Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Question: (a) Determine each of the absolute conditions of the linear map $f: \mathbb{R} \rightarrow \mathbb{R}:$

  1. $f(x) = |x|$
  2. $f(x) = \frac{\pi}{2}$

(b) Provide both an example of a well conditioned and an ill conditioned function evaluation (both the function $f$ and the argument $x$). Do not use any of the examples from (a).

My attempt: so although this seems really simple I'm unfortunately having trouble with exactly what to do. The given definition of absolute condition $\kappa_{abs}$ is the smallest number for which: $|f(x_0) - f(x)| \leq \kappa_{abs}|x_0 - x| + o(|x_0 - x|)$. However in the one example I have it seems that the last term with $o$ is ignored (I assume it is too small...). I was thinking that perhaps for (a) 2. I could write something like: $|f(x_0) - f(x)| = |\frac{\tilde{\pi}}{2} - \frac{\pi}{2}|= \frac{1}{2}|\tilde{\pi}-\pi| \Rightarrow \kappa_{abs} = \frac{1}{2}$ ? With 1. (a) I am don't know which sort of answer would be expected...

For (b) I am guessing it must be measured in terms of relative condition, right? Since it appears that the rule is condition(not specified which) greater than 1 is ill conditioned. I know that subtraction of almost equal numbers does not do well. I thought a way to guarantee that the function would always subtract two such numbers would to be to let $f(x) = x - \frac{99x}{100}$ using the formula for relative condition of subtraction($\kappa = \frac{|x|+|y|}{|x-y|}$) I would have $$\frac{|x| + |\frac{99x}{100}|}{|x| - |\frac{99x}{100}|} = 199 > 1$$ $\Rightarrow$ This function is ill-conditioned ? For the well conditioned function I have the theorem: for $x,y > 0$ $$\frac{|(x+y) - (\tilde{x} + \tilde{y})|}{|x+y|} \leq 1\epsilon \Rightarrow \kappa = 1$$ but I don't understand how to use that here since it seems self explanatory that a positive number divided by a larger positive number is less than 1. Can I simply substitue any positive number for $y$ and let my function be $f(x) = x^{2} +10$ ?

As if it weren't obvious I am very lost with this and any help, tips, advice, etc is as usual greatly appreciated!

share|improve this question
$\tilde{\pi}$?!? It's the same $\pi$ in both $f(x_0)$ and $f(x)$, so $f(x_0)-f(x)=0$, thus $\kappa_{\mathrm{abs}}=0$. For $f(x)=|x|$, think geometrically: what's the maximum slope of the graph? –  Hans Lundmark Nov 30 '10 at 11:40
@Hans: haha sorry about that with $\pi$. I just thought that maybe it had something to do with machines not being able to store/represent $\pi$ to the same amount of places and the $\tilde{\pi}$ would have been a slightly different approximation. With $|x|$ wouldn't the slope be $=1$ ? so $\kappa_{\mathrm{abs}} = 1$ ? –  ghshtalt Nov 30 '10 at 11:53
Are you in mathematics (where the numbers are exact) or in computer science (where you have to worry about the errors in floating point representation)? If in math, your last comment that the condition is 1 is correct. And what would be |f(x)-f(x0)| if f is a constant? –  Ross Millikan Nov 30 '10 at 14:06
@Ross: I am in 'computer-oriented mathematics', and it seems that a lot of the focus so far is on error (I couldn't tell you whether or not that is exclusively 'errors in floating point representation' since I don't know exactly what that means). But I thought that the whole point of condition was to measure the effect of a mapping on inaccuracy(error), right? Does this still occur in math? If $f$ were constant $|f(x)-f(x_o)| = 0$ as Hans pointed out, correct? –  ghshtalt Nov 30 '10 at 14:44
@user3711: Yes, Hans is correct. But in a computer you cannot represent real exactly, only with some error. For your second function, you would expect pi/2 to always be represented the same way (unless you calculate it in different ways) so f(x)-f(x0) will be truly zero. But in the first, if you try to define |represented f(x)- represented f(x0)|/|real x - real x0| you could get a surprise as x gets close to x0. –  Ross Millikan Nov 30 '10 at 15:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.