Is division ill-conditioned when divisor is close to zero?
1
$begingroup$
My intuition is that division of real numbers is ill-conditioned when divisor is close to zero. Is this intuition correct?
numerical-methods arithmetic condition-number
edited Dec 8 '18 at 16:34
MJD
47.3k29213396
asked Dec 8 '18 at 16:17
gaazkamgaazkam
456314
$endgroup$
add a comment |
1
$begingroup$
My intuition is that division of real numbers is ill-conditioned when divisor is close to zero. Is this intuition correct?
numerical-methods arithmetic condition-number
edited Dec 8 '18 at 16:34
MJD
47.3k29213396
asked Dec 8 '18 at 16:17
gaazkamgaazkam
456314
$endgroup$
add a comment |
1
1
1
$begingroup$
My intuition is that division of real numbers is ill-conditioned when divisor is close to zero. Is this intuition correct?
numerical-methods arithmetic condition-number
edited Dec 8 '18 at 16:34
MJD
47.3k29213396
asked Dec 8 '18 at 16:17
gaazkamgaazkam
456314
$endgroup$
My intuition is that division of real numbers is ill-conditioned when divisor is close to zero. Is this intuition correct?
numerical-methods arithmetic condition-number
numerical-methods arithmetic condition-number
edited Dec 8 '18 at 16:34
MJD
47.3k29213396
asked Dec 8 '18 at 16:17
gaazkamgaazkam
456314
edited Dec 8 '18 at 16:34
MJD
47.3k29213396
asked Dec 8 '18 at 16:17
gaazkamgaazkam
456314
edited Dec 8 '18 at 16:34
MJD
47.3k29213396
edited Dec 8 '18 at 16:34
MJD
47.3k29213396
edited Dec 8 '18 at 16:34
MJD
47.3k29213396
47.3k29213396
asked Dec 8 '18 at 16:17
gaazkamgaazkam
456314
asked Dec 8 '18 at 16:17
gaazkamgaazkam
456314
asked Dec 8 '18 at 16:17
gaazkamgaazkam
456314
456314
add a comment |
add a comment |
1 Answer
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No. The linear system $ax=b$ where $a not = 0$ has condition number $1$. This is as good as it gets.
Addendum: There is more than one way to describe the conditioning of this problem. Initially, I choose to treat the problem as a standard linear system with a single unknown. Here the classical condition number as well as Skeel's condition numbers are all $1$. However, there is value in applying basic principles. We are interested in the perturbed equation
$$
(a + Delta a) (x + Delta x) = (b + Delta b).
$$
The relevant condition number is
$$ kappa =underset{epsilon rightarrow 0_+}{lim} , sup left { frac{1}{epsilon} frac{|Delta x|}{|x|} : Big | : (a + Delta a) (x + Delta x) = (b + Delta b), : frac{|Delta a|}{|a|} leq epsilon, : frac{|Delta b|}{|b|} leq epsilon right}$$
By direct computation we have
$$Delta x = frac{b + Delta b}{a + Delta a} - frac{b}{a} = frac{b}{a} left(frac{1 + (Delta b)/b}{1 + (Delta a)/a} - 1right)= frac{b}{a} frac{(Delta b)/b - (Delta a)/a}{1+(Delta a)/a}.$$
It follows, that
$$ frac{Delta x}{x}= frac{(Delta b)/b - (Delta a)/a}{1+(Delta a)/a}.$$
This implies that
$$ frac{1}{epsilon} left| frac{Delta x}{x} right| leq frac{2}{1 - epsilon} $$
when $epsilon < 1$. Moreover, equality is possible when $Delta a = - epsilon a$ and $Delta b = epsilon b$. It follows that $kappa = 2$. In particular, we have
$$ left| frac{Delta x}{x} right| approx 2 epsilon$$
and the approximation is good for small values of the relative error $epsilon$.
The conditioning of a problem reflects the sensitivity of the solution to small relative changes to the defining parameters. In particular, large relative changes are irrelevant. We are only interested in the limit where the relative changes tend to zero.
answered Dec 8 '18 at 22:44
Carl ChristianCarl Christian
5,4431721
$endgroup$
$begingroup$
B-but! Error of $x$ seems to grow horribly with error of $a$! Up to infinity, actually, when error of $a$ is 100%! Less extremely: when error of $a$ is 50%, error of $x$ is no less than 200%! How can this be $1$??
$endgroup$
– gaazkam
Dec 8 '18 at 22:57
$begingroup$
When error of $a$ is "only" 50%: $frac{b}{a-frac12 a}=2frac ba = 2 x$
$endgroup$
– gaazkam
Dec 8 '18 at 22:57
$begingroup$
$2x$ is hardly $xpm 50% x$
$endgroup$
– gaazkam
Dec 8 '18 at 22:59
add a comment |
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No. The linear system $ax=b$ where $a not = 0$ has condition number $1$. This is as good as it gets.
Addendum: There is more than one way to describe the conditioning of this problem. Initially, I choose to treat the problem as a standard linear system with a single unknown. Here the classical condition number as well as Skeel's condition numbers are all $1$. However, there is value in applying basic principles. We are interested in the perturbed equation
$$
(a + Delta a) (x + Delta x) = (b + Delta b).
$$
The relevant condition number is
$$ kappa =underset{epsilon rightarrow 0_+}{lim} , sup left { frac{1}{epsilon} frac{|Delta x|}{|x|} : Big | : (a + Delta a) (x + Delta x) = (b + Delta b), : frac{|Delta a|}{|a|} leq epsilon, : frac{|Delta b|}{|b|} leq epsilon right}$$
By direct computation we have
$$Delta x = frac{b + Delta b}{a + Delta a} - frac{b}{a} = frac{b}{a} left(frac{1 + (Delta b)/b}{1 + (Delta a)/a} - 1right)= frac{b}{a} frac{(Delta b)/b - (Delta a)/a}{1+(Delta a)/a}.$$
It follows, that
$$ frac{Delta x}{x}= frac{(Delta b)/b - (Delta a)/a}{1+(Delta a)/a}.$$
This implies that
$$ frac{1}{epsilon} left| frac{Delta x}{x} right| leq frac{2}{1 - epsilon} $$
when $epsilon < 1$. Moreover, equality is possible when $Delta a = - epsilon a$ and $Delta b = epsilon b$. It follows that $kappa = 2$. In particular, we have
$$ left| frac{Delta x}{x} right| approx 2 epsilon$$
and the approximation is good for small values of the relative error $epsilon$.
The conditioning of a problem reflects the sensitivity of the solution to small relative changes to the defining parameters. In particular, large relative changes are irrelevant. We are only interested in the limit where the relative changes tend to zero.
answered Dec 8 '18 at 22:44
Carl ChristianCarl Christian
5,4431721
$endgroup$
$begingroup$
B-but! Error of $x$ seems to grow horribly with error of $a$! Up to infinity, actually, when error of $a$ is 100%! Less extremely: when error of $a$ is 50%, error of $x$ is no less than 200%! How can this be $1$??
$endgroup$
– gaazkam
Dec 8 '18 at 22:57
$begingroup$
When error of $a$ is "only" 50%: $frac{b}{a-frac12 a}=2frac ba = 2 x$
$endgroup$
– gaazkam
Dec 8 '18 at 22:57
$begingroup$
$2x$ is hardly $xpm 50% x$
$endgroup$
– gaazkam
Dec 8 '18 at 22:59
add a comment |
0
$begingroup$
No. The linear system $ax=b$ where $a not = 0$ has condition number $1$. This is as good as it gets.
Addendum: There is more than one way to describe the conditioning of this problem. Initially, I choose to treat the problem as a standard linear system with a single unknown. Here the classical condition number as well as Skeel's condition numbers are all $1$. However, there is value in applying basic principles. We are interested in the perturbed equation
$$
(a + Delta a) (x + Delta x) = (b + Delta b).
$$
The relevant condition number is
$$ kappa =underset{epsilon rightarrow 0_+}{lim} , sup left { frac{1}{epsilon} frac{|Delta x|}{|x|} : Big | : (a + Delta a) (x + Delta x) = (b + Delta b), : frac{|Delta a|}{|a|} leq epsilon, : frac{|Delta b|}{|b|} leq epsilon right}$$
By direct computation we have
$$Delta x = frac{b + Delta b}{a + Delta a} - frac{b}{a} = frac{b}{a} left(frac{1 + (Delta b)/b}{1 + (Delta a)/a} - 1right)= frac{b}{a} frac{(Delta b)/b - (Delta a)/a}{1+(Delta a)/a}.$$
It follows, that
$$ frac{Delta x}{x}= frac{(Delta b)/b - (Delta a)/a}{1+(Delta a)/a}.$$
This implies that
$$ frac{1}{epsilon} left| frac{Delta x}{x} right| leq frac{2}{1 - epsilon} $$
when $epsilon < 1$. Moreover, equality is possible when $Delta a = - epsilon a$ and $Delta b = epsilon b$. It follows that $kappa = 2$. In particular, we have
$$ left| frac{Delta x}{x} right| approx 2 epsilon$$
and the approximation is good for small values of the relative error $epsilon$.
The conditioning of a problem reflects the sensitivity of the solution to small relative changes to the defining parameters. In particular, large relative changes are irrelevant. We are only interested in the limit where the relative changes tend to zero.
answered Dec 8 '18 at 22:44
Carl ChristianCarl Christian
5,4431721
$endgroup$
$begingroup$
B-but! Error of $x$ seems to grow horribly with error of $a$! Up to infinity, actually, when error of $a$ is 100%! Less extremely: when error of $a$ is 50%, error of $x$ is no less than 200%! How can this be $1$??
$endgroup$
– gaazkam
Dec 8 '18 at 22:57
$begingroup$
When error of $a$ is "only" 50%: $frac{b}{a-frac12 a}=2frac ba = 2 x$
$endgroup$
– gaazkam
Dec 8 '18 at 22:57
$begingroup$
$2x$ is hardly $xpm 50% x$
$endgroup$
– gaazkam
Dec 8 '18 at 22:59
add a comment |
0
0
0
$begingroup$
No. The linear system $ax=b$ where $a not = 0$ has condition number $1$. This is as good as it gets.
Addendum: There is more than one way to describe the conditioning of this problem. Initially, I choose to treat the problem as a standard linear system with a single unknown. Here the classical condition number as well as Skeel's condition numbers are all $1$. However, there is value in applying basic principles. We are interested in the perturbed equation
$$
(a + Delta a) (x + Delta x) = (b + Delta b).
$$
The relevant condition number is
$$ kappa =underset{epsilon rightarrow 0_+}{lim} , sup left { frac{1}{epsilon} frac{|Delta x|}{|x|} : Big | : (a + Delta a) (x + Delta x) = (b + Delta b), : frac{|Delta a|}{|a|} leq epsilon, : frac{|Delta b|}{|b|} leq epsilon right}$$
By direct computation we have
$$Delta x = frac{b + Delta b}{a + Delta a} - frac{b}{a} = frac{b}{a} left(frac{1 + (Delta b)/b}{1 + (Delta a)/a} - 1right)= frac{b}{a} frac{(Delta b)/b - (Delta a)/a}{1+(Delta a)/a}.$$
It follows, that
$$ frac{Delta x}{x}= frac{(Delta b)/b - (Delta a)/a}{1+(Delta a)/a}.$$
This implies that
$$ frac{1}{epsilon} left| frac{Delta x}{x} right| leq frac{2}{1 - epsilon} $$
when $epsilon < 1$. Moreover, equality is possible when $Delta a = - epsilon a$ and $Delta b = epsilon b$. It follows that $kappa = 2$. In particular, we have
$$ left| frac{Delta x}{x} right| approx 2 epsilon$$
and the approximation is good for small values of the relative error $epsilon$.
The conditioning of a problem reflects the sensitivity of the solution to small relative changes to the defining parameters. In particular, large relative changes are irrelevant. We are only interested in the limit where the relative changes tend to zero.
answered Dec 8 '18 at 22:44
Carl ChristianCarl Christian
5,4431721
$endgroup$
No. The linear system $ax=b$ where $a not = 0$ has condition number $1$. This is as good as it gets.
Addendum: There is more than one way to describe the conditioning of this problem. Initially, I choose to treat the problem as a standard linear system with a single unknown. Here the classical condition number as well as Skeel's condition numbers are all $1$. However, there is value in applying basic principles. We are interested in the perturbed equation
$$
(a + Delta a) (x + Delta x) = (b + Delta b).
$$
The relevant condition number is
$$ kappa =underset{epsilon rightarrow 0_+}{lim} , sup left { frac{1}{epsilon} frac{|Delta x|}{|x|} : Big | : (a + Delta a) (x + Delta x) = (b + Delta b), : frac{|Delta a|}{|a|} leq epsilon, : frac{|Delta b|}{|b|} leq epsilon right}$$
By direct computation we have
$$Delta x = frac{b + Delta b}{a + Delta a} - frac{b}{a} = frac{b}{a} left(frac{1 + (Delta b)/b}{1 + (Delta a)/a} - 1right)= frac{b}{a} frac{(Delta b)/b - (Delta a)/a}{1+(Delta a)/a}.$$
It follows, that
$$ frac{Delta x}{x}= frac{(Delta b)/b - (Delta a)/a}{1+(Delta a)/a}.$$
This implies that
$$ frac{1}{epsilon} left| frac{Delta x}{x} right| leq frac{2}{1 - epsilon} $$
when $epsilon < 1$. Moreover, equality is possible when $Delta a = - epsilon a$ and $Delta b = epsilon b$. It follows that $kappa = 2$. In particular, we have
$$ left| frac{Delta x}{x} right| approx 2 epsilon$$
and the approximation is good for small values of the relative error $epsilon$.
The conditioning of a problem reflects the sensitivity of the solution to small relative changes to the defining parameters. In particular, large relative changes are irrelevant. We are only interested in the limit where the relative changes tend to zero.
answered Dec 8 '18 at 22:44
Carl ChristianCarl Christian
5,4431721
edited Dec 9 '18 at 22:01
answered Dec 8 '18 at 22:44
Carl ChristianCarl Christian
5,4431721
answered Dec 8 '18 at 22:44
Carl ChristianCarl Christian
5,4431721
answered Dec 8 '18 at 22:44
Carl ChristianCarl Christian
5,4431721
5,4431721
$begingroup$
B-but! Error of $x$ seems to grow horribly with error of $a$! Up to infinity, actually, when error of $a$ is 100%! Less extremely: when error of $a$ is 50%, error of $x$ is no less than 200%! How can this be $1$??
$endgroup$
– gaazkam
Dec 8 '18 at 22:57
$begingroup$
When error of $a$ is "only" 50%: $frac{b}{a-frac12 a}=2frac ba = 2 x$
$endgroup$
– gaazkam
Dec 8 '18 at 22:57
$begingroup$
$2x$ is hardly $xpm 50% x$
$endgroup$
– gaazkam
Dec 8 '18 at 22:59
add a comment |
$begingroup$
B-but! Error of $x$ seems to grow horribly with error of $a$! Up to infinity, actually, when error of $a$ is 100%! Less extremely: when error of $a$ is 50%, error of $x$ is no less than 200%! How can this be $1$??
$endgroup$
– gaazkam
Dec 8 '18 at 22:57
$begingroup$
When error of $a$ is "only" 50%: $frac{b}{a-frac12 a}=2frac ba = 2 x$
$endgroup$
– gaazkam
Dec 8 '18 at 22:57
$begingroup$
$2x$ is hardly $xpm 50% x$
$endgroup$
– gaazkam
Dec 8 '18 at 22:59
$begingroup$
B-but! Error of $x$ seems to grow horribly with error of $a$! Up to infinity, actually, when error of $a$ is 100%! Less extremely: when error of $a$ is 50%, error of $x$ is no less than 200%! How can this be $1$??
$endgroup$
– gaazkam
Dec 8 '18 at 22:57
$begingroup$
B-but! Error of $x$ seems to grow horribly with error of $a$! Up to infinity, actually, when error of $a$ is 100%! Less extremely: when error of $a$ is 50%, error of $x$ is no less than 200%! How can this be $1$??
$endgroup$
– gaazkam
Dec 8 '18 at 22:57
$begingroup$
When error of $a$ is "only" 50%: $frac{b}{a-frac12 a}=2frac ba = 2 x$
$endgroup$
– gaazkam
Dec 8 '18 at 22:57
$begingroup$
When error of $a$ is "only" 50%: $frac{b}{a-frac12 a}=2frac ba = 2 x$
$endgroup$
– gaazkam
Dec 8 '18 at 22:57
$begingroup$
$2x$ is hardly $xpm 50% x$
$endgroup$
– gaazkam
Dec 8 '18 at 22:59
$begingroup$
$2x$ is hardly $xpm 50% x$
$endgroup$
– gaazkam
Dec 8 '18 at 22:59
add a comment |
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