Maybe this is a silly question but does the normal form $\dot{x}=\mu\pm x^2$ of the fold bifurcation has something to do with the normal form by Dulac and Poincaré or are this completely different things?
The Poincaré-Dulac normal form is that we can write $f(x,\mu)$ as the sum of some linear part $Ax$ and a sum of resonant monomials, i.e. $$ f(x,\mu)=Ax+\sum_{(m,\lambda)=\lambda_k}^{\infty}b_{mk}(\mu)x^m. $$ where $\lambda_k=(m,\lambda)$ are the resonances.
Now, if we have $\dot{x}=f(x,\mu)$ fulfilling the conditions to be a fold birfurcation, what is the Poincaré-Dulac normal form of $f(x,\mu)$ and does it coincide with $\mu\pm x^2$?
Edit:
I think, first of all, one can show that $\alpha\pm x^2$ is a normal form, i.e. the easiest form. Then one can show that this is the Dulac-normal form (or some truncated form of it).
So let's suppose that we have shown that we can ease $\dot{x}=f(x,\alpha)$ to the form $\dot{x}=\alpha\pm x^2$. This is shown in some books.
Then for $\alpha=0$, we have the eigenvalue $\lambda=0$, and have infinitely many resonances: $\lambda=m\lambda$ for $m\geq 2$. If we choose $N=2$, one can use the truncated Dulac normal form, where $\alpha$ is the linear summand and $x^2$ is the resonant monomial. All other summands of the truncated normal form are $0$.
For $\alpha<0$, we have eigenvalues $\pm\sqrt{-\alpha}$ and we have - as far as I see - resonances with $\lvert m\rvert\geq 3$. So we can use that for $N=2$ we do not have resonances. A theorem tells us that we can write it as $Ax+R(x)$, where $Ax$ is the linear summand and $R(x)$ is a polonomial of resonant monomials of order at most N. So here, $Ax=\alpha, R(x)=x^2$.
So the normal form $\dot{x}=\alpha\pm x^2$ is the truncated Dulac normal form for $N=2$?