Newton’s method, question 7

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

Let $x^{*}$ be the $x$ -coordinate of the point on the curve $y = e^{x}$ closest to the point $(2, 0)$ . Write the equation $D'(x) = 0$ , where $D(x)$ is the squared distance from $(x, e^{x})$ to $(2, 0)$ , and estimate $x^{*}$ with one step of Newton’s method on the function $D'$ itself, from $x_{0} = 0$ .

Your answer

A fraction, or a decimal to three decimal places.

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Theory UBC Math 100 §10.1Key results

Worked solution

Strategy

The closest point minimises the squared distance, so it is a zero of that distance’s derivative. Newton’s method finds zeros, so it is applied to the derivative, not to the distance.

Working

$D(x) = (x - 2)^{2} + \left(e^{x}\right)^{2} = (x - 2)^{2} + e^{2x}$
The squared distance is smallest where the distance is, and has no square root to differentiate.
$D'(x) = 2(x - 2) + 2e^{2x}, \qquad D''(x) = 2 + 4e^{2x} > 0$
D is concave up, so its one critical point is the minimum.
$D'(x^{*}) = 0 \iff h(x^{*}) = x^{*} - 2 + e^{2x^{*}} = 0$
Halving D′ changes neither its zeros nor the Newton steps.
$h(0) = -2 + 1 = -1, \qquad h'(x) = 1 + 2e^{2x}, \quad h'(0) = 3$
Value and slope at x₀ = 0. The chain rule gives the 2 in 2e²ˣ.
$x_{1} = 0 - \frac{-1}{3} = \frac{1}{3}$
The minimiser is 0.2731 to four places; one more step gives 0.276.

Answer

x^{*} - 2 + e^{2x^{*}} = 0, \qquad x_{1} = \frac{1}{3}

Check

D′(0) = −2 and D′(1/3) ≈ 0.56 have opposite signs, so the minimiser lies between 0 and 1/3, nearer 1/3.

Where this goes wrong

Applying Newton’s method to the squared distance instead of to its derivative, which gives 5/2. The distance is never zero; it is its derivative that vanishes at the closest point, so the derivative is the function to iterate.