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I am trying to prove that the uncertainty of the equation $$Q = xy$$ is equal to $$\frac{\Delta Q}{Q_0} = \frac{\Delta x}{x} + \frac{\Delta y}{y}.$$

However what I am getting is $$\Delta Q =\sqrt{(y\Delta X)^ 2 + (x \Delta y)^²} $$ and I am stuck there.. How to continue to prove that it is equal to the second equation that I have written?

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  • $\begingroup$ Possible duplicate by OP: physics.stackexchange.com/q/382376/2451 $\endgroup$
    – Qmechanic
    Jan 27, 2018 at 20:54
  • $\begingroup$ Hi Green P. Welcome to Phys.SE. Please don't repost a closed question in a new entry. Instead, you are supposed to edit the original question within the original entry. $\endgroup$
    – Qmechanic
    Jan 27, 2018 at 20:58

2 Answers 2

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If I understand correctly you are trying to derive

$$\frac{\Delta Q}{Q_0} = \frac{\Delta x}{x} + \frac{\Delta y}{y}.$$

for

for $$Q = xy$$

and you are using the propagation of error general formula in order to do that:

$$\Delta Q =\sqrt{( \frac {\partial Q}{\partial x}\Delta x)^2+(\frac {\partial Q}{\partial y}\Delta y)^2} $$

But you are getting

$$\Delta Q =\sqrt{(y\Delta X)^ 2 + (x \Delta y)^²} $$

I think two points are relevant here.

1) as @SuperCioca points out you should divide by Q to get $${\Delta Q \over Q}=\sqrt{({y\Delta X \over Q})^ 2 + ({x \Delta y\over Q})^²} $$ which simplifies to $${\Delta Q \over Q}=\sqrt{({\Delta X \over x})^ 2 + ({\Delta y\over y})^²} $$

2) It is impossible to get the error propagation expression

$$\frac{\Delta Q}{Q_0} = \frac{\Delta x}{x} + \frac{\Delta y}{y}.$$

from the general formula - The expression that you have been asked to derive is a simplification that is sometimes used because it is simpler that the `correct' version of that you have derived. This expression you are trying to derive is considered simpler because it does not involve squaring and taking the square root. From another point of view the expression that you are trying to derived gives the 'maximum' possible error given errors in x and y of $\Delta$x and $\Delta$y. It is easier to show that $\Delta$Q = $\Delta$x + $\Delta$y is the maximum possible error for Q=x+y -- although the more correct version for the error in Q=x+y is a $\sqrt{ (\Delta x^2 +\Delta y^2)}$.

So in essence the expression you want to derive is a 'maximum possible error', but the general formula give a 'statistically likely error'.


For maximum possible error...

if $$Q=xy $$ $$Q + \Delta Q= (x+\Delta x)(y+\Delta y)$$ $$Q + \Delta Q= xy +y\Delta x + x\Delta y + \Delta x\Delta y $$ now xy = Q so $$ \Delta Q= y\Delta x + x\Delta y + \Delta x\Delta y $$ Now $\Delta x\Delta y$ should be small enough to ignore.... and we divide by Q $$ {\Delta Q \over Q}= {y\Delta x\over Q} + {x\Delta y \over Q} $$ hence $$ {\Delta Q \over Q}= {\Delta x\over x} + {\Delta y \over y} $$

But as I hope is clear by now this formula is not as good as the one you derived from the general error formula

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  • $\begingroup$ The expression you are trying to derive is considered simpler because it does not involve squaring and taking the square root. - From another point of view the expression that you are trying to derived gives the 'maximum' possible error given errors in x and y of $\Delta$x and $\Delta$y. It is easier to prove that $\Delta$Q = $\Delta$x + $\Delta$y is the maximum possible error for Q=x+y -- although the more correct version is again $\sqrt{(\Delta x^2 +\Delta y^2)^{1/2}}$ --- I have updated the post $\endgroup$
    – tom
    Jan 27, 2018 at 21:49
  • $\begingroup$ Ok... for $Q=x+y$ then if $\Delta x=1$ and $\Delta y=1$ then the maximum possible $\Delta Q$ is $\Delta x + \Delta y=2$ . This is easy to show for $Q=x+y$. It is a little bit harder, but possible to show it for $Q=xy$ - you need to maybe substitute... in - I will put a little edit in perhaps $\endgroup$
    – tom
    Jan 28, 2018 at 12:53
  • $\begingroup$ answer edited -- hope this is all clear now... $\endgroup$
    – tom
    Jan 28, 2018 at 13:02
  • $\begingroup$ Great, glad that the answer was helpful $\endgroup$
    – tom
    Jan 28, 2018 at 16:52
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  • The formula is

$$\left ( \frac{\Delta Q}{Q_0} \right)^2 = \left ( \frac{\Delta x}{x} \right )^2 + \left ( \frac{\Delta y}{y} \right )^2. $$

  • Your other, more general formula, is correct.

In your last step, divide by $Q_0 = xy$.

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  • $\begingroup$ Yes it's wrong. It's correct for one variable though... since you can just square root it. $\endgroup$
    – SuperCiocia
    Jan 27, 2018 at 20:34
  • $\begingroup$ The 'correct' formula is the quadratic one, generalised with the use of partial derivatives like here: physics.stackexchange.com/questions/59628/…. I say "correct" because this is all statistics, so all prone to what you decide to define you uncertainty as. You can use your formula if you want, but you have to state it in order for people to know how you got it from. Otherwise everyone will assume you used the standard one, derived by means of a Taylor expansion. $\endgroup$
    – SuperCiocia
    Jan 27, 2018 at 21:22
  • $\begingroup$ My equation is the standard one, derived with a Taylor expansion. Whenever you quote uncertainties, people will assume you used that one. Yours seems a simplification of the more general formula. To be honest, error analysis is all a bunch of statistics anyway, so as long as you quote the formula that you used to get the error (so that people can reproduce it) it's fine. $\endgroup$
    – SuperCiocia
    Jan 30, 2018 at 0:24

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