Solution. \(f(2) = -17<0<2 = f(3)\), so there is a root in \((2,3)\). We need \(2^{-N}<10^{-4}\), that is \(N\geq 14\). Fourteen iterations give \(p_{14} = 2.92401\), and \(\sqrt [3]{25} = 2.9240177\). The first steps are \(p_1 = 2.5\) (\(f<0\)), \(p_2 = 2.75\) (\(f<0\)), \(p_3 = 2.875\) (\(f<0\)), \(p_4 = 2.9375\) (\(f>0\)), \(p_5 = 2.90625\) (\(f<0\)).

Solution. With \(f(x) = x^2-6\), \(p_n = p_{n-1}-\frac {p_{n-1}^2-6}{2p_{n-1}} = \frac 12\left (p_{n-1}+\frac {6}{p_{n-1}}\right )\), which is Heron’s iteration (Example 2.3.13). The iterates are \(3.5\), \(2.607143\), \(2.454256\), \(2.449494\), against \(\sqrt 6 = 2.449490\). The first step overshoots because \(p_0 = 1\) is far from the root; from then on the convergence is quadratic and from above.

Solution. \(f'(x) = -\frac {1}{x^2}\), so \(p_n = p_{n-1}-\frac {1/p_{n-1}-a}{-1/p_{n-1}^2} = p_{n-1}(2-ap_{n-1})\). With \(a = 7\): \(0.13\), \(0.1417\), \(0.14284777\), \(0.1428571422\), against \(\frac 17 = 0.1428571429\). Writing \(e_n = 1-ap_n\) gives \(e_n = e_{n-1}^2\), so the iteration converges exactly when \(|1-ap_0|<1\), that is \(0<p_0<\frac 2a\), and then quadratically.

Solution. On \([0,1]\), \(g\) is increasing with \(g(0) = 0.6\) and \(g(1) = 0.8\), so it maps \([0,1]\) into itself, and \(|g'(x)| = \frac {2x}{5}\leq 0.4 = k\). By the Fixed-Point Theorem there is a unique fixed point, \(p = \frac {5-\sqrt {13}}{2} = 0.6972\). With \(|p_1-p_0| = 0.2\), bound (2.4) requires \(\frac {0.4^n}{0.6}(0.2)<10^{-5}\), that is \(n>11.4\), so \(12\) iterations suffice. (Near \(p\) the slope is \(g'(p) = 0.279\), so fewer are needed in practice.)

Solution. With \(f(x) = e^{-x}-x\) the iterates are \(0.61270\), \(0.56384\), \(0.56717036\), \(0.5671433066\), \(0.5671432904\), with errors \(4.6\times 10^{-2}\), \(3.3\times 10^{-3}\), \(2.7\times 10^{-5}\), \(1.6\times 10^{-8}\), \(7.9\times 10^{-14}\). The exponents of the errors, about \(1.3, 2.5, 4.6, 7.8, 13.1\), grow by a factor of about \(1.7\), close to the theoretical \(1.618\).

Solution. With \(f = (x-p)^mq\), as in the proof of Theorem 2.5.3, \(\frac {g(x)-p}{x-p} = 1-\frac {mq(x)}{mq(x)+(x-p)q'(x)}\longrightarrow 1-1 = 0\), so \(g'(p) = 0\) and Theorem 2.3.11 gives quadratic convergence. For \(1-\cos x\) the root \(0\) is double (\(f(0) = f'(0) = 0\), \(f''(0) = 1\)). Plain Newton gives \(0.454\), \(0.223\), \(0.111\), \(0.0554\), \(0.0277\): halving each time. The modified iteration \(p_n = p_{n-1}-2\frac {1-\cos p_{n-1}}{\sin p_{n-1}}\) gives \(-9.3\times 10^{-2}\), \(6.6\times 10^{-5}\), \(1.3\times 10^{-12}\).

Solution. The iterates cycle \(0, 1, 0, 1,\dots \) (Example 2.4.6). The real root is \(-1.7693\), and \(f\) has a local minimum at \(x = \sqrt {2/3}\) and a local maximum at \(-\sqrt {2/3}\); starting values to the right of the maximum are pulled towards the minimum, where the tangent is nearly horizontal. The remedy is to bracket first: \(f(-2) = -2<0<2 = f(-1)\), and a few bisection steps on \([-2,-1]\) give a starting value from which Newton converges quadratically (for example \(p_0 = -1.5\) gives \(-1.8421\), \(-1.7728\), \(-1.769301\), \(-1.7692924\)).

Solution. \(\Delta p_n = c\lambda ^n(\lambda -1)\) and \(\Delta ^2p_n = c\lambda ^n(\lambda -1)^2\), so \(\hat p_n = p+c\lambda ^n-\frac {c^2\lambda ^{2n}(\lambda -1)^2}{c\lambda ^n(\lambda -1)^2} = p+c\lambda ^n-c\lambda ^n = p\). Aitken’s method assumes exactly this geometric model of the error, and removes the error completely when the model is exact; for other linearly convergent sequences the model holds only asymptotically, which is why the improvement grows with \(n\).