3
$\begingroup$

I am fairly new to teaching error propagation to students but came across a question I couldn’t answer myself. Say we have a set of data that produces a parabola. In order to linearize the data and get a straight line we plot the $x$-data as $x^2$. My question is how do I deal with the uncertaintities in the x-data?

$\endgroup$
8
  • $\begingroup$ Do you have error in both the x and y data? $\endgroup$ Feb 14, 2019 at 3:37
  • $\begingroup$ In some cases yes. $\endgroup$ Feb 14, 2019 at 5:27
  • 1
    $\begingroup$ Related. If you're serious about uncertainty analysis, this kind of "linearization" may introduce more problems than it solves. $\endgroup$
    – rob
    Feb 14, 2019 at 6:11
  • 1
    $\begingroup$ Wold Cross Validated be a better home for this question? $\endgroup$
    – Qmechanic
    Feb 14, 2019 at 7:19
  • 1
    $\begingroup$ All of this is good feedback. Thank you. The reality is the new curriculum I have to deliver has a strong focus on linearization and error propagation. $\endgroup$ Mar 28, 2019 at 2:25

1 Answer 1

1
$\begingroup$

Whether or not this is the best way to deal with uncertainty propagation may be up for debate, but one common way that instructors teach uncertainty propagation with regard to linear fitting is simply to sweep uncertainties in individual values under the rug. The goodness-of-fit of the line is used to determine uncertainties in the best-fit slope and y-intercept.

The uncertainty in the slope can be found from the $R^2$ value using $$ \sigma_m = m\sqrt{\frac{(1/R^2)-1}{n-2}}, $$ where $m$ and $\sigma_m$ are the slope and uncertainty in the slope respectively, and $n$ is the number of data points used to make the fit. Students should be reminded (several times!) to be careful - $R^2$ is already squared! Don't square it again.

The uncertainty in y-intercept can be found using the above result: $$ \sigma_b = \sigma_m \sqrt{\frac{\sum x^2}{n}}. $$

While the $R^2$ value should account for the precision of data (i.e. more precise data will tend to have a higher $R^2$ value -- especially when $n$ is big), this approach completely disregards errors that affect the accuracy of the data. For example, if you used a very low-quality ruler that had the potential to be 20% too short or too long, but got really precise results, the $R^2$ value (and hence $\sigma_m$ would be high even though the results might be quite biased!

If anyone else has a better method that is appropriate for intro/sophomore-level physics students, I would love to hear the answer as well! On the other hand, perhaps my method is better than I gave it credit for being, and if anyone can prove that, I'm similarly all ears.

$\endgroup$
1
  • $\begingroup$ How do you get that first formula? $\endgroup$
    – jvriesem
    Mar 10, 2022 at 2:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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