Arithmetic Functions

Arithmetic Function

An arithmetic function

f

is a function such that

dom (f) = ℕ_{1}

Divisor Counting Function

Let

n \in ℕ_{1}

and define

τ (n) : = | {d \in ℕ_{1} : d | n} |

Equation for the Divisor Counting Function

Suppose that

n \in ℕ_{2}

so that

n = \prod_{i \in ℕ_{1}} p_{i}^{α_{i}}

then

τ (n) = \prod_{i \in ℕ} (α_{i} + 1)

Divisor Sum Function

Let

n \in ℕ_{1}

and define

σ (n) = \sum {d \in ℕ_{1} : d | n}

the above notation is a sum fold

Equation for the Divisor Sum Function

Suppose that

n \in ℕ_{2}

so that

n = \prod_{i = 1}^{k} p_{i}^{α_{i}}

then

σ (n) = \prod_{i = 1}^{k} \frac{p_{i}^{α_{i}} - 1}{p_{i} - 1}

Multiplicative

Suppose that

f : ℕ_{1} \to C \subseteq ℂ

is arithmetic, then we say that

f

is multiplicative if for every coprime

n, m \in ℕ_{1}

we have

f (n m) = f (n) f (m)

Multiplicative Functions that map 1 to 0 Are the Constant 0 Function

Suppose that

f : ℕ_{1} \to C

is multiplicative then if

f (1) = 0

then we have

f (n) = 0

for every

n \in ℕ_{1}

Let

n \in ℕ_{1}

then

f (n) = f (1 \cdot n) = f (1) f (n) = 0

as needed.

Non-Zero Multiplicative Functions map 1 to 1

Suppose that

f : ℕ_{1} \to C

is non-zero multiplicative then

f (1) = 1

We know that

f (1 \cdot 1) = f (1) \cdot f (1)

so we have

f (1) = f (1) f (1)

since

f (1) \in C

f (1) = 0

then

f

would be the constant zero function which is a contradiction therefore

f (1) \neq 0

so that it has a multiplicative inverse which allows us to conclude that

f (1) = 1

by cancelling

f (1)

from both sides

Totally Multiplicative

Suppose that

f : ℕ_{1} \to C \subseteq ℂ

is arithmetic, then we say that

f

is totally multiplicative if for

n, m \in ℕ_{1}

we have

f (n m) = f (n) f (m)

The Base Function is Totally Multiplicative

As per title.

Notice that this is the same as multiplicative but without the coprime condition.

When the Constant Function is Multiplicative

The constant function is multiplicative if and only if

c \in {0, 1}

The Product of Two Multiplicative Functions is Multiplicative

Suppose that

f, g : ℕ_{1} \to C

be multiplicative then so is

f \cdot g

Nowhere Zero

We say that a function is nowhere zero if

0 \notin im (f)

The Quotient of Two Multiplicative Functions is Multiplicative

Suppose that

f, g : ℕ_{1} \to C

be multiplicative and that

g

is nowhere zero then

\frac{f}{g}

is multiplicative

Two Multiplicative Functions are Equal if they Agree on Prime Powers

Suppose that

f, g : ℕ_{1} \to C

are multiplicative such that

f (1) = g (1)

then if for any

p \in ℙ

and

k \in ℕ_{1}

we have

f (p^{k}) = g (p^{k})

then

f = g

Sum of Reciprocals of Divisors Equation

Show that

\frac{σ (n)}{n} = \sum_{d | n} \frac{1}{d}

Since both functions are multiplicative, then we confirm that the align at

1

which they do. Then let

p \in ℙ

and

k \in ℕ_{1}

then we have

\begin{matrix} \sum_{d | p^{k}} \frac{1}{d} & = \sum_{j = 0}^{k} \frac{1}{p^{j}} \\ = \frac{1}{p^{k}} (\sum_{j = 0}^{k} p^{k}) \\ = \frac{σ (p^{k})}{p^{k}} \end{matrix}

therefore the two functions are equal

The Sum of a Multiplicative Function over Divisors of a Number is Multiplicative

Let

f : ℕ_{1} \to C

be multiplicative and set

F (n) = \sum_{d | n} f (d)

is multiplicative

Dirichlet Convolution

Suppose

f, g : ℕ \to C

are arithmetic then the Dirichlet Convolution of

f

and

g

denoted

f * g

is defined as

(f * g) (n) : = \sum_{d | n} f (d) g (\frac{n}{d})

The Dirichlet Convolution of two Multiplicative Functions is Multiplicative

Let

f, g : ℕ_{1} \to C

be multiplicative, then

f * g

is multiplicative

The Dirichlet Convolution Identity

Suppose we define

ϵ : ℕ_{1} \to ℂ

so that

ϵ (n) : = {\begin{matrix} 1 & if n = 1 \\ 0 & otherwise \end{matrix}

The Dirichlet Identity is Completely Multiplicative

As per title.

The Dirichlet Identity is an Identity

Suppose that

f

is an arithmetic function, then

f * ϵ = ϵ * f = f