Difference between revisions of "Multinomial theorem"
Borishaase (talk | contribs) (Multinomial theorem) |
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Cases <math>k \in \{1, 2\}</math> are clear. [[w:Mathematical induction|<span class="wikipedia">Induction step</span>]] from <math>k</math> to <math>\grave{k}</math> by summarising the last two summands for <math>p \in {}^{\omega}\mathbb{N}</math>: | Cases <math>k \in \{1, 2\}</math> are clear. [[w:Mathematical induction|<span class="wikipedia">Induction step</span>]] from <math>k</math> to <math>\grave{k}</math> by summarising the last two summands for <math>p \in {}^{\omega}\mathbb{N}</math>: | ||
− | <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}{ | + | <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>]]: | 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>]]: |
Revision as of 02:48, 8 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
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
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]:
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: