# Propagation of uncertainty

I have a problem to understand the concept of propagation of uncertainty. To be honest there are two issues am confused about. (1) Do we need to use sum of uncertainties or sum of squares? (2) Derivation from Taylor approximation produces approach slightly different formula for uncertainties, including an extra terms, and I would be happy if someone could explain the difference.

TL;DR

The first problem is related to sum of uncertainties. I followed the youtube tutorials by Prof. Glenn Stark (Part 1 and 2 and Part3) which explains the concept quite well. According to the video, formula for the error propagation is:

$f(x_1,\dots,x_n)\ = f(.)$

$$\delta f(.)\ =\ \left|\frac{\partial f(.)}{\partial x_1}|_{x_{10}, \dots,x_{n0}}\right|\delta x_1\ +\ \dots\ +\ \left|\frac{\partial f(.)}{\partial x_n}|_{x_{10}, \dots,x_{n0}}\right|\delta x_n\ .$$

After deriving partial derivatives, usually the final equation has the form:

$$\frac{\delta f(.)}{F}\ =\ C_1\frac{\delta x_1}{x_1}\ +\ \dots\ +\ C_n\frac{\delta x_n}{x_n}\ .$$

I found another video (here) which states that we should sum squared values of uncertainties!

The second problem is related to derivation using Taylor approximation:

\begin{align} f(x) \approx f(x_0)+f'(x_0)(x-x_0) \end{align}

Using this approach, I ended up having the same form of uncertainties but with an extra component, where two uncertainties are multiplied. As an example, X is equal $X=A/B^2$. Uncertainties are propagating as follows:

$$X=\frac{A + \delta A}{(B + \delta B)^2}=\frac{A + \delta A}{B^2}\left(1 + \frac{\delta B}{B}\right)^{-2}$$

The term in denominator can be approximated using Taylor series:

$$\left(1 + \frac{\delta B}{B}\right)^{-2} \approx 1 - 2\frac{\delta B}{B}$$

Then approximation of uncertainties associated with X can be derived by multiplying two terms together:

\begin{align} X\ &\approx\ \frac{A + \delta A}{B^2} \left(1 - \frac{2\delta B}{B} \right) \\ &\approx\ \frac{A}{B^2} + \frac{\delta_A}{B^2} - \left(\frac{2A\delta_A}{B^3} + \frac{\delta_A\delta_B}{B^3} \right) \end{align}

The result is similar to example two but this approach generates an extra term $\delta_A\delta_B/B^3$ which I don't understand.

I provided two example to show how I understand uncertainties.

Example 1:

$A = wh$

\begin{align} \delta A\ &=\ \left| \frac{\partial A}{\partial w}|_{w_0,h_0} \right| \delta w\ +\ \left| \frac{\partial A}{\partial h}|_{w_0,h_0} \right| \delta h \\ &=\ h_0 \delta w\ +\ w_0 \delta h\ . \\ \end{align}

Dividing by $A_0\ =\ w_0h_0$:

$$\frac{\delta A}{A_0}\ =\ \frac{\delta h}{h_0}\ +\ \frac{\delta w}{w_0},\quad C_1=C_2=1\ .$$

Example 2:

$g = \frac{4\pi^2L}{T^2}$

\begin{align} \delta g\ &=\ \left| \frac{\partial g}{\partial L}|_{L_0,T_0} \right| \delta L\ +\ \left| \frac{\partial g}{\partial T}|_{L_0,T_0} \right| \delta T \\ &=\ \frac{4\pi^2}{T^2} \delta L \ +\ \frac{8\pi^2L}{T^3} \delta T\ . \end{align}

Dividing by $g_0\ =\ 4\pi^2L_0/T_0^2$:

$$\frac{\delta g}{g_0}\ =\ \frac{\delta g}{L_0}\ +\ \frac{2\delta T}{T_0},\quad C_1=1,\ C_2=2\ .$$