Approximating derivative as parementer decreases 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!

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
