First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 4. NUMERICAL QUADRATURE AND DIFFERENTIATION 171
x 0 +2h. Then, we obtain
f ′ (x 0 )= 1
2h [−3f(x 0)+4f(x 0 + h) − f(x 0 +2h)] + h2
3 f (3) (ξ 0 ) (4.11)
f ′ (x 0 + h) = 1
2h [−f(x 0)+f(x 0 +2h)] − h2
6 f (3) (ξ 1 ) (4.12)
f ′ (x 0 +2h) = 1
2h [f(x 0) − 4f(x 0 + h)+3f(x 0 +2h)] + h2
3 f (3) (ξ 2 ). (4.13)
It turns out that the first and third equations ((4.11) and(4.13)) are equivalent. To see
this, first substitute x 0 by x 0 − 2h in the third equation to get (ignoring the error term)
f ′ (x 0 )= 1
2h [f(x 0 − 2h) − 4f(x 0 − h)+3f(x 0 )] ,
and then set h to −h in the right-hand side to get 1 [−f(x 2h 0 +2h)+4f(x 0 + h) − 3f(x 0 )],
which gives us the first equation.
Therefore we have only two distinct equations, (4.11) and(4.12). We rewrite these
equations below with one modification: in (4.12), we substitute x 0 by x 0 − h. We then
obtain two different formulas for f ′ (x 0 ):
f ′ (x 0 )= −3f(x 0)+4f(x 0 + h) − f(x 0 +2h)
2h
f ′ (x 0 )= f(x 0 + h) − f(x 0 − h)
2h
+ h2
3 f (3) (ξ 0 ) → three-point endpoint formula
− h2
6 f (3) (ξ 1 ) → three-point midpoint formula
The three-point midpoint formula has some advantages: it has half the error of the endpoint
formula, and it has one less function evaluation. The endpoint formula is useful if one does
not know the value of f on one side; a situation that may happen if x 0 is close to an endpoint.
Example 84. The following table gives the values of f(x) =sinx. Estimate f ′ (0.1),f ′ (0.3)
using an appropriate three-point formula.
x f(x)
0.1 0.09983
0.2 0.19867
0.3 0.29552
0.4 0.38942
Solution. To estimate f ′ (0.1), we set x 0 =0.1, and h =0.1. Note that we can only use the
three-point endpoint formula.
f ′ (0.1) ≈ 1 (−3(0.09983) + 4(0.19867) − 0.29552) = 0.99835.
0.2