Landau's derivation of a free particle's kinetic energy- expansion of a function? I was reading a bit of Landau and Lifshitz's Mechanics the other day and ran into the following part, where the authors are about to derive the kinetic energy of a free particle. They use the fact that the Lagrangian of this particle must be the same (or at most, differ by the total time derivative of a function of co-ordinates and time) in different inertial frames.

We have $L'=L(v'^2)=L(v^2+2\mathbf{v}\cdot\boldsymbol{\epsilon}+\boldsymbol{\epsilon}^2)$. Expanding this expression in powers of $\boldsymbol{\epsilon}$ and neglecting terms above the first order, we obtain 
  $$L(v'^2)=L(v^2)+\frac{\partial L(v^2)}{\partial (v^2)}2\mathbf{v}\cdot\boldsymbol{\epsilon}.$$

I think I'm ok with all the physics in this section. What I don't get is just the part I quoted above (so maybe this post is better suited for the math site, but since this is book is so physics-y I thought I'd post it here). My math is pretty rusty, so I'm not really sure- how do the authors expand the function to arrive at the above expression? It reminds me a bit of a Taylor expansion, but not very much. What's the process used to arrive at it?  
 A: It is a Taylor expansion.  You might be a little freaked out because they are treating $L$ as a function of $v^2$, but that doesn't matter, consider
$$ L(x + \delta) \sim L(x) + \frac{\partial L}{\partial x} \delta + O(\delta^2) $$
but take $x=v^2$ and $\delta = 2 \boldsymbol{v} \cdot \boldsymbol{\epsilon} + \epsilon^2 $,
$$
\begin{align*}
 L(v'^2) &= L( v^2 + 2 \boldsymbol{v} \cdot \boldsymbol{\epsilon} + \epsilon^2) \\
   &\sim L(v^2) + \frac{\partial L}{\partial v^2} ( 2 \boldsymbol{v} \cdot \boldsymbol{\epsilon} + \epsilon^2) + O(\epsilon^2) \\
 &\sim L(v^2) + \frac{\partial L}{\partial v^2}  2 \boldsymbol{v} \cdot \boldsymbol{\epsilon}  + O(\epsilon^2) 
\end{align*}
$$
We can drop the other $\epsilon^2 \frac{\partial L}{\partial v^2}$ term since we are only interested in terms of first order.
Taylor Expansion Generally
If its the taylor expansion you're having trouble with, that's pretty easy to show as well.  Consider $f(x + \epsilon)$ with $\epsilon$ small, we might want to express the value of $f(x + \epsilon)$ as a power series in $\epsilon$, so we look for something of the form
$$ f(x + \epsilon) = a_0 + a_1  \epsilon + a_2 \epsilon^2 + \cdots $$
How do we determine the coefficients, well we can find $a_0$ by taking the limit as $\epsilon \to 0$, obtaining
$$ f(x) = a_0 $$
Great, but how would we get the $a_1$ coefficient?  Well, just take a derivative of both sides
$$ f'(x + \epsilon) = a_1 + 2 a_2 \epsilon + \cdots $$
and just take the limit again, obtaining
$$ f'(x) = a_1 $$
Doing that process once more you get
$$ f''(x) = 2 a_2 $$
So, so far we've got
$$ f(x + \epsilon) = f(x) + f'(x) \epsilon + \frac 12 f''(x) \epsilon^2 + \cdots $$
If you think generally, you should be able to convince yourself that in general, for the $n$th term, we'll have
$$ f^{(n)}(x) = n! a_n $$
giving us the general Taylor series result
$$ f(x + \epsilon) = \sum_{i=0}^\infty  \frac{1}{n!} f^{(n)}(x) \epsilon^n $$ 
