Dirichlet product of derivative arithmetic with an arithmetic function multiplicative
aa r X i v : . [ m a t h . G M ] A ug D IR IC HLET PRODUCT OF DER IVATIVE AR ITHMETIC WITH ANAR ITHMETIC FUNC TION MULTIPLICATIVE
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REPRINT
Es-said En-naoui [email protected]
August 21, 2019 A BSTRACT
We define the derivative of an integer to be the map sending every prime to 1 and satisfying theLeibniz rule. The aim of this article is to calculate the Dirichlet product of this map with a functionarithmetic multiplicative.
Barbeau [1] defined the arithmetic derivative as the function δ : N → N , defined by the rules :1. δ ( p ) = 1 for any prime p ∈ P := { , , , , . . . , p i , . . . } .2. δ ( ab ) = δ ( a ) b + aδ ( b ) for any a, b ∈ N (the Leibnitz rule) .Let n a positive integer , if n = Q si =1 p α i i is the prime factorization of n , then the formula for computing the arithmeticderivative of n is (see, e.g., [1, 3]) giving by : δ ( n ) = n s X i =1 α i p i = n X p α || n αp (1)A brief summary on the history of arithmetic derivative and its generalizations to other number sets can be found, e.g.,in [4] .First of all, to cultivate analytic number theory one must acquire a considerable skill for operating with arithmeticfunctions. We begin with a few elementary considerations. Definition 1 (arithmetic function) . An arithmetic function is a function f : N −→ C with domain of definition the setof natural numbers N and range a subset of the set of complex numbers C . Definition 2 (multiplicative function) . A function f is called an multiplicative function if and only if : f ( nm ) = f ( n ) f ( m ) (2) for every pair of coprime integers n , m . In case (2) is satisfied for every pair of integers n and m , which are notnecessarily coprime, then the function f is called completely multiplicative . Clearly , if f are a multicative function , then f ( n ) = f ( p α ) . . . f ( p α s s ) , for any positive integer n such that n = p α . . . p α s s , and if f is completely multiplicative , so we have : f ( n ) = f ( p ) α . . . f ( p s ) α s . Example 3.
Let n ∈ N ∗ , This is the same classical arithmetic functions used in this article :1. Identity function : The function defined by Id ( n ) = n for all positive integer n. irichlet product of derivative arithmetic with an arithmetic function multiplicative A PREPRINT The Euler phi function : ϕ ( n ) = n P k =1 gcd ( k,n )=1 .3. The number of distinct prime divisors of n : ω ( n ) = P p | n .4. The Mobiuse function : µ ( n ) = if n = 10 if p | n f or some prime p ( − ω ( n ) otherwise5. number of positive divisors of n defined by : τ ( n ) = P d | n .6. sum of divisors function of n defined by : σ ( n ) = P d | n d . Now ,if f, g : N −→ C are two arithmetic functions from the positive integers to the complex numbers, the Dirichletconvolution f ∗ g is a new arithmetic function defined by: ( f ∗ g )( n ) = X d | n f ( d ) g ( nd ) = X ab = n f ( a ) g ( b ) (3)where the sum extends over all positive divisors d of n , or equivalently over all distinct pairs ( a, b ) of positive integerswhose product is n .In particular, we have ( f ∗ g )(1) = f (1) g (1) , ( f ∗ g )( p ) = f (1) g ( p ) + f ( p ) g (1) for any prime p and for any powerprime p m we have : ( f ∗ g )( p m ) = m X j =0 f ( p j ) g ( p m − j ) (4)This product occurs naturally in the study of Dirichlet series such as the Riemann zeta function. It describes themultiplication of two Dirichlet series in terms of their coefficients: (cid:18) X n ≥ (cid:0) f ∗ g (cid:1) ( n ) n s (cid:19) = (cid:18) X n ≥ f ( n ) n s (cid:19)(cid:18) X n ≥ g ( n ) n s (cid:19) (5)with Riemann zeta function or is defined by : ζ ( s ) = X n ≥ n s These functions are widely studied in the literature (see, e.g., [5, 6, 7]).Now before to past to main result we need this propriety , if f and g are multiplicative function , then f ∗ g ismultiplicative. In this section we give the new result of Dirichlet product between derivative arithmetic and an arithmetic functionmultiplicative f , and we will give the relation between τ and the derivative arithmetic . Theorem 4.
Given a multiplicative function f , and lets n and m two positive integers such that gcd ( n, m ) = 1 , Thenwe have : ( f ∗ δ )( nm ) = (cid:0) Id ∗ f (cid:1) ( n ) . (cid:0) f ∗ δ (cid:1) ( m ) + (cid:0) Id ∗ f (cid:1) ( m ) . (cid:0) f ∗ δ (cid:1) ( n ) (6)2irichlet product of derivative arithmetic with an arithmetic function multiplicative A PREPRINT
Proof.
Lets n and m two positive integers such that gcd ( n, m ) = 1 , and let f an arithmetic function multiplicative ,then we have : ( f ∗ δ )( nm ) = X d | nm f (cid:0) nmd (cid:1) δ ( d ) = X d | n d | m f ( nmd d ) δ ( d d ) = X d | n d | m f ( nd ) f ( md ) (cid:18) d δ ( d ) + d δ ( d ) (cid:19) = X d | n d | m (cid:18) d f ( nd ) f ( md ) δ ( d ) + d f ( md ) f ( nd ) δ ( d ) (cid:19) = (cid:18) X d | n d f ( nd ) (cid:19)(cid:18) X d | m f ( md ) δ ( d ) (cid:19) + (cid:18) X d | m d f ( md ) (cid:19)(cid:18) X d | n f ( nd ) δ ( d ) (cid:19) = (cid:0) Id ∗ f (cid:1) ( n ) . (cid:0) f ∗ δ (cid:1) ( m ) + (cid:0) Id ∗ f (cid:1) ( m ) . (cid:0) f ∗ δ (cid:1) ( n ) Lemma 5.
For any natural number n , if n = Q si =1 p α i i is the prime factorization of n, then : (cid:0) f ∗ δ (cid:1) ( n ) = (cid:0) Id ∗ f (cid:1) ( n ) s X i =1 (cid:0) f ∗ δ (cid:1) ( p α i i ) (cid:0) Id ∗ f (cid:1) ( p α i i ) (7) Proof.
Let n a positive integer such that n = p α . . . p α s s and let f an arithmetic function , Then : (cid:0) f ∗ δ (cid:1) ( n ) = (cid:0) f ∗ δ (cid:1) ( p α . . . p α s s )= (cid:0) Id ∗ f (cid:1) ( p α . . . p α s s ) . (cid:0) f ∗ δ (cid:1) ( p α ) + (cid:0) Id ∗ f (cid:1) ( p α ) . (cid:0) f ∗ δ (cid:1) ( p α . . . p α s s )= (cid:0) Id ∗ f (cid:1) ( n ) . (cid:0) f ∗ δ (cid:1) ( p α ) (cid:0) Id ∗ f (cid:1) ( p α ) + (cid:0) Id ∗ f (cid:1) ( p α ) . (cid:20)(cid:0) Id ∗ f (cid:1) ( p α . . . p α s s ) . (cid:0) f ∗ δ (cid:1) ( p α )++ (cid:0) Id ∗ f (cid:1) ( p α ) . (cid:0) f ∗ δ (cid:1) ( p α . . . p α s s ) (cid:21) = (cid:0) Id ∗ f (cid:1) ( n ) (cid:0) f ∗ δ (cid:1) ( p α ) (cid:0) Id ∗ f (cid:1) ( p α ) + (cid:0) Id ∗ f (cid:1) ( n ) (cid:0) f ∗ δ (cid:1) ( p α ) (cid:0) Id ∗ f (cid:1) ( p α ) ++ (cid:0) Id ∗ f (cid:1) ( p α ) (cid:0) Id ∗ f (cid:1) ( p α ) (cid:0) f ∗ δ (cid:1) ( p α . . . p α s s ) ... = (cid:0) Id ∗ f (cid:1) ( n ) (cid:0) f ∗ δ (cid:1) ( p α ) (cid:0) Id ∗ f (cid:1) ( p α ) + (cid:0) Id ∗ f (cid:1) ( n ) (cid:0) f ∗ δ (cid:1) ( p α ) (cid:0) Id ∗ f (cid:1) ( p α ) + . . . + (cid:0) Id ∗ f (cid:1) ( n ) (cid:0) f ∗ δ (cid:1) ( p α s s ) (cid:0) Id ∗ f (cid:1) ( p α s s )= (cid:0) Id ∗ f (cid:1) ( n ) (cid:20) (cid:0) f ∗ δ (cid:1) ( p α ) (cid:0) Id ∗ f (cid:1) ( p α ) + . . . + (cid:0) f ∗ δ (cid:1) ( p α s s ) (cid:0) Id ∗ f (cid:1) ( p α s s ) (cid:21) = (cid:0) Id ∗ f (cid:1) ( n ) s X i =1 (cid:0) f ∗ δ (cid:1) ( p α i i ) (cid:0) Id ∗ f (cid:1) ( p α i i ) an other prof by induction on s that if n = Q si =1 p α i i then ( f ∗ δ )( n ) = (cid:0) Id ∗ f (cid:1) ( n ) s P i =1 (cid:0) f ∗ δ (cid:1) ( p αii ) (cid:0) Id ∗ f (cid:1) ( p αii ) . Proof.
Consider n ∈ N and express n = Q si =1 p α i i where all p i are distinct .where s = 1 , it is clear that ( f ∗ δ )( n ) = (cid:0) Id ∗ f (cid:1) ( n ) P i =1 (cid:0) f ∗ δ (cid:1) ( p αii ) (cid:0) Id ∗ f (cid:1) ( p αii ) = (cid:0) Id ∗ f (cid:1) ( p α ) (cid:0) f ∗ δ (cid:1) ( p α ) (cid:0) Id ∗ f (cid:1) ( p α ) = (cid:0) f ∗ δ (cid:1) ( p α ) . A PREPRINT
Assume that n = Q si =1 p α i i , then we have : (cid:0) id ∗ δ (cid:1) ( n.p α s +1 s +1 ) = (cid:0) Id ∗ f (cid:1) ( p α s +1 s +1 ) . (cid:0) f ∗ δ (cid:1) ( n ) + (cid:0) Id ∗ f (cid:1) ( n ) . (cid:0) f ∗ δ (cid:1) ( p α s +1 s +1 )= (cid:0) Id ∗ f (cid:1) ( p α s +1 s +1 ) . (cid:0) Id ∗ f (cid:1) ( n ) s X i =1 (cid:0) f ∗ δ (cid:1) ( p α i i ) (cid:0) Id ∗ f (cid:1) ( p α i i ) + (cid:0) Id ∗ f (cid:1) ( p α s +1 s +1 ) . (cid:0) Id ∗ f (cid:1) ( n ) (cid:0) f ∗ δ (cid:1) ( p α s +1 s +1 ) (cid:0) Id ∗ f (cid:1) ( p α s +1 s +1 )= (cid:0) Id ∗ f (cid:1) ( n.p α s +1 s +1 ) s X i =1 (cid:0) f ∗ δ (cid:1) ( p α i i ) (cid:0) Id ∗ f (cid:1) ( p α i i ) + (cid:0) Id ∗ f (cid:1) ( n.p α s +1 s +1 ) (cid:0) f ∗ δ (cid:1) ( p α s +1 s +1 ) (cid:0) Id ∗ f (cid:1) ( p α s +1 s +1 )= (cid:0) Id ∗ f (cid:1) ( n.p α s +1 s +1 ) (cid:20) s X i =1 (cid:0) f ∗ δ (cid:1) ( p α i i ) (cid:0) Id ∗ f (cid:1) ( p α i i ) + (cid:0) f ∗ δ (cid:1) ( p α s +1 s +1 ) (cid:0) Id ∗ f (cid:1) ( p α s +1 s +1 ) (cid:21) = (cid:0) Id ∗ f (cid:1) ( n.p α s +1 s +1 ) s +1 X i =1 (cid:0) f ∗ δ (cid:1) ( p α i i ) (cid:0) Id ∗ f (cid:1) ( p α i i ) Proposition 6.
Let f a function arithmetic multiplicative , and δ the derivative arithmetic , then we have : (cid:0) Id ∗ δ (cid:1) ( n ) = 12 τ ( n ) δ ( n ) (8) Proof.
Since ( Id ∗ Id )( n ) = P d | n nd d = n P d | n nτ ( n ) .and : ( Id ∗ δ )( p α ) = α P j =1 δ ( p j ) Id ( p α − j ) = α P j =1 jp j − p α − j = α ( α + 1) p α − .Then for every a positive integer n such that n = p α . . . p α s s , we have : (cid:0) Id ∗ δ (cid:1) ( n ) = (cid:0) Id ∗ Id (cid:1) ( n ) s X i =1 (cid:0) Id ∗ δ (cid:1) ( p α i i ) (cid:0) Id ∗ Id (cid:1) ( p α i i )= nτ ( n ) s X i =1 12 α i ( α i + 1) p α i − i p α i i τ ( p α i i )= nτ ( n ) s X i =1 12 α i ( α i + 1) p α i − i p α i i ( α i + 1)= 12 nτ ( n ) s X i =1 α i p i = 12 τ ( n ) δ ( n ) So by the proposition 6 , and the equality 5 we have this relation between arithmetic derivative and the function τ : ζ ( s − X n ≥ δ ( n ) n s = X n ≥ δ ( n ) τ ( n ) n s (9)Let defined the new function arithmetic called En-naoui function , by : Φ ϕ ( n ) = n X p | n (cid:18) − p (cid:19) (10)Then we have this equality related between 8 arithmetic function : (cid:0) µ ∗ δ (cid:1) ( n ) = ϕ ( n ) (cid:18) δ ( n ) − ω ( n ) + B ( n ) + Φ ϕ ( n ) n + (cid:0) B ∗ Id (cid:1) ( n ) σ ( n ) (cid:19) . (11)with B is the arithmetic function defined by : B ( n ) = P p α || n αp . In next article i will prove this equality, just I need tosubmit an article about this new function . 4irichlet product of derivative arithmetic with an arithmetic function multiplicative A PREPRINT
I would like to thank S.Allo for the opportunity to collaborate on my research and for his continued mentorship, andfriendship, and I’m indebted to the Professor Zouhaïr Mouayn to endorse me to submit my article here.I also extendmy deepest appreciation to my family for opening their minds, hearts, and homes to me and my work.
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