HookThe equation you cannot solve — that still points the telescope
In 1609 Johannes Kepler published the equation that governs where a planet is in its orbit at a given time: \(M=E-e\sin E\), where \(M\) is set by the clock, \(e\) is the orbit's eccentricity, and \(E\) is the angle he actually needed. There is a problem that has never gone away in four centuries: you cannot rearrange that equation to make \(E\) the subject. No amount of algebra isolates \(E\), because it sits both inside and outside a sine. Kepler solved it by hand, guessing and correcting; today the flight software that points the James Webb Space Telescope and keeps every GPS satellite on schedule solves the very same equation millions of times a day — not with algebra, but by iterating to a numerical answer good enough for the job.
That is the whole premise of Chapter 9. Most equations that matter in science and engineering have no tidy closed-form solution, so we find the root approximately, to whatever accuracy is required, and we watch carefully for the ways the process can betray us. You will locate a root by spotting a change of sign, close in on it with fixed-point iteration (and read its behaviour off a cobweb or staircase diagram), accelerate with the Newton-Raphson method, learn precisely how each method can fail, estimate an area no formula gives you with the trapezium rule, and apply all of it to problems in context. The running example below is the cubic \(x^3-2x-5=0\) — the very equation Newton chose in the 1660s to demonstrate his method, whose root is about \(2.0946\).
ModelLocating a root by change of sign — and when it lies to you
The starting move needs only one idea from earlier maths: if \(f\) is continuous on \([a,b]\) and \(f(a)\) and \(f(b)\) have opposite signs, then somewhere between them the graph must cross the axis, so there is a root in \((a,b)\). This is the change-of-sign method, and it is how you justify that a root exists before you go hunting for its value. The word 'continuous' is load-bearing and must be stated — the argument is about the graph having no breaks.
The examinable subtlety is how it fails. First, a sign change guarantees an odd number of roots in the interval, not exactly one — three roots also flip the sign. Second, no sign change does not prove there is no root: a repeated root (an even multiplicity), where the curve touches the axis and turns back, keeps \(f(a)\) and \(f(b)\) the same sign despite a root sitting between them. Third, if \(f\) has a discontinuity — think \(\dfrac{1}{x-2}\), which flips sign across \(x=2\) without ever being zero — the method reports a 'root' that does not exist. Whenever you use a sign change, you are implicitly promising the function is continuous across the interval.
Show that \(f(x)=x^3-2x-5\) has a root between \(x=2\) and \(x=3\).
Evaluate at the ends: \(f(2)=2^3-2(2)-5=8-4-5=-1\), and \(f(3)=3^3-2(3)-5=27-6-5=16\). One is negative and one is positive — a change of sign. Since \(f\) is a polynomial it is continuous everywhere, in particular on \([2,3]\), so by the change-of-sign argument there is at least one root \(\alpha\) with \(2<\alpha<3\). Note the wording carefully: 'at least one', because in principle three roots could lie in the interval; and the sentence only works because we noted \(f\) is continuous — drop that clause and the mark goes with it. The root turns out to be near \(2.0946\), which the next two blocks close in on.
MechanismFixed-point iteration — and the cobweb and staircase that reveal it
To find the root's value, rearrange \(f(x)=0\) into the form \(x=g(x)\), then feed a starting guess through repeatedly: \(x_{n+1}=g(x_n)\). If the sequence settles down, its limit \(\alpha\) satisfies \(\alpha=g(\alpha)\), which is the root. The same equation can be rearranged several ways, and this is where care is needed: some rearrangements converge to the root and some diverge away from it. The test is the gradient — iteration converges near the root when \(|g'(\alpha)|<1\), and the flatter \(g\) is there, the faster it homes in.
The behaviour is made visible by drawing \(y=x\) and \(y=g(x)\) on the same axes and bouncing between them. A staircase diagram appears when \(g'(\alpha)\) is positive: the iterates step towards (or away from) the root from one side, like climbing stairs. A cobweb diagram appears when \(g'(\alpha)\) is negative: the iterates spiral inward (or outward), alternating sides of the root. Convergence looks like the steps or the spiral tightening onto the intersection of the two lines; divergence looks like them flying apart. Being able to sketch which one occurs, and say whether it converges, is a standard exam ask.
Use the rearrangement \(x=\sqrt[3]{2x+5}\) with \(x_0=2\) to estimate the root of \(x^3-2x-5=0\).
From \(x^3=2x+5\) we get \(x=(2x+5)^{1/3}\), so \(x_{n+1}=(2x_n+5)^{1/3}\). Iterate from \(x_0=2\):
\[x_1=(2(2)+5)^{1/3}=9^{1/3}=2.0801,\quad x_2=(2(2.0801)+5)^{1/3}=2.0923,\] \[x_3=(2(2.0923)+5)^{1/3}=2.0942,\quad x_4=2.0945.\]
The iterates rise steadily towards \(2.0946\) from below — a staircase, because \(g'(x)=\tfrac{2}{3}(2x+5)^{-2/3}\) is small and positive near the root, comfortably satisfying \(|g'(\alpha)|<1\). Had we instead rearranged to \(x=\tfrac{x^3-5}{2}\), the derivative \(g'(x)=\tfrac{3x^2}{2}\approx 6.6\) at the root would exceed \(1\), and the iteration would have diverged — same equation, useless rearrangement.
MechanismNewton-Raphson — the tangent that races to the root
Fixed-point iteration crawls; Newton-Raphson sprints. The idea is geometric: from a guess \(x_n\), follow the tangent to the curve down to where it hits the \(x\)-axis, and use that crossing as your next guess. Writing the tangent line and solving for its root gives the formula
\[x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.\]
When it works, it works spectacularly — the number of correct digits roughly doubles each step (quadratic convergence), which is why it lives inside calculators and solvers. But it needs the derivative, and it has sharp failure modes you must be able to name. If \(f'(x_n)=0\) at or near an iterate, the tangent is horizontal and never meets the axis — the formula divides by zero and blows up. A starting guess near a turning point can fling the next iterate far away, sometimes converging to a different root than intended or diverging entirely. And if the tangent repeatedly lands on the same two values, the iteration can cycle forever. Newton-Raphson is fast but not foolproof; a bad start is punished hard.
Apply Newton-Raphson to \(f(x)=x^3-2x-5\) from \(x_0=2\).
Differentiate: \(f'(x)=3x^2-2\). The iteration is \(x_{n+1}=x_n-\dfrac{x_n^3-2x_n-5}{3x_n^2-2}\). Step one, \(x_0=2\): \(f(2)=-1\), \(f'(2)=3(4)-2=10\), so \(x_1=2-\dfrac{-1}{10}=2.1\). Step two, \(x_1=2.1\): \(f(2.1)=9.261-4.2-5=0.061\), \(f'(2.1)=3(4.41)-2=11.23\), so \(x_2=2.1-\dfrac{0.061}{11.23}=2.09457\). Step three lands on \(x_3=2.094551\), correct to six decimal places. Compare with the previous block: fixed-point iteration needed four steps to reach three-decimal accuracy, while Newton-Raphson reached six in three — the doubling of correct digits per step in action. The catch to remember: starting instead at \(x_0=0\), where the curve is near-flat, would have sent the first tangent off to a wildly different value.
MechanismThe trapezium rule — area when there is no antiderivative
Some integrals cannot be evaluated exactly — there is simply no elementary antiderivative — yet you still need the area. The trapezium rule estimates it by slicing \([a,b]\) into \(n\) strips of equal width \(h=\dfrac{b-a}{n}\), reading off the \(y\)-values \(y_0,y_1,\dots,y_n\) at the strip edges, and approximating each strip by a trapezium rather than a rectangle:
\[\int_a^b y\,dx\approx \frac{h}{2}\Big[\,y_0+y_n+2(y_1+y_2+\dots+y_{n-1})\,\Big].\]
The structure is worth memorising as 'half the width, times the ends plus twice the middles'. Because each strip is topped by a straight chord instead of the true curve, the estimate is biased, and you can predict the direction. Where the curve is convex (\(f''>0\), bending upward) the chords sit above the curve, so the rule overestimates; where the curve is concave (\(f''<0\), bending downward) the chords sit below, so it underestimates. Being able to state, with a reason, whether your estimate is an over- or under-estimate is a routine follow-up mark — and using more strips (smaller \(h\)) always tightens the estimate.
Estimate \(\displaystyle\int_0^4 \sqrt{x}\,dx\) using the trapezium rule with \(4\) strips, and say whether it over- or under-estimates.
With \(n=4\) strips on \([0,4]\), the width is \(h=\dfrac{4-0}{4}=1\), and the \(x\)-values are \(0,1,2,3,4\). Compute the heights \(y=\sqrt{x}\): \(y_0=0,\ y_1=1,\ y_2=1.4142,\ y_3=1.7321,\ y_4=2\). Apply the rule:
\[\int_0^4\sqrt{x}\,dx\approx \frac{1}{2}\Big[0+2+2(1+1.4142+1.7321)\Big]=\frac{1}{2}\big[2+8.2926\big]=5.146.\]
The exact value is \(\displaystyle\int_0^4 x^{1/2}\,dx=\left[\tfrac{2}{3}x^{3/2}\right]_0^4=\tfrac{2}{3}(8)=5.333\). Our estimate \(5.146\) is smaller — an underestimate — exactly as predicted, because \(y=\sqrt{x}\) is concave (it bends downward), so every chord lies below the curve. Quoting that reason, not just the number, is what earns the justification mark.
CaseNumerical methods in context — from car speedometers to interest rates
The final spec point drops the abstraction: real problems arrive as data or as equations no formula cracks, and you deploy these tools on them. A velocity-time table from a car's telemetry has no equation at all, yet the distance travelled is the area under it — so the trapezium rule turns speedometer readings into a distance estimate (and, because a car accelerating hard traces a convex curve, you can even say whether that estimate is high or low). A savings or mortgage problem produces a high-degree polynomial in the interest rate with no algebraic solution — so a spreadsheet finds the rate by Newton-Raphson, which is precisely how the 'IRR' function works. And a designed component whose stress equation mixes polynomials with trig functions is solved for its safe load by iteration.
The examiner's extra demand in context is judgement: choose a sensible method, state the accuracy you are working to, and comment on reliability. A change of sign confirms a root exists; an iteration or Newton step finds it; and you confirm the final answer to the required accuracy by checking a sign change across the rounding interval. Never present a numerical answer as exact — it is an approximation, and saying so, with its accuracy, is part of the mathematics.
A car's speed is logged every \(2\) seconds as it accelerates: at \(t=0,2,4,6,8\) s the speed is \(0,\ 6,\ 14,\ 24,\ 30\) m/s. Estimate the distance travelled in the \(8\) seconds.
Distance is the area under the speed-time graph, and we only have five readings, so use the trapezium rule with \(h=2\) s and heights \(y_0=0,\ y_1=6,\ y_2=14,\ y_3=24,\ y_4=30\):
\[\text{distance}\approx \frac{2}{2}\Big[0+30+2(6+14+24)\Big]=1\times\big[30+2(44)\big]=30+88=118\ \text{m}.\]
So the car covers roughly \(118\) m. Because the speed is rising ever more steeply for most of the interval — a broadly convex trace — the straight chords sit slightly above the true curve early on, so this is likely a small overestimate; with readings only every \(2\) s it is an approximation, and logging speed every half-second would tighten it. Stating the method, the estimate, and its limitation together is the full contextual answer.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
For a change-of-sign question, write both function values, state that they have opposite signs, and — the clause candidates forget — say that \(f\) is continuous on the interval; without it the deduction is not valid and the mark is lost. When asked to show a root is accurate to a given number of decimal places, evaluate \(f\) at the two ends of the rounding interval (e.g. \(2.0945\) and \(2.0955\) for '\(2.095\) to 3 d.p.'), show the sign change, and conclude — an iteration alone does not prove the accuracy.
In iteration and Newton-Raphson questions, quote the formula, then show every iterate to the accuracy asked, keeping full calculator precision between steps and only rounding at the end. State the Newton-Raphson formula with \(f\) and \(f'\) written out; a slip in \(f'\) poisons every line after it. For the trapezium rule, write \(h\), lay the \(y\)-values out in a small table, and reproduce the \(\frac{h}{2}[\text{ends}+2(\text{middles})]\) structure explicitly so a single arithmetic slip does not cost the method marks — then justify any over- or under-estimate by naming the curve's concavity, not by comparing with an exact value you were not given. Throughout, present numerical answers as approximations to a stated accuracy: 'exact' is the wrong word for everything in this chapter.