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Newton-Raphson and the trapezium rule

Two workhorses finish the unit. Newton-Raphson rides tangent lines down to a root, gaining digits at a startling rate when it behaves and sailing off to nowhere when it does not, and the trapezium rule prices an area with straight-topped strips whose error direction a sketch reveals in advance.

Year 13EDEXCEL 9MA0 9.3-9.5

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

IN THIS TOPIC

  • Run Newton-Raphson, understanding each step as a tangent sliding to the axis.
  • Know when the method fails: a horizontal tangent has no axis-cut to offer.
  • Estimate definite integrals by the trapezium rule and read the error's direction from a sketch.

WHAT YOU PROBABLY THINK

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 to the axis: the crossing point is the next estimate. As a formula, from the booklet,

xn+1 = xn − f(xn)f'(xn)IN THE FORMULAE BOOKLET

and on a well-behaved curve the convergence is ferocious, roughly doubling the correct digits each step.

Newton-Raphson on the cubic x cubed minus x minus 1: the tangent at x nought equals 1.5 slides down to cut the axis at 1.3478, the tangent there cuts at 1.3252, and two steps have landed within a whisker of 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. The plain iteration of the previous lesson took twice as many steps for the same equation, which is why this method earns its booklet entry.

That headline lie dies at any stationary point. Where f'(xn) is zero or nearly so, the tangent runs horizontal, meets the axis nowhere or miles away, and the formula divides by (almost) nothing: starting this cubic at x = 1/√3, a turning point, kills the method on step one. A starting value on the wrong side of a turning point can also send the slide to a different root than the one intended.

YOUR TURN

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 with it to 4 decimal places; the method is, among other things, how calculators find square roots.

Pricing an area numerically

Some integrands have no antiderivative the course can write down, and some are only known as a table of readings. The trapezium rule prices the area anyway: slice into n strips of width h, join the tops with straight chords, and add the trapezia,

T = h2(y0 + 2(y1 + … + yn−1) + yn)IN THE FORMULAE BOOKLET

with h = (b − a)/n. Ends once, middles twice, is the shape of the bracket, because each interior ordinate walls two trapezia.

The trapezium rule with four strips under the log curve from 1 to 2: chords under a curve that bends downwards cut the corners, so the estimate 0.3837 sits just below the true area 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 predicts the direction before the arithmetic exists.

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. The curve bends downwards, every chord sits below it, and the sketch called the direction of the error in advance. Doubling to 8 strips tightens the estimate to 0.3856.

TRY IT UNSEEN

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 bends upwards, so the chords ride above it and the estimate is an overestimate; the true value is 1.1478. The reason is worth a mark and takes one sentence about concavity.

THE EXAM BIT

  • 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, the tangent is horizontal, is the whole mark.
  • Trapezium bracket: ends once, middles doubled; the commonest slip is a wrong h from miscounting strips versus ordinates.
  • Over- or underestimate comes from the bend of the curve; sketch, then say which side of the curve the chords sit.
  • Questions state when an iterative method is required; unprompted exact methods still earn full credit elsewhere.

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.

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, the doubling in action.

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

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

CHECK YOUR PROGRESS

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  • Run Newton-Raphson, understanding each step as a tangent sliding to the axis.
  • Know when the method fails: a horizontal tangent has no axis-cut to offer.
  • Estimate definite integrals by the trapezium rule and read the error's direction from a sketch.

No animated video for this topic yet; these notes stand alone.