THE SOLUTION OF EQUATIONS BY 

 OPERATIVE DIVISION.— Part II 



By SIR RONALD ROSS, K.C.B., F.R.S., D.Sc. 



CONTENTS OF PART II 



IV. (i) Descending Division and Examples A to D— (2) The General 

 Descending Invert— (3) Examples and Corollaries E to N — 

 (4) Formal Algebra. 

 V. (1) The Methods of D ary and Newton— (2) The General Theorem 

 of Solution by Iteration — (3) Solution by Operative Division 

 gives the Algebraic Expression of Solution by Iteration — 

 (4) Proof by Taylor's Theorem— (5) The Curve [0 +/]". 



VI. (1) We have now studied operative division when the divisor 

 is of the form + b, or + bo 2 + co 3 + . . . , that is when the 

 divisor commences with the first power of o ; and in these cases 

 the powers of in the quotient are integral if they are integral 

 in the first dividend. But how shall we carry out the division 

 if the divisor commences with other powers of 0, as for example 

 when we wish to divide by o 3 — 20 or by o 2 -f bo 3 + co 4 ? 

 We proceed in exactly the same way except that instead of 

 expanding the divisor for integral powers in order to obtain 

 each successive subtrahend, we may now have to expand it by 

 negative or fractional powers. If the dividend (either the first 

 or any successive dividend) commences with o'' and the divisor 

 commences with o n , then the corresponding power of in the 

 quotient must be such that when it operates on 0" it converts 



it into o r — that is, it must be o~, whatever numbers positive, 

 negative, integral, or fractional n and r may be. Then, in 



r 



order to obtain the corresponding dividend, o» must operate 

 on the whole divisor ; that is, the dividend will be the ex- 

 pansion of the (-)th power of the divisor by the binomial or the 



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