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Old Nov 13th 2017, 10:51 AM   #1
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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!
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Old Nov 13th 2017, 11:37 AM   #2
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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.
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Old Nov 13th 2017, 11:52 AM   #3
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Originally Posted by studiot View Post
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?
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Calculating Uncertainy-img_1810.jpg  
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Old Nov 13th 2017, 01:01 PM   #4
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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.
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Old Nov 14th 2017, 04:13 AM   #5
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Originally Posted by HallsofIvy View Post
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?
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Old Nov 14th 2017, 06:41 AM   #6
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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)
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Old Nov 14th 2017, 07:32 AM   #7
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Originally Posted by Glavien View Post
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,

$\displaystyle \delta X = -2X \left[\left(\frac{\delta A}{A}\right)^2 + \left(\frac{\delta B}{B}\right)^2\right]^{1/2}$
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Old Nov 14th 2017, 04:44 PM   #8
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So many responses and they're all wrong. Uncertainty in x, i.e. <x>, 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.
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Old Nov 15th 2017, 03:51 AM   #9
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Originally Posted by benit13 View Post
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.
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Old Nov 15th 2017, 06:28 AM   #10
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Originally Posted by Pmb View Post
So many responses and they're all wrong. Uncertainty in x, i.e. <x>, 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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