1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    mathman

    User Avatar
    Science Advisor

    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

    pam

    User Avatar

    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

    jtbell

    User Avatar

    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

    User Avatar
    Science Advisor
    Education Advisor

    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook