Binomial Series: Products, Partial Fractions and Approximations

A2 · P3 · 12 min

Once you can expand a single bracket like $(1 + 2x)^{-3}$ , the exam asks you to do something with it: expand a product such as $(1 + x)\sqrt{1 - 2x}$ , a quotient such as $\dfrac{1 + 2x}{(1 - x)^2}$ , or a whole rational function by first splitting it into partial fractions. Each piece has its own range of validity, and the combined expansion is valid only where every piece is. The classic P3 question is "express in partial fractions, hence obtain the expansion up to $x^2$ ", worth 8 to 10 marks across two parts.

Products: expand, multiply, truncate

To expand a product, expand each factor separately to the required power of $x$ , then multiply, keeping only the terms up to that power. There is no need to work out any term you will throw away.

Expanding a product or quotient

Rewrite any quotient as a product: $\dfrac{p(x)}{(1 + x)^2} = p(x)(1 + x)^{-2}$ .
Expand each non-polynomial factor up to the highest power required.
Multiply, collecting only terms up to that power. A grid of "this term times that term" stops you missing products.
State the range of validity: the intersection of the ranges of every series used, which is the narrowest one.

A product

Expand $(1 + x)\sqrt{1 - 2x}$ in ascending powers of $x$ up to and including the term in $x^2$ , and state the set of values of $x$ for which the expansion is valid.

Solution

First, with $n = \tfrac{1}{2}$ and $u = -2x$ :

\sqrt{1 - 2x} = 1 + \tfrac{1}{2}(-2x) + \frac{\tfrac{1}{2}\left(-\tfrac{1}{2}\right)}{2}(-2x)^2 + \cdots = 1 - x - \tfrac{1}{2}x^2 + \cdots

Multiply by $(1 + x)$ , keeping terms up to $x^2$ :

(1 + x)\left(1 - x - \tfrac{1}{2}x^2\right) = 1 - x - \tfrac{1}{2}x^2 + x - x^2 + \cdots = 1 - \tfrac{3}{2}x^2 + \cdots

The $x$ terms cancel, so the coefficient of $x$ is $0$ .

Valid for $|2x| < 1$ , i.e. $|x| < \tfrac{1}{2}$ . (The factor $1 + x$ is a polynomial and imposes no condition.)

A quotient

Expand $\dfrac{1 + 2x}{(1 - x)^2}$ up to and including the term in $x^3$ , and state the range of validity.

Solution

$\dfrac{1 + 2x}{(1 - x)^2} = (1 + 2x)(1 - x)^{-2}$ . With $n = -2$ , $u = -x$ :

(1 - x)^{-2} = 1 + (-2)(-x) + \frac{(-2)(-3)}{2}x^2 + \frac{(-2)(-3)(-4)}{6}(-x)^3 + \cdots = 1 + 2x + 3x^2 + 4x^3 + \cdots

Multiply by $(1 + 2x)$ :

	$1$	$2x$	$3x^2$	$4x^3$
$\times 1$	$1$	$2x$	$3x^2$	$4x^3$
$\times 2x$	$2x$	$4x^2$	$6x^3$	(not needed)

Collect: $1 + (2 + 2)x + (3 + 4)x^2 + (4 + 6)x^3$ .

\frac{1 + 2x}{(1 - x)^2} = 1 + 4x + 7x^2 + 10x^3 + \cdots, \qquad |x| < 1.

A product needing a factor taken out

Expand $(2 - x)(1 + 3x)^{-1/3}$ up to and including the term in $x^2$ .

Solution

With $n = -\tfrac{1}{3}$ and $u = 3x$ :

(1 + 3x)^{-1/3} = 1 + \left(-\tfrac{1}{3}\right)(3x) + \frac{\left(-\tfrac{1}{3}\right)\left(-\tfrac{4}{3}\right)}{2}(3x)^2 + \cdots = 1 - x + \tfrac{2}{9} \cdot 9x^2 + \cdots = 1 - x + 2x^2 + \cdots

Multiply:

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

Valid for $|3x| < 1$ , i.e. $|x| < \tfrac{1}{3}$ .

Watch out

When multiplying, every term in the first bracket must meet every term in the second. A common slip in the quotient example is to forget $2x \times 3x^2 = 6x^3$ and get $4x^3$ as the final $x^3$ term. A grid prevents this.

Using partial fractions first

A fraction like $\dfrac{11 - x - x^2}{(2 + x)(1 - x)^2}$ could be expanded as a product of three series, but that is slow and error-prone. Splitting it into partial fractions gives three single-bracket expansions to add.

Expanding via partial fractions

Express the function in partial fractions.
Write each term as a constant times a bracket to a power, with the bracket in the form $(1 + kx)$ : for example $\dfrac{1}{2 + x} = \tfrac{1}{2}\left(1 + \tfrac{x}{2}\right)^{-1}$ and $\dfrac{3}{(1 - x)^2} = 3(1 - x)^{-2}$ .
Expand each to the required power.
Add like terms.
Validity: the narrowest of the individual ranges.

Partial fractions then expansion

Let $f(x) = \dfrac{11 - x - x^2}{(2 + x)(1 - x)^2}$ .

(a) Express $f(x)$ in partial fractions.

(b) Hence obtain the expansion of $f(x)$ in ascending powers of $x$ , up to and including the term in $x^2$ , and state the set of values of $x$ for which it is valid.

Solution

(a) $f(x) \equiv \dfrac{A}{2 + x} + \dfrac{B}{1 - x} + \dfrac{C}{(1 - x)^2}$ , so

11 - x - x^2 \equiv A(1 - x)^2 + B(2 + x)(1 - x) + C(2 + x).

$x = 1$ : $9 = 3C$ , so $C = 3$ . $\quad x = -2$ : $11 + 2 - 4 = 9A$ , so $A = 1$ . $\quad x^2$ coefficients: $-1 = A - B$ , so $B = 2$ .

f(x) = \frac{1}{2 + x} + \frac{2}{1 - x} + \frac{3}{(1 - x)^2}.

(b) Expand each term:

\frac{1}{2 + x} = \tfrac{1}{2}\left(1 + \tfrac{x}{2}\right)^{-1} = \tfrac{1}{2}\left(1 - \tfrac{x}{2} + \tfrac{x^2}{4} - \cdots\right) = \tfrac{1}{2} - \tfrac{1}{4}x + \tfrac{1}{8}x^2 - \cdots

\frac{2}{1 - x} = 2(1 - x)^{-1} = 2 + 2x + 2x^2 + \cdots

\frac{3}{(1 - x)^2} = 3(1 - x)^{-2} = 3(1 + 2x + 3x^2 + \cdots) = 3 + 6x + 9x^2 + \cdots

Add:

f(x) = \left(\tfrac{1}{2} + 2 + 3\right) + \left(-\tfrac{1}{4} + 2 + 6\right)x + \left(\tfrac{1}{8} + 2 + 9\right)x^2 + \cdots = \frac{11}{2} + \frac{31}{4}x + \frac{89}{8}x^2 + \cdots

The first expansion is valid for $|x| < 2$ ; the other two for $|x| < 1$ . All three hold when $|x| < 1$ .

Tip

A quick check: the constant term of the expansion must equal $f(0)$ . Here $f(0) = \dfrac{11}{2 \times 1} = \dfrac{11}{2}$ , which matches.

A quadratic factor

For a term like $\dfrac{x + 2}{x^2 + 2}$ , take out the $2$ from the denominator and expand $\left(1 + \tfrac{x^2}{2}\right)^{-1}$ in powers of $x^2$ , then multiply by the numerator.

Linear and quadratic factors

Given that $\dfrac{5x^2 + x + 6}{(x - 1)(x^2 + 2)} = \dfrac{4}{x - 1} + \dfrac{x + 2}{x^2 + 2}$ , find the expansion of this function in ascending powers of $x$ up to and including the term in $x^3$ , and state the range of validity.

Solution

First term. To get a bracket starting with $1$ , write $x - 1 = -(1 - x)$ :

\frac{4}{x - 1} = -4(1 - x)^{-1} = -4(1 + x + x^2 + x^3 + \cdots) = -4 - 4x - 4x^2 - 4x^3 - \cdots

Second term:

\frac{x + 2}{x^2 + 2} = (x + 2) \cdot \tfrac{1}{2}\left(1 + \tfrac{x^2}{2}\right)^{-1} = \tfrac{1}{2}(x + 2)\left(1 - \tfrac{x^2}{2} + \cdots\right)

= \tfrac{1}{2}\left(2 + x - x^2 - \tfrac{1}{2}x^3 + \cdots\right) = 1 + \tfrac{1}{2}x - \tfrac{1}{2}x^2 - \tfrac{1}{4}x^3 + \cdots

Add:

-3 - \frac{7}{2}x - \frac{9}{2}x^2 - \frac{17}{4}x^3 + \cdots

Validity: $|x| < 1$ for the first series and $\left|\tfrac{x^2}{2}\right| < 1$ , i.e. $|x| < \sqrt{2}$ , for the second. Both hold for $|x| < 1$ .

Watch out

$\dfrac{4}{x - 1}$ is not $4(1 + x)^{-1}$ or $4(1 - x)^{-1}$ . Factor out the $-1$ : $x - 1 = -(1 - x)$ , so the term is $-4(1 - x)^{-1}$ . Getting the sign wrong here flips every coefficient of that piece.

Unknown constants

A missing term

In the expansion of $(1 + ax)(1 - 2x)^{-1/2}$ in ascending powers of $x$ , the coefficient of $x^2$ is zero. Find $a$ , and the coefficient of $x$ .

Solution

(1 - 2x)^{-1/2} = 1 + \left(-\tfrac{1}{2}\right)(-2x) + \frac{\left(-\tfrac{1}{2}\right)\left(-\tfrac{3}{2}\right)}{2}(-2x)^2 + \cdots = 1 + x + \tfrac{3}{2}x^2 + \cdots

(1 + ax)\left(1 + x + \tfrac{3}{2}x^2\right) = 1 + (1 + a)x + \left(\tfrac{3}{2} + a\right)x^2 + \cdots

The $x^2$ coefficient is zero, so $a = -\tfrac{3}{2}$ . The coefficient of $x$ is then $1 + a = -\tfrac{1}{2}$ .

Approximations with a clever choice of $x$

The skill is to choose a small $x$ that turns the expression into something related to the number you want.

Approximating a surd

(a) Show that, for small $x$ , $\sqrt{\dfrac{1 + x}{1 - x}} \approx 1 + x + \tfrac{1}{2}x^2 + \tfrac{1}{2}x^3$ .

(b) By substituting $x = 0.02$ , find an approximation to $\sqrt{51}$ , giving your answer to 5 decimal places.

Solution

(a) $\sqrt{\dfrac{1 + x}{1 - x}} = (1 + x)^{1/2}(1 - x)^{-1/2}$ .

(1 + x)^{1/2} = 1 + \tfrac{1}{2}x - \tfrac{1}{8}x^2 + \tfrac{1}{16}x^3 + \cdots

(1 - x)^{-1/2} = 1 + \tfrac{1}{2}x + \tfrac{3}{8}x^2 + \tfrac{5}{16}x^3 + \cdots

Multiply, collecting up to $x^3$ :

$x$ : $\tfrac{1}{2} + \tfrac{1}{2} = 1$
$x^2$ : $\tfrac{3}{8} + \tfrac{1}{4} - \tfrac{1}{8} = \tfrac{1}{2}$
$x^3$ : $\tfrac{5}{16} + \tfrac{1}{2} \cdot \tfrac{3}{8} - \tfrac{1}{8} \cdot \tfrac{1}{2} + \tfrac{1}{16} = \tfrac{5}{16} + \tfrac{3}{16} - \tfrac{1}{16} + \tfrac{1}{16} = \tfrac{1}{2}$

So the expression $\approx 1 + x + \tfrac{1}{2}x^2 + \tfrac{1}{2}x^3$ , valid for $|x| < 1$ .

(b) With $x = 0.02$ : $\dfrac{1.02}{0.98} = \dfrac{102}{98} = \dfrac{51}{49}$ , so $\sqrt{\dfrac{51}{49}} = \dfrac{\sqrt{51}}{7}$ .

\frac{\sqrt{51}}{7} \approx 1 + 0.02 + 0.0002 + 0.000004 = 1.020204

\sqrt{51} \approx 7 \times 1.020204 = 7.141428 = 7.14143 \ \text{(5 d.p.)}.

(The true value is $7.1414284\ldots$ .)

Common mistakes

Watch out

Stating only one range of validity. When several series are combined, the answer is valid only where all of them are. Give the narrowest range, and say why.

Watch out

Expanding too few terms of a factor. To get the $x^3$ term of a product of two series, each series needs terms up to $x^3$ , because $1 \times x^3$ contributes.

Watch out

Adding instead of multiplying. $(1 + x)\sqrt{1 - 2x}$ is a product; $\dfrac{2}{1 - x} + \dfrac{3}{(1 - x)^2}$ is a sum. Read the structure before expanding.

Exam tip

"Hence" after partial fractions means you must expand the partial fractions, not the original fraction. Using another method can score zero for that part.
Show each separate expansion with its own simplified coefficients before adding; a method mark is usually given for "expanding one term correctly up to $x^2$ ".
Check the constant term against $f(0)$ : it takes five seconds and catches sign errors in the partial fractions.
For an approximation, show the substitution, the link to the required number (such as $\sqrt{51/49} = \sqrt{51}/7$ ), and give the accuracy asked for.

Summary

Products: expand each factor to the required power, multiply, discard higher powers.
Quotients: rewrite as a product with a negative power.
Rational functions: partial fractions first, then expand each term and add.
Bring each term to the form $c(1 + kx)^n$ before expanding; watch $\dfrac{1}{x - a} = -\dfrac{1}{a}\left(1 - \tfrac{x}{a}\right)^{-1}$ .
The combined expansion is valid where every series used is valid.
Check: the constant term equals $f(0)$ .

Practice

Question

Expand $\dfrac{3 + x}{(1 + x)^2}$ in ascending powers of $x$ up to the term in $x^2$ , and state the range of validity.
Expand $\dfrac{1 + x}{\sqrt{4 - x}}$ up to the term in $x^2$ , and state the range of validity.
Expand $\dfrac{\sqrt{1 - x}}{(1 + x)^2}$ up to the term in $x^2$ .
(a) Express $\dfrac{4x^2 + 5x + 3}{(2x + 1)(x + 1)^2}$ in partial fractions. (b) Hence expand it up to the term in $x^2$ and state the range of validity.
Given $\dfrac{5x^2 - 3x + 1}{(2x + 1)(x^2 + 1)} = \dfrac{3}{2x + 1} + \dfrac{x - 2}{x^2 + 1}$ , expand the function up to the term in $x^3$ .
The expansion of $\dfrac{a + bx}{1 + 2x}$ in ascending powers of $x$ begins $3 - 5x$ . Find $a$ and $b$ , and the coefficient of $x^2$ .
Express $\dfrac{2x^2 + 5x + 1}{(x + 1)(x + 2)}$ in partial fractions, and hence expand it up to the term in $x^2$ , stating the range of validity.
In the expansion of $\dfrac{(1 + ax)^{1/2}}{1 - 3x}$ in ascending powers of $x$ , the coefficient of $x$ is zero. Find $a$ , the coefficient of $x^2$ , and the range of validity.
Expand $(8 + x)^{1/3}$ up to the term in $x^2$ , and use it to find an approximation to $\sqrt[3]{8.24}$ , giving your answer to 4 decimal places.

Answers

$(1 + x)^{-2} = 1 - 2x + 3x^2 - \cdots$ . Then $(3 + x)(1 - 2x + 3x^2) = 3 - 6x + 9x^2 + x - 2x^2 = 3 - 5x + 7x^2$ . Valid for $|x| < 1$ .
$(4 - x)^{-1/2} = \tfrac{1}{2}\left(1 - \tfrac{x}{4}\right)^{-1/2} = \tfrac{1}{2}\left(1 + \tfrac{x}{8} + \tfrac{3x^2}{128}\right)$ . Multiply by $(1 + x)$ : $\tfrac{1}{2}\left(1 + \tfrac{9}{8}x + \left(\tfrac{3}{128} + \tfrac{1}{8}\right)x^2\right) = \tfrac{1}{2} + \tfrac{9}{16}x + \tfrac{19}{256}x^2$ . Valid for $|x| < 4$ .
$\sqrt{1 - x} = 1 - \tfrac{1}{2}x - \tfrac{1}{8}x^2$ and $(1 + x)^{-2} = 1 - 2x + 3x^2$ . Product: $x$ : $-2 - \tfrac{1}{2} = -\tfrac{5}{2}$ ; $x^2$ : $3 + 1 - \tfrac{1}{8} = \tfrac{31}{8}$ . Answer $1 - \tfrac{5}{2}x + \tfrac{31}{8}x^2$ , valid for $|x| < 1$ .
(a) $\dfrac{6}{2x + 1} - \dfrac{1}{x + 1} - \dfrac{2}{(x + 1)^2}$ (see the partial fractions note). (b) $6(1 + 2x)^{-1} = 6 - 12x + 24x^2$ ; $-(1 + x)^{-1} = -1 + x - x^2$ ; $-2(1 + x)^{-2} = -2 + 4x - 6x^2$ . Sum: $3 - 7x + 17x^2$ . Valid for $|x| < \tfrac{1}{2}$ . (Check: $f(0) = \tfrac{3}{1} = 3$ .)
$3(1 + 2x)^{-1} = 3 - 6x + 12x^2 - 24x^3$ . $(x - 2)(1 + x^2)^{-1} = (x - 2)(1 - x^2) = -2 + x + 2x^2 - x^3$ . Sum: $1 - 5x + 14x^2 - 25x^3$ , valid for $|x| < \tfrac{1}{2}$ .
$(a + bx)(1 - 2x + 4x^2 - \cdots) = a + (b - 2a)x + (4a - 2b)x^2 + \cdots$ . So $a = 3$ , $b - 6 = -5$ , $b = 1$ . Coefficient of $x^2$ : $12 - 2 = 10$ .
$2 - \dfrac{2}{x + 1} + \dfrac{1}{x + 2}$ . $-2(1 + x)^{-1} = -2 + 2x - 2x^2$ ; $\tfrac{1}{2}\left(1 + \tfrac{x}{2}\right)^{-1} = \tfrac{1}{2} - \tfrac{1}{4}x + \tfrac{1}{8}x^2$ . Total: $\tfrac{1}{2} + \tfrac{7}{4}x - \tfrac{15}{8}x^2$ , valid for $|x| < 1$ .
$(1 + ax)^{1/2} = 1 + \tfrac{a}{2}x - \tfrac{a^2}{8}x^2$ and $(1 - 3x)^{-1} = 1 + 3x + 9x^2$ . Coefficient of $x$ : $3 + \tfrac{a}{2} = 0$ , so $a = -6$ . Coefficient of $x^2$ : $9 + 3 \cdot \tfrac{a}{2} - \tfrac{a^2}{8} = 9 - 9 - \tfrac{36}{8} = -\tfrac{9}{2}$ . Validity: $|6x| < 1$ and $|3x| < 1$ , so $|x| < \tfrac{1}{6}$ .
$(8 + x)^{1/3} = 2\left(1 + \tfrac{x}{8}\right)^{1/3} = 2\left(1 + \tfrac{x}{24} - \tfrac{x^2}{576}\right) = 2 + \tfrac{x}{12} - \tfrac{x^2}{288}$ , valid for $|x| < 8$ . With $x = 0.24$ : $2 + 0.02 - 0.0002 = 2.0198$ (true value $2.01980\ldots$ ).

Products: expand, multiply, truncate

Using partial fractions first

A quadratic factor

Unknown constants

Approximations with a clever choice of xxx

Common mistakes

Summary

Practice

Console

Approximations with a clever choice of $x$