# Binomial approximation

The binomial approximation is useful for approximately calculating powers of sums of a small number and 1. It states that if ${\displaystyle |x|<1}$ and ${\displaystyle |\alpha x|}$${\displaystyle 1}$ where ${\displaystyle x}$ and ${\displaystyle \alpha }$ are real or complex numbers, then

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

If either ${\displaystyle x}$ or ${\displaystyle \alpha }$ are complex then the absolute value denotes the modulus of the complex number.

The benefit of this approximation is that ${\displaystyle \alpha }$ is converted from a power to multiplicative factor. This can greatly simplify mathematical expressions (see example) and is a common tool in physics.[1]

This approximation can be proven several ways including the binomial theorem and ignoring the terms beyond the first two.

By Bernoulli's inequality, the left-hand side of this relation is greater than or equal to the right-hand side whenever ${\displaystyle x>-1}$ and ${\displaystyle \alpha \geq 1}$.

## Derivation using linear approximation

The function

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

is a smooth function for x near 0. Thus, standard linear approximation tools from calculus apply: one has

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

and so

${\displaystyle f'(0)=\alpha .}$

Thus

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

## Derivation using Taylor Series

The function

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

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

{\displaystyle {\begin{aligned}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{aligned}}}

If ${\displaystyle |x|<1}$ and ${\displaystyle |\alpha x|}$${\displaystyle 1}$, then the terms in the series become progressively smaller and it can be truncated to

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

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

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

### Example simplification

Consider the following expression where ${\displaystyle a}$ and ${\displaystyle b}$ are real but ${\displaystyle a}$${\displaystyle b}$.

${\displaystyle {\frac {1}{\sqrt {a+b}}}-{\frac {1}{\sqrt {a-b}}}}$

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

{\displaystyle {\begin{aligned}{\frac {1}{\sqrt {a+b}}}-{\frac {1}{\sqrt {a-b}}}&={\frac {1}{\sqrt {a}}}{\Big (}{\big (}1+{\frac {b}{a}}{\big )}^{-1/2}-{\big (}1-{\frac {b}{a}}{\big )}^{-1/2}{\Big )}\\&\approx {\frac {1}{\sqrt {a}}}{\Big (}{\big (}1+{\big (}-{\frac {1}{2}}{\big )}{\frac {b}{a}}{\big )}-{\big (}1-{\big (}-{\frac {1}{2}}{\big )}{\frac {b}{a}}{\big )}{\Big )}\\&\approx {\frac {1}{\sqrt {a}}}{\Big (}1-{\frac {b}{2a}}-1-{\frac {b}{2a}}{\Big )}\\&\approx -{\frac {b}{a{\sqrt {a}}}}\end{aligned}}}

Evidently the expression is linear in ${\displaystyle b}$ when ${\displaystyle a}$${\displaystyle b}$ which is otherwise not obvious from the original expression.

### Example keeping the quadratic term

Consider the expression:

${\displaystyle (1+\epsilon )^{n}-(1-\epsilon )^{-n}}$

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

{\displaystyle {\begin{aligned}(1+\epsilon )^{n}-(1-\epsilon )^{-n}&\approx (1+n\epsilon )-(1-(-n)\epsilon )\\&\approx (1+n\epsilon )-(1+n\epsilon )\\&\approx 0\end{aligned}}}.

While the expression is small, it is not exactly zero. It is possible to extract a nonzero approximate solution by keeping the quadratic term in the Taylor Series, i.e. ${\displaystyle (1+x)^{\alpha }\approx 1+\alpha x+{\frac {1}{2}}\alpha (\alpha -1)x^{2}}$ so now,

{\displaystyle {\begin{aligned}(1+\epsilon )^{n}-(1-\epsilon )^{-n}&\approx (1+n\epsilon +{\frac {1}{2}}n(n-1)\epsilon ^{2})-(1+(-n)(-\epsilon )+{\frac {1}{2}}(-n)(-n-1)(-\epsilon )^{2})\\&\approx (1+n\epsilon +{\frac {1}{2}}n(n-1)\epsilon ^{2})-(1+n\epsilon +{\frac {1}{2}}n(n+1)\epsilon ^{2})\\&\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{aligned}}}

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

## References

1. ^ For example calculating the multipole expansion. Griffiths, D. (1999). Introduction to Electrodynamics (Third ed.). Pearson Education, Inc. pp. 146–148.