6.1 The Mystery of Archimedes' Number
6.1.1 The Key to all Puzzles If you have worked your way through Chapters 1 - 5 will now be rewarded. We are ready to explain the puzzles of Chapter 1. In fact, this book has been written for the sake of one short conclusion: If you want to approximate a real number with high accuracy by a fraction with a sufficiently simple denominator, you should replace it by convergents.
We will now solve in this manner Archimedes' problem and the problem of the calendar.
Note that Christian Huygens encounetered continued fractions when trying to approximate real numbers by sufficiently simple fractions. He wanted to construct a model of the Solar system in which planets were modelled by gear-wheels. In order to reproduce revolution periods with sufficiently high accuracy wheels, he had to have staggeringly large numbers of teeth. Huygens looked for and found a general method for solving such problems: Substitution of much smaller numbers for the large ones, reproducing their ratios as accurately as possible. In this way, he invented continued fractions as an auxiliary tool and discovered many of their properties, although Raffael Bombelli in Italy had operated with these fractions (in a more superficial way) one hundred years earlier.
N. N. Luzin used to say in such cases that even chips and shavings are valuable in the laboratory of a great scientist.
6.1.2. The Secret of Archimedes' Number In order to obtain approximations to the number p, expand it into a continued fraction. Take its decimal approximation with a high accuracy, for example,
3.14159265 = 314159265/100 000 000
and apply Euclid's algorithm:
p = [3; 7,15,1,288,1, ··· ].
Now calculate the convergents by the method of 3.1.4:
And this is all, it is as simple as that. This table exposes Archimedes' secret as well as that of Metius, The table demonstrates:
Can we really say that Archimedes and Metius are at last exposed? That they employed continued fractions, Archimedes using the convergent p1/q1 and Metius the convergent p3/q3? No, we cannot, at least not as far as Archimedes is concerned.
It should be clear to you that the solved problem is one of mathematics, not of history. We have demonstrated how one could have come to an approximation of p by the fraction 22/7, but this does not mean that Archimedes used this approach. In fact, it cannot be ruled out that he used the continued fractions algorithm. This conjecture is supported by two arguments: 1. This is the most natural approach when decimals had not yet been invented. 2. the ancients preferred fractions with unity in the numerator. Only such fractions were in use in Egypt and Babylon, other fractions gaining recognition only very slowly. Nevertheless, these are merely speculative arguments and would be rejected in any court of law. No direct evidence has been found. In order to evaluate p, Archimedes calculated the perimeters of inscribed and circumscribed regular polygons, using the duplication formula. We do not know how Archimedes extracted roots, for he only has presented final results. Regarding this subject, historians have been unable to reach a universally acceptable conclusion .
The advantages of fractions with the denominator 7 can be discovered empirically by comparison with fractions with different denominators. But Metius (or rather, Antoniszoon) could not act this way. It would have been hardly possible to find the complicated fraction 355/113 without a theory. There is virtually no doubt that Antoniszoon resorted to continued fractions. It is perfectly clear why he stopped with the convergent 355/113. In fact, this is the last acceptable fraction. The next one is 102595/32657 and is so cumbersome that it cannot have any practical significance.
6.2 The Solution to the Calendar Problem
6.2.1 The Use of Continued Fractions Consider first of all how we ourselves would solve the problem of alternation of ordinary and leap years, We would represent the duration of the year by a continued fraction
1 year = 365 days 5 hours 48 minutes 46 seconds=[365; 4, 7, 1, 3, 5, 64) days.
Note 1 The number p is irrational and represented by a non-terminating continued fraction. The length of the year is an empirical quantity. All empirical quantities involve errors, so it would be meaningless to consider them to be rational or irrational. The length of the year, given above, is the adopted value and must be treated as exact. It is given by a terminating continued fraction.
Note 2. You do not have to express the length of the year by a decimal in fractions of one day (analogous to what was done in the case of p) if you wish to represent this duration by a continued fraction. The calculations are carried out as follows (dropping the integral component):
Find several convergents of the continued fraction representing the length of the year. You can omit the integral part, because you must remember that each year contains 365 full days:
Each cnlumn yields a solution to the calendar problem. For example, the first column allots to the year an approximate length of 365¼ days. This duration is achieved by makings one year in four a leap year. In general, the third row gives the length of the cycle (or period) and the second row the number of leap years per cycle. For example, the secong column yields the solution: Seven leap years in a 29-years cycle. This corresponds to the average duration of the year of 3657/29 days. This is a more accurate pattern than 365¼, but it is more complicated.
6.2.2 How to Choose a Calendar It is now clear that we are offered only four alternatives. In order to avoid a misunderstanding, we have to realize that there exist a very large number of calendars in the world. There exist solar and lunar calendars. Different people use different starting points for years, different numbers of months per year (12 or 13), different (and tremendously diverse) starting dates for the year and different celebration dates. We will not enbrace here the entire variety of these distinct features and pick up a single aspect, namely, the average length of the year. There are only four, acceptably simple and exact possibilities. They are given by the first four columns of the table above. Combinations arising from the fifth, etc. columns are far too complicated. The possible—and acceptable—options are listed in Table 3.
The minus sign in the Error column indicates that the average duration of the year is greater than the true value.
The first option is the Julian calendar.
The second option is inexpedient, It is as complicated as the third option and much less accurate.
The third option (8 leap years in a 33-years cycle) was proposed in 1079 by the great Persian and Tajik scholar and poet Omar Khayyam.
The fourth option is exceptionally accurate. The error of 1 sec is of no practical significance. This calendar has therefore been proposed, for example, in 1864 by the Russian astronomer Medler for its introduction in Russia at the beginning of the 20th century. It demanded only one correction to the Gregorian calendar: Jumping the leap year every 128 years (i.e. treating this year as an ordinary year). Indeed, the Julian calendar contains 32 leap years during a 128-years cycle.
However, this calendar has not been enacted in Russia nor anywhere else. The likely reasons are that the 128-years period is not rounded-off and that people are strongly accustomed to the existing calendar.
6.2.3 The Secret of Pope Gregory XIII The last section did not solve the mystery of Pope Gregory XIII, the Gregorian calendar is not included among the four options in the table above. For this reason, having solved the mathematical problem, we shall devote some time to the historical problem. What were the arguments behind the decision of Pope Gregory XIII (or rather, of the commission he had set up)?
Take a look at a very appealing hypothesis: The Pope preferred the ratio 31:128, but wanted to replace the period of 128 years by something more convenient and chose for this 400 years. If 128 years contain 31 leap years, how many leap years are contained in 400 years? The proportion
31/128 = x/400
yields x = 96.875 » 97. This is precisely the Gregorian calendar: 97 leap years per a 400 years cycle.
Quite conclusive, isn't it? Unfortunately, it, is wrong!
When reasoning a historical case, including an event in the history of science, you must avoid ascribing our modern way of reasoning to the scholars of yore. Quite the opposite must be done; you ought to try and penetrate their world of ideas and knowledge. Furthermore, speculative arguments of the type it could quite likely have been like this are not popular with historians. You have to find documents and ascertain that it was so and not differently. A great deal is known about the Gregorian calendar reform, including the names of the scholars who were in the commission given the job of planning the reform.
The flaw in the above speculative reasoning is this: The duration of the year was not known at the time of Pope Gregory XIII as accurately as it is known these days. The Pope's commission used the astronomical tables compiled by the Academy of Toledo at the order of King Alphonse X (the Wise) of Castile (1221-1284). The length of the year in these tables was
1 year = 365 days 5 hours 49 minutes 16 seconds.
Converted into a continued fraction, it gives
1 year = [365: 4, 8, 7, 2, 2, 17].
Its convergents (with the integral part omitted) are
1/4, 8/33, 57/ 235.
Pope Gregory's commission could not be aware of the ratio 31/128, whatever method it chose.
As has already been mentioned, the average duration of the year in the Gregorian calendar is 365 days 5 hours 49 minutes 12 seconds and is thus 27 seconds longer than the modern duration. However, this is the modern attitude, while the Pope believed that his year was 4 seconds shorter than the duration believed in modern times. Hence it turns out that the Pope's commission could have been very much satisfied with the accuracy it had achieved.
Note that nothing indicates that the papal commission employed continued fractions; they were unknown in Europe at that time. Rather, the commission reached its decision by a method of trial and error. This could have occurred quite easily. According to the Tables of Alphonse X, the Julian year was longer than the true year by 10m44s. How many years would it take to accumulate an error of one full day? Divide 24 hours by 10m44s:
24h/10m44s = 86·400/644 » 134.
Hence a leap year must be overlooked once in every 134 years. But this would not be suitable, because the coming 134th year may not be a leap year. But 134»400/3, whence you overlook 3 leap years during 400 years. This yields the Gregorian calendar!