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An Exploration of the Arithmetic Derivative Alaina Sandhu Research Midterm Report, Summer 2006 Under Supervision: Dr . An Exploration of the Arithmetic Derivative Alaina Sandhu Research Midterm Report, Summer 2006 Under Supervision: Dr. McCallum, Ben Levitt, Cameron McLeman 1 Introduction The Arithmetic Derivative is a recently defined function of whose properties directly relate to some of the most well known conjectures within Number Theory. The arithmetic derivative of an integer is defined to be the map which sends every prime integer to 1 and that satisfies the Leibnitz rule, for all a, b ∈ Z,(ab)0 = a0b + ab0. From this definition, we can conjecture that for any a ∈ N, there exists a solution to the differential equation n0 = 2a. A proof of Goldbach’s conjecture would validate this statement, as the derivative of the product of 0 two primes (p1 · p2) is the sum. Our goal is to explore the properties of the arithmetic derivative and propose further conjectures with regard to the nature of the function as well as its implications for the Twin Prime Conjecture and Goldbach Conjecture, among others. 2 Definition The arithmetic derivative n0 of any natural number is defined as follows: • p0 = 1 for any prime p. • (ab)0 = a0b + ab0 for any a, b ∈ N (Leibniz rule). • 00 = 0. To illustrate this, 150 = (3 · 5)0 = 30 · 5 + 3 · 50 = 1 · 5 + 3 · 1 = 8. We now examine the derived formula and ensure the function is well-defined. 1 Qk ni Theorem 1 For any natural number n, if n = i=1 pi is the prime factorization of n, then k X ni n0 = n . (1) p i=1 i Qk 0 Pk 1 First we prove by induction on k that if n = pi, then n = n . i=1 i=1 pi Qk Proof. Consider n ∈ N and express n = i=1 pi, where all pi are not necessarily distinct. 0 Qk When k = 1, it is clear that n = 1. Assume that n = i=1 pi. Then, 0 0 0 (npk+1) = n pk+1 + n(pk+1) k X 1 = n + np k+1 p i=1 i k+1 X 1 = np k+1 p i=1 i Qk ni Now, if we express n = i=1 pi , our formula becomes n0 = n Pk ni i=1 pi Now, we check that our formula is consistent with the Leibniz rule for the product of two Qk ai Qk bi natural numbers; Let a = i=1 pi and b = i=1 pi . Then, k k ! k ! X ai + bi X ai X bi ab = a b + a b p p p i=1 i i=1 i i=1 i . This definition maintains some natural properties, such as (nk)0 = knk−1n0. Also, because 10 = (1 · 1)0 = 10 · 1 + 1 · 10 = 2 · 10,, the derivative of 1 is defined as 10 = 0 . 2.1 Bounds for the Arithmetic Derivative Theorem 2 The only positive integer which satisfies n0 = 0 is n = 1. Proof. This is a direct result of the definition of our function. Theorem 3 The only solutions to n0 = 1 in natural numbers are prime numbers. Proof. A composite number can be expressed as the product of prime numbers, of which the derivative of (by the product rule) is the sum of at least two positive integers, which is greater than 1. Theorem 4 [1] For any positive integer n n log n n0 ≤ 2 . (2) 2 2 If n is composite, √ n0 ≥ 2 n. (3) Furthermore, if n is a product of k factors larger than 1, then 0 k−1 n ≥ kn k . (4) Qk ni (Proof taken directly from [1]) Proof. If n = i=1 pi , then k k Y ni X n ≥ 2 ⇒ log2 n ≥ ni. i=1 i=1 Using Theorem 1, this inequality translates to k Pk X ni n ni n log n n0 = n ≤ i=1 ≤ 2 . p 2 2 i=1 i If n = n1n2n3 ··· nk then, according to the Leibnitz rule, 0 0 0 0 0 n = n1n2n3 ··· nk + n1n2n3 ··· nk + n1n2n3 ··· nk + ... + n1n2n3 ··· nk ≥ n2n3n4 ··· nk + n1n3n4 ··· nk + n1n2n4 ··· nk + ... + n1n2 ··· nk−1 = 1 k 1 1 1 1 1 1 −1 k−1 n + + ... + ≥ n · k · ··· = k · n · n k = k · n k . n1 n2 nk n1 n2 nk Here we have replaced the arithmetic mean by the geometric mean. 3 Properties of the Arithmetic Derivative While the arithmetic derivative has only recently been defined, there has been some signifi- cant research published on various properties of this function, notably ”How to Differentiate a Number” by Ufnarovski and Ahlander [1]. We will now address the theorems and properties relevant to our research, and discuss the related questions we seek to resolve. 3.1 Solutions to n0 = n Numerical derivations have shown positive integers of the form pp (where p is prime) solve the differential equation n0 = n. X p (pp)0 = pp = pp (5) p This unique property of pp is useful in the following theorems. Theorem 5 [1] If n = pp · m for some prime p and natural m > 1, then n0 = pp(m + m0) (k) and limk→∞ n = ∞. 3 Proof. The derivative of n = pp · m, n0 = (pp)0 · m + pp · m0 = pp(m + m0) > n and further proof by induction shows that n(k) ≥ n + k. Theorem 6 [1] Let pk be the highest power of prime p that divides the natural number n. If 0 < k < p, then pk−1 is the highest power of p that divides n0. Furthermore, each derivative n, n0, n00, . , n(k) is distinct. Proof. Let n = pkm. Then n0 = kpk−1m + pkm0 = pk−1(km + pm0), and because k < p, the inside term is not divisible by p. We see that indeed, the only solutions to this equation are integers of the form pp. Theorem 7 [1] For n ∈ N, n0 = n if and only n = pp, where p is a prime. Furthermore, there is an infinite number of solutions to the equation. Proof. Assume n0 = n. Then by Theorem 6, if p | n at least pp | n or else it would contradict n0 = n. Conversely, assume n = pp. As seen in (5), n0 = pp = n. These properties lead to the conjecture that n(k) = n has only numbers of the form pp as solutions. As we will discuss later, we do not expect the function n(k) to cycle for any arbitrary n not of the form pp. 3.2 Solutions to n0 = a As previously discussed in Section 2.1, we have specific solutions to the differential equations n0 = 0 and n0 = 1. Now we turn our attention to the existence of solutions to the equation n0 = a, where a > 1. First, we observe that there can only be a finite number of solutions to n0 = a because a2 all potential solutions are bounded above by 4 . One direct application of n0 = a concerns the Goldbach Conjecture, which states that every even number larger than 3 is the sum of two distinct prime numbers. In terms of our function, we can restate it as follows: Conjecture 1 For any a ∈ N, there exists a solution to the equation n0 = 2a. The Goldbach Conjecture allows us to represent 2a = p1 + p2, so if we take the derivative 0 0 0 of the product (p1 · p2) = (p1) (p2) + p1(p2) = p2 + p1 = 2a. Additionally, we find some solutions for n0 = a where a is odd: Theorem 8 For any prime, p, and a ∈ N which can be expressed a = p + 2, 2p is a solution to the n0 = a. Proof. (2p)0 = 20p + 2p0 = p + 2. It is useful to explore the properties of the function I(a) which calculates all positive integers n such that n0 = a and the corresponding function i(n) which denotes the number of such solutions. After implementing the program within Pari, initial explorations supported 4 our above conjecture that the each even integer through 100, had at least one solution to n0 = a. We observed that a substantial amount of prime numbers (through 1000) did not have any anti-derivatives, which may prove useful when considering bounds for the function i(n) and I(n). Theorem 9 [1] The function i(n) is unbounded for n > 1. (This proof was taken directly from [1]) Proof. Suppose that i(n) < C for all n > 1 for some constant C. Then 2n X i(k) < 2Cn k=2 for any n. But for any two primes p, q the product pq belongs to I(p + q) thus 2n X X 0 π(n)(π(n) + 1) π(n)2 i(k) > 1 = > , 2 2 k=2 p≤q≤n where P0 means that the sum runs over the primes, and π(n) is the number of primes not exceeding n. This leads to the inequality π(n)2 √ 2Cn > ⇒ π(n) < 2 Cn, 2 n which contradicts the known asymptotic behavior π(n) ≈ ln n . We are going to use this proof as a starting point towards a stronger statement regarding the infinitude of i(n) for n > 1. We would like to yield some asymptotic relationship for the function. 3.3 Dynamics of n(k) as k → ∞ It appears that for any natural number, the trajectory of its n-th derivative can follow any three paths: Conjecture 2 For any n ∈ N, one of the following could happen: • n(k) = 0 as k → ∞ • n(k) = ∞ as k → ∞ • n = pp for some prime p., and thus n(k) = pp as k → ∞ We would like to focus on the proofs of the former two points, as well as prove that not only are cycles infinite, but they never repeat themselves.
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