So suppose you have a value $A$ with an absolute uncertainty $\pm a$ and another value $B$ with absolute uncertainty $b$, it is easy to prove that the rule for dealing with the uncertainties when adding these values like so: $$(A\pm a)+(B\pm b)=(A+B)\pm (a+b)$$ However the rule for dealing with uncertainties when multiplying the values is: $$(A\pm a)\times(B\pm b)=(A\times B)\pm \left[\left(\frac{a}{A}\cdot100\right)+\left(\frac{b}{B}\cdot100\right)\right]\%$$ my question is how would one go about proving this fact or is it just useful convention with no logical proof. If it is the latter why is useful to deal with uncertainties this way. Thanks in advance :)
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1Do you mean $A\times B$ instead of $A+B$ for the RHS of last equation? – Karn Watcharasupat Dec 27 '17 at 16:21
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Your last formula cannot be correct, since putting $a = b = 0$ we get $A\times B = A+B$. – JMJ Dec 27 '17 at 16:22
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yes is do I have changed it now – cal Dec 27 '17 at 16:37
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why do you have the 100's in the answer? – JMJ Dec 27 '17 at 16:40
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in physics when you multiply two measurements with uncertainties you add the percentage uncertainties – cal Dec 27 '17 at 16:47
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the last equation is incorrect, need to multiply the part after +/- with (A X B). The forms in the answers are correct – ten1o Jan 26 '21 at 14:06
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The rule is $(A \pm a) \times (B \pm b)=AB \pm Ab \pm Ba \pm ab$, which you can verify by the distributive principle. (Note that: the case where all plus-minus's in the product are minus is not realizable). We often write this as $(A \pm a) \times (B \pm b)=AB(1 \pm \frac bB \pm \frac aA \pm \frac {ab}{AB})$ which has the idea you are remembering that the fractional errors add. Then if the fractional errors are small we say $\frac {ab}{AB}$ is small twice and can be ignored, giving the more famous $$(A \pm a) \times (B \pm b)=AB\left(1 \pm \frac aA \pm \frac bB\right)$$
EEH
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Ross Millikan
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tbh this question may belong on the physics board as it come from the practice of adiing the percentage unceartainty when multiplying measurements. Apologise if the equation you gave me is the same as this but I cant see how it is so does this mean that the formula I gave is just convention and has no proof? – cal Dec 27 '17 at 16:47
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@ten1o: because $a$ and $b$ are uncertanties and are positive. The measured value can be low or high, so we use the $\pm$ sign. – Ross Millikan Jan 26 '21 at 15:02
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But the general formula includes only the plus sign. a is taken to be either -a or a and b as either -b or b, multiplying these values gives us two positive values and one negative value, which equals one positive value. – ten1o Jan 26 '21 at 15:32
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That is a reasonable convention. I have usually seen the uncertainty defined as positive. So you might have a measurement that is given as $65\pm1$. The maximum error is given as positive, but the measurement can be $64$. Certainly to get the extremes of the range you want $+$ in the final set of parentheses. – Ross Millikan Jan 26 '21 at 15:36
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Suppose $A, B$ are the variables and $a, b$ are their absolute uncertainties respectively. We could then calculate absolute uncertainty of $AB$ as follows:
$$\max(AB)= (A+a)(B+b)=AB + Ab + Ba + ab$$ $$\min(AB)= (A-a)(B-b)=AB - Ab - Ba + ab$$
So absolute uncertainty of $AB = \frac{\max(AB) - \min(AB)}2 = \frac{2Ab + 2Ba}2 = Ab + Ba$
% uncertainty of $AB$ = absolute / $AB = b/B + a/A =$ %uncertainty of $A$ + % uncertainty of $B$.
