Let $F(x):\mathbb{R}^n\rightarrow \mathbb{R}^n$. Newton's method is: $x_{k+1} := x_k + d_k$, where $d_k$ is computed to satisfy $F'(x_k)d_k = -F(x_k)$.
If the error at the current step is $e_k = x^* - x_k$, then why is $\|e_k\|$ proportional to $\|F(x_k)\|$?
It makes sense since $F(x_k)\rightarrow 0$ iff $\|e_k\|\rightarrow 0$, but I just can't see why they're proportional, something isn't clicking.
If $x_k$ is close to the solution $x^*$, the Jacobian of $F$ should not change very much and the linear approximation to $F$ about $x^*$ should be accurate, so $e_k$ should be approximately equal to $-DF(x^*)^{-1}F(x_k)$. Then $\frac{\| e_k \|}{\| F(x_k) \|}$ will be approximately bounded by the smallest and largest singular values of $DF(x^*)^{-1}$.