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How can I propagate error through an equation such as the following:

$$A \cdot B \cdot C = D$$

where $A$ and $B$ have errors of $\pm0.2$ and $\pm0.5$ respectively but $C$ has an error of a factor of $3$ (between $\frac{1}{3}$ and $3$ times $C$). So, what will $D$ have for error here and how do I repeat the process in the future?

Thanks for any help!

Would the best way to this be: $A(\pm0.2)\cdot B(\pm0.5)\cdot C(\pm C\cdot\frac{2}{3}) = D (\pm X)$ and then report the min as $D$ minus $X$. Then take $A(\pm0.2)\cdot B(\pm0.5)\cdot C(\pm C\cdot2) = D (\pm Y)$ and then report the max as $D$ plus $Y$? Note: that provides a different answer than simply multiplying each min and then each max.

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    If you want a precise result, you may be looking for [interval arithmetic](https://en.wikipedia.org/wiki/Interval_arithmetic), in which you compute the smallest and largest _true_ value of $D$ and read the error bounds off of that. – 2017-02-02
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    for first-order calculations, remember these two rules: (1) the absolute error of a summation or subtraction is the sum of the absolute errors of the terms, and (2) the relative error of a product or division is the sum of the relative errors of the factors. The 2nd rule ignores the contribution of 2nd-order error (products of two or more error terms together). For your equation, it says that $$\frac{\Delta D}D \approx \frac{\Delta A}A +\frac{\Delta B}B +\frac{\Delta C}C$$ – 2017-02-03
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    Henning, what you said makes sense but is it best to get the min and max from all three and multiply for min and max range or propagate error through A and B and then multiple by min and max of C to get a D range? They provide two different answers. – 2017-02-09
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    Paul I don't see how your method works for combining error reported in different manners. – 2017-02-09

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