Stronger version of Taylor's Theorem Show that Taylor's Theorem may be strengthened as follows:
Let $f$ be a continuous real-valued function on the closed interval in R of extremities $a$ and $b$. That is, (n + 1) times differentiable on the open interval with these same extremities and suppose that $\lim_{x \to a} f'(x), \lim_{x \to a} f''(x), \dots, \lim_{x \to a} f^{(n)}(x)$ exist and that $f',f'',\dots,f^{(n)}$ are bounded.
Then,
$$f(b) = f(a) + (\lim_{x \to a} f'(x)) \frac{(b - a)}{1!} + \dots | (\lim_{x \to a} f^{(n)}(x)) \frac{(b - a)^n}{n!} + f^{(n + 1)}(c) \frac{(b - a)^{n + 1}}{(n + 1)!}$$
for some c between a and b.
I'm not too familiar with Taylor's Theorem in an Analysis sense. We briefly talked about it in class, but we sort of moved on quickly. This was said to be a "cool problem" to complete, so I would like to see it. Can anyone help me with this one? Thank you!
 A: Theorem: Let $a,b\in\mathbb{R}$ with $a<b$. Let $n\geq 0$ be an integer. Let $f:[a,b]\rightarrow\mathbb{R}$ be a function
that satisfies the following conditions:
(1) $f$ is continuous on $[a,b]$,
(2) $f^{(k)}$ exists on $(a,b)$ for $k=1,2,\ldots,n+1$.
(3) $\lim_{x\rightarrow a+}f^{(k)}(x)$ exists for $k=1,2,\ldots,n$.
(4) $\lim_{x\rightarrow b-}f^{(k)}(x)(b-x)^{k}=0$ for $k=1,2,\ldots,n.$
(Remark: Note that condition (4) is implied by the condition that
$f^{(k)}$ is bounded on $(a,b)$. However, the condition that $f^{(k)}$
is bounded on $(a,b)$ is too strong and we can replace it by a much
weaker condition.)
Denote $f^{(k)}(a+)=\lim_{x\rightarrow a+}f^{(k)}(x)$. Then there
exists $\xi\in(a,b)$ such that
$$
f(b)=f(a)+\sum_{k=1}^{n}f^{(k)}(a+)\cdot\frac{(b-a)^{k}}{k!}+f^{(n+1)}(\xi)\cdot\frac{(b-a)^{n+1}}{(n+1)!}.
$$
Proof: Define $\theta:[a,b]\rightarrow\mathbb{R}$ by
$$
\theta(t)=\begin{cases}
f(b)-f(t)-\sum_{k=1}^{n}f^{(k)}(t)\cdot\frac{(b-t)^{k}}{k!}, & \mbox{if }t\in(a,b)\\
f(b)-f(a)-\sum_{k=1}^{n}f^{(k)}(a+)\cdot\frac{(b-a)^{k}}{k!}, & \mbox{if }t=a\\
0, & \mbox{if }t=b
\end{cases}.
$$
Clearly $\theta$ is continuous on $(a,b)$. By condition (3), $\theta(a)=\lim_{t\rightarrow a+}\theta(t).$
By condition (4), $\theta(b)=\lim_{t\rightarrow b-}\theta(t)$. In
short, $\theta$ is continuous on $[a,b]$. It is also clear that
$\theta$ is differentiable on $(a,b)$. By direct calculation (observe
the telescope cancellation), we have
$$
\theta'(t)=-f^{(n+1)}(t)\cdot\frac{(b-t)^{n}}{n!}.
$$
Define $g:[a,b]\rightarrow\mathbb{R}$ by $g(t)=(b-t)^{n+1}$. Clearly
$g$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $g'(t)\neq0$
for all $t\in(a,b)$. By Cauchy's mean-value theorem, there exists
$\xi\in(a,b)$ such that
\begin{eqnarray*}
\frac{\theta(b)-\theta(a)}{g(b)-g(a)} & = & \frac{\theta'(\xi)}{g'(\xi)}\\
 & = & -f^{(n+1)}(\xi)\cdot\frac{(b-\xi)^{n}}{n!}\cdot\frac{1}{-(n+1)(b-\xi)^{n}}\\
 & = & f^{(n+1)}(\xi)\cdot\frac{1}{(n+1)!}.
\end{eqnarray*}
On the other hand,
\begin{eqnarray*}
\frac{\theta(b)-\theta(a)}{g(b)-g(a)} & = & -\frac{f(b)-f(a)-\sum_{k=1}^{n}f^{(k)}(a+)\cdot\frac{(b-a)^{k}}{k!}}{-(b-a)^{n+1}}.
\end{eqnarray*}
Rearranging terms, we obtain,
$$
f(b)=f(a)+\sum_{k=1}^{n}f^{(k)}(a+)\cdot\frac{(b-a)^{k}}{k!}+f^{(n+1)}(\xi)\cdot\frac{(b-a)^{n+1}}{(n+1)!}.
$$
