How many Gauss points are required if the Gauss quadrature rule should provide the exact value of the integral $I=\int_{-1}^1f(x)dx$ for $f(x)=(x^2-1)^2$?

I am really not sure what theorem to use to solve this problem. What I can think of is a theorem about Gaussian quadrature with orthogonal polynomials as follows:

If a polynomial $p$ of degree $n+1$ is orthogonal to all polynomials of lower degree on the interval $[a,b]$ then it has $n+1$ distinct roots $x_i$ with $a<x_0<\ldots<x_n<b$ and if one uses these roots to determine the weights $A_i$ in the approximate integration formula $\int_{a}^{b}f(x)dx\approx\sum_{i=0}^{n}A_if(x_i)$ so that it is exact for all polynomials of degree up to $n$, then it is in fact exact for all polynomials of degree up to $2n+1$

But somehow I still cannot relate this theorem to the problem.

Could anyone please lend some help?


  • $\begingroup$ You have it: Gauss Ian quadrature uses exactly those nodes and weights. Then your polynomial is degree 4 so... $\endgroup$ – Ian Apr 27 '16 at 10:54
  • $\begingroup$ @Ian so it has 5 distinct roots. So 5 Gauss points? $\endgroup$ – user71346 Apr 27 '16 at 10:59
  • $\begingroup$ No, think about that $2n+1$ thing... $\endgroup$ – Ian Apr 27 '16 at 13:07
  • $\begingroup$ @Ian. If $n$ is 3, then $2(3)+1=7$ points? But $2n+1$ is the degree of the polynomials, not the number of Gauss points? $\endgroup$ – user71346 Apr 27 '16 at 13:32
  • $\begingroup$ With $n+1$ points you exactly integrate polynomials of degree $2n+1$, that is the important message of your paragraph above. $\endgroup$ – Ian Apr 27 '16 at 13:39

As far is I know the correct formula for determining the number of Gauss points is given by:

$p + 1 = 2n$


$p = 2n-1$

where p is the degree of the polynomial and n are the number of Gauss points.

Since your problem involves a fourth degree polynomial, you need 5/2 gauss points. This problem would therefore require 3 integration points instead of 2:

$(4+1)/2 = 5/2$

I hope this might solve your problem. I tried it out on a simple fourth order polynomial which gave me the exact answer.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.