# Calculating observational error for complex expressions

While doing an experiment I've came upon the need to calculate the error of my value that was calculated using observed values with known observational error. While I know how to calculate the error for expressions with two observed values:

$$\frac{\mathrm{d}F}{F}=\sqrt{\left(\frac{\mathrm{d}x}{x}\right)^2+\left(\frac{\mathrm{d}y}{y}\right)^2} \quad \text{for } F=x y \text{ or } F=\frac{x}{y}$$

Is there a formula for more complex expressions, like $F=xyz$, of $F=\frac{xy}{z}$ and such, or am I forced to calculate the value of $x y$ and its error and then use it with $z$ and its error?

-
Assuming independence of these errors, it would be $\frac{\mathrm{d}F}{F}=\sqrt{\left(\frac{\mathrm{d}x}{x}\right)^2+\left(\frac{ \mathrm{d}y}{y}\right)^2 + \left(\frac{\mathrm{d}z}{z}\right)^2}$ – Sasha Jan 30 '12 at 18:42
Write it as the answer so I could accept it. – Ilya Melamed Jan 30 '12 at 19:10

Let $w = F(x_1, x_2, \ldots, x_n)$ represent the relationship between actual measured quantities $x_k$ and the quantity of interest $w$. Assuming $F$ is a analytic function, we get $$\Delta(F) \approx \sum_{k=1}^n \frac{\partial F}{\partial x_k} \Delta x_k$$ Is we further assume that sources of error $\Delta x_k$ for each measured quantity $x_k$ are normally distributed with zero mean and independent, we get $$\mathbb{E}\left( (\Delta F)^2 \right) \approx \sum_{k=1}^n \sum_{\ell=1}^n \frac{\partial F}{\partial x_k} \frac{\partial F}{\partial x_\ell} \mathbb{E}\left( \Delta x_k \Delta x_\ell\right)$$ Due to independence, for $k \not= \ell$, $\mathbb{E}(\Delta x_k \Delta x_\ell) = \mathbb{E}(\Delta x_k) \mathbb{E}(\Delta x_\ell) = 0$, thus $$\mathbb{E}((\Delta F)^2) = \sum_{k=1}^n \left( \frac{\partial F}{\partial x_k} \right)^2 \mathbb{E}((\Delta x_k)^2)$$
In the case of $F$ that is a simple product of powers $F=x_1^{p_1} \cdots x_n^{p_n}$, $$\frac{\partial F}{\partial x_k} = p_k \frac{F}{x_k}$$