What are Numerical Methods in the context of Edexcel International A Levels Pure Mathematics 3?

Numerical methods in Edexcel International A Levels Pure Mathematics 3 refer to techniques used to find approximate solutions to mathematical problems that may not be solvable analytically. These methods are crucial for solving equations, integration, and differentiation when exact solutions are difficult or impossible to obtain.

How is the Newton-Raphson method used to find roots of equations in Numerical Methods?

The Newton-Raphson method is an iterative numerical technique used to find approximate roots of a real-valued function. Starting with an initial guess, the method uses the function's derivative to iteratively converge to a root. It is particularly useful in Pure Mathematics 3 for solving equations where analytical solutions are not feasible.

What is the significance of error analysis in Numerical Methods for Pure Mathematics 3?

Error analysis in Numerical Methods is crucial for understanding the accuracy and reliability of approximate solutions. In Pure Mathematics 3, students learn to estimate and minimize errors, ensuring that numerical solutions are as close as possible to the true values. This involves understanding truncation and rounding errors in calculations.

Can you explain the use of the trapezium rule in Numerical Integration?

The trapezium rule is a numerical method used to approximate the definite integral of a function. By dividing the area under a curve into trapezoids rather than rectangles, it provides a more accurate approximation. This method is particularly useful in Pure Mathematics 3 for integrating functions that are difficult to integrate analytically.

Why is the iterative method important in solving equations numerically?

Iterative methods are important in numerical solutions because they provide a systematic approach to approximating solutions of equations. These methods, such as the Newton-Raphson method, allow students in Pure Mathematics 3 to handle complex equations by refining guesses until a satisfactory level of accuracy is achieved, making them essential tools in numerical analysis.

Why is "Quoting $f(a) f(b) < 0 \Rightarrow$ root, without mentioning continuity" a common mistake in Numerical methods?

Why it happens: Continuity feels obvious for polynomials. How to avoid it: ALWAYS state 'continuous'. The change-of-sign argument relies on the Intermediate Value Theorem, which requires continuity. Without it, sign change does not guarantee a root. Source: WMA13/01 examiner reports — flagged annually

Why is "Writing $x_{n+1} = x_{n} + \dfrac{f(x_{n})}{f'(x_{n})}$ (plus instead of minus)" a common mistake in Numerical methods?

Why it happens: Misremembering the formula. How to avoid it: Derive each time: the tangent at (x_n, f(x_n)) has equation y - f(x_n) = f'(x_n)(x - x_n). Setting y = 0 and solving: x = x_n - f(x_n)f'(x_n). MINUS.

Why is "Picking a rearrangement that diverges" a common mistake in Numerical methods?

Why it happens: Choosing g(x) without checking |g'()| < 1. How to avoid it: Before iterating, evaluate g' at (or near) the suspected root. If |g'| ≥ 1, REARRANGE differently. Multiple valid rearrangements always exist; pick the convergent one.

Why is "Quoting the final root without intermediate iterations" a common mistake in Numerical methods?

Why it happens: Calculator does it in one button press. How to avoid it: Mark scheme awards M marks for each iteration. Always write x_2, x_3, explicitly, to the requested precision.

Why is "Rounding intermediate iterations to 2 d.p., then quoting 5 d.p. final" a common mistake in Numerical methods?

Why it happens: Wanting tidy intermediate values. How to avoid it: Carry FULL calculator precision through intermediate iterations. Only round the FINAL answer to the requested precision. Mark schemes deduct for incorrect rounding propagation. Source: WMA13/01 mark schemes — strict on precision

Pure Mathematics 3Edexcel International A Levels Mathematics (XMA01-YMA01)

Numerical methods

Q: Why is "Quoting $f(a) f(b) < 0 \Rightarrow$ root, without mentioning continuity" a common mistake in Numerical methods?

Why it happens: Continuity feels obvious for polynomials. How to avoid it: ALWAYS state 'continuous'. The change-of-sign argument relies on the Intermediate Value Theorem, which requires continuity. Without it, sign change does not guarantee a root. Source: WMA13/01 examiner reports — flagged annually

Q: Why is "Writing $x_{n+1} = x_{n} + \dfrac{f(x_{n})}{f'(x_{n})}$ (plus instead of minus)" a common mistake in Numerical methods?

Why it happens: Misremembering the formula. How to avoid it: Derive each time: the tangent at (x_n, f(x_n)) has equation y - f(x_n) = f'(x_n)(x - x_n). Setting y = 0 and solving: x = x_n - f(x_n)f'(x_n). MINUS.

Q: Why is "Picking a rearrangement that diverges" a common mistake in Numerical methods?

Why it happens: Choosing g(x) without checking |g'()| < 1. How to avoid it: Before iterating, evaluate g' at (or near) the suspected root. If |g'| ≥ 1, REARRANGE differently. Multiple valid rearrangements always exist; pick the convergent one.

Q: Why is "Quoting the final root without intermediate iterations" a common mistake in Numerical methods?

Why it happens: Calculator does it in one button press. How to avoid it: Mark scheme awards M marks for each iteration. Always write x_2, x_3, explicitly, to the requested precision.

Q: Why is "Rounding intermediate iterations to 2 d.p., then quoting 5 d.p. final" a common mistake in Numerical methods?

Why it happens: Wanting tidy intermediate values. How to avoid it: Carry FULL calculator precision through intermediate iterations. Only round the FINAL answer to the requested precision. Mark schemes deduct for incorrect rounding propagation. Source: WMA13/01 mark schemes — strict on precision

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Numerical Methods Study Notes — Edexcel IAL Mathematics P3 (WMA13/01, 2018 spec — 2026 onwards)

Locating roots by sign change. Fixed-point iteration $x_{n+1} = g(x_{n})$ with convergence condition $|g'(\alpha)| < 1$ near the fixed point. Newton–Raphson method $x_{n+1} = x_{n} - \dfrac{f(x_{n})}{f'(x_{n})}$ — fast quadratic convergence when it works, failures when the tangent goes wrong. Staircase / cobweb diagrams visualise the iteration behaviour.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XMA01-YMA01 syllabus (2026 onwards).

P3 8.1 — Locate roots of $f(x) = 0$ by considering changes of sign of $f$ in an interval.
P3 8.2 — Use fixed-point iteration $x_{n+1} = g(x_{n})$ to approximate a root; describe the convergence condition.
P3 8.3 — Apply the Newton–Raphson method; identify when it fails and describe failure modes.

Locating roots: sign change

Two opposite-sign values + continuity ⇒ root in between.

Theorem (Intermediate Value). If $f$ is continuous on $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs (i.e. $f(a) f(b) < 0$ ), then there exists $c \in (a, b)$ with $f(c) = 0$ .

Standard exam exposition.

To show that $f(x) = 0$ has a root in $(a, b)$ :

Compute $f(a)$ and $f(b)$ — show they are on opposite sides of zero.
State that $f$ is continuous on $[a, b]$ .
Conclude by the change-of-sign principle.

Worked example. Show $x^{3} - 5x + 1 = 0$ has a root in $(0, 1)$ .

$f(x) = x^{3} - 5x + 1$ .
$f(0) = 1 > 0$ . $f(1) = 1 - 5 + 1 = -3 < 0$ .
$f$ is a polynomial, so continuous on $[0, 1]$ .
By the change-of-sign principle, $f$ has a root in $(0, 1)$ .

Interval bisection (refinement). Once a root is bracketed in $(a, b)$ , bisect:

Compute $m = (a + b)/2$ and $f(m)$ .
If $f(a) f(m) < 0$ , root lies in $(a, m)$ ; otherwise in $(m, b)$ .
Repeat to narrow the interval.

This is reliable but SLOW — typically only used to set up a starting value for a faster method (fixed-point or Newton–Raphson).

Pitfalls.

Continuity matters. Two opposite-sign values do NOT imply a root if $f$ has a discontinuity in between (e.g. a vertical asymptote).
Number of roots. Sign change guarantees AT LEAST one root, not exactly one. Two or more roots in $(a, b)$ are consistent with one sign change at the endpoints.

$f(a) f(b) < 0$ + $f$ continuous ⇒ root in $(a, b)$ .
Always state continuity explicitly.
Bisection refines the interval.
Sign change guarantees at least one root, not exactly one.

Fixed-point iteration

Rearrange to $x = g(x)$ and iterate. Convergence needs $|g'(\alpha)| < 1$ .

Setup. Rearrange $f(x) = 0$ as $x = g(x)$ for some function $g$ . A root of $f$ corresponds to a FIXED POINT of $g$ (a point where $g(\alpha) = \alpha$ ).

Iteration. Pick $x_{1}$ near the suspected root. Compute $x_{2} = g(x_{1}), \quad x_{3} = g(x_{2}), \quad \ldots$

If the sequence converges, it converges to a fixed point.

Convergence condition. Near a fixed point $\alpha$ , the iteration converges if $|g'(\alpha)| < 1.$

If $|g'(\alpha)| > 1$ , the iteration diverges from $\alpha$ regardless of how close you start.

Multiple rearrangements. Every $f(x) = 0$ has many valid rearrangements $x = g(x)$ . Some will converge, others diverge. Pick one with $|g'(\alpha)| < 1$ .

Worked example. $f(x) = x^{3} - 5x + 1 = 0$ has a root near $0.2$ .

Rearrangement 1: $5x = x^{3} + 1 \Rightarrow x = \dfrac{x^{3} + 1}{5}$ , so $g(x) = \dfrac{x^{3} + 1}{5}$ . $g'(x) = \dfrac{3x^{2}}{5}$ . At $\alpha \approx 0.2$ : $g'(\alpha) \approx 0.024$ . $|g'| \ll 1$ — converges fast. ✓

Rearrangement 2: $x^{3} = 5x - 1 \Rightarrow x = (5x - 1)^{1/3}$ . $g'(x) = \dfrac{5}{3(5x - 1)^{2/3}}$ . At $\alpha \approx 0.2$ : $g'(\alpha) = \dfrac{5}{3(0)^{2/3}}$ — UNDEFINED! Different problem here.

Iterate Rearrangement 1 with $x_{1} = 0.2$ : $x_{2} = \dfrac{0.008 + 1}{5} = 0.2016, \quad x_{3} \approx 0.2016, \quad x_{4} \approx 0.2016.$

Converged to $\alpha \approx 0.2016$ in just two steps.

Convergence speed. $|g'(\alpha)|$ small ⇒ fast convergence. $|g'(\alpha)|$ close to $1$ ⇒ slow convergence.

Sign of $g'$ and pattern.

$g'(\alpha) > 0$ : STAIRCASE (iterates approach from one side).
$g'(\alpha) < 0$ : COBWEB (iterates alternate above/below the root).

Fixed point = root of $f(x) = 0$ via $x = g(x)$ .
Convergence condition: $|g'(\alpha)| < 1$ .
Multiple rearrangements possible; pick a convergent one.
Staircase if $g'(\alpha) > 0$ ; cobweb if $< 0$ .

Newton–Raphson method

Tangent-line approximation. Fast — when it works.

Formula. $x_{n+1} = x_{n} - \dfrac{f(x_{n})}{f'(x_{n})}.$

Geometric derivation. Tangent to $y = f(x)$ at $(x_{n}, f(x_{n}))$ : $y - f(x_{n}) = f'(x_{n})(x - x_{n}).$ Setting $y = 0$ (where the tangent meets the $x$ -axis): $-f(x_{n}) = f'(x_{n})(x - x_{n}) \Rightarrow x = x_{n} - \dfrac{f(x_{n})}{f'(x_{n})}$ . This is the next iterate.

Quadratic convergence. If the iteration converges, the error roughly squares each step: $|x_{n+1} - \alpha| \approx C |x_{n} - \alpha|^{2}.$

So if $x_{n}$ has error $0.01$ , $x_{n+1}$ has error roughly $0.0001$ . Each iteration roughly DOUBLES the number of correct digits — far faster than fixed-point iteration.

Worked example. $f(x) = x^{3} - 5x + 1$ , root near $0.2$ , $x_{1} = 0.5$ .

$f'(x) = 3x^{2} - 5$ .
$f(0.5) = 0.125 - 2.5 + 1 = -1.375$ . $f'(0.5) = 0.75 - 5 = -4.25$ .
$x_{2} = 0.5 - \dfrac{-1.375}{-4.25} = 0.5 - 0.3235 = 0.1765$ .
$f(0.1765) \approx 0.123$ . $f'(0.1765) \approx -4.906$ .
$x_{3} = 0.1765 - \dfrac{0.123}{-4.906} \approx 0.2016$ .

Converged to $4$ d.p. in two steps.

When NR fails.

$f'(x_{n}) = 0$ : horizontal tangent, division by zero in the formula.
$f'(x_{n})$ very small: tangent step is huge, iterate jumps far from the root.
Oscillation: iterate bounces between two values without converging.
Diverges to a different root: if the curve curves away, the iterate can land on a different basin of attraction.

Geometric interpretation of failure. Sketch the curve and the tangent at $x_{n}$ . If the tangent meets the $x$ -axis far from the root (or on the wrong side of a turning point), NR will misbehave.

Worked example of failure. $f(x) = x^{3} - 5x + 1$ . From $x_{1} = -1$ : $f(-1) = -1 + 5 + 1 = 5$ , $f'(-1) = -2$ . NR gives $x_{2} = -1 - 5/(-2) = -1 + 2.5 = 1.5$ . The iteration jumped to a different region — and from $x_{2} = 1.5$ it may not converge to the root near $0.2$ (it may instead converge to the root near $-2.13$ or $2.13$ , missing the target).

Practical advice. Always SKETCH $y = f(x)$ before applying NR, to choose $x_{1}$ wisely.

$x_{n+1} = x_{n} - \dfrac{f(x_{n})}{f'(x_{n})}$ .
Quadratic convergence near a root.
Tangent meets $x$ -axis at $x_{n+1}$ .
Failures: $f'(x_{n}) = 0$ , oscillation, jumping far.

Cobweb and staircase diagrams

Visualise the fixed-point iteration via $y = g(x)$ and $y = x$ .

Setup. Plot $y = g(x)$ and $y = x$ on the same axes. Fixed points are intersections.

Drawing the iteration.

Start at $x = x_{1}$ on the $x$ -axis.
Go VERTICALLY up to $y = g(x_{1})$ on the curve.
Go HORIZONTALLY to the line $y = x$ , hitting $(g(x_{1}), g(x_{1}))$ — i.e. $(x_{2}, x_{2})$ .
Drop vertically to the $x$ -axis at $x = x_{2}$ .
Repeat.

Staircase ( $g'(\alpha) > 0$ ). Iterates approach $\alpha$ from one side. The path looks like a staircase climbing or descending.

Cobweb ( $g'(\alpha) < 0$ ). Iterates alternate above and below $\alpha$ . The path looks like a spiral / web.

Convergence vs divergence.

$|g'(\alpha)| < 1$ : iteration spirals/staircase IN — CONVERGES.
$|g'(\alpha)| > 1$ : iteration spirals/staircase OUT — DIVERGES.

$g'(\alpha)$	Pattern	Convergence
$0 < g'(\alpha) < 1$	Staircase	Converges
$g'(\alpha) > 1$	Staircase	Diverges
$-1 < g'(\alpha) < 0$	Cobweb	Converges
$g'(\alpha) < -1$	Cobweb	Diverges

Worked example. $g(x) = \dfrac{x^{3} + 1}{5}$ . $g'(x) = \dfrac{3x^{2}}{5} > 0$ for $x \neq 0$ . At $\alpha \approx 0.2016$ : $g' \approx 0.024 < 1$ . STAIRCASE, CONVERGES. ✓

Use in exams. P3 occasionally asks for a small sketch of the cobweb/staircase diagram. Draw two iterations (vertical-horizontal-vertical-horizontal) clearly.

Plot $y = g(x)$ AND $y = x$ together.
Vertical to curve, horizontal to line, vertical, horizontal.
$g'(\alpha) > 0$ ⇒ staircase. $g'(\alpha) < 0$ ⇒ cobweb.
$|g'(\alpha)| < 1$ ⇒ converges. $> 1$ ⇒ diverges.

How it’s examined

Numerical Methods is a $9$ - $13$ -mark question on every WMA13/01 paper. Standard format: (a) sign-change root location ( $3$ marks); (b) rearrange and iterate ( $4$ marks — show all $x_{2}, x_{3}, x_{4}$ ); (c) Newton–Raphson with $x_{1}$ given ( $4$ marks — show $x_{2}$ and $x_{3}$ ); (d) describe a failure of NR or interpret a cobweb diagram ( $2$ - $3$ marks). Marks lost: missing continuity statement (B mark), missing iteration values, incorrect rounding ( $5$ d.p. then quoting only $3$ s.f.). Always carry full calculator precision; round only at the final step.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA13/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Numerical methods

Step-by-step solutions to past-paper-style questions on numerical methods, written exactly the way a tutor would explain them at the board.

1Locate a root by sign change (3 marks, P3)
Core• Adapted from WMA13/01 January 2024 Q5(a)• sign change, root location
Question
Show that the equation $x^{3} - 5x + 1 = 0$ has a root $\alpha$ in the interval $(0, 1)$ .
Step-by-step solution
1. Step 1
  Define $f(x) = x^{3} - 5x + 1$ . Evaluate at the endpoints.
  $f(0) = 0 - 0 + 1 = 1 > 0$
2. Step 2
  And:
  $f(1) = 1 - 5 + 1 = -3 < 0$
3. Step 3
  $f$ is continuous (polynomial) on $[0, 1]$ and changes sign — by the change-of-sign principle there is at least one root in $(0, 1)$ .
Answer
$f(0) = 1 > 0$ and $f(1) = -3 < 0$ , and $f$ is continuous, so by the change-of-sign principle a root lies in $(0, 1)$ .
Examiner tip
M1 for evaluating both endpoints (or specifying the function). A1 for the two numerical values. A1 for stating BOTH the change of sign AND the continuity argument. Just quoting $f(0) > 0$ and $f(1) < 0$ without mentioning continuity loses the final A. (Even better: write 'continuous' explicitly.)
2Fixed-point iteration: rearrange and iterate (6 marks, P3)
Extended• Adapted from WMA13/01 June 2024 Q5• fixed-point iteration, convergence
Question
The equation $x^{3} - 5x + 1 = 0$ has a root $\alpha$ in $(0, 1)$ . (a) Rearrange the equation into the form $x = g(x) = \dfrac{x^{3} + 1}{5}$ . (b) Using $x_{1} = 0.2$ , apply the iteration $x_{n+1} = \dfrac{x_{n}^{3} + 1}{5}$ to find $x_{2}, x_{3}, x_{4}$ to 4 d.p.
Step-by-step solution
1. Step 1
  (a) From $x^{3} - 5x + 1 = 0$ , isolate $5x$ : $5x = x^{3} + 1$ , so $x = \dfrac{x^{3} + 1}{5}$ . ✓
2. Step 2
  (b) Iterate starting from $x_{1} = 0.2$ .
  $x_{2} = \dfrac{0.2^{3} + 1}{5} = \dfrac{0.008 + 1}{5} = \dfrac{1.008}{5} = 0.2016$
3. Step 3
  Next iteration:
  $x_{3} = \dfrac{0.2016^{3} + 1}{5} = \dfrac{0.00819 + 1}{5} \approx 0.2016$
4. Step 4
  Continue:
  $x_{4} \approx 0.2016 \text{ (4 d.p.)}$
5. Step 5
  The sequence has converged to $\alpha \approx 0.2016$ . Convergence justified by $|g'(\alpha)| = |3\alpha^{2}/5| \approx 0.024 \ll 1$ .
Answer
$x_{2} \approx 0.2016$ , $x_{3} \approx 0.2016$ , $x_{4} \approx 0.2016$ ; root $\alpha \approx 0.2016$ .
Examiner tip
(a) M1 algebraic rearrangement. A1 final $g(x)$ . (b) M1 substitute $x_{1}$ . A1 $x_{2}$ . A1 $x_{3}$ . A1 $x_{4}$ . Always show ALL iterations to the requested precision — even if they look identical. Mark scheme penalises missing iterations.
3Newton–Raphson method (5 marks, P3)
Extended• Adapted from WMA13/01 January 2023 Q8• Newton-Raphson
Question
$f(x) = x^{3} - 5x + 1$ , with a root $\alpha$ in $(0, 1)$ . Use the Newton–Raphson method with $x_{1} = 0.5$ to find $x_{2}$ and $x_{3}$ to 5 d.p.
Step-by-step solution
1. Step 1
  $f(x) = x^{3} - 5x + 1$ , $f'(x) = 3x^{2} - 5$ . Newton–Raphson:
  $x_{n+1} = x_{n} - \dfrac{f(x_{n})}{f'(x_{n})}$
2. Step 2
  At $x_{1} = 0.5$ : $f(0.5) = 0.125 - 2.5 + 1 = -1.375$ . $f'(0.5) = 0.75 - 5 = -4.25$ .
  $x_{2} = 0.5 - \dfrac{-1.375}{-4.25} = 0.5 - 0.32353 = 0.17647$
3. Step 3
  Round: $x_{2} \approx 0.17647$ .
4. Step 4
  At $x_{2} = 0.17647$ : $f(0.17647) \approx 0.00550 - 0.88235 + 1 = 0.12315$ . $f'(0.17647) \approx 0.09343 - 5 = -4.90657$ .
  $x_{3} = 0.17647 - \dfrac{0.12315}{-4.90657} \approx 0.17647 + 0.02510 = 0.20157$
5. Step 5
  So $x_{3} \approx 0.20157$ (consistent with the root $\alpha \approx 0.2016$ ).
Answer
$x_{2} \approx 0.17647$ , $x_{3} \approx 0.20157$ .
Examiner tip
M1 quote NR formula. A1 each correct evaluation of $f, f'$ . A1 $x_{2}$ . A1 $x_{3}$ . NR converges much faster than fixed-point iteration but requires $f'(x_{n}) \neq 0$ . Always carry 5+ d.p. through intermediate steps to avoid rounding loss.
4Explain when Newton–Raphson fails (3 marks, P3)
Extended• Adapted from WMA13/01 June 2024 Q5(c)• Newton-Raphson, failure analysis
Question
Explain why Newton–Raphson cannot be used to find the root of $f(x) = x^{3} - 5x + 1$ if you start from $x_{1} = -1$ . (Hint: compute $f'(-1)$ .)
Step-by-step solution
1. Step 1
  $f'(x) = 3x^{2} - 5$ . At $x = -1$ : $f'(-1) = 3 - 5 = -2$ . Not zero — so the iteration is defined. (The hint is misleading; check if a turning point is nearby.)
2. Step 2
  Actually, look at where $f'$ vanishes: $3x^{2} - 5 = 0 \Rightarrow x = \pm \sqrt{5/3} \approx \pm 1.291$ . At any iterate close to these values $f'$ is small and the Newton step is huge.
3. Step 3
  Starting from $x_{1} = -1$ , NR overshoots (because $f'(-1) = -2$ is small in magnitude and $f(-1) = -1 + 5 + 1 = 5$ is large) and the iteration diverges or jumps to a different root.
4. Step 4
  General failure modes: (i) $f'(x_{n}) = 0$ (division by zero — tangent is horizontal); (ii) the tangent at $x_{n}$ crosses far from the root; (iii) oscillation between two values.
Answer
NR fails / behaves poorly when $f'(x_{n}) = 0$ (or is very small) because the next iterate is $x_{n} - f(x_{n})/f'(x_{n})$ , which diverges. From $x_{1} = -1$ , the tangent step lands far from the target root.
Examiner tip
B1 identify the role of $f'(x_{n})$ . B1 explain the geometric reason (horizontal tangent / large jump). B1 link to the specific starting value. P3 questions on NR failure expect a clear geometric explanation, not just 'it diverges'.
5Interpret a cobweb / staircase diagram (4 marks, P3)
Extended• Adapted from WMA13/01 June 2023 Q4• fixed point, cobweb
Question
Sketch the iteration $x_{n+1} = g(x_{n})$ where $g(x) = \dfrac{x^{3} + 1}{5}$ , starting from $x_{1} = 0.2$ . Indicate the staircase diagram converging to the fixed point.
Step-by-step solution
1. Step 1
  Set axes: plot $y = g(x) = \dfrac{x^{3} + 1}{5}$ and the line $y = x$ on the same axes. Fixed points = intersections.
2. Step 2
  At $x_{1} = 0.2$ on the $x$ -axis. Go vertically up to $y = g(0.2) \approx 0.2016$ . Then horizontally to $y = x$ , hitting $(0.2016, 0.2016)$ .
3. Step 3
  Drop vertically to the $x$ -axis at $x_{2} = 0.2016$ . Repeat: vertical to $g$ , horizontal to $y = x$ . Each step is tiny because $|g'(0.2016)| \approx 0.024 \ll 1$ — fast convergence.
4. Step 4
  Staircase (both steps moving in the same direction) vs cobweb (alternating): which one appears depends on the sign of $g'$ . Here $g'(x) = 3x^{2}/5 > 0$ , so STAIRCASE.
Answer
Staircase converges to $\alpha \approx 0.2016$ because $0 < g'(\alpha) < 1$ (positive — staircase; magnitude less than $1$ — converges).
Examiner tip
B1 plot $y = g(x)$ and $y = x$ . M1 first step. M1 second step. A1 identify staircase vs cobweb. Positive derivative gives staircase; negative gives cobweb. Magnitude $< 1$ converges; $> 1$ diverges. Memorise this $2 \times 2$ matrix of behaviours.

Key Formulae — Numerical methods

The formulae you need to memorise for numerical methods on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Change-of-sign principle
$f \text{ continuous on } [a, b], \;\; f(a) f(b) < 0 \;\Rightarrow\; \exists \, c \in (a, b) \text{ with } f(c) = 0$
$f$
continuous function
$(a, b)$
interval where the sign changes
When to use
Whenever you need to LOCATE a root. Two evaluations with opposite sign + continuity = root exists. NOT in the IAL formula booklet — state the conditions in words.
Example
$f(0) = 1$ , $f(1) = -3$ , $f$ continuous polynomial $\Rightarrow$ root in $(0, 1)$ .
Fixed-point iteration
$x_{n+1} = g(x_{n})$
$g(x)$
rearrangement of $f(x) = 0$ into $x = g(x)$
$x_{1}$
initial guess
When to use
When you can rearrange $f(x) = 0$ as $x = g(x)$ . Sequence converges to a root $\alpha$ if $|g'(\alpha)| < 1$ . NOT in the formula booklet.
Example
$x^{3} - 5x + 1 = 0 \Rightarrow x = \dfrac{x^{3} + 1}{5}$ ; iterate.
Newton–Raphson method
$x_{n+1} = x_{n} - \dfrac{f(x_{n})}{f'(x_{n})}$
$x_{n}$
current iterate
$f, f'$
function and its derivative
When to use
When you have $f(x) = 0$ and $f'(x)$ is easy to compute. Quadratic convergence near the root. Fails if $f'(x_{n}) = 0$ (horizontal tangent) or if the iteration oscillates. NOT in the IAL formula booklet — memorise.
Example
$f(x) = x^{3} - 5x + 1$ , $x_{1} = 0.5 \Rightarrow x_{2} \approx 0.176$ , $x_{3} \approx 0.202$ .
Convergence condition (fixed-point iteration)
$|g'(\alpha)| < 1 \Rightarrow \text{iteration converges near } \alpha$
$\alpha$
the fixed point (root)
$g'$
derivative of the iterator $g$
When to use
Justifying convergence of a fixed-point scheme. If $|g'(\alpha)| > 1$ , the iteration DIVERGES from $\alpha$ no matter how close you start. NOT in the formula booklet.
Example
$g(x) = \dfrac{x^{3} + 1}{5}$ , $g'(\alpha) = \dfrac{3\alpha^{2}}{5} \approx 0.024 \Rightarrow$ converges.

Key Definitions and Keywords — Numerical methods

Definitions to memorise and the exact keywords mark schemes credit for numerical methods answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Root of an equation
Examiner keyword
A value $\alpha$ such that $f(\alpha) = 0$ . Equivalently a zero of $f$ . Roots may be real or complex; in P3 we work with real roots.
Change-of-sign / interval bisection
Examiner keyword
A method of locating a root by finding two values $a$ , $b$ with $f(a) f(b) < 0$ — implies (by continuity) that a root lies between them.
Fixed point of $g$
Examiner keyword
A value $\alpha$ with $g(\alpha) = \alpha$ . Geometrically the intersection of $y = g(x)$ with the line $y = x$ . Equivalently a root of $f(x) = x - g(x) = 0$ .
Newton–Raphson method
Examiner keyword
An iteration $x_{n+1} = x_{n} - \dfrac{f(x_{n})}{f'(x_{n})}$ that converges quadratically to a root of $f$ (when it converges). Geometrically: take the tangent at $(x_{n}, f(x_{n}))$ and let $x_{n+1}$ be where the tangent meets the $x$ -axis.
Staircase / cobweb diagram
A graphical representation of the fixed-point iteration on the same axes as $y = g(x)$ and $y = x$ . Staircase (steps in one direction) when $g'(\alpha) > 0$ ; cobweb (alternating) when $g'(\alpha) < 0$ .

Common Mistakes and Misconceptions — Numerical methods

The traps other students keep falling into on numerical methods questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Quoting $f(a) f(b) < 0 \Rightarrow$ root, without mentioning continuity
WMA13/01 examiner reports — flagged annually
Why it happens
Continuity feels obvious for polynomials.
How to avoid it
ALWAYS state 'continuous'. The change-of-sign argument relies on the Intermediate Value Theorem, which requires continuity. Without it, sign change does not guarantee a root.
✕Writing $x_{n+1} = x_{n} + \dfrac{f(x_{n})}{f'(x_{n})}$ (plus instead of minus)
Why it happens
Misremembering the formula.
How to avoid it
Derive each time: the tangent at $(x_{n}, f(x_{n}))$ has equation $y - f(x_{n}) = f'(x_{n})(x - x_{n})$ . Setting $y = 0$ and solving: $x = x_{n} - \dfrac{f(x_{n})}{f'(x_{n})}$ . MINUS.
✕Picking a rearrangement that diverges
Why it happens
Choosing $g(x)$ without checking $|g'(\alpha)| < 1$ .
How to avoid it
Before iterating, evaluate $g'$ at (or near) the suspected root. If $|g'| \geq 1$ , REARRANGE differently. Multiple valid rearrangements always exist; pick the convergent one.
✕Quoting the final root without intermediate iterations
Why it happens
Calculator does it in one button press.
How to avoid it
Mark scheme awards M marks for each iteration. Always write $x_{2}, x_{3}, \ldots$ explicitly, to the requested precision.
✕Rounding intermediate iterations to 2 d.p., then quoting 5 d.p. final
WMA13/01 mark schemes — strict on precision
Why it happens
Wanting tidy intermediate values.
How to avoid it
Carry FULL calculator precision through intermediate iterations. Only round the FINAL answer to the requested precision. Mark schemes deduct for incorrect rounding propagation.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Numerical methods — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Numerical methods

Short Study Notes in the form of Slides

Short Study Notes — Numerical methods

Detailed Notes

What you’ll learn

How it’s examined

1Locate a root by sign change (3 marks, P3)

2Fixed-point iteration: rearrange and iterate (6 marks, P3)

3Newton–Raphson method (5 marks, P3)

4Explain when Newton–Raphson fails (3 marks, P3)

5Interpret a cobweb / staircase diagram (4 marks, P3)

Change-of-sign principle

Fixed-point iteration

Newton–Raphson method

Convergence condition (fixed-point iteration)

Root of an equation

Change-of-sign / interval bisection

Fixed point of ggg

Newton–Raphson method

Staircase / cobweb diagram

✕Quoting f(a)f(b)<0⇒f(a) f(b) < 0 \Rightarrowf(a)f(b)<0⇒ root, without mentioning continuity

✕Writing xn+1=xn+f(xn)f′(xn)x_{n+1} = x_{n} + \dfrac{f(x_{n})}{f'(x_{n})}xn+1​=xn​+f′(xn​)f(xn​)​ (plus instead of minus)

✕Picking a rearrangement that diverges

✕Quoting the final root without intermediate iterations

✕Rounding intermediate iterations to 2 d.p., then quoting 5 d.p. final

Practice questions

Past paper style quiz

Video lesson

Numerical methods — frequently asked questions

Fixed point of $g$

✕Quoting $f(a) f(b) < 0 \Rightarrow$ root, without mentioning continuity

✕Writing $x_{n+1} = x_{n} + \dfrac{f(x_{n})}{f'(x_{n})}$ (plus instead of minus)