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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