$$
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\Aex}{{A_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\tp}{\thinspace .}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
\newcommand{\x}{\boldsymbol{x}}
\newcommand{\X}{\boldsymbol{X}}
\renewcommand{\u}{\boldsymbol{u}}
\renewcommand{\v}{\boldsymbol{v}}
\newcommand{\e}{\boldsymbol{e}}
\newcommand{\f}{\boldsymbol{f}}
\newcommand{\dfc}{\alpha} % diffusion coefficient
\newcommand{\Ix}{\mathcal{I}_x}
\newcommand{\Iy}{\mathcal{I}_y}
\newcommand{\Iz}{\mathcal{I}_z}
\newcommand{\If}{\mathcal{I}_s} % for FEM
\newcommand{\Ifd}{{I_d}} % for FEM
\newcommand{\Ifb}{{I_b}} % for FEM
\newcommand{\sequencei}[1]{\left\{ {#1}_i \right\}_{i\in\If}}
\newcommand{\basphi}{\varphi}
\newcommand{\baspsi}{\psi}
\newcommand{\refphi}{\tilde\basphi}
\newcommand{\psib}{\boldsymbol{\psi}}
\newcommand{\sinL}[1]{\sin\left((#1+1)\pi\frac{x}{L}\right)}
\newcommand{\xno}[1]{x_{#1}}
\newcommand{\Xno}[1]{X_{(#1)}}
\newcommand{\xdno}[1]{\boldsymbol{x}_{#1}}
\newcommand{\dX}{\, \mathrm{d}X}
\newcommand{\dx}{\, \mathrm{d}x}
\newcommand{\ds}{\, \mathrm{d}s}
$$
The least squares method; calculations
$$
\begin{align*}
E(c_0) &= (\e,\e) = (\f - \u, (\f - \u) = (\f - c_0\psib_0, \f - c_0\psib_0)\\
&= (\f,\f) - 2c_0(\f,\psib_0) + c_0^2(\psib_0,\psib_0)
\end{align*}
$$
$$
\begin{equation}
\frac{\partial E}{\partial c_0} = -2(\f,\psib_0) + 2c_0 (\psib_0,\psib_0) = 0
\tag{1}
\end{equation}
$$
$$
c_0 = \frac{(\f,\psib_0)}{(\psib_0,\psib_0)} = \frac{3a + 5b}{a^2 + b^2}
$$
Observation to be used later: the vanishing derivative (1)
can be alternatively written as
$$ (\e, \psib_0) = 0 $$
