Proof of Taylor's Theorem There are several proofs of Taylor's Theorem, which
lead to various forms of the remainder. The following proof is instructive
because it leads to two different forms of the remainder and it relies on the
Fundamental Theorem of Calculus, integration by parts, and the Mean Value
Theorem for Integrals. Assume \(f\) has at least \(n+1\) continuous derivatives on
an interval containing \(a\).
a. Show that the Fundamental Theorem of Calculus can be written in the form
$$f(x)=f(a)+\int_{a}^{x} f^{\prime}(t) d t$$
b. Use integration by parts \(\left(u=f^{\prime}(t), d v=d t\right)\) to show
that \right.
$$f(x)=f(a)+(x-a) f^{\prime}(a)+\int_{a}^{x}(x-t) f^{\prime \prime}(t) d t$$
c. Show that \(n\) integrations by parts give
$$\begin{aligned}
f(x)=& f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2
!}(x-a)^{2}+\cdots \\
&+\frac{f^{(n)}(a)}{n !}(x-a)^{n}+\underbrace{\int_{a}^{x}
\frac{f^{(n+1)}(t)}{n !}(x-t)^{n} d t}_{R_{n}(x)}
\end{aligned}$$
d. Challenge: The result in part (c) has the form \(f(x)=p_{n}(x)+R_{n}(x),\)
where \(p_{n}\) is the \(m\) th-order Taylor polynomial and \(R_{n}\) is a new form
of the remainder, known as the integral form of the remainder. Use the Mean
Value Theorem for Integrals (Section 5.4 ) to show that \(R_{n}\) can be
expressed in the form
$$R_{n}(x)=\frac{f^{(n+1)}(c)}{(n+1) !}(x-a)^{n+1}$$
where \(c\) is between \(a\) and \(x\)
