Difference between revisions of "Multinomial theorem"
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For <math>\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>\tbinom{m}{n} := \widetilde{n_1! ... {n}_k!}m!</math>, it holds that | For <math>\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>\tbinom{m}{n} := \widetilde{n_1! ... {n}_k!}m!</math>, it holds that | ||
− | <div style="text-align:center;"><math>(1{\ | + | <div style="text-align:center;"><math>(1{\upharpoonleft}_k^Tz)^m={\LARGE{\textbf{+}}}_{|n|=m}{\tbinom{m}{n}z^n}.</math></div> |
==== Proof: ==== | ==== Proof: ==== | ||
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> where <math>\tbinom{m}{n} = \tbinom{m}{n_1, ...,n_{\acute{k}},p}\tbinom{p}{n_k, n_{\grave{k}}}</math> and <math>p=n_k+n_{\grave{k}}</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> where <math>\tbinom{m}{n} = \tbinom{m}{n_1, ...,n_{\acute{k}},p}\tbinom{p}{n_k, n_{\grave{k}}}</math> and <math>p=n_k+n_{\grave{k}}</math>: | ||
− | <div style="text-align:center;"><math>\left.{({1{\ | + | <div style="text-align:center;"><math>\left.{({1{\upharpoonleft}_{\grave{k}}^Tz})^m}\right |_{\zeta_{k}=z_k+z_{\grave{k}}}=\left.{\LARGE{\textbf{+}}}_{|n|=m}{\tbinom{m}{n}z^n}\right |_{{\eta}_k!={n_k!}{n_{\grave{k}}!}} = {\LARGE{\textbf{+}}}_{|n|=m}{\tbinom{m}{n}z^n}</math> resp. from <math>m</math> to <math>\grave{m}</math></div>: |
− | <div style="text-align:center;"><math>(1{\ | + | <div style="text-align:center;"><math>(1{\upharpoonleft}_{k}^T z)^{\grave{m}} =\grave{m}{\uparrow}_{0}^{z_j}\left.{(1{\upharpoonleft}_{k}^T z)}^m\right |_{z_j=\zeta}{{\downarrow}\zeta}+\left.(1{\upharpoonleft}_{k}^T z)^{\grave{m}} \right |_{z_j=0}=\left.\grave{m}{\uparrow}_{0}^{z_j}{\LARGE{\textbf{+}}}_{|n|=m}{\tbinom{m}{n}z^{n}}\right |_{z_j=\zeta}{{\downarrow}\zeta}+\left.(1{\upharpoonleft}_{k}^T z)^{\grave{m}} \right |_{z_j=0}={\LARGE{\textbf{+}}}_{|\grave{n}|=\grave{m}}{\tbinom{\grave{m}}{\grave{n}}z^{\grave{n}}}.\square</math></div> |
=== General Leibniz formula === | === General Leibniz formula === |
Latest revision as of 18:24, 1 May 2024
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 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{ \grave{k} }[/math] where [math]\displaystyle{ \tbinom{m}{n} = \tbinom{m}{n_1, ...,n_{\acute{k}},p}\tbinom{p}{n_k, n_{\grave{k}}} }[/math] and [math]\displaystyle{ p=n_k+n_{\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