Hyperoperation: Introduction to the Theory and Potential Solutions

Hyperoperation: Introduction To The Theory And Potential Solutions Item Type text; Electronic Thesis Authors Dalthorp, Mark Publisher The University of Arizona. Rights Copyright © is held by the author. Digital access to this material is made possible by the University Libraries, University of Arizona. Further transmission, reproduction or presentation (such as public display or performance) of protected items is prohibited except with permission of the author. Download date 24/09/2021 15:26:55 Item License http://rightsstatements.org/vocab/InC/1.0/ Link to Item http://hdl.handle.net/10150/632817 HYPEROPERATION: INTRODUCTION TO THE THEORY AND POTENTIAL SOLUTIONS By MARK DALTHORP ____________________ A Thesis Submitted to The Honors College In Partial Fulfillment of the Bachelors degree With Honors in Mathematics THE UNIVERSITY OF ARIZONA M A Y 2019 Approved by: ____________________________ Dr. Douglas Pickrell Department of Mathematics Hyperoperation Mark Dalthorp May 1, 2019 Abstract We investigate the sequence of hyperoperations, which begins with addition, multiplication, and exponentiation. The nth hyperoperation is defined by iterating the previous operation. We address the problem of extending hyperoperations to non-integer values. We show existence of an analytic solution, and present several approaches to construction. 1 Introduction Our motivation is the observation that the arithmetic operations of addition, multiplication, and exponentiation can be thought of as a sequence, since multiplication and exponentiation are iterated forms of addition and multiplication, respectively: a · b = a + a + ::: + a (1) | {z } b times ab = a · a · ::: · a (2) | {z } b times One could easily imagine extending this and defining an operation to represent iterated exponentiation: :::a a ? b = aa (3) | {z } b times Since exponentiation is not commutative, there are two possible conventions for this. The standard way to think about (a:::a) b−1 this is right-nested parentheses a(a ), because the other direction is relatively easy to understand (((aa)a):::)a = a(a ). The iterative perspective on addition, multiplication, and our new operation ?, can be seen as a recursion. For example, observe that a · b = a + a + ::: + a = a + (a + a + ::: + a) = a + a · (b − 1). We have in general | {z } | {z } b times b−1 times a · b = a + (a · (b − 1)) (4) ab = a · (ab−1) (5) a ? b = aa?(b−1) (6) where each operation ∗ satisfies a ∗ 1 = a. It is fairly natural to continue this idea, defining a new operation to compound ?, and so on. We define a sequence of hyperoperation recursively for n > 1 by: a ∗1 b = a + b (7) a ∗n 1 = a (8) a ∗n b = a ∗n−1 (a ∗n (b − 1)) (9) This definition only applies to integer values of a and b. It is straightforward to see that ∗2 and ∗3 are multiplication and exponentiation, respectively. To see this definition in action, let us consider 2 ∗5 3 a simple example for n > 3. 2 ∗5 3 = 2 ∗4 (2 ∗5 2) = 2 ∗4 (2 ∗4 (2 ∗5 1)) (10) = 2 ∗4 (2 ∗4 2) = 2 ∗4 (2 ∗3 (2 ∗4 1)) (11) 2 = 2 ∗4 (2 ∗3 2) = 2 ∗4 (2 ) = 2 ∗4 4 (12) = 2 ∗3 (2 ∗4 3) = 2 ∗3 (2 ∗3 (2 ∗4 2)) (13) = 2 ∗3 (2 ∗3 (2 ∗3 (2 ∗4 1))) (14) = 2 ∗3 (2 ∗3 (2 ∗3 2)) (15) (22) 2 4 = 2 = 2(2 ) = 216 (16) = 65536 (17) 1 We know that multiplication and exponentiation have natural extensions to non-integer values. This means we can make sense of a ∗4 b for noninteger a (but not b). What about for non-integer b? Our goal in this paper is to find a sequence of hyperoperations satisfying the recursion of (9-7), but are real-analytic functions in both arguments for a; b > 1. A solution to this recursion is not unique, because given any 1-periodic function θ(x) such that θ(0) = 0, then the operation a∗^nb = a ∗n (b + θ(b)) will also satisfy the same recursion, because a∗^nb = a ∗n (b + θ(b)) = a ∗n−1 (a ∗n (b − 1 + θ(b))) = a ∗n−1 (a ∗n (b − 1 + θ(b − 1))) = a ∗n−1 (a∗^n(b − 1)) Thus, once we have constructed a solution, we actually have an infinite family of solutions. Ideally, our solution would also be monotonic in both variables. However, constructing a solution is actually more difficult than might be expected. In terms of construction, we will focus on ∗4, since this is the first difficult operation, and we can use the same techniques to construct the higher-order hyperoperations. Other authors have constructed analytic forms of a ∗4 b on limited domains, specifically for a 2 (1; e1=e) or in (e1=e; 1). Ideally, we would be able to construct a "universal" solution, i.e. one that is analytic on the whole region from 1 to 1. Though I do not yet have a solution, herein I offer several promising approaches. In this paper, we will briefly discuss the general theory of iteration of analytic functions, because this provides a framework to think about hyperoperation. Then, we will look at properties that will hold for any nice solution. Finally, we will examine the known non-universal constructions of hyperoperations, and include some of the approaches to finding a universal solution. 2 Theory of function iteration The most straightforward way to approach the problem of solving the hyperoperation recursion is by looking at functional iteration, since they are defined by iteration. 2.1 Preliminary Definitions Definition 1. For arbitrary f whose image is a subset of its range, we use the following notation for iterates of f f ◦n(x) = f(f(:::f(f (x)):::)) (18) | {z } n iterations f ◦−n(x) = f −1(f −1(:::f −1(f −1(x)):::)) (19) | {z } n iterations f ◦0(x) = x (20) ◦n+m ◦n ◦m ◦b Note that f = f ◦ f . Also notice, if f(b) = a ∗n b for some fixed a, we will have f (1) = a ∗n+1 b, so if we can find an analytic way to define analytic non-integer iterations we will have found an analytic solution to the hyperoperator equation. Definition 2. A point c is a fixed point of f if f(c) = c. A fixed point c is attracting if 0 < jf 0(c)j < 1, repelling if jf 0(c)j > 1, superattracting if jf 0(c)j = 0, and neutral or parabolic if jf 0(c)j = 1. We will mostly be concerned with attracting and repelling fixed points of functions, as these are the easiest to analyze. ◦n Proposition 1. For all x in a neighborhood of an attracting or superattracting fixed point c, limn!1 f (x) = c, and f ◦n(x) − c ∼ κf 0(c)n for some constant κ (dependant on f, x, and c, but independent of n). The proof of this proposition is simple if we expand f into the first term of its Taylor series. Also, notice that if c is a repelling fixed point of f, it is an attracting fixed point of f −1. This will be useful for understanding the iteration of functions around repelling fixed points. 2.2 The Schroeder Function One of the easiest ways to solve for non-integer iteration is by using a function to control iteration, for example, given a function f, the Schroeder Function for f is a function h satisfying h(f(x)) = λh(x) (21) i.e. h is an eigenfunction of the composition operator by f. One could imagine defining Schroeder functions for arbitrary f : A ! A, but for this paper we will only consider the case of Schroeder functions for f a complex analytic function. The iteration of f can be determined by f ◦n(x) = h−1(λnh(x)), which opens up the possibility for non-integer iteration of f, since the right hand side does not necessitate and integer n. The following theorem tells us how to find Schroeder functions: 2 Theorem 1. If c is an attracting or repelling fixed point of an analytic function f :Ω ! C, where Ω is a connected open set containing c, then there exists a unique function h analytic in a neighborhood of c such that h(f(x)) = λh(x) and h0(c) = 1. In this case, λ = f 0(c). Also, if c is an attracting fixed point of f then h is given by 1 ! f ◦n(x) − c Z x Y f 0(f ◦k(t)) h(x) = lim = dt (22) n!1 f 0(c)n f 0(c) c k=0 and in this case h extends to the entire basin of attraction of c Proof. First, we deal with the attracting case. We begin by showing that (22) solves (21): If h is defined by the limit definition (which converges in a neighborhood of c by Proposition 1), we have f ◦n(f(x)) − c f ◦n+1(x) − c f ◦n+1(x) − c h(f(x)) = lim = lim = f 0(c) lim = f 0(c)h(x) (23) n!1 f 0(c)n n!1 f 0(c)n n!1 f 0(c)n+1 How we show that this function is analytic and equal to the integral expression: Observe that using the fact that f ◦n(x) ! c exponentially, the product inside the integral converges absolutely, and we can show that it converges uniformly on any compact subset of the basin of attraction of c, and hence the product defines an analytic function on the basin of attraction of c.

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