[Untitled]

French title: Binome

Link to original French version in ARTFL: page 2:258

Category of knowledge (if given): Algebra

Author (if known): D’Alembert, Jean Le Rond

English Title: Binomial

Translator: Gilles Carrier

Translator email: g.carrier007@gmail.com

Faculty supervisor and email (if applicable): N/A

BINOMIAL (Algebra) is a quantity composed of two parts, or two terms linked by the + or - signs. See Mononomial. Thus, a + c and 5-3 are binomials.

If an algebraic quantity has three parts, such as a + b + c, it is called a trinomial. If it has more, it is called a quadrinomial, etc. and in general multinomial. See Trinome.

Mr. Newton has given a method for raising a generic binomial a + b, to any power m, whose exponent is an integer or a fractional number, positive or negative.

Here is this formula¹:

$(a + b)^{m} = a^{m} + m.a^{(m - 1)}b + \ \frac{m\ .\ (m - 1)}{2}a^{(m - 2)}b^{2}\ + \frac{m\ .\ (m - 1)\ (m - 2)}{2.3}a^{(m - 3)}b^{3} + etc.$

A mere inspection of the terms reveals the law better than a long speech.

One can see that when m is an integer, this sequence is reduced to a finite number of terms. For example, let m=2, since m-2 = 0, therefore all the terms following the first three will be = 0, since they will each be multiplied by m-2.

M. le Marquis de l'Hôpital², in his Traité des sections coniques³, Book X., has demonstrated this formula for the case where m is an integer. M. l'Abbé de Molieres⁴ has also demonstrated this in his Éléments de mathématiques. Finally, we find a demonstration of this by the combinations in the Éléments d’algèbre⁵ by M. Clairaut⁶.

When m is a negative number or a fraction, the sequence is infinite, and therefore it represents the value of (a + b)^m only in the case where it is convergent, that is, where each term is greater than the next. See Serie or Suite. See also Convergent, Divergent, etc.

For example, let’s have an imperfect square aa + b, from which we must extract the square root. All we have to do is raise aa + b to the power 1/2 since finding the square root, or to raise to the power of 1/2, is the same thing. See Exponent. Thus, we will have:

${(aa + b)}^{\frac{1}{2}} = {aa}^{\frac{1}{2}} + \ \frac{1}{2} \times b \times {aa}^{\left( \frac{1}{2\ } - 1 \right)} + \ \frac{1}{2} \times \frac{1}{2} - 1 \times \frac{b^{2} \times aa^{\left( \frac{1}{2} - 2 \right)}}{2},\ etc.$

$= a + \ \frac{b}{2a}\ - \frac{bb}{8a^{3}},\ etc.$

a formula or an infinite series which will approach more and more to the root sought.

Similarly, if we want to extract the cubic root of (a³+b), we will have to raise this quantity to the power 1/3. We will then find

${(aa + b)}^{\frac{1}{3}} = a + \ \frac{b}{3a^{2}} - \ \frac{b^{2}}{9a^{3}},\ etc.$

and so on for with others. But these infinite series are good only in so far as they are convergent.

Let n be the rank occupied by any term in the sequence of the binomial a + b raised to any power m; we will find that this term is to the next as 1 is to $\left( \frac{b}{a} \times \frac{m - n + 1}{n} \right)$ from which it follows that for the series to be convergent. That is to say, for the terms to always decrease, the term b.(m-n+1) must always be smaller than na.

Thus, in order to find the approximate [square] root of (aa + b) using the previous formula, b(½-n+1), taken positively, must be smaller than naa, n being any integer.

Similarly, to extract the [square] root of (a³ + b) using this formula, b.(1/3-n+1), taken positively, must always be smaller than na³. (O)

Part of bnomial