Approximation of the first derivative by writing Taylor expansions What is the most accurate approximation you can write for $f'(x)$ using the values $f(x-2h)$, $f(x)$ and $f(x+4h)$? What is the order of this approximation?
Note: I know how to approximate $f'(x)$ with these given values by using Taylor expansion but I couldn't understand how to find the most accurate one.
 A: Let's try to find the best linear approximation for $f'(x)$ using $f(x-2h)$, $f(x)$, $f(x+4h)$.  It'll have the form 
$$\begin{align}
f'(x) &\approx \frac{\alpha f(x-2h) + \beta f(x) + \gamma f(x+4h)}{h}\end{align}$$
We can use Taylor expansions to try to minimise the error
$$\begin{align}
f(x-2h)&= f(x) - 2hf'(x) + 2h^2f''(x) - \frac{4}{3}h^3f'''(x) + \mathcal{O}(h^4) \\
f(x)&= f(x)  \\
f(x+4h)&= f(x)+4hf'(x) + 8h^2f''(x) + \frac{32}{3}h^3f'''(x) + \mathcal{O}(h^4)\end{align}$$
The numerator of the approximation is now
$$\begin{align}
(\alpha + \beta + \gamma)f(x)+(-2\alpha + 4\gamma)hf'(x)+(2\alpha+8\gamma)h^2 f''(x) + \left(-\frac{4}{3}\alpha + \frac{32}{3}\gamma\right)h^3 f'''(x)
\end{align}$$
For the error to behave at all nicely as $h\to0$, we must have $\alpha+\beta+\gamma=0$.  For the approximation to be accurate in the limit $h\to0$, we must have $-2\alpha + 4\gamma=1$.  We still have one free parameter, since there are two equations and three unknowns, so let us say $2\alpha+8\gamma=0$ to make the approximation most accurate.  Solving these three equations gives 
$$\alpha=-\frac{1}{3}, \quad \beta=\frac{1}{4}, \quad \gamma=\frac{1}{12}$$
Our approximation is $$\begin{align}
f'(x) &\approx \frac{-\frac{1}{3} f(x-2h) + \frac{1}{4} f(x) + \frac{1}{12} f(x+4h)}{h}\end{align}$$
The approximation is second-order accurate, since the $h^2$ term disappears and the $h^3$ term doesn't.
A: You are asked to guess that the "most accurate" expansion is the one where as many initial terms as possible of the Taylor expansion are canceled out.
For this, you will form a linear combination of the expansions of $f(x-2h), f(x)$ and $f(x+4h)$, such that


*

*the constant term (coefficient of $f$) vanishes,

*the linear term (in front of $f'$) gets the coefficient $1$,

*as many higher order coefficients as possible vanish.


As you have three unknowns, you can fulfill three conditions and cancel up to the quadratic term. The approximation is up to the second order.
