$$
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$$
Deriving an exact numerical solution; calculations (1)
$$
u^n = IA^n = I\exp{(\tilde\omega \Delta t\, n)}=I\exp{(\tilde\omega t)} =
I\cos (\tilde\omega t) + iI\sin(\tilde \omega t)
\tp
$$
$$
\begin{align*}
[D_tD_t u]^n &= \frac{u^{n+1} - 2u^n + u^{n-1}}{\Delta t^2}\\
&= I\frac{A^{n+1} - 2A^n + A^{n-1}}{\Delta t^2}\\
&= I\frac{\exp{(i\tilde\omega(t+\Delta t))} - 2\exp{(i\tilde\omega t)} + \exp{(i\tilde\omega(t-\Delta t))}}{\Delta t^2}\\
&= I\exp{(i\tilde\omega t)}\frac{1}{\Delta t^2}\left(\exp{(i\tilde\omega(\Delta t))} + \exp{(i\tilde\omega(-\Delta t))} - 2\right)\\
&= I\exp{(i\tilde\omega t)}\frac{2}{\Delta t^2}\left(\cosh(i\tilde\omega\Delta t) -1 \right)\\
&= I\exp{(i\tilde\omega t)}\frac{2}{\Delta t^2}\left(\cos(\tilde\omega\Delta t) -1 \right)\\
&= -I\exp{(i\tilde\omega t)}\frac{4}{\Delta t^2}\sin^2(\frac{\tilde\omega\Delta t}{2})
\end{align*}
$$