Mobius Inversion and Dirichlet Convolution

Mobius Inversion and Dirichlet Convolution

Mobius¨ inversion revisited 1 Arithmetic functions and Dirichlet series >0 Suppose f : N ! C is an arithmetic function. In that case, we may form X f(n) L(f; s) := ; ns n>0 we will usually ignore convergence issues and refer to this as a formal Dirichlet series. If f and g are arithmetic functions, then we may define the arithmetic function f + g by the formula f + g(n) = f(n) + g(n). In that case, observe that the formula: L(f + g; s) = L(f; s) + L(g; s) holds. While we can multiply arithmetic functions as well via (fg)(n) = f(n)g(n), the result is not as closely connected with L-functions. Instead, consider the product of L-functions: 1 1 X f(n) X g(n) L(f; s) · L(g; s) = ( )( ): ns ns n=1 n=1 If we multiply out and group terms, f(2) f(3) g(2) g(3) (f(1) + + + ··· )(g(1) + + + ··· ) 2s 3s 2s 3s f(1)g(2) + f(2)g(1) f(1)g(3) + g(1)f(3) f(1)g(4) + f(2)g(2) + g(1)f(4) =(f(1)g(1) + + + + ··· ) 2s 3s 4s Extrapolating from this, the coefficients of L(f; s) · L(g; s) take the form: X f(a)g(b): ab=n This defines a new rule for composing arithmetic functions. Definition 1.1. If f and g are two arithmetic functions, then their Dirichlet convolution is the arith- metic function defined by X n X (f ∗ g)(n) = f(d)g( ) = f(a)g(b): d djn ab=n 1 2 1.1 The ring of arithmetic functions under convolution Example 1.2. We have used this notation already. If f is an arithmetic function, then we defined X X F (n) := f(d) = f(d) · 1: djn djn If we write 1 for the function that takes the constant value 1 on every positive integer, then in terms of convolution, this is the formula F := f ∗ 1. Example 1.3. We introduced the Mobius function µ(n). We then proved that µ ∗ 1 = χ1; where χ1 was the characteristic function of 1, i.e., it took the value 1 if n = 1 and was zero otherwise. Note that, by definition, f ∗ g = g ∗ f, i.e., Dirichlet convolution is a commutative composition. 1.1 The ring of arithmetic functions under convolution Now that we have introduced a new binary operation on the set of arithmetic functions, it is natural to ask whether the new operation gives rise to a commutative (unital) ring. Thus, let R be the set of arithmetic functions. Addition of arithmetic functions as defined above with additive identity 0 satisfies the axioms for the addition in a ring. Let us consider (R; +; 0; ∗). In order to check that ∗ defines a commutative unital ring, we need a prospective unit for ∗, we need to check ∗ is associative, and we want to check that ∗ distributes over +. Proposition 1.4. The function χ1 is a multiplicative unit for ∗, i.e., for any arithmetic function f, f ∗ χ1 = f = χ1 ∗ f. Proof. Since ∗ is commutative, it suffices to check one of these statements. In that case, consider: X n f(d)χ ( ): 1 d djn n n Since χ1( d ) = 0 unless d = 1, i.e., n = d, this sum collapses to give f(n), which is what we wanted to show. Proposition 1.5. The operation ∗ is associative, i.e., if f; g and h are three arithmetic functions, then (f ∗ g) ∗ h = f ∗ (g ∗ h). Proof. We just compute: X f ∗ g = f(a)g(b): ab=n Then, X (f ∗ g) ∗ h = f(a)g(b)h(c): abc=n Computing in the other direction is similar and yields the result. 3 1.1 The ring of arithmetic functions under convolution Proposition 1.6. The operation ∗ distributes over +, i.e., given three arithmetic functions f; g and h, f ∗ (g + h) = f ∗ g + f ∗ h. Proof. Again, we just compute: X X X X f∗(g+h) = f(a)(g+h)(b) = f(a)(g(b)+h(b)) = f(a)g(b)+ f(a)h(b) = f∗g+f∗h; ab=n ab=n ab=n ab=n as claimed. We can now give a very simple proof of Mobius¨ inversion. Theorem 1.7. If f is an arithmetic function, and F = f ∗ 1, then F ∗ µ = f. Proof. Indeed, F ∗ µ = (f ∗ 1) ∗ µ by definition. Then, by associativity of ∗, (f ∗ 1) ∗ µ = f ∗ (1 ∗ µ): Now, 1 ∗ µ = χ1 by the formula we proved last time, so f ∗ (1 ∗ µ) = f ∗ χ1 = f; where the last equality holds because χ1 is a multiplicative unit for ∗. Remark 1.8. It’s worth pointing out that this proof is not so different than the first proof we gave in class (the one in your book), but I think it is much easier to remember. Theorem 1.9. If f and g are multiplicative arithmetic functions, then f ∗ g is also multiplicative. Proof. Suppose m and n are coprime integers, then X mn (f ∗ g)(mn) = f(d)g( ): d djmn Now, if djmn, then d = d1d2 with d1jm and d2jn. In other words, X mn X mn f(d)g( ) = f(d1d2)g( ): d d1d2 djmn d1jm;d2jn Since f and g are multiplicative, and d1 and d2 are automatically coprime, X mn X m n f(d1d2)g( ) = f(d1)f(d2)g( )g( ); d1d2 d1 d2 d1jm;d2jn d1jm;d2jn the latter expression can be rewritten as a double sum giving exactly what we want. Corollary 1.10. If F = f ∗ 1 is multiplicative, then so is f. Proof. If F = f ∗ 1, then by Mobius¨ inversion F ∗ µ = f. Since µ is multiplicative, it follows from the preceding result that F ∗ µ is also multipliative, which is what we wanted to prove..

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    3 Page
  • File Size
    -

Download

Channel Download Status
Express Download Enable

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

  • Not to be reproduced or distributed without explicit permission.
  • Not used for commercial purposes outside of approved use cases.
  • Not used to infringe on the rights of the original creators.
  • If you believe any content infringes your copyright, please contact us immediately.

Support

For help with questions, suggestions, or problems, please contact us