Applications of Binomial Expansion (Edexcel IGCSE Further Pure Maths)

Revision Note

Written by: Roger B

Reviewed by: Dan Finlay

Applications of Binomial Expansion

How can I use the binomial expansion with more complex expressions?

You may be asked to find a series expansion for an expression like $\frac{1 + x}{3 + 2 x}$
- Rewrite as a product, $\frac{1 + x}{3 + 2 x} = (1 + x) {(3 + 2 x)}^{- 1}$

Find the binomial expansion of ${(3 + 2 x)}^{- 1}$

${(3 + 2 x)}^{- 1} = \frac{1}{3} - \frac{2}{9} x + \frac{4}{27} x^{2} - \frac{8}{81} x^{3} + . . .$

Note this has only been expanded up to the $x^{3}$ term
Multiply that expansion by $(1 + x)$ and simplify

$\begin{array}{rcl} (1 + x) {(3 + 2 x)}^{- 1} & = & (1 + x) (\frac{1}{3} - \frac{2}{9} x + \frac{4}{27} x^{2} - \frac{8}{81} x^{3} + . . .) \\ = & \frac{1}{3} - \frac{2}{9} x + \frac{4}{27} x^{2} - \frac{8}{81} x^{3} + \frac{1}{3} x - \frac{2}{9} x^{2} + \frac{4}{27} x 3 - \frac{8}{81} x^{4} + . . . \\ = & \frac{1}{3} + \frac{1}{9} x - \frac{2}{27} x^{2} + \frac{4}{81} x^{3} - \frac{8}{81} x^{4} + . . . \\ = & \frac{1}{3} + \frac{1}{9} x - \frac{2}{27} x^{2} + \frac{4}{81} x^{3} - . . . \end{array}$

This is only valid up to the $x^{3}$ term
To get more terms we would have to start with more terms for ${(3 + 2 x)}^{- 1}$
$- \frac{8}{81} x^{4}$ is not the correct $x^{4}$ term for $\begin{array}{rcl} (1 + x) {(3 + 2 x)}^{- 1} \end{array}$ as there are more $x^{4}$ terms that were not found
Use the same process to find the expansion for something like $\frac{1 - 2 x}{\sqrt{2 + 3 x}}$
- Rewrite as a product, $\frac{1 - 2 x}{\sqrt{2 + 3 x}} = (1 - 2 x) {(2 + 3 x)}^{- \frac{1}{2}}$
- Find the binomial expansion of ${(2 + 3 x)}^{- \frac{1}{2}}$
- Multiply the expansion by $(1 - 2 x)$ and simplify

How can I use the binomial expansion to estimate a value?

The binomial expansion can be used to find estimates or approximations
- When $|x| < 1$ , higher powers of $x$ will be very small
- So even the first 3 or 4 terms of an expansion can form a good approximation
- The more terms used the closer the approximation will be to the true value
  - Also the closer to zero $x$ is, the better the approximation will be
For example, find an approximation for $\sqrt{0.96}$ using the expansion of ${(1 - x)}^{\frac{1}{2}}$
- Compare the value you are approximating to the expression being expanded
  - ${(1 - x)}^{\frac{1}{2}} = 0 . 96^{\frac{1}{2}}$
- Find the value of $x$ to use by solving the appropriate equation
  - $\begin{array}{rcl} 1 - x & = & 0.96 \\ x & = & 0.04 \end{array}$

Substitute this value of $x$ into the binomial expansion of ${(1 - x)}^{\frac{1}{2}}$
- ${(1 - x)}^{\frac{1}{2}} = 1 - \frac{1}{2} x - \frac{1}{8} x^{2} - . . .$
- So $\sqrt{0.96} \approx 1 - \frac{1}{2} (0.04) - \frac{1}{8} {(0.04)}^{2} = 0.9798$
- The true value of $\sqrt{0.96}$ is $0.97979589 . . .$
On the exam this is often used to approximate square roots
- It can also be used to approximate other things
- For example approximate the fraction $\frac{8}{125}$ using the binomial expansion of $\frac{1}{{(3 - x)}^{3}}$
  - $\frac{1}{{(3 - x)}^{3}} = \frac{8}{125} \Rightarrow \frac{1}{3 - x} = \frac{2}{5} \Rightarrow x = 0.5$
  - So substitute $x = 0.5$ into the expansion
Always check that the value of $x$ is within the interval of convergence for the expansion
- If $x$ is outside the interval of convergence then the approximation is not reliable

How can I use the binomial expansion with calculus?

A complete binomial expansion is exactly equal to the function it represents

This means that it is valid to differentiate or integrate a binomial expansion
- These will always be powers of $x$ derivatives or integrals

For example, the function $f (x) = \frac{1 + x}{3 + 2 x}$
- We saw above that $\begin{array}{rcl} \frac{1 + x}{3 + 2 x} & = & \frac{1}{3} + \frac{1}{9} x - \frac{2}{27} x^{2} + \frac{4}{81} x^{3} - . . . \end{array}$
- We can differentiate that:
  - $f^{'} (x) = \frac{1}{9} - \frac{4}{27} x + \frac{4}{27} x^{2} - . . .$
- Or integrate it
  - $\int f (x) d x = \int \begin{array}{rcl} (\frac{1}{3} + \frac{1}{9} x - \frac{2}{27} x^{2} + \frac{4}{81} x^{3} - . . .) \end{array} d x = \frac{1}{3} x + \frac{1}{18} x^{2} - \frac{2}{81} x^{3} + \frac{1}{81} x^{4} - . . . + c$
This can be used to find estimates or approximations
- For example to estimate $\int_{0}^{0.5} \frac{1 + x}{3 + 2 x} d x$
- Integrate the binomial expansion (as we just did above)
  - $\int_{0}^{0.5} \frac{1 + x}{3 + 2 x} d x \approx {[\frac{1}{3} x + \frac{1}{18} x^{2} - \frac{2}{81} x^{3} + \frac{1}{81} x^{4}]}_{0}^{0.5} = \frac{77}{432} = 0.178240 . . .$
  - The true value of the integral is 0.178079...
Always check that any values of $x$ you use are within the interval of convergence for the expansion
- This includes the integration limits if you are approximating a definite integral
- If any $x$ values are outside the interval of convergence then the approximation is not reliable

How can I find the percentage error of an approximation?

Use the following formula
- $percentage error = (\frac{v_{E} - v_{A}}{v_{E}}) \times 100 %$
  - $ν_{E}$ is the exact value
  - $ν_{A}$ is the approximated value
- The exact value must be in the denominator!
Percentage errors are usually given as positive values
- If the formula gives you a negative value, you can just remove the minus sign
- But you will usually get the marks for a correct positive or negative answer

Examiner Tips and Tricks

When substituting values of $x$ into a binomial expansion
- Always make sure they are within the interval of convergence
- If they are not then you may have made a mistake earlier in the question

Worked Example

The binomial expansion of $\frac{1}{\sqrt{9 - 3 x}}$ is $\frac{1}{3} + \frac{1}{18} x + \frac{1}{72} x^{2} + . . .$ , with interval of convergence $- 3 < x < 3$ .

(a) Use the expansion to estimate the value of $\frac{1}{\sqrt{10}}$ , giving your answer as a fraction.

Find the value of $x$ you need to use

$\begin{array}{rcl} \frac{1}{\sqrt{9 - 3 x}} & = & \frac{1}{\sqrt{10}} \\ 9 - 3 x & = & 10 \\ 3 x & = & - 1 \\ x & = & - \frac{1}{3} \end{array}$

That is within the interval of convergence $- 3 < x < 3$ , so we can use it to find approximation

Substitute it into the expansion

$\frac{1}{\sqrt{10}} \approx \frac{1}{3} + \frac{1}{18} (- \frac{1}{3}) + \frac{1}{72} {(- \frac{1}{3})}^{2} = \frac{1}{3} - \frac{1}{54} + \frac{1}{648} = \frac{205}{648}$

$\frac{1}{\sqrt{10}} \approx \frac{205}{648}$

(b) Find the percentage error, to 3 decimal places, of your approximation from the actual value.

Use $percentage error = (\frac{v_{E} - v_{A}}{v_{E}}) \times 100 %$

Make sure the exact value is in the denominator!

$percentage error = (\frac{\frac{1}{\sqrt{10}} - \frac{205}{648}}{\frac{1}{\sqrt{10}}}) \times 100 = - 0.041191 . . .$

That is negative because the approximated value is greater than the exact value

Percentage errors are usually given as positive numbers, so remove the minus sign

Round to 3 decimal places

$0.041 % (3 d . p .)$

Last updated: 20 October 2024

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Previous:General Binomial ExpansionNext:Radian Measure