Interpolation in the Integral

There are three integrals

G1(1)=0

G2(1)=0

G3(1)=0

And

G1 and G2 are defined at points 3,5,7,…,N, so that the first two lines of can be used for Richardson’s extrapolation to find

Interpolating G_a

When G_a is found in , the value of a(x) was implicitly found at the points 3,5,7,…

Linearly interpolating in a(x)/N² = G_a(i)-G1(i) to the value i+1 yields

The term in v’’ which is dropped is of order 1/N² times the error in G_a which is of order 1/N⁴ è a code segment of the form

DO I=1,N,2

DGI=GA(I)-G1(I)

DGIP2=GA(I+1)-G1(I+2)

DGIP1=DGI+(X(I+1)-X(I))*(DGIP2-DGI)/(X(I+2)-X(I))

GA(I+1)=G1(I+1)+DGIP1

ENDDO

Note that the bulk of the variation with x is included in G1 and hence in G_a and only the error estimate has been interpolated.

All three G’s are defined at i=7,13,19, …, N

At these locations all three G’s can be used to define

The error can then be determined as

This values is expected to be of order 1/N⁸ so that accuracy in interpolating this is not an issue. The code segment is simply

DO I=1,N,6

SIGMAP=(SIGMA(I+6)-SIGMA(I))/ (X(I+2)-X(I))

DO J=1,5

SIGMA(I+J)=SIGMA(I)+(X(I+J)-X(I))*SIGMAP

ENDDO