so this is something we both agree on:

$L(x+a) - L(x) = a\frac{\partial L}{\partial x} + \frac{a^2}{2!}\frac{\partial^2 L}{\partial x^2} + \frac{a^3}{3!}\frac{\partial^3 L}{\partial x^3} + ... = \dot x \frac{\partial f}{\partial x} + \frac{\partial f}{\partial t}$

The idea is that it must work for every a. what this means is that for the same L and f, you change/plug in different $a$, but same L and f and it still should work. That's the beauty of homogeneity criterion. What this means is that unless higher order derivatives are zero, our criterion won't hold true. why ? because once you …

Even then, my ideas of those "effective moral" I've mentioned is mostly aligned with what we'd call "good" or "right" today. To make an example, since that's
what you've mentioned: I, the person who you're talking to now, obviously consider genocides wrong. If this society's "New Thing" appeared, my effective moral would not change and
I'd still deem it as wrong (in that effective sense). Or, to consider a different scenario, if this version of me that is writing were sent to a far past with different ideas, I would still

when we have $L(x+a) - L(x) = a\frac{\partial L}{\partial x} + \frac{a^2}{2!}\frac{\partial^2 L}{\partial x^2} + \frac{a^3}{3!}\frac{\partial^3 L}{\partial x^3} + ... = \dot x \frac{\partial f}{\partial x} + \frac{\partial f}{\partial t}$
, shouldn't this work for every $a$ ? meaning that the equation equality must hold true if $a=3$, if $a=2$, if $a=5$, so by pliugging in, if I put $a=2$ in the equation, LHS and RHS must be equal, if I put $a=3$ in the equation, LHS and RHS must again be equal