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Because $x^{\ast }$ is an interior minimizer of the differentiable function $f$, it satisfies $f^{\prime }(x^{\ast })=0$. Apply the Taylor theorem with Lagrange remainder to $f^{\prime }$ about $x_n$. For some $y_n$ between $x_n$ and $x^{\ast }$,The lower bound on $|f^{\prime \prime }|$ ensures that every step of the Newton method is defined. Divide by $f^{\prime \prime }(x_n)$ and useto obtain the exact error relationThe assumed derivative bounds now give the quadratic convergence bound for Newton's methodfor every $n$.

Newton's method for minimization usesFor a quantitative local bound, suppose on a convex neighbourhood containing the iterates thatwith $m>0$, and let $x^{\ast }$ be the minimizer. The integral form of the gradient and the Hessian Lipschitz bound give the quadratic convergence bound for Newton's methodIf $q=M\Vert x_0-x^{\ast }\Vert /(2m)<1$, induction yieldsSince $f(x)-f(x^{\ast })\le L\Vert x-x^{\ast }{\Vert }^2/2$,