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 General Physics General Physics Help Forum Nov 13th 2017, 09:51 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, 10:37 AM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 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, 10:52 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?   Nov 13th 2017, 12:01 PM #4 Senior Member   Join Date: Aug 2010 Posts: 434 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 12:11 PM.   Nov 14th 2017, 03:13 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, 05:41 AM #6 Senior Member   Join Date: Jun 2016 Location: England Posts: 1,005 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, 06:32 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, 03:44 PM #8 Physics Team   Join Date: Apr 2009 Location: Boston's North Shore Posts: 1,576 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 04:14 PM.   Nov 15th 2017, 02:51 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, 05:28 AM   #10
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 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.  Tags calculating, uncertainy Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post helpmeout Kinematics and Dynamics 0 Oct 9th 2015 08:59 AM wisezeta Energy and Work 1 Jan 16th 2010 01:24 PM christina Kinematics and Dynamics 3 Nov 10th 2009 10:49 PM hana1 Electricity and Magnetism 2 Jun 20th 2009 05:40 AM natasha Kinematics and Dynamics 1 Aug 5th 2008 06:08 PM