# Explanation on Taylor expansions needed

1. Jun 24, 2011

### Stalafin

I have a question about Taylor expanding functions. For both cases I can't get my head around why things are the way they are. I just don't see how one would perform Taylor expansions like that.

The first:
The starting point of a symmetry operations is the following expansion:
$$f(r+a) = f(r) + \left(a\cdot\frac{\partial}{\partial r}\right) f(r) + \frac{1}{2} \left(a\cdot\frac{\partial}{\partial r}\right)^2 f(r) + \frac{1}{3!} \left(a\cdot\frac{\partial}{\partial r}\right)^2 f(r) + \ldots$$

The second:
This one comes from Landau's book, first chapter, fourth section:
Given the equation $L = L(v^{\prime 2})$ where $\boldsymbol{v}^\prime = \boldsymbol{v} + \boldsymbol{\varepsilon}$ and $\boldsymbol{v}^2 = v^2$ s.t. $L(v^{\prime 2}) = L(v^2 + 2\boldsymbol{v}\boldsymbol{\varepsilon} + \boldsymbol{\varepsilon}^2)$: why does expanding in powers of $\boldsymbol{\varepsilon}$ and neglecting terms above first order lead to:
$$L(v^{\prime 2}) = L(v^2)+\frac{\partial L}{\partial v^2} 2\boldsymbol{v}\cdot \boldsymbol{\varepsilon}$$

Some insight into why this is would be greatly appreciated. :-)

2. Jun 24, 2011

### Stalafin

Okay, the first and the second one are equivalent if I set $r=v^2$, $a=2\boldsymbol{v}\boldsymbol{\varepsilon}$ and disregard terms $\boldsymbol{\varepsilon}^2$. I still do not quite see how to get the first equation done properly.

For a normal Taylor expansion (up to first order), we have:
$$f(x) = f(c) + (x-c) \frac{\operatorname{d}}{\operatorname{d}x}f(c) + \ldots$$

If I now identify $x=r+a$ and $c=a$ I recover most of the terms I need, with the exception of $\frac{\operatorname{d}}{\operatorname{d}r+a}f(r)$.

Suggestions?

3. Jun 24, 2011

### micromass

Hi Stalafin!

This is an abuse of notation and an abuse that I don't like at all. Basically, this equation has two kind of r's: an r that represents a value and an r that represents a variable.

$$f(r+a) = f(r) + a\cdot\frac{\partial}{\partial r}f(r) + \frac{1}{2} a^2\cdot\frac{\partial^2}{\partial r^2}f(r) + \frac{1}{3!} a^3\cdot\frac{\partial^3}{\partial r^3} f(r) + \ldots$$
Because the notation $\frac{\partial}{\partial f}$ sucks (because r is used as a variable there), we will replace it with a better one:
$$f(r+a) = f(r) + af^\prime(r) + \frac{1}{2} a^2f^{\prime\prime}(r) + \frac{1}{3!} a^3f^{\prime\prime\prime}(r) + \ldots$$