Binomial Theorem/General Binomial Theorem
Theorem
Let $\alpha \in \R$ be a real number.
Let $x \in \R$ be a real number such that $\size x < 1$.
Then:
\(\displaystyle \paren {1 + x}^\alpha\) | \(=\) | \(\displaystyle \sum_{n \mathop = 0}^\infty \frac {\alpha^{\underline n} } {n!} x^n\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{n \mathop = 0}^\infty \dbinom \alpha n x^n\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {n!} \paren {\prod_{k \mathop = 0}^{n - 1} \paren {\alpha - k} } x^n\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle 1 + \alpha x + \dfrac {\alpha \paren {\alpha - 1} } {2!} x^2 + \dfrac {\alpha \paren {\alpha - 1} \paren {\alpha - 2} } {3!} x^3 + \cdots\) |
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:
- $\displaystyle \map f x = \sum_{n \mathop = 0}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \paren {\alpha - k} } {n!} x^n$
Then:
\(\displaystyle \frac 1 R\) | \(=\) | \(\displaystyle \lim_{n \mathop \to \infty} \frac {\size {\alpha \paren {\alpha - 1} \dotsm \paren {\alpha - n} } } {\paren {n + 1}!} \frac {n!} {\size {\alpha \paren {\alpha - 1} \dotsm \paren {\alpha - n + 1} } }\) | Radius of Convergence from Limit of Sequence | ||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \lim_{n \mathop \to \infty} \frac {\size {\alpha - n} } {n + 1}\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle 1\) |
Thus for $\size x < 1$, Power Series is Differentiable on Interval of Convergence applies:
- $\displaystyle D_x \map f x = \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \paren {\alpha - k} } {n!} n x^{n - 1}$
This leads to:
\(\displaystyle \paren {1 + x} D_x \map f x\) | \(=\) | \(\displaystyle \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \paren {\alpha - k} } {\paren {n - 1}!} x^{n - 1} + \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \paren {\alpha - k} } {\paren {n - 1}!} x^n\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \alpha + \sum_{n \mathop = 1}^\infty \paren {\frac {\prod \limits_{k \mathop = 0}^n \paren {\alpha - k} } {n!} + \frac {\prod \limits_{k \mathop = 0}^{n - 1} \paren {\alpha - k} } {\paren {n - 1}!} } x^n\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \alpha + \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^n \paren {\alpha - k} } {\paren {n - 1}!} \paren {\frac 1 n + \frac 1 {\alpha - n} } x^n\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \alpha + \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^n \paren {\alpha - k} } {\paren {n - 1}!} \, \frac \alpha {n \paren {\alpha - n} } x^n\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \alpha \paren {1 + \sum_{n \mathop = 1}^\infty \frac {\prod \limits_{k \mathop = 0}^{n - 1} \paren {\alpha - k} } {n!} x^n}\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \alpha \, \map f x\) |
Gathering up:
- $\paren {1 + x} D_x \map f x = \alpha \, \map f x$
Thus:
- $\map {D_x} {\dfrac {\map f x} {\paren {1 + x}^\alpha} } = -\alpha \paren {1 + x}^{-\alpha - 1} \map f x + \paren {1 + x}^{-\alpha} D_x \map f x = 0$
So $\map f x = c \paren {1 + x}^\alpha$ when $\size x < 1$ for some constant $c$.
But $\map f 0 = 1$ 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 $t = 0$:
\(\displaystyle \sum_k \dbinom {\alpha - n \times 0} n \dfrac \alpha {\alpha - n \times 0} z^n\) | \(=\) | \(\displaystyle x^\alpha\) | |||||||||||
\(\displaystyle \leadsto \ \ \) | \(\displaystyle \sum_n \dbinom \alpha n \dfrac \alpha \alpha z^n\) | \(=\) | \(\displaystyle \left({1 + z}\right)^\alpha\) | ||||||||||
\(\displaystyle \leadsto \ \ \) | \(\displaystyle \sum_n \dbinom \alpha n z^n\) | \(=\) | \(\displaystyle \left({1 + z}\right)^\alpha\) |
$\blacksquare$
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
- 1937: Eric Temple Bell: Men of Mathematics ... (previous) ... (next): Chapter $\text{VI}$: On the Seashore
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 20$: Binomial Series: $20.4$
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $\text{F} \ (15)$
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.9$: Generating Functions: $(19)$