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Propagation of Error/Uncertainty

  1. Jul 24, 2010 #1
    I'm trying to get an intuitive sense for errors and picked some random numbers:

    x = 2.5 +/- 0.01
    find f(x) = x³

    d f(x) / dx = 3x²
    d f(x) = 3x² dx
    = 3(2.5)² 0.01
    = 0.1875

    What I don't get is why f(x - Δx) ≠ f(x) - 0.1875 and why f(x + Δx) ≠ f(x) + 0.1875

    Where did I go wrong in my method for finding the uncertainty value? Thanks
     
  2. jcsd
  3. Jul 24, 2010 #2
    I found a webpage which lists:

    [tex]Q = a^n \Rightarrow \frac{\Delta Q}{Q} = |n| \frac{\Delta a}{a}[/tex]

    Applying the method I used in OP here,

    [tex]\frac{\delta Q}{\delta a} = n a^{n-1}\\
    \delta Q = na^{n-1}\delta a\\
    \therefore \frac{\delta Q}{Q} = \frac{na^{n-1}\delta a}{a^n} = n \frac{\delta a}{a}[/tex]

    So it looks like my method of deriving the uncertainty is correct.

    Working out f(x) = x^6; x = 25 +/- 1 I get do,

    f(25 - 1) = f(24) = 191102976
    f(25 + 1) = f(26) = 308915776

    now using the identity df(x) = 6 25^5 * 1 = 58593750

    25^6 + 58593750 ≠ (25+1)^6
    25^6 - 58593750 ≠ (25-1)^6

    Any ideas???
     
    Last edited: Jul 24, 2010
  4. Jul 24, 2010 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, you can think of a small error as an "differential" and approximate errors in functions of the measurement using the derivatives.

    If [itex]y= x^2[/itex] then [itex]dy/dx= 2x[/itex] so that [itex]dy= 2xdx[/itex]. You could also say, then, that
    [tex]\frac{dy}{y}= \frac{2xdx}{y}= \frac{2xdx}{x^2}= 2\frac{dx}{x}[/tex]
    so that the "relative error", the actual error in the measurement divided by the measurement, is multiplied by 2.

    More generally, if f(x,y)= xy, where x and y are independent measurements, then [itex]df= ydx+ xdy[/itex] and so
    [tex]\frac{df}{f}= \frac{ydx+ xdy}{xy}= \frac{dx}{x}+ \frac{dy}{y}[/tex].

    This is equivalent to the old engineering "rule of thumb": "When measurements are added, their errors add, when measurements are multiplied, their relative errors add".
     
  5. Jul 24, 2010 #4
    Thank you for your informative post.

    But what I still don't get is what they are showing?

    If I make a measurement of x = 2 +/- 1 otherwise written as x = [1,3]

    Then y = x² would be [1,9]

    Using the rule derived dy = y (2 dx / x) = 2 x dx = 2*2*1 = 4

    y = 2² +/- 4 = [0,8] ≠ [1,9]

    So what does this represent intuitively?

    Is this only an approximation? It doesn't look like it should be
     
    Last edited: Jul 24, 2010
  6. Jul 24, 2010 #5

    Office_Shredder

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    Gold Member

    Using the derivative is basically a way of saying "We're going to assume our function really looks like a line, and use the slope of that line to figure out what the error is".

    The bigger your interval, the more room for error as the tangent line becomes a worse and worse approximation.

    Notice that [0,8] is the range of the tangent line at x=2 over the interval[1,3]
     
  7. Jul 24, 2010 #6
    Thank you! Of course...
     
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