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((Short description|Approximation of powers of some binomials))
((distinguish|Binomial distribution#Normal approximation))
((refimprove|date=February 2016))


The '''binomial approximation''' is useful for approximately calculating [[exponentiation|powers]] of sums of 1 and a small number ''x''.  It states that

: <math> (1 + x)^\alpha \approx 1 + \alpha x.</math>

It is valid when <math>|x|<1</math> and <math>|\alpha x| \ll 1</math> where <math>x</math> and <math>\alpha</math> may be [[real number|real]] or [[complex number]]s.

The benefit of this approximation is that <math>\alpha</math> is converted from an exponent to a multiplicative factor.  This can greatly simplify mathematical expressions (as in [[#Example|the example below]]) and is a common tool in physics.<ref>For example calculating the [[multipole expansion]]. ((cite book |last=Griffiths |first=D.  |year=1999 |title=Introduction to Electrodynamics |publisher=Pearson Education, Inc. |edition=Third |pages=146–148))</ref>

The approximation can be proven several ways, and is closely related to the [[binomial theorem]].  By [[Bernoulli's inequality]], the left-hand side of the approximation is greater than or equal to the right-hand side whenever <math>x>-1</math> and <math>\alpha \geq 1</math>.

== Derivations ==
=== Using linear approximation ===
The function
:<math> f(x) = (1 + x)^{\alpha}</math>
is a [[smooth function]] for ''x'' near 0.  Thus, standard [[linear approximation]] tools from [[calculus]] apply: one has
:<math> f'(x) = \alpha (1 + x)^{\alpha - 1}</math>
and so
:<math> f'(0) = \alpha.</math>
Thus
:<math> f(x) \approx f(0) + f'(0)(x - 0) = 1 + \alpha x.</math>

By [[Taylor's theorem]], the error in this approximation is equal to <math display="inline"> \frac{\alpha(\alpha - 1) x^2}{2} \cdot (1 + \zeta)^{\alpha - 2}</math> for some value of <math>\zeta</math> that lies between 0 and ((mvar|x)).  For example, if <math> x < 0 </math> and <math>\alpha \geq 2</math>, the error is at most <math display="inline"> \frac{\alpha(\alpha - 1) x^2}{2}</math>.  In [[Little O notation|little o notation]], one can say that the error is <math>o(|x|)</math>, meaning that <math display="inline"> \lim_{x \to 0} \frac{\textrm{error)){|x|} = 0</math>.

=== Using Taylor series ===
The function 
:<math> f(x) = (1+x)^\alpha </math>
where <math>x</math> and <math>\alpha</math> may be real or complex can be expressed as a [[Taylor series]] about the point zero.

:<math>\begin{align}
f(x) &= \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\\
f(x) &= f(0) + f'(0) x + \frac{1}{2} f''(0) x^2 + \frac{1}{6} f'''(0) x^3 + \frac{1}{24} f^{(4)}(0) x^4 + \cdots\\
(1+x)^{\alpha} &= 1 + \alpha x + \frac{1}{2} \alpha (\alpha-1) x^2 + \frac{1}{6} \alpha (\alpha-1)(\alpha-2)x^3 + \frac{1}{24} \alpha (\alpha-1)(\alpha-2)(\alpha-3)x^4 + \cdots
\end{align}</math>

If <math>|x| < 1</math> and <math>|\alpha x| \ll 1</math>, then the terms in the series become progressively smaller and it can be truncated to
:<math>(1+x)^\alpha \approx 1 + \alpha x .</math>

This result from the binomial approximation can always be improved by keeping additional terms from the Taylor series above.  This is especially important when <math>|\alpha x|</math> starts to approach one, or when evaluating a more complex expression where the first two terms in the Taylor series cancel ([[#Quadratic example|see example]]).

Sometimes it is wrongly claimed that <math>|x| \ll 1</math> is a sufficient condition for the binomial approximation.  A simple counterexample is to let <math>x=10^{-6}</math> and <math>\alpha=10^7</math>.  In this case <math>(1+x)^\alpha > 22,000</math> but the binomial approximation yields <math>1 + \alpha x = 11</math>. For small <math>|x|</math> but large  <math>|\alpha x|</math>, a better approximation is:

: <math> (1 + x)^\alpha \approx e^{\alpha x} .</math>

== Example ==
The binomial approximation for the [[square root]], <math>\sqrt{1+x} \approx 1+x/2</math>, can be applied for the following expression,
:<math>	\frac{1}{\sqrt{a+b)) - \frac{1}{\sqrt{a-b)) </math>
where <math>a</math> and <math>b</math> are real but <math>a \gg b</math>.

The mathematical form for the binomial approximation can be recovered by factoring out the large term <math>a</math> and recalling that a square root is the same as a power of one half.

:<math>\begin{align}
 \frac{1}{\sqrt{a+b)) - \frac{1}{\sqrt{a-b)) &= \frac{1}{\sqrt{a)) \left(\left(1+\frac{b}{a}\right)^{-1/2} - \left(1-\frac{b}{a}\right)^{-1/2}\right)\\
 &\approx\frac{1}{\sqrt{a)) \left(\left(1+\left(-\frac{1}{2}\right)\frac{b}{a}\right) - \left(1-\left(-\frac{1}{2}\right)\frac{b}{a}\right)\right) \\
 &\approx\frac{1}{\sqrt{a)) \left(1-\frac{b}{2a} - 1 -\frac{b}{2a}\right) \\
 &\approx -\frac{b}{a \sqrt{a))
\end{align}</math>

Evidently the expression is linear in <math>b</math> when <math>a \gg b</math> which is otherwise not obvious from the original expression.

==Generalization==
((further|Binomial series))

While the binomial approximation is linear, it can be generalized to keep the quadratic term in the Taylor series:
:<math> (1+x)^\alpha \approx 1 + \alpha x + (\alpha/2) (\alpha-1) x^2</math> 

Applied to the square root, it results in:
:<math>\sqrt{1+x} \approx 1 + x/2 - x^2 / 8.</math>

===Quadratic example===
Consider the expression:
:<math>	(1 + \epsilon)^n - (1 - \epsilon)^{-n} </math>
where <math>|\epsilon|<1</math> and <math>|n \epsilon| \ll 1</math>.  If only the linear term from the binomial approximation is kept <math>(1+x)^\alpha \approx 1 + \alpha x</math> then the expression unhelpfully simplifies to zero
:<math>\begin{align}
(1 + \epsilon)^n - (1 - \epsilon)^{-n} &\approx (1+ n \epsilon) - (1 - (-n) \epsilon)\\
&\approx (1+ n \epsilon) - (1 + n \epsilon)\\
&\approx 0 .
\end{align}</math>

While the expression is small, it is not exactly zero. 
So now, keeping the quadratic term:
:<math>\begin{align}
(1+\epsilon)^n - (1 - \epsilon)^{-n}&\approx \left(1+ n \epsilon + \frac{1}{2} n (n-1) \epsilon^2\right) - \left(1 + (-n)(-\epsilon) + \frac{1}{2} (-n) (-n-1) (-\epsilon)^2\right)\\
&\approx \left(1+ n \epsilon + \frac{1}{2} n (n-1) \epsilon^2\right) - \left(1 + n \epsilon + \frac{1}{2} n (n+1) \epsilon^2\right)\\
&\approx \frac{1}{2} n (n-1) \epsilon^2 - \frac{1}{2} n (n+1) \epsilon^2\\
&\approx \frac{1}{2} n \epsilon^2 ((n-1) - (n+1)) \\
&\approx - n \epsilon^2 
\end{align}</math>

This result is quadratic in <math>\epsilon</math> which is why it did not appear when only the linear terms in <math>\epsilon</math> were kept.

==References==
((Reflist))

[[Category:Factorial and binomial topics]]
[[Category:Approximations]]

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