Explanation on Taylor expansions needed

[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:
<br /> L(v^{\prime 2}) = L(v^2)+\frac{\partial L}{\partial v^2} 2\boldsymbol{v}\cdot \boldsymbol{\varepsilon}<br />


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

 

[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:
<br /> f(x) = f(c) + (x-c) \frac{\operatorname{d}}{\operatorname{d}x}f(c) + \ldots<br />

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?

 

[micromass]

Hi Stalafin!

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[/tex]

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.

Rewriting your equation gives us
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

But this is exactly the Taylor series! So the original formula was correct (but abusive).