9 Numerical methods — notes on one page

Every Edexcel A-level Maths 9MA0 specification point in 9, read straight through: 5 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

Checked against Edexcel 9MA0 section 9. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

9.1 · Locate roots of f(x) = 0 by considering changes of sign of f(x) in an interval on which f(x) is sufficiently well behaved; understand how change of sign methods can fail.

If a continuous function has values of opposite sign at the endpoints of an interval, the intermediate value theorem guarantees at least one root inside that interval.
Evaluate $f(a)$ and $f(b)$ accurately, record their signs, and conclude that a root lies in $(a,b)$ only after checking that $f$ is continuous there.
For example, $f(1)<0$ and $f(2)>0$ for a continuous $f$ , so the graph must cross the $x$ -axis at least once between $1$ and $2$ .
A sign change across a discontinuity need not contain a root, while a repeated root can touch the axis without changing sign; a common error is to treat the sign test as an equivalence.

Worked example

For $f(x)=x^3+x-5$ , show that there is exactly one root in $(1,2)$ .

1.Calculate $f(1)=1+1-5=-3$ and $f(2)=8+2-5=5$ .
2.Continuity gives at least one root in $(1,2)$ .
3.Also $f'(x)=3x^2+1$ , which is positive for every real $x$ , so $f$ is strictly increasing and cannot cross the axis more than once.

Answer: $f(1)=-3$ and $f(2)=5$ , so a root lies in $(1,2)$ .; Since $f'(x)=3x^2+1>0$ , $f$ is strictly increasing and the root is unique.

Common mistakes

Don't conclude that an even-multiplicity root cannot exist because endpoint values have the same sign.
Don't claim a root from a sign change without checking continuity, or claim uniqueness without a monotonicity argument.

Exam tip. To establish exactly one root, combine a sign change on a continuous interval with a separate uniqueness argument.

9.2 · Solve equations approximately using simple iterative methods; be able to draw associated cobweb and staircase diagrams.

A fixed-point iteration rewrites an equation as $x=g(x)$ and generates approximations using $x_{n+1}=g(x_n)$ from a chosen starting value.
Calculate successive values without premature rounding; on a graph of $y=g(x)$ and $y=x$ , move vertically to $y=g(x)$ and horizontally to $y=x$ to display each iteration.
Near a fixed point $\alpha$ , iterations usually converge when $|g'(\alpha)|<1$ : a positive gradient gives a staircase pattern and a negative gradient gives a cobweb pattern.
Different rearrangements of the same equation can converge at different rates or diverge; a common error is to assume that obtaining $x=g(x)$ automatically produces a useful recurrence.

A staircase diagram alternates between

y=g(x)

and

y=x

to display successive fixed-point iterates.

Worked example

Use $x_{n+1}=\sqrt{7-x_n}$ with $x_0=2$ to calculate $x_1$ , $x_2$ and $x_3$ to $4$ decimal places. State the equation satisfied by any limiting value.

1.Repeated substitution gives $x_1=\sqrt5=2.236067\ldots$ , $x_2=\sqrt{7-2.236067\ldots}=2.182643\ldots$ and $x_3=\sqrt{7-2.182643\ldots}=2.194847\ldots$ .
2.If the sequence tends to $L$ , then $L=\sqrt{7-L}$ , so $L^2+L-7=0$ .

Answer: $x_1=2.2361$ , $x_2=2.1826$ and $x_3=2.1948$ .; A limiting value satisfies $x^2+x-7=0$ .

Common mistakes

Don't draw horizontal moves to $y=g(x)$ and vertical moves to $y=x$ , reversing the cobweb construction.
Don't calculate iterates and fail to state the fixed-point equation or assess convergence from the iteration graph.

Exam tip. Show each substituted iterate to the requested accuracy, then set the limiting value equal to the iteration function.

9.3 · Solve equations using the Newton-Raphson method and other recurrence relations of the form xₙ₊₁ = g(xₙ); understand how such methods can fail.

Newton-Raphson replaces the curve locally by its tangent, giving $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ for a root of $f(x)=0$ .
Differentiate first, substitute the current approximation into both $f$ and $f'$ , and retain extra calculator digits until the requested final accuracy.
For instance, applying Newton-Raphson to $f(x)=x^2-a$ gives $x_{n+1}=\frac12\left(x_n+\frac{a}{x_n}\right)$ .
The method can fail when $f'(x_n)=0$ , when a tangent sends the next value away from the desired root, or when the values enter a cycle; a common error is to continue without checking the iterates.
Set $f(x)=x^3+x-4$ , so $f'(x)=3x^2+1$ .

Newton–Raphson uses the tangent at

x_n

; its intercept with the

x

-axis is the next iterate

x_{n+1}

Worked example

Derive a Newton-Raphson recurrence for $x^3+x-4=0$ . Starting with $x_0=1.3$ , find $x_1$ and $x_2$ to $5$ decimal places.

1.Set $f(x)=x^3+x-4$ , so $f'(x)=3x^2+1$ .
2.Substitution in the Newton-Raphson formula gives $x_{n+1}=x_n-\frac{x_n^3+x_n-4}{3x_n^2+1}$ .
3.Using $x_0=1.3$ gives $x_1=1.382866\ldots$ , and substituting this unrounded value gives $x_2=1.378806\ldots$ .

Answer: $x_{n+1}=x_n-\frac{x_n^3+x_n-4}{3x_n^2+1}$ .; $x_1=1.38287$ and $x_2=1.37881$ .

Common mistakes

Don't choose a starting value where $f'(x_0)=0$ , making the first Newton step undefined.
Don't use the Newton formula with the derivative evaluated at the new iterate rather than the current one.

Exam tip. Write the Newton recurrence explicitly, retain guard digits between iterations and round only the reported values.

9.4 · Understand and use numerical integration of functions, including the use of the trapezium rule and estimating the approximate area under a curve and limits that it must lie between.

The trapezium rule approximates an integral by replacing the curve over each strip with a straight chord and adding the resulting trapezium areas.
For ordinates $y_0,\ldots,y_n$ at equal spacing $h$ , use $\frac{h}{2}[y_0+y_n+2(y_1+\cdots+y_{n-1})]$ and include each interior ordinate twice.
A convex curve with $f''(x)>0$ lies below its chords, so the trapezium rule overestimates its integral; more, narrower strips usually improve the estimate.
Numerical area is not automatically the signed integral: a common error is to leave contributions below the $x$ -axis negative when the question asks for total area.

The trapezium rule joins consecutive ordinates with straight segments; convexity indicates whether this polygonal boundary lies above or below the curve.

Worked example

Use four equal strips and the trapezium rule to estimate $\int_0^2(x^2+1)\,dx$ . State whether the estimate is an overestimate or an underestimate, with a reason.

1.Here $h=0.5$ and the ordinates at $x=0,0.5,1,1.5,2$ are $1,1.25,2,3.25,5$ .
2.Thus the estimate is $\frac{0.5}{2}[1+5+2(1.25+2+3.25)]=0.25(19)=4.75$ .
3.Since $f''(x)=2>0$ , each chord lies above the curve, so the result is an overestimate.

Answer: $4.75$ ; It is an overestimate because $f''(x)=2>0$ , so the curve is convex.

Common mistakes

Don't apply weight two to the endpoints and weight one to the interior ordinates, reversing the trapezium-rule coefficients.
Don't use the number of ordinates as the number of strips, giving the wrong strip width and endpoint weights.

Exam tip. For the trapezium rule, calculate strip width from the interval and use half weight only on the two endpoints.

9.5 · Use numerical methods to solve problems in context.

A contextual numerical model translates a physical condition, such as a target value or accumulated quantity, into a root, recurrence or numerical integral.
Define the function and units before applying the method, choose an interval or starting value that is meaningful in the context, and state the requested accuracy.
For a threshold $M(t)=C$ , solve $M(t)-C=0$ numerically; for a varying rate $q(t)$ , approximate the accumulated change with $\int q(t)\,dt$ .
A calculator value is not a complete contextual answer: a common error is to omit units, ignore the model's domain or round in a way that makes a safety decision invalid.

Worked example

A culture is modelled by $P(t)=200e^{0.3t}$ , where $t$ is in hours. Use Newton-Raphson on $f(t)=200e^{0.3t}-1000$ with $t_0=5$ to find $t_1$ and $t_2$ . Hence estimate when the culture reaches $1000$ cells.

1.Differentiate to get $f'(t)=60e^{0.3t}$ .
2.Use $t_{n+1}=t_n-\frac{200e^{0.3t_n}-1000}{60e^{0.3t_n}}$ .
3.Substitution of $t_0=5$ gives $t_1=5.385502\ldots$ , and the next substitution gives $t_2=5.364857\ldots$ .
4.The positive time is therefore about $5.365$ hours.

Answer: $t_1=5.38550$ and $t_2=5.36486$ .; The culture reaches $1000$ cells after approximately $5.365$ hours.

Common mistakes

Don't select a root that satisfies the equation but lies outside the interval allowed by the context.
Don't carry out a numerical iteration but give no contextual time, units or check that the root is meaningful.

Exam tip. After numerical iteration, translate the root back into the context and report a sensible accuracy with units.