Numerical methods in contextEdexcel A-Level Maths: Revision notes

Many real problems end in an equation that cannot be solved by algebra, such as $x^3-4x-7=0$ or $t=5e^{-0.4t}$ (a polynomial or $t$ mixed with an exponential, or $\theta$ mixed with a trigonometric function). The first job is to model the situation: define the variable, write the relationship and move everything to one side to make $f(x)=0$. Example: an open box from a $30$ cm by $20$ cm sheet with squares of side $x$ cm cut from the corners has volume $x(30-2x)(20-2x)=4x^3-100x^2+600x$. For volume $900$ cm$^3$ this gives $f(x)=x^3-25x^2+150x-225=0$. Keep the domain from the context in mind: here $0<x<10$, since the width $20-2x$ must be positive.

If $f$ is continuous on $[a,b]$ and $f(a)$ and $f(b)$ have opposite signs, there is at least one root of $f(x)=0$ in $[a,b]$. A full answer gives: the values (accurate enough to show their sign), the statement that $f$ is continuous, and the conclusion. Example: $f(x)=x^3-4x-7$ has $f(2)=-7$ and $f(3)=8$, so there is a root in $[2,3]$. To find a root to $n$ decimal places, trap it in an interval of width $10^{-n}$ whose end points round differently, such as $2.295$ and $2.305$ for $2.30$. A sign change can fail to locate a root when $f$ is discontinuous, for example $f(x)=\frac{1}{x}$ on $[-1,1]$, and two roots in an interval can hide the sign change altogether.

Rearrange $f(x)=0$ into $x=g(x)$ and use the iterative formula $x_{n+1}=g(x_n)$ from a starting value $x_0$. If the sequence converges, it converges to a root $\alpha$ where $\alpha=g(\alpha)$. The same equation can be rearranged in several ways. From $x^3-4x-7=0$: $x=\sqrt[3]{4x+7}$, or $x=\sqrt{4+\frac{7}{x}}$, or $x=\frac{x^3-7}{4}$. With $x_0=2$, the first gives $x_1=\sqrt[3]{15}=2.466$, $x_2=2.564$, $x_3=2.584$, ... converging to $2.588$ (3 d.p.). Show each step of a rearrangement in a 'show that' question.

The sequence $x_{n+1}=g(x_n)$ converges to $\alpha$ when $|g'(\alpha)|<1$. A positive gradient gives a staircase that approaches $\alpha$ from one side; a negative gradient gives a cobweb that spirals in alternately above and below. For the pond, $\theta-\sin\theta=0.6$ rearranges to $\theta_{n+1}=0.6+\sin\theta_n$. Here $g'(\theta)=\cos\theta$ is close to $0$ near $\theta\approx1.6$, so it converges quickly: $1.5\to1.5975\to1.5996\to1.5996$. A rearrangement such as $x=\frac{x^3-7}{4}$ for the break-even equation has a large gradient near the root and diverges.

Worked example: the concentration equation $t=5e^{-0.4t}$, with $f(t)=t-5e^{-0.4t}$.

1. $f(2)=-0.247$ and $f(3)=1.494$: sign change, so $2<\alpha<3$.
2. $f(2.1)=-0.059$ and $f(2.2)=0.126$, so $2.1<\alpha<2.2$.
3. Iterate $t_{n+1}=5e^{-0.4t_n}$ from $t_0=2$: $t_1=2.247$, $t_2=2.036$, $t_3=2.215$ ... the values oscillate (a cobweb) and close in on $\alpha\approx2.13$.
4. Interpret: the two concentrations are equal about $2.13$ hours after the start. Always give the final answer in the units and with the meaning of the context.