Newton’s method, question 5

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

The equation $\cos x = x$ has exactly one solution, the zero of $f(x) = \cos x - x$ . Use one step of Newton’s method on $f$ from $x_{0} = \dfrac{\pi}{4}$ to estimate the solution. Give $x_{1}$ exactly.

Your answer

An exact expression, or a decimal to four decimal places.

How sure are you?

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Theory UBC Math 100 §10.1Key results

Worked solution

Strategy

A solution of cos x = x is a zero of f(x) = cos x − x, which Newton’s method can find. π/4 is a convenient start because sine and cosine are known exactly there.

Working

$f(x) = \cos x - x, \qquad f'(x) = -\sin x - 1$
A solution of cos x = x is a zero of f.
$f\!\left(\tfrac{\pi}{4}\right) = \frac{\sqrt{2}}{2} - \frac{\pi}{4}, \qquad f'\!\left(\tfrac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} - 1$
cos(π/4) = sin(π/4) = √2/2.
$x_{1} = \frac{\pi}{4} - \frac{\frac{\sqrt{2}}{2} - \frac{\pi}{4}}{-\frac{\sqrt{2}}{2} - 1} = \frac{\pi}{4} + \frac{\frac{\sqrt{2}}{2} - \frac{\pi}{4}}{\frac{\sqrt{2}}{2} + 1}$
The update formula. The minus signs cancel; this is already a calculator-ready answer.
$x_{1} = \frac{\frac{\sqrt{2}}{2}\left(\frac{\pi}{4} + 1\right)}{\frac{\sqrt{2}}{2} + 1} = \frac{\pi + 4}{4(\sqrt{2} + 1)}$
Optional tidying: over a common denominator the π/4 terms cancel, then divide top and bottom by √2/2.
$x_{1} = \frac{(\pi + 4)(\sqrt{2} - 1)}{4} \approx 0.7395$
The solution is 0.7391 to four places, so one step lands close.

Answer

x_{1} = \frac{\pi + 4}{4(\sqrt{2} + 1)} = \frac{(\pi + 4)(\sqrt{2} - 1)}{4} \approx 0.7395

Check

cos(0.7395) ≈ 0.7388, so x₁ misses solving cos x = x by less than 0.001.

Where this goes wrong

Differentiating cos x as sin x, so that f′(π/4) = √2/2 − 1 and x₁ comes out near 0.518. The derivative of cos x is −sin x, which makes f′ negative everywhere here.