# 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 $\alpha \geq 1$.

## 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
    $f'(x)=\alpha (1+x)^{\alpha -1}$
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 α(α−1)x$^2$/2 ⋅(1+ζ)$^{α−2}${\textstyle {\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 α(α−1)x$^2$/2 {\textstyle {\frac {\alpha (\alpha -1)x^{2}}{2}}}. In little o notation, one can say that the error is $o(|x|)$, meaning that lim$_{x→0}$ error/|x|=0{\textstyle \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.
    ${\begin{aligned}f(x)&=\sum _{n=0}^{\infty }{\frac {f^{(n)}(0)}{n!}}x^{n}\\f(x)&=f(0)+f'(0)x+{\frac {1}{2}}f''(0)x^{2}+{\frac {1}{6}}f'''(0)x^{3}+{\frac {1}{24}}f^{(4)}(0)x^{4}+\cdots \\(1+x)^{\alpha }&=1+\alpha x+{\frac {1}{2}}\alpha (\alpha -1)x^{2}+{\frac {1}{6}}\alpha (\alpha -1)(\alpha -2)x^{3}+{\frac {1}{24}}\alpha (\alpha -1)(\alpha -2)(\alpha -3)x^{4}+\cdots \end{aligned}}$
If $|x|<1$ 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.
    ${\begin{aligned}{\frac {1}{\sqrt {a+b}}}-{\frac {1}{\sqrt {a-b}}}&={\frac {1}{\sqrt {a}}}\left(\left(1+{\frac {b}{a}}\right)^{-1/2}-\left(1-{\frac {b}{a}}\right)^{-1/2}\right)\\&\approx {\frac {1}{\sqrt {a}}}\left(\left(1+\left(-{\frac {1}{2}}\right){\frac {b}{a}}\right)-\left(1-\left(-{\frac {1}{2}}\right){\frac {b}{a}}\right)\right)\\&\approx {\frac {1}{\sqrt {a}}}\left(1-{\frac {b}{2a}}-1-{\frac {b}{2a}}\right)\\&\approx -{\frac {b}{a{\sqrt {a}}}}\end{aligned}}$
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
    ${\begin{aligned}(1+\epsilon )^{n}-(1-\epsilon )^{-n}&\approx (1+n\epsilon )-(1-(-n)\epsilon )\\&\approx (1+n\epsilon )-(1+n\epsilon )\\&\approx 0.\end{aligned}}$
While the expression is small, it is not exactly zero. So now, keeping the quadratic term:
    ${\begin{aligned}(1+\epsilon )^{n}-(1-\epsilon )^{-n}&\approx \left(1+n\epsilon +{\frac {1}{2}}n(n-1)\epsilon ^{2}\right)-\left(1+(-n)(-\epsilon )+{\frac {1}{2}}(-n)(-n-1)(-\epsilon )^{2}\right)\\&\approx \left(1+n\epsilon +{\frac {1}{2}}n(n-1)\epsilon ^{2}\right)-\left(1+n\epsilon +{\frac {1}{2}}n(n+1)\epsilon ^{2}\right)\\&\approx {\frac {1}{2}}n(n-1)\epsilon ^{2}-{\frac {1}{2}}n(n+1)\epsilon ^{2}\\&\approx {\frac {1}{2}}n\epsilon ^{2}((n-1)-(n+1))\\&\approx -n\epsilon ^{2}\end{aligned}}$
This result is quadratic in $\epsilon$ which is why it did not appear when only the linear terms in $\epsilon$ were kept.