Polynomial approximation for $f$ induces an approximation to $\sqrt f$? Assume  $f:[0,1] \rightarrow \mathbb{R}$ satisfies $f(t)\geq 0, f(0)=0$ 
I am looking for a machinery, which given a polynomial approximation of $f$ of a certain degree, determines the highest order of polynomial approximation which can be achieved for $\sqrt{f}$. (Based on the coefficients of the approximation for $f$).
Also, I would like an algorithmic way to determine the coefficients of the approxiamtion for the $\sqrt{f}$.
Example: assume $f(t)= at^2+bt^3+ct^4 + R(t)$ where $(\frac{R(t)}{t^4}) \stackrel{t\rightarrow 0}{\longrightarrow} 0, R(0)=0$.
Note that for such an $f$, $\sqrt{f(t)}$ is differntiable at $0$ and $\frac{d}{dt}\sqrt{f(t)}|_{t=0}=\sqrt a$. 
A particular question: For which such functions $f$ (that is for which coefficents $a,b,c$) there exists a polynomial approximation up to third order for $\sqrt{f(t)}$ ? Does the answer depends only on $a,b,c$?
That is there exists $e,f$ such that:
$\sqrt{f(t)}= \sqrt a t+et^2+ft^3 + \tilde R(t)$ where $(\frac{\tilde R(t)}{t^3}) \stackrel{t\rightarrow 0}{\longrightarrow} 0$.
Is there a simple way to deduce the values of $e,f$ as functions of $a,b,c$? (other than writing an expression for the approximation , squaring it, comparing coefficients etc)
Note:
We cannot use the result on Taylor's polynomials for composition of functions, since  $g(t)=\sqrt t$ is not differentiable in $0$.
Also, it is interesting to find what changes if we require existence of approximation for $\sqrt{f(t)}$ up to lower order than 3? do we get more functions for which this holds? What changes if we assume existence of approximations of different orders for the original function?
Lastly, note that the existence of $n$-order approximation does not imply $f$ is $n$ times differentiable. If we do assume some  degree of differentiability, does it help?

I will try a standard technique
for fiddling with power series.
If
$g(t)
=\sqrt{f(t)}
$,
then,
differentiating,
$g'(t)
=\frac {f'(t)}{2\sqrt{f(t)}}
=\frac {f'(t)}{2g(t)}
$
so
$2g'(t)g(t)
= f'(t)
$.
If 
$f(t)
=\sum_{n \ge 0} a_n t^n
$
and
$g(t)
=\sum_{n \ge 0} b_n t^n
$,
then
$f'(t)
=\sum_{n \ge 1} na_n t^{n-1}
=\sum_{n \ge 0} (n+1)a_{n+1} t^{n}
$
and
$g'(t)
=\sum_{n \ge 1} nb_n t^{n-1}
=\sum_{n \ge 0} (n+1)b_{n+1} t^{n}
$.
Multiplying the series
for $g(t)$ and $g'(t)$,
and equating this
to the series for $f'(t)$
you get a recurrence
which will allow
the coefficients $b_n$
to be iteratively computed.
I will leave the details to you.
This technique
can be used to get
$e^{f(t)}$,
$f^{\alpha}(t)$
for real $\alpha$,
and similar results.
