real function ppvalu (break, coef, l, k, x, jderiv )
c from * a practical guide to splines * by c. de boor
calls interv
calculates value at x of jderiv-th derivative of pp fct from pp-repr
c
c****** i n p u t ******
c break, coef, l, k.....forms the pp-representation of the function f
c to be evaluated. specifically, the j-th derivative of f is
c given by
c
c (d**j)f(x) = coef(j+1,i) + h*(coef(j+2,i) + h*( ... (coef(k-1,i) +
c + h*coef(k,i)/(k-j-1))/(k-j-2) ... )/2)/1
c
c with h = x - break(i), and
c
c i = max( 1 , max( j , break(j) .le. x , 1 .le. j .le. l ) ).
c
c x.....the point at which to evaluate.
c jderiv.....integer giving the order of the derivative to be evaluat-
c ed. a s s u m e d to be zero or positive.
c
c****** o u t p u t ******
c ppvalu.....the value of the (jderiv)-th derivative of f at x.
c
c****** m e t h o d ******
c the interval index i , appropriate for x , is found through a
c call to interv . the formula above for the jderiv-th derivative
c of f is then evaluated (by nested multiplication).
c
integer jderiv,k,l, i,m,ndummy
real break(l+1),coef(k,l),x, fmmjdr,h
ppvalu = 0.
fmmjdr = k - jderiv
c derivatives of order k or higher are identically zero.
if (fmmjdr .le. 0.) go to 99
c
c find index i of largest breakpoint to the left of x .
call interv ( break, l+1, x, i, ndummy )
c
c Evaluate jderiv-th derivative of i-th polynomial piece at x .
h = x - break(i)
m = k
9 ppvalu = (ppvalu/fmmjdr)*h + coef(m,i)
m = m - 1
fmmjdr = fmmjdr - 1.
if (fmmjdr .gt. 0.) go to 9
99 return
end