$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\tp}{\thinspace .}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
\newcommand{\x}{\boldsymbol{x}}
\newcommand{\dfc}{\alpha} % diffusion coefficient
\newcommand{\Ix}{\mathcal{I}_x}
\newcommand{\Iy}{\mathcal{I}_y}
\newcommand{\If}{\mathcal{I}_s} % for FEM
\newcommand{\Ifd}{{I_d}} % for FEM
\newcommand{\basphi}{\varphi}
\newcommand{\baspsi}{\psi}
\newcommand{\refphi}{\tilde\basphi}
\newcommand{\xno}[1]{x_{#1}}
\newcommand{\dX}{\, \mathrm{d}X}
\newcommand{\dx}{\, \mathrm{d}x}
\newcommand{\ds}{\, \mathrm{d}s}
$$
Newton's method
Write the nonlinear algebraic equation as
$$ F(u) = 0 $$
Newton's method: linearize \( F(u) \) by two terms from the Taylor series,
$$
\begin{align*}
F(u) &= F(u^{-}) + F^{\prime}(u^{-})(u - u^{-}) + {\half}F^{\prime\prime}(u^{-})(u-u^{-})^2
+\cdots\\
& \approx F(u^{-}) + F^{\prime}(u^{-})(u - u^{-}) \equiv \hat F(u)
\end{align*}
$$
The linear equation \( \hat F(u)=0 \) has the solution
$$ u = u^{-} - \frac{F(u^{-})}{F^{\prime}(u^{-})}$$
Note that \( \hat F \) in Picard and Newton are different!
