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(2) Corresponding to any linear factor x b that occurs more than once, assume partial fractions of the form

B,

+

B2

B2

+

+ .....

x-b ( x − b ) 2 ̄ (x—b)3

one for each repetition of the factor.

(3) Corresponding to a quadratic factor x2+px+q, assume

a partial fraction

Px+Q x2+px+q

All the coefficients A, B, P and Q are constants.

Example 1: For example, suppose we wish to break up

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The two members of this identity represent the same function, and so have the same values for any value of x. If we let x=2, and again let x=-3, we obtain:

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Note that in order to determine one of the coefficients, we give x a value that will cause the other coefficient to drop out.

Example 2: As a second example, we will break up

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In order to find B1, substitute the values of A and B, and simplify:

x2+2x+2=2(x+1)2+В ̧x(x+1) −x,

− x2−x=B1x(x+1).

As one form of the function is divisible by x(x+1), the other must be; dividing the identity, we find

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Example 3: The following example illustrates the use of the imaginary i=V-1: Separate

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substitute and simplify, dividing out x; then

−x=B1(x2+1)+(Px+Q)x.

Let x=0; then

0=B1;

substitute, simplify, and divide out x; then

Let x=i; then

-1=Px+Q.

-1=Pi+Q,

whence P=0, Q=-1, as the real and imaginary parts must be separately equal.

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It must be observed that these methods apply only to fractions of which the numerator is of lower degree than the denominator; i. e., to proper fractions. An improper fraction must first be separated by division into an integral expression plus a proper fraction.

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137. Theorem of Undetermined Coefficients.-The following principle is often useful:

If two polynomials are identical, the coefficient of any power of their variable is the same in both.

This may be seen as follows: Given

a-bx+cx2+ dx3 +.... = A + Bx+Cx2+Dx3 +.....

Being expressions for the same function, and therefore equal for any value of x, these polynomials are equal when x=0; hence

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Proceeding in this way, we see the truth of the proposition above. Note that after dividing out x from any one of these identities, we still have an identity; for x is not identically zero; i. e., zero is not the only value x can have. (See Art. 36.)

138. Partial Fractions by Comparison of Coefficients.-This theorem can be applied to partial fractions as follows: Separate

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or

1=B1x(x2+1)+B2(x2+1)+(Px+Q)x2,

1 = (B1+P)x+(В2+Q) x2+B1x+В2.

2

Comparing coefficients, B1+P=0, B2+Q=0, B1=0, B2=1, whence

B1=0, B2=1, P=0, Q= −1.

Examples.

Separate into partial fractions:

x2-1

1.

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(Compare coefficients.)

(Put x=-1 in your identity, simplify the iden

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