Error Propagation: Explaining AB Calculation

[NotStine]

Recently I came across an example for working out error propagation, and I'm having trouble following the steps:

A = 100 $$\pm$$ 1%
B = 10 $$\pm$$ 1%

AB = (100 $$\pm$$ 1%).(10 $$\pm$$ 1%)

=$$\left\{1000 \pm \left[\left(100.1\%\right) \pm \left(10.1\%\right)\right]\right\}$$ // get confused here, how does this happen?

= 1000 $$\pm$$ 1.1
= 998.9 - 1001.1
Can somebody please explain to me how the section I marked above appears.

Thank you.

 

[HallsofIvy]

It doesn't happen- what you have is wrong. If $A= 100\pm 1\%$, so that A lies between 99 and 101, and $B= 10\pm 1\%$, so that B lies between 9.9 and 10.1, then AB lies between 99*9.9= 980.1 101*10.1= 1020.1. We could write those as $1000\pm (20\mp .1)$. 100*10= 1000 is not the midpoint of that interval: it is 1000+ 20+.1 and 1000- 20+ .1

We can think of it this way: the differential of AB is d(AB)= A(dB)+ B(dA). Dividing both sides by AB, dAB/(AB)= dA/A+ dB/B. In this problem dA/A= dB/B= .01 so approximately, dAB/AB= 2(.01)= 2%.

This is an example of an old mechanic's "rule of thumb": when two measurements add, the error of their sum is the sum of their errors; when two measurements multiply, the error of their product is the sum of their relative (or percentage) errors.

 

[NotStine]

I see.

Thank you very much HallsofIvy. This was bugging me for a long time.

 

[elibj123]

There is a general formula to calculate error propagation, which is statistically more accurate:

If you want to calculate the error of a function $$y=f(x_{1},x_{2},x_{3},...)$$
With $$x_{i}$$ being measurements with given errors $$\delta x_{i}$$
Then the error of y is given by:

$$\delta y=\sqrt{\sum_{i=1}^{N}\left| \frac{\partial f}{\partial x_{i}}\delta x_{i}\right|^{2}}$$


F.e. here y=f(A,B)=AB

$$f_{A}(A,B)=B; f_{B}(A,B)=A$$

so $$\delta (AB)=\sqrt{B^{2}\delta A^{2}+A^{2}\delta B^{2}}$$