CUBIC EQUATIONS BY HELP OF RECURRING CHAIN-FRACTIONS. 323 



The elimination now gives 



S3{ r s^3 _ qrd 2 e + ( g2 _ 2pr)d 2 f+prde 2 - (pq + 3r)dcf+ (p 2 + 2q)df 2 + re 3 - qe 2 f+pef 2 +/ 3 } 



- S 2 {3r 3 Dd 2 - qr(2T>de + Yd 2 ) + (q 2 - 2pr)(2Vdf+ Yd 2 ) +pr(J)e 2 + 2Yde) - (pq + 3r) 



(Def+ Ydf+ Yde) + (p 2 + 2q)(Df 2 + 2Ydf ) + 3rYe 2 - q(2Yef+ Ye 2 ) +p(Yf 2 + 2Yef) 

 + 3F/ 2 } 



+ S!{3r3D 2 d- qr(D 2 e + 2BYd) + {q 2 - 2pr)(D 2 f+ 2DYd) +pr(2BEe + Y 2 d) - (pq + 3r) 

 (DE/+ DFe + YYd) + [p 2 + 2 2 )(2DF/+ Y 2 d) + 3rY 2 c - q(E 2 f+ 2YYe) +p 

 (2EF/+F 2 e) + 3F 2 /} 



-S°{r 3 D 3 -jrD 2 E + (q 2 - 2pr)D 2 Y+prJ)Y 2 -(pq + 3r)DEF + (^ 2 + 2 ? )DF 2 + rE 3 - 2 E 2 F 

 + pEF 2 + F 3 } = (4) 



It may be interesting to apply this method to some problems in geometry ; 

 and we may take the construction of the heptagon as a first example. 



The ratio of the diagonal to the side of a regular pentagon is given by the 

 well-known series 0, 1, 1 , 1, 2, 3, 5, 8, 13, 21, &c, in which each term 

 is the sum of the two preceding, this being a progression of the second order 

 having the multipliers q = 1 , p — 1 . 



The relation between the long diagonal, a , and the side, b , of a heptagon 

 is easily shown to be 



a*-2a 2 b-ab 2 + V = 0, 



which gives at once the multipliers r — 1, q—+l, p — 2, whence the pro- 

 gression 



, , 1 , 2 , 5 , 11 , 25 , 56 , 126 , 283 , 636 , &c. , 



each new term being the double of the last found together with the difference 

 between the preceding terms. The convergence here is slow ; to make it more 

 rapid we may write a = 2b + c , and so get the equation 



-¥ + 3b 2 c + 4bc 2 + c 3 = 0. 

 This gives the multipliers r = l , y = 4, p — 3 ; whence the progression 



1 2 7 29 117 474 & 

 0' 0' 1' 3' 13' 52 ' 211' °" 



