 Solution

First control the denominator. The mean value theorem applied to $f^{\prime }$ yields, for $0<x\le 2000$,Thus $f$ is strictly increasing on $[0,2000]$ and all proposed Newton steps within that interval are defined.

Assume $0\le x_n<x^{\ast }$. Then $f(x_n)<0$ and $f^{\prime }(x_n)>0$, so $x_{n+1}>x_n$. Apply Taylor's theorem instead at the root, expanding about $x_n$:After rearranging and substituting the Newton step, the signed error satisfiesConsequently $x_n<x_{n+1}<x^{\ast }$. Starting from $x_0=0$, mathematical induction proves this for every finite $n$, establishing that the signed errors are negative and strictly increasing. Since $f^{\prime \prime }({\xi }_n)>-1/4000$ and $2f^{\prime }(x_n)>1$,Writing $E_n=-{ϵ}_n$ gives $0<E_{n+1}\le E_n^2/4000$. Since $E_0=x^{\ast }<2000$, this also implies $E_n\le 4000(E_0/4000{)}^{2^n}\to 0$, proving quadratic convergence. This is the concave Newton iteration from below.