Binomial Theorem/General Binomial Theorem

Theorem

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

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

Then:

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

where $\alpha^{\underline k}$ denotes the falling factorial.

That is:

$\displaystyle \left({1 + x}\right)^\alpha = 1 + \alpha x + \frac {\alpha \left({\alpha - 1}\right)} {2!} x^2 + \frac {\alpha \left({\alpha - 1}\right) \left({\alpha - 2}\right)} {3!} x^3 + \cdots$

Proof

Let $R$ be the radius of convergence of the power series $\displaystyle f \left({x}\right) = \sum_{n=0}^\infty \frac {\prod \limits_{k=0}^{n-1} \left({\alpha - k}\right)} {n!} x^n$.

Then by Radius of Convergence from Limit of Sequence, we have:

$\displaystyle \frac 1 R = \lim_{n \to \infty} \frac {\left|{\alpha \left({\alpha - 1}\right) \cdots \left({\alpha - n}\right)}\right|} {\left({n+1}\right)!} \frac {n!} {\left|{\alpha \left({\alpha - 1}\right) \cdots \left({\alpha - n + 1}\right)}\right|}$

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac 1 R\)	\(=\)	\(\displaystyle \lim_{n \to \infty} \frac {\mid {\alpha \left({\alpha - 1}\right) \cdots \left({\alpha - n}\right)}\mid} {\left({n+1}\right)!} \frac {n!} {\mid {\alpha \left({\alpha - 1}\right) \cdots \left({\alpha - n + 1}\right)}\mid}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{n \to \infty} \frac {\mid {\alpha - n}\mid} {n+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Thus for $\left|{x}\right| < 1$ we can use Power Series Differentiable on Interval of Convergence and get:

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

This leads us to:

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

Gathering up what we've got, that is: $\left({1 + x}\right) D_x f \left({x}\right) = \alpha f \left({x}\right)$.

Thus we get $\displaystyle D_x \left({\frac {f \left({x}\right)} {\left({1 + x}\right)^\alpha}}\right) = -\alpha \left({1 + x}\right)^{-\alpha - 1} f \left({x}\right) + \left({1 + x}\right)^{-\alpha} D_x f \left({x}\right) = 0$

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 $c = 1$.

$\blacksquare$

Historical Note

The General Binomial Theorem was announced by Isaac Newton in 1676. However, he had no real proof. Euler made an incomplete attempt in 1774, but the full proof had to wait for Gauss to provide it in 1812.

Sources

• Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (1968): $\S 1.2.6 \ \text{F} \ (13), \ (15)$, Exercise $15$
• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $20.4$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 15.10 \ (1)$