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
 A: 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
\[ 
  f'''(x) = \frac{-f(x) + 3f(x+h) - 3f(x+2h) + f(x+3h)}{h^3} + O(h).
\] 
A: 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,
A: Starting with $$f'(x)h\approx f(x+h)-f(x),$$ 
we get
$$\begin{align}f''(x)h^2&\approx f'(x+h)h-f'(x)h \\ &\approx f(x+2h)-f(x+h)-(f(x+h)-f(x))\\ &= f(x+2h)-2f(x+h)+f(x)\end{align}$$
Repeat: 
$$\begin{align}f'''(x)h^3&\approx f''(x+h)h-f''(x)h \\&\approx (f(x+3h)-2f(x+2h)+f(x+h))-(f(x+2h)-2f(x+h)+f(x))\\ &= f(x+3h)-3f(x+2h)+3f(x+h)-f(x)\end{align}$$
You can go on as long as you like... The coefficients are binomial with alternating signs.
Perhaps this approach does not quality as using the Taylor expansion, but it seems less painful. 
