Calculating Error in Quadrature Equation: y=10^x with dx and dy Values

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SUMMARY

The discussion focuses on calculating the error in the function y=10^x given an error in x, denoted as dx. The correct method to find the associated error in y, dy, is through differentiation, resulting in dy = 10^x * ln(10) * dx. This approach provides a more accurate error estimation compared to previous incorrect methods. The equation is particularly relevant for calculating cosmological distances using the formula d=10^((m(v)-M(v)+5)/5), where errors in m(v) and M(v) are combined using the square root of the sum of their squares.

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Jack_O
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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).
 
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Jack_O said:
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:
 
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:
 
Jack_O said:
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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