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FP1.3 - Numerical solution of equations

Syllabus
2019
Topic
—
Level
AS

Approximate a root numerically

Numerical root methods approximate a solution of f(x)=0f(x)=0 when exact algebra is unavailable or inconvenient. First locate a root: if ff is continuous on [a,b][a,b] and f(a)f(b)<0f(a)f(b)<0, then at least one root lies between aa and bb.

Method Next approximation or interval Main control
interval bisection evaluate the midpoint m=(a+b)/2m=(a+b)/2 and keep the half whose endpoint values have opposite signs preserves a sign-change bracket; width halves each step
linear interpolation x≈a−f(a)(b−a)f(b)−f(a)\displaystyle x\approx a-\frac{f(a)(b-a)}{f(b)-f(a)} uses the xx-intercept of the chord through the two endpoint values
Newton-Raphson xn+1=xn−f(xn)f′(xn)\displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} uses the tangent at xnx_n; needs f′(xn)≠0f'(x_n)\ne0

For f(x)=x3−x−1f(x)=x^3-x-1, f(1)=−1f(1)=-1 and f(2)=5f(2)=5, so continuity gives a root in [1,2][1,2]. Since f(1.5)=0.875f(1.5)=0.875, one bisection gives [1,1.5][1,1.5]; since f(1.25)=−0.296875f(1.25)=-0.296875, a second gives [1.25,1.5][1.25,1.5]. The interval width is now 0.250.25.

Using x0=1.3x_0=1.3 for the same function, f′(x)=3x2−1f'(x)=3x^2-1. One Newton-Raphson step gives x1=1.3−1.33−1.3−13(1.3)2−1=1.325307….x_1=1.3-\frac{1.3^3-1.3-1}{3(1.3)^2-1}=1.325307\ldots. Keep unrounded values inside the calculation, then round the requested approximation only at the end.

Bisection is slower but keeps a certified bracket. Linear interpolation usually improves a bracket with one straight-line estimate. Newton-Raphson can converge quickly from a suitable starting value, but it does not preserve a bracket and a poor start or a near-zero derivative may send the iteration away from the intended root.

A sign change plus continuity proves at least one root, not exactly one. Always show the function values that select a bisection half, use the stated interval or starting approximation, differentiate the actual function for Newton-Raphson, and report the precision requested rather than rounding every intermediate value.