Graduate Theses, Dissertations, and Problem Reports 2008 Pierce-Engel hybrid expansions Andrea Sutyak West Virginia University Follow this and additional works at: https://researchrepository.wvu.edu/etd Recommended Citation Sutyak, Andrea, "Pierce-Engel hybrid expansions" (2008). Graduate Theses, Dissertations, and Problem Reports. 2718. https://researchrepository.wvu.edu/etd/2718 This Dissertation is protected by copyright and/or related rights. It has been brought to you by the The Research Repository @ WVU with permission from the rights-holder(s). You are free to use this Dissertation in any way that is permitted by the copyright and related rights legislation that applies to your use. For other uses you must obtain permission from the rights-holder(s) directly, unless additional rights are indicated by a Creative Commons license in the record and/ or on the work itself. This Dissertation has been accepted for inclusion in WVU Graduate Theses, Dissertations, and Problem Reports collection by an authorized administrator of The Research Repository @ WVU. For more information, please contact [email protected]. Pierce-Engel Hybrid Expansions Andrea Sutyak Dissertation submitted to the Eberly College of Arts and Sciences at West Virginia University in partial fulfillment of the requirements for the degree of Doctor of Philosophy in Mathematics Michael E. Mays, PhD., chair J. Goldwasser, PhD. H.W. Gould, M.A. K. Subramani, PhD. J. Wojciechowski, PhD. Department of Mathematics Morgantown, West Virginia 2008 Keywords: Pierce expansion, Engel expansion Copyright 2008 Andrea Sutyak ABSTRACT Pierce-Engel Hybrid Expansions Andrea Sutyak Pierce and Engel expansions are representations of numbers between 0 and 1 as sums of unitary fractions (of alternating signs in the case of Pierce) whose denominators are built multiplicatively, choosing the successive factors greedily. We show some results for Pierce expansions, and investigate the idea of hybrid expansions, which are built similarly but without regard to the signs of the terms. Dedication Trust in the Lord with all your heart and lean not on your own understanding: in all your ways acknowledge Him, and He will make your paths straight. Proverbs 2: 5-6 First and foremost, I thank God for not only opening paths for me, but giving me the strength to walk them. Many thanks to Dr. Mays, whose time and insights have proven immensely valuable. I would also like to thank the other members of my committee for their comments and support. I would be remiss not to extend my gratitude to Professor Michael Berry: PB - YFAGHMMTMTYWEK. Finally, my eternal thanks to my husband Mark and to my family, who have lovingly and patiently supported me through this endeavor. iii Contents 1 Pierce and Engel Expansions 1 1.1 Introduction.................................... 1 1.2 NotableResults .................................. 3 2 Hybrid Expansions 15 2.1 Introduction.................................... 15 2.2 NotableResults .................................. 21 2.3 InfiniteExpansions ............................... 30 3 Appendix A: Maple Code 39 4 Appendix B: Values of H(b, a) 44 iv 1 Pierce and Engel Expansions 1.1 Introduction The Pierce expansion of a number 0 < x ≤ 1 is the unique way of writing 1 1 1 x = − + −··· , q1 q1q2 q1q2q3 where the sequence qn is a strictly increasing sequence of positive integers. Similarly, the Engel expansion is the unique representation 1 1 1 x = + + + ··· , q1 q1q2 q1q2q3 where qn is an increasing sequence of positive integers. a The terms of the sequences qn for the respective expansions of a rational number b can be found using variations on the Euclidean algorithm, or division algorithm. Without loss of generality, hereafter we will assume that a ≤ b. The well-known algorithm is as follows: THE EUCLIDEAN ALGORITHM: Given a pair of positive integers, b, a, there exist a unique quotient q and a unique remainder r with 0 ≤ r < a such that b = aq + r. Iterating the division algorithm, we get a sequence of expressions b = aq1 + r1 a = r1q2 + r2 1 r1 = r2q3 + r3 . rn−2 = rn−1qn +0, where each ri satisfies 0 ≤ ri < ri−1 < a. (The greatest common divisor of (b, a) arises as the last non-zero remainder, namely rn−1). The quotients produced by this algorithm also a generate the continued fraction expansion of b . By changing the way that we iterate this algorithm, we find a tool for devising the a quotients necessary for the Pierce expansion of b , iterating for each pair (b, ri) as follows: THE PIERCE ALGORITHM: b = aq1 + r1 b = r1q2 + r2 b = r2q3 + r3 . b = rn−1qn +0, where each ri satisfies 0 ≤ ri <ri−1 < a. [3] a The sequence qi produces the unique Pierce expansion for the quotient b , given a pair of a positive integers b, a. This sequence terminates if and only if b is rational. [4] The algorithm for finding the Engel expansion of a rational number is also closely related to the division algorithm, with the exception that the quotients are chosen so that the remainder is negative (with magnitude less than that of a.) THE ENGEL ALGORITHM: 2 b = aq1 − r1 b = r1q2 − r2 b = r2q3 − r3 . b = rn−1qn − 0, where each ri satisfies 0 ≤ ri <ri−1 < a. It should be immediately clear that for either of these algorithms, any two equivalent a ka fractions b and kb will yield precisely the same sequence of quotients qi, and thus have the a same expansion. As a consequence, we may assume that b is in lowest terms. 1.2 Notable Results Pierce expansions, while not unexplored, still hold a certain mystery. Perhaps the most intriguing question revolves around predicting the length of the expansion. As in the work of Erd¨os and Shallit [1], we define a function P (b, a) such that for each pair b, a, P (b, a) is equal to the number of terms in the Pierce expansion (or, equivalently, the number of steps for the Pierce algorithm to terminate). As an example, consider b =7, a = 4: 7 = 4(1)+3 7 = 3(2)+1 7 =1(7)+0. 3 Since the algorithm terminates after three steps, we say P (7, 4) = 3. The expansion also 4 1 1 1 1 1 1 has three terms: 7 = 1 − 1·2 + 1·2·7 = 1 − 2 + 14 . Below is a short table of values for P (b, a) for 1 ≤ a ≤ b ≤ 25. b|a 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 1 1 2 1 1 3 121 4 1121 5 12321 6 111221 7 1223321 8 11213221 9 121232321 10 1122133221 11 1234235432 1 12 1111213222 2 1 13 1222323433 3 2 1 14 1122321343 3 2 2 1 15 1212122332 3 2 3 2 1 161121222133 3 2 3 2 2 1 171232344235 5 4 3 4 3 2 1 181112213213 4 2 3 3 2 2 2 1 191223425323 4 6 3 5 4 3 3 2 1 201121123221 3 3 4 3 2 2 3 2 2 1 211212221322 3 3 4 2 3 3 3 2 3 2 1 221122232432 1 3 4 5 3 4 3 3 3 2 2 1 231234453454 2 3 5 6 5 4 6 55 4 3 2 1 24111121212221333232322221 251222123242323435343233321 Some natural first questions are to wonder if it is possible to use known values of P (b, a) to predict the value for certain other pairs, to consider the properties of the sequence P (b, a), and so forth. The following observations are relevant in responding to such inquiries. Theorem 1.1. For fixed b > 0, if maxcP (b, c) = n, then for each j ≤ n, ∃a such that P (b, a)= j. Proof. Since maxcP (b, c)= n, ∃d such that P (b, d)= n. Then applying the Pierce algorithm, 4 we have b = dq1 + r1 b = r1q2 + r2 b = r2q3 + r3 . b = rn−1qn +0, from which it is clear that P (b, r1)= n − 1, P (b, r2)= n − 2, . P (b, rn−1)=1. b Theorem 1.2. If P (b, a) = n where a < b − a (or equivalently, a < 2 ), then P (b, b − a) = n +1. Proof. Apply the Pierce algorithm to the pair (b, b − a): b b =(b − a) · ⌊ ⌋ + a. b − a 5 b (Note that ⌊ b−a ⌋ = 1.) This is merely the first step. To compute P (b, b − a), the problem falls to finding P (b, a). Thus P (b, b − a)= n + 1. Recalling that equivalent fractions share a Pierce expansion, the two theorems below consider some simple congruence classes. Theorem 1.3. For b> 3, b (i) P (b, 2 )=1 if b is even. b (ii) P (b, ⌊ 2 ⌋)=2 if b is odd. b (iii) P (b, ⌈ 2 ⌉)=3 if b is odd. b b Proof. (i) If b is even, the result is obvious: b = 2 · 2+0, so P (b, 2 )=1. If b is odd, then b b = ⌊ ⌋· 2+1 2 b = 1(b)+0. and b b − 1 b = ⌈ ⌉· 1+ 2 2 b − 1 b = · 2+1 2 b = 1(b)+0. b b Theorem 1.4. If b ≡ 1 (mod k) and k ≥ 2, then P (b, ⌊ k ⌋)=2. Proof. b ≡ 1 (mod k) implies that b = km + 1 for some non-negative integer m. Then km+1 b b = ⌊ k ⌋· k +1, or b = 1(b) + 0 Thus P (b, ⌊ k ⌋)=2. 6 b b Theorem 1.5. If b ≡ 2 (mod k) and k ≥ 3, then P (b, ⌊ k ⌋)=2 if b is even, 3 if b is odd. Proof. b ≡ 2 (mod k) implies that b = km + 2 for some non-negative integer m.