# Propagation of uncertainty when summation symbol is involved

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:

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

• The basis for your conclusion "get a huge uncertainty" is unclear. Have you carried out the calculations for small values of $N,$ such as $N=1,2,3$? You will learn much from that. – whuber Nov 8 '18 at 16:01

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 $$R = \frac{(n_1d_1^3 \pm 0.09n_1d_1^3) + (n_2d_2^3\pm0.09n_2d_2^3)}{(n_1d_1^2 \pm 0.06n_1d_1^2) + (n_2d_2^2\pm0.06n_2d_2^2)}$$

$$R = \frac{(n_1d_1^3+n_2d_2^3) \pm \sqrt{(0.09n_1d_1^3)^2 + (0.09n_2d_2^3)^2 }} {(n_1d_1^2+n_2d_2^2) \pm \sqrt{(0.06n_1d_1^2)^2 + (0.06n_2d_2^2)^2 }}$$
$$R=\frac{ \left[ \sum_{i=1}^Nn_id_i^3\right] \pm \left[ 0.09\sqrt{\sum_{i=1}^N(n_id_i^3)^2 } \right] } { \left[ \sum_{i=1}^Nn_id_i^2\right] \pm \left[ 0.06\sqrt{\sum_{i=1}^N(n_id_i^2)^2 } \right] }$$
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. $$R = \frac{A}{B} \pm \frac{A}{B}\left( \frac{u_A}{A} + \frac{u_B}{B} \right)$$