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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?

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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 $-hf''(\epsilon)$

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  • $\begingroup$ is it possible to do this without using taylor expansion? because this quesiton did not specify to use taylor (other questions of mine did specify so that's what i did for other questions but thanks.) $\endgroup$ – TomWang Feb 12 '18 at 10:30
  • $\begingroup$ I do not know of a way to do it. $\endgroup$ – Satish Ramanathan Feb 12 '18 at 10:56
  • $\begingroup$ can you take a look at my other question here if you have time (if this is not allowed then i apologize) because the other question also involves interpolation and taylor math.stackexchange.com/questions/2647115/… $\endgroup$ – TomWang Feb 12 '18 at 11:38

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