Binomial approximation

Short description: Approximation of powers of some binomials

Not to be confused with Binomial distribution § Normal 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 $\alpha \ge 1$.

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 has

$f^{\prime }(x)=\alpha (1+x{)}^{\alpha -1}$

and so

$f^{\prime }(0)=\alpha .$

Thus

$f(x)\approx f(0)+f^{\prime }(0)(x-0)=1+\alpha x.$

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 $\alpha \ge 2$, 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 $\underset{x\to 0}{\lim }\frac{\text{<mrow data-mjx-texclass="ORD">error</mrow>}}{|x|}=0$.

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.

$\begin{array}{rl}f(x)& =\sum _{n=0}^{\mathrm{\infty }}\frac{f^{(n)}(0)}{n!}{x}^n\\ f(x)& =f(0)+f^{\prime }(0)x+\frac{1}{2}{f}^{″}(0)x^2+\frac{1}{6}{f}^{‴}(0)x^3+\frac{1}{24}{f}^{(4)}(0)x^4+\cdots \\ (1+x{)}^{\alpha }& =1+\alpha x+\frac{1}{2}\alpha (\alpha -1)x^2+\frac{1}{6}\alpha (\alpha -1)(\alpha -2)x^3+\frac{1}{24}\alpha (\alpha -1)(\alpha -2)(\alpha -3)x^4+\cdots \end{array}$

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.

$\begin{array}{rl}\frac{1}{\sqrt{a+b}}-\frac{1}{\sqrt{a-b}}& =\frac{1}{\sqrt{a}}({(1+\frac{b}{a})}^{-1/2}-{(1-\frac{b}{a})}^{-1/2})\\ & \approx \frac{1}{\sqrt{a}}((1+(-\frac{1}{2})\frac{b}{a})-(1-(-\frac{1}{2})\frac{b}{a}))\\ & \approx \frac{1}{\sqrt{a}}(1-\frac{b}{2a}-1-\frac{b}{2a})\\ & \approx -\frac{b}{a\sqrt{a}}\end{array}$

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 a quadratic approximation keeping the second 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:

$(1+ϵ{)}^n-(1-ϵ{)}^{-n}$

where $|ϵ|<1$ and $|nϵ|\ll 1$. 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

$\begin{array}{rl}(1+ϵ{)}^n-(1-ϵ{)}^{-n}& \approx (1+nϵ)-(1-(-n)ϵ)\\ & \approx (1+nϵ)-(1+nϵ)\\ & \approx 0.\end{array}$

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

$\begin{array}{rl}(1+ϵ{)}^n-(1-ϵ{)}^{-n}& \approx (1+nϵ+\frac{1}{2}n(n-1){ϵ}^2)-(1+(-n)(-ϵ)+\frac{1}{2}(-n)(-n-1)(-ϵ{)}^2)\\ & \approx (1+nϵ+\frac{1}{2}n(n-1){ϵ}^2)-(1+nϵ+\frac{1}{2}n(n+1){ϵ}^2)\\ & \approx \frac{1}{2}n(n-1){ϵ}^2-\frac{1}{2}n(n+1){ϵ}^2\\ & \approx \frac{1}{2}n{ϵ}^2((n-1)-(n+1))\\ & \approx -n{ϵ}^2\end{array}$

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