Applications of Binomial Expansion (Edexcel IGCSE Further Pure Maths): Revision Note

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  $\frac{1+x}{3+2x}$

  • Rewrite as a product,  $\frac{1+x}{3+2x}=\left(1+x\right){\left(3+2x\right)}^{-1}$

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

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

• Note this has only been expanded up to the ${\mathit{x}}^{\mathbf{3}}$ term

• Multiply that expansion by $\left(1+x\right)$ and simplify

$\begin{array}{rcl}\left(1+x\right){\left(3+2x\right)}^{-1}& =& \left(1+x\right)\left(\frac{1}{3}-\frac{2}{9}x+\frac{4}{27}{x}^2-\frac{8}{81}{x}^3+...\right)\\ & =& \frac{1}{3}-\frac{2}{9}x+\frac{4}{27}{x}^2-\frac{8}{81}{x}^3+\frac{1}{3}x-\frac{2}{9}{x}^2+\frac{4}{27}x3-\frac{8}{81}{x}^4+...\\ & =& \frac{1}{3}+\frac{1}{9}x-\frac{2}{27}{x}^2+\frac{4}{81}{x}^3-\frac{8}{81}{x}^4+...\\ & =& \frac{1}{3}+\frac{1}{9}x-\frac{2}{27}{x}^2+\frac{4}{81}{x}^3-...\end{array}$

• This is only valid up to the $x^3$ term

• To get more terms we would have to start with more terms for ${\left(3+2x\right)}^{-1}$

• $-\frac{8}{81}{x}^4$ is not the correct $x^4$ term for $\begin{array}{rcl}& & \left(1+x\right){\left(3+2x\right)}^{-1}\end{array}$ as there are more $x^4$ terms that were not found

• Use the same process to find the expansion for something like  $\frac{1-2x}{\sqrt{2+3x}}$

  • Rewrite as a product,  $\frac{1-2x}{\sqrt{2+3x}}=\left(1-2x\right){\left(2+3x\right)}^{-\frac{1}{2}}$

  • Find the binomial expansion of  ${\left(2+3x\right)}^{-\frac{1}{2}}$  

  • Multiply the expansion by $\left(1-2x\right)$ 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 $\left|x\right|<1$, 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 ${\left(1-x\right)}^{\frac{1}{2}}$  

  • 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 $\mathit{x}$ 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 ${\left(1-x\right)}^{\frac{1}{2}}$

  • ${\left(1-x\right)}^{\frac{1}{2}}=1-\frac{1}{2}x-\frac{1}{8}{x}^2-...$

  • So $\sqrt{0.96}\approx 1-\frac{1}{2}\left(0.04\right)-\frac{1}{8} {\left(0.04\right)}^2=0.9798$

  • 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 $\frac{1}{{\left(3-x\right)}^3}$

    • $\frac{1}{{\left(3-x\right)}^3}=\frac{8}{125}\Rightarrow \frac{1}{3-x}=\frac{2}{5}\Rightarrow x=0.5$

    • 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 $\mathit{x}$ derivatives or integrals

• For example, the function  $\mathrm{f}\left(x\right)=\frac{1+x}{3+2x}$

  • We saw above that  $\begin{array}{rcl}\frac{1+x}{3+2x}& =& \frac{1}{3}+\frac{1}{9}x-\frac{2}{27}{x}^2+\frac{4}{81}{x}^3-...\end{array}$

  • We can differentiate that:

    • ${\mathrm{f}}^{\text{'}}\left(x\right)=\frac{1}{9}-\frac{4}{27}x+\frac{4}{27}{x}^2-...$

  • Or integrate it

    • $\int \mathrm{f}\left(x\right)\mathrm{d}x=\int \begin{array}{rcl}& & \left(\frac{1}{3}+\frac{1}{9}x-\frac{2}{27}{x}^2+\frac{4}{81}{x}^3-...\right)\end{array}\mathrm{d}x=\frac{1}{3}x+\frac{1}{18}{x}^2-\frac{2}{81}{x}^3+\frac{1}{81}{x}^4-...+c$

• This can be used to find estimates or approximations

  • For example to estimate ${\int }_0^{0.5}\frac{1+x}{3+2x}dx$

  • Integrate the binomial expansion (as we just did above)

    • ${\int }_0^{0.5}\frac{1+x}{3+2x}dx\approx {\left[\frac{1}{3}x+\frac{1}{18}{x}^2-\frac{2}{81}{x}^3+\frac{1}{81}{x}^4\right]}_0^{0.5}=\frac{77}{432}=0.178240...$

    • 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

  • $\mathrm{percentage}\mathrm{error}=\left(\frac{v_E-v_A}{v_E}\right)\times 100\%$

    • ${\nu }_E$  is the exact value

    • ${\nu }_A$ 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