+++ /dev/null
- real function bvalue ( t, bcoef, n, k, x, jderiv )
-c from * a practical guide to splines * by c. de boor
-calls interv
-c
-calculates value at x of jderiv-th derivative of spline from b-repr.
-c the spline is taken to be continuous from the right, EXCEPT at the
-c rightmost knot, where it is taken to be continuous from the left.
-c
-c****** i n p u t ******
-c t, bcoef, n, k......forms the b-representation of the spline f to
-c be evaluated. specifically,
-c t.....knot sequence, of length n+k, assumed nondecreasing.
-c bcoef.....b-coefficient sequence, of length n .
-c n.....length of bcoef and dimension of spline(k,t),
-c a s s u m e d positive .
-c k.....order of the spline .
-c
-c w a r n i n g . . . the restriction k .le. kmax (=20) is imposed
-c arbitrarily by the dimension statement for aj, dl, dr below,
-c but is n o w h e r e c h e c k e d for.
-c
-c x.....the point at which to evaluate .
-c jderiv.....integer giving the order of the derivative to be evaluated
-c a s s u m e d to be zero or positive.
-c
-c****** o u t p u t ******
-c bvalue.....the value of the (jderiv)-th derivative of f at x .
-c
-c****** m e t h o d ******
-c The nontrivial knot interval (t(i),t(i+1)) containing x is lo-
-c cated with the aid of interv . The k b-coeffs of f relevant for
-c this interval are then obtained from bcoef (or taken to be zero if
-c not explicitly available) and are then differenced jderiv times to
-c obtain the b-coeffs of (d**jderiv)f relevant for that interval.
-c Precisely, with j = jderiv, we have from x.(12) of the text that
-c
-c (d**j)f = sum ( bcoef(.,j)*b(.,k-j,t) )
-c
-c where
-c / bcoef(.), , j .eq. 0
-c /
-c bcoef(.,j) = / bcoef(.,j-1) - bcoef(.-1,j-1)
-c / ----------------------------- , j .gt. 0
-c / (t(.+k-j) - t(.))/(k-j)
-c
-c Then, we use repeatedly the fact that
-c
-c sum ( a(.)*b(.,m,t)(x) ) = sum ( a(.,x)*b(.,m-1,t)(x) )
-c with
-c (x - t(.))*a(.) + (t(.+m-1) - x)*a(.-1)
-c a(.,x) = ---------------------------------------
-c (x - t(.)) + (t(.+m-1) - x)
-c
-c to write (d**j)f(x) eventually as a linear combination of b-splines
-c of order 1 , and the coefficient for b(i,1,t)(x) must then be the
-c desired number (d**j)f(x). (see x.(17)-(19) of text).
-c
- integer jderiv,k,n, i,ilo,imk,j,jc,jcmin,jcmax,jj,kmax,kmj,km1
- * ,mflag,nmi,jdrvp1
- parameter (kmax = 20)
- real bcoef(n),t(n+k),x, aj(kmax),dl(kmax),dr(kmax),fkmj
- bvalue = 0.
- if (jderiv .ge. k) go to 99
-c
-c *** Find i s.t. 1 .le. i .lt. n+k and t(i) .lt. t(i+1) and
-c t(i) .le. x .lt. t(i+1) . If no such i can be found, x lies
-c outside the support of the spline f , hence bvalue = 0.
-c (The asymmetry in this choice of i makes f rightcontinuous, except
-c at t(n+k) where it is leftcontinuous.)
- call interv ( t, n+k, x, i, mflag )
- if (mflag .ne. 0) go to 99
-c *** if k = 1 (and jderiv = 0), bvalue = bcoef(i).
- km1 = k - 1
- if (km1 .gt. 0) go to 1
- bvalue = bcoef(i)
- go to 99
-c
-c *** store the k b-spline coefficients relevant for the knot interval
-c (t(i),t(i+1)) in aj(1),...,aj(k) and compute dl(j) = x - t(i+1-j),
-c dr(j) = t(i+j) - x, j=1,...,k-1 . set any of the aj not obtainable
-c from input to zero. set any t.s not obtainable equal to t(1) or
-c to t(n+k) appropriately.
- 1 jcmin = 1
- imk = i - k
- if (imk .ge. 0) go to 8
- jcmin = 1 - imk
- do 5 j=1,i
- 5 dl(j) = x - t(i+1-j)
- do 6 j=i,km1
- aj(k-j) = 0.
- 6 dl(j) = dl(i)
- go to 10
- 8 do 9 j=1,km1
- 9 dl(j) = x - t(i+1-j)
-c
- 10 jcmax = k
- nmi = n - i
- if (nmi .ge. 0) go to 18
- jcmax = k + nmi
- do 15 j=1,jcmax
- 15 dr(j) = t(i+j) - x
- do 16 j=jcmax,km1
- aj(j+1) = 0.
- 16 dr(j) = dr(jcmax)
- go to 20
- 18 do 19 j=1,km1
- 19 dr(j) = t(i+j) - x
-c
- 20 do 21 jc=jcmin,jcmax
- 21 aj(jc) = bcoef(imk + jc)
-c
-c *** difference the coefficients jderiv times.
- if (jderiv .eq. 0) go to 30
- do 23 j=1,jderiv
- kmj = k-j
- fkmj = float(kmj)
- ilo = kmj
- do 23 jj=1,kmj
- aj(jj) = ((aj(jj+1) - aj(jj))/(dl(ilo) + dr(jj)))*fkmj
- 23 ilo = ilo - 1
-c
-c *** compute value at x in (t(i),t(i+1)) of jderiv-th derivative,
-c given its relevant b-spline coeffs in aj(1),...,aj(k-jderiv).
- 30 if (jderiv .eq. km1) go to 39
- jdrvp1 = jderiv + 1
- do 33 j=jdrvp1,km1
- kmj = k-j
- ilo = kmj
- do 33 jj=1,kmj
- aj(jj) = (aj(jj+1)*dl(ilo) + aj(jj)*dr(jj))/(dl(ilo)+dr(jj))
- 33 ilo = ilo - 1
- 39 bvalue = aj(1)
-c
- 99 return
- end
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