How many Gauss points are required to provide exact value for the Gauss quadrature rule
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 $int_{a}^{b}f(x)dxapproxsum_{i=0}^{n}A_if(x_i)$ 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?
Thanks.
numerical-methods
asked Apr 27 '16 at 10:02
user71346user71346
1,81231540
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
|
show 2 more comments
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 $int_{a}^{b}f(x)dxapproxsum_{i=0}^{n}A_if(x_i)$ 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?
Thanks.
numerical-methods
asked Apr 27 '16 at 10:02
user71346user71346
1,81231540
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
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
|
show 2 more comments
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 $int_{a}^{b}f(x)dxapproxsum_{i=0}^{n}A_if(x_i)$ 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?
Thanks.
numerical-methods
asked Apr 27 '16 at 10:02
user71346user71346
1,81231540
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 $int_{a}^{b}f(x)dxapproxsum_{i=0}^{n}A_if(x_i)$ 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?
Thanks.
numerical-methods
numerical-methods
asked Apr 27 '16 at 10:02
user71346user71346
1,81231540
asked Apr 27 '16 at 10:02
user71346user71346
1,81231540
edited Apr 27 '16 at 10:51
user71346
asked Apr 27 '16 at 10:02
user71346user71346
1,81231540
asked Apr 27 '16 at 10:02
user71346user71346
1,81231540
asked Apr 27 '16 at 10:02
user71346user71346
1,81231540
1,81231540
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
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
|
show 2 more comments
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
|
show 2 more comments
1 Answer
1
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votes
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.
answered Feb 12 '18 at 16:40
Frits RooyackersFrits Rooyackers
111
add a comment |
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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.
answered Feb 12 '18 at 16:40
Frits RooyackersFrits Rooyackers
111
add a comment |
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.
answered Feb 12 '18 at 16:40
Frits RooyackersFrits Rooyackers
111
add a comment |
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.
answered Feb 12 '18 at 16:40
Frits RooyackersFrits Rooyackers
111
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.
answered Feb 12 '18 at 16:40
Frits RooyackersFrits Rooyackers
111
answered Feb 12 '18 at 16:40
Frits RooyackersFrits Rooyackers
111
answered Feb 12 '18 at 16:40
Frits RooyackersFrits Rooyackers
111
answered Feb 12 '18 at 16:40
Frits RooyackersFrits Rooyackers
111
111
add a comment |
add a comment |
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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