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Newton-Raphson and the trapezium rule

Two methods finish the unit. Newton-Raphson uses successive tangent intersections with the x-axis to approximate a root, converging quickly on a well-behaved curve but failing to converge where the tangent is flat. The trapezium rule approximates a definite integral with straight-topped strips, and a sketch shows in advance whether the estimate is too big or too small.

Year 13EDEXCEL 9MA0 9.3, 9.4, 9.5

Builds on Locating roots and iteration and Tangents, turning points and curve behaviour.

IN THIS TOPIC

  • Run Newton-Raphson, reading each step as a tangent sliding to the axis.
  • Explain the method's failure when the tangent is horizontal or nearly so.
  • Estimate definite integrals by the trapezium rule with a stated number of strips.
  • Decide from the bend of the curve whether the estimate is too big or too small.

COMMON MISCONCEPTION

Newton-Raphson always finds the root, whatever the starting value.

The tangent slide

Newton-Raphson turns differentiation into a root-finder. Stand at an estimate xn, draw the tangent to the curve there, and follow it down to the axis. Where it lands is your next estimate. The booklet prints it as

xn+1=xn−f(xn)f′(xn)x_{n+1} = x_{n} − \frac{f(x_{n})}{f'(x_{n})}IN THE FORMULAE BOOKLET

and on a well-behaved curve the convergence is very fast, roughly doubling the number of correct digits at each step.

Newton-Raphson on x cubed minus x minus 1 from 1.5: tangents cut the axis at 1.3478 then 1.3252, closing fast on the root 1.3247x₀ = 1.5x₁ = 1.3478tangent in, axis-cut out, repeatthen x₂ = 1.3252, x₃ = 1.3247y = x³ − x − 1
FIG. 1Two tangent slides on x³ − x − 1: from 1.5 the tangent cuts at 1.3478, from there at 1.3252, and 1.3247 arrives one step later.

WORKED EXAMPLE

Digits by the doubling

Use Newton-Raphson on f(x) = x3 − x − 1 with x0 = 1.5.

f(1.5) = 0.875 and f'(1.5) = 5.75, so x1 = 1.5 − 0.875/5.75 = 1.3478.

Repeating: x2 = 1.3252, x3 = 1.3247, and the digits have stopped moving at 4 decimal places.

The root is 1.3247. Three steps. The plain iteration in the previous lesson needed roughly twice as many on the same equation, and that speed is why this method gets its own booklet entry.

Newton-Raphson fails at any stationary point. Where f'(xn) is zero or close to it, the tangent runs flat, meets the axis nowhere or miles away, and the formula divides by almost nothing. Start this cubic at x = 1/√3, one of its turning points, and the method fails at the first step. A starting value on the wrong side of a turning point can also slide you off to a different root than the one the question wanted.

GUIDED PRACTICE

A square root by tangents

Apply Newton-Raphson to f(x) = x2 − 7 with x0 = 2.5, finding x1 and x2, before opening the working.

Show the working

f' = 2x, so the formula reads xn+1 = xn − (xn2 − 7)/(2xn).

x1 = 2.5 − (−0.75)/5 = 2.65, and x2 = 2.6458 to 4 decimal places.

√7 = 2.64575… , so two steps already agree to 4 decimal places. This is, among other things, how your calculator finds square roots.

Estimating an area numerically

Some integrands have no antiderivative this course can write down, and some functions arrive as nothing but a table of readings. The trapezium rule estimates the area anyway. Slice into n strips of width h, join the tops of the strips with straight chords, and add up the trapezia.

T=h2(y0+2(y1+…+yn−1)+yn)T = \frac{h}{2}(y_{0} + 2(y_{1} + … + y_{n−1}) + y_{n})IN THE FORMULAE BOOKLET

with h = (b − a)/n. Ends once, middles twice, is the shape of that bracket, because each interior ordinate is a side of two trapezia at once. Count ordinates and you get n + 1 of them for n strips, and miscounting there is the single commonest error in the topic.

The trapezium rule with four strips under the log curve from 1 to 2: chords cut the corners, so the estimate 0.3837 sits just below the true 0.386312y = ln xfour trapezia: 0.3837true area 0.3863: the chords cut the corners
FIG. 2Four strips under ln x from 1 to 2: the chords dip below the bending curve, so 0.3837 undershoots the true 0.3863. The sketch gives the direction of the error before any arithmetic.

WORKED EXAMPLE

Under a known curve, as a test

Estimate ∫12 ln x dx with 4 strips, and compare with the exact value.

h = 0.25, and the five ordinates are 0, 0.2231, 0.4055, 0.5596, 0.6931.

T = (0.25/2)(0 + 0.6931 + 2(0.2231 + 0.4055 + 0.5596)) = 0.3837.

The parts lesson gave the exact answer, 2 ln 2 − 1 = 0.3863, so the rule is under by 0.0026. This curve is concave, every chord sits below it, and a sketch called the direction of the error before any arithmetic happened. Doubling to 8 strips tightens the estimate to 0.3856.

INDEPENDENT PRACTICE

Over or under, then the number

Using 4 strips, estimate ∫01 √(1 + x2) dx, and state with a reason whether your estimate is too big or too small.

Show the working

The ordinates at 0, 0.25, 0.5, 0.75, 1 are 1, 1.0308, 1.1180, 1.25, 1.4142.

T = (0.25/2)(1 + 1.4142 + 2(1.0308 + 1.1180 + 1.25)) = 1.1515.

This curve is convex, so the chords ride above it and the estimate is an overestimate. The true value is 1.1478. That reason is worth a mark and takes one sentence.

ASSESSMENT FOCUS

  • Quote the Newton-Raphson formula, substitute visibly, and give iterates to more places than the final answer needs.
  • If f'(x₀) = 0 the method fails at once, and saying why, that the tangent is horizontal and never meets the axis, is what earns the mark.
  • Trapezium bracket: ends once, middles doubled. Wrong h from miscounting strips against ordinates is the commonest slip in the topic.
  • Over or under comes from the bend of the curve. Sketch it, then say which side of the curve the chords sit on.
  • Questions state when an iterative method is required. Where they do, an exact method does not answer the question asked, however correct it is.
  • Round only at the end. Rounding each iterate as you go can shift the final digit and lose the accuracy mark.

CHECK YOURSELF

Show that x = e−x has a root between 0.5 and 0.6, and find it to 5 decimal places using Newton-Raphson on f(x) = x − e−x with x0 = 0.6.

Show a hint

f(0.5) and f(0.6) first; f' = 1 + e−x.

Show the answer

f(0.5) = −0.1065 < 0 and f(0.6) = 0.0512 > 0 with f continuous, so a root lies between them.

x1 = 0.6 − 0.0512/1.5488 = 0.56695, and x2 = x3 = 0.56714.

The root is 0.56714. Two tangent slides pinned five decimal places, which is the doubling in action.

Newton-Raphson slides tangents to the axis and doubles its correct digits, until a flat tangent makes it fail.

The trapezium rule adds straight-topped strips, ends once and middles twice, and the curve's bend determines the sign of the error.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

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Question workbook

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Answer book

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7 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the newton-raphson and the trapezium rule questions page.

CHECK YOUR PROGRESS

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  • Run Newton-Raphson, reading each step as a tangent sliding to the axis.
  • Explain the method's failure when the tangent is horizontal or nearly so.
  • Estimate definite integrals by the trapezium rule with a stated number of strips.
  • Decide from the bend of the curve whether the estimate is too big or too small.

Open the full revision checklist to see every objective in the course in one place.