Binomial Theorem
Theorem
Integral Index
Let $X$ be one of the set of numbers $\N, \Z, \Q, \R, \C$.
Let $x, y \in X$.
Then:
| \(\ds \forall n \in \Z_{\ge 0}: \ \ \) | \(\ds \paren {x + y}^n\) | \(=\) | \(\ds \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds x^n + \binom n 1 x^{n - 1} y + \binom n 2 x^{n - 2} y^2 + \binom n 3 x^{n - 3} y^3 + \cdots\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds x^n + n x^{n - 1} y + \frac {n \paren {n - 1} } {2!} x^{n - 2} y^2 + \frac {n \paren {n - 1} \paren {n - 3} } {3!} x^{n - 3} y^3 + \cdots\) |
where $\dbinom n k$ is $n$ choose $k$.
Ring Theory
Let $\struct {R, +, \odot}$ be a ringoid such that $\struct {R, \odot}$ is a commutative semigroup.
Let $n \in \Z: n \ge 2$.
Then:
- $\ds \forall x, y \in R: \odot^n \paren {x + y} = \odot^n x + \sum_{k \mathop = 1}^{n - 1} \binom n k \paren {\odot^{n - k} x} \odot \paren {\odot^k y} + \odot^n y$
where $\dbinom n k = \dfrac {n!} {k! \ \paren {n - k}!}$ (see Binomial Coefficient).
If $\struct {R, \odot}$ has an identity element $e$, then:
- $\ds \forall x, y \in R: \odot^n \paren {x + y} = \sum_{k \mathop = 0}^n \binom n k \paren {\odot^{n - k} x} \odot \paren {\odot^k y}$
General Binomial Theorem
Let $\alpha \in \R$ be a real number.
Let $x \in \R$ be a real number such that $\size x < 1$.
Then:
| \(\ds \paren {1 + x}^\alpha\) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \frac {\alpha^{\underline n} } {n!} x^n\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \dbinom \alpha n x^n\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 0}^\infty \frac 1 {n!} \paren {\prod_{k \mathop = 0}^{n - 1} \paren {\alpha - k} } x^n\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 + \alpha x + \dfrac {\alpha \paren {\alpha - 1} } {2!} x^2 + \dfrac {\alpha \paren {\alpha - 1} \paren {\alpha - 2} } {3!} x^3 + \cdots\) |
where:
- $\alpha^{\underline n}$ denotes the falling factorial
- $\dbinom \alpha n$ denotes a binomial coefficient.
Multiindices
Let $\alpha$ be a multiindex, indexed by $\left\{{1, \ldots, n}\right\}$ such that $\alpha_j \ge 0$ for $j = 1, \ldots, n$.
Let $x = \left({x_1, \ldots, x_n}\right)$ and $y = \left({y_1, \ldots, y_n}\right)$ be ordered tuples of real numbers.
Then:
- $\displaystyle \left({x + y}\right)^\alpha = \sum_{0 \mathop \le \beta \mathop \le \alpha} {\alpha \choose \beta} x^\beta y^{\alpha - \beta}$
where $\displaystyle {n \choose k}$ is a binomial coefficient.
Extended Binomial Theorem
Let $r, \alpha \in \C$ be complex numbers.
Let $z \in \C$ be a complex number such that $\cmod z < 1$.
Then:
- $\ds \paren {1 + z}^r = \sum_{k \mathop \in \Z} \dbinom r {\alpha + k} z^{\alpha + k}$
where $\dbinom r {\alpha + k}$ denotes a binomial coefficient.
Abel's Generalisation
- $\displaystyle \left({x + y}\right)^n = \sum_k \binom n k x \left({x - k z}\right)^{k - 1} \left({y + k z}\right)^{n - k}$
Hurwitz's Generalisation
- $\displaystyle \paren {x + y}^n = \sum x \paren {x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}^{\epsilon_1 + \cdots + \epsilon_n - 1} \paren {y - \epsilon_1 z_1 - \cdots - \epsilon_n z_n}^{n - \epsilon_1 - \cdots - \epsilon_n}$
where the summation ranges over all $2^n$ choices of $\epsilon_1, \ldots, \epsilon_n = 0$ or $1$ independently.
Examples
Cube of Sum
- $\paren {x + y}^3 = x^3 + 3 x^2 y + 3 x y^2 + y^3$
Cube of Difference
- $\paren {x - y}^3 = x^3 - 3 x^2 y + 3 x y^2 - y^3$
Fourth Power of Sum
- $\paren {x + y}^4 = x^4 + 4 x^3 y + 6 x^2 y^2 + 4 x y^3 + y^4$
Fourth Power of Difference
- $\paren {x - y}^4 = x^4 - 4 x^3 y + 6 x^2 y^2 - 4 x y^3 + y^4$
Fifth Power of Sum
- $\paren {x + y}^5 = x^5 + 5 x^4 y + 10 x^3 y^2 + 10 x^2 y^3 + 5 x y^4 + y^5$
Fifth Power of Difference
- $\paren {x - y}^5 = x^5 - 5 x^4 y + 10 x^3 y^2 - 10 x^2 y^3 + 5 x y^4 - y^5$
Sixth Power of Sum
- $\paren {x + y}^6 = x^6 + 6 x^5 y + 15 x^4 y^2 + 20 x^3 y^3 + 15 x^2 y^4 + 6 x y^5 + y^6$
Sixth Power of Difference
- $\paren {x - y}^6 = x^6 - 6 x^5 y + 15 x^4 y^2 - 20 x^3 y^3 + 15 x^2 y^4 - 6 x y^5 + y^6$
Power of $11$: $11^4$
- $11^4 = \left({10 + 1}\right)^4 = 14 \, 641$
Binomial Theorem: $\paren {1 + x}^7$
- $\paren {1 + x}^7 = 1 + 7 x + 21 x^2 + 35 x^3 + 35 x^4 + 21 x^5 + 7 x^6 + x^7$
Square Root of 2
- $\sqrt 2 = 2 \paren {1 - \dfrac 1 {2^2} - \dfrac 1 {2^5} - \dfrac 1 {2^7} - \dfrac 5 {2^{11} } - \cdots}$
Also known as
This result is also known as the binomial formula.