Two Newton iterations for a square root

HardCalculus Programming & DSA~5m

Newton's method solves $f(x)=0$ via $x_{k+1} = x_k - \dfrac{f(x_k)}{f'(x_k)}$ . To approximate $\sqrt{2}$ , apply it to $f(x) = x^2 - 2$ starting from $x_0 = 1$ . What is $x_2$ (the estimate after two iterations)? Give a decimal to four places.

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For $f=x^2-2$ the update simplifies to $x_{k+1} = x_k/2 + 1/x_k$ .
$x_0=1 \to x_1 = 3/2 \to x_2 = 17/12$ .

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1.4167 (± 0.0005)

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Two Newton iterations for a square root

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