# What type of expansion is this?

1. Feb 27, 2012

### Chuck88

When I am reading the paper about Rayleigh instability, I found this type of expanding method.

$$\sqrt{1+(\frac{2\pi\delta}{\lambda})^2 \cos^2(\frac{2\pi x}{\lambda})} = 1 + \frac{1}{2}(\frac{2\pi\delta}{\lambda})^2\cos^2 (\frac{2\pi x}{\lambda}) + \cdots$$

Can someone tell me what type of expansion is this?

2. Feb 27, 2012

### AlephZero

It's the Binomial series expansion of $(1+x)^{1/2}$

3. Feb 27, 2012

### Chuck88

If we suppose that $f(x) = (\frac{2\pi\delta}{\lambda})^2 \cos^2(\frac{2\pi x}{\lambda})$, is it right that we could take derivative with respect to $f(x)$ to get the Taylor Expansion? The first order derivative I mean is presented below:

$$\frac{d(1 + f(x))^{\frac{1}{2}}}{df(x)} = \frac{1}{2} \frac{1}{\sqrt{1 + f(x)}}$$

4. Feb 27, 2012

### AlephZero

I'm not sure where that is leading to. I meant
$(1+x)^n = 1 + n x + n(n-1)x^2 / 2! + n(n-1)(n-2)x^3/3! + \dots$
where x is the trig function and n = 1/2.

5. Feb 27, 2012

### Chuck88

OK. Now I comprehend. Thanks a lot.