Scaling forcing function before numerical integration of linear ODE
$begingroup$
Suppose that we have a linear ode which we want to integrate numerically:
$$
L{y}=f(t)
$$
where $L$ is a linear differential operator (and some initial conditions), by using a method of order $mathcal{O}(h^n)$ -- $h$ being the step size of the method.
Is it true that if we scale up the forcing function $f(t)$ with a factor $K$ before integrating, and then rescale down the solution $y$ by $frac{1}{K}$ we have increased the precision of the method? The argument being that the order of the method is still the same and only dependent on the step size. So when we scale down the error is reduced.
ordinary-differential-equations numerical-methods soft-question
asked Jan 7 at 20:31
AmbeshAmbesh
1,4351140
$endgroup$
add a comment |
$begingroup$
Suppose that we have a linear ode which we want to integrate numerically:
$$
L{y}=f(t)
$$
where $L$ is a linear differential operator (and some initial conditions), by using a method of order $mathcal{O}(h^n)$ -- $h$ being the step size of the method.
Is it true that if we scale up the forcing function $f(t)$ with a factor $K$ before integrating, and then rescale down the solution $y$ by $frac{1}{K}$ we have increased the precision of the method? The argument being that the order of the method is still the same and only dependent on the step size. So when we scale down the error is reduced.
ordinary-differential-equations numerical-methods soft-question
asked Jan 7 at 20:31
AmbeshAmbesh
1,4351140
$endgroup$
add a comment |
$begingroup$
Suppose that we have a linear ode which we want to integrate numerically:
$$
L{y}=f(t)
$$
where $L$ is a linear differential operator (and some initial conditions), by using a method of order $mathcal{O}(h^n)$ -- $h$ being the step size of the method.
Is it true that if we scale up the forcing function $f(t)$ with a factor $K$ before integrating, and then rescale down the solution $y$ by $frac{1}{K}$ we have increased the precision of the method? The argument being that the order of the method is still the same and only dependent on the step size. So when we scale down the error is reduced.
ordinary-differential-equations numerical-methods soft-question
asked Jan 7 at 20:31
AmbeshAmbesh
1,4351140
$endgroup$
Suppose that we have a linear ode which we want to integrate numerically:
$$
L{y}=f(t)
$$
where $L$ is a linear differential operator (and some initial conditions), by using a method of order $mathcal{O}(h^n)$ -- $h$ being the step size of the method.
Is it true that if we scale up the forcing function $f(t)$ with a factor $K$ before integrating, and then rescale down the solution $y$ by $frac{1}{K}$ we have increased the precision of the method? The argument being that the order of the method is still the same and only dependent on the step size. So when we scale down the error is reduced.
ordinary-differential-equations numerical-methods soft-question
ordinary-differential-equations numerical-methods soft-question
asked Jan 7 at 20:31
AmbeshAmbesh
1,4351140
asked Jan 7 at 20:31
AmbeshAmbesh
1,4351140
asked Jan 7 at 20:31
AmbeshAmbesh
1,4351140
asked Jan 7 at 20:31
AmbeshAmbesh
1,4351140
asked Jan 7 at 20:31
AmbeshAmbesh
1,4351140
1,4351140
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Imagine the function $y_s$ is the actual solution to the problem
$$
L{y_s} = f(t) tag{1a}
$$
And $hat{y}$ is an approximation, the error is measured as
$$
epsilon(n) = |y_s(t_n) - hat{y}_n| tag{2a}
$$
Now, since $L$ is linear, if you multiply both sides of (1a) by $K$, then
$$
L{ Ky_s } = K f(t) tag{1b}
$$
And the approximated solution is $hat{y}^{K}$ so the error is
$$
epsilon^K(n) = |K y_s(t_n) - hat{y}^{K}| = Kleft|y_s(t_n) - frac{hat{y}^{K}}{K}right| = K epsilon(n) tag{2b}
$$
So the error also scales as $K$. That is: if you multiply by $K$ and then divide the result, the accuracy does not increase
answered Jan 7 at 21:07
caveraccaverac
14.6k31130
$endgroup$
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
$begingroup$
Imagine the function $y_s$ is the actual solution to the problem
$$
L{y_s} = f(t) tag{1a}
$$
And $hat{y}$ is an approximation, the error is measured as
$$
epsilon(n) = |y_s(t_n) - hat{y}_n| tag{2a}
$$
Now, since $L$ is linear, if you multiply both sides of (1a) by $K$, then
$$
L{ Ky_s } = K f(t) tag{1b}
$$
And the approximated solution is $hat{y}^{K}$ so the error is
$$
epsilon^K(n) = |K y_s(t_n) - hat{y}^{K}| = Kleft|y_s(t_n) - frac{hat{y}^{K}}{K}right| = K epsilon(n) tag{2b}
$$
So the error also scales as $K$. That is: if you multiply by $K$ and then divide the result, the accuracy does not increase
answered Jan 7 at 21:07
caveraccaverac
14.6k31130
$endgroup$
add a comment |
$begingroup$
Imagine the function $y_s$ is the actual solution to the problem
$$
L{y_s} = f(t) tag{1a}
$$
And $hat{y}$ is an approximation, the error is measured as
$$
epsilon(n) = |y_s(t_n) - hat{y}_n| tag{2a}
$$
Now, since $L$ is linear, if you multiply both sides of (1a) by $K$, then
$$
L{ Ky_s } = K f(t) tag{1b}
$$
And the approximated solution is $hat{y}^{K}$ so the error is
$$
epsilon^K(n) = |K y_s(t_n) - hat{y}^{K}| = Kleft|y_s(t_n) - frac{hat{y}^{K}}{K}right| = K epsilon(n) tag{2b}
$$
So the error also scales as $K$. That is: if you multiply by $K$ and then divide the result, the accuracy does not increase
answered Jan 7 at 21:07
caveraccaverac
14.6k31130
$endgroup$
add a comment |
$begingroup$
Imagine the function $y_s$ is the actual solution to the problem
$$
L{y_s} = f(t) tag{1a}
$$
And $hat{y}$ is an approximation, the error is measured as
$$
epsilon(n) = |y_s(t_n) - hat{y}_n| tag{2a}
$$
Now, since $L$ is linear, if you multiply both sides of (1a) by $K$, then
$$
L{ Ky_s } = K f(t) tag{1b}
$$
And the approximated solution is $hat{y}^{K}$ so the error is
$$
epsilon^K(n) = |K y_s(t_n) - hat{y}^{K}| = Kleft|y_s(t_n) - frac{hat{y}^{K}}{K}right| = K epsilon(n) tag{2b}
$$
So the error also scales as $K$. That is: if you multiply by $K$ and then divide the result, the accuracy does not increase
answered Jan 7 at 21:07
caveraccaverac
14.6k31130
$endgroup$
Imagine the function $y_s$ is the actual solution to the problem
$$
L{y_s} = f(t) tag{1a}
$$
And $hat{y}$ is an approximation, the error is measured as
$$
epsilon(n) = |y_s(t_n) - hat{y}_n| tag{2a}
$$
Now, since $L$ is linear, if you multiply both sides of (1a) by $K$, then
$$
L{ Ky_s } = K f(t) tag{1b}
$$
And the approximated solution is $hat{y}^{K}$ so the error is
$$
epsilon^K(n) = |K y_s(t_n) - hat{y}^{K}| = Kleft|y_s(t_n) - frac{hat{y}^{K}}{K}right| = K epsilon(n) tag{2b}
$$
So the error also scales as $K$. That is: if you multiply by $K$ and then divide the result, the accuracy does not increase
answered Jan 7 at 21:07
caveraccaverac
14.6k31130
answered Jan 7 at 21:07
caveraccaverac
14.6k31130
answered Jan 7 at 21:07
caveraccaverac
14.6k31130
answered Jan 7 at 21:07
caveraccaverac
14.6k31130
14.6k31130
add a comment |
add a comment |
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