11—Numerical Analysis 326
This is known as Simpson’s rule.
Simpson’s Rule
Before applying this last result, I’ll go back and derive it in a more systematic way, putting it into the form you’ll
see most often.
Integrate Taylor’s expansion over a symmetric domain to simplify the algebra:
∫h
−h
dxf(x) = 2hf(0) +
2
6
h^3 f′′(0) +
2
120
h^5 f′′′′(0) +···.
I’ll try to approximate this by a three point formulaα(−h) +βf(0) +γf(h)whereα,β, andγ, are unknown.
Because of the symmetry of the problem, you can anticipate thatα=γ, but let that go for now and it will come
out of the algebra.
αf(−h) +βf(0) +γf(h) =
α
[
f(0)−hf′(0) +
1
2
h^2 f′′(0)−
1
6
h^3 f′′′(0) +
1
24
h^4 f′′′′(0) +···
]
+βf(0)
+γ
[
f(0) +hf′(0) +
1
2
h^2 f′′(0) +
1
6
h^3 f′′′(0) +
1
24
h^4 f′′′′(0) +···
]
You now determine the three constants by requiring that the two series for the same integral agree to as
high an order as is possible for any f.
2 h=α+β+γ
0 =−αh+γh
1
3
h^3 =
1
2
(α+γ)h^2
=⇒ α=γ=h/ 3 , β= 4h/ 3
and so,
∫h
−h
dxf(x)≈
h
3
[
f(−h) + 4f(0) +f(h)