162. Alternative proof of Taylor’s Theorem

We shall now give the alternative form of the proof of Taylor’s Theorem to which we alluded in § 147.

Let $f(x)$ be a function whose first $n$ derivatives are continuous, and let $$F_n(x)=f(b)–f(x)–(b–x)f^{\prime }(x)–\cdots –\frac{(b–x{)}^{n-1}}{(n–1)!}{f}^{(n-1)}(x).$$

Then $$F_n^{\prime }(x)=-\frac{(b–x{)}^{n-1}}{(n–1)!}{f}^{(n)}(x),$$ and so $$F_n(a)=F_n(b)–{\int }_a^b{F}_n^{\prime }(x)dx=\frac{1}{(n–1)!}{\int }_a^b(b–x{)}^{n-1}{f}^{(n)}(x)dx.$$ If now we write $a+h$ for $b$, and transform the integral by putting $x=a+th$, we obtain $$\begin{array}{}\text{(1)}& f(a+h)=f(a)+h{f}^{\prime }(a)+\cdots +\frac{h^{n-1}}{(n–1)!}{f}^{(n-1)}(a)+R_n,\end{array}$$ where $$\begin{array}{}\text{(2)}& R_n=\frac{h^n}{(n–1)!}{\int }_0^1(1–t{)}^{n-1}{f}^{(n)}(a+th)dt.\end{array}$$

Now, if $p$ is any positive integer not greater than $n$, we have, by Theorem (9) of § 160, $$\begin{array}{rl}{\int }_0^1(1–t{)}^{n-1}{f}^{(n)}(a+th)dt& ={\int }_0^1(1–t{)}^{n-p}(1–t{)}^{p-1}{f}^{(n)}(a+th)dt\\ & =(1–\theta {)}^{n-p}{f}^{(n)}(a+\theta h){\int }_0^1(1–t{)}^{p-1}dt,\end{array}$$ where $0<\theta <1$. Hence $$\begin{array}{}\text{(3)}& R_n=\frac{(1–\theta {)}^{n-p}{f}^{(n)}(a+\theta h)h^n}{p(n–1)!}.\end{array}$$

If we take $p=n$ we obtain Lagrange’s form of $R_n$ (§ 148). If on the other hand we take $p=1$ we obtain Cauchy’s form, viz. $$\begin{array}{}\text{(4)}& R_n=\frac{(1–\theta {)}^{n-1}{f}^{(n)}(a+\theta h)h^n}{(n–1)!}.\end{array}$$