Binomial approximation explained

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

( 1 + x) ^{\alpha} ≈ 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.^[1]

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 .

Derivations

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 hasf'(x) = \alpha ( 1 + x) ^{\alpha} and sof'( 0 ) = \alpha. Thusf(x) ≈ f( 0 ) + f'( 0 )(x - 0 ) = 1 + \alpha x.

By Taylor's theorem, the error in this approximation is equal to $\backslash frac\; \backslash cdot\; (1\; +\; \backslash zeta)^$ for some value of

\zeta that lies between 0 and . For example, if x < 0 and \alpha \geq 2 , the error is at most $\backslash frac$. In little o notation, one can say that the error is o(|x|) , meaning that $\backslash lim\_\; \backslash frac$ = 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 ^{inf ty} _{n =0} (f ^{(n)} ( 0 ))/(n!) x ^{n\\ f(x)} & = f( 0 ) + f'( 0 ) x + (1)/(2) f''( 0 ) x ^{2} + (1)/(6) f'''( 0 ) x ^{3} + (1)/(24) f ^{(4)} ( 0 ) x ^{4} + … \\ ( 1+ x) ^{\alpha} & = 1 + \alpha x + (1)/(2) \alpha (\alpha -1 ) x ^{2} + (1)/(6) \alpha (\alpha -1 )(\alpha -2 )x ^{3} + (1)/(24) \alpha (\alpha -1 )(\alpha -2 )(\alpha -3 )x ^{4} + … \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} ≈ 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} ≈ e ^{\alpha} .

Example

The binomial approximation for the square root,

\sqrt{ 1+ x} ≈ 1+ x/ 2 , can be applied for the following expression,(1)/(\sqrt{a + b) } - \frac 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} (1)/(\sqrt{a + b) } - \frac &= \frac \left(\left(1+\frac\right)^ - \left(1-\frac\right)^\right)\\ &\approx\frac \left(\left(1+\left(-\frac\right)\frac\right) - \left(1-\left(-\frac\right)\frac\right)\right) \\ &\approx\frac \left(1-\frac - 1 -\frac\right) \\ &\approx -\frac\end

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} ≈ 1 + \alpha x + (\alpha/ 2 ) (\alpha -1 ) x ^{2}

Applied to the square root, it results in:

\sqrt{ 1+ x} ≈ 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} ≈ 1 + \alpha x then the expression unhelpfully simplifies to zero\begin{align} ( 1 + \epsilon) ^{n} - ( 1 - \epsilon) ^{-n} & ≈ ( 1+ n \epsilon) - ( 1 - ( - n) \epsilon)\\ & ≈ ( 1+ n \epsilon) - ( 1 + n \epsilon)\\ & ≈ 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} & ≈ \left( 1+ n \epsilon + (1)/(2) n (n -1 ) \epsilon ^{2\right)} - \left( 1 + ( - n)( - \epsilon) + (1)/(2) ( - n) ( - n -1 ) ( - \epsilon) ^{2\right)\\ & ≈} \left( 1+ n \epsilon + (1)/(2) n (n -1 ) \epsilon ^{2\right)} - \left( 1 + n \epsilon + (1)/(2) n (n +1 ) \epsilon ^{2\right)\\ & ≈} (1)/(2) n (n -1 ) \epsilon ^{2} - (1)/(2) n (n +1 ) \epsilon ^{2\\ & ≈} (1)/(2) n \epsilon ^{2} ((n -1 ) - (n +1 )) \\ & ≈ - 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.

Notes and References

1. For example calculating the multipole expansion. Book: Griffiths, D. . 1999 . Introduction to Electrodynamics . Pearson Education, Inc. . Third . 146–148.