Suppose you have two measured (independent) physical quantities $x$ and $y$ with relative errors $r_x := \frac{\delta x}{x}$ and $r_y := \frac{\delta y}{y}$, where $\delta x$ and $\delta y$ are the corresponding absolute errors. Now you want to calculate $z = xy$ (or $z = \frac{x}{y}$).

Usually the relative error $r_z$ is calculated as

$$ r_z = \sqrt{r_x^2 + r_y^2} $$

Sometimes (usually in lower level or high school courses) it is said that you have to take just the sum of the relative errors, i.e.

$$ r_z = r_x + r_y $$

For example for the product this seems easy to derive: $$ \begin{align} (x + \delta x)\cdot (y + \delta y) = xy + x \delta y + y \delta x + \delta x \delta y \\ (x - \delta x)\cdot (y - \delta y) = xy - x \delta y - y \delta x + \delta x \delta y \\ \end{align} $$

Subtracting both equations you get for the right side: $2(x\delta y + y \delta x)$ and half of it seems to be a good measure for the absolute value of $\delta z$. So you get for the relative error $r_z$

$$ r_z = \frac{\delta z}{z} = \frac{x\delta y + y \delta}{xy} = \frac{\delta y}{y} + \frac{\delta x}{x} = r_y + r_x $$

What's wrong with this reasoning?

What is the correct formula and why? Are there different domains of application of both formulas?

  • $\begingroup$ It wouldn't be unheard of to assume the product of the errors $\delta x \delta y$ is negligible, but this assumes that both are small numbers relative to x and y $\endgroup$
    – tpg2114
    Sep 7 '15 at 12:56
  • $\begingroup$ @EmilioPisanty: You're right (and I closed the other one myself as dupe of that), I chose the dupe link too quickly. $\endgroup$
    – ACuriousMind
    Sep 7 '15 at 13:10
  • 1
    $\begingroup$ "What's wrong with this reasoning?" Intuitively, what's wrong is that it assumes the errors all have the same sign, whereas in most cases you need to assume that the errors are uncorrelated. Sum ten uncorrelated noise terms together: what's the probability that they're all in the same direction? Its $2^{-9}\approx 0.002$. Even with two terms, the probability that they are both of the same sign is only a half. Half of the time, they will mitigate one another. $\endgroup$ Sep 7 '15 at 13:28

Browse other questions tagged or ask your own question.