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Mar 26, 2021 at 17:30 comment added Jan Bohr On $M=\mathbb{R}$ we take $x(t)=t$, such that $X=\{0\}$. Now, if a smooth function $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfies $f(0)=0$, then by exactness of the SES we can write $f(t)=t g(t)$ for another smooth function $g:\mathbb{R}\rightarrow \mathbb{R}$. Inductively, if $f(t)=O(t^k)$, then $f(t) = t^k g(t)$ for a smooth function $g:\mathbb{R}\rightarrow\mathbb{R}$. This applies to the remainder in the Taylor series, where $g$ might a priori only be bounded.
Mar 24, 2021 at 21:53 comment added Zach Teitler Can you please expand a bit on how the $M=\mathbb{R}$ case corresponds to remainders in Taylor series?
S Aug 25, 2020 at 10:38 history answered Jan Bohr CC BY-SA 4.0
S Aug 25, 2020 at 10:38 history made wiki Post Made Community Wiki by Jan Bohr