CHARACTERIZING COMPLETELY MULTIPLICATIVE FUNCTIONS BY GENERALIZED MÖBIUS FUNCTIONS

where α∈R, and n=∏p primepp denotes the prime factorization of n. This function is called the generalized Möbius function because μ1 = μ, the wellknown Möbius function. Note that μ0 = I, the identity function with respect to Dirichlet convolution, μ−1 = ζ, the arithmetic zeta function and μα+β = μα∗μβ; α, β being real numbers. Recall that an arithmetic function f is said to be completely multiplicative if f(1)≠ 0 and f(mn)= f(m)f(n) for allm andn. As a tool to characterize completely multiplicative functions, Apostol [1] or Apostol [2, Problem 28(b), page 49], it is known that for a multiplicative function f , f is completely multiplicative if and only if


Introduction. Hsu
, see also Brown et al. [3], introduced a very interesting arithmetic function where α ∈ R, and n = p prime p νp (n) denotes the prime factorization of n.This function is called the generalized Möbius function because µ 1 = µ, the wellknown Möbius function.Note that µ 0 = I, the identity function with respect to Dirichlet convolution, µ −1 = ζ, the arithmetic zeta function and µ α+β = µ α * µ β ; α, β being real numbers.Recall that an arithmetic function f is said to be completely multiplicative if f (1) ≠ 0 and f (mn) = f (m)f (n) for all m and n.As a tool to characterize completely multiplicative functions, Apostol [1] or Apostol [2,Problem 28(b), page 49], it is known that for a multiplicative function f , f is completely multiplicative if and only if Our first objective is to extend this result to µ α .
Theorem 1.1.Let f be a nonzero multiplicative function and α a nonzero real number.Then f is completely multiplicative if and only if In another direction, Haukkanen [5] proved that if f is a completely multiplicative function and α a real number, then f α = µ −α f .Here and throughout, all powers refer to Dirichlet convolution; namely, for positive integral α, define f α := f * ••• * f (α times) and for real α, define f α = Exp(α Log f ), where Exp and Log are Rearick's operators [9].Our second objective is to establish the converse of this result.There is an additional hypothesis, referred to as condition (NE) which appears frequently.By condition (NE), we refer to the condition that: if α is a negative even integer, then assume that f (p −α−1 ) = f (p) −α−1 for each prime p. Theorem 1.2.Let f be a nonzero multiplicative function and Because of the different nature of the methods, the proof of Theorem 1.2 is divided into two cases, namely, α ∈ Z and α ∈ Z.As applications of Theorem 1.2, we deduce an extension of Corollary 3.2 in [11] and a modified extension of [7, Theorem 4.1(i)].

Proof of Theorem
follows easily from Haukkanen's theorem [5].To prove the other implication, it suffices to show that f (p k ) = f (p) k for each prime p and nonnegative integer k.This is trivial for k = 0, 1.Assuming f (p j ) = f (p) j for j = 0, 1,...,k−1, we proceed by induction to settle the case j = k > 1.From hypothesis, we get (2.1) (2.2) Simplifying and using induction hypothesis, we get From Riordan [10, identity (5), page 8], the coefficient of f (p) k on the right-hand side is equal to and the desired result follows.
Remark 2.1.(1) To prove the "only if" part of Theorem 1.1, instead of using Haukkanen's result, a direct proof based on [1, Theorem 4(a)] can be done as follows: (2) To prove the "if" part of Theorem 1.1, instead of using [10, identity (5)], a selfcontained proof can be done as follows: from (1+z

Proof of Theorem 1.2.
The proof of Theorem 1.2 is much more involved and we treat the integral and nonintegral cases separately.This is because the former can be settled using only elementary binomial identities, while the proof of the latter, which is also valid for integral α, makes use of Rearick logarithmic operator, which deems nonelementary to us.Proposition 3.1.Let f be a nonzero multiplicative function and r a positive integer where p is a prime and k a nonnegative integer.This clearly holds for k = 0, 1.As an induction hypothesis, assume this holds for 0, 1,...,k− 1 (≥ 1).From we get, using induction hypothesis, Simplifying, we arrive at We have, for k ≥ 2, Using induction hypothesis and [10, identity (5)], the right-hand expression is For positive integers r and k (≥ 2), observe that r + (−1) k r k = 0 if and only if k = r − 1 and k is odd.The conclusion hence follows.
Remark 3.4.In the case of r being a positive even integer, without an additional assumption on f (p r −1 ), Proposition 3.3 fails to hold as seen from the following example.
Take r = 4.For each prime p, set and for k ≥ 4, define f (p k ) by the relation Define other values of f by multiplicativity, namely, (3.9) p i prime, a i nonnegative integer.This particular function satisfies µ 4 f = f −4 and is multiplicative, but not completely multiplicative.Now for the case of nonintegral index, we need one more auxiliary result.For more details about the Rearick logarithm, see [8,9].Lemma 3.5.Let f be an arithmetic function, p a prime, k a positive integer, and let Log denote the Rearick logarithmic operator defined by Proof.From hypothesis, we have and so Log Now proceed by induction noting, as in the lemma of Carroll [4], that f (1) = 1 and (3.15) Now for the final case, we prove the following proposition.
Proposition 3.6.Let f be multiplicative and Proof.As before, we proceed by induction on nonnegative integer k to show that f (p k ) = f (p) k the result being trivial for k = 0, 1.
Let D be the log-derivation on the ring of arithmetic functions (cf.[7,8,9]).Since where Log denotes the Rearick logarithmic operator mentioned in Lemma 3.5, then taking derivation D on both sides of µ −α f = f α and evaluating at p k , we get that is, (3.18) Using induction hypothesis and the lemma, we have and so with the aid of [10, identity ( 5)], we get Since α ∈ R − Z, then the coefficients on both sides are the same nonzero real number, which immediately yields the desired conclusion.
The following corollaries are immediate consequences of Theorem 1.2 and the main theorem in [5].

Call for Papers
Thinking about nonlinearity in engineering areas, up to the 70s, was focused on intentionally built nonlinear parts in order to improve the operational characteristics of a device or system.Keying, saturation, hysteretic phenomena, and dead zones were added to existing devices increasing their behavior diversity and precision.In this context, an intrinsic nonlinearity was treated just as a linear approximation, around equilibrium points.Inspired on the rediscovering of the richness of nonlinear and chaotic phenomena, engineers started using analytical tools from "Qualitative Theory of Differential Equations," allowing more precise analysis and synthesis, in order to produce new vital products and services.Bifurcation theory, dynamical systems and chaos started to be part of the mandatory set of tools for design engineers.
This proposed special edition of the Mathematical Problems in Engineering aims to provide a picture of the importance of the bifurcation theory, relating it with nonlinear and chaotic dynamics for natural and engineered systems.Ideas of how this dynamics can be captured through precisely tailored real and numerical experiments and understanding by the combination of specific tools that associate dynamical system theory and geometric tools in a very clever, sophisticated, and at the same time simple and unique analytical environment are the subject of this issue, allowing new methods to design high-precision devices and equipment.
Authors should follow the Mathematical Problems in Engineering manuscript format described at http://www .hindawi.com/journals/mpe/.Prospective authors should submit an electronic copy of their complete manuscript through the journal Manuscript Tracking System at http:// mts.hindawi.com/according to the following timetable:

. 4 ) 1 r − 1 −Remark 3 . 2 .Proposition 3 . 3 .
Since r ≥ 2, then k+r −r ≠ 0, and we have the result.The case r = 1 is excluded for µ −1 f = ζf is always equal to f , and so the assumption is empty.The case r = 0 is excluded becauseI = f 0 = µ 0 f = If holds for any arithmetic function f with f (1) = 1.Let f be a nonzero multiplicative function and −α = r a positive integer.Assuming condition (NE), if f −r = µ r f , then f is completely multiplicative.Proof.As in Proposition 3.1, we show by induction that f (p k ) = f (p) k for prime p and nonnegative integer k, noting that it holds trivially for k = 0, 1.The main assumption of the theorem gives µ r f * f r = I.(3.5) and the converse is true provided condition (NE) holds.Corollary 3.8 (cf.[7, Theorem 4.1(i)]).Let α ∈ R − {0, 1} and f a nonzero multiplicative function.If f is completely multiplicative, then