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Data Representations, Significant Figures, Precision, Convergence Tolerances, Uncertainty, etc. |
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| Significant Figures | Significant Figures
Most numbers have uncertainties associated with them. For example, if I say I weigh 168 lbs, what I am really saying is: 167.5 lbs < my weight < 168.5 lbs (Eq. 1) I don't say that I weigh 168.000 lbs, since that would imply extraordinarily precise knowledge of my weight: 167.0005 lbs < my weight < 168.0005 lbs (Eq. 2) Eq. 2 might be true, but it might not be true: my scale is not good enough to tell. But I think my scale is good enough that I can be sure about Eq. 1. We say that 168 has three significant figures (i.e. three digits in the number are known to be correct), but 168.000 has six significant figures. Non-zero digits always count toward the number of significant figures; zeroes count except where they are only setting the scale. Here are some examples: 120000. has only two significant digits 120000.0 has seven significant digits 0.000012 has only two significant digits 1.000012 has seven significant digits 0.000120 has three significant digits Note that calculators and computers often spew out a large number of digits in their output, which may or may not all be significant. (To put it more strongly: most of the digits printed in typical computer program outputs are meaningless.) It is the responsibility of the programmer and user to ensure that these numbers are interpreted correctly. Relationship Between Significant Figures and Uncertainty Estimates Knowing a number to three significant figures means that the relative uncertainty in that number is < 1%; if you know a number to six significant figures, the relative uncertainty is less than 0.001 %. Some definitions: Relative uncertainty = max(|Reported Value – True Value|/|True Value|) Absolute uncertainty = max(|Reported Value| - |True Value|) Eqs.(3) Almost always you do not know the True Value, and the uncertainties you report (by how many significant figures you write down) are only estimates. I think my scale is accurate to within a pound, and that I read the number correctly off the dial, and that my weight has not changed by more than half a pound since a I weighed myself, so I confidently say I weigh 168 lbs (three significant figures). But after a big meal, I am not so sure, so I may instead report that I weigh 170 lbs. (two significant figures) i.e. 165 lbs < my true weight < 175 lbs. Only rarely are you absolutely sure that the true value lies in a certain range; what is more often meant is that there is a very high probability that the true value lies in the stated range. Sometimes one knows enough about sources of the uncertainty and the statistics of the measurements of a value that one can assign a numerical probability that the true value lies in the stated range. Note that if the estimated absolute uncertainty is 0.5 lbs, and the corresponding estimate relative uncertainty 0.5/168 = 0.3%, we can say that my weight is known to a precision of 0.5 lbs, or to a precision of 0.3%. Carrying Significant Figures Through Arithmetic Multiplication and Division If I weigh 168 lbs, what is my weight in grams? With great effort, it has been determined that there are 453.5923 grams/lb. (seven significant digits!). So I type on my calculator 168 lbs x 453.5923 grams/lb = 76203.5064 grams but of course I cannot report this value as my weight, since it implies I know my weight to an absolute precision of 50 micrograms (about 1000x less of the weight of a drop of sweat). Instead, I should report a value in grams that has about the same (0.3%) relative uncertainty as the initial 168 lbs: my weight = 7.6 x 104 grams, i.e. 75500 grams = 166.4 < my true weight < 76500 grams = 168.7 lbs Eq.(4) which is true if Eq. 1 is true. Note that significant figures are only a rough way to convey uncertainty information; the lower bound in our reported uncertainty range is a pound less in Eq. 4 than the original estimate in Eq. 1. The rough rule for multiplication and division is to count the number of significant digits of the two numbers involved. Use the smaller number of digits when reporting the answer. Addition and Subtraction If I weigh 168 lbs, and I lose 0.1 lbs, how much do I weigh now? 167.5 lbs < my true original weight < 168.5 lbs 167.4 lbs < my true weight after dieting < 168.4 lbs which we would still write as 168 lbs. So 168 – 0.1 = 168 Very disappointing. Of course this sort of thing easily leads to numerical problems, e.g. if I lose 0.1 lbs a day, how much do I weigh after a month? What if I get into my 1600 lb car? 1550 lbs < true weight of car < 1650 lbs combining with Eq. 1: 1717.5 < true weight of car plus me < 1818.5 which one would normally write as 1800 lbs, or more precisely as 1768 +/- 51 lbs. The rough rule for addition and subtraction is to line up the numbers vertically, and keep only columns where every digit is significant. Here the zeroes in 1600 are not significant 1600 so we round 168 (to 200): 1600 A particularly problematic case is when you have two numbers which are each known precisely, but when you subtract them the result has only a few significant figures. For example, I use a precision balance to weigh a sample sample weight with container = 1.34568 grams and then I heat the sample to drive a chemical reaction which releases a gas. Now I measure weight again: product weight with container = 1.34564 grams So the weight of material which was lost as a gas is 1.34568 grams which is only determined to one significant figure! (>10% relative error). |