How Is Uncertainty Calculated for (a+b)/(c+d) When a=b=c=d?

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SUMMARY

The uncertainty of the expression \(\frac{a+b}{c+d}\) when \(a=b=c=d\) and \(\sigma_a = \sigma_b = \sigma_c = \sigma_d\) is calculated as \(\frac{\sigma_a}{a}\). The derivation involves using the formula for the uncertainty of sums and ratios, leading to \(\sigma_{a+b} = \sqrt{2} \sigma_a\) and \(\sigma_{\frac{a+b}{c+d}} = \frac{\sqrt{2} \sqrt{2} \sigma_a}{2a}\), which simplifies to \(\frac{\sigma_a}{a}\). This result highlights that only relative errors are significant in ratio calculations, not absolute uncertainties.

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Homework Statement


what is the uncertainty of \frac{a+b}{c+d} if a=b=c=d and σ_a = σ_b =σ_c=_d

Homework Equations



σ_{a+b}=√(σ_a^2+σ_b^2)

σ_{\frac{a}{b}}=√((\frac{σ_a}{a})^2+(\frac{σ_b}{b}^2))

The Attempt at a Solution


since a=b=c=d

<br /> σ_{a+b}=√2 σ_a

σ_{\frac{a}{b}}=√2 \frac{σ_a}{a}

so σ_{\frac{a+b}{c+d}} = \frac{√2 √2 σ _a}{2a} = \frac{σ_a}{a}

is this correct?!

Thanks
 
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The result looks good to me!
 
Great thanks :) any idea why it is the same as the uncertainty in a single measurement... Just doesn't seem right to me! Xxx
 
It is NOT the same as the uncertainty in a single measurement! That would be ##\sigma_a##. Since you are evaluating a ratio, only relative errors matter. The factors ##\sqrt 2## and 2 just happen to cancel.

You can repeat the exercise with ##{a+b+c}\over e+f+g## and see what happens...
 

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