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Quick quadrature question.

  1. Apr 20, 2009 #1
    I have an equation in the form y=10^x, i have an error in x, dx. I am unsure how to find the associated error in y. dy=10^dx gives to small an error and dy=x-dx doesn't seem logical as a smaller error in x gives a bigger error in y.

    (The actual equation is d=10^((m(v)-M(v)+5)/5), i have already have errors for m(v) and M(v) and have combined them by using sqrt(dM(v)^2+dm(v)^2), the equation is used for calculating cosmoligical distances).
     
  2. jcsd
  3. Apr 20, 2009 #2

    tiny-tim

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    Hi Jack_O! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    Hint: just differentiate … if y = 10x, then dy = (what)dx ? :wink:
     
  4. Apr 20, 2009 #3
    Hi tiny-tim, doesn't bode well that i had to look up the differential of 10x:frown:

    Anyhoo i know get my error in y as 10xln(10)dx, which gives me a much more reasonable answer, thanks for your help:smile:
     
  5. Apr 20, 2009 #4

    tiny-tim

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    Hi Jack_O! :smile:

    Quick trick: 10x = (eln10)x = exln10 :wink:
     
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