Propagation of uncertainty when summation symbol is involved

Question

I am unsure how to estimate the propagation of uncertainty when there is a summation symbol involved.

I have the formula (its used to calculate the sauter mean diameter but I will give a simpler example here):

$$ R = \frac{\sum_{i=1}^{N}n_{i}d_{i}^{3}}{\sum_{i=1}^{N}n_{i}d_{i}^{2}} $$

For the benefit of clarity, lets say I have taken the weight (to the nearest integer) of 5000 individuals, so $N$ = 5000. $d_{i}$ is the weight of each individual and $n_{i}$ is the total number of people with weight $d_{i}$.

The fractional uncertainty in $d_{i}$ is 3%. I am not sure what the uncertainty is in $R$. I can estimate the uncertainty of $d_{i}^3$ using the advice here:

http://ipl.physics.harvard.edu/wp-uploads/2013/03/PS3_Error_Propagation_sp13.pdf

But if I just then proceed to carry out the methods when variables are multiplied/ divided, I would surely get a huge uncertainty?

Answer 1

A go at it here:

I'm assume all $n$'s and $d$'s are independent, and that there is no error associated with $n$.

Let's assume $N=2$.

$$ R = \frac{n_1d_1^3 + n_2d_2^3}{n_1d_1^2 + n_2d_2^2} $$

When raising a value with an uncertainty to a power without an uncertainty, the following applies: $$ Z=A^p \\ u_Z = \left | pA^{p-1} \right | u_A $$

A 3% error on $d$ means that the uncertainty of any $d$ is given as $u_d=0.03d$. It follows from this that the uncertainty on $Z=d^p$ is given by $$ u_Z = pd^{p-1}0.03d = 0.03pd^p $$

This can all go back into our original equation $$ R = \frac{n_1(d_1^3 \pm 0.09d_1^3) + n_2(d_2^3\pm0.09d_2^3)}{n_1(d_1^2 \pm 0.06d_1^2) + n_2(d_2^2\pm0.06d_2^2)} $$ multiplying in the $n$'s $$ R = \frac{(n_1d_1^3 \pm 0.09n_1d_1^3) + (n_2d_2^3\pm0.09n_2d_2^3)}{(n_1d_1^2 \pm 0.06n_1d_1^2) + (n_2d_2^2\pm0.06n_2d_2^2)} $$

From your link to uncertainty propagation, when adding uncertainties, we add them in quadrature

$$ R = \frac{(n_1d_1^3+n_2d_2^3) \pm \sqrt{(0.09n_1d_1^3)^2 + (0.09n_2d_2^3)^2 }} {(n_1d_1^2+n_2d_2^2) \pm \sqrt{(0.06n_1d_1^2)^2 + (0.06n_2d_2^2)^2 }} $$

We can look at this and turn it back into sum notation

$$ R=\frac{ \left[ \sum_{i=1}^Nn_id_i^3\right] \pm \left[ 0.09\sqrt{\sum_{i=1}^N(n_id_i^3)^2 } \right] } { \left[ \sum_{i=1}^Nn_id_i^2\right] \pm \left[ 0.06\sqrt{\sum_{i=1}^N(n_id_i^2)^2 } \right] } $$

We can simplify this as $$ R=\frac{A \pm u_A}{B \pm u_B} $$

Now, $A$ and $B$ are not independent. Therefore, their uncertainties add as the relative errors, not the squares of the relative errors. $$ R = \frac{A}{B} \pm \frac{A}{B}\left( \frac{u_A}{A} + \frac{u_B}{B} \right) $$

Hope this helps.