Wikipedia:Reference desk/Archives/Mathematics/2011 February 21

= February 21 =

Adjusting for leap years
Earth's year is about 365.242... days long; to correct for this over a 400-year period we need three rules (every fourth year is a leap year, except centuries are not leap years, except years divisible by 400 are leap years). It got me wondering: What is the worst-case fraction in terms of the number of rules needed to adjust over a given period of time? Are we quite lucky with the fraction 365.242... that nature has given us? 86.160.219.91 (talk) 00:36, 21 February 2011 (UTC)


 * Interesting question. The three leap-year rules you cite mean that the average length of a year in the Gregorian calendar is (365 + 1/4 − 1/100 + 1/400) = 365.2425 days. This excess fraction, 0.2425 or 97/400, could be represented in a single "rule," I suppose, by specifying "Out of every 400 years, the following 97 shall be leap years: …," but obviously that's not in the spirit of the question. If we restrict the rules to be of the form, "Every nth year is an exception to the previous rules," as you've written the rules for the Gregorian calendar, then the excess fraction of a day in the average length of a year will be expressible as an alternating sum of unit fractions. This is similar to Egyptian fractions, except that we are subtracting as well as adding. I don't know the answer to your question, but there has been a lot written about Egyptian fractions, so perhaps if you dig around a bit in that area you'll find something. —Bkell (talk) 00:48, 21 February 2011 (UTC)
 * Correction: I said that, given a particular form of the rules, the average length of a year will be an alternating sum of unit fractions. That will only be true if each level of "exceptions" is entirely included in the previous level. In the Gregorian calendar, for example, every year divisible by 4 is a leap year, except years divisible by 100 are not leap years (all such years are divisible by 4), except years divisible by 400 are leap years (all such years are divisible by 100)—the three levels of "exceptions" here are all nested inside each other, so the average length of a year is an alternating sum of unit fractions, as I mentioned above. But this wouldn't be true for a hypothetical calendar that used rules such as the following:
 * Every year divisible by 3 is a leap year.
 * Every year divisible by 5 is an exception to rule 1.
 * In this strange calendar, rule 2 not only takes away some leap years from rule 1 (such as years 15 and 30), but also adds some new leap years (like years 5 and 10). So it's not true that the average length of a year in this calendar is 365 + 1/3 − 1/5. —Bkell (talk) 00:55, 21 February 2011 (UTC)

Skipping just one leap year every 128 years would be very accurate: 365.2421875. Compare that with the mean tropical year of 365.2421897 days. That would amount to three skipped leap years every 284 years, as opposed to the present (Gregorian) calendar's three every 400 years. Michael Hardy (talk) 01:27, 21 February 2011 (UTC)


 * And even if what we're aiming for is 365.2425 exactly, the Gregorian rule does not appear to be "optimal" from a purely number-theoretic viewpoint. We would get the same average year length by the rules
 * Every year divisible by 4 is leap ...
 * except that years divisible by 132 are not leap ...
 * except that years divisible by 13,200 are leap.
 * which might be argued to be a superior way to approximate .2425, particularly if one considers the precision achieved by truncated rulesets. On the other hand, it is not at all clear how such choices ought to be compared. The Gregorian rules have the large pragmatic advantage that the divisions they involve are easy to carry out in base 10, but it is not easy to account for this formally without ending in rampant ad-hoccery. –Henning Makholm (talk) 02:13, 21 February 2011 (UTC)


 * Note also how attempting to generalize the leap-year rules leads to expression of the form
 * $$0.2425 = \frac{1-\frac{1-\frac{1}{4}}{25}}{4} = \frac{1-\frac{1-\frac{1}{100}}{33}}{4} = \frac{1-\frac{1-\frac{1-\frac{1}{5}}{2}}{20}}{4} = \frac{1-\frac{1-\frac{1-\frac{1-\frac{1-\frac{1-\frac{1}{25}}{6}}{3}}{2}}{16}}{32}}{4}$$
 * which look somewhat like parallel-universe counterparts of continued fractions, but much less well-behaved... –Henning Makholm (talk) 02:38, 21 February 2011 (UTC)
 * These look like Engel expansions. Michael Hardy (talk) 06:23, 21 February 2011 (UTC)


 * Interesting; thanks for the reference. What we have here appears to be Pierce expansions, except for the rule that successive denominators must increase (which I think corresponds to by vague notion of optimality). –Henning Makholm (talk) 13:21, 22 February 2011 (UTC)


 * Answering part of the original question, the "worst fraction to approximate in this way" can be taken to be the Pierce expansion with the lowest period for any particular number of rules. This must be the one with successive divisors 2, 3, 4, ..., which happens to produce the power series for exp(-1)=0.3678794..., so in this sense at least 365.3678794... would be the worst fraction to approximate. –Henning Makholm (talk) 15:22, 22 February 2011 (UTC)


 * The fraction that the Gergorian calendar is approximating is not the mean tropical year of 365.2421897 days, but the mean time between Vernal equinoxes which is 365.242374 days at present but is gradually increasing. John Herschel, 150 years ago, proposed making multiples of 4000 years not leap years which would make the average year 365.24225 days, but this was never adopted because, by the year 4000, the Vernal Equinox year (if we are still using it as we have been doing for many thousands of years) might have lengthened to 365.2425 days.    D b f i r s   21:37, 22 February 2011 (UTC)