Consider some Hamiltonian system $$ \begin{cases}\dot{y}=\frac{\partial H(y,p)}{\partial p}\\\dot{p}=-\frac{\partial H(y,p)}{\partial p}\end{cases}, $$ where $H(y,p)$ is a given smooth function at least $C^2$. Then it is well known that $H(y,p)$ is preserved by the system, i.e. $$ \frac{d}{dt}H(y,p)=\frac{\partial H}{\partial y}\dot{y}+\frac{\partial H}{\partial p}\dot{p}=0.~~~(*) $$
Today, in the lecture we had that $(*)$ only holds if the dimension $n$ of vector $y$ and the dimension $m$ of vector $p$ are equal, i.e. $m=n$.
As an explanation of that, it was written on the board that $$ H(y,p)=H(y_0,p_0)+\frac{\partial H}{\partial y_1}(y_1-y_{01})+\frac{\partial H}{\partial y_2}(y_2-y_{02})+\cdots+\frac{\partial H}{\partial y_n}(y_n-y_{0n})+\frac{\partial H}{\partial p_1}(p_1-p_{01})+\cdots+\frac{\partial H}{\partial p_m}(p_m-p_{0m})+\mathcal{O}_2.~~~~(**) $$
I think, this is the Taylor series at $(y_0,p_0)$, where $$ y_0=(y_{01},\ldots,y_{0n}),~~~~~p_0=(p_{01},...,p_{0m}). $$
But why does $(**)$ tell me that we need to have $m=n$ in order to have $(*)$?