Talk:Propagation of uncertainty

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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)[reply]

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)[reply]

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)[reply]

Linear Combinations section had following formula:

\sigma _{f}^{2}=\sum _{i}^{n}\sum _{j}^{n}a_{i}\Sigma _{ij}^{x}a_{j}=\mathbf {a\Sigma ^{x}a^{t}}

I changed this to

\sigma _{f}^{2}=\sum _{i}^{n}\sum _{j}^{n}a_{i}\Sigma _{ij}^{f}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)[reply]

f=aA^b 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+σ_f=a(A+σ_A)^b and discard all terms with powers of σ_A greater than 1, because they should be negligible (assuming the uncertainty is small compared to the absolute value). This gives σ_f = abA^b-1σ_A = abfσ_A/A. For negative b we again start with f+σ_f=a(A+σ_A)^b, then we split it as f+σf= aA^b(1+σ_A/A)^b. We can then expand this as per http://mathworld.wolfram.com/NegativeBinomialSeries.html, to give f+σf=aA^b(1+bσ_A/A), again assuming terms with σ_A to the power two or more are negligible. substituting in for f and rearranging gives σ_f=abA^b-1σ_A - the same result as for positive b!