Talk:Propagation of uncertainty

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Derivative is not partial in Example calculation: Inverse tangent function[edit]

If f(x) = \arctan(x) then you cannot take \frac{\partial f}{\partial x}, because f only depends on one variable. It should be \frac{df}{dx} — Preceding unsigned comment added by Finkeltje (talkcontribs) 15:31, 1 February 2013 (UTC)

It's been fixed already. Feel free to correct the page directly instead of just reporting it. Fgnievinski (talk) 18:38, 2 November 2014 (UTC)

Linear Combinations section had following formula:[edit]

\sigma^2_f= \sum_i^n \sum_j^n a_i \Sigma^x_{ij} a_j= \mathbf{a \Sigma^x a^t}

I changed this to

\sigma^2_f= \sum_i^n \sum_j^n a_i \Sigma^f_{ij} a_j= \mathbf{a \Sigma^f a^t}

f=aAb is incorrect[edit]

The example for f=aA+/-b is incorrect. The derivation which gives this result is based on the f=AB example used b-1 times. However the f=AB example assumes A and B are independent. When considering powers they are not. Instead for positive b we need to do a binomial expansion of f+σf=a(A+σA)b and discard all terms with powers of σA greater than 1, because they should be negligible (assuming the uncertainty is small compared to the absolute value). This gives σf = abAb-1σA = abfσA/A. For negative b we again start with f+σf=a(A+σA)b, then we split it as f+σf= aAb(1+σA/A)b. We can then expand this as per http://mathworld.wolfram.com/NegativeBinomialSeries.html, to give f+σf=aAb(1+bσA/A), again assuming terms with σA to the power two or more are negligible. substituting in for f and rearranging gives σf=abAb-1σA - the same result as for positive b!

I have not yet made the change to the page but I will if nobody objects

Philrosenberg (talk) 14:30, 16 September 2014 (UTC)