<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] As for $1 \le i \le 3$
\[ f(x+ih) = f(x) + ihf'(x) + \frac 12 {i^2}{h^2}f''(x) + \frac 16{i^3}{h^3}f'''(x) + O(h^4)
\]
So we have for your right hand side
\begin{align*}
   f(x)\bigl(A + B + C + D)\\
{} + f'(x)\bigl(B + 2C + 3D\bigr)h\\
{} + f''(x)\frac 12\bigl(B + 4C + 9D\bigr)h^2\\
{} + f'''(x)\frac 16 \bigl(B + 8C + 27D)h^3\\
{} + O(h^4)
\end{align*}
So we want to have
\begin{align*}
  A + B + C + D &= 0\\
      B + 2C + 3D &= 0\\
      B + 4C + 9D &= 0
\end{align*}
Wlog we can let $A = 1$. Then our System gives 
\begin{align*}
      B + C + D   &= -1\\
          C + 2D &= 1\\
         3C + 8D &= 1
\end{align*}
hence
\begin{align*}
      B + C + D   &= -1\\
          C + 2D &= 1\\
              2D &= -2
\end{align*}
so $D = -1$, $C = 3$, $B = -3$. Then $B + 8C + 27D = -3+24-27 = -6$. That gives
\[ 
  f'''(x) = \frac{-f(x) + 3f(x+h) - 3f(x+2h) + f(x+3h)}{h^3} + O(h).
\] 
</s>