Numerical methods in contextAQA A-Level Maths: Revision notes

Many problems cannot be solved analytically, meaning by exact algebra. Equations such as $e^x=3x$ and $d+\ln d=4$ mix a variable with its exponential or logarithm, and integrals such as $\int e^{-x^2}\,dx$ have no simple antiderivative. Numerical methods give a solution to a required level of accuracy instead of an exact one. In practice they are used in science, engineering and finance wherever the exact value is out of reach.

Roots of $f(x)=0$: a change of sign or interval bisection is reliable but slow; fixed-point iteration $x_{n+1}=g(x_n)$ needs a rearrangement with $|g'(\alpha)|<1$; Newton-Raphson $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ is usually fastest but needs $f'$ and fails if $f'(x_n)=0$. Areas and integrals: the trapezium rule with enough strips. For $f(\theta)=\theta-0.5\sin\theta-1$, $f'(\theta)=1-0.5\cos\theta$ and $\theta_0=1.5$ gives $\theta_1=1.4987$ and $\theta_2=1.498701$ in only two steps.

First form the equation from the context, then rewrite it as $f(x)=0$ with defined units. For $d+\ln d=4$, $f(d)=d+\ln d-4$, $f(2.9)=-0.0353$ and $f(3)=0.0986$, so $2.9<d<3$. The iteration $d_{n+1}=4-\ln d_n$ with $d_0=3$ gives $d_1=2.901$ and $d_2=2.935$, which settle on the root. Always check that the answer makes sense: depths and widths must be positive, and the units must be stated.

Stop when successive iterates agree to the required number of decimal places, but this alone is not a proof. To show a root is $2.46$ to 2 d.p., show a change of sign over $[2.455,2.465]$. For $f(x)=x^3-4x-5$: $f(2.455)=-0.0237<0$ and $f(2.465)=0.1179>0$, so $\alpha=2.46$ (2 d.p.). For integrals, compare estimates with different numbers of strips, or compute the percentage error $\frac{\text{estimate}-\text{exact}}{\text{exact}}\times100$ when an exact value is known. For $\int_0^1e^{-x^2}\,dx$, 4 strips give $0.743$ against $0.7468$, an error of $0.51\%$.

Be ready to say why a method fails or is poor. A rearrangement fails if $|g'(\alpha)|>1$: for $x^3-4x-5=0$, $x_{n+1}=\frac{x_n^3-5}{4}$ has $g'(x)=\frac{3x^2}{4}\approx4.5$ at the root, and from $x_0=2.5$ it gives $2.656$, $3.435$, diverging; the rearrangement $x_{n+1}=\sqrt[3]{4x_n+5}$ converges. Quote the numbers and link them to the behaviour. State the limits of a method too: bisection is slow, Newton-Raphson can fail near stationary points, and the trapezium rule is only as accurate as the number of strips allows.