<s>[INST] a representation of $f''$ only using terms of $f$ and constants. Using Taylor formulare How can I prove the following result? 
$
f''\left( x \right) \approx Af\left( x \right) + Bf\left( {x + h} \right) + Cf\left( {x + 2h} \right) + Df\left( {x + 3h} \right)\,\,h > 0
$
Obviously I have to consider the taylor expansion and "playing" with the terms, but that's my problem . Sorry for asking this stupid things
 [/INST] Write
\begin{align}
f(x+h) & = f(x) + hf'(x) + \frac{h^2}2 f''(x) + \frac{h^3}{6}f'''(\zeta_1) \\\
f(x+2h) & = f(x) + 2hf'(x) + \frac{(2h)^2}2 f''(x) + \frac{(2h)^3}{6}f'''(\zeta_2) \\\
f(x+3h) & = f(x) + 3hf'(x) + \frac{(3h)^2}2 f''(x) + \frac{(3h)^3}{6}f'''(\zeta_3). \\\
\end{align}
Then after you work out the equations, you get
$$
\begin{align}
A f(x) + B f(x+h) & + C f(x+2h) + D f(x+3h) \\\
& = \\\
(A+B+C+D)f(x) + (B+2C + 3D)hf'(x) &+ \left( \frac B2 + 2C + \frac 92D \right)h^2 f''(x) + \text{error term}
\end{align}
$$
where the error term contains the third derivatives. So just solve for
$$
\begin{align}
A+B+C+D = 0 \\\
B+2C + 3D = 0 \\\
B +2C + 9D/2 = 1/h^2 \\\
\end{align}
$$
and you will find four functions of $h$ ($A(h)$, $B(h)$, $C(h)$ and $D(h)$) such that your approximation is as good as the upper bound you can put on the third derivative error term.
Hope that helps,
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