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Worked solution

Partial fractions before a local expansion

Express f(x)=3x2−3x+4(2x−1)(x2+3)f(x)=\dfrac{3x^2-3x+4}{(2x-1)(x^2+3)} in partial fractions and hence obtain f(x)f(x) as a series expansion in ascending powers of xx up to and including the term in x2x^2. marks

Find the range of values of xx for which the expansion is valid. mark

[2009/HCI/I/7(modified)]

CJC · Review problems · Question 1

Answer

f(x)=1/(2x−1)+(x−1)/(x2+3)=−4/3−(5/3)x−(35/9)x2+O(x3)f(x)=1/(2x-1)+(x-1)/(x^2+3)=-4/3-(5/3)x-(35/9)x^2+O(x^3), valid for −1/2<x<1/2-1/2<x<1/2.

Find the partial fractions

Write f(x)=A/(2x−1)+(Bx+C)/(x2+3)f(x)=A/(2x-1)+(Bx+C)/(x^2+3). Then 3x2−3x+4=A(x2+3)+(Bx+C)(2x−1)3x^2-3x+4=A(x^2+3)+(Bx+C)(2x-1). Using x=1/2x=1/2 gives 13/4=(13/4)A13/4=(13/4)A, so A=1A=1. At x=0x=0, 4=3−C4=3-C, so C=−1C=-1; at x=1x=1, 4=4+B−14=4+B-1, so B=1B=1. Hence

f(x)=12x−1+x−1x2+3.\begin{aligned}&f(x)\\ & =\frac1{2x-1}+\frac{x-1}{x^2+3}.\end{aligned}

How did this part go?

Expand and collect through x²

12x−1=−(1−2x)−1=−1−2x−4x2+⋯ .\begin{aligned}&\frac1{2x-1}\\ & =-(1-2x)^{-1}\\ & =-1-2x-4x^2+\cdots.\end{aligned}

Also

x−1x2+3=x−13(1+x23)−1=13(x−1+x23−x33+⋯ ).\begin{aligned}&\frac{x-1}{x^2+3}\\ & =\frac{x-1}{3}\left(1+\frac{x^2}{3}\right)^{-1}\\ & =\frac13\left(x-1+\frac{x^2}{3}-\frac{x^3}{3}+\cdots\right).\end{aligned}

Therefore

f(x)=−43−53x−359x2+O(x3).\begin{aligned}&f(x)\\ & =-\frac43-\frac53x-\frac{35}{9}x^2\\ &{}+O(x^3).\end{aligned}

How did this part go?

Intersect the validity intervals

The first series requires ∣2x∣<1|2x|<1. The second requires ∣x2/3∣<1|x^2/3|<1, or ∣x∣<3|x|<\sqrt3. Both hold for

−12<x<12.-\frac12<x<\frac12.

How did this part go?

Common mistakes

The valid interval is the intersection of both expansion conditions.

Solution notes

The source bracket contains +x²/3, confirmed with a higher-resolution page crop. Native extraction conflates the superscript and denominator; this is an extraction artifact, not a source mathematical error. The negative x³/3 term is an added intermediate expansion beyond the requested quadratic order.

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