Lagrangian for 2 inertial frames where only Speed is different by small amount

Question

In Landau & Liftshitz’s book p.5, they go ahead and writes down lagrangians for 2 different inertial frames. They say that Lagrangian is a function of $v^2$.

So in one frame, we got $L(v^2)$.

In another frame (the speed relative to the first frame is $\epsilon$ so its actual speed is $v+\epsilon$

So lagrangian for the second reference frame is $L(v^2 + 2v\epsilon + \epsilon^2)$.

Then, the book proceeds to mention that we need to expand this in powers of $\epsilon$ and then neglect the terms above first order, we obtain:

$L(v^2) + \frac{\partial L}{\partial v^2}2v\epsilon$ (This is what I don't get).

We know taylor is given by: $p(x) = f(a) + f'(a)(x-a) + ....$

I tried doing this in respect to $\epsilon$. We know $\epsilon$ is super small, so our taylor is $f(0) + f'(a)(x-0) = L(v^2) + \frac{\partial L}{\partial \epsilon}(2v+2\epsilon)|(\epsilon=0)(\epsilon-0) = L(v^2) + \frac{\partial L}{\partial \epsilon}2v\epsilon$ (note that I don't have $\frac{\partial L}{\partial v^2}$)

If I try doing taylor with respect to $v^2$, then I don't get $L(v^2)$ as the first part because $f(a) != L(v^2)$

Where am I making a mistake?

Update

A couple of things that confuse me.

1. $\frac{d}{d\epsilon}L(v^2 + 2v\epsilon + \epsilon^2)|_{\epsilon = 0} = \frac{d}{d\epsilon}L(x)|_{x = v^{2}}$ If we imagined $x = v^2 + 2v\epsilon + \epsilon^2$, why do we say $|x=v^2$ only ?
2. $[\frac{d}{dx} L(x)]|_{x = v^2} = \frac{d}{dv^2} L(v^2)$ (here the left pat says differentiate L with x and then put $v^2$ instead of $x$. how is this the same as the right part ?

Answer 1

It's a question of understanding notation.

$f'(a) \equiv [f'](a)$ means the [derivative of the function] that is then (evaluated at the point a). Notice that the stuff in [...] makes no reference to the label that you give the function's argument. But it might be helpful to invent an auxiliary argument, $x$, say and evaluate it like this:

$f'(a) = [\frac{d}{dx} f(x)]|_{x = a}$.

In your case, you haven't applied the chain rule correctly. In your approach, $L'(a)$ should be $\frac{d}{d\epsilon}L(v^2 + 2v\epsilon + \epsilon^2)|_{\epsilon = 0}$ which you can evaluate as follows: let $x = v^2 + 2v\epsilon + \epsilon^2$ so that

$\frac{d}{d\epsilon}L(v^2 + 2v\epsilon + \epsilon^2)|_{\epsilon = 0} = \frac{d}{d\epsilon}L(x)|_{x = v^{2}} = \frac{d}{dx}L(x)|_{x = v^2} \times \frac{dx}{d\epsilon}|_{\epsilon = 0} = \frac{d}{dx}L(x)|_{x= v^{2}}\times (2v + 2\epsilon)_{\epsilon = 0} = 2vL'(v^{2}).$

where $L'(v^2)$ can be calculated by pretending momentarily that the argument of $L$ is $x$, say, differentiating and then setting $x = v^2$:

$L'(v^2) = [\frac{d}{dx} L(x)]|_{x = v^2}$

The only reason to write it like this is to clarify what the derivative means. In any case, your function depends on $v^2$ so differentiation wrt its argument is differentiation wrt $v^2$:

$L'(v^2) = \frac{d}{dv^2} L(v^2)$.

At the end, multiply by $(\epsilon - 0) = \epsilon$.