legendre.cpp File Reference

compute legendre polynomial at specific x. This is inspired from the p_polynomial_value package, written by John Burkardt (see below for full documentation). Here, we apply the function differently: we assume previous two evaluations are provided, and we compute the level n evaluation: More...

#include "./types.h"
#include "../Exceptions/exceptions.h"
#include "../MemOps/MemOps.h"
#include "./recast.h"

Go to the source code of this file.

Functions
IssmDouble *	p_polynomial_value (int m, int n, IssmDouble *x)

Detailed Description

compute legendre polynomial at specific x. This is inspired from the p_polynomial_value package, written by John Burkardt (see below for full documentation). Here, we apply the function differently: we assume previous two evaluations are provided, and we compute the level n evaluation:

Definition in file legendre.cpp.

Function Documentation

p_polynomial_value()

IssmDouble* p_polynomial_value	(	int	m,
		int	n,
		IssmDouble *	x
	)

Definition at line 142 of file legendre.cpp.

                                                             {
 
   /******************************************************************************{{{/
   Purpose:
 
   P_POLYNOMIAL_VALUE evaluates the Legendre polynomials P(n,x).
 
   Discussion:
 
   P(n,1) = 1.
   P(n,-1) = (-1)^N.
   | P(n,x) | <= 1 in [-1,1].
 
   The N zeroes of P(n,x) are the abscissas used for Gauss-Legendre
   quadrature of the integral of a function F(X) with weight function 1
   over the interval [-1,1].
 
   The Legendre polynomials are orthogonal under the inner product defined
   as integration from -1 to 1:
 
   Integral ( -1 <= X <= 1 ) P(I,X) * P(J,X) dX
   = 0 if I =/= J
   = 2 / ( 2*I+1 ) if I = J.
 
   Except for P(0,X), the integral of P(I,X) from -1 to 1 is 0.
 
   A function F(X) defined on [-1,1] may be approximated by the series
   C0*P(0,x) + C1*P(1,x) + ... + CN*P(n,x)
   where
   C(I) = (2*I+1)/(2) * Integral ( -1 <= X <= 1 ) F(X) P(I,x) dx.
 
   The formula is:
 
   P(n,x) = (1/2^N) * sum ( 0 <= M <= N/2 ) C(N,M) C(2N-2M,N) X^(N-2*M)
 
   Differential equation:
 
   (1-X*X) * P(n,x)'' - 2 * X * P(n,x)' + N * (N+1) = 0
 
   First terms:
 
   P( 0,x) =      1
   P( 1,x) =      1 X
   P( 2,x) = (    3 X^2 -       1)/2
   P( 3,x) = (    5 X^3 -     3 X)/2
   P( 4,x) = (   35 X^4 -    30 X^2 +     3)/8
   P( 5,x) = (   63 X^5 -    70 X^3 +    15 X)/8
   P( 6,x) = (  231 X^6 -   315 X^4 +   105 X^2 -     5)/16
   P( 7,x) = (  429 X^7 -   693 X^5 +   315 X^3 -    35 X)/16
   P( 8,x) = ( 6435 X^8 - 12012 X^6 +  6930 X^4 -  1260 X^2 +   35)/128
   P( 9,x) = (12155 X^9 - 25740 X^7 + 18018 X^5 -  4620 X^3 +  315 X)/128
   P(10,x) = (46189 X^10-109395 X^8 + 90090 X^6 - 30030 X^4 + 3465 X^2-63)/256
 
   Recursion:
 
   P(0,x) = 1
   P(1,x) = x
   P(n,x) = ( (2*n-1)*x*P(n-1,x)-(n-1)*P(n-2,x) ) / n
 
   P'(0,x) = 0
   P'(1,x) = 1
   P'(N,x) = ( (2*N-1)*(P(N-1,x)+X*P'(N-1,x)-(N-1)*P'(N-2,x) ) / N
 
   Licensing:
 
   This code is distributed under the GNU LGPL license.
 
   Modified:
 
   08 August 2013
 
   Author:
 
      John Burkardt
 
         Reference:
 
         Milton Abramowitz, Irene Stegun,
               Handbook of Mathematical Functions,
               National Bureau of Standards, 1964,
               ISBN: 0-486-61272-4,
               LC: QA47.A34.
 
                  Daniel Zwillinger, editor,
               CRC Standard Mathematical Tables and Formulae,
               30th Edition,
               CRC Press, 1996.
 
                  Parameters:
 
                  Input, int M, the number of evaluation points.
 
                  Input, int N, the highest order polynomial to evaluate.
                  Note that polynomials 0 through N will be evaluated.
 
                  Input, double X[M], the evaluation points.
 
                  Output, double P_POLYNOMIAL_VALUE[M*(N+1)], the values of the Legendre
                  polynomials of order 0 through N.
   }}}*/
 
   int i,j;
 
   if(n<0) return NULL;
 
   IssmDouble* v = xNew<IssmDouble>(m*(n+1));
 
   for ( i = 0; i < m; i++ ) v[i*(n+1)+0] = 1.0;
   if ( n < 1 ) return v;
 
   for ( i = 0; i < m; i++ ) v[i*(n+1)+1] = x[i];
 
   for ( i = 0; i < m; i++ ) {
      for ( j = 2; j <= n; j++ ) {
 
         v[j+i*(n+1)] = ( ( IssmDouble ) ( 2 * j - 1 ) * x[i] * v[(j-1)+i*(n+1)]
               - ( IssmDouble ) (     j - 1 ) *        v[(j-2)+i*(n+1)] )
            / ( IssmDouble ) (     j     );
 
      }
   }
 
   return v;
}