find coefficients of a numerical integration formula
$begingroup$
we have this formula for calculating integral
$int_{0}^{h} xf(x) dx = (h)^2 [A_0 f(0) + A_1f(h) + A_2f(2h)] + R \$
the goal is to find $A_0,A_1, A_2$ and $R$, when $f^{primeprimeprime}$ is constant.
I tried to get to this formula by taking integral of Taylor Expansion of $xf(x)$ but it didn't work, or I messed up in it.
integration numerical-methods
asked Jan 8 at 16:47
SajadSajad
103
$endgroup$
add a comment |
$begingroup$
we have this formula for calculating integral
$int_{0}^{h} xf(x) dx = (h)^2 [A_0 f(0) + A_1f(h) + A_2f(2h)] + R \$
the goal is to find $A_0,A_1, A_2$ and $R$, when $f^{primeprimeprime}$ is constant.
I tried to get to this formula by taking integral of Taylor Expansion of $xf(x)$ but it didn't work, or I messed up in it.
integration numerical-methods
asked Jan 8 at 16:47
SajadSajad
103
$endgroup$
add a comment |
$begingroup$
we have this formula for calculating integral
$int_{0}^{h} xf(x) dx = (h)^2 [A_0 f(0) + A_1f(h) + A_2f(2h)] + R \$
the goal is to find $A_0,A_1, A_2$ and $R$, when $f^{primeprimeprime}$ is constant.
I tried to get to this formula by taking integral of Taylor Expansion of $xf(x)$ but it didn't work, or I messed up in it.
integration numerical-methods
asked Jan 8 at 16:47
SajadSajad
103
$endgroup$
we have this formula for calculating integral
$int_{0}^{h} xf(x) dx = (h)^2 [A_0 f(0) + A_1f(h) + A_2f(2h)] + R \$
the goal is to find $A_0,A_1, A_2$ and $R$, when $f^{primeprimeprime}$ is constant.
I tried to get to this formula by taking integral of Taylor Expansion of $xf(x)$ but it didn't work, or I messed up in it.
integration numerical-methods
integration numerical-methods
asked Jan 8 at 16:47
SajadSajad
103
asked Jan 8 at 16:47
SajadSajad
103
asked Jan 8 at 16:47
SajadSajad
103
asked Jan 8 at 16:47
SajadSajad
103
asked Jan 8 at 16:47
SajadSajad
103
103
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
The third derivative $f$ being constant means that $f$ is a cubic, so let $f(x)=ax^3+b^2x+cx+d$. Plug that into your equation. The result needs to be true for all $a,b,c,d$. You will get simultaneous equations for $A_0,A_1,A_2$. Note that $f(x)=0$ tells us that $R=0$. It will fail because you have four equations in three unknowns.
answered Jan 8 at 16:53
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
so it's ok that for R to be 0? cause I did this too but I thought it's wrong :))
$endgroup$
– Sajad
Jan 8 at 16:58
$begingroup$
Yes, $R$ has to be $0$. If it weren't, you could let $h$ go to zero and have the left be very small, smaller than $R$. Some of the $A$s are $0$ in formulas like this as well. There is no solution to the problem as stated. You need one more point and one more $A$ on the right to make it work, or to demand that $f''(x)$ is constant so you only have a quadratic.
$endgroup$
– Ross Millikan
Jan 8 at 17:01
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The third derivative $f$ being constant means that $f$ is a cubic, so let $f(x)=ax^3+b^2x+cx+d$. Plug that into your equation. The result needs to be true for all $a,b,c,d$. You will get simultaneous equations for $A_0,A_1,A_2$. Note that $f(x)=0$ tells us that $R=0$. It will fail because you have four equations in three unknowns.
answered Jan 8 at 16:53
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
so it's ok that for R to be 0? cause I did this too but I thought it's wrong :))
$endgroup$
– Sajad
Jan 8 at 16:58
$begingroup$
Yes, $R$ has to be $0$. If it weren't, you could let $h$ go to zero and have the left be very small, smaller than $R$. Some of the $A$s are $0$ in formulas like this as well. There is no solution to the problem as stated. You need one more point and one more $A$ on the right to make it work, or to demand that $f''(x)$ is constant so you only have a quadratic.
$endgroup$
– Ross Millikan
Jan 8 at 17:01
add a comment |
$begingroup$
The third derivative $f$ being constant means that $f$ is a cubic, so let $f(x)=ax^3+b^2x+cx+d$. Plug that into your equation. The result needs to be true for all $a,b,c,d$. You will get simultaneous equations for $A_0,A_1,A_2$. Note that $f(x)=0$ tells us that $R=0$. It will fail because you have four equations in three unknowns.
answered Jan 8 at 16:53
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
so it's ok that for R to be 0? cause I did this too but I thought it's wrong :))
$endgroup$
– Sajad
Jan 8 at 16:58
$begingroup$
Yes, $R$ has to be $0$. If it weren't, you could let $h$ go to zero and have the left be very small, smaller than $R$. Some of the $A$s are $0$ in formulas like this as well. There is no solution to the problem as stated. You need one more point and one more $A$ on the right to make it work, or to demand that $f''(x)$ is constant so you only have a quadratic.
$endgroup$
– Ross Millikan
Jan 8 at 17:01
add a comment |
$begingroup$
The third derivative $f$ being constant means that $f$ is a cubic, so let $f(x)=ax^3+b^2x+cx+d$. Plug that into your equation. The result needs to be true for all $a,b,c,d$. You will get simultaneous equations for $A_0,A_1,A_2$. Note that $f(x)=0$ tells us that $R=0$. It will fail because you have four equations in three unknowns.
answered Jan 8 at 16:53
Ross MillikanRoss Millikan
297k23198371
$endgroup$
The third derivative $f$ being constant means that $f$ is a cubic, so let $f(x)=ax^3+b^2x+cx+d$. Plug that into your equation. The result needs to be true for all $a,b,c,d$. You will get simultaneous equations for $A_0,A_1,A_2$. Note that $f(x)=0$ tells us that $R=0$. It will fail because you have four equations in three unknowns.
answered Jan 8 at 16:53
Ross MillikanRoss Millikan
297k23198371
edited Jan 8 at 17:01
answered Jan 8 at 16:53
Ross MillikanRoss Millikan
297k23198371
answered Jan 8 at 16:53
Ross MillikanRoss Millikan
297k23198371
answered Jan 8 at 16:53
Ross MillikanRoss Millikan
297k23198371
297k23198371
$begingroup$
so it's ok that for R to be 0? cause I did this too but I thought it's wrong :))
$endgroup$
– Sajad
Jan 8 at 16:58
$begingroup$
Yes, $R$ has to be $0$. If it weren't, you could let $h$ go to zero and have the left be very small, smaller than $R$. Some of the $A$s are $0$ in formulas like this as well. There is no solution to the problem as stated. You need one more point and one more $A$ on the right to make it work, or to demand that $f''(x)$ is constant so you only have a quadratic.
$endgroup$
– Ross Millikan
Jan 8 at 17:01
add a comment |
$begingroup$
so it's ok that for R to be 0? cause I did this too but I thought it's wrong :))
$endgroup$
– Sajad
Jan 8 at 16:58
$begingroup$
Yes, $R$ has to be $0$. If it weren't, you could let $h$ go to zero and have the left be very small, smaller than $R$. Some of the $A$s are $0$ in formulas like this as well. There is no solution to the problem as stated. You need one more point and one more $A$ on the right to make it work, or to demand that $f''(x)$ is constant so you only have a quadratic.
$endgroup$
– Ross Millikan
Jan 8 at 17:01
$begingroup$
so it's ok that for R to be 0? cause I did this too but I thought it's wrong :))
$endgroup$
– Sajad
Jan 8 at 16:58
$begingroup$
so it's ok that for R to be 0? cause I did this too but I thought it's wrong :))
$endgroup$
– Sajad
Jan 8 at 16:58
$begingroup$
Yes, $R$ has to be $0$. If it weren't, you could let $h$ go to zero and have the left be very small, smaller than $R$. Some of the $A$s are $0$ in formulas like this as well. There is no solution to the problem as stated. You need one more point and one more $A$ on the right to make it work, or to demand that $f''(x)$ is constant so you only have a quadratic.
$endgroup$
– Ross Millikan
Jan 8 at 17:01
$begingroup$
Yes, $R$ has to be $0$. If it weren't, you could let $h$ go to zero and have the left be very small, smaller than $R$. Some of the $A$s are $0$ in formulas like this as well. There is no solution to the problem as stated. You need one more point and one more $A$ on the right to make it work, or to demand that $f''(x)$ is constant so you only have a quadratic.
$endgroup$
– Ross Millikan
Jan 8 at 17:01
add a comment |
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