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# Arithmetic logarithmic derivative

(Redirected from Number logarithmic derivative)

## Definition

Considering a natural number's prime factorization

${\displaystyle n=\prod _{i=1}^{\omega (n)}{p_{i}}^{e_{i}},\,}$

where ${\displaystyle \scriptstyle \{p_{1},\,\ldots ,\,p_{\omega (n)}\}\,}$ are the distinct prime factors of ${\displaystyle \scriptstyle n\,}$, ω(n) is the number of distinct prime factors of ${\displaystyle \scriptstyle n\,}$ and ${\displaystyle \scriptstyle \{e_{1},\,\ldots ,\,e_{\omega (n)}\}\,}$ are positive integers,

the arithmetic logarithmic derivative of ${\displaystyle \scriptstyle n\,}$ is defined as

${\displaystyle ld(n)\equiv (\log n)'\equiv {\frac {n'}{n}}=\sum _{i=1}^{\omega (n)}{\frac {e_{i}}{p_{i}}},\,}$

where ${\displaystyle \scriptstyle n'\,}$ is the arithmetic derivative of ${\displaystyle \scriptstyle n\,}$.

### Arithmetic logarithmic derivative of zero

The arithmetic logarithmic derivative of a zero is undefined

${\displaystyle ld(0)\equiv {\frac {0'}{0}}={\frac {0}{0}}\,}$, which is undefined.

### Arithmetic logarithmic derivative of units

The arithmetic logarithmic derivative of a unit ${\displaystyle \scriptstyle u\,}$ is

${\displaystyle ld(u)\equiv {\frac {u'}{u}}={\frac {0}{u}}=0.\,}$

### Arithmetic logarithmic derivative of primes

The arithmetic logarithmic derivative of a prime ${\displaystyle \scriptstyle p\,}$ is

${\displaystyle ld(p)\equiv {\frac {p'}{p}}={\frac {1}{p}}.\,}$

## Properties

### Arithmetic logarithmic derivative of a product

For any nonzero integer ${\displaystyle \scriptstyle n\,}$

${\displaystyle ld(-n)\equiv {\frac {(-n)'}{(-n)}}={\frac {-n'}{-n}}={\frac {n'}{n}}\equiv ld(n)\,}$
${\displaystyle ld({\tfrac {1}{n}})\equiv {\frac {({\tfrac {1}{n}})'}{({\tfrac {1}{n}})}}=n\cdot {\bigg (}{\frac {-n'}{n^{2}}}{\bigg )}=-{\frac {n'}{n}}\equiv -ld(n).\,}$

The arithmetic logarithmic derivative of a product has the property

${\displaystyle ld(mn)\equiv {\frac {(mn)'}{mn}}={\frac {m'n+mn'}{mn}}={\frac {m'}{m}}+{\frac {n'}{n}}\equiv ld(m)+ld(n),\,}$

or

${\displaystyle {\frac {(mn)'}{mn}}={\frac {m'}{m}}+{\frac {n'}{n}}\,}$

where ${\displaystyle \scriptstyle m\,}$ and ${\displaystyle \scriptstyle n\,}$ are any nonzero integers.

Thus

${\displaystyle ld{\bigg (}\prod _{i=1}^{k}n_{i}{\bigg )}=\sum _{i=1}^{k}ld(n_{i})\,}$

where the ${\displaystyle \scriptstyle n_{i}\,}$ are any nonzero integers.

#### Arithmetic logarithmic derivative of powers

Also

${\displaystyle ld(n^{k})=k\cdot ld(n)\,}$

where ${\displaystyle \scriptstyle n\,}$ is any nonzero integer and ${\displaystyle \scriptstyle k\,}$ is any integer (for ${\displaystyle \scriptstyle k\,=\,0\,}$ we get ${\displaystyle \scriptstyle ld(1)\,=\,ld(n^{0})\,=\,0\cdot ld(n)\,=\,0\,}$, which is the wanted result.)

### Arithmetic logarithmic derivative of a quotient

The arithmetic logarithmic derivative of a quotient has the property

${\displaystyle ld{\bigg (}{\frac {n}{d}}{\bigg )}\equiv {\frac {({\tfrac {n}{d}})'}{\tfrac {n}{d}}}={\frac {\frac {n'd-d'n}{d^{2}}}{\tfrac {n}{d}}}={\frac {d}{n}}\cdot {\bigg (}{\frac {n'd-d'n}{d^{2}}}{\bigg )}={\frac {n'd-d'n}{nd}}={\frac {n'}{n}}-{\frac {d'}{d}}\equiv ld(n)-ld(d),\,}$

or

${\displaystyle {\frac {({\tfrac {n}{d}})'}{\tfrac {n}{d}}}={\frac {n'}{n}}-{\frac {d'}{d}}\,}$

where ${\displaystyle \scriptstyle n\,}$ and ${\displaystyle \scriptstyle d\,}$ are any nonzero integers.