the solution to this equation is an algebraic function which can be represented by the Puiseux series: $x=\sum_{j=1}^\infty c_j\ep^{\f{j}{p}}$ in some neighborhood of zero. Thus $x=c_1\ep^{\f{1}{p}}(1+O(\ep))$ as $\ep\to 0$. Now from (\[algeq\]) one gets: $$c_1^m\ep^{\f{m}{p}}(1+O(\ep))+c_1\ep^{1+\f{1}{p}}(1+O(\ep))=z_0\ep.$$ Thus $p=m$, $c_1=z_0^{\f{1}{m}}$ and $x(\ep)=z_0^{\f{1}{m}}\ep^{\f{1}{m}}(1+O(\ep))$. For $\ep=0$ one gets the solution $y=0$. Therefore $$\label{algeq1}
|x(\ep)-y|\sim \ep^{\f{1}{m}},\q \ep\to 0,$$ where $m$ can be chosen arbitrary large.

Below in Propositions \[prop1\] and  \[prop2\] some sufficient conditions are given that allow one to obtain the estimates for $||x(t)-y||$. \[prem\]

[**Proof**]{} [**of Theorem \[t3\]**]{} Choosing $\mu(t)=\f{\lambda}{\ep(t)}$, where $\lambda$ is a constant, from conditions (\[ric1\]) and (\[xdot\]) one gets the following inequalities: $$\label{rho1}
N_2\le\lambda\left(1-\f{|\dot{\ep}(t)|}{\ep(t)}\right),$$ $$\label{rho2}
2||z_0-y||\f{|\dot{\ep}(t)|}{\ep^2(t)}\le\f{1}{\lambda}
\left(1-\f{|\dot{\ep}(t)|}{\ep(t)}\right),\q \lambda||z_0-x(0)||<\ep(0).$$ Choose $$\label{chlam}
\lambda:=\f{N_2}{1-C_\ep}.$$ It follows from (\[rhoc\]) that (\[rho1\]) holds. From (\[rho\]) one gets: $$\label{ad1}
\ep(0)>\f{N_2||z_0-y||}{1-C_\ep}.$$ On the other hand from (\[toobad\]) it follows that $$\label{ad2} 
||z_0-x(0)||\le ||z_0-y||.$$ Thus from (\[ad1\]), (\[ad2\]) and (\[chlam\]) one obtains the second inequality in (\[rho2\]). Using (\[rho\]) once again, one gets: $$\label{ad3} 
\ep(0)\geq \f{2N_2C_\ep||z_0-y||}{(1-C_\ep)^2}.$$ This inequality and (\[rhoc\]) imply: $$\label{ad4}
\lambda=\f{N_2}{1-C_\ep}\le 
\f{1-C_\ep}{2||z_0-y||\f{|\dot{\ep}(t)|}{\ep^2(t)}}.$$ By (\[rhoc\]) one gets $$\label{ad5}
\f{N_2}{1-C_\ep}\le\f{1-\f{|\dot{\ep}(t)|}{\ep(t)}}{2||z_0-y||
\f{|\dot{\ep}(t)|}{\ep^2(t)}}.$$ The first inequality in (\[rho2\]) is equivalent to (\[ad5\]) for $\lambda$ chosen in (\[chlam\]). Therefore one gets inequality (\[lap\]) by applying Theorem \[t1\], while the second relation 