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Propagation of Uncertainties using Partial Differentials and w/ and w/o Probability

  1. Sep 19, 2006 #1
    Assume we have the function [tex]z = x\sin y[/tex]
    Our best guest for our measurement is x=1.0 and y=2.0. The uncertainty in x is 0.05. The uncertainty in y is 0.10.

    We want to calculate the final uncertainty as the initial uncertainties propagate through the function.

    ***** Method 1 *****
    In Calculus III we find the propagation of uncertainties in multivariable functions using the following method:

    [tex]
    dz = \frac{{\partial z}}{{\partial x}}dx + \frac{{\partial z}}{{\partial y}}dy
    [/tex]

    So the uncertainty would be

    [tex]
    \begin{array}{l}
    dz = \sin \left( y \right)dx + x\cos \left( y \right)dy \\
    dz = \sin \left( {2.0} \right)\left( {0.05} \right) + \left( {1.0} \right)\cos \left( {2.0} \right)\left( {0.10} \right) \\
    dz = 0.0039 \\
    \end{array}
    [/tex]


    ***** Method 2 *****

    According to
    http://www.amazon.com/Introduction-Error-Analysis-Uncertainties-Measurements/dp/093570275X

    It says we should use this formula to calculate the propagated uncertainty:

    [tex]
    \delta z = \sqrt {\left( {\frac{{\partial z}}{{\partial x}}dx} \right)^2 + \left( {\frac{{\partial z}}{{\partial y}}dy} \right)^2 }
    [/tex]

    Using this method the uncertainty is

    [tex]
    \begin{array}{l}
    \delta z = \sqrt {\left[ {\sin \left( {2.0} \right)\left( {0.05} \right)} \right]^2 + \left[ {\left( {1.0} \right)\cos \left( {2.0} \right)\left( {0.10} \right)} \right]^2 } \\
    \delta z = 0.062 \\
    \end{array}
    [/tex]

    The uncertainty in method 2 is nearly 16 times larger than the uncertainty in method 1.
    I am assuming method 2 represents the uncertainty better than method 1.

    My question is: What is method 2 taking into account that method 1 isnt? Why does method 2 represent the uncertainty better than method 1?
     
    Last edited: Sep 19, 2006
  2. jcsd
  3. Sep 20, 2006 #2

    mathman

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    Science Advisor
    Gold Member

    To reconcile the 2 approaches, I suggest you modify method 1 to use absolute value for both terms and then add. This would bring them closer.

    Method 2 is the usual statistical approach, since errors can be negative or positive.
     
  4. Sep 21, 2006 #3

    Chronos

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    Science Advisor
    Gold Member
    2015 Award

    I believe the law of averages would rapidly assert itself in this scenario. The Chi squared probability is the most reliable method, IMO.
     
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