If we substitute any numbers whatever for a and r, the two members of this equation will be identical. The equation is true for every value which can be assigned to a and r.

Assume

в заб

=

· a 6 X 32-6
2-1

- 192-6=186

= 6, r = 2. art+ar+ar2+ar+a=96 +48 +24+ 12 + 6 = 186 The equations A, B and C are called Analytical equations; they are true for every value of the letters. The Algebraic equations which we have used in the solution of questions, are true only for some particular values of the letters.

1. Develop the following compound terms, or find their equivalents in single terms:

2. What analytical equations can be formed with the preceding compound terms?

SECTION XCIX.

Indeterminate Coefficients.

The equivalent single terms of any compound term may be found, by what is called the method of indeterminate coeffi

cients.

Here A, B, C, D, &c. are the indeterminate coefficients; they are unknown, and it is proposed to find them; so that the equation may be analytical.

Clear the equation of fractions, and collect together the terms containing the same powers of x, and we have

· x = A+ (B—A) x + (C −B) x2 + (D—C) x3, &c. This last must be an analytical equation; that is, it must be true whatever x may be.

It is evident that we can make the equation true by making the terms containing the same powers of x, in each member, equal to each other.

Now we have a term, A, in the second member, which does not contain x; but there is no term in the first member which does not contain x; hence we make (1) A=0. We have (B-A) x or (B—A) times x, in the

second member, and once x in the first; hence we put

We have (CB) x2, or (C—B) times x2, in the second member; but, we have no 2 in the first member; hence we put

From equations (1,) (2,) (3,) (4,) we obtain

A=0

B=1

C=1

D=1&c.

Put these values of A, B, C, D, &c. in place of them, in equation (A,) and we have

(B)

− = x + x2+x3 + xa, &c. 1-x

1st. In equation (B) we have only written four terms of the second member; but it is evident that the series does not end here, nor does it in fact terminate at all; and hence it is called an infinite series.

2nd. The exponents of x increase every succeeding term; hence it is called a series according to the ascending powers of x.

3rd. If the coefficients of x decrease every succeeding term, the terms are called a series according to the descending powers of x.

4th. If x were a fraction, the terms would become continu

ally smaller: thus, if x, x+x2 + x2 + x2, &c.={+} ++, &c.; it would then be called a converging series.

5th. If x were a whole number, greater than 1, the terms would continually increase; and it would then be called a diverging series.

2. Find the equivalent series of the following, in which a, b, c, g, are known numbers:

Indeterminate Coefficients.

1. Find the equivalent single terms of the following:

(a + x)}

This may be done by the rule for extracting the root of

We may obtain this same equation by the method of indeterminate coefficients.

Assume (A) (a + x)=A+Bx+Çx2+Dx3, &c.
Squaring both members, we have

a+x=A2+2AB x + (2 AC+B2) x2 + (2 AD+2 BC) x3,

&c.

Putting the terms which do not contain x, and those which contain the same powers of x, in the two members of the equation equal to each other,

have

Substituting these values of A, B, C, D, in equation (A,) we

2. If a 16 and x = 1; what will equation (A') become?

Ans. (17)3=4+÷—sta + 1875 &c.

Or (17)3 =4.2017, nearly.

We see, that the series expressing the root of 17, is very converging. Hence, four of the first terms give nearly the root of 17; as the rest of the terms are very small.

3. If a = 16 and x = 1; what will equation (B') become?

Ans. (15)=4—;+3+7—18375, &c.

Or (15)=3.117, nearly.

Equations (A') and (B') furnish a convenient method of finding the approximate roots of numbers.

Let it be proposed to find the approximate root of 26. Take the greatest square number in 26, which is 25; call it a; and make x=1. Then substitute these values of a and x in equation (A',) and the second member of the equation will give the approximate root of 26. Recollect that x must be much smaller than a, or the series will not be very converging; and, a few terms will not give the root sufficiently exact. 4. Find, by this method, the root of 65; of 83; of 8, 37, 35, 48, 50, 101, 145, 143.

5. Find the series of single terms equivalent to

(3) (q x — x2)3 — x3 (a—x)3 (6) (ax+x2)3 = x* (a+x)}

=

1. Given, (1) y=x+2x2 + 4x4, &c., to find a series of y: powers of y equal to x, which is called reversing the series. Assume (2) x=By+Cy+Dy3, &c."

Putting this value of x, in place of x, in equation (1,) we

have

(3) y=By+ (2 B2 + C) y2 + (4 B C + D) y3, &c.

Putting the terms containing the same powers of y, in the first and second members of this equation, equal to each other, and we have