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 $int_{a}^{b}f(x)dxapproxsum_{i=0}^{n}A_if(x_i)$ 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
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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