Is the problem well conditioned or not $x=1-e^p$.
$begingroup$
For large numbers $|p|gg1$ we want to compute $x=1-e^p$.
Is the problem for $|p|gg 1$ well conditioned.
We have
$$kappa = frac{|x'|}{|x|}cdot |p|=frac{|-e^p|}{|1-e^p|}cdot |p|$$
If $p>0$ we have
$$kappa=frac{e^p}{-1+e^p} cdot p=frac{pe^p}{e^p-1}approx p$$ so the problem is ill conditioned.
If $p<0$
$$kappa=frac{e^p}{1-e^p}cdot -p=frac{-p e^p}{1-e^p}approx -p e^p to 0 ~(pto-infty),$$
so the problem is well conditioned.
I am not sure if my work is correct.
proof-verification numerical-methods condition-number
asked Jan 13 at 10:35
orangeorange
729316
$endgroup$
|
show 5 more comments
$begingroup$
For large numbers $|p|gg1$ we want to compute $x=1-e^p$.
Is the problem for $|p|gg 1$ well conditioned.
We have
$$kappa = frac{|x'|}{|x|}cdot |p|=frac{|-e^p|}{|1-e^p|}cdot |p|$$
If $p>0$ we have
$$kappa=frac{e^p}{-1+e^p} cdot p=frac{pe^p}{e^p-1}approx p$$ so the problem is ill conditioned.
If $p<0$
$$kappa=frac{e^p}{1-e^p}cdot -p=frac{-p e^p}{1-e^p}approx -p e^p to 0 ~(pto-infty),$$
so the problem is well conditioned.
I am not sure if my work is correct.
proof-verification numerical-methods condition-number
asked Jan 13 at 10:35
orangeorange
729316
$endgroup$
$begingroup$
It's when $|p|ll 1$ that the problem is ill-conditioned. Then $e^papprox 1$ and so the subtraction loses precision.
$endgroup$
– TonyK
Jan 13 at 10:56
$begingroup$
We are assuming $p$ is large in absolute value always. Be it negative or positive.
$endgroup$
– orange
Jan 13 at 10:58
$begingroup$
Then the problem is not ill-conditioned.
$endgroup$
– TonyK
Jan 13 at 11:51
$begingroup$
I fixed the obvious error in the derivative but otherwise I still have a large condition number for positive $p$.
$endgroup$
– orange
Jan 13 at 11:57
$begingroup$
The function $$y=frac{left|-e^xright|}{left|1-e^xright|}left|xright|$$ explodes at $+infty$ and tends to $0$ at $-infty$
$endgroup$
– orange
Jan 13 at 12:02
|
show 5 more comments
$begingroup$
For large numbers $|p|gg1$ we want to compute $x=1-e^p$.
Is the problem for $|p|gg 1$ well conditioned.
We have
$$kappa = frac{|x'|}{|x|}cdot |p|=frac{|-e^p|}{|1-e^p|}cdot |p|$$
If $p>0$ we have
$$kappa=frac{e^p}{-1+e^p} cdot p=frac{pe^p}{e^p-1}approx p$$ so the problem is ill conditioned.
If $p<0$
$$kappa=frac{e^p}{1-e^p}cdot -p=frac{-p e^p}{1-e^p}approx -p e^p to 0 ~(pto-infty),$$
so the problem is well conditioned.
I am not sure if my work is correct.
proof-verification numerical-methods condition-number
asked Jan 13 at 10:35
orangeorange
729316
$endgroup$
For large numbers $|p|gg1$ we want to compute $x=1-e^p$.
Is the problem for $|p|gg 1$ well conditioned.
We have
$$kappa = frac{|x'|}{|x|}cdot |p|=frac{|-e^p|}{|1-e^p|}cdot |p|$$
If $p>0$ we have
$$kappa=frac{e^p}{-1+e^p} cdot p=frac{pe^p}{e^p-1}approx p$$ so the problem is ill conditioned.
If $p<0$
$$kappa=frac{e^p}{1-e^p}cdot -p=frac{-p e^p}{1-e^p}approx -p e^p to 0 ~(pto-infty),$$
so the problem is well conditioned.
I am not sure if my work is correct.
proof-verification numerical-methods condition-number
proof-verification numerical-methods condition-number
asked Jan 13 at 10:35
orangeorange
729316
asked Jan 13 at 10:35
orangeorange
729316
edited Jan 13 at 11:56
orange
asked Jan 13 at 10:35
orangeorange
729316
asked Jan 13 at 10:35
orangeorange
729316
asked Jan 13 at 10:35
orangeorange
729316
729316
$begingroup$
It's when $|p|ll 1$ that the problem is ill-conditioned. Then $e^papprox 1$ and so the subtraction loses precision.
$endgroup$
– TonyK
Jan 13 at 10:56
$begingroup$
We are assuming $p$ is large in absolute value always. Be it negative or positive.
$endgroup$
– orange
Jan 13 at 10:58
$begingroup$
Then the problem is not ill-conditioned.
$endgroup$
– TonyK
Jan 13 at 11:51
$begingroup$
I fixed the obvious error in the derivative but otherwise I still have a large condition number for positive $p$.
$endgroup$
– orange
Jan 13 at 11:57
$begingroup$
The function $$y=frac{left|-e^xright|}{left|1-e^xright|}left|xright|$$ explodes at $+infty$ and tends to $0$ at $-infty$
$endgroup$
– orange
Jan 13 at 12:02
|
show 5 more comments
$begingroup$
It's when $|p|ll 1$ that the problem is ill-conditioned. Then $e^papprox 1$ and so the subtraction loses precision.
$endgroup$
– TonyK
Jan 13 at 10:56
$begingroup$
We are assuming $p$ is large in absolute value always. Be it negative or positive.
$endgroup$
– orange
Jan 13 at 10:58
$begingroup$
Then the problem is not ill-conditioned.
$endgroup$
– TonyK
Jan 13 at 11:51
$begingroup$
I fixed the obvious error in the derivative but otherwise I still have a large condition number for positive $p$.
$endgroup$
– orange
Jan 13 at 11:57
$begingroup$
The function $$y=frac{left|-e^xright|}{left|1-e^xright|}left|xright|$$ explodes at $+infty$ and tends to $0$ at $-infty$
$endgroup$
– orange
Jan 13 at 12:02
$begingroup$
It's when $|p|ll 1$ that the problem is ill-conditioned. Then $e^papprox 1$ and so the subtraction loses precision.
$endgroup$
– TonyK
Jan 13 at 10:56
$begingroup$
It's when $|p|ll 1$ that the problem is ill-conditioned. Then $e^papprox 1$ and so the subtraction loses precision.
$endgroup$
– TonyK
Jan 13 at 10:56
$begingroup$
We are assuming $p$ is large in absolute value always. Be it negative or positive.
$endgroup$
– orange
Jan 13 at 10:58
$begingroup$
We are assuming $p$ is large in absolute value always. Be it negative or positive.
$endgroup$
– orange
Jan 13 at 10:58
$begingroup$
Then the problem is not ill-conditioned.
$endgroup$
– TonyK
Jan 13 at 11:51
$begingroup$
Then the problem is not ill-conditioned.
$endgroup$
– TonyK
Jan 13 at 11:51
$begingroup$
I fixed the obvious error in the derivative but otherwise I still have a large condition number for positive $p$.
$endgroup$
– orange
Jan 13 at 11:57
$begingroup$
I fixed the obvious error in the derivative but otherwise I still have a large condition number for positive $p$.
$endgroup$
– orange
Jan 13 at 11:57
$begingroup$
The function $$y=frac{left|-e^xright|}{left|1-e^xright|}left|xright|$$ explodes at $+infty$ and tends to $0$ at $-infty$
$endgroup$
– orange
Jan 13 at 12:02
$begingroup$
The function $$y=frac{left|-e^xright|}{left|1-e^xright|}left|xright|$$ explodes at $+infty$ and tends to $0$ at $-infty$
$endgroup$
– orange
Jan 13 at 12:02
|
show 5 more comments
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$begingroup$
It's when $|p|ll 1$ that the problem is ill-conditioned. Then $e^papprox 1$ and so the subtraction loses precision.
$endgroup$
– TonyK
Jan 13 at 10:56
$begingroup$
We are assuming $p$ is large in absolute value always. Be it negative or positive.
$endgroup$
– orange
Jan 13 at 10:58
$begingroup$
Then the problem is not ill-conditioned.
$endgroup$
– TonyK
Jan 13 at 11:51
$begingroup$
I fixed the obvious error in the derivative but otherwise I still have a large condition number for positive $p$.
$endgroup$
– orange
Jan 13 at 11:57
$begingroup$
The function $$y=frac{left|-e^xright|}{left|1-e^xright|}left|xright|$$ explodes at $+infty$ and tends to $0$ at $-infty$
$endgroup$
– orange
Jan 13 at 12:02