10
$\begingroup$
Around[100.123456, 0.12]

gives

100.12±0.12

But

Around[100.123456, 0.42]

gives

100.1±0.4

why not 100.12±0.42. What is the rule for Around to show number of significant digits in uncertainty?

Update

I found it seems that Around take 35 as a boundary for uncertainty, no matter what the value is. So

Around[1, 0.35]

gives

1.00±0.35

and

Around[1, 0.36]

gives

1.0±0.4

But the problem is I did not see rule like this in the definite guide GUM(Guide to the Expression of Uncertainty in Measurement). Is there other reference has this rule for uncertainty report?

$\endgroup$
0

2 Answers 2

7
$\begingroup$

TL;DR

The number of significant figures displayed both for the value and its uncertainty depends on the value of the uncertainty. The details are not documented, but there seems to be a threshold around $0.35* 10^n$

Scope

As pointed out by @JimB, this is a display-only issue, the internal representation of the values is not changed.

Wolfram usually does not document the details of the internal implementations of Mathematica. So the scope of the answer necessarily is restricted to what is documented and what can be observed. I will not engage on speculation.

Documentation

The documentation for Around reads

Around[x,δ] displays with one or two digits of the uncertainty δ shown; x is shown with the same number of digits to the right of the decimal point as is shown in δ.

enter image description here

Other display formats exist too

Around[x,δ] is typically displayed as x±δ. If δ is very small compared to x, as in Around[1.2345678,0.0000012], it is instead displayed in a form like 1.23456(78±12).

enter image description here

That is the extent of the explanation provided, details of the implementation are not explained.

Expected behaviour

The criteria should be to keep the Precision and Accuracy coherent. There is no point in having many decimal points that are within the uncertainty of the value.

The number of significant figures (precision) should be dependent upon the uncertainty of the value (accuracy).

Observed behaviour

One can observe that once is decided the number of significant digits on the value, the uncertainty is also displayed with the same number of significant digits, as expected and discussed in the previous title.

The seems to be a preferred threshold

TableForm[
    Around[100.11111111111, #]&/@{0.035, 0.036, 0.35,0.36,3.5, 3.6 }
]

enter image description here

$\endgroup$
5
  • $\begingroup$ Hi, rhermans. Thank you for answering. But this does not solve my question. Can you for example show why Around[100.123456, 0.42] should round uncertainty to 0.4 instead of 0.42 $\endgroup$
    – matheorem
    Jul 6 at 15:21
  • 1
    $\begingroup$ @matheorem There is the "display" of Around[100.123456, 0.42] and what Mathematica stores internally which is used in future calculations: Around[100.123456, 0.42] // FullForm gets you Around[100.12345, 0.42] so in that sense it does do what is advertised. (Sorry, I don't know how to get the backticks to display properly.) $\endgroup$
    – JimB
    Jul 6 at 15:34
  • $\begingroup$ @JimB Thanks for comment. I know the FullForm has everything original. But the display form is also important. I found NIST, GUM documentation almost always use two significant digits. $\endgroup$
    – matheorem
    Jul 6 at 15:39
  • 1
    $\begingroup$ @matheorem I don't know what else do you expect an answer to contain. Wolfram does not reveal the details of their implementations. $\endgroup$
    – rhermans
    Jul 6 at 15:53
  • $\begingroup$ You say that it "depends on the relative value of the uncertainty"; I don't think so, there doesn't seem to be any relative dependency. It's simply that uncertainty values below $0.355\times10^n$ (with $n\in\mathbb{Z}$) show with two significant figures, and values above $0.355\times10^n$ show with one significant figure. As to why they chose $0.355$ as the cutoff: no idea (I would have chosen $1/\sqrt{10}\approx0.316$ as we're in the decimal system). $\endgroup$
    – Roman
    Jul 6 at 16:09
4
$\begingroup$

What I have been taught (and what Mathematica seems to do as well, albeit with a higher threshold) is that you use two significant digits for the uncertainty until a certain threshold (I was told to use 25, Mathematica seems to use 35). The idea is that an uncertainty of 12 has an implied error of 0.5 (i.e. you don't know the third digit), same goes for 90, where the implied error is also 0.5. The "problem" now is that 90 ± 0.5 has a much smaller relative error than 12 ± 0.5. For this reason, one removes one digit from the uncertainty above a certain threshold - if you always show two digits, you (arbitrarily) chose 100 as the threshold, and Mathematica (arbitrarily) chose 35.

$\endgroup$

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.