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Worked solution

Expand a rational function using partial fractions

Express f(x)=x2+4x+3(2x+1)(x2+1)f(x)=\dfrac{x^2+4x+3}{(2x+1)(x^2+1)} in partial fractions. Hence obtain its expansion in ascending powers of xx up to and including x3x^3. Find also the range of values of xx for which the expansion is valid.

[2012/SRJC/Promo/5]

CJC · Discussion · Question 3

Answer

f(x)=12x+1+2x2+1=3−2x+2x2−8x3+O(x4)f(x)=\dfrac1{2x+1}+\dfrac2{x^2+1}=3-2x+2x^2-8x^3+O(x^4), for ∣x∣<1/2|x|<1/2.

Find the partial fractions

Write f(x)=A2x+1+Bx+Cx2+1f(x)=\dfrac A{2x+1}+\dfrac{Bx+C}{x^2+1}. Then x2+4x+3=A(x2+1)+(Bx+C)(2x+1)x^2+4x+3=A(x^2+1)+(Bx+C)(2x+1). At x=−1/2x=-1/2, 5/4=(5/4)A5/4=(5/4)A, so A=1A=1. At x=0x=0, 3=1+C3=1+C, so C=2C=2. At x=1x=1, 8=2+3(B+2)8=2+3(B+2), so B=0B=0. Hence f(x)=12x+1+2x2+1f(x)=\dfrac1{2x+1}+\dfrac2{x^2+1}.

How did this part go?

Combine geometric expansions

(1+2x)−1=1−2x+4x2−8x3+O(x4)(1+2x)^{-1}=1-2x+4x^2-8x^3+O(x^4) and 2(1+x2)−1=2−2x2+O(x4)2(1+x^2)^{-1}=2-2x^2+O(x^4). Therefore f(x)=3−2x+2x2−8x3+O(x4)f(x)=3-2x+2x^2-8x^3+O(x^4).

How did this part go?

Intersect the validity conditions

Require ∣2x∣<1|2x|<1 and ∣x2∣<1|x^2|<1. Their intersection is −1/2<x<1/2-1/2<x<1/2.

How did this part go?

Common mistakes

The combined expansion must satisfy both validity conditions.

Read the hint