Английская Википедия:Binomial approximation

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The binomial approximation is useful for approximately calculating 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 or complex numbers.

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 the example below) and is a common tool in physics.

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 Шаблон:Mvar. 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, 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 (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

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.