Numerical Methods — A-Level Mathematics Pure 3 Study Guide

For: A-Level Mathematics candidates sitting Paper 3 (Pure Mathematics 3).

Covers: sign-change root location, iterative method convergence, the $|g^{\prime }(x)|<1$ convergence condition, identification of divergent or oscillatory iterations, and application of these techniques to find roots to specified accuracy.

You should already know: A-Level Mathematics Pure 1 (functions, calculus, trig).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Mathematics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Numerical Methods?

Numerical methods are algorithmic, approximation-based techniques used to solve mathematical problems that have no closed-form algebraic solution, such as $x=e^{-x}$ or high-degree polynomial equations. Unlike algebraic methods that return exact solutions, numerical methods generate increasingly precise estimates of the solution, up to a user-specified level of accuracy. This topic accounts for 8-12 marks per A-Level Mathematics Paper 3, usually in 1-2 structured questions, making it a high-yield, low-effort area to master for your exam.

2. Locating roots — sign-change argument

A root $\alpha$ of a function $f(x)$ is a value where $f(\alpha )=0$, corresponding to the x-intercept of the graph $y=f(x)$. The sign-change argument is the first step in root finding, used to narrow down the interval where the root lies:

Sign-change rule: If $f(x)$ is a continuous function (no breaks, jumps, or asymptotes) on the interval $[a,b]$, and $f(a)$ and $f(b)$ have opposite signs, then there is at least one root of $f(x)=0$ in the open interval $(a,b)$.

Examiners explicitly require you to state the continuity condition to earn full marks for this step, even if the function is obviously continuous (e.g., a polynomial).

Worked Example

Show that $f(x)=x^3-x-2$ has a root between $x=1$ and $x=2$:

1. Calculate $f(1)=1^3-1-2=-2$ (negative)
2. Calculate $f(2)=2^3-2-2=4$ (positive)
3. State: $f(x)$ is a polynomial, so it is continuous on $[1,2]$. The opposite signs of $f(1)$ and $f(2)$ confirm there is a root in $(1,2)$.

3. Iterative methods — convergence of $x_{n+1}=g(x_n)$

Once you have a rough interval for the root, iterative methods are used to generate increasingly precise estimates of the root. The process works as follows:

1. Rearrange the original equation $f(x)=0$ into the form $x=g(x)$, where $g(x)$ is a continuous function.
2. Choose an initial guess $x_0$ near the root (usually taken from the sign-change interval).
3. Use the recurrence relation $x_{n+1}=g(x_n)$ to generate a sequence of estimates $x_1,x_2,x_3,...$

The sequence is convergent if ${\lim }_{n\to \infty }{x}_n=\alpha$, the true root, meaning each subsequent estimate gets closer to $\alpha$. If the sequence does not approach $\alpha$, it is non-convergent.

Worked Example

For $x^3-x-2=0$, two possible rearrangements to $x=g(x)$ form are:

1. $g_1(x)=\sqrt[3]{x+2}$
2. $g_2(x)=x^3-2$

Testing with initial guess $x_0=1.5$:

• For $g_1(x)$: $x_1=\sqrt[3]{1.5+2}\approx 1.518$, $x_2=\sqrt[3]{1.518+2}\approx 1.521$, $x_3\approx 1.521$. The sequence converges to ~1.521.
• For $g_2(x)$: $x_1=1.5^3-2=1.375$, $x_2=1.375^3-2\approx 0.609$, $x_3\approx -1.774$. The sequence moves further away from the root, so it diverges.

The choice of rearrangement directly determines if the iteration converges.

4. Convergence condition $|g^{\prime }(x)|<1$ near the root

The convergence of $x_{n+1}=g(x_n)$ depends on the gradient of $g(x)$ near the root. We can derive the condition using the Mean Value Theorem:

• For the true root $\alpha$, $\alpha =g(\alpha )$ by definition.
• The error at step $n$ is $e_n=x_n-\alpha$. Using the Mean Value Theorem: $$x_{n+1}-\alpha =g(x_n)-g(\alpha )=g^{\prime }(c)(x_n-\alpha )$$ where $c$ is a value between $x_n$ and $\alpha$.
• Taking absolute values: $|e_{n+1}|=|g^{\prime }(c)||e_n|$

For the error to get smaller with each iteration, $|g^{\prime }(c)|<1$ for all $c$ near the root $\alpha$. This is the core convergence condition you will be tested on in almost every numerical methods question.

Worked Example

Verify that $g_1(x)=\sqrt[3]{x+2}$ converges to the root $\alpha \approx 1.521$ of $x^3-x-2=0$:

1. Calculate the derivative: $g_1^{\prime }(x)=\frac{1}{3}(x+2{)}^{-2/3}=\frac{1}{3(x+2{)}^{2/3}}$
2. Substitute $x=\alpha \approx 1.521$: $g_1^{\prime }(\alpha )=\frac{1}{3(1.521+2{)}^{2/3}}\approx \frac{1}{3\times 2.31}\approx 0.144<1$
3. State: The absolute value of the derivative is less than 1 near the root, so the iteration converges.

For $g_2(x)=x^3-2$, $g_2^{\prime }(x)=3x^2$, so $g_2^{\prime }(\alpha )\approx 3\times 2.31=6.93>1$, which explains why it diverges.

5. Spotting divergent or oscillatory iterations

Non-convergent sequences fall into two categories, and you can identify their behavior using both the sign and magnitude of $g^{\prime }(\alpha )$ without calculating multiple iterations:

1. Divergent sequences: Values of $x_n$ move further away from $\alpha$, trending to $\pm \infty$. This occurs when $|g^{\prime }(\alpha )|>1$.
2. Oscillatory sequences: Values of $x_n$ bounce between values on either side of $\alpha$. This occurs when $g^{\prime }(\alpha )$ is negative. There are two subtypes:

• Oscillatory convergence: $|g^{\prime }(\alpha )|<1$, so the oscillations get smaller over time and approach $\alpha$.
• Oscillatory divergence: $|g^{\prime }(\alpha )|>1$, so the oscillations get larger over time and move away from $\alpha$.

Worked Example

For $g(x)=\cos (x)$, the root of $x=\cos (x)$ is $\alpha \approx 0.739$:

1. $g^{\prime }(x)=-\sin (x)$, so $g^{\prime }(\alpha )\approx -\sin (0.739)\approx -0.673$
2. $|g^{\prime }(\alpha )|<1$, so the sequence converges, and the negative derivative means it converges oscillatory. Testing with $x_0=0$: $x_1=1$, $x_2\approx 0.540$, $x_3\approx 0.857$, $x_4\approx 0.654$, which bounces around 0.739 while getting closer.

6. Application: find roots of equations to specified accuracy

To find a root to a specified level of accuracy, follow this exam-standard process:

1. Use the sign-change rule to find an initial interval for the root.
2. Generate iterative terms using the given $x_{n+1}=g(x_n)$ formula, keeping 4-5 decimal places for intermediate values to avoid rounding error.
3. Stop when two consecutive terms round to the same value at the required accuracy.
4. Verify with a final sign-change test: confirm the root lies in an interval of width less than $0.5\times 10^{-d}$ for $d$ decimal places, or the equivalent threshold for significant figures. This proves that all values in the interval round to the same final answer.

Worked Example

Find the root of $x^3-x-2=0$ to 2 decimal places:

1. We already know the root is in $(1.5,1.6)$.
2. Iterative terms with $g_1(x)=\sqrt[3]{x+2}$: $x_1\approx 1.518$, $x_2\approx 1.521$, $x_3\approx 1.521$.
3. Sign-change test: $f(1.521)=(1.521{)}^3-1.521-2\approx -0.003$, $f(1.522)\approx 0.003$. The interval $(1.521,1.522)$ has width $0.001<0.005$ (the threshold for 2 decimal places), so the root is $1.52$ to 2 decimal places.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting to state continuity when using the sign-change argument. Why students do it: They assume the sign change alone is enough. Correct move: Always explicitly state the function is continuous on the interval, and explain why (e.g., "polynomial, so continuous everywhere") to earn full marks.
• Wrong move: Calculating $g^{\prime }(x)$ incorrectly when checking convergence. Why students do it: Rushing differentiation, especially for fractional powers or quotients. Correct move: Write out the derivative step-by-step, substitute values across the full interval given in the question, and explicitly confirm $|g^{\prime }(x)|<1$ for all x in the interval.
• Wrong move: Stopping iterations as soon as two terms match the required accuracy, without the final sign-change test. Why students do it: They think matching consecutive terms is sufficient proof. Correct move: Always perform the sign-change test on the interval of uncertainty to confirm the root lies inside, as required by Cambridge mark schemes.
• Wrong move: Rounding intermediate iteration values too early. Why students do it: To speed up calculation. Correct move: Keep at least 4-5 decimal places for all intermediate terms, only round your final answer to the required accuracy.
• Wrong move: Confusing the sign of $g^{\prime }(x)$ for oscillatory behavior. Why students do it: They mix up magnitude and sign rules. Correct move: Remember magnitude of $g^{\prime }(x)$ determines convergence/divergence, sign determines monotonic/oscillatory behavior.

8. Practice Questions (A-Level Mathematics P3 Style)

Question 1

(a) Show that the equation $e^x-3x=0$ has a root $\alpha$ in the interval $[1,2]$. [2 marks] (b) The equation is rearranged to the form $x=g(x)=\ln (3x)$. Show that this iterative form will converge to $\alpha$ for values of $x$ near $\alpha$. [3 marks]

Solution

(a) Let $f(x)=e^x-3x$. $f(1)=e-3\approx -0.282$ (negative), $f(2)=e^2-6\approx 1.389$ (positive). $f(x)$ is the sum of an exponential and linear function, so it is continuous on $[1,2]$. The opposite signs confirm a root in $(1,2)$. (b) Calculate $g^{\prime }(x)=\frac{d}{dx}\ln (3x)=\frac{1}{x}$. For $x$ near the root $\alpha \approx 1.512$, $g^{\prime }(\alpha )=\frac{1}{1.512}\approx 0.66$, so $|g^{\prime }(x)|<1$ near the root, confirming convergence.

Question 2

The iterative formula $x_{n+1}=4-\frac{1}{x_n^2}$ is used to find a root of $x^3-4x^2+1=0$, with initial value $x_0=3.8$. (a) Calculate the first 4 iterations of the sequence, giving values to 4 decimal places. [3 marks] (b) Determine if the sequence is convergent, divergent, or oscillatory, justifying your answer. [2 marks]

Solution

(a) $x_0=3.8$ $x_1=4-\frac{1}{3.8^2}\approx 4-0.0693=3.9307$ $x_2=4-\frac{1}{3.9307^2}\approx 4-0.0647=3.9353$ $x_3=4-\frac{1}{3.9353^2}\approx 4-0.0646=3.9354$ $x_4\approx 3.9354$ (b) $g^{\prime }(x)=\frac{2}{x^3}$. At the root $\alpha \approx 3.9354$, $g^{\prime }(\alpha )\approx \frac{2}{61}\approx 0.03<1$, and $g^{\prime }(\alpha )$ is positive, so the sequence converges monotonically.

Question 3

Find the root of $x-\cos (x)=0$ to 3 significant figures, using the iterative formula $x_{n+1}=\cos (x_n)$ with initial value $x_0=0$. [5 marks]

Solution

Iterative terms (kept to 4 decimal places): $x_0=0$, $x_1=1.0000$, $x_2=0.5403$, $x_3=0.8576$, $x_4=0.6543$, $x_5=0.7935$, $x_6=0.7014$, $x_7=0.7640$, $x_8=0.7221$, $x_9=0.7504$, $x_{10}=0.7314$, $x_{11}=0.7442$, $x_{12}=0.7356$, $x_{13}=0.7414$, $x_{14}=0.7375$, $x_{15}=0.7401$, $x_{16}=0.7384$, $x_{17}=0.7395$, $x_{18}=0.7388$ Sign change test: Let $f(x)=x-\cos (x)$. $f(0.739)\approx 0.739-0.740=-0.001$, $f(0.740)\approx 0.740-0.739=+0.001$. The interval $[0.739,0.740]$ has width $0.001<0.005$, so the root is $0.739$ to 3 significant figures.

9. Quick Reference Cheatsheet

10. What's Next

The numerical root-finding techniques you have learned in this guide are a foundational part of A-Level Mathematics Paper 3, and they also extend directly to later topics in the syllabus, including numerical solution of differential equations (Euler's method) and numerical integration (trapezium rule). The approximation and error-checking principles you practice here will also help you in applied topics like mechanics and statistics, where exact analytical solutions are often unavailable.

If you are stuck on any convergence proof, iteration calculation, or sign-change test, you can ask Ollie our AI tutor for personalized walkthroughs, extra practice questions, and feedback on your working directly on the homepage. You can also find more A-Level Mathematics Paper 3 topic guides and full mock papers on the OwlsPrep platform to refine your exam preparation.

Aligned with the Cambridge International AS & A Level Mathematics 9709 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.