Locating roots by change of sign, iterative methods, the Newton-Raphson method, and the trapezium rule for numerical integration.

What this dot point is asking

WJEC wants you to locate roots by the change-of-sign method, to use iterative formulae $x_{n+1} = g(x_n)$ to converge on a root, to apply the Newton-Raphson method, and to estimate a definite integral with the trapezium rule. These methods handle equations and integrals that have no neat closed-form answer, which is most of the realistic ones.

The answer

Locating roots by change of sign

If $f$ is continuous and $f(a)$ and $f(b)$ have opposite signs, then $f(x) = 0$ has at least one root between $a$ and $b$.

Iterative methods

Rearranging $f(x) = 0$ into the form $x = g(x)$ gives an iteration $x_{n+1} = g(x_n)$. Starting from an estimate $x_0$, repeated application converges to a root when the iteration is well chosen. A staircase or cobweb diagram on $y = x$ and $y = g(x)$ shows the convergence.

The Newton-Raphson method

Newton-Raphson uses the tangent at the current estimate to find the next, usually converging much faster than a simple iteration.

The trapezium rule

The trapezium rule approximates the area under a curve by dividing it into vertical strips of equal width $h$ and treating each strip as a trapezium.

For a curve that is concave up, the trapezium rule overestimates the area; for a concave-down curve it underestimates.

Examples in context

Example 1. Estimating an awkward integral. To estimate $\displaystyle\int_0^2 \sqrt{1 + x^3}\,dx$ with $4$ strips, $h = 0.5$ and you evaluate $\sqrt{1 + x^3}$ at $x = 0, 0.5, 1, 1.5, 2$. The trapezium rule combines these as $\tfrac{0.5}{2}[y_0 + y_4 + 2(y_1 + y_2 + y_3)]$, giving an approximation where no elementary antiderivative exists. The method makes the intractable integral computable.

Example 2. Iteration convergence. Rearranging $x^3 + x - 1 = 0$ as $x = \dfrac{1}{x^2 + 1}$ and iterating from $x_0 = 0.5$ gives $x_1 = 0.8$, $x_2 \approx 0.61$, $x_3 \approx 0.728$, converging towards the root near $0.682$. The fixed-point iteration homes in on the solution step by step.

Try this

Q1. Show $f(x) = x^2 - 2$ has a root between $1$ and $2$. [2 marks]

• Cue. $f(1) = -1$, $f(2) = 2$; sign change and $f$ continuous, so a root lies between.

Q2. Apply one Newton-Raphson step to $f(x) = x^2 - 5$ from $x_0 = 2$. [3 marks]

• Cue. $f' = 2x$: $x_1 = 2 - \dfrac{4 - 5}{4} = 2 + 0.25 = 2.25$.

Q3. State how many ordinates are used in the trapezium rule with $5$ strips. [1 mark]

• Cue. $5 + 1 = 6$ ordinates.