Editing Binomial theorem (section)

== Applications ==

=== Multiple-angle identities ===
For the [[complex numbers]] the binomial theorem can be combined with [[de Moivre's formula]] to yield [[List of trigonometric identities#Multiple-angle formulae|multiple-angle formulas]] for the [[sine]] and [[cosine]]. According to De Moivre's formula,
<math display="block">\cos\left(nx\right)+i\sin\left(nx\right) = \left(\cos x+i\sin x\right)^n.</math>

Using the binomial theorem, the expression on the right can be expanded, and then the real and imaginary parts can be taken to yield formulas for {{math|cos(''nx'')}} and {{math|sin(''nx'')}}. For example, since
<math display="block">\left(\cos x + i\sin x\right)^2 = \cos^2 x + 2i \cos x \sin x - \sin^2 x
= (\cos^2 x-\sin^2 x) + i(2\cos x\sin x),</math>
But De Moivre's formula identifies the left side with <math>(\cos x+i\sin x)^2 = \cos(2x)+i\sin(2x)</math>, so
<math display="block">\cos(2x) = \cos^2 x - \sin^2 x \quad\text{and}\quad\sin(2x) = 2 \cos x \sin x,</math>
which are the usual double-angle identities. Similarly, since
<math display="block">\left(\cos x + i\sin x\right)^3 = \cos^3 x + 3i \cos^2 x \sin x - 3 \cos x \sin^2 x - i \sin^3 x,</math>
De Moivre's formula yields
<math display="block">\cos(3x) = \cos^3 x - 3 \cos x \sin^2 x \quad\text{and}\quad \sin(3x) = 3\cos^2 x \sin x - \sin^3 x.</math>
In general,
<math display="block">\cos(nx) = \sum_{k\text{ even}} (-1)^{k/2} {n \choose k}\cos^{n-k} x \sin^k x</math>
and
<math display="block">\sin(nx) = \sum_{k\text{ odd}} (-1)^{(k-1)/2} {n \choose k}\cos^{n-k} x \sin^k x.</math>There are also similar formulas using [[Chebyshev polynomials]].

=== Series for ''e'' ===
The [[e (mathematical constant)|number {{mvar|e}}]] is often defined by the formula
<math display="block">e = \lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n.</math>

Applying the binomial theorem to this expression yields the usual [[infinite series]] for {{mvar|e}}. In particular:
<math display="block">\left(1 + \frac{1}{n}\right)^n = 1 + {n \choose 1}\frac{1}{n} + {n \choose 2}\frac{1}{n^2} + {n \choose 3}\frac{1}{n^3} + \cdots + {n \choose n}\frac{1}{n^n}.</math>

The {{mvar|k}}th term of this sum is
<math display="block">{n \choose k}\frac{1}{n^k} = \frac{1}{k!}\cdot\frac{n(n-1)(n-2)\cdots (n-k+1)}{n^k}</math>

As {{math|''n'' → ∞}}, the rational expression on the right approaches {{math|1}}, and therefore
<math display="block">\lim_{n\to\infty} {n \choose k}\frac{1}{n^k} = \frac{1}{k!}.</math>

This indicates that {{mvar|e}} can be written as a series:
<math display="block">e=\sum_{k=0}^\infty\frac{1}{k!}=\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots.</math>

Indeed, since each term of the binomial expansion is an [[Monotonic function|increasing function]] of {{mvar|n}}, it follows from the [[monotone convergence theorem]] for series that the sum of this infinite series is equal to&nbsp;{{mvar|e}}.

=== Probability ===
The binomial theorem is closely related to the probability mass function of the [[negative binomial distribution]]. The probability of a (countable) collection of independent Bernoulli trials <math>\{X_t\}_{t\in S}</math> with probability of success <math>p\in [0,1]</math> all not happening is
:<math> P\biggl(\bigcap_{t\in S} X_t^C\biggr) = (1-p)^{|S|} = \sum_{n=0}^{|S|} {|S| \choose n} (-p)^n.</math>
An upper bound for this quantity is <math> e^{-p|S|}.</math><ref>{{Cite book |title=Elements of Information Theory |chapter=Data Compression |last1=Cover |first1=Thomas M. |author1-link=Thomas M. Cover |last2=Thomas |first2=Joy A. |author2-link=Joy A. Thomas |date=1991 |publisher=Wiley |isbn=9780471062592 |at=Ch. 5, {{pgs|78–124}} |doi=10.1002/0471200611.ch5}}<!-- a specific page number would be helpful. previously this citation noted p. 320 but that's not in this chapter. --> </ref>