Loading [MathJax]/extensions/TeX/boldsymbol.js
\newcommand{\uex}{{u_{\small\mbox{e}}}}
\newcommand{\vex}{{v_{\small\mbox{e}}}}
\newcommand{\half}{\frac{1}{2}}
\newcommand{\tp}{\thinspace .}
\newcommand{\Oof}[1]{\mathcal{O}(#1)}
Taylor series inserted in the backward difference approximation
\begin{align*}
[D_t^-u]^n - u'(t_n) &= \frac{u(t_n) - u(t_{n-1})}{\Delta t} - u'(t_n)\\
&= \frac{u(t_n) - (u(t_n) - u'(t_n)\Delta t + {\half}u''(t_n)\Delta t^2 + \Oof{\Delta t^3} )}{\Delta t}\\
&\quad -u'(t_n)\\
&= -{\half}u''(t_n)\Delta t + \Oof{\Delta t^2} )
\end{align*}
Result:
\begin{equation}
R^n = -{\half}u''(t_n)\Delta t + \Oof{\Delta t^2} )
\tp
\end{equation}
The difference approximation is of
first order in \Delta t . It is exact for linear \uex .
