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Please compute f'(1.4) from the table of data given in problem 40 of
Chapter 5, using:

- first-order forward differences, with h=0.1
- first-order forward differences, with h=0.2
- first-order backward differences, with h=0.1
- center differences, with h=0.1
- center differences, with h=0.2


f'(1.4), using first-order forward differences, with h=0.1:

2.352 - 2.151
-------------	=  2.010
 1.5 - 1.4

f'(1.4), using first-order forward differences, with h=0.2:

2.577 - 2.151
-------------	=  2.130
 1.6 - 1.4

f'(1.4), using first-order backward differences, with h=0.1:

2.151 - 1.971
-------------	=  1.800
 1.4 - 1.3

f'(1.4), using center differences, with h=0.1:

2.352 - 1.971
-------------	=  1.905
 1.5 - 1.3

f'(1.4), using center differences, with h=0.2:

2.577 - 1.811
-------------	=  1.915
 1.6 - 1.2

(I did these quickly; please let me know if you find arithmetic
errors)

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Use center differences and the extrapolation technique of section 5.4
to get f'(1.4) from the table of data given in problem 40 of Chapter 5
to an O(h^4) accuracy:

The key idea here is that the 1.905 answer is "better" and the 1.915
answer is "worse" -- simply because the h used to get the latter was
larger.  Moreover, we know how MUCH better, because we know how the
error of the method depends on h --- as O(h^2), in this case.

(See the formulas on the bottom of page 373 if you need to look up a
derivative error order.)

To do the first round of extrapolation, then, we simply stuff those
three values into equation 5.21:

better = 1.905 + (1.905 - 1.915)
                 ---------------
                     4 - 1           (since n=2)

       = 1.902

This answer has O(h^4) error.

We could continue this process, as the book does in Table 5.4, by
using h=0.4 to calculate f'(1.4) and then using equation 5.21 with
that value as "worse" and 1.915 as "better," but the problem
requirements only call for O(h^4).

We couldn't go any further, incidentally, given the available data,
because we'd need f(0.6) and f(2.2) to calculate f(1.4) using h=0.8.