E X P L A N A T I O N S
		            -----------------------

	                 Method of Weigting Coefficients
		         to integrate a discrete function
			 F(x) from x(0) to x(N)


    We are looking here for the weigting coefficients A(i), i=0..N,
  (N is the number of points), such as the error E given by:
	
         b
	E = Sum f(x) dx - [A0 f(x0) + A1 f(x1) + ... + An f(xn)]
	     a

  is null for all distinct points X(i), i=0..N, when f(x) is a polynomial Pn(x)
  of degree <= N.

  Note that the Xi's are not necessarily equally spaced.

    By successively taking as polynomials 1, x, x^2, x^3...x^n as f(x), the in-
  tegration leads to a linear equations system:
	
    (a=x(0), b=x(N) )
	
                     b
    for f(x)=1 ==>  Sum 1 dx = A0 + A1 + ... + An = b - a
                     a

                     b
    for f(x)=x ==>  Sum x dx = A0 x0 + A1 x1 + ... + An xn = (b^2 - a^2)/2
                     a

                       b
    for f(x)=x^2 ==>  Sum x^2 dx = A0 x0^2 + A1 x1^2 + ... + An xn^2 = (b^3 
	               a
                                                           - a^3)/3
    - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

                       b
    for f(x)=x^n ==>  Sum x^n dx = A0 x0^n + A1 x1^n + ... + An xn^n = (b^n+1
                       a
                                                           - a^n+1)/(n+1)

    This linear system, to calculate the weighting coefficients, A(i), i=0..n,
  can be written in matrix form as follows:

	| 1    1    1    ... 1    | | A0 |   |  b - a              |
	| x0   x1   x2   ... xn   | | A1 |   | (b^2-a^2)/2         |
 	| x0^2 x1^2 x2^2 ... xn^2 | | A2 | = | (b^3-a^3)/3         |
	| ----------------------- | | -- |   | ------------------- |
	| x0^n x1^n x2^n ... xn^n | | An |   | (b^n+1-a^n+1)/(n+1) |

    The determinant of the left matrix, of size (n+1)(n+1), never equals zero if
  the Xi's are all distinct (determinant of Vandermonde) and the linear system
  has always a solution.
	
	Let us see how it works on an example:
	
	Let be the function f(x) = ch(x) = (e^x + e^-x)/2  (hyperbolic cosinus)
	and the given points:

			x			y
		-----------------------
			0			1
			1			1.5431
			2			3.7622

	                        2
	Let us approximate I = Sum ch(x) dx by: A0 f(0) + A1 f(1) + A2 f(2).
	                        0

	We obtain the linear system:

	| 1   1   1 | | A0 |   |  2-0    |
	| 0   1   2 | | A1 | = | (4-0)/2 |
	| 0   1   4 | | A2 |   | (8-0)/2 |
	
	The roots are: A0 = 1/3, A1 = 4/3 and A2 = 1/3.
	
	So the integral approximation is:
	
	I = 1/3 x 1 + 4/3 x 1.5431 + 1/3 x 3.7612 = 3.6449
						    ======
												
	The exact solution is sh(2) - sh(0) = 3.6269.
	
	The error is 3.6269 - 3.6449 = -0.018
	
	
	NOTE: When the Xi's are equally spaced, this method is
	      the same as the formulas of Newton-Cotes.


    Reference: "Méthodes de calcul numérique - Tome 1
                By Claude Nowakowski, PS1 1981" [BIBLI 07]. 
	
    Translated by J-P Moreau
	
	
	Sum is here for the integral sign
	^ is here for the power sign.	
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End of file dinteg.txt