1 ` uq k q a k`1 p k q k ` up k 1 q k a k`1 p k q k ` upp k q k 1 p 1q k 1 u k q a k`1 p k q k ` up k q k 1 ` p 1q k u k`1 p k q k`1 ` p 1q k u k`1 With the calculations for k `, it is nearly identical to show that p k` q k`1 p k`1 q k` ` p 1q k`1 u k` Thus, by the principle of mathematical induction, () holds for n P N Proof of () will also follow from induction For the base case, and x ra 0, x 1 s a 0 ` u{ x 1 x ra 0, a 1, x s a 0 ` u{ a 1 a 0x 1 ` u x 1 u{ ` x p 0x 1 ` up 1 q 0 x 1 ` uq 1 a 0a 1 x ` ux ` a 0 u a 1 x ` u p 1x ` up 0 q 1 x ` uq 0 Now assume that () holds for some k, k P N With this we can show that the relation holds for both k ` 1 and k ` In a similar manner to the proof of (1), and for the same reasoning that allows it, we can replace an x k`1 term with a term using both a k and x k`1 That is, x ra 0, a 1,, x k`1 s a 0, a 1,, a k ` u{ j a 0 ` u{ x k`1 a 1 ` ` p kpa k ` u{ x k`1 q ` up k p kx k`1pa k ` uq ` up k x k`1 q k pa k ` u{ x k`1 q ` uq k q k x k`1 pa k ` uq ` uq k x k`1 x k`1pa k p k 1 ` up k q ` up k 1 x k`1 pa k q k 1 ` uq k q ` uq k 1 p kx k`1 ` up k 1 q k x k`1 ` uq k 1 u{ a k ` u{ x k`1 Therefore, the relation holds for k ` 1, and a similar result follows for k ` Thus, by the principle of mathematical induction, () holds for n P N It is conenient to hae a table of some conergent alues on hand to aoid the computations each time they are needed

10 n p n q n a a 0 a 1 ` u a 1 a 0 a 1 a ` ua 0 ` ua a 1 a ` u a 0 a 1 a a ` ua 0 a 1 ` ua 0 a ` ua a ` u a 1 a a ` ua 1 ` ua Table 1: Values of p n and q n for ra 0, a 1, s u{ with small n Corollary For n P N and the conergents, pn q n, of a continued fraction using numerator u, (i) q n 1 (ii) gcdpp n, q n q u n Proof For piq, note q 1 a 1 q 0 `uq 1 a 1, and a 1 Assume that for some k, q k 1 Then q k`1 a k`1 q k ` uq k 1 Since a k`1 q k and q k 1, then q k`1 For piiq, let d gcdpp n, q n q Then d p n and d q n, so d p n q n 1 p n 1 q n By (), d p 1q n u n, so d u n The relations shown aboe can be demonstrated by extending a preious example Example We calculated 0 r,,,, s { By (1), and pp n q p, 1, 6, 10, 0q pq n q p1,,, 8, 81q Subsequently, ˆpn q n ˆ 1, 1, 6, 10 8, 0 ˆ 81 1, 1, 6, 10 8, 0 Continued Fraction Expansions This section focuses on methods for using expansions that hae already been calculated We first turn our sights to finding continued fraction expansions that are ery similar to one another As obsered, a gien expansion may hae terminating or periodic behaior, or een appear to go on foreer without becoming periodic Definition Two continued fraction expansions are equialent if they hae the same form (ie, periodic, terminating, etc) with the same lengths (eg, periodic with period length ), and eentually become the same expansion from some point on 8

11 Example Let r We calculate the following 1 r0,, s r 0,,, r 1 r0, 1, s r 0, 1,, r 8 1 r, 1, s r 8, 1,, r Note how these share certain characteristics The obserations to make are that 1 1`, 8 ` 1, `, and 8 ` Following from the preious example, we shall now show two results that will assist us in finding equialent continued fractions Lemma 6 For integers u,, x, and y, if then for any positie integer i Proof Notice that px`yiq y x y x ` yi y x y ra 0, a 1, a, s u{, ra 0 ` i, a 1, a, s u{ ` i The result follows immediately by construction Lemma For integers u,, x, and xj ` k with j ě 1 and 0 ď k ă x, if then for any positie integer i Proof Let x 0 x xj`k x xj ` k r0, a 1, a, s u{, x xpj ` iq ` k r0, a 1 ` ui, a, s u{ with j ě 1 and 0 ă k ă x, then gcdpx, xj ` kq 1 Suppose that By the algorithm that we introduced earlier, a 0 0, x 1 x 0 r0, a 1, a, s u{ r0, a 1, x s u{ upmj ` kq m, a 1 tx 1 u, x u{ x 1 tx 1 u, are the beginning pieces for constructing the expansion Then we shall do a similar construction for where i is a positie integer With the same algorithm, x xpj`iq`k a 1 0 0, x 1 upmpj ` iqq ` kq 1 x 1 ` uim m m a 1 1 tx 1 ` uiu a 1 ` ui, x 1 ` ui,

12 and thus, Therefore, x 1 u{ x 1 1 a 1 1 x r0, a xpj`iq`k 1 ` ui, a, s u{ u{ x 1 ` ui a 1 ui u{ x 1 tx 1 u With the results of the preious lemmas, the following can be shown Theorem 8 With numerator u and integers x and y, when considering continued fraction expansions under the same algorithm, x y x pmod yq y x pmod yq y pmod px pmod yqqq Proof By definition, x pmod yq x ` yi for some i The first equialence follows from Lemma 6 Now let l x pmod yq Then l ă y, and we can rewrite y as y lj ` k for some j ě 1 and 0 ă k ă l Next, y pmod lq y ` li lpj ` iq ` k for some i The second equialence follows from Lemma One application of Theorem 8 is the ability to classify continued fractions more efficiently, especially if we are only looking for general properties Example We wish to know whether there are any periodic continued fraction expansions for numbers of the form 1, where y is an integer, with numerator We calculate y and 1 r0, s { 1 r0,, s { By Theorem 8, eery number 1 will hae one of the aboe forms, so no periodic continued y fraction expansions arise from the form x 1 y Example We would like to determine all unique period lengths that continued fraction expansions hae for numerator when we expand all x where x, y ă 1000 Calculations y show that becomes periodic with period length under the max algorithm We now can sae significant time by no longer needing to compute the expansions for,,,,, etc, since they are all equialent (each will be periodic with period length ) 6 As demonstrated, time can be saed by calculating some continued fraction expansions and then no longer needing to do the same for a large class of others, possibly infinite The next results are extensions of classic work on simple continued fractions Theorem For integers u,, x, and y, if 0 ă x ă y and x y r0, a 1, a, a, s u{, then uy x ra 1, a, a, s u{ 10

13 Proof With 0 ă x ă y and x 0 x y, tx 0u 0, so x 1 u{ x{y 0 uy x Theorem gies the ability to extract a piece of a continued fraction, as demonstrated in the next example Example After calculating r1,,,,,, s {, we wish to pull off the front of the expansion Using Theorem 8 and Theorem, we know r0,,,,,, s { and 1 r,,,,, s 8 { We can continue in this fashion, repeatedly applying the two theorems, finding expansions such as r,,, s { Theorem 10 If pn q n (i) for ď k ď n, (ii) if a 0 0, for 0 ď k ď n, (iii) if a 0 0, for ď k ď n, ra 0, a 1, a,, a n s u{ and k P N, then q k q k 1 ra k,, a, a 1 s u{ p k p k 1 ra k,, a, a 1, a 0 s u{ p k p k 1 ra k,, a, a, a s u{ Additionally, if a i for all 0 ď i ď n, then (i) for 1 ď k ` 1 ď n, q k`1 q k a k`1,, a, a 1 u{ () if a 0 0, for 1 ď k ` 1 ď n, p k`1 p k a k`1,, a, a 1, a 0 u{ (i) if a 0 0, for ď k ` 1 ď n, p k`1 p k a k`1,, a, a, a u{ Proof The proof of piq follows by induction We are interested in an index on k and for k, q q 1 a q 1 ` uq 0 q 1 a ` uq 0 q 1 a ` u a 1 a ` u{ a 1 ra, a 1 s u{, which completes the base case Now assume that the result holds for some positie integer l That is, Then, q l ra l,, a, a 1 s q u{ a l ` u{ u{ u{ l 1 a l 1 ` ` a ` a 1 q l` a l`q l`1 ` uq l a l` ` uq l u{ a l` ` q l`1 q l`1 q l`1 pq l`1 {q l q{ u{ a l` ` ppa l`1 q l ` uq l 1 q{q l q{ a u{ l` ` a l`1 ` puq l 1 q{pq l q u{ a l` ` a l` ` u{ a l`1 ` a l`1 ` ` u{ q l {q l 1 u{ a 1 ra l`,, a, a 1 s u{, as desired for l `, which proes the result The proofs for piiq, piiiq, piq, pq, and piq are ery similar

14 Example We calculate 1 r,,,, s { Furthermore, pq n q p1, 6, 1, 8, q Then, by Theorem 10i, 1 6 r, s { and 8 r,,, s { Example We calculate r6,, s { Furthermore, Then, by Theorem 10, 6, 6 Periodic Results 1 General Periodicity pp n q p6,, 6q { r6, s { As mentioned earlier, some continued fraction expansions appear to eentually become periodic Here, we go into further detail about what this entails and find special results Periodic continued fraction expansions consist of a leading tail and an orbit, both of which can ary in length These hae the form ra 0, a 1,, a k, a k`1,, a k`n 1, a k`n 1 s u{ with the part under the line repeating indefinitely Periodicity can be proen, as demonstrated next Example We wish to compute the continued fraction expansion of 1 with numerator and the max algorithm When calculated, there are sequences pa n q p0,, 1,, 10, q and px n q p, 10, 0,, 10, q At that point, we are able to proe periodicity The term reappears in the px 8 nq sequence, so the steps will repeat at that point forward, causing period length Indeed, 1 0,, 1,, 10 { Remark A remark must be made about the preious example the algorithm for choosing terms should be consistent For instance, if the max choice is made at one point and the min choice at another, the expansion will not necessarily hae periodic behaior as the steps for choosing the terms may not follow the same pattern There is slight difficulty in classifying periodic expansions when we hae to work with the tails It would be helpful to extract the periodic part in the sense of taking the alue of the orbit into a new continued fraction By doing this, we will hae one that is purely periodic with period length n of the form ra 0, a 1,, a n 1 s u{ Preiously, we showed how Theorem could be used to extract the end of an expansion Example Calculations show that 1,,,, 1,,, but we would like to work with 0 { an expansion that is purely periodic with period length Applying ideas from Theorem 8 p q and Theorem twice, we arrie at,, 1,, pp0 q p qq 8 { 1

15 Going further, a purely periodic continued fraction can be used in a recursie manner Recall that periodic behaior occurs when a alue in the sequence px n q reappears Thus, for a purely periodic expansion of x with a period defined by a 0, a 1,, a n 1, we can write x ra 0, a 1,, a n 1, xs u{ With this form of a periodic continued fraction at hand, preious results can be used to derie seeral properties First, a known fact of simple continued fractions is that if x is a real number and the expansion of x is eentually periodic, then x is a quadratic surd [, p8] This is not, in general, true for continued fractions with rational numerators, as we hae demonstrated Howeer, if x ra 0, a 1,, a k, a k`1,, a k`n 1, a k`n 1 s u{, and we let b ra k, a k`1,, a k`n 1, a k`n 1 s u{, then () tells us or b p n 1b ` up n q n 1 b ` uq n b p n 1b ` up n q n 1 b ` uq n That is, b satisfies a quadratic equation, so b is either a rational number or a quadratic surd Since x ra 0, a 1,, bs u{, x can be simplified and will be either a rational number or quadratic surd as well For the next property, we can go on to say more about expansions of rational numbers using a rational numerator Theorem 1 If a continued fraction expansion of x, a rational number, with numerator u is purely periodic with period length n, then there exists an integer solution to where p n q n and p n 1 q n 1 puq n ` p n 1 q u n s, if n is een (1) puq n ` p n 1 q ` u n s, if n is odd () are the n and n 1 conergents, respectiely, of x and s is an integer Proof Assume that we hae a purely periodic continued fraction expansion x ra 0, a 1,, a n 1, xs u{ where n is een Recall the x n ersion of () for x, so x p n 1x n ` up n q n 1 x n ` uq n p n 1x ` up n q n 1 x ` uq n, which can be rearranged; then apply the quadratic formula to q n 1 x ` puq n p n 1 qx up n 0

16 Since x is rational, the discriminant of the quadratic must be a perfect square Here, the discriminant is and we can simplify by utilizing (), then u q n ` uq n 1 p n uq n p n 1 ` p n 1, u q n ` uq n p n 1 ` p 1q n u n ` p n 1 puq n ` p n 1 q u n With een n, this results gies (1) By a ery similar process for the case when n is odd, we arrie at () With these equations for periodicity, we take note of an obseration about odd period lengths, which can now be proed Theorem If there exists integers x and y such that the expansion of x y r has odd period length, then for some u, w P N, r u w with numerator Proof Suppose that we hae a continued fraction with odd period length n and numerator u Then there must also exist a purely periodic continued fraction with the same period length and numerator Since is an integer, and we know that integers can be decomposed into a square part and a square-free part, we can write w 1 where w is an integer and 1 is a square-free integer Substitute this into () and we know there exists an integer solution to puq n ` w 1 p n 1 q ` w 1 u n s () Now suppose that there is a prime number p such that p ą and p 1 Since 1 is square-free, we know p 1 If p w, since w is a perfect square, then p m w for some m Next, following from Corollary i, since w 1 q n, then p p m p m`1 q n With these facts, we know p m`1 puq n ` w 1 p n 1 q, and since gcdpw 1, uq 1, we know p m`1 w 1 u n, which ultimately gies the result that p m`1 puq n `w 1 p n 1 q `w 1 u n Howeer, s is a perfect square so the prime decomposition under the fundamental theorem of arithmetic must hae all primes to an een power We hae shown p m`1 s, a contradiction Now instead suppose that 1, and consider two cases First, if m w for some m ą 0, then pm`1q m` puq n ` w 1 p n 1 q and m m` w 1 u n This implies that m` s Since s is a perfect square, we find a contradiction On the other hand, if gcdpw, q 1, then puq n ` w 1 p n 1 q and 8 w 1 u n In that case, () can be diided by as ˆuqn ` w 1 p n 1 ` w u n s Well-known properties of squares tell us that both sides of the aboe must be 0 pmod q or 1 pmod q This is not the case, howeer, as w u n pmod q Thus, 1 is neither nor diisible by an odd prime, so 1 1 and w 1 w 1

17 As preiously shown, Theorem can be used to slightly alter periodic behaior To further the possibilities of altering expansions, we set out to determine reersals of purely periodic continued fractions First, we need to proe a useful fact about conergents that we will make use of Lemma If x ra 0, a 1,, a n 1, a n, a n`1, s u{ and x 1 ra n, a n 1,, a 1, a 0, b 1, b, s u{, then for p k q k, the k conergent of x, and p1 k, the k conergent of x 1, q 1 k p n p 1 n, () q n 1 qn 1, 1 () # p 1 q n n 1, if n is een, p 1 (6) n 1 if n is odd, # qn 1 pn 1, if n is een, () if n is odd p n 1 Proof For conenience and conciseness, it is possible to think of p n and q n as polynomials in the ariables u,, a 0, a 1, For example, let q pa 0, a 1, a q a 1 a ` u, then q 1 q pa, a 1, a 0 q This way, we aoid the prime notation and write p n pa 0, a 1,, a n q p n pa n, a n 1,, a 0 q, q n 1 pa 0, a 1,, a n q q n 1 pa n, a n 1,, a 0 q, # pn 1 pa n, a n 1,, a 0 q, if n is een, q n pa 0, a 1,, a n q p n 1 pa n, a n 1,, a 0 q if n is odd, # pn 1 pa 0, a 1,, a n q, if n is een, q n pa n, a n 1,, a 0 q p n 1 pa 0, a 1,, a n q if n is odd There are seeral important details to notice about p n and q n First, one can see from the formulas that p n depends only on the first n`1 arguments, that is, p n pa 0, a 1,, a n, a n`1, q p n pa 0, a 1,, a n q Similarly, q n does not depend on the first argument nor arguments past n ` 1, that is, q n 1 pa 0, a 1,, a n, a n`1, q q n 1 pa 1,, a n 1 q (one must take care with the fact that this cannot be applied repeatedly, that is, if q n 1 pa 0, a 1,, a n, a n`1, q q n 1 pa 1,, a n 1 q, it is not necessarily true that q n 1 pa 1,, a n 1 q q n 1 pa,, a n 1 q) By using these facts, we can reconsider the aboe formulas and aim to proe p n pa 0, a 1,, a n q p n pa n, a n 1,, a 0 q, (8) q n 1 pa 0, a 1,, a n 1 q q n 1 pa n, a n 1,, a 1 q, () # pn 1 pa n, a n 1,, a 1 q, if n is een, q n pa 0, a 1,, a n q (10) p n 1 pa n, a n 1,, a 1 q if n is odd, # pn 1 pa 0, a 1,, a n 1 q, if n is een, q n pa n, a n 1,, a 1 q () p n 1 pa 0, a 1,, a n 1 q if n is odd 1

18 Again using the same facts introduced aboe, we find seeral additional formulas that will be of use, # pn 1 pa 1,, a n q, if n is een, q n pa 0, a 1,, a n q p n 1 pa 1,, a n q if n is odd, $ & u p n pa 0, a 1,, a n q a 0 q n pa 0, a 1,, a n q ` q n 1pa 1,, a n q, % uq n 1 pa 1,, a n q if n is een, if n is odd (1) () When iewed as four formulas, (1) and () can be proed by induction To start, see that q 0 pa 0 q 1 p 1 pa 0 q and q 1 pa 0, a 1 q a 1 p 0 pa 1 q as base cases For een n, assume that q n 1 pa 0, a 1,, a n 1 q p n pa 1,, a n 1 q and q n pa 0, a 1,, a n q p n 1 pa 1,, a n q, then q n`1 q n`1 pa 0, a 1,, a n, a n`1 q a n`1 q n pa 0, a 1,, a n, a n`1 q ` uq n 1 pa 0, a 1,, a n, a n`1 q a n`1 q n pa 0, a 1,, a n q ` uq n 1 pa 0, a 1,, a n 1 q a n`1 p n 1 pa 1,, a n q ` up n pa 1,, a n 1 q p n pa 1,, a n, a n`1 q, as desired for (1) The case for odd n follows almost the same steps After showing base cases and assuming induction hypotheses, () can be proed in a similar manner Now we can proe (8) through (), also using induction The bases are triial, as we hae shown in similar situations Assume that each equation holds up to n Starting with the een n case of (8), we hae p n`1 pa 0, a 1,, a n, a n`1 q a n`1 p n pa 0,, a n q ` up n 1 pa 0,, a n 1 q a n`1 a 0 q n pa 0,, a n q ` ua n`1 q n 1 pa 1,, a n 1 q ` ua 0 q n 1 pa 0,, a n 1 q ` u q n pa 1,, a n 1 q a n`1 a 0 q n pa n`1,, a 1 q ` ua n`1 q n 1 pa n,, a q ` ua 0 q n 1 pa n,, a 1 q ` u q n pa n 1,, a q a n`1 q n`1 pa n`1,, a 1, a 0 q ` uq n pa n,, a 0 q p n`1 pa n`1, a n,, a, a 1, a 0 q, as desired The case for odd n of (8) follows in a similar manner The proofs for (), (10), and () are omitted for breity, but use similar strategies By proing those equations, the relations that we originally introduced hold as well Theorem If x y ra 0, a 1,, a k 1 s u{ and x1 y 1 where p k q k and p k 1 q k 1 x 1 ra k 1,, a 1, a 0 s u{, then xq k 1, y 1 yp k if k is een (1) x 1 xq k 1, y 1 yp k if k is odd (1) are the k and k 1 conergents of x y, respectiely

19 Proof Assume that we hae x y ra 0, a 1,, a k 1 s u{ a 0, a 1,, a k 1, x j y u{ where k is een We can apply the x n ersion of () to get x y p x k 1 ` up y k x q k 1 ` uq y k Through multiplication, p k 1 x y ` up k q k 1 x y ` uq k q k 1 p k q k 1 p k p k 1 xq k 1 yp k ` uq k 1 xq q k 1 k 1 yp k ` uq k 1q k p k xq k 1 yp k x y 1 1 q k 1 1 p k 1 p k xq k 1 yp k ` uq k 1 xq q k 1 k 1 yp k ` uq k 1q k p k 1 1 q k 1 xq k 1 yp k ` uq k 1 1 p k 1 xq p k p k 1 k yp k ` uq k Suppose that x1 y 1 ra k 1,, a 1, a 0 s u{ By the results of Lemma, xq k 1 yp k p1 Thus, x1 y 1 xq k 1 yp k ra k 1,, a 1, a 0 s u{ The proof for odd k is nearly identical xq k 1 k 1 qk 1 xq k 1 yp k ` up 1 k 1 yp k ` uqk 1 The utility of the preious theorem can be demonstrated through example Example We preiously discoered By Theorem, 8,, 1,, { q 8p 8 86,, 1,, { Note that this does not, in some cases, gie results through the same algorithm Here, for example,,, 1,, 8 with the max algorithm while,,, 1,, with 8 { 86 { the same algorithm With these results, we hae tools to find and confirm periodic continued fraction expansions Howeer, a question still remains about how many different purely periodic expansions can be made for a particular numerator Theorem If there exists x 0 y 0 ra 0, a 1,, a k 1 s u{

20 for some numerator u and integer k ě, then for n, the number of purely periodic expansions with period length k, k n Furthermore, if a i a k 1 i for some 0 ď i ď k 1 (that is, the period of x 0 y 0 is not symmetric), then k n In particular, for 0 ď l ă k pmod kq, there are k expansions of the form and there are k expansions of the form Proof Start with an expansion of x 0 y 0 By utilizing Theorem 8 and Theorem, x l y l ra l,, a k 1, a 0,, a l 1 s u{ x l y l ra l,, a 0, a k 1,, a l`1 s u{ with numerator u, and we will construct more Suppose x 0 y 0 ra 0, a 1,, a k 1 s u{ uy px a 0 yq x 1 y 1 ra 1, a,, a k 1, a 0 s u{ Continuing with the same method, we are able to construct expansions that are purely periodic that use each a i as a leading term This makes k expansions of this form with period length k Next, the application of Theorem to x 0 y 0 gies x 1 0 y 1 0 ra k 1,, a 1, a 0 s u{ Then, similarly to aboe, we can repeatedly apply Theorem 8 and Theorem to find k expansions that are purely periodic with period length k, which use each a i as leading term, but in reerse order Period Length 1 With tools at hand to examine periodic continued fractions, we next gie special attention to period length 1, where we can extend ideas that applied to any period length By using (), the numerators that can produce expansions with periodic length 1 are classified by soling s puq 1 ` p 0 q ` u a 0 ` u () Example For numerator, see that ` 6 This shows the existence of a continued fraction which has period length 1, specifically with a 0 Repeatedly applying examples to () shows a pattern, which brings us to the following theorem 18

21 Theorem 6 There exists rational x ra 0 s u{, a purely periodic continued fraction with period length 1, if and only if the following conditions hold 1 w for some w P N, w ă u There exists m 1 ą m ą 0 such that u n pm 1 m q and w k m 1 n k m for some 0 ď k ď n Then, a 0 pk m 1 n k m q w and x a 0 ` k m 1` n k m w k m 1 w Proof Suppose that we hae a continued fraction of the form x ra 0 s u{ Following directly from Theorem, w for some w P N, so we rewrite () as w a 0 ` w u w pw a 0 ` uq ps 1 q where s 1 is an integer Let s s1 w in order to eliminate the w factor, and thus there must be an integer solution to This can be rearranged as and factored as w a 0 ` u s s w a 0 u, ps wa 0 qps ` wa 0 q u () Suppose u n m 1 m for some n, m 1, m P N where m 1 ą m, m 1, and m Now we hae ps wa 0 qps ` wa 0 q n` m 1 m Let x s wa 0 and y s ` wa 0, then xy n` m 1 m and y x wa 0 Since y x, it must be that x and y Therefore, we may suppose that y k`1 m 1 and x n k`1 m for some 0 ď k ď n So, and after diiding by we find wa 0 y x k`1 m 1 n k`1 m, wa 0 k m 1 n k m Since a 0 is an integer, w k m 1 n k m and a 0 k m 1 n k m w Finally, we will sole for x as x ra 0 s u{w ùñ x a 0 ` u{w x ùñ x a 0 x u{w 0

22 The the quadratic formula gies Substitute a 0 and u to find x x k m 1 n k m w k m 1 n k m w a 0 b a 0 ` u w k m 1 n k m k m 1` n k m w w b p k m 1 n k m q w ` n` m 1 m w b p k m 1` n k m q w The subtractie case gies a negatie number, so we shall only consider the additie case Therefore, x k m 1 ` k m 1 w k m 1 w The conerse can be proed by construction For numerator u, suppose that w for some w P N, w ă u Also suppose that there exists m 1 ą m ą 0 such that u n pm 1 m q and w k m 1 n k m for some 0 ď k ď n Let a 0 k m 1 n k m and x x w 0 k m 1 For w the next step, x 1 u w x 0 a 0 n m 1 m w k m 1 k m 1 n k m w w n m 1 m w n k m w k m 1 w x 0 Thus, we can construct x ra 0 s u{, a purely periodic continued fraction with period length 1 Recall the discussion of numerator, and apply our new result Example Let r By Theorem 6, a 0 p 1q and x ` p`1q Indeed, we confirm with direct computation that To go further, use Theorem 8 to find r examples such as and so on 1 0, r,, r, 10 1,,,, r, Other Results and Conjectures For continued fractions with integer numerators, certain facts can be shown such as the existence of arbitrarily long terminating expansions and infinitely many distinct periodic 0

23 expansions [1, Theorem 1 on p6] This relies on seeral expansion techniques [1, Lemma 110 on p6], such as for N ě and k ě 0, ı N pn 1q k, N and ı N pn 1q 8 The hope is to make an analogue for expansions with rational numerators Howeer, these proe to be slightly more complicated, especially considering the ariety of algorithms that could be used Seeral specific cases hae been discoered, and left as propositions Proposition 1 For m, n, u, P N and the max algorithm, (i) u mn`n rmn, m ` 1s u{, where ą n (ii) u m 1 r1,, m, m 1s m u{, where m ą (iii) u m r1,, m, m 1s m u{, where m ą (i) u 6m 1 r,, m, 1m s m u{, where m ą 1 Proof The construction of piq follows N N a 0 t mn ` n u tmn ` n u mn x 1 pmn ` nq{ n{ m ` 1 The construction of piiq follows a 0 t m 1 m u t 1 m u 1 a 1 t m 1 m 1 u a t m m ` 1 u tm ` 1 m m u m The construction of piiiq follows a 0 t m m u t m u 1 a 1 t m m u a t m 6m ` u tm ` m m u m x 1 x x x 1 x x pm 1q{m pm 1q{m m 1 m 1 pm 1q{m m m ` 1 1{pm 1q m pm 1q{m m 1 1{m pm q{m pm q{m m m pm q{m m 6m ` {pm q m pm q{m m 1 {pmq 1

24 The construction of piq follows a 0 t 6m 1 m u t 1 m u a 1 t 6m 1 m 1 u a t 1m 8m ` 1 u tm ` 1 m m u m x 1 x x p6m 1q{pmq pm 1q{pmq 6m 1 m 1 p6m 1q{pmq 1m 8m ` 1 {pm 1q m p6m 1q{pmq 1m 1{pmq Proposition For any u, P N and the min algorithm, u r0, 1s u{ Proof Select a 0 t u u t u u 0 Then, x 1 u{ u{ 1 For conjectures about the behaior of continued fraction expansions with rational numerators, there are a number of interesting properties that show up that could not be proed at the time of this writing In particular, these conjectures were made while collecting data oer a large sample of expansions, and are mentioned in Appendix A1 Following that, more conjectures could be made about the data in Appendix A Future Work From the onset, there were seeral inquiries regarding continued fractions, and more hae spawned along the way One of the desires for this work was to find a method for proing that any expansion of a rational number terminates for a certain numerator (such as ) This is one of the most basic properties of simple continued fractions and appears to carry oer to some cases here See A1 for lists of numerators that appear to hae this property with the max and min algorithms We hae discoered some that work for both algorithms, which includes,,,, 10, and 1 The most that can be said up to this point is that none of these will produce continued fraction expansions that are periodic with odd period length Future work could perhaps show that eery rational number will indeed terminate with these numerators This could inole a method of showing a bound on the algorithm remainders or showing a strictly decreasing sequence Furthermore, it is within the realm of possibility that the list of numerators that appear to terminate all rational numbers could be added to by computing further That is, oer the set of rational numbers, calculate the expansions to a greater length than is done here Perhaps this could show that some expansions with unknown behaior actually do terminate

25 Another possibility for future work is to look into other algorithms for computing continued fraction expansions The main focus in this work was to apply either the max or min algorithm, yet there are numerous other possibilities For example, one could compute continued fractions by alternating each step between the min and max algorithms This could produce patterns that were not seen here There is also the possibility that eery rational expansion could terminate if the correct algorithm is chosen One could intelligently choose the max, min, or some other choice at each step Howeer, this is all speculation and requires inestigation To conclude, there is certainly further work to be done on the subject of continued fractions A number of interesting patterns and properties hae been discoered for rational numerators, but questions remain These discoeries could hae impact on other fields of mathematics, and that is an intriguing prospect for the future

26 References [1] Anselm, M and Weintraub, H A Generalization of Continued Fractions Journal of Number Theory 1 (0): -60 [] Olds, C D Continued Fractions New York: Random House, 6

27 A Data Collection This section is included to look at some of the data and trends that hae been found, with the hope of inspiring future work A1 Terminating Rationals As discussed, eery simple continued fraction expansion of a rational number terminates We hae shown that this is not the case for expansions with rational numerators Howeer, there appears to be some numerators for which all rational expansions terminate, based on haing large samples that do just that Here we list all numerators u (u, ă 100) for which all expansions of x (x, y ă 00) terminate in less than 100 steps with the y (i) max algorithm:,,,, 8, 8,, 10, 10,,, 1, 1, 1, 1, 1, 1,,, 18, 18, 18,, 0, 0, 0 1,,,,,,,, 6, 6,,, 8,,, 0, 0,,,,,,,,, 6, 6, 6, 8, 8,,,, 0, 0, 0,,,,,,,,, 8, 8, 8, 8, 8, 8, 0, 1,,,,,,,,,,,,, 6, 6,, 8, 60, 60, 60, 60, 60, 60, 6, 1 6, 6, 6, 6, 6, 6, 6, 66, 66, 66, 6, 68, 68, 68, 6, 6, 0, 0,,,,,,,, 6, 6,,, 8, 8, 80, 80, 80, 8, 8, 8, 8, 8, 8, 8, 86, 86, 8, 8, 88, 88, 0, 0, 0, 0, 0,,,,,,, 6, 6, 6, 6, 8,, 1, (ii) min algorithm:,,,, 10, 1 Next, we examine seeral particular numerators listed aboe to see patterns in the lengths of the continued fraction expansions For numerators that appear to make all rational numbers terminate, there seems to be seeral patterns 1 The max and min algorithms produce different behaior in regard to frequency of expansion length Expansions of odd length are relatiely more common than een lengths The cures of odd and een lengths appear to differ by a factor of approximately (where u is the numerator that was used) The min algorithm appears to produce expansion lengths of larger mean and larger standard deiation Examples of these patterns are shown in the following figures

28 Frequency Length Figure 1: Terminating lengths of x y (x, y ă 1000) with numerator and the max algorithm Frequency Length Figure : Terminating lengths of x y (x, y ă 1000) with numerator and the min algorithm 6

29 Frequency Length Figure : Terminating lengths of x y (x, y ă 1000) with numerator and the max algorithm Frequency Length Figure : Terminating lengths of x y (x, y ă 1000) with numerator and the min algorithm

30 Frequency Length Figure : Terminating lengths of x y (x, y ă 1000) with numerator and the max algorithm A Periodic Rationals Unlike simple continued fractions, those with rational numerators may be periodic For numerators that can be used to create periodic expansions, there are two data sets of interest First is the frequency of lengths for those expansions that do terminate Second is the frequency of tail lengths for the expansions that become periodic Seeral examples of these are shown below 8

31 Frequency Length Figure 6: Terminating lengths of x y (x, y ă 1000) with numerator and the max algorithm Frequency Length Figure : Tail lengths of x y (x, y ă 1000) with numerator and the max algorithm

32 Frequency Length Figure 8: Terminating lengths of x y (x, y ă 1000) with numerator 6 and the max algorithm Frequency Length Figure : Tail lengths of x y (x, y ă 1000) with numerator 6 and the max algorithm 0

33 A Continued Fraction Rational Numerators Below is the complete dataset that was collected for a sample of continued fraction numerators The collected data includes, for each numerator, the portion of expansions that terminate, the portion of expansions that are eentually periodic, the period lengths that were discoered, and the portion of expansions that neither terminated nor became periodic Table : Numerators u (u, ă ) and the corresponding percentage of expansions that are terminating, periodic, or unknown oer a sample of all x (x, y ă 00, 1181 y total) in less than 00 steps with the max algorithm u Terminating % Periodic % Period Lengths Unknown % , 1, , 0 0, ,,, ,, ,, , , 6, ,, Continued on next page 1

34 Table Continued from preious page u Terminating % Periodic % Period Lengths Unknown % 0 6,, 6, 8,, , , , 6, 8,, ,, 6, 8, 1, 1, 18, 0,,,, 8,, ,, , , 6, 1,, , 8,,, 0, ,,, ,, 6, 10, 1, , 8, 10, 1, , , , 10, Continued on next page

35 Table Continued from preious page u Terminating % Periodic % Period Lengths Unknown % , , 8, 10,, ,,, 6, ,, 6, 10, 1,, 18, , 8,, 18, 0, ,, 6, , ,, 6, 10, 1, 1, ,, ,, 8, ,, 6, 8, 1,, , 1, ,,, 6, 8, 10, 1,,,,,, , 8, 10, 0 Continued on next page

36 Table Continued from preious page u Terminating % Periodic % Period Lengths Unknown % ,, , 8, ,, 6, 8, 10, , ,, , , , ,,,,, 6, , , 6, 10, 1, 1, 6,, , ,, , , 6 1 6, 8, 1, , Continued on next page

37 Table Continued from preious page u Terminating % Periodic % Period Lengths Unknown % 1 88, 10, , , , 8, , , 8, 1, , , 10,, ,, 6, 1, , 1, , 1, 18,, , 1 1, , 6, 8, , ,

38 Table : Numerators u (u, ă ) and the corresponding percentage of expansions that are terminating, periodic, or unknown oer a sample of all x (x, y ă 00, 1181 y total) in less than 00 steps with the min algorithm u Terminating % Periodic % Period Lengths Unknown % , 1,, 8, , 0 1 0, 6, 1, 18, 8, 0, 66, 6 8, ,, 6, 10,, 18, , , 1,, , 10, 80, , 1,, , 1, 18, 0, , 6, 10, ,, , 6, 8, , 0, ,, 6, 10, 8 Continued on next page 6

39 Table Continued from preious page u Terminating % Periodic % Period Lengths Unknown % , , 6, 1, 18,,, , 6, 8, 10, 1,, , , , ,, 6, 10, 1, , , ,, 6, 8, ,, , 10, , , 1, 1, 0, ,, , , 8, 10, 1,,, , 010 Continued on next page

40 Table Continued from preious page u Terminating % Periodic % Period Lengths Unknown % 1 06, , 1,,, , , 10, ,,, 6, 8, 10,, , , 1,, 8, , 6, 8,, 0, , 8, 10, , , , 6, 8, , , 1, , 10 Continued on next page 8

41 Table Continued from preious page u Terminating % Periodic % Period Lengths Unknown % ,, ,, 8,, , , 6, 1, 0, , , , 1, , , 18, , 10, , , , Continued on next page

42 Table Continued from preious page u Terminating % Periodic % Period Lengths Unknown % , , 10, , B Mathematica Code The following code can be pasted as seen and executed in a Mathematica notebook There are three functions which encompass the computations needed for this work, each of which are described in more detail using code comments CFRational - Create a single continued fraction expansion MultiXSingleR - Collect data about a single numerator oer a sample of continued fraction expansions MultiR - Collect data oer a sample of numerators, each oer a sample of continued fraction expansions (* CFRational Computes a continued fraction for a rational number x/y \ with a numerator u/ Arguments: 0