In my nonlinear dynamics lecture, it has been stated that:
In general, close to a bifurcation, we expand $f(x,\nu)$ up to linear in $\nu-\nu_c$ and up to first non-zero term in $x-x_*$, $\newcommand{\pd}[2]{\frac{\partial#1}{\partial#2}}$ \begin{multline} \dot x=\underbrace{f(x_*,\nu_c)}_{=0}+\underbrace{\pd{f}{x}\big|_{...}(x-x_*)}_{=0}+\pd{f}{\nu}\big|_{...}(\nu-\nu_c) \\ +\pd{^2f}{x\partial\nu}\big|_{...}(x-x_*)(\nu-\nu_c) +\frac12\pd{^2f}{x^2}\big|_{...}(x-x_*)^2 +\frac12\pd{^2f}{\nu^2}\big|_{...}(\nu-\nu_c)^2 \\ +\frac16\pd{^3f}{x^3}\big|_{...}(x-x_*)^3+\dots \end{multline} enter image description here \begin{array}{cc|ccl} \pd{f}{\nu}&\pd{^2f}{x\partial\nu} & \pd{^2f}{x^2} & \pd{^3f}{x^3} & \\ \hline \ne 0&&\ne 0&&\begin{aligned} \dot x &= a(\nu-\nu_c)+b(x-x_*)^2\\ \dot x &= \nu-x^2\\ &\text{saddle-node} \end{aligned}\\ \hline =0&\ne 0&\ne 0&& \begin{aligned} \dot x &= a(\nu-\nu_c)(x-x_*)+b(x-x_*)^2\\ \dot x &= \nu x-x^2\\ &\text{transcritical} \end{aligned}\\ \hline =0&\ne 0&=0&\ne 0& \begin{aligned} \dot x &= a(\nu-\nu_c)(x-x_*)+b(x-x_*)^3\\ \dot x &= \nu x-x^3\\ &\text{pitchfork} \end{aligned}\\ \end{array}
For a given system of the form $$\dot x = f(x, \nu),$$ where $\nu$ is the bifurcation parameter, at a bifurcation point, the stability of the fixed point is mainly determined by highest order nonzero terms in the Taylor series expansion of $f$ wrt $x$ and $\nu$.
However, is there a proof of this fact?
Intuitively, the higher-order terms should have less effect, but this doesn't necessarily mean that they will not effect the resulting stability. After all, we will be ignoring infinitely any terms.