DECOMPOSITION OF RATIONAL FRACTIONS.

369. By means of the properties of identical equations, a fraction may often be separated into two or more partial fractions, whose denominators shall be simpler than the given denominator. In every such case, the given fraction is the sum of the partial fractions; hence its denominator will be a commoTM multiple of the denominators of the partial fractions.

By inspection, we perceive that

x2-7x+10= (x − 5) (x − 2).

Since the first member is simply the sum of the two fractions in the second member, this is obviously an identical equation. Clearing of fractions and uniting terms,

in which 31 in the first member, and (2A + 5B) in the second, may be considered as coefficients of 20. Now according to 368, III, the coefficients of the like powers of x in the two members must be equal; therefore,

From these equations we readily obtain

A 3, and B = 5;

whence, from equation (1), we have

It should be observed that equations (3) and (4) are the equations of condition, which must exist in order that equation (1) shall be true for all values of x.

7x2 + x = (2A + B) x + (B − A) ;

transposing all the terms to the first member,

7x2 + (1 − 2A — B) x + (A − B) = 0 . . (2).

If this equation be possible, it must be an identical equation ; and as one member is zero, the coefficients of the different powers of x must be separately equal to zero (368, IV); and we shall have

7=0,

which is absurd. Hence, we infer that the fraction cannot be separated into partial fractions, having numerators independent of x. Again, assume

7x2+x=(2A + B) x2 + (B - A) x ... (2);

equating the coefficients of like powers of x,

From this example we learn that if we assume an impossible form for the partial fractions, the fact will be made apparent by some absurdity in the equations of condition.

NOTE.-If the given denominator consists of three or more factors, there will be three or more partial fractions. But there will always be as many equations of condition as there are numerators to be determined.

370. It has been shown (89, 4) that my is exactly divisible by xy, if m is a positive whole number. The form of the quotient is as follows:

-3

+xm−2 y + xm¬3 y2 + xm−4 y3 + + ym-1,

....

the number of terms in the quotient being equal to m.

Now suppose = y; then each term will become xm-1, and since there are m terms, we have the formula,

The subscript equation, y = x, is used to indicate the condition

under which the first member of (A) will be equal to the second.

371. We will

the value of m.

now show that this formula is true, whatever be There will be two cases:

Now suppose x=y, then zu; and since r and s are positive whole numbers, we have from (1),

Hence, the formula is true when the exponent is positive and fractional.

2d. When m is negative, and either integral or fractional.

Suppose the exponent of x and y to be m; we shall have

xm

x-my-m (xn

X y

Now suppose x=y; then, whether m be integral or fractional, we shall have, from the principles already established,

BINOMIAL THEOREM.

372. The Binomial Theorem has for its object the development of a binomial with any exponent, into a series. This theorem is expressed by an equation, called the Binomial Formula.

373. It is required to expand (a+x)" into a series, n being any real quantity, positive or negative, entire or fractional.

We observe that

a + x = a(1+2); therefore (a + x)" = a" (1 + 2)".

an

Hence, if we first expand (1 + 2)", 2)", and then multiply the

result by a", we shall have the expansion of (a + x)”.

Let us now assume the equation,

(1 + z)" = A + Bz+ Cz2 + Dz3 + Ez1 + . . . .

z)”

in which A, B, C, D, etc., are independent of z.

(1),

We are to find

the values of these coefficients which will render equation (1) true

for all possible values of z.

Suppose z = 0; then from equation (1), we have A = 1.

Hence, the assumed development becomes

(1+2)=1+ Bz + Cz2+ Dz3 +Ez1+....

for all values of z. Put zu; then

(1 + u)" = 1+ Bu+ Cu2 + Du3 + Eu2 + . . . .

Subtracting (3) from (2), and dividing the result by z — u,

Let P=1+%, and Q=1+u; then P-Q = z — u.
Equation (4) now becomes

(2)

(3).

(4);

Now suppose zu; then P= Q. And by the Residual Formula (370), we shall have