Confusion about Numerical Stability
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Numerical Analysis is giving me some trouble. In specific, I'm highly confused by our definition of numerical stability. I'm hoping some of you can help me clear my confusion.
Definition. An algorithm $hat{f}$ for a problem $f$ is called stable if$$ | hat{f}(x) - f(tilde{x}) |_{text{rel}} = mathcal{O}(varepsilon_{text{mach}})$$ for some suitably chosen $tilde{x}$ with $|x-tilde{x} |_{text{rel}} = mathcal{O}(varepsilon_{text{mach}})$.
So what do I do in practice to prove stability/instability? It may be my unfamiliarity with Landau notation in numerical purposes. Let me try to have a guess.
Proving Stability: Show that there exists a constant $c > 0$ such that: There exists a constant $C > 0$ such that there exists a $tilde{x}$ with $|x-tilde{x} |_{text{rel}} < c varepsilon_{text{mach}}$ for each sufficiently small $varepsilon_{text{mach}} > 0$ such that$$| hat{f}(x) - f(tilde{x}) |_{text{rel}} < C varepsilon_{text{mach}}$$
Do I read the Landau symbols correctly?
Proving Instability: Show that such $c > 0$ does not exist.
Is that how you would prove stability in practice? For instance, would you start by defining such constants to prove that $x mapsto log(1 oplus operatorname{fl}(x))$ is instable?
EDIT: Perhaps this seems more convenient: Take $tilde{x} = (1+c varepsilon_{text{mach}})x$ and show that the difference of functions also scales in $varepsilon_{text{mach}}$. As for instability, anytime we do that, the result will not decrease as quickly.
I guess that's how one should go about that?
numerical-linear-algebra stability-theory
asked Jan 20 at 18:34
KezerKezer
1,422621
$endgroup$
add a comment |
$begingroup$
Numerical Analysis is giving me some trouble. In specific, I'm highly confused by our definition of numerical stability. I'm hoping some of you can help me clear my confusion.
Definition. An algorithm $hat{f}$ for a problem $f$ is called stable if$$ | hat{f}(x) - f(tilde{x}) |_{text{rel}} = mathcal{O}(varepsilon_{text{mach}})$$ for some suitably chosen $tilde{x}$ with $|x-tilde{x} |_{text{rel}} = mathcal{O}(varepsilon_{text{mach}})$.
So what do I do in practice to prove stability/instability? It may be my unfamiliarity with Landau notation in numerical purposes. Let me try to have a guess.
Proving Stability: Show that there exists a constant $c > 0$ such that: There exists a constant $C > 0$ such that there exists a $tilde{x}$ with $|x-tilde{x} |_{text{rel}} < c varepsilon_{text{mach}}$ for each sufficiently small $varepsilon_{text{mach}} > 0$ such that$$| hat{f}(x) - f(tilde{x}) |_{text{rel}} < C varepsilon_{text{mach}}$$
Do I read the Landau symbols correctly?
Proving Instability: Show that such $c > 0$ does not exist.
Is that how you would prove stability in practice? For instance, would you start by defining such constants to prove that $x mapsto log(1 oplus operatorname{fl}(x))$ is instable?
EDIT: Perhaps this seems more convenient: Take $tilde{x} = (1+c varepsilon_{text{mach}})x$ and show that the difference of functions also scales in $varepsilon_{text{mach}}$. As for instability, anytime we do that, the result will not decrease as quickly.
I guess that's how one should go about that?
numerical-linear-algebra stability-theory
asked Jan 20 at 18:34
KezerKezer
1,422621
$endgroup$
$begingroup$
I'd say you're right. If it helps, my "laymen" definition of stability is: small perturbations in the data imply small perturbations in the result. But (surely it's abundantly obvious) keep in mind that if $x$ is a vector, you need a norm of the difference $| x - bar{x} |$ to be small so the formula for $bar{x}$ at the bottom works only in the scalar case.
$endgroup$
– VorKir
Jan 21 at 19:54
$begingroup$
Oh right, true, thanks. Yeah, I do know the intuition for stability, though I‘m having trouble working with it.
$endgroup$
– Kezer
Jan 23 at 8:35
add a comment |
$begingroup$
Numerical Analysis is giving me some trouble. In specific, I'm highly confused by our definition of numerical stability. I'm hoping some of you can help me clear my confusion.
Definition. An algorithm $hat{f}$ for a problem $f$ is called stable if$$ | hat{f}(x) - f(tilde{x}) |_{text{rel}} = mathcal{O}(varepsilon_{text{mach}})$$ for some suitably chosen $tilde{x}$ with $|x-tilde{x} |_{text{rel}} = mathcal{O}(varepsilon_{text{mach}})$.
So what do I do in practice to prove stability/instability? It may be my unfamiliarity with Landau notation in numerical purposes. Let me try to have a guess.
Proving Stability: Show that there exists a constant $c > 0$ such that: There exists a constant $C > 0$ such that there exists a $tilde{x}$ with $|x-tilde{x} |_{text{rel}} < c varepsilon_{text{mach}}$ for each sufficiently small $varepsilon_{text{mach}} > 0$ such that$$| hat{f}(x) - f(tilde{x}) |_{text{rel}} < C varepsilon_{text{mach}}$$
Do I read the Landau symbols correctly?
Proving Instability: Show that such $c > 0$ does not exist.
Is that how you would prove stability in practice? For instance, would you start by defining such constants to prove that $x mapsto log(1 oplus operatorname{fl}(x))$ is instable?
EDIT: Perhaps this seems more convenient: Take $tilde{x} = (1+c varepsilon_{text{mach}})x$ and show that the difference of functions also scales in $varepsilon_{text{mach}}$. As for instability, anytime we do that, the result will not decrease as quickly.
I guess that's how one should go about that?
numerical-linear-algebra stability-theory
asked Jan 20 at 18:34
KezerKezer
1,422621
$endgroup$
Numerical Analysis is giving me some trouble. In specific, I'm highly confused by our definition of numerical stability. I'm hoping some of you can help me clear my confusion.
Definition. An algorithm $hat{f}$ for a problem $f$ is called stable if$$ | hat{f}(x) - f(tilde{x}) |_{text{rel}} = mathcal{O}(varepsilon_{text{mach}})$$ for some suitably chosen $tilde{x}$ with $|x-tilde{x} |_{text{rel}} = mathcal{O}(varepsilon_{text{mach}})$.
So what do I do in practice to prove stability/instability? It may be my unfamiliarity with Landau notation in numerical purposes. Let me try to have a guess.
Proving Stability: Show that there exists a constant $c > 0$ such that: There exists a constant $C > 0$ such that there exists a $tilde{x}$ with $|x-tilde{x} |_{text{rel}} < c varepsilon_{text{mach}}$ for each sufficiently small $varepsilon_{text{mach}} > 0$ such that$$| hat{f}(x) - f(tilde{x}) |_{text{rel}} < C varepsilon_{text{mach}}$$
Do I read the Landau symbols correctly?
Proving Instability: Show that such $c > 0$ does not exist.
Is that how you would prove stability in practice? For instance, would you start by defining such constants to prove that $x mapsto log(1 oplus operatorname{fl}(x))$ is instable?
EDIT: Perhaps this seems more convenient: Take $tilde{x} = (1+c varepsilon_{text{mach}})x$ and show that the difference of functions also scales in $varepsilon_{text{mach}}$. As for instability, anytime we do that, the result will not decrease as quickly.
I guess that's how one should go about that?
numerical-linear-algebra stability-theory
numerical-linear-algebra stability-theory
asked Jan 20 at 18:34
KezerKezer
1,422621
asked Jan 20 at 18:34
KezerKezer
1,422621
edited Jan 20 at 18:40
Kezer
asked Jan 20 at 18:34
KezerKezer
1,422621
asked Jan 20 at 18:34
KezerKezer
1,422621
asked Jan 20 at 18:34
KezerKezer
1,422621
1,422621
$begingroup$
I'd say you're right. If it helps, my "laymen" definition of stability is: small perturbations in the data imply small perturbations in the result. But (surely it's abundantly obvious) keep in mind that if $x$ is a vector, you need a norm of the difference $| x - bar{x} |$ to be small so the formula for $bar{x}$ at the bottom works only in the scalar case.
$endgroup$
– VorKir
Jan 21 at 19:54
$begingroup$
Oh right, true, thanks. Yeah, I do know the intuition for stability, though I‘m having trouble working with it.
$endgroup$
– Kezer
Jan 23 at 8:35
add a comment |
$begingroup$
I'd say you're right. If it helps, my "laymen" definition of stability is: small perturbations in the data imply small perturbations in the result. But (surely it's abundantly obvious) keep in mind that if $x$ is a vector, you need a norm of the difference $| x - bar{x} |$ to be small so the formula for $bar{x}$ at the bottom works only in the scalar case.
$endgroup$
– VorKir
Jan 21 at 19:54
$begingroup$
Oh right, true, thanks. Yeah, I do know the intuition for stability, though I‘m having trouble working with it.
$endgroup$
– Kezer
Jan 23 at 8:35
$begingroup$
I'd say you're right. If it helps, my "laymen" definition of stability is: small perturbations in the data imply small perturbations in the result. But (surely it's abundantly obvious) keep in mind that if $x$ is a vector, you need a norm of the difference $| x - bar{x} |$ to be small so the formula for $bar{x}$ at the bottom works only in the scalar case.
$endgroup$
– VorKir
Jan 21 at 19:54
$begingroup$
I'd say you're right. If it helps, my "laymen" definition of stability is: small perturbations in the data imply small perturbations in the result. But (surely it's abundantly obvious) keep in mind that if $x$ is a vector, you need a norm of the difference $| x - bar{x} |$ to be small so the formula for $bar{x}$ at the bottom works only in the scalar case.
$endgroup$
– VorKir
Jan 21 at 19:54
$begingroup$
Oh right, true, thanks. Yeah, I do know the intuition for stability, though I‘m having trouble working with it.
$endgroup$
– Kezer
Jan 23 at 8:35
$begingroup$
Oh right, true, thanks. Yeah, I do know the intuition for stability, though I‘m having trouble working with it.
$endgroup$
– Kezer
Jan 23 at 8:35
add a comment |
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$begingroup$
I'd say you're right. If it helps, my "laymen" definition of stability is: small perturbations in the data imply small perturbations in the result. But (surely it's abundantly obvious) keep in mind that if $x$ is a vector, you need a norm of the difference $| x - bar{x} |$ to be small so the formula for $bar{x}$ at the bottom works only in the scalar case.
$endgroup$
– VorKir
Jan 21 at 19:54
$begingroup$
Oh right, true, thanks. Yeah, I do know the intuition for stability, though I‘m having trouble working with it.
$endgroup$
– Kezer
Jan 23 at 8:35