Applications of Binomial Expansion (Edexcel IGCSE Further Pure Maths)

Revision Note

Roger B

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  fraction numerator 1 plus x over denominator 3 plus 2 x end fraction

    • Rewrite as a productfraction numerator 1 plus x over denominator 3 plus 2 x end fraction equals open parentheses 1 plus x close parentheses open parentheses 3 plus 2 x close parentheses to the power of negative 1 end exponent

  • Find the binomial expansion of  open parentheses 3 plus 2 x close parentheses to the power of negative 1 end exponent

open parentheses 3 plus 2 x close parentheses to the power of negative 1 end exponent equals 1 third minus 2 over 9 x plus 4 over 27 x squared minus 8 over 81 x cubed plus...

  • Note this has only been expanded up to the bold italic x to the power of bold 3 term

  • Multiply that expansion by open parentheses 1 plus x close parentheses and simplify

table row cell open parentheses 1 plus x close parentheses open parentheses 3 plus 2 x close parentheses to the power of negative 1 end exponent end cell equals cell open parentheses 1 plus x close parentheses open parentheses 1 third minus 2 over 9 x plus 4 over 27 x squared minus 8 over 81 x cubed plus... close parentheses end cell row blank equals cell 1 third minus 2 over 9 x plus 4 over 27 x squared minus 8 over 81 x cubed plus 1 third x minus 2 over 9 x squared plus 4 over 27 x 3 minus 8 over 81 x to the power of 4 plus... end cell row blank equals cell 1 third plus 1 over 9 x minus 2 over 27 x squared plus 4 over 81 x cubed minus 8 over 81 x to the power of 4 plus... end cell row blank equals cell 1 third plus 1 over 9 x minus 2 over 27 x squared plus 4 over 81 x cubed minus... end cell end table

  • This is only valid up to the x cubed term

  • To get more terms we would have to start with more terms for open parentheses 3 plus 2 x close parentheses to the power of negative 1 end exponent

  • negative 8 over 81 x to the power of 4 is not the correct x to the power of 4 term for table row blank blank cell open parentheses 1 plus x close parentheses open parentheses 3 plus 2 x close parentheses to the power of negative 1 end exponent end cell end table as there are more x to the power of 4 terms that were not found

  • Use the same process to find the expansion for something like  fraction numerator 1 minus 2 x over denominator square root of 2 plus 3 x end root end fraction

    • Rewrite as a productfraction numerator 1 minus 2 x over denominator square root of 2 plus 3 x end root end fraction equals open parentheses 1 minus 2 x close parentheses open parentheses 2 plus 3 x close parentheses to the power of negative 1 half end exponent

    • Find the binomial expansion of  open parentheses 2 plus 3 x close parentheses to the power of negative 1 half end exponent  

    • Multiply the expansion by open parentheses 1 minus 2 x close parentheses 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 open vertical bar x close vertical bar less than 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 xis, the better the approximation will be

  • For example, find an approximation for square root of 0.96 end root using the expansion of open parentheses 1 minus x close parentheses to the power of 1 half end exponent  

    • Compare the value you are approximating to the expression being expanded

      • left parenthesis 1 space minus space x right parenthesis to the power of 1 half end exponent space equals space 0.96 to the power of 1 half end exponent

    • Find the value of bold italic x to use by solving the appropriate equation

      • table attributes columnalign right center left columnspacing 0px end attributes row blank blank blank row cell 1 space minus space x space end cell equals cell space 0.96 end cell row cell space x space end cell equals cell space 0.04 end cell end table

  • Substitute this value of x into the binomial expansion of open parentheses 1 minus x close parentheses to the power of 1 half end exponent

    • open parentheses 1 minus x close parentheses to the power of 1 half end exponent equals 1 minus 1 half x minus 1 over 8 x squared minus...

    • So square root of 0.96 end root almost equal to 1 minus 1 half open parentheses 0.04 close parentheses minus 1 over 8 blank open parentheses 0.04 close parentheses squared equals 0.9798

    • The true value of square root of 0.96 end root 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 8 over 125 using the binomial expansion of 1 over open parentheses 3 minus x close parentheses cubed

      • 1 over open parentheses 3 minus x close parentheses cubed equals 8 over 125 space space space rightwards double arrow space space space fraction numerator 1 over denominator 3 minus x end fraction equals 2 over 5 space space space rightwards double arrow space space space x equals 0.5

      • So substitute x equals 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 bold italic x derivatives or integrals

  • For example, the function  straight f open parentheses x close parentheses equals fraction numerator 1 plus x over denominator 3 plus 2 x end fraction

    • We saw above that  table row cell fraction numerator 1 plus x over denominator 3 plus 2 x end fraction end cell equals cell 1 third plus 1 over 9 x minus 2 over 27 x squared plus 4 over 81 x cubed minus... end cell end table

    • We can differentiate that:

      • straight f to the power of apostrophe open parentheses x close parentheses equals 1 over 9 minus 4 over 27 x plus 4 over 27 x squared minus...

    • Or integrate it

      • integral straight f open parentheses x close parentheses space straight d x space equals space integral table row blank blank cell open parentheses 1 third plus 1 over 9 x minus 2 over 27 x squared plus 4 over 81 x cubed minus... close parentheses end cell end table space straight d x space equals 1 third x plus 1 over 18 x squared minus 2 over 81 x cubed plus 1 over 81 x to the power of 4 minus... plus c

  • This can be used to find estimates or approximations

    • For example to estimate integral subscript 0 superscript 0.5 end superscript space fraction numerator 1 plus x over denominator 3 plus 2 x end fraction d x

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

      • integral subscript 0 superscript 0.5 end superscript space fraction numerator 1 plus x over denominator 3 plus 2 x end fraction d x almost equal to open square brackets 1 third x plus 1 over 18 x squared minus 2 over 81 x cubed plus 1 over 81 x to the power of 4 close square brackets subscript 0 superscript 0.5 end superscript equals 77 over 432 equals 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 space error equals open parentheses fraction numerator v subscript E minus v subscript A over denominator v subscript E end fraction close parentheses cross times 100 percent sign

      • nu subscript E  is the exact value

      • nu subscript 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  fraction numerator 1 over denominator square root of 9 blank minus blank 3 x end root end fraction is 1 third plus 1 over 18 x plus 1 over 72 x squared plus..., with interval of convergence  negative 3 less than x less than 3.

(a) Use the expansion to estimate the value of fraction numerator 1 over denominator square root of 10 end fraction, giving your answer as a fraction.

Find the value of xyou need to use

table row cell fraction numerator 1 over denominator square root of 9 minus 3 x end root end fraction end cell equals cell fraction numerator 1 over denominator square root of 10 end fraction end cell row cell 9 minus 3 x end cell equals 10 row cell 3 x end cell equals cell negative 1 end cell row x equals cell negative 1 third end cell end table

That is within the interval of convergence  negative 3 less than x less than 3, so we can use it to find approximation

Substitute it into the expansion

fraction numerator 1 over denominator square root of 10 end fraction almost equal to 1 third plus 1 over 18 open parentheses negative 1 third close parentheses plus 1 over 72 open parentheses negative 1 third close parentheses squared equals 1 third minus 1 over 54 plus 1 over 648 equals 205 over 648

fraction numerator bold 1 over denominator square root of bold 10 end fraction bold almost equal to bold 205 over bold 648

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

Use  percentage space error equals open parentheses fraction numerator v subscript E minus v subscript A over denominator v subscript E end fraction close parentheses cross times 100 percent sign

Make sure the exact value is in the denominator!

percentage space error equals open parentheses fraction numerator begin display style fraction numerator 1 over denominator square root of 10 end fraction end style minus begin display style 205 over 648 end style over denominator begin display style fraction numerator 1 over denominator square root of 10 end fraction end style end fraction close parentheses cross times 100 equals negative 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

bold 0 bold. bold 041 bold percent sign bold space begin bold style stretchy left parenthesis 3 space d. p. stretchy right parenthesis end style

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Roger B

Author: Roger B

Expertise: Maths

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Dan Finlay

Author: Dan Finlay

Expertise: Maths Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.