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fraction of the form

x-a

x-b, occurring twice in the denominator there correspond two partial fractions, and If x-b occurs three times,

B

and so on.

To any

x-b there is an additional fraction, quadratic factor x2+px+q there corresponds a partial fraction Pr+Q of the form ; if the factor x2+px+q occurs twice, x2+px + q there is a second partial fraction,

Rx+S
(x2+px+q)2 ́

and so on.

Here the quantities A, B, C, D,.. P, Q, R, S are all independent of x.

We shall make use of these results in the examples that follow.

A

; to any factor of the first degree, as

whence

Example 2. Resolve

с

(x-b)2

Assume

Example 1. Separate

into partial fractions.

Since the denominator 2x2+x-6=(x+2) (2x-3), we assume

B

5–11 A
+
2x2+x-6 x+2 2x-3

where A and B are quantities independent of x whose values have to be determined.

Clearing of fractions,

5 −11 2x2+x-6

[ocr errors]

=

5x-11=4(2x-3)+B(x+2).

Since this equation is identically true, we may equate coefficients of like powers of x; thus

2A+B=5, −3A+2B=−11;

A=3, B=-1.

5 −11 3 2x2+x-6x+2

......

=

mx + n

(x−α)(x+b)

mx+n

(x−α)(x+b)

[ocr errors]

=

A

B

+
x-α x+b

•'. mx+n=A(x+b)+B(x−a)

.(1).

We might now equate coefficients and find the values of A and B, but it is simpler to proceed in the following manner:

1 2x-3

into partial fractions.

Since A and B are independent of x, we may give to x any value

we please.

In (1) put x-a=0, or x=a; then

A=

putting x+b=0, or x=—b, B=

...

Assume

mx+n

(x− a)(x+b)

Example 3. Resolve

[ocr errors]

=

ma+n
a+b
mb-n

a+b

1

a+b (

into partial fractions.

23x-11.x2

A

B

+

C +

(2x−1)(3+x)(3−x) 2x-1 3+x 3

... 23x-11x2=A(3+x)(3−x)+B(2x−1)(3−x)

=

Example 5. Resolve

Assume

23x-11x2

(2x-1)(9-x2)

[ocr errors]

;

[ocr errors]

=

=

ma+n

x-a

+C(2x−1)(3+x).

By equating the coefficients of like powers of x, or putting in succession 2x-1=0, 3+x=0, 3-x=0, we find that

1

4

A=1, B=4, C=-1.
23x-11x2
+
(2x-1)(9-x2) 2x-1 3+x
3x2+x-2
(x-2)2(1-2x)
A

Example 4. Resolve

into partial fractions.

3x2+x-2

B

с

Assume

+

+

(x-2)2(1—2x) 1-2x x-2 (x − 2)2 3

.. 3x2+x-2=A(x-2)2+B(1−2x) (x-2)+ C(1-2x).

Let 1-2x=0, then

A=

3

let x-2=0, then

C=-4.

To find B, equate the coefficients of x2; thus

5

3

3-A-2B; whence B=

——

+mb-n).

1

3x2+x-2
(x-2)2 (1—2x) 3(1-2x)
42-19x
(x2+1)(x-4)

[ocr errors][merged small]

5

4

3(x-2) (x-2)2

into partial fractions. Ax+B+

42-19x

с

;

(x2+1)(x-4) x2+1 x-4
.. 42-19x=(4x+B) (x-4)+ C(x2+1).

(1);

Let x=4, then

C=-2;

equating coefficients of x2, 0=A+ C, and A=2; equating the absolute terms, 42=−4B+ C, and B=-11, 42-19x 2x-11 2

(x2+1)(x-4) x2+1 X- -4

and

464. In all the preceding examples the numerator has been of lower dimensions than the denominator; if this is not the case, we divide the numerator by the denominator until a remainder is obtained which is of lower dimensions than the denominator.

Example. Resolve

By division,

6x3+5x2-7
3x2-2x-1

6x3+5x2-7
3x2-2x-1

=

=

==

=2x+3+

8x-4
5
1
+
3x2-2x-1 3x+1 X- -1
6x3+5x2-7
3x2-2x-1

=2x+3+

into partial fractions.

8x-4 3x2-2x-1

5

1

+

3x+1 X- 1

465. We shall now explain how resolution into partial fractions may be used to facilitate the expansion of a rational fraction in ascending powers of x.

Example. Find the general term of

panded in a series of ascending powers of x. By Ex. 4, Art. 463, we have

3x2+x-2
(x-2)2(1-2x)

1

5

3x2+x-2
(x-2)2(1-2x) 3(1-2x) 3(x-2)

4

(x-2)2
4

1

5

+
3(1-2x)' 3(2-x)

=-(1-2)+(-)-(1-1)

(2-x)2

when ex

Hence the (r+1)th or general term of the expansion is

2r 5

r+

(- 3 + 1 - 1 - - 11 )

6 2r 2+

"

466. The following example sufficiently illustrates the method to be pursued when the denominator contains a quadratic factor.

Example. Expand

and find the general term.

Assume

1.

Let 1 + x = 0,
equating the absolute terms, 7
equating the coefficients of x2, 0 =
7+ x
3 4-3x
(1 + x)(1 + x2)

+
1 + x 1 + x2

1

7+ x

(1 + x)(1 + x2)

7+ x

A

=

+

(1 + x)(1 + x2) 1 + x

4.

6.

[ocr errors]

· ·. 7 + x = A(1 + x2) + (Bx + C)(1 + x).
then A = 3;

7x-1

5x+6x2

in ascending powers of x.

Resolve into partial fractions:

[blocks in formation]

A+ C, whence C = 4;
A+ B, whence B = 3.

2.

Bx + C

1 + x2

......

To find the coefficient of x":

(1) If is even, the coefficient of x in the second series

g0

is 4(-1); therefore in the expansion the coefficient of x is

g"

3 + 4( − 1)2.

(2) If r is odd, the coefficient of x in the second series is

r-1

r+1

— 3(− 1) 2, and the required coefficient is 3(-1) 2 — 3. ·

x210x+13

(* − 1)(x2 – 5x+6)

9

(x − 1)(x+2)2

;

EXAMPLES XLIV. d.

46 + 13 x 12x2 - 11x - 15

5.

[ocr errors]

1...

7.

•+(− 1)2x2+.....}.

[blocks in formation]

8.

12.

10.

15.

17.

26 x2 + 208 x

(x2 + 1)(x+5)

1+ 3x
1+11x+28x2

19.

3 x3 8 x2+10

(x − 1)4

21.

[ocr errors]

Find the general term of the following expressions when expanded in ascending powers of x.

13.

2 x 4

(1 − x2)(1 − 2 x)

3+2x-x2

(1 + x)(1 − 4 x)2

2x + 1

(x-1)(x2+1)

9.

11.

1

(1 − ax)(1 − bx)(1 − cx)

5x+6

(2 + x)(1 − x)

16.

18.

11x+5

(x-3)(x2 + 2 x − 5)

20.

22.

2 x2

5x3+6x +52

(x2 − 1)(x + 1)3

[blocks in formation]

4+3x+2x2

(1-x)(1 + x2x2)

4+7x
(2+3x)(1 + x)2

1−x+2x2

(1 - x)8

3- 2x2
(2-3x+x2)2

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