= Solution
A local minimizer of a <differentiable function> on $\mathbb R^N$ necessarily satisfies $\boxed{\nabla f(x_*)=0}$. A sufficient condition for a strict local minimum is $\boxed{\nabla f(x_*)=0,\quad\nabla^2f(x_*)\text{ positive definite}}$. Indeed, the second-order <Taylor expansion> has a strictly positive quadratic term that dominates its $o(\|x-x_*\|^2)$ remainder near $x_*$.
For <Newton's method>, solve $H_ns_n=-g_n$ with $H_n=\nabla^2f(x_n)$, $g_n=\nabla f(x_n)$, and set $x_{n+1}=x_n+s_n$. A positive-definite Hessian makes this a <descent direction>; general Hessians may need modification or a separate <line search> for a globally useful method.
The sequence is q-quadratically convergent if there is $C<\infty$ such that $\|x_{n+1}-x_*\|\leq C\|x_n-x_*\|^2$ eventually. Sufficient local hypotheses are $\nabla f(x_*)=0$, a nonsingular $H_*=\nabla^2f(x_*)$, a <Lipschitz continuous> Hessian near $x_*$, and $x_0$ sufficiently close. For convergence to a local minimum take $H_*$ positive definite. Near $x_*$, inverse Hessians are uniformly bounded, and integration along the error segment gives
$$
x_{n+1}-x_*=H_n^{-1}\int_0^1[H_n-\nabla^2f(x_*+t(x_n-x_*))](x_n-x_*)\,dt.
$$
A Hessian Lipschitz constant $L_H$ yields $\|x_{n+1}-x_*\|\leq\tfrac12L_H\|H_n^{-1}\|\|x_n-x_*\|^2$. Choosing a sufficiently small invariant neighborhood proves <quadratic convergence>. Continuous second derivatives alone do not guarantee this quadratic rate; the Lipschitz hypothesis supplies the needed remainder bound.
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