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Error propagation when you take the inverse?

  1. Feb 7, 2008 #1
    Say something is a value +/- .05. What happens when you take the inverse of the value? For example, 30 V +/- .05 V. 1/V...what would the error be?
  2. jcsd
  3. Feb 7, 2008 #2


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    This is a math question. 1/(x+y)=1/(x(1+y/x)).=.(1/x)(1-y/x)=1/x-y/x2.

    The assumption is|y|<<|x|, .=. means approx =

    I'll let you do the arithmetic.
  4. Feb 8, 2008 #3


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    When you take the inverse, use % error. That is the same for the inverse as for the original.
  5. Sep 8, 2011 #4
    Sorry, I have the same qns but i don't get what both of you are saying, elaborate with example? thanks
  6. Sep 8, 2011 #5


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    Staff: Mentor

    In the original question, the error in V is 0.05 V or (0.05/30)*100% = 0.1667%.

    1/V = 0.0333 V^{-1}. The error in this is also 0.1667%, or about 0.0000556 V^{-1}.
  7. Sep 8, 2011 #6

    Andy Resnick

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    The uncertainty in any function of one variable is [itex]\delta y = \left|\frac{dy}{dx}\right| \delta x[/itex]. If y = x^n (in your case n = -1), then [itex]\frac{\delta y}{|y|} = |n| \frac{\delta x}{|x|} [/itex]. For your case, the error is unchanged.

    Taylor's book "An introduction to error analysis" is well worth reading.
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