Roots of Nonlinear Equations Questions and Answers

Answer: A

The order of convergence is \(\frac{1 + \sqrt{5}}{2}\), about 1.618.

For the secant method near a simple root, the errors satisfy \(e_{n+1} \approx C e_{n} e_{n-1}\). Seeking \(|e_{n+1}| \approx A|e_{n}|^{p}\) and substituting gives \(p^{2} = p + 1\), whose positive root is the golden ratio:

\[ p = \frac{1 + \sqrt{5}}{2} \approx 1.618 \]

So the secant method is superlinear but slower than Newton's quadratic method. \(\frac{2}{1 + \sqrt{5}} \approx 0.618\) is the reciprocal, less than 1, which is not an order of convergence; \(\frac{1+\sqrt{5}}{3} \approx 1.08\) and \(\frac{3}{1+\sqrt{5}} \approx 0.93\) do not solve \(p^{2} = p + 1\).

Answer: B

The sequence converges to \(\sqrt{2}\) whenever \(x_{0} \gt \sqrt{2/3}\).

The scheme is Newton's method for \(x^{2} - 2 = 0\): \(x - \frac{x^{2}-2}{2x} = \frac{x}{2} + \frac{1}{x}\). For any \(x_{0} \gt 0\), the AM-GM inequality gives

\[ \begin{aligned} x_{1} &= \frac{x_{0}}{2} + \frac{1}{x_{0}} \\ &\geq 2\sqrt{\frac{1}{2}} = \sqrt{2} \end{aligned} \]

From then on the terms stay at or above \(\sqrt{2}\) and decrease, so they converge to the positive fixed point \(\sqrt{2}\). Every \(x_{0} \gt \sqrt{2/3}\) is positive, so the statement holds.

The map is odd, so a negative start converges to \(-\sqrt{2}\). That makes 'for all \(x_{0} \neq 0\)' false, and also the (-1, 1) option, which includes negative starts. The sequence does not diverge for any non-zero start.

Answer: A

S1 is TRUE and S2 is FALSE.

A fixed-point scheme \(x_{k+1} = g(x_{k})\) converges from every start in an interval if g maps the interval into itself and \(|g'| \lt 1\) there.

S1: \(g(x) = \sqrt{10/(4 + x)}\) is decreasing, with \(g(1) = \sqrt{2} \approx 1.414\) and \(g(1.5) \approx 1.348\), so it maps the interval into itself; and \(|g'(x)| = \sqrt{10}/(2(4 + x)^{3/2}) \le 0.15\). So it converges: S1 is true.

S2: \(g(x) = \tfrac{1}{2}\sqrt{10 - x^{3}}\) is also decreasing, with \(g(1) = 1.5\) and \(g(1.5) \approx 1.287\), so it maps (1, 3/2) into itself, and \(|g'(x)| = 3x^{2}/(4\sqrt{10 - x^{3}})\) is at most about 0.66 there. It converges for every start in the interval, so it never diverges and S2 is false.

Answer: B

The first approximation is 1/3.

Newton-Raphson: \(x_{1} = x_{0} - f(x_{0})/f'(x_{0})\), with \(f(x) = x^{3} + 3x - 1\) and \(f'(x) = 3x^{2} + 3\).

\[ \begin{aligned} x_{1} &= 0 - \frac{f(0)}{f'(0)} \\ &= 0 - \frac{-1}{3} = \frac{1}{3} \end{aligned} \]

As a check, \(f(1/3) = 1/27 \gt 0\) and \(f(0) = -1\), so the root lies just below 1/3, and the next iterate moves only slightly. 1/2 would be the bisection midpoint, and 1/4 and 1/5 do not come from one Newton step.