With given:
$k_1 = 37.1\pm0.3$
$k_2 = 9.87\pm0.11$
$k_3 = 6.052\pm0.016$
estimate the absolute and relative error for $w = k_1 \cdot k_2^2 \cdot k_3^3$ and round both $w$ and the error in such a way not to lose any precise figures.

So my attempt looks like this:

We know the absolute errors for the three variables so we can calculate the absolute error for our w with: $\Delta w = 0,3 \cdot (0.11)^2 \cdot (0.016)^3 = 1.486848 \cdot 10^{-8}$

The value of the $w$ itself for our variables is $w = k_1 \cdot k_2^2 \cdot k_3^3 = 801133.6485723691$

Hence, the relative error for w is $\delta w = \frac{\Delta w}{w} \cdot 100\% = 1.8559300344575253 \cdot 10^{-12} \%$.

And as for the rounding: the "precise figure" is my translation as I couldn't find the exact thing I mean on Wikipedia. So by that I mean: we say that a rounded number has n precise significant figures if the absolute error of the number isn't higher than $0.5\cdot10^{-n}$. So for example $t=0.1132$ such that $\Delta t = 0.0001$ has 3 precise figures since $0.0001<0.5\cdot10^3$. I hope it clears things up a bit...

So for the rounding: the absolute error is of form $0.<seven-zeroes>1486<...>$ so we can round without loss of precise figures to the form of $801133.6485723$.

However, I don't have the slightest idea how to round the errors $\delta w, \Delta w$ to not lose any precise figures. I mean: if I need the absolute error of a value to determine how many precise figures it has, how can I do it if I don't know the error of the errors?

Could you please guide ma and tell me if my thinking is correct and - if not - help me understand the problem? I heartily thank you in advance :)


You can't just multiply the errors to get the error of the product. Let us just look at the error in $v=k_2^2$. We are given that $k_2=9.87\pm0.11$. The minimum $k_2$ can be is $9.75$, in which case $k_2^2= 95.0625$. The largest $k_2$ can be is $9.98$ with square $99.6004$. This means $k_2^2$ can range over $99.6004-95.0625=4.5379$, much greater than $0.11^2=0.0121$

What is happening here? If we are asked to find $v=(k_2\pm \Delta k_2)^2$ we can expand it to $v=k_2^2 \pm 2k_2\Delta k_2+(\Delta k_2)^2$. You were calculating the third term as the error, but the second is much larger. In fact, we often ignore the third term and we would write $v=k_2^2 \pm 2k_2 \Delta k_2$ or $v=97.4169\pm 2.1714$ This shows that the last digits are spurious. I'm not sure how your book would round this "so as not to lose any significant figures". Certainly you don't know even the ones digit exactly, so maybe what is desired is $97 \pm 2$, but you do know it is greater than $97.2$ and it might be almost as large as $99.6$

For the more general case, we can follow the product and power rules for derivatives and write $\Delta w=\Delta k_1k_2^2k_3^3=k_2^2k_3^3\Delta k_1+2k_1k_2k_3^3 \Delta k_2+3k_1k_2^2k_3^2\Delta k_3$. If you divide both sides by $w$ you get the convenient $\frac {\Delta w}w=\frac {\Delta k_1}{k_1}+2\frac {\Delta k_2}{k_2}+3\frac {\Delta k_3}{k_3}$

  • $\begingroup$ Thank you very much. Then, the absolute error is $\Delta w = 37.4\cdot 9.98^{2}\cdot 6.068^{3} - 36.8 \cdot 9.76^2 \cdot 6.036^3 = 61385.03477967705$, is that right? However, I don't know how to calculate the relative error, then. It seems I can't do it as long as I don't know any of the numbers for sure and so it happens that every one of them is $\pm$. The formula you provided in the end confirms it. Does that mean I can't determine the relative error or round it to some precise digits and the absolute error is the best I could do? $\endgroup$ – Straightfw Mar 6 '13 at 16:14
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    $\begingroup$ That is twice the absolute error, because you have subtracted the min from the max. The relative error is just the absolute error divided by the central value: about $\frac {30200}{801134}\approx 0.038$. The last expression in my original post (with the $2$ corrected to $w$) is precisely the relative error. Note that the relative error (in this approximation)is the sum of the relative errors of the factors times their powers. this gives us a good check on the relative error: the relative error on $k_1$ is about $1\%$ and on $k_2$ is $1\%$, so they contribute a total of $3\%$ to the $\endgroup$ – Ross Millikan Mar 6 '13 at 16:27
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    $\begingroup$ overall. The relative error on $k_3$ is much smaller, so we can neglect it. In practice, these error values are usually quite rough. We would say the relative error on $w$ is $4\%$ and not worry about whether the exact ends match with the calculation. $\endgroup$ – Ross Millikan Mar 6 '13 at 16:28
  • $\begingroup$ Thank you, now I understand. What about the rounding so that we don't lose any precise digits, though? I mean: we can see that the absolute error is going to have 17 significant figures and we can't determine the relative error's number of significant figures (too much digits after the comma). Also, $w$ has 13 significant figure, but $w$ has only one precise digit ($30200<50000$) and we can round it to $801133.65\pm0.00857$. Is it OK? $\endgroup$ – Straightfw Mar 6 '13 at 16:46
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    $\begingroup$ @Straightfw: we have $w$ is in the range $7980641$ to $8042026$. So in one sense you don't know any digits for sure. I would report this as $w=8.01\pm 0.03$ (though in fact it could go slightly over my top limit). We do know that if we round to two places we will get $8.0 E6$. Maybe you are supposed to give $8000000 \pm 100000$ but I don't know. I think errors shouldn't have more than two significant digits (almost always only one) and the value shouldn't have digits past the least significant of the error. But you will have to check what the book wants. $\endgroup$ – Ross Millikan Mar 6 '13 at 16:57

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