AppliedNumeric answer

Newton’s method for a square root · Part 2 of 3

You compute $\sqrt{10}$ with Newton’s method on $f(x) = x^2 - 10$ , which gives the iteration $x_{k+1} = \tfrac{1}{2}\left(x_k + 10/x_k\right)$ , starting from $x_0 = 3$ .

The error $e_k = x_k - \sqrt{10}$ obeys $e_{k+1} = e_k^2/(2x_k)$ exactly. Given $e_2 \approx 3.04 \times 10^{-6}$ , what is $e_3$ , in units of $10^{-12}$ , to two decimals?

Answer with a number. Fractions, powers and expressions like 23/6 or C(52,5) are read correctly in practice.

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Root finding, with implied volatility as the worked case

The error ek=xk−10e_k = x_k - \sqrt{10}ek​=xk​−10​ obeys ek+1=ek2/(2xk)e_{k+1} = e_k^2/(2x_k)ek+1​=ek2​/(2xk​) exactly. Given e2≈3.04×10−6e_2 \approx 3.04 \times 10^{-6}e2​≈3.04×10−6, what is e3e_3e3​, in units of 10−1210^{-12}10−12, to two decimals?

More numerical methods questions

The error $e_k = x_k - \sqrt{10}$ obeys $e_{k+1} = e_k^2/(2x_k)$ exactly. Given $e_2 \approx 3.04 \times 10^{-6}$ , what is $e_3$ , in units of $10^{-12}$ , to two decimals?