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Iteration and Newton-RaphsonAQA A-Level Maths: Flashcards

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What is an iteration $x_{n+1}=g(x_n)$?

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What is an iteration xn+1=g(xn)x_{n+1}=g(x_n)?
A rule that produces each term from the previous one, starting from x0x_0.
How do you turn f(x)=0f(x)=0 into an iteration?
Rearrange to x=g(x)x=g(x) and use xn+1=g(xn)x_{n+1}=g(x_n).
If an iteration converges to α\alpha, what is true of α\alpha?
α=g(α)\alpha=g(\alpha), so α\alpha is a root of the original equation.
State the Newton-Raphson formula.
xn+1=xn−f(xn)f′(xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}
Where does the Newton-Raphson formula come from?
The tangent to y=f(x)y=f(x) at xnx_n meets the xx-axis at xn+1x_{n+1}.
Newton-Raphson for f(x)=x2−7f(x)=x^2-7?
xn+1=xn−xn2−72xnx_{n+1}=x_n-\frac{x_n^2-7}{2x_n}
What does a staircase diagram show?
Monotone convergence, when 0<g′(α)<10<g'(\alpha)<1.
What does a cobweb diagram show?
Convergence with terms alternating either side of the root, when −1<g′(α)<0-1<g'(\alpha)<0.
When does a fixed-point iteration diverge?
When ∣g′(α)∣>1|g'(\alpha)|>1, so the error grows at each step.
When does Newton-Raphson fail completely?
When f′(xn)=0f'(x_n)=0: the tangent is horizontal and the formula divides by zero.
Other ways Newton-Raphson can go wrong?
It converges to a different root, or oscillates, when x0x_0 is near a stationary point or far from the root.
How do you know when to stop iterating?
When successive values agree to the required number of decimal places; then check with a sign change.

Exam questions on Iteration and Newton-Raphson

  1. The equation x3+x−3=0x^3+x-3=0 has a root α\alpha between 1 and 2. A student uses the iteration xn+1=3−xn3x_{n+1}=\sqrt[3]{3-x_n} with x0=1x_0=1.
    Find x2x_2 and x3x_3, giving each to 4 decimal places.2 marks
  2. The Newton-Raphson method is used to approximate 7\sqrt7 by solving f(x)=0f(x)=0, where f(x)=x2−7f(x)=x^2-7, with starting value x0=3x_0=3.
    Find x2x_2, giving your answer to 4 decimal places.2 marks
  3. The equation f(x)=0f(x)=0, where f(x)=x3−2x−5f(x)=x^3-2x-5, has a root α\alpha close to 2. The Newton-Raphson method is used with x0=2x_0=2.
    Show that the Newton-Raphson iteration is xn+1=xn−xn3−2xn−53xn2−2x_{n+1}=x_n-\frac{x_n^3-2x_n-5}{3x_n^2-2}, and hence show that x1=2.1x_1=2.1.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).