Binomial approximation

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


 * $$ (1 + x)^\alpha \approx 1 + \alpha x.$$

It is valid when $$|x|<1$$ and $$|\alpha x| \ll 1$$ where $$x$$ and $$\alpha$$ may be real or complex numbers.

The benefit of this approximation is that $$\alpha$$ 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 $$x>-1$$ and $$\alpha \geq 1$$.

Using linear approximation
The function
 * $$ f(x) = (1 + x)^{\alpha}$$

is a smooth function for x near 0. Thus, standard linear approximation tools from calculus apply: one has
 * $$ f'(x) = \alpha (1 + x)^{\alpha - 1}$$

and so
 * $$ f'(0) = \alpha.$$

Thus
 * $$ f(x) \approx f(0) + f'(0)(x - 0) = 1 + \alpha x.$$

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

Using Taylor series
The function
 * $$ f(x) = (1+x)^\alpha $$

where $$x$$ and $$\alpha$$ may be real or complex can be expressed as a Taylor series about the point zero.


 * $$\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}$$

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

This result from the binomial approximation can always be improved by keeping additional terms from the Taylor series above. This is especially important when $$|\alpha x|$$ 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 $$|x| \ll 1$$ is a sufficient condition for the binomial approximation. A simple counterexample is to let $$x=10^{-6}$$ and $$\alpha=10^7$$. In this case $$(1+x)^\alpha > 22,000$$ but the binomial approximation yields $$1 + \alpha x = 11$$. For small $$|x|$$ but large $$|\alpha x|$$, a better approximation is:


 * $$ (1 + x)^\alpha \approx e^{\alpha x} .$$

Example
The binomial approximation for the square root, $$\sqrt{1+x} \approx 1+x/2$$, can be applied for the following expression,
 * $$	\frac{1}{\sqrt{a+b}} - \frac{1}{\sqrt{a-b}} $$

where $$a$$ and $$b$$ are real but $$a \gg b$$.

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


 * $$\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}$$

Evidently the expression is linear in $$b$$ when $$a \gg b$$ 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:
 * $$ (1+x)^\alpha \approx 1 + \alpha x + (\alpha/2) (\alpha-1) x^2$$

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

Quadratic example
Consider the expression:
 * $$	(1 + \epsilon)^n - (1 - \epsilon)^{-n} $$

where $$|\epsilon|<1$$ and $$|n \epsilon| \ll 1$$. If only the linear term from the binomial approximation is kept $$(1+x)^\alpha \approx 1 + \alpha x$$ then the expression unhelpfully simplifies to zero
 * $$\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}$$

While the expression is small, it is not exactly zero. So now, keeping the quadratic term:
 * $$\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}$$

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