From: jeanfi61@aol.com (Jeanfi61)
Subject: Re: Cubic polynomial
Date: 07 Jul 1999 01:21:00 GMT
Newsgroups: sci.math
Keywords: elementary example of Gauss quadrature

neenag@cableol.co.uk
wrote:

> Find constants a_1, a_2, u_1 and u_2,
> such that whenever P is a cubic polynomial,

> 1
> int P(t) dt = a_1 P(u_1)  + a_2 P(u_2)
> -1


A correct answer has been already given in this thread
(put P(X)=1, X, X^2, X^3).


Just a comment:
The problem is called in literature "Gaussian quadrature", or "Gauss-Legendre
quadrature".
In fact, if you fix an integer n, you can find a unique
(a_1,...,a_n; u_1,...,u_n) such that :

Int[P(t) ; -1 <= t <= 1]=a_1P(u_1)+...+a_nP(u_n)
for all polynomial P with degree(P)<2n.

To find the a_j, u_j, consider the infinite sequence (L_k)_k of Lengendre
polynomials, defined by :
L_0=1, L_1=X,
L_{k+1}=(2k+1/k+1) X*L_k -(k/k+1)L_{k-1}, for all k>1.

Then let u_1<u_2<...<u_n be the zeros of L_n (which happen to be real, and in the
interval ]-1;1[), and :
a_j=-2/[ (n+1)*P'_n(u_i)*P_{n+1}(u_i) ].


For example, if you want to integrate cubic polynomials, take n=2 ; then
L_2=(3x^2-1)/2.

So u_1=-u_2=sqrt[1/3].
a_1=a_2=1.

You can find many more things in books on numerical analysis, for example an
estimation of the error when you integrate something else than a polynomial of
degree <2n.
But the proof though not hard, is not quite elementary, because you have to
deal with systems of nonlinear equations (2n equations in 2n unknowns).