Binomial approximation

Short description: Approximation of powers of some binomials

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

[math]\displaystyle{ (1 + x)^\alpha \approx 1 + \alpha x. }[/math]

It is valid when [math]\displaystyle{ |x|\lt 1 }[/math] and [math]\displaystyle{ |\alpha x| \ll 1 }[/math] where [math]\displaystyle{ x }[/math] and [math]\displaystyle{ \alpha }[/math] may be real or complex numbers.

The benefit of this approximation is that [math]\displaystyle{ \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]\displaystyle{ x\gt -1 }[/math] and [math]\displaystyle{ \alpha \geq 1 }[/math].

Derivations

Using linear approximation

The function

[math]\displaystyle{ 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]\displaystyle{ f'(x) = \alpha (1 + x)^{\alpha - 1} }[/math]

and so

[math]\displaystyle{ f'(0) = \alpha. }[/math]

Thus

[math]\displaystyle{ 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]\displaystyle{ \frac{\alpha(\alpha - 1) x^2}{2} \cdot (1 + \zeta)^{\alpha - 2} }[/math] for some value of [math]\displaystyle{ \zeta }[/math] that lies between 0 and x. For example, if [math]\displaystyle{ x \lt 0 }[/math] and [math]\displaystyle{ \alpha \geq 2 }[/math], the error is at most [math]\displaystyle{ \frac{\alpha(\alpha - 1) x^2}{2} }[/math]. In little o notation, one can say that the error is [math]\displaystyle{ o(|x|) }[/math], meaning that [math]\displaystyle{ \lim_{x \to 0} \frac{\textrm{error}}{|x|} = 0 }[/math].

Using Taylor series

The function

[math]\displaystyle{ f(x) = (1+x)^\alpha }[/math]

where [math]\displaystyle{ x }[/math] and [math]\displaystyle{ \alpha }[/math] may be real or complex can be expressed as a Taylor series about the point zero.

[math]\displaystyle{ \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]\displaystyle{ |x| \lt 1 }[/math] and [math]\displaystyle{ |\alpha x| \ll 1 }[/math], then the terms in the series become progressively smaller and it can be truncated to

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

[math]\displaystyle{ (1 + x)^\alpha \approx e^{\alpha x} . }[/math]

Example

The binomial approximation for the square root, [math]\displaystyle{ \sqrt{1+x} \approx 1+x/2 }[/math], can be applied for the following expression,

[math]\displaystyle{ \frac{1}{\sqrt{a+b}} - \frac{1}{\sqrt{a-b}} }[/math]

where [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math] are real but [math]\displaystyle{ a \gg b }[/math].

The mathematical form for the binomial approximation can be recovered by factoring out the large term [math]\displaystyle{ a }[/math] and recalling that a square root is the same as a power of one half.

[math]\displaystyle{ \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]\displaystyle{ b }[/math] when [math]\displaystyle{ a \gg b }[/math] which is otherwise not obvious from the original expression.

Generalization

Further information: Binomial series

While the binomial approximation is linear, it can be generalized to keep the quadratic term in the Taylor series:

[math]\displaystyle{ (1+x)^\alpha \approx 1 + \alpha x + (\alpha/2) (\alpha-1) x^2 }[/math]

Applied to the square root, it results in:

[math]\displaystyle{ \sqrt{1+x} \approx 1 + x/2 - x^2 / 8. }[/math]

Quadratic example

Consider the expression:

[math]\displaystyle{ (1 + \epsilon)^n - (1 - \epsilon)^{-n} }[/math]

where [math]\displaystyle{ |\epsilon|\lt 1 }[/math] and [math]\displaystyle{ |n \epsilon| \ll 1 }[/math]. If only the linear term from the binomial approximation is kept [math]\displaystyle{ (1+x)^\alpha \approx 1 + \alpha x }[/math] then the expression unhelpfully simplifies to zero

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \epsilon }[/math] which is why it did not appear when only the linear terms in [math]\displaystyle{ \epsilon }[/math] were kept.