GREGORIAN CALENDAR. Abstract

Transcription

1 Bulletin of the Marathwada Mathematical Society Vol. 12, No. 1, June 2011, Pages GREGORIAN CALENDAR S.R.Joshi 8, Karmayog, Tarak Colony, Opp. to Ram-krishna Ashram, Beed bye pass Road, Aurangabad , M. S., India. pranav.coatings@gmail.com and U.S. Muley, Dept.of Mathematics, Govt. College of Arts and Science, Aurangabad, , M. S., India. Abstract After giving a brief history of the Gregorian calendar, a few formulae for finding the day (Sunday to Saturday) on a given date are derived, which can be used in obtaining the calendar of any given year. It has been shown that these formulae remain valid up to the end of 32nd century approximately, in the present context of the rules about deciding a leap year. A modified formula for finding the day, which will also work after is obtained by adding one more rule in the existing rules for deciding a leap year. 1 INTRODUCTION The dictionary meaning of the word calendar is the table of dates, months, and days in a given year. Different calendars are followed in different countries depending on the culture, tradition and astronomical practices of the countries [7, 8]. Even in one country different date systems are followed in different parts of the country. For example in India 30 different date systems are used in 30 different parts [8]. The current national calendar of India was set up in 1957 and adopted on March 22, 1957 by the Calendar Reform committee that formalized a lunar-solar calendar in which leap years coincide with the Gregorian calendar, which is an internationally accepted civil calendar [2]. Unfortunately this national calendar is not followed in many parts of India. 99 percent people of India don t know even that there is a national calendar of their country. In this article, we give a short history of the Gregorian calendar and explain 65

2 66 S.R.Joshi and U.S.Muley how it is prepared for a given year using the formulae to be derived. Different formulae are available in deciding the day for a given date. All these formulae mainly depend on the following five things. (i) The definition of a year (by a year we mean the tropical year or the solar year) (ii) Every year is divided into 12 months, January to December having fixed number of days. (iii) The seven days of a week are repeated periodically with period 7. (iv) The concept of the leap year, and (v) The day on January 1, After deriving a few formulae for finding the day for a given date, we shall show that these formulae remain valid up to December 31, 3199 approximately. We shall also obtain a modified formula for finding the day of a given date for which one has to add one more rule in the existing rules for deciding a leap year. 2 HISTORY OF THE GREGORIAN CALENDAR [1,2,6,8] The calendar that is used now a days in almost all countries over the world is the Gregorian calendar and it is an internationally accepted civil calendar [2]. It was introduced by Pope Gregory XIII, after whom the calendar was named. The Gregorian calendar year differs from the solar year by 26 seconds approximately. By a solar year we mean the number of days taken by the Earth to go round the Sun in its elliptical orbit. This solar year is then qualified with adjectives like tropical, sidereal, eclipse, etc. depending on the reference point for recknowing a revolution. For example the tropical solar year is the time interval between two successive passages of the sun, as observed from the earth, through the vernal equinox point on the ecliptic. For the purpose of Gregorian calendar the length of a tropical solar year is used. The length of a tropical solar year is given by [4, 7]; i.e. yr = 365 days, 5 hours, 48 minutes, seconds (approximately) yr = (approximately). (2.1) For the Hindu Sauramana calendar on which the Makar Sankranti,Tamil new year s day etc are based, the sidereal solar year is used. Its length is days. There is some debate among astronomers and other parties as to what actually constitutes the solar year, and these figures can vary by several seconds.[7] Before 25th March,1752, the calendar that was followed is known as Julian calendar. There were 12 months in a year, January to December. There were 7 days in a week (Sunday to Saturday) and they were used to occur periodically as is the case today with one exception. The exception was that a year was considered as

3 Gregorian Calendar 67 a leap year whenever the year number is divisible by 4 even if it is divisible by any multiple of 100. By a leap year we mean that year in which an extra day is added to the calendar in order to synchronize it with the seasons. Secondly every year was used to begin on 25th March instead of 1st January. For example after March 24, 1673, the next day would be March 25, 1674, advancing the year only. Due to this, a somewhat confusing phenomena occurred, and for the present day imaginations, comprehension requires a close attention. Gregorian calendar is a compilation of usages dating from the days of Julius Caesar, when the Greek astronomer and mathematician Sosigenes devised for Rome, a calendar, [2] only essentially familiar to us as our present day calendar. It was during the late 16th century, that the work of the 6th century Anglosaxon monk Bede, was submitted to Pope Gregory XIII, who accepted the calculations and made the decision to issue a more accurate calendar which ultimately was accepted. The Gregorian calendar was adopted by most Roman catholic countries. Many protestant countries did not accept the new calendar until 18th century. Britain and America accepted the Gregorian calendar for themselves and all their possessions were effective from March 25,1772. But a remarkable 2-step change was made which is little realized today. Before following the Gregorian calendar, the British made the following two decisions in the form of a decree.[2] (D2.1) It was decreed that year 1752 should end with December 31st and not be carried on to the next March 25. (D2.2) It was decreed that the arrival of September 2, 1752, should be called September 14, For the sake of clarity, we explain; the period of January 1,... March 24, 1751 was the end of an epoch. The year 1752 began on March 25th and ended with December 31, The earlier days of 1752 never existed (83 days). Also the deleted days of September, 2-13 (11 days) of 1752 never existed. Thus the year 1752 was a very short year; 72 days shorter. Due to omission of 11 days the vernal equinox was resorted to the 21st March.[8] Because of the statements (D2) cited above, the day September 2, 1752 existed (Wednesday) and this was followed by Thursday September 14, 1752 (the order of the days was not disturbed). The 11 days from September 3 to September 13,1752 did not exist. Because of this, protestors give the slogan Give us back our 11 days even today. From astronomical and mathematical point of view this slogan given by protestors due to blind belief has no meaning. Of course, dates may be recomputed from the old to the new calendars. For instance, George Washington s birth day was February 11, 1731, as far as his mother was concerned. Today we recon his birthday as February 21, 1732 according to Gregorian calendar. In order to convert from old Julian calendar to the new Gregorian calendar, one must add 10 to 13 days to the old date, and sometimes change the year one extra when the date considered falls within the period January 1-March

4 68 S.R.Joshi and U.S.Muley 24. This is very important for those interested in genealogy and historical research. Documented dates before March 25,1752 do not necessarily always coordinate with a stated period of time. Further, references to any new years day before 1752, in Great Britain meant March 25th. 3 FORMULAE FOR OBTAINING THE DAY OF A GIVEN DATE The problem of obtaining the Gregorian calendar for a given year depends on the problem of knowing the day of a given date. For this purpose let d-m-y be a given date where d, m, y are all positive integers satisfying 1 d 31 and 1 m 12,. We shall assume that y is a 4-digit number abce,where a, b, c are nonnegative integers and e > 0. Let fr be the fractional part of the value of the year yr given by the equation (2.1) i.e. fr = If k is a positive integer and x,y are any two integers we say that x is congruent to y modulo k if the difference (x - y) is divisible by k and express this by writing x y (mod k). Throughout this paper we assume k = 7. Hence for brevity by x y we mean x y (mod 7). It is clear that if r is the remainder when an integer n is divided by 7 then n r (mod 7). Because of this if a certain day occurs on a given date say dt1 and if a date dt2 occurs after or before n days then the same day occurs on dt2, whenever n is a multiple of 7. For simplicity we shall denote the date 1st January, 0001 by the symbol D 0. Now the problem of finding the day on date d-m-y depends on the value of yr given by (2.1), the day on D 0, and the number of days passed before d-m-y and up to D 0. Note that the value of the yr is not an integer. As a first approximation we take the value of the the yr as i.e. 365 days and a quarter of a day. In practice we take the value of an year as a whole number 365, and after a period of four years we take the value of a year as 366 days calling that year as a leap year, taking 29 days in February instead of 28. Note that the value is a little more than the actual value of yr by an amount of df = yr = (in days). This difference may cause difficulties in counting the actual days passed after D 0. If we multiply the difference df by 400 we get This means we have taken 3 days (again approximately) more in counting the leap years within a period of 400 years. If we neglect this difference for ever, after a long period, seasons in a year will not take place in specified months, as we observe and experience today. To overcome this difficulty a provision is made in the Gregorian calendar in deciding a leap year. The present rules about leap years are as follows. (L1) If y = abce is a year and the number ce (10c + e) is divisible by 4 and ce 0, then the year y is supposed to be a leap year. (L2) If ce = 0 and ab is divisible by 4 (i.e. y is divisible by 400) then also the year is supposed to be a leap year. For example the years 344, 2368, 1200, 2000, 2800 are all leap years. But

5 Gregorian Calendar 69 the years 345, 1946, 1800, 2100, are not leap years according to the rules (L1) and (L2). In order to find the day on D 0 we shall assume that the Gregorian calendar existed in the past years up to D 0. This assumption and the rules (L1) and (L2) will help us to find the number of days passed from D 0 to d-m-y. For this we shall consider that date on which the day is known. For example we know that the day on was Monday. Further it can be verified that the total number of days passed from D 0 to (including both the days) is This number is exactly divisible by 7 and hence using the congruency relation modulo 7, we conclude that the day on D 0 must be Monday. Hence the problem of finding the day on D 0 is solved. Let T be the total number of days passed from the date d-m-y to D 0. Let r be the remainder when T is divided by 7. Clearly there are 7 values of r, 0 to 6. From these values of r and because the day on D 0 is Monday, the day on d-m-y is decided from the following table. T able 3.1 Remainder r Day Sun Mon Tue Wed Thu Fri Sat Note that if T is replaced by T - 7n, where n is any integer, then the same remainders are obtained when T - 7n is divided by 7 as are obtained for T. In practice it is not convenient to find the value of T and then decide the day on d-m-y, sice it takes time for calculation. There are shortcuts to decide the day. These are discussed below in the form of formulae. Formula 3.1 :Every non-leap year contains 365 days and Hence if a particular day occurs on 1st January of such a year, then the next day occurs on 1st January of the next year. For example the day on was Friday and the day on was Saturday. Hence as far as the calculation of a day is concerned, advancing a non-leap year is equivalent to advancing a day, because (mod 7). (3.1) Similarly advancing a leap year is equivalent to advancing two days, since (mod 7). (3.2) Now for the date d-m-y, the number of complete years passed from this date to D 0 is (y-1). Among these years some years are leap, obtained by the rules (L1) and (L2) mentioned earlier. If l is the number of leap years passed, then it can be observed that l = [y/4] [3(ab + 1)/4]. (3.3) where by [x] we mean the integer part of the real number x. Note that in a period of 400 years there are [400/4] - 3 i.e. 97 leap years and not 400, because of the rules

6 70 S.R.Joshi and U.S.Muley (L1) and (L2). Hence if there are d 1 number of days passed before 1st January,y, to D 0, then by using (3.1), (3.2) and (3.3) we get d 1 (y 1) + [y/4] [3(ab + 1)/4]. (3.4) Let d 2 be the number of days from 1st January, y, to d-m-y and d 3 be the number of days from 1st January, y to the end of (m-1)th month of the year y. Then clearly d 2 = d + d 3. (3.5) Note that d 3 is 0 if the month number m is 1 (for January). There are 12 values of d 3 and these are fixed since the days of months are fixed. These 12 numbers are 0,31,59 (31+28),90 ( ), 120, 151, 182, 212, 243, 273, 304, and 334. It is sufficient to take the remainders of these numbers after dividing by 7, since days are repeated periodically. It is observed that the remainders are 0,3,3,6,1,4,2,5,0,3,and 5 respectively. We shall denote these remainders by the symbol cm and call them as month codes. These are given in the following table. T able 3.2 Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec cm Note that we have supposed that there are 28 days in February whether the year y is leap or not leap. Further d 3 is congruent to one of these 12 month codes depending on the value of m i.e. d 3 cm (3.6) Now T d 1 + d 2 d 1 + d + d 3. Hence using (3.4) and (3.6), we get T d + cm + (y 1) + [y/4] [3(ab + 1)/4] (3.7) The formula (3.7) for finding the day on d-m-y is valid for all dates except when y is a leap year and when the month is January (m = 1) or February (m = 2). This is so because the 12 month codes given in Table 3.1 are obtained by taking 28 days of February irrespective of the nature of the year. For this exceptional case the number 1 is to be subtracted from the right hand side of the equation (3.7) and calculations are to be carried out. The formula (3.7) can be verified for any date from D 0 to 31st December We shall explain the reason for this after discussing two more formulae for finding the day on a given date. Formula 3.2 :The formula (3.7) can be modified further so that it will take less time in finding the day on a given date d-m-y. According to the rules (L1) and (L2) about the leap years there are 97 leap years in a period of 400 years. Also

7 Gregorian Calendar 71 an advancement of one non-leap year is equivalent to advancing a day. Hence in a period of 400 years the total days advanced is equal to 497. This number is exactly divisible by 7. Hence after every gap of 400 years the days are repeated in the same order. For example if the day on 15th August, 1547 is Friday, then, the day on 15th August 1947 or 2347 is also Friday. This fact can be verified for any two dates differing by 400 years with the help of the formula (3.7). Note that we have assumed that the Gregorian calendar existed in the past years also up to D 0. Hence for a given date d-m-y if the corresponding year number > 400, then it can be reduced by subtracting a suitable multiple of 400 from the number y. So there is no loss of generality if we assume that y < 400. Let y = abce, with a = 0. With this assumption we have ab = b, y = 100b + ce, [y/4] = 25b + [ce/4] and [3(ab + 1)/4] = [3(b+1)/4]. Note that there are only 4 possible values of b viz 0,1,2 and 3, which decide the dates in the first, second, third and fourth century respectively. With these changes the formula (3.7) can be written in the form, where T x 1 + x 2, (3.8) x 1 = d + cm + ce + [ce/4], and x 2 = 125b 1 [3(b + 1)/4]. (3.9) Since x 2 depends on b and b has only 4 values (0,1,2,3), x 2 has 4 values viz - 1, 123, 247 and 371 respectively. These four numbers are congruent to 6,4,2 and 0 respectively modulo 7. We shall call these 4 numbers as century codes and we denote them by a common variable say c 0. These four code numbers are corresponding to the 4 values of b (0,1,2,3) which are also the remainders when the century part ab of y is divided by 4 even if a is nonzero. The four values of c 0 are given in the following table. T able3 Remainder Code c Note that x 2 c 0. Hence using (3.8) and (3.9), we get from (3.7) T d + cm + ce + [ce/4] + c 0 (3.10) This is the second formula for deciding the day on a given date. This formula is much more simpler than the Formula 3.1 from calculation point of view. It does not require any kind of calculator. Only thing is that one has to remember the 12 month codes (values of cm) given in Table 3.2 and the century codes c 0 given in Table 3.2 Further as in case of Formula 1, here also, one has to subtract 1 from the right hand side of (3.10) when one wants to find the day on a date which falls in January or February and when the year y is a leap year. Readers may verify the validity of the formula by considering any date for which the corresponding day is known.

8 72 S.R.Joshi and U.S.Muley Formula 3.3: In [7] on page 264 the following formula is given to find the day on a given date. T d + [2.6M 0.2] + ce + [ce/4] + [ab/4] 2ab (1 + L)[M/11](mod 7), (3.11) where M is the month number beginning with March and not with January and the values of M (beginning with March and ending with February) are 1,2,3,...,12, and L is 1 or 0 according as the year y = abce is leap or not. In [7] it is also said that this formula does not work before the year 1582, in the sense that the days obtained by this formula and the those obtained by using old calendars do not always coincide. Though the derivation of this formula is not given in [7](authors of this paper don t know the derivation part), it can be verified that the formula gives the same results as obtained by Formulae 3.1 and MODIFIED FORMULA As remarked at the end of Formula 1, the Formulae 2 and 3 also do not remain valid after 32th century. This is due to the assumption that there are only 97 leap years in a period of 400 years. The difference of 3 years (400/4-97) is obtained by considering the product 400( yr) = 3.124, where yr is the value of a year given by the relation (2.1). This product is greater than 3 by the amount If we consider a period of 3200 years( more approximate period is 3300 years ) then according to our assumptions there are 97 8 i.e. 776 leap years. But if we multiply the difference ( yr) by 3200 we get which is approximately equal to 25. This means there are 3200/4-25 i.e. 775 leap years instead of 776. Thus according to the value of yr and the present rules (L1) and (L2) about the leap years we are taking one extra leap year in a period of 3200 years. This further implies that there will be 100 extra leap years in a period of years. This will create difficulties in experiencing seasons, in the sense that seasons will not occur in specified months as we observe today. Of course this situation will start occurring after slowly as centuries will pass. To overcome this difficulty the rulers of different nations all over the world will have to take help from mathematicians and astronomers to add the following rule about leap years. (L3) If y is a multiple of 3200, then y is supposed to be a non-leap year, though y is divisible by 400. The problem of deciding the rules about leap years can also be handled by considering the convergents of the rational number yr expressed as a continued fraction [7]. On pages the authors of [5] have considered the value of yr as yr = 365 days seconds (this value differs from the value given by (2.1) by just 0.784

9 Gregorian Calendar 73 seconds ) i.e. yr = This yr can also be expressed as a continued fraction. It is yr = (365; 4, 7, 1, 3, 5, 64). The consecutive convergents are given by p 0 q 0 = 365, p 1 q 1 = (365, 4) = 365 days 6 hours, p 2 q 2 = (365; 4,7) = 365 days 5 hours 47 minutes and 35 seconds, p 3 q 3 = [365; 4, 7, 1] = 365 days 5 hours 49 minutes and 45 seconds etc.it can be observed that p 0 are the lower approximates to the q 0, p 2 q 2, p 4 q 4 actual values of the yr and the errors are 5 hours 48 min 46 sec, 1 min 11 sec and 1 sec respectively. The errors go on decreasing and if yr = p 4 q 4 approximately then, the error is just 1 second. Further the convergents p 1 q 1, p 3 q 3 are upper approximates to yr. The errors are 11 min 14 sec and 19 sec respectively. Here also the errors p are decreasing. For practical purposes one year = 365 days i.e. 0 q 0. Clearly this is much lower estimate than the exact value of yr. If we take yr = p 1 q 1 then we have to account for 6 hours time each year. In four years the error accounted is 24 hours. Hence to compensate this error it is a practice to add one day to every 4th year and call it a leap year. If yr = p 2 q 2 approximately, we again commit an upward mistake of 11 min. 14 sec. each year. In a century the accumulated mistake is 18 hours 43 min 20 sec. To compensate this error we take every 4th century as a leap year. The story does not end here since there is a mistake of 2 hour 53 min. 20 sec. in every 400 years. To compensate this mistake every 32nd century is to be exempted from leap year. For more details about convergents related to leap years readers may refer [5]. It is clear that the rule (L3) is applicable only when y The rule (L3) along with (L1) and (L2) can be used to obtain the following modified formula for deciding the day on a given date and which will hold up to approximately. Let d-m-y be a given date, where y = pqabce is a six digit number. Let c 1 = [pqab/32]. Then the number c 1 indicates the number of leap years to be subtracted according the rule (L3). Hence the equation (3.11) is changed as T d + cm + ce + [ce/4] + c 0 c 1 (4.1) This is an extended formula which can be used to find the day on any date d-m-y up to and the Formula 2 is a particular case in the sense that it coincides with Formula 3 whenever y Lastly we give below an algorithm depending on the modified formula for deciding the day on a given date d-m-y. Algorithm (1) Input the date in the form d-m-y,where y = pqabce, so that d,m,y are all positive integers. (2) If y then goto (14) (3) If d 0 or d [d] or m 0 or m [m] or y 0 or y [y] then print date inserted is invalid. Go to (1). (4) If m = 1,2,3,...12, then let cm = 0,3,3,6,1,4,6,2,5,0,3,5 respectively. (5) y 1 = 100 [y/100 [y/100]] and y 2 = [y 1 /4].

10 74 S.R.Joshi and U.S.Muley (6) r = pqab - 4 [pqab/4] (7) c 1 = [pqab/32]. (8) If r = 0, 1, 2 or 3 then c 0 = 6, 4, 2 or 0. (9) T = d + cm + y 1 + y 2 + c 0 c 1 (10) days = T - 7 [T/7]. (11) If y is a leap year and m = 1 or 2, then days = days -1. (12) If days = 0,1,2,...,6, then DAY = Sunday, Monday,...,Saturday respectively. (13) Print the day on d-m-y is DAY. (14) Print the day on d-m-y can not be obtained since y (15) End. 5 CONCLUDING REMARKS We have obtained a formula for finding the day on a given date which holds true up to Note that the number we have considered in the derivation of Modified Formula is chosen for practical purpose in order to handle the Rule (L3) as it has been done for handling the rules (L1) and (L2). An integer near to but greater than can also be considered. But the rule (L3) will become unpracticable to handle. Further one can not obtain a universal formula for finding the day on any given date. This is so because the value of yr given by (2.2) is not exact. In fact yr is an irrational number since the length of an ellipse depends on π, an irrational number. Hence there is no end to the extension of any modified formula. ACKNOWLEDGEMENT The authors are very much thankful to Dr. S. Balchandra Rao, Hon. Director, Bharatiya vidya Bhavan s Gandhi Centre of Science and Human Values, Bangalore, for providing useful information in the preparation of this article. References [1] The New Caxton Encyclopedia, International Learning Systems Corp.(Athens,NewYork),(1969) [2] Lawrence A. Crowl, The British Switch to the Gregorian Calendar, crowl@cs.orst.edu,(october 4,1955) [3] Subhamoy Das, The Hindu Calendar System,About.comGuide, Google Search, [4] Sudhir Date, Anantkalin Dindarshika (Marathi), Sakal-Daily, Maharatra, India, ( ).

11 Gregorian Calendar 75 [5] S.G. Deo, D.Y.Kasture, H.V. Kumbhojkar and V.G.Tikeker, Popular Lectures on Mathematics, Universities Press(India), Pvt. Ltd.(2009). [6] Google Search, hinduism.about.com/od/history/a/calendar.htm, [7] Ivan Niven and Herbert S. Zuckerman, Wiley Eastern Limited,(1976). [8] Robert Wilde, Introduction of the Gregorian Calendar, About.comGuide, Google Search,