is lagrange error bound always true? The Next CEO of Stack OverflowQuestion regarding upper bound of fixed-point functionHermite Interpolation of $e^x$. Strange behaviour when increasing the number of derivatives at interpolating points.Runge function error second factorFind the error boundPolynomial Interpolation and Error BoundLagrange Interpolating Polynomials - Error BoundError Bound for Lagrange InterpolatingError bound of Lagrange interpolatingError Expectations for Composite Simpson's RuleLagrange error bound
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is lagrange error bound always true?
The Next CEO of Stack OverflowQuestion regarding upper bound of fixed-point functionHermite Interpolation of $e^x$. Strange behaviour when increasing the number of derivatives at interpolating points.Runge function error second factorFind the error boundPolynomial Interpolation and Error BoundLagrange Interpolating Polynomials - Error BoundError Bound for Lagrange InterpolatingError bound of Lagrange interpolatingError Expectations for Composite Simpson's RuleLagrange error bound
$begingroup$
I am asking this question because I am currently working with the function $f$ defined by $f(x) = 1/(1+x^2)$ and interpolating it at $n+1$ for the interval $[-2,2]$. I have found that it theoretically is less.
My finding are as follow:
beginarrayc
hline
textValue of $n$ & textMaximum error (approximation) & textMaximum error (Theoretical) \
hline
2 & 0.305573 & 0.299488 \
hline
4 & 0.1618 & 0.0950007 \
hline
8 & 0.0992211 & 0.0167722 \
hline
18& 0.0717574 & 0.000890209 \
hline
endarray
Could there be cases where Lagrange bound is not satisfied?
numerical-methods
edited Mar 28 at 2:32
Brian
1,220116
asked Mar 28 at 1:53
KbiirKbiir
556
$endgroup$
add a comment |
$begingroup$
I am asking this question because I am currently working with the function $f$ defined by $f(x) = 1/(1+x^2)$ and interpolating it at $n+1$ for the interval $[-2,2]$. I have found that it theoretically is less.
My finding are as follow:
beginarrayc
hline
textValue of $n$ & textMaximum error (approximation) & textMaximum error (Theoretical) \
hline
2 & 0.305573 & 0.299488 \
hline
4 & 0.1618 & 0.0950007 \
hline
8 & 0.0992211 & 0.0167722 \
hline
18& 0.0717574 & 0.000890209 \
hline
endarray
Could there be cases where Lagrange bound is not satisfied?
numerical-methods
edited Mar 28 at 2:32
Brian
1,220116
asked Mar 28 at 1:53
KbiirKbiir
556
$endgroup$
$begingroup$
Would we be teaching it if it were not true? (humor)
$endgroup$
– Brian
Mar 28 at 2:04
$begingroup$
You do not say you are approximating with Taylor polynomials and do not explicitly say the center of your expansion, but you should. Also, the tag "lagrange-multiplier" does not apply to your question. Finally, since I find the maximum of the true error for $n = 2$ is $16/5$ (occurring at $x = 2$) and the maximum of the Lagrange error is a little less than $50$ for $n = 2$ (occurring at $x = 2$), perhaps you should show your work for the true error and the error bound for at least one value of $n$.
$endgroup$
– Eric Towers
Mar 28 at 2:09
add a comment |
$begingroup$
I am asking this question because I am currently working with the function $f$ defined by $f(x) = 1/(1+x^2)$ and interpolating it at $n+1$ for the interval $[-2,2]$. I have found that it theoretically is less.
My finding are as follow:
beginarrayc
hline
textValue of $n$ & textMaximum error (approximation) & textMaximum error (Theoretical) \
hline
2 & 0.305573 & 0.299488 \
hline
4 & 0.1618 & 0.0950007 \
hline
8 & 0.0992211 & 0.0167722 \
hline
18& 0.0717574 & 0.000890209 \
hline
endarray
Could there be cases where Lagrange bound is not satisfied?
numerical-methods
edited Mar 28 at 2:32
Brian
1,220116
asked Mar 28 at 1:53
KbiirKbiir
556
$endgroup$
I am asking this question because I am currently working with the function $f$ defined by $f(x) = 1/(1+x^2)$ and interpolating it at $n+1$ for the interval $[-2,2]$. I have found that it theoretically is less.
My finding are as follow:
beginarrayc
hline
textValue of $n$ & textMaximum error (approximation) & textMaximum error (Theoretical) \
hline
2 & 0.305573 & 0.299488 \
hline
4 & 0.1618 & 0.0950007 \
hline
8 & 0.0992211 & 0.0167722 \
hline
18& 0.0717574 & 0.000890209 \
hline
endarray
Could there be cases where Lagrange bound is not satisfied?
numerical-methods
numerical-methods
edited Mar 28 at 2:32
Brian
1,220116
asked Mar 28 at 1:53
KbiirKbiir
556
edited Mar 28 at 2:32
Brian
1,220116
asked Mar 28 at 1:53
KbiirKbiir
556
edited Mar 28 at 2:32
Brian
1,220116
edited Mar 28 at 2:32
Brian
1,220116
edited Mar 28 at 2:32
Brian
1,220116
1,220116
asked Mar 28 at 1:53
KbiirKbiir
556
asked Mar 28 at 1:53
KbiirKbiir
556
asked Mar 28 at 1:53
KbiirKbiir
556
556
$begingroup$
Would we be teaching it if it were not true? (humor)
$endgroup$
– Brian
Mar 28 at 2:04
$begingroup$
You do not say you are approximating with Taylor polynomials and do not explicitly say the center of your expansion, but you should. Also, the tag "lagrange-multiplier" does not apply to your question. Finally, since I find the maximum of the true error for $n = 2$ is $16/5$ (occurring at $x = 2$) and the maximum of the Lagrange error is a little less than $50$ for $n = 2$ (occurring at $x = 2$), perhaps you should show your work for the true error and the error bound for at least one value of $n$.
$endgroup$
– Eric Towers
Mar 28 at 2:09
add a comment |
$begingroup$
Would we be teaching it if it were not true? (humor)
$endgroup$
– Brian
Mar 28 at 2:04
$begingroup$
You do not say you are approximating with Taylor polynomials and do not explicitly say the center of your expansion, but you should. Also, the tag "lagrange-multiplier" does not apply to your question. Finally, since I find the maximum of the true error for $n = 2$ is $16/5$ (occurring at $x = 2$) and the maximum of the Lagrange error is a little less than $50$ for $n = 2$ (occurring at $x = 2$), perhaps you should show your work for the true error and the error bound for at least one value of $n$.
$endgroup$
– Eric Towers
Mar 28 at 2:09
$begingroup$
Would we be teaching it if it were not true? (humor)
$endgroup$
– Brian
Mar 28 at 2:04
$begingroup$
Would we be teaching it if it were not true? (humor)
$endgroup$
– Brian
Mar 28 at 2:04
$begingroup$
You do not say you are approximating with Taylor polynomials and do not explicitly say the center of your expansion, but you should. Also, the tag "lagrange-multiplier" does not apply to your question. Finally, since I find the maximum of the true error for $n = 2$ is $16/5$ (occurring at $x = 2$) and the maximum of the Lagrange error is a little less than $50$ for $n = 2$ (occurring at $x = 2$), perhaps you should show your work for the true error and the error bound for at least one value of $n$.
$endgroup$
– Eric Towers
Mar 28 at 2:09
$begingroup$
You do not say you are approximating with Taylor polynomials and do not explicitly say the center of your expansion, but you should. Also, the tag "lagrange-multiplier" does not apply to your question. Finally, since I find the maximum of the true error for $n = 2$ is $16/5$ (occurring at $x = 2$) and the maximum of the Lagrange error is a little less than $50$ for $n = 2$ (occurring at $x = 2$), perhaps you should show your work for the true error and the error bound for at least one value of $n$.
$endgroup$
– Eric Towers
Mar 28 at 2:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is important to mention the exact location of the interpolation nodes. The interpolation error using nodes $x_0, cdots, x_n in [-2,2]$ is given by
$$
e_n(x) = dfracf^(n+1)(xi)(n+1)!prod_i=0^n(x-x_i),
$$
So the error can be bounded by
$$
frac(n+1)!.
$$
Depending on the location of the nodes, the term $ | prod_i=0^n(x-x_i)|_infty$ can behave very differently (you should look into Runge's example and the use of Chebyshev nodes).
answered Mar 28 at 8:55
PierreCarrePierreCarre
1,695212
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
It is important to mention the exact location of the interpolation nodes. The interpolation error using nodes $x_0, cdots, x_n in [-2,2]$ is given by
$$
e_n(x) = dfracf^(n+1)(xi)(n+1)!prod_i=0^n(x-x_i),
$$
So the error can be bounded by
$$
frac(n+1)!.
$$
Depending on the location of the nodes, the term $ | prod_i=0^n(x-x_i)|_infty$ can behave very differently (you should look into Runge's example and the use of Chebyshev nodes).
answered Mar 28 at 8:55
PierreCarrePierreCarre
1,695212
$endgroup$
add a comment |
$begingroup$
It is important to mention the exact location of the interpolation nodes. The interpolation error using nodes $x_0, cdots, x_n in [-2,2]$ is given by
$$
e_n(x) = dfracf^(n+1)(xi)(n+1)!prod_i=0^n(x-x_i),
$$
So the error can be bounded by
$$
frac(n+1)!.
$$
Depending on the location of the nodes, the term $ | prod_i=0^n(x-x_i)|_infty$ can behave very differently (you should look into Runge's example and the use of Chebyshev nodes).
answered Mar 28 at 8:55
PierreCarrePierreCarre
1,695212
$endgroup$
add a comment |
$begingroup$
It is important to mention the exact location of the interpolation nodes. The interpolation error using nodes $x_0, cdots, x_n in [-2,2]$ is given by
$$
e_n(x) = dfracf^(n+1)(xi)(n+1)!prod_i=0^n(x-x_i),
$$
So the error can be bounded by
$$
frac(n+1)!.
$$
Depending on the location of the nodes, the term $ | prod_i=0^n(x-x_i)|_infty$ can behave very differently (you should look into Runge's example and the use of Chebyshev nodes).
answered Mar 28 at 8:55
PierreCarrePierreCarre
1,695212
$endgroup$
It is important to mention the exact location of the interpolation nodes. The interpolation error using nodes $x_0, cdots, x_n in [-2,2]$ is given by
$$
e_n(x) = dfracf^(n+1)(xi)(n+1)!prod_i=0^n(x-x_i),
$$
So the error can be bounded by
$$
frac(n+1)!.
$$
Depending on the location of the nodes, the term $ | prod_i=0^n(x-x_i)|_infty$ can behave very differently (you should look into Runge's example and the use of Chebyshev nodes).
answered Mar 28 at 8:55
PierreCarrePierreCarre
1,695212
answered Mar 28 at 8:55
PierreCarrePierreCarre
1,695212
answered Mar 28 at 8:55
PierreCarrePierreCarre
1,695212
answered Mar 28 at 8:55
PierreCarrePierreCarre
1,695212
1,695212
add a comment |
add a comment |
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$begingroup$
Would we be teaching it if it were not true? (humor)
$endgroup$
– Brian
Mar 28 at 2:04
$begingroup$
You do not say you are approximating with Taylor polynomials and do not explicitly say the center of your expansion, but you should. Also, the tag "lagrange-multiplier" does not apply to your question. Finally, since I find the maximum of the true error for $n = 2$ is $16/5$ (occurring at $x = 2$) and the maximum of the Lagrange error is a little less than $50$ for $n = 2$ (occurring at $x = 2$), perhaps you should show your work for the true error and the error bound for at least one value of $n$.
$endgroup$
– Eric Towers
Mar 28 at 2:09