Determine the error term for the formula $f′(x) ≈ \frac 1 4 h*[f(x + 3h) − f(x − h)]$
The solution is −hf′′(ξ).
My attempt is to use the fact that for one-sided formula $f(x)≈ \frac 1 h [ f (x + h) − f (x)]$ the error term is $\frac {−1} 2 h f′′(ξ )$ how do i transform the formula in the question into the standard one sided formula?
Using taylor expansion for (x-h) and (x+3h), you get
$$f(x - h) = f(x) - hf'(x) + \frac{h^2}{2}f''(x) + O(h^3)\tag 1$$
$$f(x + 3h) = f(x) + 3hf'(x) + \frac{9h^2}{2}f''(x) + O(h^3)\tag2$$
Subtract (1) from (2)
$$f(x + 3h)- f(x-h) = 4hf'(x) + \frac{8h^2}{2}f''(x) $$
Thus $$f'(x) = \frac{f(x + 3h)- f(x-h)}{4h} - \frac{4h^2}{4h}f''(x)$$
Thus $$f'(x) = \frac{f(x + 3h)- f(x-h)}{4h} - hf''(x)\tag 3$$
We also know that
$f′(x) ≈ \frac {1} {4h}*[f(x + 3h) − f(x − h)]\tag4$
Comparing (3) and (4), there exists some epsilon such that
$$f′(x) = \frac {1} {4h}*[f(x + 3h) − f(x − h)] -hf''(\epsilon)$$
The error term is $-hf''(\epsilon)$