Binomial Series Derivation
Formulas
$|x| < 1$ implies $\alpha \in \mathbb{C}$, then $$ \begin{align*} (1 + x )^{\alpha} =& \sum_{k=0}^{\infty} \binom{\alpha}{k} x^{k} \\ =& 1 + \alpha x + \dfrac{\alpha (\alpha-1)}{2!}x^{2} + \dfrac{\alpha (\alpha-1)(\alpha-2)}{3!}x^{3} + \cdots \end{align*} $$
Negative Binomial Series
$$ \begin{align*} (1 - x)^{-\alpha} &= \sum\limits_{k=0}^{\infty} \binom{\alpha + k - 1}{k} x^{k} \\ &= 1 + \alpha x + \dfrac{\alpha(\alpha+1)}{2!} x^{2} + \dfrac{\alpha(\alpha+1)(\alpha+2)}{3!} x^{3} + \cdots \end{align*} $$
Description
Known as Newton’s binomial theorem, it can be seen as a generalization of binomial expansion to infinity and complex numbers. Meanwhile, using $x,y$, the familiar form can be simply derived as follows.
$$ \begin{align*} && \left( 1 + {{y} \over {x}} \right)^{\alpha} =& \sum_{k=0}^{\infty} \binom{\alpha}{k} \left( \dfrac{y}{x} \right)^{k} \\ \implies && x^{-\alpha} \left( x + y \right)^{\alpha} =& \sum_{k=0}^{\infty} \binom{\alpha}{k} y^{k} x^{-k} \\ \implies && \left( x + y \right)^{\alpha} =& \sum_{k=0}^{\infty} \binom{\alpha}{k} x^{\alpha-k} y^{k} \end{align*} $$
Derivation
Strategy: Create a function $g(\alpha)$ based on $\alpha$ and then show that $F$ is continuous. If a continuous function satisfies $g( \alpha + \beta ) = g( \alpha ) g( \beta )$, it inherits properties related to the exponential function, leading us to deduce $(1 + x)^\alpha$ on the left side.
Let’s define $g$ as $g ( \alpha ) := \sum_{k=0}^{\infty} \binom{\alpha}{k} x^{k}$.
$$ \lim_{k \to \infty} \left| {{ \binom{\alpha}{k+1} x^{k+1} } \over { \binom{\alpha}{k} x^{k} }} \right| = \lim_{k \to \infty} \left| {{ \alpha - k } \over { k + 1 }} \right| | x | = | x | < 1 $$
By the Ratio Test, $F$ absolutely and uniformly converges on $\mathbb{C}$, and therefore, is continuous on $\mathbb{C}$.
Product of Two Power Series: If the convergence interval of $f(x) : = \sum_{k=0}^{\infty} a_{k} x^{k}$ and $g(x) : = \sum_{k=0}^{\infty} b_{k} x^{k}$ is $(-r,r)$ and $c_{k} := \sum_{j=0}^{k} a_{j} b_{k-j}$, then $\sum_{k=0}^{\infty} c_{k} x^{k}$ converges to $f(x)g(x)$ on the convergence interval $(-r,r)$.
Properties of Binomial Coefficients: $\sum_{j=0}^{k} \binom{\alpha}{k-j} \binom{\beta}{j} = \binom{\alpha + \beta}{k}$
$$ \begin{align*} g(\alpha) g(\beta) =& \sum_{k=0}^{\infty} \binom{\alpha}{k} x^{k} \sum_{k=0}^{\infty} \binom{\beta}{k} x^{k} \\ =& \sum_{k=0}^{\infty} \sum_{j=0}^{k} \binom{\alpha}{k-j} \binom{\beta}{j} x^{k} \\ =& \sum_{k=0}^{\infty} \binom{ \alpha + \beta}{k} x^{k} \\ =& g (\alpha + \beta ) \end{align*} $$
Properties of Continuous Homomorphisms: If continuous function $g : \mathbb{R} \to ( 0 , \infty )$ satisfies $g(\alpha + \beta) = g(\alpha) g(\beta)$ for all $\alpha, \beta \in \mathbb{R}$, then $g(\alpha) = \left( g(1) \right)^\alpha$
Therefore, $g(1) = \sum_{k=0}^{\infty} \binom{1}{k} = 1 + x$
$$ (1 + x )^{\alpha} = g ( \alpha ) = \sum_{k=0}^{\infty} \binom{\alpha}{k} x^{k} $$
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