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What type of expansion is this?

by Chuck88
Tags: expansion
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Chuck88
#1
Feb27-12, 09:56 AM
P: 37
When I am reading the paper about Rayleigh instability, I found this type of expanding method.

[tex]
\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
[/tex]

Can someone tell me what type of expansion is this?
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AlephZero
#2
Feb27-12, 10:18 AM
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It's the Binomial series expansion of ##(1+x)^{1/2}##
Chuck88
#3
Feb27-12, 10:27 AM
P: 37
Quote Quote by AlephZero View Post
It's the Binomial series expansion of ##(1+x)^{1/2}##
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:

[tex]
\frac{d(1 + f(x))^{\frac{1}{2}}}{df(x)} = \frac{1}{2} \frac{1}{\sqrt{1 + f(x)}}
[/tex]

AlephZero
#4
Feb27-12, 01:31 PM
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What type of expansion is this?

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.
Chuck88
#5
Feb27-12, 07:30 PM
P: 37
Quote Quote by AlephZero View Post
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.
OK. Now I comprehend. Thanks a lot.


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