# [SOLVED]Binomial expansion

#### dwsmith

##### Well-known member
Use the binomial expansion to give the approximation $\frac{1}{\sqrt{1 - A^2u^2}}\approx 1 + \frac{1}{2}A^2u^2$

How can this be done?
Using the definition for (x - y), we have
$$(x - y)^n = \sum_{k = 1}^{n}(-1)^k\binom{n}{k}x^{n - k}y^{k}$$
but $n\notin\mathbb{Z}$.

#### dwsmith

##### Well-known member
Re: binomial expansion

Use the binomial expansion to give the approximation $\frac{1}{\sqrt{1 - A^2u^2}}\approx 1 + \frac{1}{2}A^2u^2$

How can this be done?
Using the definition for (x - y), we have
$$(x - y)^n = \sum_{k = 1}^{n}(-1)^k\binom{n}{k}x^{n - k}y^{k}$$
but $n\notin\mathbb{Z}$.
It has to be the Taylor series.

#### Opalg

##### MHB Oldtimer
Staff member
Re: binomial expansion

Use the binomial expansion to give the approximation $\frac{1}{\sqrt{1 - A^2u^2}}\approx 1 + \frac{1}{2}A^2u^2$

How can this be done?
Using the definition for (x - y), we have
$$(x - y)^n = \sum_{k = 1}^{n}(-1)^k\binom{n}{k}x^{n - k}y^{k}$$
but $n\notin\mathbb{Z}$.
It is the generalised ("Newton") binomial expansion $(1+x)^s = 1 + sx + \frac{s(s-1)}{2!}x^2 + \frac{s(s-1)(s-2)}{3!}x^3 + \ldots$, which is valid for any real number $s$, provided that $|x|<1$ (because it is an infinite series and you need that condition in order for it to converge). In this case you take $s = -\frac12$ and $x = -A^2u^2$.