Is there any method for computation of the coefficients $A_i$, $B_i$, $u\in[-\pi/2,\pi/2],v\in[-\pi,\pi]$, of the inverse series to
$$X=F(u,v)=a_1u^0v^1+a_2u^2v^1+a_3u^2v^3,$$ $$Y=G(u,v)=b_1uv^0+b_2uv^2+b_3u^3v+b_4u^2v^3,$$
in the form of
$$u=A_1X^0Y^0+A_2X^0Y+A_3XY^0+A_4XY+...,$$ $$v=B_1X^0Y^0+B_2X^0Y+B_3XY^0+B_4XY+B_5X^2Y+...,$$
or, only a numerical solution of the problem
$$min[(F(u,v)-X)^2 + (G(u,v)-Y)^2]$$
exists? For example, the Gauss-Newton method $h=-(J^TJ)^{-1}J^Tf$ may be used, where $$J=\left[\begin{array}{cc} \partial F(u,v)/\partial u & \partial F(u,v)/\partial v\\ \partial G(u,v)/\partial u & \partial G(u,v)/\partial v \end{array}\right].$$
Thanks for your help.
Recall that a univarite formal power series can be inverted \begin{eqnarray*} f(x)&=&x+a_1 x^2+ a_2 x^3 +\cdots \\ f^{-1}(x) &=& x-a_1 x^2 +(2a_1^2 -a_2)x^3+ \cdots \end{eqnarray*} A $2$ dimensional analogue can be done in your case, let $a_1=1$ and $b_1=1$ for the sake of simplicity and this can be achieved by rescaling $u$ and $v$. Your function is $ T:(v,u) \rightarrow (X=F(u,v),Y=G(u,v))$ \begin{eqnarray*} X&=&v+a_2 u^2 v + a_3 u^2 v^3 \\ Y&=&u +b_2 uv^2+b_3u^3v +b_4 u^2 v^3. \end{eqnarray*} The inverse function $ T:(X,Y) \rightarrow (v=V(X,Y),u=U(X,Y))$ is (up to third order) \begin{eqnarray*} v=X-a_2 X Y^2 +\cdots \\ u= Y-b_2 X^2 Y +\cdots \end{eqnarray*} Now add fourth order terms ( $\alpha_u X^4+ \beta_u X^3 Y + \gamma_u X^2 Y^2 +\delta_u X Y^3 +\epsilon_u Y^4$) to these series and use $ T \circ S (X,Y)=(X,Y)$ to calculate the next terms.