The binomial expansion for rational powers

The generalised binomial expansion writes $(1+x)^n$ as a power series when $n$ is rational, including fractional and negative values:

$$\boxed{(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots}$$

If $n$ is not a non-negative integer, this series does not terminate and is valid only when

$$\boxed{|x|<1.}$$

This restriction is part of the answer whenever an expansion is requested.

Prerequisites

You should be able to:

• apply the laws of indices
• manipulate fractions and factorise expressions
• use the binomial expansion for positive integer powers
• understand what it means for an infinite geometric series to converge
• use an inequality to describe an interval

The general term and how to generate coefficients

For rational $n$, define

$$\binom nr=\frac{n(n-1)(n-2)\cdots(n-r+1)}{r!}.$$

Then

$$(1+x)^n=\sum_{r=0}^{\infty}\binom nr x^r, \qquad |x|<1.$$

The first few terms are

$$T_{r+1}=\binom nr x^r,$$

so the first term has $r=0$. For $r\geq 0$, consecutive terms satisfy

$$\boxed{T_{r+2}=T_{r+1}\times\frac{n-r}{r+1}x.}$$

This multiplier is often the quickest and safest way to continue an expansion.

Worked example 1: a fractional power

Expand $(1+x)^{1/2}$ up to and including the term in $x^4$.

Substitute $n=\frac12$:

$$\begin{aligned} (1+x)^{1/2} &=1+\frac12x +\frac{\frac12(-\frac12)}{2}x^2 +\frac{\frac12(-\frac12)(-\frac32)}{6}x^3\\ &\quad+\frac{\frac12(-\frac12)(-\frac32)(-\frac52)}{24}x^4+\cdots\\ &=1+\frac12x-\frac18x^2+\frac1{16}x^3-\frac5{128}x^4+\cdots, \qquad |x|<1. \end{aligned}$$

The factors in the numerator descend by $1$:

$$\frac12,\ -\frac12,\ -\frac32,\ -\frac52,\ldots$$

They do not repeatedly halve.

Worked example 2: a negative integer power

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

Here $n=-3$ and the small quantity replacing $x$ in the formula is $-2x$:

$$\begin{aligned} (1-2x)^{-3} &=1+(-3)(-2x) +\frac{(-3)(-4)}{2}(-2x)^2\\ &\quad+\frac{(-3)(-4)(-5)}{6}(-2x)^3+\cdots\\ &=1+6x+24x^2+80x^3+\cdots. \end{aligned}$$

The condition is

$$|-2x|<1\quad\Longrightarrow\quad |x|<\frac12.$$

The validity condition uses the whole small term

For $(1-2x)^{-3}$, the formula’s small term is $-2x$, not $x$. Therefore the condition is $|-2x|<1$, not merely $|x|<1$.

Rewriting into the form $(1+u)^n$

The formula applies directly to $(1+u)^n$. If the constant term inside the bracket is not $1$, factor it out first.

$$(a+bx)^n=a^n\left(1+\frac ba x\right)^n.$$

The validity condition is then

$$\left|\frac ba x\right|<1.$$

Worked example 3: factor out the constant

Expand $(4+x)^{-1/2}$ up to the term in $x^3$, and state the interval of validity.

First make the constant inside the bracket equal to $1$:

$$(4+x)^{-1/2} =4^{-1/2}\left(1+\frac{x}{4}\right)^{-1/2} =\frac12\left(1+\frac{x}{4}\right)^{-1/2}.$$

Using $n=-\frac12$ and $u=\frac{x}{4}$,

$$\begin{aligned} (1+u)^{-1/2} &=1-\frac12u+\frac38u^2-\frac5{16}u^3+\cdots. \end{aligned}$$

Hence

$$\begin{aligned} (4+x)^{-1/2} &=\frac12\left(1-\frac12\frac{x}{4} +\frac38\frac{x^2}{16} -\frac5{16}\frac{x^3}{64}+\cdots\right)\\ &=\frac12-\frac{x}{16}+\frac{3x^2}{256}-\frac{5x^3}{2048}+\cdots. \end{aligned}$$

For validity,

$$\left|\frac{x}{4}\right|<1 \quad\Longrightarrow\quad \boxed{-4<x<4}.$$

Do not lose the outside factor

The factor $4^{-1/2}=\frac12$ multiplies every term of the expansion, not just the first term.

Self-check 1

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

Answer

With $n=-2$ and $u=3x$,

$$\begin{aligned} (1+3x)^{-2} &=1-2(3x)+\frac{(-2)(-3)}2(3x)^2 +\frac{(-2)(-3)(-4)}6(3x)^3+\cdots\\ &=1-6x+27x^2-108x^3+\cdots. \end{aligned}$$

Also,

$|3x|<1\quad\Longrightarrow\quad -\frac13<x<\frac13.$

Finding a particular coefficient

The term containing $u^r$ is

$$T_{r+1}=\frac{n(n-1)\cdots(n-r+1)}{r!}u^r.$$

When $u$ contains $x$, include its coefficient and power before extracting the coefficient of $x^r$.

Worked example 4: find one coefficient directly

Find the coefficient of $x^4$ in $(1-3x)^{2/3}$.

The $x^4$ term corresponds to $r=4$:

$$\begin{aligned} T_5 &=\frac{\frac23(-\frac13)(-\frac43)(-\frac73)}{4!}(-3x)^4\\ &=-\frac7{243}\cdot81x^4\\ &=-\frac73x^4. \end{aligned}$$

Therefore the coefficient is

$$\boxed{-\frac73}.$$

A useful sign check is to count signs in both the binomial coefficient and $(-3)^4$ separately.

Worked example 5: a coefficient in a product

Find the coefficient of $x^3$ in

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

First expand only as far as needed:

$$(1-x)^{-1/2} =1+\frac12x+\frac38x^2+\frac5{16}x^3+\cdots.$$

An $x^3$ term in the product can arise in two ways:

$$1\times\frac5{16}x^3, \qquad 2x\times\frac38x^2.$$

Thus the coefficient is

$$\frac5{16}+2\left(\frac38\right) =\frac5{16}+\frac{12}{16} =\boxed{\frac{17}{16}}.$$

Binomial approximations

If $|x|$ is small, higher powers $x^2,x^3,\ldots$ are usually much smaller than $x$. Truncating the expansion can therefore give an efficient approximation:

$$(1+x)^n\approx1+nx$$

for a first order approximation, or

$$(1+x)^n\approx1+nx+\frac{n(n-1)}2x^2$$

for a second order approximation.

The condition $|x|<1$ guarantees convergence, but it does not guarantee that a short truncation is highly accurate. Accuracy improves when $|x|$ is closer to $0$ and when more terms are retained.

Worked example 6: approximate a square root

Use the first four terms of a binomial expansion to approximate $\sqrt{1.04}$.

Set $x=0.04$ in

$$(1+x)^{1/2}=1+\frac12x-\frac18x^2+\frac1{16}x^3+\cdots.$$

Then

$$\begin{aligned} \sqrt{1.04} &\approx1+\frac12(0.04)-\frac18(0.04)^2+\frac1{16}(0.04)^3\\ &=1+0.02-0.0002+0.000004\\ &=\boxed{1.019804}. \end{aligned}$$

Squaring the result gives approximately $1.04000016$, which is consistent with the requested value.

Worked example 7: rewrite a number strategically

Use a binomial expansion to approximate $\dfrac1{\sqrt{3.96}}$.

Choose the nearby convenient value $4$:

$$3.96=4(0.99)=4(1-0.01).$$

Therefore

$$\frac1{\sqrt{3.96}} =\frac12(1-0.01)^{-1/2}.$$

Using

$$(1+x)^{-1/2}=1-\frac12x+\frac38x^2-\frac5{16}x^3+\cdots$$

with $x=-0.01$ gives

$$\begin{aligned} \frac1{\sqrt{3.96}} &\approx\frac12\left(1+0.005+0.0000375+0.0000003125\right)\\ &=\boxed{0.50251890625}. \end{aligned}$$

The important modelling choice was to create a small quantity, $-0.01$. Writing $3.96=3(1+0.32)$ would converge, but its early terms would usually give a poorer approximation.

Self-check 2

1. Use terms up to $x^2$ in $(1+x)^{-1}$ to approximate $1/1.02$.
2. Explain why the same truncation is expected to work better for $1/1.02$ than for $1/1.8$.

Answers

Since

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

putting $x=0.02$ gives

$$\frac1{1.02}\approx1-0.02+0.02^2=\boxed{0.9804}.$$

For $1/1.02$, $|x|=0.02$; for $1/1.8$, $|x|=0.8$. The omitted powers decrease much faster when $|x|=0.02$, so the same number of retained terms gives a much smaller error.

Combining two infinite expansions

Sometimes a rational expression is best handled by expanding its factors separately and then multiplying, retaining only terms that can affect the required order.

Worked example 8: multiply truncated series

Find the expansion of

$$\frac{1}{(1-x)\sqrt{1+x}}$$

up to and including the term in $x^2$.

Use

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

and

$$(1+x)^{-1/2}=1-\frac12x+\frac38x^2+\cdots.$$

Multiplying and collecting only powers up to $x^2$:

$$\begin{aligned} (1+x+x^2)\left(1-\frac12x+\frac38x^2\right) &=1+\left(1-\frac12\right)x\\ &\quad+\left(1-\frac12+\frac38\right)x^2+\cdots\\ &=1+\frac12x+\frac78x^2+\cdots. \end{aligned}$$

Both original expansions require $|x|<1$, so the product expansion is valid for

$$\boxed{|x|<1.}$$

Common misconceptions

Four errors to avoid

1. Using factorial notation for a fractional top number. Write $n(n-1)\cdots$, since an expression such as $(\frac12)!$ is not part of the A-level formula.
2. Forgetting that the series is infinite. A non-integer or negative power does not produce a final term, so include $+\cdots$.
3. Using the wrong validity condition. For $(1+u)^n$, require $|u|<1$, then solve this inequality for $x$.
4. Expanding before normalising the bracket. Rewrite $(a+bx)^n$ as $a^n(1+\frac ba x)^n$ and keep the factor $a^n$.

Mixed self-check

1. Write the first four terms of $(1-4x)^{1/3}$ and state its interval of validity.
2. Find the coefficient of $x^3$ in $(2+x)^{-2}$.
3. Find the coefficient of $x^2$ in $(1+x)^3(1-2x)^{-1}$.

Answers

1. With $n=\frac13$ and $u=-4x$,

$$\begin{aligned} (1-4x)^{1/3} &=1+\frac13(-4x) +\frac{\frac13(-\frac23)}2(-4x)^2\\ &\quad+\frac{\frac13(-\frac23)(-\frac53)}6(-4x)^3+\cdots\\ &=1-\frac43x-\frac{16}{9}x^2-\frac{320}{81}x^3+\cdots. \end{aligned}$$

The condition $|-4x|<1$ gives $-\frac14<x<\frac14$.

2. Rewrite

$$(2+x)^{-2}=\frac14\left(1+\frac x2\right)^{-2}.$$

The coefficient of $u^3$ in $(1+u)^{-2}$ is

$$\frac{(-2)(-3)(-4)}6=-4.$$

After substituting $u=x/2$ and multiplying by $1/4$, the coefficient is

$$\frac14(-4)\left(\frac12\right)^3=\boxed{-\frac18}.$$

3. Up to $x^2$,

$$(1+x)^3=1+3x+3x^2+\cdots$$

and

$$(1-2x)^{-1}=1+2x+4x^2+\cdots.$$

The coefficient of $x^2$ is

$$1(4)+3(2)+3(1)=\boxed{13}.$$

Summary

For rational $n$,

$$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2 +\frac{n(n-1)(n-2)}{3!}x^3+\cdots, \qquad |x|<1.$$

To use it reliably:

1. rewrite the expression as $A(1+u)^n$
2. substitute $n$ and the complete expression $u$
3. simplify one term at a time
4. state and solve $|u|<1$
5. retain only as many terms as the question requires

Next, use these expansions in series in modelling or revisit the general term for finite binomial expansions. The approximation ideas also support later work with numerical methods.