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Unformatted text preview: ARITHMETICAL FUNCTIONS II: CONVOLUTION AND INVERSION PETE L. CLARK 1. Sums over divisors, convolution and Mobius Inversion The proof of the multiplicativity of the functions k , easy though it was, actu ally establishes a more general result. Namely, suppose that f is a multiplicative function, and define a new function F = d f as F ( n ) = X d  n F ( d ) . For instance, if we start with the function f ( n ) = n k , then F = k . Note that f ( n ) = n k is (in fact completely) multiplicative. The generalization of the proof is then the following Proposition 1. If f is a multiplicative function, then so is F ( n ) = d  n f ( d ) . Proof: If n 1 and n 2 are coprime, then F ( n 1 n 2 ) = d  n 1 n 2 F ( d ) = X d 1  n 1 , d 2  n 2 f ( d 1 d 2 ) = X d 1  n 1 , d 2  n 2 f ( d 1 ) f ( d 2 ) = ( X d 1  n 1 f ( d 1 ))( X d 2  n 2 f ( d 2 )) = F ( n 1 ) F ( n 2 ) . Note that the process does not preserve completely multiplicative functions. It turns out that the operation f 7 F is of general interest; it gives rise to a certain kind of duality among arithmetic functions. Slightly less vaguely, some times f is simple and F is more complicated, but sometimes the reverse takes place. Definition: Define the function by (1) = 1 and ( n ) = 0 for all n > 1. Note that is multiplicative. Also write for the function n 7 n . Proposition 2. a) For all n > 1 , d  n ( d ) = 0 . b) For all n Z + , d  n ( n ) = n . In other words, the sum over the divisors of the Mobius function is , and the sum over the divisors of is . Proof: a) Write n = p a 1 1 p a r r . Then d  ( n ) = ( 1 ,..., r ) ( p 1 1 p r r ), where the i are 0 or 1. Thus X d  n ( d ) = 1 r + r 2  r 3 + ... + ( 1) r r r = (1 1) r = 0 . For part b) we take advantage of the fact that since is multiplicative, so is the sum over its divisors. Therefore it is enough to verify the identity for a prime power 1 2 PETE L. CLARK p a , and as usual this is significantly easier: X d  p a ( p a ) = a X i =0 ( p i ) = 1 + a X i =1 ( p i p i 1 ) = 1 + ( p a 1) = p a , where we have cancelled out a telescoping sum. This indicates that the Mobius function is of some interest. We can go further by asking the question: suppose that F = d f is multiplicative; must f be mul tiplicative? Well, the first question is to what extent f is determined by its divisor sum func tion: if F = d f = d g = G , must f = g ? If so, is there a nice formula which gives f in terms of F ? Some calculations: f (1) = F (1); for any prime p , F ( p ) = f (1) + f ( p ), so f ( p ) = F ( p ) F (1); F ( p 2 ) = f (1) + f ( p ) + f ( p 2 ) = F ( p ) + f ( p 2 ), so f ( p ) = F ( p 2 ) F ( p ); indeed f ( p n ) = F ( p n ) F ( p n 1 )....
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 Spring '11
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 Number Theory

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