Is division ill-conditioned when divisor is close to zero?












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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







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    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







    share|cite|improve this question











    $endgroup$















      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 16:34









      MJD

      47.3k29213396




      47.3k29213396










      asked Dec 8 '18 at 16:17









      gaazkamgaazkam

      456314




      456314






















          1 Answer
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          $begingroup$

          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.




          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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











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          1 Answer
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          $begingroup$

          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.




          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          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.




          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          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.




          share|cite|improve this answer











          $endgroup$



          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.





          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 '18 at 22:01

























          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


















          • $begingroup$
            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


















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