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Propagation of uncertainties

  1. Oct 15, 2008 #1
    1. The problem statement, all variables and given/known data
    Given that [tex] f=\frac{\bar u \bar v}{\bar u +\bar v}[/tex]

    show that

    [tex]e_f=f^2({\frac{e_u}{\bar u^2} + \frac{e_v}{\bar v^2}) [/tex]

    where [tex]e[/tex] refers to the error. ok so I added up the fractional uncertainties

    2. Relevant equations

    3. The attempt at a solution
    and I got this


    after some simplifying, I got to this,


    and then I realized that I could never get the answer, however, if this term was negative,
    [tex]\frac{e_u+e_v}{uv}[/tex], i would get the answer perfectly, but how can it be negative??? Problem is even in division, shouldn't the fractional uncertianties add up??
  2. jcsd
  3. Oct 15, 2008 #2
    any help?
  4. Oct 16, 2008 #3
    e_f is the maximum error, I have no idea why the term should be negative
  5. Oct 16, 2008 #4


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    I didn't notate it properly, but the derivatives are partial derivatives. Treat df as ef, du as eu, dv as ev
  6. Oct 16, 2008 #5
    Im sorry, I dont understand your post. how is my working affected by this equation?
  7. Oct 16, 2008 #6


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    I don't follow your working at all, so I'm not sure if it's a right way that I don't know about.

    Are you familiar with differentiation?
  8. Oct 16, 2008 #7
    yeh but not with partial derivatives,

    What I did was to add up the fractional uncertainties.

    I tried evaluating your equation, but I arrived at an ugly equation that doesnt fit the final bill.
  9. Oct 16, 2008 #8


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    OK, that's good. Partial differentiation is exactly like ordinary differentiation. So for examle df(u,v)/du just means normal differentiation of f with respect to u, holding v constant. It may look ugly, but it should work out.

    So first evaluate df(u,v)/du.
  10. Oct 16, 2008 #9
    ok, I tried working out the coefficient of e_u according to your equation. Which is the partial derivative of f wrt u.


    unfortunately, it evaluaes out to be

    [tex] f \times (\frac{v}{u(u+v)})[/tex]

    which isn't what the question asked for. Have I differentiated wrongly?
  11. Oct 16, 2008 #10


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    Try a different rearrangement, maybe take the common denominator to be (u+v)2
  12. Oct 16, 2008 #11
    I have already combined the two terms in my previous post, but i got [tex]f \times (\frac{v}{u(u+v)}) [/tex] for the coefficient of [tex]e_u[/tex] when it should have been
    [tex] \frac{f}{u^2}[/tex]
  13. Oct 16, 2008 #12
    ok sorry, i got the answer, i have to factorise another f out again. THANKS FOR YOUR TIME, IM VERY VERY GRATEFUL!!!!
  14. Oct 16, 2008 #13


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    Wait - you have to do it for df(u,v)/dv also!
  15. Oct 16, 2008 #14
    yeh it's the same for the other term, it should evaluate out to be the same.
  16. Oct 16, 2008 #15


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    And you understand what I meant in #4?
  17. Oct 16, 2008 #16
    yeh i do,

    df=partial derivative of f wrt u times du + partial derivative of f wrt v times dv


    from df du and dv to ef eu and ev is an approximation for a small change.
  18. Oct 16, 2008 #17


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    OK, great (assuming what I said is correct :rolleyes:) :smile:
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