# Binomial approximation

The binomial approximation is useful for approximately calculating powers of numbers close to 1. It states that if $x$ is a real number close to 0 and $\alpha$ is a real number, then

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

This approximation can be obtained by using the binomial theorem and ignoring the terms beyond the first two.

The left-hand side of this relation is always greater than or equal to the right-hand side for $x>-1$ and $\alpha$ a non-negative integer, by Bernoulli's inequality.

## Derivation using linear approximation

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

When x = 0:

$f'(0) = \alpha.$

Using linear approximation:

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

## Derivation using Mellin transform

$M(p) = \int^\infty_0 (1+\alpha x)^{-\gamma}x^{p-1}dx$

Let $y=\alpha x\,$

$M(p) = \alpha^{-p}\int^\infty_0 (1+y)^{-\gamma}y^{p-1}dy$

Let $y=z/(1-z)$

$M(p) = \alpha^{-p}\int^1_0(1-z)^{\gamma-p-1}z^{p-1} dz$

$= \alpha^{-p}B(\gamma-p,p)\,$

$= \alpha^{-p}\frac{\Gamma(\gamma-p)\Gamma(p)}{\Gamma(\gamma)}.$

Using the inverse Mellin transform:

$(1+\alpha x)^{-\gamma}=\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}(x\alpha)^{-p}\frac{\Gamma(\gamma-p)\Gamma(p)}{\Gamma(\gamma)}dp$

Closing this integral to the left, which converges for $|\alpha x|<1\,$, we get:

$(1+\alpha x)^{-\gamma}=\Sigma_{n=0}^{\infty}(\alpha x)^n \frac{(-1)^n}{n!}\frac{\Gamma(\gamma+n)}{\Gamma(\gamma)}$

$=1-\alpha x \gamma+(1/2)(\alpha x)^2 (\gamma+1)\gamma-...\,$