Part I. Section II. Chapter 13. “Of the Resolution of Negative Powers.”

370 We have already shown, that $$\frac{1}{a}$$ may be expressed by $$a^{-1}$$; we may therefore express $$\dfrac{1}{a+b}$$ also by $$(a+b)^{-1}$$; so that the fraction $$\dfrac{1}{a+b}$$ 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 $$\dfrac{1}{a+b}$$ 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,

$\frac{n}{1}=-1; \quad \frac{n-1}{2}=-1; \quad \frac{n-2}{3}=-1; \quad \frac{n-3}{4}=-1, \quad \textrm{etc.}$

And, for the powers of $$a$$, we have

$a^n = a^{-1}=\dfrac{1}{a}; \quad a^{n-1} = a^{-2} = \dfrac{1}{a^2}; \quad a^{n-2} = \frac{1}{a^3}; \quad a^{n-3} = \frac{1}{a^4}, \quad \textrm{etc.}:$

so that

$(a+b)^{-1} = \dfrac{1}{a+b} = \dfrac{1}{a} - \dfrac{b}{a^2} + \dfrac{b^2}{a^3} - \dfrac{b^3}{a^4} +\dfrac{b^4}{a^5} - \dfrac{b^5}{a^6}, \; \textrm{etc.}$

which is the same series that we found before by division.

372 Farther, $$\dfrac{1}{(a+b)^2}$$ being the same with $$(a+b)^{-2}$$, let us reduce this quantity also to an infinite series. For this purpose, we must suppose $$n=-2$$, and we shall first have, for the coefficients,

$\frac{n}{1}=-\frac{2}{1}; \quad \frac{n-1}{2}=-\frac{3}{2}; \quad \frac{n-2}{3}=-\frac{4}{3}; \quad \frac{n-3}{4} = -\frac{5}{4}; \; \textrm{etc.};$

and, for the powers of $$a$$, we obtain

$a^n=\dfrac{1}{a^2}; \quad a^{n-1}=\dfrac{1}{a^3}; \quad a^{n-2} = \dfrac{1}{a^4}; \quad a^{n-3} = \dfrac{1}{a^5}, \quad \textrm{etc.}$

We have therefore

$(a+b)^{-2} = \dfrac{1}{(a+b)^2} = \dfrac{1}{a^2} - \dfrac{2\cdot b}{1\cdot a^3} + \dfrac{2\cdot 3\cdot b^2}{1\cdot 2\cdot a^4} -\dfrac{2\cdot 3\cdot 4\cdot b^3}{1\cdot 2\cdot 3\cdot a^5} +\dfrac{2\cdot 3\cdot 4\cdot 5\cdot b^4}{1\cdot 2\cdot 3\cdot 4\cdot a^6}, \; \textrm{etc.}$

Now,

$\frac{2}{1}=2; \quad \dfrac{2\cdot 3}{1\cdot 2}=3; \quad \dfrac{2\cdot 3\cdot 4}{1\cdot 2\cdot 3}=4; \quad \dfrac{2\cdot 3\cdot 4\cdot 5}{1\cdot 2\cdot 3\cdot 4}=5, \quad \textrm{etc.}$

and consequently,

$\dfrac{1}{(a+b)^2} = \dfrac{1}{a^2} - 2\cdot \dfrac{b}{a^3} +3 \cdot \dfrac{b^2}{a^4} - 4\cdot \dfrac{b^3}{a^5} + 5\cdot \dfrac{b^4}{a^6} -6\cdot \dfrac{b^5}{a^7} +7\cdot \dfrac{b^6}{a^8}, \; \textrm{etc.}$

373 Let us proceed, and suppose $$n=-3$$, and we shall have a series expressing the value of $$\dfrac{1}{(a+b)^3}$$, or of $$(a+b)^{-3}$$. Here the coefficients will be

$\frac{n}{1} = -\frac{3}{1}; \quad \frac{n-1}{2}=-\frac{4}{2}; \quad \frac{n-2}{3} = -\frac{5}{3}, \quad \textrm{etc.}$

which gives

\begin{align} \dfrac{1}{(a+b)^3}&= \dfrac{1}{a^3} - \dfrac{3\cdot b}{1\cdot a^4} + \dfrac{3\cdot 4\cdot b^2}{1\cdot 2\cdot a^5} - \dfrac{3\cdot 4\cdot 5\cdot b^3}{1\cdot 2\cdot 3\cdot a^6}+\dfrac{3\cdot 4\cdot 5\cdot 6\cdot b^4}{1\cdot 2\cdot 3\cdot 4\cdot a^7}+\textrm{etc.}\\ &=\dfrac{1}{a^3}-3\cdot \dfrac{b}{a^4}+6\cdot \dfrac{b^2}{a^5} -10\cdot \dfrac{b^3}{a^6} +15\cdot\dfrac{b^4}{a^7} -21\cdot \dfrac{b^5}{a^8} +28\cdot \dfrac{b^6}{a^9}+\textrm{etc.} \end{align}

If we now make $$n=-4$$; we shall have for the coefficients

$\frac{n}{1} = -\frac{4}{1}; \quad \frac{n-1}{2}=-\frac{5}{2}; \quad \frac{n-2}{3} = -\frac{6}{3}; \quad \frac{n-3}{4} = -\frac{7}{4}, \quad \textrm{etc.}$

And for the powers,

$a^n=\dfrac{1}{a^4}; \quad a^{n-1}=\dfrac{1}{a^5}; \quad a^{n-2} = \dfrac{1}{a^6}; \quad a^{n-3} = \dfrac{1}{a^7}; \quad a^{n-4} = \dfrac{1}{a^8},$

whence we obtain,

\begin{align} \dfrac{1}{(a+b)^4} &= \dfrac{1}{a^4} - \dfrac{4b}{1\cdot a^5} - \dfrac{4\cdot 5\cdot b^2}{1\cdot 2\cdot a^6}- \dfrac{4\cdot 5\cdot 6\cdot b^3}{1\cdot 2\cdot 3\cdot a^7}+\textrm{etc.}\\ &= \dfrac{1}{a^4} - 4\dfrac{b}{a^5} + 10\dfrac{b^2}{a^6}- 20\dfrac{b^3}{a^7} + 35\dfrac{b^4}{a^8} -56\dfrac{b^5}{a^9}+\textrm{etc.} \end{align}

374 The different cases that have been considered enable us to conclude with certainty, that we shall have, generally, for any negative power of $$a+b$$;

$\dfrac{1}{(a+b)^m} = \dfrac{1}{a^m} - \dfrac{m\cdot b}{a^{m+1}} +\dfrac{m\cdot (m-1)\cdot b^2}{2\cdot a^{m+2}} -\dfrac{m\cdot (m-1)\cdot (m-2)\cdot b^3}{2\cdot 3\cdot a^{m+3}}, + \textrm{etc.}$

And, by means of this formula, we may transform all such fractions into infinite series, substituting fractions also, or fractional exponents, for $$m$$, in order to express irrational quantities.

375 The following considerations will illustrate the subject still farther: for we have seen that

$\dfrac{1}{a+b} = \dfrac{1}{a} - \dfrac{b}{a^2} + \dfrac{b^2}{a^3} - \dfrac{b^3}{a^4} + \dfrac{b^4}{a^5} - \dfrac{b^5}{a^6}+ \textrm{etc.}$

If, therefore, we multiply this series by $$a+b$$, the product ought to be = 1; and this is found to be true, as will be seen by performing the multiplication:

 $$\dfrac{1}{a}$$ $$-\dfrac{b}{a^2}$$ $$+\dfrac{b^2}{a^3}$$ $$-\dfrac{b^3}{a^4}$$ $$+\dfrac{b^4}{a^5}$$ $$-\dfrac{b^5}{a^6}$$ + etc. $$a$$ $$+b$$ 1 $$-\dfrac{b}{a}$$ $$+\dfrac{b^2}{a^2}$$ $$-\dfrac{b^3}{a^3}$$ $$+\dfrac{b^4}{a^4}$$ $$-\dfrac{b^5}{a^5}$$ + etc. $$+\dfrac{b}{a}$$ $$-\dfrac{b^2}{a^2}$$ $$+\dfrac{b^3}{a^3}$$ $$-\dfrac{b^4}{a^4}$$ $$+\dfrac{b^5}{a^5}$$ + etc.

where all the terms but the first cancel each other.

376 We have also found, that

$\dfrac{1}{(a+b)^2} = \dfrac{1}{a^2} - \dfrac{2b}{a^3} + \dfrac{3b^2}{a^4} - \dfrac{4b^3}{a^5} +\dfrac{5b^4}{a^6} - \dfrac{6b^5}{a^7} + \textrm{etc.}$

And if we multiply this series by $$(a+b)^2$$, the product ought to be equal to 1. Now, $$(a+b)^2 = a^2+2ab+b^2$$, and

 $$\dfrac{1}{a^2}$$ $$-\dfrac{2b}{a^3}$$ $$+\dfrac{3b^2}{a^4}$$ $$-\dfrac{4b^3}{a^5}$$ $$+\dfrac{5b^4}{a^6}$$ $$-\dfrac{6b^5}{a^7}$$ + etc. $$a^2$$ $$+2ab$$ $$+b^2$$ 1 $$-\dfrac{2b}{a}$$ $$+\dfrac{3b^2}{a^2}$$ $$-\dfrac{4b^3}{a^3}$$ $$+\dfrac{5b^4}{a^4}$$ $$-\dfrac{6b^5}{a^5}$$ + etc. $$+\dfrac{2b}{a}$$ $$-\dfrac{4b^2}{a^2}$$ $$+\dfrac{6b^3}{a^3}$$ $$-\dfrac{8b^4}{a^4}$$ $$+\dfrac{10b^5}{a^5}$$ + etc. $$+\dfrac{b^2}{a^2}$$ $$-\dfrac{2b^3}{a^3}$$ $$+\dfrac{3b^4}{a^4}$$ $$-\dfrac{4b^5}{a^5}$$ + etc.

which gives 1 for the product, as the nature of the thing required.

377 If we multiply the series which we found for the value of $$\dfrac{1}{(a+b)^2}$$, by $$a+b$$ only, the product ought to answer to the fraction $$\dfrac{1}{a+b}$$, or be equal to the series already found, namely,

$\dfrac{1}{a}-\dfrac{b}{a^2}-\dfrac{b^2}{a^3}-\dfrac{b^3}{a^4}+\dfrac{b^4}{a^5}+ \textrm{etc.}$

and this is the actual multiplication will confirm.

 $$\dfrac{1}{a^2}$$ $$-\dfrac{2b}{a^3}$$ $$+\dfrac{3b^2}{a^4}$$ $$-\dfrac{4b^3}{a^5}$$ $$+\dfrac{5b^4}{a^6}$$ + etc. $$a$$ $$+b$$ $$\dfrac{1}{a}$$ $$-\dfrac{2b}{a^2}$$ $$+\dfrac{3b^2}{a^3}$$ $$-\dfrac{4b^3}{a^4}$$ $$+\dfrac{5b^4}{a^5}$$ + etc. $$+\dfrac{b}{a^2}$$ $$-\dfrac{2b^2}{a^3}$$ $$+\dfrac{3b^3}{a^4}$$ $$-\dfrac{4b^4}{a^5}$$ + etc. $$\dfrac{1}{a}$$ $$-\dfrac{b}{a^2}$$ $$+\dfrac{b^2}{a^3}$$ $$-\dfrac{b^3}{a^4}$$ $$+\dfrac{b^4}{a^5}$$ + etc.

as required.

Editions

1. Leonhard Euler. Elements of Algebra. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822.
2. Leonhard Euler. Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange. Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.