Comparing Numerical Methods

This lesson brings together all the numerical methods covered in the Edexcel A-Level Mathematics specification (9MA0) — fixed-point iteration, Newton-Raphson, and bisection/sign change — and compares their strengths, weaknesses, and appropriate uses. You also need to understand error bounds.

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Overview of the Three Methods

Feature	Sign Change / Bisection	Fixed-Point Iteration	Newton-Raphson
Formula	Midpoint m = (a+b)/2, check sign	x_{n+1} = g(x_n)	x_{n+1} = x_n - f(x_n)/f'(x_n)
Requires	f(x) only	A rearrangement x = g(x)	f(x) and f'(x)
Convergence speed	Slow (linear)	Moderate (linear)	Fast (quadratic)
Reliability	Very reliable	Variable	Usually good
Always converges?	Yes (if continuous and sign change exists)	Only if	g'(α)
Error bounds	Easy to determine	Harder to determine	Harder to determine

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Bisection / Sign Change Method

Advantages

• Always works (for continuous functions with a sign change) — guaranteed convergence.
• Easy to understand and implement.
• Predictable error bounds — after n steps, error ≤ (b - a)/2ⁿ.
• Does not require differentiation.

Disadvantages

• Slow — each step only halves the interval (linear convergence).
• Requires a known interval containing the root.
• Cannot detect repeated roots (where f touches but does not cross the x-axis).
• Misses roots when two roots are close together in the same interval.

When to Use

• When reliability is more important than speed.
• When you need guaranteed error bounds.
• When you cannot differentiate f(x).
• As a starting method to narrow down an interval before using a faster method.

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Fixed-Point Iteration

Advantages

• Simple iteration formula: just evaluate g(x) repeatedly.
• Does not require differentiation.
• Can be adapted to converge to a specific root by choosing the right rearrangement.
• Graphical interpretation (staircase/cobweb diagrams) aids understanding.

Disadvantages

• Not always convergent — depends on |g'(α)| < 1.
• Finding a suitable rearrangement can be difficult.
• Convergence is linear — can be slow.
• Different rearrangements converge to different roots (or may diverge).

When to Use

• When differentiation is difficult or impossible.
• When a natural rearrangement x = g(x) presents itself.
• When you want to visualise convergence with staircase/cobweb diagrams.

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Newton-Raphson Method

Advantages

• Fast convergence — quadratic convergence means the number of correct decimal places roughly doubles each iteration.
• Works well for most well-behaved functions.
• Widely used in professional mathematics and engineering.

Disadvantages

• Requires f'(x) — you must be able to differentiate the function.
• Can fail near turning points (f'(x_n) ≈ 0 causes overshoot).
• Can converge to the wrong root if the starting value is poor.
• Can diverge or oscillate for certain starting values.
• No easy way to predict error bounds.

When to Use

• When speed matters and you can differentiate f(x).
• When you have a good initial estimate.
• When the function is smooth and well-behaved near the root.

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Speed of Convergence — Practical Comparison

Consider finding a root of f(x) = x³ - 2 (the cube root of 2, α ≈ 1.2599).

Bisection (starting from [1, 2]):

Iteration	Interval Width	Decimal Places Correct
5	0.03125	~1
10	0.000977	~3
15	0.0000305	~4
20	0.00000095	~6

20 iterations for 6 decimal places.

Newton-Raphson (starting from x₀ = 1):

Iteration	x_n	Decimal Places Correct
1	1.3333	~1
2	1.2639	~2
3	1.25992	~4
4	1.2599210	~7

4 iterations for 7 decimal places. The speed advantage is dramatic.

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Error Bounds

Bisection Error Bound

After n bisection steps on the interval [a, b]:

|error| ≤ (b - a) / 2ⁿ

This is a guaranteed upper bound on the error.

Example: Starting from [1, 2], after 10 bisections: |error| ≤ 1/2¹⁰ = 1/1024 ≈ 0.000977

How Many Bisection Steps for a Given Accuracy?

To achieve an error less than ε, you need: (b - a)/2ⁿ < ε 2ⁿ > (b - a)/ε n > log₂((b - a)/ε)

Example: For [1, 2] with accuracy 10⁻⁶: n > log₂(10⁶) = 6 log₂(10) ≈ 6 × 3.322 = 19.93

So you need at least 20 bisection steps.

Fixed-Point Iteration Error Estimate

There is no simple guaranteed bound, but as a rough guide, the error after n iterations is approximately proportional to |g'(α)|ⁿ.

Newton-Raphson Error Estimate

The error approximately squares each iteration (quadratic convergence). If the error at step n is ε, the error at step n + 1 is roughly Cε² for some constant C.

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Choosing the Right Method

Decision Framework

Question 1: Can you differentiate f(x)?

• Yes → Consider Newton-Raphson (fast) or bisection (reliable).
• No → Use bisection or fixed-point iteration.

Question 2: Do you need guaranteed convergence?

• Yes → Use bisection (provided a sign change exists).
• No → Newton-Raphson is usually the fastest choice.

Question 3: Do you need it to converge to a specific root?

• Yes → Choose starting value carefully (all methods); bisection lets you target a specific interval.
• No → Any method works.

Question 4: Is speed important?

• Yes → Newton-Raphson.
• No → Bisection is simplest and most reliable.

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Exam-Style Questions

Typical Question Types

1. Apply a named method — Use Newton-Raphson / iteration / bisection for a given number of steps.
2. Compare two methods — Explain which is faster, more reliable, or more appropriate for a given problem.
3. Explain why a method fails — Describe situations where Newton-Raphson diverges, or iteration diverges.
4. Find error bounds — Calculate the maximum error after n steps of bisection.
5. Choose a method — Given a problem, justify which numerical method you would use.

Example Exam Question

"Explain one advantage and one disadvantage of the Newton-Raphson method compared to the bisection method."

Advantage: Newton-Raphson converges much faster (quadratic vs linear convergence), so fewer iterations are needed for a given accuracy.

Disadvantage: Newton-Raphson can diverge or converge to the wrong root if the starting value is poor, whereas bisection always converges provided a sign change exists.

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Combining Methods

In practice, a common strategy is:

1. Sign change/bisection to find an approximate interval containing the root.
2. Newton-Raphson to refine the root quickly once a good starting value is known.

This combines the reliability of bisection with the speed of Newton-Raphson.

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Summary Table

Method	Speed	Reliability	Requirements	Error Bounds
Bisection	Slow (halving)	Excellent	f(x) continuous, sign change	Easy: (b-a)/2ⁿ
Fixed-point iteration	Moderate	Variable	x = g(x) with	g'(α)
Newton-Raphson	Fast (quadratic)	Good but can fail	f(x) and f'(x)	Approximate only

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Common Mistakes

1. Stating Newton-Raphson always converges — It does not. It can diverge or oscillate.
2. Claiming bisection is fast — It is reliable but slow.
3. Forgetting convergence conditions — Fixed-point iteration needs |g'(α)| < 1.
4. Not justifying method choice — In comparison questions, give specific reasons (speed, reliability, requirements).
5. Confusing the methods — Make sure you can clearly state each formula and when to use it.

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Summary

• Bisection is reliable but slow — guaranteed convergence, error ≤ (b - a)/2ⁿ.
• Fixed-point iteration is simple but convergence depends on the rearrangement.
• Newton-Raphson is fast (quadratic convergence) but can fail near turning points or with poor starting values.
• Choose based on: whether you can differentiate, whether you need guaranteed convergence, and how fast you need the answer.
• In practice, combine methods: use sign change to find an interval, then Newton-Raphson to refine.

A-Level Deep Dive: Comparing Numerical Methods

Spec mapping

Edexcel 9MA0-02 specification section 13 — Numerical methods: "Locate roots of $f(x) = 0$f(x)=0 by considering changes of sign of $f(x)$f(x) in an interval of $x$x on which $f(x)$f(x) is sufficiently well behaved; solve equations approximately using simple iterative methods, including recurrence relations of the form $x_{n+1} = g(x_n)$xn+1​=g(xn​); solve equations using the Newton-Raphson method and other recurrence relations of the form $x_{n+1} = x_n - f(x_n)/f'(x_n)$xn+1​=xn​−f(xn​)/f′(xn​); understand how such methods can fail; use numerical integration of functions, including the use of the trapezium rule." Section 13 sits in the Year 2 Pure content of Paper 2 and is structurally synoptic — every numerical-methods question rests on differentiation (section 9), integration (section 10) or graph-sketching (section 4) underneath. The Edexcel formula booklet provides the trapezium rule ($\int_a^b y\,dx \approx \tfrac{1}{2}h\{y_0 + y_n + 2(y_1 + y_2 + \dots + y_{n-1})\}$∫ab​ydx≈21​h{y0​+yn​+2(y1​+y2​+⋯+yn−1​)}) but not the Newton-Raphson formula — that must be memorised as $x_{n+1} = x_n - f(x_n)/f'(x_n)$xn+1​=xn​−f(xn​)/f′(xn​).

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Worked example with full mark scheme

Question (8 marks): The equation $f(x) = x^3 + 2x - 5 = 0$f(x)=x3+2x−5=0 has a single real root $\alpha$α in the interval $[1, 2]$[1,2].

(a) Show that $\alpha$α lies in the interval $[1.3, 1.4]$[1.3,1.4]. (2)

(b) Using the iteration $x_{n+1} = \sqrt[3]{5 - 2x_n}$xn+1​=35−2xn​​ with $x_0 = 1.3$x0​=1.3, find $x_1$x1​, $x_2$x2​ and $x_3$x3​ to 4 decimal places. (3)

(c) Apply the Newton-Raphson method with $x_0 = 1.3$x0​=1.3 to obtain a single iterate $x_1$x1​ to 4 d.p., and compare the speed of convergence of the two methods to the root $\alpha = 1.3284$α=1.3284 (4 d.p.). (3)

Solution with mark scheme:

(a) Evaluate $f$f at the endpoints:

$f(1.3) = 2.197 + 2.6 - 5 = -0.203$f(1.3)=2.197+2.6−5=−0.203 $f(1.4) = 2.744 + 2.8 - 5 = 0.544$f(1.4)=2.744+2.8−5=0.544

M1 — both values computed correctly (or one value with an explicit sign-change reference). A1 — explicit conclusion that $f(1.3) < 0$f(1.3)<0 and $f(1.4) > 0$f(1.4)>0, that $f$f is continuous on $[1.3, 1.4]$[1.3,1.4] (a polynomial), so by the intermediate value theorem there exists $\alpha \in (1.3, 1.4)$α∈(1.3,1.4) with $f(\alpha) = 0$f(α)=0. The continuity statement is required for the A1 — examiners deduct for "sign change so root", with no mention of continuity.

(b) Apply the recurrence:

$x_1 = \sqrt[3]{5 - 2(1.3)} = \sqrt[3]{2.4} = 1.3389$x1​=35−2(1.3)​=32.4​=1.3389 $x_2 = \sqrt[3]{5 - 2(1.3389)} = \sqrt[3]{2.3222} = 1.3242$x2​=35−2(1.3389)​=32.3222​=1.3242 $x_3 = \sqrt[3]{5 - 2(1.3242)} = \sqrt[3]{2.3516} = 1.3298$x3​=35−2(1.3242)​=32.3516​=1.3298

M1 — correct substitution into the recurrence at $x_0$x0​. A1 — $x_1 = 1.3389$x1​=1.3389 to 4 d.p. A1 — $x_2$x2​ and $x_3$x3​ both correct to 4 d.p. The iterates are oscillating around $\alpha$α, halving the error roughly each step.

(c) Newton-Raphson requires $f'(x) = 3x^2 + 2$f′(x)=3x2+2, so $f'(1.3) = 3(1.69) + 2 = 7.07$f′(1.3)=3(1.69)+2=7.07. Then:

$x_1 = 1.3 - \dfrac{-0.203}{7.07} = 1.3 + 0.02871 = 1.3287$x1​=1.3−7.07−0.203​=1.3+0.02871=1.3287

M1 — correct $f'(x)$f′(x) and substitution into the Newton-Raphson formula. A1 — $x_1 = 1.3287$x1​=1.3287 to 4 d.p.

A1 (comparison) — after one Newton-Raphson step the error is $|1.3287 - 1.3284| = 0.0003$∣1.3287−1.3284∣=0.0003; after three iterations of the cube-root recurrence the error is $|1.3298 - 1.3284| = 0.0014$∣1.3298−1.3284∣=0.0014. Newton-Raphson reaches 4 d.p. accuracy in one step where the fixed-point iteration needs four or five — Newton-Raphson exhibits quadratic convergence (error roughly squares each step) versus the linear convergence of the cube-root recurrence (error multiplied by a fixed factor each step).

Total: 8 marks (M3 A5).

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Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): The function $f$f is defined by $f(x) = e^x - 3x$f(x)=ex−3x for $x \in \mathbb{R}$x∈R.

(a) Show that the equation $f(x) = 0$f(x)=0 has a root $\beta$β in the interval $[0.6, 0.7]$[0.6,0.7]. (2)

(b) Use the trapezium rule with five strips to estimate $\int_0^1 (e^x - 3x)\,dx$∫01​(ex−3x)dx, giving your answer to 3 decimal places. (3)

(c) State, with a brief reason, whether the trapezium-rule estimate over- or under-estimates the true value of the integral. (1)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1b) — $f(0.6) = e^{0.6} - 1.8 = 1.8221 - 1.8 = 0.0221$f(0.6)=e0.6−1.8=1.8221−1.8=0.0221 and $f(0.7) = e^{0.7} - 2.1 = 2.0138 - 2.1 = -0.0862$f(0.7)=e0.7−2.1=2.0138−2.1=−0.0862.
• A1 (AO2.1) — sign change and continuity statement ($f$f is continuous as the difference of an exponential and a linear function), giving $\beta \in (0.6, 0.7)$β∈(0.6,0.7) by IVT.

(b)

• M1 (AO1.1a) — $h = (1 - 0)/5 = 0.2$h=(1−0)/5=0.2 and tabulating $y_0, y_1, \dots, y_5$y0​,y1​,…,y5​ at $x = 0, 0.2, 0.4, 0.6, 0.8, 1.0$x=0,0.2,0.4,0.6,0.8,1.0.
• M1 (AO1.1b) — substituting into $T = \tfrac{1}{2}h\{y_0 + y_5 + 2(y_1 + y_2 + y_3 + y_4)\}$T=21​h{y0​+y5​+2(y1​+y2​+y3​+y4​)}.
• A1 (AO1.1b) — $T \approx 0.224$T≈0.224 to 3 d.p. (computed values: $y_0 = 1$y0​=1, $y_1 = 0.6214$y1​=0.6214, $y_2 = 0.2918$y2​=0.2918, $y_3 = 0.0221$y3​=0.0221, $y_4 = -0.1745$y4​=−0.1745, $y_5 = -0.2817$y5​=−0.2817; $T = 0.1\{0.7183 + 2(0.7608)\} = 0.1 \cdot 2.2399 = 0.2240$T=0.1{0.7183+2(0.7608)}=0.1⋅2.2399=0.2240).

(c)

• B1 (AO2.4) — $f''(x) = e^x > 0$f′′(x)=ex>0 for all $x$x, so $f$f is convex; the trapezium rule joins chords above a convex curve, hence over-estimates the integral.

Total: 6 marks split AO1 = 4, AO2 = 2. The Year 2 character of section 13 shows in part (c): the AO2 mark rewards reasoning about convexity that links section 9 (second derivative) to section 13 (numerical integration). This is the synoptic AO2 step that distinguishes A from A*.

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Synoptic links

Connects to:

1. Section 9 — Differentiation: Newton-Raphson is differentiation in disguise — $x_{n+1} = x_n - f(x_n)/f'(x_n)$xn+1​=xn​−f(xn​)/f′(xn​) is the $x$x-intercept of the tangent line at $(x_n, f(x_n))$(xn​,f(xn​)). Failure of Newton-Raphson at stationary points (where $f'(x_n) = 0$f′(xn​)=0 and the tangent is horizontal) is a section-9 fact applied to a section-13 algorithm.

2. Section 10 — Integration: the trapezium rule estimates $\int_a^b f(x)\,dx$∫ab​f(x)dx when $f$f has no elementary antiderivative — for example $\int_0^1 e^{-x^2}\,dx$∫01​e−x2dx (the Gaussian, with no closed form). Symbolic integration and numerical integration are complementary, not rival.

3. Section 4 — Graphs and transformations: sign-change methods require sketching $y = f(x)$y=f(x) to confirm a single root in the interval (multiple roots breaking the IVT logic). A* candidates always sketch before locating.

4. Section 6 — Exponentials and logarithms: $e^x$ex and $\ln x$lnx have no elementary inverses for many compound expressions, so equations like $e^x = 3x$ex=3x are routinely solved numerically. Numerical methods are the only tool for transcendental equations beyond basic forms.

5. Section 8 — Sequences and series: an iterative recurrence $x_{n+1} = g(x_n)$xn+1​=g(xn​) is a sequence; convergence to a fixed point requires $|g'(\alpha)| < 1$∣g′(α)∣<1 at the root, which is the section-8 limit-of-sequence concept applied to a section-13 algorithm.

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Mark-scheme literacy

Numerical-methods questions on 9MA0-02 split AO marks across all three categories more evenly than algebra topics: