TO THE COMPARISON OF SEVERAL MAGNITUDES. 61 



if we continue this operation, using always the same multiplier, or a circulating 

 set of multipliers, the fractions so resulting converge to the root of a quadratic 

 equation. If we should assume three fractions, and combine fixed multiples of 

 their members, so as to form a progression of the third order, as we may call it, 

 to what value do the terms of this progression converge ? 



In a paper read by me some time ago to the Society, it was shown that the 

 convergence in this case is toward the root of a cubic equation; and that the 

 same arrangement may be extended to the still higher orders ; as examples of 

 the application of this method, two cases may be cited. 



If we begin with the two fractions q > ^ and form a progression by adding 



to the double of each member of the last, the corresponding member of the 

 preceding, we form the well-known progression 



1 1 3 7 17 41 99 o 

 ' 1 ' 2 ' 5 ' f2 ' 29 ' 70 ' ' 



which converges toward the ratio of the diagonal to the side of a square. 



If, beginning with the same pair, we form a progression by taking the sums 

 of the members of the last and of the penult, we obtain 



1 1 2 3 5 8 13 21 34 o 

 6' I* 1' 2' 3' 5' "8 ' 13' 21' ' 



which converges toward the ratio of the diagonal of a regular pentagon to its 

 side. In this case, the numerator of one fraction becomes the denominator of 

 the succeeding, so that it is unnecessary to write both progressions. These 

 are familiar examples of quadratic roots. 



Let us now assume three terms, 0, 0, 1, and continue a progression by 

 adding to the double of the last term the difference between the two previous 

 ones, thus — 



0, 0, 1, 2, 5, 11, 25, 56, 126, 283, 636, 1429, 3211, 7215, 16212, 36428, 



81853, 183922, &c, 



and we obtain an approximation to the ratio of the long diagonal to the side of 

 a regular heptagon. Thus, if the side of the heptagon be 283, its longest 

 diagonal is almost exactly 636. 



Or again if, assuming the same three terms 0, 0, 1, we form a series by 

 deducting the antepenult from the triple of the last term, thus — 



0, 0, 1, 3, 9, 26, 75, 216, 622, 1791, 5157, 14849, 42756, 123111, 354484, &c, 



we obtain an approximation to the ratio of the long diagonal of an enneagon to 

 its side. 



I have shown that, in progressions of this kind, that is, where the numerator 



