Taylor Series of a function not showing the right derivatives

Question

The function is the following $$ f[t, y + \epsilon \eta , \dot y + \epsilon \dot \eta)$$ How is a Taylor Expansion done in a way that is in powers of $\epsilon$? $$ f[t, y + \epsilon \eta , \dot y + \epsilon \dot \eta)\approx a_0 + \epsilon a_1 + \mathcal O[\epsilon ^2) $$

If doing a Taylor expansion around $\epsilon$ it should be $$ f[t, y + \epsilon \eta , \dot y + \epsilon \dot \eta)\approx f[t,y,\dot y) + \epsilon (\partial_\epsilon f)|_{\epsilon=0} + \mathcal O[\epsilon ^2)$$

When I searched in the topic of Lagrangians it was like: $$ f[t, y + \epsilon \eta , \dot y + \epsilon \dot \eta)\approx f[t,y,\dot y) + \epsilon ( \eta \partial_y f + \dot \eta \partial_{\dot y} f) + \mathcal O[\epsilon ^2) \; \; \tag{3}$$

Doing Chain Rule $$a_1 = \frac{\partial f}{\partial ( y +\epsilon \eta)} \frac{\partial ( y +\epsilon \eta)}{\partial \epsilon} + \frac{\partial f}{\partial (\dot y +\epsilon \dot \eta)} \frac{\partial (\dot y +\epsilon \dot\eta)}{\partial \epsilon} $$

$$a_1=\frac{\partial f}{\partial ( y +\epsilon \eta)} \eta +\frac{\partial f}{\partial (\dot y +\epsilon \dot \eta)} \dot \eta$$

But why the derivatives on $(3)$ do not show the $\epsilon$ and $\eta$?

Answer 1

I think you are missing subtlety here, I have been confused by this in the past.

A little background is necessary first. When studying the effects of perturbations of the Lagrangian $f(t,y,\dot y)$ around a given trajectory $y(t)$, one often replaces $$ y(t) \rightarrow y(t)+\delta y(t) $$ where $\delta y$ is supposed to be small. One way to express that $\delta y$ is small, is to say $$\delta y(t) = \epsilon \eta(t), $$ where $\eta$ is any function of time, and $\epsilon\ll 1$ is not a constant. One can make the perturbation as small as required by changing the $\epsilon$, while keeping $\eta$ the same. This way of splitting the "size" and the "shape" of the perturbation has the added benefit of clearly separating effect that are linear, quadratic (and so on) in the perturbation.

Ok, to your problem. I'll omit the time dependence on $f$ as it just rides along. We want to relate $f(y+\epsilon \eta, \dot y + \epsilon \dot \eta)$ to $f(y, \dot y)$. To first order in $\epsilon$. $f$ is a function of 2 variables. Let us denote $\partial_1 f(a,b)$ and $\partial_2 f(a,b)$ the partial derivatives with respect to the first and second argument evaluated at $(y,\dot y) = (a,b)$. Then by Taylor: $$f(y+\epsilon \eta, \dot y + \epsilon \dot \eta) = f(y,\dot y) + \epsilon\eta\partial_1f(y,\dot y) + \epsilon\dot\eta\partial_2f(y,\dot y) + \mbox{higher order terms}$$ where by higher order, I mean terms proportional to $\partial_1^nf\partial_2^mf\times(\epsilon \eta)^n(\epsilon\dot\eta)^m$, with $m+n>1$. Note that the formula above will naturally lead to an expansion in powers of $\epsilon$ as each higher order term comes with powers of $\epsilon$.

This is the way to understand formula (3). In particular, the term $\epsilon\eta\partial_yf$ is just shorthand for $\epsilon\eta\partial_1f(y,\dot y)$. The shorthand is useful, but it can be confusing.

We can also see why (3) agrees with your $a_1$, even though you are also using shorthand notation. Your $$\frac{\partial f}{\partial(y+\epsilon\eta)}$$ is none other than $\partial_1 f$ aka $\partial_y$!