forward - backward differencing = central differencing From Taylor series, we can derive:
Forward Differencing Formula: 
$$ f'(x_{i}) = \frac{f(x_{i+1}) - f(x_{i})}{h} - \frac{f''(x_{i})h}{2!} $$
$$\tag 1 f'(x) \approx \dfrac{f(x+h)-f(x)}{h}$$   (1)
Backward Differencing Formula: 
$$ f'(x_{i}) = \frac{f(x_{i}) - f(x_{i-1})}{h} + \frac{f''(x_{i})h}{2!} $$
$$\tag 1 f'(x) \approx \dfrac{f(x)-f(x-h)}{h}$$  (2)
I know that in order to calculate the central differencing formula, we subtract (2) from (1). However, I am unable to do the subtraction properly. I must obtain an error term consisting of 1/6 after the subtraction, along with the central differencing formula. However, I just get 2f(x+h).... which is wrong.
A step by step subtraction would be really helpful. 
 A: First we must assume that $f$ is three times continuously differentiable, $f\in C^3$, then we can calculate the Taylor series of $f$ around $x_i$, 
$$
f(x_i + h) = f(x_i) + h f'(x_i) + \frac{h^2}{2} f''(x_i) + \frac{h^3}{6} f'''(\xi_+)
\tag{I} \label{1}
$$
with $\xi_+ \in [x_i, x_i+h]$. In addition we approximate in the other direction
$$
f(x_i - h) = f(x_i) - h f'(x_i) + \frac{h^2}{2} f''(x_i) - \frac{h^3}{6} f'''(\xi_-)
\tag{II}\label{2}
$$
with $\xi_- \in [x_i, x_i - h]$. Now we subtract \eqref{2} from \eqref{1} 
$$
f(x_i + h) - f(x_i - h) = 2 h f'(x_i) +  \frac{h^3}{6} f'''(\xi_+)  + \frac{h^3}{6} f'''(\xi_-)
$$
which we can rewrite to 
$$
f'(x_i) = \frac{f(x_i + h) - f(x_i - h)}{2h}  -  \frac{h^2}{12} f'''(\xi_+)  - \frac{h^2}{12} f'''(\xi_-)
\tag{III} \label{3}
$$
Now here comest the kicker. As the function $f'''(x)$ is continuous, we can use the Intermediate Value Theorem to conclude that $f'''(x)$ must attain all values between 
$f'''(\xi_-)$ and $f'''(\xi_+)$. Specifically it must attain the mean of these values. Mathematically put, there must exists $\xi \in [x_i - h, x_i + h]$ such that
$$
f'''(\xi) = \frac{f'''(\xi_-) + f'''(\xi_+)}{2} 
$$
which means we can rewrite \eqref{3} as
$$
f'(x_i) = \frac{f(x_i + h) - f(x_i - h)}{2h}  -  \frac{h^2}{6} f'''(\xi)
$$
for $\xi \in [x_i - h, x_i + h]$. 
