Newton’s method, question 10

Topic: Newton’s method
Difficulty: Challenging
Suggested time: 7 min
Marks: 4
Theory: UBC Math 100 §10.1Key results

Newton’s method for a root of $f(x) = x^{3} - 7x$ is started at some $x_{0} > 0$ . For exactly one such $x_{0}$ , the first step lands on $x_{1} = -x_{0}$ , and from then on the estimates alternate $x_{0}, -x_{0}, x_{0}, \ldots$ for ever without approaching a root. Find that $x_{0}$ .

Your answer

An exact value, or a decimal to three places.

How sure are you?

SureNot sureGuessing

Theory UBC Math 100 §10.1Key results

Worked solution

Strategy

Write the condition x₁ = −x₀ as an equation in x₀ and solve it. Then use the symmetry of f: because f is odd, the step from −x₀ mirrors the step from x₀, so the estimates swap sides for ever.

Working

$x_{1} = x_{0} - \frac{x_{0}^{3} - 7x_{0}}{3x_{0}^{2} - 7}$
One Newton step, with f′(x) = 3x² − 7.
$x_{1} = -x_{0} \iff 2x_{0}\bigl(3x_{0}^{2} - 7\bigr) = x_{0}^{3} - 7x_{0} \iff 5x_{0}^{3} = 7x_{0}$
The step exists only where f′(x₀) = 3x₀² − 7 ≠ 0; there, multiply through by it and collect terms.
$x_{0} > 0:\quad x_{0}^{2} = \frac{7}{5}, \qquad x_{0} = \sqrt{\frac{7}{5}} = \frac{\sqrt{35}}{5} \approx 1.183$
Divide by x₀, which is not 0.
$f(-x) = -f(x), \quad f'(-x) = f'(x) \quad\Longrightarrow\quad x_{2} = -x_{1} = x_{0}$
The step from −x₀ is the mirror image of the step from x₀, so the estimates alternate and approach no root.

Answer

x_{0} = \sqrt{\dfrac{7}{5}} = \dfrac{\sqrt{35}}{5}

Check

With x₀² = 7/5, f(x₀) = x₀(7/5 − 7) = −28x₀/5 and f′(x₀) = 21/5 − 7 = −14/5, so the step f/f′ is exactly 2x₀ and lands on −x₀.

Where this goes wrong

Adding f(x₀)/f′(x₀) instead of subtracting it, which gives √3. Newton’s method moves by −f/f′, so the condition for landing on −x₀ is 2x₀ f′(x₀) = f(x₀).