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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?

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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 $$\Delta x = \frac{b + \Delta b}{a + \Delta a} - \frac{b}{a} = \frac{b}{a} \left(\frac{1 + (\Delta b)/b}{1 + (\Delta a)/a} - 1\right)= \frac{b}{a} \frac{(\Delta b)/b - (\Delta a)/a}{1+(\Delta a)/a}.$$ 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.

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  • $\begingroup$ 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%: $\frac{b}{a-\frac12 a}=2\frac ba = 2 x$ $\endgroup$ – gaazkam Dec 8 '18 at 22:57
  • $\begingroup$ $2x$ is hardly $x\pm 50\% x$ $\endgroup$ – gaazkam Dec 8 '18 at 22:59

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