Numerical Methods

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Numerical Methods

Edexcel International A Level (IAL) Maths: Pure 3

The Big Idea: When equations are too difficult to solve using normal algebraic methods, we can find approximate solutions by either testing where a function changes sign across an interval, or by rearranging the equation and repeatedly feeding answers back in until we converge on the solution.

Quick Summary

Change of Sign A root exists between a and b if f(a) and f(b) have opposite signs (for continuous functions). Use bounds to prove accuracy.

Why Sign Change Fails Large intervals, multiple roots, discontinuities, and functions that touch rather than cross the axis can fool the method.

Iteration Basics Rearrange the equation into the form xₙ₊₁ = g(xₙ) and substitute each answer back in repeatedly until the value stabilises.

Convergence & Diagrams If |g'(x)| < 1 at the root, iteration converges. The sign of g'(x) determines whether you get a staircase or cobweb pattern.

1. Change of Sign Method

What Does "Change of Sign" Mean?

A change of sign happens when a function's output goes from positive to negative (or vice versa) as you move from one point to another. This tells us something fundamental about continuous functions: they must cross the x-axis at least once in between.

Think of it like a pencil line: if you draw from a point above the axis to a point below it without lifting your pencil, you must cross the line. That crossing point is a root.

The Sign Change Rule:
If a continuous function f(x) satisfies f(a) × f(b) < 0, then there is at least one root between a and b.

How to Use It

The method is simple but powerful:

Calculate f(a) and f(b) by substituting the numbers into your function.
Check the signs. If one is positive and one is negative, there's a root between them.
State your conclusion: "Since f(a) is negative and f(b) is positive, there is a root between a and b."

Example: Show that f(x) = 27 − 5tan(x) has a root of 1.39 correct to 2 decimal places.

Part a) Show 0.7 < p < 0.8:
f(0.7) = 3 × 0.7ᵉ − 5 = −0.7711... (negative)
f(0.8) = 3 × 0.8ᵉ − 5 = 0.3412... (positive)
Sign change → root exists between 0.7 and 0.8 ✓

Part b) Show p = 0.771 to 3 d.p.:
Now find upper and lower bounds that round to 0.771:
f(0.7705) = −0.0052... (negative)
f(0.7715) = 0.0062... (positive)
All values between 0.7705 and 0.7715 round to 0.771
Therefore p = 0.771 to 3 d.p. ✓

Key Requirements for the Method:

The function must be continuous in your interval (no sudden jumps or asymptotes).
The interval must be small enough that only one root sits inside.
You're looking for roots where the graph actually crosses the x-axis, not just touches it.

Finding Bounds for Accuracy

every value rounds to the same number.

1.39 f(1.385) f(1.395) 1.39

Practice Question 1

f(x) = x³ − 4x + 1

Solution:

Concepts Checklist

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Also in the full note

2. When the Change of Sign Method Fails
3. x = g(x) Iteration
Exam Tips & Common Mistakes
Failure Case 1: Large Intervals with Multiple Roots
Failure Case 2: Discontinuities and Asymptotes
Failure Case 3: Functions That Touch but Don't Cross
What is Iteration?
How to Rearrange an Equation into Iterative Form

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Numerical Methods