Most equations cannot be solved exactly. Most areas under curves cannot be integrated by formula. Numerical methods get a good enough answer anyway — by approximating, then improving.
Chord gradients replace differentiation, when there is no formula
The trapezium rule replaces integration, when the integral is intractable
| Word | Meaning |
|---|---|
| Root | A solution of $\mathrm{f}(x) = 0$. |
| Change of sign | $\mathrm{f}(a)$ and $\mathrm{f}(b)$ have opposite signs, so a root lies between. |
| Iteration | Repeating a formula, feeding each answer back in. |
| Converge | The iterates settle towards a fixed value. |
| Diverge | The iterates run away, without settling. |
| Chord | A straight line joining two points on a curve. |
| Trapezium rule | Estimating an area with trapezium-shaped strips. |
| Over-estimate | An approximation larger than the true value. |
| Ordinate | One of the $y$-values at a strip boundary. |
then a root lies between $a$ and $b$
where $h = \dfrac{b-a}{n}$. This formula will be provided in the examination.
Show that $x^3 + x - 3 = 0$ has a root between $1$ and $2$, and find it to $1$ decimal place.
- NM1
NM3Change of SignLocating a root between two values, and decimal search. - NM2
NM3Iterative Methods$x_{n+1} = \mathrm{g}(x_n)$, convergence, and when iteration fails. - NM4
NM5Gradients from ChordsEstimating a tangent's gradient, and improving the estimate. - NM6
NM7
NM8Estimating AreasRectangular strips, the trapezium rule, and over- or under-estimates. - NM9Numerical Methods in ContextApplying these techniques to real data and real problems.
Change of sign
Opposite signs at the ends $\Rightarrow$ a root between.
State everything
Both values, both signs, the conclusion.
Continuity
Essential — asymptotes can fake a sign change.
Iteration
$x_{n+1} = \mathrm{g}(x_n)$ with a starting value.
ANS key
The fastest way to iterate.
Full accuracy
Never round mid-iteration.
Chord gradient
$\dfrac{\mathrm{f}(x+h)-\mathrm{f}(x)}{h}$.
Improve it
Make $h$ smaller.
Trapezium rule
Ends once, middles doubled, all $\times \tfrac{h}{2}$.
Over or under?
Decide from the curvature, and say why.
Show that $x^3 - 5 = 0$ has a root between $1$ and $2$.
▶ Show solution
$\mathrm{f}(1) = 1 - 5 = -4$ (negative)
$\mathrm{f}(2) = 8 - 5 = 3$ (positive)
The sign changes and $\mathrm{f}$ is continuous, so a root lies between $1$ and $2$.
$\mathrm{f}(x) = x^3 + 2x - 8$. Evaluate $\mathrm{f}(1)$ and $\mathrm{f}(2)$ and state what you conclude.
▶ Show solution
$\mathrm{f}(1) = 1 + 2 - 8 = -5$
$\mathrm{f}(2) = 8 + 4 - 8 = 4$
Opposite signs, so a root lies in $(1, 2)$.
Use the iteration $x_{n+1} = \sqrt{\dfrac{6}{x_n}}$ with $x_1 = 2$ to find $x_2$ and $x_3$.
▶ Show solution
$x_2 = \sqrt{3} = 1.73205$
$x_3 = \sqrt{6 \div 1.73205} = \sqrt{3.46410} = 1.86121$
Estimate the gradient of $y = x^2$ at $x = 3$ using the chord from $x = 3$ to $x = 3.1$.
▶ Show solution
$\dfrac{3.1^2 - 3^2}{0.1} = \dfrac{9.61 - 9}{0.1} = \dfrac{0.61}{0.1} = 6.1$
(The exact gradient is $6$, so the estimate is slightly high.)
Use the trapezium rule with two strips to estimate $\displaystyle\int_0^2 x^2 \,\mathrm{d}x$.
▶ Show solution
$h = 1$; ordinates at $x = 0, 1, 2$ are $0, 1, 4$.
$\approx \tfrac12(1)\Big[(0 + 4) + 2(1)\Big] = \tfrac12(6) = 3$
(The exact value is $\tfrac83 = 2.667$, so this over-estimates.)
How many ordinates are needed for the trapezium rule with $5$ strips?
▶ Show solution
$6$ ordinates — always one more than the number of strips.
Explain how to improve a chord estimate of a gradient.
▶ Show solution
Use a smaller interval $h$, so the second point is closer to the first.
The chord then lies closer to the tangent, so its gradient is a better approximation.
Explain how to improve a trapezium rule estimate.
▶ Show solution
Use more strips (a smaller $h$).
Each trapezium's straight top then hugs the curve more closely, so less area is wrongly included or excluded.
Give an example showing that a change of sign does not always mean there is a root.
▶ Show solution
Take $\mathrm{f}(x) = \dfrac{1}{x}$.
$\mathrm{f}(-1) = -1$ and $\mathrm{f}(1) = 1$ — the sign changes.
But $\dfrac1x$ is never zero, so there is no root in $(-1, 1)$.
The sign change happens because the function is discontinuous at $x = 0$, jumping from $-\infty$ to $+\infty$ rather than passing through zero. This is why continuity must always be checked.
$\mathrm{f}(x) = x^3 - 2x - 5$.
(a) Show that a root lies between $2$ and $3$. (b) Use a decimal search to find it to $1$ d.p. (c) Show that the equation can be written as $x = \sqrt[3]{2x+5}$, and use the iteration $x_{n+1} = \sqrt[3]{2x_n+5}$ with $x_1 = 2$ to find the root to $3$ d.p. (d) Compare the two methods.
▶ Show solution
(a) $\mathrm{f}(2) = 8 - 4 - 5 = -1$ (negative)
$\mathrm{f}(3) = 27 - 6 - 5 = 16$ (positive)
Sign change with $\mathrm{f}$ continuous, so a root lies in $(2, 3)$.
(b) $\mathrm{f}(2.1) = 9.261 - 4.2 - 5 = 0.061$ (positive)
So the root is in $(2.0, 2.1)$. Test the midpoint:
$\mathrm{f}(2.05) = 8.615 - 4.1 - 5 = -0.485$ (negative)
The root is in $(2.05, 2.1)$, so it rounds to $\mathbf{2.1}$ to $1$ d.p.
(c) $x^3 - 2x - 5 = 0 \;\Rightarrow\; x^3 = 2x + 5 \;\Rightarrow\; x = \sqrt[3]{2x+5}$ ✓
$x_1 = 2$
$x_2 = \sqrt[3]{9} = 2.080084$
$x_3 = \sqrt[3]{9.160168} = 2.092351$
$x_4 = \sqrt[3]{9.184701} = 2.094217$
$x_5 = \sqrt[3]{9.188434} = 2.094501$
$x_6 = \sqrt[3]{9.189001} = 2.094544$
The iterates are settling, so the root is $\mathbf{2.095}$ to $3$ d.p.
(d) The decimal search is guaranteed to work and gives a bracket containing the root, so the accuracy is certain — but it needs a new function evaluation for every extra digit, and progress is slow.
The iteration converged to $3$ decimal places in only five steps, and with the ANS key each step is one button press. But it offers no guarantee: a different rearrangement, such as $x = \tfrac{x^3-5}{2}$, would diverge from the same start.
In practice the two are used together: a change of sign to prove the root exists and locate it roughly, then an iteration to refine it quickly.