= Solution
<Taylor's theorem> gives a finite <Taylor polynomial> of degree $n$, consisting of the first $n+1$ terms, and a remainder:
$$
\boxed{f(x)=\sum_{j=0}^{n}\frac{f^{(j)}(a)}{j!}(x-a)^j+R_{n+1}(x).}
$$
In the <Taylor theorem with Lagrange remainder> form, assuming the required <derivatives> exist on an <interval> containing $a$ and $x$,
$$
\boxed{R_{n+1}(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1},}
$$
for some $\xi$ strictly between $a$ and $x$ when $x\ne a$; at $x=a$ the remainder is zero. In particular, if $|f^{(n+1)}|\leq M$ on that segment, then $|R_{n+1}(x)|\leq M|x-a|^{n+1}/(n+1)!$. This is a finite expansion; it does not assume equality with an infinite <Taylor series>.
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