Talk:Propagation of uncertainty

Derivative is not partial in Example calculation: Inverse tangent function
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 (talk • contribs) 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)

True, but it would actually be more instructive in this article to use the partials to provide the propagated error of arctan(y/x) where the standard deviations of x and y are known or estimated. 193.17.11.20 (talk) 14:23, 17 September 2018 (UTC)

Linear Combinations section had following formula:

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

I changed this to
 * $$\sigma^2_f= \sum_i^n \sum_j^n a_i \Sigma^f_{ij} a_j= \mathbf{a \Sigma^f a^t}$$

Can we please not have a single equation where a single sign is both a variable and an operator ? It almost seems like someone want to make it unclear on purpose — Preceding unsigned comment added by 128.252.25.45 (talk) 21:48, 5 February 2019 (UTC)

f=aAb is incorrect
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)

Note also that the derivative of Ab is not bAb-1 but ln(A)Ab (https://en.wikipedia.org/wiki/Derivative). — Preceding unsigned comment added by 193.136.48.112 (talk) 09:07, 5 August 2015 (UTC)

Propagation of errors resulting from algebraic manipulations
, currently a redirect to this page, has been nominated for deletion. The discussion at Redirects for discussion/Log/2017 April 16 needs input from those with knowledge of the subject. Thryduulf (talk) 13:38, 17 April 2017 (UTC)

Why don't we include this figure in the article?
The figure was recently removed without explaination. 2A01:C23:783B:B600:B48D:E0C5:7819:B0DD (talk) 10:52, 10 January 2020 (UTC)