Suppose that $$f(x)$$ is a function all of whose differential coefficients are continuous in an interval $${[a – \eta, a + \eta]}$$ surrounding the point $$x = a$$. Then, if $$h$$ is numerically less than $$\eta$$, we have $f(a + h) = f(a) + hf'(a) + \dots + \frac{h^{n-1}}{(n – 1)!} f^{(n-1)}(a) + \frac{h^{n}}{n!} f^{(n)}(a + \theta_{n} h),$ where $$0 < \theta_{n} < 1$$, for all values of $$n$$. Or, if $S_{n} = \sum_{0}^{n-1} \frac{h^{\nu}}{\nu!} f^{(\nu)}(a),\quad R_{n} = \frac{h^{n}}{n!} f^{(n)}(a + \theta_{n} h),$ we have $f(a + h) – S_{n} = R_{n}.$

Now let us suppose, in addition, that we can prove that $$R_{n} \to 0$$ as $$n \to \infty$$. Then $f(a + h) = \lim_{n\to\infty} S_{n} = f(a) + hf'(a) + \frac{h^{2}}{2!} f”(a) + \dots.$

This expansion of $$f(a + h)$$ is known as Maclaurin’s Series. When $$a = 0$$ the formula reduces to $f(h) = f(0) + hf'(0) + \frac{h^{2}}{2!} f”(0) + \dots,$ which is known as . The function $$R_{n}$$ is known as Lagrange’s form of the remainder.

The reader should be careful to guard himself against supposing that the continuity of all the derivatives of $$f(x)$$ is a sufficient condition for the validity of Taylor’s series. A direct discussion of the behaviour of $$R_{n}$$ is always essential.

Example LVI

1. Let $$f(x) = \sin x$$. Then all the derivatives of $$f(x)$$ are continuous for all values of $$x$$. Also $$|f^{n}(x)| \leq 1$$ for all values of $$x$$ and $$n$$. Hence in this case $$|R_{n}| \leq h^{n}/n!$$, which tends to zero as $$n \to \infty$$ (Ex. XXVII 12) whatever value $$h$$ may have. It follows that $\sin(x + h) = \sin x + h\cos x – \frac{h^{2}}{2!}\sin x – \frac{h^{3}}{3!}\cos x + \frac{h^{4}}{4!}\sin x + \dots,$ for all values of $$x$$ and $$h$$. In particular $\sin h = h – \frac{h^{3}}{3!} + \frac{h^{5}}{5!} – \dots,$ for all values of $$h$$. Similarly we can prove that $\cos(x + h) = \cos x – h\sin x – \frac{h^{2}}{2!}\cos x + \frac{h^{3}}{3!} \sin x + \dots,\quad \cos h = 1 – \frac{h^{2}}{2!} + \frac{h^{4}}{4!} – \dots.$

2. The Binomial Series. Let $$f(x) = (1 + x)^{m}$$, where $$m$$ is any rational number, positive or negative. Then $$f^{(n)}(x) = m(m – 1) \dots (m – n + 1) (1 + x)^{m-n}$$ and Maclaurin’s Series takes the form $(1 + x)^{m} = 1 + \binom{m}{1}x + \binom{m}{2}x^{2} + \dots.$

When $$m$$ is a positive integer the series terminates, and we obtain the ordinary formula for the Binomial Theorem with a positive integral exponent. In the general case $R_{n} = \frac{x^{n}}{n!} f^{(n)}(\theta_{n}x) = \binom{m}{n}x^{n}(1 + \theta_{n}x)^{m-n},$ and in order to show that Maclaurin’s Series really represents $$(1 + x)^{m}$$ for any range of values of $$x$$ when $$m$$ is not a positive integer, we must show that $$R_{n} \to 0$$ for every value of $$x$$ in that range. This is so in fact if $$-1 < x < 1$$, and may be proved, when $$0\leq x < 1$$, by means of the expression given above for $$R_{n}$$, since $$(1 + \theta_{n}x)^{m-n} < 1$$ if $$n > m$$, and $$\dbinom{m}{n} x^{n} \to 0$$ as $$n \to \infty$$ (Ex. XXVII. 13). But a difficulty arises if $$-1 < x < 0$$, since $$1 + \theta_{n}x < 1$$ and $$(1 + \theta_{n}x)^{m-n} > 1$$ if $$n > m$$; knowing only that $$0 < \theta_{n} < 1$$, we cannot be assured that $$1 + \theta_{n}x$$ is not quite small and $$(1 + \theta _{n}x)^{m-n}$$ quite large.

In fact, in order to prove the Binomial Theorem by means of Taylor’s Theorem, we need some different form for $$R_{n}$$, such as will be given later (§ 162).