Weierstrass approximation theorem. Approximation of |x|. I'm studying a proof of the Weierstrass approximation theorem that requires an uniform approximation using polynomials of the function |x| in the interval $[-1,1]$ i.e. we need a sequence of polynomials that converge to |x| in the supreme norm. One way to do this is to define for $t\in[0,1]$ the following sequence:
First set $P_0(t)=0$ for all t, and then using induction define $$P_{n+1}(t)=P_n(t)+\frac{1}{2}(t-P_n(t)^2).$$
The next step would be to prove using induction that for all $t\in[0,1]$, $$0\leq\sqrt{t}-P_n(t)\leq\frac{2\sqrt{t}}{2+n\sqrt{t}},$$ but i have not been able to do this. Can someone help me???
 A: Proof by induction. Consider:
\begin{align}
\sqrt{t}-P_{n+1}(t) &= \sqrt{t}-P_n(t)-\frac{1}{2}(t−P_n(t)^2) \\
&=(\sqrt{t}-P_n(t))(1-\frac{1}{2}(\sqrt{t}+P_n(t)))
\end{align}
Now by induction hypothesis:
\begin{align}
\sqrt{t}-P_{n+1}(t) & \le\frac{2\sqrt{t}}{2+n\sqrt{t}}(1-\frac{1}{2}(\sqrt{t}+P_n(t)))
\end{align}
Now the last term, still by induction, is majored by
$$
1-\frac{1}{2}(\sqrt{t}+P_n(t))=1-\sqrt{t}+\frac{1}{2}(\sqrt{t}-P_n(t))\le1-\sqrt{t}+\frac{\sqrt{t}}{2+n\sqrt{t}}
$$
Putting thing together we have:
\begin{align}
\sqrt{t}-P_{n+1}(t) & \le\frac{2\sqrt{t}}{2+n\sqrt{t}}(1-\sqrt{t}+\frac{\sqrt{t}}{2+n\sqrt{t}}) \\
& =\frac{2\sqrt{t}}{2+(n+1)\sqrt{t}}\underbrace{\left[\frac{2+(n+1)\sqrt{t}}{2+n\sqrt{t}}(1-\sqrt{t}+\frac{\sqrt{t}}{2+n\sqrt{t}})\right]}_{g_n(t)}
\end{align}
Now we must prove that $g_n(t)\le 1$. You can compute $g_n(t)$ derivative, but please use a formal calculus tool like Maple or Mathematica... You get:
$$
g'_n(t)=-\frac{n^3 t+n^2 \left(t+6 \sqrt{t}\right)+n \left(6 \sqrt{t}+8\right)+4}{2
   \left(n \sqrt{t}+2\right)^3}
$$
which is always negative hence its maximal value is for $t=0$ and $g_n(t=0)=1$. We get our expected result:
$$
\sqrt{t}-P_{n+1}(t) \le\frac{2\sqrt{t}}{2+(n+1)\sqrt{t}}
$$
This proof is not very nice (someone will certainly gives another one which is nicer).

Now, there is a much more elegant way for this problem.
1/ It s obvious that $0\le P_n(t)$. Now take $P_n(t)\le \sqrt{t}$ as induction hypothesis. This is true for $n=0$, as $P_0=0$. 
Now as before write:
\begin{align}
\sqrt{t}-P_{n+1}(t) &= \sqrt{t}-P_n(t)-\frac{1}{2}(t−P_n(t)^2) \\
&=(\sqrt{t}-P_n(t))(1-\frac{1}{2}(\sqrt{t}+P_n(t)))
\end{align}
By induction hypothesis $P_n(t)\le\sqrt{t}$, hence $\frac{1}{2}(\sqrt{t}+P_n(t))\le 1$ which also implies $0 \le 1-\frac{1}{2}\sqrt{t}+P_n(t))$.
By consequence
$$
\sqrt{t}-P_{n+1}(t) = (\sqrt{t}-P_n(t))(1-\frac{1}{2}(\sqrt{t}+P_n(t))) \ge 0
$$
hence $P_{n+1}(t)\le\sqrt{t}$ which is what we wanted to prove.
2/ Now back to: $P_{n+1}(t)=P_n(t)+\frac{1}{2}(t-P_n(t)^2)$, observe that:
$$
P_{n+1}(t)-P_n(t)=\frac{1}{2}(t-P_n(t)^2)\ge 0
$$
hence your sequence $P_n$ is increasing.
3/ Observe that $\forall t$, the sequence $P_n(t)$ is increasing and bounded, hence it converges point wise. Now, as $[0,1]$ is compact, by the Dini theorem you can conclude that the convergence is also uniform. Now the limit is obtained solving $p(t)=p(t)-(t-p(t)^2)$ which gives $p(t)=\sqrt{t}$. Q.E.D
