The cubic-regularized quadratic model

Let $f:{\mathbb{R}}^n\to \mathbb{R}$ be twice differentiable at $x$, let $g:=\mathrm{\nabla }f(x)$, let $B\in {\mathbb{R}}^{n\times n}$ be symmetric, and let $\sigma >0$. The cubic-regularized quadratic model is

$m_{\sigma }(s):=f(x)+⟨g,s⟩+\frac{1}{2}{s}^{T}Bs+\frac{\sigma }{3}\mathrm{\parallel }s{\mathrm{\parallel }}_2^3,s\in {\mathbb{R}}^n,$

using the Euclidean conventions of The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$, the gradient and Hessian conventions of The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case and The Hessian matrix and critical points of a scalar field, and the induced matrix norm of The matrix norm induced by a published vector p-norm. Its predicted reduction is $m_{\sigma }(0)-m_{\sigma }(s)$. Define the actual reduction by $\mathrm{ared}(s):=f(x)-f(x+s)$ and, when the predicted reduction is positive, define the ratio ${\rho }_{\sigma }(s):=\frac{\mathrm{ared}(s)}{m_{\sigma }(0)-m_{\sigma }(s)}\mathrm{.}$ When $B={\mathrm{\nabla }}^{2}f(x)$, this is the exact-Hessian ARC model; approximation error in a different $B$ requires separate bounds. The definition alone gives no convergence or evaluation-complexity guarantee.