Could someone check my solution for finding constant of a difference quotient? So the question was, 
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be three times differentiable and $f'''$ is bounded, find constants $a,b,c$ such that $$f''(x) = \lim_{h\rightarrow 0} \frac{af(x-h)+bf(x)+cf(x+2h)}{h^2}$$
My Solution: 
Expand $f
(x-h)$ and $f(x+2h)$ by Taylors theorem and solve for $f''(x)$:
$f(x-h) = f(x) - hf'(x) + 0.5h^2f''(x)$ so $f''(x) = 2f(x-h)-2f(x)+2hf'(x)$ and putting in the limit definition of $f'(x)$ we get $$f''(x) = \frac{2f(x-h)-4f(x)+2f(x+h)}{h^2}$$
Doing the same for $f(x+2h)$ we get $$\frac{4f''(x)}{3} =\frac{f(x+2h)+f(x)-2f(x+h)}{h^2}$$
Can I subtract $\frac{1}{3}$ of first one from the second to get $$f''(x) = \frac{f(x+2h)+\frac{7}{3} f(x) - \frac{2}{3} f(x-h) - \frac{8}{3} f(x+h)}{h^2}$$
And also can I say that $-f(x+h) = f(x-h)$? 
 A: We find the first three  terms, in powers of $h$, of the  Taylor expansion of the numerator.
The first term is
$$af(x)+bf(x)+cf(x).\tag{1}$$
The second term is 
$$\left(-af'(x)+2cf'(x)\right)h.\tag{2}$$
The third term is
$$\frac{af''(x)+4cf''(x)}{2}h^2.\tag{3}$$
We will arrange for the first two terms to vanish, and for the third term to be $f''(x)h^2+o(h^2)$. This can be done by letting $a+b+c=0$, $-a+2c=0$, and $\frac{a+4c}{2}=1$. When we solve this system of equations, we get $a=2/3$, $b=-1$, and $c=1/3$.
A: It is probably not of much interest to students when some drudge of a mathematician hastens to point out the real source of some of the assigned problems.  So downvote if this puts you to sleep.
The idea of the problem is not simply an exercise in L'Hopital's rule or Taylor's theorem or a malicious attempt at a tricky problem.  There is real interest in finding similar expressions: can I find a characterization of the $n$th derivative of a function that does not require the computation of all the intermediate derivatives?
Consider a possible $n$th derivative ("P" for proposed?)
$$
PD_n f(x) =   \lim_{h\rightarrow 0} \frac{ \sum_{i=1}^n a_i f(x+ b_i h) )}{h^n} .
$$
The first and most elementary of the difficulties facing us is to determine conditions on the $a_i$ and the $b_i$ so that
whenever the $n$th derivative does exist it would agree with $PD_n f(x)$.
The solution to this part of the problem is that a necessary condition is that the numbers  must satisfy the $n+1$ equations 
$$\sum_{i=1}^n a_ib_i^ r = 0  \ \ r=0,1,2,3,..., n-1$$
and
$$\sum_{i=1}^n a_ib_i^ n = n!   $$
This is obtained essentially by the same methods already given in the other answer.
For the problem posed here $n=2$, $b_1=-1$, $b_2=0$, $b_3=2$
so the three  equations  are what we have already seen in the answer provided:
$$ \begin{array}{2}
  a_1+a_2+a_3= 0 \\
  -a_1 +0 +   2a_3 =0 \\
  a_1+0 + 2^2a_3 = 2!
 \end{array}
 $$
There are lots of other choices more famous than this one.  For example the symmetric Riemann derivatives have been much studied.  The third order looks like this: 
$$SRD_3 F(x) =  \lim_{h\to 0}\frac{ F(x+3h)-3F(x+h)+3F(x-h)-F(x-3h)}{8h^3}.$$
I can leave this by encouraging a dip into the literature.  J. Marshall Ash has long been interested in problems of this nature (originating from his studies under Zygmund at Chicago).  This paper   will get you started:

J. Marshall Ash, Generalizations of the Riemann derivative.
  Trans. Amer. Math. Soc. 126 1967 181–199.

