Atomic Structure Theory

2.4 Quadrature Rules (rint) 47
where ɛk(ξ) is an error function evaluated at some point x = ξ on the integration
interval. The endpoint coefficients a1, a2, ···ak, are chosen to insure
that the integration is exact for polynomials of degree (0, 1, ··· ,k− 1). The
modified trapezoidal rule is also exact for polynomials of degree k, for odd
values of k only. There is, consequently, no gain in accuracy for a rule based
on an even value of k compared with one based on the next lower odd value.
Therefore, we concentrate on rules of form (2.89) with odd values of k.
It is relatively simple to determine the weights ai. To this end, we examine
the difference ∆l between the value given by the integration rule for a polynomial
f(x) =x l and the exact value of the integral (mh) l+1 /(l + 1). Let us
consider the example k =3:
∆0 = h (2a1 +2a2 +2a3 − 5)
∆1 = h 2
a1 + a2 + a3 − 5
m
2
∆2 = h 3
2a2 +8a3 − 10 + −2a2 − 4a3 + 37
m + a1 + a2 + a3 −
6
5
m
2
2
∆3 = h 4
(3a2 +12a3− 15) m + −3a2 − 6a3 + 37
m
4
2
+ a1 + a2 + a3 − 5
m
2
3
. (2.90)
It can be seen that ∆l is a polynomial of degree m l ; the lowest power of
m in ∆l is m for odd l and 1 for even l. The coefficient of m i in ∆l is
proportional to the coefficient of m i−1 in ∆l−1. Therefore, if ∆k−1 vanishes,
then all ∆l, l=0···k − 2 automatically vanish. Thus, for odd values of k, we
require that the k coefficients of m l ,l=0···k − 1 in the expression for ∆k−1
vanish. This leads to a system of k equations in k unknowns. For our example
k = 3, we obtain the following set of equations on requiring coefficients of m l
in ∆2 to vanish:
The solution to these equations is
a1 + a2 + a3 = 5
2 ,
2a2 +4a3 = 37
6 ,
2a2 +8a3 =10.
(a2, a2, a3) =
9 28 23
, , .
24 24 24
It should be noted that if ∆k−1 vanishes, ∆k will also vanish! Values of al,
l =1···k represented as ratios al = cl/d with a common denominator, are
tabulated for k =1, 3, 5, 7 in Table 2.2. The above analysis for the case k =1
leads precisely to the trapezoidal rule.