Numerical methods quadratic interpolation errors
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Show that the truncation error of quadratic interpolation in an equidistant table is bounded by $$frac{h^3}{9cdot3^{0.5}}max f''' (x)$$
I have gotten to nothing because i don't seem to understand quadratic interpolation errors unlike linear interpolation errors.
numerical-methods
edited Dec 11 '18 at 9:27
LutzL
60k42057
asked Dec 11 '18 at 6:26
Kevin GachathiKevin Gachathi
61
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add a comment |
$begingroup$
Show that the truncation error of quadratic interpolation in an equidistant table is bounded by $$frac{h^3}{9cdot3^{0.5}}max f''' (x)$$
I have gotten to nothing because i don't seem to understand quadratic interpolation errors unlike linear interpolation errors.
numerical-methods
edited Dec 11 '18 at 9:27
LutzL
60k42057
asked Dec 11 '18 at 6:26
Kevin GachathiKevin Gachathi
61
$endgroup$
$begingroup$
I have edited, partially heavily, your question. Please check that the corrections conform with your original task, especially that the subscript "$m$" was originally three dashes of the third derivative.
$endgroup$
– LutzL
Dec 11 '18 at 9:28
$begingroup$
Write down the interpolation formula and apply Taylor's theorem.
$endgroup$
– Yves Daoust
Dec 11 '18 at 10:02
add a comment |
$begingroup$
Show that the truncation error of quadratic interpolation in an equidistant table is bounded by $$frac{h^3}{9cdot3^{0.5}}max f''' (x)$$
I have gotten to nothing because i don't seem to understand quadratic interpolation errors unlike linear interpolation errors.
numerical-methods
edited Dec 11 '18 at 9:27
LutzL
60k42057
asked Dec 11 '18 at 6:26
Kevin GachathiKevin Gachathi
61
$endgroup$
Show that the truncation error of quadratic interpolation in an equidistant table is bounded by $$frac{h^3}{9cdot3^{0.5}}max f''' (x)$$
I have gotten to nothing because i don't seem to understand quadratic interpolation errors unlike linear interpolation errors.
numerical-methods
numerical-methods
edited Dec 11 '18 at 9:27
LutzL
60k42057
asked Dec 11 '18 at 6:26
Kevin GachathiKevin Gachathi
61
edited Dec 11 '18 at 9:27
LutzL
60k42057
asked Dec 11 '18 at 6:26
Kevin GachathiKevin Gachathi
61
edited Dec 11 '18 at 9:27
LutzL
60k42057
edited Dec 11 '18 at 9:27
LutzL
60k42057
edited Dec 11 '18 at 9:27
LutzL
60k42057
60k42057
asked Dec 11 '18 at 6:26
Kevin GachathiKevin Gachathi
61
asked Dec 11 '18 at 6:26
Kevin GachathiKevin Gachathi
61
asked Dec 11 '18 at 6:26
Kevin GachathiKevin Gachathi
61
61
$begingroup$
I have edited, partially heavily, your question. Please check that the corrections conform with your original task, especially that the subscript "$m$" was originally three dashes of the third derivative.
$endgroup$
– LutzL
Dec 11 '18 at 9:28
$begingroup$
Write down the interpolation formula and apply Taylor's theorem.
$endgroup$
– Yves Daoust
Dec 11 '18 at 10:02
add a comment |
$begingroup$
I have edited, partially heavily, your question. Please check that the corrections conform with your original task, especially that the subscript "$m$" was originally three dashes of the third derivative.
$endgroup$
– LutzL
Dec 11 '18 at 9:28
$begingroup$
Write down the interpolation formula and apply Taylor's theorem.
$endgroup$
– Yves Daoust
Dec 11 '18 at 10:02
$begingroup$
I have edited, partially heavily, your question. Please check that the corrections conform with your original task, especially that the subscript "$m$" was originally three dashes of the third derivative.
$endgroup$
– LutzL
Dec 11 '18 at 9:28
$begingroup$
I have edited, partially heavily, your question. Please check that the corrections conform with your original task, especially that the subscript "$m$" was originally three dashes of the third derivative.
$endgroup$
– LutzL
Dec 11 '18 at 9:28
$begingroup$
Write down the interpolation formula and apply Taylor's theorem.
$endgroup$
– Yves Daoust
Dec 11 '18 at 10:02
$begingroup$
Write down the interpolation formula and apply Taylor's theorem.
$endgroup$
– Yves Daoust
Dec 11 '18 at 10:02
add a comment |
1 Answer
1
active
oldest
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$begingroup$
The interpolation error for the quadratic polynomial interpolating at the points $x_{k-1},x_k,x_{k+1}$ and their tabulated values is bounded by
$$
|(x-x_{k-1})(x-x_k)(x-x_{k+1})|frac{1}6max_{zin[x_{k-1},x_{k+1}]}|f'''(z)|
$$
If $x$ is parametrized as $x_k+sh$ with $sin[-frac12,frac12]$, then this bound becomes
$$
|s-s^3|frac{h^3}6max_{zin[x_{k-1},x_{k+1}]}|f'''(z)|
$$
The local maxima of the first factor in $[-1,1]$ are located at $s=pmfrac1{sqrt3}$, which should result in your bound.
But these $s$ are outside the interval $[-frac12,frac12]$. To get an approximate value for the point $x_k+frac1{sqrt3}h$ from the table, one would use the interpolation polynomial based on the triple centered at $x_{k+1}$. Thus the maximum error for the more realistic procedure is at $s=pmfrac12$.
answered Dec 11 '18 at 9:42
LutzLLutzL
60k42057
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The interpolation error for the quadratic polynomial interpolating at the points $x_{k-1},x_k,x_{k+1}$ and their tabulated values is bounded by
$$
|(x-x_{k-1})(x-x_k)(x-x_{k+1})|frac{1}6max_{zin[x_{k-1},x_{k+1}]}|f'''(z)|
$$
If $x$ is parametrized as $x_k+sh$ with $sin[-frac12,frac12]$, then this bound becomes
$$
|s-s^3|frac{h^3}6max_{zin[x_{k-1},x_{k+1}]}|f'''(z)|
$$
The local maxima of the first factor in $[-1,1]$ are located at $s=pmfrac1{sqrt3}$, which should result in your bound.
But these $s$ are outside the interval $[-frac12,frac12]$. To get an approximate value for the point $x_k+frac1{sqrt3}h$ from the table, one would use the interpolation polynomial based on the triple centered at $x_{k+1}$. Thus the maximum error for the more realistic procedure is at $s=pmfrac12$.
answered Dec 11 '18 at 9:42
LutzLLutzL
60k42057
$endgroup$
add a comment |
$begingroup$
The interpolation error for the quadratic polynomial interpolating at the points $x_{k-1},x_k,x_{k+1}$ and their tabulated values is bounded by
$$
|(x-x_{k-1})(x-x_k)(x-x_{k+1})|frac{1}6max_{zin[x_{k-1},x_{k+1}]}|f'''(z)|
$$
If $x$ is parametrized as $x_k+sh$ with $sin[-frac12,frac12]$, then this bound becomes
$$
|s-s^3|frac{h^3}6max_{zin[x_{k-1},x_{k+1}]}|f'''(z)|
$$
The local maxima of the first factor in $[-1,1]$ are located at $s=pmfrac1{sqrt3}$, which should result in your bound.
But these $s$ are outside the interval $[-frac12,frac12]$. To get an approximate value for the point $x_k+frac1{sqrt3}h$ from the table, one would use the interpolation polynomial based on the triple centered at $x_{k+1}$. Thus the maximum error for the more realistic procedure is at $s=pmfrac12$.
answered Dec 11 '18 at 9:42
LutzLLutzL
60k42057
$endgroup$
add a comment |
$begingroup$
The interpolation error for the quadratic polynomial interpolating at the points $x_{k-1},x_k,x_{k+1}$ and their tabulated values is bounded by
$$
|(x-x_{k-1})(x-x_k)(x-x_{k+1})|frac{1}6max_{zin[x_{k-1},x_{k+1}]}|f'''(z)|
$$
If $x$ is parametrized as $x_k+sh$ with $sin[-frac12,frac12]$, then this bound becomes
$$
|s-s^3|frac{h^3}6max_{zin[x_{k-1},x_{k+1}]}|f'''(z)|
$$
The local maxima of the first factor in $[-1,1]$ are located at $s=pmfrac1{sqrt3}$, which should result in your bound.
But these $s$ are outside the interval $[-frac12,frac12]$. To get an approximate value for the point $x_k+frac1{sqrt3}h$ from the table, one would use the interpolation polynomial based on the triple centered at $x_{k+1}$. Thus the maximum error for the more realistic procedure is at $s=pmfrac12$.
answered Dec 11 '18 at 9:42
LutzLLutzL
60k42057
$endgroup$
The interpolation error for the quadratic polynomial interpolating at the points $x_{k-1},x_k,x_{k+1}$ and their tabulated values is bounded by
$$
|(x-x_{k-1})(x-x_k)(x-x_{k+1})|frac{1}6max_{zin[x_{k-1},x_{k+1}]}|f'''(z)|
$$
If $x$ is parametrized as $x_k+sh$ with $sin[-frac12,frac12]$, then this bound becomes
$$
|s-s^3|frac{h^3}6max_{zin[x_{k-1},x_{k+1}]}|f'''(z)|
$$
The local maxima of the first factor in $[-1,1]$ are located at $s=pmfrac1{sqrt3}$, which should result in your bound.
But these $s$ are outside the interval $[-frac12,frac12]$. To get an approximate value for the point $x_k+frac1{sqrt3}h$ from the table, one would use the interpolation polynomial based on the triple centered at $x_{k+1}$. Thus the maximum error for the more realistic procedure is at $s=pmfrac12$.
answered Dec 11 '18 at 9:42
LutzLLutzL
60k42057
edited Dec 12 '18 at 9:09
answered Dec 11 '18 at 9:42
LutzLLutzL
60k42057
answered Dec 11 '18 at 9:42
LutzLLutzL
60k42057
answered Dec 11 '18 at 9:42
LutzLLutzL
60k42057
60k42057
add a comment |
add a comment |
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$begingroup$
I have edited, partially heavily, your question. Please check that the corrections conform with your original task, especially that the subscript "$m$" was originally three dashes of the third derivative.
$endgroup$
– LutzL
Dec 11 '18 at 9:28
$begingroup$
Write down the interpolation formula and apply Taylor's theorem.
$endgroup$
– Yves Daoust
Dec 11 '18 at 10:02