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Legendre polynomials in the reverse direction

  1. Apr 1, 2015 #1
    I have just written a program to calculate Legendre Polynomials, finding for Pl+1 using the recursion (l+1)Pl+1 + lPl-1 - (2l+1).x.Pl=0 That is working fine.
    The next section of the problem is to investigate the recursive polynomial in the reverse direction. I would solve this for Pl-1 in terms of Pl and Pl+1 - could someone please confirm if that is the correct way to 'reverse'? Also I would use the Rodriques formula to get the 2 starting values of Pn and Pn-1, it's a complex formula, is there a better way to get those starting values? Thanks for all help.
     
  2. jcsd
  3. Apr 2, 2015 #2
    Hi again - I found enough to convince me that solving for Pl-1 is the correct way to calculate in the reverse direction. I also decided against using the Rodriguez formula, since I work out Pl and Pl+1 in the forward program, I can just use those as starting values for the reverse direction. All seemed good. However what I expect in the reverse direction, is to get P(x) for l=1, which should be always x. I figured that the last 2 P values correspond to L_input and L_input - 1, so loop down from L_input - 2 down to l=1 - see program code below.
    I have spent hours with the debugger and pouring over the code, can't spot the bug or mistake in approach. Help please?
    Code (Text):

    program LegendreDual
       implicit none
       integer, parameter   ::   rep=60
       integer, parameter   ::   dp = selected_real_kind(15, 307)
       real(kind=dp)     ::   X,P,P_minus,P_plus,saveP
       integer         ::   j, L_input
       character(len=rep )   ::   stars=REPEAT('*',rep)
       !----- header ----  
       print *, " "
       print *, 'Program LegendreDual starts'
       print *, stars
       !----- Initialise ----  
       print *, " "
       print *,'Enter  l, X (l .LT.  0 to stop)'
       read *, L_input, X
       print *, " "
      while (L_input.GE.0) do
           print 20, 'L input','X Input','P'
         If (L_input.EQ.0) then
         P=0
      else if (L_input.EQ.1) then
           P=X
      else
         P_minus=1
      P=X
      Do j=1, L_input-1,+1           ! loop for fwd recursion
         P_plus=(((2*j)+1)*X*P)-(j*P_minus)
         P_plus=P_plus/(j+1)
             saveP=P
         P_minus=P                 ! roll values
         P=P_plus
           end do
         Print 50, L_input, X, P  
      P=saveP                   ! P from fwd; we can use P_plus and P
    !-----------------------               ! to start the backward recursion
          Do j=L_input-2,1,-1               ! loop for reverse recursion,
         P_minus=(((2*j)+1)*X*P)-((j+1)*P_plus)   ! using P(n+1) and P just found
         P_minus=P_minus/j
         P=P_minus                   ! roll values
         P_plus=P
           end do    
      end if
      Print 50, j, X, P             ! only print once if l=0 or 1
         print *, stars  
         print *, " "
         print *,'Enter  l, X (l .LT.  0 to stop)'
         read *, L_input, X
      end do
       print *, " "
    !--- admin ----
       print *, 'User selected end'
       print *, stars  
    20   format (3X, A, 3X, A, 18X,A)
    50   format (3X, I3, 5X, F10.5, 5X, F25.5)
    end program LegendreDual
     
     
  4. Apr 10, 2015 #3
    Please feel free to close or delete this thread and sorry for any inconvenience
     
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