Numerical Methods
Numerical Methods
When equations cannot be solved algebraically, numerical methods provide systematic ways to find approximate solutions. A-Level covers the change of sign method, iteration, and Newton-Raphson.
Locating Roots: Change of Sign
If f(x) is continuous on [a, b] and f(a) and f(b) have opposite signs, then there is at least one root in the interval (a, b).
Worked Example: Show that x³ - 3x + 1 = 0 has a root between x = 1 and x = 2.
f(1) = 1 - 3 + 1 = -1 (negative)
f(2) = 8 - 6 + 1 = 3 (positive)
Since f(1) < 0 and f(2) > 0 and f is continuous, there is a sign change, so a root lies in (1, 2).
Limitations: The method fails when:
- The curve touches the x-axis without crossing (a repeated root)
- There are two roots in the interval (sign changes cancel)
- The function is discontinuous (e.g. 1/x between -1 and 1 has a sign change but no root)
Fixed-Point Iteration
Rearrange f(x) = 0 into the form x = g(x), then use the iterative formula:
xₙ₊₁ = g(xₙ)
Starting from an initial estimate x₀, compute x₁, x₂, x₃, ... until the values converge to a fixed point.
Worked Example: Find a root of x³ - 3x + 1 = 0 using the rearrangement x = (x³ + 1)/3, starting from x₀ = 1.5.
x₁ = (1.5³ + 1)/3 = (3.375 + 1)/3 = 4.375/3 = 1.4583
x₂ = (1.4583³ + 1)/3 = (3.1013 + 1)/3 = 1.3671
x₃ = (1.3671³ + 1)/3 = (2.5544 + 1)/3 = 1.1848
The values are converging toward a root (approximately 1.532 after more iterations with a better rearrangement).
Convergence of Iteration
Not every rearrangement converges. The iteration xₙ₊₁ = g(xₙ) converges to a root α if:
|g'(α)| < 1
Graphically, the iteration converges when the curve y = g(x) crosses y = x with a gradient between -1 and 1 at the intersection. If |g'(α)| > 1, the iterations diverge (move away from the root).
A cobweb diagram (when g'(α) < 0) or staircase diagram (when g'(α) > 0) illustrates the iterative process.
The Newton-Raphson Method
This method uses tangent lines to find successively better approximations. Starting from x₀, the formula is:
xₙ₊₁ = xₙ - f(xₙ) / f'(xₙ)
This is usually much faster than simple iteration — it typically doubles the number of correct decimal places with each step (quadratic convergence).
Worked Example: Use Newton-Raphson to find a root of f(x) = x³ - 3x + 1, starting from x₀ = 2.
f(x) = x³ - 3x + 1, so f'(x) = 3x² - 3.
x₀ = 2: f(2) = 3, f'(2) = 9. x₁ = 2 - 3/9 = 2 - 1/3 = 1.6667
x₁ = 1.6667: f(1.6667) = 4.6297 - 5.0001 + 1 = 0.6296. f'(1.6667) = 8.3335 - 3 = 5.3335.
x₂ = 1.6667 - 0.6296/5.3335 = 1.6667 - 0.1180 = 1.5487
x₂ = 1.5487: f(1.5487) = 3.7155 - 4.6461 + 1 = 0.0694. f'(1.5487) = 7.1943 - 3 = 4.1943.
x₃ = 1.5487 - 0.0694/4.1943 = 1.5487 - 0.0165 = 1.5321
The root is approximately 1.532 (to 3 d.p.).
When Newton-Raphson Fails
The method can fail or behave poorly when:
- f'(xₙ) = 0 at some iteration (division by zero — the tangent is horizontal)
- The starting point is near a local maximum or minimum rather than near a root
- The function has a discontinuity near the root
- The tangent overshoots to a different region of the curve (bad starting point)
Comparing the Methods
| Method | Speed | Requirements | Pitfalls | ||
|---|---|---|---|---|---|
| Change of sign | Slow (bisection halves interval each step) | Continuity | Misses repeated roots; multiple roots cancel | ||
| Fixed-point iteration | Moderate | Suitable rearrangement with | g' | < 1 | Wrong rearrangement diverges |
| Newton-Raphson | Fast (quadratic convergence) | f(x) and f'(x) | Fails at turning points; sensitive to x₀ |
Exam Tips
- For change of sign questions, always evaluate f(x) at both endpoints, state the sign change, and state the conclusion ("therefore a root lies between...").
- In iteration questions, show your working to at least 4 decimal places unless told otherwise.
- For Newton-Raphson, show f(xₙ) and f'(xₙ) at each step — method marks depend on this.
- If asked to explain why an iteration fails, discuss |g'(x)| at the root — if it exceeds 1, the method diverges.
- Always give your final answer to the degree of accuracy specified in the question.