Is This Uncertainty Propagation Correct?

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SUMMARY

The forum discussion centers on the correctness of uncertainty propagation in mathematical expressions. The first calculation presented, involving the sum of two variables divided by the product of two others, is confirmed as correct. The second calculation, which attempts to propagate uncertainty for the reciprocal of a variable, is deemed incorrect. The correct approach involves adding the relative uncertainties of the numerator and denominator to determine the overall uncertainty.

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lapo3399
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For the following calculation, would the uncertainty propagate as I have estimated?


\frac{\left(a \pm \Delta a \right) + \left(b \pm \Delta b \right)}{\left(c \pm \Delta c \right) \cdot \left(d \pm \Delta d \right)} = \frac{a+b}{cd} \pm \frac{a+b}{cd} \left( \frac{\Delta a + \Delta b}{a + b} + \frac{\Delta c}{c} + \frac{\Delta d}{d} \right)

Thanks.
 
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Yes, looks good to me.
 
Thanks!

One more quick question - is the following uncertainty propagation correct also?

\frac{1}{ \left( r \pm \Delta r \right)} = \frac{1}{r} \pm \frac{1}{ \Delta r}

Thanks again.
 
lapo3399 said:
Thanks!

One more quick question - is the following uncertainty propagation correct also?

\frac{1}{ \left( r \pm \Delta r \right)} = \frac{1}{r} \pm \frac{1}{ \Delta r}

Thanks again.

That doesn't look right to me. To get the relative uncertainty of the fraction... add the relative uncertainty of the top (0)... to the relative uncertainty of the bottom...

So the relative uncertainty of the fraction seems to be \frac{\Delta r}{r} so the absolute uncertainty would be \frac{1}{r}\times \frac{\Delta r}{r}
 

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