Solution. Here \(\mathbf {F} = \mathbf {X}\) for every \(\boldsymbol \theta \), and \(\mathbf {e}^{(0)} = \mathbf {y} - \mathbf {X}\boldsymbol \theta ^{(0)}\). The step is \(\boldsymbol \delta = (\mathbf {X}^\top \mathbf {X})^{-1}\mathbf {X}^\top (\mathbf {y} - \mathbf {X}\boldsymbol \theta ^{(0)}) = \hat {\boldsymbol \beta } - \boldsymbol \theta ^{(0)}\), so \(\boldsymbol \theta ^{(1)} = \hat {\boldsymbol \beta }\). The linear approximation is exact, so the first step is the answer.

Solution. (a) With \(g_i = e^{-\theta _2x_i}\), \[ \hat \theta _1(\theta _2) = \frac {\sum e^{-\theta _2x_i}y_i}{\sum e^{-2\theta _2x_i}} , \qquad S_p(\theta _2) = \sum y_i^2 - \frac {\bigl (\sum e^{-\theta _2x_i}y_i\bigr )^2}{\sum e^{-2\theta _2x_i}} . \]

(b) \(\hat \theta _1(0.3) = 23.7162/1.18313 = 20.045\) and \[ S_p(0.3) = 475.57 - \frac {23.7162^2}{1.18313} = 475.57 - 475.399 = 0.171 , \] a little above the minimum \(0.154\) at \(\theta _2 = 0.3047\). Since \(S_p\) is the difference of two nearly equal numbers, the sums must be carried to many figures.

Solution. By Theorem 12.5.1, with \(t\) as the non-linear parameter, \(\hat t = 4\), where the residual sum of squares is smallest (a finer search between \(3\) and \(5\) could refine it). The fitted model rises at \(1.973\) per unit of \(x\) up to \(x = 4\) and at \(1.973 - 1.468 = 0.505\) per unit afterwards. Note that the usual \(t\)-tests treat \(t\) as known; since it was chosen from the data, they are somewhat optimistic.

Solution. (a) \(\partial f/\partial \theta _1 = 1\), \(\partial f/\partial \theta _2 = -e^{-\theta _3x}\), \(\partial f/\partial \theta _3 = \theta _2xe^{-\theta _3x}\).

(b) For fixed \(\theta _3\) the model is linear in \((\theta _1, \theta _2)\), with columns \(\mathbf {1}\) and \(-e^{-\theta _3\mathbf {x}}\). So \(S_p(\theta _3)\) is the residual sum of squares from regressing \(y\) on \(e^{-\theta _3x}\) with an intercept, a function of one variable to be minimised.

(c) The ceiling is \(\theta _1 \approx 10\), and at \(x = 0\), \(f = \theta _1 - \theta _2 \approx 3\), so \(\theta _2 \approx 7\). For \(\theta _3\), find the \(x\) at which the curve has covered half the distance, \(y \approx 6.5\); there \(e^{-\theta _3x} = \frac 12\), so \(\theta _3 \approx \log 2/x\).

Solution. With \(t_{0.025,4} = 2.776\): \(9.938 \pm 2.776(0.328) = 9.938 \pm 0.911\), from \(9.03\) to \(10.85\). The Lineweaver–Burk value \(8.968\) lies outside the interval: the linearised fit underestimates the maximum rate, for the reason given in the example.