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Arithmetic Subderivatives: P-Adic Discontinuity and Continuity

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Arithmetic Subderivatives: P-Adic Discontinuity and Continuity 1 2 Journal of Integer Sequences, Vol. 23 (2020), 3 Article 20.7.3 47 6 23 11 Arithmetic Subderivatives: p-adic Discontinuity and Continuity Pentti Haukkanen and Jorma K. Merikoski1 Faculty of Information Technology and Communication Sciences FI-33014 Tampere University Finland [email protected] [email protected] Timo Tossavainen Department of Arts, Communication and Education Lulea University of Technology SE-97187 Lulea Sweden [email protected] Abstract In a previous paper, we proved that the arithmetic subderivative DS is discontinuous at any rational point with respect to the ordinary absolute value. In the present paper, we study this question with respect to the p-adic absolute value. In particular, we show that DS is in this sense continuous at the origin if S is finite or p∈ / S. 1 Introduction Let 0 =6 x ∈ Q. There exists a unique sequence (νp(x))p∈P of integers (with only finitely many nonzero terms) such that 1Corresponding author. 1 x = (sgn x) pνp(x). (1) Y p∈P Here P stands for the set of primes, and sgn x = x/|x|. Define that sgn0 = 0 and νp(0) = ∞ for all p ∈ P. In addition to the ordinary axioms of ∞, we state that 0 · ∞ = 0. Then (1) holds also for x = 0. We recall the basic properties of the p-adic order νp. Proposition 1. For all x,y ∈ Q, (a) νp(x)= ∞ if and only if x = 0; (b) νp(xy)= νp(x)+ νp(y); (c) νp(x + y) ≥ min(νp(x),νp(y)); (d) νp(x + y) = min(νp(x),νp(y)) if νp(x) =6 νp(y). Proof. Properties (a) and (b) are trivial. For (c) and (d), see, e.g., [2, Proposition 2.4]. Throughout this paper, we let a ∈ Q, p,q ∈ P, p =6 q, and ∅= 6 S ⊆ P. The arithmetic subderivative [11, 8, 9] of x ∈ Q with respect to S, a.k.a. the arithmetic type derivative [4] is ν (x) D (x)= x p . S X p p∈S The arithmetic partial derivative [10, 7] of x with respect to p is Dp(x) = D{p}(x). The arithmetic derivative [12, 3, 13] of x is D(x)= DP(x). Clearly, D (x)= D (x),D(x)= D (x). S X p X p p∈S p∈P The function DS is very strongly discontinuous at any a [8, Theorem 4] with respect to the ordinary absolute value. But do we succeed better if we use the p-adic absolute value of x, defined by 1 |x|p = ? pνp(x) (In particular, |0|p =1/∞ = 0.) We recall the basic properties of |·|p. Proposition 2. For all x,y ∈ Q, (a) |x|p =0 if and only if x = 0; (b) |xy|p = |x|p|y|p; 2 (c) |x + y|p ≤ max(|x|p, |y|p); (d) |x + y|p = max(|x|p, |y|p) if |x|p =6 |y|p. Proof. This proposition is equivalent to Proposition 1. Let us write 1 νp(x) νp(x) νp(x) x = x p = x|x|pp = µp(x)p , pνp(x) where x µp(x)= x|x|p = . (2) pνp(x) Proposition 3. For all x,y ∈ Q, (a) νp(µp(x))=0 if x =06 , νp(µp(0)) = νp(0) = ∞; (b) |µp(x)|p =1 if x =06 , |µp(0)|p = 0; (c) µp(xy)= µp(x)µp(y); (d) Dp(µp(x))=0. Proof. Trivial. The p-adic distance |x − y|p is smaller the larger νp(x − y) is. We say that a function Q → Q is p-adically continuous, in short p-continuous, if it is continuous with respect to |·|p. So we ask: Is DS p-continuous at some a? We study this question by considering sequences (xi) of rational numbers. If |xi − a|p → 0, equivalently νp(xi − a) → ∞, then (xi) converges p-adically, in short p-converges, to a. Let xi →p a denote this convergence. Sections 2–3 are introductory. We present in Section 2 a “light version” of Dirichlet’s theorem on arithmetic progressions. We study p-convergence in Section 3. Sections 4–7 contain our main results. We prove in Section 4 that DS is p-continuous at a = 0 if S is finite or p∈ / S. We also prove that Dp is p-continuous also at a =6 0. On the other hand, we show in Section 5 that Dq can be (and conjecture that it always is) p-discontinuous at a =6 0. In Section 6, we extend the results of Section 5 to DS when S is finite. Although Section 5 is only a special case of Section 6, we find it instructive to present it separately. We complete our paper with the conclusion in Section 7. 2 “Poor man’s theorem on arithmetic progressions” Throughout this section, a,b ∈ Z with gcd(a,b) = 1. As suggested by Graham et al. [5], we let a ⊥ b denote that gcd(a,b)=1. See also [6, 4]. We recall Dirichlet’s theorem on arithmetic progressions. 3 Theorem 4. If b> 0, then the set T = {a + nb : n ∈ Z+} (3) contains infinitely many primes. Proof. See, e.g., [1, Theorem 7.9]. Corollary 5. If b =6 0, then the set (3) contains infinitely many primes or their additive inverses. Proof. Trivial. Theorem 4 is advanced, while our paper is elementary. We do not need the full force of this theorem, and we want to use only elementary methods. Therefore we apply, instead of Theorem 4, the following “poor man’s theorem on arithmetic progressions.” It is elementary but strong enough for us. (Remember that ∅= 6 S ⊆ P throughout.) Theorem 6. If S is finite and b =06 , then the set (3) contains infinitely many numbers that are not divisible by any element of S. Proof. If a = 0, then b = ±1, since otherwise a 6⊥ b, a contradiction. Therefore T = Z+ or T = Z−, and the claim is trivially true. Now assume that a =6 0. Let S = {p1,...,ph,q1,...,qk}, p1,...,ph ∤ a, q1,...,qk | a. (Either the pi or the qi can be missing. Clearly, pi =6 qj for all i, j.) We show that the numbers a + nb apply when n goes through the set N = {mp1 ··· ph : m ∈ Z+, q1,...,qk ∤ m}. Write c = p1 ··· ph. (If the pi are missing, then the “empty product” c = 1.) Let x ∈ T with n ∈ N, that is, x = a + mcb, q1,...,qk ∤ m. (4) Each pi | c but pi ∤ a, so pi ∤ x. Each qi | a but qi ∤ mcb. (Clearly, qi ∤ m,c. If qi | b, then a 6⊥ b, a contradiction.) Therefore also qi ∤ x. Consequently, s ∤ x for all s ∈ S. Because there are infinitely many numbers (4), the claim follows. 4 3 Convergence Continuity is usually proved by the “ε − δ technique”, while discontinuity is often proved using suitable sequences. For consistency, we use sequences also in proving continuity. To that end, we need a characterization of p-convergence. Proposition 7. Let (xi) be a sequence of rational numbers. If a =6 0, then the following conditions are equivalent. (a) xi →p a; (b) µp(xi) →p µp(a) and there is i0 ∈ Z+ such that νp(xi0 )= νp(xi0+1)= ··· = νp(a). If a =0, then (b) ⇒ (a) but not conversely. Proof. Case 1: a =6 0. (a) ⇒ (b): If i0 does not exist, then (xi) has a subsequence (xik ) whose each term satisfies νp(xik ) =6 νp(a). Consequently, Prop.1(d) a6=0 νp(xik − a) = min(νp(xik ),νp(a)) ≤ νp(a) < ∞, k ≥ 1. Hence νp(xik − a) 6→ ∞, implying xi 6→p a, a contradiction. Therefore i0 exists, i.e., νp(a) xi = µp(xi)p , i ≥ i0. Now, for i ≥ i0, we have νp(a) xi − a =(µp(xi) − µp(a))p , (5) and further (5),Prop. 1(b) (a) νp(µp(xi) − µp(a)) + νp(a) = νp(xi − a) → ∞, verifying µp(xi) →p µp(a). (b) ⇒ (a): Since (b) νp(a) νp(a) νp(a) xi − a = µp(xi)p − µp(a)p =(µp(xi) − µp(a))p , i ≥ i0, we have Prop. 1(b) (b) νp(xi − a) = νp(µp(xi) − µp(a)) + νp(a) → ∞, verifying (a). Case 2: a = 0. (b) ⇒ (a). Since νp(xi0 )= νp(xi0+1)= ··· = νp(0) = ∞, it follows that xi0 = xi0+1 = ··· = 0. Therefore xi →p 0. i (a) 6⇒ (b). If xi = p , then xi →p 0, but µp(xi)=1 →p 1 =0=6 µp(0). 5 Proposition 8. A function F : Q → Q is p-continuous at a if and only if any sequence (xi) of rational numbers satisfying xi →p a satisfies F (xi) →p F (a). Proof. Proceed as in proving the corresponding property of the ordinary continuity. There are three formally different ways to consider p-convergence. First, use νp every- where. Second, use |·|p everywhere. Third, use either νp or |·|p, depending on the situation. We follow the first way. 4 The cases of DS, a =0, and Dp, a arbitrary We begin with a lemma that may be interesting on its own. Lemma 9. Let S be finite and y ∈ Q. Assume that {q ∈ P | νq(y) =06 }⊆ S. Factorize y = qνq(y) = u(y)w(y), Y q∈S where u(y)= qνq(y)−1, w(y)= q. Y Y q∈S q∈S Then DS(y)= u(y)v(y), where v(y)= ν (y) r. X q Y q∈S r∈S\{q} Proof. We have D (y)= D (qνq(y)) rνr(y) = ν (y)qνq(y)−1 rνr(y) S X S Y X q Y q∈S r∈S\{q} q∈S r∈S\{q} = ν (y)qνq(y)−1 rνr(y)−1r = ν (y) rνr(y)−1 r , X q Y X q Y Y q∈S r∈S\{q} q∈S r∈S r∈S\{q} verifying the claim.
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