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The Newton-Raphson processEdexcel International A Level Further Maths: Revision notes

Section 1

The idea

The Newton-Raphson method improves an estimate of a root of f(x)=0f(x)=0 using the tangent to the curve. If xnx_n is an approximation, the tangent at (xn,f(xn))(x_n,f(x_n)) has gradient f′(xn)f'(x_n) and meets the xx-axis at xn+1x_{n+1}. Gradient =f(xn)−0xn−xn+1=\frac{f(x_n)-0}{x_n-x_{n+1}}, so xn+1=xn−f(xn)f′(xn).x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}. It needs ff to be differentiable and f′(xn)≠0f'(x_n)\neq0. In IAL FP1 the functions are those from P1 and P2, such as polynomials, ln⁡x\ln x, exe^x and trigonometric functions (in radians).

Key termsNewton-Raphsontangentiteration
Common mistake

Forgetting the minus sign or the xnx_n: xn+1=xn−ff′x_{n+1}=x_n-\frac{f}{f'}, not ff′\frac{f}{f'} on its own.

Section 2

Worked example

Take f(x)=x3−2x−5f(x)=x^3-2x-5 with x0=2x_0=2. Then f′(x)=3x2−2f'(x)=3x^2-2.

  • f(2)=−1f(2)=-1, f′(2)=10f'(2)=10, so x1=2+0.1=2.1x_1=2+0.1=2.1.
  • f(2.1)=0.061f(2.1)=0.061, f′(2.1)=11.23f'(2.1)=11.23, so x2=2.1−0.00543=2.0946x_2=2.1-0.00543=2.0946.

The values settle quickly. Newton-Raphson usually converges much faster than bisection, roughly doubling the number of correct digits each step once close to the root. Use the ANS key on a calculator to keep full accuracy between iterations.

Key termsconvergence
Exam tip

Differentiate ff first, write the formula for xn+1x_{n+1}, then substitute x0x_0. Show f(x0)f(x_0) and f′(x0)f'(x_0) in your working.

Section 3

Deriving an iteration formula

Questions often ask you to show a specific form. For f(x)=cos⁡x−xf(x)=\cos x-x (radians), f′(x)=−sin⁡x−1f'(x)=-\sin x-1, so xn+1=xn−cos⁡xn−xn−sin⁡xn−1=xn+cos⁡xn−xn1+sin⁡xn.x_{n+1}=x_n-\frac{\cos x_n-x_n}{-\sin x_n-1}=x_n+\frac{\cos x_n-x_n}{1+\sin x_n}. Starting from x0=1x_0=1: x1=0.7504x_1=0.7504 and x2=0.7391x_2=0.7391. For f(x)=x2−7f(x)=x^2-7 the formula simplifies to xn+1=xn2+72xnx_{n+1}=\frac{x_n^2+7}{2x_n}.

Key termsradians
Common mistake

Leaving the calculator in degree mode when ff involves sin⁡x\sin x or cos⁡x\cos x.

Section 4

When Newton-Raphson fails

The method can fail or go to the wrong root:

  • f′(xn)=0f'(x_n)=0: the tangent is horizontal, so there is no xn+1x_{n+1}. For f(x)=x3−3x+1f(x)=x^3-3x+1, f′(1)=0f'(1)=0, so x0=1x_0=1 fails.
  • x0x_0 too far away, or near a stationary point: the first step can fling the estimate far away, or to another root. For f(x)=x3−4x−2f(x)=x^3-4x-2, x0=0x_0=0 converges to −0.539-0.539, not to the largest root 2.2142.214.
  • Not differentiable at the root, or ff undefined at an iterate.

Choose x0x_0 close to the root, found from a sign change or sketch, and where f′f' is not small.

Key termsstationary point
Exam tip

If the question says the method fails, look for f′(x0)=0f'(x_0)=0 or a start close to a turning point.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on The Newton-Raphson process

  1. The equation x3−2x−5=0x^3-2x-5=0 has a root α\alpha close to 22. Let f(x)=x3−2x−5f(x)=x^3-2x-5 and use the Newton-Raphson method with x0=2x_0=2.
    Find x2x_2, giving your answer to 4 decimal places.2 marks
  2. Let f(x)=x3−3x+1f(x)=x^3-3x+1. The equation f(x)=0f(x)=0 has a root α\alpha between 00 and 11.
    Taking x0=13x_0=\frac13, find x1x_1 to 4 decimal places.2 marks
  3. Let f(x)=cos⁡x−xf(x)=\cos x-x, where xx is in radians. The equation f(x)=0f(x)=0 has a root α\alpha near 0.70.7.
    Show that the Newton-Raphson iteration for this equation can be written as xn+1=xn+cos⁡xn−xn1+sin⁡xnx_{n+1}=x_n+\frac{\cos x_n-x_n}{1+\sin x_n}.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).