Newton’s method, question 8

Topic: Newton’s method
Difficulty: Challenging
Suggested time: 5 min
Marks: 3
Theory: UBC Math 100 §10.1Key results

The function $f(x) = x^{3} - x$ has roots $-1$ , $0$ and $1$ . Newton’s method is started at $x_{0} = \tfrac{1}{2}$ . Which root do the estimates approach, and why?

−1: the tangent at 1/2 has the small slope −1/4, so it crosses the x-axis far away, at exactly x₁ = −1, and every later estimate stays there.
0: the first step moves left from 1/2, and Newton’s method homes in on the nearest root in the direction it first moves, so the estimates settle on 0.
1: the first step is x₁ = 1/2 + (−3/8)/(−1/4) = 2, just past the root 1, and from there the estimates decrease steadily towards the root 1.
1: with f′(x) = 3x², the tangent at 1/2 has slope 3/4, so it crosses the x-axis at exactly x₁ = 1, and every later estimate stays there.

How sure are you?

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Theory UBC Math 100 §10.1Key results

Worked solution

Strategy

Take the first step exactly. The slope at 1/2 is small, which is the warning that the step will be long.

Working

$f\!\left(\tfrac{1}{2}\right) = \tfrac{1}{8} - \tfrac{1}{2} = -\tfrac{3}{8}, \qquad f'(x) = 3x^{2} - 1, \quad f'\!\left(\tfrac{1}{2}\right) = -\tfrac{1}{4}$
Value and slope at the start.
$x_{1} = \tfrac{1}{2} - \frac{-3/8}{-1/4} = \tfrac{1}{2} - \tfrac{3}{2} = -1$
The step is 3/2 long, three times the distance from 1/2 to either neighbouring root.
$f(-1) = 0 \implies x_{2} = -1 - \frac{0}{f'(-1)} = -1$
f′(−1) = 2 is not zero, so the step is zero and every estimate from x₁ on is −1.
Near 1/√3 ≈ 0.58, where f′ = 0, the tangent lines are nearly flat, so a start there is thrown a long way: here past the root 0 to the root −1.

Answer

x_{1} = -1, \quad \text{so the estimates converge to the root } -1

Check

The tangent at x = 1/2 simplifies to y = −(x + 1)/4, which crosses the axis at −1, itself a root of x³ − x.

Where this goes wrong

Assuming Newton’s method finds the root nearest the start. The step is set by the tangent line, and a shallow tangent can carry the estimate past a nearby root to a distant one.