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Here is the question and below my understanding/attempt:

Let $f(x) = \arctan (x)$. Use the derivative approximation: $f'(x) = \frac{8f(x+h) - 8f(x-h) - f(x+2h) +f(x-2h)}{12h} $ to approximate $f'(\frac14\pi)$ using $h^{-1}$ = 2, 4, 8 . Try to take h small enough that the rounding error effect begins to dominate the mathematical error. For what value of h does this begin to occur? (You may have to restrict yourself to working in single precision.)

So, from what I gather I just I have to do $\frac{8 \times \arctan(\frac\pi4 + \frac12) -8\times\arctan(\frac\pi4 - \frac12)..........}{12h}$

And then the same for $h^{-1}$= 4 and 8. How does the decreasing h come into play?

Thanks!

  • 0
    Possible duplicate of [Derivative approximation](http://math.stackexchange.com/questions/2109407/derivative-approximation) – 2017-01-23

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As $h$ decreases, the results should get closer to the exact answer.

The error of such approximations usually behaves like $c h^m$ for some real $c$ and integer $m$.

Therefore, as you go from $h$ to $h/2$, the error would go approximately from $ch^m$ to $c(h/2)^m =ch^m/2^m$, so it would decrease by a factor of $2^m$.

Considering this further can lead to the concept of Richardson extrapolation. See here for an explanation:

https://en.wikipedia.org/wiki/Richardson_extrapolation