fortran - Polynomial Interpolation with derivatives -

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i made post issue few weeks ago , edited i'm not sure if it's still active. since still bothering me i'm making new 1 in better form, hope.

for assignment supposed interpolate function values , derivative values in newton form. can on paper i'm lost when comes turning fortran code. have looked @ different source codes either opened file or asked input, given 5 sets of values , have use those. able write small program interpolation without derivatives. tried add derivatives it. in code below "yp" derivatives

program main implicit none  double precision , allocatable , dimension (:,:) :: nt double precision , allocatable , dimension (:) :: xnodes, yval, yp double precision :: x, evalnewton integer :: i,n,k  n = 7 allocate ( xnodes (0:n), yval (0:n), yp (0:n), nt (0:n, 0:n) )  xnodes = (/ 1.32d0, 2.47d0, 5.81d0, 6.83d0, 7.0d0 /) yval = (/ 5.63d0, 6.11d0, 8.12d0, 4.33d0, 6.15d0 /) yp = (/ 1.29d0, 2.21d0, 1.48d0, -3.61d0, 3.11d0 /)  call computent (n, xnodes, yval, yp, nt)  = 0,n     x = xnodes(i)     print*, xnodes(i), yval(i), yp(i), evalnewton (n, xnodes, nt, x) end  open (unit = 4, file = 'z', status = 'replace')  = 0,100     x = dble(i) * 1.0d-1     write (4,*) x, evalnewton (n, xnodes, nt, x) end  close (4)  deallocate (xnodes, yval, yp)  stop end   subroutine computent (n, xnodes, yval, yp, nt) implicit none  integer :: i, j, n, k, top, bot double precision :: nt (0:n, 0:n) double precision xnodes (0:n), yval (0:n), yp(0:n) double precision :: x, evalnewton  nt = 0.0d0  = 0,n     nt (i,0) = yval (i)  end  k = 1,n     top = k     bot = 0       = k,n         nt (i,k) = (nt (i,k-1) - nt (i-1,k-1))/(xnodes(top) - xnodes(bot))         top = top + 1         bot = bot + 1     end       print*, 'column number', k     j = 0,n         print*, nt (j,k)     end  end  return end   double precision function evalnewton (n, xnodes, nt, x) implicit none  integer :: i, j, n, k, top, bot double precision :: nt(0:n, 0:n) double precision xnodes (0:n), yval(0:n), yp(0:n) double precision :: x  evalnewton = nt(n,n) = n, 1, -1     evalnewton = evalnewton * (x - xnodes (i-1)) + nt (i-1, i-1) end   return end

to honest added "yp" everywhere, had "yval" , hoped mystery solver magic. have not figured out yet add program include derivatives in calculations. nice if explain me next steps take.

edit: added minor changes computent thought solve issue: = 0,n nt (i*2,0) = yval (i) nt (i*2+1,0) = yval (i) end do

do k = 1,n     top = k      bot = 0         = k,n         nt (i,k) = (nt (i,k-1) - nt (i-1,k-1))/(xnodes(top) - xnodes(bot))         if (nt(i,k) == 0)             nt(i,k) = yp(i/2)         end if         top = top + 2         bot = bot + 2     end       print*, 'column number', k     j = 0,n         print*, nt (j,k)     end

now first number 2.420...e^-322, , number messing rest of calculations. ideas how rid of it?

i'm pretty figured out small issue. have answer in case else needs it.

program main implicit none  double precision , allocatable , dimension (:,:) :: yt double precision , allocatable , dimension (:) :: xnodes, yval, yp, xt double precision :: x, evalnewton integer :: i,n,k  n = 9 allocate ( xnodes (0:n), yval (0:n), yp (0:n), yt (0:n, 0:n), xt (0:n) )  xnodes = (/ 1.32d0, 2.47d0, 5.81d0, 6.83d0, 7.0d0 /) yval = (/ 5.63d0, 6.11d0, 8.12d0, 4.33d0, 6.15d0 /) yp = (/ 1.29d0, 2.21d0, 1.48d0, -3.61d0, 3.11d0 /)  call computent (n, xnodes, yval, yp, yt, xt)  = 0,n     x = xnodes(i)     print*, xnodes(i), yval(i), yp(i), evalnewton (n, xnodes, yt, x) end  open (unit = 4, file = 'z', status = 'replace')  = 0,100     x = dble(i) * 1.0d-1     write (4,*) x, evalnewton (n, xnodes, yt, x) end  close (4)  deallocate (xnodes, yval, yp, yt, xt)  stop end   subroutine computent (n, xnodes, yval, yp, yt, xt) implicit none  integer :: i, j, n, k, top, bot double precision :: yt (0:n, 0:n) double precision xnodes (0:n), yval (0:n), yp(0:n), xt (0:n) double precision :: x, evalnewton  xt = 0.0d0 yt = 0.0d0  = 0,n     xt(i*2) = xnodes (i)     xt(i*2+1) = xnodes (i)     yt(i*2,0) = yval (i)     yt(i*2+1,0) = yval (i)   end  k = 1,n      = k,n         yt(i,k) = (yt(i,k-1) - yt(i-1,k-1))/(xt(i) - xt(i-1))         if (xt(i) - xt(i-1) == 0)             yt(i,k) = yp(i/2)         end if      end      print*, 'column number', k     j = 0,n         print*, yt (j,k)     end  end  return end   double precision function evalnewton (n, xnodes, yt, x) implicit none  integer :: i, j, n, k, top, bot double precision :: yt(0:n, 0:n) double precision xnodes (0:n), yval(0:n), yp(0:n) double precision :: x  evalnewton = yt(n,n) = n, 1, -1     evalnewton = evalnewton * (x - xnodes (i-1)) + yt (i-1, i-1) end   return end

the small issue error message when run program , file not created

error message: a.out: malloc.c:2372: sysmalloc: assertion `(old_top == (((mbinptr) (((char *) &((av)->bins[((1) - 1) * 2])) - __builtin_offsetof (struct malloc_chunk, fd)))) && old_size == 0) || ((unsigned long) (old_size) >= (unsigned long)((((__builtin_offsetof (struct malloc_chunk, fd_nextsize))+((2 *(sizeof(size_t))) - 1)) & ~((2 *(sizeof(size_t))) - 1))) && ((old_top)->size & 0x1) && ((unsigned long) old_end & pagemask) == 0)' failed.

program received signal sigabrt: process abort signal.

backtrace error:


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