Binomial Series Derivation
📂Analysis Binomial Series Derivation ∣ x ∣ < 1 |x| < 1 ∣ x ∣ < 1 α ∈ C \alpha \in \mathbb{C} α ∈ C ( 1 + x ) α = ∑ k = 0 ∞ ( α k ) x k = 1 + α x + α ( α − 1 ) 2 ! x 2 + α ( α − 1 ) ( α − 2 ) 3 ! x 3 + ⋯
\begin{align*}
(1 + x )^{\alpha} =& \sum_{k=0}^{\infty} \binom{\alpha}{k} x^{k}
\\ =& 1 + \alpha x + \dfrac{\alpha (\alpha-1)}{2!}x^{2} + \dfrac{\alpha (\alpha-1)(\alpha-2)}{3!}x^{3} + \cdots
\end{align*}
( 1 + x ) α = = k = 0 ∑ ∞ ( k α ) x k 1 + αx + 2 ! α ( α − 1 ) x 2 + 3 ! α ( α − 1 ) ( α − 2 ) x 3 + ⋯
Negative Binomial Series ( 1 − x ) − α = ∑ k = 0 ∞ ( α + k − 1 k ) x k = 1 + α x + α ( α + 1 ) 2 ! x 2 + α ( α + 1 ) ( α + 2 ) 3 ! x 3 + ⋯
\begin{align*}
(1 - x)^{-\alpha} &= \sum\limits_{k=0}^{\infty} \binom{\alpha + k - 1}{k} x^{k} \\
&= 1 + \alpha x + \dfrac{\alpha(\alpha+1)}{2!} x^{2} + \dfrac{\alpha(\alpha+1)(\alpha+2)}{3!} x^{3} + \cdots
\end{align*}
( 1 − x ) − α = k = 0 ∑ ∞ ( k α + k − 1 ) x k = 1 + αx + 2 ! α ( α + 1 ) x 2 + 3 ! α ( α + 1 ) ( α + 2 ) x 3 + ⋯
Description Known as Newton’s binomial theorem , it can be seen as a generalization of binomial expansion to infinity and complex numbers. Meanwhile, using x , y x,y x , y
( 1 + y x ) α = ∑ k = 0 ∞ ( α k ) ( y x ) k ⟹ x − α ( x + y ) α = ∑ k = 0 ∞ ( α k ) y k x − k ⟹ ( x + y ) α = ∑ k = 0 ∞ ( α k ) x α − k y k
\begin{align*}
&& \left( 1 + {{y} \over {x}} \right)^{\alpha} =& \sum_{k=0}^{\infty} \binom{\alpha}{k} \left( \dfrac{y}{x} \right)^{k}
\\ \implies && x^{-\alpha} \left( x + y \right)^{\alpha} =& \sum_{k=0}^{\infty} \binom{\alpha}{k} y^{k} x^{-k}
\\ \implies && \left( x + y \right)^{\alpha} =& \sum_{k=0}^{\infty} \binom{\alpha}{k} x^{\alpha-k} y^{k}
\end{align*}
⟹ ⟹ ( 1 + x y ) α = x − α ( x + y ) α = ( x + y ) α = k = 0 ∑ ∞ ( k α ) ( x y ) k k = 0 ∑ ∞ ( k α ) y k x − k k = 0 ∑ ∞ ( k α ) x α − k y k
Derivation Strategy: Create a function g ( α ) g(\alpha) g ( α ) α \alpha α F F F g ( α + β ) = g ( α ) g ( β ) g( \alpha + \beta ) = g( \alpha ) g( \beta ) g ( α + β ) = g ( α ) g ( β ) ( 1 + x ) α (1 + x)^\alpha ( 1 + x ) α
Let’s define g g g g ( α ) : = ∑ k = 0 ∞ ( α k ) x k g ( \alpha ) := \sum_{k=0}^{\infty} \binom{\alpha}{k} x^{k} g ( α ) := ∑ k = 0 ∞ ( k α ) x k
lim k → ∞ ∣ ( α k + 1 ) x k + 1 ( α k ) x k ∣ = lim k → ∞ ∣ α − k k + 1 ∣ ∣ x ∣ = ∣ x ∣ < 1
\lim_{k \to \infty} \left| {{ \binom{\alpha}{k+1} x^{k+1} } \over { \binom{\alpha}{k} x^{k} }} \right| = \lim_{k \to \infty} \left| {{ \alpha - k } \over { k + 1 }} \right| | x | = | x | < 1
k → ∞ lim ( k α ) x k ( k + 1 α ) x k + 1 = k → ∞ lim k + 1 α − k ∣ x ∣ = ∣ x ∣ < 1
By the Ratio Test , F F F C \mathbb{C} C C \mathbb{C} C
Product of Two Power Series : If the convergence interval of f ( x ) : = ∑ k = 0 ∞ a k x k f(x) : = \sum_{k=0}^{\infty} a_{k} x^{k} f ( x ) := ∑ k = 0 ∞ a k x k g ( x ) : = ∑ k = 0 ∞ b k x k g(x) : = \sum_{k=0}^{\infty} b_{k} x^{k} g ( x ) := ∑ k = 0 ∞ b k x k ( − r , r ) (-r,r) ( − r , r ) c k : = ∑ j = 0 k a j b k − j c_{k} := \sum_{j=0}^{k} a_{j} b_{k-j} c k := ∑ j = 0 k a j b k − j ∑ k = 0 ∞ c k x k \sum_{k=0}^{\infty} c_{k} x^{k} ∑ k = 0 ∞ c k x k f ( x ) g ( x ) f(x)g(x) f ( x ) g ( x ) ( − r , r ) (-r,r) ( − r , r )
Properties of Binomial Coefficients : ∑ 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 )
g ( α ) g ( β ) = ∑ k = 0 ∞ ( α k ) x k ∑ k = 0 ∞ ( β k ) x k = ∑ k = 0 ∞ ∑ j = 0 k ( α k − j ) ( β j ) x k = ∑ k = 0 ∞ ( α + β k ) x k = g ( α + β )
\begin{align*}
g(\alpha) g(\beta) =& \sum_{k=0}^{\infty} \binom{\alpha}{k} x^{k} \sum_{k=0}^{\infty} \binom{\beta}{k} x^{k}
\\ =& \sum_{k=0}^{\infty} \sum_{j=0}^{k} \binom{\alpha}{k-j} \binom{\beta}{j} x^{k}
\\ =& \sum_{k=0}^{\infty} \binom{ \alpha + \beta}{k} x^{k}
\\ =& g (\alpha + \beta )
\end{align*}
g ( α ) g ( β ) = = = = k = 0 ∑ ∞ ( k α ) x k k = 0 ∑ ∞ ( k β ) x k k = 0 ∑ ∞ j = 0 ∑ k ( k − j α ) ( j β ) x k k = 0 ∑ ∞ ( k α + β ) x k g ( α + β )
Properties of Continuous Homomorphisms : If continuous function g : R → ( 0 , ∞ ) g : \mathbb{R} \to ( 0 , \infty ) g : R → ( 0 , ∞ ) g ( α + β ) = g ( α ) g ( β ) g(\alpha + \beta) = g(\alpha) g(\beta) g ( α + β ) = g ( α ) g ( β ) α , β ∈ R \alpha, \beta \in \mathbb{R} α , β ∈ R g ( α ) = ( g ( 1 ) ) α g(\alpha) = \left( g(1) \right)^\alpha g ( α ) = ( g ( 1 ) ) α
Therefore, g ( 1 ) = ∑ k = 0 ∞ ( 1 k ) = 1 + x g(1) = \sum_{k=0}^{\infty} \binom{1}{k} = 1 + x g ( 1 ) = ∑ k = 0 ∞ ( k 1 ) = 1 + x
( 1 + x ) α = g ( α ) = ∑ k = 0 ∞ ( α k ) x k
(1 + x )^{\alpha} = g ( \alpha ) = \sum_{k=0}^{\infty} \binom{\alpha}{k} x^{k}
( 1 + x ) α = g ( α ) = k = 0 ∑ ∞ ( k α ) x k
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