Solution. The first step uses the chord from \((1,-5)\) to \((2,14)\): \[p_1 = 2-14\cdot \frac {2-1}{14-(-5)} = 2-\frac {14}{19} = 1.263157895,\] and \(f(p_1) = -1.6022<0\), so the new bracket is \([1.263157895, 2]\). Continuing:

After \(10\) steps the error is \(4.2\times 10^{-7}\), compared with \(9.7\times 10^{-4}\) for bisection. Two features of the table stand out: the right end point \(b_n = 2\) never changes, and each error is almost exactly a quarter of the previous one. The next theorem explains both.

Proof. Uniqueness. A root exists by the Intermediate Value Theorem. If there were two roots \(p<q\) in \((a,b)\), write \(p = \theta a+(1-\theta )q\) with \(\theta \in (0,1)\). Convexity gives \(f(p)\leq \theta f(a)+(1-\theta )f(q) = \theta f(a)<0\), contradicting \(f(p) = 0\).

The right end point is fixed. Suppose \(a_n<p<b_n = b\) with \(f(a_n)<0\). Since \(f''>0\), the graph of \(f\) on \([a_n,b]\) lies below the chord joining its end points. At the point \(p_n\) where the chord is zero, therefore, \(f(p_n)<0\). So \(p_n\) lies to the left of \(p\), the sign test keeps \(b_{n+1} = b\) and moves the left end point to \(a_{n+1} = p_n\). Also \(p_n>a_n\), since the chord is negative at \(a_n\). By induction \(a<p_1<p_2<\dots <p\).

Convergence. The sequence \(\{p_n\}\) is increasing and bounded above by \(p\), so it converges to some \(q\leq p\). Since \(p_{n+1} = b-f(b)\frac {b-p_n}{f(b)-f(p_n)}\) and \(f\) is continuous, letting \(n\longrightarrow \infty \) gives \(q = b-f(b)\frac {b-q}{f(b)-f(q)}\), which simplifies to \(f(q)(b-q) = 0\). As \(q\leq p<b\), \(f(q) = 0\), and by uniqueness \(q = p\).

Rate. Write \(e_n = p_n-p\). From the second form of (2.2) with \(a_n = p_n\), \(b_n = b\), \[e_{n+1} = \frac {p_nf(b)-bf(p_n)}{f(b)-f(p_n)}-p = \frac {e_nf(b)-(b-p)f(p_n)}{f(b)-f(p_n)} .\] Dividing by \(e_n\) and using \(\frac {f(p_n)}{e_n} = \frac {f(p_n)-f(p)}{p_n-p}\longrightarrow f'(p)\) and \(f(p_n)\longrightarrow 0\), \[\frac {e_{n+1}}{e_n} = \frac {f(b)-(b-p)\frac {f(p_n)}{e_n}}{f(b)-f(p_n)}\longrightarrow \frac {f(b)-(b-p)f'(p)}{f(b)} .\] Since \(f'' > 0\), \(f'\) is increasing. By the Mean Value Theorem, \(f(p)-f(a) = f'(\zeta )(p-a)\) for some \(\zeta \in (a,p)\), so \(f'(p)>f'(\zeta ) = \frac {-f(a)}{p-a}>0\); and \(f(b) = f(b)-f(p) = f'(\eta )(b-p)\) for some \(\eta \in (p,b)\), with \(f'(\eta )>f'(p)\). Hence \(0<(b-p)f'(p)<f(b)\), and \(0<\lambda <1\). □

Solution. Here \(f\) is convex, \(f(1.3) = 12.79\) is large and \(f\) is very flat on most of \([0,1]\). The chords are almost horizontal, so the fixed end point \(b = 1.3\) pulls the iterates only slowly towards \(1\):

False position needs \(52\) iterations to reach an error below \(10^{-5}\), while bisection needs \(14\). The theorem predicts the asymptotic ratio \(\lambda = 1-\frac {0.3\times 10}{12.79} = 0.765\), compared with bisection’s guaranteed factor \(\frac 12\). False position is faster only when the fixed end point is not too far from the root.

Solution. For \(x^{10}-1\) the iterates are \[0.0943,\ 0.1818,\ 0.3330,\ 0.5634,\ 0.8464,\ 1.0749,\ 0.9455,\ 0.9828,\ 1.0041,\ 0.99968,\ 0.9999941,\] reaching an error below \(10^{-5}\) after \(11\) iterations instead of \(52\); from the sixth step on they fall alternately on both sides of the root. For the test problem the errors after steps \(4\) to \(7\) are \(1.5\times 10^{-4}\), \(9.0\times 10^{-7}\), \(8.9\times 10^{-7}\) and \(3.9\times 10^{-13}\). The Illinois method still brackets the root at every step, so it keeps the safety of bisection, and its convergence is superlinear (of order about \(1.44\)).