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I am confused about the uncertainty equations calculations.

If I have the equation X1 * Y1 = X2 * Y2 and I want to find the uncertainty of X1 , I convert the equation to X1 = (X2 * Y2)/(Y1) . Now, I would know how to calculate the uncertainty if it was a straight division or multiplication uncertainty but this equation is both. How do I go about calculating the uncertainty if the equation contains both multiplication and division? Do I just do the multiplication uncertainty first and then the division uncertainty?

Thanks any advice is appreciated.

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You have three variables, for simplicity let's call them, x, y, z each with a (standard deviation) uncertainty $\sigma_x, \sigma_y, \sigma_z$. The formula you want to find the error on is then $f=xy/z$ which will have a standard deviation $\sigma_f$ The method that always works is the the formula that gives the square of the uncertainty $$ \sigma_f ^2 = \left( \frac{\partial f}{\partial x}\right)^2_{y,z}.\sigma_x^2+\left( \frac{\partial f}{\partial y}\right)^2_{x,z}.\sigma_y^2+\left( \frac{\partial f}{\partial z}\right)^2_{x,y}.\sigma_z^2$$

which looks horrid and intimidating but is very easy to use as it only involves taking the derivative of f with x, y and z in turn, while keeping the other variables constant.

In your example $\partial f/\partial x = y/z$, $\partial f/\partial y =x/z$ and $\partial f/\partial z =-xy/z^2$. Putting the parts together gives $$\sigma_f^2= \frac{y^2}{z^2}\sigma_x^2 + \frac{x^2}{z^2}\sigma_y^2 +\frac{x^2y^2}{z^4}\sigma_z^2 $$

You can simplify this a bit but you should have all the values to plug into this. Don't forget to take the square root at the end.

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  • $\begingroup$ That method doesn't "always" work. Like mine it ignores the covariances, $\sigma_{x2y2}, \sigma_{x2y1}, \sigma_{y1y2}, \sigma_{x2y2y1}$ $\endgroup$ – MaxW Feb 8 '17 at 18:03
  • $\begingroup$ As long as the variables are un-correlated this method is ok, and I suspect that this is the case in the OP's question. You would need to know all the cross-correlation sigmas of course and then its best to use matrices to do the calculation. $\endgroup$ – porphyrin Feb 8 '17 at 21:30
  • $\begingroup$ multiply all the terms in the last equation by $$\dfrac{f^2z^2}{x^2y^2}$$ which is equal to 1. then you'll get $$\dfrac{\sigma_f^2}{f^2} = \dfrac{\sigma_x^2}{x^2} + \dfrac{\sigma_y^2}{y^2} +\dfrac{\sigma_z^2}{z^2} $$ $\endgroup$ – MaxW Feb 8 '17 at 23:01
  • $\begingroup$ nice, I didn't try to simplify this but I'll add thus to the answer if that's ok with you? I could also add covariance bit so that there is complete answer for others to use. $\endgroup$ – porphyrin Feb 9 '17 at 9:39
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$\%\text{RSD}_{X1} = \sqrt{\%\text{RSD}^2_{X2} + \%\text{RSD}^2_{Y2} + \%\text{RSD}^2_{Y1}}$

For multiplication and division the error propagates as the % relative standard deviation.

This ignores the covariances which means that there is no interaction between the variables.

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