The sequence of functions $u_{n+1}(t)=u_n(t)+\frac{1}{2}(t-u_n^2(t))$ approximates $\sqrt{t}$ from below

Question

The square root function on $[0,1]$ is approximated by a sequence of functions (polynomials) defined on $[0,1]$. The induction hypothesis is that you have functions such that

$$0 = u_0\leq u_1\leq ... \leq u_n \leq\sqrt{} \leq 1$$

Through induction it is then said that you can use

$$u_{n+1}(t)=u_n(t)+\frac{1}{2}(t-u_n^2(t))$$

($t$ is the number you want to calculate the square root of) and that $u_{n+1}>u_n$ and $u_{n+1}\leq \sqrt{}$ . I've proved the first statement but after many subsitutions I still can't seem to figure out how to prove that

$u_{n+1}\leq \sqrt{}$

Answer 1

You want to prove that $u+\frac12(t-u^2)\le \sqrt{t}$ when $0\le u\le \sqrt{t}$. The function $f(u)=u+\frac12(t-u^2)$ is increasing because $f'(u)=1-u\ge 0$. Since $f(\sqrt{t})=\sqrt{t}$, the conclusion follows.