= Solution
The <intermediate value theorem> gives a root $r\in(a,b)$, unique because $f'>0$ makes $f$ <strictly increasing>. We show that <Newton root-finding iteration>, started at $b$, always stays to the right of $r$. If $x>r$, the <mean value theorem> gives $f(x)=f'(\xi)(x-r)$ for some $\xi\in(r,x)$. Since $f''>0$, $f'$ is <strictly increasing>, so $0<f'(\xi)<f'(x)$. It follows that
$$
0<\frac{f(x)}{f'(x)}<x-r,\qquad \boxed{r<F(x)<x.}
$$
Induction gives $r<x_{n+1}<x_n\le b$, so the <bounded monotone sequence theorem> supplies a limit $\alpha\ge r$. The continuous positive $f'$ is bounded away from zero on $[a,b]$, making $F$ continuous. Passing to the limit in $x_{n+1}=F(x_n)$ gives $\alpha=F(\alpha)$ and hence $f(\alpha)=0$. Uniqueness implies \b[$\boxed{\alpha=r}$].
For the rate, apply the <Taylor theorem with Lagrange remainder> to $f$ between $r$ and $x_n$:
$$
0=f(r)=f(x_n)-f'(x_n)(x_n-r)+\frac12f''(\xi_n)(x_n-r)^2,\qquad \xi_n\in(r,x_n).
$$
With $e_n=x_n-r>0$, rearrangement gives
$$
\boxed{e_{n+1}=\frac{f''(\xi_n)}{2f'(x_n)}e_n^2,\qquad
\frac{e_{n+1}}{e_n^2}\longrightarrow\frac{f''(r)}{2f'(r)}>0.}
$$
Thus the <monotone Newton convergence for a strictly convex function> has \b[quadratic convergence]. More concretely, if $K=\max_{[a,b]}f''/[2\min_{[a,b]}f']$, then $e_{n+1}\le Ke_n^2$. Once $Ke_N<1$, induction gives $e_{N+j}\le K^{-1}(Ke_N)^{2^j}$, exhibiting the repeated squaring of small errors.
Back to article page
