Binomial Theorem/General Binomial Theorem

From ProofWiki
Jump to: navigation, search

Theorem

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

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


Then:

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

where $\alpha^{\underline n}$ 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:

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

By Radius of Convergence from Limit of Sequence:

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

Thus for $\left|{x}\right| < 1$, Power Series 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:

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

Gathering up:

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

Thus:

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

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

Retrieved from "https://proofwiki.org/w/index.php?title=Binomial_Theorem/General_Binomial_Theorem&oldid=183583"