From: spellucci@mathematik.tu-darmstadt.de (Peter Spellucci)
Subject: Re: Gauss Quadrature
Date: 13 Apr 2000 16:57:17 GMT
Newsgroups: sci.math.num-analysis
Summary: [missing]

In article <38F5D17F.41C6@pb.izm.fhg.de>,
 El Houssine Touh <touh@pb.izm.fhg.de> writes:
|> Hi all, 
|> 
|> i have some questions about Gauss-Quadrature: 
|> 
|> 1- if the Gauss rule is: 
|> 
|> \Int_a^b w(x)f(x)dx = \sum_{i=1}^N w_i f(x_i)
|> 
|> where w(x)present the weight function, w_i present the weights and x_i
|> the absissa. 
|> 
|> How can we prove that x_i \in [a,b] ?
you first show that the x_i are the roots of the associated orthogonal
polynomial of degree n. this is easy using for q in P_{2n-1}
     q = v*p_n + r with degree v,w <n
where p_n is this polynomial.
then you show that orthogonal polynomials with respect to a weight function on an 
interval have only real roots, single and inside the interval. the proof is by
contradiction. you assume that p_n has less than n real roots (counting
multiplicity) in the interval and set
   r(x)=prod(x_x_i) x_i those real zeroes. hence w has degree less than n
and hence by orthogonality 
    \int_ r(x)w(x)p_n(x) dx = 0
but r(x)p_n(x)w(x) >=0 and not identical 0 in the interval, hence the 
    \int 
 should be >0 .

|> 
|> 
|> 2- I we have 2 Gauss rules for the weight function w(x) with different
|> number of quadrature points N and M. 
|> 
|> If N > M is that right that the N-rule better than M-rule ? 
this is quite unclear. what do you mean by "better"?
it has order 2N the other order 2M. the order is better, but the 
specific error for a specific function might be larger depending on the behaviour
of the function in question.


|> 
|> How to prove it? 
|> 
|> 
|> 3- from wich parameters depend the optimal number of the quadrature
|> points?
|> 
|>   -from the weight function? 
|> 
|>   - from the intervall of the integration? 
what please is the optimal number of points?
If the function is continuous, then the rule converges to the true integral 
for n to infinity. you have in mind some complexity measure for the integrand
or what? Of course, in practice one has to decide how much work to invest in such
an approximate integral, and, given some weight function of work involved 
try to minimize this weight function under the side condition of some 
precision to be obtained. the outcome then will depend on everything,
in first line the integrand itself and of course also the type of weight function
and the interval.
hope that helps
peter