Is This Uncertainty Propagation Correct?

AI Thread Summary
The discussion centers on the correctness of uncertainty propagation in two calculations. The first calculation, involving the sum and product of variables, is confirmed to be accurate. However, the second calculation regarding the uncertainty of a reciprocal is deemed incorrect; the correct approach involves adding the relative uncertainties of the numerator and denominator. The absolute uncertainty should be calculated as the product of the reciprocal and the relative uncertainty of the variable. Overall, the participants clarify the methods for correctly propagating uncertainty in these mathematical expressions.
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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