Physics Help Forum Calculating Uncertainy

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 Nov 13th 2017, 10:51 AM #1 Junior Member   Join Date: Nov 2017 Posts: 4 Calculating Uncertainy If I want to calculate the uncertainty in X given that X = 2/(A*B)^2, how would I do it? I know the uncertainties in A and B and want to find out the uncertainty in X. Thanks!
 Nov 13th 2017, 11:37 AM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 708 I recommend splitting A and B up so that X = {sqrt(2)/ A^2} * {sqrt(2)/ B^2} You can then do the partial differentiation to get dX/dA and dX/dB and then combine them to form the total variation by the product formulae.
Nov 13th 2017, 11:52 AM   #3
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 Originally Posted by studiot I recommend splitting A and B up so that X = {sqrt(2)/ A^2} * {sqrt(2)/ B^2} You can then do the partial differentiation to get dX/dA and dX/dB and then combine them to form the total variation by the product formulae.
I haven't done partial differentiation before. I have these equations:

Can I use the first equation to calculate uncertainty in A*B then use the 3rd equation to calculate uncertainty in (A*B)^-2?
Attached Thumbnails

 Nov 13th 2017, 01:01 PM #4 Senior Member   Join Date: Aug 2010 Posts: 272 You don't need to use "partial differentiation". Instead, think of A and B as being functions of some other variable, say, x, and use the "product rule" and the"chain rule". With $\displaystyle f(x)= 2A^{-2}B^{-2}$, $\displaystyle \frac{df}{dx}= -4A^{-3}B^{-2}\frac{dA}{dx}- 4A^{-2}B^{-3}\frac{dB}{dx}$. In terms of differentials, $\displaystyle df=-4A^{-3}B^{-2}dA- 4A^{-2}B^{-3}dB$. Now take "dA" and "dB" to be the "uncertainties" in A and B and use that formula to calculate "df" the "uncertainty" in f. For example, if A is measured to be 2 with uncertainty dA= 0.1 and B were measured to be 3 with uncertainty dB= 0.2, then $\displaystyle df= =-\frac{4}{(8)(9)}(0.1)- \frac{4}{(4)(27)}(-0.2)= -0.005555- 0.185185= -0.190740$. Last edited by HallsofIvy; Nov 13th 2017 at 01:11 PM.
Nov 14th 2017, 04:13 AM   #5
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 Originally Posted by HallsofIvy You don't need to use "partial differentiation". Instead, think of A and B as being functions of some other variable, say, x, and use the "product rule" and the"chain rule". With $\displaystyle f(x)= 2A^{-2}B^{-2}$, $\displaystyle \frac{df}{dx}= -4A^{-3}B^{-2}\frac{dA}{dx}- 4A^{-2}B^{-3}\frac{dB}{dx}$. In terms of differentials, $\displaystyle df=-4A^{-3}B^{-2}dA- 4A^{-2}B^{-3}dB$. Now take "dA" and "dB" to be the "uncertainties" in A and B and use that formula to calculate "df" the "uncertainty" in f. For example, if A is measured to be 2 with uncertainty dA= 0.1 and B were measured to be 3 with uncertainty dB= 0.2, then $\displaystyle df= =-\frac{4}{(8)(9)}(0.1)- \frac{4}{(4)(27)}(-0.2)= -0.005555- 0.185185= -0.190740$.
Thanks you for the help! Can I not use the equations attached in my previous post?

 Nov 14th 2017, 06:41 AM #6 Senior Member     Join Date: Jun 2016 Location: England Posts: 366 My simplistic approach would (as you suggest) follow the equations in your attachment in a step by step way. X = 2/(A*B)^2 let Q=A*B then (from your attachment): dQ/Q=sqrt[(dA^2/A)+(dB^2)/B] let Z=Q^2 then (from your attachment): dZ/Z=2(dQ/Q) thus: dZ/Z=2{sqrt[(dA^2/A)+(dB^2)/B]} we can see that: X=2/Z again from your attachment we have that: dX/X = dZ/Z (Studiot / HallsofIvy please correct me if this simple approach is not correct) __________________ ~\o/~
Nov 14th 2017, 07:32 AM   #7
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 Originally Posted by Glavien If I want to calculate the uncertainty in X given that X = 2/(A*B)^2, how would I do it? I know the uncertainties in A and B and want to find out the uncertainty in X. Thanks!
Let $\displaystyle u = AB$,

$\displaystyle \frac{\delta u}{u} = \left[\left(\frac{\delta A}{A}\right)^2 + \left(\frac{\delta B}{B}\right)^2\right]^{1/2}$

Let $\displaystyle X = \frac{2}{u^2} = 2u^{-2}$,

$\displaystyle \frac{\delta X}{X} = -2 \frac{\delta u}{u} = -2 \left[\left(\frac{\delta A}{A}\right)^2 + \left(\frac{\delta B}{B}\right)^2\right]^{1/2}$

Therefore,

$\displaystyle \delta X = -2X \left[\left(\frac{\delta A}{A}\right)^2 + \left(\frac{\delta B}{B}\right)^2\right]^{1/2}$

 Nov 14th 2017, 04:44 PM #8 Physics Team   Join Date: Apr 2009 Location: Boston's North Shore Posts: 1,286 So many responses and they're all wrong. Uncertainty in x, i.e. , is a standard deviation which is found by integrating but the value is a function of the wave function. See: https://en.wikipedia.org/wiki/Uncertainty_principle Last edited by Pmb; Nov 14th 2017 at 05:14 PM.
Nov 15th 2017, 03:51 AM   #9
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 Originally Posted by benit13 I don't think this is a QM question. The uncertainties for A and B are known and the OP wants to calculate the uncertainty in X, which can be calculated using propagation of errors in quadrature. I posted a specific answer yesterday but it got blocked Here it is again on the off-chance it doesn't get blocked: Let $\displaystyle u = AB$, $\displaystyle \frac{\delta u}{u} = \left[\left(\frac{\delta A}{A}\right)^2+\left(\frac{\delta B}{B}\right)^2\right]^{1/2}$ Let $\displaystyle x = \frac{2}{(AB)^2} = \frac{2}{u^2} = 2 u^{-2}$ Therefore, $\displaystyle \frac{\delta x}{x} = -2 \frac{\delta u}{u} = -2 \left[\left(\frac{\delta A}{A}\right)^2+\left(\frac{\delta B}{B}\right)^2\right]^{1/2}$
Thank you that all makes sense, exactly what I was looking for.

Nov 15th 2017, 06:28 AM   #10
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 Originally Posted by Pmb So many responses and they're all wrong. Uncertainty in x, i.e. , is a standard deviation which is found by integrating but the value is a function of the wave function. See: https://en.wikipedia.org/wiki/Uncertainty_principle
In other words, you did not understand the question.

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