8th 6x9+5 59X7+6 419X2+59 897 X 4+419 897 2843 4007 = 2843X4+1328 12700 = 12700X6+2843 79043 Original fraction, 24939X8+4007_203519 As a third and practical illustration, let it be required to express, in terms of a day, the difference between the solar year (365 d. 5 h. 48 min. 49 sec.) and 365 days. The number of seconds in the difference is 20,929, and there are 86,400 seconds in 24 hours. In terms of a day, therefore, the difference is expressed by the fraction 20929. Dealing with this fraction in 86400 the manner already explained, we obtain the following results: Quotients, 4 7 I 3 I 16 I Approximations, I 15 I 7 8 31 39 655 694 1349 20929 4 29 33 128 161 2704 2865 5569 86400 So that the difference between the solar year and 365 days amounts to nearly 1 day in 4 years, to a little more than 7 days in 29 years, to somewhat less than 8 days in 33 years,* to very little more than 39 days in 161 years, &c. Continued fractions suggest a means---different from that already explained-for the extraction of square roots. Let it be required, for example, to extract 17. This root lies between 4 and 5. Regarding 4 as the "known" part of the root, and putting u for the "unknown" part, we have 4+u=√ (4+u)2 = 17; 42+2×4×u+u2=17; 16+8 x u + u2. 8xu+u2= (17-16=)1; (8+u) xu=1;u= I I I 17; = 17; 8+u 8+8+2 I 8+ *It is an interesting fact that the addition of 8 days to every 33 years (of 365 days each) was proposed by the Persian astronomers nearly seven centuries ago. † Each of these fractions is formed from the preceding one by the substitution of I for u. Here we have an interminate series of approximations—the continued fraction being interminate. Without, however, proceeding farther than the third approximation, we obtain a sufficiently accurate result: u= ='123; √17=4+u=4123. 65 528 = Next, let it be required to extract 41. This root lies between 6 and 7. Regarding 6 as the known part of the root, and putting x for the unknown part, we have 6+x=√ŢI ; (6+x)2=41; 62+2×6×x+x2=41; 36+12×x+x2=41 ; 5 12 X x+x2=(41-36=) 5; (12+x)×x=5;x= = 12+x thus find x= 745 and converting this expression into an ordinary fraction, we = 5 149 12+ 12 Continued fractions enable us to express any number as a power of any other number. Let us suppose the fact to have Each of these fractions is formed from the preceding one by the 5 substitution of for x. 12+x been, in some way, ascertained that 409 is the 292nd power of the 87th root of 6: 292 409=687. Putting a for the 87th root of 6, we have 409=a292 Dividing 409 by 6 as often as possible, It thus appears that, as often as 6 can be divided into 409, so often, exactly, can 87 be subtracted from 292. Substituting 67 for a, we have It thus appears that 1·8935185 can be divided into 6 as often -and only as often-as 31 can be subtracted from 87. Substituting 1-8935185 for b, we have 1-6734477 1-8935185## 1-67344771.893518525 Putting c for the 25th root of 1.6734477, It thus appears that 16734477 can be divided once only into 1.8935185, and that 25 can be subtracted once only from 31. Substituting 1'6734477's for c, we have 11315074 16734477's 113150741.6734477° 25 1131507416734477 Putting d for the 6th root of 1.1315074, 1.6734477=d25 It thus appears that 11315074 can be divided into 1.6734477 as often-and only as often-as 6 can be subtracted from 25. Substituting 11315074 for d, we have Now, as often as I can be subtracted from 6, so often, exactly, ought 10208985 to be contained as factor in 1'1315074; and this we find to be the case: *This last quotient would be exactly 1 if 409 were exactly the 292nd power of the 87th root of 6 ; but the index 292 is only an approximation. 87 |