Arithmetic logarithmic derivative

Definition

Considering a natural number's prime factorization

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

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

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

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

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

Arithmetic logarithmic derivative of zero

The arithmetic logarithmic derivative of a zero is undefined

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

Arithmetic logarithmic derivative of units

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

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

Arithmetic logarithmic derivative of primes

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

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

Properties

Arithmetic logarithmic derivative of a product

For any nonzero integer $\scriptstyle n\,$

$ld(-n)\equiv {\frac {(-n)'}{(-n)}}={\frac {-n'}{-n}}={\frac {n'}{n}}\equiv ld(n)\,$

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

The arithmetic logarithmic derivative of a product has the property

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

or

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

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

Thus

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

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

Arithmetic logarithmic derivative of powers

Also

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

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

Arithmetic logarithmic derivative of a quotient

The arithmetic logarithmic derivative of a quotient has the property

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

or

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

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

See also

• Arithmetic derivative