1.4.4 Validity of the expansion (A-level only)

What you'll learn

Why some binomial expansions are infinite and only valid for certain values of $x$ .
How to rewrite $a+bx)^n$ in the standard form $1+u)^n$ .
How to use $∣bxa∣<1\left|\frac{bx}{a}\right|<1$ to find the interval of validity.
How validity affects approximations made using the first few terms of an expansion.

Prerequisite: the general binomial expansion

For any real or rational power $n$ , the general binomial expansion begins

(1+u)n=1+nu+n(n−1)2!u2+n(n−1)(n−2)3!u3+⋯(1+u)^n = 1+nu+\frac{n(n-1)}{2!}u^2 +\frac{n(n-1)(n-2)}{3!}u^3+\cdots

Here, $u$ is the expression being added to 1. The dots show that the expansion may continue indefinitely.

If $n$ is a non-negative integer, such as 2 or 5, one of the factors in the coefficients eventually becomes zero. The expansion therefore terminates and is a finite polynomial.

If $n$ is not a non-negative integer, for example $−12-\frac12$ , $13\frac13$ or $- 2$ , the expansion generally continues forever.

Definition

Infinite binomial series

An infinite binomial series is the non-terminating expansion of $1+u)^n$ when the power $n$ is not a non-negative integer.

Example

Recognising a non-terminating expansion

Expand $(1+u)−12(1+u)^{-\frac12}$ up to and including the term in $u^3$ .

Substitute $n=−12n=-\frac12$ into the general expansion:
$(1+u)−12=1−12u+(−12)(−32)2u2+(−12)(−32)(−52)6u3+⋯(1+u)^{-\frac12} = 1-\frac12u +\frac{\left(-\frac12\right)\left(-\frac32\right)}{2}u^2 +\frac{\left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\right)}{6}u^3 +\cdots$
Simplify each coefficient:
$(−12)(−32)2=38,(−12)(−32)(−52)6=−516.\frac{\left(-\frac12\right)\left(-\frac32\right)}{2} =\frac38, \qquad \frac{\left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\right)}{6} =-\frac5{16}.$
Therefore,
$(1+u)−12=1−12u+38u2−516u3+⋯(1+u)^{-\frac12} = 1-\frac12u+\frac38u^2-\frac5{16}u^3+\cdots$
This does not terminate, so its validity must be considered.

What does “valid” mean?

An infinite series does not necessarily approach a finite value for every value of its variable.

A series is convergent when the sum of its terms approaches a finite limit. It is divergent when its partial sums do not approach a finite limit.

For the general binomial series of $1+u)^n$ , the expansion is guaranteed to converge to the original expression when

∣ u ∣ < 1.

The notation $∣ u ∣$ means the absolute value or magnitude of $u$ . Therefore, $∣ u ∣ < 1$ is equivalent to

- 1 < u < 1.

Key Idea

Validity condition

The infinite binomial expansion of $1+u)^n$ is valid for $∣ u ∣ < 1$ . Before giving an interval of validity, identify the entire expression playing the role of $u$ .

The inequality is strict, so the endpoints $u = - 1$ and $u = 1$ are not included in the standard interval of validity. Although some particular series may behave differently at an endpoint, OCR expects the condition $∣ u ∣ < 1$ here.

Rewriting $a+bx)^n$

The general formula applies directly to an expression beginning with 1. To expand $a+bx)^n$ , first factor out the constant $a$ :

a+bx=a(1+bxa).a+bx=a\left(1+\frac{bx}{a}\right).

Therefore,

(a+bx)n=an(1+bxa)n.(a+bx)^n = a^n\left(1+\frac{bx}{a}\right)^n.

The expression replacing $u$ is

u=bxa.u=\frac{bx}{a}.

So the expansion is valid when

∣bxa∣<1.\left|\frac{bx}{a}\right|<1.

Provided $a≠0a\neq0$ and $b≠0b\neq0$ , this can also be written as

∣x∣<∣ab∣.|x|<\left|\frac{a}{b}\right|.

Equivalently, the interval is

−∣ab∣<x<∣ab∣.-\left|\frac{a}{b}\right|<x<\left|\frac{a}{b}\right|.

Number line showing that the expansion is valid strictly between minus the absolute value of a over b and the absolute value of a over b

Common Mistake

Using the unfactored term

For $a+bx)^n$ , the condition is not generally $∣ b x ∣ < 1$ . After factoring out $a$ , the quantity replacing $u$ is $bxa\frac{bx}{a}$ , so the condition is $∣bxa∣<1\left|\frac{bx}{a}\right|<1$ .

Example

Finding an interval of validity

Find the interval of validity of the binomial expansion of $(2−3x)−12(2-3x)^{-\frac12}$ .

Factor out 2 so that the bracket begins with 1:
$(2−3x)−12=2−12(1−32x)−12.(2-3x)^{-\frac12} = 2^{-\frac12}\left(1-\frac32x\right)^{-\frac12}.$
Identify the expression replacing $u$ :
$u=−32x.u=-\frac32x.$
Apply the condition $∣ u ∣ < 1$ :
$∣−32x∣<1.\left|-\frac32x\right|<1.$
Use $∣−32x∣=32∣x∣\left|-\frac32x\right|=\frac32|x|$ and solve:
$32∣x∣<1⇒∣x∣<23.\frac32|x|<1 \quad\Rightarrow\quad |x|<\frac23.$
Hence the expansion is valid for
$−23<x<23.-\frac23<x<\frac23.$

Expanding and stating validity together

A complete answer may require both the expansion and its interval of validity. The interval comes from the expression substituted for $u$ , not from the number of terms you calculate.

Example

Expanding with a validity condition

Expand $4+x)^{-1}$ up to and including the term in $x^3$ , and state the values of $x$ for which the expansion is valid.

Factor out 4:
$(4+x)−1=14(1+x4)−1.(4+x)^{-1} = \frac14\left(1+\frac{x}{4}\right)^{-1}.$
Use the expansion
$(1+u)−1=1−u+u2−u3+⋯(1+u)^{-1}=1-u+u^2-u^3+\cdots$
with $u=x4u=\frac{x}{4}$ :
$).(4+x)^{-1} = \frac14\left( 1-\frac{x}{4}+\frac{x^2}{16}-\frac{x^3}{64}+\cdots \right).$
Multiply every term by $14\frac14$ :
$(4+x)−1=14−x16+x264−x3256+⋯(4+x)^{-1} = \frac14-\frac{x}{16}+\frac{x^2}{64}-\frac{x^3}{256}+\cdots$
Apply the validity condition to $u=x4u=\frac{x}{4}$ :
$∣x4∣<1⇒∣x∣<4.\left|\frac{x}{4}\right|<1 \quad\Rightarrow\quad |x|<4.$
Therefore, the stated expansion is valid for
$- 4 < x < 4.$

Tip

A quick interval check

For an expansion centred at $x = 0$ , the interval obtained from $∣bxa∣<1\left|\frac{bx}{a}\right|<1$ should be symmetric about zero.

Validity and approximation

Using only the first few terms of an infinite expansion gives a truncated series. This provides an approximation rather than an exact equality.

Values of $x$ must lie inside the interval of validity. In general, the approximation is better when $∣bxa∣\left|\frac{bx}{a}\right|$ is small, because later powers then decrease rapidly.

Example

Checking whether an approximation is valid

Suppose the expansion of $(2−3x)−12(2-3x)^{-\frac12}$ is used to estimate its value when $x=12x=\frac12$ .

The expansion is valid when
$−23<x<23.-\frac23<x<\frac23.$
Compare the proposed value with this interval:
$−23<12<23.-\frac23<\frac12<\frac23.$
Therefore, $x=12x=\frac12$ lies inside the interval, so the infinite expansion is valid there. However,
$∣−32⋅12∣=34,\left|-\frac32\cdot\frac12\right|=\frac34,$
which is fairly close to 1. A short truncation may therefore converge less quickly than it would for a value of $x$ nearer zero.

Common Mistake

Inside the interval does not mean exact

The condition $∣bxa∣<1\left|\frac{bx}{a}\right|<1$ guarantees convergence of the complete infinite series. If you retain only a few terms, your result is still an approximation.

Exam technique

In the exam

Factor out the constant term so that the bracket is in the form $1+u)^n$ .
Identify the whole expression corresponding to $u$ , including its sign and any numerical factor.
Apply $∣ u ∣ < 1$ , solve the inequality carefully, and exclude both endpoints.
State the final answer as either an absolute-value inequality or an open interval.
Check that the interval is symmetric about zero for an expression of the form $a+bx)^n$ .

Self review

Check yourself

What expression replaces $u$ when $5-2x)^n$ is written in the form $5^n(1+u)^n$ ?
For which values of $x$ is the infinite binomial expansion of $(3+4x)−12(3+4x)^{-\frac12}$ valid?
Why can a truncated binomial expansion be valid at a particular value of $x$ but still give only an approximation?

What you'll learn

Why some binomial expansions are infinite and only valid for certain values of $x$ .
How to rewrite $a+bx)^n$ in the standard form $1+u)^n$ .
How to use $∣bxa∣<1\left|\frac{bx}{a}\right|<1$ to find the interval of validity.
How validity affects approximations made using the first few terms of an expansion.

Prerequisite: the general binomial expansion

For any real or rational power $n$ , the general binomial expansion begins

(1+u)n=1+nu+n(n−1)2!u2+n(n−1)(n−2)3!u3+⋯(1+u)^n = 1+nu+\frac{n(n-1)}{2!}u^2 +\frac{n(n-1)(n-2)}{3!}u^3+\cdots

Here, $u$ is the expression being added to 1. The dots show that the expansion may continue indefinitely.

If $n$ is a non-negative integer, such as 2 or 5, one of the factors in the coefficients eventually becomes zero. The expansion therefore terminates and is a finite polynomial.

If $n$ is not a non-negative integer, for example $−12-\frac12$ , $13\frac13$ or $- 2$ , the expansion generally continues forever.

Definition

Infinite binomial series

An infinite binomial series is the non-terminating expansion of $1+u)^n$ when the power $n$ is not a non-negative integer.

Example

Recognising a non-terminating expansion

Expand $(1+u)−12(1+u)^{-\frac12}$ up to and including the term in $u^3$ .

Substitute $n=−12n=-\frac12$ into the general expansion:
$(1+u)−12=1−12u+(−12)(−32)2u2+(−12)(−32)(−52)6u3+⋯(1+u)^{-\frac12} = 1-\frac12u +\frac{\left(-\frac12\right)\left(-\frac32\right)}{2}u^2 +\frac{\left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\right)}{6}u^3 +\cdots$
Simplify each coefficient:
$(−12)(−32)2=38,(−12)(−32)(−52)6=−516.\frac{\left(-\frac12\right)\left(-\frac32\right)}{2} =\frac38, \qquad \frac{\left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\right)}{6} =-\frac5{16}.$
Therefore,
$(1+u)−12=1−12u+38u2−516u3+⋯(1+u)^{-\frac12} = 1-\frac12u+\frac38u^2-\frac5{16}u^3+\cdots$
This does not terminate, so its validity must be considered.

What does “valid” mean?

An infinite series does not necessarily approach a finite value for every value of its variable.

A series is convergent when the sum of its terms approaches a finite limit. It is divergent when its partial sums do not approach a finite limit.

For the general binomial series of $1+u)^n$ , the expansion is guaranteed to converge to the original expression when

∣ u ∣ < 1.

The notation $∣ u ∣$ means the absolute value or magnitude of $u$ . Therefore, $∣ u ∣ < 1$ is equivalent to

- 1 < u < 1.

Key Idea

Validity condition

The infinite binomial expansion of $1+u)^n$ is valid for $∣ u ∣ < 1$ . Before giving an interval of validity, identify the entire expression playing the role of $u$ .

Rewriting $a+bx)^n$

The general formula applies directly to an expression beginning with 1. To expand $a+bx)^n$ , first factor out the constant $a$ :

a+bx=a(1+bxa).a+bx=a\left(1+\frac{bx}{a}\right).

Therefore,

(a+bx)n=an(1+bxa)n.(a+bx)^n = a^n\left(1+\frac{bx}{a}\right)^n.

The expression replacing $u$ is

u=bxa.u=\frac{bx}{a}.

So the expansion is valid when

∣bxa∣<1.\left|\frac{bx}{a}\right|<1.

Provided $a≠0a\neq0$ and $b≠0b\neq0$ , this can also be written as

∣x∣<∣ab∣.|x|<\left|\frac{a}{b}\right|.

Equivalently, the interval is

−∣ab∣<x<∣ab∣.-\left|\frac{a}{b}\right|<x<\left|\frac{a}{b}\right|.

Number line showing that the expansion is valid strictly between minus the absolute value of a over b and the absolute value of a over b

Common Mistake

Using the unfactored term

Example

Finding an interval of validity

Find the interval of validity of the binomial expansion of $(2−3x)−12(2-3x)^{-\frac12}$ .

Factor out 2 so that the bracket begins with 1:
$(2−3x)−12=2−12(1−32x)−12.(2-3x)^{-\frac12} = 2^{-\frac12}\left(1-\frac32x\right)^{-\frac12}.$
Identify the expression replacing $u$ :
$u=−32x.u=-\frac32x.$
Apply the condition $∣ u ∣ < 1$ :
$∣−32x∣<1.\left|-\frac32x\right|<1.$
Use $∣−32x∣=32∣x∣\left|-\frac32x\right|=\frac32|x|$ and solve:
$32∣x∣<1⇒∣x∣<23.\frac32|x|<1 \quad\Rightarrow\quad |x|<\frac23.$
Hence the expansion is valid for
$−23<x<23.-\frac23<x<\frac23.$

Expanding and stating validity together

A complete answer may require both the expansion and its interval of validity. The interval comes from the expression substituted for $u$ , not from the number of terms you calculate.

Example

Expanding with a validity condition

Expand $4+x)^{-1}$ up to and including the term in $x^3$ , and state the values of $x$ for which the expansion is valid.

Factor out 4:
$(4+x)−1=14(1+x4)−1.(4+x)^{-1} = \frac14\left(1+\frac{x}{4}\right)^{-1}.$
Use the expansion
$(1+u)−1=1−u+u2−u3+⋯(1+u)^{-1}=1-u+u^2-u^3+\cdots$
with $u=x4u=\frac{x}{4}$ :
$).(4+x)^{-1} = \frac14\left( 1-\frac{x}{4}+\frac{x^2}{16}-\frac{x^3}{64}+\cdots \right).$
Multiply every term by $14\frac14$ :
$(4+x)−1=14−x16+x264−x3256+⋯(4+x)^{-1} = \frac14-\frac{x}{16}+\frac{x^2}{64}-\frac{x^3}{256}+\cdots$
Apply the validity condition to $u=x4u=\frac{x}{4}$ :
$∣x4∣<1⇒∣x∣<4.\left|\frac{x}{4}\right|<1 \quad\Rightarrow\quad |x|<4.$
Therefore, the stated expansion is valid for
$- 4 < x < 4.$

Tip

A quick interval check

For an expansion centred at $x = 0$ , the interval obtained from $∣bxa∣<1\left|\frac{bx}{a}\right|<1$ should be symmetric about zero.

Validity and approximation

Using only the first few terms of an infinite expansion gives a truncated series. This provides an approximation rather than an exact equality.

Values of $x$ must lie inside the interval of validity. In general, the approximation is better when $∣bxa∣\left|\frac{bx}{a}\right|$ is small, because later powers then decrease rapidly.

Example

Checking whether an approximation is valid

Suppose the expansion of $(2−3x)−12(2-3x)^{-\frac12}$ is used to estimate its value when $x=12x=\frac12$ .

The expansion is valid when
$−23<x<23.-\frac23<x<\frac23.$
Compare the proposed value with this interval:
$−23<12<23.-\frac23<\frac12<\frac23.$
Therefore, $x=12x=\frac12$ lies inside the interval, so the infinite expansion is valid there. However,
$∣−32⋅12∣=34,\left|-\frac32\cdot\frac12\right|=\frac34,$
which is fairly close to 1. A short truncation may therefore converge less quickly than it would for a value of $x$ nearer zero.

Common Mistake

Inside the interval does not mean exact

The condition $∣bxa∣<1\left|\frac{bx}{a}\right|<1$ guarantees convergence of the complete infinite series. If you retain only a few terms, your result is still an approximation.

Exam technique

In the exam

Factor out the constant term so that the bracket is in the form $1+u)^n$ .
Identify the whole expression corresponding to $u$ , including its sign and any numerical factor.
Apply $∣ u ∣ < 1$ , solve the inequality carefully, and exclude both endpoints.
State the final answer as either an absolute-value inequality or an open interval.
Check that the interval is symmetric about zero for an expression of the form $a+bx)^n$ .

Self review

Check yourself

What expression replaces $u$ when $5-2x)^n$ is written in the form $5^n(1+u)^n$ ?
For which values of $x$ is the infinite binomial expansion of $(3+4x)−12(3+4x)^{-\frac12}$ valid?
Why can a truncated binomial expansion be valid at a particular value of $x$ but still give only an approximation?

What you'll learn

Prerequisite: the general binomial expansion

What does “valid” mean?

Rewriting $a+bx)^n$

Expanding and stating validity together

Validity and approximation

Review 1.4.4 Validity of the expansion (A-level only) by teaching Genie

1.4.4 Validity of the expansion (A-level only)

What you'll learn

Prerequisite: the general binomial expansion

What does “valid” mean?

Rewriting $a+bx)^n$

Expanding and stating validity together

Validity and approximation

Review 1.4.4 Validity of the expansion (A-level only) by teaching Genie

1.4.4 Validity of the expansion (A-level only)

What you'll learn

Prerequisite: the general binomial expansion

What does “valid” mean?

Rewriting (a+bx)n(a+bx)^n(a+bx)n

Expanding and stating validity together

Validity and approximation

Review 1.4.4 Validity of the expansion (A-level only) by teaching Genie

1.4.4 Validity of the expansion (A-level only) Revision Guide

1.4.4 Validity of the expansion (A-level only)

What you'll learn

Prerequisite: the general binomial expansion

What does “valid” mean?

Rewriting (a+bx)n(a+bx)^n(a+bx)n

Expanding and stating validity together

Validity and approximation

Review 1.4.4 Validity of the expansion (A-level only) by teaching Genie

1.4.4 Validity of the expansion (A-level only) Revision Guide

Rewriting $a+bx)^n$

Rewriting $a+bx)^n$