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Error Propagation

  1. Jan 20, 2010 #1
    Recently I came across an example for working out error propagation, and I'm having trouble following the steps:

    A = 100 [tex]\pm[/tex] 1%
    B = 10 [tex]\pm[/tex] 1%

    AB = (100 [tex]\pm[/tex] 1%).(10 [tex]\pm[/tex] 1%)

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

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

    Thank you.
     
    Last edited: Jan 20, 2010
  2. jcsd
  3. Jan 20, 2010 #2

    HallsofIvy

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    Science Advisor

    It doesn't happen- what you have is wrong. If [itex]A= 100\pm 1\%[/itex], so that A lies between 99 and 101, and [itex]B= 10\pm 1\%[/itex], 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 [itex]1000\pm (20\mp .1)[/itex]. 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.
     
  4. Jan 20, 2010 #3
    I see.

    Thank you very much HallsofIvy. This was bugging me for a long time.
     
  5. Jan 20, 2010 #4
    There is a general formula to calculate error propagation, which is statistically more accurate:

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

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


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

    [tex]f_{A}(A,B)=B; f_{B}(A,B)=A[/tex]

    so [tex]\delta (AB)=\sqrt{B^{2}\delta A^{2}+A^{2}\delta B^{2}}[/tex]
     
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