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Kuzmin's Zero Measure Extraordinary Kuzmin’s Zero Measure Extraordinary Set Matthew Michelini Mathematics Department Princeton University April 9, 2004 My special thanks to my parents who have made my Princeton experience pos- sible. This thesis and more importantly graduation would not have been possible if not for the guidance, patience, and unfaltering faith of both Ramin Takloo- Bighash and Steven Miller. Because of Dr. Takloo-Bighash and Dr. Miller, I conquered the impossible. 1 Contents 1 Preliminaries 1 1.1 Notation . 1 1.2 Definitions . 2 1.3 Finding the Quotients of a Continued Fraction and Uniqueness . 2 1.4 Properties of Convergents . 4 1.5 Infinite Continued Fractions . 8 1.6 Advantages and Disadvantages of Continued Fractions . 12 2 Convergence and Approximation 14 2.1 Discussion and Bounds . 14 2.2 Convergents as Best Approximations . 15 2.3 Absolute Difference Approximation . 16 2.4 Liouville’s Theorem . 23 3 Kuzmin’s Theorem and Levy’s Improved Bound 25 3.1 The Gaussian Problem . 25 3.2 Intervals of Rank n . 26 3.2.1 Definition and Intuition . 26 3.2.2 Prob(an = k)........................ 29 3.3 Kuzmin’s Theorem . 31 3.3.1 Notation and Definitions . 31 3.3.2 Necessary Lemmas . 34 3.3.3 Proof of Main Result . 38 3.3.4 Kuzmin’s Result . 45 3.4 Levy’s Refined Results . 47 3.5 Experimental Results for Levy’s Constants . 48 3.5.1 Motivation . 48 3.5.2 Problems in Estimating A and λ . 50 2 3.5.3 Method . 52 3.5.4 Results . 57 4 Bounded Coefficients 61 4.1 Known Theory . 61 4.2 Motivation . 67 4.3 Results . 70 4.4 Possible Theoretical Explanations of the Results . 72 4.4.1 Kuzmin’s Measure Zero Set . 73 5 Conclusion 84 6 Appendix A 85 3 Abstract In this paper, I will reconstruct Khintchine’s presentation of Kuzmin’s Theo- rem but with vastly more details and explanations. I will then use this formulation to give a method of approximating the absolute positive constants A and λ in Levy’s error term: n 1 o ln 1 + k(k+2) A n −λ(n−1) µ(E ) − < e . k ln 2 k(k + 1) I conducted a numerical experiment to estimate these constants given certain con- ditions and will present the results in this paper. Finally, there exists some guiding theory to describe the zero measure set of α ∈ [0, 1] that does not obey Kuzmin’s Theorem, but the existing theory has never been fully summarized in a single exposition. I will provide this summary, and I will present two additional sets, for which theory suggests Kuzmin’s Theorem does not hold. Chapter 1 Preliminaries 1.1 Notation A continued fraction is the representation of a number α ∈ R and is of the form: 1 α = a + (1.1.1) 0 1 a + 1 1 a2 + . 1 .. + an If the continued fraction is infinite, then the expansion will not terminate with an like the expansion above does. Throughout this thesis we will also represent a finite continued fraction with [a0; a1, . , an] and an infinite continued fraction with [a0; a1,...]. Where appropriate, I will make the following notational distinctions: x = [a0; a1,...] is the value of the continued fraction of arbitrary length, but α is the actual number being represented by the continued fraction, or the number to which the continued fraction x converges. In other words, we attempt to rep- resent α by a rational number, x = [a0; a1,...]. For example, if α = π, then x = [3; 7, 15, 1, 292, 1,...] is the value of the continued fraction expansion of arbitrary length that ultimately converges to π. 1 1.2 Definitions Many of the following definitions are found in [MT]: Definition 1.2.1. Coefficients of a Continued Fraction: If x = [a0; a1,...], then the ai are the digits or coefficients. Definition 1.2.2. Positive Continued Fraction: A continued fraction [a0; a1,...] is positive if each ai > 0. Definition 1.2.3. Simple Continued Fraction: A continued fraction is simple if all ai are positive integers. Definition 1.2.4. Convergents of a Continued Fraction: Let x = [a0; a1,...], and if x = [a ; a , . , a ] = pn , then pn is the nth quotient or convergent. n 0 1 n qn qn Property 1.2.5. Let k ≥ 2, then the following is an increasing sequence for even values of k and a decreasing for odd values of k: p p + p p + 2p p + a p p k−2 , k−2 k−1 , k−2 k−1 ,..., k−2 k k−1 = k (1.2.2) qk−2 qk−2 + qk−1 qk−2 + 2qk−1 qk−2 + akqk−1 qk Definition 1.2.6. Intermediate Fractions: The fractions standing between pk−2 qk−2 and pk are called intermediate fractions. qk Let us define: sk = [a0; a1, . , ak], (1.2.3) or a section comprised of the first k coefficients of a continued fraction. Cor- respondingly, rk is the remainder of the continued fraction beginning with the coefficient ak: rk = [ak; ak+1,...], (1.2.4) where rk terminates at an if the continued fraction expansion is finite, or x = [a0; a1, . , an] = [a0; a1, . , ak−1, rk]. If rk exists, then rk ≥ 1 because ak ≥ 1 for all k. 1.3 Finding the Quotients of a Continued Fraction and Uniqueness Although there exist other methods to determine a continued fraction’s coef- ficient values, these algorithms build on the same idea behind the Lang-Trotter 2 method [LT]. However, the “old-fashioned” method utilizes the Euclidean algo- rithm and implementing this method yields x’s unique continued fraction expan- sion [MT]. Assume that x is represented by a simple continued fraction: 1 x = a + (1.3.5) 0 1 a + 1 1 a + 2 . .. To find a0, we take the greatest integer less than x (denoted by [x]) to be a0’s value. Hence, the remainder of the continued fraction expansion represents the value x − [x]: 1 x − [x] = , (1.3.6) 1 a + 1 1 a + 2 . .. where by taking the reciprocal, we determine a1’s value: 1 1 x = = a + 1 x − [x] 1 1 a + 2 1 a3 + ... ⇒ [x1] = a1. (1.3.7) 1 We repeat this process for x2 i.e. x2 = and all subsequent xi until our x1−[x1] expansion repeats, terminates, or a desired n is reached. Since we assumed that x’s continued fraction representation was simple, the only modification needed if x < 0 is to allow a0 < 0, then every other coefficient value is a natural number. Assuming that a continued fraction possesses at least m coefficients, it is also important to reiterate that ai ≥ 1 for 1 ≤ i ≤ m because 0 ≤ xi − [xi] < 1. If xi − [xi] = 0 then the remainder is 0 and the expansion terminates. From this traditional method of finding coefficients, we note a very important property: 3 Property 1.3.1. Let α be a rational number with a finite continued fraction ex- 1 pansion of length n, then α = [a0; a1, . , an] = [a0; a1, . , ak−1] + , where ai rk is a positive integer for all i ≤ n. For expositions on this see [MT]. We will now show that if x has an infinite continued fraction expansion, then the expansion is unique. 0 0 0 0 Theorem 1.3.2. Let x = [a0; a1, a2, a3,...] = [a0; a1, a2, a3,...] be two continued 0 fraction expansions for x, then ai = ai for all i. Thus, a continued fraction expansion is unique. Proof: This proof is found in [Ki]. Let the conditions of the theorem hold, 0 0 0 0 namely x = [a0; a1, a2, a3,...] = [a0; a1, a2, a3,...], where the expansions can be finite or infinite. Let i = 0, then since both expansions represent x, we have a0 = 0 0 0 [x] and a 0 = [x], which implies a0 = a0. Assume ai = ai for all i ≤ n. Then 0 0 by Theorem 1.4.2, we have pi = pi and qi = qi for all i ≤ n. From the definition of rk and Theorem 1.4.8, we have x = [a0; a1, . , an] = [a0; a1, . , ak−1, rk], which implies: 0 0 0 0 pnrn+1 + pn−1 pnrn+1 + pn−1 pnrn+1 + pn−1 x = = 0 0 0 = 0 ; (1.3.8) qnrn+1 + qn−1 qnrn+1 + qn−1 qnrn+1 + qn−1 0 0 0 therefore, rn+1 = rn+1. But we know that an+1 = [rn+1] and an+1 = [rn+1] so 0 an+1 = an+1, and the two expansions are identical.2 From this proof we conclude if α’s continued fraction expansion terminates with a coefficient an = 1, then two possible continued fraction representations 0 exist for α: one representation ends with an = 1 and the other ends with an−1 = an−1+1. Only in this case is it possible for a number to have two distinct continued fraction representations. 1.4 Properties of Convergents Property 1.4.1. Let [a0; a1, . , an] be a continued fraction, then [MT]: 1 1. [a0; a1, . , an] = [a0; a1, . , an−1 + ] an 2. [a0; a1, . , an] = [a0; a1, . , am−1, [am, . , an]]. (1.4.9) 4 The following theorem will be integral in answering certain questions involv- ing measure and continued fractions, especially with regard to a continued frac- tion’s tail or interval of uncertainty. Theorem 1.4.2. For any m ∈ {2, . , n} and am a positive integer, we have 1. p0 = a0, p1 = a0a1 + 1, and pm = ampm−1 + pm−2 2.
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