Define $$x\_{k+1}(t)=\frac{3x^4\_k(t)+6(1-t)x\_k^2(t)-(1-t)^2}{8x\_k^3(t)},$$
 with $x\_0(t)=1$. It is not difficult to see $x\_k(t)$ converges to $\sqrt{1-t}$, whose (Maclaurin expansion) has negative coefficients unless the first one. Let 


$$x\_k(t)=\sum\limits\_{i =0}^{\infty}c\_{k,i}t^{i}, \mid t\mid<1$$


Is it true $c\_{k,i}\le0$ for all $k\ge 1, i\ge 1$?



A classical analysis problem
This is not the first time I am fighting with positivity (nonnegativity).
But this problem looks not natural enough for a standard technique, and
it seems to me that the resulting sequence $x\_k(t)$ is always between
$\sqrt{1-t}$ and $1$, in the sense that the expansions of $x\_k(t)-\sqrt{1-t}$
and $1-x\_k(t)$ have nonnegative coefficients only.


I can only give a partial solution. After the change
$$
z\_k(t)=1-\frac{x\_k(t)}{\sqrt{1-t}}
$$
the recursion for $x\_k(t)$ translates into
$$
z\_{k+1}=\frac{z\_k^3(4-3z\_k)}{8(1-z\_k)^3}
\quad\text{for}\quad k=0,1,2,\dots
$$
with the initial data
$$
z\_0=1-\frac1{\sqrt{1-t}}=-\frac12t+O(t^2).
$$
The recursion implies $z\_{k+1}=z\_k^3/2+O(z\_k^4)$, therefore
by induction on $k$ we obtain
$$
z\_k=-\frac{t^{3^k}}{2^{(3^{k+1}-1)/2}}+\text{higher terms}.
$$
This implies that
$$
x\_k(t)=\sqrt{1-t}-\sqrt{1-t}z\_k(t)
$$
agrees with $\sqrt{1-t}$ up to the term $t^{3^k}$. In particular,
as the expansion of $\sqrt{1-t}$ involves negative coefficients besides
the constant term, we conclude that $c\_{k,i}<0$ for $i=1,2,\dots,3^k$
in the expansion
$$
x\_k(t)=1+\sum\_{k=1}^\infty c\_{k,i}t^i.
$$


