0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – whuber Nov 8 '18 at 16:01
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.