Find formula for $f''(0)$ using $f(0), f(h), f(2h), f'(h)$ I want to find an approximation for $f''(0)$ using the values of: $f(0), f(h), f(2h), f'(h)$
Usually I would use unknown coefficients and work the formula $$Af(0) + Bf(h) +Cf(2h)+Df'(h)$$
using taylor's expansion constraints. The problem here, that in the expansion of $f'(h)$, there is a difference in the multiplication by $h$: $f'(h) = f'(0) + hf''(0) + ...$
unlike $f(h) = f(0) + hf'(0) + \frac{h^2}{2}f''(0) + ...$
Moreover, I don't know the value of $f'(0)$
Help would be appreciated
 A: You can still do as you say, you'll just get some coefficients that depend on $h$ in different ways:
Let's expand a few Taylor series':
\begin{align*}
f(h)&=f(0)+hf'(0)+h^2/2f''(0)+\mathcal{O}(h^3)\\
f(2h)&=f(0)+2hf'(0)+2h^2f''(0)+\mathcal{O}(h^3)\\
f'(h)&=f'(0)+hf''(0)+h^2/2f'''(0)+\mathcal{O}(h^3).
\end{align*}
 Combining everything, we have 
\begin{align*}
&Af(0)+Bf(h)+Cf(2h)+Df'(0)\\
&=Af(0)\\
&+B(f(0)+hf'(0)+h^2/2f''(0)+\mathcal{O}(h^3))\\
&+C(f(0)+2hf'(0)+2h^2f''(0)+\mathcal{O}(h^3))\\
&+D(f'(0)+hf''(0)+h^2/2f'''(0)+\mathcal{O}(h^3)).
\end{align*} We want this to be $\approx f''(0).$ So, we require that \begin{align*} A+B+C&=0\\ Bh+2Ch+D&=0\\
\frac{Bh^2}{2}+2Ch^2+Dh&=1\end{align*} These conditions will guarantee that we have terms of the form $f''(0)+\text{higher-order terms}.$ We need some more conditions to uniquely determine $A,B,C,D.$ It's natural to require that that $f'''(0)$ terms vanish (just expand the Taylor series' from the beginning out a bit more). I suspect that you'll find that your approximation is second-order.
Alternatively, you could require exactness for polynomials $\{x^n:0\leq n\leq N\}$ up to a sufficiently high order $N$.
EDIT: If you impose that condition that I mentioned earlier, you get $A=\frac{4}{h^2}, B=-\frac{2}{h^2}, C=-\frac{2}{h^2}, D=\frac{6}{h}$, as mentioned by @eyeballfrog below, which will yield a second-order accurate approximation. 
