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Two Problems on Cantor Set Arithmetic

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Two Problems on Cantor Set Arithmetic Two Problems on Cantor Set Arithmetic Lauren Chen under the direction of Yuqiu Fu Department of Mathematics Massachusetts Institute of Technology Research Science Institute July 30, 2019 Abstract We find the cardinalities of the solution sets to the polynomial equations c = a + b and c = a − b on variants of the Cantor set. We also compute examples for the equation c = ab. A previous theorem states f(C × C) = [0; 1] for the Cantor set C where f(x; y) = x2y. Our log 2 second problem generalizes this to f = xαy for α in the range ≤ α ≤ 2. We also log 3=2 explore the case when α is greater than 2. We consider the expansion of f(Cn × Cn) for a few small n, where Cn is the nth iteration of the Cantor set, to find intervals of α > 2 such that f(C × C) does not cover the entire interval [0; 1]. Summary The numbers we typically use are in base 10, meaning they contain the digits 0 through 9. Similarly, numbers in base 3 only contain the digits 0, 1, and 2. The Cantor set is the set of all numbers from [0; 1] that can be written in base 3 without 1. We study two problems. In the first problem, we consider three polynomial equations in base 3 where all numbers involved do not include one of the digits 0, 1, or 2. The second problem generalizes a theorem about the equation u = x2y in the Cantor set to u = xαy. 1 Introduction P1 −k The Cantor set C is defined as all numbers that can be expressed in the form k=1 ak3 ; ak 2 f0; 2g. In other words, it is the set of points in the interval [0; 1] that can be written in ternary without 1. It is produced by repeatedly removing the middle third of each of the remaining segments, starting with the interval [0; 1]. Denote these iterations Cn, as seen in Figure 1. Figure 1: Iterations of the Cantor set Cn. The Cantor set satisfies C = limn!1 Cn. Pj k We define the set C(A) as fa 2 Z≥0ja = k=0 ak3 for ak 2 A, j 2 Z≥0g, where A is a two-element subset of the set f0; 1; 2g. C(A) is an analog of the Cantor set. Numbers in both sets consist of only two distinct numbers from the set f0; 1; 2g in their ternary representation. Because the Cantor set has been more extensively studied, the connection between the sets reveals some potential directions of exploration for our own problem. N We define the set C(A)N to be fa 2 C(A): a < 3 g. We want to find the cardinality of the set SN := f(a; b) 2 C(A)N × C(A)N : 9 c 2 C(A) s.t. P (a; b) = cg for some polynomial P . Note that SN depends on the polynomial P . We will specify which polynomial SN refers to in each section. We consider variations of this problem, such as changing A to be a different two-element subset of f0; 1; 2g. Our first problem is the discrete version of the problem of finding the Minkowski dimension of the solution set [1]. Motivation for the problem thus arises from this connection. We also explore a second problem on the Cantor set itself. In 2017, Athreya, Reznick, and Tyson [2] proved that every element u of [0; 1] can be written in the form u = x2y, where 1 x; y are elements of the Cantor set C. We investigate the range of possible values for α in u = xαy for which the result still holds. In Sections 2 and 3, we find formulas for the cardinalities of the solution sets for P (a; b) = a + b and P (a; b) = a − b, respectively. In Section 4, we compute the cardinalities jSN j for some small N for P (a; b) = ab. In Section 5, we form a conjecture for the lower bound on α in u = xαy based on the proof by Athreya, Reznick, and Tyson for α = 2. Section 6 generalizes the result to a larger range for α < 2, and Section 7 explores what happens for α > 2. 2 jSN j for P (a; b) = a + b For our first problem, we find formulas for the cardinalities of the solution sets to polyno- mial equations on the Cantor set variant C(A). We first consider the case where P (a; b) = a+b and C(f0; 1g). Theorem 2.1. When SN := f(a; b) 2 C(A)N × C(A)N : 9 c 2 C(A) s.t. a + b = cg and Pj k N C(A) := fa 2 Z≥0 : a = k=0 ak3 for ak 2 f0; 1g, j 2 Z≥0g, jSN j = 3 . PN−1 i PN−1 i PN−1 i Proof. Let a = i=0 ai3 and b = i=0 bi3 with ai; bi 2 f0; 1g, so c = a + b = i=0 ci3 where ci 2 f0; 1; 2g. We want ci to also be 0; 1 for all i so c 2 C(A). For each i 2 f0; 1; :::; N − 1g, there are four cases: When ai = 0 and bi = 0, we have ai + bi = 0. When ai = 0 and bi = 1, we have ai + bi = 1. When ai = 1 and bi = 0, we have ai + bi = 1. When ai = 1 and bi = 1, we have ai + bi = 2. Thus, ai + bi does not exceed 2 for any i, so ci = ai + bi and each of the N sums is N independent from the others. Each sum can be any of the first three cases, so jSN j = 3 . We now consider the case where P (a; b) = a+b and A is the set f1; 2g, following a similar method of proof. 2 Theorem 2.2. When SN := f(a; b) 2 C(A)N × C(A)N : 9 c 2 C(A) s.t. a + b = cg and 3N + 1 C(A) := fa 2 : a = Pj a 3k for a 2 f1; 2g, j 2 g, jS j = . Z≥0 k=0 k k Z≥0 N 2 PN−1 i PN−1 i Proof. As before, we let a = i=0 ai3 and b = i=0 bi3 with ai; bi 2 f1; 2g, so c = a+b = PN−1 i i=0 ci3 where ci 2 f0; 1; 2g. We find a recursive formula for jSN j. There are four cases: When ai = 1 and bi = 1, we have ai + bi = 2. When ai = 1 and bi = 2, we have ai + bi = 3. When ai = 2 and bi = 1, we have ai + bi = 3. When ai = 2 and bi = 2, we have ai + bi = 4. When x0 = y0 = 1, there are jSN−1j such ordered pairs. The second and third cases are not possible when i = 0 because ci would then equal 0, which is not part of our solution set. When the last case holds for i = 0, the sum a0 + b0 causes a regroup and increases z1 by 1. If this happens, a1 and b1 have to be one of the last three cases. Then by the same reasoning, N−1 all following ai and bi must be one of the last three cases. Thus, there are 3 ordered pairs N−1 with i = 0 as the last case. Thus, jSN j = 3 + jSN−1j with S0 = 1. We can rewrite the 3N + 1 recursive formula in closed form as jS j = . N 2 3 jSN j for P (a; b) = a − b For P (a; b) = a − b, we allow negative integers in C(A). By symmetry, jSN j is equal for a > b and a < b. Thus, it suffices to consider when a > b. Define Sm;n to be the solution set for N N S S S a and b with exactly m and n digits, respectively. Note that SN = Sm;n = Sm;n. m;n≤N m=1 n=1 Theorem 3.1. When S := f(a; b) 2 C(A)N × C(A)N : 9 c 2 C(A) s.t. a − b = cg and P k n−1 m−n C(A) := fa 2 Z≥0 : a = k=0 ak3 for ak 2 f0; 1g, j 2 Z≥0g, jSm;nj = 3 2 . PN−1 i PN−1 i Proof. As before, we let a = i=0 ai3 and b = i=0 bi3 with ai; bi 2 f0; 1g, so c = a−b = PN−1 i i=0 ci3 where ci 2 f0; 1; 2g. We first consider the four cases: 3 When ai = 1 and bi = 1, we have ai − bi = 0. When ai = 0 and bi = 1, we have ai − bi = −1. When ai = 1 and bi = 0, we have ai − bi = 1. When ai = 0 and bi = 0, we have ai − bi = 0. If m = n, we assume am = bn = 1. Then i = 0; 1; :::; m − 1 has to be case 3 or 4, so there are 2m possibilities. Now let m > n. In case 2, the −1 is equivalent to a ci = 2 and subtracting 1 from ci+1. This cannot happen because we cannot have any ci = 2. Thus, i = 0; 1; :::; n − 2 can be cases 1, 3, or 4, so there are 3n−1 ways to assign a case to i = 0; 1; :::; n − 2. We can assume that bn is 1, so i = n − 1 has to be case 1.
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