So I was working on ΔS (it is called action in physics but the question I am about to present is purely mathematical).
$S=∫L dt => ΔS=Δ(∫(\frac{1}{2}mv^2 -V)dt)= Δ(∫\frac{1}{2}mv^2 dt)$
The problem I face here is if we were to assume that t interval is small. Then could we say
$$ΔS= Δ(∫\frac{1}{2}mv^2 dt)=\frac{1}{2}mv^2Δt$$
Is this acceptable?
Note: I am reminded that this question may not be seen as "purely mathematical" and decided to wrote this note to avoid confusion regarding terms related to physics. However, this note will get a bit more mathematical towards the end. I asked this question here because I think my problem is more mathematical than physical. m is mass and v is velocity. It is assumed that they are constants. ΔS is "change in action" and is defined as ∫L dt with L being $L=\frac{1}{2}mv^2 -V$. This "change" actually represents deviation from the action of the classical path a particle may follow.
V (potential term) in my first definition of ΔS is not important. What I am trying to show here is that ΔS can be represented as $\frac{1}{2}mv^2Δt$. Let's visualise the issue that I am having.
When $Δt<<1$, it is immediate to see that $∫\frac{1}{2}mv^2 dt=\frac{1}{2}mv^2Δt$
So we have $Δ(\frac{1}{2}mv^2Δt)=\frac{1}{2}mv^2Δ(Δt)$. This is a serious issue for me. I felt like we can overcome this by saying $Δt=t'$ and thus we have $$Δ(\frac{1}{2}mv^2Δt)=\frac{1}{2}mv^2Δ(Δt)=\frac{1}{2}mv^2Δt'=ΔS$$
But I need $Δt'\equiv Δt$ and that feels wrong in a mathematical way. It works while making commentaries on a solution but I want to be able to show it mathematically.
If $Δt'\equiv Δt$ somehow works than ΔS becomes $$ΔS=\frac{1}{2}mv^2Δt=\frac{1}{2}m(\frac{Δx}{Δt})^2Δt=\frac{1}{2}m\frac{(Δx)^2}{Δt}=\frac{1}{2}m\frac{\eta^2}{\epsilon}$$ Don't worry about $\eta$ and $\epsilon$. They are just a symbol. They still show changes in the respective "variables". That's not changed.