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EnglandMathsSyllabus dot point

When you cannot solve an equation or integrate a function exactly, how do you find a reliable approximate answer?

Locating roots by sign change, iterative methods including fixed point iteration and the Newton-Raphson method, the conditions under which they succeed or fail, and the trapezium rule for approximating definite integrals.

A focused answer to the AQA A-Level Mathematics numerical methods content, covering locating roots by sign change, fixed point iteration, the Newton-Raphson method and its failures, and the trapezium rule for definite integrals.

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  1. What this dot point is asking
  2. Locating roots by sign change
  3. Fixed point iteration
  4. The Newton-Raphson method
  5. The trapezium rule
  6. Why we need numerical methods, and how to present them

What this dot point is asking

AQA wants you to locate roots of an equation using a change of sign, use iterative formulae including the Newton-Raphson method, understand when these methods succeed or fail, and use the trapezium rule to approximate a definite integral and judge whether it over- or under-estimates. The questions reward precise reasoning: stating continuity, justifying convergence, and explaining the direction of error.

Locating roots by sign change

A single sign change guarantees an odd number of roots in the interval (at least one), not exactly one; if the curve dips and returns, there could be more. To confirm a root to a given accuracy, evaluate ff at the two ends of a narrow bracketing interval and show the sign change.

Fixed point iteration

Rearrange the equation into the form x=g(x)x = g(x), then iterate xn+1=g(xn)x_{n+1} = g(x_n) from a starting value. The sequence may converge to a root, shown neatly on a staircase or cobweb diagram, provided the gradient of gg near the root has magnitude less than one. Different rearrangements of the same equation can converge, diverge, or converge to different roots, so the choice of gg matters.

The Newton-Raphson method

The method fails if f(xn)f'(x_n) is zero or near zero (the tangent is nearly horizontal and the next estimate shoots far away), or if the starting value is near a turning point so that the tangent points away from the desired root.

The trapezium rule

The rule over-estimates the area when the curve bends upwards (concave up), because the straight tops of the trapezia lie above the curve, and under-estimates when the curve bends downwards. Increasing the number of strips reduces the error. Justifying the direction of error by referring to the concavity of the curve is an examined point.

Why we need numerical methods, and how to present them

Many equations that arise in practice, such as x35x3=0x^3 - 5x - 3 = 0 or x=cosxx = \cos x, have no neat algebraic solution, and many functions, such as 1+x3\sqrt{1 + x^3}, have no elementary antiderivative. Numerical methods give a usable approximation in these cases, and the exam rewards a tidy, reproducible presentation. For root finding, state the function, show the sign change with continuity, then either iterate or apply Newton-Raphson, recording each estimate to the requested accuracy. For the trapezium rule, tabulate the ordinates clearly before substituting, since a single mis-evaluated ordinate ruins the answer.

To confirm a root to a stated number of decimal places, evaluate the function at the two ends of the rounding interval. For instance, to show a root is 2.492.49 to two decimal places, check that f(2.485)f(2.485) and f(2.495)f(2.495) have opposite signs; the sign change inside that interval guarantees the rounded root. This "interval bisection of the rounding band" argument is the standard way AQA expects you to justify an accuracy claim, rather than simply asserting that successive iterates agree.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20197 marksPaper 1, Section B. The equation x35x3=0x^3 - 5x - 3 = 0 has a root α\alpha between 22 and 33. (a) Show that the root lies between 22 and 33. (b) Use the Newton-Raphson method with x0=2.5x_0 = 2.5 to find x1x_1 and x2x_2, giving your answers to four decimal places. (c) State one circumstance in which the Newton-Raphson method might fail.
Show worked answer →

For (a), with f(x)=x35x3f(x) = x^3 - 5x - 3, f(2)=8103=5f(2) = 8 - 10 - 3 = -5 and f(3)=27153=9f(3) = 27 - 15 - 3 = 9; opposite signs and ff continuous, so a root lies between 22 and 33. For (b), f(x)=3x25f'(x) = 3x^2 - 5. Then x1=2.5f(2.5)f(2.5)=2.515.62512.5318.755=2.50.12513.75=2.4909x_1 = 2.5 - \frac{f(2.5)}{f'(2.5)} = 2.5 - \frac{15.625 - 12.5 - 3}{18.75 - 5} = 2.5 - \frac{0.125}{13.75} = 2.4909. Repeating from x1x_1 gives x22.4909x_2 \approx 2.4909 (converged). For (c), the method fails if f(xn)=0f'(x_n) = 0 or is near zero (a near-horizontal tangent), or if the starting value is too far from the root. Markers reward the sign-change argument with continuity stated, accurate iteration, and a valid failure condition.

AQA 20216 marksPaper 1, Section A. Use the trapezium rule with four strips to estimate 021+x3dx\int_0^2 \sqrt{1 + x^3} \, dx, giving your answer to three decimal places. State, with a reason, whether your estimate is likely to be an over-estimate or an under-estimate.
Show worked answer →

With 44 strips on [0,2][0, 2], h=0.5h = 0.5 and ordinates at x=0,0.5,1,1.5,2x = 0, 0.5, 1, 1.5, 2: y0=1=1y_0 = \sqrt{1} = 1, y1=1.125=1.0607y_1 = \sqrt{1.125} = 1.0607, y2=2=1.4142y_2 = \sqrt{2} = 1.4142, y3=4.375=2.0917y_3 = \sqrt{4.375} = 2.0917, y4=9=3y_4 = \sqrt{9} = 3. The trapezium rule gives 0.52[1+3+2(1.0607+1.4142+2.0917)]=0.25[4+2(4.5666)]=0.25×13.13323.283\frac{0.5}{2}[1 + 3 + 2(1.0607 + 1.4142 + 2.0917)] = 0.25[4 + 2(4.5666)] = 0.25 \times 13.1332 \approx 3.283. The curve is concave up (increasing gradient), so the straight trapezium tops lie above the curve, making this an over-estimate. Markers reward correct ordinates, the rule applied with only interior ordinates doubled, and a justified over- or under-estimate.

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