Elements of Algebra

CHAP. XIII. OF ALGEBRA. 123 CHAP. XIII. *Of the Resolution of Negative Powers.* 370. We have already shewn, that $\frac{1}{a}$ may be expressed by $a^{-1}$; we may therefore express $\frac{1}{a+b}$ also by $(a+b)^{-1}$; so that the fraction $\frac{1}{a+b}$ may be considered as a power of $a+b$, namely, that power whose exponent is $-1$; from which it follows, that the series already found as the value of $(a+b)^n$ extends also to this case. 371. Since, therefore $\frac{1}{a+b}$ is the same as $(a+b)^{-1}$, let us suppose, in the general formula, [Art. 361.] $n = -1$; and we shall first have, for the coefficients, $\frac{n}{1} = -1$; $\frac{n-1}{2} = -1$; $\frac{n-2}{3} = -1$; $\frac{n-3}{4} = -1$, \ldots And, for the powers of $a$, we have $a^n = a^{-1} = \frac{1}{a}$; $a^{n-1} = a^{-2} = \frac{1}{a^2}$; $a^{n-2} = a^{-3} = \frac{1}{a^3}$; $a^{n-3} = a^{-4} = \frac{1}{a^4}$, \ldots: so that $(a+b)^{-1} = \frac{1}{a+b} = \frac{1}{a} - \frac{b}{a^2} + \frac{b^2}{a^3} - \frac{b^3}{a^4} + \frac{b^4}{a^5} - \frac{b^5}{a^6} + \ldots$ which is the same series that we found before by division. 372. Farther, $\frac{1}{(a+b)^2}$ being the same with $(a+b)^{-2}$, let any degree whatever by approximation; where he demonstrates this general formula, $\sqrt[m]{a^m \pm b} = \frac{m-2}{m-1}a + \sqrt{\frac{a^2}{(m-1)^2} \pm \frac{2b}{(m^2-m)a^{m-1}}}$. Those who have not an opportunity of consulting the Philosophical Transactions, will find the formation and the use of this formula explained in the new edition of Leçons Elementaires de Mathematiques by M. D'Abbé de la Caille, published by M. L'Abbé Marie. F. T. See also Dr. Hutton's Math. Dictionary.