I am working modulo high order terms that is I neglect in all expressions terms of order more than $2$.
I have the transformation $\theta = f(\psi)$ defined by $$ \theta_1 = \alpha_1 \psi_1^2 + 2\beta_1 \psi_1 \psi_2 + \gamma_1 \psi_2^2 $$ $$ \theta_2 = \alpha_2 \psi_1^2 + 2\beta_2 \psi_1 \psi_2 + \gamma_2 \psi_2^2 $$
I would like to write the inverse $\psi = f^{-1}(\theta)$ but I fail in the computations.
My point is that first the inverse could be not defined ? I have two parabolas surfaces in $\psi_1,\psi_2$ so they don't span all the real $\theta_i,\,i=1,2$... but since we neglect high order terms can we define the inverse ?
The inverse cannot be defined as a formal power series because the tangent plane at the origin is 'vertical', i.e. the first derivative of $f^{-1}(\theta)$ would be infinitely large. Take for example $\beta_{1,2} = 0$ and $\gamma_1 = 0 = \alpha_2$, then the inverse function would be \begin{align} \psi_1 &= \pm \sqrt{\alpha_1 \theta_1},\\ \psi_2 &= \pm \sqrt{\gamma_2 \theta_2}, \end{align} neither of which can be expressed in a power series in $\theta_{1,2}$, since $\lim_{\theta_i \to 0} \frac{\text{d} \psi_i}{\text{d} \theta_i} = \pm \infty$.
To get a workable expression, you could use the Ansatz \begin{equation} \psi_i = \sqrt{\theta_2} \left(a_0 + a_1 \frac{\theta_1}{\theta_2} + a_2 \frac{\theta_1^2}{\theta_2^2} + \ldots\right), \end{equation} which seemed to work for me.