Multinomial theorem
Theorem (binomial series)
From [math]\displaystyle{ alpha \in {}^{(\nu)}\mathbb{C}, \binom{\alpha}{n}:=\widetilde{n!}\alpha\acute{\alpha}...(\overset{\scriptsize{\grave{}}}{\alpha}-n) }[/math] and [math]\displaystyle{ \left|\binom{\alpha}{\overset{\scriptsize{\grave{}}}{m}}/\binom{\alpha}{m}\right|<1 }[/math] for all [math]\displaystyle{ m \ge \nu }[/math] and [math]\displaystyle{ \binom{\alpha}{0}:=1 }[/math], it follows for [math]\displaystyle{ z \in \mathbb{D}^\ll }[/math] or [math]\displaystyle{ z \in {}^{(\omega)}\mathbb{C} }[/math] for [math]\displaystyle{ \alpha \in {}^{(\omega)}\mathbb{N} }[/math] the Taylor series centred on 0 that
Multinomial theorem
For [math]\displaystyle{ \zeta \in {}^{(\omega)}\mathbb{C}, z \in {}^{(\omega)}\mathbb{C}^{k}, k \in {}^{(\omega)}\mathbb{N}_{\ge 2}, m, n_j \in {}^{\omega}\mathbb{N}^{*}, |n| := {\LARGE{\textbf{+}}}_{j=1}^{k}{n_j}, z^n := {\times}_{j=1}^{k}{{z_j}^{n_j}} }[/math] and [math]\displaystyle{ \tbinom{m}{n} := \widetilde{n_1! ... {n}_k!}m! }[/math], it holds that
Proof:
Cases [math]\displaystyle{ k \in \{1, 2\} }[/math] are clear. Induction step from [math]\displaystyle{ k }[/math] to [math]\displaystyle{ \overset{\scriptsize{\grave{}}}{k} }[/math] where [math]\displaystyle{ \tbinom{m}{n} = \tbinom{m}{n_1, ...,n_{\acute{k}},p}\tbinom{p}{n_k, n_{\overset{\scriptsize{\grave{}}}{k}}} }[/math] and [math]\displaystyle{ p=n_k+n_{\overset{\scriptsize{\grave{}}}{k}} }[/math]:
:
General Leibniz formula
Putting [math]\displaystyle{ {\downarrow}^n := {\downarrow}_1^{n_1}...{\downarrow}_k^{n_k} }[/math] and [math]\displaystyle{ {\downarrow}_j^{n_j} := {\downarrow}^{n_j}/{\downarrow}{z_j}^{n_j} }[/math], it follows for [math]\displaystyle{ j, k, m, n \in {}^{(\omega)}\mathbb{N} }[/math] and differentiable [math]\displaystyle{ f = f_1\cdot...\cdot f_k \in {}^{(\omega)}\mathbb{C} }[/math] from the multinomial theorem that