Mathematical Induction

Problem 159: Using mathematical induction, show that for $n > 1$

(1) $\begin{equation*} \frac{d^n}{dx^n} \ln x = (-1)^{n-1} \frac{(n-1)!}{x^n}. \end{equation*}$

Solution: Since we will use mathematical induction to prove (1), we need to prove the base case and the inductive case. That is,

Base Case: Let $n = 2$ . Then

(2) $\begin{align*} \frac{d^n}{dx^n} \ln x & = (-1)^{n-1} \frac{(n-1)!}{x^n} \\ \frac{d^2}{dx^2} \ln x & = (-1)^{2-1} \frac{(1)!}{x^2} \\ -\frac{1}{x^2} & = - \frac{1}{x^2}.\end{align*}$

Inductive Case: We assume that (1) holds for $n = k$ . We will show that it also holds for $n = k+1$ . That is,

(3) $\begin{align*} \frac{d^{k+1}}{dx^{k+1}} \ln x & = (-1)^{k+1-1} \frac{(k+1-1)!}{x^{k+1}} \\ & = \frac{(-1)^{k} k!}{x^{k+1}} \\ & = \frac{(-1)^{k}k (k-1)!}{x^k \cdot x} \\ & = - \frac{k}{x} \cdot \underbrace{\frac{(-1)^{k-1}(k-1)!}{x^k}}_{= \ (1)} \\ & = -\frac{k}{x} \cdot \frac{d^k}{dx^k} \ln x \\ \frac{1}{x} \frac{d^k}{dx^{k}} \ln x & = - \frac{k}{x} \cdot \frac{d^k}{dx^k}. \ln x\end{align*}$

It must be the case that $k = -1$ so that it proves (1).

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