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Uncertainty Derivations/Calculation

  1. Mar 3, 2014 #1
    1. The problem statement, all variables and given/known data
    If F = aXn = f +- f +δf where a is a constant, show f = xn and [itex]\frac{δf}{f}[/itex] = [itex]\frac{nδx}{x}[/itex].

    X = x +- δx

    x refers to the average and δx refers to uncertainty in x.


    2. Relevant equations
    The power rule for error propagation shows that the uncertainty is multiplied n times (where n is the power raised).


    3. The attempt at a solution
    I'm having trouble showing that f = xn. Through the use of algebraic manipulation, I was able to get a(x+δx)n = f + δf. I then made the assumption to ignore the constant a and by deduction say x = 5 +- 0.5, set f = xn because it is continuously multiplied by whatever the function x is to the nth power. The second part is easier- mainly I just took the differential δf = n*xn-1δx. This simplifies to [itex]\frac{δf}{f}[/itex] = n[itex]\frac{δx}{x}[/itex]
     
  2. jcsd
  3. Mar 3, 2014 #2

    BvU

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    Hello t and welcome to PF. If I want to help, I have to be able to read what you've written. Could you proofread your stuff and explain what isn't completely universally known language ?

    I recognize some aspects from error propagation theory, but I find your jargon hard to understand.
     
  4. Mar 3, 2014 #3
    Hello,

    I am supposed to show that f = xn given that F = aXn = f ± δf where a is a constant. X is equal to x ± δx. In plain English, X is equal to the average x, plus or minus the uncertainty in x (δx).

    I am having trouble showing that f = xn. I have already derived δf/f using the method shown above.

    Thanks again!
     
  5. Mar 3, 2014 #4

    BvU

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    Yes, it still looks weird that a should disappear from f. Is that really how it is presented to you ?
    I mean if F = 1000 X2 and X = 5.000 ± 0.001 there is no way that f can be x2
     
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