[SOLVED] Quadrature formula help

Nov 2009
2
0
Hello,

I have to determine a quadrature formula of this shape:



with maximum degree of precision.

I can't find any examples to start me off. I looked at some of the polynomials they use in examples to start off (Legendre, Laguerre etc.) but can't find one that matches, or even an example like this...

Can anyone walk me through the solution, or at least start me off (what's the maximum precision, which polynomial do I start with)

Thanks!
 
Feb 2010
16
4
Berlin
Hey Tommy,

I'm not on this site so often; I hope this thread isn't too cold that posting suggestions here is still worth while.

The function \(\displaystyle W(x)=x^4 - 1\) in the integral is a weight function. Notice that it is positive everywhere on the interval \(\displaystyle (1, 2)\).

First you need to find the first few orthogonal polynomials for this weight, that is, polynomials \(\displaystyle p_0, p_1, \ldots\) satisfying

\(\displaystyle \int_1^2W(x)p_m(x)p_n(x)~dx = 0, \quad m\neq n\),

and satisfying \(\displaystyle \deg p_n = n\). You can do this by applying Gram-Schmidt to the polynomial basis \(\displaystyle 1, x, x^2,\ldots\). You can also do this by hand, since it only involves definite integrals of polynomials.

The quadrature rule you want has three "abscissae" \(\displaystyle x_1, x_2, x_3\), which will be zeros of the polynomial \(\displaystyle p_3\) that you find. This will give a quadrature rule of maximum precision.

Note: \(\displaystyle x_1, x_2, x_3 \in (1,2)\). If not, you made a mistake somewhere!

It does not matter if you normalize the \(\displaystyle p_n\) when calculating. Scaling a polynomial by a constant does not affect the location of its zeros. However the coefficients

\(\displaystyle k_n = \int_1^2 W(x)p_n^2(x)~dx\)

are relevant to the weights \(\displaystyle A_1, A_2, A_3\) in the quadrature rule. There is a formula; also for \(\displaystyle R(f)\); if you want, I can look around and maybe find a reference online.

Yours,
Jerry
 
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