$f(x) = \tan^{-1}(x)$ || Is there any easy method to show that the sequence [$f^n(0)$] is unbounded?

Question

$f(x) = \tan^{-1}(x)$

$f^n(0)$ denotes the $n^{th}$ derivative of $f(x)$ at the point $0$.

Is there any easy method to show that the sequence [$f^n(0)$] is unbounded?

Answer 1

Direct method [no Taylor series here].

First we know that $$ f’(x) = \frac 1 {1+x^2} \implies (1 + x^2) f’(x) = 1. $$ Now take the $n$-th derivative: by the Leibniz rule, $$ (1 + x^2)f^{(n+1)}(x) + 2nx f^{(n)}(x) + n(n-1) f^{(n-1)}(x) = 0. $$ Plug $x = 0$ into the equation above, we have $$ f^{(n+1)} (0) + n(n-1) f^{(n-1)} (0) = 0. $$ Since $f^{(0)} (0)= 0, f^{(1)} (0) = 1$, $f^{(2n)} (0) = 0$, and $$ f^{(2n+1)}(0) = - 2n (2n-1) f^{(2n-1)}(0) = 2n(2n-1)(2n-2)(2n-3)f^{(2n-3)}(0) = \cdots = (2n)! (-1)^n. $$

This method is not as quick as the one using Taylor series, but feasible.