The Newton-Raphson Method

This lesson covers the Newton-Raphson method for finding roots of equations — as required by the Edexcel A-Level Mathematics specification (9MA0). You need to know the formula, understand how it works, recognise when it fails, and compare it with other numerical methods.

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The Formula

The Newton-Raphson method uses the following iteration formula:

x_{n+1} = x_n - f(x_n) / f'(x_n)

Starting from an initial estimate x₀, this generates a sequence x₁, x₂, x₃, ... that (usually) converges rapidly to a root of f(x) = 0.

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How It Works — The Geometric Idea

At each step, the method:

1. Evaluates f(x_n) and f'(x_n) at the current estimate.
2. Draws the tangent line to y = f(x) at the point (x_n, f(x_n)).
3. Finds where this tangent crosses the x-axis — this gives x_{n+1}.

The tangent at (x_n, f(x_n)) has equation: y - f(x_n) = f'(x_n)(x - x_n)

Setting y = 0 (x-axis crossing): -f(x_n) = f'(x_n)(x - x_n) x = x_n - f(x_n)/f'(x_n)

This is the Newton-Raphson formula.

Key Insight: The method replaces the curve with its tangent at each step. Since the tangent is a good local approximation to the curve, the x-intercept of the tangent is close to the root.

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Worked Example 1

Use the Newton-Raphson method to find a root of f(x) = x³ - 2x - 5, starting from x₀ = 2.

Step 1: Find f'(x) = 3x² - 2.

Step 2: Iterate.

n	x_n	f(x_n)	f'(x_n)	x_{n+1} = x_n - f/f'
0	2	8 - 4 - 5 = -1	12 - 2 = 10	2 - (-1/10) = 2.1
1	2.1	9.261 - 4.2 - 5 = 0.061	13.23 - 2 = 11.23	2.1 - 0.061/11.23 = 2.09457
2	2.09457	≈ 0.000089	≈ 11.155	≈ 2.09455

The method converges very quickly. After just 2 iterations, we have the root correct to 5 significant figures: x ≈ 2.0946.

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Worked Example 2

Find a root of f(x) = eˣ - 3x using x₀ = 2.

f'(x) = eˣ - 3

n	x_n	f(x_n)	f'(x_n)	x_{n+1}
0	2	e² - 6 = 7.389 - 6 = 1.389	e² - 3 = 4.389	2 - 1.389/4.389 = 1.6835
1	1.6835	e^(1.6835) - 5.0505 = 5.383 - 5.051 = 0.332	e^(1.6835) - 3 = 2.383	1.6835 - 0.332/2.383 = 1.5441
2	1.5441	e^(1.5441) - 4.6323 = 4.684 - 4.632 = 0.052	4.684 - 3 = 1.684	1.5441 - 0.052/1.684 = 1.5132
3	1.5132	≈ 0.0015	≈ 1.541	≈ 1.5122

Root ≈ 1.512 (3 d.p.)

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Rate of Convergence

The Newton-Raphson method has quadratic convergence — this means that the number of correct decimal places roughly doubles with each iteration. This makes it much faster than simple fixed-point iteration.

Method	Convergence Type	Typical Speed
Fixed-point iteration	Linear	Slow
Newton-Raphson	Quadratic	Fast
Bisection	Linear	Reliable but slow

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When Newton-Raphson Fails

The method can fail in several situations:

1. f'(x_n) = 0 (Horizontal Tangent)

If f'(x_n) = 0, the formula involves division by zero. The tangent is horizontal and never crosses the x-axis.

2. Poor Starting Value

A bad choice of x₀ can cause the method to:

• Converge to the wrong root (a different root from the one you wanted)
• Oscillate between values without converging
• Diverge (values getting larger and larger)

3. Stationary Points Near the Root

If f(x) has a turning point near the root, the tangent may overshoot badly, leading to divergence.

4. Repeated Root

If f(x) has a repeated root (e.g., f(x) = (x - 2)²), the convergence slows from quadratic to linear.

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Comparison with Fixed-Point Iteration

Feature	Newton-Raphson	Fixed-Point Iteration
Formula	x_{n+1} = x_n - f(x_n)/f'(x_n)	x_{n+1} = g(x_n)
Requires	f(x) and f'(x)	A rearrangement x = g(x)
Speed	Fast (quadratic convergence)	Slower (linear convergence)
Robustness	Can fail near turning points	Depends on
Ease of use	Need to differentiate f(x)	Need to find a suitable g(x)

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Using Newton-Raphson to Solve Equations

Example: Solve x = tan x near x = 4.5

Let f(x) = x - tan x, so f'(x) = 1 - sec²x.

Note: This equation has infinitely many solutions. We are finding the one near x = 4.5.

n	x_n	f(x_n)	f'(x_n)	x_{n+1}
0	4.5	4.5 - tan(4.5) = 4.5 - 4.6373 = -0.1373	1 - sec²(4.5) = 1 - 22.498 = -21.498	4.5 - (-0.1373)/(-21.498) = 4.4936
1	4.4936	4.4936 - 4.4937 = -0.0001	≈ -21.16	4.4934

Root ≈ 4.493 (3 d.p.)

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Applying Newton-Raphson in Exam Questions

Typical exam questions will ask you to:

1. Show that a given equation can be written as f(x) = 0.
2. Apply Newton-Raphson with a given x₀ for a specified number of iterations.
3. Show your working clearly (f(x_n), f'(x_n), and the formula).
4. Give your answer to a specified number of decimal places or significant figures.
5. Explain why the method might fail for a different starting value.

Exam Tip: Always show the formula and your substitutions clearly. The examiner needs to see your method, not just the final answer.

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Common Mistakes

1. Wrong derivative — Make sure f'(x) is correct. A wrong derivative gives a completely wrong iteration.
2. Calculator errors — Use your calculator carefully and write down intermediate values.
3. Not enough iterations — Continue until the answer stabilises to the required accuracy.
4. Ignoring failure conditions — If the question asks when the method might fail, think about horizontal tangents, poor starting values, and turning points.

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Summary

• Newton-Raphson formula: x_{n+1} = x_n - f(x_n)/f'(x_n).
• Geometrically, it finds where the tangent at x_n crosses the x-axis.
• Convergence is quadratic — roughly doubles correct decimal places per iteration.
• The method requires f(x) and f'(x).
• It can fail when f'(x_n) = 0, the starting value is poor, or there are turning points near the root.
• Faster than fixed-point iteration but less robust.

A-Level Deep Dive: The Newton-Raphson Method

Spec mapping

Edexcel 9MA0-02 specification section 13 — Numerical methods, sub-strands 13.3 and 13.4 covers the Newton-Raphson method for finding approximate roots of equations of the form $f(x) = 0$f(x)=0. Understand how such methods can fail (refer to the official specification document for exact wording). This sub-strand sits inside Year 2 Pure Mathematics (Paper 2) and is consistently examined as a structured 8-12 mark question. Newton-Raphson connects directly to section 9 (Differentiation, since the iteration formula $x_{n+1} = x_n - f(x_n)/f'(x_n)$xn+1​=xn​−f(xn​)/f′(xn​) requires fluent computation of $f'(x)$f′(x)), section 10 (Integration, where Newton's method appears as the inverse operation in solving $F(x) = c$F(x)=c), and section 13.1-13.2 (Locating roots and fixed-point iteration, the alternative numerical methods). The Edexcel formula booklet does reproduce the iteration formula, but candidates are expected to derive it geometrically when the question demands explanation. Calculator usage is permitted on Paper 2, but full algebraic working for $f'(x)$f′(x) is required.

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Worked example with full mark scheme

Question (8 marks):

(a) The equation $x^3 - 2x - 5 = 0$x3−2x−5=0 has a single real root $\alpha$α in the interval $[2, 3]$[2,3]. Using the Newton-Raphson method with starting value $x_0 = 2$x0​=2, find $x_1$x1​, $x_2$x2​ and $x_3$x3​, giving each iterate to four decimal places. (5)

(b) State, with a reason, whether your value of $x_3$x3​ gives $\alpha$α correct to three decimal places. (3)

Solution with mark scheme:

(a) Step 1 — set up the iteration formula.

With $f(x) = x^3 - 2x - 5$f(x)=x3−2x−5, differentiate to get $f'(x) = 3x^2 - 2$f′(x)=3x2−2. The iteration is

$x_{n+1} = x_n - \dfrac{x_n^3 - 2x_n - 5}{3x_n^2 - 2}$xn+1​=xn​−3xn2​−2xn3​−2xn​−5​

M1 — correct $f'(x)$f′(x) stated and the Newton-Raphson formula written with $f$f and $f'$f′ substituted. Common slip: writing $f'(x) = 3x^2$f′(x)=3x2 (forgetting the derivative of $-2x$−2x) — that propagates through every iterate and loses every subsequent A mark.

Step 2 — compute $x_1$x1​.

At $x_0 = 2$x0​=2: $f(2) = 8 - 4 - 5 = -1$f(2)=8−4−5=−1 and $f'(2) = 12 - 2 = 10$f′(2)=12−2=10.

$x_1 = 2 - \dfrac{-1}{10} = 2 + 0.1 = 2.1$x1​=2−10−1​=2+0.1=2.1

A1 — $x_1 = 2.1000$x1​=2.1000 to four decimal places.

Step 3 — compute $x_2$x2​.

At $x_1 = 2.1$x1​=2.1: $f(2.1) = 9.261 - 4.2 - 5 = 0.061$f(2.1)=9.261−4.2−5=0.061 and $f'(2.1) = 13.23 - 2 = 11.23$f′(2.1)=13.23−2=11.23.

$x_2 = 2.1 - \dfrac{0.061}{11.23} = 2.1 - 0.005432... = 2.094568...$x2​=2.1−11.230.061​=2.1−0.005432...=2.094568...

To four decimal places, $x_2 = 2.0946$x2​=2.0946.

A1 — $x_2 = 2.0946$x2​=2.0946.

Step 4 — compute $x_3$x3​.

At $x_2 = 2.094568$x2​=2.094568: $f(x_2) \approx (2.094568)^3 - 2(2.094568) - 5 \approx 9.18814 - 4.18914 - 5 \approx 0.00018$f(x2​)≈(2.094568)3−2(2.094568)−5≈9.18814−4.18914−5≈0.00018 (the value is now extremely small — quadratic convergence in action). $f'(x_2) \approx 3(2.094568)^2 - 2 \approx 13.16182 - 2 \approx 11.16182$f′(x2​)≈3(2.094568)2−2≈13.16182−2≈11.16182.

$x_3 = 2.094568 - \dfrac{0.00018}{11.16182} \approx 2.094568 - 0.0000161 \approx 2.094552$x3​=2.094568−11.161820.00018​≈2.094568−0.0000161≈2.094552

To four decimal places, $x_3 = 2.0946$x3​=2.0946.

M1 A1 — correct method applied at each stage; correct value of $x_3$x3​ to four decimal places.

(b) Step 1 — name the test.

A change-of-sign argument is required. Evaluate $f(2.0945)$f(2.0945) and $f(2.0955)$f(2.0955) — the endpoints of the interval that round to $2.095$2.095 — to test whether $\alpha$α definitely lies in this interval.

M1 — recognising that confirming "correct to three decimal places" requires a sign change on $[2.0945, 2.0955]$[2.0945,2.0955], not merely a small residual at $x_3$x3​.

Step 2 — compute the two values.

$f(2.0945) = (2.0945)^3 - 2(2.0945) - 5 \approx 9.18815 - 4.18900 - 5 \approx -0.00085$f(2.0945)=(2.0945)3−2(2.0945)−5≈9.18815−4.18900−5≈−0.00085.

$f(2.0955) = (2.0955)^3 - 2(2.0955) - 5 \approx 9.20132 - 4.19100 - 5 \approx 0.01032$f(2.0955)=(2.0955)3−2(2.0955)−5≈9.20132−4.19100−5≈0.01032.

A1 — both values correct, with sign change observed: $f(2.0945) < 0$f(2.0945)<0 and $f(2.0955) > 0$f(2.0955)>0.

Step 3 — conclude.

Since $f$f is continuous on $[2.0945, 2.0955]$[2.0945,2.0955] and changes sign, by the intermediate value theorem there is a root $\alpha$α in this interval. Hence $\alpha = 2.095$α=2.095 to three decimal places, and $x_3 = 2.0946$x3​=2.0946 rounds to the same value, so $x_3$x3​ is correct to three decimal places.

A1 — fully justified conclusion citing continuity and the change-of-sign / intermediate value reasoning.

Total: 8 marks (M3 A5).

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Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): The equation $g(x) = e^x - 4x = 0$g(x)=ex−4x=0 has a root $\beta$β in $[2, 3]$[2,3].

(a) Show that the Newton-Raphson iteration applied to $g$g gives the formula

$x_{n+1} = \dfrac{(x_n - 1)e^{x_n}}{e^{x_n} - 4}$xn+1​=exn​−4(xn​−1)exn​​

(3)

(b) Taking $x_0 = 2.2$x0​=2.2, find $x_1$x1​ to four decimal places. (2)

(c) Explain briefly why taking $x_0 = \ln 4$x0​=ln4 would cause the Newton-Raphson method to fail at the first iteration. (1)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — differentiating: $g'(x) = e^x - 4$g′(x)=ex−4 correctly stated.
• M1 (AO1.1b) — substituting into $x_{n+1} = x_n - g(x_n)/g'(x_n)$xn+1​=xn​−g(xn​)/g′(xn​) and writing $x_n - \dfrac{e^{x_n} - 4x_n}{e^{x_n} - 4}$xn​−exn​−4exn​−4xn​​.
• A1 (AO2.1) — combining over the common denominator and simplifying the numerator: $x_n(e^{x_n} - 4) - (e^{x_n} - 4x_n) = x_n e^{x_n} - 4x_n - e^{x_n} + 4x_n = (x_n - 1)e^{x_n}$xn​(exn​−4)−(exn​−4xn​)=xn​exn​−4xn​−exn​+4xn​=(xn​−1)exn​. The cancellation of $\pm 4x_n$±4xn​ is the AO2.1 reasoning step.

(b)

• M1 (AO1.1b) — correct substitution $x_0 = 2.2$x0​=2.2 into the simplified formula: $x_1 = (1.2)e^{2.2}/(e^{2.2} - 4)$x1​=(1.2)e2.2/(e2.2−4).
• A1 (AO1.1b) — $x_1 \approx (1.2)(9.0250)/(9.0250 - 4) \approx 10.8300/5.0250 \approx 2.1552$x1​≈(1.2)(9.0250)/(9.0250−4)≈10.8300/5.0250≈2.1552 to 4 d.p.

(c)

• B1 (AO2.4) — at $x_0 = \ln 4$x0​=ln4, $e^{x_0} = 4$ex0​=4, so $g'(x_0) = e^{x_0} - 4 = 0$g′(x0​)=ex0​−4=0. The denominator of the iteration is zero and $x_1$x1​ is undefined (the tangent is horizontal and never crosses the x-axis).

Total: 6 marks split AO1 = 4, AO2 = 2. Edexcel typically awards a single AO2.4 reasoning mark for spotting the failure case in part (c) — short answer, but pinpoint accuracy required.

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Synoptic links

Connects to:

1. Section 9 — Differentiation: every Newton-Raphson question is also a differentiation question. The iteration cannot proceed without $f'(x)$f′(x), and questions deliberately choose $f$f that requires the chain rule, product rule or implicit differentiation. Errors in $f'$f′ propagate through every iterate.

2. Section 13.1 — Locating roots by change of sign: Newton-Raphson finds a root quickly but does not prove the root exists. Examiners pair the methods deliberately: locate-by-sign-change establishes existence on an interval, Newton-Raphson refines the value, and a second sign-change check verifies the final accuracy claim.

3. Section 13.2 — Fixed-point iteration $x_{n+1} = F(x_n)$xn+1​=F(xn​): Newton-Raphson is itself a fixed-point iteration with $F(x) = x - f(x)/f'(x)$F(x)=x−f(x)/f′(x). The convergence condition $|F'(x)| < 1$∣F′(x)∣<1 near the root is automatically satisfied for Newton-Raphson when $f'(\alpha) \neq 0$f′(α)=0 — which is why it converges quadratically rather than linearly.

4. Section 6 — Logarithms and exponentials: equations like $e^x = kx$ex=kx or $\ln x = mx$lnx=mx have no closed-form solution and demand numerical methods. Newton-Raphson is the standard tool, and the differentiation $\frac{d}{dx}e^x = e^x$dxd​ex=ex keeps the iteration formula tractable.

5. Calculator computation: the Casio fx-991EX SOLVE function and most graphic calculator equation solvers run Newton-Raphson internally. Understanding the algorithm explains why calculators occasionally fail on equations with stationary points near roots and why a sensible starting value matters.

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Mark-scheme literacy

Newton-Raphson questions on 9MA0-02 split AO marks roughly:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–65%	Stating $f'(x)$f′(x), substituting into the iteration formula, computing iterates to specified accuracy, performing sign-change checks
AO2 (reasoning / interpretation)	25–35%	Justifying convergence claims, identifying failure modes ($f'(x_n) = 0$f′(xn​)=0, divergence, oscillation), explaining the geometric interpretation, deducing accuracy by intermediate value theorem
AO3 (problem-solving)	5–15%	Selecting an appropriate starting value, sketching tangent lines to demonstrate failure, modelling a real-world equation as $f(x) = 0$f(x)=0 for numerical solution