Approximation formula for third derivative, is my approach right? 
Derive by using Taylor approximation up to 4th degree (in $h$) of $f$ in $x_0 \pm h$, $x_0\pm 2h$ at $x_0$, an formula for approximation of $f'''(x_0)$ with an error term of order $h^2$.

Could someone verify my solution, would the following be right?
Is there some restraint on $\xi_1, \xi_2, \xi_3, \xi_4$? (as to where these values reside?)
For a certain $\xi_1, \xi_2, \xi_3, \xi_4$
$$
\begin{align}
f(x_0+h) & = f(x_0) + f'(x_0)h+f''(x_0)\dfrac{h^2}{2}+ f'''(x_0)\dfrac{h^3}{6} + f^{(4)}(x_0)\dfrac{h^4}{24} + f^{(5)}(\xi_1)\dfrac{h^5}{120}\\
f(x_0-h) & = f(x_0) - f'(x_0)h+f''(x_0)\dfrac{h^2}{2}- f'''(x_0)\dfrac{h^3}{6} + f^{(4)}(x_0)\dfrac{h^4}{24} - f^{(5)}(\xi_2)\dfrac{h^5}{120}\\
f(x_0+2h) & = f(x_0) + f'(x_0)2h+f''(x_0)2h^2+ f'''(x_0)\dfrac{4h^3}{3} + f^{(4)}(x_0)\dfrac{2h^4}{3} + f^{(5)}(\xi_3)\dfrac{4h^5}{15}\\
f(x_0-2h) & = f(x_0) - f'(x_0)2h+f''(x_0)2h^2- f'''(x_0)\dfrac{4h^3}{3} + f^{(4)}(x_0)\dfrac{2h^4}{3} - f^{(5)}(\xi_4)\dfrac{4h^5}{15}\\
\end{align}$$
Then by multiplying and adding this can be rewritten for a certain $\tilde \xi$:
$$f(x_0+2h)-f(x_0-2h)-2[f(x_0+h)-f(x_0-h)] =  f'''(x_0)2h^3 + \tilde \xi h^5 $$ 
Such that for a certain $\xi$:
$$\dfrac{1}{2h^3} \left( f(x_0+2h)-f(x_0-2h)-2[f(x_0+h)-f(x_0-h)]\right) + \xi h^2 =  f'''(x_0) $$ 

I checked that your formula gives indeed an approximation to $f'''$ that is correct to order $h^2$
