Mathematics 9709 · Pure Mathematics 3

Partial fractions with binomial expansions

Here the two ideas come together. This is the main reason you learn partial fractions at all.
A messy fraction will not expand on its own. But split it into partial fractions first and each simple piece is ready for the binomial.
Expand each piece, then add them up term by term.

Split, expand, add. Split the function into partial fractions, expand each piece with the rational-n binomial (§7.3) to the term you need, then add them up. For the combined validity, take the part that both ranges share. The smaller bound always wins.

Split, expand each, combine

Three steps, in order:

Split into pieces like $\dfrac{A}{1-ax}$ , then write each as $A(1-ax)^{-1}$ .
Expand each one with the rational-n binomial (§7.3) up to the term you need.
Add the series term by term, and give the validity $|x|<\min\!\big(\tfrac1a,\tfrac1b\big)$ (the smaller band wins).

Why the smaller bound always wins

Each piece has its own band, say $|x|<\tfrac12$ and $|x|<1$ . The combined sum only works where both pieces work. The part those two bands share is just the smaller one, $|x|<\tfrac12$ . So take the part they share, not both put together. The smaller bound always wins.

Worked example

Textbook: expand $\dfrac{10x+1}{(1-2x)(1+x)}$ up to $x^3$ , with validity

Partial fractions (distinct linear): $\dfrac{10x+1}{(1-2x)(1+x)}=\dfrac{4}{1-2x}-\dfrac{3}{1+x}$ .
Expand each as a power: $4(1-2x)^{-1}=4(1+2x+4x^2+8x^3)$ and $3(1+x)^{-1}=3(1-x+x^2-x^3)$ .
Subtract term by term: $(4-3)+(8+3)x+(16-3)x^2+(32+3)x^3$ .
Validity: $(1-2x)^{-1}$ needs $|x|<\tfrac12$ , $(1+x)^{-1}$ needs $|x|<1$ ; the tighter wins.

Answer

1+11x+13x^2+35x^3+\cdots,\quad |x|<\tfrac12

Where the marks go

Common mistake

The combined range is the part the bands share: with bands

|x|<\tfrac12

and

|x|<1

, the answer is

|x|<\tfrac12

, not

|x|<1

. Writing the bigger band (or both put together) is the usual lost mark here.

Common mistake

A repeated factor gives an

(a+x)^{-2}

piece (§7.4): pull its

a

out before expanding, and don't forget the

a^n

in front. The squared piece is where most marks are lost in these combined questions.

Now you try

Split $\dfrac{24x+13}{(1-2x)(2+x)^2}$ into partial fractions, then expand it up to $x^2$ . (9709/31 Oct/Nov 2023 Q10)

Your turn— tap to reveal the worked answer (9709/31 Oct/Nov 2023 Q10)

Find the constants first. Repeated factor, so use $\dfrac{A}{1-2x}+\dfrac{B}{2+x}+\dfrac{C}{(2+x)^2}$ . Multiply up: $24x+13=A(2+x)^2+B(1-2x)(2+x)+C(1-2x)$ .

Put $x=\tfrac12$ : $25=A(\tfrac{25}{4})\Rightarrow A=4$ ; put $x=-2$ : $-35=C(5)\Rightarrow C=-7$ ; match $x^2$ : $A-2B=0\Rightarrow B=2$ . So $\dfrac{4}{1-2x}+\dfrac{2}{2+x}-\dfrac{7}{(2+x)^2}$ .

Now expand each up to $x^2$ ( $4(1-2x)^{-1}$ , $(1+\tfrac{x}{2})^{-1}$ , $-\tfrac74(1+\tfrac{x}{2})^{-2}$ ) and add. The smallest band is $|x|<\tfrac12$ :

Answer

\tfrac{13}{4}+\tfrac{37}{4}x+\tfrac{239}{16}x^2+\cdots,\quad |x|<\tfrac12

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Partial fractions with binomial expansions

Here the two ideas come together. This is the main reason you learn partial fractions at all.
A messy fraction will not expand on its own. But split it into partial fractions first and each simple piece is ready for the binomial.
Expand each piece, then add them up term by term.

Split, expand each, combine

Three steps, in order:

Split into pieces like $\dfrac{A}{1-ax}$ , then write each as $A(1-ax)^{-1}$ .
Expand each one with the rational-n binomial (§7.3) up to the term you need.
Add the series term by term, and give the validity $|x|<\min\!\big(\tfrac1a,\tfrac1b\big)$ (the smaller band wins).

Why the smaller bound always wins

Worked example

Textbook: expand $\dfrac{10x+1}{(1-2x)(1+x)}$ up to $x^3$ , with validity

Partial fractions (distinct linear): $\dfrac{10x+1}{(1-2x)(1+x)}=\dfrac{4}{1-2x}-\dfrac{3}{1+x}$ .
Expand each as a power: $4(1-2x)^{-1}=4(1+2x+4x^2+8x^3)$ and $3(1+x)^{-1}=3(1-x+x^2-x^3)$ .
Subtract term by term: $(4-3)+(8+3)x+(16-3)x^2+(32+3)x^3$ .
Validity: $(1-2x)^{-1}$ needs $|x|<\tfrac12$ , $(1+x)^{-1}$ needs $|x|<1$ ; the tighter wins.

Answer

1+11x+13x^2+35x^3+\cdots,\quad |x|<\tfrac12

Where the marks go

Common mistake

The combined range is the part the bands share: with bands

|x|<\tfrac12

and

|x|<1

, the answer is

|x|<\tfrac12

, not

|x|<1

. Writing the bigger band (or both put together) is the usual lost mark here.

Common mistake

A repeated factor gives an

(a+x)^{-2}

piece (§7.4): pull its

a

out before expanding, and don't forget the

a^n

in front. The squared piece is where most marks are lost in these combined questions.

Now you try

Split $\dfrac{24x+13}{(1-2x)(2+x)^2}$ into partial fractions, then expand it up to $x^2$ . (9709/31 Oct/Nov 2023 Q10)

Your turn— tap to reveal the worked answer (9709/31 Oct/Nov 2023 Q10)

Find the constants first. Repeated factor, so use $\dfrac{A}{1-2x}+\dfrac{B}{2+x}+\dfrac{C}{(2+x)^2}$ . Multiply up: $24x+13=A(2+x)^2+B(1-2x)(2+x)+C(1-2x)$ .

Put $x=\tfrac12$ : $25=A(\tfrac{25}{4})\Rightarrow A=4$ ; put $x=-2$ : $-35=C(5)\Rightarrow C=-7$ ; match $x^2$ : $A-2B=0\Rightarrow B=2$ . So $\dfrac{4}{1-2x}+\dfrac{2}{2+x}-\dfrac{7}{(2+x)^2}$ .

Now expand each up to $x^2$ ( $4(1-2x)^{-1}$ , $(1+\tfrac{x}{2})^{-1}$ , $-\tfrac74(1+\tfrac{x}{2})^{-2}$ ) and add. The smallest band is $|x|<\tfrac12$ :

Answer

\tfrac{13}{4}+\tfrac{37}{4}x+\tfrac{239}{16}x^2+\cdots,\quad |x|<\tfrac12