With given:

$\displaystyle k_1 = 37.1\pm0.3$

$\displaystyle k_2 = 9.87\pm0.11$

$\displaystyle k_3 = 6.052\pm0.016$

estimate the absolute and relative error for $\displaystyle w = k_1 \cdot k_2^2 \cdot k_3^3$ and round both $\displaystyle w$ 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: $\displaystyle \Delta w = 0.3 \cdot (0.11)^2 \cdot (0.016)^3 = 1.486848 \cdot 10^{-8}$

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

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

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 $\displaystyle 0.5\cdot10^{-n}$. So for example $\displaystyle t=0.1132$ such that $\displaystyle \Delta t = 0.0001$ has 3 precise figures since $\displaystyle 0.0001<0.5\cdot10^3$. I hope it clears things up a bit...

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

However, I don't have the slightest idea how to round the errors $\displaystyle \delta w, \Delta w$ 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