I read a part of the wikipedia page of Newton's method where it is shown that the order of convergence for Newton's method of finding a root is quadratic, i.e. $$ |\varepsilon_{n+1}| = \dfrac{f''(\xi_n)}{2|f'(x_n)|} \cdot \varepsilon_n^2 \quad , $$ where $\varepsilon_n = \alpha - x_n$, $\alpha$ is the root, $x_n$ is the "nth guess" and $\xi_n$ is in between $x_n$ and $\alpha$. It is stated that the equation shows that
...the order of convergence is at least quadratic if the following conditions are satisfied:
- $f'(x) \neq 0$; for all $x \in I$ where $I$ is the interval $[\alpha - r, \alpha + r]$ for some $r \geq |\alpha - x_0|$;
- $f''(x)$ is continuous, for all $x \in I$;
- $x_0$ is sufficiently close to the root $\alpha$.
The term sufficiently close in this context means the following:
a. Taylor approximation is accurate enough such that we can ignore higher order terms;
b. $\dfrac{1}{2} \left| \dfrac{f''(x_n)}{f'(x_n)} \right| < C\left| \dfrac{f''(\alpha)}{f'(\alpha)} \right|$ for some $C < \infty$;
c. $C \left | \dfrac{f''(\alpha)}{f'(\alpha)} \right | \varepsilon_n < 1$,
for $n \in \mathbb{Z}$, $n \geq 0$ and $C$ satisfying condition b.
I wonder if condition 2. is required and in that case why?
The only hint at why this continuity of the second derivative is required is that the article claims that it ensures Taylor's theorem can be applied to write $f(x_n)$ as a Taylor series of order $1$ with the remainder in Lagrange form. However, the wikipedia page on Taylor's theorem does not require the second derivative be continuous, only that $f$ is $2$-times differentiable and some other conditions. So I don't understand where the requirement of a continuous second derivative may come from.