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


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