$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\Aex}{{A_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\tp}{\thinspace .}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
\newcommand{\x}{\boldsymbol{x}}
\newcommand{\X}{\boldsymbol{X}}
\renewcommand{\u}{\boldsymbol{u}}
\renewcommand{\v}{\boldsymbol{v}}
\newcommand{\e}{\boldsymbol{e}}
\newcommand{\f}{\boldsymbol{f}}
\newcommand{\dfc}{\alpha} % diffusion coefficient
\newcommand{\Ix}{\mathcal{I}_x}
\newcommand{\Iy}{\mathcal{I}_y}
\newcommand{\Iz}{\mathcal{I}_z}
\newcommand{\If}{\mathcal{I}_s} % for FEM
\newcommand{\Ifd}{{I_d}} % for FEM
\newcommand{\Ifb}{{I_b}} % for FEM
\newcommand{\sequencei}[1]{\left\{ {#1}_i \right\}_{i\in\If}}
\newcommand{\basphi}{\varphi}
\newcommand{\baspsi}{\psi}
\newcommand{\refphi}{\tilde\basphi}
\newcommand{\psib}{\boldsymbol{\psi}}
\newcommand{\sinL}[1]{\sin\left((#1+1)\pi\frac{x}{L}\right)}
\newcommand{\xno}[1]{x_{#1}}
\newcommand{\Xno}[1]{X_{(#1)}}
\newcommand{\xdno}[1]{\boldsymbol{x}_{#1}}
\newcommand{\dX}{\, \mathrm{d}X}
\newcommand{\dx}{\, \mathrm{d}x}
\newcommand{\ds}{\, \mathrm{d}s}
$$
The least squares method
Idea: find \( c_0,\ldots,c_N \) such that \( E= ||\e||^2 \) is minimized, \( \e=\f-\u \).
$$
\begin{align*}
E(c_0,\ldots,c_N) &= (\e,\e) = (\f -\sum_jc_j\psib_j,\f -\sum_jc_j\psib_j)
\nonumber\\
&= (\f,\f) - 2\sum_{j=0}^Nc_j(\f,\psib_j) +
\sum_{p=0}^N\sum_{q=0}^N c_pc_q(\psib_p,\psib_q)
\end{align*}
$$
$$
\begin{equation*}
\frac{\partial E}{\partial c_i} = 0,\quad i=0,\ldots,N
\end{equation*}
$$
After some work we end up with a linear system
$$
\begin{align}
\sum_{j=0}^N A_{i,j}c_j &= b_i,\quad i=0,\ldots,N\\
A_{i,j} &= (\psib_i,\psib_j)\\
b_i &= (\psib_i, \f)
\end{align}
$$
