복소수에 대해 일반화된 이항 계수
📂복소해석 복소수에 대해 일반화된 이항 계수 정의 복소수 α ∈ C \alpha \in \mathbb{C} α ∈ C 이항 계수 binomial Coefficient 라고 한다.
( α k ) : = { α ( α − 1 ) ⋯ ( α − k + 1 ) k ! , k ∈ N 1 , k = 0
\binom{\alpha}{k} := \begin{cases} \displaystyle {{ \alpha ( \alpha - 1 ) \cdots ( \alpha - k + 1 ) } \over { k! }} & , k \in \mathbb{N}
\\ 1 & ,k=0
\end{cases}
( k α ) := ⎩ ⎨ ⎧ k ! α ( α − 1 ) ⋯ ( α − k + 1 ) 1 , k ∈ N , k = 0
설명 원래 이항계수 는 α ∈ N \alpha \in \mathbb{N} α ∈ N 자연수 일 필요가 없다. 당장 생각할 수 있는 음의 정수는 물론 실수, 심지어는 복소수로 확장 가능하다.
정리 ∑ j = 0 k ( α k − j ) ( β j ) = ( α + β k )
\sum_{j=0}^{k} \binom{\alpha}{k-j} \binom{\beta}{j} = \binom{\alpha + \beta}{k}
j = 0 ∑ k ( k − j α ) ( j β ) = ( k α + β )
증명 전략: 수학적 귀납법 과 지루한 계산 뿐이다.
k = 0 k = 0 k = 0 1 ⋅ 1 = 1
1 \cdot 1 = 1
1 ⋅ 1 = 1 k = 1 k = 1 k = 1 α ⋅ 1 + 1 ⋅ β = α + β
\alpha \cdot 1 + 1 \cdot \beta = \alpha + \beta
α ⋅ 1 + 1 ⋅ β = α + β k ≥ 1 k \ge 1 k ≥ 1 ∑ j = 0 k ( α k − j ) ( β j ) = ( α + β k ) \displaystyle \sum_{j=0}^{k} \binom{\alpha}{k-j} \binom{\beta}{j} = \binom{\alpha + \beta}{k} j = 0 ∑ k ( k − j α ) ( j β ) = ( k α + β ) ( α + β k + 1 ) = ( α + β k ) α + β − k k + 1 = ∑ j = 0 k ( α k − j ) ( β j ) ( α − k + j k + 1 + β − j k + 1 ) = ∑ j = 0 k [ k − j + 1 k + 1 ( α k − j + 1 ) ( β j ) + j + 1 k + 1 ( α k − j ) ( β j + 1 ) ] = ( α k + 1 ) + ∑ j = 1 k ( k − j + 1 k + 1 + j k + 1 ) ( α k − j + 1 ) ( β j ) + ( β k + 1 ) = ( α k + 1 ) + ∑ j = 1 k ( α k − j + 1 ) ( β j ) + ( β k + 1 ) = ∑ j = 0 k + 1 ( α ( k + 1 ) − j ) ( β j )
\begin{align*}
\binom{ \alpha + \beta }{ k + 1 } =& \binom{ \alpha + \beta }{ k } {{ \alpha + \beta - k } \over { k + 1 }}
\\ =& \sum_{j=0}^{k} \binom{ \alpha }{ k - j } \binom{ \beta }{ j } \left( {{ \alpha - k + j } \over { k + 1 }} + {{ \beta - j } \over { k + 1 }} \right)
\\ =& \sum_{j=0}^{k} \left[ {{ k - j + 1 } \over { k + 1 }} \binom{ \alpha }{ k - j + 1 } \binom{ \beta }{ j } + {{ j + 1 } \over { k + 1 }} \binom{ \alpha }{ k - j } \binom{ \beta }{ j + 1 } \right]
\\ =& \binom{ \alpha }{ k + 1 } + \sum_{j=1}^{k} \left( {{ k - j + 1 } \over { k + 1 }} + {{ j } \over { k + 1 }} \right) \binom{ \alpha }{ k - j + 1 } \binom{ \beta }{ j } + \binom{ \beta }{ k + 1 }
\\ =& \binom{ \alpha }{ k + 1 } + \sum_{j=1}^{k} \binom{ \alpha }{ k - j + 1 } \binom{ \beta }{ j } + \binom{ \beta }{ k + 1 }
\\ =& \sum_{j=0}^{k+1} \binom{ \alpha }{ (k + 1) - j } \binom{ \beta }{ j }
\end{align*}
( k + 1 α + β ) = = = = = = ( k α + β ) k + 1 α + β − k j = 0 ∑ k ( k − j α ) ( j β ) ( k + 1 α − k + j + k + 1 β − j ) j = 0 ∑ k [ k + 1 k − j + 1 ( k − j + 1 α ) ( j β ) + k + 1 j + 1 ( k − j α ) ( j + 1 β ) ] ( k + 1 α ) + j = 1 ∑ k ( k + 1 k − j + 1 + k + 1 j ) ( k − j + 1 α ) ( j β ) + ( k + 1 β ) ( k + 1 α ) + j = 1 ∑ k ( k − j + 1 α ) ( j β ) + ( k + 1 β ) j = 0 ∑ k + 1 ( ( k + 1 ) − j α ) ( j β )
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