91Ó°ÊÓ

Consider the following four methods for calculating \(2^{1 / 4}\), the fourth root of 2 . (a) Rank them for speed of convergence, from fastest to slowest. Be sure to give reasons for your ranking. (A) Bisection Method applied to \(f(x)=x^{4}-2\) (B) Secant Method applied to \(f(x)=x^{4}-2\) (C) Fixed-Point Iteration applied to \(g(x)=\frac{x}{2}+\frac{1}{x^{3}}\) (D) Fixed-Point Iteration applied to \(g(x)=\frac{2 x}{3}+\frac{2}{3 x^{3}}\) (b) Are there any methods that will converge faster than all above suggestions?

Prove that Newton's Method applied to \(f(x)=a x+b\) converges in one step.

Use Newton's Method to produce a quadratically convergent method for calculating the \(n\)th root of a positive number \(A\), where \(n\) is a positive integer. Prove quadratic convergence.

Consider the Fixed-Point Iteration \(x \rightarrow g(x)=x^{2}-0.24\). (a) Do you expect Fixed-Point Iteration to calculate the root \(-0.2\), say, to 10 or to correct decimal places, faster or slower than the Bisection Method? (b) Find the other fixed point. Will FPI converge to it?

Which of the following three Fixed-Point Iterations converge to \(\sqrt{2}\) ? Rank the ones that converge from fastest to slowest. (A) \(x \longrightarrow \frac{1}{2} x+\frac{1}{x}\) (B) \(x \rightarrow \frac{2}{3} x+\frac{2}{3 x}\) (C) \(x \longrightarrow \frac{3}{4} x+\frac{1}{2 x}\)

Assume that \(g\) is a continuously differentiable function and that the Fixed- Point Iteration \(g(x)\) has exactly three fixed points, \(-3,1\), and 2. Assume that \(g^{\prime}(-3)=2.4\) and that FPI started sufficiently near the fixed point 2 converges to 2 . Find \(g^{\prime}(1)\).