Binomial Theorem/General Binomial Theorem

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Theorem

Let $\alpha \in \R$ be a real number.

Let $x \in \R$ be a real number such that $\left|{x}\right| < 1$.


Then:

\(\displaystyle \left({1 + x}\right)^\alpha\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \frac {\alpha^{\underline n} } {n!} x^n\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \dbinom \alpha n x^n\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {n!} \left({\prod_{k \mathop = 0}^{n - 1} \left({\alpha - k}\right)}\right) x^n\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 1 + \alpha x + \dfrac {\alpha \left({\alpha - 1}\right)} {2!} x^2 + \dfrac {\alpha \left({\alpha - 1}\right) \left({\alpha - 2}\right)} {3!} x^3 + \cdots\) $\quad$ $\quad$

where:

$\alpha^{\underline n}$ denotes the falling factorial
$\dbinom \alpha n$ denotes a binomial coefficient.


Proof 1

Let $R$ be the radius of convergence of the power series:

$\displaystyle f \left({x}\right) = \sum_{n \mathop = 0}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \left({\alpha - k}\right)} {n!} x^n$


Then:

\(\displaystyle \frac 1 R\) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \frac {\left\vert{\alpha \left({\alpha - 1}\right) \cdots \left({\alpha - n}\right)}\right\vert} {\left({n + 1}\right)!} \frac {n!} {\left\vert{\alpha \left({\alpha - 1}\right) \cdots \left({\alpha - n + 1}\right)}\right\vert}\) $\quad$ Radius of Convergence from Limit of Sequence $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to \infty} \frac {\left\vert{\alpha - n}\right\vert} {n + 1}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 1\) $\quad$ $\quad$

Thus for $\left|{x}\right| < 1$, Power Series is Differentiable on Interval of Convergence applies:

$\displaystyle D_x f \left({x}\right) = \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \left({\alpha - k}\right)} {n!} n x^{n - 1}$


This leads to:

\(\displaystyle \left({1 + x}\right) D_x f \left({x}\right)\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^{n-1} \left({\alpha - k}\right)} {\left({n - 1}\right)!} x^{n - 1} + \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \left({\alpha - k}\right)} {\left({n - 1}\right)!} x^n\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \alpha + \sum_{n \mathop = 1}^\infty \left({\frac {\prod \limits_{k \mathop = 0}^n \left({\alpha - k}\right)} {n!} + \frac {\prod \limits_{k \mathop = 0}^{n - 1}\left({\alpha - k}\right)} {\left({n - 1}\right)!} }\right)x^n\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \alpha + \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^n \left({\alpha - k}\right)} {\left({n - 1}\right)!} \left({\frac 1 n + \frac 1 {\alpha - n} }\right)x^n\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \alpha + \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^n \left({\alpha - k}\right)} {\left({n - 1}\right)!} \ \frac \alpha {n \left({\alpha - n}\right)} x^n\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \alpha \left({1 + \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \left({\alpha - k}\right)} {n!} x^n}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \alpha f \left({x}\right)\) $\quad$ $\quad$

Gathering up:

$\left({1 + x}\right) D_x f \left({x}\right) = \alpha f \left({x}\right)$

Thus:

$D_x \left({\dfrac {f \left({x}\right)} {\left({1 + x}\right)^\alpha}}\right) = -\alpha \left({1 + x}\right)^{-\alpha - 1} f \left({x}\right) + \left({1 + x}\right)^{-\alpha} D_x f \left({x}\right) = 0$

So $f \left({x}\right) = c \left({1 + x}\right)^\alpha$ when $\left|{x}\right| < 1$ for some constant $c$.

But $f \left({0}\right) = 1$ and hence $c = 1$.

$\blacksquare$


Proof 2

From Sum over k of r-kt choose k by r over r-kt by z^k:

$\displaystyle \sum_n \dbinom {\alpha - n t} k \dfrac \alpha {\alpha - n t} z^n = x^\alpha$

where:

$z = x^{t + 1} - x^t$
$x = 1$ for $z = 0$.

Setting $t = 0$:

\(\displaystyle \sum_k \dbinom {\alpha - n \times 0} n \dfrac \alpha {\alpha - n \times 0} z^n\) \(=\) \(\displaystyle x^\alpha\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_n \dbinom \alpha n \dfrac \alpha \alpha z^n\) \(=\) \(\displaystyle \left({1 + z}\right)^\alpha\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_n \dbinom \alpha n z^n\) \(=\) \(\displaystyle \left({1 + z}\right)^\alpha\) $\quad$ $\quad$

$\blacksquare$


Historical Note

The General Binomial Theorem was first conceived by Isaac Newton during the years $1665$ to $1667$ when he was living in his home in Woolsthorpe.

He announced the result formally, in letters to Henry Oldenburg on $13$th June $1676$ and $24$th October $1676$ but did not provide a proper proof (at that time the need for the appropriate level of rigor had not been recognised).

Leonhard Paul Euler made an incomplete attempt in $1774$, but the full proof had to wait for Carl Friedrich Gauss to provide it in $1812$.

This was, in fact, the first time anything about infinite summations was proved adequately.


Sources