   Chapter 8, Problem 11PS

Chapter
Section
Textbook Problem

Partial Fraction Decomposition Suppose the denominator of a rational function can be factored into distinct linear factors D ( x ) = ( x − c 1 ) ( x − c 2 ) ⋯ ( x − c n ) vfor a positive integer n and distinct real numbers c 1 , c 2 , ... , c n . If N is a polynomial of degree less than n, show that N ( x ) D ( x ) = P 1 x − c 1 + P 2 x − c 2 + ⋯ + P n x − c n where P k = N ( c k ) / D ′ ( c k ) for k = 1 ,   2 , ⋯ , n . Note that this is the partial fraction decomposition of N ( x ) / D ( x ) .

To determine

To prove: The relation N(x)D(x)=P1xc1+P2xc2+...+Pnxcn where D(x)=(xc1)(xc2)...(xcn)

Explanation

Given:

We can factor the denominator of a rational function into distinct linear factors.

D(x)=(xc1)(xc2)...(xcn)

Where c1,c2,...cn are distinct linear factors.

Here, Pk=N(ck)D(ck) where N is a polynomial degree less than n and k=1,2,3,...,n, which is partial fraction decomposition of N(x)D(x).

Formula used:

Product Rule of Differentiation states that for functions u and v

ddx(uv)=uv+vu

Proof:

Assume N(x)D(x)=P1xc1+P2xc2+...+Pnxcn to be true.

Then take LCM

N(x)(xc1)(xc2)...(xcn)={P1(xc2)(xc3)...(xcn)+P2(xc1)(xc3)...(xcn)+...+Pn(xc1)(xc2)...(xcn1)}(xc1)(xc2)...(xcn)N(x)={P1(xc2)(xc3)...(xcn)+P2(xc1)(xc3)...(xcn)+...+Pn(xc1)(xc2)...(xcn1)}

Let x=c1.

Therefore,

N(c1)=P1(c1c2)(c1c3)...(c1cn)P1=N(c1)(c1c2)(c1c3)...(c1cn)

Similarly, let x=c2

N(c2)=P2(c2c1)(c2c3)..

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