Difference between revisions of "Multinomial theorem"

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m (General Leibniz formula)
m (Proof:)
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<div style="text-align:center;"><math>({(1,z^T){\underline{1}}_{\grave{k}})}^m=\sum_{{(n_0,n)\underline{1}}_{\grave{k}}=m}{\binom{m}{n_0,n}z^n}</math> because of <math>\binom{m}{n_1,...,n_{k-2},p} \binom{p}{n_{\grave{k}},n_k} = \binom{m}{n_1,...,n_k}.</math></div>
 
<div style="text-align:center;"><math>({(1,z^T){\underline{1}}_{\grave{k}})}^m=\sum_{{(n_0,n)\underline{1}}_{\grave{k}}=m}{\binom{m}{n_0,n}z^n}</math> because of <math>\binom{m}{n_1,...,n_{k-2},p} \binom{p}{n_{\grave{k}},n_k} = \binom{m}{n_1,...,n_k}.</math></div>
  
Induction step from <math>m</math> to <math>\grave{m}</math> for <math>\grave{n} := n+(1,0, ... ,0)</math> by individual [[w:Integral|<span class="wikipedia">Integration</span>]]:
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Induction step from <math>m</math> to <math>\grave{m}</math> for <math>\grave{n} := n+(1,0, ... ,0)</math> by individual [[w:Integral|<span class="wikipedia">integration</span>]]:
  
 
<div style="text-align:center;"><math>{\grave{m}\int_{0}^{\check{z}_1}{\left({\underline{1}}_k^Tz\right)^mdz_1}=\left({\underline{1}}_k^Tz\right)^{\grave{m}}-\left.\left({\underline{1}}_k^Tz\right)^{\grave{m}}\right|_{{\check{z}}_1=0}=\sum_{{\grave{n}}{\underline{1}}_k=\grave{m}}\binom{\grave{m}}{\grave{n}}{\check{z}}^{\grave{n}}}.\square</math></div>
 
<div style="text-align:center;"><math>{\grave{m}\int_{0}^{\check{z}_1}{\left({\underline{1}}_k^Tz\right)^mdz_1}=\left({\underline{1}}_k^Tz\right)^{\grave{m}}-\left.\left({\underline{1}}_k^Tz\right)^{\grave{m}}\right|_{{\check{z}}_1=0}=\sum_{{\grave{n}}{\underline{1}}_k=\grave{m}}\binom{\grave{m}}{\grave{n}}{\check{z}}^{\grave{n}}}.\square</math></div>

Revision as of 00:27, 9 February 2022

Theorem (binomial series)

From [math]\displaystyle{ \alpha \in {}^{(\nu)}\mathbb{C}, \binom{\alpha}{n}:=\widehat{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=\sum\limits_{n=0}^{\omega}{\binom{\alpha}{n}z^n}.\square }[/math]

Multinomial theorem

For [math]\displaystyle{ z, \check{z} \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} := \widehat{n_1! ... {n}_k!}m!\;(k \ge 2) }[/math], it holds that

[math]\displaystyle{ \left({\underline{1}}_k^Tz\right)^m=\sum\limits_{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=\sum_{{(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}\int_{0}^{\check{z}_1}{\left({\underline{1}}_k^Tz\right)^mdz_1}=\left({\underline{1}}_k^Tz\right)^{\grave{m}}-\left.\left({\underline{1}}_k^Tz\right)^{\grave{m}}\right|_{{\check{z}}_1=0}=\sum_{{\grave{n}}{\underline{1}}_k=\grave{m}}\binom{\grave{m}}{\grave{n}}{\check{z}}^{\grave{n}}}.\square }[/math]

General Leibniz formula

For [math]\displaystyle{ \partial^n := \partial_1^{n_1}...\partial_k^{n_k} }[/math] and [math]\displaystyle{ \partial_j^{n_j} := \partial^{n_j}/\partial{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{ \partial^mf = \sum\limits_{n\underline{1}_k=||m||_1}{\binom{||m||_1}{n}\partial^nf}.\square }[/math]

See also