§4. Finding one pair roots of the equation performed through the continued fractions

With the help of the rule of continued fractions we should find one pair root of this inexact equation:

(4.1)

For example, we have an equation like this:

(4.2)

Before solving it let’s convert the fraction into a continued form. And it is clear that:

_(4.3)

Now let’s find the second convergent fraction from the end of the last continued fraction (it is clear that the first convergent fraction from the end is equal to ). It certain that it is of the value . As the found convergent fraction is at the even place, its value is greater than , and the difference is in the value of . Therefore an expression

_(4.4)

is performed. In order to make the equation (4.4) similar to the inexact equation, let’s increase all the members of the last equation by 13 and write it in the following form: or

(4.5)

Now let’s compare the equation (4.5) and the given equation, if we take _{instead of and 39 instead of} , they are appeared to be mutually exact. Consequently, one pair root of the equation is found; it is: . That is why now it is possible to find all the roots of the equation: .

They will be found through these chains:

As it is no conditional value for number , if we make expression as a condition of the upper chain, chains of roots appeared in a convenient form like this:

Example 2: We have equation . Let’s convert into the continued fraction. So it will be as this:

Let’s find a convergent fraction second from the end, it is . As the taken convergent is at the uneven place, its meaning is less than . That is why:

Hereof expression is performed. If we multiple members of the last equation by 5, we can find one pair roots of the equation given through expression . They are:

Consequently, the value of all roots of the given equation is found from these chains:

As the number has a conditional value, if we change for , previous chains will have the following form:

Chapter II

§1. Continued fraction and calendar

As we know 1 year 1 = 365 days 5 hours 48 minutes 46 seconds = 365,242199...days. Days are measured as a circadian period and fractions of the sun years conform to its length. It is possible to make a calendar of 365, 242199…=365;4,7,1,3,5,…days.

The first convergent fraction 36514 conforms to Julian calendar: every four years it is a leap year. Let’s find the difference among the followings,

365,25-365,242199 = 0,007801 days = 0,187224 hours = 11,23344 minutes = 11 minutes 14 seconds, i.e. the average length of the year is greater than that period for 11 minutes 14 seconds. The third convergent fraction 365;4,7,1=365833 is a basis of the Persian calendar. It was suggested in 1079 by mathematician, astronomer and poet Omar Khayyam. Here if we find the difference as 365833-365,242199 = 0,000225242 days = 0,005406 hours = 0,324348 minutes = 19,5 seconds, then our error will be 19 seconds in one year, but if we divide years into a 33-year cycle, the cycle inside the leap year coming after every four years is counted seven times and the eighth is not convenient in the fifth one, i.e. ІІІ-ІІІ-ІІІ-ІІІ-ІІІ-ІІІ-ІІІ-ІІІ-ІІІ-ІІІ-ІІІ (here sticks express years and dashes express leap years).

In 1864 Russian astronomer Johann Heinrich Medler suggested that the fourth convergent fraction 365, 24199…=36531128 gave one more calendar. Meddler suggested this in XXth century in Russia. For this it was necessary to pass 1 leap year every 128 years, because according to Julian calendar there are 32 leap years in 128 years. This calendar was accepted, because number 128 is not round. What is our error here?

36531128-365,242199 = -0,0000115 days = -0,000276 hours = -0,01656 minutes = -0,99 seconds = -1second!

Therefore this calendar has an error of 1 second.

In 1582 Pope Gregory ХІІІ made reformation correcting inaccuracy of Julian calendar. Also the the sequence of common and leap years, if the number is ended with two zeroes and the number is divided into four, then it is a leap year.

The error has become 10 days (years 1700, 1800, 1900) since the Christ’s new year. According to Julian calendar 13 days does not fit. And now let’s count the length of Gregorian calendar. According to Julian calendar if 100 years of 400 years are leap and according to Gregorian calendar they are 97, that is why the average length of Gregorian year is 36597400 days = 365,2425 days = 365 days 5 hours 49 minutes 12 seconds, it is greater 26 seconds than the current period. There are different types of moon-and-sun calendars in states of South-Eastern Asia, Iran and Israel.