From: spellucci@mathematik.tu-darmstadt.de (Peter Spellucci)
Subject: Re: Gaussian Quadrature
Date: 20 Sep 2000 08:40:34 GMT
Newsgroups: sci.math.num-analysis
Summary: [missing]

In article <20000919221801.13551.00000059@ng-fe1.aol.com>,
 professorgauss@aol.com (Professor Gauss) writes:
|> I understand that n-point Gaussian quadrature is exact for polynomials up to
|> degree 2n-1.  What I am looking for is a lucid explanation as to why this is
|> so.  I know that the technique is related to the roots of Legendre polynomials,
|> but how this works is not obvious to me.
|> 
theorem : if t_i are n pairwise distinct points in [a,b] and you define

    w_i = integral over [a,b} of i-th Lagrange-polynomial with respect to 
          the grid {t_1,..,t_n}   
then
    sum over i {1 to n} w_i f(t_i) = integral over [a,b] f(t) dt   (*) 
if f is a polynomial of degree <= n-1.
proof : obvious, since 
    sum over i {1 to n} f(t_i)*La_i(t) == f(t) .  (I write La_i for the 
    Lagrange polynomial in order to discern it from the L_i, the Legendre)
    La_i(t) = product over j (1 to n , j not = i) (t-t_j)/(t_i-t_j)

(w.l.o.g. let [a,b] = [-1,1] ; otherwise transform [a,b] to [-1,1] in the
 obvious manner ) 
Now let t_i be the roots of the n-th Legendre polynomial L_i(t) on [-1,1] .
(that these are simple, real and within [-1,1] follows from the orthogonality
property of the L_i (by definition), using a proof by contradiction).

the Legendre polynomials have the property 

  integral over [-1,1] (L_i(t))^2 dt =1 
  integral over [-1,1] (L_i(t)*L_j(t)) dt =0 if i not= j   (**)

Let f be an arbitrary polynomial of degree <= 2n-1 and write, using polynomial
division     
    f(t) = L_n(t)*psi(t) + phi(t) ,   (dec)

 with phi and psi polynomials of degree <=n-1.
then 
    psi(t) = sum over j {0 to n-1} L_j(t)*beta_j for some beta_j .
Hence 
    integral over [-1,1] L_n(t)*psi(t) dt =0  from (**)
From  the definition of t_i also 
    sum over i {1 to n} w_i*L_n(t_i)*psi(t_i) = 0
Hence from (dec)
   integral over [-1,1] f(t) dt = integral over [-1,1] phi(t) dt =
   sum over i {1 to n } phi(t_i)w_i  (from (*))
   =
   sum over i {1 to n} f(t_i)*w_i (from (dec) and the definition of the t_i) 
done.
(the uniqueness of the rule (up to numbering of the nodes) 
follows by a similar argument).

hope this helps
peter