Multinomial theorem

Theorem (binomial series)

From [math]\displaystyle{ \alpha \in {}^{(\nu)}\mathbb{C}, \binom{\alpha}{n}:=\widetilde{n!}\alpha\acute{\alpha}...(\grave{\alpha}-n) }[/math] and [math]\displaystyle{ \left|\binom{\alpha}{\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 by derivating

[math]\displaystyle{ {\grave{z}}^\alpha={+}_{n=0}^{\omega}{\binom{\alpha}{n}z^n}.\square }[/math]

Multinomial theorem

For [math]\displaystyle{ z, \zeta \in {}^{(\omega)}\mathbb{C}^{k}, n^T \in {}^{(\omega)}\mathbb{N}^{k}, k, m \in {}^{\omega}\mathbb{N}^{*}, z^n := z_1^{n_1} ... z_k^{n_k} }[/math] and [math]\displaystyle{ \binom{m}{n} := \widetilde{n_1! ... {n}_k!}m!\;(k \ge 2) }[/math], it holds that

[math]\displaystyle{ \left({\underline{1}}_k^Tz\right)^m={+}_{n\underline{1}_k=m}{\binom{m}{n}z^n}. }[/math]

Proof:

Cases [math]\displaystyle{ k \in \{1, 2\} }[/math] are clear. Induction step from [math]\displaystyle{ k }[/math] to [math]\displaystyle{ \grave{k} }[/math] by summarising the last two summands for [math]\displaystyle{ p \in {}^{\omega}\mathbb{N} }[/math]:

[math]\displaystyle{ ({(1,z^T){\underline{1}}_{\grave{k}})}^m={+}_{{(n_0,n)\underline{1}}_{\grave{k}}=m}{\binom{m}{n_0,n}z^n} }[/math] because of [math]\displaystyle{ \binom{m}{n_1,...,n_{k-2},p} \binom{p}{n_{\grave{k}},n_k} = \binom{m}{n_1,...,n_k}. }[/math]

Induction step from [math]\displaystyle{ m }[/math] to [math]\displaystyle{ \grave{m} }[/math] for [math]\displaystyle{ \grave{n} := n+(1,0, ... ,0) }[/math] by individual integration:

[math]\displaystyle{ {\grave{m}{\uparrow}_{0}^{\zeta_1}{\left({\underline{1}}_k^Tz\right)^m{\downarrow}z_1}=\left.\left({\underline{1}}_k^T\zeta\right)^{\grave{m}}\right|_0^{{\zeta}_1}={+}_{{\grave{n}}{\underline{1}}_k=\grave{m}}\binom{\grave{m}}{\grave{n}}z^{\grave{n}}}.\square }[/math]

General Leibniz formula

For [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{ m^T, n^T \in {}^{(\omega)}\mathbb{N}^{k}, j, k \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

[math]\displaystyle{ {\downarrow}^mf = {+}_{n\underline{1}_k=||m||_1}{\binom{||m||_1}{n}{\downarrow}^nf}.\square }[/math]

See also

• List of mathematical symbols
• Multinomial theorem