Use partial fractions for distant coefficients · source reference

Printed question

Express $\frac{2x^2+3x-1}{(x-1)(x^2+1)}$ in partial fractions.

Hence find the series expansion of $\frac{2x^2+3x-1}{(x-1)(x^2+1)}$ in ascending powers of $x$, up to and including the term in $x^4$. Find the coefficient of $x^{2008}$. [6]

[2008/HCI/Prelim/I/6]

Printed working

$$\begin{aligned}\frac{2x^2+3x-1}{(x-1)(x^2+1)}&=\frac A{x-1}+\frac{Bx+C}{x^2+1},\\2x^2+3x-1&=A(x^2+1)+(Bx+C)(x-1).\end{aligned}$$

Let $x=1$: $4=2A\Rightarrow A=2$. Let $x=0$: $-1=2-C\Rightarrow C=3$. Compare coefficients of $x^2$: $2=2+B\Rightarrow B=0$.

$2/(x-1)+3/(x^2+1)=-2(1-x)^{-1}+3(1+x^2)^{-1}$.

$-2(1+x+x^2+x^3+x^4+\cdots)+3(1-x^2+x^4-\cdots)=1-2x-5x^2-2x^3+x^4+\cdots$.

Coefficient of $x^{2008}=x^{2(1004)}$ is $-2+3=1$.

Teacher note 1

For expansion of $(1-x)^{-1}$, coefficient of $x^{2008}$ is $-2(1)=-2$. For $(1+x^2)^{-1}=1+(-1)x^2+(-1)^2(x^2)^2+(-1)^3(x^2)^3+\cdots$, coefficient of $x^{2008}$ is $3(-1)^{1004}=3$.

Mathematical review

Write $F(x)=A/(x-1)+(Bx+C)/(x^2+1)$. Substitution $x=1$ gives $A=2$, $x=0$ gives $C=3$, and comparison of $x^2$ gives $B=0$.

Thus $F=-2(1-x)^{-1}+3(1+x^2)^{-1}=1-2x-5x^2-2x^3+x^4+O(x^5)$, valid for $|x|<1$. For $x^{2008}$, the first series contributes $-2$ and the second $3(-1)^{1004}=3$. Total coefficient: $1$.

Provenance

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