Numerical methods

Bisection is started on $[1, 2]$ to find $\sqrt 2$ as a root of $x^2 - 2$. After $n$ halvings we report the midpoint of the current interval. What is the smallest $n$ for which Theorem 3 guarantees an error of at most $10^{-6}$?

Newton's method for $x^2 - 2$ is $x_{n+1} = \frac12\left(x_n + \frac{2}{x_n}\right)$. Starting from $x_0 = 1$, compute exactly $x_1 = \frac32$, $x_2 = \frac{17}{12}$ and $x_3 = \frac{p}{q}$ in lowest terms. Enter the numerator $p$.

For large $x$ (say $x = 10^{12}$), which formula computes $\sqrt{x + 1} - \sqrt{x}$ most accurately in floating point?

Newton's method for $f(x) = \sqrt[3]{x}$ (the real cube root) gives $x_{n+1} = -2x_n$ and diverges from every $x_0 \ne 0$, although $f(0) = 0$. Which hypothesis of Theorem 4 fails?

Apply Simpson's rule with $n = 2$ strips ($h = 2$) to $\int_0^4 x^3\,dx$. What value does it give?

Which statements are true? (Choose all that apply.)

In binary64 arithmetic, approximately which step $h$ minimises the total error of the forward difference $\frac{f(x + h) - f(x)}{h}$ for a typical smooth function with $f$, $f''$ of moderate size?

With IsBinary64 x meaning "$x$ is a positive binary64 number", which Lean statement says that $2^{-52}$ is the gap between $1$ and the next larger binary64 number?

Write newton(f, df, x0, tol=1e-12, max_iter=50) that runs Newton's method and returns (x, n): the first iterate with abs(f(x)) <= tol and the number of steps taken. If a derivative is exactly 0, or max_iter steps pass without success, return (None, n).