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Uncertainty Propagation Argument

  1. Aug 31, 2011 #1
    A while back, one of my undergraduate physics professors gave an argument for why the uncertainty in a function or quantity F is given by


    [itex]\Delta F [/itex] = [itex]\sqrt{^{N}_{i-1}\sum(\frac{\partial F}{\partial x_{i}})^{2}(\Delta x_{i})^{2}}[/itex]


    He argued to think of a right triangle and think of c=[itex]\sqrt{a^{2}+b^{2}}[/itex]

    The uncertainty in the length of side c, would be calculate in a similar method, however it would be

    [itex]\Delta c [/itex] =[itex]\sqrt{(da)^{2}+(db)^{2}}[/itex] which via chain rule would be

    [itex]\Delta c [/itex] =[itex]\sqrt{(\frac{\partial c}{\partial a}\Delta a)^{2}+(\frac{\partial c}{\partial b}\Delta b)^{2}}[/itex]


    He then argued that in a function of several variables, those variables can be thought of as perpendicular to each other in the same way that (a) and (b) are in the right triangle (because they pertain to degrees of freedom), this is why we call for the sum in quadrature.


    I know that it can be derived from normal distribution, however is the argument above correct reasoning?

    Thank you to all

    Alex Z
     
  2. jcsd
  3. Aug 31, 2011 #2
    The summation should read (i=1 to i=N) in the first equation.




    Maybe this is not the correct place for this question???


    Looking back at my first post, I think I might have incorrectly worded his arguement.


    Suppose c is some function of a and b, c is not necessarily the hypotenuse of the triangle. c = f(a,b)

    Then because, c has two degrees of freedom, these act like perpendicular vectors (right triangle).

    So their uncertainties should combine in quadrature. as I've shown above.

    ....

    It sounds like hand waiving, if that made a sound.
     
    Last edited: Aug 31, 2011
  4. Sep 2, 2011 #3
    Anyone?
     
  5. Sep 2, 2011 #4
    It's going to depend on how the variables are related. For example, take the function
    F = a/b + c
    Increasing A or C makes F larger, but in different ways. The only way I know of to compute a margin of error for F is to use maximum and minimum values of a, b, and c and see what range you come up with. Using the above example lets say that
    a = 25 +/- 1
    b = 5 +/- 1
    c = 5 +/- 1

    so F could be anywhere between 24/6 + 4 and 26/4 + 6. Anywhere from 8 to 12.5, or 10.25 +/- 2.25
     
  6. Sep 2, 2011 #5
    I don't think you understand my question.


    The formula that I've provided (for the uncertainty in c) is correct, im not asking if it is. I'm asking about the validity of its derivation by the argument given.
     
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