Symmetry Condition in Noether's Theorem

Question

Suppose $q = \{q_1,\cdots, q_i\}$ is a coordinate system for Lagrangian $L(q,\dot{q},t)$. In this text by David Morin, on page 16 in chapter 6, it states that a symmetry is a transformation of the Lagrangian coordinates, $q'_i = q_i + \varepsilon\cdot K_i(q)$, $\varepsilon$ small and where $K_i(q)$ is a function on the set $q$, that does not lead to a first-order change in the Lagrangian.

It later stated that not having a first-order change in the Lagrangian means that $$\frac{dL}{d \varepsilon}=\sum_{i}\left(\frac{\partial L}{\partial q_i}\frac{\partial q_i}{\partial \varepsilon} + \frac{\partial L}{\partial \dot q_i}\frac{\partial \dot q_i}{\partial \varepsilon} \right) = 0.\tag{6.59}$$

Is an equivalent condition for symmetry given by requiring the Taylor expansion of $L(q',\dot{q}')$ about $L(q,\dot{q})$ to lack first order terms in $\varepsilon$? This comes from thinking that $L(q',\dot{q}')$ can be written as

$$L(q',\dot{q}') = L(q,\dot{q}) + \frac{dL}{d\varepsilon}\varepsilon + O(\varepsilon^2),$$

which would show that the two are equivalent, but I feel like I'm glossing over some details.

Answer 1

I don't think there is the statement that $K_i(q)$ has to be linear in $q$. This is certainly in a lot of examples. The simplest is a translation symmetry $q'=q+\epsilon$.

Going to your question though, this is certainly equivalent to asking that the Taylor expansion of $L$ around $q$ has no first order in $\varepsilon$. However, your Taylor expansion of the Lagrangian is wrong. It is simply $$L(q')=L(q)+ \epsilon\frac{\text{d}L}{\text{d}\epsilon}+\mathcal O(\epsilon^2)$$.