The Newton-Raphson Method — Edexcel A-level Mathematics (Pure)

SPEC 9.3 • NUMERICAL METHODS

Newton-Raphson Method

A powerful root-finding algorithm using tangent lines

The Problem We're Solving

We want to solve equations of the form f(x) = 0.

What We Know So Far

Change of sign — locates where roots exist

Fixed point iteration — finds roots using x = g(x)

Now we introduce a faster, more powerful alternative:

Key Idea

Use the tangent line at each point to "jump" closer to the root

Why Newton-Raphson?

Newton-Raphson typically converges much faster than fixed point iteration — often doubling the number of correct decimal places with each step!

Geometric Interpretation

The Process

Start at point x₀, go up to the curve at (x₀, f(x₀))

Draw the tangent line at this point

Find where the tangent crosses the x-axis → this gives x₁

x₁ is closer to the root α — repeat to get even closer!

The Newton-Raphson Formula

Newton-Raphson Iteration

x_{n + 1} = x_{n} - \frac{f ( x _{n} )}{f ^{'} ( x _{n} )}

Requirements

f(x) must be differentiable

f'(xₙ) must not equal zero (can't divide by zero)

Need both f(x) AND f'(x) to apply this method

Key Difference from Fixed Point

Fixed point iteration uses xₙ₊₁ = g(xₙ) where you choose g(x).

Newton-Raphson always uses the same formula — you just need f(x) and f'(x).

Method Steps

1
Differentiate: Find f'(x) from f(x)
2
Choose x₀: Pick a starting value near the expected root
3
Calculate: xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ)
4
Repeat: Until desired accuracy is achieved
5
Verify: Check that f(xₙ) ≈ 0

Common Exam Mistake

Don't forget Step 1! Many students try to apply the formula without first finding f'(x).

Worked Example01

Use Newton-Raphson to find the root of f(x) = x³ − 2 near x = 1. Perform 3 iterations.

•

1. f(x) = x³ − 2, so f'(x) = 3x²

•

2. Formula: xₙ₊₁ = xₙ − (xₙ³ − 2)/(3xₙ²)

•

3. x₀ = 1

x₁ = 1 − (1 − 2)/(3) = 1 − (−1/3) = 1.3333...

•

4. x₂ = 1.3333 − (1.3333³ − 2)/(3 × 1.3333²)

= 1.3333 − (2.3704 − 2)/(5.3333)

= 1.3333 − 0.0694 = 1.2639

•

5. x₃ = 1.2639 − (1.2639³ − 2)/(3 × 1.2639²)

= 1.2639 − (0.0186)/(4.7927) = 1.2599

Answerx₃ = 1.2599 (converging to ∛2 ≈ 1.25992...)

Worked Example02

Use Newton-Raphson with x₀ = 1 to find a root of f(x) = eˣ − 3x. Give x₂ to 4 decimal places.

•

1. f(x) = eˣ − 3x, so f'(x) = eˣ − 3

•

2. Formula: xₙ₊₁ = xₙ − (eˣⁿ − 3xₙ)/(eˣⁿ − 3)

•

3. x₀ = 1

f(1) = e¹ − 3 = 2.7183 − 3 = −0.2817

f'(1) = e¹ − 3 = −0.2817

x₁ = 1 − (−0.2817)/(−0.2817) = 1 − 1 = 0

•

4. x₁ = 0

f(0) = e⁰ − 0 = 1

f'(0) = e⁰ − 3 = −2

x₂ = 0 − (1)/(−2) = 0 + 0.5 = 0.5

Answerx₂ = 0.5000

When Newton-Raphson Fails

Newton-Raphson doesn't always work. There are three main failure modes:

Failure Mode 1: Horizontal Tangent

f'(xₙ) = 0 or very small

The tangent is horizontal and never crosses the x-axis, or gives a huge jump away from the root. This causes division by zero or divergence.

Failure Mode 2: Cycling

Iteration oscillates between values

The sequence bounces back and forth without converging to a single value.

Failure Mode 3: Divergence

Values get further from all roots

The sequence tends to infinity instead of converging.

Exam Tip

Questions often ask you to explain why Newton-Raphson fails. Always check f'(x₀) first!

Worked Example03

Show that Newton-Raphson applied to f(x) = x^(1/3) with x₀ = 1 diverges.

•

1. f(x) = x^(1/3) = ∛x, so f'(x) = (1/3)x^(−2/3) = 1/(3∛x²)

•

2. Formula: xₙ₊₁ = xₙ − ∛xₙ / [1/(3∛xₙ²)]

= xₙ − 3∛xₙ × ∛xₙ² = xₙ − 3xₙ = −2xₙ

•

3. x₀ = 1

x₁ = −2(1) = −2

x₂ = −2(−2) = 4

x₃ = −2(4) = −8

x₄ = −2(−8) = 16

AnswerThe sequence |xₙ| → ∞ as n → ∞. The iteration diverges to infinity, alternating in sign.

Choosing Starting Values

The success of Newton-Raphson depends heavily on the choice of x₀.

Good Strategies:

Use change of sign to locate approximate root position

Sketch the graph to identify root locations

Start reasonably close to the expected root

Check that f'(x₀) is not zero or very small

What to Avoid:

Starting at a stationary point (f'(x₀) = 0)

Starting far from any root

Starting where the curve is nearly flat

If It Diverges

If your iteration diverges or oscillates, try a different starting value. There's no guarantee Newton-Raphson will work from any given x₀.

Practice

Question 1: Use Newton-Raphson to find the positive root of x² − 7 = 0, starting with x₀ = 3. Give x₂ to 4 decimal places.

f(x) = x² − 7, f'(x) = 2x

x₀ = 3

x₁ = 3 − (9 − 7)/(6) = 3 − 2/6 = 3 − 0.3333 = 2.6667

x₂ = 2.6667 − (7.1113 − 7)/(5.3333) = 2.6667 − 0.0209 = 2.6458

Answer: x₂ = 2.6458 (√7 ≈ 2.6458...)

Practice

Question 2: The equation x³ + x − 1 = 0 has a root near x = 0.7. Use Newton-Raphson to find x₂ to 4 decimal places.

f(x) = x³ + x − 1, f'(x) = 3x² + 1

x₀ = 0.7

f(0.7) = 0.343 + 0.7 − 1 = 0.043

f'(0.7) = 3(0.49) + 1 = 2.47

x₁ = 0.7 − 0.043/2.47 = 0.7 − 0.0174 = 0.6826

f(0.6826) = 0.3183 + 0.6826 − 1 = 0.0009

f'(0.6826) = 3(0.4659) + 1 = 2.3978

x₂ = 0.6826 − 0.0009/2.3978 = 0.6826 − 0.0004 = 0.6823

Answer: x₂ = 0.6823

Summary

Newton-Raphson Formula

x_{n + 1} = x_{n} - \frac{f ( x _{n} )}{f ^{'} ( x _{n} )}

Exam Essentials

Always find f'(x) first before applying the formula

Check f'(x₀) ≠ 0 before starting

Be ready to explain why iteration might fail

Compare convergence rates between methods