#    

```Interpolation in the Integral
There are three integrals
G1(1)=0
G2(1)=0
(1.1)
G3(1)=0
And
G1(i )    x j  x j 1  f j  f j 1  i  2,3, 4,5..., N
i
j 2
G 2  i     x j  x j  2  f j  f j  2  i  3,5, 7,9..., N (1.2)
i
j 3
G3  i     x j  x j  2  f j  f j  2  i  4, 7,10,..., N
i
j 4
As shown in BLI_integration.htm,
c1
1
G  x   G1 x   2  O 4
N
N
4c1
1
G  x   G 2  x   2  O 4 (1.3)
N
N
9c1
1
G  x   G3  x   2  O 4
N
N
G1 and G2 are defined at points 3,5,7,…,N, so that the first two lines of (1.3) can be used for Richardson’s
extrapolation to find
Ga  i    4G1 i   G 2  i   / 3 i  1,3,5,..., N (1.4)
Interpolating Ga
  x   x
Ga ( x)  G1 x  

(1.5)
2
4
N
N
When Ga is found in (1.4), the value of (x) was implicitly found at the points 3,5,7,…
Linearly interpolating in (x)/N2 = Ga(i)-G1(i) to the value i+1 yields
 Ga  i  2   G1 i  2     Ga  i   G1 i   
2
Ga  i  1  G1 i  1  Ga  i   G1 i    xi 1  xi  
  xi 1  xi  v ''
xi  2  xi
2
The term in v’’ which is dropped is of order 1/N times the error in Ga which is of order 1/N4  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 Ga and only the error estimate has been
interpolated.
Finding the error estimate
All three G’s are defined at i=7,13,19, …, N
At these locations all three G’s can be used to define
Ga  i    4G1 i   G 2  i   / 3 

Gb  i   9G1 i   G3  i   / 8  i  7,13,19,..., N (1.6)

Gc  x   9G 2  i   4G3(i )  / 5
The error can then be determined as
 G  i   G  i    G i   G i    G i   G i  

2

2
i 
a
b
2
a
c
b
2
c
3
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
ENDDO
```