Abstract

CHAPTER 1. OVERVIEW 24
rule. The problem still remains that since y is not a grid point, the function g(x) is
not evaluated at x = y. To solve this problem, we use Taylor expansions to get that
and
g(xv) =g(y) − h1g ′ (y)+O(h 2 1)
g(xv+1) =g(y)+h2g ′ (y)+O(h 2 2)
We note that since h1 ≤ ∆x and h2 ≤ ∆x, thenO(h 2 1)=O(h 2 2)=O(∆x 2 ). Further,
since h1 + h2 =∆x, then by using the above Taylor expansion, we have that
h2 h1
g(xv)+
∆x ∆x g(xv+1) =g(y)+O(∆x 2 )
This gives us an estimate of g(y) using known function values that is second-order
accurate with respect to ∆x. Substituting this back into the approximation of the
integral gives
y
0
g(x)dx ≈ ∆x
2 g(x1)+∆x
v−1
which is the same as
i=2
g(xi)+ ∆x
2
y v+1
g(x)dx ≈ g(xi)wi
0
h1 h2 h1
g(xv)+ (g(xv)+ g(xv)+
2 ∆x ∆x g(xv+1))
i=1