Analytic Number Theory/Arithmetic functions

In this chapter, we shall set up the basic theory of arithmetic functions. This theory will be seen in action in later chapters, but in particular in chapter 9. Definitions 0 100% #FFFFFF Definition 2.1: An arithmetical function is a function . 0 100% #FFFFFF Definition 2.2 (important arithmetical functions): The Kronecker delta: Euler's totient function: Möbius' -function: The von Mangoldt function: The monomials: The number of distinct prime divisors: , The sum of prime factors with multiplicity: , The Liouville function: Arithmetic function Exercises Exercise 2.1.1: Compute , and . Exercise 2.1.2: Compute . Hint: . Exercise 2.1.3: Compute up to three decimal places. Hint: Use a Taylor expansion. Exercise 2.1.4: Prove that for each and . The convolution and the ring of arithmetic functions 0 100% #FFFFFF Definition 2.3: Let be arithmetical functions. Then the convolution of and is defined to be the function . Dirichlet convolution In the following theorem, we show that the arithmetical functions form an Abelian monoid, where the monoid operation is given by the convolution. Further, since the sum of two arithmetic functions is again an arithmetic function, the arithmetic functions form a commutative ring. In fact, as we shall also see, they form an integral domain. 0 100% #FFFFFF Theorem 2.4 (Abelian monoid properties of the arithmetical functions): The convolution is commutative, i. e. . The convolution is associative, i. e. . The function from definition 2.2 is an identity for the convolution, i. e. . Proof: 1.: , where is a bijection from the set of divisors of to itself. 2.: , where the last equality follows from the identity function being a bijection. But and hence associativity. 3.: 0 100% #FFFFFF Theorem 2.5: The ring of arithmetic functions is an integral domain. Proof: Let be arithmetic functions, and let be minimal such that , . Then . We shall now determine the units of the ring of arithmetic functions. 0 100% #FFFFFF Theorem 2.6: Let be an arithmetic function. Then is invertible (with respect to convolution) if and only if . Proof: Assume first . Then for any arithmetic function , . Assume now . Then , given by the recursive formula , , is an inverse (and thus the inverse) of , since and for inductively Exercises Exercise 2.2.1: Exercise 2.2.2: Multiplicative functions 0 100% #FFFFFF Definition 2.7: An arithmetical function is called multiplicative iff it satisfies , and . 0 100% #FFFFFF Theorem 2.8: Let be multiplicative arithmetical functions. Then is multiplicative. Proof: Let . Then , since the function is a bijection from the divisors of to the Cartesian product of the divisors of and the divisors of ; this is because multiplication is the inverse: , . To rigorously prove this actually is an exercise in itself. But due to the multiplicativity of and , . Furthermore, . Since is multiplicative, we conclude that the multiplicative functions form an Abelian submonoid of the arithmetic functions with convolution. Unfortunately, we do not have a subring since the sum of two multiplicative functions is never multiplicative (look at ). 0 100% #FFFFFF Theorem 2.9: Let be a multiplicative function such that converges absolutely. Then . Proof: Let be the ordered sequence of all prime numbers. For all we have due to the multiplicativity of . For each , we successively take , ..., and then . It follows from the definitions and the rule that the right hand side converges to . We claim that . Indeed, choose such that . Then by the fundamental theorem of arithmetic, there exists an and such that . Then we have by the triangle inequality for , and arbitrary that From this easily follows the claim. It is left to show that the product on the left is independent of the order of multiplication. But this is clear since if the sequence is enumerated differently, the argument works in just the same way and the left hand side remains the same. 0 100% #FFFFFF Definition 2.10: An arithmetical function is called strongly multiplicative iff it satisfies , and . Equivalently, a strongly multiplicative function is a monoid homomorphism . 0 100% #FFFFFF Theorem 2.11: Let be a strongly multiplicative function such that converges absolutely. Then . Proof: Due to theorem 2.9, we have . Due to strong multiplicativity and the geometric series, the latter expression equals . Exercises Exercise 2.3.1: Let be an arithmetic function such that for all , and let . Prove that the function is multiplicative. Bell series 0 100% #FFFFFF Definition 2.12: Let be an arithmetic function. Then for a prime the Bell series modulo is the formal power series . Examples 2.13: We shall here compute the Bell series for some important arithmetic functions. We note that in general for a completely multiplicative function , we have . In particular, in this case the Bell series defines a function. 1. The Kronecker delta: 2. Euler' totient function (we use lemma 9.?): 3. The Möbius function: 4. The von Mangoldt function: 5. The monomials: 6. The number of distinct prime divisors: 7. The number of prime divisors including multiplicity: 8. The Liouville function: 0 100% #FFFFFF Theorem 2.14 (compatibility of Bell series and convolution): Let arithmetic functions, and be a prime. Then . Proof: In case of multiplicativity, we have the following theorem: 0 100% #FFFFFF Theorem 2.15 (Uniqueness theorem): Let be multiplicative functions. Then . Proof: is pretty obvious; : as formal power series is equivalent to saying . If now , then due to the multiplicativity of and . In chapter 9, we will use Bell series to obtain equations for number-theoretic functions. Derivatives 0 100% #FFFFFF Definition 2.16: Let be an arithmetic function. Then the derivative of is defined to be the function . 0 100% #FFFFFF Theorem 2.17 (rules for the derivative): Let arithmetic functions. We have the following rules: if invertible, i.e. Note that is not the inverse function of (this wouldn't make much sense anyway since arithmetic can not be surjective, since is uncountable), but rather the convolution inverse. Proof: 1. is easily checked. 2.: 3. We have and . Hence, by 2. . Convolving with and using yields the desired formula. Note that a chain rule wouldn't make much sense, since arithmetic may map anywhere but to and thus doesn't make a lot of sense in general. Further reading terrytao.wordpress.com: Derived multiplicative functions