Talk:Taylor's Theorem/One Variable

Invocation of Questionable template

I don't understand, if $f$ is $n$ times differentiable at $x$, then its $n$th derivative is continuous at $x$ by definition, is it not? --prime mover (talk) 07:07, 30 August 2020 (UTC)

Counterexample:

$f(x) = \begin{cases}

x^2 \sin(1/x) \quad (x\neq 0) \\
0 \quad (x=0)

\end{cases}$ Then

$f'(x) = \begin{cases}

2x \sin(1/x) - \cos(1/x) \quad (x\neq 0) \\
0 \quad (x=0)

\end{cases}$ is not continuous at $x=0$. -- Sawasawa (talk) 10:07, 30 August 2020 (UTC)

What we need is for a page (or suite of pages) which detail the precise conditions under which this theorem holds and demonstrates why relaxing any of those conditions renders the theorem invalid. The work from which the original statement came deilvered the statement in a different form from this, and the subtleties of exactly how continuous and derivable the function is were glossed over. As of now, nobody has attempted a comprehensive exposition of Taylor's Theorem. This is long overdue. --prime mover (talk) 10:46, 30 August 2020 (UTC)