11—Numerical Analysis 317− 3 k −k 0 k 3 kf(k) =f(0) +kf′(0) +1
2
k^2 f′′(0) +1
6
k^3 f′′′(0) +···. (2)I want to isolatef(0)from this, so take
f(k) +f(−k) = 2f(0) +k^2 f′′(0) +1
12
k^4 f′′′′(0) +···f(3k) +f(− 3 k) = 2f(0) + 9k^2 f′′(0) +81
12
k^4 f′′′′(0) +···.The biggest term after thef(0)is ink^2 f′′(0), so I’ll eliminate this.
[
f(3k) +f(− 3 k)]
− 9
[
f(k)−f(−k)]
≈− 16 f(0) +[
81
12
−
9
12
]
k^4 f′′′′(0)f(0)≈1
16
[
−f(− 3 k) + 9f(−k) + 9f(k)−f(3k)]
−
[
−
3
8
k^4 f′′′′(0)]
. (3)
The error estimate is then− 3 h^4 f′′′′(0)/ 128.
To apply this, take the same example as before,f(x) = 2xatx=. 5
21 /^2 ≈
1
16
[
− 2 −^1 + 9. 20 + 9. 21 − 22
]
=
45
32
= 1. 40625 ,
and the error is 1. 40625 − 1 .41421 =−. 008 , a tenfold improvement over the previous interpolation despite the
fact that the function changes markedly in this interval and you shouldn’t expect interpolation to work very well
here.