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CHAPTER XLIV.

UNDETERMINED COEFFICIENTS.

453. The method of Undetermined Coefficients is often employed in developing algebraic expressions. It consists in assuming that the expression is equal to a series in which the coefficients are at first unknown, but determined by subsequent operations. The method is based on the following principles:

454. If the infinite series an+a,x+a,x2+azx8+ ...... is equal to zero for every finite value of x for which the series is convergent, then each coefficient must be equal to zero identically.

Let the series be denoted by S, and let S, stand for the expression a, +2.9x +Azx2 + ...... ; then S=a,+s, and therefore, by hypothesis, a, +äsn=0 for all finite values of x. But since s is convergent, S, cannot exceed some finite limit; therefore by taking a small enough xS, may be made as small as we please. In this case the limit of S is ao; but S is always zero, therefore

ao must be equal to zero identically. Removing the term do, we have xSq=0 for all finite values of X; that is, a, +a ,X +agu2+ vanishes for all finite values of x.

Similarly, we may prove in succession that each of the coefficients ay, A2, A3,

is equal to zero identically. 455. If two infinite series are convergent and equal to one another for every finite value of the variable, the coeficients of like powers of the variable in the two series are equal. Suppose that the two series are denoted by

ao++ax?+ 0,28+ ...... and

4,+A42++4 3:28+ ......; then the expression

Qo-A,+(a, -A1.2+(az- A2).x2+(az - Az)x3 + ...

vanishes for all values of x within the assigned limits; therefore by the last article

Qo-A,=0, 0, -A,=0, 0, - A2=0, 2,- A,=0), ...... that is,

a,= A, a,= Ay ay= A ,, ag=A 3, ......; hence the coefficients of like powers of the variable are equal, which proves the proposition.

456. The principle of Art. 455 is equally true of finite series, for we may regard them as infinite series in which the missing terms are to be supplied with zero coefficients.

EXPANSION OF FRACTIONS INTO SERIES.

457. Expand

2+22

in a series of ascending powers of x. 1+2-x2

2 +22 Let

= A + Bx + Cxo+D.x3 + ...,

1+2--22 where A, B, C, D, ... are constants whose values are to be determined; then

2+22=(1+2— 22)(A + B.c + C22 +0.23 +...)
=A+B+ C122+ D 28
A
B

(
- A

-B

Equating the coefficients of like powers of 2, we have
A=2; B+A=0, C+B-A=1, D+C-B=0,

... B= -2; .. C=5; ... D= -7; thus

2+x2

-=2-23+5x2—7x3 + ...

1+2 — x2 458. Both numerator and denominator should always be arranged with reference to the ascending or descending powers of the same quantity; then dividing the first term of the numerator by the first term of the denominator det the form of the expansion.

2 Example. Expand in a series of ascending powers of x.

2x2 – 3x3

Dividing 2', of the first term of the numerator, by x?, of the first term of the denominator, we obtain -2 for the first term of the expansion; therefore we assume

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Equating the coefficients of like powers of x, we have
A=1;
2B-3A=0, 2C-3B=0, 2D-3C=0,

9
.. B=;

27 .: C=

.: D=

;

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1.

Expand to four terms in ascending powers of x:
1+ 2x
1-8x
1+x

3+x
2.
3.

4.
1-2-32 1-3-622

2+x+x?

2--2-22 1 2–2x+3x2

1 6.

7. 1+ aa- ax2 — 303

4+2+x2

2x - 3x2 10.

1 + x + x2 9. 3x2 + x3

XC+23+x4

5.

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8.

b-ax

EXPANSION OF SURDS INTO SERIES.

459. Expand V1 + x in ascending powers of x.

Let V1+x= A + Bx + Cx2 + D28+ Ex4. By squaring both sides of the equation, we have 1+x= A2+2 AB | 2+ B2 X2 + 2AD | 23 + C'2 +2AC +2BC +2BD

+2AE

x4

.. C=

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Equating the coefficients of like powers of x, we have A=1; 2AB=1, B2+2AC=0, 2AD+2BC=0, 1

1 B:

i

16 C2+2BD+2AE=0,

5

i

128

x x35x4 thus

+
2 8 16 128

.. E=

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V1+x=1+

t...

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460. To revert a series y=ax+bx2 +cQ3 + ... is to express x in a series of ascending powers of y. Revert the series y=x— 2x2 + 3x3 — 4x4+.

(1). Assume x= Ay+By2 + Cy8+ Dyt +...

Substituting in this equation the value of y as given in (1), we have

x= A(x – 2x2 + 3x3 – 4.24 + ...) = A(x 2x2 +3.28 — 4x4 + ...) +B(x 2x2 + 3x3 — 4x4 + ...)2=B (x2 + 4x4 — 4x3 + 6x4 + ...) +C(x-2x2 + 3x3–4x4 +...)3=C(28-6x4 - ...

+D(2-222 +3.23 -4.74 + ...)+=D(24+... Equating the coefficients of like powers of x A=1; B-2A =0, C4B+34 =0, D-6C+10B-4A=0, .:. B=2; .. C=5;

.:. D=14. Hence

x=y+2y2 + 5y +14y4+...

461. If the series be y=1+2x+3x2+4x3+... put y-1=z; then

z=2x+3x2+4x3+... Assume x=Az+Bz2 + C23+... and complete the work as in Art. 460, after which replace z by its value y-1.

EXAMPLES XLIV. C.

Revert each of the following series to four terms: 1. y=x+x2 + x3 +34 +...

2. y=x+3x2+5x3+734 + ... 22 2003 24

22 23 х4 3. y=x

t..
4. y=1+3+

+.
2 4 8

2 6 24 25 27 5. Y=X

+... 6. y=ax+bx2 + cx3 +d34 +... 7

+

+

+

PARTIAL FRACTIONS.

462. A group of fractions connected by the signs of addition and subtraction is reduced to a more simple form by being collected into one single fraction whose denominator is the lowest common denominator of the given fractions. But the converse process of separating a fraction into a group of simpler, or partial, fractions is often required. For example, if we wish to

3 – 5x expand

in a series of ascending powers of x, we 1-4x + 3x2 might use the method of Art. 457, and so obtain as many terms as we please. But if we wish to find the general term of the series this method is inapplicable, and it is simpler to express

1 2 the given fraction in the equivalent form +

Each 1

1- 3x of the expressions (1-x)-1 and (1-3 x)-! can now be expanded by the Binomial Theorem, and the general term obtained.

463. We shall now give some examples illustrating the decomposition of a rational fraction into partial fractions. For a fuller discussion of the subject the reader is referred to treatises on Higher Algebra, or the Integral Calculus. In these works it is proved that any rational fraction may be resolved into a series of partial fractions; and that to any factor of the first degree, as -a, in the denominator there corresponds a partial

-X

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