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The problem says to show that the relative error for division on a computer is

\begin{align}\textrm{Rel}\left(\frac{x_{A}}{y_{A}}\right)&=\frac{\textrm{Rel}(x_{A})-\textrm{Rel}(y_{A})}{1-\textrm{Rel}(y_{A})}\\ &\approx \textrm{Rel}(x_{A})-\textrm{Rel}(y_{A})\end{align}

provided that the relative error of $y_{A}$ is small compared to one.

I know that $$\textrm{Rel}(x_{A})=\frac{x_{T}-x_{A}}{x_{T}}$$

and $x_{A}=x_{T}(1-e_{x})$ with $e_{x}$ being the error.

but I'm really not sure how to proceed from here.

Edit again: I emailed the professor and he sent out a class-wide email totally rearranging it so that's probably where confusion stems from. This is the new and actual problem.

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    $\begingroup$ What is $x_B$...? $\endgroup$
    – Ron Gordon
    Commented Feb 13, 2013 at 14:41
  • $\begingroup$ And what is $x_T$ in the definition of relative error? $\endgroup$ Commented Feb 13, 2013 at 14:45
  • $\begingroup$ @Ross: I think it is the denominator in his division operation. But perhaps the OP can verify this. $\endgroup$
    – Ron Gordon
    Commented Feb 13, 2013 at 14:48
  • $\begingroup$ @rlgordonma: I see it there, but then the definition of relative error in $x_A$ doesn't make sense as it shouldn't refer to what you will divide it by. $\endgroup$ Commented Feb 13, 2013 at 14:50
  • $\begingroup$ @RossMillikan: I see what you mean. So the OP needs to figure out this stuff for himself. Perhaps he could have stated things better by saying that $x_T = x_A (1 + \epsilon)$. But then again, this doesn't make a heck of a lot of sense either. $\endgroup$
    – Ron Gordon
    Commented Feb 13, 2013 at 15:24

1 Answer 1

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Your notation is a complete mess (I made a correction, but it's still all wrong). You cannot start to trying proving something if you cannot make sense of that something. Try, for example, to work out a numeric example, to get some consistent notation.

I'll try. I define $e_X = (x_T- x_A)/x_T$, so $x_A = x_T (1-e_X)$ ($x_T$ is the true value, $x_A$ the approximate or actual value, $e_X$ the relative error).

Then $$z_A=\frac{x_A}{y_A}=\frac{x_T(1-e_X)}{y_T(1-e_Y)} = z_T \frac{1-e_X}{1-e_Y} \approx z_T (1-e_X)(1+e_Y) \approx z_T (1 - e_X + e_Y)$$

where the last approximations assumes $e_Y \ll 1$. But the sign of the relative error is immaterial, hence $ e_Z= e_X + e_Y$.

Update: following the revised question:

$$z_A = z_T \frac{1-e_X}{1-e_Y} = z_T (1 - e_Z) \implies e_Z = 1 - \frac{1-e_X}{1-e_Y} = \frac{e_X - e_Y}{1-e_Y}$$

Again, the denominator tends to 1 if $e_Y$ is small, and the numerator should be writen as $e_X + e_Y$ if we are computing propagation of errors.

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  • $\begingroup$ Since $e_X$ and $e_Z$ are $<<1,$ shouldn't $e_Z=e_X+e_Y$ as $e_Xe_Y$ is smaller? $\endgroup$ Commented Feb 13, 2013 at 19:33
  • $\begingroup$ @RossMillikan: Of course, fixed, thanks. $\endgroup$
    – leonbloy
    Commented Feb 13, 2013 at 20:03
  • $\begingroup$ I think I follow this actually, thank you. I've edited the problem to show the revision my professor made, is the answer you gave still applicable? $\endgroup$ Commented Feb 13, 2013 at 21:01
  • $\begingroup$ Deleted a couple comments because it was beginning to feel like spamming you. After the arrow you have ez=(error) and I understand that step, but when you go from that to the very last (ex-ey)/(1-ey) I am missing how that transition is made. Something tells me it is very easy algebra but my brain is fried right now. Care to explain quick? $\endgroup$ Commented Feb 13, 2013 at 22:05
  • $\begingroup$ Basic algebra, common denominator $ 1 - (1-A)/(1-B) = [(1-B) - (1-A)]/(1-B) = (A-B)/(1-B)$ $\endgroup$
    – leonbloy
    Commented Feb 14, 2013 at 0:02

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