# What is the reasoning behind the deviration of propogation of uncertainty?

When considering the uncertainty of a signal which is determined by multiple inputs propagation of uncertainty states that for measurement $$y = f(x), x=\{x_1, x_2,..., x_N\}$$ uncertainty in $y$ is $$u_y=\sum_i \frac{\partial f}{\partial x_i} u_{x_i}$$ My question is how do we get to this rule. I know that it can be obtained by a taylor expansion of $f_i$ but I don't see why this is a sensible/logical thing to do.

• Please define uncertainty. What your formula looks like is just a step along the definition of derivative of $y$ with respect to $x$: $$\Delta y = \sum_i \Delta f_i(x) = \sum_i \frac{\Delta f_i}{\Delta x}\Delta x$$ – Dilip Sarwate May 1 '14 at 17:24
• @DilipSarwate The uncertainty of a measurement is basically the standard deviation (or a multiple) that you would expect if you repeated the measurement multiple times. See wikipedia for more about propagation – nivag May 2 '14 at 8:25
• Do you actually mean $y(x_1,\ldots,x_n)=\sum_if_i(x_1,\ldots,x_n)$? – Matt L. May 2 '14 at 9:16
• @MattL. Not quite, but I definitely didn't mean what I had written. Edited this now, hopefully it makes more sense – nivag May 2 '14 at 9:25
• If the uncertainty does mean standard deviation as you claim, then your formula is sheer nonsense. It might make sense of uncertainty means variance. – Dilip Sarwate May 3 '14 at 1:52