On highest degree of precision of numerical integration scheme that comes from interpolating polynomial
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Let $x_1,...,x_n$ be distinct points in $[a,b]$ and $l_i(x):=prod_{kne i}dfrac {x-x_k}{x_i-x_k} $. Let $w_i=int_a^b l_i(x)dx$. For every $f in C[a,b]$, let $I_n(f):=sum_{i=1}^n w_i f(x_i)$.
If $I_n(P)=int_a^b P(t) dt$ for every polynomial $P(x)$ of degree $le m$, then how to prove that $m le 2n-1$ ?
(Please do not quote any big theorems, I am trying to prove that any numerical intergation formula that comes from interpolation at $n$-points, has degree of precision at most $2n-1$)
numerical-methods interpolation lagrange-interpolation numerical-calculus
asked Jan 12 at 5:02
user521337user521337
1,1791417
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add a comment |
$begingroup$
Let $x_1,...,x_n$ be distinct points in $[a,b]$ and $l_i(x):=prod_{kne i}dfrac {x-x_k}{x_i-x_k} $. Let $w_i=int_a^b l_i(x)dx$. For every $f in C[a,b]$, let $I_n(f):=sum_{i=1}^n w_i f(x_i)$.
If $I_n(P)=int_a^b P(t) dt$ for every polynomial $P(x)$ of degree $le m$, then how to prove that $m le 2n-1$ ?
(Please do not quote any big theorems, I am trying to prove that any numerical intergation formula that comes from interpolation at $n$-points, has degree of precision at most $2n-1$)
numerical-methods interpolation lagrange-interpolation numerical-calculus
asked Jan 12 at 5:02
user521337user521337
1,1791417
$endgroup$
$begingroup$
Do you already know anything about Gaussian quadrature?
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– Ian
Jan 12 at 5:18
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@Ian: yes I know Gaussian quadrature and I know that it has degree of precision $2n-1$ ... now I am trying to prove that it is the best we can do ...
$endgroup$
– user521337
Jan 12 at 5:20
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One way to do it is very similar to how you prove Gaussian quadrature has positive weights: construct a polynomial of degree $2n$ which is everywhere nonnegative and which vanishes at all $n$ nodes. There's a somewhat obvious way to do this...
$endgroup$
– Ian
Jan 12 at 15:38
add a comment |
$begingroup$
Let $x_1,...,x_n$ be distinct points in $[a,b]$ and $l_i(x):=prod_{kne i}dfrac {x-x_k}{x_i-x_k} $. Let $w_i=int_a^b l_i(x)dx$. For every $f in C[a,b]$, let $I_n(f):=sum_{i=1}^n w_i f(x_i)$.
If $I_n(P)=int_a^b P(t) dt$ for every polynomial $P(x)$ of degree $le m$, then how to prove that $m le 2n-1$ ?
(Please do not quote any big theorems, I am trying to prove that any numerical intergation formula that comes from interpolation at $n$-points, has degree of precision at most $2n-1$)
numerical-methods interpolation lagrange-interpolation numerical-calculus
asked Jan 12 at 5:02
user521337user521337
1,1791417
$endgroup$
Let $x_1,...,x_n$ be distinct points in $[a,b]$ and $l_i(x):=prod_{kne i}dfrac {x-x_k}{x_i-x_k} $. Let $w_i=int_a^b l_i(x)dx$. For every $f in C[a,b]$, let $I_n(f):=sum_{i=1}^n w_i f(x_i)$.
If $I_n(P)=int_a^b P(t) dt$ for every polynomial $P(x)$ of degree $le m$, then how to prove that $m le 2n-1$ ?
(Please do not quote any big theorems, I am trying to prove that any numerical intergation formula that comes from interpolation at $n$-points, has degree of precision at most $2n-1$)
numerical-methods interpolation lagrange-interpolation numerical-calculus
numerical-methods interpolation lagrange-interpolation numerical-calculus
asked Jan 12 at 5:02
user521337user521337
1,1791417
asked Jan 12 at 5:02
user521337user521337
1,1791417
asked Jan 12 at 5:02
user521337user521337
1,1791417
asked Jan 12 at 5:02
user521337user521337
1,1791417
asked Jan 12 at 5:02
user521337user521337
1,1791417
1,1791417
$begingroup$
Do you already know anything about Gaussian quadrature?
$endgroup$
– Ian
Jan 12 at 5:18
$begingroup$
@Ian: yes I know Gaussian quadrature and I know that it has degree of precision $2n-1$ ... now I am trying to prove that it is the best we can do ...
$endgroup$
– user521337
Jan 12 at 5:20
$begingroup$
One way to do it is very similar to how you prove Gaussian quadrature has positive weights: construct a polynomial of degree $2n$ which is everywhere nonnegative and which vanishes at all $n$ nodes. There's a somewhat obvious way to do this...
$endgroup$
– Ian
Jan 12 at 15:38
add a comment |
$begingroup$
Do you already know anything about Gaussian quadrature?
$endgroup$
– Ian
Jan 12 at 5:18
$begingroup$
@Ian: yes I know Gaussian quadrature and I know that it has degree of precision $2n-1$ ... now I am trying to prove that it is the best we can do ...
$endgroup$
– user521337
Jan 12 at 5:20
$begingroup$
One way to do it is very similar to how you prove Gaussian quadrature has positive weights: construct a polynomial of degree $2n$ which is everywhere nonnegative and which vanishes at all $n$ nodes. There's a somewhat obvious way to do this...
$endgroup$
– Ian
Jan 12 at 15:38
$begingroup$
Do you already know anything about Gaussian quadrature?
$endgroup$
– Ian
Jan 12 at 5:18
$begingroup$
Do you already know anything about Gaussian quadrature?
$endgroup$
– Ian
Jan 12 at 5:18
$begingroup$
@Ian: yes I know Gaussian quadrature and I know that it has degree of precision $2n-1$ ... now I am trying to prove that it is the best we can do ...
$endgroup$
– user521337
Jan 12 at 5:20
$begingroup$
@Ian: yes I know Gaussian quadrature and I know that it has degree of precision $2n-1$ ... now I am trying to prove that it is the best we can do ...
$endgroup$
– user521337
Jan 12 at 5:20
$begingroup$
One way to do it is very similar to how you prove Gaussian quadrature has positive weights: construct a polynomial of degree $2n$ which is everywhere nonnegative and which vanishes at all $n$ nodes. There's a somewhat obvious way to do this...
$endgroup$
– Ian
Jan 12 at 15:38
$begingroup$
One way to do it is very similar to how you prove Gaussian quadrature has positive weights: construct a polynomial of degree $2n$ which is everywhere nonnegative and which vanishes at all $n$ nodes. There's a somewhat obvious way to do this...
$endgroup$
– Ian
Jan 12 at 15:38
add a comment |
1 Answer
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Consider $f(x)=(x-x_1)^2(x-x_2)^2cdots (x-x_n)^2$.
This $f$ is nonnegative everywhere and positive at all but finitely many points, so its integral over any nontrivial interval is positive. On the other hand, any integration rule based on the values at the $n$ points $x_1,x_2,dots,x_n$ must return zero, so it can't integrate $f$ exactly. That's a polynomial of degree $2n$, and thus no linear combination of values at $n$ points can find the integral accurately for all polynomials of degree up to $2n$.
answered Jan 12 at 6:06
jmerryjmerry
13.6k1629
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add a comment |
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1 Answer
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$begingroup$
Consider $f(x)=(x-x_1)^2(x-x_2)^2cdots (x-x_n)^2$.
This $f$ is nonnegative everywhere and positive at all but finitely many points, so its integral over any nontrivial interval is positive. On the other hand, any integration rule based on the values at the $n$ points $x_1,x_2,dots,x_n$ must return zero, so it can't integrate $f$ exactly. That's a polynomial of degree $2n$, and thus no linear combination of values at $n$ points can find the integral accurately for all polynomials of degree up to $2n$.
answered Jan 12 at 6:06
jmerryjmerry
13.6k1629
$endgroup$
add a comment |
$begingroup$
Consider $f(x)=(x-x_1)^2(x-x_2)^2cdots (x-x_n)^2$.
This $f$ is nonnegative everywhere and positive at all but finitely many points, so its integral over any nontrivial interval is positive. On the other hand, any integration rule based on the values at the $n$ points $x_1,x_2,dots,x_n$ must return zero, so it can't integrate $f$ exactly. That's a polynomial of degree $2n$, and thus no linear combination of values at $n$ points can find the integral accurately for all polynomials of degree up to $2n$.
answered Jan 12 at 6:06
jmerryjmerry
13.6k1629
$endgroup$
add a comment |
$begingroup$
Consider $f(x)=(x-x_1)^2(x-x_2)^2cdots (x-x_n)^2$.
This $f$ is nonnegative everywhere and positive at all but finitely many points, so its integral over any nontrivial interval is positive. On the other hand, any integration rule based on the values at the $n$ points $x_1,x_2,dots,x_n$ must return zero, so it can't integrate $f$ exactly. That's a polynomial of degree $2n$, and thus no linear combination of values at $n$ points can find the integral accurately for all polynomials of degree up to $2n$.
answered Jan 12 at 6:06
jmerryjmerry
13.6k1629
$endgroup$
Consider $f(x)=(x-x_1)^2(x-x_2)^2cdots (x-x_n)^2$.
This $f$ is nonnegative everywhere and positive at all but finitely many points, so its integral over any nontrivial interval is positive. On the other hand, any integration rule based on the values at the $n$ points $x_1,x_2,dots,x_n$ must return zero, so it can't integrate $f$ exactly. That's a polynomial of degree $2n$, and thus no linear combination of values at $n$ points can find the integral accurately for all polynomials of degree up to $2n$.
answered Jan 12 at 6:06
jmerryjmerry
13.6k1629
answered Jan 12 at 6:06
jmerryjmerry
13.6k1629
answered Jan 12 at 6:06
jmerryjmerry
13.6k1629
answered Jan 12 at 6:06
jmerryjmerry
13.6k1629
13.6k1629
add a comment |
add a comment |
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$begingroup$
Do you already know anything about Gaussian quadrature?
$endgroup$
– Ian
Jan 12 at 5:18
$begingroup$
@Ian: yes I know Gaussian quadrature and I know that it has degree of precision $2n-1$ ... now I am trying to prove that it is the best we can do ...
$endgroup$
– user521337
Jan 12 at 5:20
$begingroup$
One way to do it is very similar to how you prove Gaussian quadrature has positive weights: construct a polynomial of degree $2n$ which is everywhere nonnegative and which vanishes at all $n$ nodes. There's a somewhat obvious way to do this...
$endgroup$
– Ian
Jan 12 at 15:38