I'm following the paper "Cordova 2014, Comparative Study of two compact finite difference methods".
It states:
Given a discretization of a line by $x_j = -1 + jh$, where $j = \{ 0, 1, \ldots, N \}$ and $Nh=2$, we have a second order central Finite difference to the first derivative given by:
$$ v'(x_j) - \frac{v_{j+1}-v_{j-1}}{2h} = - \frac{h^2}{6} v'''(x_j) - \frac{h^4}{120} v^{(5)}(x_j) + \cdots $$ ( in fact the paper says $v^{(4)}$ here but I think that the correct is $v^{(5)}$, since adding forward and backwards FD we keep the even terms, and subtracting we keep the odd terms, and as correctly stated in comment the sign should changed in RHS).
That is ok, we get by taking the forward and backwards FD and subtract both.
But then it states:
"A similar Taylor expansion for v′(x) shows the second order difference between v′(xj) and the average value of neighbors as
$$ v'(x_j) - \frac{v'_{j+1}-v'_{j-1}}{2} = \frac{h^2}{2} v'''(x_j) + \frac{h^4}{24} v^{(4)}(x_j) + \cdots $$"
Then it concludes:
"A simple reduction of above shows that the following equality holds to $O(h^4)$.
$$ \frac{1}{3}V'_{j-1} + \frac{4}{3}V'_{j} + \frac{1}{3}V'_{j+1} = \frac{v_{j+1}-v_{j-1}}{h} $$
where discrete values $V$ have reaplaced the evaluations of the exact $v'(x)$."
Of course I "believe" in the author, but I would like to know:
How to find the second equation;
With them, how to reach the third one.
Just for reference (taken from Cordova et al, A COMPARATIVE STUDY OF TWO COMPACT FINITE DIFFERENCES METHODS: STANDARD VS.MIMETIC)
From the fundamental theorem of calculus you get $$ v_{j+1}-v_{j-1}=\int_{x_{j-1}}^{x_{j+1}}v'(x)\,dx $$ which you can approximate using the Simpson method as $$ v_{j+1}-v_{j-1}=\frac{2h}6(v'_{j-1}+4v'_j+v'_{j+1})+O((x_{j+1}-x_{j-1})h^4). $$
That said, the second cited formula is dubious. There is a sign mismatch in the mean value, and the right side should be sign reversed. $$ \frac{2v'_j-v'_{j-1}-v'_{j+1}}2=v'_j-\frac{v'_{j-1}+v'_{j+1}}2=-\frac{h^2}2v'''_j-\frac{h^4}{24}v^{(4)}_j-...\tag{II} $$ again by simply inserting the Taylor expansions for $v'(x\pm h)$.
Similarly the first formula should be $$ v'_j-\frac{v_{j+1}-v_{j-1}}{2h}=-\frac{h^2}6v_j'''+O(h^4)\tag{I} $$
After that just combine both equations to eliminate the third derivative terms to again get the method step formula, $(I)-\frac13\,(II)$ gives \begin{align} \left(v'_j-\frac{v_{j+1}-v_{j-1}}{2h}\right)-\left(\frac{v'_j}3-\frac{v'_{j-1}+v'_{j+1}}6\right)&=-\frac{h^2}6v_j'''-\frac13\left(-\frac{h^2}2v'''_j\right)+O(h^4)\\ \frac{2v'_j}3+\frac{v'_{j-1}+v'_{j+1}}6&=\frac{v_{j+1}-v_{j-1}}{2h}+O(h^4) \end{align}