Mathematics · OCR H240 · Topic 1.09

Numerical methods

Finding approximate answers when exact ones are not possible: sign changes, iteration, Newton–Raphson and the trapezium rule.

  • 5 sub-topics
  • Taught in Year 13
  • 61 practice questions
  • 5 animated lessons
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What the specification says

Every sub-topic OCR can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 1.09a-bSign change methodsY13
  2. 1.09c-eFormal iterative methods: iteration, cobweb and staircase diagramsY13
  3. 1.09c-eFormal iterative methods: the Newton-Raphson methodY13
  4. 1.09fNumerical integrationY13
  5. 1.09gUse numerical methods in contextY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 61 questions in this topic, free, no account needed.

Numerical methods

Not needed for Year 12Easy

Locating a root · Year 13 · 2 marks

f(x)=x3−2x−5\mathrm{f}(x) = x^3 - 2x - 5
  1. Show that f(x)=0\mathrm{f}(x) = 0 has a root between x=2x = 2 and x=3x = 3.
    (2)
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  1. Work out both values.f(2)=−1\mathrm{f}(2) = -1, f(3)=16\mathrm{f}(3) = 16
  2. Mention both the sign change and continuity.There is a change of sign and f\mathrm{f} is continuous, so there is a root between 2 and 3

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Numerical methods

Not needed for Year 12Medium

Rearranging for an iteration · Year 13 · 4 marks

  1. (a)
    Show that x3−2x−5=0x^3 - 2x - 5 = 0 can be written as x=2+5xx = \sqrt{2 + \dfrac{5}{x}}.
    (2)
  2. (b)
    Use xn+1x_{n+1}=2+5xn{}= \sqrt{2 + \dfrac{5}{x_n}} with x0=2x_0 = 2 to find x1x_1, x2x_2 and x3x_3.
    (2)

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Part (a)2 marks

  1. Divide by xx.x3x^3=2x+5{}= 2x + 5⇒x2{}\Rightarrow x^2=2+5x{}= 2 + \dfrac{5}{x}
  2. Take the positive square root (the root is positive)

Part (b)2 marks

  1. x1=2.1213x_1 = 2.1213, x2=2.0873x_2 = 2.0873
  2. The values swing either side of the root as they close in.x3=2.0965x_3 = 2.0965

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Numerical methods

Not needed for Year 12Hard

When Newton–Raphson fails · Year 13 · 5 marks

f(x)=x3−3x+1\mathrm{f}(x) = x^3 - 3x + 1
  1. (a)
    Explain why the Newton–Raphson method fails with x0=1x_0 = 1.
    (2)
  2. (b)
    Use x0=1.5x_0 = 1.5 to find x1x_1.
    (3)
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Part (a)2 marks

  1. f′(1)=3−3=0\mathrm{f}'(1) = 3 - 3 = 0
  2. The tangent is horizontal, so it never meets the xx-axis (we would divide by zero)

Part (b)3 marks

  1. f(1.5)=−0.125\mathrm{f}(1.5) = -0.125, f′(1.5)=3.75\mathrm{f}'(1.5) = 3.75
  2. x1=1.5−−0.1253.75x_1 = 1.5 - \dfrac{-0.125}{3.75}
  3. =1.5333= 1.5333

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