homework and exercises - Trouble with Landau & Lifshitz's expansion of the Lagrangian with respect to $epsilon$ and $v$

Hello I have a quick question on what I have been reading in Landau & Lifshitz's book on classical mechanics. I am in the very beginning of the book and I am having trouble with his derivation on the Lagrangian of a free particle. At the very top of page 7, he equates two Lagrangian by

$$\mathrm{L}'=\mathrm{L}(v'^2)=\mathrm{L}(v^2+2v \cdot \epsilon+ \epsilon^2).$$

Then he goes to to say that you can expand this in powers of $ \epsilon$ and neglect all terms above the first order to get

$$\mathrm{L}(v'^2)=\mathrm{L}(v^2)+\frac{\partial \mathrm{L}}{\partial v^2}2v \cdot \epsilon.$$

How do you expand in terms of $\epsilon$ to get this result? This Lagrangian seems to me to be a function of three variables and I am looking at the multivariable Taylor expansion and I am not seeing how Landau could get his result.

Answer

Landau defines $v'\equiv v+\epsilon$. Then

$$ L(v'^2)= L[ (v+\epsilon)^2] = L(v^2+2v\cdot\epsilon+\epsilon^2)\sim L(v^2+2v\cdot \epsilon) $$

In the last step we have omited the second order terms in $\epsilon$.

So we have $L(v^2+2v\cdot\epsilon)$ which is a function like $f(a+x)$. We expand it as

$$ f(a+x)\sim f(a)+f'(a)x $$

at first order. In our case, we have $a=v^2$ and $x=2v\cdot\epsilon$, so

$$ L(v'^2)\sim L(v^2 + 2v\cdot\epsilon) \sim L(v^2) + \frac{\partial L}{\partial v^2} 2v\cdot\epsilon$$