How many Gauss points are required to provide exact value for the Gauss quadrature rule

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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 $int_a^bf(x)dxapproxsum_i=0^nA_if(x_i)$ 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?



Thanks.







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  • You have it: Gauss Ian quadrature uses exactly those nodes and weights. Then your polynomial is degree 4 so...
    – Ian
    Apr 27 '16 at 10:54











  • @Ian so it has 5 distinct roots. So 5 Gauss points?
    – user71346
    Apr 27 '16 at 10:59










  • No, think about that $2n+1$ thing...
    – Ian
    Apr 27 '16 at 13:07










  • @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?
    – user71346
    Apr 27 '16 at 13:32










  • With $n+1$ points you exactly integrate polynomials of degree $2n+1$, that is the important message of your paragraph above.
    – Ian
    Apr 27 '16 at 13:39














up vote
0
down vote

favorite












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 $int_a^bf(x)dxapproxsum_i=0^nA_if(x_i)$ 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?



Thanks.







share|cite|improve this question






















  • You have it: Gauss Ian quadrature uses exactly those nodes and weights. Then your polynomial is degree 4 so...
    – Ian
    Apr 27 '16 at 10:54











  • @Ian so it has 5 distinct roots. So 5 Gauss points?
    – user71346
    Apr 27 '16 at 10:59










  • No, think about that $2n+1$ thing...
    – Ian
    Apr 27 '16 at 13:07










  • @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?
    – user71346
    Apr 27 '16 at 13:32










  • With $n+1$ points you exactly integrate polynomials of degree $2n+1$, that is the important message of your paragraph above.
    – Ian
    Apr 27 '16 at 13:39












up vote
0
down vote

favorite









up vote
0
down vote

favorite











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 $int_a^bf(x)dxapproxsum_i=0^nA_if(x_i)$ 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?



Thanks.







share|cite|improve this question














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 $int_a^bf(x)dxapproxsum_i=0^nA_if(x_i)$ 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?



Thanks.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 27 '16 at 10:51

























asked Apr 27 '16 at 10:02









user71346

1,80131336




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  • You have it: Gauss Ian quadrature uses exactly those nodes and weights. Then your polynomial is degree 4 so...
    – Ian
    Apr 27 '16 at 10:54











  • @Ian so it has 5 distinct roots. So 5 Gauss points?
    – user71346
    Apr 27 '16 at 10:59










  • No, think about that $2n+1$ thing...
    – Ian
    Apr 27 '16 at 13:07










  • @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?
    – user71346
    Apr 27 '16 at 13:32










  • With $n+1$ points you exactly integrate polynomials of degree $2n+1$, that is the important message of your paragraph above.
    – Ian
    Apr 27 '16 at 13:39
















  • You have it: Gauss Ian quadrature uses exactly those nodes and weights. Then your polynomial is degree 4 so...
    – Ian
    Apr 27 '16 at 10:54











  • @Ian so it has 5 distinct roots. So 5 Gauss points?
    – user71346
    Apr 27 '16 at 10:59










  • No, think about that $2n+1$ thing...
    – Ian
    Apr 27 '16 at 13:07










  • @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?
    – user71346
    Apr 27 '16 at 13:32










  • With $n+1$ points you exactly integrate polynomials of degree $2n+1$, that is the important message of your paragraph above.
    – Ian
    Apr 27 '16 at 13:39















You have it: Gauss Ian quadrature uses exactly those nodes and weights. Then your polynomial is degree 4 so...
– Ian
Apr 27 '16 at 10:54





You have it: Gauss Ian quadrature uses exactly those nodes and weights. Then your polynomial is degree 4 so...
– Ian
Apr 27 '16 at 10:54













@Ian so it has 5 distinct roots. So 5 Gauss points?
– user71346
Apr 27 '16 at 10:59




@Ian so it has 5 distinct roots. So 5 Gauss points?
– user71346
Apr 27 '16 at 10:59












No, think about that $2n+1$ thing...
– Ian
Apr 27 '16 at 13:07




No, think about that $2n+1$ thing...
– Ian
Apr 27 '16 at 13:07












@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?
– user71346
Apr 27 '16 at 13:32




@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?
– user71346
Apr 27 '16 at 13:32












With $n+1$ points you exactly integrate polynomials of degree $2n+1$, that is the important message of your paragraph above.
– Ian
Apr 27 '16 at 13:39




With $n+1$ points you exactly integrate polynomials of degree $2n+1$, that is the important message of your paragraph above.
– Ian
Apr 27 '16 at 13:39










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



$p + 1 = 2n$



or



$p = 2n-1$



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.






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    1 Answer
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    1 Answer
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    up vote
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    down vote













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



    $p + 1 = 2n$



    or



    $p = 2n-1$



    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.






    share|cite|improve this answer
























      up vote
      0
      down vote













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



      $p + 1 = 2n$



      or



      $p = 2n-1$



      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.






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









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



        $p + 1 = 2n$



        or



        $p = 2n-1$



        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.






        share|cite|improve this answer












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



        $p + 1 = 2n$



        or



        $p = 2n-1$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 12 at 16:40









        Frits Rooyackers

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