Questions Regarding the Fundamental Rules of Error Analysis

[JC2000]

(A)My textbook states the following :

Finally, remember that intermediate results in a multi-step computation should be calculated to one more significant figure in every measurement than the number of digits in the least precise measurement. These should be justified by the data and then the arithmetic operations may be carried out; otherwise rounding errors can build up. For example, the reciprocal of 9.58, calculated (after rounding off) to the same number of significant figures (three) is 0.104, but the reciprocal of 0.104 calculated to three significant figures is 9.62. However, if we had written 1/9.58 = 0.1044 and then taken the reciprocal to three significant figures, we would have retrieved the original value of 9.58.
This example justifies the idea to retain one more extra digit (than the number of digits in the least precise measurement) in intermediate steps of the complex multi-step calculations in order to avoid additional errors in the process of rounding off the numbers.

(1) The example explaining the italicised statement is rather confusing. My understanding of the statement is that we need to include one extra significant figure than what the relevant rule requires us to round off to, for intermediate calculations. For instance while solving $(4.338 + 4.835*3.88/3.0)$, using the rule for multiplication (which tells that the result should retain as many significant figures as there are in the original number with least significant figures) the result of 3.88/3.0 must be taken as 1.29 instead of 1.3. Similarly, for 4.835*1.29 we use 6.237 instead of 6.24(?). Lastly, when adding (6.237 + 4.338) we follow the rule for addition as it is and get 10.575.

(2) Is my understanding correct?

(3) Should 10.575 be rounded to 2 significant figures?

(B) 5.87446 has to be rounded to three significant figures.

(4) Do you start by rounding the rightmost figure and then inwards (gives 5.88) or do you simply ignore all the figures except the fourth and simply round that one (gives 5.87)?

 

[BvU]

(1) Point is you don't want to introduce a shift in outcome solely through repeated rounding. Risk isn't all that great nowadays with calulators and computers, but the principle remains standing.

His example is begging for a mentioning of an exception: number of significant digits for results around a power of ten. It is logical to consider 958 as three significant digits (one digit $\approx$ 1/1000 but then 1043 is also 'three digits' and should not be written as 1040 (some multimeter manufacturers call it 3½ digits).
(In short: err on the generous side if the first digit is a 1)

In your example the 3.0 would be 2 digits, so calculating in three would give
3.88/3 = 1.29 $\qquad$ 1.29 * 4.84 = 6.26 $\qquad$ 6.26 + 4.34 = 10.60
(again, I added the 0 because the first digit is a 1)
2 $\rightarrow$ 2½ digits for the result yields 10.6

All this only on the basis of numbers of digits, where the 3.0 by convention is in the range [2.95,3.05] so some 2%. The other errors are negligible compared to that. So 6.26 $\pm$ 0.13 $\rightarrow$ 10.6 $\pm$ 0.13

10.6 without further info suggests a number in the range [10.55.,10.65] which is slightly optimistic;
but with 11 as reported result the 1½ digits suggest 5% and that would be shortchanging the work done to get the 3.0.

All this changes again if the actual (or estimated) uncertainties of the numbers are given !

(2) You're doing well. give yourself some time to pick up experience (and make errors -- pun intended)

(3) I hope you agree my two cents to argue that reporting: 11 would be too rough

(4) You definitely come from the left. There are some puritans who want to round off fives in a way that the preceding number becomes even. But I never obeyed that one (easier to let the calculator come up with more digits and make the cut at 5000000000 -- so much for good behaviour ).

 

[JC2000]

His example is begging for a mentioning of an exception: number of significant digits for results around a power of ten. It is logical to consider 958 as three significant digits (one digit ≈≈\approx 1/1000 but then 1043 is also 'three digits' and should not be written as 1040 (some multimeter manufacturers call it 3½ digits).
(In short: err on the generous side if the first digit is a 1)

What you mean is for numbers close to 1 or 10 or 100 and so on it is better to carry over an extra significant digit (hence the terminology involving 1/2 as in 3½ digits)?

My impression is that this 'rule' was put in there simply to cover the exceptions so can I safely ignore it in other circumstances?

In your example the 3.0 would be 2 digits, so calculating in three would give
3.88/3 = 1.29 \qquad 1.29 * 4.84 = 6.26 \qquad 6.26 + 4.34 = 10.60
(again, I added the 0 because the first digit is a 1)
2 →→\rightarrow 2½ digits for the result yields 10.6

If I were to solve this example applying the rule above only for the 'cases with 1' I would do it as follows :

3.88/3.0 = 1.29 (Since the result is close to 1 and hence rounding off with an extra significant figure?)

1.29 * 4.835 = 6.24 (Since the 1.29 as three significant figures I rounded off the answer to three figures. Here I do not understand (a) why you chose to round off 4.835 before multiplying and (b) how the result was rounded up to 6.26. )

4.338 + 6.24 = 10.58 (When adding the rule for significant figures mentions that you round off so as to 'retain as many decimal places as there are in the number with least decimal places.' So do you round off the result or do you round of the numbers that you are adding so as to match the decimal digits, as you seem to have done by rounding off 4.338 to 4.34, or do both give the same results)

(4) You definitely come from the left. There are some puritans who want to round off fives in a way that the preceding number becomes even. But I never obeyed that one (easier to let the calculator come up with more digits and make the cut at 5000000000 -- so much for good behaviour ).

From the link you helpfully provided : If a number has to be rounded to a certain number of significant figures/ decimal places, you start with the digit to the right of the last significant figure need and round off ( ignoring the other digits, if any).

My book also states the same with regard to round off fives.

Though I don't understand how disregarding this works.

 

[JC2000]

...On a related note when should you round off before doing an operation or after?

For instance when dividing 4.237 by 2.5 :

One way is to divide to get 1.6948 but since only 2 significant figures are required 4 and 8 are ignored, and the result is 1.7

Or, since the result has to have two significant figures, round off 4.237 to 4.2 and 4.2/2.5 gives 1.68 which also gives 1.7.

Which method is correct, since I don't think both would result in the same answer always.

 

[Vanadium 50]

In this age of calculators, intermediate rounding makes little sense. Round at the end.

 

[BvU]

Since the result is close to 1 and hence rounding off with an extra significant figure?

3.0 is two significant figures, so we have to calculate intermediary result to 3 digits. I forgot to apply the other rule: to add an extra digit if the first is a 1. So 1.293

But when looking at 4.835*3.88/3.0 you could also say the whole term is 6.253 and then round off to 6.25

Simplest rule: if it makes no difference in the final result you can round off but the only way to know that is to carry sufficent digits. Matter of experience and gut feeling. Here at PF we often see things like: a length of e.g 5 m processed to a result in 11 digits. I try to carefully object, friendly but consistently .

Re #4: Agree with Vanadium but don't bother to type in 11 digit numbers if you see a rounding off to 2 or 3 coming up.

 

[JC2000]

Just to state precisely what I have understood/ Precisely ask what I have not understood ( I want to ignore the role of calculators and computers so as to understand the nuts and bolts of this). (I also intend to change my username to Neanderthal. Apologies for asking the same questions again.):

• Use an extra digit (than the number of digits in the least precise measurement) in intermediate steps. (a) Is there a add on effect of this (since an extra digit would have to be taken on for each intermediate step?)? OR (b) Only when dividing by powers of one use an extra digit in the result(?).

• Point is you don't want to introduce a shift in outcome solely through repeated rounding. Risk isn't all that great nowadays with calculators and computers, but the principle remains standing.

  If so : (c) Can I follow (b) always, using the rules for operations involving significant figures as they are, without adding extra figures, unless dividing by powers of one? Since

  3.0 is two significant figures, so we have to calculate intermediary result to 3 digits. I forgot to apply the other rule: to add an extra digit if the first is a 1. So 1.293

  seems to suggest carrying an extra digit in the presence of 1 anywhere?
• Regarding rounding off when performing operations : (d) Rounding measurements before the operation always give the same result as rounding the result of the operation (1.29*4.84 where 4.835 is rounded to three significant digits giving 6.2436 which is rounded off to 6.24 as compared to 1.29*4.835 = 6.23715 which is rounded off to three digits giving 6.24)(OR as BvU stats it is a matter of experience/ gut feeling in which case is it safer to only round the final result?

• 6.26 + 4.34 = 10.60
  (again, I added the 0 because the first digit is a 1)
  2 →→\rightarrow 2½ digits for the result yields 10.6

  (e)Here did you add the zero because the rule for addition states that the decimal places in the result must be the same as the decimal places in the least precise measurement?
• (f)Is assuming uncertainty $\pm 2%$ by convention?

Thank you for your patience.

 

[BvU]

Just to state precisely what I have understood/ Precisely ask what I have not understood ( I want to ignore the role of calculators and computers so as to understand the nuts and bolts of this). (I also intend to change my username to Neanderthal. Apologies for asking the same questions again

I appreciate your tenacity in trying to do this perfectly -- but keep in mind that error propagation and error handling in general can't be cast in concrete. It's not really nuts and bolts. So don't go Neanderthal, you'll need common sense at all times.

To illustrate: statistics teaches that the relative error in an estimate of sigma is around $\ 1\over\sqrt N\$ with N the number of measurements (the size of the sample population). So if someone quotes an error in three digits you can be almost sure it's nonsense.

In addition, errors come in two flavours: statistical and systematic. Statistics is noise, rounding errors and the like - more measurements can reduce statistical errors. Systematic errors are common to all measurements (or a subset), e.g. due to imperfect calibration of measurement tools. Repeating with the same instrument doesn't improve the systematic error.

Conscientious experimenters report both errors to the best of their conscience: $A\pm\sigma_{\text stat}\pm\sigma_{\text syst}$. Handling error propagation with something like that is topnotch (The Particle Data Group are really good at it. So are the guys that do fitting of fundamental constants.)

Now I suspect you have a slight handicap in that your textbook adds errors linearly, which, as I tried to explain, is a slightly pessimistic mode of error propagation.

Remember the common sense I mentioned. I'll try concise answers to your questions but it's not cast in concrete and others may differ.
(a) no add-on effect
(b) powers of one ? They are all 1 ! First digit 1 ? Use common sense !
(c) never ever hide behind rules. As the pirate says: they are only guidelines
(d) when in doubt, round off later
(e) no. I wrote why I added the 0
(f) can't follow this one -- does it come from your book ?

only on the basis of numbers of digits, where the 3.0 by convention is in the range [2.95,3.05]

I sure hope I'm not the only one cherishing this convention . I didn't invent it, it was taught to me, so I'll be ok in that respect.

On a side note; my common sense says never treat a result like 2.0853763 in that way unless you have good evidence otherwise !

And I always laugh myself silly about results in the news like '64.3 percent of respondents agree': it means they have asked 14 people and 9 agreed. But very likely 12 out of 14 misinterpreted the question.

 

[JC2000]

Thanks a lot for the detailed response, I am a lot clearer about how to approach this, thanks to the perspective you offered.

Regarding (a), (b) What I meant was, should I implement the rule on carrying an extra digit only when numbers close to powers of 10 are involved in multiplication/division (958 or 9.58 or 1.29)? What you said about the rules being guidelines tells me that until I develop experience with this, it would be safer to carry over an extra digit for all intermediate operations involving multiplication/division.

Regarding (f) :

All this only on the basis of numbers of digits, where the 3.0 by convention is in the range [2.95,3.05] so some 2%. The other errors are negligible compared to that. So 6.26 ±±\pm 0.13 →→\rightarrow 10.6 ±±\pm 0.13

My interpretation is that since the error/ uncertainty was not mentioned you chose $\pm 2$% by convention...

 

[gmax137]

My interpretation is that since the error/ uncertainty was not mentioned you chose ±2±2\pm 2% by convention...

Well that's the thing about significant figures, it isn't a percentage concept. If you say the value is 3.0 that means it is less than/equal to 3.05; if it were higher than that, you would have told us it was 3.1. Likewise on the low end; the 3.0 reported value means actual is greater than/equal to 2.95.

If you report 6.0 it's the same: actual is less than/equal to 6.05 so now the "percentage" (0.05/6) is closer to 1%.

You could evaluate the example
( 4.338 + 4.835 ∗ 3.88 / 3.0 )

by adjusting each value to minimize or maximize the result:

Max:
( 4.3385 + 4.8355 ∗ 3.885 / 2.95 )=10.70661
Min:
( 4.3375 + 4.8345 ∗ 3.875 / 3.05 )=10.47969

Statistical approaches refine this, basically by saying "what are the chances that all of the measurements are biased in the same direction?"

 

[Friend of Kalina]

This is why we need to go back to slide rules. Precision was always fixed by the physical dimensions of the rule. The internal precision was always maintained, but at the end, you couldn't read it any closer than the tolerance of the markings, so restraining yourself to the number of digits in the original numbers was easy.

:)

I think the idea here is that you retain enough digits through the calculation that when you round to the significant digits of the least precise value in the calculation, the answer doesn't change. The text says that carrying 1 extra digit is enough. If you carried 12 extra digits, and ended up rounding to n significant digits, you would get the same value as carrying 1 extra digit. If you carry NO extra digits, the answer may change. All you're trying to to is not change the number through intermediate rounding, which throws away some of the initial precision of the data. The FINAL result, of course, is limited to the number of significant digits in the least accurate component.

as for (4) Do you start by rounding the rightmost figure and then inwards (gives 5.88) or do you simply ignore all the figures except the fourth and simply round that one (gives 5.87)?

You start from the left. 5.87446 is less than 5.875, and rounds to 5.87. Sequentially rounding from the right cannot make it larger than 5.875.