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

$M(p)=\int _{0}^{\infty }(1+\alpha x)^{-\gamma }x^{p-1}dx$

Let $y=\alpha x\,$

$M(p)=\alpha ^{-p}\int _{0}^{\infty }(1+y)^{-\gamma }y^{p-1}dy$

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

$M(p)=\alpha ^{-p}\int _{0}^{1}(1-z)^{\gamma -p-1}z^{p-1}dz$

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

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

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:

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

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