I am reading through the proof of the following theorem:
The proof follows the the proof which can be found in Arnold's book about Classical mechanics.
What I don't understand is in the following part of the proof:
Question What is $O(s^2 t,st^2)$? Shouldn't it be $O(s^2+t^2)$, since we have a Taylor expansion of order two?
Attempt of an answer
- I think from Corollary C.16 (below) we have that it follows that $|\Delta(s,t,x)-T_{2}|\in O(||(s,t)||^{3})$. Going back to the definition of $O$ I get $|\Delta(s,t)-T_{2}|\leq M|(s^2+t^2)^{\frac{3}{2}}|\leq|(s^2+t^2)^{3}|=M|s^3+3ts^2+3st^2+t^3|\leq M|3ts^2+3st^2|=3M|ts^2+st^2|$. Hence, $\Delta(s,t)-T_{2}\in O(st^2,ts^2)\iff \Delta(s,t)=T_{2}+O(st^2,ts^2)$
- To show that $\Delta(s,t)=st\frac{\partial^{2}\Delta}{\partial s\partial t}|_{(0,0)}+O(st^2,ts^2)$, I would go back to the definition of the partial derivative. Then I would say the fact that $\Delta(0,t,x)=\Delta(s,0,x)=0$ implies that both $\frac{\partial\Delta}{\partial s}(0,0)=0$ and $\frac{\partial\Delta}{\partial t}(0,0)=0$. Therefore $\frac{\partial^{2} \Delta}{\partial s^{2}}(0,0)=\frac{\partial^{2}\Delta}{\partial t^{2}}(0,0)=0$. For the mixed term I get $\frac{\partial^{2}\Delta}{\partial s\partial t} (0,0):=\lim_{h_{1},h_{2}\to 0}\frac{\Delta(h_{1},h_{2})}{h_{1}h_{2}}$. Of course having in mind that itterated limits are not necessarily the same. But since $\Delta$ is cont. partially differentiable, I guess it is okay.
The largest terms of the error term, $O(s^2 t,st^2)$, come from the cube, $(s+t)^3$. The reason that $s^3$ and $t^3$ are omitted is that they will always vanish under the limit. This is not true in general for the middle terms (recall from multivariable calculus that convergence to a point for 2 (or more) independent variables can sometimes work for some trajectories, but not others; directional limits may not agree or not even exist for all trajectories, even when some do exist).