# Notes on Chapter 2: Arithmetical Functions and Dirichlet Multiplication

By Susam Pal on 26 Mar 2021

## § 2.11: The inverse of a completely multiplicative function

### Completely Multiplicative Function

$f(mn) = f(m) f(n) \text{ for all } m, n.$

### Identity function (Dirichlet product)

$I(n) = \begin{cases} 1 & \text{ if } n = 1, \\ 0 & \text{ if } n > 1. \end{cases}$

\begin{align*} f(n)I(n) & = \begin{cases} 1 \cdot 1 & \text{ if } n = 1, \\ f(n) \cdot 0 & \text{ if } n > 1. \end{cases} \\ & = \begin{cases} 1 & \text{ if } n = 1, \\ 0 & \text{ if } n > 1. \end{cases} \\ & = I(n). \end{align*}

### Möbius function for prime powers

$\mu(1) = 1, \qquad \mu(p) = -1, \qquad \mu(p^2) = \mu(p^3) = \dots = 0.$

### Third equation in the proof of Theorem 2.17

\begin{align*} \sum_{d \mid p^a} \mu(d) f(d) f\left(\frac{p^a}{d}\right) & = \sum_{d = 1, p, p^2, \dots, p^a} \mu(d) f(d) f\left(\frac{p^a}{d}\right) \\ & = \begin{aligned}[t] & \mu(1) f(1) f\left( \frac{p^a}{1} \right) + \mu(p) f(p) f\left( \frac{p^a}{p} \right) \\ & + \underbrace{\mu(p^2) f(p^2) f\left( \frac{p^a}{p^2} \right) + \dots + \mu(p^a) f(p^a) f\left( \frac{p^a}{p^a} \right)}_{=\,0} \end{aligned} \\ & = \mu(1) f(1) f(p^a) + \mu(p) f(p) f(p^{a - 1}) \\ & = f(p^a) - f(p) f(p^{a - 1}). \end{align*}

### Final step in the proof of Theorem 2.17

\begin{align*} f(p^a) & = f(p)f(p^{a - 1}) \\ & = f(p)f(p)f(p^{a - 2}) \\ & = \dots \\ & = \underbrace{f(p)f(p)f(p) \dots f(p)}_{a \text{ times}} \\ & = \left( f(p) \right)^a. \end{align*}

### How the completely multiplicative property works for prime powers

$f(mn) = f(m)f(n) \text{ whenever } (m, n) = 1.$

$f(p_1^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k}) = f(p_1^{\alpha_1}) f(p_2^{\alpha_2}) \dots f(p_k^{\alpha_k}).$

### Euler totient function as Dirichlet product

$\varphi(n) = \sum_{d \mid n} \mu(d) \frac{n}{d} = \sum_{d \mid n} \mu(d) N\left(\frac{n}{d}\right) = (\mu * N)(n).$

### Proof of Theorem 2.18

Let $$f$$ be multiplicative. We want to show that $\sum_{d \mid n} \mu(d) f(d) = \prod_{p \mid n} (1 - f(p)).$ Note the following: $g(n) = \sum_{d \mid n} \mu(d) f(d) = \sum_{d \mid n} (\mu f) (d) u\left( \frac{n}{d} \right) = (\mu f) * u.$ The functions $$\mu$$ and $$f$$ are multiplicative. Thus $$\mu f$$ is multiplicative. Thus $$(\mu f) * u$$ is multiplicative. Therefore $g(n) = g(p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}) = g(p_1^{a_1}) g(p_2^{a_2}) \dots g(p_k^{a_k}).$ But \begin{align*} g(p_i^{a_i}) & = \sum_{d \mid p_i^{a_i}} \mu(d) f(d) \\ & = \mu(1) f(1) + \mu(p_i) f(p_i) + \underbrace{\mu(p_i^2) f(p_i^2) + \dots + \mu(p_i^{a_i}) f(p_i^{a_i})}_{=\,0} \\ & = 1 - f(p). \end{align*} From the two equations above, we get \begin{align*} g(n) & = g(p_1^{a_1}) g(p_2^{a_2}) \dots g(p_k^{a_k}) \\ & = (1 - f(p_1)) (1 - f(p_2)) \dots (1 - f(p_k)) \\ & = \prod_{p \mid n} (1 - f(p)). \end{align*}

## § 2.15: Formal power series

### Product of formal power series

\begin{align*} A(x)B(x) & = \left( \sum_{n=0}^{\infty} a(n) x^n \right) \left( \sum_{n=0}^{\infty} b(n) x^n \right) \\ & = \left( a(0) + a(1)x + a(2)x^2 + \dots \right) \left( b(0) + b(1)x + b(2)x^2 + \dots \right) \\ & = a(0)b(0) + \Bigl( a(0)b(1) + a(1)b(0) \Bigr) x + \Bigl( a(0)b(2) + a(1)b(1) + a(2)b(0) \Bigr) x^2 + \dots \\ & = \sum_{k=0}^0 a(k)b(n - k) + \sum_{k=0}^1 a(k)b(1 - k)x + \sum_{k=0}^2 a(k)b(2 - k)x^2 + \dots \\ & = \sum_{n=0}^{\infty} \sum_{k=0}^n a(k)b(n - k). \end{align*}

### Commutativity of product of formal power series

$A(x)B(x) = \sum_{n=0}^{\infty} \underbrace{\left\{ \sum_{k=0}^{n} a(k) b(n - k) \right\}}_{c(n)} x^n.$ $B(x)A(x) = \sum_{n=0}^{\infty} \underbrace{\left\{ \sum_{k=0}^{n} a(n - k) b(k) \right\}}_{c'(n)} x^n.$ $c(3) = a(0)b(3) + a(1)b(2) + a(2)b(1) + a(3)b(0).$ $c'(3) = a(3)b(0) + a(2)b(1) + a(1)b(2) + a(0)b(3).$

### Distributivity of multiplication over addition in formal power series

$A(x)\Bigl(B(x) + C(x)\Bigr) = A(x)B(x) + A(x)C(x).$ $\Bigl(B(x) + C(x)\Bigr)A(x) = B(x)A(x) + C(x)A(x).$ \begin{align*} A(x)\Bigl(B(x) + C(x)\Bigr) & = \left( \sum_{n=0}^{\infty} a(n) x^n \right) \left( \sum_{n=0}^{\infty} \Bigl( b(n) + c(n) \Bigr) x^n \right) \\ & = \sum_{n=0}^{\infty} \Bigl\{ \sum_{k=0}^{n} a(k) \Bigl( b(n - k) + c(n - k) \Bigr) \Bigr\} x^n. \end{align*} $A(x)B(x) + A(x)C(x) = \sum_{n=0}^{\infty} \sum_{k=0}^n a(k) b(n - k) x^n + \sum_{n=0}^{\infty} \sum_{k=0}^n a(k) c(n - k) x^n.$

### Determining coefficients of inverse of power series

\begin{align*} A(x)B(x) & = \sum_{n=0}^{\infty} \Bigl( \sum_{k=0}^{n} a(k) b(n - k) \Bigr) x^n \\ & = \Bigl( a(0) b(0) \Bigr) x^0 + \Bigl( a(0) b(1) + a(1) b(0) \Bigr) x^1 + \Bigl( a(0) b(2) + a(1) b(1) + a(2) b(0) \Bigr) x^2 + \dots \\ & = 1. \end{align*}

### Inverse of geometric series

\begin{align*} A(x) & = 1 + ax + (ax)^2 + (ax)^3 + \dots, \\ B(x) & = 1 - ax. \end{align*} \begin{align*} A(x) B(x) & = \Bigl( 1 + ax + (ax)^2 + (ax)^3 + \dots \Bigr) (1 - ax) \\ & = \Bigl( 1 + ax + (ax)^2 + (ax)^3 + \dots \Bigr) - \Bigl( (ax) - (ax)^2 - (ax)^3 - \dots \Bigr) = 1. \end{align*}

## § 2.16: The Bell series of an arithmetical function

### Bell series of $$f$$ modulo $$p$$

$f_p(x) = \sum_{n=0}^{\infty} f(p^n) x^n = f(1) + f(p) x + f(p^2) x^2 + f(p^3) x^3 + \dots$

### Theorem 2.24: Uniqueness theorem

\begin{align*} f(n) & = f(p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}) = f(p_1^{a_1}) f(p_2^{a_2}) \dots f(p_k^{a_k}), \\ \\ g(n) & = g(p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}) = g(p_1^{a_1}) g(p_2^{a_2}) \dots g(p_k^{a_k}). \\ \end{align*}

### Example 1: MÃ¶bius function

\begin{align*} \mu_p(x) = \sum_{n=0}^{\infty} \mu(p^n) x^n & = \mu(1) + \mu(p) x + \mu(p^2) x^2 + \mu(p^3) x^3 + \dots \\ & = 1 - x + 0 + 0 + \dots \\ & = 1 - x. \end{align*}

## § 2.17: Bell series and Dirichlet multiplication

### Power series multiplication

$A(x) = \sum_{n=0}^{\infty} a(n) x^n, \quad B(x) = \sum_{n=0}^{\infty} b(n) x^n, \quad A(x) B(x) = \sum_{n=0}^{\infty} \underbrace{\sum_{k=0}^n a(k) b(n - k)}_{c(n)} x^n.$

### Relationship with Dirichlet multiplication

$(f * g)_p(x) = f_p(x) g_p(x).$ $f_p(x) = \sum_{n=0}^{\infty} f(p^n) x^n, \quad g_p(x) = \sum_{n=0}^{\infty} g(p^n) x^n, \quad f_p(x) g_p(x) = \sum_{n=0}^{\infty} \sum_{k=0}^n f(p^k) g(p^{n-k}) x^n.$ $h = f * g = \sum_{d \mid n} f(d) g\left( \frac{n}{d} \right).$ $h_p(x) = \sum_{n=0}^{\infty} h(p^n) x^n = \sum_{n=0}^{\infty} \sum_{d \mid p^n} f(d) g\left( \frac{p^n}{d} \right) x^n = \sum_{n=0}^{\infty} \sum_{k=0}^{n} f(p^k) g(p^{n-k}) x^n.$

Some steps of Example 1: $I(n)= \mu^2(n) * \lambda(n) \implies I_p(x) = \mu_p^2(x) \lambda_p(x)$ $I_p(x) = \mu_p^2(x) \lambda_p(x) \iff 1 = \mu_p^2(x) \cdot \frac{1}{1 + x} \iff \mu_p^2(x) = 1 + x.$

Some steps of Example 2: \begin{align*} \frac{1}{1 - p^{\alpha}} \cdot \frac{1}{1 - x} & = \frac{1}{1 - x - p^{\alpha}x + p^{\alpha}x^2} \\ & = \frac{1}{1 - (1 + p^{\alpha})x + p^{\alpha}x^2} \\ & = \frac{1}{1 - \sigma_{\alpha}(p)x + p^{\alpha}x^2}. \end{align*} Note that $$\sigma_{\alpha}(n) = \sum_{d\,\mid\,n} d^{\alpha},$$ so $\sigma_{\alpha}(p) = \sum_{d\,\mid\,p} d^{\alpha} = 1^{\alpha} + p^{\alpha} = 1 + p^{\alpha}.$

Some steps of Example 3: Showing That $$f(n) = 2^{\nu(n)}$$ is multiplicative: $f(n) = 2^{\nu(n)}.$ $f(p_1^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k}) = 2^{\nu(p_1^{\alpha_1} p_2^{\alpha_2} \dots p_k^{\alpha_k})} = 2^k.$ $f(p_1^{\alpha_1}) f(p_2^{\alpha_2}) \dots f(p_k^{\alpha_k}) = 2^{\nu(p_1^{\alpha_1})} 2^{\nu(p_2^{\alpha_2})} \dots 2^{\nu(p_k^{\alpha_k})} = \underbrace{2 \cdot 2 \cdot \dots \cdot 2}_{k \text{ times}}. = 2^k.$

## § 2.18: Derivatives of arithmetical functions

### Ordinary derivative

$(f + g)' = f' + g'.$ $(fg)' = f'g + fg'.$ $\left( f^{-1} \right)' = \frac{-f'}{f^2} = -f' \cdot (f \cdot f)^{-1}.$

### Similarities in special derivative

$f'(n) = f(n) \log n.$ $(f + g)' = f' + g'.$ $(f * g)' = f' * g + f * g'.$ $\left( f^{-1} \right)' = -f' * (f * f)^{-1}.$