Maths · Functions, Sequences and Rates of Change
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Iteration
Using iterative formulae, rearranging equations, and finding roots by a change of sign and trial and improvement.
Learning Objectives
- 1Use an iterative formula \(x_{n+1} = f(x_n)\) to find the next terms from a starting value.
- 2Rearrange an equation into an iterative formula.
- 3Use a change of sign to show that a root lies between two values.
- 4Find a root to a given accuracy by trial and improvement, and find the limit of an iteration.
Solving equations by repeating a process
Many equations cannot be solved exactly by a simple method, but their solutions can be found as accurately as you like by repeating a calculation. Iteration starts with a guess, puts it into a formula, and uses the answer as the next guess. If the values settle down towards a single number, that number is a solution of the equation. Iteration is a Higher tier topic on every board. On a non-calculator paper the numbers are chosen to be simple, so that you can work with fractions and easy decimals.
An iterative formula
The formula \(x_{n+1} = 3 - \dfrac{2}{x_n}\) takes the current value \(x_n\) and gives the next value \(x_{n+1}\). Starting from \(x_0 = 4\), the first answer is \(x_1 = 3 - \dfrac{2}{4} = 2.5\).
Using an iterative formula
- Start The first value is given, such as \(x_0 = 4\).
- Step Substitute the current value into the right-hand side.
- Next value The answer becomes \(x_{1}\), then \(x_{2}\), and so on.
- Keep going Write down every value, so that a pattern is clear.
Using a formula
\(x_{n+1} = 3 - \dfrac{2}{x_n}\) and \(x_0 = 4\). Work out \(x_1\) and \(x_2\).
Show the solutionHide the solution
- 1 First step \(x_1 = 3 - \dfrac{2}{4} = 3 - 0.5 = 2.5\).
- 2 Second step \(x_2 = 3 - \dfrac{2}{2.5} = 3 - 0.8 = 2.2\).
- 3 Pattern The values are getting closer to 2.
- 4 Next value \(x_3 = 3 - \dfrac{2}{2.2} = 2.0909\ldots\), still moving towards 2.
Answer\(x_1 = 2.5\) and \(x_2 = 2.2\)
The values settle down
Each step goes up to the curve, across to the line \(y = x\), and up to the curve again. The steps get smaller, and the values move towards \(x = 2\), where the curve meets the line. A point where the curve meets \(y = x\) is a solution of \(x = 3 - \dfrac{2}{x}\).
Where the iteration ends up
- Limit The number the values approach, here \(2\).
- Why it works At the limit, \(x_{n+1} = x_n\).
- Equation So the limit \(a\) satisfies \(a = 3 - \dfrac{2}{a}\).
- Solving Multiply by \(a\): \(a^2 = 3a - 2\), so \(a^2 - 3a + 2 = 0\) and \(a = 1\) or \(a = 2\).
Rearranging into an iterative formula
An iterative formula comes from rearranging an equation so that \(x\) is on one side.
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Start
\(x^2 - 3x + 2 = 0\).
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Divide by x
\(x - 3 + \dfrac{2}{x} = 0\).
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Rearrange
\(x = 3 - \dfrac{2}{x}\).
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Iterative form
\(x_{n+1} = 3 - \dfrac{2}{x_n}\).
A change of sign
A root of \(f(x) = 0\) lies between two values if \(f\) changes sign between them.
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Method
Work out \(f\) at both values. If one is negative and the other is positive, there is a root between them.
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Condition
The graph must be a continuous curve, with no gaps.
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Narrowing down
Try a value in the middle, and keep the interval where the sign changes.
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To 1 decimal place
Test the midpoint of the last interval, so that you know which way to round.
Trial and improvement
\(f(x) = x^3 + x - 3\). Show that \(f(x) = 0\) has a root between 1 and 2, and find this root to 1 decimal place.
Show the solutionHide the solution
- 1 Test 1 and 2 \(f(1) = 1 + 1 - 3 = -1\) and \(f(2) = 8 + 2 - 3 = 7\). The sign changes, so there is a root between 1 and 2.
- 2 Narrow down \(f(1.2) = 1.728 + 1.2 - 3 = -0.072\), and \(f(1.3) = 2.197 + 1.3 - 3 = 0.497\).
- 3 Interval The root is between 1.2 and 1.3.
- 4 Midpoint \(f(1.25) = 1.953125 + 1.25 - 3 = 0.203\ldots\) is positive, so the root is between 1.2 and 1.25, and rounds to 1.2.
AnswerThe root is 1.2 to 1 decimal place
Test yourself
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1
What does \(x_{n+1}\) mean in an iterative formula?
Show answerHide answer
The next value, found from \(x_n\).
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2
What is true at the limit of an iteration?
Show answerHide answer
\(x_{n+1} = x_n\).
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3
How do you show a root lies between two values?
Show answerHide answer
Show that \(f\) changes sign between them.
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4
Why do you test the midpoint when rounding?
Show answerHide answer
To see which side of the midpoint the root is, and so how to round.
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5
How do you find the limit of an iteration exactly?
Show answerHide answer
Replace \(x_{n+1}\) and \(x_n\) with \(a\) and solve.
Exam technique: iteration
Show every value, and write why you reach your conclusion.
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Write each term
Show \(x_1\), \(x_2\), and so on, so that errors can be spotted.
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Limit equation
Replace \(x_n\) and \(x_{n+1}\) with \(a\) and solve for \(a\).
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Choose the right root
The question will usually tell you the values are positive, or give a start value.
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Conclusion
Write that the sign changes, so there is a root between the two values.
Summary and exam focus
- Iteration repeats a formula, using each answer as the next input.
- The limit \(a\) satisfies \(a = f(a)\).
- A change of sign shows a root between two values, for a continuous function.
- Rearrange an equation to get an iterative formula.
Exam focus
\(x_{n+1} = 1 + \dfrac{6}{x_n}\) converges to a positive number \(a\). Show that \(a^2 - a - 6 = 0\), and find the value of \(a\). (4 marks) (4 marks)
At the limit \(a = 1 + \dfrac{6}{a}\), so \(a^2 = a + 6\) and \(a^2 - a - 6 = 0\). Factorise: \((a - 3)(a + 2) = 0\), and \(a\) is positive, so \(a = 3\).
Key terms
The words this lesson expects you to use. Each one is linked from the first place it appears above.
- Iteration
- A process that is repeated, using each output as the next input.
- Iterative formula
- A formula that gives the next value from the current one.
- Limit
- The value that the terms of an iteration approach.
- Root
- A solution of an equation \(f(x) = 0\).
- Change of sign
- The function is negative at one value and positive at another.
- Continuous
- A graph with no breaks or gaps.
- Trial and improvement
- Testing values to find a solution to a given accuracy.
- Converge
- To get closer and closer to a value.
- Starting value
- The first input of an iteration.
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