Theory of errors and confusion with roundings

**Glyper** · March 6th 2013, 03:51 AM

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

**BobP** · March 6th 2013, 04:08 AM

Your calculation for the maximum absolute error, (you can't know what the actual absolute error is), is wrong.

The calculated value for w is $37.1\times 9.87^{2}\times 6.052^{3}.$

The maximum possible value for w would be $37.4\times 9.98^{2}\times 6.068^{3}.$

The maximum absolute error is the difference between these.

**Glyper** · March 6th 2013, 05:33 AM

Thank you. But if we're to calculate the maximum absolute error, shouldn't we then subtract $37.4\times 9.98^{2}\times 6.068^{3} - 36.8 \times 9.76^2 \times 6.036^3$ which is maximum possible value minus minimum possible value? Why are we using the "center" of our possible spectrum for this (which is $37.1\times 9.87^{2}\times 6.052^{3}$ )?

**BobP** · March 6th 2013, 08:16 AM

When dealing with 'multiplication' problems, it tends to be the relative errors that are important. If you calculate the two possible values for the relative error calculated in this way, (that is (maxposs(w) - calculated(w))/calculated(w) and (calculated(w) - minposs(w))/calculated(w)), you should find that they are pretty much the same. And, given that we are dealing with a whole lot of rounded numbers and that this is the maximum possible relative error that we are talking about, the difference will not be significant.
It is more usual to calculate the relative error in a different way.
For your problem it would be
$|\frac{\Delta w}{w}| \leq |\frac{0.3}{37.1}|+2|\frac{0.11}{9.87}|+3|\frac{0. 016}{6.052}|\approx 0.0383.$

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