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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 α 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 α a non-negative integer, by Bernoulli's inequality.

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-...\,


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