A-Level · Mathematics · AQA · Mark scheme decoded
AQA A-Level Mathematics: Iterative Methods and Newton-Raphson — mark scheme explained
The short answer
In AQA A-Level Mathematics, the topic of numerical methods includes solving equations approximately using simple iterative methods. This involves understanding how to use recurrence relations and graphical techniques like cobweb and staircase diagrams. Additionally, you will learn about the Newton-Raphson method, a powerful technique for finding roots of equations, and understand its limitations.
The question
Use the simple iterative method to solve the equation x 3 - 2x - 5 = 0. Start with an initial guess of x 0 = 2 and iterate until the difference between successive terms is less than 0.001. [Paraphrased for study — not reproduced from any exam paper.]
Mark scheme, decoded
What each mark is really for — in plain English — and the wording trap that loses it.
- S1
Rewrite the equation as x = g(x): x = (2x + 5) 1/3
- S2
Use the recurrence relation: x n+1 = (2x n + 5) 1/3
- S3
Start with x 0 = 2
- S4
Calculate x 1 : x 1 = (2 × 2 + 5) 1/3 ≈ 2.080
- S5
Calculate x 2 : x 2 = (2 × 2.080 + 5) 1/3 ≈ 2.094
- S6
Calculate x 3 : x 3 = (2 × 2.094 + 5) 1/3 ≈ 2.096
- S7
Check the difference: |x 3 - x 2 | = |2.096 - 2.094| = 0.002 (not less than 0.001)
- S8
Calculate x 4 : x 4 = (2 × 2.096 + 5) 1/3 ≈ 2.097
- S9
Check the difference: |x 4 - x 3 | = |2.097 - 2.096| = 0.001 (less than 0.001)
- S10
The sequence converges to approximately x ≈ 2.097
Model answer
Worked through, with each step tagged to the mark it earns.
- S1
Rewrite the equation as x = g(x): x = (2x + 5) 1/3
- S2
Use the recurrence relation: x n+1 = (2x n + 5) 1/3
- S3
Start with x 0 = 2
- S4
Calculate x 1 : x 1 = (2 × 2 + 5) 1/3 ≈ 2.080
- S5
Calculate x 2 : x 2 = (2 × 2.080 + 5) 1/3 ≈ 2.094
- S6
Calculate x 3 : x 3 = (2 × 2.094 + 5) 1/3 ≈ 2.096
- S7
Check the difference: |x 3 - x 2 | = |2.096 - 2.094| = 0.002 (not less than 0.001)
- S8
Calculate x 4 : x 4 = (2 × 2.096 + 5) 1/3 ≈ 2.097
- S9
Check the difference: |x 4 - x 3 | = |2.097 - 2.096| = 0.001 (less than 0.001)
- S10
The sequence converges to approximately x ≈ 2.097
Final answer: x ≈ 2.097
Common mistakes
- Using the wrong recurrence relation for a given equation. — Double-check the rearrangement and ensure it is in the correct form x = g(x).
- Not checking for convergence criteria. — Always check the difference between successive terms to ensure it is below a specified tolerance level.
- Using an inappropriate initial guess. — Choose an initial guess close to the expected root, possibly by sketching the function or using a calculator.
- Failing to recognize when the Newton-Raphson method fails. — Understand the limitations of the method and check for these conditions before applying it.
- Incorrectly plotting cobweb or staircase diagrams. — Practice drawing these diagrams step by step and ensure each line is drawn accurately according to the rules.
- Not using the derivative correctly in the Newton-Raphson method. — Always compute and use the correct derivative when applying the Newton-Raphson formula.
- Failing to check for multiple roots in the Newton-Raphson method. — Plot the function and identify all possible roots before applying the method. Choose an initial guess close to the desired root.
- Not understanding the difference between cobweb and staircase diagrams. — Understand that a cobweb diagram uses vertical and horizontal lines alternately, while a staircase diagram uses steps. Practice drawing both types of diagrams.
Where the marks go
- Full worked solution (all marking points)5 marks