Cliste School
Get started

Algebra

Iteration (Higher)

Year 10 · Year 11

  • ✓By the end of this lesson students will be able to understand what an iterative formula is and how to use it.
  • ✓By the end of this lesson students will be able to generate a sequence of approximations for the root of an equation using an iterative formula.
  • ✓By the end of this lesson students will be able to use iteration to find an approximate solution to an equation given a starting value.
  • ✓By the end of this lesson students will be able to recognise when an iterative process converges to a root.

Key concepts

What is Iteration?

Iteration is a numerical method of repeatedly applying a process to generate a sequence of approximations that get closer and closer to the exact solution of a problem. In mathematics, it is often used to find roots of equations that cannot be solved algebraically.

Iterative Formulae

An iterative formula is a rule of the form x_{n+1} = f(x_n) which uses the previous term (x_n) to calculate the next term (x_{n+1}) in a sequence. A starting value, x_0, is always given to begin the process.

x_{n+1} = f(x_n)
Solving Equations Numerically using Iteration

To solve an equation f(x) = 0 using iteration, we first rearrange it into the form x = g(x). This g(x) then becomes the iterative formula x_{n+1} = g(x_n). We start with an initial estimate x_0 and substitute it into the formula to find x_1, then use x_1 to find x_2, and so on, until the values converge to a desired degree of accuracy. This method is particularly useful for equations that are difficult or impossible to solve directly.

x_{n+1} = g(x_n) (where g(x) is a rearrangement of f(x) = 0)

Key facts to remember

  • 1Iteration is a numerical method for finding approximate solutions to equations, especially those that cannot be solved algebraically.
  • 2An iterative formula is typically written in the form x_{n+1} = f(x_n), where x_n is the current approximation and x_{n+1} is the next.
  • 3A starting value, x_0, is always required to initiate the iterative process.
  • 4To solve an equation f(x) = 0 using iteration, it must first be rearranged into the form x = g(x) to create the iterative formula x_{n+1} = g(x_n).
  • 5Always use the full calculator value from the previous step when calculating the next term to maintain accuracy throughout the process.
  • 6The iterative process stops when successive approximations agree to the required degree of accuracy.
  • 7Not all rearrangements of an equation will lead to a convergent iterative process for a given starting value; some may diverge.

Worked examples

Example 1

Use the iterative formula x_{n+1} = \sqrt{3x_n + 1} with x_0 = 2 to find the values of x_1, x_2, and x_3. Give your answers to 3 decimal places.

IGiven x_0 = 2 and the formula x_{n+1} = \sqrt{3x_n + 1}.
IIFor n = 0: x_1 = \sqrt{3(x_0) + 1} = \sqrt{3(2) + 1} = \sqrt{6 + 1} = \sqrt{7}
IIIx_1 = 2.645751311... \approx 2.646 (to 3 d.p.)
IVFor n = 1: x_2 = \sqrt{3(x_1) + 1} = \sqrt{3(2.645751311...) + 1} = \sqrt{7.937253933... + 1} = \sqrt{8.937253933...}
Vx_2 = 2.98952577... \approx 2.990 (to 3 d.p.)
VIFor n = 2: x_3 = \sqrt{3(x_2) + 1} = \sqrt{3(2.98952577...) + 1} = \sqrt{8.96857731... + 1} = \sqrt{9.96857731...}
VIIx_3 = 3.1573049... \approx 3.157 (to 3 d.p.)

Answer

x_1 = 2.646, x_2 = 2.990, x_3 = 3.157

Always use the full calculator value from the previous step for subsequent calculations to maintain accuracy, only rounding the final answer or specific intermediate steps as requested.

Example 2

Show that the equation x^3 - 3x - 5 = 0 has a root between x = 2 and x = 3. Use the iterative formula x_{n+1} = \sqrt[3]{3x_n + 5} with x_0 = 2.5 to find the root correct to 2 decimal places.

ILet f(x) = x^3 - 3x - 5.
IITo show a root exists between x = 2 and x = 3, evaluate f(x) at these points:
IIIf(2) = (2)^3 - 3(2) - 5 = 8 - 6 - 5 = -3
IVf(3) = (3)^3 - 3(3) - 5 = 27 - 9 - 5 = 13
VSince f(2) is negative and f(3) is positive, there is a change of sign, indicating that a root lies between x = 2 and x = 3.
VIGiven x_0 = 2.5 and the formula x_{n+1} = \sqrt[3]{3x_n + 5}.
VIIx_0 = 2.5
VIIIx_1 = \sqrt[3]{3(2.5) + 5} = \sqrt[3]{7.5 + 5} = \sqrt[3]{12.5} = 2.3223945...
9x_2 = \sqrt[3]{3(2.3223945...) + 5} = \sqrt[3]{6.9671835... + 5} = \sqrt[3]{11.9671835...} = 2.2879505...
10x_3 = \sqrt[3]{3(2.2879505...) + 5} = \sqrt[3]{6.8638515... + 5} = \sqrt[3]{11.8638515...} = 2.2798938...
11x_4 = \sqrt[3]{3(2.2798938...) + 5} = \sqrt[3]{6.8396814... + 5} = \sqrt[3]{11.8396814...} = 2.2779816...
12x_5 = \sqrt[3]{3(2.2779816...) + 5} = \sqrt[3]{6.8339448... + 5} = \sqrt[3]{11.8339448...} = 2.2775330...
13x_6 = \sqrt[3]{3(2.2775330...) + 5} = \sqrt[3]{6.8325990... + 5} = \sqrt[3]{11.8325990...} = 2.2774238...
14x_7 = \sqrt[3]{3(2.2774238...) + 5} = \sqrt[3]{6.8322714... + 5} = \sqrt[3]{11.8322714...} = 2.2773970...
15Rounding to 2 decimal places:
16x_5 \approx 2.28
17x_6 \approx 2.28
18x_7 \approx 2.28
19Since x_5, x_6, and x_7 all round to 2.28 (2 d.p.), the root is 2.28.

Answer

The root is 2.28 (2 d.p.).

The change of sign method is a fundamental way to demonstrate the existence of a root within an interval before applying iteration.

Example 3

The equation x^2 - 5x + 3 = 0 can be rearranged to form the iterative formula x_{n+1} = \frac{3}{5-x_n}. Use this formula with x_0 = 0.5 to find the root correct to 3 decimal places.

IGiven x_0 = 0.5 and the formula x_{n+1} = \frac{3}{5-x_n}.
IIx_0 = 0.5
IIIx_1 = \frac{3}{5-0.5} = \frac{3}{4.5} = 0.6666666...
IVx_2 = \frac{3}{5-0.6666666...} = \frac{3}{4.3333333...} = 0.6923076...
Vx_3 = \frac{3}{5-0.6923076...} = \frac{3}{4.3076923...} = 0.6963953...
VIx_4 = \frac{3}{5-0.6963953...} = \frac{3}{4.3036046...} = 0.6970405...
VIIx_5 = \frac{3}{5-0.6970405...} = \frac{3}{4.3029594...} = 0.6971488...
VIIIx_6 = \frac{3}{5-0.6971488...} = \frac{3}{4.3028511...} = 0.6971666...
9x_7 = \frac{3}{5-0.6971666...} = \frac{3}{4.3028333...} = 0.6971695...
10x_8 = \frac{3}{5-0.6971695...} = \frac{3}{4.3028304...} = 0.6971700...
11Rounding to 3 decimal places:
12x_6 \approx 0.697
13x_7 \approx 0.697
14x_8 \approx 0.697
15Since x_6, x_7, and x_8 all round to 0.697 (3 d.p.), the root is 0.697.

Answer

The root is 0.697 (3 d.p.).

It is crucial to continue iterating until successive values agree to the required degree of accuracy, not just one value.

Common mistakes

  • ✗Rounding intermediate values too early in the calculation, which can lead to an inaccurate final answer.
  • ✗Using the original equation f(x) = 0 instead of the iterative formula x_{n+1} = g(x_n) for the iterative steps.
  • ✗Incorrectly rearranging the equation f(x) = 0 into the form x = g(x).
  • ✗Not checking if the successive approximations have converged to the required degree of accuracy before stating the final answer.
  • ✗Confusing the notation x_n (the current term) with x_{n+1} (the next term) during calculations.

Exam tips

  • ★Always show your working for each iteration (e.g., x_1 = ..., x_2 = ..., etc.) as marks are usually awarded for these steps.
  • ★Use the 'ANS' button on your calculator to store the previous result and use it in the next calculation. This saves time and ensures maximum accuracy.
  • ★When asked to show that a root exists between two values, use the change of sign method with the original function f(x), clearly stating the sign change and conclusion.
  • ★Read the question carefully for the required degree of accuracy for the final answer (e.g., 2 decimal places, 3 significant figures). Ensure your final answer meets this requirement.
  • ★If an iterative process diverges (values get further away from a root), it means that particular rearrangement or starting value does not work for finding that specific root. Do not force convergence.

Ready to practise?

Try a problem on this topic

Snap a photo or type a question — get step-by-step working instantly.

Solve a problem →← Maths curriculum