For Taylor expansion, we have
\begin{equation}
f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{f^{''}(x_0)}{2 !}(x-x_0)^{2}+\cdots+\frac{f^{(N)}(x_0)}{N !}(x-x_0)^{N} + R_N,
\end{equation}
where $R_N$ is the remainder term of this $N$-th order Taylor polynomial.
However, I feel a bit confusion that whether we have
\begin{equation}
\left|\frac{f^{(N)}(x_0)}{N !}(x-x_0)^{N}\right| > |R_n|.
\end{equation}
I don't know how to prove it.
Best Answer
$R_n$ and the last term are not related in any specific way, it depends on the specific function. The only thing that is true in general is that $R_n$ is an infinitesimal of order higher than the order of the last term, that is $$ R_n(x)=o(|x-x_0|^n) $$ as $x\to x_0$. In other words, when $f^n(x_0)\neq 0$ you have $R_n(x)<< \textrm{last term}$ in the asymptotic sense.