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Learn7357Paper 2 · Pure and MechanicsPure content (sections A–J)Newton–Raphson, bisection and interpolation
Pure content (sections A–J) · Subtopic 59 of 66

Newton–Raphson, Bisection and Interpolation

Slide down the tangent to the axis and repeat: the Newton–Raphson method roughly doubles the number of correct digits at every step. Also the two simpler interval methods in Edexcel FP1, and the ways all of them can go wrong. For AQA and Edexcel Further Pure 1; not examined in Cambridge 9709.

Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2Edexcel YFM01 WFM01Edexcel 9FM0 OptionAQA 7357 Paper 1AQA 7357 Paper 2≈ 40 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Use the Newton–Raphson formula xn+1=xn−f(xn)f′(xn)x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)} and explain it with a tangent.
  • Recognise when Newton–Raphson fails or finds an unexpected root.
  • Narrow down a root by interval bisection.
  • Estimate a root by linear interpolation (Edexcel FP1).
Pause and think

Near a root, a smooth curve looks almost like its tangent. If you replaced the curve by its tangent at x1x_1, where would you predict the root to be?

Section 2 of 14

Core idea

Start hereRide the tangent down

Standing on a curve, you want to reach where it crosses the axis. Pretend the curve is its tangent line, slide down the tangent to the axis, then repeat from the new point. Each step roughly doubles the number of correct digits. This is how calculators find square roots.

The big idea

Newton–Raphson: xn+1=xn−f(xn)f′(xn)x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}. Fast, but it fails when f′(xn)≈0f'(x_n)\approx0 (near turning points).

1. Derivation

Newton–Raphson on y = x² − 2, starting at 3
0.511.522.53−224680xyy = x² − 2x₀ = 3x₁ = 1.833x₂
3→1.8333→1.4621→1.4150→1.414213\to1.8333\to1.4621\to1.4150\to1.41421. After four steps it matches 2\sqrt2 to five decimal places.
Where the tangent meets the axis
  1. y−f(xn)=f′(xn)(x−xn)y-f(x_n)=f'(x_n)(x-x_n)
    Tangent at xnx_n.
  2. 0−f(xn)=f′(xn)(x−xn)0-f(x_n)=f'(x_n)(x-x_n)
    Set y=0y=0.
  3. x=xn−f(xn)f′(xn)x=x_n-\frac{f(x_n)}{f'(x_n)}
    Divide by f′(xn)f'(x_n); add xnx_n.
Worked example2\sqrt2 from f(x)=x2−2f(x)=x^2-2, x0=3x_0=3

Find 2\sqrt2 to 5 d.p.

Goal
Approximation.
Strategy
f′(x)=2xf'(x)=2x.
  1. xn+1=xn−xn2−22xnx_{n+1}=x_n-\frac{x_n^2-2}{2x_n}
    Formula.
  2. 1.83333, 1.46212, 1.41500, 1.414211.83333,\ 1.46212,\ 1.41500,\ 1.41421
    Iterate.
Answer1.414211.41421.
✓ Check1.414212=1.999991.41421^2=1.99999 ✓.
Worked examplex3−2x−5=0x^3-2x-5=0, x0=2x_0=2

Find the root.

Goal
Root to 5 d.p.
Strategy
f′=3x2−2f'=3x^2-2.
  1. x1=2−−110=2.1x_1=2-\frac{-1}{10}=2.1
    Step 1.
  2. x2=2.1−0.06111.23=2.09457x_2=2.1-\frac{0.061}{11.23}=2.09457
    Step 2.
  3. x3=2.09455x_3=2.09455
    Step 3.
Answer2.094552.09455.
✓ CheckAgrees with the iteration in the previous lesson ✓.
Edge caseFailure If f′(xn)=0f'(x_n)=0 the tangent is horizontal and never meets the axis. Near a turning point it shoots far away.

2. Linear interpolation

Linear interpolation for x³ − 2x − 5 on [2, 3]
22.22.42.62.8351015xyy = f(x)chordf(3) = 16f(2) = −1estimate 2.059
By similar triangles the chord crosses at 2+11+16=2.0592+\dfrac{1}{1+16}=2.059, close to the true root 2.09462.0946.
Worked examplex3−2x−5x^3-2x-5 on [2,3][2,3]

Estimate the root.

Goal
One estimate.
Strategy
Join the end points with a chord; similar triangles.
  1. f(2)=−1, f(3)=16f(2)=-1,\ f(3)=16
    Ends.
  2. x−21=3−x16⇒16x−32=3−x\frac{x-2}{1}=\frac{3-x}{16}\Rightarrow16x-32=3-x
    Similar triangles.
  3. x=3517=2.059x=\tfrac{35}{17}=2.059
    Solve.
Answer2.0592.059.
✓ CheckCruder than Newton–Raphson, as expected ✓.
Pause and think

N–R formula for f(x)=ex−3xf(x)=e^x-3x?

Summary

  • xn+1=xn−f/f′x_{n+1}=x_n-f/f'.
  • Fast; fails near stationary points.
Key words
Newton–Raphson
Tangent-based root finding.
HistoryThe same cubic, 350 years ago. Isaac Newton described his method around 1669 and illustrated it with the equation y3−2y−5=0y^3-2y-5=0, the very cubic used throughout this topic. Joseph Raphson published a simpler, purely algebraic version in 1690, and the familiar formula with f′(x)f'(x) was written down later by Thomas Simpson in 1740.
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