Is This Uncertainty Propagation Correct?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 2K views
lapo3399
Messages
53
Reaction score
0
For the following calculation, would the uncertainty propagate as I have estimated?


[tex]\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)[/tex]

Thanks.
 
Physics news on Phys.org
Thanks!

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

[tex]\frac{1}{ \left( r \pm \Delta r \right)} = \frac{1}{r} \pm \frac{1}{ \Delta r}[/tex]

Thanks again.
 
lapo3399 said:
Thanks!

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

[tex]\frac{1}{ \left( r \pm \Delta r \right)} = \frac{1}{r} \pm \frac{1}{ \Delta r}[/tex]

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 [tex]\frac{\Delta r}{r}[/tex] so the absolute uncertainty would be [tex]\frac{1}{r}\times \frac{\Delta r}{r}[/tex]