Page:EB1911 - Volume 01.djvu/646

 that an equation is not altered by multiplying both members by the same number. Suppose, for instance, that (x+1)＝(x−2). Here each member is a number, and the equation may, by the commutative law for multiplication, be written $2(x+1)⁄5$＝$4(x−2)⁄3$. This means that, whatever unit A we take, $2(x+1)⁄5$ A and $4(x−2)⁄3$ A are equal. We therefore take A to be 15, and find that 6(x+1)＝20(x−2). Thus, if we have an equation P＝Q, where P and Q are numbers involving fractions, we can clear of fractions, not by multiplying P and Q by a number m, but by applying the equal multiples P and Q to a number m as unit. If the P and Q of our equation were quantities expressed in terms of a unit A, we should restate the equation in terms of a unit A/m, as explained in §§ 18 and 21 (i.) (a).

(vii.) One result of the rule of transposition is that we can transpose all the terms in x to one side of equation, and all the terms not containing x to the other. An equation of the form ax＝b, where a and b do not contain x, is the standard form of simple equation.

(viii.) The quadratic equation is the equation of two expressions, monomial or multinomial, none of the terms involving any power of x except x and x2. The standard form is usually taken to be

ax2+bx+c＝0,

from which we find, by transformation,

(2 ax+b)2＝b2–4ac,

and thence x＝$√{b^{2} − 4ac} − b⁄2a$.

This only gives one root. As to the other root, see § 47 (iii.).

39. Fractional Expressions.—An equation may involve a fraction of the form $P⁄Q$, where Q involves x.

(i.) If P and Q can (algebraically) be written in the forms RA and SA respectively, where A may or may not involve x, then $P⁄Q$＝$RA⁄SA$＝$R⁄S$, provided A is not＝0.

(ii.) In an equation of the form $P⁄Q$＝$U⁄V$, the expressions P, Q, U, V are usually numerical. We then have $P⁄Q$ QV＝$U⁄V$ QV, or PV＝UQ, as in § 38 (vi.). This is the rule of cross-multiplication.

(iii.) The restriction in (i.) is important. Thus $x^{2}−1⁄x^{2} + x − 2$＝$(x−1) (x+1)⁄(x−1) (x+2)$ is equal to $x+1⁄x+2$, except when x＝1. For this latter value it becomes $0⁄0$, which has no direct meaning, and requires interpretation (§ 61).

40. Powers of a Binomial.—We know that (A+a)2＝A2 + 2Aa + a2. Continuing to develop the successive powers of A+a into multinomials, we find that (A+a)³＝A³ + 3A2a + 3Aa2 + a³, &c.; each power containing one more term than the preceding power, and the coefficients, when the terms are arranged in descending powers of A, being given by the following table:—

where the first line stands for (A+a)0＝1. A0a0, and the successive numbers in the (n+1)th line are the coefficients of Ana0, An−1a1,. . . A0an in the n+1 terms of the multinomial equivalent to (A+a)n.

In the same way we have (A−a)2＝A2−2Aa+a2, (A−a)3＝A3−3A2a + 3Aa2−a3,. . ., so that the multinomial equivalent to (A − a)n has the same coefficients as the multinomial equivalent to (A+a)n, but with signs alternately + and −.

The multinomial which is equivalent to (A ± a)n, and has its terms arranged in ascending powers of a, is called the expansion of (A ± a)n.

41. The binomial theorem gives a formula for writing down the coefficient of any stated term in the expansion of any stated power of a given binomial.

(i.) For the general formula, we need only consider (A+a)n. It is clear that, since the numerical coefficients of A and of a are each 1, the coefficients in the expansions arise from the grouping and addition of like terms (§ 37 (ii.)). We therefore determine the coefficients by counting the grouped terms individually, instead of adding them. To individualize the terms, we replace (A+a) (A+a) (A+a) … by (A+a) (B+b) (C+c) …, so that no two terms are the same; the “like” -ness which determines the placing of two terms in one group being the fact that they become equal (by the commutative law) when B, C, … and b, c, .... are each replaced by A and a respectively.

Suppose, for instance, that n＝5, so that we take five factors (A+a) (B+b) (C+c) (D+d) (E+e) and find their product. The coefficient of A2a3 in the expansion of (A+a)5 is then the number of terms such as ABcde, AbcDe, AbCde,..., in each of which there are two large and three small letters. The first term is ABCDE, in which all the letters are large; and the coefficient of A2a3 is therefore the number of terms which can be obtained from ABCDE by changing three, and three only, of the large letters into small ones.

We can begin with any one of the 5 letters, so that the first change can be made in 5 ways. There are then 4 letters left, and we can change any one of these. Then 3 letters are left, and we can change any one of these. Hence the change can be made in 3. 4. 5 ways.

If, however, the 3. 4. 5 results of making changes like this are written down, it will be seen that any one term in the required product is written down several times. Consider, for instance, the term AbcDe, in which the small letters are bce. Any one of these 3 might have appeared first, any one of the remaining 2 second, and the remaining 1 last. The term therefore occurs 1. 2. 3 times. This applies to each of the terms in which there are two large and three small letters. The total number of such terms in the multinomial equivalent to (A+a) (B+b) (C+c) (D+d) (E+e) is therefore (3. 4. 5) ÷ (1. 2. 3); and this is therefore the coefficient of A2a3 in the expansion of (A+a)5.

The reasoning is quite general; and, in the same way, the coefficient of An−rar in the expansion of (A+a)n is {(n−r+1) (n−r+2)  (n−1)n} ÷ {1. 2. 3 r}. It is usual to write this as a fraction, inverting the order of the factors in the numerator. Then, if we denote it by n(r), so that

we have

where n(0), introduced for consistency of notation, is defined by

This is the binomial theorem for a positive integral index.

(ii.) To verify this, let us denote the true coefficient of An−rar by $n(n−1)...(n−r+1)⁄1. 2. 3...r$, so that we have to prove that $$＝n(r), where n(r) is defined by (1); and let us inspect the actual process of multiplying the expansion of (A+a)n−1 by A+a in order to obtain that of (A+a)n. Using detached coefficients (§ 37 (v.)), the multiplication is represented by the following:—

Now suppose that the formula (2) has been established for every power of A+a up to the (n−1)th inclusive, so that $$＝(n−1)(r), $$＝(n−1)(r−1). Then $$, the coefficient of An−rar in the expansion of (A+a)n, is equal to (n−1)(r)+(n−1)(r−1). But it may be shown that (r being > 0)