### Part I. Section II. Chapter 12. “Of the Expression of Irrational Powers by Infinite Series.”

361 As we have shewn the method of finding any power of the root $$a + b$$, however great the exponent may be, we are able to express, generally, the power of $$a + b$$ whose exponent is undetermined; for it is evident that if we represent that exponent by $$n$$, we shall have by the rule already given (Art. 348 and the following):

\begin{align} (a+b)^n &= a^n + \frac{n}{1}a^{n-1}b+\frac{n}{1}\cdot \frac{n-1}{2} a^{n-2}b^2\\ &+\frac{n}{1}\cdot \frac{n-1}{2} \cdot \frac{n-2}{3} a^{n-3}b^3\\ &+\frac{n}{1}\cdot \frac{n-1}{2} \cdot \frac{n-2}{3} \cdot \frac{n-3}{4} a^{n-4}b^4 +\textrm{etc.} \end{align}

362 If the same power of the root $$a-b$$ were required, we need only change the signs of the second, fourth, sixth, etc. terms, and should have

\begin{align} (a-b)^n &= a^n - \frac{n}{1}a^{n-1}b+\frac{n}{1}\cdot \frac{n-1}{2} a^{n-2}b^2\\ &-\frac{n}{1}\cdot \frac{n-1}{2} \cdot \frac{n-2}{3} a^{n-3}b^3\\ &+\frac{n}{1}\cdot \frac{n-1}{2} \cdot \frac{n-2}{3} \cdot \frac{n-3}{4} a^{n-4}b^4 -\textrm{etc.} \end{align}

363 These formulas are remarkably useful, since they serve also to express all kinds of radicals; for we have shown that all irrational quantities may assume the form of powers whose exponents are fractional, and that $$\sqrt[2]{\vphantom{a}}a=a^{\frac{1}{2}}$$, $$\sqrt[3]{\vphantom{a}}a=a^{\frac{1}{3}}$$, $$\sqrt[4]{\vphantom{a}}a=a^{\frac{1}{4}}$$, etc.: we have, therefore,

$\sqrt[2]{\vphantom{(a+b)}}(a+b)=(a+b)^{\frac{1}{2}}; \; \sqrt[3]{\vphantom{(a+b)}}(a+b)=(a+b)^{\frac{1}{3}}; \; \sqrt[4]{\vphantom{(a+b)}}(a+b)=(a+b)^{\frac{1}{4}}, \; \textrm{etc.}$

Consequently, if we wish to find the square root of $$a+b$$, we have only to substitute for the exponent $$n$$ the fraction ½ in the general formula, Art. 361, and we shall have first, for the coefficients,

$\frac{n}{1}=\frac{1}{2}; \; \frac{n-1}{2}=-\frac{1}{4}; \; \frac{n-2}{3}=-\frac{3}{6}; \; \frac{n-3}{4}=-\frac{5}{8}; \; \frac{n-4}{5}=-\frac{7}{10}; \; \frac{n-5}{6}=-\frac{9}{12}.$

Then $$a^n=a^{\frac{1}{2}}=\surd a$$ and $$a^{n-1}=\dfrac{1}{\surd a}$$; $$a^{n-2}=\dfrac{1}{a\surd a}$$; $$a^{n-3}=\dfrac{1}{a^2 \surd a}$$, etc. or we might express those powers of $$n$$ in the following manner: $$a^n=\surd a$$; $$a^{n-1}=\dfrac{\surd a}{a}$$; $$a^{n-2}=\dfrac{a^n}{a^2}=\dfrac{\surd a}{a^2}$$; $$a^{n-3}=\dfrac{a^n}{a^3} = \dfrac{\surd a}{a^3}$$; $$a^{n-4}=\dfrac{a^n}{a^4} = \dfrac{\surd a}{a^4}$$, etc.

364 This being laid down, the square root of $$a + b$$ may be expressed in the following manner:

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

365 If $$a$$ therefore be a square number, we may assign the value of $$\surd a$$, and, consequently, the square root of $$a+b$$ may be expressed by an infinite series, without any radical sign.

Let, for example, $$a=c^2$$, we shall have $$\surd a=c$$; then

$\surd(c^2+b) = c + \frac{1}{2} \cdot \dfrac{b}{c} - \frac{1}{8} \cdot \dfrac{b^2}{c^3} +\frac{1}{16} \cdot \dfrac{b^3}{c^5} - \frac{5}{128} \cdot \dfrac{b^4}{c^7}, \; \textrm{etc.}$

We see, therefore, that there is no number, whose square root we may not extract in this manner; since every number may be resolved into two parts, one of which is a square represented by $$c^2$$. If, for example, the square root of 6 be required, we make 6 = 4 + 2, consequently, $$c^2=4$$, $$c = 2$$, $$b = 2$$, whence results

√6 = 2 + ½ - ¹⁄₁₆ + ¹⁄₆₄ - ⁵⁄₁₀₂₄ + etc.

If we take only the two leading terms of this series, we shall have 2½ = ⁵⁄₂, the square of which, ²⁵⁄₄, is ¼ greater than 6; but if we consider three terms, we have 2⁷⁄₁₆ = ³⁹⁄₁₆, the square of which, ¹⁵²¹⁄₂₅₆, is still ¹⁵⁄₂₅₆ too small.

366 Since, in this example, ⁵⁄₂ approaches very nearly to the true value of √6, we shall take for 6 the equivalent quantity ²⁵⁄₄ - ¼; thus $$c^2=\frac{25}{4}$$; $$c=\frac{5}{2}$$; $$b=\frac{1}{4}$$; and calculating only the two leading terms, we find

$\surd 6 = \frac{5}{2}+\frac{1}{2}\cdot \dfrac{-\frac{1}{4}}{\frac{5}{2}} = \frac{5}{2} - \frac{1}{2} \cdot \dfrac{\frac{1}{4}}{\frac{5}{2}} = \frac{5}{2} - \frac{1}{20} = \frac{49}{20},$

the square of which fraction being ²⁴⁰¹⁄₄₀₀, it exceeds the square of √6 only by ¹⁄₄₀₀.

Now, making 6 = ²⁴⁰¹⁄₄₀₀ - ¹⁄₄₀₀, so that $$c=\frac{49}{20}$$ and $$b=-\frac{1}{400}$$; and still taking only the two leading terms, we have

$\surd 6 = \frac{49}{20} + \frac{1}{2} \cdot \dfrac{-\frac{1}{400}}{\frac{49}{20}} =\frac{49}{20} - \frac{1}{2} \cdot \dfrac{\frac{1}{400}}{\frac{49}{20}} = \frac{49}{20} -\frac{1}{1960} = \frac{4801}{1960},$

the square of which is ²³⁰⁴⁹⁶⁰¹⁄₃₈₄₁₆₀₀; and 6, when reduced to the same denominator, is = ²³⁰⁴⁹⁶⁰⁰⁄₃₈₄₁₆₀₀; the error therefore is only ¹⁄₃₈₄₁₆₀₀.

367 In the same manner, we may express the cube root of $$a+b$$ by an infinite series; for since $$\surd[3]{\vphantom{(a+b)}}(a+b) = (a+b)^{\frac{1}{3}$$, we shall have in the general formula, $$n=\frac{1}{2}$$, and for the coefficients,

$\frac{n}{1}=\frac{1}{3}; \; \frac{n-1}{2}=-\frac{1}{3}; \; \frac{n-2}{3}=-\frac{5}{9}; \; \frac{n-3}{4} = -\frac{2}{3}; \; \frac{n-4}{5}= - \frac{11}{15}; \quad \textrm{etc.}$

and, with regard to the powers of $$a$$, we shall have

$a^n = \sqrt[3]{\vphantom{a}}a; \; a^{n-1} = \dfrac{\sqrt[3]{\vphantom{a}}a}{a}; \; a^{n-2} = \dfrac{\sqrt[3]{\vphantom{a}}a}{a^2}; \; a^{n-3} = \dfrac{\sqrt[3]{\vphantom{a}}a}{a^3}; \quad \textrm{etc.}$

then

$\sqrt[3]{\vphantom{(a+b)}} = \sqrt[3]{\vphantom{a}} + \frac{1}{3} \cdot b \dfrac{\sqrt[3]{\vphantom{a}}a}{a} -\frac{1}{9} \cdot b^2 \dfrac{\sqrt[3]{\vphantom{a}}a}{a^2} +\frac{5}{81} \cdot b^3 \dfrac{\sqrt[3]{\vphantom{a}}a}{a^3} -\frac{10}{243} \cdot b^4 \dfrac{\sqrt[3]{\vphantom{a}}a}{a^4}, \quad \textrm{etc.}$

368 If $$a$$ therefore be a cube, or $$a=c^3$$, we have $$\sqrt[3]{\vphantom{a}}=c$$, and the radical signs will vanish; for we shall have

$\sqrt[3]{\vphantom(c^3+b)}} (c^3+b) = c + \frac{1}{3} \cdot \frac{b}{c^2}-\frac{1}{9} \cdot \frac{b^2}{c^5} +\frac{5}{81} \codt \frac{b^3}{c^8} - \frac{10}{243} \cdot \frac{b^4}{c^11} +\textrm{etc.}$

369 We have therefore arrived at a formula, which will enable us to find, by approximation, the cube root of any number; since every number may be resolved into two parts, as $$c^3+b$$, the first of which is a cube.

If we wish, for example, to determine the cube root of 2, we represent 2 by 1 + 1, so that $$c=1$$ and $$b=1$$; consequently,

$\sqrt[3]{\vphantom{2}} 2 = 1 + \frac{1}{3} - \frac{1}{9} + \frac{5}{81}, \textrm{etc.}$

The two leading terms of this series make 1⅓ = ⁴⁄₃, the cube of which ⁶⁴⁄₂₇ is too great by ¹⁰⁄₂₇: let us therefore make 2 = ⁶⁴⁄₂₇ - ¹⁰⁄₂₇, we have $$c=\frac{4}{3}$$ and $$b=-\frac{10}{27}$$, and consequently,

$\sqrt[3]{\vphantom{2}} 2 = \frac{4}{3} + \frac{1}{3} \cdot \dfrac{-\frac{10}{27}}{\frac{16}{9}}:$

two two terms give ⁴⁄₃ - ⁵⁄₇₂ = ⁹¹⁄₇₂, the cube of which is ⁷⁵³⁵⁷¹⁄₃₇₃₂₄₈: but, 2 = ⁷⁴⁶⁴⁹⁶⁄₃₇₃₂₄₈, so that the error is ⁷⁰⁷⁵⁄₃₇₃₂₄₈; and in this way we might still approximate, the faster in proportion as we take a greater number of terms.

#### 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.