Mixing
Coefficients of numerical approximation of differentiation
$begingroup$
there is given
$y^prime(frac{a+b}{2}) = alpha y(a) + beta y(b) + gamma y^{prime prime}(a) + delta y^{prime prime}(b) quad quad (dagger)$
we want to find $alpha, beta, gamma, delta$ such that ($dagger$) is as accurate as possible for high degree polynomials and what is error when $|a-b| rightarrow 0$.
I actually don't know where I should start from, any tip?
numerical-methods
asked Jan 9 at 14:55
SajadSajad
103
$endgroup$
add a comment |
$begingroup$
there is given
$y^prime(frac{a+b}{2}) = alpha y(a) + beta y(b) + gamma y^{prime prime}(a) + delta y^{prime prime}(b) quad quad (dagger)$
we want to find $alpha, beta, gamma, delta$ such that ($dagger$) is as accurate as possible for high degree polynomials and what is error when $|a-b| rightarrow 0$.
I actually don't know where I should start from, any tip?
numerical-methods
asked Jan 9 at 14:55
SajadSajad
103
$endgroup$
add a comment |
$begingroup$
there is given
$y^prime(frac{a+b}{2}) = alpha y(a) + beta y(b) + gamma y^{prime prime}(a) + delta y^{prime prime}(b) quad quad (dagger)$
we want to find $alpha, beta, gamma, delta$ such that ($dagger$) is as accurate as possible for high degree polynomials and what is error when $|a-b| rightarrow 0$.
I actually don't know where I should start from, any tip?
numerical-methods
asked Jan 9 at 14:55
SajadSajad
103
$endgroup$
there is given
$y^prime(frac{a+b}{2}) = alpha y(a) + beta y(b) + gamma y^{prime prime}(a) + delta y^{prime prime}(b) quad quad (dagger)$
we want to find $alpha, beta, gamma, delta$ such that ($dagger$) is as accurate as possible for high degree polynomials and what is error when $|a-b| rightarrow 0$.
I actually don't know where I should start from, any tip?
numerical-methods
numerical-methods
asked Jan 9 at 14:55
SajadSajad
103
asked Jan 9 at 14:55
SajadSajad
103
asked Jan 9 at 14:55
SajadSajad
103
asked Jan 9 at 14:55
SajadSajad
103
asked Jan 9 at 14:55
SajadSajad
103
103
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
You have four adjustable parameters, so should expect to be able to be exact for up to third degree. Let $y=cx^3+dx^2+ex+f$ and plug it and its derivatives into the formula. You get one equation from the terms in each of $c,d,e,f$, so have four equations in four unknowns. There will be a lot of symmetry, so the system will be easy to solve.
It is even a little easier if you assume $a=-b$ so the center point is $0$. The result will be the same but some of the terms will disappear.
answered Jan 9 at 15:02
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
"You have four adjustable parameters, so should expect to be able to be exact for up to third degree", it's not clear to me why is that true
$endgroup$
– Sajad
Jan 9 at 15:05
$begingroup$
Because a cubic has four parameters. If you tried with a quartic there would be five equations in four unknowns and you would be unlikely to have a solution.
$endgroup$
– Ross Millikan
Jan 9 at 15:07
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
You have four adjustable parameters, so should expect to be able to be exact for up to third degree. Let $y=cx^3+dx^2+ex+f$ and plug it and its derivatives into the formula. You get one equation from the terms in each of $c,d,e,f$, so have four equations in four unknowns. There will be a lot of symmetry, so the system will be easy to solve.
It is even a little easier if you assume $a=-b$ so the center point is $0$. The result will be the same but some of the terms will disappear.
answered Jan 9 at 15:02
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
"You have four adjustable parameters, so should expect to be able to be exact for up to third degree", it's not clear to me why is that true
$endgroup$
– Sajad
Jan 9 at 15:05
$begingroup$
Because a cubic has four parameters. If you tried with a quartic there would be five equations in four unknowns and you would be unlikely to have a solution.
$endgroup$
– Ross Millikan
Jan 9 at 15:07
add a comment |
$begingroup$
You have four adjustable parameters, so should expect to be able to be exact for up to third degree. Let $y=cx^3+dx^2+ex+f$ and plug it and its derivatives into the formula. You get one equation from the terms in each of $c,d,e,f$, so have four equations in four unknowns. There will be a lot of symmetry, so the system will be easy to solve.
It is even a little easier if you assume $a=-b$ so the center point is $0$. The result will be the same but some of the terms will disappear.
answered Jan 9 at 15:02
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
"You have four adjustable parameters, so should expect to be able to be exact for up to third degree", it's not clear to me why is that true
$endgroup$
– Sajad
Jan 9 at 15:05
$begingroup$
Because a cubic has four parameters. If you tried with a quartic there would be five equations in four unknowns and you would be unlikely to have a solution.
$endgroup$
– Ross Millikan
Jan 9 at 15:07
add a comment |
$begingroup$
You have four adjustable parameters, so should expect to be able to be exact for up to third degree. Let $y=cx^3+dx^2+ex+f$ and plug it and its derivatives into the formula. You get one equation from the terms in each of $c,d,e,f$, so have four equations in four unknowns. There will be a lot of symmetry, so the system will be easy to solve.
It is even a little easier if you assume $a=-b$ so the center point is $0$. The result will be the same but some of the terms will disappear.
answered Jan 9 at 15:02
Ross MillikanRoss Millikan
295k23198371
$endgroup$
You have four adjustable parameters, so should expect to be able to be exact for up to third degree. Let $y=cx^3+dx^2+ex+f$ and plug it and its derivatives into the formula. You get one equation from the terms in each of $c,d,e,f$, so have four equations in four unknowns. There will be a lot of symmetry, so the system will be easy to solve.
It is even a little easier if you assume $a=-b$ so the center point is $0$. The result will be the same but some of the terms will disappear.
answered Jan 9 at 15:02
Ross MillikanRoss Millikan
295k23198371
answered Jan 9 at 15:02
Ross MillikanRoss Millikan
295k23198371
answered Jan 9 at 15:02
Ross MillikanRoss Millikan
295k23198371
answered Jan 9 at 15:02
Ross MillikanRoss Millikan
295k23198371
295k23198371
$begingroup$
"You have four adjustable parameters, so should expect to be able to be exact for up to third degree", it's not clear to me why is that true
$endgroup$
– Sajad
Jan 9 at 15:05
$begingroup$
Because a cubic has four parameters. If you tried with a quartic there would be five equations in four unknowns and you would be unlikely to have a solution.
$endgroup$
– Ross Millikan
Jan 9 at 15:07
add a comment |
$begingroup$
"You have four adjustable parameters, so should expect to be able to be exact for up to third degree", it's not clear to me why is that true
$endgroup$
– Sajad
Jan 9 at 15:05
$begingroup$
Because a cubic has four parameters. If you tried with a quartic there would be five equations in four unknowns and you would be unlikely to have a solution.
$endgroup$
– Ross Millikan
Jan 9 at 15:07
$begingroup$
"You have four adjustable parameters, so should expect to be able to be exact for up to third degree", it's not clear to me why is that true
$endgroup$
– Sajad
Jan 9 at 15:05
$begingroup$
"You have four adjustable parameters, so should expect to be able to be exact for up to third degree", it's not clear to me why is that true
$endgroup$
– Sajad
Jan 9 at 15:05
$begingroup$
Because a cubic has four parameters. If you tried with a quartic there would be five equations in four unknowns and you would be unlikely to have a solution.
$endgroup$
– Ross Millikan
Jan 9 at 15:07
$begingroup$
Because a cubic has four parameters. If you tried with a quartic there would be five equations in four unknowns and you would be unlikely to have a solution.
$endgroup$
– Ross Millikan
Jan 9 at 15:07
add a comment |
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