Iteration Revision Notes
Subject: Further Mathematics | Level: GCSE | Exam Board: OCR
Iteration is a powerful numerical method that allows you to find approximate solutions to equations that cannot be solved algebraically. By repeatedly applying an iterative formula of the form x_{n+1} = g(x_n), you systematically converge on roots to a specified degree of accuracy. This topic is essential for OCR GCSE Further Mathematics and typically appears as structured 4-6 mark questions testing both algebraic manipulation and precise calculator technique.
Revision Notes & Key Concepts
Revision Podcast Transcript
ITERATION IN GCSE FURTHER MATHEMATICS - 10-MINUTE EDUCATIONAL PODCAST Speaker: Female educator, warm and engaging tone [INTRO - 1 MINUTE] Hello and welcome to this study session on Iteration - one of the most powerful numerical methods you'll encounter in GCSE Further Mathematics. I'm here to guide you through everything you need to know to confidently tackle iteration questions in your OCR exam. Now, you might be wondering: what exactly is iteration, and why should I care? Well, imagine you're trying to solve an equation like x cubed minus x minus one equals zero. You can't factorise it easily, and the quadratic formula won't help because it's not a quadratic. This is where iteration comes to the rescue. It's a systematic method that lets you find approximate solutions to equations that would otherwise be impossible to solve by hand. The beauty of iteration is that it's like a mathematical treasure hunt. You make an educated guess, apply a formula repeatedly, and watch as your answers get closer and closer to the true solution. It's satisfying, it's powerful, and it's absolutely essential for your exam. So let's dive in. [CORE CONCEPTS - 5 MINUTES] Let's start with the fundamentals. Iteration is all about turning an equation of the form f of x equals zero into what we call an iterative formula: x subscript n plus one equals g of x subscript n. This might look intimidating at first, but think of it as a recipe. You start with an ingredient - your initial value x naught - and you follow the same steps over and over until you get the result you want. The first crucial step is rearranging your equation. If you're given something like x cubed equals x plus one, you need to isolate x on one side. You might rearrange it to x equals the cube root of x plus one. This becomes your iterative formula: x subscript n plus one equals the cube root of x subscript n plus one. The key here is algebraic manipulation - and this is where many candidates lose marks. You must show every single step clearly. Don't skip lines, don't work backwards, and definitely don't just write down the final formula without justification. Once you have your iterative formula, you need a starting value - we call this x naught. Sometimes the question will give you this starting value, like "use x naught equals two." Other times, you might need to estimate it from a graph or use your mathematical intuition. A good starting value helps the iteration converge faster, but even a rough estimate will usually get you there eventually. Now comes the repetitive part - and this is where your calculator becomes your best friend. You substitute x naught into your formula to get x one. Then you substitute x one to get x two. Then x two to get x three, and so on. Here's the golden rule that will save you marks: use the ANS key on your calculator. Press equals, then immediately type your formula using ANS as the variable, and keep pressing equals. This maintains full precision throughout your calculations and prevents rounding errors. Let me give you a concrete example. Suppose we're solving x cubed minus x minus one equals zero, and we've rearranged it to x equals the cube root of x plus one. If we start with x naught equals one, we get: x one equals the cube root of one plus one, which is the cube root of two, approximately one point two six. Then x two equals the cube root of one point two six plus one, which is approximately one point three one two. Then x three is approximately one point three two two. Notice how the values are getting closer together? That's convergence in action. You keep going until consecutive values agree to the required degree of accuracy. If the question asks for three significant figures, you need to continue until at least the first three digits stop changing. Typically, you'll need to calculate four or five iterations and write them all down to at least five decimal places. This shows the examiner you understand the process and aren't just guessing. One more thing about convergence: not all rearrangements work. Some iterative formulas diverge, meaning the values get further apart rather than closer together. OCR will always give you a formula that converges, but you should be aware that the way you rearrange the equation matters. If you're ever asked to suggest your own rearrangement, think carefully about which form will bring values together rather than push them apart. [EXAM TIPS & COMMON MISTAKES - 2 MINUTES] Now let's talk about how to maximise your marks in the exam. Iteration questions typically appear as four to six mark questions, and they follow a predictable structure. You'll usually be asked to show that an equation rearranges to a given form, then use iteration to find a root to a specified accuracy. For the "show that" part, you must show every algebraic step. Start with the original equation, clearly state what operation you're performing, and arrive at the required form. Examiners are looking for logical progression - if you jump straight to the answer, you'll lose method marks even if you're correct. For the iteration part, write down at least your first three or four values to five decimal places. Label them clearly: x naught, x one, x two, x three. Then state your final answer to the required accuracy. If the question asks for three significant figures, write something like "therefore the root is one point three two to three significant figures." Now, the mistakes. The biggest one? Rounding too early. If you round x one to three significant figures, then use that rounded value to calculate x two, your errors compound and you'll never reach the correct answer. Keep full precision until the very end. Second mistake: incorrect algebra when rearranging. Watch out for negative signs, especially when you're moving terms across the equals sign. And if you're dividing by x, remember you can't do that if x might be zero. Third mistake: stopping too soon. Just because x two and x three look similar doesn't mean they've converged to three significant figures. Always calculate one or two extra iterations to be sure. And finally, not using the ANS key. I can't stress this enough - it's the difference between a correct answer and one that's slightly off due to rounding errors. [QUICK-FIRE RECALL QUIZ - 1 MINUTE] Let's test your understanding with some quick-fire questions. Pause after each one and see if you can answer. Question one: What is the standard form of an iterative formula? Answer: x subscript n plus one equals g of x subscript n. Question two: Why must you use the ANS key on your calculator? Answer: To maintain full precision and avoid cumulative rounding errors. Question three: How many decimal places should you write down for your intermediate iterations? Answer: At least five decimal places. Question four: What does it mean when an iterative sequence converges? Answer: The values get closer and closer together, approaching the true root. Question five: In a "show that" question, can you work backwards from the answer? Answer: No - you must show logical forward progression from the starting equation. [SUMMARY & SIGN-OFF - 1 MINUTE] Let's wrap up. Iteration is a powerful method for solving equations that can't be solved algebraically. The process is straightforward: rearrange your equation into the form x subscript n plus one equals g of x subscript n, choose a starting value, and repeatedly apply the formula until your values converge to the required accuracy. Remember the key exam techniques: show all algebraic steps in rearrangement questions, use the ANS key to maintain precision, write down multiple iterations to at least five decimal places, and don't round until you state your final answer. Iteration questions are highly structured and predictable, which means they're a fantastic opportunity to secure marks if you follow the method carefully. Practice a few examples, get comfortable with your calculator, and you'll find these questions become routine. Thank you for listening, and best of luck with your revision. Keep practising, stay precise, and remember - iteration is just a systematic treasure hunt for solutions. You've got this!
Key Terms & Definitions
- Iteration
- A numerical method for finding approximate solutions to equations by repeatedly applying a formula to generate a sequence of values that converge toward the true root.
- Iterative Formula
- A formula of the form x_{n+1} = g(x_n) that defines each term in a sequence based on the previous term, derived by rearranging an equation f(x) = 0.
- Convergence
- The property of an iterative sequence where successive terms get progressively closer together, approaching a limiting value (the root of the equation).
- Starting Value (x₀)
- The initial value used to begin an iterative sequence, often provided in the question or estimated from a graph.
- Root
- A value of x for which f(x) = 0; the solution to the equation. In iteration, the root is the limiting value toward which the sequence converges.
- Precision
- The number of digits or decimal places to which a value is stated. Full calculator precision (typically 10-12 digits) must be maintained during iteration to avoid cumulative rounding errors.
- Rearrangement
- The algebraic process of transforming an equation f(x) = 0 into the form x = g(x), which then becomes the iterative formula x_{n+1} = g(x_n).
- Significant Figures
- The number of meaningful digits in a number, counting from the first non-zero digit. Used to specify the required accuracy of a final answer.
Worked Examples
Worked Example
Question: The equation x³ - 4x - 2 = 0 has a root between x = 2 and x = 3. (a) Show that the equation can be rearranged to x = ∛(4x + 2). (b) Use the iterative formula x_{n+1} = ∛(4x_n + 2) with x₀ = 2 to find the root correct to 2 decimal places.
Solution: **Part (a) [2 marks]** Step 1: Start with the original equation x³ - 4x - 2 = 0 Step 2: Add 4x and 2 to both sides x³ = 4x + 2 Step 3: Take the cube root of both sides x = ∛(4x + 2) ✓ Therefore, the equation rearranges to the required form. **Part (b) [3 marks]** Using x_{n+1} = ∛(4x_n + 2) with x₀ = 2: x₁ = ∛(4×2 + 2) = ∛10 = 2.15443 x₂ = ∛(4×2.15443 + 2) = ∛10.61772 = 2.19685 x₃ = ∛(4×2.19685 + 2) = ∛10.78740 = 2.20906 x₄ = ∛(4×2.20906 + 2) = ∛10.83624 = 2.21235 x₅ = ∛(4×2.21235 + 2) = ∛10.84940 = 2.21323 x₆ = ∛(4×2.21323 + 2) = ∛10.85292 = 2.21346 The values have converged to 2.21 (to 2 d.p.) **Final answer: x = 2.21 to 2 decimal places**
Worked Example
Question: The equation x² - 5x + 3 = 0 has a root near x = 1. Use the iterative formula x_{n+1} = (x_n² + 3)/5 with x₀ = 1 to find this root correct to 3 significant figures. Show your working clearly.
Solution: Using x_{n+1} = (x_n² + 3)/5 with x₀ = 1: x₀ = 1 x₁ = (1² + 3)/5 = 4/5 = 0.80000 x₂ = (0.8² + 3)/5 = (0.64 + 3)/5 = 3.64/5 = 0.72800 x₃ = (0.728² + 3)/5 = (0.529984 + 3)/5 = 3.529984/5 = 0.70600 x₄ = (0.706² + 3)/5 = (0.498436 + 3)/5 = 3.498436/5 = 0.69969 x₅ = (0.69969² + 3)/5 = (0.489566 + 3)/5 = 3.489566/5 = 0.69791 x₆ = (0.69791² + 3)/5 = (0.487079 + 3)/5 = 3.487079/5 = 0.69742 x₇ = (0.69742² + 3)/5 = (0.486395 + 3)/5 = 3.486395/5 = 0.69728 The values have stabilised to 0.697 (to 3 s.f.) **Final answer: x = 0.697 to 3 significant figures**
Worked Example
Question: A student uses the iterative formula x_{n+1} = √(2x_n + 7) with x₀ = 3 to solve an equation. (a) Write down the values of x₁, x₂, and x₃ to 5 decimal places. (b) To what value does the sequence appear to be converging? Give your answer to 2 decimal places.
Solution: **Part (a) [2 marks]** Using x_{n+1} = √(2x_n + 7) with x₀ = 3: x₁ = √(2×3 + 7) = √13 = 3.60555 x₂ = √(2×3.60555 + 7) = √14.21110 = 3.76971 x₃ = √(2×3.76971 + 7) = √14.53942 = 3.81306 **Part (b) [2 marks]** Continuing the iteration: x₄ = √(2×3.81306 + 7) = √14.62612 = 3.82418 x₅ = √(2×3.82418 + 7) = √14.64836 = 3.82709 x₆ = √(2×3.82709 + 7) = √14.65418 = 3.82785 x₇ = √(2×3.82785 + 7) = √14.65570 = 3.82805 The sequence is converging to **3.83 to 2 decimal places**
Practice Questions
Question: The equation x² + x - 7 = 0 has a root between 2 and 3. Show that this equation can be rearranged to x = √(7 - x).
Answer:
Question: Use the iterative formula x_{n+1} = √(7 - x_n) with x₀ = 2 to find a root of x² + x - 7 = 0 correct to 1 decimal place.
Answer:
Question: The equation x³ - 2x - 5 = 0 has a root near x = 2. (a) Show that the equation can be rearranged to x = ∛(2x + 5). (b) Use the iterative formula x_{n+1} = ∛(2x_n + 5) with x₀ = 2 to find the root correct to 3 significant figures.
Answer:
Question: A student uses the iterative formula x_{n+1} = (x_n³ + 3)/4 with x₀ = 1.5 to solve the equation 4x - x³ - 3 = 0. Explain why the student's sequence of values is likely to diverge rather than converge.
Answer:
Question: The iterative formula x_{n+1} = √(3x_n + 1) is used with x₀ = 2. (a) Calculate x₁, x₂, and x₃ to 4 decimal places. (b) By considering your answers, suggest to what value the sequence is converging. (c) Verify your answer by substituting it into the original equation x² - 3x - 1 = 0.
Answer:


