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Quick quadrature question.

  • Thread starter Jack_O
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  • #1
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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).
 

Answers and Replies

  • #2
tiny-tim
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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.
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:
 
  • #3
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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:
 
  • #4
tiny-tim
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Hi tiny-tim, doesn't bode well that i had to look up the differential of 10x:frown:
Hi Jack_O! :smile:

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

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