11—Numerical Analysis 328Gaussian Integration
If the integrand is known at all points of the interval and not just at discrete locations as for tabulated or
experimental data, there is more freedom that you can use to gain higher accuracy even though using just a two
point formula: ∫
h
−hf(x)dx≈α[
f(β) +f(−β)]
.
I could try picking two arbitrary points, not symmetrically placed, in the interval, but the previous experience with
Simpson’s rule indicates that the result will come out as indicated. (Though it’s easy to check what happens if
you pick two general points in the interval.)
2 hf(0) +1
3
h^3 f′′(0) +1
60
h^5 f′′′′(0) +···=α[
2 f(0) +β^2 f′′(0) +1
12
β^4 f′′′′(0) +···]
To make this an equality through the low orders implies
or2 h= 2α
α=h1
3
h^3 =αβ^2β=h/√
3.
(24)
with an error term
1
60
h^5 f′′′′(0)−1
12
.^1
9
h^5 f′′′′(0) =1
135
h^5 f′′′′(0),and ∫
h
−hf(x)dx≈h[
f(
h/√
3
)
+f(
−h/√
3
)]
+
1
135
h^5 f′′′′(0). (25)With only two points, this expression yields an accuracy equal to the three point Simpson formula.
Notice that the two points found in this way are roots of a certain quadratic
(
x−
1
√
3
)(
x+1
√
3
)
=x^2 − 1 / 3 ,which is proportional to
3
2
x^2 −1
2
=P 2 (x), (26)