<s>[INST] 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 $-f(x+h) = f(x-h)$? 
 [/INST] 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 $c=1/3$.
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