the existence of a real polynomial satisfying the following property It is easy to verify that 
$$ \frac{t}{2}\leq \frac{1}{4}+\frac{1}{4}t^2\leq \frac{t}{1-(1-t^2)^2}-\frac{t}{2}
\quad \quad  0<t\leq1$$
I want to ask if there exist a real polynomial $h(t)$ such that$$ \frac{t}{2}\leq h(t^2)\leq \frac{t}{1-(1-t^2)^n}-\frac{t}{2} \quad \quad  0<t\leq1$$
when the positive integer $n\geq3?$
 A: Basically this is asking for an even polynomial $f=2h$ on $[-1,1]$
such that $f(t) \geq |t|$ but $f(t)-t \ll (1-t)^n$ as $t \rightarrow 1$
from below.  There's a standard construction of a uniform approximation to 
$|t|$ on $[-1,1]$: truncate the Taylor expansion
$$
(1-x)^{1/2} =
1 - \frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16} - \frac{5x^4}{128} - \cdots
$$
before the $x^n$ term to obtain some polynomial $P_n(x)$, 
and let $f_n(t) = P_n(1-t^2)$.
Because each nonconstant term in the Taylor expansion has
a negative coefficient, we have $P_n(x) > \sqrt{1-x}$ for all $x \in (0,1]$.
Thus $f_n(t) > \sqrt{1-(1-t^2)} = |t|$ for $t \in [0,1)$.  For $t$ near $1$,
the desired $f_n(t) = t + O((1-t)^n)$ follows from
$P_n(x) = \sqrt{1-x^2} + O(x^n)$.  If the $O$-constant 
happens to be too large for the desired application, 
use $f_{n'}(t)$ for some $n' > n$; as $n$ increases, 
this must succeed eventually, because $f_n(t)$ decreases to $|t|$ 
as $n \rightarrow \infty$ and the convergence is uniform in $t \in [-1,1]$.
