Dr. Rankine on Rational Approximations to the Circle. 425 



will be equal to each other, and to the diameter of the circle 

 X sin 6 ; and n + 1 of the half-sides of the circumscribed polygon 



will be equal to each other, and to the diameter x • „ » . ; the 



remaining side of the inscribed polygon and the remaining pair 

 of half-sides of the circumscribed polygon will have for their 

 arithmetical ratios to the diameter respectively sin {n-\-\)6 and 



—fz — , but will be negative ; which means that as the 



A 



first side and the (n + l)th side of the inscribed polygon, and the 



corresponding parts of the circumscribed polygon, will overlap 



each other, the closing side of the one and pair of half-sides of 



the other must run backwards, and must be subtracted in the 



computation of perimeters. 



Thus the ratios of the perimeters to the diameter will now be 



inscribed polygon , nN . n . . -, N/1 ,„ x 



diaJL = («+l)sm0-sm (» + !)*, . (7) 



circ tt msc^edjoIygon =(B + 1)taue _ tan( _ ?i _ 1) ^ _ (g) 



both of which ratios are rational fractions, and the difference 



between them is less than — of the greater of them. Q. E. F. 



m 



10. Proposition IV. Problem. — In solving the previous 

 problem, to ensure that the denominator of the ratio of the peri- 

 meter of the circumscribed polygon to the diameter shall contain 

 any given factor (f ) . 



In applying Lemma I., make either a— b or a + b a multiple 

 of the given factor ; that is to say, having chosen an arbitrary 

 value of b } make 



a = kf±b, . (9) 



and determine the multiplier k by trial, so as to satisfy the con- 

 dition expressed by equation (2), viz. 



fl 9 =(A/+5) s >(2i»-l)i 9 (10) 



a 2 — b 2 

 Then cos#= 9 , L9 will contain the given factor f. which will 

 a? -f b l D J 



therefore be a factor in the denominator of tan 0, and of the tan- 

 gent of every odd multiple of 0, as may be proved by considering 

 the results of dividing equation (4) by equation (3). Then, in 

 solving the problem of Proposition III., use the fii*st solution if n 

 is odd, and the second solution if n is even ; and the denominator 

 of the ratio of the perimeter of the circumscribed polygon to the 

 diameter will contain the given factor. Q. E. F. 



11. It is easy to see that there are an endless variety of 

 Phil. Mag. S. 4. Vol. 29. No. 198. June 1865. 2 F 



