Significant figures calculations

[mg11]

Hi,

I am currently taking a college-level Physics course and ran into some troubling questions regarding significant figures. I wasn’t sure which forum I should post this to but the math forum seemed like the most adequate one.

So here goes…

I have no problem understanding the basic rules for calculating significant figures as long as one operation is performed. My problem starts when solving a physics equation requires performing multiple operations.

Consider the following calculation:

(3.5)*(4.5)/(2.0)

Method A (performing one operation at a time and adjusting the result to the correct number of sig. figs.)

First, we calculate 3.5*4.5 which gives us 15.75
The result has to have 2 sig. figs. So we round it and get 16.
Next we perform 16/2.0 which gives us 8.0

Problem: There is a sig. fig. calculator at http://calculator.sig-figs.com/ which seems to be using this method. This method seems better than method B because it does not violate the basic rules for calculating sig. figs. However, it seems to violate some basic algebraic rules. For example, consider the following calculation:

(5.0+5.0)/(11.0)

Option 1

We first perform 5.0+5.0 and get 10.0
Then we perform 10.0/11.0 and get 0.909090…. so we round it to 0.909

Option 2

Algebraically speaking, we are allowed to write the equation as:

(5.0/11.0)+(5.0/11.0)

We perform 5.0/11.0 and get 0.45454545…… so we round it to 0.45
Then we perform 5.0/11.0 and get 0.45454545…… so again we round it to 0.45
We perform 0.45+0.45 and get 0.90

Problem: we got two different answers with different numbers of sig. figs. simply by using some “legal” algebra.

Method B (entering the entire calculation into a calculator and adjusting the result to the correct number of sig. figs.)

We use our calculator to perform the entire calculation (3.5)*(4.5)/(2.0) and get 7.875
Again, the result should have 2 sig. figs. so we round it and get 7.9 (vs. 8.0 using method A).

Problem: My physics instructor uses this method but it seems to violate the basic rules for performing sig. figs. calculations (as they appear in my textbook). When solving equations that don’t include addition/subtraction we know that the result must have the same number of sig. figs. as the operand with the least sig. figs. However, when we use addition and subtraction we might end up with any number of sig. figs. so using this method will cause us to lose track of how many sig. figs. have to be in the result.

So as you can see, both methods present some problems which I cannot settle. I would love to get an explanation on what is the right method for performing these calculations.

Thanks!

 

[Mute]

I would do the entire calculation and round to the correct number of sig figs at the very end of the calculation. If I calculate something that I report to the desired number of sig figs, but then need to use the number in a subsequent calculation, I would use the unrounded number, do the calculation, and adjust sig figs at the end.

The fact that your method A gives you two different results based on how you parse the equation might be reason enough to use method B instead. If you want more reason, consider the following example:

Say we want to calculate the area of an annulus (a circle with a circle missing in the middle). The outer radius is 1.0 and the inner radius is 0.99. The area is

A = pi*(1.0^2-0.99^2).

If you use method A, pi*1.0^2 = 3.1 to 2 sig figs, 0.99^2 = 0.9801 = 0.98, pi*0.98 = 3.079... = 3.1 to 2 sig figs, so you get

A = 0...

However, you know the area isn't zero! Doing the calculation without adjusting sig figs until the end, we get

A = 0.0625... -> 0.062.

So, I probably would not use method A.

 

[hotvette]

I agree with Mute - do the calculations using unrounded numbers and round in the end. If you were in a Computer Science course studying the computational effects of significant digits the advice might be different.

I remember the days when calculators first showed up in the classroom (yeah, I'm dating myself) and my Engineering professors went nuts when we provided answers to 8 or 10 decimal places on calculations that probably had an accuracy no better than 5-10%. One of my professors refused to let calculators in the classroom (which I quietly applaud to this day).

 

[mg11]

Mute,

Thank you for taking the time to answer. Your answer makes a lot of sense but I still have some difficulty accepting it. While using method B will always give the same result (on a calculator) it is still sensitive to parsing.

Lets bring back this calculation: (5.0+5.0)/11.0

Whether I punch in (5+5)/11 or 5/11+6/11 in my calculator I will get the same result but as I try to determine significant figures I will still get 3 sig figs for (5.0+5.0)/11.0 and 2 sig figs for 5.0/11.0+6.0/11.0 depending on parsing.

Another problem we encounter using method B, is that in some cases we will have to evaluate some intermediate calculations to know the correct number of sig figs.

Take the following case for example:

49/100+51/100

when we punch the whole thing to a calculator we obviously get 1. Now, when we go to evaluate sig figs:

We can see that 49/100 should give 1 sig fig and the same goes for 51/100 but how many figures will be in the sum 49/100+51/100?
To know that we have to actually calculate. Not only do we have to calculate but we have to apply the basic rules to these intermediate calculations because if we try to do 0.49+0.51 without adjusting sig figs we might think that the answer has 3 sig figs (1.00).
So we round the intermediate calculations and get 0.5+0.5=1.0 giving us 2 sig figs which (I think) is the right number of sig figs.
So we will be forced to implement "some of" method A to get the right number of sig figs in the final result.

 

[Mark44]

I remember the days when calculators first showed up in the classroom (yeah, I'm dating myself) and my Engineering professors went nuts when we provided answers to 8 or 10 decimal places on calculations that probably had an accuracy no better than 5-10%. One of my professors refused to let calculators in the classroom (which I quietly applaud to this day).

They can too often be a poor substitute for thinking. One of the physics instructors at the CC where I used to teach was working with a young woman student in his class. She had arrived at an answer that was too large by a factor of 10. When he told he this, she immediately went to her calculator to get the right answer. At this point he put his hand over the calculator to get her to think a moment about what she was doing.

 

[Mute]

Mute,

Thank you for taking the time to answer. Your answer makes a lot of sense but I still have some difficulty accepting it. While using method B will always give the same result (on a calculator) it is still sensitive to parsing.

Lets bring back this calculation: (5.0+5.0)/11.0

Whether I punch in (5+5)/11 or 5/11+6/11 in my calculator I will get the same result but as I try to determine significant figures I will still get 3 sig figs for (5.0+5.0)/11.0 and 2 sig figs for 5.0/11.0+6.0/11.0 depending on parsing.

Another problem we encounter using method B, is that in some cases we will have to evaluate some intermediate calculations to know the correct number of sig figs.

Take the following case for example:

49/100+51/100

when we punch the whole thing to a calculator we obviously get 1. Now, when we go to evaluate sig figs:

We can see that 49/100 should give 1 sig fig and the same goes for 51/100 but how many figures will be in the sum 49/100+51/100?
To know that we have to actually calculate. Not only do we have to calculate but we have to apply the basic rules to these intermediate calculations because if we try to do 0.49+0.51 without adjusting sig figs we might think that the answer has 3 sig figs (1.00).
So we round the intermediate calculations and get 0.5+0.5=1.0 giving us 2 sig figs which (I think) is the right number of sig figs.
So we will be forced to implement "some of" method A to get the right number of sig figs in the final result.

In method B you still have to round your answer to the correct number of significant figures, but you do not round until you have your final answer. You will of course pick up digits beyond what you want in the end, but discarding them before you reach your final answer can introduce round-off errors into your calculation, as shown by the area of the annulus calculation example I gave above. Generally, the more times you have to round your numbers, the more errors you introduce, and these compound themselves.

If you have a calculation with a large number of operations in it, rounding after every operation is sure to introduce errors that accumulate with every subsequent operation, such that your final answer is going to be off due to all the rounding you've been doing. (Nevermind the fact that rounding after every operation is performed is going to be computationally expensive for calculations in which large numbers of operations need to be performed).

 

[AlephZero]

"Significant figures" is not a very precise way to measure how accurately you are giving a result.

Think about "2 significant figures" for example. Suppose the value is close to 100. The difference between 99 and 100 is about 1%, but the difference between 100 and 110 is 10%.

That is the reason for your different results from 10/11 and 5/11 + 5/11.

Mute is right, the best method is to do all the intermediate calculations with carrying some "extra" significant figures, and round the final answer.

At some stage you will learn a better way to handle the errors than just counting significant figures, that deals with the range of error for each quantitiy separately (for example a measurement of "5" might really mean "5 plus or minus 0.2", which you can't describe accurately as a number of significant figures).

The real point of "significant figures" is to stop you giving results to a ridiculously low or high precision compared with the accuracy of the input data. They aren't an "exact" science, as you have figured out for yourself.

 

[mg11]

The real point of "significant figures" is to stop you giving results to a ridiculously low or high precision compared with the accuracy of the input data. They aren't an "exact" science, as you have figured out for yourself.

Indeed, significant figures are very far from being exact science. In fact, I think they're pretty far from being science at all. I did some calculations of my own to check the validity of significant figures. I performed similar calculations using sig figs on one side and precise error notation on the other. I discovered that sig figs were extremely bad at representing the level of uncertainty even after performing just one or two operations. I'm no great mathematician but I think I can prove that sig figs are an infinitely bad way of tracking uncertainty.

Consider the following:

The number 1.0 represents some (uncertain!) level of uncertainty. Let's call that uncertainty "a". So that 1±0.1 for example becomes 1±a

If we use interval arithmetic (http://en.wikipedia.org/wiki/Interval_arithmetic) to add a series of 1's we get:

(1±a)+(1±a)+(1±a)+(1±a)+(1±a)+...+(1±a) = (n±na)

If we do the same for sig figs we get:

1.0+1.0+1.0+1.0+1.0+...+1.0 = n.0

suppose we add up a million 1's in this way using sig fig arithmetic. we will end up with the number 1000000.0 which suggests an uncertainty of the same "a" that we started with. While in reality the uncertainty is now "na". If we started with a=0.1 for example, the uncertainty will now be 1000000±100000.

If you take a long enough chain of 1's you can show that sig fig arithmetic is inaccurate for ANY given uncertainty "a". I have found similar proofs for subtraction, multiplication, and division as well.

In conclusion, I find that the use of sig fig arithmetic is completely idiotic and pointless. Why push students to use pseudo-scientific methods just for the sake of aesthetics? Why not ask students to always give a precision of two or three decimal values in their answers instead?

I became interested in this subject because my community college physics instructor is being very unfair with his grading. He is correcting me for things which I know to be true and deducting a lot of points for things such as forgetting to put units in equations and the use of sig figs. I wanted to have the truth on my side when I dispute his grading but only ended up with an understanding of how ridiculous sig fig arithmetic is.

The only thing I can do now is figure out what he expects me to do and do it simply for the sake of not losing points...

 

[Mute]

Indeed, significant figures are very far from being exact science. In fact, I think they're pretty far from being science at all. I did some calculations of my own to check the validity of significant figures. I performed similar calculations using sig figs on one side and precise error notation on the other. I discovered that sig figs were extremely bad at representing the level of uncertainty even after performing just one or two operations. I'm no great mathematician but I think I can prove that sig figs are an infinitely bad way of tracking uncertainty.

Consider the following:

The number 1.0 represents some (uncertain!) level of uncertainty. Let's call that uncertainty "a". So that 1±0.1 for example becomes 1±a

If we use interval arithmetic (http://en.wikipedia.org/wiki/Interval_arithmetic) to add a series of 1's we get:

(1±a)+(1±a)+(1±a)+(1±a)+(1±a)+...+(1±a) = (n±na)

If we do the same for sig figs we get:

1.0+1.0+1.0+1.0+1.0+...+1.0 = n.0

suppose we add up a million 1's in this way using sig fig arithmetic. we will end up with the number 1000000.0 which suggests an uncertainty of the same "a" that we started with. While in reality the uncertainty is now "na". If we started with a=0.1 for example, the uncertainty will now be 1000000±100000.

If you take a long enough chain of 1's you can show that sig fig arithmetic is inaccurate for ANY given uncertainty "a". I have found similar proofs for subtraction, multiplication, and division as well.

In conclusion, I find that the use of sig fig arithmetic is completely idiotic and pointless. Why push students to use pseudo-scientific methods just for the sake of aesthetics? Why not ask students to always give a precision of two or three decimal values in their answers instead?

I became interested in this subject because my community college physics instructor is being very unfair with his grading. He is correcting me for things which I know to be true and deducting a lot of points for things such as forgetting to put units in equations and the use of sig figs. I wanted to have the truth on my side when I dispute his grading but only ended up with an understanding of how ridiculous sig fig arithmetic is.

The only thing I can do now is figure out what he expects me to do and do it simply for the sake of not losing points...

You have to be careful with your interpretations here. First, in your calculation that yields 1000000 +/- 100000, you haven't properly represented the number to display the correct number of significant figures. Not all of the zeros in the million are significant. If you write the result using scientific notation you can appropriately express the result to two significant figures:

1.0 x10^6 +/- 0.1 x 10^6,

The fact that I only wrote 1.0 indicates that only the first zero after the 1 in the million was significant, and we know this based on the error you calculated. For the pure sig fig calculation, writing the result to two sig figs requires you to write

1.0 x 10^6.

How do you know to write the answer as two sig figs without an error calculation? If your original 1.0 is given to 2 sig figs but with no error attached, then the fact that it is only reported to 2 sig figs tells you there is an uncertainty in the last digit. Since the uncertainty is not reported, it could be anywhere from 0.0 to 0.9. You didn't account for this in the pure sig fig calculation, but it's there, and so you would get a similar result to the one where you included an error explicitly.

In practice, we must do an error calculation to determine how many digits are significant - your result cannot be more accurate than your uncertainty in it. Doing a sig fig calcuation without errors is just to get you used to the idea of what sig figs are and how they are represented. A number reported to some specified number of sig figs necessarily implies there is some error in the last digit. That means that, in your original example, where you get 0.90 versus 0.91, in both of those values there is an uncertainty in the last digit - that digit could be anywhere from 0 to 9, so your reported values in either case might as well be 0.90 +/- 0.09 and 0.91 +/- 0.09. You'll notice these values overlap within their errors.

Since it's so computationally expensive to round to the appropriate number of sig figs at every step, you might as well just do it at the end. (Plus, I suppose it's possible that rounding at every step will introduce exponentially growing errors, but I'm not sure - keeping everything to the appropriate number of sig figs based on errors, the roundoff error might only introduce errors on the order of the last significant digit).

 

[Borek]

Since it's so computationally expensive to round to the appropriate number of sig figs at every step, you might as well just do it at the end.

Never round down intermediate results. Report them as rounded, but for further calculations use them in full precision (or at least use several guard digits).

 

[mg11]

You have to be careful with your interpretations here. First, in your calculation that yields 1000000 +/- 100000, you haven't properly represented the number to display the correct number of significant figures. Not all of the zeros in the million are significant. If you write the result using scientific notation you can appropriately express the result to two significant figures:

1.0 x10^6 +/- 0.1 x 10^6,

The fact that I only wrote 1.0 indicates that only the first zero after the 1 in the million was significant, and we know this based on the error you calculated.

I didn’t use the right notation here but it wasn’t because I didn’t understand the meaning of the result. The bottom line here, as we seem to agree, is that the uncertainty is very high (±0.1 x 10^6) when we take the actual uncertainty factor into account.

For the pure sig fig calculation, writing the result to two sig figs requires you to write

1.0 x 10^6.

How do you know to write the answer as two sig figs without an error calculation? If your original 1.0 is given to 2 sig figs but with no error attached, then the fact that it is only reported to 2 sig figs tells you there is an uncertainty in the last digit. Since the uncertainty is not reported, it could be anywhere from 0.0 to 0.9. You didn't account for this in the pure sig fig calculation, but it's there, and so you would get a similar result to the one where you included an error explicitly.

What you’re saying may be true but, unfortunately, it does not result from applying sig fig arithmetic rules as they are presented in my textbook or in any web sources. Check out Wikipedia for example: http://en.wikipedia.org/wiki/Significance_arithmetic.

The rule for addition/subtraction as it appears in my textbook is:

“When numbers are added or subtracted, the number of decimal places in the result should equal the smallest number of decimal places of any term in the sum”

The result 1.0 x 10^6 as you wrote it will not be reached by using sig fig arithmetic as it is presented to students. Therefore a chain of 1.0+1.0+…+1.0 would eventually give a value of 1000000.0 and not 1.0 x 10^6. That is the whole problem with sig fig arithmetic as I am claiming. And my “proof” still seems valid to me.

In practice, we must do an error calculation to determine how many digits are significant - your result cannot be more accurate than your uncertainty in it. Doing a sig fig calcuation without errors is just to get you used to the idea of what sig figs are and how they are represented. A number reported to some specified number of sig figs necessarily implies there is some error in the last digit. That means that, in your original example, where you get 0.90 versus 0.91, in both of those values there is an uncertainty in the last digit - that digit could be anywhere from 0 to 9, so your reported values in either case might as well be 0.90 +/- 0.09 and 0.91 +/- 0.09. You'll notice these values overlap within their errors.

The results were not 0.91 and 0.90. They were 0.909 and 0.90. So if I follow your example these two values would actually represent 0.909±0.009 and 0.90±0.09 or if we use interval arithmetic notation [0.9 , 0.918] and [0.81 , 0.99].

Now let’s compare that to what we would get using interval arithmetic. The original calculation was (5.0+5.0)/(11.0) we’ll use an error range of ±0.9 for this calculation as well.

[4.1 , 5.9]+[4.1 , 5.9] = [8.2 , 11.8]

[8.2 , 11.8] / [10.1 , 11.9] = [~0.6890756 , ~1.168316]

We find that the real lower bound is ~0.69 vs. 0.9 and 0.81 using sig fig arithmetic and the real upper bound is ~1.17 vs. 0.918 and 0.99 using sig fig arithmetic. This is what's known as [incredibly] Superfluous Precision.

Let’s see what the % error is for one of these pairs.

(0.9-0.69)/0.69*100 = ~30.43% error

~30.43% error after performing just two operations is pretty lousy in my book.

Since it's so computationally expensive to round to the appropriate number of sig figs at every step, you might as well just do it at the end. (Plus, I suppose it's possible that rounding at every step will introduce exponentially growing errors, but I'm not sure - keeping everything to the appropriate number of sig figs based on errors, the roundoff error might only introduce errors on the order of the last significant digit).

As I showed you before, when addition and/or subtraction are involved, you cannot always know what the appropriate number of sig figs will be at the end without making some intermediate calculations AND adjusting them to the correct number of sig figs, at least not according to the rules of sig fig arithmetic as I know them. And I do believe that doing that will introduce exponentially growing errors, that is exactly what I find ridiculous about sig fig arithmetic. Although, if you look at my example you will see that sig fig arithmetic is going to produce incredibly far-fetched results no matter how we do it.