Binomial Theorem

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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 $\set {1, \ldots, n}$ such that $\alpha_j \ge 0$ for $j = 1, \ldots, n$.

Let $x = \tuple {x_1, \ldots, x_n}$ and $y = \tuple {y_1, \ldots, y_n}$ be ordered tuples of real numbers.

Then:

$\ds \paren {x + y}^\alpha = \sum_{0 \mathop \le \beta \mathop \le \alpha} \dbinom \alpha \beta x^\beta y^{\alpha - \beta}$

where $\dbinom n k$ 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

$\ds \paren {x + y}^n = \sum x \paren {x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}^{\epsilon_1 + \cdots + \epsilon_n - 1} \paren {y - \epsilon_1 z_1 - \cdots - \epsilon_n z_n}^{n - \epsilon_1 - \cdots - \epsilon_n}$

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.


Also see