Numerical solution of equations

A-Level Maths 9709 · Topic 3.6 revision notes

Everything from the 2 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.

Locating roots by a change of sign
Iterative methods

1Locating roots by a change of sign

Syllabus 3.6

Write the equation as \(f(x) = 0\). If \(f\) is continuous on \([a, b]\) and \(f(a)\) and \(f(b)\) have opposite signs, then there is at least one root between \(a\) and \(b\). To show a root is \(\alpha\) correct to 2 d.p., show the sign changes between \(\alpha - 0.005\) and \(\alpha + 0.005\).

Fig. 1: y = ln x and y = 3 − x cross exactly once, so ln x = 3 − x has one root α, between 2 and 3 (α ≈ 2.21)

Worked example: Graphs and a sign change

By sketching suitable graphs, show that the equation \(\ln x = 3 - x\) has exactly one real root. Verify by calculation that this root lies between 2 and 3.

Sketch \(y = \ln x\) (increasing, through \((1, 0)\)) and \(y = 3 - x\) (decreasing straight line). An increasing curve and a decreasing line cross exactly once (Fig. 1).
Let \(f(x) = \ln x + x - 3\), continuous for \(x > 0\).
\(f(2) = \ln 2 - 1 = -0.307\) and \(f(3) = \ln 3 = 1.099\).
The sign changes and \(f\) is continuous, so the root lies between 2 and 3.

Answer: One root, \(2 < \alpha < 3\)

Rearrange to f(x) = 0, then test the sign.

Sketch both sides of the equation to count roots and see roughly where they are.
Rearrange to \(f(x) = 0\) and check \(f\) is continuous on the interval.
Evaluate \(f(a)\) and \(f(b)\), in radians for trig, and quote the values.
Conclude: "sign change, and \(f\) continuous, so a root lies between \(a\) and \(b\)".
Accuracy: test the rounding bounds, e.g. 1.505 and 1.515 for 1.51.

Watch out!

In a sketch-and-count question the marks are for two correct graph shapes in the right places AND a stated conclusion ("one point of intersection, so one root"). A sketch with no statement does not earn the final mark.

Compare with zero, not with each other: you must rearrange to \(f(x) = 0\) first. Checking that \(\ln 2 < 3 - 2\) and \(\ln 3 > 3 - 3\) is fine, but only if you say clearly that the sign of \(\ln x - (3 - x)\) has changed.

No sign change does NOT mean no root. \(f(x) = (x-1)^2\) has a root at \(x = 1\) but is positive on both sides; and two roots between \(a\) and \(b\) cancel out the sign change. A sketch protects you from both traps.

Revisit it: Locating roots by a change of sign lesson · worked practice

2Iterative methods

Syllabus 3.6

If the sequence given by \(x_{n+1} = F(x_n)\) converges to \(\alpha\), then \(\alpha = F(\alpha)\): the limit is a root of the equation \(x = F(x)\). Iterate until successive values agree to the accuracy required, then give the root to that accuracy.

Fig. 1: a cobweb diagram for x(n+1) = cos x(n) starting at x1 = 0.2: go up to the curve, across to y = x, up or down to the curve again; the path spirals in to the crossing point α ≈ 0.739, the root of x = cos x

Worked example: Find the exact limit

The sequence defined by \(x_{n+1} = \dfrac{2x_n}{3} + \dfrac{4}{3x_n^2}\), \(x_1 = 1.5\), converges to \(\alpha\). Find the exact value of \(\alpha\).

At the limit, \(\alpha = \dfrac{2\alpha}{3} + \dfrac{4}{3\alpha^2}\).
Subtract \(\dfrac{2\alpha}{3}\): \(\dfrac{\alpha}{3} = \dfrac{4}{3\alpha^2}\).
Multiply by \(3\alpha^2\): \(\alpha^3 = 4\).

Answer: \(\alpha = \sqrt[3]{4}\)

Rearrange, iterate, stop, check.

Rearrange the equation into \(x = F(x)\), or check that a given \(F\) comes from it.
Iterate with ANS, writing each value to one more decimal place than needed.
Stop when successive values agree to the required accuracy.
Exact limit: put \(x_{n+1} = x_n = \alpha\) and solve.
Failure: values that run away or oscillate more widely mean the rearrangement does not converge.

Watch out!

Keep \(x_{n+1}\) and \(x_n\) distinct while iterating, but set them BOTH equal to \(\alpha\) (or \(x\)) when finding the equation. Writing \(\alpha = \dfrac{2x}{3} + \ldots\) with a mixture of letters goes nowhere.

Do not round each iterate before feeding it back in: use the full calculator value (ANS). Rounding early can stall the sequence or change the final digit.

A start close to the root does not guarantee convergence: the rearrangement decides. In an exam, if your values run away, recheck the arithmetic and calculator mode (radians!) before concluding that the iteration fails.

Revisit it: Iterative methods lesson · worked practice