Number-Theoretic Function

Transcription

1 Chapter 1 Number-Theoretic Function 1.1 The function τ and σ Definition Given a positive integer n, let τ(n) denote the number of positive divisor of n and σ(n) denote the sum of these divisor. Definition Any function whose domain of definition is the set of positive integers is called number theoretic function. Note σ(n), τ(n) are non theoretic function. n Theorem If n = p k i i = p k 1 1 p k p kr r is the prime factorization of n > 1, then a) τ(n) = (k 1 + 1)(k 2 + 1) (k r + 1). b) σ(n) = ( pk pk2+1 p 1 )( p 2 1 ) ( pkr+1 r 1 ). p r 1 Proof. Any divisor d of n is of the form d = p α 1 1 p α p αr r where 0 α i k i, there are k choices for the exponent α 1, k choices for the exponent α 2,..., k r +1 choices for the exponent α r, hence there are (k 1 +1)(k 2 +1) (k r +1) possible divisor of n. τ(n) = (k 1 + 1)(k 2 + 1) (k r + 1) = To evaluate σ(n) consider the product r (k i + 1). (1 + p 1 + p p k 1 1 )(1 + p 2 + p p k 2 2 ) (1 + p r + p 2 r p kr r )...(1) each positive divisor of n is one of the terms in the expansion of this product (1), so that 1

2 σ(n) = ( pk 1 +1 = 1 1 pk2+1 p 1 )( p 2 1 ( pk i+1 i 1 r p i 1 ). ) ( pkr+1 r 1 ) p r 1 Example n = = 376 τ(n) = (3 + 1)(1 + 1) = 8. σ(n) = ( 24 1 ) (472 1) = = (47 1) Definition A number theoretic functions f is called multiplicative number theoretic functions if f(mn) = f(m) f(n), where gcd(m, n) = 1. Theorem The functions τ and σ are both multiplicative functions. Proof. If gcd(m, n) = 1, then the result true if m = n = 1. Assume m > 1, n > 1, if m = p k 1 1 p kr r, n = q j 1 1 q js s, p i q i for 1 i r, 1 i s, then mn = p k 1 1 p kr r q j 1 1 qs js τ(mn) = (k 1 + 1)(k 2 + 1) (k r + 1)(j 1 + 1)(j 2 + 1) (j s + 1) also σ(mn) = = τ(m) τ(n). r i 1 s p i 1 ) ( pk i+1 = σ(m) σ(n). j=1 j 1 p j 1 ) ( pk j+1 Definition f(d) means the value f(d) runs over all the positive divisors of the positive integer n. Example d 20 f(d) = f(1) + f(2) + f(4) + f(5) + f(10) + f(20). special cases: τ(n) = σ(n) = 1 = number of divisor. d sum of divisor. Example τ(10) = d 10 σ(10) = d 10 d = = 18 1 = = 4 2

3 Lemma If gcd(m, n) = 1, then the set of positive divisor of mn consists of all products d 1 d 2, where d 1 n, d 2 n and gcd(d 1, d 2 ) = 1, further more, these products are all distinct. Theorem If f is a multiplicative function and F is defined by F (n) = f(d); then F is also multiplicative. Proof. Let g.c.d(m, n) = 1, then F (mn) = f(d) = f(d 1 d 2 ) d 1 n,d 2 n Since every divisor of mn can be uniquely written as a product of a divisor d 1 of m and a divisor d 2 of n, where g.c.d(d 1, d 2 ) = 1. By the definition of a multiplicative functions we get f(d 1 d 2 ) = f(d 1 )f(d 2 ) F (mn) = Example Let m = 8, n = 3 d 1 n,d 2 n f(d 1 )f(d 2 ) = f(d 1 ) f(d 2 ) d 1 n d 2 n = F (m) F (n) F (8 3) = f(d) d 24 = f(1) + f(2) + f(3) + f(4) + f(6) + f(8) + f(12) + f(24) = f(1 1)+f(2 1)+f(1 3)+f(4 1)+f(2 3)+f(8 1)+f(4 3)+f(8 3) = f(1) f(1) + f(2) f(1) + f(1) f(3) + f(4) f(1) + f(2) f(3)+ f(8) f(1) + f(4) f(3) + f(8) f(3) = [f(1) + f(2) + f(4) + f(8)][f(1) + f(3)] = f(d) f(d) d 8 d 3 = F (8) F (3). 3

4 Definition We say that n is not a square free if p 2 n, p is prime. And n is a square free if n = p 1 p 2 p r 1.2 The Möbius inversion formula Definition For a positive integer n, define µby the rules 1, if n = 1 µ = 0, if p 2 n for some p ( 1) r, if n = p 1 p 2 p r, where the p i are distinct primes Example µ(30) = µ(2 3 5) = ( 1) 3 = 1, µ(1) = 1, µ(2) = 1, µ(4) = 0, µ(6) = µ(2 3) = ( 1) 2 = 1. Note If p is a prime number, it is clear that µ(p) = 1; also, µ(p k ) = 0 for k 2. Theorem The function µ is a multiplicative function. Proof. We want to show that µ(mn) = µ(m) µ(n), where g.c.d(m, n) = 1. If either p 2 m or p a prime, then p 2 mn. Hence, µ(mn) = µ(m) µ(n), and the formula holds. Let m = p 1 p 2 p r, n = q 1 q 2 q s, where p i q j for 1 i r, 1 j s, then µ(mn) = µ(p 1 p 2 p r ) µ(q 1 q 2 q s ) = ( 1) r+s = ( 1) r ( 1) s = µ(m) µ(n). Note If n = 1, then µ(d) = 1, µ(d) = 1 suppose n > 1 and put F (n) = µ(d) let n = p k, the positive divisor of p k are k + 1 integers (1, p, p 2,..., p k ), so that F (n) = F (p k ) = d p k µ(d) = µ(1) + µ(p) + µ(p 2 ) µ(p k ) = µ(1) + µ(p) = 1 + ( 1) = 0 If n = p k 1 1 p k 2 2 p kr r if p i p j for 1 i r, 1 j r, g.c.d(p i, p j ) = 1. F (n) = µ(d) is multiplicative (since µ(d) is multiplicative) F (n) = F (p k 1 1 )F (p k 2 2 ) F (p kr r ) = 0 Theorem { For each positive integer n 1 1, if n = 1 µ(d) = 0, if n > 1 where d runs through the positive divisors of n. 4

5 Example n = 10, The divisor of n are 1, 2, 5, 10 µ(d) = µ(1) + µ(2) + µ(5) + µ(10) d/10 = µ(1) + µ(2) + µ(5) + µ(10) = 1 + ( 1) + ( 1) + 1 = 0 Theorem (Möbius inversion formula) let F and f be tow number-theoretic functions related by the formula F (n) = f(d) then f(n) = µ(d)f () = µ(n d)f (d) Proof. let d = n d, as d ranges over all positive divisors of n, so dose d. (1) claim: µ(d)f () = and c d = n d iff c n and d n c. (µ(d) c ( n d ) f(c)) = k Z n = d k, k = n d n = d n d If c n c n ( because c n and d n) d d n also, s Z n d = cs n d = ds d n d Because of this, the last expression in (1) becomes (2) ( µ(d)f(c)) c d ( µ(d)f(c)) = (f(c)µ(d)) c d c n d ( n c ) = (f(c) µ(d)). c n d ( n c ) By theorem 6.6 µ(d) = 0 except when n = 1 d d ( n c ) That is when n = c, in this case it is equal to 1. so the right hand side of equation(2) simplifies to c n (f(c) d ( n c ) µ(d)) = f(c) 1 = f(n) c=n 5

6 Example Let n = 10, to illustrate how the double sum in (2) is turned. In this instance we find that d 10 c ( 10 d ) (µ(d)f(c)) = µ(1)[f(1) + f(2) + f(5) + f(10)] + µ(2)[f(1) + f(5)]+ µ(5)[f(1) + f(2)] + µ(10)[f(1)] = f(1)[µ(1) + µ(2) + µ(5) + µ(10)+]+ f(2)[µ(1) + µ(5)+] + f(5)[µ(1) + µ(2)+] + f(10)[µ(1)+] = (f(c)µ(d)) c 10 d ( 10 c ) To see how Möbius inversion works in a particular case, we remind the reader that the functions τ and σ may both be described as sum functions τ(n) = 1 and σ(n) = d By theorem = ( µ( n d )τ(d)) and n = ( µ( n d )σ(d)) valid for all n 1. Theorem If F is a multiplicative function and F (n) = f(d), then f is also multiplicative Proof. Let m and n be relatively prime positive integers. we recalled that any divisor d of mn can be uniquely written as d = d 1 d 2, where d 1 m, d 2 n and gcd(d 1, d 2 ) = 1. Thus, using the inversion formula. f(mn) = µ(d)f ( mn d ) = µ(d 1 d 2 )F ( mn ) d 1 d 2 d mn d 1 m,d 2 n = µ(d 1 )µ(d 2 )F ( m )F ( n ) d 1 d 2 d 1 m,d 2 n = µ(d 1 )F ( m ) µ(d 2 )F ( n ) d 1 d 2 d 1 m d 2 n = f(m)f(n) 6