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 $F_{n}(x) = f(b) – f(x) – (b – x)f'(x) – \dots – \frac{(b – x)^{n-1}}{(n – 1)!} f^{(n-1)}(x).$

Then $F_{n}'(x) = -\frac{(b – x)^{n-1}}{(n – 1)!} f^{(n)}(x),$ and so $F_{n}(a) = F_{n}(b) – \int_{a}^{b}F_{n}'(x)\, dx = \frac{1}{(n – 1)!} \int_{a}^{b} (b – x)^{n-1} f^{(n)}(x)\, dx.$ If now we write $$a + h$$ for $$b$$, and transform the integral by putting $$x = a + th$$, we obtain $\begin{equation*} f(a + h) = f(a) + hf'(a) + \dots + \frac{h^{n-1}}{(n – 1)!} f^{(n-1)}(a) + R_{n}, \tag{1} \end{equation*}$ where $\begin{equation*} R_{n} = \frac{h^{n}}{(n – 1)!} \int_{0}^{1} (1 – t)^{n-1} f^{(n)}(a + th)\, dt. \tag{2} \end{equation*}$

Now, if $$p$$ is any positive integer not greater than $$n$$, we have, by Theorem (9) of § 160, \begin{aligned} \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{aligned} where $$0 < \theta < 1$$. Hence $\begin{equation*} R_{n} = \frac{(1 – \theta)^{n-p} f^{(n)}(a + \theta h)h^{n}}{p(n – 1)!}. \tag{3} \end{equation*}$

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. $\begin{equation*} R_{n} = \frac{(1 – \theta)^{n-1} f^{(n)}(a + \theta h) h^{n}}{(n – 1)!}. \tag{4} \end{equation*}$