Nth Derivative of Natural Logarithm by Reciprocal
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Theorem
- $\dfrac {\d^n} {\d x^n} \dfrac {\ln x} x = \paren {-1}^{n + 1} n! \dfrac {H_n - \ln x} {x^{n + 1} }$
where $H_n$ denotes the $n$th harmonic number:
- $H_n = \ds \sum_{r \mathop = 1}^n \dfrac 1 r = 1 + \dfrac 1 2 + \dfrac 1 3 + \cdots + \dfrac 1 r$
Proof
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
- $\dfrac {\d^n} {\d x^n} \dfrac {\ln x} x = \paren {-1}^{n + 1} n! \dfrac {H_n - \ln x} {x^{n + 1} }$
Basis for the Induction
$\map P 1$ is the case:
- $\dfrac \d {\d x} \dfrac {\ln x} x = \paren {-1}^n n! \dfrac {H_n - \ln x} {x^{n + 1} }$
\(\ds \dfrac \d {\d x} \dfrac {\ln x} x\) | \(=\) | \(\ds \dfrac {1 - \ln x} {x^2}\) | Derivative of Logarithm over Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^2 \times 1! \times \dfrac {H_1 - \ln x} {x^{1 + 1} }\) |
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
- $\dfrac {\d^k} {\d x^k} \dfrac {\ln x} x = \paren {-1}^{k + 1} k! \dfrac {H_k - \ln x} {x^{k + 1} }$
from which it is to be shown that:
- $\dfrac {\d^{k + 1} } {\d x^{k + 1} } \dfrac {\ln x} x = \paren {-1}^{k + 2} \paren {k + 1}! \dfrac {H_{k + 1} - \ln x} {x^{k + 2} }$
Induction Step
This is the induction step:
\(\ds \dfrac {\d^{k + 1} } {\d x^{k + 1} } \dfrac {\ln x} x\) | \(=\) | \(\ds \dfrac \d {\d x} \paren {-1}^{k + 1} k! \dfrac {H_k - \ln x} {x^{k + 1} }\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{k + 1} k! \map {\dfrac \d {\d x} } {\dfrac {H_k} {x^{k + 1} } - \dfrac {\ln x} {x^{k + 1} } }\) | Linear Combination of Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{k + 1} k! \paren {\dfrac {-\paren {k + 1} H_k} {x^{k + 2} } - \dfrac {1 - \paren {k + 1} \ln x} {x^{k + 2} } }\) | Derivative of Power, Derivative of Logarithm over Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{k + 2} \paren {k + 1}! \paren {\dfrac {H_k} {x^{k + 2} } + \dfrac 1 {\paren {k + 1} x^{k + 2} } - \dfrac {\ln x} {x^{k + 2} } }\) | extracting $-\paren {k + 1}$ as a factor | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{k + 2} \paren {k + 1}! \paren {\paren {H_k + \dfrac 1 {\paren {k + 1} } } \dfrac 1 {x^{k + 2} } - \dfrac {\ln x} {x^{k + 2} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{k + 2} \paren {k + 1}! \paren {\dfrac {H_{k + 1} } {x^{k + 2} } - \dfrac {\ln x} {x^{k + 2} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^{k + 2} \paren {k + 1}! \dfrac {H_{k + 1} - \ln x} {x^{k + 2} }\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \Z_{>0}: \dfrac {\d^n} {\d x^n} \dfrac {\ln x} x = \paren {-1}^{n + 1} n! \dfrac {H_n - \ln x} {x^{n + 1} }$
$\blacksquare$