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Mathematics for AVT Dr Ian Drumm Week 6

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A linear factor in the denominator gives rise to a single partial fraction ... Proper fractions with quadratic factors. Where your denominator can't be factorised. ... – PowerPoint PPT presentation

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Title: Mathematics for AVT Dr Ian Drumm Week 6


1
Mathematics for AVTDr Ian DrummWeek 6
  • Aims
  • Partial Fractions
  • Objectives
  • Polynomials - factorising
  • Equating Coefficients, Proper and Improper
    Fractions,
  • Partial Fractions

2
Polynomials
  • Equations of the form
  • For example
  • Must be non-negative whole number powers
  • Degree of polynomial is highest power
  • Cubic polynomial has a degree of 3

3
Quadratic Equations
  • A quadratic equation is and equation of the form
  • Solving finding roots
  • Factorisation
  • Completing the Square
  • Using a formula

4
Factorisation split into factors to find roots
(remember multiply by first coefficient and hence
express using z4x)
5
Completing the square
  • It can be shown that
  • For examplecomplete square of
  • We know k8 hence
  • Hence roots

6
Solution by formula
  • Given quadratic equation of the form
  • Solutions of the form

7
Types of roots
8
Factorising cubic equations Equating coefficients
  • Factorise
  • given a factor is
  • Write in terms of a, b and c hence find unknowns

9
Verifying Solution
  • Substitute for x
  • If we know x4 is a root, we can write polynomial
    as

10
Solving polynomial graphically
  • Roots where y0
  • Use computer graphing package (e.g. matlab,
    excel, etc.)

11
Partial Fractions
  • Splitting complicated fractions into simpler
    ones, particularly useful for integration
  • For example

12
Proper and improper fractions
  • Degree of numerator n is highest power in the
    numerator
  • Degree of denominator d is highest power in
    denominator
  • Proper if dgtn, otherwise improper if d n
  • For example n3gtd2 hence improper

13
Linear factors
  • A linear factor in the
    denominator gives rise to a single partial
    fraction
  • For example the expression here has two linear
    factors to give two partial fractions
  • Hence we can find the unknowns A and B

14
Find partial fractions(proper fraction)
  • Split denominator into linear factors
  • Multiply partial fractions by linear factors
  • Choose value of x to eliminate a term, e.g. x-1
  • Hence can rearrange and compare coefficients to
    find A

15
Proper Fractions with repeated linear factors
  • A repeated linear factor
  • in the denominator produces two partial
    fractions
  • For example

16
Proper fractions with quadratic factors
  • Where your denominator cant be factorised.
  • A quadratic factor in
    denominator gives partial fraction
  • For example

17
Find partial fractionImproper Fraction
  • Extra polynomial added to partial fractions that
    arise
  • Subtract degree of numerator from degree of
    denominator, hence polynomial to degree of n-d
  • For examplen2 d1 hence add
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