Second order accurate numerical approximation for first derivative Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Taylor expansion: first derivative approximation with third orderBoundary Conditions for a Finite Difference Approximation of a Sixth Derivative“Symmetric” numerical computation of second derivativeNumerical differentiation (approximation with three supporting points )Second order one-sided finite difference approximation to a partial derivative4th order accurate difference formula less accurate than 2nd order formula?Approximation of the first derivative by writing Taylor expansionsDerivation of fourth-order accurate formula for the second derivativeApproximation of $log(x)$ for very small $x$Second order approximation of first derivative gives odd results
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Second order accurate numerical approximation for first derivative
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Taylor expansion: first derivative approximation with third orderBoundary Conditions for a Finite Difference Approximation of a Sixth Derivative“Symmetric” numerical computation of second derivativeNumerical differentiation (approximation with three supporting points )Second order one-sided finite difference approximation to a partial derivative4th order accurate difference formula less accurate than 2nd order formula?Approximation of the first derivative by writing Taylor expansionsDerivation of fourth-order accurate formula for the second derivativeApproximation of $log(x)$ for very small $x$Second order approximation of first derivative gives odd results
$begingroup$
We just learnt about one sided and centred difference approximations in class and we have been given a problem to find $a_0, a_1$ and $a_2$ in the below numerical approximation for a first derivative in order to make the approximation second order accurate:$$F'(x) approx frac1Delta x[a_0F(x + Delta x) + a_1F(x + 2 Delta x) + a_2F(x + 3 Delta x)].$$
Would it be right to say that we want the RHS to be in this form: $$fracF(x + Delta x) - F(x -Delta x)2 Delta x$$ and so, somehow, we'd manipulate the Taylor expansion of each $F(...)$ term in the first derivative approximation equation?
derivatives numerical-methods
asked Apr 2 at 10:56
FrancisFrancis
393
$endgroup$
add a comment |
$begingroup$
We just learnt about one sided and centred difference approximations in class and we have been given a problem to find $a_0, a_1$ and $a_2$ in the below numerical approximation for a first derivative in order to make the approximation second order accurate:$$F'(x) approx frac1Delta x[a_0F(x + Delta x) + a_1F(x + 2 Delta x) + a_2F(x + 3 Delta x)].$$
Would it be right to say that we want the RHS to be in this form: $$fracF(x + Delta x) - F(x -Delta x)2 Delta x$$ and so, somehow, we'd manipulate the Taylor expansion of each $F(...)$ term in the first derivative approximation equation?
derivatives numerical-methods
asked Apr 2 at 10:56
FrancisFrancis
393
$endgroup$
add a comment |
$begingroup$
We just learnt about one sided and centred difference approximations in class and we have been given a problem to find $a_0, a_1$ and $a_2$ in the below numerical approximation for a first derivative in order to make the approximation second order accurate:$$F'(x) approx frac1Delta x[a_0F(x + Delta x) + a_1F(x + 2 Delta x) + a_2F(x + 3 Delta x)].$$
Would it be right to say that we want the RHS to be in this form: $$fracF(x + Delta x) - F(x -Delta x)2 Delta x$$ and so, somehow, we'd manipulate the Taylor expansion of each $F(...)$ term in the first derivative approximation equation?
derivatives numerical-methods
asked Apr 2 at 10:56
FrancisFrancis
393
$endgroup$
We just learnt about one sided and centred difference approximations in class and we have been given a problem to find $a_0, a_1$ and $a_2$ in the below numerical approximation for a first derivative in order to make the approximation second order accurate:$$F'(x) approx frac1Delta x[a_0F(x + Delta x) + a_1F(x + 2 Delta x) + a_2F(x + 3 Delta x)].$$
Would it be right to say that we want the RHS to be in this form: $$fracF(x + Delta x) - F(x -Delta x)2 Delta x$$ and so, somehow, we'd manipulate the Taylor expansion of each $F(...)$ term in the first derivative approximation equation?
derivatives numerical-methods
derivatives numerical-methods
asked Apr 2 at 10:56
FrancisFrancis
393
asked Apr 2 at 10:56
FrancisFrancis
393
asked Apr 2 at 10:56
FrancisFrancis
393
asked Apr 2 at 10:56
FrancisFrancis
393
asked Apr 2 at 10:56
FrancisFrancis
393
393
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The formula you mention is a centered difference, but you are required to obtain a forward difference. You just need to use Taylor's formula and compute the coefficients that cancel out the lower order terms.
$$
F(x+Delta x)= F(x)+F'(x) Delta x + fracF''(x)2 (Delta x)^2 + O((Delta x)^3)
$$
$$
F(x+ 2 Delta x)= F(x)+F'(x) 2 Delta x + fracF''(x)2 (2 Delta x)^2 + O((Delta x)^3)
$$
$$
F(x+3Delta x)= F(x)+F'(x) 3Delta x + fracF''(x)2 (3Delta x)^2 + O((Delta x)^3)
$$
so you see that
beginalign*
F'(x)-frac1Delta x & (a_0 F(x+Delta x)+a_1 F(x+2Delta x) + a_2 F(x+3 Delta x))\
=& frac1Delta xleft(F'(x) Delta x - a_0 (F(x)+F'(x) Delta x + fracF''(x)2 (Delta x)^2 + O((Delta x)^3))right.\
& - a_1 (F(x)+F'(x) 2 Delta x + fracF''(x)2 (2 Delta x)^2 + O((Delta x)^3))\
& left.- a_2 (F(x)+F'(x) 3Delta x + fracF''(x)2 (3Delta x)^2 + O((Delta x)^3))right)\
=& frac1Delta xleft( -(a_0+a_1+a_2)F(x)+(1-a_0-2a_1-3a_2)F'(x)Delta xright.\
& left.-(a_0+4a_1+9a_2)F''(x) (Delta x)^2/2 + O((Delta x)^3)right)
endalign*
To get the desired accuracy you just have to set $a_0, a_1, a_2$ such that
$$
a_0+a_1+a_2=0, quad a_0+2a_1+3a_2 = 1, quad a_0+4a_2+9a_2=0,
$$
that is
$$
a_0 = -frac 52, quad a_1 =4, quad a_2=-frac 32
$$
and the formula becomes
$$
F'(x) approx frac12Delta x left(-5 F(x+Delta x)+8F(x+2 Delta x) - 4F(x+3 Delta x) right)
$$
answered Apr 4 at 8:04
PierreCarrePierreCarre
2,2781215
$endgroup$
add a comment |
$begingroup$
$a_0f(x+Delta x) = a_0f(x) + a_0Delta xf'(x) + a_0frac12Delta x^2f'(x) + a_0frac16Delta x^3f'''(xi_1)$
$a_1f(x+2Delta x) = a_1f(x) + a_12Delta xf'(x) +
a_12Delta x^2f''(x) + a_1frac43Delta x^3f'''(xi_2)$
$a_2f(x+3Delta x) = a_2f(x) + a_23Delta xf(x) + a_2frac94Delta x^2f''(x) + a_2frac92Delta x^3f'''(xi_3) $
this implies the augmented matrix:
$left(beginarrayc
1 & 1 & 1 & 0\
1 & 2 & 3 & 1 \
frac12 & 2 & frac94 & 0
endarrayright)$
do you see why?
answered Apr 4 at 5:36
GeauxMathGeauxMath
1114
$endgroup$
add a comment |
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2 Answers
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2 Answers
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votes
$begingroup$
The formula you mention is a centered difference, but you are required to obtain a forward difference. You just need to use Taylor's formula and compute the coefficients that cancel out the lower order terms.
$$
F(x+Delta x)= F(x)+F'(x) Delta x + fracF''(x)2 (Delta x)^2 + O((Delta x)^3)
$$
$$
F(x+ 2 Delta x)= F(x)+F'(x) 2 Delta x + fracF''(x)2 (2 Delta x)^2 + O((Delta x)^3)
$$
$$
F(x+3Delta x)= F(x)+F'(x) 3Delta x + fracF''(x)2 (3Delta x)^2 + O((Delta x)^3)
$$
so you see that
beginalign*
F'(x)-frac1Delta x & (a_0 F(x+Delta x)+a_1 F(x+2Delta x) + a_2 F(x+3 Delta x))\
=& frac1Delta xleft(F'(x) Delta x - a_0 (F(x)+F'(x) Delta x + fracF''(x)2 (Delta x)^2 + O((Delta x)^3))right.\
& - a_1 (F(x)+F'(x) 2 Delta x + fracF''(x)2 (2 Delta x)^2 + O((Delta x)^3))\
& left.- a_2 (F(x)+F'(x) 3Delta x + fracF''(x)2 (3Delta x)^2 + O((Delta x)^3))right)\
=& frac1Delta xleft( -(a_0+a_1+a_2)F(x)+(1-a_0-2a_1-3a_2)F'(x)Delta xright.\
& left.-(a_0+4a_1+9a_2)F''(x) (Delta x)^2/2 + O((Delta x)^3)right)
endalign*
To get the desired accuracy you just have to set $a_0, a_1, a_2$ such that
$$
a_0+a_1+a_2=0, quad a_0+2a_1+3a_2 = 1, quad a_0+4a_2+9a_2=0,
$$
that is
$$
a_0 = -frac 52, quad a_1 =4, quad a_2=-frac 32
$$
and the formula becomes
$$
F'(x) approx frac12Delta x left(-5 F(x+Delta x)+8F(x+2 Delta x) - 4F(x+3 Delta x) right)
$$
answered Apr 4 at 8:04
PierreCarrePierreCarre
2,2781215
$endgroup$
add a comment |
$begingroup$
The formula you mention is a centered difference, but you are required to obtain a forward difference. You just need to use Taylor's formula and compute the coefficients that cancel out the lower order terms.
$$
F(x+Delta x)= F(x)+F'(x) Delta x + fracF''(x)2 (Delta x)^2 + O((Delta x)^3)
$$
$$
F(x+ 2 Delta x)= F(x)+F'(x) 2 Delta x + fracF''(x)2 (2 Delta x)^2 + O((Delta x)^3)
$$
$$
F(x+3Delta x)= F(x)+F'(x) 3Delta x + fracF''(x)2 (3Delta x)^2 + O((Delta x)^3)
$$
so you see that
beginalign*
F'(x)-frac1Delta x & (a_0 F(x+Delta x)+a_1 F(x+2Delta x) + a_2 F(x+3 Delta x))\
=& frac1Delta xleft(F'(x) Delta x - a_0 (F(x)+F'(x) Delta x + fracF''(x)2 (Delta x)^2 + O((Delta x)^3))right.\
& - a_1 (F(x)+F'(x) 2 Delta x + fracF''(x)2 (2 Delta x)^2 + O((Delta x)^3))\
& left.- a_2 (F(x)+F'(x) 3Delta x + fracF''(x)2 (3Delta x)^2 + O((Delta x)^3))right)\
=& frac1Delta xleft( -(a_0+a_1+a_2)F(x)+(1-a_0-2a_1-3a_2)F'(x)Delta xright.\
& left.-(a_0+4a_1+9a_2)F''(x) (Delta x)^2/2 + O((Delta x)^3)right)
endalign*
To get the desired accuracy you just have to set $a_0, a_1, a_2$ such that
$$
a_0+a_1+a_2=0, quad a_0+2a_1+3a_2 = 1, quad a_0+4a_2+9a_2=0,
$$
that is
$$
a_0 = -frac 52, quad a_1 =4, quad a_2=-frac 32
$$
and the formula becomes
$$
F'(x) approx frac12Delta x left(-5 F(x+Delta x)+8F(x+2 Delta x) - 4F(x+3 Delta x) right)
$$
answered Apr 4 at 8:04
PierreCarrePierreCarre
2,2781215
$endgroup$
add a comment |
$begingroup$
The formula you mention is a centered difference, but you are required to obtain a forward difference. You just need to use Taylor's formula and compute the coefficients that cancel out the lower order terms.
$$
F(x+Delta x)= F(x)+F'(x) Delta x + fracF''(x)2 (Delta x)^2 + O((Delta x)^3)
$$
$$
F(x+ 2 Delta x)= F(x)+F'(x) 2 Delta x + fracF''(x)2 (2 Delta x)^2 + O((Delta x)^3)
$$
$$
F(x+3Delta x)= F(x)+F'(x) 3Delta x + fracF''(x)2 (3Delta x)^2 + O((Delta x)^3)
$$
so you see that
beginalign*
F'(x)-frac1Delta x & (a_0 F(x+Delta x)+a_1 F(x+2Delta x) + a_2 F(x+3 Delta x))\
=& frac1Delta xleft(F'(x) Delta x - a_0 (F(x)+F'(x) Delta x + fracF''(x)2 (Delta x)^2 + O((Delta x)^3))right.\
& - a_1 (F(x)+F'(x) 2 Delta x + fracF''(x)2 (2 Delta x)^2 + O((Delta x)^3))\
& left.- a_2 (F(x)+F'(x) 3Delta x + fracF''(x)2 (3Delta x)^2 + O((Delta x)^3))right)\
=& frac1Delta xleft( -(a_0+a_1+a_2)F(x)+(1-a_0-2a_1-3a_2)F'(x)Delta xright.\
& left.-(a_0+4a_1+9a_2)F''(x) (Delta x)^2/2 + O((Delta x)^3)right)
endalign*
To get the desired accuracy you just have to set $a_0, a_1, a_2$ such that
$$
a_0+a_1+a_2=0, quad a_0+2a_1+3a_2 = 1, quad a_0+4a_2+9a_2=0,
$$
that is
$$
a_0 = -frac 52, quad a_1 =4, quad a_2=-frac 32
$$
and the formula becomes
$$
F'(x) approx frac12Delta x left(-5 F(x+Delta x)+8F(x+2 Delta x) - 4F(x+3 Delta x) right)
$$
answered Apr 4 at 8:04
PierreCarrePierreCarre
2,2781215
$endgroup$
The formula you mention is a centered difference, but you are required to obtain a forward difference. You just need to use Taylor's formula and compute the coefficients that cancel out the lower order terms.
$$
F(x+Delta x)= F(x)+F'(x) Delta x + fracF''(x)2 (Delta x)^2 + O((Delta x)^3)
$$
$$
F(x+ 2 Delta x)= F(x)+F'(x) 2 Delta x + fracF''(x)2 (2 Delta x)^2 + O((Delta x)^3)
$$
$$
F(x+3Delta x)= F(x)+F'(x) 3Delta x + fracF''(x)2 (3Delta x)^2 + O((Delta x)^3)
$$
so you see that
beginalign*
F'(x)-frac1Delta x & (a_0 F(x+Delta x)+a_1 F(x+2Delta x) + a_2 F(x+3 Delta x))\
=& frac1Delta xleft(F'(x) Delta x - a_0 (F(x)+F'(x) Delta x + fracF''(x)2 (Delta x)^2 + O((Delta x)^3))right.\
& - a_1 (F(x)+F'(x) 2 Delta x + fracF''(x)2 (2 Delta x)^2 + O((Delta x)^3))\
& left.- a_2 (F(x)+F'(x) 3Delta x + fracF''(x)2 (3Delta x)^2 + O((Delta x)^3))right)\
=& frac1Delta xleft( -(a_0+a_1+a_2)F(x)+(1-a_0-2a_1-3a_2)F'(x)Delta xright.\
& left.-(a_0+4a_1+9a_2)F''(x) (Delta x)^2/2 + O((Delta x)^3)right)
endalign*
To get the desired accuracy you just have to set $a_0, a_1, a_2$ such that
$$
a_0+a_1+a_2=0, quad a_0+2a_1+3a_2 = 1, quad a_0+4a_2+9a_2=0,
$$
that is
$$
a_0 = -frac 52, quad a_1 =4, quad a_2=-frac 32
$$
and the formula becomes
$$
F'(x) approx frac12Delta x left(-5 F(x+Delta x)+8F(x+2 Delta x) - 4F(x+3 Delta x) right)
$$
answered Apr 4 at 8:04
PierreCarrePierreCarre
2,2781215
answered Apr 4 at 8:04
PierreCarrePierreCarre
2,2781215
answered Apr 4 at 8:04
PierreCarrePierreCarre
2,2781215
answered Apr 4 at 8:04
PierreCarrePierreCarre
2,2781215
2,2781215
add a comment |
add a comment |
$begingroup$
$a_0f(x+Delta x) = a_0f(x) + a_0Delta xf'(x) + a_0frac12Delta x^2f'(x) + a_0frac16Delta x^3f'''(xi_1)$
$a_1f(x+2Delta x) = a_1f(x) + a_12Delta xf'(x) +
a_12Delta x^2f''(x) + a_1frac43Delta x^3f'''(xi_2)$
$a_2f(x+3Delta x) = a_2f(x) + a_23Delta xf(x) + a_2frac94Delta x^2f''(x) + a_2frac92Delta x^3f'''(xi_3) $
this implies the augmented matrix:
$left(beginarrayc
1 & 1 & 1 & 0\
1 & 2 & 3 & 1 \
frac12 & 2 & frac94 & 0
endarrayright)$
do you see why?
answered Apr 4 at 5:36
GeauxMathGeauxMath
1114
$endgroup$
add a comment |
$begingroup$
$a_0f(x+Delta x) = a_0f(x) + a_0Delta xf'(x) + a_0frac12Delta x^2f'(x) + a_0frac16Delta x^3f'''(xi_1)$
$a_1f(x+2Delta x) = a_1f(x) + a_12Delta xf'(x) +
a_12Delta x^2f''(x) + a_1frac43Delta x^3f'''(xi_2)$
$a_2f(x+3Delta x) = a_2f(x) + a_23Delta xf(x) + a_2frac94Delta x^2f''(x) + a_2frac92Delta x^3f'''(xi_3) $
this implies the augmented matrix:
$left(beginarrayc
1 & 1 & 1 & 0\
1 & 2 & 3 & 1 \
frac12 & 2 & frac94 & 0
endarrayright)$
do you see why?
answered Apr 4 at 5:36
GeauxMathGeauxMath
1114
$endgroup$
add a comment |
$begingroup$
$a_0f(x+Delta x) = a_0f(x) + a_0Delta xf'(x) + a_0frac12Delta x^2f'(x) + a_0frac16Delta x^3f'''(xi_1)$
$a_1f(x+2Delta x) = a_1f(x) + a_12Delta xf'(x) +
a_12Delta x^2f''(x) + a_1frac43Delta x^3f'''(xi_2)$
$a_2f(x+3Delta x) = a_2f(x) + a_23Delta xf(x) + a_2frac94Delta x^2f''(x) + a_2frac92Delta x^3f'''(xi_3) $
this implies the augmented matrix:
$left(beginarrayc
1 & 1 & 1 & 0\
1 & 2 & 3 & 1 \
frac12 & 2 & frac94 & 0
endarrayright)$
do you see why?
answered Apr 4 at 5:36
GeauxMathGeauxMath
1114
$endgroup$
$a_0f(x+Delta x) = a_0f(x) + a_0Delta xf'(x) + a_0frac12Delta x^2f'(x) + a_0frac16Delta x^3f'''(xi_1)$
$a_1f(x+2Delta x) = a_1f(x) + a_12Delta xf'(x) +
a_12Delta x^2f''(x) + a_1frac43Delta x^3f'''(xi_2)$
$a_2f(x+3Delta x) = a_2f(x) + a_23Delta xf(x) + a_2frac94Delta x^2f''(x) + a_2frac92Delta x^3f'''(xi_3) $
this implies the augmented matrix:
$left(beginarrayc
1 & 1 & 1 & 0\
1 & 2 & 3 & 1 \
frac12 & 2 & frac94 & 0
endarrayright)$
do you see why?
answered Apr 4 at 5:36
GeauxMathGeauxMath
1114
answered Apr 4 at 5:36
GeauxMathGeauxMath
1114
answered Apr 4 at 5:36
GeauxMathGeauxMath
1114
answered Apr 4 at 5:36
GeauxMathGeauxMath
1114
1114
add a comment |
add a comment |
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