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I want to ask a question about the multiplication of two variables and the error associated with the process.

I was learning about calculating errors in the physical labs at school and was taught that the following formula was used to calculate errors:

$\frac{\triangle Z}{Z} = \frac{\triangle A}{A} + \frac{\triangle B}{B}$

This made sense, given that we are in essence calculating percentage uncertainties and these are added when calculating the uncertainty of Z.

However, in my current undergraduate lab, we square each term:

$(\frac{\triangle Z}{Z})^2 = (\frac{\triangle A}{A})^2 + (\frac{\triangle B}{B})^2$

and I cannot find any reference to explain the purpose of squaring the values.

Why do we square each term as above?

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    $\begingroup$ Do you assume that the uncertainties of $A$ and $B$ are correlated or independent? $\endgroup$ – Faded Giant Feb 23 at 17:14
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    $\begingroup$ It is complicated. See the examples in the Wikipedia article on "Propagation of uncertainty." $\endgroup$ – MaxW Feb 23 at 17:20
  • $\begingroup$ We are told the measurements of A and B are independent. $\endgroup$ – vik1245 Feb 23 at 17:20
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    $\begingroup$ @vik1245 Possibly you stumble over «variance». For the later, if you have a function $z$ which is the result of two mutually independent functions $a$ and $b$, you access the variance of $z$ via the partial derivatives: $\mathrm{var}(z) = (\frac{\partial{}z}{\partial{}a})^2 \cdot \mathrm{var}(a) + (\frac{\partial{}z}{\partial{}b})^2 \cdot \mathrm{var}(b)$. Using the squares here ensures that the variances will add, regardless of the sign of either summand, hence for either $z = a + b$, or $z = a - b$. So the second equation comes closer to the principle of determining the variance. $\endgroup$ – Buttonwood Feb 23 at 21:24
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    $\begingroup$ @vik1245 ... as the $\Delta{}b$, for example, could be negative, hence rendering the fraction $\frac{\Delta{}b}{b}$ negative -- just depending on the order of minuend and subtrahend. This could lead to the (erroneous) result that the error in $z$ decreases. $\endgroup$ – Buttonwood Feb 23 at 21:32

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