Given the error propagation of (one dimensional) Newton's method over interval $I$ $$ |\epsilon_{n+1}|\leq M\epsilon_{n}^{2} $$ with $$ M=\sup_{x\in I}\Big|\frac{f''(x)}{f'(x)}\Big| $$ I was wondering what the error propagation is in a multi variable setting for $x=(x_{1},...,x_{m})$ with $m$ dimensions?
In a one dimensional setting, Taylor's expansion implies for $f(a)=0$ and $\Delta x_{n}=a-x_{n}$ with $x_{n}\rightarrow a$ as $n\rightarrow\infty$, such that $$ 0=f(a)=f(x_{n})+f'(x_{n})\Delta x_{n}+\frac{1}{2}f''(\zeta_{n})\Delta x_{n}^{2} $$ with $\zeta_{n}$ between $x_{n}$ and $a$. Next, dividing both sides by $f'(x_{n})$ and filling in $x_{n+1}=x_{n}-\frac{f''(x_{n})}{f'(x_{n})}$ according to Newton's method, one finds $$ \epsilon_{n+1}=\frac{-f''(x_{n})}{2f'(x_{n})}\epsilon_{n} $$
In a multi variable setting, Newton's method states $$ x_{n+1}=x_{n}-H^{-1}(x_{n})g(x_{n}) $$ with $H(x_{n})$ the Hessian and $g(x_{n})=\nabla f(x_{n})$ the gradient of $f$.
Essentially, I was wondering if an above propagation of the error also exists in a multivariable setting as dividing both sides by the gradient is not possible, neither is taking the inverse of the gradient.