Finite difference : relationship involving gamma Given the following PDE, 
$$
\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial^2V}{\partial S^2}=0
$$
and its finite difference approximation, 
$$
\frac{V_n^{m+1}-V_n^m}{\Delta t} + \frac{1}{2}\sigma^2S_n^2\Gamma_n^m=0 \qquad (*)
$$
where
$$
\Gamma_n^m=\frac{V_{n+1}^{m}-2V_n^m + V_{n-1}^m}{\Delta S^2}
$$
How do i show by considering suitable linear combination of (*) evaluated at mesh points $S_{n-1}$, $S_n$, $S_{n+1}$ that, 
$$
\frac{\Gamma_n^{m+1}-\Gamma_n^m}{\Delta t} + \frac{1}{2}\sigma^2S_n^2\frac{\Gamma_{n+1}^m-2\Gamma_n^m+\Gamma_{n-1}^m}{\Delta S^2}+2\sigma^2S_n\frac{\Gamma_{n+1}^m-\Gamma_{n-1}^m}{2\Delta S}+\sigma^2\frac{\Gamma_{n+1}^m+\Gamma_{n-1}^m}{2}=0
$$
 A: Basically we need to show that
$$
D (S_n^2 \Gamma_n) = S_n^2 D \Gamma_n + 4S_n \frac{\Gamma_{n+1} - \Gamma_{n-1}}{2\Delta S} + \Gamma_{n+1} + \Gamma_{n-1}
$$
where $D$ is second order finite difference derivative operator
$$
D u_n \equiv \frac{u_{n+1} -2 u_n+ u_{n-1}}{\Delta S^2}
$$
Note, that $D$ can be written as
$$
D = D_+ D_- = D_- D_+
$$
where
$$
D_+ u_n = \frac{u_{n+1} - u_n}{\Delta S}\\
D_- u_n = \frac{u_{n} - u_{n-1}}{\Delta S}\\
$$
are one-sided finite difference first order derivative operators. Really,
$$
D_+D_- u_n = D_+\left( \frac{u_{n} - u_{n-1}}{\Delta S}\right) = 
\frac{u_{n+1} - u_{n}}{\Delta S^2} - \frac{u_{n} - u_{n-1}}{\Delta S^2}
= D u_n.
$$
Same for the inverse order.
Next,
$$
D_{\pm} (a_n b_n) = \pm\frac{a_{n\pm 1}b_{n \pm 1} -a_nb_n}{\Delta S}
= \pm\frac{a_{n\pm 1}b_{n \pm 1} -a_{n\pm 1}b_n+a_{n\pm 1}b_n-a_nb_n}{\Delta S} =
a_{n \pm 1} D_\pm b_n + b_n D_\pm a_n.
$$
Thus 
$$D_\pm S_n^2 = S_{n \pm 1} D_\pm S_n + S_n D_\pm S_n = S_n + S_{n \pm 1}\\
D_\pm \Gamma_n S_n^2 = \Gamma_{n \pm 1} (S_n + S_{n\pm 1}) + S_n^2 D_\pm \Gamma_n.
$$
This far we found that 
$$
D_+ \Gamma_n S_n^2 = \Gamma_{n + 1} (S_n + S_{n+1}) + S_n^2 D_+ \Gamma_n.
$$
Let's apply $D_-$ to that
$$
D_- D_+ \Gamma_n S_n^2 = D_-\Gamma_{n + 1} (S_n + S_{n+1})
+ D_- S_n^2 D_+ \Gamma_n.
$$
Starting with the $D_- S_n^2 D_+ \Gamma_n$ we have
$$
D_- S_n^2 D_+ \Gamma_n = S_{n-1}^2 D_-D_+ \Gamma_n + (S_n + S_{n-1})D_+\Gamma_n = 
S_{n-1}^2 D \Gamma_n + (S_n + S_{n-1})D_+ \Gamma_n.
$$
Next
$$
D_-\Gamma_{n + 1} (S_n + S_{n+1}) = \Gamma_n D_-(S_n+S_{n+1}) + (S_n+S_{n+2}) D_-\Gamma_{n+1} =
 2\Gamma_n + (S_n + S_{n+1}) D_+\Gamma_n.
$$
So far we have
$$
D S_n^2 \Gamma_n = 2\Gamma_n + (S_{n-1} + 2S_n + S_{n+1}) D_+ \Gamma_n + S_{n-1}^2 D\Gamma_n.
$$
Computing the same expression by reversing the $D_+D_-$ order, we have
$$
D S_n^2 \Gamma_n = 2\Gamma_n + (S_{n-1} + 2S_n + S_{n+1}) D_- \Gamma_n + S_{n+1}^2 D\Gamma_n.
$$
Also, since $S_{n\pm 1} = S_{n}\pm \Delta S$,
$$
D S_n^2 \Gamma_n = 2\Gamma_n + 4S_n D_+ \Gamma_n + (S_{n}-\Delta S)^2 D\Gamma_n\\
D S_n^2 \Gamma_n = 2\Gamma_n + 4S_n D_- \Gamma_n + (S_{n}+\Delta S)^2 D\Gamma_n.
$$
Halfsumming the equations finally we have
$$
D S_n^2 \Gamma_n = 2\Gamma_n + 4S_n \frac{\Gamma_{n+1} - \Gamma_{n-1}}{2\Delta S} + (S_n^2 + \Delta S^2)D\Gamma_n = \\ =
S_n^2 D\Gamma_n + 4S_n \frac{\Gamma_{n+1} - \Gamma_{n-1}}{2\Delta S} +
2 \Gamma_n + \Delta S^2\frac{\Gamma_{n-1} - 2\Gamma_n + \Gamma_{n+1}}{\Delta S^2}.
$$
Rest should be obvious now.
A: Actually I've managed to figure out that we just need linear combinations of 1, -2, 1 for (*) evaluated at $S_{n-1}$, $S_n$, $S_{n+1}$ respectively and we will get the required expression.
