ZIMSEC A Level · Pure Mathematics · Form 6
Locating roots
If f(x) is continuous and f(a) and f(b) have opposite signs, a root lies between a and b.
Iteration
Rearrange f(x) = 0 into x = g(x) and use xₙ₊₁ = g(xₙ). The iteration converges if |g′(x)| < 1 near the root.
Worked example: Use xₙ₊₁ = ∛(2xₙ + 5) with x₀ = 2 to find a root of x³ − 2x − 5 = 0 to 3 d.p.
x₁ = ∛9 = 2.0801; x₂ = ∛9.1602 = 2.0924; x₃ = 2.0942; x₄ = 2.0945; x₅ = 2.0946
Root ≈ 2.095 (check: f(2.0945) < 0 and f(2.0955) > 0)
Newton–Raphson method
xₙ₊₁ = xₙ − f(xₙ)/f′(xₙ). It converges quickly if the starting value is close to the root, but fails near stationary points.
The trapezium rule
∫ from a to b of y dx ≈ (h/2)[y₀ + 2(y₁ + y₂ + … + yₙ₋₁) + yₙ], where h = (b − a)/n.
It overestimates if the curve is concave up (convex) and underestimates if it is concave down.
Check yourself
- Show that x³ + x − 3 = 0 has a root between 1 and 2.
- Write the Newton–Raphson formula for f(x) = x² − 7.
- Use the trapezium rule with 2 strips to estimate ∫ from 0 to 2 of x² dx.
Show answers
- f(1) = −1 < 0 and f(2) = 7 > 0; sign change, so a root lies between 1 and 2.
- xₙ₊₁ = xₙ − (xₙ² − 7)/(2xₙ)
- h = 1: ½[0 + 2(1) + 4] = 3 (exact value 8/3 – an overestimate)
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