GCSE · Maths · Edexcel · Spec 1MA1 · Higher

Iterative methods

Put a guess into a formula, put the answer back in, and keep going. With the right formula, you walk straight to a solution that algebra can't reach.

Iteration

Catch a solution between two values

The formula xₙ₊₁ = ∛(21 − 4xₙ) takes one value of x and gives back the next. Start with x₀ = 2, feed each answer straight back in, and you get the six values below. Place each one on the line, then check.

x₀ — x₀ = 2 (the start)x₁ — x₁ = 2.3513x₂ — x₂ = 2.2634x₃ — x₃ = 2.2860x₄ — x₄ = 2.2802x₅ — x₅ = 2.2817

The full method, one step at a time

Problem

Use iteration to find a solution of x³ + 4x − 21 = 0, correct to 3 significant figures. Start with x₀ = 2.

Does it converge?

Sort the runs of iterations

Where does each run of iterations belong?

Still to sort

Converging (0)

The gaps between successive iterations are getting smaller.

Where the line is: The direction doesn't matter. Up, down or zigzag, what counts is shrinking gaps.

Not converging (0)

The gaps between successive iterations are getting bigger.

Where the line is: A run can zigzag and still not converge, if each swing is wider than the one before.

Can't use this x₀ (0)

The formula can't even be worked out for this starting value.

Where the line is: This is about the starting value, not the gaps. You never get a second value to compare.

6 of 6 still to sort.

Work out the gap between each value and the next. Are the gaps getting smaller or bigger?

Watch out: Values that go up every time can still converge. So can values that zigzag. The only question is whether the gaps are getting smaller.

Spot the slip

Where does this answer lose accuracy?

Use xₙ₊₁ = ∛(8 − xₙ) with x₀ = 2 to find x₁, x₂ and x₃, each correct to 4 decimal places.

A student's answer — which line goes wrong?

x₁ = ∛(8 − 2) = ∛6 = 1.817120…, so x₁ = 1.8171
To keep things quick, I'll use x₁ = 1.82 as the next input.
x₂ = ∛(8 − 1.82) = ∛6.18 = 1.835113…, so x₂ = 1.8351
x₃ = ∛(8 − 1.835113…) = 1.833616…, so x₃ = 1.8336
Answer: x₁ = 1.8171, x₂ = 1.8351, x₃ = 1.8336

Exam line: Don't round a single iteration and call it the answer, either. Keep going until successive iterations round to the same value.

Where the sign flips

x: 2. y = x³ + 4x − 21: -5

slide across where the curve meets y = 0

Watch out: A change of sign only works on an equation arranged to equal 0. Rearrange first, then substitute.

Is it proof?

How far can you trust a change of sign?

You substitute two values of x into an equation arranged to equal 0 and compare the signs of the results.

Which of these is closest to what you think right now?

If the two results have different signs, there must be a solution between them.If the two results have the same sign, there can't be a solution between them.A change of sign shows a solution when the graph crosses the axis between the two values, and no change of sign doesn't rule one out.

How sure are you?

I'm sureI think soI'm guessing

If the two results have different signs, there must be a solution between them.

This is a common misconception

Only if the graph actually crosses the x-axis between them. Try y = 1/x. At x = −1 you get −1 and at x = 1 you get 1, so the sign changes. But the graph of y = 1/x never touches the x-axis. It jumps across at an asymptote instead. For straight lines, parabolas and cubics, a change of sign does mean the graph crosses.

If the two results have the same sign, there can't be a solution between them.

This is a common misconception

There still might be. For y = (x − 2)², x = 1 and x = 3 both give 1, yet x = 2 is a solution. The curve just touches the axis there (a repeated root), so the sign never changes. And for y = x² − 5x + 6, x = 1 and x = 4 both give 2, but two solutions, 2 and 3, are hiding between them.

A change of sign shows a solution when the graph crosses the axis between the two values, and no change of sign doesn't rule one out.

This matches the evidence

That's it. For lines, parabolas and cubics, a change of sign does mean the graph crosses. Either way, choose values close together or look at the graph. Then you won't be fooled by a jump, a touch or a pair of hidden solutions.

Different signs point to a solution when the graph crosses in between. Same signs don't prove there isn't one. Keep your values close together, and if you're unsure, sketch the graph.

WHAT YOU'VE LEARNED

A quick recap of today's lesson.

xₙ₊₁ = g(xₙ)

Use each answer as the next input, and let the values close in on a solution you can't find with algebra.

What you need to know

Iteration means doing the same calculation again and again, with each output becoming the next input.
In xₙ₊₁ = g(xₙ), xₙ is the input and xₙ₊₁ is the output. x₀ is the starting value, x₁ is the first iteration, and so on.
To make an iterative formula, rearrange the equation to get x on its own on one side, with x still on the other side.
The iterations converge when the gaps between successive values keep getting smaller, whichever direction the values move.
A change of sign between the bounds of a rounded answer confirms it to that accuracy.

The big picture

Some equations, like x³ + 4x − 21 = 0, can't be solved with ordinary algebra. Iteration gets round this. You rearrange the equation into a formula like xₙ₊₁ = ∛(21 − 4xₙ), start with a value x₀, and feed each output back in as the next input. If the gaps between the iterations shrink, the values converge on a solution. You state it once successive iterations round to the same value, and you check it by showing the sign changes between its bounds.

Key points

1Iteration: the output of each step becomes the input for the next.

2One equation can be rearranged into many different iterative formulas. A formula might find one solution, several, or none, depending on the formula and on x₀.

3Show x₀ substituted in the first iteration, then use the Ans key so the exact previous output is always the next input.

4Converging means the gaps between successive iterations are getting smaller. If the gaps are growing, it isn't converging.

5Some starting values can't be used at all. It's safest to start reasonably close to a solution.

6Don't round a single iteration. State the answer once successive iterations round to the same value, and if in doubt, do more iterations.

7To check a rounded solution, substitute its lower and upper bounds into the equation arranged to equal 0 and look for a change of sign.

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Worked example

Problem

Iteration suggests that x = 2.09 is a solution of x³ − 2x − 5 = 0, correct to 3 significant figures. Show that this is correct.

⚠ Watch out

Rounding one iteration and calling it the answer. For example, with xₙ₊₁ = ∛(21 − 4xₙ), x₁ = 2.3513 would give 2.35, which is wrong. Only state a value once successive iterations round to it, and always put the exact previous output back in.

🧠

Memory hook

Feed it back, watch the gaps, trap it between the bounds.

✓

Check yourself

Iterating a formula gives 1.52, 1.47, 1.495, 1.4875. Is it converging? Work out the gaps: 0.05, 0.025, 0.0075. They're shrinking, so yes. The values zigzag, so the solution lies between successive values.

Flashcards

(14)

What is iteration?

Repeating the same calculation, with each output used as the next input.

In xₙ₊₁ = g(xₙ), what are xₙ and xₙ₊₁?

xₙ is the input (the value you have now). xₙ₊₁ is the output (the next value).

What does x₀ stand for?

The starting value, the first input. x₁ is the first iteration.

How do you turn an equation into an iterative formula?

Rearrange it so x is on its own on one side and x still appears on the other. Then write xₙ₊₁ on the left and xₙ on the right.

Can one equation give more than one iterative formula?

Yes, many. Different formulas (and different x₀) may find one solution, more than one, or none.

How do you generate iterations quickly on a calculator?

Enter x₀ and press =. Type the formula with Ans wherever xₙ appears, then press = repeatedly.

Why must you never type a rounded value back in?

It loses accuracy in the next iteration. The Ans key keeps the exact previous output.

How can you tell iterations are converging?

The gaps between successive iterations get smaller. The values can go up, go down or zigzag.

What if the gaps between iterations keep growing?

The iteration is not converging. Try a different formula or a starting value closer to the solution.

Why can't xₙ₊₁ = √xₙ start at x₀ = −4?

√(−4) has no real value. For this formula, x₀ must be positive.

When can you state a converging value to a given accuracy?

When successive iterations round to the same value, so the digits are fixed.

Why does substituting a rounded solution not give exactly 0?

Because the solution has been rounded. It's close to the true solution, not equal to it.

What are the bounds of x = 1.83 to 3 significant figures?

1.825 and 1.835.

Does a change of sign always mean a solution?

Only if the graph crosses the x-axis between the values. y = 1/x changes sign at an asymptote without crossing.

Tap any card to flip it, or use Study as deck to go through them one at a time. In the full lesson these run as a spaced-repetition deck — you rate each card Hard, Good or Easy and the tricky ones keep coming back until they stick.

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Catch a solution between two values

Sort the runs of iterations

Where does this answer lose accuracy?

Where the sign flips

How far can you trust a change of sign?

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