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客观分析 拉格朗日插值 FORTRAN程序 LAGINT.F ! !根据给定点的函数值，应用拉格朗日插值公式，计算指定插值点处的 ！函数值。 ！在指定点的前后各取四个结点，用七次拉格朗日插值公式。 ！X，Y为输入参数，各有N个点 ！T为指定插入点 ！Z为输出参数，是插值点T处的近似值。 SUBROUTIONE LAGINT（X，Y，N ，T，Z） DIMENSION X（N），Y（N） DOUBLE PRECISION X，Y，T，X，S Z=0.0 IF (N.LE.0) RETURN IF(N.EQ.1) THEN Z=Y(1) ENDIF IF(N.EQ.2) THEN Z=(Y(1)*(T-X(2))-Y(2)*(T-X(1)))/(X(1)-X(2)) RETURN ENDIF I=1 IF(X(1) .LT. T) I=I+1 IF(I .LE. N) GOTO 10 ENDIF K=I-4 IF(K .LT. 1) K=1 M=I+3 IF( M .GT. N) M=N DO 30 I=K,M S=1.0 DO 20 J=K,M IF(J .NE. I) THEN S=S*(T-X(J))/(X(I)-X(J)) ENDIF 20 CONTINUE Z=Z+S*Y(I) 30 CONTINUE RETURN END 样条插值 FORTRAN程序 SPLINE.F ! SPLINE.FOR -- Natural Cubic Spline ! Call SplineCalculate with the two arrays of control points (and ! the number of elements in each array). To interpolate the spline, ! Call SplineEvaluate for each X for which you want a Y along the ! splines curve. ! subroutine SplineCalculate( inx, ina, numx ) !ms$if .not. defined(LINKDIRECT) !ms$attributes dllexport :: SplineCalculate !ms$endif implicit none real*4 ina(*), inx(*) integer numx integer MAXPOINTS parameter( MAXPOINTS = 1000 ) integer nxs real*4 x(0:MAXPOINTS) real*4 a(0:MAXPOINTS), b(0:MAXPOINTS) real*4 c(0:MAXPOINTS), d(0:MAXPOINTS) common /SplineData/ a, b, c, d, nxs, x integer i real*4 alpha(0:MAXPOINTS), l(0:MAXPOINTS) real*4 mu(0:MAXPOINTS), z(0:MAXPOINTS) real*4 h(0:MAXPOINTS) ! ===== Step 1 nxs = numx-1 do i = 0, nxs x(i) = inx(i+1) a(i) = ina(i+1) end do do i = 0, nxs-1 h(i) = x(i+1) - x(i) end do x(nxs+1) = 40000 ! ===== Step 2 do i = 1, nxs-1 alpha(i) = 3.0 * (a(i+1) * h(i-1)-a(i) * (x(i+1)-x(i-1)) + a(i-1) * h(i)) / (h(i-1) * h(i)) end do ! ===== Step 3 l(0) = 1.0 mu(0) = 0.0 z(0) = 0.0 ! ===== Step 4 do i = 1, nxs-1 l(i) = 2.0 * (x(i+1)-x(i-1))-h(i-1) * mu(i-1) mu(i) = h(i) / l(i) z(i) = (alpha(i)-h(i-1) * z(i-1)) / l(i) end do ! ===== Step 5